Rational Numbers to Linear Equations 1470456753, 9781470456757

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Rational Numbers to Linear Equations Hung-Hsi Wu

Rational Numbers to Linear Equations

Rational Numbers to Linear Equations Hung-Hsi Wu

2010 Mathematics Subject Classification. Primary 97-01, 97-00, 97D99, 97-02, 00-01, 00-02.

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Library of Congress Cataloging-in-Publication Data Cataloging-in-Publication Data has been applied for by the AMS. See http:www.loc.gov/publish/cip/.

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established to ensure permanence and durability. Visit the AMS home page at https://www.ams.org/ 10 9 8 7 6 5 4 3 2 1

25 24 23 22 21 20

Dedicated to the memory of David M. Collins

Contents Contents of the Companion Volumes and Structure of the Chapters

ix

Preface

xi

To the Instructor

xxvii

To the Pre-Service Teacher

xli

Prerequisites

xlv

Some Conventions

xlvii

Chapter 1. Fractions Overview of Chapters 1 and 2 1.1. Definition of a fraction 1.2. Equivalent fractions 1.3. Adding and subtracting fractions 1.4. Multiplying fractions 1.5. Dividing fractions 1.6. Complex fractions 1.7. Percent, ratio, and rate problems 1.8. Appendix: The basic laws

1 1 4 19 32 43 54 68 72 86

Chapter 2. Rational Numbers 2.1. The rational numbers 2.2. Adding rational numbers 2.3. The vectorial representation of addition 2.4. Multiplying rational numbers 2.5. Dividing rational numbers 2.6. Comparing rational numbers 2.7. FASM

89 90 91 99 105 112 121 133

Chapter 3. The Euclidean Algorithm 3.1. The reduced form of a fraction 3.2. The fundamental theorem of arithmetic

137 137 147

Chapter 4. Basic Isometries and Congruence Overview of Chapters 4 and 5 4.1. The basic vocabulary, Part 1 4.2. The basic vocabulary, Part 2 4.3. Transformations of the plane 4.4. The basic isometries: Rotations

157 157 164 180 199 216

vii

viii

CONTENTS

4.5. The basic isometries: Reflections and translations 4.6. Congruence, SAS, and ASA 4.7. A brief pedagogical discussion of proofs

229 239 252

Chapter 5. Dilation and Similarity 5.1. The fundamental theorem of similarity 5.2. Dilation 5.3. Similarity

255 255 268 283

Chapter 6. Symbolic Notation and Linear Equations 6.1. Symbolic expressions 6.2. Solving linear equations in one variable 6.3. Setting up coordinate systems 6.4. Lines in the plane and their slopes 6.5. The graphs of linear equations in two variables 6.6. Parallelism and perpendicularity 6.7. Simultaneous linear equations

297 298 322 331 337 351 363 373

Glossary of Symbols

385

Bibliography

387

Index

391

Contents of the Companion Volumes and Structure of the Chapters Algebra and Geometry [Wu2020b] Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter

1: 2: 3: 4: 5: 6: 7: 8:

Linear Functions Quadratic Functions and Equations Polynomial and Rational Functions Exponential and Logarithmic Functions Polynomial Forms and Complex Numbers Basic Theorems of Plane Geometry Ruler and Compass Constructions Axiomatic Systems

Pre-Calculus, Calculus, and Beyond [Wu2020c] Chapter Chapter Chapter Chapter Chapter Chapter Chapter

1: 2: 3: 4: 5: 6: 7:

Trigonometry The Concept of Limit The Decimal Expansion of a Number Length and Area 3-Dimensional Geometry and Volume Derivatives and Integrals Exponents and Logarithms, Revisited

ix

x

CONTENTS OF THE COMPANION VOLUMES AND STRUCTURE OF THE CHAPTERS

Structure of the chapters in this volume and its two companion volumes (RLE= Rational Numbers to Linear Equations, A&G = Algebra and Geometry, PCC = Pre-Calculus, Calculus, and Beyond)

Preface The really vital importance of definition is not, I venture to think, sufficiently emphasized even in good textbooks.. . . they form the premises from which the rest of the algebraic theorems are to be derived by a process of logical deduction. George A. Gibson ([Gibson, page 3])

A nation’s mathematics education is only as good as its mathematics teachers. The ongoing crisis in school mathematics education (cf. [RAGS]) therefore raises the question: what have we done wrong in the preparation of mathematics teachers? The answer is plenty: our longstanding neglect of the mathematical education of teachers has come home to roost. This neglect manifests itself in K–12, where we fail to ensure that correct mathematics is taught to students—especially future teachers—and we compound this neglect by failing to provide the needed corrective measures in universities for pre-service teachers to repair their mathematical mis-education in K–12 (cf. [Wu2011b]). Thus, through no fault of their own, the mathematics teachers of our nation are put in the untenable position of teaching from a position of weakness: they do not possess the needed knowledge of mathematics to carry out their basic duties. The present volume is the fourth of six volumes whose collective goal is to provide the needed mathematical backing for a full-scale attack on the crisis in the mathematical education of mathematics teachers in K–12. This volume is the first of three—the other two volumes being [Wu2020b] and [Wu2020c]—that give a systematic and grade-level-appropriate exposition of the mathematics of grades 9– 12 (excluding probability1 and statistics), together with some essential background information about rational numbers. This is the mathematical content knowledge that we believe, as of 2020, all high school mathematics teachers need for their teaching and all mathematics educators2 interested in high school mathematics need for their research. The previous three volumes—the volume on the mathematics of grades K–6 ([Wu2011a]) and the two volumes on the mathematics of grades 6–8 ([Wu2016a] and [Wu2016b])—have already been published. We hope that these six volumes will serve the dual purpose of revamping the mathematical education in the universities of pre-service mathematics teachers and future mathematics educators on the one hand, and on the other, offering textbook publishers a detailed blueprint on how to introduce mathematics into school textbooks that is 1 There 2 We

is, however, an exposition of finite probability in Section 1.10 of [Wu2016a]. use the term "mathematics educators" to refer to university faculty in schools of

education. xi

xii

PREFACE

both correct and learnable. These six volumes will also shore up the critical mathematical backgrounds of supervisors of mathematics and mathematics professional developers. There has been no lack of books on all or parts of school mathematics—the mathematics of K–12—in the education literature. We have chosen to add another 2,500 pages (the approximate total length of these six volumes) to the already voluminous literature because we believe these volumes provide a first attempt at solving two of the central problems in mathematics education: whether school mathematics can be made to respect the integrity of mathematics and how much mathematics a mathematics teacher or a mathematics educator needs to know. School mathematics that respects mathematical integrity We will address the former problem first. These six volumes give a detailed confirmation of the fact that school mathematics—while maintaining its fidelity to the progression of the standard school mathematics curriculum from kindergarten to grade 12—can be made to respect the integrity of mathematics. Such a confirmation has been a long time coming. In the following pages, we will explain what mathematical integrity is and why it is important to have an exposition of school mathematics that respects mathematical integrity.3 At first glance, it seems absurd that there would be any need to discuss whether school mathematics respects mathematical integrity. Is not school mathematics, by its very name, part of mathematics and, as such, does it not follow that school mathematics carries the integrity inherent in the subject? This is a misconception about school mathematics that we must confront without delay. School mathematics is in fact not part of mathematics if mathematics is understood to be what working mathematicians do or what is taught to math majors in college mathematics departments. Rather, school mathematics is an engineered version of mathematics—in the sense of mathematical engineering introduced in [Wu2006]—in the same way that civil engineering is an engineered version of Newtonian mechanics. Mathematical engineering customizes the abstractions of mathematics for consumption by K–12 students. For example, a fraction in mathematics is a straightforward concept: it is an element of the quotient field of the integral domain of integers. Fortunately, no one suggests that we tell this to ten-year-olds. Mathematical engineering intervenes at this point to recast the concept of fractions so that fractions can be understood by elementary students (see [Wu1998]). There are many such examples all through the K–12 curriculum, e.g., negative numbers, slope of a line, geometric measurements (length, area, and volume), congruence, similarity, exponential functions, logarithms, axioms of plane geometry, etc. The engineering that is needed to make these abstract concepts learnable by school students is therefore substantial at times. Now there is good engineering, but there is also bad engineering, and the question is whether good mathematical engineering has been put in the service of school mathematics. Unhappily, the answer is not always. In fact, school

3 It would be legitimate to also inquire why it has taken so long for someone to try to meet this obvious need.

PREFACE

xiii

mathematics and mathematical integrity parted ways at least five decades ago, and our schools have been plagued by products of very bad mathematical engineering ever since. Before proceeding further, we first explain what mathematical integrity is because this concept is coming into focus,. We say a mathematical exposition has mathematical integrity if it embodies the following five qualities: (a) Definitions: Every concept is clearly and precisely defined so that there is no ambiguity about what is being discussed. (See the quote from Gibson at the beginning of this preface.) (b) Precision: All statements are precise, especially the hypotheses that guarantee the validity of a mathematical assertion, the reasoning in a proof, and the conclusions that follow from a set of hypotheses. (c) Reasoning: All statements4 other than the unavoidable basic assumptions are supported by reasoning.5 (d) Coherence: The basic concepts and skills are logically interwoven to form a single fabric, and the interconnections among them are consistently revealed. (e) Purposefulness: The mathematical purpose behind every concept and skill is clearly brought out so as to leave no doubt about why it is where it is. These we call the Fundamental Principles of Mathematics. A fuller discussion of these principles will be found on pp. xxviii–xxxiii in the To the Instructor section on pp. xxvii ff. below, but two things need to be said right away. First, the role of definitions in school mathematics has been misunderstood, and misrepresented, in the education literature thus far, so that—to educators—the emphasis on definitions may seem to be misplaced. One will find a more balanced presentation about definitions on pp. xxix–xxx. Next, there is no difference between reasoning and proof in a mathematical context, and what is generally called problem solving in the education literature is part of what is known as theorem proving in mathematics.6 Overall, it should not be difficult to see—and these three volumes will bear witness to this fact—that these five fundamental principles are what make mathematics transparent, in the sense that everything is on the table and no guesswork or privileged knowledge is needed for its decoding. They are also the qualities that make mathematics accessible to all students and learnable by all students. If we want mathematics learning to take place in schools, it is incumbent on us to teach school mathematics that is consistent with these fundamental principles. But to return to the discussion of school mathematics of the past five decades, we have to begin by asking what is school mathematics? This is in fact the question that these six volumes ([Wu2011a], [Wu2016a], . . . , [Wu2020c]) attempt to answer, but short of that, we will have to say school mathematics is the common content of most of the mathematics textbooks in K–12 and most of the college textbooks aimed at the professional development of mathematics teachers and mathematics educators (compare the review of school textbooks in Appendix B of Chapter 3 4 With

the exception of a few standard ones such as the fundamental theorem of algebra. reasoning supports even those assumptions because there are reasons why we want to assume them. 6 Compare the discussion on pp. xxxvi-xxxvii. 5 Intuitively,

xiv

PREFACE

in [NMAP2]). If this strikes readers as too vague, they will be relieved to know that there is in fact an amazing consistency among these textbooks.7 For example, a fraction is thought of as a piece of pizza or a part-of-a-whole, although neither conveys the message to students that a fraction is a number that they have to use for extensive computations. Consequently, with such a "definition" of a fraction, the arithmetic operations on fractions cannot be defined and their computational algorithms cannot be proved.8 For another example, the concept of the slope of a line in the coordinate plane is defined in most of these textbooks by taking two preassigned points on the line to form the rise-over-run. But why is this rise-over-run equal to the rise-over-run with respect to another pair of points on the same line? Almost all textbooks insinuate that this equality is obviously true and not worth fussing about. And so on. In general, school mathematics, as defined collectively by these textbooks, is antithetical to mathematical integrity in that it lacks clarity (due to a general absence of definitions and a pervasive lack of precision in its articulation), mostly asks for rote memorization as its default mode of learning (due to the pervasive absence of reasoning), is incoherent (due to its neglect of the inherent logical structure of mathematics), and traverses the curriculum in a listless and pro forma manner (due to its failure to recognize the mathematical purpose behind each topic). We call the content of these standard school mathematics textbooks TSM (Textbook School Mathematics).9 TSM is recognized, consciously or subconsciously, by teachers and educators to be unlearnable, and it is this unlearnability that emboldens countless sensible adults to proclaim, often with pride, "I am not good in math!" There is a far more pernicious fallout from TSM, however, and it is the effect TSM has on mathematics teachers and educators. These teachers and educators have learned only TSM in K–12, but as of 2020, institutions of higher learning do not provide courses to help future teachers and educators to replace their knowledge of TSM with school mathematics with mathematical integrity. Consequently, all that most teachers can do when they go back to teach in K–12 is trot out the TSM they are familiar with, and all that most educators can do when they begin their research is to fall back on the TSM they were taught. So the next generation also learns TSM, and this is the vicious cycle that has rendered school mathematics synonymous with TSM for at least the past five decades. Most educators may have suspected that there must be more to school mathematics than TSM, but without access to an exposition of school mathematics with mathematical integrity, their suspicion remains just that, a suspicion. Back in 1985, Lee Shulman lamented in his well-known address to the AERA about "the absence of focus on subject matter among the various research paradigms for the study of teaching" ([Shulman, page 6]). Shulman was talking about all disciplines, but from the standpoint of these six volumes ([Wu2011a], [Wu2016a], . . . , [Wu2020c]), we gain a clear perspective on how this neglect of the subject matter may have come about in mathematics. We speculate that mathematics educators 7 When I first had the opportunity to sample a wide range of the available K–12 textbooks for the first time around the year 2000, I was convinced that the publishers were in collusion and simply agreed to copy each other. 8 Remember: "Ours is not to reason why. Just invert and multiply." 9 For a more extended discussion of TSM, see To the Instructor on pp. xxvii ff. as well as [Wu2014] and [Wu2018a]. Note that TSM provides a new window into the phenomenon known as math phobia.

PREFACE

xv

may have chosen not to pay any attention to mathematical content because, since school mathematics was apparently nothing more than TSM, they saw nothing in the subject matter of school mathematics worthy of their serious attention. To change mathematics educators’ perception of the subject matter in mathematics, we have to give them access to a fully detailed exposition of school mathematics with mathematical integrity. The omnipresence of TSM in the last half-century created the unmistakable impression that perhaps at least some of the travesties in TSM are necessarily endemic to school mathematics. Under the circumstances, it was not easy to imagine that school mathematics might have anything to do with mathematical integrity. But two things happened around 1990. In 1989, NCTM (National Council of Teachers of Mathematics) launched its school mathematics education reform by proclaiming that school mathematics could be made to respect mathematical integrity. Without a detailed exposition of school mathematics that respects mathematical integrity to back up its claim, NCTM was of course going out on a limb. Then in 1994, Alan Schoenfeld made a scholarly statement with the clear implication that, while a school mathematics curriculum with mathematical integrity was certainly possible, we did not have it yet. What he wrote was, "Proof is not a thing separable from mathematics, as it appears to be in our curricula . . . . And I believe it can be imbedded in our curricula, at all levels" ([Schoenfeld1994, page 76]). Schoenfeld’s statement was prompted in part by the debates surrounding the NCTM reform. Note that beyond affirming his belief in the fundamental article of faith underlying the NCTM reform, he stated openly that, indeed, this article of faith had not yet been confirmed. We will return to Schoenfeld’s statement below, but before proceeding further, we will make a few comments about the NCTM reform. The foundational documents of the NCTM reform are the two sets of standards: the 1989 [NCTM1989] and the 2000 [PSSM]. Although NCTM did not have an explicit recognition of the concept of TSM, the 1989 reform was undoubtedly a revolt against the stranglehold of TSM on school mathematics education. NCTM declared in essence that mathematical integrity must be part of school mathematics. For example, [NCTM1989] states that one of the reform’s goals is that students "become mathematically literate" (page 6). [PSSM] states that "a mathematics curriculum should be coherent" (page 15), "should focus on important mathematics" (page 15), and "reasoning and proof should be a consistent part of students’ mathematical experience in prekindergarten through grade 12" (page 56). With the hindsight of thirty years, we can see all too clearly the obstacles that confronted the NCTM reform. With students, teachers, and educators completely immersed in TSM, the clarion call for coherence, reasoning, and proof might as well have been stated in a foreign language. Most of them had no conception of what those words meant. We have to remember that, for example, what little "proof" TSM has to offer resides only in the course in high school geometry, and even there, proofs are mainly taught by rote (see [Schoenfeld1988]). Back in 1989, there was no detailed point-bypoint exposition of school mathematics that could provide a roadmap to show how mathematical integrity can be introduced into school mathematics. There were no school mathematics textbooks to replace the TSM-infested ones.10 Most fatally, NCTM made no commitment to a massive and long-term professional development 10 In the year 2020, most of us can calmly look back and see that the reform curricula that were published post-1989 were essentially different incarnations of TSM.

xvi

PREFACE

program to explain to teachers what mathematical integrity in their daily teaching could look like.11 With these three strikes against the NCTM reform before it stepped up to the plate, the reform faced an insurmountable credibility crisis. Its visionary declaration about what school mathematics education could be and ought to be almost instantly became nothing more than appealing rhetoric. The need for a detailed exposition of school mathematics with mathematical integrity could not have been more urgent. There is another way that having a detailed exposition of school mathematics with mathematical integrity would have helped with the reform. Both [NCTM1989] and [PSSM] did try to provide some mathematical details about the curriculum they envisioned and, in so doing, made some missteps. For example, page 96 of [NCTM1989] suggests that the addition of fractions—inscrutable as it is in TSM— has to be approached gingerly, and neither [NCTM1989] nor [PSSM] points out the profound error of using the least common denominator for the addition of fractions (see page 41 below for an explanation of this error). Students’ difficulty with the multiplication and division of fractions is duly noted in [NCTM1989] and [PSSM], but again there is no substantive suggestion on how to get them out of the predicament. Either document could have pointed to the need for a proof of the area formula for a rectangle with fractional sides; such a proof would add immeasurably to students’ knowledge and confidence in reasoning and proof about fraction multiplication and about the concept of area (see pp. 49ff. below for such a proof). But the fact remains that neither did. One suggestion on the division of fractions is made on page 219 of [PSSM], but it confuses the division of fractions with the concept of division-with-remainder (see Section 7.2 of [Wu2011a] for a careful discussion of the latter). The difficulties of the concept of slope for teachers (and students) are notorious (see, e.g., page 126 of [Stump] or [Newton-Poon2]), but neither [NCTM1989] nor [PSSM] seems to recognize that the concept of slope as it is known in TSM is not properly defined and therefore a new approach is called for (see pp. 338–354 in this volume), and so on. These and many other missteps could have been avoided had a detailed exposition of school mathematics with mathematical integrity been available. Some twenty years later, 2010 saw the release of CCSSM, the Common Core State Standards for Mathematics ([CCSSM]). CCSSM calls for a focused and coherent curriculum that stresses both conceptual understanding and procedural fluency (pp. 3–4 and 6 of [CCSSM]). In addition, it also asks for precision and clear definitions (page 7, loc. cit.). The most pronounced difference between CCSSM and the NCTM reform lies in the specificity of the standards in CCSSM: they are much more explicit in specifying the progression of mathematical topics through the grades and, even more importantly, in steering the curriculum away (most of the time) from the defective practices abounding in TSM. Because of the latter, most of the standards in CCSSM look different from the traditional standards (including those briefly sketched out in the NCTM documents [NCTM1989] and [PSSM]). This is especially true for the standards on fractions, finite decimals, rational numbers (along the lines of Chapters 1 and 2 in [Wu2016a], similar to Chapters 1 and 2 in this volume), part of beginning algebra (along the lines of Chapters 1 and 4 11 The absence of this commitment was no accident. To carry out this kind of professional development on a large scale, the need for something like these six volumes ([Wu2011a], . . . , [Wu2020c]) to serve as a guide would be absolute.

PREFACE

xvii

in [Wu2016b], similar to Chapter 6 of this volume), and middle and high school geometry (along the lines of Chapters 4 and 5 of [Wu2016a], similar to Chapters 4 and 5 in this volume). However, in the apparent absence of a detailed account of what a CCSSM curriculum would look like,12 the specificity of the curricular deviations in CCSSM turns out to be more of a political liability than an asset. Many people immediately put CCSSM and the NCTM reform on the same footing. Their perception was that these two movements represented what happens when a bunch of wannabes pontificate about school mathematics education without knowing what they are talking about. A conspicuous example they cited is CCSSM’s approach to the geometry curriculum in middle school and high school using reflections, rotations, and translations as the basic building blocks. Such a change is necessitated by the inherently flawed TSM geometry curriculum based on an uninformed interpretation of the work of Euclid some twenty-three centuries ago (see pp. 157–164 below for a more detailed explanation, and see Chapter 8 of [Wu2020b] for a comprehensive one). Instead, CCSSM calls for a nuanced two-step process to introduce reflections, rotations, and translations as the foundational building blocks of the school geometry curriculum. Standards 8.G on page 55 of [CCSSM] describe how, in grade 8, these transformations can be used informally (but correctly) in heuristic arguments to develop students’ intuition about transformations (as detailed in Chapters 4 and 5 of [Wu2016a]). Then in high school, these transformations are precisely defined to be used for formal proofs (as in Chapters 4 and 5 of this volume and Chapters 6 and 7 of [Wu2020b]). But not having such details available back in 2010, many critics, educators, and teachers immediately predicted the impending doom of this effort by CCSSM by citing the failures of putative similar experiments in other nations. Moreover, they also predicted (not entirely incorrectly) the almost certain confusion among teachers who would try to implement this new curriculum, and they regarded as inevitable the disappearance of proofs from CCSSM high school geometry. In the absence of a detailed exposition that shows how to navigate and implement the Common Core geometry standards with mathematical integrity,13 such misunderstanding led inevitably to harsh criticism (see, e.g., [Milgram-Wurman, pp. 4–5] and [Phelps-Milgram, page 10 and footnote 15 on page 41]). So CCSSM ends up facing the same wide credibility gap that plagued the NCTM reform twenty years ago. It did not help that the CCSSM agenda also left out the critical component of professional development for teachers, thereby creating the same sense of bewilderment in classrooms across the land (see [Education Week], [Loewus1], [Loewus2], and [Sawchuk]). It would seem that CCSSM is repeating the same mistake as the NCTM reform by not taking seriously the need to offer sustained, large-scale professional development for teachers to help with its implementation. With the 12 The situation surrounding the release of [CCSSM] is complicated. A detailed exposition of the part of the CCSSM curriculum mentioned above—fractions, finite decimals, rational numbers, parts of algebra, and middle school geometry—in the form of [Wu2010a] and [Wu2010b] in fact predated CCSSM (they were drafts for [Wu2016a] and [Wu2016b], respectively). However, the existence of these documents was not made widely known. 13 Note, however, that many curricular details on geometry were soon provided in [Wu2012], [Wu2013], [Wu2016a], and [Wu2016b], but they were not made widely known. A detailed CCSSMaligned high school geometry curriculum, in existence since 2007, will appear in the second volume of this three-volume set, [Wu2020b].

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publication of these three volumes, at least one complete exposition of school mathematics with mathematical integrity—an exposition that is also consistent in the main with CCSSM—will be available to provide the needed guidance for this kind of professional development, but will these volumes be too little too late? Only time will tell. The recent drive towards mathematical integrity It should be abundantly clear from the foregoing discussion that any real improvement in school mathematics education requires us to rethink the mathematical education of teachers and educators. In particular, the destructive presence of TSM can no longer be ignored. These six volumes have been written with the express intent of encouraging and supporting such rethinking. In the last few years, several books have made a concerted effort to promote the introduction of mathematical integrity into school mathematics, e.g., [MET2], [NCTM2009], [MUST], and the sixteen volumes in the NCTM series Developing Essential Understanding (e.g., [Ellis-Bieda-Knuth]). In a book entitled We Reason & We Prove for All Mathematics, Arbaugh et al. respond directly to Schoenfeld’s belief in the possibility of imbedding proofs in all levels of K–12 (quoted on page xv) by flatly stating that, in their volume, they "will provide guidance about how to make reasoning-and-proving a reality in your classroom" ([Arbaugh et al., page x]). These developments are welcome because their willingness to directly address the content of K–12 mathematics represents a giant step forward in school mathematics education at a time when many are still clinging to the idea that integrating fun, engaging activities into the classroom—while leaving TSM intact—is the way to improve school mathematics education. Nevertheless, we must also add a word of caution at this juncture of school mathematics education concerning the effectiveness of "providing guidance" in small doses, quite apart from the quality of the guidance itself. As of 2020, we have to face the unpleasant truth that, because of the longstanding malfeasance of the education establishment, most people in school mathematics education have been immersed in TSM, and only TSM, for their entire lives. Consequently, most end up being deficient in a detailed knowledge of the inner workings of mathematics on the one hand and in a coherent view of mathematics as a whole on the other. An example of the former is the chronic failure to recognize that, without precise definitions, correct reasoning (= proof) is unattainable. Another example of the same is the fact that a proof must not be confused with a heuristic argument, no matter how attractive that heuristic argument may be. Examples of the lack of a global, coherent view of mathematics abound in TSM, but we will limit our discussion to only three of them. The first is the lack of awareness of the overall hierarchical structure of mathematics; e.g., in order to move forward mathematically in a mathematical development, one may only use results already proved earlier. There is no better illustration of this lack than the "proof" in TSM of equivalent fractions using fraction multiplication14 —one that is universally taught in TSM. Such a "proof" should be recognized for what it is: totally anti-mathematical. Here, the details are, step by step, impeccable, but the flagrant mathematical error lies in using a fact—about fraction multiplication that 14 This

is the reasoning that

2 3

=

2×4 3×4

because

2 3

=

2 3

×1=

2 3

×

4 4

=

2×4 . 3×4

PREFACE

xix

can only be proved later in the development of fractions—to justify a foundational result about fractions that is needed almost as soon as a fraction is defined. (See pp. 270–271 in [Wu2011a] for further discussions of this error.) A second example is the role of congruence in school mathematics. In TSM, the concept of the congruence of two arbitrary figures is not well-defined, and only the congruence of triangles is used for proofs in high school geometry (ASA, SAS, SSS, etc.). Moreover, congruence seems to have little relevance to daily life. TSM does not mention that, without the fundamental assumption that lengths, areas, and volumes remain the same for congruent geometric figures, it is impossible to derive any area or volume formulas (in particular, not even the area formulas for parallelograms and triangles). This realization makes it imperative that, in teaching the area formula for a triangle in grade 6 or 7 (for example), teachers make an effort to bring out the important role that the concept of congruence plays in geometric measurements (see Section 5.3 in [Wu2016b]). The same realization also impacts the geometry curriculum in high school: the TSM treatment of congruence as the "congruence of triangles" à la Euclid will have to be upgraded so that it can make sense of the "congruence of any two geometric figures" (see the Overview of Chapters 4 and 5 on pp. 157ff.). Such an upgrade is needed, for example, for the study of quadratic functions (see Section 2.1 of [Wu2020b]). This glaring defect in the TSM treatment of a foundational concept like congruence is in fact one of the main reasons necessitating the overhaul of the TSM geometry curriculum in middle and high school (see Chapters 4 and 5 of [Wu2016a], Chapters 4 and 5 in this volume, and Chapters 6 and 7 of [Wu2020b]). This kind of longitudinal coherence of the school mathematics curriculum, so vital for students’ mathematics learning and on such a detailed level, is unlikely to be brought up in the context of providing general guidance piecemeal. A final example of the lack of a global, coherent view of mathematics in TSM is the transition in the middle school curriculum √ from rational numbers to real numbers due to the emergence of numbers such as 2, π, etc. TSM makes believe that the introduction of irrational numbers and their arithmetic operations into the school curriculum can be done surreptitiously and informally, without any explicit mathematical discussion. The resulting mathematical errors and their consequences for teachers and educators are profound. See the example on page xxxi below, among many Also see the discussion of the incorrectness of the √ examples. √ √ such equality 2 + 3 = 5 on pp. 207ff. of [MUST] which makes no mention of the fact that the arithmetic operations on irrational numbers are never given a serious and explicit discussion with ninth graders. What is needed for the purpose of helping students make this transition is something like the FASM (Fundamental Assumption of School Mathematics) stated on page 133, but unhappily, nothing like FASM has ever appeared in TSM. To address such deficiencies at both ends of the school mathematics spectrum, it would be reasonable to argue that a systematic exposure of teachers and mathematics educators to a complete exposition of the mathematics—one that honors mathematical integrity—over several grades at the very least15 will be the only effective cure (see [NMAP1, Recommendation 19 on page xxi]).

15 Teachers need to know where their students come from and where they are headed, curriculumwise.

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We have just given a partial explanation of why these six volumes (this volume, together with [Wu2011a], [Wu2016a], [Wu2016b], [Wu2020b], and [Wu2020c]) require 2,500 pages of detailed mathematical discussions to confirm the fact that school mathematics can be made to respect mathematical integrity. Because of the corrosive effects of TSM that have pervaded and degraded school mathematics for so long, we are obliged to rebuild school mathematics from the ground up. In these six volumes, we take nothing for granted. For example, we pay special attention to the need for correct definitions as the basis for reasoning and proofs; we want to drive home the point that once a definition of a concept (such as a fraction) is given, then every subsequent assertion about this concept has to be based on the definition, and on the definition alone. Every statement in these volumes, from whole numbers to calculus, is carefully proved.16 The intended goal of this effort is to clarify, cumulatively, the mathematical meaning of the declarative statement, "A implies B", as a purely deductive process that begins with the hypothesis A and arrives at the conclusion B. This is in contrast with the common practice in TSM of "explaining" something by telling a story, by drawing an analogy, or by offering an attractive pattern or heuristic argument. These six volumes take an entirely different tack: they show, consistently, how to verify "A implies B" in mathematics by moving from A to B on the basis of definitions, explicit assumptions, or theorems with the help of logic. These volumes do so—we emphasize—from the first page to the last because we believe that the way to teach is not to pontificate but to lead by example. This process of acculturating teachers (and ultimately their students) to reasoning and proof does not have to be rigid or formal, especially in the early grades (see, e.g., Sections 4.2 or Section 6.2 of [Wu2011a]), but the essential elements of logical deduction must be put in place and maintained ab initio to preserve the integrity of mathematics. We also go into extensive detail about such seemingly pedestrian topics as the proper use of symbols (Sections 6.1 and 6.2 on pp. 298ff.), the meaning of an equation, and what it means to solve an equation (see pp. 322–324), with the hope that the long years of obfuscation in TSM with such jargon as "variables" and "symbolic manipulations for solving an equation" will be brought to a merciful end. We hope that the foregoing discussion has made the case for the critical need for a thorough-going exposition of school mathematics with mathematical integrity. Incidentally, the only reason we have made repeated references in this whole discussion to the same six volumes by the present author is that there is no comparable exposition at the moment. It is in fact our hope that the publication of these six volumes will encourage others to come up with their own ways of replacing TSM across K–12 with a development of school mathematics that respects mathematical integrity. How much mathematics teachers need to know Knowing what school mathematics with mathematical integrity looks like enables us to face up to the second problem in mathematics education that was mentioned on page xii: how much mathematics a mathematics teacher or a mathematics educator needs to know. For teachers, this problem has a long history; see, 16 With the usual disclaimer that there are a very few theorems that we must intentionally assume without proof.

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e.g., [Ball], [Ball-McDiarmid], [Begle1972], [Goldhaber-Brewer], and [Monk]. We can speculate that, because the school curriculum has been dominated by TSM for so long and the flaws in TSM are so pronounced and extensive,17 mathematics educators were reluctant to prescribe the content knowledge teachers need in terms of TSM on the one hand, and they were uncertain about what to prescribe on the other. After all, there was simply no available exposition of school mathematics with mathematical integrity. Now that these six volumes are available, it is possible to make a first attempt at describing the minimum knowledge that teachers and educators in elementary, middle, and high school, respectively, need to be effective in their work (again, see Recommendation 19 on p. xxi of [NMAP1]). Those in elementary school mathematics education18 should know the equivalent of [Wu2011a] minus Chapters 23, 31, 37, 41, and 42; they should also have some acquaintance with the equivalent of Chapters 4 and 5 of [Wu2016a] and Chapters 1 and 2 of [Wu2016b]. Those in middle school mathematics education should know the equivalent of [Wu2016a] and [Wu2016b] and have some acquaintance with the equivalent of Part 1 of [Wu2011a] and Chapters 4 and 5 of this volume. Those in high school mathematics education should know the equivalent of this volume, [Wu2020b], and [Wu2020c]. In addition, because pre-service teachers and educators interested in high school mathematics are typically math majors in college, they are expected to know something about linear algebra, i.e., vector spaces and matrices. Those who intend to teach calculus or do research on the teaching of calculus should also know something about Taylor’s theorem and the Taylor series expansions of standard elementary functions such as sine, cosine, exponential function, and logarithm; they should also know some multi-variable calculus. Now consider the teaching and learning of fractions and (finite) decimals. While education researchers of the past five decades were no doubt aware of the simple treatment of fractions in abstract algebra, their uncritical acceptance of TSM misled them into believing that, for elementary school students, one can do no better than teaching fractions as pieces of pizzas or some variation thereof. Consequently, they focussed their research on the teaching and learning of fractions, for the most part, on tweaking the TSM model of fractions-as-pizzas—with no thought given to helping students learn about fractions as numbers or learn to reason their way through the arithmetic of fractions.19 As a result, education research on fractions has focussed on increasing children’s experiential and informal familiarity with fractions 17 Regardless

of the fact that the term TSM was coined only in 2011. strongly believe that the mathematics of elementary school should be taught by mathematics teachers. See [Wu2009]. 19 Unhappily, TSM also claims some professional mathematicians among its victims: these mathematicians have come to believe that teaching fractions in schools can lead to nothing more than "confusion and memorization". See, for example, [DeTurck]. 18 We

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based on the pizza model rather than on increasing children’s mathematical knowledge of fractions based on a correct definition of a fraction. If it had tried to do the latter, it would have rejected the absurd pizza model from the outset (see, e.g., pp. 33–35 of [Wu2008] for a brief discussion of the relevant literature). The same body of education research has also tried to make sense—unsuccessfully of course— of other anti-mathematical practices, such as treating decimals as a different kind of number, adding and subtracting fractions using the least common denominator, or teaching the multiplication and division of fractions without precise definitions. Only recently have researchers become aware of a more reasonable foundation for fractions (initiated in [Wu1998] and expanded in [Wu2011a]; abbreviated versions are given in Chapter 1 of [Wu2016a] and Chapter 1 of this volume)20 that puts the study of fractions on the number line, emphasizes the concept of a fraction as a number for arithmetic computations, and makes sense of (finite) decimals as a special collection of fractions. There is still some distance to go in this direction, such as honoring the definition of a fraction by using it in every situation, e.g., for multiplication, for division, for understanding ratios, etc. We eagerly look forward to a change along these lines in the education research on fractions and decimals in the years to come (cf. [Siegler et al.]). School textbooks Better school mathematics education requires not only more effective teachers but also textbooks that contain only school mathematics with mathematical integrity. Our discussion thus far has been all about getting more effective teachers but nothing about getting better textbooks. This is not because we believe textbooks are less important, but since most school textbooks are published by the major publishers, there is little that people in academia can do to convince publishers to abandon their bottom-line mentality and write better textbooks (cf. [Keeghan]). However, there are now several online curricula written more or less in accordance with CCSSM and, according to some reports, a few seem to be showing promise.21 As we said at the beginning, we hope these six volumes under discussion can serve as a blueprint for better school textbooks. But let us add a few caveats in this regard. First of all, these six volumes are certainly not student textbooks: they are written specifically for adults (teachers and educators, maybe some curious parents). Nevertheless, their mathematical content has been carefully customized (i.e., engineered) for use in the appropriate grades, at least as far as the mathematical level of sophistication is concerned, so that after some straightforward pedagogical modifications and embellishments, they can be expanded into student textbooks. An example of how such an expansion may be realized will appear before long, we hope, in the form of a student textbook for grade 8 that will be posted on the author’s homepage, https://math.berkeley.edu/~wu/. At the very least, we believe these six volumes taken together can serve as a detailed guide for textbook 20 This approach to fractions and decimals—as presented in [Wu2016a]—served as a blueprint for the fractions and decimals standards of [CCSSM]. Because this volume is for consumption by high school teachers and mathematics educators, what is in Chapter 1 is more brief and slightly more sophisticated than its counterparts in [Wu2011a] and [Wu2016a]. 21 It is uncertain whether any of the textbook evaluation agencies is aware of the importance of having mathematical integrity in mathematics textbooks.

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publishers on how to write school mathematics textbooks across K–12 that respect both the standard curricular sequence and mathematical integrity. For this purpose, textbook writers should take note that there are several major departures from the standard school curriculum in this volume and [Wu2020b] and [Wu2020c]. Briefly, they are the following: (1) The conversion of fractions to infinite decimals and geometric measurements (length, area, and volume) are two topics typically taught in middle school, but in these volumes they appear in the third volume, [Wu2020c], after the introduction of limits (see Chapters 3–5 of [Wu2020c]). Fortunately, the procedural aspect of the conversion of fractions to decimals is addressed (and partially explained) in Section 1.5 (pp. 54ff.) of this volume, and there is an intuitive discussion of geometric measurements in Chapter 5 of [Wu2016a] which is actually adequate for (a somewhat superficial) use in a high school classroom. The main reason for these two departures is that it is impossible to make sense of infinite decimals and geometric measurements without the use of limits. Our teachers’ and educators’ critical need for some real understanding of the subtleties of both topics accounts for this departure from the norm. In any case, any adaptation of Chapters 3-5 of [Wu2020c] for student textbooks will require selective omissions. (2) The presentation of high school geometry in these volumes deviates from the traditional one. The concept of congruence is defined in terms of the tangible, accessible concepts of reflections, rotations, and translations in the plane, and similarity is defined in terms of congruence and the equally tangible and accessible concept of dilation. A detailed explanation is given in the Overview of Chapters 4 and 5 on pp. 157ff. as well as in Section 4.7 on pp. 252ff. and Chapter 8 of the second volume, [Wu2020b]. Because CCSSM has since adopted this approach to middle and high school geometry, no defense of this deviation will be necessary here. (3) These three volumes propose a different progression of geometry from middle school to high school, as follows. In grade 8, teach enough informal geometry to get to the concept of similar triangles, the angle-angle similarity criterion, and the proof of the Pythagorean theorem before embarking on introductory algebra in high school. Then in the high school geometry course, revisit the topic of similar triangles, but this time from a more formal standpoint. Again, see the Overview of Chapters 4 and 5 on pp. 157ff. for an explanation. (This departure from the standard sequencing has also been adopted by CCSSM.) The presentation of the curricular shift described in (3) will be given in Chapters 4 and 5 of this volume, but with a mild twist. Because the informal geometry

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(proposed for grade 8) has already been treated in detail in [Wu2016a], the geometry in Chapters 4 and 5 of the present volume will be the formal high school counterpart of the informal geometry in [Wu2016a]. The exposition of the main body of plane geometry (geometry of the triangle and the circle along with constructions with ruler and compass) then resumes in Chapters 6 and 7 of the second volume, [Wu2020b], after we have finished discussing the standard topics of secondyear algebra. Final thoughts We call special attention to the fact that the third and last of these three volumes, [Wu2020c], is essentially an introduction to mathematical analysis, customized specifically for consumption by prospective mathematics teachers and educators.22 It is likely that this material will also benefit beginning math majors in college. We should also address an obvious question that probably has been on readers’ minds all along; namely, why does this volume on high school mathematics begin with the middle school topics of fractions and rational numbers? Nothing need be said about the obvious relevance of these topics to mathematics educators, but we owe high school teachers an explanation of why we consider these topics to be an integral part of their content knowledge. It is a fact—though hidden in TSM—that rational numbers, rather than real numbers, are the backbone of the mathematics in grades 5–12. Unfortunately, because of TSM, students in all grades seem to have trouble with fractions and, consequently, with rational numbers. Given the hierarchical structure of mathematics, it is not surprising that students’ inability to learn algebra can often be traced back to their weakness in the foundational subjects of fractions and rational numbers. This was pointed out in the National Mathematics Advisory Panel Report (see page 18 of [NMAP1]). Indeed, the story has been told many times that even students in honors sections of Algebra 2 plead with their teachers to give them instructions on fractions. So, to be effective in teaching the standard topics of high school mathematics, high school teachers must have a TSM-free working knowledge of fractions and rational numbers as well. A final reflection: Earlier, we quoted Lee Shulman’s lament about "the absence of focus on subject matter among the various research paradigms for the study of teaching" (see page xiv). These six volumes have now redefined the meaning of this subject matter for school mathematics. We hope mathematics educators will discover through these volumes that the mathematics underlying school mathematics, when presented correctly, is no longer meaningless like TSM and is worthy of their best efforts to learn it. Moreover, the subject matter, thus redefined, will have repercussions on "the study of teaching". As school mathematics becomes more learnable by all students, and therefore more teachable by all teachers, pedagogy will have to focus—not on how to render the incomprehensible23 palatable—but on how to facilitate the normal process of learning so that all students can learn how to reason critically and correctly.

22 Here as well as elsewhere in these three volumes, we are engaging in serious mathematical engineering in the sense of [Wu2006]. 23 That is, TSM.

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But for all that, it will be necessary to first make school mathematics that respects mathematical integrity an integral part of mathematics education research. This then harks back to Lee Shulman’s lament. It is our belief and our hope that school mathematics education will improve when mathematics education research begins to address, not TSM, but school mathematics with mathematical integrity. Acknowledgements The drafts of this volume and its companion volumes, [Wu2020b] and [Wu2020c], have been used since 2006 in the mathematics department at the University of California at Berkeley as textbooks for a three-semester sequence of courses, Math 151–153, that was created for pre-service high school teachers.24 The two people most responsible for making these courses a reality were the two chairs of the mathematics department in those early years: Calvin Moore and Ted Slaman. I am immensely indebted to them for their support. I should not fail to mention that, at one point, Ted volunteered to teach an extra course for me in order to free me up for the writing of an early draft of these volumes. Would that all of us had chairs like him! Mark Richards, then Dean of Physical Sciences, was also behind these courses from the beginning. His support not only meant a lot to me personally, but I suspect that it also had something to do with the survival of these courses in a research-oriented department. It is manifestly impossible to write three volumes of teaching materials without generous help from students and friends in the form of corrections and suggestions for improvement. I have been fortunate in this regard, and I want to thank them all for their critical contributions: Richard Askey,25 David Ebin, Emiliano Gómez, Larry Francis, Ole Hald, Sunil Koswatta, Bob LeBoeuf, Gowri Meda, Clinton Rempel, Ken Ribet, Shari Lind Scott, Angelo Segalla, and Kelli Talaska. Dick Askey’s name will be mentioned in several places in these volumes, but I have benefitted from his judgment much more than what those explicit citations would seem to indicate. I especially appreciate the fact that he shared my belief early on in the corrosive effect of TSM on school mathematics education. David Ebin and Angelo Segalla taught from these volumes at SUNY Stony Brook and CSU Long Beach, respectively, and I am grateful to them for their invaluable input from the trenches. I must also thank Emiliano Gómez, who has taught these courses more times than anybody else with the exception of Ole Hald. Some of his deceptively simple comments have led to much soul-searching and extensive corrections. Bob LeBoeuf put up with my last-minute requests for help, and he showed what real dedication to a cause is all about. Section 1.9 of the third volume ([Wu2020c]) on the importance of sine and cosine could not have been written without special help from Professors Thomas Kailath and Julius O. Smith III of Stanford University, as well as from my longtime collaborator Robert Greene of UCLA. I am grateful to them for their uncommon courtesy.

24 Since the fall of 2018, this three-semester sequence has been pared down to a two-semester sequence. A partial study of the effects of these courses on pre-service teachers can be found in [Newton-Poon1]. 25 Sadly, Dick passed away on October 9, 2019.

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Last but not least, I have to single out two names for my special expression of gratitude. Larry Francis has been my editor for many years, and he has pored over every single draft of these manuscripts with the same meticulous care from the first word to the last. I want to take this opportunity to thank him for the invaluable help he has consistently provided me. Ole Hald took it upon himself to teach the whole Math 151–153 sequence—without a break—several times to help me improve these volumes. That he did, in more ways than I can count. His numerous corrections and suggestions, big and small, all throughout the last nine years have led to many dramatic improvements. My indebtedness to him is too great to be expressed in words. Hung-Hsi Wu Berkeley, California February 2, 2020

To the Instructor These three volumes (the other two being [Wu2020b] and [Wu2020c]) have been written expressly for high school mathematics teachers and mathematics educators.1 Their goal is to revisit the high school mathematics curriculum, together with relevant topics from middle school, to help teachers better understand the mathematics they are or will be teaching and to help educators establish a sound mathematical platform on which to base their research. In terms of mathematical sophistication, these three volumes are designed for use in upper division courses for math majors in college. Since their content consists of topics in the upper end of school mathematics (including one-variable calculus), these volumes are in the unenviable position of straddling two disciplines: mathematics and education. Such being the case, these volumes will inevitably inspire misconceptions on both sides. We must therefore address their possible misuse in the hands of both mathematicians and educators. To this end, let us briefly review the state of school mathematics education as of 2020. The phenomenon of TSM For roughly the last five decades, the nation has had a de facto national school mathematics curriculum, one that has been defined by the standard school mathematics textbooks. The mathematics encoded in these textbooks is extremely flawed.2 We call the body of knowledge encoded in these textbooks TSM (Textbook School Mathematics; see page xiv). We will presently give a superficial survey of some of these flaws,3 but what matters to us here is the fact that institutions of higher learning appear to be oblivious to the rampant mathematical mis-education of students in K–12 and have done very little to address the insidious presence of TSM in the mathematics taught to K–12 students over the last 50 years. As a result, mathematics teachers are forced to carry out their teaching duties with all the misconceptions they acquired from TSM intact, and educators likewise continue to base their research on what they learned from TSM. So TSM lives on unchallenged. These three volumes are the conclusion of a six-volume series4 whose goal is to correct the universities’ curricular oversight in the mathematical education of 1 We

use the term "mathematics educators" to refer to university faculty in schools of education. statements about curriculum and textbooks do not take into account how much the quality of school textbooks and teachers’ content knowledge may have evolved recently with the advent of CCSSM (Common Core State Standards for Mathematics) ([CCSSM]) in 2010. 3 Detailed criticisms and explicit corrections of these flaws are scattered throughout these volumes. 4 The earlier volumes in the series are [Wu2011a], [Wu2016a], and [Wu2016b]. 2 These

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teachers and educators by providing the needed mathematical knowledge to break the vicious cycle of TSM. For this reason, these volumes pay special attention to mathematical integrity (as defined on page xiii) and transparency, so that every concept is precisely defined and every assertion is completely explained5 and so that the exposition here is as close as possible to what is taught in a high school classroom. TSM has appeared in different guises; after all, the NCTM reform (see page xv ff.) was largely ushered in around 1989. But beneath the surface its essential substance has stayed remarkably constant (compare [Wu2014]). TSM is characterized by a lack of clear definitions, faulty or nonexistent reasoning, pervasive imprecision, general incoherence, and a consistent failure to make the case about why each standard topic in the school curriculum is worthy of study. Let us go through each of these issues in some detail. (1) Definitions. In TSM, correct definitions of even the most basic concepts are usually not available. Here is a partial list: fraction, multiplication of fractions, division of fractions, one fraction being bigger or smaller than another, finite decimal, infinite decimal, mixed number, ratio, percent, rate, constant rate, negative number, the four arithmetic operations on rational numbers, congruence, similarity, length of a curve, area of a planar region, volume of a solid, expression, equation, graph of a function, graph of an inequality, half-plane, polygon, interior angle of a polygon, regular polygon, slope of a line, parabola, inverse function, etc. Consequently, students are forced to work with concepts whose mathematical meaning is at best only partially revealed to them. Consider, for example, the concept of division. TSM offers no precise definition of division for whole numbers, fractions, rational numbers, real numbers, or complex numbers. If it did, the division concept would become much more learnable because it is in fact the same for all these number systems (thus we also witness the incoherence of TSM). The lack of a definition for division leads inevitably to the impossibility of reasoning about the division of fractions, which then leads to "ours is not to reason why, just invert-and-multiply". We have here a prime example of the convergence of the lack of definitions, the lack of reasoning, and the lack of coherence. The reason we need precise definitions is that they create a level playing field for all learners, in the sense that each person—including the teacher—has all the needed information about a given concept from the very beginning and this information is the same for everyone. This eliminates any need to spend time looking for "tricks", "insider knowledge", or hidden agendas. The level playing field makes every concept accessible to all learners, and this fact is what the discussion of equity in school mathematics education seems to have overlooked thus far. To put this statement in context, think of TSM’s definition of a fraction as a piece of pizza: even elementary students can immediately see that there is more to a fraction than just being a piece of pizza. For example, " 58 miles of dirt road" has nothing to do with pieces of a pizza. The credibility gap between what students are made to learn and what they subconsciously recognize to be false disrupts the learning process, often fatally. 5 In other words, every theorem is completely proved. Of course there are a few theorems that cannot be proved in context, such as the fundamental theorem of algebra.

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In mathematics, there can be no valid reasoning without precise definitions. Consider, for example, TSM’s proof of (−2)(−3) = 2 × 3. Such a proof requires that we know what −2 is, what −3 is, what properties these negative integers are assumed to possess, and what it means to multiply (−2) by (−3) so that we can use them to justify this claim. Since TSM does not offer any information of this kind, it argues instead as follows: 3 · (−3), being 3 copies of −3, is equal to −9, and likewise, 2 · (−3) = −6, 1 · (−3) = −3, and of course 0 · (−3) = 0. Now look at the pattern formed by these consecutive products: 3 · (−3) = −9,

2 · (−3) = −6,

1 · (−3) = −3,

0 · (−3) = 0.

Clearly when the first factor decreases by 1, the product increases by 3. Now, when the 0 in the product 0 · (−3) decreases by 1 (so that 0 becomes −1), the product (−1)(−3) ceases to make sense. Nevertheless, TSM urges students to believe that the pattern must persist no matter what so that this product will once more increase by 3 and therefore (−1)(−3) = 3. By the same token, when the −1 in (−1)(−3) decreases by 1 again (so that −1 becomes −2), the product must again increase by 3 for the same reason and (−2)(−3) = 6 = 2 × 3, as desired. This is what TSM considers to be "reasoning". TSM goes further. Using a similar argument for (−2)(−3) = 2 × 3, one can show that (−a)(−b) = ab for all whole numbers a and b. Now, TSM asks students to take another big leap of faith: if (−a)(−b) = ab is true for whole numbers a and b, then it must also be true when a and b are arbitrary numbers. This is how TSM "proves" that negative times negative is positive. Slighting definitions in TSM can also take a different form: the graph of a linear inequality ax + by ≤ c is claimed to be a half-plane of the line ax + by = c, and the "proof" usually consists of checking a few examples. Thus the points (0, 0), (−2, 0), and (1, −1) are found to lie below the line defined by x + 3y = 2 and, since they all satisfy x + 3y ≤ 2, it is believable that the "lower half-plane" of the line x + 3y = 2 is the graph of x + 3y ≤ 2. Further experimentation with other points below the line defined by x + 3y = 2 adds to this conviction. Again, no reasoning is involved and, more importantly, neither "graph of an inequality" nor "half-plane" is defined in such a discussion because these terms sound so familiar that TSM apparently believes no definition is necessary. At other times, reasoning is simply suppressed, such as when the coordinates of the vertex of the graph of ax2 + bx + c are peremptorily declared to be   −b 4ac − b2 , . 2a 4a End of discussion. Our emphasis on the importance of definitions in school mathematics compels us to address a misconception about the role of definitions in school mathematics education. To many teachers and educators, the word "definition" connotes something tedious and nonessential that students must memorize for standardized tests. It may also conjure an image of cut-and-dried, top-down instruction that begins with a rigid and unmotivated definition and continues with the definition’s formal and equally unmotivated appearance in a chain of logical arguments. Understandably, most educators find this scenario unappetizing. Their response is that, at least in school mathematics, the definition of a concept should emerge at the end—but not at the beginning—of an extended intuitive discussion of the hows and whys of

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the concept.6 In addition, the so-called conceptual understanding of the concept is believed to lie in the intuitive discussion but not in the formal definition itself, the latter being nothing more than an afterthought. These two opposite conceptions of definition ignore the possibility of a middle ground: one can state the precise definition of a concept at the beginning of a lesson to set the tone of the subsequent mathematical discussion and exploration, which is to show students that this is all they will ever need to know about the concept as far as doing mathematics is concerned. Such transparency—demanded by the mathematical culture of the past century (cf. [Quinn])—is what is most sorely missing in TSM, which consistently leaves students in doubt about what a fraction is or might be, what a negative number is, what congruence means, etc. In this middle ground, a definition can be explored and explained in intuitive terms in the ensuing discussion on the one hand and, on the other, put to use in proofs—in its precise formulation—to show how and why the definition is absolutely indispensable to any kind of reasoning concerning the concept. With the consistent use of precise definitions, the line between what is correct and what is intuitive but maybe incorrect (such as the TSM-proof of negative times negative is positive) becomes clearly drawn. It is the frequent blurring of this line in TSM that contributes massively to the general misapprehension in mathematics education about what a proof is (part of this misapprehension is described in, e.g., [NCTM2009], [Ellis-Bieda-Knuth], and [Arbaugh et al.]). These three volumes (this volume, [Wu2020b], and [Wu2020c]) will always take a position in the aforementioned middle ground. Consider the definition of a fraction, for example: it is one of a special collection of points on the number line (page 10). This is the only meaning of a fraction that is needed to drive the fairly intricate mathematical development of fractions, and, for this reason, the definition of a fraction as a certain point on the number line is the one that will be unapologetically used all through these three volumes. To help teachers and students feel comfortable with this definition, we give an extensive intuitive discussion of why such a definition for a fraction is necessary on pp. 4–10. This intuitive discussion, naturally, opens the door to whatever pedagogical strategy a teacher wants to invest in it. Unlike in TSM, however, this definition is not given to be forgotten. On the contrary, all subsequent discussions about fractions will refer to this precise definition (but not to the intuitive discussion that preceded it) and, of course, all the proofs about fractions will also depend on this formal definition because mathematics demands no less. Students need to learn what a proof is and how it works; the exposition here tries to meet this need by (gently) laying bare the fact that reasoning in proofs requires precise definitions. As a second example, we give the definition of the slope of a line only after an extensive intuitive discussion on pp. 338–346 about what slope is supposed to measure and how we may hope to measure it. Again, the emphasis is on the fact that this definition of slope is not the conclusion, but the beginning of a long logical development that goes from page 346 to the end of Chapter 6 on page 383, and into trigonometry (relation with the tangent function), calculus (definition of the derivative), and beyond. 6 Proponents of this approach to definitions often seem to forget that, after the emergence of a precise definition, students are still owed a systematic exposition of mathematics using the definition so that they can learn about how the definition fits into the overall logical structure of mathematics.

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(2) Reasoning. Reasoning is the lifeblood of mathematics, and the main reason for learning mathematics is to learn how to reason. In the context of school mathematics, reasoning is important to students because it is the tool that empowers them to explore on their own and verify for themselves what is true and what is false without having to take other people’s words on faith. Reasoning gives them confidence and independence. But when students have to accustom themselves to performing one unexplained rote skill after another, year after year, their ability to reason will naturally atrophy. Many students find it more expedient to stop asking why and simply take any order that comes their way sight unseen just to get by.7 One can only speculate on the cumulative effect this kind of mathematics "learning" has on those students who go on to become teachers and mathematics educators. (3) Precision. The purpose of precision is to eliminate errors and minimize misconceptions, but in TSM students learn at every turn that they should not believe exactly what they are told but must learn to be creative in interpreting it. For example, TSM preaches the virtue of using the theorem on equivalent fractions to simplify fractions and does not hesitate to simplify a rational expression in x as follows: x2 + 3 (x − 1)(x2 + 3) = . x(x − 1) x This looks familiar because "canceling the same number from top and bottom" is exactly what the theorem on equivalent fractions is supposed to do. Unfortunately, this theorem only guarantees a ca = bc b when a, b, and c are whole numbers (b and c understood to be nonzero). In the 2 previous rational expression, however, none of (x−1), √ (x +3), and x is necessarily a whole number because x could be, for example, 5. Therefore, according to TSM, students in algebra should look back at equivalent fractions and realize that the theorem on equivalent fractions—in spite of what it says—can actually be applied to "fractions" whose "numerators" and "denominators" are not whole numbers. Thus TSM encourages students to believe that "nothing needs to be taken precisely and one must be flexible in interpreting what one learns". This extrapolation-happy mindset is the opposite of what it takes to learn a precise subject like mathematics or any of the exact sciences. For example, we cannot allow students to believe that the domain of definition of log x is [0, ∞) since [0, ∞) is more or less the same as (0, ∞). Indeed, the presence or absence of the single point "0" is the difference between true and false. Another example of how a lack of precision leads to misconceptions is the statement that "β 0 = 1", where β is a nonzero number. Because TSM does not use precise language, it does not—or cannot—draw a sharp distinction between a heuristic argument, a definition, and a proof. Consequently, it has misled numerous students and teachers into believing that the heuristic argument for defining β 0 to be 1 is in fact a "proof" that β 0 = 1. The same misconception persists for negative exponents (e.g., β −n = 1/β n ). The lack of precision is so pervasive in TSM that there is no end to such examples. 7 There

is consistent anecdotal evidence from teachers in the trenches that such is the case.

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(4) Coherence. Another reason why TSM is less than learnable is its incoherence. Skills in TSM are framed as part of a long laundry list, and the lack of definitions for concepts ensures that skills and their underlying concepts remain forever disconnected. Mathematics, on the other hand, unfolds from a few central ideas, and concepts and skills are developed along the way to meet the needs that emerge in the process of unfolding. An acceptable exposition of mathematics therefore tells a coherent story that makes mathematics memorable. For example, consider the fact that TSM makes the four standard algorithms for whole numbers four separate rote-learning skills. Thus TSM hides from students the overriding theme that the Hindu-Arabic numeral system is universally adopted because it makes possible a simple, algorithmic procedure for computations; namely, if we can carry out an operation (+, −, ×, or ÷) for single-digit numbers, then we can carry out this operation for all whole numbers no matter how many digits they have (see Chapter 3 of [Wu2011a]). The standard algorithms are the vehicles that bridge operations with single-digit numbers and operations on all whole numbers. Moreover, the standard algorithms can be simply explained by a straightforward application of the associative, commutative, and distributive laws. From this perspective, a teacher can explain to students, convincingly, why the multiplication table is very much worth learning; this would ease one of the main pedagogical bottlenecks in elementary school. Moreover, a teacher can also make sense of the associative, commutative, and distributive laws to elementary students and help them see that these are vital tools for doing mathematics rather than dinosaurs in an outdated school curriculum. If these facts had been widely known during the 1990s, the senseless debate on whether the standard algorithms should be taught might not have arisen and the Math Wars might not have taken place at all. TSM also treats whole numbers, fractions, (finite) decimals, and rational numbers as four different kinds of numbers. The reality is that, first of all, decimals are a special class of fractions (see pp. 14ff.), whole numbers are part of fractions, and fractions are part of rational numbers. Moreover, the four arithmetic operations (+, −, ×, and ÷) in each of these number systems do not essentially change from system to system. There is a smooth conceptual transition at each step of the passage from whole numbers to fractions and from fractions to rational numbers; see Parts 2 and 3 of [Wu2011a] or Sections 2.2, 2.4, and 2.5 in this volume. This coherence facilitates learning: instead of having to learn about four different kinds of numbers, students basically only need to learn about one number system (the rational numbers). Yet another example is the conceptual unity between linear functions and quadratic functions: in each case, the leading term—ax for linear functions and ax2 for quadratic functions—determines the shape of the graph of the function completely, and the studies of the two kinds of functions become similar as each revolves around the shape of the graph (see Section 2.1 of [Wu2020b]). Mathematical coherence gives us many such storylines, and a few more will be detailed below. (5) Purposefulness. In addition to the preceding four shortcomings—a lack of clear definitions, faulty or nonexistent reasoning, pervasive imprecision, and general incoherence—TSM has a fifth fatal flaw: it lacks purposefulness. Purposefulness is what gives mathematics its vitality and focus: the fact is that a mathematical investigation, at any level, is always carried out with a specific goal in mind. When a mathematics textbook reflects this goal-oriented character of mathematics, it

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propels the mathematical narrative forward and facilitates its learning by making students aware of where the discussion is headed, and why. Too often, TSM lurches from topic to topic with no apparent purpose, leading students to wonder why they should bother to tag along. One example is the introduction of the absolute value of a number. Many teachers and students are mystified by being saddled with such a "frivolous" skill: "just kill the negative sign", as one teacher put it. Yet TSM never tries to demystify his concept. (For an explanation of the need to introduce absolute value, see, e.g., the discussion on pp. 130ff.). Another is the seemingly inexplicable replacement √ of the square root and cube root symbols of a positive √ number b, i.e., b and 3 b, by rational exponents, b1/2 and b1/3 , respectively (see, e.g., Section 4.2 of [Wu2020b]). Because TSM teaches the laws of exponents as merely "number facts", it is inevitable that it would fail to point out the purpose of this change of notation, which is to shift focus from the operation of taking roots to the properties of the exponential function bx for a fixed positive b. A final example is the way TSM teaches estimation completely by rote, without ever telling students why and when estimation is important and therefore worth learning. Indeed, we often have to make estimates, either because precision is unattainable or unnecessary, or because we purposely use estimation as a tool to help achieve precision (see [Wu2011a, Section 10.3]). To summarize, if we want students to be taught mathematics that is learnable, then we must discard TSM and replace it with the kind of mathematics that possesses these five qualities: Every concept has a clear definition. Every statement is precise. Every assertion is supported by reasoning. Its development is coherent. Its development is purposeful. We have come across them before on page xiii: these are the Fundamental Principles of Mathematics (also see Section 2.1 in [Wu2018a]). TSM consistently violates all five fundamental principles. Because of the dominance of TSM for at least the past half-century, most students come out of K–12 knowing only TSM but not mathematics that respects these fundamental principles. To them, learning mathematics is not about learning how to reason or distinguish true from false but about memorizing facts and tricks to get correct answers. Faced with this crisis, what should be the responsibility of institutions of higher learning? Should it be to create courses for future teachers and educators to help them systematically replace their knowledge of TSM with mathematics that is consistent with the five fundamental principles? Or should it be, rather, to leave TSM alone but make it more palatable by helping teachers infuse their classrooms with activities that suggest visions of reasoning, problem solving, and sense making? As of this writing, an overwhelming majority of the institutions of higher learning are choosing the latter alternative. At this point, we return to the earlier question about some of the ways both university mathematicians and educators might misunderstand and misuse these three volumes.

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Potential misuse by mathematicians First, consider the case of mathematicians. They are likely to scoff at what they perceive to be the triviality of the content in these volumes: no groups, no homomorphisms, no compact sets, no holomorphic functions, and no Gaussian curvature. They may therefore be tempted to elevate the level of the presentation, for example, by introducing the concept of a field and show that, when two fractions symbols m/n and k/ (with whole numbers m, n, k, , and n = 0,  = 0) satisfying m = nk are identified, and when + and × are defined by the usual formulas, the fraction symbols form a field. In this elegant manner, they can efficiently cover all the standard facts in the arithmetic of fractions in the school curriculum.8 This is certainly a better way than defining fractions as points on the number line to teach teachers and educators about fractions, is it not? Likewise, mathematicians may find finite geometry to be a more exciting introduction to axiomatic systems than any proposed improvements on the high school geometry course in TSM. The list goes on. Consequently, pre-service teachers and educators may end up learning from mathematicians some interesting mathematics, but not mathematics that would help them overcome the handicap of knowing only TSM. Mathematicians may also engage in another popular approach to the professional development of teachers and educators: teaching the solution of hard problems. Because mathematicians tend to take their own mastery of fundamental skills and concepts for granted, many do not realize that it is nearly impossible for teachers who have been immersed in thirteen years or more of TSM to acquire, on their own, a mastery of a mathematically correct version of the basic skills and concepts. Mathematicians are therefore likely to consider their major goal in the professional development of teachers and educators to be teaching them how to solve hard problems. Surely, so the belief goes, if teachers can handle the "hard stuff", they will be able to handle the "easy stuff" in K–12. Since this belief is entirely in line with one of the current slogans in school mathematics education about the critical importance of problem solving, many teachers may be all too eager to teach their students the extracurricular skills of solving challenging problems in addition to teaching them TSM day in and day out. In any case, the relatively unglamorous content of these three volumes (this volume, [Wu2020b], and [Wu2020c])—designed to replace TSM—will get shunted aside into supplementary reading assignments. At the risk of belaboring the point, the focus of these three volumes is on showing how to replace teachers’ and educators’ knowledge of TSM in grades 9–12 with mathematics that respects the fundamental principles of mathematics. Therefore, reformulating the mathematics of grades 9–12 from an advanced mathematical standpoint to obtain a more elegant presentation is not the point. Introducing novel elementary topics (such as Pick’s theorem or the 4-point affine plane) into the mathematics education of teachers and educators is also not the point. Rather, the point in year 2020 is to do the essential spadework of revisiting the standard 9–12 curriculum—topic by topic, along the lines laid out in these three volumes— showing teachers and educators how the TSM in each case can be supplanted by mathematics that makes sense to them and to their students. For example, since most pre-service teachers and educators have not been exposed to the use of precise 8 This is my paraphrase of a mathematician’s account of his professional development institute around year 2000.

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definitions in mathematics, they are unlikely to know that definitions are supposed to be used, exactly as written, no more and no less, in logical arguments. One of the most formidable tasks confronting mathematicians is, in fact, how to change educators’ and teachers’ perception of the role of definitions in reasoning. As illustration, consider how TSM handles slope. There are two ways, but we will mention only one of them.9 TSM pretends that, by defining the slope of a line L using the difference quotient with respect to two pre-chosen points P and Q on L,10 such a difference quotient is a property of the line itself (rather than a property of the two points P and Q). In addition, TSM pretends that it can use "reasoning" based on this defective definition to derive the equation of a line when (for example) its slope and a given point on it are prescribed. Here is the inherent danger of thirteen years of continuous exposure to this kind of pseudoreasoning: teachers cease to recognize that (a) such a definition of slope is defective and (b) such a defective definition of slope cannot possibly support the purported derivation (= proof) of the equation of a line. It therefore comes to pass that— as a result of the flaws in our education system—many teachers and educators end up being confused about even the meaning of the simplest kind of reasoning: "A implies B". They need—and deserve—all the help we can give so that they can finally experience genuine mathematics, i.e., mathematics that is based on the fundamental principles of mathematics. Of course, the ultimate goal is for teachers to use this new knowledge to teach their own students so that those students can achieve a true understanding of what "A implies B" means and what real reasoning is all about. With this in mind, we introduce in Section 6.4 (pp. 337ff.) the concept of slope by discussing what slope is supposed to measure (an example of purposefulness) and how to measure it, which then leads to the formulation of a precise definition. With the availability of the AA-criterion for triangle similarity (Theorem G22 on page 288), we then show how this definition leads to the formula for the slope of a line as the difference quotient of the coordinates of any two points on the line (the "rise-over-run"). Having this critical flexibility to compute the slope—plus an earlier elucidation of what an equation is (pp. 322–324)—we easily obtain the equation of a line passing through a given point with a given slope, with correct reasoning this time around (see pp. 357ff.). Of course the same kind of reasoning can be applied to similar problems when other reasonable geometric data are prescribed for the line. By guiding teachers and educators systematically through the correction of TSM errors on a case-by-case basis, we believe they will gain a new and deeper understanding of school mathematics. Ultimately, we hope that if institutions of higher learning and the education establishment can persevere in committing themselves to this painstaking work, the students of these teachers and educators will be spared the ravages of TSM. If there is an easier way to undo thirteen years and more of mis-education in mathematics, we are not aware of it. A main emphasis in using these three volumes should therefore be on providing patient guidance to teachers and educators to help them overcome the many handicaps inflicted on them by TSM. In this light, we can say with confidence that, for 9 A second way is to define a line to be the graph of a linear equation y = mx + b and then define the slope of this line to be m. This is the definition of a line in advanced mathematics, but it is so profoundly inappropriate for use in K–12 that we will just ignore it. 10 This is the "rise-over-run".

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now, the best way for mathematicians to help educate teachers and educators is to firm up their mathematical foundations. Let us repair the damage TSM has done to their mathematics content knowledge by helping them to acquire a knowledge of school mathematics that is consistent with the fundamental principles of mathematics. Potential misuse by educators Next, we address the issue of how educators may misuse these three volumes. Educators may very well frown on the volumes’ insistence on precise definitions and precise reasoning and their unremitting emphasis on proofs while, apparently, neglecting problem solving, conceptual understanding, and sense making. To them, good professional development concentrates on all of these issues plus contextual learning, student thinking, and communication with students. Because these three volumes never explicitly mention problem solving, conceptual understanding, or sense making per se (or, for that matter, contextual learning or student thinking), their content may be dismissed by educators as merely skills-oriented or technical knowledge for its own sake and, as such, get relegated to reading assignments outside of class. They may believe that precious class time can be put to better use by calling on students to share their solutions to difficult problems or by holding small group discussions about problem-solving strategies. We believe this attitude is also misguided because the critical missing piece in the contemporary mathematical education of teachers and educators is an exposure to a systematic exposition of the standard topics of the school curriculum that respects the fundamental principles of mathematics. Teachers’ lack of access to such a mathematical exposition is what lies at the heart of much of the current education crisis. Let us explain. Consider problem solving. At the moment, the goal of getting all students to be proficient in solving problems is being pursued with missionary zeal, but what seems to be missing in this single-minded pursuit is the recognition that the body of knowledge we call mathematics consists of nothing more than a sequence of problems posed, and then solved, by making logical deductions on the basis of precise definitions, clearly stated hypotheses, and known results.11 This is after all the whole point of the classic two-volume work [Pólya-Szegö], which introduces students to mathematical research through the solutions to a long list of problems. For example, the Pythagorean theorem and its many proofs are nothing more than solutions to the problem posed by people from diverse cultures long ago: "Is there any relationship among the three sides of a right triangle?" There is no essential difference between problem solving and theorem proving in mathematics. Each time we solve a problem, we in effect prove a theorem (trivial as that theorem may sometimes be). The main point of this observation is that if we want students to be proficient in problem solving, then we must give them plenty of examples of gradeappropriate proofs all through (at least) grades 4–12 and engage them regularly 11 It is in this light that the previous remark about the purposefulness of mathematics can be better understood: before solving a problem, one should know why the problem was posed in the first place. Note that, for beginners (i.e., school students), the overwhelming emphasis has to be on solving problems rather than the more elusive issue of posing problems.

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in grade-appropriate theorem-proving activities. If we can get students to see, day in and day out, that problem solving is a way of life in mathematics and if we also routinely get them involved in problem solving (i.e., theorem proving), students will learn problem solving naturally through such a long-term immersion. In the process, they will get to experience that, to solve problems, they need to have precise definitions and precise hypotheses as a starting point, know the direction they are headed towards before they make a move (sense making), and be able to make deductions from precise definitions and known facts. Definitions, sense making, and reasoning will therefore come together naturally for students if they learn mathematics that is consistent with the five fundamental principles. We make the effort to put problem solving in the context of the fundamental principles of mathematics because there is a danger in pursuing problem solving per se in the midst of the TSM-induced corruption of school mathematics. In a generic situation, teachers teach TSM and only pay lip service to "problem solving", while in the best case scenario, teachers keep TSM intact while teaching students how to solve problems on a separate, parallel track outside of TSM. Lest we forget, TSM considers "out of a hundred" to be a correct definition of percent, expands the product of two linear polynomials by "FOILing", and assumes that in any problem about rate, one can automatically assume that the rate is constant ("Lynnette can wash 95 cars in 5 days. How many cars can Lynnette wash in 11 days?"), etc. In this environment, it is futile to talk about (correct) problem solving. Until we can rid school classrooms of TSM, the most we can hope for is having teachers teach, on the one hand, definition-free concepts with a bag of tricks-sans-reasoning to get correct answers and, on the other hand, reasoning skills for solving a separate collection of problems for special occasions. In other words, two parallel universes will co-exist in school mathematics classrooms. So long as TSM continues to reign in school classrooms, most students will only be comfortable doing one-step problems and any problem-solving ability they possess will only be something that is artificially grafted onto the TSM they know. If we want to avert this kind of bipolar mathematics education in schools, we must begin by providing teachers with a better mathematical education. Then we can hope that teachers will teach mathematics consistent with the fundamental principles of mathematics12 so that students’ problem-solving abilities can evolve naturally from the mathematics they learn. It is partly for this reason that the six volumes under discussion13 choose to present the mathematics of K–12 with explanations (= proofs) for all the skills. In particular, these three volumes on the mathematics of grades 9–12 provide proofs for every theorem. (At the same time, they also caution against certain proofs that are simply too long or too tedious to be presented in a high school classroom.) The hope is that when teachers and educators get to experience firsthand that every part of school mathematics is suffused with reasoning, they will not fail to teach reasoning to their own students as a matter of routine. Only then will it make sense to consider problem solving to be an integral part of school mathematics.

12 And, of course, to also get school textbooks that are unsullied by TSM. However, it seems likely as of 2020 that major publishers will hold onto TSM until there are sufficiently large numbers of knowledgeable teachers who demand better textbooks. See the end of [Wu2015]. 13 These three volumes, together with [Wu2011a], [Wu2016a], and [Wu2016b].

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The importance of correct content knowledge In general, the idea is that if we give teachers and educators an exposition of mathematics that makes sense and has built-in conceptual understanding and reasoning, then we can hope to create classrooms with an intellectual climate that enables students to absorb these qualities as if by osmosis. Perhaps an analogy can further clarify this issue: if we want to teach writing, it would be more effective to let students read good writing and learn from it directly rather than to let them read bad writing and simultaneously attend special sessions on the fine points of effective written communication. If we want school mathematics to be suffused with reasoning, conceptual understanding, and sense making, then we must recognize that these are not qualities that can stand apart from mathematical details. Rather, they are firmly anchored to hard-and-fast mathematical facts. Take proofs (= reasoning), for example. If we only talk about proofs in the context of TSM, then our conception of what a proof is will be extremely flawed because there are essentially no correct proofs in TSM. For starters, since TSM has no precise definitions, there can be no hope of finding a completely correct proof in TSM. Therefore, when teaching from these three volumes,14 it is imperative to first concentrate on getting across to teachers and educators the details of the mathematical reformulation of the school curriculum. Specifically, we stress the importance of offering educators a valid alternative to TSM for their future research. Only then can we hope to witness a reconceptualization—in mathematics education—of reasoning, conceptual understanding, problem solving, etc., on the basis of a solid mathematical foundation. Reasoning, conceptual understanding, and sense making are qualities intrinsic to school mathematics that respects the fundamental principles of mathematics. We see in these three volumes a continuous narrative from topic to topic and from chapter to chapter to guide the reader through this long journey. The sense making will be self-evident to the reader. Moreover, when every assertion is backed up by an explanation (= proof), reasoning will rise to the surface for all to see. In their presentation of the natural unfolding of mathematical ideas, these volumes also routinely point out connections between definitions, concepts, theorems, and proofs. Some connections may not be immediately apparent. For example, in Section 6.1 of this volume (page 310), we explicitly point out the connection between Mersenne primes and the summation of finite geometric series. Other connections span several grades: there is a striking similarity between the proofs of the area formula for rectangles whose sides are fractions (Theorem 1.7 on pp. 48ff.), the ASA congruence criterion (Theorem G9 on pp. 245ff.), the SSS congruence criterion (Theorem G28 in Section 6.2 of [Wu2020b]), the fundamental theorem of similarity (Theorem G10 in Section 6.4 of [Wu2020b]), and the theorem about the equality of angles on a circle subtending the same arc (Theorem G52 in Section 6.8 of [Wu2020b]). All these proofs are achieved by breaking up a complicated argument into two or more clearcut steps, each involving simpler arguments. In other words, they demonstrate how to reduce the complex to the simple, so prospective teachers and educators can learn from such instructive examples about the fine art of problem solving (= reasoning). 14 As

well as from the other three volumes, [Wu2011a], [Wu2016a], and [Wu2016b]).

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The foregoing unrelenting emphasis on mathematical content should not lead readers to believe that these three volumes deal with mathematics at the expense of pedagogy. To the extent that these volumes are designed to promote better teaching in the schools, they do not sidestep pedagogical issues. Extensive pedagogical comments are offered whenever they are called for, and they are clearly displayed as such; see, for example, pp. 16, 23, 37, 40, 53, 119, 261, 263, etc. Nevertheless, our most urgent task—the fundamental task—in the mathematical education of teachers and educators as of 2020 has to be the reconstruction of their mathematical knowledge base. This is not about judiciously tinkering with what teachers and educators already know or tweaking their existing knowledge here and there. Rather, it is about the hard work of replacing their knowledge of TSM with mathematics that is consistent with the fundamental principles of mathematics from the ground up. The primary goal of these three volumes is to give a detailed exposition of school mathematics in grades 9–12 to help educators and teachers achieve this reconstruction.

To the Pre-Service Teacher In one sense, these three volumes are just textbooks, and you may feel you have gone through too many textbooks in your life to need any fresh advice. Nevertheless, we are going to suggest that you approach these volumes with a different mindset than what you may have used with other textbooks, because you will soon be using the knowledge you gain from these volumes to teach your students. Reading other textbooks, you would likely congratulate yourself if you could achieve mastery over 90% of the material. That would normally guarantee an A. More is at stake with these volumes, however, because they directly address what you will need to know in order to write your lessons. Ask yourself whether a mathematics teacher whose lessons are correct only 90% of the time should be considered a good teacher. To be blunt, such a teacher would be a near disaster. So your mission in reading these volumes should be to achieve nothing short of total mastery. You are expected to know this material 100%. To the extent that the content of these three volumes is just K–12 mathematics, this is an achievable goal. This is the standard you have to set for yourself. Having said that, we also note explicitly that many Mathematical Asides are sprinkled all through the text, sometimes in the form of footnotes. These are comments—usually from an advanced mathematical perspective—that try to shed light on the mathematics under discussion. The above reference to "total mastery" does not include these comments. You should approach these volumes differently in yet another respect. Students’ typical attitude towards a math course is that if they can do all the homework problems, then most of their work is done. Think back on your calculus courses or any of the math courses when you were in school, and you will understand how true this is. But since these volumes are designed specifically for teachers, your emphasis cannot be limited to merely doing the homework assignments because your job will be more than just helping students to do homework problems. When you stand in front of a class, what you will be talking about, most of the time, will not be the exercises at the end of each section but the concepts and skills in the exposition proper.1 For example, very likely you will soon have to convince a class on geometry why the Pythagorean theorem is correct. There are two proofs of this theorem in these volumes, one in Chapter 5 of this volume and the other in Chapter 4 of [Wu2020c]. Yet on neither occasion is it possible to assign a problem that asks for a proof of this theorem. There are problems that can assess whether you know

1 I will be realistic and acknowledge that there are teachers who use class time only to drill students on how to get the right answers to exercises, often without reasoning. But one of the missions of these three volumes is to steer you away from that kind of teaching. See To the Instructor on pp. xxvii ff.

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enough about the Pythagorean theorem to apply it, but how do you assess whether you know how to prove the theorem when the proofs have already been given in the text? It is therefore entirely up to you to achieve mastery of everything in the text itself. One way to check is to pick a theorem at random and ask yourself: Can I prove it without looking at the book? Can I explain its significance? Can I convince someone else why it is worth knowing? Can I give an intuitive summary of the proof? These are questions that you will have to answer as a teacher. To the extent possible, these volumes try to provide information that will help you answer questions of this kind. I may add that the most taxing part of writing these volumes was in fact to do it in a way that would allow you, as much as possible, to adapt them for use in a school classroom with minimal changes. (Compare, for example, To the Instructor on pp. xxvii ff.) There is another special feature of these volumes that I would like to bring to your attention: these volumes are essentially school textbooks written for teachers, and as such, you should read them with the eyes of a school student. When you read Chapter 1 of this volume on fractions, for instance, picture yourself in a sixth-grade classroom and therefore, no matter how much abstract algebra you may know or how well you can explain the construction of the quotient field of an integral domain, you have to be able to give explanations in the language of sixth-grade mathematics (i.e., to sixth graders). Similarly, when you come to Chapter 6, you are developing algebra from the beginning, so even the use of symbols will be an issue (it is in fact the key issue; see Section 6.1 on pp. 298ff.). Therefore, be very deliberate and explicit when you introduce a symbol, at least for a while. The major conclusions in these volumes, as in all mathematics books, are summarized into theorems. Depending on the author’s (and other mathematicians’) whims, theorems are sometimes called propositions, lemmas, or corollaries as a way of indicating which theorems are deemed more important than others. Roughly speaking, a proposition is not regarded to be as important as a theorem, a lemma is conceptually less important than a proposition, and a corollary is supposed to follow immediately from the theorem or proposition to which it is attached. (Incidentally, a formula or an algorithm is just a theorem.) This idiosyncratic classification of theorems started with Euclid around 300 BC, and it is too late to do anything about it now. The main concepts of mathematics are codified into definitions. Definitions are set in boldface in these volumes when they appear for the first time; a few truly basic ones are even individually displayed in a separate paragraph, but most of the definitions are embedded in the text itself, so you should watch out for them. The statements of the theorems, and especially their proofs, depend on the definitions, and proofs are the guts of mathematics. Please note that when I said above that I expect you to know everything in these volumes, I was using the word "know" in the way mathematicians normally use the word. They do not use it to mean simply "know the statement by heart". Rather, to know a theorem, for instance, means know the statement by heart, know its proof, know why it is worth knowing, know what its potential implications are, and finally, know how to apply it in new situations. If you know anything short of this, how can you expect to be able to answer your students’ questions? At the very least, you should know by heart all the theorems and definitions as well as the main ideas of each proof because, if you do not, it will be futile to talk about the

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other aspects of knowing. Therefore, a preliminary suggestion to help you master the content of these volumes is for you to copy out the statements of every definition, theorem, proposition, lemma, and corollary, along with page references so that they can be examined in detail when necessary, and also to form the habit of summarizing the main idea(s) of each proof. These are good study habits. When it is your turn to teach your students, be sure to pass along these suggestions to them. You should also be aware that reading a mathematics book is not the same as reading a gossip magazine. You can probably flip through one of the latter in an hour or less. But in these volumes, there will be many passages that require slow reading and re-reading, perhaps many times. I cannot single out those passages for you because they will be different for different people. We do not all learn the same way. What you can take for granted, however, is that mathematics books make for exceedingly slow reading. (Nothing good comes easy.) Therefore if you get stuck, time and time again, on a sentence or two in these volumes, take heart, because this is the norm in mathematics learning.

Prerequisites In terms of the mathematical development of this volume, only a knowledge of whole numbers, 0, 1, 2, 3, . . . , is assumed. Thus along with place value, you are assumed to know the four arithmetic operations, their standard algorithms, and the concept of division-with-remainder and how it is related to the long division algorithm.1 Division-with-remainder assigns to each pair of whole numbers b (the dividend) and d (the divisor), where d = 0, another pair of whole numbers q (the quotient) and r (the remainder), so that b = qd + r

where 0 ≤ r < d.

Some subtle points about the concept of division among whole numbers will be briefly recalled at the beginning of Section 1.5 on page 54. A detailed exposition of the concept of "division" among whole numbers is given in Chapter 7 of [Wu2011a]. Note that 0 is included among the whole numbers. A knowledge of negative numbers, particularly integers, is not assumed. Negative numbers will be developed ab initio in Chapter 2.

Because every assertion in these three volumes (this volume, together with [Wu2020b] and [Wu2020c]) will be proved, students should be comfortable with mathematical reasoning. It is hoped that as they progress through the volumes, all students will become increasingly at ease with proofs. In terms of the undergraduate curriculum, readers of this volume—as a rule of thumb—should have already taken the usual two years of college calculus or their equivalents.

1 Unfortunately, a correct exposition of this topic is difficult to come by. Try Chapter 7 of [Wu2011a].

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Some Conventions • Each chapter is divided into sections. Titles of the sections are given at the beginning of each section as well as in the table of contents. Each section (with few exceptions) is divided into subsections; a list of the subsections in each section—together with a summary of the section in italics—is given at the beginning of each section. • When a new concept is first defined, it appears in boldface but is not often accorded a separate paragraph of its own. For example: A subset R in a plane is called convex if given any two points A, B in R, . . . (p. 172). You will have to look for many definitions in the text proper. (However, not all boldfaced words or phrases signify new concepts to be defined, because boldface fonts are sometimes used for emphasis.) • When a new notation is first introduced, it also appears in boldface. For example: The congruence notation ABC ∼ = A B  C  will be understood to mean . . . (p. 245). • Equations are labeled with numbers inside parentheses, and the first digit of the label indicates the chapter in which the equation can be found. For example, the "(1.17)" in the sentence "Thus (1.17) implies that . . . " means the 17th labeled equation in Chapter 1. • Exercises are located at the end of each section. • Bibliographic citations are labeled with the name of the author(s) inside square brackets, e.g., [Ginsburg]. The bibliography begins on page 387. • In the index, if a term is defined on a certain page, that page will be in italics. For example, the item division-with-remainder, 15, 137, 139 means that the term "division-with-remainder" appears in a significant way on all three pages, but the definition of the term is on page 139.

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CHAPTER 1

Fractions Overview of Chapters 1 and 2 These two chapters give a systematic, though somewhat terse, exposition of rational numbers1 that is suitable for classroom use in grades 5–7. The reason for the brevity is that a more leisurely exposition written for middle school teachers has been given in [Wu2016a]; you can consult the latter for more details if necessary. Such an exposition will be important to high school teachers because most students come to high school with a very defective understanding of fractions. They need help with fractions if they are to have any hope of understanding their high school math classes. High school teachers who do not possess a knowledge of fractions that makes sense mathematically and is accessible to school students will be tremendously handicapped when they try to communicate with their students. There is a good reason why students have trouble with fractions: TSM2 , in the guise of "school mathematics", has not taught them much about fractions that makes sense to them. TSM does not even tell them what a fraction is beyond putting forth the elusive metaphor of "a part of a whole" or "a piece of pizza", and it certainly shows no interest in providing understandable mathematical explanations. For example, why use the least common denominator when adding fractions, and why invert and multiply when dividing fractions? If we do not teach learnable mathematics, then nonlearning will be the inevitable consequence. Unfortunately, the nonlearning of fractions is a major stumbling block in the learning of algebra3 and, therefore, in the learning of high school mathematics as a whole. This is one reason that this volume—addressing the teaching of high school mathematics as it does—is obliged to begin with an exposition of fractions and rational numbers that is both mathematically correct and compatible with the school curriculum. We want to maximize the chances that high school teachers will be able to do their share to help students make a successful first step in their lifelong journey of a thousand miles. From the perspective of advanced mathematics, the rational numbers are among the simplest mathematical structures. As sometimes happens, however, what is mathematically simple may not be simple enough for consumption by school students. Let us briefly recall how the set of rational numbers, Q, is handled in abstract algebra. We start with the integers, Z, and Q is the set of all equivalence classes of 1 The term "rational numbers" is being used correctly here to denote the number system consisting of fractions and negative fractions. Unfortunately, this term has been incorrectly used in the education literature to mean fractions only, but not negative fractions. 2 See page xiv for a definition of TSM. 3 The importance of fractions to the learning of algebra is beginning to be recognized. See, for example, Recommendation 4 on page xvii of the report of the National Mathematics Advisory Panel ([NMAP1]). See also the article [Wu2018b].

1

2

1. FRACTIONS

ordered pairs of integers with respect to a natural equivalence relation (essentially the cross-multiplication algorithm; see page 23). Then addition + and multiplication × are introduced in Q in a way that guarantees that Q becomes a field, i.e., + and × are associative, commutative, and distributive, and so that every element has an additive inverse and every nonzero element has a multiplicative inverse. An order relation ≥ is introduced into Q and those elements ≥ 0 are what we call the fractions. For the purposes of doing mathematics, the fact that Q is an ordered field is essentially all that matters. Now there are many reasons why this way of dealing with the rational numbers is not suitable for use in grades 5–7, which is where the most substantial parts of fractions and rational numbers are systematically taught in schools. The major reason is that students around the age of twelve do not have the mathematical maturity to deal with ordered pairs of integers or equivalence classes. They also have little or no conception of why a field might be interesting or important, and they couldn’t care less whether or not fractions can be added or multiplied. They come to the concept of a fraction through everyday usage, such as two-thirds of a glass of orange juice, and they come to the addition and multiplication of fractions through their experience with whole numbers, so that + connotes "putting things together" and × signifies "repeated addition". Negative fractions, essential as they are, are even further from students’ intuitive understanding. The needs of the school classroom therefore dictate that fractions and rational numbers be introduced in a way that respects and extends students’ prior learning experiences. It would be futile to try to impose on them a mathematical worldview that is entirely foreign to them. What the foregoing discussion hints at is that school mathematics education is not the teaching of straightforward mathematics as we know it in universities but must be, rather, a variant of it. If engineering is the customization of abstract scientific principles to meet human needs, then mathematics education is mathematical engineering:4 it customizes abstract university mathematics to meet the needs of K–12 students. In the case of fractions, mathematical engineering customizes the abstract concept of Q to meet the needs of students around the age of twelve. What is essential to engineering is that, while it can customize scientific principles, it must not violate them. Likewise, the customization of abstract mathematics must be also consonant with mathematics in terms of reasoning and precision. The tension between the need to be faithful to the fundamental principles of mathematics5 and the need to make the mathematics grade-level appropriate will be the overriding theme in this and the two companion volumes, [Wu2020b] and [Wu2020c]. For the case at hand, this tension highlights the fact that these two chapters are nothing but the presentation of an "engineered version of Q". This chapter will start with fractions, while negative numbers will be the subject of the following one. We want to point out a glaring failing of TSM that these two chapters will redress. On a technical level, rational numbers are important because they constitute the number system of choice in school mathematics. School mathematics even views real numbers (rational and irrational numbers) through the lens of rational numbers. To illustrate this point, consider the following simple operation 4 For 5 See

a fuller discussion, see [Wu2006]. page xiii.

OVERVIEW OF CHAPTERS 1 AND 2

with irrational numbers: (1.1)

3

√ √ √ 2 (4 × 2) + ( 3 5) 5 √ √ + = . 4 3 4 3 √

In TSM, one does not explain what √2 and 45 are, much less the meaning of 3 √ √ adding the numbers on the left, and the same can be said about the product 3 5 on the right. Let us not forget that in teaching fractions, much time is spent on the meaning and the skills of the addition and multiplication of fractions. With this in mind, how can we account for √ the fact that the addition and multiplication of more mysterious numbers such as 3 and π are passed over in silence? The answer, which TSM √ fails to provide, is that the conceptual complexity of irrational numbers such as 3 and π is beyond the level of K–12 and therefore the computation in (1.1) has to be carried out by making a formal analogy with a similar computation involving rational numbers. More precisely, since the identity a c ad + bc + = b d bd is valid for all rational numbers a, b, c, d (b, d = 0),6 a fundamental assumption in school mathematics is that this identity will remain valid for all real numbers a, b, c, d, with b, d = 0 (see page 6√for the definition of a real number). Such being the √ case, we may let a = 2, b = 3, c = 5, and d = 4 to get the previous equality (1.1). TSM seems never to discuss how teachers and their students should deal with real numbers. Its main instructional strategy seems to be one that exploits students’ passive willingness to accept orders from authority figures about what to do, reasonable or not, and since this strategy has been implemented since kindergarten, TSM has a high expectation of compliance. Therefore, having drilled students on formal computations with fractions—whose numerators and denominators are whole numbers—TSM is comfortable asking students to simply believe that the same computations can also be carried out when the numerators and denominators are real numbers. No explanation given, of course. In order to make school mathematics learnable, it is essential that we stop such underhanded maneuvers and be explicit about what we ask students to assume. This is what we have come to call the Fundamental Assumption of School Mathematics (FASM). It will be discussed at length on pp. 133ff. Although nothing like FASM exists in TSM, FASM plays a pivotal role throughout these three volumes. (The proof of FASM is given in Section 2.1 of the third volume in this series, [Wu2020c].) Finally, in reading these two chapters, please keep in mind that the emphasis here will not be on individual facts or skills. For example, it is taken for granted that you are entirely comfortable with identifying the fractions and rational numbers as certain points on the x-axis (called "the number line" in school mathematics) and are fluent in the four arithmetic operations with rational numbers. Rather, the emphasis will be on the reconstruction of familiar facts about fractions and rational numbers into a new body of knowledge that is logical and coherent, so that it can 6 The fact that this identity is true when a, b, c, d are whole numbers is perhaps well known, but TSM does not explain its validity when a, b, c, d are fractions, much less when they are rational numbers. However, one can find such an explanation on page 118 of Section 2.5.

4

1. FRACTIONS

provide a framework for you to communicate with high school students who need help. You can be certain that, as of 2020, you will often be called upon to explain fractions and rational numbers to your students. We hope that you will take the time and trouble to become thoroughly familiar with these two chapters so that you will be able to make sense of fractions and rational numbers when explaining them to your students. This will be an important first step toward improving school mathematics education. 1.1. Definition of a fraction After a brief review of the unsatisfactory state of the teaching of fractions in TSM, we introduce the number line, explain informally why fractions should be a specific collection of points on the number line, and then give the formal definition. The definition has some inherent subtleties, including the need for precision in the specification of the unit, and these are duly pointed out. We then single out an important subclass of fractions: the decimals. The fact that decimals are by definition fractions is a main feature of this volume. Leaving the past behind (p. 4) The number line (p. 5) An informal discussion (p. 7) The formal definition of a fraction (p. 9) Miscellaneous comments on the definition (p. 10) Decimal fractions and decimals (p. 14) Locating fractions on the number line (p. 15) Leaving the past behind Reasoning in mathematics requires precise definitions for each and every concept used.7 We need a definition of a fraction, not only because this is what mathematics demands, but also because students need a precise mental image for fractions to update the mental image of their fingers for whole numbers. Because 7 3 or 13 , it is incumbent on us to there is no natural image for fractions such as 11 create one for students. They cannot go through life knowing a fraction only as a piece of pizza, as TSM basically forces them to do. They will have to use fractions to compute percents for sales tax and volumes of solids for gardening chores even when no pizza is in sight. Beyond pizzas, the most common definition TSM has to offer for fractions is "parts of a whole".8 For students, the difficulty with the conception of a fraction as "parts of a whole" is multifaceted: (1) The concept of a "whole" is elusive. TSM never defines what a "whole" is. It is many things, and thus a moving target. A concept this nebulous cannot serve as a solid foundation for learning fractions. 7 Mathematical

Aside: Except undefined concepts that are part of the foundational axioms. should point out the linguistic coincidence that the word "whole" appears in both phrases, "whole numbers" and "parts of a whole". Be careful not to confuse the two. Fortunately, we will basically have no occasion to refer to "parts of a whole" in this volume beyond this informal discussion. 8 We

1.1. DEFINITION OF A FRACTION

5

(2) A fraction is a number that you compute with, but TSM does not explain in what sense "parts of a whole" is a "number" and how to compute with "parts of a whole". For example, how does "parts of a whole" logically lead to invert and multiply in the division of fractions? (3) "Parts of a whole" is at least two things: the "parts" and then there is the "whole". It is difficult to conceptualize a single number as two separate pieces. (4) There is a psychological issue. Since "the whole" means "the whole thing", how can you have more than the whole thing as in the case of 32 ? There are additional definitions of a fraction either as a "division" or a "ratio" in TSM, or even as all three things; i.e., a fraction is "parts of a whole" and a "division" and a "ratio". Theorem 1.4 on page 29 shows that indeed a fraction can be interpreted as a division, but only after "division" between arbitrary whole numbers has been carefully defined. Not before. The division interpretation of a fraction is therefore something to prove rather than something to decree by fiat as part and parcel of what a fraction is. The concept of "ratio" will be seen to be one that can be defined precisely after the division of fractions has been put in place (page 75). Thus, to define a fraction as a "ratio" is to define one abstruse concept in terms of another that is even more abstruse. Bottom line: TSM’s approach to fractions is unlearnable from the beginning. The number line We will use the number line to formulate a definition of fractions (the exposition of this chapter goes back to [Wu1998] and [Wu2002]; compare 3.NF on page 24 of [CCSSM]). The fractions will be a particular collection of points on the number line. The definition will be unambiguous, and the geometric nature of this collection of points will make it a very accessible mental image of fractions for students. The number line is the name that school mathematics gives to what is called in mathematics the real line, i.e., the x-axis. Take a line which is (usually chosen to be) horizontal and pick a point to designate as 0. The line being horizontal, one can distinguish between the left direction and the right direction on the line. We choose another point on the line to the right of 0 and designate it as 1.9 Once a choice of the two points 0 and 1 has been fixed on this line, the line is called the number line. 0 1 If a and b are two points on the number line so that a is to the left of b, then the segment from a to b consists of all the points between a and b, together with the points a and b themselves; the notation for this segment is [a, b]. (Mathematical Aside: In calculus, a segment is called a closed bounded interval.) a

b

9 By convention, 1 is to the right of 0. It could have been to the left of 0, of course, if the convention had so dictated.

6

1. FRACTIONS

The points a and b are called the endpoints of [a, b]; a is the left endpoint and b the right endpoint. The segment [0, 1] will be called the unit segment, and its right endpoint 1 will be called the unit on the number line. We will have to make precise the common notion of "equal parts" on the number line. To this end, we have to be able to decide if two segments have the same length. Two segments [a, b] and [c, d] are said to be of the same length if, by sliding one segment along the number line until their left endpoints a and c coincide, then their right endpoints b and d also coincide. For convenience, we will also express the length of [a, b] as the distance between a and b. Mathematical Aside: We intentionally use the suggestive term of "sliding" for the comparison of two segments because we are setting the stage for teaching fractions in the upper elementary grades. The mathematical terminology for "slide" is "translate". Therefore we have implicitly introduced the concept of a translation into the study of fractions (see page 234 for the definition of translation). Formally, → we are translating [a, b] along the number line from a to c (or, along the vector − ac), and we say [a, b] and [c, d] have the same length if this translation also maps b to d. This approach to fractions therefore assumes a knowledge of Euclidean geometry. There is no logical difficulty here as Euclidean geometry can be developed without reference to numbers if we so wish (see [Hilbert]). In terms of pedagogy, this exposition is set at the right level too because the amount of geometric knowledge that is implicitly assumed of school students is nothing more than what any ten-year-old would naturally take for granted. Back to the number line on which the points 0 and 1 have been chosen. We choose another point to the right of 1 so that the distance between that point and 1 is the same as the distance between 0 and 1. We designate that point as 2. Then we choose another point to the right of 2 so that the distance between that point and 2 is the same as the distance between 0 and 1. We designate that point as 3, and so on. In this way, we get an infinite sequence of equidistant points to the right of 0 (i.e., points on the line so that the distance between any two consecutive points is the same) to which we have attached the whole numbers, N, {0, 1, 2, 3, . . .}. Think of the number line as an infinite ruler, as shown: 0

1

2

3

4

We have seen that the choice of a 0 and a 1 on a horizontal line naturally leads to a sequence of equidistant points to the right of 0. For this reason, we will also refer to a horizontal line with an infinite sequence of equidistant points identified with N on its right side as the number line. By definition, a number, or more precisely, a real number is just a point on the number line.10 Mathematical Aside: It is worth pointing out that implicit in the above definition of the sequence of whole numbers, N, on the number line is the assumption that there is a way to tell the "distance" between any two points on the number line. Thus 2 is the point to the right of 1 so that the distance between 0 and 1 is the same as the distance between 1 and 2, 3 is the point to 10 Mathematical Aside: We are in effect introducing coordinates in the line, in the same way that we will later introduce coordinates in the plane in Section 6.3. By defining a point on the number line as a number, we are adopting the usual practice of identifying a point with its coordinate(s) once the coordinate system has been fixed.

1.1. DEFINITION OF A FRACTION

7

the right of 2 so that the distance between 0 and 1 is the same as the distance between 2 and 3, and so on. We will revisit this scenario when we have to introduce the distance function in the plane on page 185. We pause to note that we have just defined explicitly what a number is, namely, a point on the number line. This may not seem like much until you recall that, in TSM, the word number is bandied about repeatedly and yet nobody can say precisely what a number is. A segment [a, b] is said to have length c, where c is a point on the number line, if the segments [a, b] and [0, c] have the same length. Thus by definition, the segment [0, c] has length c. In particular, the unit segment [0, 1] has length 1 and, for this reason, we sometimes refer to 1 as the unit length or unit distance. Let it be observed that, insofar as c is just a symbol representing a point on the number line, to say [a, b] has length c means very little at the moment. However if c happens to be a whole number, say 7, then to say [a, b] has length 7 gives us a good idea of how long [a, b] is (because we know what 7 times longer than [0, 1] is). The next order of business is to expand our pool of "standard numbers" by naming more numbers on the number line, namely, the fractions. At this juncture, we can make contact with the common notion of a "whole" again. The advantage of using the number line to define fractions is that we are in effect fixing the "whole" once and for all: it will always be the length of the unit segment [0, 1]. No ambiguity, no guessing. At the same time, the number line has the flexibility to accommodate any kind of a "whole": if we want to talk about dividing a pizza, let the unit 1 on the number line be the area of a circular pizza (always assuming the pizza has uniform thickness so that only the area of the pizza matters in the division), but if we want to consider the distance of a car from a starting point, let the unit 1 be one mile. In the former case, the number 3 then stands for the total area of 3 pizzas, and in the latter case the number 3 will stand for 3 miles. We emphasize: contrary to what TSM says, the "whole" is not the unit segment [0, 1], but the length of the unit segment [0, 1]. The precision here is everything. TSM has misled students into believing that the "whole" in "parts of a whole" is an object, e.g., a segment, a square, a pizza, a glass of water, etc., whereas the correct statement is that the whole is, respectively, the length of a segment, the area of a square, the area of a pizza, the volume of a glass of water, etc. The sloppiness of the language in TSM is a main reason for much of students’ nonlearning, as we shall see. An informal discussion With the length of [0, 1] as the "whole", let us see informally where the fractions with denominator equal to 3 (i.e., 13 , 23 , 33 , 43 , etc.) should be placed on the number line. We will assume that we can divide a given segment into any number of segments of equal length (this is something intuitive and believable and, in any case, we will show how to do it in the Pedagogical Comments on page 16). The fraction 13 is one-third of the whole (= the length of [0, 1]) and, therefore, 13 is the length of one part when we divide the length of [0, 1] into 3 equal parts, i.e., 3 segments of equal length. If we divide also each of [0, 1], [1, 2], [2, 3], . . . into 3 segments of equal length, then these division points, together with the whole numbers, form

8

1. FRACTIONS

an infinite sequence of equidistant points, to be called the sequence of thirds. Then 13 is the length of any one of the short segments below, where short segment refers to a segment between consecutive points in the sequence of thirds: 0

1

2

3

Any of the following thickened short segments is "one part when the whole is divided into 3 equal parts" and is therefore a legitimate representation of 13 : 0

1

2

3

The existence of these multiple representations of 13 cries out for clarification. To this end, we introduce the following standard representation of 13 , namely, the short segment whose length is equal to 1/3 and whose left endpoint is 0. See the thickened segment below. 0

1

2

3

1 3

With respect to the standard representation, we observe that the length of this segment determines its right endpoint, and the right endpoint determines the length of this segment. Therefore, we may as well identify the standard representation of 1 1 3 with its right endpoint. Then the fraction 3 becomes identified with a point on the number line. In like manner, the fraction 53 , being the length of 5 of these short segments, has the following standard representation (see the thickened segment below). For a similar reason, we identify the standard representation of 53 with its right endpoint and proceed to denote the latter by 53 , as shown: 0

1

2

3

5 3 m 3

In general then, a fraction for a nonzero whole number m has the standard representation consisting of m adjoining short segments abutting 0, and we identify this standard representation of m 3 with its right endpoint. When m = 0, we agree to identify 03 with the point 0. Now each fraction with denominator 3 is identified with one and only one of the points in the sequence of thirds, as shown: 0 0 3

1 1 3

2 3

3 3

2 4 3

5 3

6 3

3 7 3

8 3

9 3

10 3

11 3

Note that by our convention, 03 is just 0. In terms of the sequence of thirds, each fraction m 3 is easily located: the point m is the m-th point to the right of 0. Thus if we ignore the denominator, which is 3 3, then the naming of the points in the sequence of thirds is no different from the naming of the whole numbers. Of course the consideration of fractions with denominator equal to 3 extends to fractions with other denominators. For example, replacing 3 by 5, then we get

1.1. DEFINITION OF A FRACTION

9

the sequence of fifths, which is a sequence of equidistant points obtained by dividing each of [0, 1], [1, 2], [2, 3], . . . into 5 equal parts. The first 11 fractions with denominator equal to 5 are now shown to be identified with points in the sequence of fifths, as shown: 0 1 2 0 5

2 5

1 5

3 5

4 5

5 5

6 5

7 5

8 5

9 5

10 5

11 5

Finally, if we consider all the fractions with denominator equal to n, then we would be led to the sequence of n-ths, which is the sequence of equidistant points resulting from dividing each of [0, 1], [1, 2], [2, 3], . . . into n equal parts. The fraction m n is then the m-th point to the right of 0 in this sequence. The formal definition of a fraction We will now begin the formal presentation of fractions. We say a segment [a, b] is divided into n equal parts if [a, b] is expressed as the union of n adjoining, nonoverlapping segments of the same length. (The union of a collection of sets is the totality of all the points each of which belongs to at least one of the given sets. For example, the union of the segments [0, 1] and [1, 3] is the segment [0, 3].) We implicitly assume that it is possible to divide a segment into a given number of equal parts. (Those who feel uneasy about the possibility of dividing a given segment into n equal parts for any whole number n may wish to skip to the Pedagogical Comments on page 16 to get some assurance on how to get it done.) Divide each of the line segments [0, 1], [1, 2], [2, 3], [3, 4], . . . into 3 equal parts. The totality of division points, including the whole numbers, forms a sequence of equidistant points (see p. 6 for the definition), to be called the sequence of thirds. We now attach symbols to points in the sequence of thirds, as follows. Starting from left to right, the fraction 03 is, by definition, the first point in the sequence, which is 0. For a nonzero whole number m, m is the m-th point in the sequence to 3 1 the right of 0. Thus 3 is the first point to the right of 0, 23 is the second point, 33 is the third point, etc. Note that 33 coincides with 1, 63 coincides with 2, 93 coincides with 3, and in general, 3m 3 coincides with m for any whole number m. Here is the picture: 0 1 2 3 etc. 0 3

1 3

2 3

3 3

4 3

5 3

6 3

7 3

8 3

9 3

10 3

1 The fraction m 3 is called the m-th multiple of 3 . Note that the way we have 1 just introduced the multiples of 3 on the number line is exactly the same way that the multiples of 1 (i.e., the whole numbers) were introduced on the number line. By doing to 13 exactly what we did to the number 1 in putting the whole numbers on the number line, we obtain all the whole-number multiples of 13 , i.e., all the m 3 , where m ∈ N. (The symbol "∈" means belonging to or belongs to.) In general, if a nonzero n ∈ N is given, we introduce a new collection of points on the number line in the following way. Divide each of the line segments [0, 1], [1, 2], [2, 3], [3, 4], . . . into n equal parts; then these division points (including the

10

1. FRACTIONS

whole numbers) form an infinite sequence of equidistant points on the number line, to be called the sequence of n-ths. Starting from left to right, 0 is denoted by 0 1 . The first point in the sequence to the right of 0 is denoted by n , the second n 2 3 point by n , the third by n , etc., and the m-th point in the sequence to the right 1 of 0 is denoted by m , to be called the m-th multiple of n . The sequence of n 1 n-ths therefore consists of all the whole-number multiples of n , i.e., those m points denoted by n for some whole number m. Definition. The collection of all the sequences of n-ths, as n runs through the nonzero whole numbers 1, 2, 3, . . . , is called the fractions. For a nonzero whole number m, the m-th point to the right of 0 in the sequence of n-ths is denoted by m n. The number m is called the numerator and n is called the denominator of the symbol m n . By the traditional abuse of language, it is common to say that m and n are the numerator and denominator, respectively, of the fraction m .11 By n 0 convention, 0 is denoted by n for any n. This definition of fractions accords a special status to those fractions denoted by n1 for a nonzero whole number n: the fractions are the union of all the wholenumber multiples of n1 for some nonzero whole number n. We call these n1 ’s the unit fractions. The thing to keep in mind is that we first identify a subcollection of points as fractions before affixing any symbols to them. The meaning of the unit determines how fractions are interpreted. If the unit 1 stands for 1 pound, then 3 will be interpreted as 3 pounds, but if 1 stands for 1 mile, 13 12 Thus any reasoning with fractions on the then 13 4 will stand for 4 miles, etc. number line can be interpreted as reasoning with a specific real-world situation once the unit 1 has been specified: 1 pound, 1 cc, 1 meter, etc. This kind of flexibility only comes with the abstraction of putting numbers on the number line and is one reason for defining fractions as points on the number line. In the future, we will relieve the tedium of always having to say that the denominator n of a fraction m n is nonzero by simply not mentioning it. Miscellaneous comments on the definition (A) It is self-evident that

n n

= 1,

2n n

= 2,

3n n

= 3,

4n n

= 4, and in general,

kn = k, for all whole numbers k, n, where n = 0. n Note that we have followed the convention of denoting the product of the whole numbers k and n by kn rather than k×n. Now, letting n = 1 and k = 1, respectively, we get k n = k and = 1 for all whole numbers k, n (n = 0). 1 n (1.2)

11 The 100% correct statement is of course that "m is the numerator of the symbol which denotes the fraction that is the m-th point of the sequence of n-ths, and n is the denominator of this symbol." Needless to say, almost no one in real life—inside or outside math class—ever talks like this. 12 Of course, 13 miles would be called "3 and a quarter miles" in everyday conversation, for 4 a reason to be explained on page 36.

1.1. DEFINITION OF A FRACTION

11

(B) For the study of fractions, the need for precision about what the unit is cannot be overstated. On one level, it is impossible to say which point is what fraction until the unit segment is fixed, i.e., the two points 0 and 1. Thus the following fixed point P on the given horizontal line is either 24 or 2, depending on which point is chosen to be the unit 1: 0 0

P 1

1

P

On a second level, without a precise statement about what the unit stands for, it would be impossible to say what "equal parts" means, and without that, the ambiguity would likely lead to nonlearning (see Exercise 4 on page 18). For example, if the unit 1 "stands for a cup of water" (as is commonly done in TSM), does 13 mean a third of the volume of the liquid in the cup or a third of the liquid by height (imagine that the cup is the usual curved shape and not a right circular cylinder)? Or, if the unit 1 "stands for a ham", does 13 mean a third of the meat or a third of the ham by weight including the bone? We can also give a slightly different example to expose the error resulting from such ambiguity: suppose the unit ("the whole") is a pizza and we ask what fraction is represented by putting one of the four pieces on the left below together with one of the eight pieces on the right below: '$'$ @ @

@ @ &%&% When the answer of 38 is not forthcoming, a common conclusion is that students "do not have the necessary conceptual understanding of a fraction". However, if students are taught that the "whole" is a pizza, they may very well think of 1 as the shape of the pizza, so each fraction becomes a shape and they naturally would not know how to put two shapes together to get a fraction. Students would be more likely to "get it" if, instead of saying that "the pizza represents 1", we tell them that the area of the pizza (ignoring its depth) represents 1.13 Then (assuming they know what area is) they would have a better chance of seeing that the area of a piece on the left above is equal to twice the area of a piece on the right, so that the answer to the question should be 38 . (C) We have been talking about the number line, but in a literal sense this way of speaking is incorrect. A different choice of the line or even a different choice of the positions of the numbers 0 and 1 would lead to a different number line. What is true, however, is that anything done on one number line can be done on any other in exactly the same way,14 and therefore we may—and do—identify all of them. Now it makes sense to speak of the number line. 13 The need for precision about the unit exposes the common fallacy that introducing students to fractions via pizzas is good pedagogy. The area of a curvilinear figure like the disk is too sophisticated for children. The length of a segment or the area of a rectangle is a better alternative. 14 Mathematical Aside: All number lines are similar to each other in the sense of the definition on page 284 so that they can be identified via similarity. Algebraically, the real numbers form a complete ordered field, and since all complete ordered fields are isomorphic, we can also identify them via isomorphism.

12

1. FRACTIONS

(D) Although a fraction is formally a point on the number line, the informal discussion above makes it clear that on an intuitive level, a fraction m n is just the segment [0, m ]. So in the back of our minds, the segment image should never go n away completely, and this fact is reflected in the language we now introduce. First, we give a definition. Definition. The concatenation of two segments L1 and L2 on the number line is the line segment obtained by putting L1 and L2 along the number line so that the right endpoint of L1 coincides with the left endpoint of L2 . L1

L2

Thus the segment [0, m n ] is the concatenation of exactly m segments each of m length n1 , namely, [0, n1 ], [ n1 , n2 ], . . . , [ m−1 n , n ] (recall the concept of the length of a m 1 1 segment on p. 7). Because we identify [0, n ] with the point m n , and [0, n ] with n , it m is natural to adopt the following suggestive terminology. We say n is m copies of 1 to mean that the segment [0, m n ] is the concatenation of exactly m segments each n of length n1 . More generally, by m copies of k we mean the segment obtained by concatenating m segments each of length k .

(E) In school mathematics, the meaning of the equal sign is a subject that is much discussed (see, e.g., Chapter 2 of [Carpenter et al.]), mainly because the meaning of equality is never made clear. In addition, you will find later on the traditional use of the word equivalent for fractions when equal is meant. Such a cavalier attitude towards "equality" only adds to the confusion. For this reason, we make explicit the fact that, by definition, two fractions k and m are equal n (or, equivalent), in symbols, k = m , if they are the same point on the number n

k line. We have already seen in equation (1.1), for example, that kn n = 1 = k for any n, k ∈ N. Incidentally, because there is no definition for fractions in TSM, TSM has no precise meaning for the equality of two fractions k and m n. (F) The definition of a fraction as a point on the number line allows us to make precise the concept of order among fractions, i.e., the concept that one fraction is smaller than (or bigger than) another. First, consider the case of whole numbers. The way we put the whole numbers on the number line, a whole number m is smaller than (or less than) another whole number n (in symbols, m < n) if m is to the left of n (thus [0, m] is shorter than [0, n]). We expand on this fact by defining, for two points A and B on the number line, A is smaller than B if A is to the left of B. In symbols, A < B. We call this an inequality. In particular, if A and B are fractions, A < B means A is to the left of B. A B

Thus, 43 < 53 because, in the sequence of thirds, the 4th member in the sequence is to the left of the 5th member of the sequence (see page 9). There are two things about the definition of A < B that are worth noting here. One is that, according to this definition, one cannot compare two given fractions until both fractions are placed on the same number line. More to the point, what

1.1. DEFINITION OF A FRACTION

13

this says is that we can compare fractions only when they refer to the same unit. The other is that in TSM, the concept of A < B between fractions is never defined beyond inscrutable statements such as "A < B when B names a greater amount than A". The reason for this omission is obvious: since there is no definition for the concept of a fraction, a fraction is an unknown object and, therefore, it is impossible to say how one unknown object A can be smaller than another unknown object B. Sometimes the inequality B > A is used in place of A < B. Then we say B is bigger (or greater) than A. This is the place to mention two related symbols. A ≤ B means either A < B or A = B. Sometimes we refer to an inequality such as A ≤ B as a weak inequality. For example, the weak inequality 12 ≤ 12 may seem odd at first glance, but when it is realized that all it says is either 12 < 12 or 1 1 1 1 2 = 2 , we have to agree that 2 ≤ 2 is a correct statement since it is correct that 1 1 2 = 2 . The other new symbol is A ≥ B, which means either A > B or A = B. Analogously, it is also correct to say that 12 ≥ 12 . For the purpose of defining A < B when two points A and B are given on the number line, all that is required is that the right-pointing direction be singled out on the line. In greater detail, since the right-pointing direction is singled out on the number line by the placement of 1 to the right of 0, the meaning of A < B is that, when going towards the right, we encounter the point A before encountering B. Thus, for the concept of order, the distance between 0 and 1 is of no consequence; it is the "direction" set by going from 0 to 1 that matters. In Chapter 4, when we consider lines in the plane that are not horizontal (see page 231), the idea of a "direction" will be seen to be crucial. (G) With the availability of the concept of order among fractions, we can revisit the concept of a segment AB, where A and B are two points on the number line and A is to the left of B (see page 5). By definition, AB consists of the two points A and B, together with all the points C between A and B, which means all the points C so that C is to the right of A and to the left of B. A

C

B

In terms of the preceding definition of order, we may rephrase AB as the collection of all the points (numbers) C so that A ≤ C ≤ B. We note that, in view of the observation at the end of remark (F), the concept of a point between two other points on the number line will make sense as soon as a right-pointing direction is specified. Since specifying a right-pointing direction is equivalent to specifying a leftpointing direction, we could have rephrased the preceding discussion in terms of a left-pointing direction. For a more elaborate discussion of this and related ideas, see the discussion of betweenness on pp. 167ff. and the discussion on pp. 231ff. k (H) A final remark has to do with the fact that a fraction such as 23 , 14 5 , or  is one number. Students are known to raise the issue of why three symbols—k, , and the "fraction bar"—are needed to denote one number. Remember that a fraction is a point on the number line, so that the symbols employed are merely means to an end; namely, they serve to indicate where each point is located on the number 14 line. Thus the symbol 14 5 says precisely that 5 is the 14-th point in the sequence of 5-ths to the right of 0. Clearly, every part of the symbol 14 5 is needed for this purpose, namely, the number 5, the number 14, and the "fraction bar" in between.

14

1. FRACTIONS

The need for 5 and 14 is obvious; the role of the "fraction bar" is to separate the 14

5 from the 14 so that, for example, one does not confuse 5 with 145. It is with the same need for separation in mind that when a fraction such as 14 5 is sometimes shown horizontally as 14/5, there is a slant bar between the 14 and the 5. This piece of information about what makes up the fraction symbol should be clearly conveyed to students in grades 5–7. Decimal fractions and decimals We now single out a special class of fractions: fractions whose denominators are of the form 10n for some positive integer n, e.g., 1489 24 58900 , , . 102 105 104 These are called the decimal fractions, but they are better known in a more common notation under a slightly different name, to be described presently. Decimal fractions were understood and used in China by about 400 AD, but most likely they were transmitted to Europe as part of the so-called Hindu-Arabic numeral system only around the twelfth century. In 1593, the German Jesuit priest C. Clavius—the Vatican astronomer who was the main architect of the Gregorian calendar—introduced the idea of writing a decimal fraction without the fraction symbol: just use the numerator and then keep track of the number of zeros in the denominator (2 in the first decimal fraction, 5 in the second, and 4 in the third of the above examples) by the use of a dot, the so-called decimal point; thus, 24 58900 1489 , 0.00024 = 5 , 5.8900 = , (1.3) 14.89 = 102 10 104 respectively (see [Ginsburg]). The rationale of the notation is clear: the number of digits to the right of the decimal point, the so-called decimal digits, keeps track of the power of 10 in the respective denominators, 2 in 14.89, 5 in 0.00024, and 4 in 5.8900. In this notation, these numbers are called finite or terminating decimals.15 Until we introduce the concept of infinite decimals (in Chapter 3 of the third volume, [Wu2020c]), we will usually omit any mention of "finite" or "terminating" and just say decimals. Notice the convention that, in order to keep 24 track of the power 5 in 10 5 , three zeros are added to the left of 24 to make sure that there are 5 digits to the right of the decimal point in 0.00024. Note also that the 0 in front of the decimal point is only for the purpose of clarity and is optional. You may be struck by the odd-looking number 5.8900, because you have probably been told by TSM that it is ok to omit the zeros at the right end of the decimal point and simply write 5.89. Before going any further with this thought, just be aware that, according to the definition of a decimal, to say 5.8900 is equal to 5.89 is to say that the following two fractions are equal: 589 58900 and . 104 102 This fact is correct, but if so, we must be able to prove that it is correct before we can use it. We will give a proof in the next section. 15 Regardless of what TSM has to say, this is the correct definition of a finite decimal, one that has been adopted by CCSSM.

1.1. DEFINITION OF A FRACTION

15

Locating fractions on the number line We conclude this section by giving some examples of locating fractions on the number line. Let us start with 43 , for example. As usual, this is the fourth point to the right of 0 in the sequence of thirds: 0

4 3

1

Activity. Can you locate the fraction

20 15 ?

How is it related to 43 ?

Next we consider the problem of locating a fraction such as 84 17 , approximately, on the number line; i.e., on the following line, where should 84 17 be placed, approximately? 0

1

2

3

4

5

6

7

8

The key idea will turn out to be the use of division-with-remainder for whole numbers.16 First, look at the multiples of 17: 0, 17, 34, 51, 68, 85, . . . . Thus the 1 1 68-th multiple of 17 is 4 (because 68 = 4 × 17), and the 85-th multiple of 17 is 5 84 68 85 (because 85 = 5 × 17). Therefore, 17 lies between 17 (= 4) and 17 (= 5) and is just 1 84 1 17 shy of 5; i.e., 17 is the point on the number line which is 17 to the left of 5. In terms of division-with-remainder, since 84 = (4 × 17) + 16, we have (4 × 17) + 16 84 = . 17 17 1 So if each step we take is of length 17 , going another 16 steps to the right of 4 will 84 get us to 17 . If we go 17 steps instead, we will get to 5. Therefore 84 17 should be quite near 5, as shown:

0

1

2

3

4

5

6

7

8

6

84 17

In general, if m n is a fraction and division-with-remainder gives m = qn + k, where q and k are whole numbers and 0 ≤ k < n, then qn + k m = , n n qn and the position of m n on the number line will be between q (= n ) and q + 1 (= (q+1)n , which is qn+n n n ). 16 Caution. Because the above reasoning gives 84 17 as 17 beyond 4, most school 84 16 textbooks would tell you that 17 = 4 17 . The latter is of course an example of k a mixed number. Similarly, the above m n is supposed to be written as q n . As a teacher, however, you should exercise self-control not to introduce the concept of 16 "Division-with-remainder" in school mathematics is what is usually called the "division algorithm" in abstract algebra texts. There is a good reason why the latter terminology is not used in school mathematics: it would be too easily confused with the "long division algorithm".

16

1. FRACTIONS

a mixed numbers or the symbol 4 16 17 at this point. If you do, what would you tell your students about the meaning of the symbol 4 16 17 ? Do not say, as TSM does, that it is 4 and 16 , because how would your students interpret the word "and" in 17 this context?17 It actually means the addition of the fraction 4 and the fraction 16 17 , as in 4 + 16 . But fraction addition has not yet been defined, so you would confuse 17 students by abusing the word "and" here. TSM has a habit of using this word "and" inappropriately. See page 35 for another flagrant example. Pedagogical Comments. It is easy to divide a given segment into 2 equal parts, 4 equal parts, 8 equal parts, etc. However, many teachers have raised the issue of how to divide a given segment into 3, 7, or in general n equal parts, where n is not a power of 2. There are two answers to this question. The first one is to cheat, but to cheat honestly. Suppose you want to show a division of a unit segment into 3 equal parts. What you do is not to start with the unit segment but to start with a short segment (say, 12 inch long) that will be your 13 (see the thickened segment below) and then duplicate it two more times. Now you declare the resulting segment with the built-in equal division into thirds to be your unit segment: 0

1 





Of course the same trick works for equal division into n equal parts for any nonzero whole number n. A second answer is to confront the issue directly: given a segment, we will show how to use plastic triangles and compass to divide it into any number of equal parts. Suppose we have to divide a given segment AB into 3 equal parts. Referring to the picture below, we draw an arbitrary ray L (see page 174 for the precise definition) issuing from A and, using a compass, mark off three points C, D, and E in succession on L so that AC, CD, and DE have equal length, the precise length of AC being irrelevant. For example, if you make each of AC, CD, and DE half an inch long, then you do not even need a compass. Join BE, and through C and D draw lines parallel to BE 18 that intersect AB at C  and D , respectively. The points C  and D are then the desired division points on AB to achieve the equidivision, i.e., AC  , C  D , and D B are of equal length, as shown:

17 To understand why this confuses students, if "4 and 16 " is all they know about 4 16 , we 17 17 3 × 12 17 ? can look ahead and ask how they can compute with something like 4 16 17 18 The use of plastic triangles for this purpose is probably well known, but if not, see pp. 20–22 of [Wu2002] or pp. 243–244 of [Wu2016a].

1.1. DEFINITION OF A FRACTION

17

 L  E       D         C              A C D B In Section 7.1 of [Wu2020b], there is a proof of why AC  , C  D , and D B are of equal length. It is clear how to modify this construction if AB is to be divided into n equal parts for any nonzero whole number n. End of Pedagogical Comments.

Exercises 1.1. In doing these and subsequent exercises, please observe the following basic rules: (a) Show your work; your explanation is as important as your answer. (b) Be clear. Get used to the idea that, as a teacher, everything you say has to be understood. (1) Indicate the approximate position of each of the following on the number 1257 77 132 line, and briefly explain why: (a) 1.24, (b) 186 11 , (c) 132 , (d) 355 , (e) 1257 . (2) Suppose the unit 1 on the number line is the area of the following region enclosed by the thickened segments, where the given square is divided into eight congruent rectangles (and therefore eight parts of equal area):19

Observe that a region whose area is 18 of the area of the given square is the fraction 15 relative to this unit. In terms of this unit, what is the fraction represented by the area of each of the following shaded regions of the same square? Give a brief explanation of your answer. (In the picture in the middle, the square is divided into eight congruent rectangles, and in the picture on the right, two copies of the same square share a common side and the square on the right is divided into four parts of equal area.)

19 We will give a precise definition of congruence in Chapter 4 and will formally discuss area in Chapter 4 of [Wu2020c]. In this chapter, we only make use of both concepts in the context of triangles and rectangles, and then only in the most superficial way. For the purpose of understanding this chapter, you may therefore take both concepts in the intuitive sense. If anything more than intuitive knowledge is needed, it will be supplied on the spot, e.g., on page 47 later on in this chapter.

18

1. FRACTIONS

@ @ @ @ @ @ (3) With the unit as in Exercise 2 above, write down the fraction representing the area of the following shaded region (assume that the top and bottom sides of the square are each divided into three segments of equal length):

(Hint: If you divide the given square horizontally into 8 congruent rectangles, then you can figure out the area of the shaded region in terms of 1 ’s of the area of the given square.) the 24 (4) It was emphasized in the text (page 11) that the concept of "equal parts" should be made precise in the teaching of fractions. This exercise gives one example to illustrate this need. There are numerous others (see, e.g., the case of Two Green Triangles on page 86 of the case book [BarnettGoldstein-Jackson]). A text on professional development claims that students’ conception of "equal parts" is fragile and is prone to errors. As illustration, it says that when a circle is presented as in the left picture below to students, '$ '$ QQ &% &% 2 they have no trouble shading 3 . However, it goes on to say that when these same students are asked to construct their own picture of 23 , we often see them create pictures with unequal pieces as in the right picture above. (a) Referring to the latter presentation of "thirds" by students, in what sense are the three pieces "equal"? In what sense are they "unequal"? (b) What would you do as a teacher to prevent students from acquiring such a misconception about "equal parts"? (5) Ellen ate 15 of a large pizza with a 10-inch diameter and Kate ate 14 of a smaller pizza with an 8-inch diameter. (Assume that all pizzas have the same thickness and that the fractions of a pizza are measured in terms of area.) Ellen told Kate that since she had eaten more pizza than Kate, 1 1 5 > 4 . (i) Did Ellen eat more pizza than Kate? (ii) Is Ellen’s assertion that 15 > 14 correct? Explain why or why not. (You are allowed to use the usual area formula for a circle.) (6) (Review remark (B) on page 11 on the importance of the unit before doing this exercise. Also make sure that you do it by a careful use of the definition of a fraction rather than by some intuition you possess that you cannot explain to your students.)

1.2. EQUIVALENT FRACTIONS

19

(a) After driving 218 miles, we have gone only two-thirds of the distance we planned to drive for the day. How many miles did we plan to drive for the day? Explain. (b) After reading 236 pages of a book, I am exactly four-fifths of the way through. How many pages are in the book? Explain. (c) Alexandra was three-quarters of the way to school after having walked 0.72 miles from home. How far is her home from school? Explain. (7) Take a pair of opposite sides of a unit square and divide each side into 7 equal parts. Join the corresponding points of division to obtain 7 thin rectangles (we will assume that these are rectangles). For the remaining pair of opposite sides, divide each into 5 equal parts and also join the corresponding points of division; these lines are perpendicular to the other 7 lines. The intersections of these 7 and 5 lines create 7×5 small rectangles which are congruent to each other (we will assume that too). What is the area of each such small rectangle, and why? (Compare page 48 below.) (8) Three segments (thickened) are on the number line, as shown:

A 3

137 25

B 4

5

C 6

7

It is known that the length of the left segment is 11 16 , that of the middle 8 23 segment is 17 , and that of the right segment is 25 . What are the fractions A, B, and C? (Caution: Remember that you have to explain your answers, and that you know nothing about "mixed numbers" until we come to this concept on page 36 below.) (9) The following is found in a certain third-grade workbook: Each of the following figures represents a fraction:

Point to two figures that have the same fractions shaded. Does this problem make sense as it stands? If so, explain your answer clearly. If not, how would you rephrase it so that it makes sense? 1.2. Equivalent fractions We prove in this section the fundamental theorem in the subject of fractions: the theorem on equivalent fractions. A first consequence of this theorem is the allimportant cross-multiplication algorithm, which is mistaken to be a "rote skill" in

20

1. FRACTIONS

the mathematics education literature. Unlike TSM, we also explicitly define the concept of k of m and use it to prove the division interpretation of a fraction, n which is erroneously considered in TSM to be part of the definition of a fraction. The fundamental theorem (p. 20) The cross-multiplication algorithm (p. 21) (p. 24) The concept of k of m n The division interpretation of a fraction (p. 28) The fundamental theorem Recall that two fractions are said to be equal (or equivalent) if they are the same point on the number line. For example, we observed in equation (1.2) on k page 10 that nk n = 1 , as both are equal to k. The following is a generalization of this simple fact. Theorem 1.1 (Theorem on equivalent fractions). Given two fractions m n and k , suppose there is a nonzero whole number c so that k = cm and  = cn. k Then m n = .

Theorem 1.1 is usually stated more briefly as follows: m cm m = for all fractions (1.4) and all whole numbers c = 0. n cn n In this form, Theorem 1.1 is sometimes called the cancellation law for fractions, because one simply "cancels" the whole number c from the numerator and the denominator. This is the justification for the usual method of reducing fractions, 3 e.g., 51 34 = 2 because we can cancel the common factor 17 from the "top and bottom" of the fraction: 17 × 3 3 51 = = . (1.5) 34 17 × 2 2 We will explain below (see page 22) why we choose to state Theorem 1.1 in this clumsy fashion. Proof. First look at a special case: why is 43 equal to 5×4 5×3 ? We have as usual the following picture: 0

1

4 3

Now suppose we further divide each of the segments between consecutive points in the sequence of thirds into 5 equal parts. Then each of the segments [0, 1], [1, 2], [2, 3], . . . is now divided into 5 × 3 = 15 equal parts and, in an obvious way, we have obtained the sequence of fifteenths on the number line: 0

1

4 3

The point 43 , being the 4-th point in the sequence of thirds, is now the 20-th point in the sequence of fifteenths because 20 = 5 × 4. The latter is by definition the 5×4 4 5×4 fraction 20 15 , i.e., 5×3 . Thus 3 = 5×3 .

1.2. EQUIVALENT FRACTIONS

21

The preceding reasoning is enough to prove the general case. Thus let k = cm k and  = cn for whole numbers c, k, , m, and n. We will prove that m n =  . In other words, we will prove equation (1.4) above. The fraction m n is the m-th point in the sequence of n-ths. Now divide each of the segments between consecutive points in the sequence of n-ths into c equal parts, so that each of [0, 1], [1, 2], [2, 3], . . . is now divided into cn equal parts. Thus the sequence of n-ths, together with the new division points, becomes the sequence of cn-ths. Simple reasoning shows that the m-th point in the sequence of n-ths must be the cm-th point in the sequence of cn-ths. This is another way of saying m cm n = cn . The proof is complete. As mentioned earlier (page 12), it is a tradition in school mathematics to say k that two fraction (symbols) k and m n are equivalent if they are equal, i.e., if  and m n

are the same point (see page 12). In this terminology, Theorem 1.1 gives a

sufficient condition for two fractions k and m n to be equivalent, and this accounts for the name of the theorem. There seems to be little awareness of the power of Theorem 1.1 on equivalent fractions in TSM. Consequently, the role played by Theorem 1.1 in the TSM curriculum is a minimal one: it is mostly used to reduce fractions. This is wrong, because Theorem 1.1 is the fundamental fact about fractions. As a first demonstration of this claim, we now use Theorem 1.1 to bring closure to the discussion on page 14 of the last section about the decimal 5.8900. Recall that we had, by definition, 58900 = 5.8900. 104 We will show that 5.8900 = 5.89 and, more generally, one can append zeros to or delete zeros from the right end of a finite decimal—to the right of the decimal digits—without changing the number. Indeed, 589 × 102 589 58900 = 2 = 2 = 5.89, 4 2 10 10 × 10 10 where the third equality makes use of the cancellation law (1.4). The reasoning is of course valid in general; e.g., 5.8900 =

1270000 127 127 × 104 = = 12.70000. = 4 10 10 × 10 105 The rest of this chapter may be said to be nothing more than an extended demonstration of the importance of Theorem 1.1. 12.7 =

The cross-multiplication algorithm We have just seen that Theorem 1.1 gives a sufficient condition for two fractions and k to be equal; there is a nonzero whole number c so that k = cm and  = cn. There is an obvious interest in such a sufficient condition because each symbol represents a point on the number line and one would like to be able to decide whether the two symbols represent the same point or not. On the other hand, the condition in Theorem 1.1 is not a necessary condition, in the sense that k the equality m n =  does not imply that k = cm and  = cn for some whole number m n

22

1. FRACTIONS

c. For example, Theorem 1.1 shows that 32 = 21 14 (as 21 = 7 × 3 and 14 = 7 × 2), so that coupled with (1.5) on page 20, we have 51 21 = . 14 34 However, there is clearly no whole number c so that c times 21 yields 51 or that the same c times 14 yields 34. It turns out that, with a mild twist, Theorem 1.1 can be used to give a necessary and sufficient condition for two fractions to be equal. Precisely: Theorem 1.2 (Cross-multiplication algorithm). A necessary and sufficient condition for two fractions k and m n to be equal is that kn = m. In the context of Theorem 1.2, it becomes clear why we chose to state Theorem 1.1 in that clumsy way: it exhibits the close relationship between the theorem on equivalent fractions and the cross-multiplication algorithm. For later needs, we pause to note that there are several different but equally valid ways to state Theorem 1.2. One way is to say that m k = if and only if kn = m.  n Another says k m = is equivalent to kn = m.  n A more symbolic way is k m = ⇐⇒ kn = m.  n No matter how the theorem is stated, all it says is that both of the following statements are valid: First, k = m n implies kn = m. Second, kn = m implies k = m n. As is well known, each is said to be the converse of the other. Proof of Theorem 1.2. Part 1. We prove k = m n implies kn = m. By Theorem 1.1, m m k m n = n . Because we are assuming  = n , we therefore have

k 

=

kn n

and

kn m = . n n 1 is equal to the m-th multiple of What this says is that the kn-th multiple of n 1 . This is possible only if kn = m. n Part 2. We next prove kn = m implies k = m n . The hypothesis implies that kn m = . n n k m By Theorem 1.1, the left side is k while the right side is m n . Thus we have  = n . The proof of Theorem 1.2 is complete.

1.2. EQUIVALENT FRACTIONS

23

Pedagogical Comments. One can see the pernicious impact of TSM on school mathematics education when the fundamental Theorem 1.2 is mistaken for a rote skill in the mathematics education literature. To the contrary, once the concept of a fraction has been clearly defined, so that the concept of the equality of two fractions can also be clearly defined, Theorem 1.2 becomes a provable theorem. Let us point out the obvious: a theorem in mathematics is never a rote skill. In this case, more can be said: this is actually an important theorem because it often provides the only reasonable method to check whether two fractions are equal; see, for example, the proof of Theorem 1.3 on page 27. Therefore, the cross-multiplication algorithm has to be an integral part of every student’s mathematical survival kit. We explicitly ask you to teach your students to be proficient in making use of this fundamental result at every opportunity. As a rather trivial application of Theo203 rem 1.2, we see that 551 247 and 91 are equal because 551 × 91 = 203 × 247. End of Pedagogical Comments. Mathematical Aside: From the vantage point of abstract algebra, the importance of Theorem 1.2 (and hence of Theorem 1.1) is manifest because the crossmultiplication algorithm is exactly the equivalence relation between ordered pairs of integers when fractions are defined as equivalence classes of such ordered pairs: (a, b) ∼ (a , b ) if and only if ab = a b. From the proof of Theorem 1.2, we can extract a very useful statement about pairs of fractions, which we call the Fundamental Fact of Fraction-Pairs (FFFP): Any two fractions may be symbolically represented as two fractions with the same denominator. Such a denominator is called a common denominator of the two fractions. In other words, there will always be some whole number q, so that these two fractions belong to the same sequence of q-ths. The reason is simple: we can simply take k q = the product of the denominators because, if the fractions are m n and  , then by Theorem 1.1, we have m k nk m = and = . n n  n The two fractions are now seen to be the m-th and nk-th members in the sequence of n-ths. That said, we should also call attention to the fact that, in some special cases, some fractions can be put on equal footing without having to multiply their denominators. For example, we can use 12 as a common denominator for 32 and 9 3 12 8 because 2 = 8 . However, knowing that the product of the denominators in question always works creates a "comfort zone" in such considerations. We can paraphrase FFFP this way: any two fractions can be put on an equal footing, in the sense that they can always be put in the same sequence of q-ths for some q so that they become directly comparable. In the notation of (1.6), if m < nk, then in the sequence of n-ths, m n (being the m-th member) is to the left of k (being the nk-th member) and is therefore the smaller of the two. An analogy is to compare 155 inches with 4 meters: one cannot get a sense of which is longer until both measurements are put on an equal footing in the sense that they are expressed in terms of the same unit, e.g., an inch. Then since 1 inch is 2.54 cm, 155 inches is 155 × 2.54 = 393.7 cm = 3.937 meters, which is shorter than 4 (1.6)

24

1. FRACTIONS

meters.20 This is how we can tell that 4 meters is longer. In the same way, given 23 15 2 5 and 57 , we may replace them with 14 21 and 21 , respectively, and conclude that 3 < 7 . There will be numerous applications of FFFP in subsequent discussions. The concept of

k 

of

m n

We will give a precise meaning to a common expression, "two-thirds of something", or more generally, " k of something", and show how to use Theorem 1.1 to compute the precise value in general. There are linguistic traps here that we would do well to avoid, as we now explain. Consider what is meant by "I ate two-thirds of a pie." A little thought would reveal that it means I looked at the pie as a circular disk by ignoring its depth, cut it into 3 parts of equal area, and then ate 2 parts. So "two-thirds of the area of the pie" is implicitly understood. Another example: what is meant by "he gave threefifths of a bag of rice to his roommate"? Most likely, he measured his bag of rice by weight and, after dividing the bag of rice into 5 equal parts by weight, he gave away 3 parts. Here "three-fifths of the weight of the rice" is again implicitly understood. These examples point to the ambiguity in the common language because a choice of the unit (area in the first and weight in the second) is imbedded in the language and the reader is implicitly expected to "get it". In mathematics, there is no room for such ambiguity. To drive home this point, consider a similar statement: "I put away three-quarters of the ham." This time, there is a lot of room for different interpretations. This could mean "three-quarters of the ham that is 24 inches long" (so I put away 18 inches of it), or "three-quarters of the whole ham that weighs 8 lbs" (so I put away 6 lbs. of the ham), or "three-quarters of the 3.2 lbs. of meat from the ham" (so I put away 2.4 lbs. of ham meat). These examples illustrate the fact that, for the purpose of doing mathematics, the "something" in " k of something" has to be a number referring to a precise unit, i.e., has to be a point on the number line where the unit is clearly specified. Because the numbers known to us up to this point are fractions, the following definition will directly refer to this "something" as a fraction m n , except in Section 1.5 in is replaced by the length of an arc, which can be [Wu2020c] for the case where m n an arbitrary positive number. k m Definition. Let k and m n be fractions. Then  of n means the total length m of k parts when the segment [0, n ] is partitioned into  equal parts.

For a reason that will become clear in the following subsection, we have intentionally used the word "partition" in place of the usual word "divide" in the preceding definition. Pedagogical Comments. The preceding definition is among the most conceptually complex ones in elementary school mathematics, but we strongly recommend

20 We freely make use of the multiplication of final decimals here because we are only giving an analogy but not giving a formal mathematical proof.

1.2. EQUIVALENT FRACTIONS

25

that extra time be spent on learning it because it is critical for understanding fraction multiplication (and therefore also for understanding fraction division). End of Pedagogical Comments. Consider, for example, the case of 1 3

of 24 7 .

This is then the length of 1 part when the segment [0, 24 7 ] is partitioned into 3 parts 3×8 24 = , so that [0, ] is 3 copies of 87 (see page 12 for the of equal length. Now, 24 7 7 7 8 meaning of "copy"). Thus 13 of 24 7 is 7 . The key point here is that the numerator 24 of 7 is divisible by 3. Next, suppose we want 2 5

of 87 .

Now we have to partition [0, 87 ] into 5 equal parts and then measure the length of 2 of those parts. But first things first: we have to partition 87 into 5 equal parts. Noting that 8 is not divisible by 5, we make use of equivalent fractions to get a fraction equivalent to 87 so that its numerator is divisible by 5; i.e., 87 = 5×8 5×7 . The

8 numerator 5 × 8 is now divisible by 5, and 87 is seen to be 5 copies of 5×7 , and we 8 conclude that if [0, 7 ] is partitioned into 5 equal parts, each part would have length 8 35 .

Two of these parts then have length 2×8 35 = Pictorially, we first locate

0

8 7

16 35 .

Thus,

2 5

of

8 7

is

in the sequence of sevenths: 8 7

1

8 7

is 8 copies of 17 . Now subdivide each of these 8 segments of length parts, as shown: 0

16 35 .

1

1 7

into 5 equal 8 7

The unit segment is now partitioned into 5 × 7 = 35 equal parts, so that the new division points furnish the initial points in the sequence of 35-ths. The segment [0, 87 ] is now partitioned into 40 equal parts by this sequence of 35-ths. If we take every 8th division point in this sequence of 35-ths, starting with 0, then we get a partition of [0, 87 ] into 5 equal parts. So the length of a part in the latter partition 8 is 35 . (Of course, what we have done is merely to reprove the theorem on equivalent fractions in the special case of 87 = 5×8 5×7 .) This way of exploiting equivalent fractions will be seen to clarify many aspects of fractions, such as the interpretation of a fraction as division (next subsection) or the concept of multiplication (Section 1.4). It also allows us to solve word problems of the following type. Example. Prema walked 25 of the distance from home to school, and there was still 49 of a mile to go. How far is her home from school? Solution. We can draw the distance from home to school on the number line, with 0 being home, the unit 1 being a mile, and S being the distance of the school

26

1. FRACTIONS

from home.21 Then it is given that, when the segment from 0 to S is partitioned into 5 equal parts, Prema was at the second division point after 0: 0

Prema

S

 4 9

 mi



For convenience, call any one of these five segments a short segment. Then the total distance from home to school is 5 times the length of a short segment, and we are given that the distance from where Prema stands to S comprises 3 short segments. We are also given that the distance from where Prema stands to S is 49 of a mile. If we can find out how long a third of 49 of a mile is, then we will know the length of a short segment and the problem will be solved. By the theorem on equivalent fractions, we can easily change 49 to an equivalent fraction whose numerator is divisible by 3; e.g., 3×4 3×4 4+4+4 4 = = = , 9 3×9 27 27 4 and this exhibits 49 as 3 copies of 27 . Therefore a third of 4 from 0 to S is thus 5 copies of 27 , which is

4 9

is

4 27 .

The total distance

20 4+4+4+4+4 = . 27 27 The distance from Prema’s home to school is therefore 20 27 miles. We would like to point out that the preceding problem is one of the standard problems on fractions which is usually given after the multiplication of fractions has been introduced, and the solution method is given out as a rote algorithm ("flip over (1 − 25 ) to multiply by 49 "). However, we now see that there is no need to use multiplication of fractions for the solution, and, in addition, the reasoning behind the present method of solution is so simple that there is no need to memorize any solution template. As a final remark on the concept of " k of m n ", we prove the following theorem that will be useful for our consideration of fraction multiplication in Section 1.4. First, we obtain a general formula for k of m n , which will be part (i) of the theorem. 51 As motivation for part (ii) of the theorem, recall the fact proven above that 21 14 = 34 . m If n is a fraction, then we expect the following to hold: 21 14

51 m of m n = 34 of n . But according to the definition of "of " on page 24, this equality asserts that if [0, m n] is divided into 14 equal parts, then the length of 21 concatenated parts would be equal to the length of 51 concatenated parts when the same segment [0, m n ] is divided M into 34 equal parts. This is not obvious. Moreover, if m = , then we also expect n N that 21 14

51 M of m n = 34 of N .

21 You may notice that the unit 1 is not shown in the picture. This is because we do not know ahead of time whether S > 1 or S < 1, so we cannot place 1 on this number line until the problem has been solved.

1.2. EQUIVALENT FRACTIONS

27

This equality clearly needs a proof, and part (ii) of the following theorem will take care of that. Finally, the notation in the theorem is worthy of a comment. We will be dealing with four fractions which will be assumed to be equal in pairs. So we use lower case k and upper case K L to denote the same fraction in part (ii) to ease M the memory load somewhat. The same is true for m n and N . K M Theorem 1.3. Let k , m n , L , and N be fractions. Then: km (i) k of m n = n . m M (ii) If k = K L and n = N , then k m K M  of n = L of N .

Proof. We first prove part (i); i.e., (1.7)

k m km of = .  n n

The left side is the length of k concatenated parts when [0, m n ] is divided into  equal parts. Because 

   m m m + m+ ··· + m = = , n n n m we see that [0, m n ] is  copies of n (see page 12 for the meaning of "copy"). Therefore m if we divide [0, m n ] into  equal parts, each part will have length n . It follows that if we concatenate k of these parts, the total length will be k

   m + m+ ··· + m km = . n n This proves (1.7). In like manner, we have K L

KM of M N = LN .

KM Hence, to prove part (ii) of the theorem, we must prove km n = LN . According to Theorem 1.2 (cross-multiplication algorithm), this would be the case if we can prove kmLN = nKM . In other words, we have to prove

(kL)(mN ) = (K)(nM ). By the assumption that k = K L and by Theorem 1.2 , we have kL = K. SimiM larly, the assumption that m n = N leads to mN = nM . Therefore (kL)(mN ) = (K)(nM ), as claimed. The proof of Theorem 1.3 is complete.

It remains to observe that equation (1.7) has a remarkable consequence. By k m the definition of " k of m n ", the relationship between  and n would not seem to be symmetric, in the sense that there is no reason to believe—strictly according to the definition on page 24—that (1.8)

m m k k of = of .  n n 

28

1. FRACTIONS

However, (1.7) tells us that, because multiplication between whole numbers is commutative, the equality (1.8) is actually correct. The division interpretation of a fraction Using the idea of " k of m n ", we now give a completely different interpretation of a fraction, the so-called division interpretation. We will prove (n = 0): m n (1.9)

= the length of one part when [0, m] is partitioned into n equal parts.

1 Recall that the original definition of m n is m copies of n (see page 12 for the m definition of copy), which means that to locate n , it suffices to consider the unit segment [0, 1], partition it into n equal parts, and concatenate m of these parts. The above statement, to the contrary, says that to locate m n , one can partition, not [0, 1] but [0, m] into n equal parts and then take the first division point to the right of 0. So the two are quite different statements.

Proof of (1.9). We observe that, by the definition of "of " on page 24, the right side of (1.9) is equal to n1 of m, i.e., n1 of m 1 , which, by Theorem 1.3(i) on 1×m m page 27, is equal to n×1 = n , the left side of (1.9). The proof of (1.9) is complete. Due to the importance of (1.9), we now give a direct proof without appealing to Theorem 1.3(i). To partition [0, m] into n equal parts, we use (1.2) on page 10 1 to express m as nm n . By the definition of a fraction, m is nm copies of n . If we 1 first group m copies of n together and call it B, then we are saying that [0, m] is n copies of B. Therefore the right side of (1.9) is equal to the length of B. Since B is m copies of n1 , its length is equal to m n by definition. This again proves (1.9). It may surprise the reader to learn that the right side of (1.9) is actually something familiar to us, at least in certain settings. Consider the special case that m is a whole-number multiple of n; let us say m = kn for some whole number k. In the right side of the equality (1.9), we want the length of one part when [0, m] is partitioned into n equal parts. Since m = kn, [0, m] is just the concatenation of kn copies of the unit segment [0, 1]. Therefore we may partition kn copies of [0, 1] (i.e., [0, m]) into n equal groups ("equal" in terms of length) where each group consists of exactly k copies of [0, 1], Thus, the right side of (1.9) is just the total length of these k copies of [0, 1], which is k. Now recall the definition of the partitive interpretation of the division m÷n, which is the number of objects in a group if m (= kn) objects is divided into n equal groups. If we interpret "object" in this case to be "[0, 1]", then the right side of (1.9) is precisely m ÷ n in the partitive sense. We repeat, if m = kn for some whole number k, then the right side of (1.9) is equal to the partitive division m ÷ n, and (1.9) becomes the statement that (1.10)

m = m ÷ n. n

Behind the symbols in (1.10) lies this statement: if m is a whole number multiple of n, then the fraction m n is the same point on the number line as the partitive division m ÷ n. Moreover, at this moment, (1.10) has no meaning when m is not a

1.2. EQUIVALENT FRACTIONS

29

whole-number multiple of n, because the division m ÷ n on the right side of (1.10) has no meaning when m is not a whole-number multiple of n. The reason m ÷ n has no meaning when m is not a whole-number multiple of n is that m ÷ n is a concept originating in whole numbers and is based on counting how many objects there are in a group of objects. In other words, while 12 ÷ 3 is the total number of objects (i.e., 4) in each group when 12 objects are partitioned into 3 equal groups, 2 ÷ 5 has no meaning in whole numbers because there is no whole number k so that when a group of 2 objects is divided into 5 equal groups, there are k objects in each group. However, if we replace counting by measuring length and allow 2 ÷ 5 to be a fraction, then we can divide [0, 2] into 5 segments of equal length where each segment has length 25 . 2 5

0 



 

4 5



 

1 

6 5

 

8 5



 



2 

This then suggests the idea that if we are willing to "expand" the meaning of m ÷ n even when m is not a whole-number multiple of n, then the right side of (1.9) would be a good starting point; i.e., we can make sense of m ÷ n for any two whole numbers (n = 0) by defining it to mean the right side of (1.9). This leads to the general concept of the division m÷n of any two whole numbers m and n (n = 0): Definition. Let m and n be any two whole numbers, n =  0. Then m ÷ n is the length of one part when [0, m] is partitioned into n equal parts. With this definition at hand, we can now rephrase (1.9) as a theorem:

Theorem 1.4. For any two whole numbers m and n, n = 0, m = m ÷ n. n

There are three observations to be made about this theorem. First, the fact that a fraction m n is equal to a division m ÷ n is called "the division interpretation of a fraction". However, there are two serious conceptual errors concerning this "interpretation" in TSM: (a) this "interpretation" is a fact that we can prove (see Theorem 1.4) rather than an ad hoc meaning one sees fit to confer on a fraction (as in TSM), and (b) the two meanings of m ÷ n when m is—or is not—a wholenumber multiple of n require a careful discussion and differentiation (which is never done in TSM). The failure described in (b) leads to students’ confusion about the very meaning of whole-number division. A second observation is that what we called the "expansion" of the meaning of m ÷ n is called, in technical language, an extension of the usual concept of whole-number division. This means that although the whole numbers m and n in the preceding concept of m ÷ n have been freed from the restriction of m being a whole-number multiple of n, yet when m is a whole-number multiple of n, this definition of m ÷ n is the same as the old one on account of the discussion preceding (1.10). The general idea of extension

30

1. FRACTIONS

(for the domain of definition of a function22 ) is standard in mathematics and will reappear several more times throughout these volumes, e.g., the extension of the concept of division from whole numbers to fractions on page 58, the extension of the concept of subtraction from fractions to rational numbers on page 96, the extension of the arithmetic operations from real numbers to complex numbers in Section 5.2 of [Wu2020b], the extension of the domain of definition of sine and cosine from [0, 90] to the number line in Section 1.2 of [Wu2020c], etc. A third and final observation about Theorem 1.4 is that the extension of the meaning of m ÷ n is inevitable because m and n are fractions after all and we will learn soon enough how to divide one fraction by another (see Section 1.5 on pp. 54). Then we will see that Theorem 1.4 gives the correct answer for the division of fractions m ÷ n (see (iii) on page 58). As a result of the division interpretation of a fraction, we will retire the division symbol "÷" from now on and use fractions m n to stand for whole-number divisions m ÷ n. Exercises 1.2. [Reminder] In doing these and subsequent exercises, please observe the following basic rules: (a) Show your work; your explanation is as important as your answer. (b) Be clear. Get used to the idea that, as a teacher, everything you say has to be understood. (1) Explain each of the following as if to an eighth grader, directly and without using Theorem 1.1 or Theorem 1.2, by drawing pictures using the number line: 5 10 28 7 12 4 = , = , and = . 12 24 20 5 27 9 (2) Reduce the following fractions to lowest terms,23 i.e., until the numerator and denominator have no common divisor > 1. (You may use a four-function calculator to test the divisibility of the given numbers by various whole numbers.) 121 157 414 969 52 , , , , . 65 143 85 299 855 (Moral: It is not easy to reduce an arbitrary fraction to lowest terms.) (3) Consider the following proof of the cancellation law (1.4) on page 20: Because c = 0, cc = 1 (by (1.2). Therefore, m m c m cm =1× = × = . n n c n cn This proof is common in TSM and is among its most egregious errors. Explain what is wrong with this proof. 22 Mathematical Aside: In greater detail, let X be the subset of N × N (where N denotes + + the nonzero whole numbers) consisting of all pairs (a, b) so that a and b are whole numbers, b = 0, and a is a whole-number multiple of b. Then the standard partitive whole-number division is the mapping, f : X → N so that f (a, b) = a ÷ b. The preceding definition of m ÷ n is the following extension of f defined by F : N × N+ → Q (where Q is the rational numbers) so that F (m, n) is the fraction m/n. 23 Theorem 3.1 on page 139 of Chapter 3 will show that every fraction can be reduced to a unique fraction in lowest terms.

1.2. EQUIVALENT FRACTIONS

31

(4) School textbooks usually present the cancellation law for fractions as follows: Given a fraction m n , suppose a nonzero whole number k divides m÷k both m and n. Then m n = n÷k .

Explain as if to a seventh grader why this is true. (5) The following points on the number line have the property that the thickened segments [A, 1], [B, 2.7], [3, C], [D, 4], [ 13 3 , E] all have the same length:

A

B

0

1

C D 6

2

3

E 4

2.7

(6)

(7)

(8)

(9)

6

5

13 3

If A = 47 , what are the values of B, C, D, E? Be careful with your explanations, because we do not know how to add or subtract fractions yet. (Rest assured that on the basis of what has been discussed in this section, you can do this exercise.) 3 5 of which number? (b) 37 is 11 of which number? (c) 11 (a) 37 is 11 5 of a 7 fraction is equal to 3 . What is this fraction? (d) A wire 314 feet long is only four-fifths of the length between two posts. How far apart are the posts? (e) Helena was three-quarters of the way to school after having walked 89 miles from home. How far is her home from school? Explain as if to a sixth-grade student how to do the following exercise: Nine students chip in to buy a 50-pound sack of rice. They are to share the rice equally by weight. How many pounds should each person get? (If you just say, "Divide 50 by 9", that won’t be good enough. You must explain what is meant by "50 divided by 9" and why the answer is 50 9 .) James gave a riddle to his friends: "I was on a hiking trail, and after walking 79 of a mile, I was 49 of the way to the end. How long is the trail?" Help his friends solve the riddle. (i) Prove that for two nonzero fractions ab and dc , ab = dc if and only if ab = dc . (ii) Prove that the following three statements are equivalent for any four whole numbers a, b, c, and d, with b = 0 and d = 0: (a)

a b

=

c d.

(b)

a a+b

=

c c+d .

(c)

a+b b

=

c+d d .

(One way is to prove that (a) implies (b) and (b) implies (a). Then prove (a) implies (c) and (c) implies (a).) 11 9 (10) Place the three fractions 13 6 , 5 , and 4 on the number line and explain how they get to be where they are. m m+b (11) For which fractions m n is it true that n = n+b , where b is a nonzero whole number? k (12) m n of a fraction is equal to  . What is this fraction?

32

1. FRACTIONS

1.3. Adding and subtracting fractions This section gives the precise definitions of fraction addition and subtraction, with special emphasis given to the fact that these definitions are conceptually identical to those among whole numbers and that the addition and subtraction formulas do not involve the least common denominator. For the purpose of defining subtraction, we will need the cross-multiplication inequality. The addition of fractions (p. 32) Applications of the addition of fractions (p. 34) The cross-multiplication inequality (p. 36) The subtraction of fractions (p. 38) The addition of fractions What is the meaning of 57 + 38 ? This simple question, incredibly, seems to have no answer in TSM.24 What is usually done in TSM, after some vague statements about giving two fractions a common denominator, is to give a formula for the sum in terms of the lowest common denominator of the two fractions in question. Now students in the upper elementary grades have some intuitive ideas about "adding numbers" because they have seen that adding whole numbers is "combining things" and they expect the same for adding fractions. The use of the lowest common denominator in the addition of fractions, however, obliterates this intuitive idea and disrupts students’ normal learning process. In this subsection, we will restore this intuition by defining the addition of fractions to be a direct extension of the usual definition of adding whole numbers. (For further comments on the addition of fractions as practiced in TSM, see the Pedagogical Comments on page 40.) Consider, for example, the addition of 4 to 7. In terms of the number line, this is just the total length of the concatenation of two segments, one of length 4 and the other of length 7, which is of course 11, as shown. (Review the meaning of length on page 6 if necessary.)

0

4 



 

11 

 7

Similarly, if we have two whole numbers m and n, then m + n is simply the length of the concatenation of the two segments of length m and n: m 

n  m+n



With the expectation that addition should mean the same thing for whole numbers and fractions, we are led to the following definition of the sum of two fractions:

24 See

page xiv of the preface for the definition of TSM.

1.3. ADDING AND SUBTRACTING FRACTIONS

33

k m Definition. Given fractions k and m n , we define their sum  + n by k m + = the length of two concatenated segments, one  n

of length

k m , followed by one of length :  n m n

k 



 k 

+

 m n

It follows directly from this definition that the addition of fractions satisfies the associative and commutative laws; see Exercise 1 on page 41 (cf. the appendix on page 86 for a summary of these laws). It is also an immediate consequence of the definition that (1.11)

k+m k m + =   

because both sides are equal to the length of k + m copies of 1 (see page 12 for the meaning of "copy"). More explicitly, the left side is the length of k copies of 1 combined with m copies of 1 and is therefore the length of k + m copies of 1 , which is exactly the right side. This tells us that, to compute the sum of two fractions with the same denominator , one adds them as one would with whole numbers, with the only difference being that, instead of adding so many copies of the unit 1, we now add so many copies of the unit fraction 1 , as above. Because of FFFP (see page 23), the general case of adding two fractions with unequal denominators is immediately reduced to the case of equal denominators; i.e., in order to add k m +  n m m where  = n, we use FFFP to rewrite k as kn n and n as n . Then we obtain the general formula for adding fractions:

kn m kn + m k m + = + = .  n n n n We emphasize that we obtain this formula from the definition of the sum of two fractions by the use of reasoning. (1.12)

Activities. If a student tells you 35 + 13 = 48 = 12 , how would you explain to him that he is wrong? (Hint: Does he know what " 53 + 13 " means? Can he use common sense to see why the sum cannot be 12 ? Up to this point, when two fractions are given, we have used the product of their denominators as a common denominator for both, i.e., a whole number so that both fractions are equivalent to a fraction with that as their denominator. Sometimes it can happen that a different, smaller common denominator is "handed to you for free". For example, when the denominator of one of the fractions is

34

1. FRACTIONS

already a multiple of the other denominator, then the bigger denominator already serves as a common denominator; e.g., 6 7 13 3 7 + = + = 4 8 8 8 8 or 13 27 130 157 27 + = + = . (1.13) 100 10 100 100 100 (Incidentally, the second example can be equally well expressed as 0.27+1.3 = 1.57. See page 14 for the decimal notation.) A slightly more sophisticated example is 7 5 25 of 12 12 + 8 . It is relatively easy to notice that 24 is the least common multiple and 8 and, as such, 24 is the least common denominator of the two fractions. Since 24 = 2 × 12 = 3 × 8, the addition can be done more simply as follows: 7 5 2×7 3×5 14 + 15 29 + = + = = . 12 8 2 × 12 3 × 8 24 24 By comparison, if we use (1.12), then we would get   5 (7 × 8) + (12 × 5) 116 7 29 + = = = 12 8 12 × 8 96 24 where the last equality in the parentheses is due to the cancellation law (page 20). k In general, suppose m n and  are given and there is a whole number D that is a common multiple of both n and , say D = n =  n; then bthe computation of the sum k + m n can make use of D as the common denominator instead of n, as follows: kn kn +  m k m  m + =  +  = .  n n n D A more interesting example to illustrate the advantage of using a simpler common denominator can be found in Exercise 13 on page 42. Applications of the addition of fractions The first application of fraction addition is the explanation of the addition algorithm for (finite) decimals. (The definition of a decimal is given on page 14.) For example, consider 4.0451 + 7.28. This algorithm calls for (α) lining up 4.0451 and 7.28 by their decimal points, (β) adding the two numbers as if they were whole numbers and getting a whole number, to be called N , and (γ) putting the decimal point back in N to get the answer to 4.0451 + 7.28. We now supply the reasoning for the algorithm (in essence, it has already been given in equation (1.13) above). First of all, we make use of the fact that zeros can be added to the right of a decimal (see page 21 for the precise statement) to rewrite the two given decimals as two decimals with the same number of decimal digits:26 40451 72800 4.0451 + 7.28 = 4.0451 + 7.2800 = + . 104 104 25 For 26 A

a definition, see Exercise 4 on page 156. little reflection would tell you that we are essentially using FFFP (page 23) here.

1.3. ADDING AND SUBTRACTING FRACTIONS

35

Then 4.0451 + 7.28

=

40451 + 72800 104

=

113251 104

(corresponds to (β))

= 11.3251

(corresponds to (γ)).

(corresponds to (α))

The reasoning is of course completely general and serves to explain the algorithm (α)–(γ) for any pair of decimals. A second application is to get the so-called complete expanded form of a (finite) decimal. For example, given 4.1297, we know it is the fraction 41297 . 104 But we have the expanded form of the whole number 41297: 41297 = (4 × 104 ) + (1 × 103 ) + (2 × 102 ) + (9 × 101 ) + (7 × 100 ). (Recall that 100 is by definition equal to 1.) We also know that, by equivalent 4 3 1 fractions, 4×10 = 4, 1×10 = 10 , etc. Thus by (1.12) on page 33, 104 104 1 2 9 7 (1.14) 4.1297 = 4 + + + 3 + 4. 10 102 10 10 This expression of 4.1297 in (1.14) as a sum of descending powers27 of 10, where the coefficients of these powers are the digits of the number itself (i.e., 4, 1, 2, 9, and 7), is called the complete expanded form of 4.1297. Equation (1.14) is the reason that we can say 4.1297 is "the sum of 4 and 1 tenth and 2 hundredths and 9 thousandths and 7 ten-thousandths". Please observe that this conclusion is a precise, logical consequence of the definition of a decimal (in equation (1.3) on page 14) and the addition formula (1.11) on page 33. Please do not make the standard TSM mistake of telling students—without first proving (1.14)—that 4.1297 "means 4 and 1 tenth and 2 hundredths and 9 thousandths and 7 ten-thousandths". You would confuse them. In the same way, a decimal 0.d1 d2 · · · dn ,28 where each dj is a single-digit number, has the following complete expanded form: d2 d1 dn + 2 + ··· + n. (1.15) 0.d1 d2 · · · dn = 10 10 10 A third application of fraction addition is to introduce the concept of mixed numbers. We have seen (on page 15) that, in order to locate fractions on the number line, it is an effective method to use division-with-remainder on the numerator. With the availability of equation (1.12) on page 33, we are now in a position to clarify the whole procedure; e.g., 187 (13 × 14) + 5 (13 × 14) 5 5 = = + = 13 + . 14 14 14 14 14 5 Thus the sum 13 + 14 , as a concatenation (see page 12 for the meaning of "con5 catenation") of two segments of lengths 13 and 14 (recall the definition of adding 1 if you think of 10 as 10−1 , etc. Negative exponents will be discussed in Section 4.1 of [Wu2020b]. 28 The notation here is unfortunate: "d d · · · d " is not the product of d , d , . . . , d . n n 1 2 1 2 27 "Descending"

36

1. FRACTIONS

fractions on page 33), clearly exhibits the fraction 187 14 as a point on the number line 5 about one-third beyond the number 13. The sum 13 + 14 is usually abbreviated 5 to 13 14 by omitting the + sign and, as such, it is called a mixed number. More generally, a mixed number is a sum n + k , where n is a whole number and k is a proper fraction (i.e., its numerator is smaller than its denominator), and it is usually abbreviated as just n k .29 This concept usually causes terror among students, probably because it is usually introduced in TSM30 before the concept 5 is then explained by of the addition of fractions is taught. Something like 13 14 5 the usual baby talk about this number being "13 and 14 ". Unfortunately, the word "and" in this context masks the concept of fraction addition, which is not yet known to students at this point. Students are therefore at a loss when they are forced to 5 . Inevitably, nonlearning follows. do computations with 13 14 It is for the reason of avoiding this pitfall that we have postponed the introduction of the concept of a mixed number until now. So just remember: a mixed number is a sum of a whole number and a proper fraction. No more, and no less. The cross-multiplication inequality We next wish to discuss the subtraction of fractions. We are handicapped by not having negative fractions at our disposal, however, so that, as in the case of subtracting whole numbers, we must first make sure that k < m n before we m k can compute n −  . For this reason, how to determine that one fraction is less than another now becomes our next concern. To this end, we will need the basic inequality in Theorem 1.5 below. Recall from page 12 that k < m n means the point k m is to the left of the point on the number line:  n k 

m n

In practice, it may not be easy to tell, for example, which of 49 or 37 is bigger. We need a general method for comparing fractions. Now FFFP comes to the rescue: we simply put both fractions in the sequence of 63-rd’s (63 = 9 × 7) and see which of the two fractions comes first. We rewrite both to have denominator 9 × 7, so that 28 3 27 4 = and = . 9 63 7 63 4 Then in the sequence of 63-rd’s, 9 is the 28-th point and 37 is the 27-th point. Therefore the latter is to the left of the former. Consequently, 49 is the bigger of the two. This reasoning is perfectly valid in general, so we have the following theorem. Theorem 1.5 (Cross-multiplication inequality). Given two fractions k k m and m n , then  < n is equivalent to kn < m. Proof. We simply follow the reasoning in the preceding special example. By kn m k m FFFP, we can rewrite k and m n as n and n , respectively. If  < n , then we 29 Discussions of fractions and decimals seem to be rife with notational problems. In this case, please note that n k is not the product of n and k . 30 See page xiv of the preface for the definition of TSM.

1.3. ADDING AND SUBTRACTING FRACTIONS

37

m 1 have kn n < n . This means the kn-th multiple of n is to the left of the m-th multiple of the same, so that kn must be smaller than m; i.e., kn < m. Conversely, 1 suppose kn < m. Then the kn-th multiple of n is to the left of the m-th multiple 1 of n , so that

m kn < . n n By the theorem on equivalent fractions, this becomes k < m n . The proof of the theorem is complete. Corollary. (i) ac < bc is equivalent to a < b. (ii)  > n is equivalent to

1 


n is equivalent to n < , which is of course identical to 1 · n < 1 ·  which, by the preceding theorem, is equivalent to 1 < n1 . The corollary is proved. Note that we have just made use of the dot in 1· to denote 1×. Pedagogical Comments. TSM generally considers both parts of this corollary to be too obvious for any proof. For part (i), it is easy to take it for granted because it "looks" obvious. However, if a teacher does not prove that, for example, 37 41 19 < 19 but simply declares that it is obvious because 37 < 41, what will this 19 teacher say to a student who claims that 19 37 < 41 because 37 < 41? Now, a teacher might object to proving something as intuitive as part (i) by appealing to a complicated theorem such as Theorem 1.5. Because of this objection, we now offer a more direct proof of part (i), as follows. The reason that ac < bc implies a < b is that, by definition, ac < cb means bc is to the right of ac . But cb is the b-th point in the sequence of 1/c and ac is the a-th point in the sequence of 1/c. Therefore what we have is that the b-th point of the sequence is to the right of the a-th point of the sequence. Since one counts the points in the sequence from left to right, this can only mean that a < b. Conversely, suppose a < b. Then the a-th point in the sequence of 1/c is to the left of the b-th point of the same sequence. This means precisely that ac < bc , as desired. Recall that we bemoaned the absence in TSM of a definition of the concept of a fraction being smaller than another on page 13 when we formalized this concept. The preceding argument is an explicit demonstration of the fact that having precise definitions facilitates learning. The need for reasoning in support of part (ii) is slightly different. Many teachers dismiss the need for any proof of this corollary, and the common thinking behind the dismissal is that, for small values of  and n, e.g.,  = 3 and n = 2, a third of a pizza is visibly smaller than a half of a pizza, and so 13 < 12 . One can even get more sophisticated by appealing to the number line and infer from the following picture that 13 < 12 : 0

1 3

1 2

2 3

1

38

1. FRACTIONS

While this kind of intuitive argument is invaluable for pointing students toward the correct conclusion, it cannot be confused with valid reasoning. After all, how would such an intuitive argument bring conviction to the claim that 1 1 > ? 8590007 8590008 This is why the correct reasoning using the cross-multiplication inequality must be taught in addition to the intuitive argument using small values of  and n. End of Pedagogical Comments. The following observations about the comparison of fractions are useful and also easy to prove. Let A, B, C, D be fractions. Then we have: (1) A < B ⇐⇒ there is a fraction C so that A + C = B. (2) A < B implies A + C < B + C for every fraction C. (3) A < B and C < D implies A + C < B + D. The proofs require nothing more than making mathematical interpretations of correct drawings on the number line. We will leave them as an exercise (see Exercise 14 on page 42). Activity. Without using any calculator or computer software, decide which of the following fractions is greater: 33333 1234567

and

33335 . 1234569

The subtraction of fractions We can now define the subtraction of fractions: suppose k ment of length m n is longer than a segment of length  . Definition.

m n

>

k ;

then a seg-

m k k If m n >  > 0, then the subtraction n −  is by definition

the length of the remaining segment when a segment of length k is taken from one end of a segment of length m n.  

m n









k  k In the picture, m n −  is the length of the thickened segment on the right. We also define m m m m − = 0 and −0= . n n n n This definition of the subtraction of fractions is clearly modeled on the subtraction of whole numbers. For example, 9 − 4 is the length of the remaining segment when a segment of length 4 is taken from a segment of length 9.

1.3. ADDING AND SUBTRACTING FRACTIONS

39

The definition of subtraction for fractions has the following pleasant (but expected) consequence: let k and m n be distinct fractions; then  k m m k (1.16) the length of the segment − . , is

 n

n



m n]

Indeed, by the definition of length on page 6, the length of [0, (respectively, [0, k ]) m k is n (respectively,  ). Therefore (1.16) is clear from the definition of subtraction:

 0

m n

 k 



k 

 m n

The reasoning used in the proof of (1.11) on p. 33, together with FFFP, gives m − nk m k − = . n  n Note that this formula makes implicit use of the preceding cross-multiplication inequality (Theorem 1.5), because the subtraction of whole numbers in the numerator of the right side of (1.17), m − nk, does not make sense unless we know m > nk; but since k < m n , Theorem 1.5 guarantees that m > nk. We wish to bring out the fact that subtraction is an alternate way of expressing k addition. Indeed, the definition of m n −  , together with the definition of adding fractions as the concatenation of segments (page 33), implies that   m k k m + − (1.18) = .  n  n

(1.17)

This is quite obvious by looking at the preceding picture. Thus we may regard m k k m n −  as the fraction that must be added to  to get n . Although this more abstract perspective seems to add nothing to the concept of subtraction, it will serve as a bridge to the definition of the division of fractions (see page 57). The subtraction of mixed numbers reveals a sidelight about subtraction that may not be entirely devoid of interest. Consider the subtraction of 17 25 − 7 34 . One can do this routinely by converting the mixed numbers into fractions: 3 85 + 2 28 + 3 87 31 87 × 4 − 5 × 31 193 2 − = − = = . 17 − 7 = 5 4 5 4 5 4 5×4 20 However, there is another way to do the computation: 3 2 3 2 17 − 7 = (17 + ) − (7 + ). 5 4 5 4 Anticipating a reasoning that will be made routine when we come to the next chapter on rational numbers (see equation (2.14) on page 97),31 we can rewrite the right side as (17 − 7) + ( 25 − 34 ). But now we are stuck because 25 < 34 , so the subtraction on the right cannot be performed according to our present definition of subtraction. Using an idea that is reminiscent of the "trading" technique in 31 Rest assured that there is no circular reasoning in appealing to a result in Chapter 2 at this juncture, because the mathematical development of Chapter 2 does not depend on this section.

40

1. FRACTIONS

the subtraction algorithm for whole numbers, we can get around this difficulty by computing as follows:     2 3 2 3 17 − 7 = 16 + 1 − 7+ 5 4 5 4   2 3 = (16 − 7) + 1 − (1.19) 5 4   13 7 3 13 − =9 . =9+ = 9+ 5 4 20 20 Again, the second line makes use of (2.14) on page 97. The whole computation looks longer than it actually is because we interrupted it with explanations. Normally, we would have done it the following way:     2 3 2 3 7 3 13 13 17 − 7 = (16 − 7) + 1 − − =9 , =9+ =9+ 5 4 5 4 5 4 20 20 which is exactly the same as before. Finally, there is a similar subtraction algorithm for finite decimals that allows finite decimals to be subtracted as if they were whole numbers provided they are aligned by their decimal points, and then the decimal point is restored in the final result at the end. The reasoning is exactly the same as the case of addition (of decimals) and will therefore be left as an exercise (Exercise 10 on page 42). Pedagogical Comments. We wish to address the issue of why TSM’s standard formula for the addition of fractions in terms of their least common denomik m nator is pedagogically unsound. Given k and m n , students are told to add  + n by first finding the least common denominator of the fractions, which is by definition the LCM32 of the denominators  and n, say B. Then if B = n =  n for some whole numbers  and n , the sum of these two fractions—according to TSM—is given by kn kn +  m  m k m + =  +  = .  n n n B First of all, TSM usually offers this formula as a definition of the sum of fractions, and this is certainly unacceptable because it bears no resemblance to the intuitive notion of addition as "combining things". But even as a formula for addition, it is no less objectionable because, as we have seen in (1.12) on page 33, it is not necessary to use the LCM of  and n when the product of the denominators, n, is both adequate and more natural as a common denominator. Finally, this formula is destructive in terms of mathematics learning because many students confuse the LCM of two whole numbers with their greatest common divisor (see page 138 for this concept) and there is no reason to complicate the learning of a simple skill by artificially inflating its complexity. The same comment applies to the use of least common denominator for the subtraction of fractions. 32 Least

common multiple. For a precise definition, see Exercise 4 on page 156.

1.3. ADDING AND SUBTRACTING FRACTIONS

41

Be sure to take every opportunity to eradicate this approach to the addition and subtraction of fractions from your teaching, because it has caused great harm to mathematics learning. End of Pedagogical Comments. Mathematical Aside: The use of the least common denominator for the definition of the addition of fractions is more than just a pedagogical disaster. From a mathematical perspective, it is conceptually incorrect. If it were necessary to find the LCM of the two denominators before the addition of two fractions could be defined, it would imply that addition cannot be performed in the field of quotients of an integral domain unless the latter has the special property that any two elements in it have an LCM. This is almost the statement that addition cannot be defined in the field of quotients of an integral domain unless the domain has the unique factorization property. However, we know that this is false because addition can be defined in the field of quotients of any domain. Exercises 1.3. (1) Use the definition of the sum of two fractions on page 33 to show that the addition of fractions satisfies the associative and commutative laws. (See (1.56) and (1.57) on page 87 for the precise statement of these two laws and the subsequent discussion for their significance.) (2) Compute (you may use a four-function calculator for the arithmetic computations; no need to simplify your answers but you have to show all your steps): 5 8 5 7 5 3 13 + 27 , (ii) 34 − 51 , (iii) 24 12 + 32 11 (i) 18 15 , (iv) 315 8 − 312 20 . (3) Compute (for the addition of three numbers, 1.8 on page 86):

2 3see  Section 5 1 7 5 4 7 13 (i) 6 + 4 + 8 , (ii) 8 + 15 + 16 , (iii) 3 + 4 − 16 . 3 − 267 (4) Compute 118 52 13 in two different ways, and check that both give the same result. (Large numbers are used on purpose. You may use a four-function calculator to do the calculations with whole numbers, but only for that purpose.) kn+m (5) (a) We have an algorithm for adding two fractions: k + m n = n . Now explain as if to an eighth grader how to obtain an algorithm for adding p three fractions k + m n + q . Make sure you can justify the algorithm.

(6)

(7) (8) (9)

1 1 1 (b) If a, b, c are nonzero whole numbers, what is ab + bc + ac ? Simplify your answer as much as possible. Show a sixth grader how to do the following problem by using the number line: I have two fractions whose sum is 18 11 and whose difference (i.e., the 1 larger one minus the smaller one) is 2 . What are the fractions? (Caution: No need to consider simultaneous linear equations.) Explain as if to a fifth grader why 1.2 is bigger than 1.1987. Explain as if to a sixth grader how to get 5.09 + 7.9287 = 13.0187. (a) Which is closer to 37 , 49 or 11 21 ? (b) Which whole number is closest to the sum 1300 312 + ? 1305 51 (Don’t forget to prove it!)

42

1. FRACTIONS

(10) State the subtraction algorithm for finite decimals, and explain why it is true. (See the discussion of the addition algorithm for finite decimals on page 34.) 7 (11) (a) 25 + 12 = ? (b) Laura ran for 35 minutes, stopped to take a rest, and then ran for another 24 minutes. How long did she run altogether, and what does this have to do with part (a)? (12) An alcohol solution mixes 5 fl. oz. of water with 24 fl. oz. of alcohol. Then 4 fl. oz. of water and 19 fl. oz. of alcohol are added to the solution. Which has a higher concentration of alcohol (which is defined to be the number (volume of the alcohol in the solution) ÷ (total volume of the solution) in the sense of (1.10) on page 28): the old solution or the new? (13) If n is a whole number, we define n! (read: n factorial) to be the product of all the whole numbers from 1 through n. Thus 5! = 1 × 2 × 3 × 4 × 5. We also define 0! to be 1. Define the so-called binomial coefficients nk for any whole number k satisfying 0 ≤ k ≤ n as   n n! . = k (n − k)! k! Then prove

(14) (15) (16)

(17)

(18)

      n n−1 n−1 = + . k k k−1

(The proper context for this exercise is Pascal’s triangle, which is discussed in Section 5.4 of [Wu2020b].) Prove each of the statements (1)–(3) on page 38 for fractions A, B, C, and D. Suppose a, b are whole numbers so that 1 < a < b. Which is bigger: a−1 a a+1 b+1 or b−1 ? Can you tell by inspection? What about and ? b a b (a) Suppose ab and dc are fractions and ab < dc . Prove that ab < a+c b+d < c . (b) Prove that between any two distinct fractions, there is another d fraction. Let ab be a nonzero fraction, with a = b. Order the following (infinite a+2 a+3 number of) fractions: ab , a+1 b+1 , b+2 , b+3 , . . . . (Caution: It makes a difference whether a < b or a > b.) In the notation of Exercise 13, observe that each fraction n! j , where n, j are whole numbers and 1 ≤ j ≤ n, is actually a whole number. Find the following sum and simplify your answer as much as possible: 1 49! 1

(19) (a) Prove

1 4

+

1 49! 2

− 25 + 16 =

+

1 60 .

1 49! 3

+ ··· +

1 49! 48

+

1 49! 49

.

(b) Generalize the following: let n be a nonzero

2 1 + n+2 is always a unit fraction, whole number; then prove that n1 − n+1 i.e., a fraction whose numerator is 1.

1.4. MULTIPLYING FRACTIONS

43

1.4. Multiplying fractions Multiplication—not division—is the most complex of the four arithmetic operations among fractions. We first give a motivation for the definition of multiplication and then dive in to prove the two most important facts about fraction multiplication: km the product formula k × m n = n and the area formula for a rectangle with fractional sides. It is sobering to realize that something as basic as the latter area formula is not proved in TSM. Then we use the product formula to explain the rule—also not explained in TSM—for the multiplication of finite decimals. Motivation for the definition of multiplication (p. 43) The formal definition of multiplication (p. 45) The product formula (p. 46) Areas of rectangles (p. 47) Three remarks (p. 51) Motivation for the definition of multiplication To facilitate student learning in school mathematics, it is of vital importance that we give students a precise definition for each concept. They have to know what they are doing. This is especially true for fraction multiplication. Such a definition is rarely, if ever, found in TSM33 or even in the education research literature. Recall that for whole numbers, multiplication is, by definition, just repeated addition: 3×5 means 5 + 5 + 5 and 4 × 17 means 17 + 17 + 17 + 17. This definition cannot be literally extended to fractions; e.g., it makes little sense to define 47 × 12 as "adding 1 4 2 to itself 7 times". This difficulty has led some educators to advocate extreme measures to achieve any kind of understanding of this concept.34 We will do mathematics the standard way by giving a precise definition of fraction multiplication and deducing precise logical consequences. Since we are not yet in possession of a definition, we must first look for one. The process of how to formulate such a precise definition is usually hidden in textbooks and, strictly speaking, not part of the logical development of mathematics. For this reason, textbooks are not obliged to give it.35 However, because the definition of fraction multiplication, while consistent with the definition of whole-number multiplication, is more sophisticated than the latter, we will give some indication of this kind of necessary "groping in the dark" to shed some light on the ultimate definition itself. We have to be perfectly clear: what is in the next paragraph is not formal mathematics, but a rather fuzzy patchwork of educated guesses and wishful thinking—more or less what takes place when one is struggling to solve a problem. The point is not whether any particular step is correct or reliable, but whether what comes out of this process turns out to be mathematically correct and consistent with common sense. You can make up your mind on that after you have read through this section. What follows in this paragraph (as announced) consists of heuristic arguments about why we should define multiplication of fractions in a particular way (essentially as given in Section 7.2 of [Wu2002]); it is not part of the logical development 33 See

page xiv of the preface for a definition of TSM. suggestion was that "multiplication of fractions is about finding multiplicative relationships between multiplicative structures". This is poetic, but not mathematically helpful. 35 In any case, when such an effort is made in TSM, it is usually a mess; e.g., the motivation for defining 5−3 as 1/53 becomes indistinguishable from a proof that 5−3 = 1/53 . 34 One

44

1. FRACTIONS

of fractions, and there is no pretense that any of what follows is logically correct. It so happens that some of the following claims or guesses will eventually be proved on the basis of the definition of multiplication on page 45, but until that happens, nothing in this paragraph should be construed as proven and therefore usable for reasoning. That said, let k and m be whole numbers; then the concept of k × m is no mystery: it is the length of k copies of m (see page 12 for the meaning of "copy"): k × m = m + m + ··· + m.    k

Now by analogy (and some wishful thinking but not by logic), we would like to m believe that for a whole number k and for a fraction m n , the multiplication k × n m "should also be" the length of k copies of n ; i.e., m m m = + ··· + . (1.20) k× n n  n k

Adding the right side, we get k×

(1.21)

km m = . n n

Next, what "should be" a reasonable definition of 1 × m n for a nonzero whole number ? It seems "reasonable" to expect that the multiplication of fractions—like the multiplication of whole numbers—obeys the associative law. Granting this, we get     1 m 1 m × × = × × .  n  n By (1.21), we have  × (1.22)

1 

=

 

= 1 and also 1 × m n =   1 m m × × = .  n n

m n.

Therefore,

We interpret (1.22) as follows: if A denotes 1 × m n , then the left side of (1.22) is equal to  copies of A, by equation (1.20).36 Thus, (1.22) says that the length of  m copies of A is equal to m n . Therefore, A is the length of one part when n is divided into  equal parts, or, (1.23)

m 1 m × = length of one part when is divided into  equal parts.  n n

Furthermore, we also observe that if m = 1 in (1.21), we get k × n1 = nk for any whole numbers k and n (n = 0). For a reason that will be clear presently, we change the notation and rewrite it as 1 k =k× .   And remembering that multiplication is assumed to obey the associative law, we finally get     1 m k m 1 m × = k× = k× × × .  n  n  n 36 A reminder: equation (1.20) is not a proven fact, only part of our wishful thinking of what the world "should" look like if we have our way.

1.4. MULTIPLYING FRACTIONS

45

Therefore, by applying (1.20) and (1.23) in succession to the last expression, we have   1 m k m × = length of k copies of ×  n  n

(1.24)

= length of k of the parts m is divided into  equal parts. when n

Now let us take stock of the situation. We have arrived at (1.24) by making a series of what we believe to be reasonable assumptions. If indeed everything turns out to be what we believe it ought to be, then (1.24) should be a reasonable definition of k × m n . At this point, this belief is nothing more than wishful thinking, but it is at least wishful thinking grounded in our collective mathematical experience. On this basis, we proceed to adopt the provisional definition (1.24). The formal definition of multiplication We now push the reset button. We will leave our heuristic discussion behind and formalize our provisional definition in (1.24) in the following definition. From this formal definition we will draw logical conclusions about fraction multiplication. is by definition Definition. The product of two fractions k × m n the length of the concatenation of k of the parts when [0, m n ] is partitioned into  equal parts. (See page 12 for the meaning of "concatenation".) Note that, according to the definition on page 24, we may rephrase the definition as (1.25)

k m k m × = of .  n  n

The preceding definition of multiplication poses a potential problem, and we should deal with it right away. We first illustrate the problem with a simple example. Consider 12 × 34 . We know that 12 = 24 and 34 = 15 20 , and therefore we expect that, no matter how fraction multiplication is defined, we should have the equality 1 3 2 15 × = × . 2 4 4 20 By (1.25), this is equivalent to asking whether 3 2 15 1 of = of . 2 4 4 20 Now if this equality turns out to be incorrect, then it would mean the definition makes no sense, or in the usual language of mathematics, is not well-defined. More generally, suppose k K =  L

and

m M = . n N

46

1. FRACTIONS

Then the question is whether the following equality is valid: k m K M × = × .  n L N By (1.25), this is equivalent to asking whether

(1.26)

k m K M of = of .  n L N By Theorem 1.3(ii) on page 27, the answer is affirmative. So we know that our definition of the product of two fractions is well-defined. The product formula For the conceptual understanding of fraction multiplication, the following theorem is all-important. k m km Theorem 1.6 (Product formula). For all fractions k and m n ,  × n = n .

Because of (1.25) above, this theorem is nothing more than a restatement of Theorem 1.3(i) on page 27. In TSM and in standard professional development materials, this formula is often presented as the definition of the product k × m n. However, without the definition on page 45, it is difficult to teach the solving of word problems involving fraction multiplication except by rote (see, e.g., Exercises 8 and 9 on page 54). As an immediate corollary of the product formula, we have Corollary. The multiplication of fractions obeys the associative, commutative, and distributive laws. (See the appendix on page 86 for a summary of these laws.) We will leave the detailed proof of the corollary to an exercise (Exercise 4 on page 53). There are two immediate consequences of our definition of fraction multiplication. The first is the interpretation of division by a (nonzero) whole number as multiplication. Recall that we defined on page 29 the concept of k ÷  for two arbitrary whole numbers k and  ( = 0) to mean the length of one part when [0, k] is partitioned into  equal parts. But from the definition of fraction multiplication (page 45), 1 × k has exactly the same meaning: the length of one part when [0, k] is partitioned into  equal parts. Therefore, 1 (1.27) k ÷  = × k for whole numbers k and ,  = 0.  We can put equation (1.27) into a more general context. (For a reason that will be obvious, we are going to change the notation in the definition on page 29 from m and n to k and , respectively.) While the definition of k ÷  on page 29 for any two whole numbers k and  ( = 0) is a generalization of the partitive division of k ÷  when k is a whole-number multiple of , this definition can itself be generalized by replacing k with a fraction A. Indeed, we now define the division of a fraction A by a nonzero whole number  to be the length of one part when [0, A] is partitioned into  equal parts. Then it follows from the definition of fraction multiplication that 1 × A is equal to A divided by  in the sense just defined. For

1.4. MULTIPLYING FRACTIONS

47

this reason: Division (of a whole number or a fraction) by a nonzero whole number  will henceforth be replaced by multiplication by 1 . Incidentally, in view of the product formula, equation (1.27) implies that k ÷  = k , which is exactly Theorem 1.4 on page 29. This then provides a different perspective on Theorem 1.4. A second immediate consequence of the definition of fraction multiplication is that if k is a whole number, then according to the definition of multiplication on k m page 45, k × m n = 1 × n = the length of the concatenation of k segments of length m n . By the definition of addition (page 33), the latter is equal to adding k copies of the fraction m n: m m m = + ··· + . (1.28) k× n n n   k

In other words, the multiplication k × retains the meaning of repeated addition. As is well known, the product formula has numerous applications. One of the simplest may be the explanation of the usual cancellation rule for fractions. For example, we have 105 135 49 × = 28 9 4 m n

because we can "cancel" the 9’s and 7’s in the numerators and denominators. The precise reasoning is the following. By the product formula, 135 49 135 × 49 × = 28 9 28 × 9 Therefore, 135 49 15 × 9 × 7 × 7 15 × 7 15 7 × = = = × , 28 9 4×7×9 4 4 1 where we have made use of equivalent fractions. The same reasoning of course k proves that for any fractions ma n and a we have the general cancellation rule for fractions: k mk ma × = . (1.29) n a n Areas of rectangles A more substantial application of the product formula is undoubtedly the following interpretation of fraction multiplication in terms of area;37 this interpretation is as basic as the definition (of fraction multiplication) itself. We will prove that the area of a rectangle is equal to the product of its sides. Let us first review some basic properties of area. We fix a unit square, i.e., a square whose sides all have length 1. The area of the unit square is by definition equal to 1. A geometric figure is by definition just a subset of the plane. We will assume that congruent geometric figures have the same area. A geometric figure S is said to be paved by 37 See the footnote on page 17 concerning the concepts of congruence and area. You may take both in the naive sense until Chapter 4 of this volume and Chapter 4 in [Wu2020c], respectively.

48

1. FRACTIONS

a collection of geometric figures S1 , . . . , Sn if the union of S1 , . . . , Sn is S and if S1 , . . . , Sn overlap only at their boundaries. A fundamental fact is that if S1 , . . . , Sn pave S, then the area of S is the sum of the areas of S1 , . . . , Sn . If the unit square is paved by n congruent pieces, then all these pieces have equal areas and, consequently, these pieces give rise to a division of the unit (area of the unit square) into n equal parts; by the definition of the fraction n1 , the area of each piece is n1 . For example, each of the following shaded regions of the unit square has area equal to 14 : @ @ @ @ @ @

The interpretation of fraction multiplication in question is the following theorem.

to

Theorem 1.7. The area of a rectangle with sides of lengths k and m n is equal k m × .  n

In school mathematics, this theorem is the basis of the statement that "area is length times width". It is sobering to realize that in TSM, there is no explanation of 38 Theorem 1.7 except (perhaps) the case of k and m When n being whole numbers. the lengths of the sides of a rectangle are numbers that may not be fractions,39 the theorem continues to hold, but the reasoning becomes more sophisticated. This issue will be considered in Section 4.4 of [Wu2020c]. Before giving the proof of Theorem 1.7, let us work out a special case: why is the area of a rectangle with sides lengths 23 and 52 equal to 23 × 52 ? First consider a simpler case: why does a rectangle with sides lengths 13 and 12 have area equal to 13 × 12 ? Such a rectangle is related to the unit square in the following way. Let one pair of opposite sides of the unit square be divided into 3 parts of equal length and the other pair into 2 halves. By joining corresponding points of the division on opposite sides, we obtain a paving of the unit square by 6 small rectangles. Each of these 6 small rectangles has side lengths 13 and 12 , and they are congruent to each other (geometrically obvious, but see Exercises 4 and 7 on page 237). 38 Note 39 In

that CCSSM ([CCSSM]) shows awareness of the significance of Theorem 1.7. other words, irrational numbers.

1.4. MULTIPLYING FRACTIONS

49

1 3 1 2

These 3 × 2 small rectangles are congruent and therefore have equal areas. The areas of these rectangles thus form an equal partition of the unit 1 (= area of unit square) into 3 × 2 equal parts. Consequently, the area of each small rectangle is the 1 . By the product formula, we have fraction 3×2 1 1 1 = × . 3×2 3 2 We have just shown that a rectangle with side lengths 13 and 12 has area 13 × 12 . Now we can tackle the case of sides with lengths 23 and 52 . Let a rectangle R with side lengths 23 and 52 be given. We want to show that its area is 23 × 52 . The key observation here is that R is paved by rectangles with side lengths 13 and 12 . Precisely, it is paved by 2 rows of 5 such rectangles—the 2 being from the numerator of 23 and the 5 being from the numerator of 52 —as shown below.

R

1 3 1 2

Since each of these 2 × 5 small rectangles has area equal to the sum of 2 × 5 of these areas; i.e., area of R =

1 3×2 ,

the area of R is

1 1 + ··· + . 3×2 3×2    (2×5)

By (1.28) on page 47, the right side is equal to (2 × 5) ×

1 2×5 2 5 = = × , 3×2 3×2 3 2

where we have used the product formula twice. Therefore Theorem 1.7 is proved for this special case. A little reflection on the preceding reasoning will reveal that the specific numbers, 23 and 52 , do not play a crucial role, and the reasoning is valid in general. Therefore we can simply write down the general proof along exactly the same lines. Proof of Theorem 1.7. The proof is broken up into two parts: Part I: the case where the lengths of the sides are unit fractions, i.e., fractions with numerators equal to 1 (see page 10) and Part II: the general case. The proof will take for granted certain geometrically obvious facts about rectangles that can be easily proved once the needed tools are available (see Exercises 4 and 7 on page 237).

50

1. FRACTIONS

Part I. We are going to prove that if a rectangle has two sides of lengths 1 and n1 , then its area is 1 × n1 . On the number line, let the unit 1 be the area of the unit square. We will show how to divide this particular unit segment into n equal parts, as follows. Partition one pair of opposite sides of the unit square into  parts of equal length and the other pair into n parts of equal length. Join the corresponding points of the division to obtain a paving of the unit square by n congruent rectangles, each having side lengths 1 and n1 , as shown: 6

 copies

?

1 



1 n

n copies

-

These n rectangles have equal areas because they are congruent. The unit square therefore has been partitioned into n equal parts—in terms of area—by these rectangles. Consider the shaded rectangle in the picture. By the definition of a fraction (in the case that the unit is the area of the unit square), its area is the 1 on this number line because it is one part when the unit (area of the fraction ×n unit square) is partitioned into n equal parts. By the product formula, its area is equal to 1 1 1 = × . ×n  n The proof of Part I is complete. Part II. We are going to prove that the area of a rectangle with sides of length k k m and m n is  × n . Let R be a rectangle with sides of length k and m n . Then R is paved by k rows of m rectangles each of which has side lengths 1 and n1 , as shown: 6

R

k copies

?

1 



1 n

m copies

-

1.4. MULTIPLYING FRACTIONS

51

1 We have just seen that each of these small rectangles has area equal to n . Since R is paved by km of these congruent small rectangles, the area of R is equal to

1 1 km k m 1 + ··· + = = × , = km × n n  n n  n km

where we have used (1.28) on page 47 and the product formula. The proof of Theorem 1.7 is now complete. Pedagogical Comments. The preceding proof of Theorem 1.7 is very instructive in that it is the first example of a proof that is achieved by neatly breaking it up into two smaller steps. First, we prove it for the case where the side lengths are unit fractions and then, on the basis of this special case, we prove with ease the general case. The general strategy of breaking a complex task into several simple tasks is basic in mathematics. While it will not work for every proof, it works often enough to make it worth learning. For example, see the discussion of the proofs of ASA on page 249, the proofs of SSS and FTS in Sections 6.2 and 6.4, respectively, of [Wu2020b], and the proof of Theorem G52 in Section 6.8 of [Wu2020b]. End of Pedagogical Comments. Three remarks We round off our discussion of the multiplication of fractions with three remarks. The first is the explanation of the usual multiplication algorithm for finite decimals. Consider for example 1.25 × 0.0067. The algorithm says to (α) multiply the two numbers as if they are whole numbers by ignoring the decimal points, (β) count the total number of decimal digits of the two decimal numbers, say p, and (γ) put the decimal point back in the whole number obtained in (α) so that it has p decimal digits. We now justify the algorithm using this example, noting at the same time that the reasoning in the general case is the same. By the product formula, we have 67 125 × 67 125 . 1.25 × 0.0067 = 2 × 4 = 2 10 10 10 × 104 Noting that 125 × 67 = 8375 and recalling that 10m × 10n = 10m+n for all nonzero whole numbers m and n, we get 1.25 × 0.0067 = =

8375 × 104

102

8375 102+4

= 0.008375

(corresponding to (α))

(corresponding to (β)) (corresponding to (γ)).

52

1. FRACTIONS

A second remark is about a different perspective on the number line in terms of multiplication. Let a number line be given; we will call this the original number line. Suppose on the original number line we make a different choice of 0 and 1 (see page 5) by designating two points A and B as the new 0 and 1, respectively. With respect to this new choice of A and B as 0 and 1, let P denote the fraction m n on the new number line, by which we mean the number line which has A and B as 0 and 1, respectively. The question is how to locate the point P . We claim that P is the point so that P and B are either both to the right of A or both to the left of A and so that m (1.30) |AP | = × |AB|, n where |AP | and |AB| are the lengths of the segments AP and AB with respect to the original number line and the fraction m n refers to the original number line. (To make sense of the right side of (1.30), we will assume that |AB| is a fraction for now. Once we have the real numbers, we can drop this restriction and the subsequent argument will make sense in general.) 25 We will prove (1.30) for the special case that m n is 7 , and it will be seen that the proof of the general case is not different. For simplicity, let B be to the right of A; the argument for the case where B is to the left of A is entirely similar. Since 25 4 4 25 7 = 3 7 = 3 + 7 and since P represents 7 on the new number line, if we replicate the segment [A, B] two more times to the right of 0, we get to the point representing the number 3 of the new number line. Now in the next segment of the same length as [A, B] but to the right of this 3, if we divide it into 7 equal parts, then 4 "small steps" from 3 to the right will get to P , as shown: A  0



B   1

P 

 



 3

By the definitions of the addition of fractions and by (1.25) on page 45, we have     4 4 25 |AP | = (3 × |AB|) + × |AB| = 3 + × |AB| = × |AB|. 7 7 7 25 This proves (1.30) when m n is equal to 7 . In TSM, the argument would go something like this: if we compare the two number lines, then 25 7 is to 1 (length of [0, 1]) as |AP | is to |AB| by proportional reasoning. Thus |AP | 25/7 = . 1 |AB|

By the cross-multiplication algorithm, we get |AP | = 25 7 × |AB| again. However, since we do not know what proportional reasoning is, this argument has no validity. A third and final remark is that there are two standard inequalities concerning multiplication that are worth knowing: if A, B, C, and D are fractions, then: (i) If A > 0, then AB < AC is equivalent to B < C. (ii) A < B and C < D imply AC < BD. Both are obvious when we interpret fraction multiplication as the area of a rectangle. See Exercise 3 on page 53.

1.4. MULTIPLYING FRACTIONS

53

Pedagogical Comments. Multiplication is probably the most misunderstood arithmetic operation among the usual four in the study of fractions. The simplicity of the product formula (Theorem 1.6 on page 46), compared with the somewhat mystifying addition formula ((1.12) on page 33) and invert-and-multiply ((1.33) on page 57 below), has misled many educators and mathematicians to believe that multiplication is the easiest of the four operations to teach. In a typical passage in TSM, it is said that, "Multiplying fractions is usually simpler than adding fractions. To find the product, you just multiply the numerators and multiply the denominators." We know that multiplication is in fact complicated. There are different ways to define multiplication, but no matter how it is done, there are complications. In this volume, it is defined in terms of the concept of "of " on page 24 and Theorem 1.3 on page 27. One also has to prove the product formula using the precise definition and then prove the area formula of a rectangle, Theorem 1.7 on page 48, to round off its conceptual foundation. There is subtlety in the reasoning of all these theorems and definitions. What is even more important is the fact that, without a solid understanding of multiplication, it is impossible to approach division, as the next section amply shows. The usual difficulty with the division of fractions is less a statement about the difficulty of the concept of division per se but more about students’ lack of a good grounding in the multiplication of fractions in TSM. Multiplication has to be taught carefully if one hopes to have any success teaching the division of fractions. End of Pedagogical Comments. Exercises 1.4. (1) Do each of the following without calculators. 1 1 1 × 2 19 × 2 19 )× (a) (12 23 × 12 23 × 12 23 ) × (2 19 7 × 4 23 ) + (2 61 × (b) ( 18

(2)

(3)

(4) (5)

(6)

(7)

7 18 )

1 26 .

7 + ( 18 × 3 16 ) .

2 (c) 8 50 × 1250 12 . (i) Prove that 123.45 × 1014 = 0.012345. (ii) Prove that 67.8901 × 103 = 67890.1. (iii) Formulate and prove generalizations of parts (i) and (ii). Prove the following for fractions A, B, C, and D (see page 52): (i) If A > 0, then AB < AC is equivalent to B < C. (ii) A < B and C < D imply AC < BD. Give a detailed proof of the corollary to Theorem 1.6 on page 46. Consider the following two numbers A and B, where A is the length of the concatenation (see page 12) of 4 parts when [0, 18 29 ] is divided into 17 equal parts. B is the length of 4 the concatenation of 18 parts when [0, 17 ] is divided into 29 equal parts. Is A equal to B? Why or why not? (a) Find a fraction q so that 28 12 = q×5 14 . Do the same for 218 17 = q×19 12 . (b) Make up a (realistic) word problem for each situation, and make sure that the problems are not the same for both. The perimeter of a rectangle is by definition the sum of the lengths of its four sides. Show that given a fraction A and a fraction L, (a) there is

54

1. FRACTIONS

a rectangle with area equal to A but with a perimeter that is bigger than L and (b) there is a rectangle with perimeter equal to L but with an area that is less than A. (8) (a) 16 12 cups of liquid would fill a punch bowl. If the capacity of the cup is 9 13 fluid ounces, what is the capacity of the punch bowl? Explain carefully. (b) The length of a rod is 18 58 times the length of a short piece that is 3 14 inches long. How long is the rod? Explain. (9) How many buckets of water would fill a container if the capacity of the bucket is 3 13 gallons and that of the container is 7 12 gallons? (Caution: Getting an answer for this exercise is easy, but explaining it logically is not.) (10) Give a proof of the distributive law for the division of whole numbers; namely, let k, m, n be whole numbers, and let n > 0. Then (m ÷ n) + (k ÷ n) = (m + k) ÷ n. (11) (This is Exercise 8 on page 31. Now do it again using the concept of fraction multiplication.) James gave a riddle to his friends: "I was on a hiking trail, and after walking 79 of a mile, I was 49 of the way to the end. How long is the trail?" (12) The difference of two given fractions is equal to 45 of the smaller one, while their sum is equal to 28 15 . What are the fractions? (Hint: Use the number line.) 1.5. Dividing fractions Understanding the division of fractions requires the understanding of the division of whole numbers and the multiplication of fractions. After a review of the former, we give the definition of fraction division that is patterned closely after the division of whole numbers. The precise definition leads immediately to the invertand-multiply rule. As another application of the precise definition, we solve a word problem (on page 58) using only reasoning based on the definition of division, but nothing involving analogies or metaphors as is the custom in TSM. The discussion in the last subsection on the division of finite decimals is only a beginning and cannot be concluded until Chapter 3 of [Wu2020c]. Division of whole numbers, revisited (p. 54) Definition of the division of fractions (p. 57) A typical word problem (p. 58) Division of finite decimals (p. 60) Division of whole numbers, revisited The overriding fact concerning the concept of division is that—in a sense to be made precise in (**) on page 57—division is an alternate, but equivalent, way of writing multiplication.40 Unfortunately, this simple fact has been corrupted in 40 Mathematical Aside: In abstract algebra, this fact is expressed incisively as follows: the division by a nonzero element x is by definition the multiplication by the multiplicative inverse of x.

1.5. DIVIDING FRACTIONS

55

TSM to read: "Division is the inverse operation of multiplication." We will give some indication below (*) on page 55 as to why this is nonsensical. We first review the division of whole numbers. We teach children that 36 9 = 4 because 4 × 9 = 36. This then is the statement that 36 divided by 9 is the whole number which, when multiplied by 9, gives 36. In symbols, we may express the foregoing as follows: 36 9 is by definition the number k which satisfies k × 9 = 36. Likewise, 84 is the whole number m which satisfies 7 m × 7 = 84, etc. In general, given any two whole numbers m and n with n = 0, the division m n is the whole number q so that qn = m, and since we only have whole numbers at our disposal, this cannot make sense unless m is a multiple of n to begin with. Notice that we say "the" whole number q, because this q is unique; i.e., there is only one such number. In elementary school, the "uniqueness" is (understandably) not emphasized, but we have to emphasize it now in anticipation of the division of fractions. Let us make sure that this uniqueness is actually true by showing that if there is another whole number q  so that q  n = m, then q  n = qn since both are equal to m. Therefore (q  − q)n = q  n − qn = 0, and since n = 0, q  − q = 0, and we have q  = q. So this q is indeed unique. The precise formulation of the concept of division among whole numbers is then the following: If m and n are whole numbers, with n = 0 and m being a multiple , is the unique of n, then the division of m by n, denoted by m n whole number q so that m = qn.41 Once we have this precise definition of division, we can assert the following: (*)

Given whole numbers m, n, and q, with n = 0 and m being a multiple of n, m n = q is equivalent to m = qn

This is the correct formulation of the confusing statement in TSM that "division is the inverse operation of multiplication" when the concept of division is introduced. The difference between (*) and the latter statement could not be more stark. (*) is a straightforward statement, after both multiplication and division have been defined, that the two simple facts, m n = q and m = qn, are equivalent. The latter statement, however, is usually made as an attempt to inform students about what "division" means. But the statement utterly fails to be informative because students have no idea of what an "inverse operation" is. Mathematical Aside: The statement "division is the inverse operation of multiplication" does not make any sense whatsoever, because an operation on the whole numbers N is a mapping F : N × N → N so that an "inverse operation" would have to be a map G : N → N × N so that F ◦ G = I and G ◦ F = I, where we have used I as a generic symbol for the identity mappings on N and N × N. Thus with F : N × N → N defined by F (m, n) = mn (multiplication of m and n), clearly there can be no such G. For example, although F (3, 8) = 24, it is also true that F (4, 6) = 24 and even F (12, 2) = 24. So how can we define G? Should G(24) be (3, 8) or (4, 6) or (12, 2)? What TSM should have said is that, with a whole number 41 This precise definition of division provides a simple explanation that division of a nonzero number by 0 has no meaning, because if it had meaning, then for a nonzero whole number m, m 0 is the whole number q so that q × 0 = m. But the last equation cannot hold because the product on the left side is equal to 0 whereas the right side m is nonzero.

56

1. FRACTIONS

m fixed and m being a multiple of a nonzero whole number n, it is always the case that m × n ÷ n = m and m ÷ n × n = m. But as we have indicated all along, TSM is simply incapable of achieving such precision. The preceding definition of division among whole numbers is important for the understanding of division among fractions because the latter is patterned after the former, with one important caveat. The definition of whole number division m ÷ n makes sense only when m is a multiple of n, but, with n fixed and n > 1, there are relatively few whole numbers m that are multiples of n. Our first task in approaching the division of fractions is to remove this restrictive condition in our prospective definition of fraction division by showing that, given a nonzero fraction B, every fraction is a (fractional) multiple of B, as the following theorem shows. Theorem 1.8. Given fractions A and B (B = 0), there is a unique fraction C, so that A = CB. Proof. If A = 0, we may let C = 0. So we will assume A = 0 from now on and let A = k and B = m n , where k, , m, n are nonzero whole numbers. Let us first prove the uniqueness of a fraction C so that A = CB; i.e., regardless of whether there is such a fraction C or not, we want to be assured that there can be at most one such fraction. So suppose A = CB. Then m k =C× .  n n kn Multiplying both sides by m yields m = C. This shows that if there is a fraction C kn . This conclusion so that A = CB, then there is only one possibility for C: C = m proves the uniqueness of C, but it says more. It also tells us how to prove existence: kn and it is clear that A = CB by virtue of the cancellation rule simply let C = m (1.29) on page 47. The theorem is proved.

The proof of Theorem 1.8 shows explicitly how to get the fraction C so that CB = A: if A = k and B = m n , then the proof gives C as C =

kn k n = × . m  m

In other words, (1.31)

if C ×

k k n m = , then C = × . n   m

This fact will be useful below. Despite the simplicity of the statement, Theorem 1.8 is conceptually subtle and may take some getting used to. As mentioned above, it says that if a fraction B is nonzero, then every fraction A is a fractional multiple of B, in the sense that A = CB for some fraction C. (Note that, since we are no longer dealing exclusively with whole numbers, the meaning of multiple has to be suitably modified. In the future, if we want to indicate that there is a whole number C so that A = CB, we will say explicitly that A is a whole number multiple of B.) Taking A = 1, then Theorem 1.8 implies that there is exactly one fraction, which we will denote by B −1 , so that B −1 B = 1. We call this B −1 the inverse (or multiplicative inverse, to be precise) of B. In fact, (1.31) shows that m n (1.32) if B = , then B −1 = . n m

1.5. DIVIDING FRACTIONS

57

For this reason, B −1 is also called the reciprocal of B in the context of fractions. Using this notation, the expression of C in equation (1.31) above can be rewritten as C = AB −1 . Thus, if A = CB, then C = AB −1 . For example, if A = 11 C that satisfies C( 23 8 ) = 5 is, according to (1.31), C =

11 5

and B =

23 8 ,

then the

8 88 11 × = . 5 23 115

Definition of the division of fractions We can now give the definition of fraction division. It is, word for word, the same as the definition of whole number division given on page 55, with the exception that, thanks to Theorem 1.8, there is no need to require that A be a fractional multiple of B. Definition. If A, B are fractions (B = 0), then the division of A by B, A , is the unique fraction C so that A = CB. denoted by B The division of A by B is also called the quotient of A divided by B. In analogy with (*) on page 55, we can also state: (**)

A Given fractions A, B, and C, with B = 0, = C is equivalent B to A = CB.

This is the precise meaning of the statement that division is an alternate, but equivalent, way of writing multiplication. At this point, we call attention to the resemblance of this definition of division to the abstract interpretation of subtraction in equation (1.18) on page 39. Remarks. We wish to make three remarks on the division of fractions. (i) Invert and multiply. Given fractions k and m n (the latter being nonzero), we claim that the following invert-and-multiply rule holds for the division of fractions: (1.33)

k  m n

=

n k × .  m

Indeed, if we denote the left side of (1.33) by C, then, by definition, C is the unique fraction so that k = C × m n . By (1.31), n k × .  m So (1.33) is proved. (We see that there is nothing mysterious about this rule: it is a simple consequence of the correct definition of division.) (ii) Division is well-defined. Given fractions k and m n (the latter being nonzero), k K m M . Now suppose we have defined the division of k by m n  = L and n = N for C=

M whole numbers k, , . . . , M , N . Then the division of K L by N is also defined. So

58

1. FRACTIONS

the question is this: does it hold that k  m n

(1.34)

=

K L M N

?

If not, then it would mean that we do not have a concept of the division of fractions (which are points on the number line), but only the division of the particular fractions symbols chosen to represent these fractions (see the discussion of (1.26) M n N on page 46). Fortunately, (1.34) is correct because m n = N implies m = M (for example, use the cross-multiplication algorithm, Theorem 1.2 on page 22). Therefore by (1.26) on page 46, we have n K N k × = × .  m L M By the invert-and-multiply rule, equation (1.33), this is equivalent to equation (1.34), as desired. (iii) Two concepts of division for whole numbers. The following discussion is best understood from the viewpoint of the concept of extension on page 29. Consider 7 and 5. On the one hand, we can consider the whole-number division 7 ÷ 5 in the sense defined on page 29. On the other hand, we can also consider the division of the fraction 7 by the fraction 5 in the sense of the preceding definition (remember that each whole number is also a fraction). Are they equal? The easy answer is yes, because both are equal to the fraction 75 , but here is a more detailed explanation. For the division as whole numbers, 7 ÷ 5 yields the fraction 75 by virtue of Theorem 1.4 on page 29. Now the division as fractions shows that, by virtue of the invert-and-multiply rule (1.33), 7 1 5 1

=

7 1 7 × = . 1 5 5

So the two are equal. Obviously, the reasoning holds in general for any two nonzero whole numbers. A typical word problem The following is a typical application of the concept of fraction division in school mathematics. We ask you to take note of the difference between the usual presentation in TSM and the one given here: we give the explicit reason why division has to be used and why—according to mathematics—the answer must be interpreted in a particular way. Example. A rod 43 38 meters long is cut into pieces which are 53 meters long. How many such pieces can we get out of the rod? If we change the numbers in this example to "if a rod 48 meters long is cut into pieces which are 2 meters long, how many such pieces can we get out of the rod?", then there would be no question that we should do the problem by dividing 48 by 2. So we will begin the discussion by following this analogy and simply divide 43 38 by 53 and see what we get: 43 38 5 3

=

1 1041 = 26 . 40 40

1.5. DIVIDING FRACTIONS

59

We have used invert-and-multiply for the preceding computation, of course. Now 1 mean? Remembering the definition of division, we see what does the answer 26 40 3 1 that the division of 43 8 by 53 being equal to 26 40 is equivalent to   1 5 1 3 5 = 26 × = 26 + 43 × 8 40 3 40 3     1 5 5 = 26 × + × (distributive law), 3 40 3 where the distributive law for fractions is stated in the corollary on page 46. The first term on the right, 26× 53 , is the length of the concatenation of 26 segments each 1 of length 53 , and the second term on the right, 40 × 53 , is the length of a segment 1 5 which is 40 of 3 , by the definition of fraction multiplication. Thus the rod can be 1 cut into 26 pieces each of 53 meters in length plus a piece that is only 40 of 53 meters. This then provides the complete answer to the problem and retroactively justifies the use of division to do the problem. Notice that the key to getting the correct answer is knowing the precise definition of division (which allowed us to convert the division into a multiplication) and knowing the distributive law (which allowed us to arrive at a correct interpretation 1 from the division). of the answer 26 40

(1.35)

You may find such an after-the-fact justification of the use of division to do the problem to be unsatisfactory. There is in fact a line of logical reasoning that leads inexorably to the conclusion that division should be used. We now present this reasoning. (Compare Exercise 8(b) on page 54.) Let there be a maximum of K copies of 53 in 43 83 , where K is a whole number. Then 43 38 − K × 53 is less than 53 (as otherwise K would not be the maximum number of such copies). Denote 43 83 − K × 53 by r (note that r is a fraction); then we may rewrite the definition of r as   5 3 5 + r, where 0 ≤ r < . 43 = K × 8 3 3 Now, by Theorem 1.8 (page 56), we may express r as a multiple of so that i.e., there is a fraction m n r =

5 ; 3

m 5 × . n 3

must be a proper fraction in the sense that m < n We notice that m n because r < 53 , so that, in view of the preceding equation, m =

3 5 3 ×r×n< × ×n=n 5 5 3

(see item (B) on page 52). Therefore, substituting this value of r into the equation (1.35) gives     3 5 5 m 43 = K× + × 8 3 n 3  m 5 × , = K+ n 3

60

1. FRACTIONS

where K + m is a mixed number because m is a proper fraction. So we n n may rewrite the preceding equation in the notation of mixed numbers:  m 5 3 × . 43 = K 8 n 3 By the definition of division, we see that K

43 3 m = 58. n 3

, then we would know the answer to If we know the mixed number K m n the problem, which is K. Therefore, the significance of the preceding equation is that, in order to find the maximum number of 53 ’s in 43 38 , we should do the division 43 83 . 5 3 1 = 40 . Recall that, by the above calculation, K = 26 and m n We have thus explained how one can give an a priori justification for the use of division to solve this problem.

Division of finite decimals We now come to the last part of the arithmetic of finite decimals: division. In principle, this is very simple because we are going to show that the division of decimals is reduced to the division of whole numbers. The following example is sufficient to illustrate the general case: the division 0.311 0.64 becomes, upon using the definition of a decimal and invert-and-multiply, 0.311 = 0.64

311 103 64 102

=

311 . 640

This reasoning is naturally valid for the division of any two finite decimals. Therefore, the division of any two finite decimals is equal to a fraction. Now a fraction is not quite the right answer to a division of two finite decimals because we want the answer to be a finite decimal. The next step is therefore to convert a given fraction to a finite decimal; i.e., given a fraction m n for some whole numbers m and n, to convert m to a finite decimal is to find whole numbers n N and k so that N m = k, n 10 where the right side is, by definition, a finite decimal (see page 14). It turns out that this is not always possible, as we now explain. The overriding fact is that a fraction in lowest terms is equal to a finite decimal if and only if its denominator is a product of 2’s and 5’s42 (Theorem 3.8 on page 152). Keeping this in mind, what we will do here is show how to convert a fraction, in three different ways, to a finite decimal if its denominator is a product of 2’s and 5’s. At the end, we will also make a passing comment on the general case of converting an arbitrary fraction to an "infinite decimal". 42 It

is understood that it could be just 2’s or just 5’s.

1.5. DIVIDING FRACTIONS

61

In one sense, the conversion is easy when the denominator is a product of 2’s and 5’s. What makes it easy is the simple observation that a power of 10 is a product of 2’s and 5’s; i.e., 10k = 2k × 5k for any whole number k (because 10 = 2 × 5). 7 , for example. Since To see how this leads to the conversion, take the fraction 125 125 = 53 and 103 = 23 × 53 , we see that we can directly change the denominator 125 to 103 by multiplying it by 23 . Therefore, by invoking equivalent fractions, we get 7 7 23 × 7 56 = 3 = 3 = 3 = 0.056. 3 125 5 2 ×5 10 Here is another example: consider the fraction 24 × 54 , we have

28 625 .

Since 625 = 54 and 104 =

28 24 × 28 448 28 = 4 = 0.0448. = 4 = 4 625 5 2 × 54 10 Finally, let us take up the fraction above,

311 640 .

We have

311 311 = 7 . 640 2 ×5 Since 107 = 27 × 57 = (27 × 5) × 56 = 640 × 56 , we see that multiplying 640 by 56 changes 640 to 107 . Therefore, using equivalent fractions again, we get 311 × 56 311 311 × 56 4859375 = = = = 0.4859375. 7 6 640 (2 × 5) × 5 107 107 There should be no difficulty at this point in converting any fraction with a denominator equal to a product of 2’s and 5’s to a finite decimal. This is not the end of the story, however. We need two more observations to round out the picture. First, there is another way to make use of the given hypothesis that the denominator of a fraction is a power of 2’s and 5’s. Thus far, we have focussed our attention on the denominator of the fraction, but we will now shift the focus to the numerator instead. We can use equivalent fractions to introduce a large power of 10 into the numerator so that we can cancel the 2’s and 5’s in the denominator to get a whole number. This statement needs to be amplified, so 7 again. Knowing in advance that 125 = 53 , let us look at an example: consider 125 3 3 3 we know that if we have 10 (= 2 × 5 = 23 × 125) in the numerator, we would be able to cancel the 125 in the denominator. Precisely, using the cancellation rule (1.29) on page 47, we have   7 × 103 1 7 × 3. (1.36) = 125 125 10    N3

The subscript 3 in the fraction we call N3 on the right refers to the fact that 103 is in its numerator. N3 is in fact a whole number because its numerator is a multiple of 125 (since 103 = 23 × 125) so that it cancels the 125 in the denominator. Thus N3 = 7 × 23 = 56 so that (1.37)

7 N3 56 = 3 = 3 = 0.056 125 10 10

62

1. FRACTIONS

7 as before. For another example, let us revisit the fraction 311 640 . Knowing 640 = 2 ×5 in advance, we will multiply the numerator 311 by 107 , as follows:   311 × 107 311 1 = (1.38) × 7. 640 640 10    N7

Again, the subscript 7 in the fraction we call N7 on the right refers to the 107 in its numerator. N7 is also a whole number because the number 107 in its numerator is a multiple of 640 (since 107 = (27 × 5) × 56 = 640 × 56 ) and this factor of 640 in the numerator of N7 cancels its denominator. Thus N7 = 311 × 56 = 4859375 so that N7 4859375 311 = 7 = = 0.4859375. (1.39) 640 10 107 5 Let us do one more example. Consider the fraction 15 32 . Because 32 = 2 , we rewrite the fraction as   15 × 105 15 1 (1.40) = × 5. 32 32 10    N5

Once again, the fraction N5 is a whole number because the 105 in its numerator is equal to 105 = 25 × 55 = 32 × 55 and the denominator 32 of N5 therefore gets canceled. Thus N5 = 15 × 55 = 46875. Consequently, N5 15 46875 = 5 = (1.41) = 0.46875. 32 10 105 The second observation builds on the first and establishes a connection between the way of converting a fraction to a finite decimal in the first observation and the traditional way that uses long division. To explain this statement in greater detail, 7 again. As we explained after (1.36), the number N3 in (1.36) is a let us look at 125 whole number (in fact N3 = 56). Therefore N3 is the quotient of the long division of 7×103 by 125. By tradition, one refers to this long division as the long division of the numerator 7 by the denominator 125, it being understood that it is actually the long division of 7 × 103 (and not 7) by the denominator 125. In TSM, the conversion of a fraction to a decimal by long division is a rote skill, and part of this rote skill is to "put the decimal point back in the quotient N3 (= 56)" in a particular way. In our setting, such a placement of the decimal point—third digit from the right—is precisely explained (and dictated) by the denominator 103 in (1.37). As a second example, consider the fraction 311 640 in (1.38). Again, since we already know that the N7 in (1.38) is a whole number (in fact, N7 = 4859375), N7 is now seen to be the quotient of the "long division of the numerator 311 by the denominator 640" (again it is understood that the dividend is actually 311 × 107 ). Equation (1.39) then shows why 311 640 is equal to the decimal obtained by placing the decimal point 7 digits from the right in the quotient N7 . Similarly, the fraction 15 32 in (1.41) is equal to the decimal obtained by placing the decimal point 5 digits from the right in the numerator N5 (= 46875), where N5 is the quotient of the long division of 15 (actually 15 × 105 ) by 32 as in (1.40). We must tie up one last loose end concerning (1.36). To express the fraction 7 as a finite decimal, do we have to use 103 as in (1.36) or can we use 10k for any 125

1.5. DIVIDING FRACTIONS

63

k ≥ 3? In other words, suppose we take any whole number k ≥ 3 and define Nk in terms of 10k as follows:   7 × 10k 7 1 = (1.42) × k. 125 125 10    Nk 7 Comparing with (1.36) on page 61, we have expressed 125 in two different ways: as 3 k N3 /10 and as Nk /10 for a whole number k ≥ 3. What is the relationship between Nk and N3 ? The simple answer is that

Nk = N3 × 10k−3 because

7 × 103 7 × 10k = × 10k−3 = N3 × 10k−3 . 125 125 In other words, Nk is the same quotient of the long division of 7 × 103 by 125 with a suitable number of zeros added to the right end. This means we can obtain the 7 by taking the quotient of the long division of 7 × 10k finite decimal equal to 125 by 125 for any k ≥ 3 and then putting the decimal point k digits from the right in this quotient. Similarly, for the fraction 311 640 , it is equal to the finite decimal obtained by taking the quotient of the long division of 311×10k (for any k ≥ 7) by 640 and then putting the decimal point k digits from the right in this quotient. A similar statement can be made about the fraction 15 32 . In fact, the uniformity of the reasoning in these three examples shows that the reasoning is valid in general. Therefore we have proved the following theorem, which is the traditional algorithm for converting a fraction to a decimal by the long division of the numerator by the denominator at least for the special case where the denominator is a product of 2’s and 5’s. Nk =

Theorem 1.9. Let m n be a fraction so that n is a product of 2’s and 5’s. Then for any sufficiently large whole number k, the long division of m × 10k by n has quotient q (a whole number) and remainder 0, and m n is equal to the finite decimal q . 10k

It is clear from the reasoning used to arrive at Theorem 1.9 that the k in Theorem 1.9 can be taken to be any whole number larger than or equal to both of the exponents of 2 and 5 when n is expressed as the product of a power of 2 times a power of 5. This latitude in the choice of k is what makes the algorithm of the conversion of a fraction to decimal so easy to use, because it says that if in doubt, one can add any number of zeros to the right of the numerator to do the long division. Such a practical consideration also brings up another observation about Theorem 1.9, namely, that there is in fact no need to consider any fraction whose denominator contains factors of both 2 and 5. This is because, if factors of both 2 and 5 are present, the denominator will be a multiple of 10 (= 2 × 5) and the factors of 10 can be "split off" from the beginning to simplify the computations. 6 Let us illustrate with the above fraction 311 640 : because 640 = 2 × (2 × 5) = 64 × 10, we have 311 311 1 = × . 640 64 10

64

1. FRACTIONS

This equality means that to obtain the decimal conversion of 311 640 , it suffices to first 311 obtain the decimal conversion of 64 and then move the decimal point of the latter decimal one digit to the left (compare Exercise 2 on page 53). The point is, of 6 course, the fact that the denominator 64 of 311 64 is a power of 2 alone (i.e., 2 ). Now the general case. Most fractions will not be equal to a finite decimal because Theorem 3.8 on page 152 says that if a fraction in lowest terms is equal to a finite decimal, then its denominator is a product of 2’s or 5’s or both. Thus in general, a fraction will be equal to an infinite decimal, whose definition can only be given by using the concept of limit; see Chapter 3 of [Wu2020c]. The fact that every fraction is equal to an infinite repeating decimal is the content of Theorem 3.8 in Section 3.4 of [Wu2020c].43 TSM usually teaches the procedure of "converting a fraction to an infinite decimal by the long division of the numerator by the denominator" entirely by rote without the slightest explanation of what an "infinite decimal" is or why the procedure involving long division is valid. Worse, the simple explanation (implicit in (1.42) on page 63) that one can attach as many zeros to the right of the numerator to continue the long division and then put back the decimal point in the quotient could have been given but never was. While it is true that the definition of an infinite decimal is beyond the scope of school mathematics, one can nevertheless approach the conversion of fractions to infinite decimals in school mathematics in a more civilized manner, as we now explain. Consider, for definiteness, the decimal conversion of 27 . Instead of saying 27 is equal to the infinite repeating decimal 0.285714, we can say instead that we can approximate 27 as closely as we like by a finite decimal. For example, suppose we want a finite decimal that is within 1/106 of 27 . Following up our success with equation (1.36) on page 61, we should at least try something like the following:   2 × 108 2 1 = × 8. 7 7 10 The hope is that the fraction in parentheses on the right will yield something that serves our purpose. We use 108 instead of 106 because we are unsure of our footing here and want to play it safe (we could even use 1010 if 108 turns out to be not good enough). By the long division of 2 × 108 by 7, we get the division-with-remainder 2 × 108 = (28571428 × 7) + 4.

(1.43) Thus, 2 7

= =

(28571428 × 7) + 4 1 × 8 = 7 10   4 1 28571428 × + . 108 7 108

Equivalently, 2 − 0.28571428 = 7





4 28571428 + 7

4 1 × 7 108

 ×

1 108

 ,

43 A finite decimal is considered to be an infinite repeating decimal in this context, with a repeating block consisting of the single digit 0.

1.5. DIVIDING FRACTIONS

65

But the right side is a nonzero fraction, so we actually have   4 1 2 × 8 . 0 < − 0.28571428 = 7 7 10 We can further simplify the right side. Since 47 < 1, inequality (i) on page 52 implies   4 1 1 1 × < 1 × 8 = 8. 7 108 10 10 Therefore we obtain 1 2 − 0.28571428 < 8 . 7 10 These inequalities can be captured in a picture: the thickened horizontal segment below represents the difference: 27 − 0.28571428.

(1.44)

0
r, then E too has C as its boundary (see Exercise 4 on page 198 again). We will call E the exterior of C. To distinguish between the two sets D and E, we introduce the following concepts: a set S in the plane is said to be bounded if it is contained in some closed disk of radius R; otherwise it is unbounded. For example, a circle of radius r is bounded, and so is an open disk or a closed disk of radius r, but any line, any ray, or any half-plane of a line is unbounded. We can also understand boundedness from a slightly different point of view, as follows. Lemma 4.12. A set S in the plane is bounded if and only if there is a point O  in the plane and a positive number R so that the distance of every point in S from O  is ≤ R. Proof. First let S be bounded. So it is contained in a closed disk D of radius R. If the center of D is O  , then the distance of each point of S from O  is ≤ R because 23 It

does not matter whether the disk is closed or open.

4.2. THE BASIC VOCABULARY, PART 2

195

each point of S is a point of D. Conversely, if a set S has the property that for some fixed positive number R, every point of S is of distance ≤ R from a fixed point O  in the plane, then S is bounded because S is contained in the closed disk of radius R centered at O  . The proof is complete. With the same notation, we see that the open disk D is bounded but the exterior E of the circle C is not. It is clear that the plane is the disjoint union of D, C, and E (see page 175 for the definition of disjoint union). There is also a property about D and E that is intuitively obvious: Any segment joining a point in the open disk D to a point in the exterior E of the circle must intersect the circle C. Like many "obvious" statements in geometry, a proof of this assertion involves subtle concepts about real numbers. (Compare the intermediate value theorem in Section 6.2 of [Wu2020c].) For this reason, we will assume this fact here without proof. A set in the plane that contains all of its boundary is called a closed set. In terms of the preceding notation, the closed disk with center O and radius r, to be denoted by D, is a closed set. Observe that D = D ∪ C, where we recall that the symbol "∪" stands for union. In this volume, we will refer to the closed disk D with radius r and center O as the closed set inside the circle C. Sometimes we say more simply that D is the inside of C. We have discussed the situation of the circle in such detail because it sets up a model for the discussion of polygons (see page 171 for the definition). The linguistic abuse of the word "circle" described on page 194 in fact spills over to "triangle", "quadrilateral", and in general, "polygon". For example, a triangle is by definition the union of the three segments consisting of the three sides, but when we speak of, e.g., "the area of the triangle", we certainly do not mean "the area of the three segments" but, rather, "the area inside those three segments", where the meaning of inside is usually understood in an intuitive and imprecise sense. We now try to shed some light on the word "inside". First, let us draw a parallel with the situation of a circle by invoking a theorem that we will not prove in this volume. Two new definitions will be needed for the statement of the theorem. A polygonal segment is a finite collection of segments A1 A2 , A2 A3 , A3 A4 , . . . , An−2 An−1 , An−1 An , with the understanding that these segments could be collinear and that there may be intersections among them. Then a region R in the plane is said to be connected if any two points in R can be joined by a polygonal segment that lies completely in R. It is easy to see that the open disk D with center O and radius r and the exterior E of the circle with center O and radius r are disjoint connected sets (see Exercise 5 on page 198). Also, recall the definition of the complement of a subset S in the plane as all the points in the plane not lying in S (see the footnote on page 181). Theorem 4.13. The complement of a polygon P consists of two nonempty planar regions B and E with the following properties: (i) B and E are both connected, B is bounded and E is unbounded, and P is their common boundary. Moreover, the plane is the disjoint union of the three sets B, E, and P.

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4. BASIC ISOMETRIES AND CONGRUENCE

(ii) A segment joining a point of B to a point of E must intersect the polygon P. In addition, suppose we have two nonempty planar regions B  and E  so that P is their common boundary and so that the plane is the disjoint union of the three sets B, E, and P. Then after a change of notation, if necessary, we have B  = B and E  = E. This theorem should remind you of the plane separation assumption (L4) on page 176. Not to belabor the point, but it should be obvious that if we replace P, B, and E in the theorem by the sets C, D, and E (that arose in the preceding discussion of the circle C), respectively, then the theorem (except for the last part) is just a summary of what we found out about the circle C. From now on, we will denote the bounded set in the theorem exclusively by B. Then we call the union of B and P the inside of P or the region enclosed by P; the region E in Theorem 4.13 will be called the exterior of P.24 It follows from Theorem 4.13 that both the inside of a polygon and the union of a polygon with its exterior are closed sets, and both have P as boundary. We will not give a proof of Theorem 4.13, to avoid getting sidetracked because it is long, and it involves technical arguments that cannot be said to be basic to the K–12 curriculum. It may be mentioned that the standard statement of Theorem 4.13 does not include the last part about when "B  = B and E  = E". It is included here because of our later needs and because it is an easy consequence of the first part of the theorem. In any case, an essentially complete and readable proof of Theorem 4.13 is given on pp. 267–269 of the classic What Is Mathematics? ([Courant-Robbins]).25 One can easily believe this theorem by looking at a few pictures; the shaded set in each of the following is the inside of the polygon in question.

From Theorem 4.13, a polygon is the boundary of the inside of the polygon, which is a closed bounded set. This motivates the following definition. Definition. A polygonal region is the inside of some polygon.

24 Caution: The "inside" of P, as defined, is a closed set, whereas the "exterior" of P is an open set. This is the reason why we use the term "inside" rather than "interior". 25 This theorem is the special case of the famous Jordan Curve Theorem when the curve in question is a polygon. One can find an elementary proof of this theorem in [Henle]. Incidentally, the Courant-Robbins volume is highly recommended as a general introduction to advanced mathematics.

4.2. THE BASIC VOCABULARY, PART 2

197

Thus a polygonal region is always a closed set, by definition, because it includes the boundary polygon. It follows that every polygon is the boundary of its polygonal region. When school mathematics talks about "the area of a polygon", what is actually meant is "the area of the polygonal region inside the polygon". Finally, we are in a position to formulate an equivalent definition of a regular polygon. We say a polygon is convex if the inside of the polygon is convex, i.e., the region enclosed by the polygon is a convex set. Then the decisive result in this connection is the following. Theorem 4.14. A polygon whose sides have the same length and whose angles have the same degree is a regular polygon if and only if it is convex. We need Theorem 4.14 because it clarifies the concept of a regular polygon. It tells us that we could have defined a regular polygon as a convex polygon whose sides have the same length and whose angles have the same degree. This would be a more direct definition in that it does not involve a circle. We should point out, however, that there is some subtlety involved in either definition. Because of our standing convention that an angle is automatically taken to be the convex set determined by the two rays, the cross on the right in the preceding picture of three polygons would be a regular 12-gon if we do not require convexity. Similarly, this 12-gon would also qualify as a regular polygon if we do not require it to be inscribed in a circle. We will not prove Theorem 4.14 in this volume but will leave the proof to the article, A characterization of regular polygons, on the author’s homepage, https://math.berkeley.edu/~wu/. This is because Theorem 4.14 is not needed in the remaining geometric discussion in these volumes and the proof is also long and sophisticated. If one is willing to spend the effort, one can define the concept of an interior angle of a polygon. When that is done, then a regular polygon could equally well be defined as a polygon whose sides all have the same length and whose interior angles have the same degree. But, again, we will not enter into those details because of the sophistication involved. A triangle is always convex (Exercise 6 on p. 198), but for a triangle (a 3-gon) to be regular, it is not necessary to assume both the equality of the lengths of the sides and the equality of the degrees of the angles. It will be seen in Chapter 6 of [Wu2020b] that a triangle is regular if either the sides have the same length or the angles have the same degree (see Theorem G26 and Exercise 2 in Section 6.2 of [Wu2020b]). For this reason, a regular 3-gon is just an equilateral triangle (which literally means a triangle with all sides of the same length). Moreover, as soon as we can show that the sum of all the angles of a convex quadrilateral is 360◦ , it will follow that a regular 4-gon must be a square. Exercise 9 in Exercises 6.8 of [Wu2020b] will show that regular n-gons do exist for any whole number n ≥ 3, but a priori this is not obvious. Pedagogical Comments. We have tried to give precise definitions for as many concepts as feasible, but we have also left a few undefined (including region and length of arc26 ), and the hope is that your intuition would fill in the gaps in the meantime. The reason for this omission is that the definitions of region, length of arc, and a host of other geometric concepts are not simple. We did make an 26 Although

length of arc will be precisely defined in Sections 4.2 and 4.3 of [Wu2020c].

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4. BASIC ISOMETRIES AND CONGRUENCE

exception to define boundary of a set and closed set because they are absolutely essential for the considerations of area in Chapter 4 of [Wu2020c]. However, we suggest that you do not make heavy weather of these two definitions in the school classroom because school students have far more pressing concerns than learning these subtle definitions. The everyday meaning of "boundary" is good enough most of the time in the school setting. Therefore, this definition is for your own conceptual clarification as a teacher: to the extent possible, these three volumes will try to convince you at every step of the way that there is no room for ambiguity in mathematics. End of Pedagogical Comments. Exercises 4.2. (1) Imagine the hands of a clock to be idealized rays emanating from the center of the clock. (a) What is the angle between the hour and minute hands precisely at 8:20 am?27 (b) At what time between 8 am and 9 am will the hour and minute hands coincide? (c) Is there any time—other than 12 am and 12 pm—that the hour, minute, and second hands all coincide?28 (2) Prove the characterization, stated on page 181, of Γ , the nonconvex angle determined by two distinct rays ROA and ROB with a common vertex O. (3) Prove on the basis of (L6) that every angle has one and only one angle bisector. (4) To do this exercise, use any theorem you know from high school geometry, but be sure to state precisely what you are using. (i) Prove that the boundary of the open (or closed) disk of radius r around a point O is the circle of radius r around O. (ii) Prove that the exterior of a circle C (see page 194) has C as its boundary. (iii) Prove that the exterior of a circle is never convex.29 (5) Prove that if C is a given circle with center O and radius r, then its open disk, closed disk, and exterior are all connected. (For the connectedness of the exterior, use any theorem you know from high school geometry, but be sure to state precisely what you are using.) (6) A triangular region in the plane is by definition the intersection of the three angles of a triangle. (a) Show that a triangular region is always convex. (b) Let T be a triangular region in the plane. Show—without invoking Theorem 4.13 on page 195—that if P ∈ T and Q is in the exterior of T , then the segment P Q intersects the boundary of T . (7) Given a circle C and a point P on C, a line LP is said to be a tangent to C at P if LP intersects C exactly at P ; i.e., LP ∩ C = {P }. Assume that every point of a circle has a tangent (a fact we will prove in Section 6.8 of [Wu2020b]) and that the circle always lies entirely in a closed half-plane of each tangent. Then prove that any disk, open or closed, is convex. (Caution: This exercise is not as easy as it seems; try to do everything according to the definitions.) 27 Exercise

due to Tony Gardiner. (b) and (c) are due to Ole Hald. 29 A disk, open or closed, is always convex, but the proof is not entirely trivial; see Section 6.8 of [Wu2020b]. 28 Both

4.3. TRANSFORMATIONS OF THE PLANE

199

(8) Assume that for any given subset S of the plane, if a segment joins a point in S and a point not in S, then the segment contains a boundary point of S.30 Then prove that any closed bounded set in the plane with a circle C as its boundary is the closed disk with the same radius and center.

4.3. Transformations of the plane The development of geometry in these volumes is built on the foundation of what we call the basic isometries (consisting of rotations, reflections, and translations) and dilations. These are examples of the general concept of transformations of the plane. This section gives a brief introduction to the generalities of transformations, with special emphasis given to those that possess an inverse. To give the generalities some substance, we zero in on the rotations by defining them precisely. Rotations fall into two categories: clockwise and counterclockwise; the meanings of the latter are intuitively clear but their definitions are cumbersome. We will use the terminology but leave the definitions to the appendix of this section (pp. 212ff.). Why transformations (p. 199) Rotations (p. 201) Generalities about transformations (p. 205) Inverse transformations (p. 208) Appendix (p. 212) Why transformations Given two segments AB and CD, how can we compare which one is longer without first getting their individual lengths? For example, suppose we have a rectangle ABCD. Do the opposite sides AB and CD have the same length? A

D

B

C

Similarly, given two angles, how can we compare which one is bigger without first getting their individual degrees? For example, if two lines L and L are parallel and they are intersected by another line, how can we tell if the angles ∠a and ∠b as shown have the same degree?

L a







b L



30 Mathematical Aside: This simple fact requires the least upper bound axiom for its proof; see Section 2.1 in [Wu2020c].

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4. BASIC ISOMETRIES AND CONGRUENCE

These questions, while seemingly silly when the figures are drawn on a piece of paper, take on a new meaning if the sides of the rectangle ABCD are a few miles apart or if the lines L and L in the case of angles are also very far apart. We are therefore confronted with a real-world situation of having to find out whether two geometric figures (two segments, two angles, or two triangles) in different parts of the plane are "the same" in some sense (e.g., same length, same degree, etc.). The traditional way of dealing with this problem in Euclidean geometry is to write down a set of axioms which abstractly guarantee that two triangles are "the same" (i.e., congruent). This is how it is usually done in TSM, and the drawback of such an approach is that, in a mathematical environment where proofs and reasoning are scarce or nonexistent, to introduce students to proofs by the opaque formalism of axioms is to invite discontent and also to ensure nonlearning. As of the last decade, the teaching of geometry in many high schools still vacillated between teaching proofs by rote via axioms from the beginning and teaching no proofs at all.31 We propose a third approach, one that is more direct and more tangible and that makes use of three standard "moves"32 to bring one figure on top of another in order to check whether two geometric figures are congruent. Even more importantly, we base proofs of theorems directly on these "moves". In this way, congruence ceases to be mysterious and abstract; it becomes a tactile concept which can be realized concretely via these standard "moves". The key issue then is what it means to "move" things around in a plane, with the understanding that the lengths of segments and degrees of angles remain unchanged in the process. Since "moving things around in a plane" is exactly where the concept of a "transformation" comes in, we first define transformations. For convenience, we denote the plane by Π. A transformation F of Π is a rule that assigns to each point P of Π a unique point F (P ) (read: "F of P ") in Π. We also say F maps P to F (P ) or, sometimes, F moves P to F (P ). Indeed, it is intuitively appealing to think of a transformation as a way of "moving" the points of the plane around. There are two extreme examples of a transformation. The first is a constant transformation: if X is a point in Π, then the transformation FX which assigns to every P of Π the same point X is called a constant transformation. Thus FX (P ) = X for every point P in Π. In this case, we can think of a constant transformation as a rule that moves every point of the plane to a single point. The other extreme is the identity transformation I which maps every P of Π to the same point P . Thus I(P ) = P for every P in Π and I does not move any point at all. To acquire some intuitive feelings for transformations in general, we need some nontrivial examples beyond the constant and identity transformations. In this and the next two sections, we will introduce three kinds of transformations that will be among the mainstays of this and subsequent volumes. They are examples of isometries: a transformation F of the plane is said to be an isometry if F preserves distance in the sense that dist(F (P ), F (Q)) = dist(P, Q) for all the points P and Q in the plane Π. Since the length of a segment is defined in terms 31 One 32 To

can get a glimpse of the general situation from the book review [Wu2004a]. be called basic isometries (page 217).

4.3. TRANSFORMATIONS OF THE PLANE

201

of the distance function (see page 185), an equivalent definition is therefore that an isometry F is a transformation that preserves length; i.e., the length of any segment P Q, where P and Q are points in the plane, is equal to the length of the segment P  Q , where P  = F (P ) and Q = F (Q ). Thus, if a picture of P , Q, P  , and Q looks like the following, then the transformation F is not an isometry because the length of P Q is visibly longer than the length of P  Q . q Q q P

qP 

q Q

On the other hand, the identity transformation is an isometry. The next subsection will introduce a class of isometries called rotations. Rotations To define rotations, we will have to make use of the concept of a clockwise or counterclockwise direction on a given circle. For reasons given in the Pedagogical Comments on page 204, we will use these two (related) concepts in an intuitive sense without a precise definition of either. This statement must be supplemented with two additional comments, however. First, in the appendix of this section on pp. 212ff., we do provide precise definitions of clockwise and counterclockwise rotations. Therefore we have in no way deviated from our stated goal of bringing precision to any discussion of school mathematics. Second, it will be seen that for the purpose of understanding clockwise or counterclockwise rotations, an intuitive understanding of these terms is sufficient. That said, let us single out the two (very plausible) properties we need about clockwise and counterclockwise rotations. For counterclockwise rotations, these are: (a) If P1 on a circle is turned φ degrees (0 < φ ≤ 360) counterclockwise to P2 , P2 is turned θ degrees counterclockwise (0 < θ ≤ 360) to P3 , and φ + θ ≤ 360 as in the left picture below, then P3 can be obtained from P1 by turning it (φ + θ) degrees counterclockwise. See the left figure below. (Compare Exercise 9 on page 216.)

P2

Q2

Q3 θ φ

θ O

P3

P1

O

φ

Q1

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4. BASIC ISOMETRIES AND CONGRUENCE

(b) Suppose Q2 on a circle is obtained from Q1 on the circle by turning φ degrees counterclockwise and Q3 on the circle is obtained from the same Q1 by turning θ degrees counterclockwise. Suppose 0 < φ, θ < 360. If φ = θ, then Q2 = Q3 . See the right figure above. (Compare the assumption of Lemma 4.10 on page 190.) The reason for the requirement in (a) that φ + θ ≤ 360 is that, at the moment, we do not know as yet what it means to turn (counterclockwise or clockwise) an angle that is greater than 360◦ . Of course, the analogs of (a) and (b) for clockwise rotations are assumed to be true as well. Definition. Let O be a point in the plane Π and let a number θ be given so that −360 ≤ θ ≤ 360. Then the rotation of θ degrees around O (sometimes we say with center O) is the transformation θ defined as follows: θ (O) = O, and if P ∈ Π and P = O, let C be the circle of radius |OP | centered at O. Then: If θ ≥ 0, θ (P ) is the point Q on C so

Q

that Q is obtained from P by turning θ degrees in the counterclockwise

O

θ

direction along C (in other words,

P

|∠QOP | = θ ◦ ).

If θ < 0, θ (P ) is the point Q on C obtained from P by turning |θ| degrees in the clockwise direction along C (in

O

θ

P

◦

other words, |∠P OQ| = |θ| ).

Q Note that the assignment of Q to the given P is unambiguous on account of Lemma 4.10 on page 190. Hence θ is well-defined (i.e., it makes sense). For an intuitive understanding of rotations, the following activity, preferably done in class, will be helpful. It gives a tactile realization of a rotation of 32 degrees. Activity. This Activity will give an idea what ρ, a counterclockwise rotation of 32◦ , does to the plane. On a piece of paper, which is our model for the plane Π, fix a point O, and then draw a geometric figure S and a point Q, as shown below in black. Place a clear transparency over this sheet of paper with figure S and points O and Q in black. With a different color (say, red), copy on the transparency the S, O, and Q right on top of the originals. In particular, the red point O on the transparency is on top of the point O on the paper.

4.3. TRANSFORMATIONS OF THE PLANE

B A

B

203

S S

Q

A

o

32 o

Q



32

O Now use a pointed object (e.g., the needle of a compass) to pin the transparency to the paper at the point O. Holding the paper fixed, rotate the transparency around O, counterclockwise by 32 degrees33 and stop. For the moment, ignore the angles ∠AQB and ∠A Q B  in the above picture and concentrate on S and Q. We will denote the red geometric figure in this new position by S  . S  is exactly where ρ has moved S. Similarly, we denote the red point Q in this new position by Q ; this point Q is where the rotation ρ has moved Q. Notice that ρ does not move O, the center of rotation. Needless to say, there is nothing special about the number 32; one should do this Activity with angles of any degree, clockwise or counterclockwise. This Activity suggests that the rotation ρ is an isometry. Indeed, if A and B are two points in the plane Π and ρ moves them to A and B  , respectively (see the preceding picture), then the Activity tells us how to locate A and B  ; namely, copy O, A, and B in red on a piece of transparency, and then rotate the transparency 32 degrees around O. The positions of the red A and red B on the transparency are the locations of A and B  , respectively. Since the distance from the red A to the red B  on the transparency is exactly the distance from A to B in Π, ρ is distancepreserving; i.e., ρ is an isometry. The Activity also suggests that ρ preserves the degrees of angles in the sense that—in the notation above—the angle ∠AQB, for example, is moved by ρ to ∠A Q B  and since the red angle ∠A Q B  is a copy of ∠AQB on the transparency, we have |∠A Q B  | = |∠AQB|. Everything we have said thus far has nothing in particular to do with "32 degrees", so what the Activity suggests is that any rotation is an isometry that also preserves degrees of angles. However, from the point of view of our mathematical development, the fact that a rotation is an isometry is something that cannot be proved but must be assumed. See page 217 and page 237. Remarks. (1) Because we will be talking about rotations of negative degrees, e.g., "a rotation of (−36) degrees", we need to further clarify the concept of degree at this point. The degree of an angle is always ≥ 0 (assumption (L6), page 188). Therefore, the concept of negative degree arises only in the context of rotations; 33 In practical terms, the way to achieve this is to pencil in an angle ∠AOB of 32 degrees somewhere on the paper (not the transparency) (with the help of a protractor) so that the ray ROA is in the counterclockwise direction of the ray ROB . Copy ∠AOB on the transparency. Then rotate the transparency counterclockwise around O until the ray ROB on the transparency is on top of the ray ROA on the paper.

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e.g., "a rotation of (−36) degrees" or "a (−36)-degree rotation" signifies that we have to rotate 36 degrees clockwise. (2) It is a curious fact—but sometimes useful nevertheless—that since we allow ourselves to use both clockwise and counterclockwise rotations, it suffices to use angles of θ degrees so that |θ| ≤ 180 in any discussion of rotations. Indeed, a rotation of 235 degrees is equal to a rotation of −125 degrees because 360 − 125 = 235, and a rotation of −235 degrees is equal to a rotation of 125 degrees. In general, if 180 < θ ≤ 360, then a rotation of θ degrees is equal to a rotation of −(360 − θ) degrees, and if −360 ≤ −θ < −180, then a rotation of −θ degrees is equal to a rotation of (360 − θ) degrees. Of course, if 180 < θ ≤ 360, we have | ± (360 − θ)| < 180. We will put this to use later. Pedagogical Comments. Although we have left the concepts of clockwise rotation and counterclockwise rotation undefined, we wish to point out that they can be defined—see the appendix on pp. 212ff.—but that we decided not to do so in the main body of the exposition for a pedagogical reason. A brief look at the appendix will reveal that the definitions are far from simple. Unless absolutely necessary, the average school student should be spared the tedium of learning something that is intuitively obvious but whose precise explanation is intricate and—it is safe to say—uninteresting. In the case of clockwise or counterclockwise rotations, we consider the precise definitions to be not absolutely necessary. Moreover, the definitions require the use of translations (see page 234). Therefore a completely correct exposition using these definitions of clockwise and counterclockwise rotations will mandate that we begin with the definition of a 180-degree rotation (for which clockwise or counterclockwise is irrelevant), prove Theorems G1 to G3 (pp. 220–224), go on to define translations, and then come back to define rotations in general, clockwise and counterclockwise. In terms of mathematics learning, it is a bad idea to break up the concept of rotation this way (regardless of how technically correct the mathematics may be). Given that introductory geometry is already overladen with a great many definitions as it is (see Sections 4.1 and 4.2), we do not believe that it is absolutely necessary to throw two more long-winded definitions into the mix. Unlike the concepts of half-lines and half-planes (see Lemma 4.5 and (L4) on pp. 173 and 176, respectively), which are truly fundamental to the whole discussion of geometry, students can easily get by with only an intuitive knowledge of clockwise or counterclockwise rotations. We therefore made the decision not to offer these precise definitions in the main body of our exposition. We have mentioned more than once (cf. the Pedagogical Comments on page 178 and page 197) that introductory geometry is full of unpleasant details, and we will have occasion to do so many more times in the future (e.g., pp. 261, 263, 277, and 292). If our goal is to optimize student learning, then it would behoove us to smooth students’ learning path by bringing out the truly essential ideas and soft-pedaling unpleasant details that are of secondary importance. At this stage of students’ mathematics learning, we believe an intuitive understanding of clockwise and counterclockwise rotations is all they need. End of Pedagogical Comments.

4.3. TRANSFORMATIONS OF THE PLANE

205

Generalities about transformations To facilitate subsequent discussions, we introduce a standard concept, that of the image of a set by a transformation F . Given a point Q of the plane, we will call F (Q) the image of Q by F or the image of Q under F . If S is a subset of the plane, then the image of S by F , denoted by F (S), is the collection of all the images of the points in S by F . Equivalently, F (S) is the collection of all the points in the plane Π which can be written as F (Q) for some point Q in S. Intuitively, F moves S to F (S). We also say F maps S to F (S). Thus the ρ(Q) and ρ(S) in the picture in the Activity of the preceding subsection are the images of the point Q and the black figure S by ρ, respectively. Likewise, for the constant transformation FX and identity transformation I, we have FX (Π) = {X}, the set consisting of the single point X, and I(Π) = Π, respectively. We will be looking at transformations that are very "well-behaved", in the following sense. First, define a transformation T to be injective, or an injection,34 if for any two distinct points P1 and P2 of Π, T assigns to them distinct points T (P1 ) and T (P2 ) of Π. Thus the constant transformation FX (for a given point X) is not injective because, if we take any two distinct points P1 and P2 of Π, then FX maps both into the same point X. On the other hand, the identity transformation I is injective because if P1 = P2 , then also I(P1 ) = I(P2 ). An isometry is injective. Indeed, let F be an isometry. Suppose P = Q; then dist(P, Q) > 0, on account of (L5)(i) on page 184. Thus dist(F (P ), F (Q)) = dist(P, Q) > 0, and therefore F (P ) = F (Q). This shows F is indeed injective. We define a transformation T to be surjective, or a surjection,35 if for every point Q of Π, Q = T (P ) for some point P of Π; i.e., for every Q in Π, there is a point P in the plane Π which is assigned to Q by T . For example, the constant transformation FX is not surjective because, if Q is a point in Π and Q = X, then there is no point P in Π so that FX (P ) = Q. But the identity transformation I is clearly surjective. Finally, a transformation T is said to be bijective, or a bijection,36 if it is both injective and surjective. It will turn out that every isometry is surjective so that, in fact, an isometry is a bijection. However, the surjectivity of an isometry is far from obvious and cannot be proved until Section 6.6 of [Wu2020b]. We remark that the common terminology for these concepts, i.e., one-to-one, onto, and one-to-one correspondence, are linguistically awkward. The suggested replacements of injective, surjective, and bijective, respectively (first made by the French group Bourbaki), are clearer and more civilized. To make sense of these new concepts, we have to look at more than the constant transformation and the identity transformation. Let us consider the rotations defined above. We claim that a rotation is always a bijection. While this is intuitively obvious, we will go through the argument carefully. Let us fix a rotation ρ around some O of θ degrees. Since a rotation of 0 degrees is the identity transformation, we may assume 0 < |θ| ≤ 360. By a remark on page 204, we may in fact assume 34 The

usual terminology is one-to-one. usual terminology is onto. 36 The usual terminology is one-to-one correspondence. 35 The

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0 < |θ| ≤ 180. Because the case of θ = 180 is obvious, we will henceforth assume 0 < |θ| < 180. ρ is injective. Let P1 and P2 be two distinct points, and we must show ρ(P1 ) = ρ(P2 ). If one of them is equal to O, say P1 = O, then ρ(P1 ) = O by the definition of a rotation, and, because P2 = O, ρ(P2 ) = O, also by the definition of a rotation. Therefore ρ(P1 ) = ρ(P2 ) in this case. We may therefore assume that both P1 and P2 are different from O. If |OP1 | = |OP2 |, then there is nothing to prove because, if P1 and P2 denote ρ(P1 ) and ρ(P2 ), respectively, then by the definition of a rotation, |OP1 | = |OP1 | = |OP2 | = |OP2 |. Therefore |OP1 | = |OP2 | so that P1 = P2 ; i.e., ρ(P1 ) = ρ(P2 ). So let us suppose |OP1 | = |OP2 |. Then both P1 and P2 lie on some circle C around O and ρ maps them to P1 and P2 , respectively, as shown: P2 P 1

φ

θ

P2

P1

O

C For definiteness, we may assume θ > 0 because the argument for the case of a negative θ is similar. Thus, we are looking at a counterclockwise rotation of θ degrees, 0 < θ < 180. Let ∠P1 OP2 denote the convex angle as usual, and let |∠P1 OP2 | = φ◦ , where 0 < φ ≤ 180 (see Lemma 4.9 on page 188). By switching the points P1 and P2 if necessary, we may assume that the counterclockwise rotation of φ◦ maps P1 to P2 , as shown in the above picture. Therefore we may characterize P2 (which is ρ(P2 )) as the point obtained from P1 , first by a counterclockwise rotation of φ degrees (which moves P1 to P2 ), followed by a counterclockwise rotation of θ degrees (which moves P2 to P2 ). By (a) on page 201, P2 is the point obtained from P1 by a counterclockwise rotation of (φ + θ) degrees. But P1 is by definition the point obtained from P1 by a counterclockwise rotation of θ degrees. Since φ > 0, (θ + φ) = θ and therefore P1 = P2 , by (b) on page 201. The proof of the injectivity of ρ is complete. ρ is surjective. Let a point Q be given. We must find a point Q so that ρ(Q) = Q . If Q = O, just let Q = O. So let Q = O, and let C be the circle with center O and radius OQ . Now rotate Q by θ degrees along C in the clockwise direction to get to a point Q, as shown below. By definition of ρ, we have ρ(Q) = Q . Hence ρ is surjective. This proves that ρ is bijective.

Q

O

θ

Q C

4.3. TRANSFORMATIONS OF THE PLANE

207

Pictorially, what a bijection T of Π does is to move the points of the plane Π in such a way that distinct points are not "lumped together" by T into the same point (injectivity) and such that the image of the plane by T , T (Π), "covers" all of Π and not just a part of it (surjectivity). The following three examples of transformations make use of coordinates, the inverse tangent function, and roots of cubic polynomials, respectively. These are concepts that will not be discussed until Chapters 6 of this volume, Chapter 1 of [Wu2020c], and Chapter 3 of [Wu2020b], respectively. Therefore, these examples are not part of the logical development in this volume. However, since these examples are likely to help you build up your geometric intuition about transformations, we put them here solely as a learning aid. Example 1. Let coordinates be introduced in the plane, so that points in the plane are now just a pair of numbers (x, y). Define the folding transformation ϕ by ϕ(x, y) = (|x|, y). Pictorially, ϕ "folds" the plane along the y-axis onto the right half-plane of the y-axis, because for every (x0 , y) to the right of the y-axis, i.e., x0 > 0, we have ϕ(x0 , y) = (x0 , y), i.e., ϕ leaves it unchanged, while for the point (−x0 , y) on the left of the y-axis (still assuming x0 > 0), ϕ(−x0 , y) = (x0 , y), i.e., ϕ "folds" (−x0 , y) onto (x0 , y) . Therefore the definition of ϕ implies that ϕ is not injective. Furthermore, ϕ is not surjective either, because, for example, the point (−1, 0) cannot be written as ϕ(x , y  ) since ϕ(x , y  ) = (|x |, y  ) no matter what x may be, so that the x-coordinate of ϕ(x , y  ) will always be ≥ 0 and can never be equal to −1 for any x . You can see easily that the image ϕ(Π) is in fact the right half-plane together with the y-axis. Example 2. Recall the inverse tangent function from trigonometry, arctan, which is defined for every number and is increasing, i.e., arctan(x) makes sense for every number x, and if x < x , then arctan(x) < arctan(x ). Furthermore, − π2 < arctan(x) < π2 for every number x. Here is the graph:

Now again assume that coordinates have been introduced in Π. We claim that the following transformation G of Π, defined by G(x, y) = (arctan(x), y), is injective but not surjective. To show G is injective, we must show that if (x1 , y1 ) = (x2 , y2 ), then G(x1 , y1 ) = G(x2 , y2 ). To this end, observe that (x1 , y1 ) = (x2 , y2 ) means x1 = x2 or y1 = y2 . First suppose x1 = x2 ; then either x1 < x2 or x1 > x2 . If x1 < x2 , then

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arctan x1 < arctan x2 (because arctan is increasing), so that G(x1 , y1 ) = (arctan x1 , y1 ) = (arctan x2 , y2 ) = G(x2 , y2 ); i.e., G(x1 , y1 ) = G(x2 , y2 ). Similarly, if x1 > x2 , then also G(x1 , y1 ) = G(x2 , y2 ). Therefore x1 = x2 implies G(x1 , y1 ) = G(x2 , y2 ). On the other hand, if y1 = y2 , then obviously (arctan x1 , y1 ) = (arctan x2 , y2 ) (the two points have different second coordinates) and hence also G(x1 , y1 ) = G(x2 , y2 ). So G is injective. On the other hand, G is not surjective on the plane, and this is because since − π2 < arctan(x) < π2 , the expression of G as G(x, y) = (arctan(x), y) means the x-coordinates of all the image points G(x, y) of G, no matter what x and y may be, lie in the open interval (− π2 , π2 ). This implies that the image G(Π) of G lies in the infinite "vertical strip" in the plane bounded between the vertical lines x = − π2 and x = π2 . In particular, the image G(Π) of the plane Π under G is not all of Π, and therefore G is not surjective. One can also see the nonsurjectivity of G directly by noting that the point (5, 0) cannot be written as G(x , y  ) for any (x , y  ), because if it were, then (5, 0) = G(x , y  ) = (arctan(x ), y  ), so that arctan(x ) = 5 and y  = 0. In particular, arctan(x ) > π2 , which is impossible. Example 3. Assume as before that we have coordinates in the plane. Define a transformation H so that H(x, y) = (x3 − 9x + 4, y). We claim that H is surjective but not injective. The failure of injectivity is easy: H(0, y) = H(3, y) = H(−3, y) = (4, y) no matter what y may be. To show surjectivity, given any point, (2, 3), for instance, we will show how to find an (x0 , y0 ) so that H(x0 , y0 ) = (2, 3); i.e., (x30 − 9x0 + 4, y0 ) = (2, 3). Consider the cubic equation x3 − 9x + 4 = 2, which is the same as x3 − 9x + 2 = 0. But we know that any polynomial (whose coefficients are real numbers) of odd degree must have a real root (Section 3.1 of [Wu2020b]). So let x0 be a real root of x3 − 9x + 2 = 0. Then with this x0 and with y0 = 3, we get H(x0 , y0 ) = (2, 3). The reasoning with (2, 3) replaced by any (x0 , y0 ) is the same. This shows H is surjective. (It can be seen from the preceding argument that, for the purpose of surjectivity, the cubic polynomial x3 − 9x + 4 could be replaced by any cubic polynomial.) We now return from the examples to the main line of our discussion. We note explicitly that at this point of the present logical development of geometry, there is no place for coordinates. Inverse transformations Bijections can be understood from a completely different angle. To this end, we will have to introduce a few more concepts. If F and G are transformations, we say the transformations F and G are equal, in symbols F = G, if F (Q) = G(Q) for every point Q ∈ Π. The composite transformation F ◦ G (sometimes also called the composition of F and G) is by definition the transformation which assigns a point P in the plane to the point F (G(P )); i.e., if P  denotes the point G(P ), then F ◦ G sends P to F (P  ). Observe that we have now introduced a new meaning to the equal sign, the equality of two transformations. This is a break from the past because, up to this point, we have only used the equal sign between two numbers or two sets. Observe also the

4.3. TRANSFORMATIONS OF THE PLANE

209

fact that this definition is completely unambiguous, so that understanding the equality of two transformations is just a routine part of learning mathematics that does not require a psychological discussion of our a priori perception of the concept of equality. For example, no matter what F is, F ◦ I = I ◦ F = F . Moreover, if FX is the constant transformation into the point X, then no matter what the transformation G is, FX ◦ G = G ◦ FX = FW , where W = G(X) so that FW is the constant transformation that assigns every point to the point G(X). As another example, the folding transformation of Example 1 on page 207 satisfies ϕ ◦ ϕ = ϕ. (Can you explain this?) Note also that the composite of two bijections is again a bijection, and the composition of two isometries is an isometry. Both are simple to verify directly (see Exercise 1 on page 215). A more revealing example is the composition of two rotations with the same center. Thus let ρ and ρ be two rotations, both with the same center O, of degrees θ and φ, respectively. One can easily see that if θ = 30 and φ = 45, then ρ ◦ ρ = ρ ◦ ρ = a rotation around O of 75◦ . Or if θ = 30 and φ = −45, then ρ ◦ ρ = ρ ◦ ρ = a rotation around O of −15◦ , i.e., a clockwise rotation of 15 degrees. In general, if ρθ and ρφ are two rotations, both with the same center O, of degrees θ and φ so that −360 ≤ θ, φ ≤ 360 and −360 ≤ θ + φ ≤ 360, then (4.1)

ρθ ◦ ρφ = ρφ ◦ ρθ = a rotation around O of (θ + φ)◦ .

Recall that the restriction of −360 ≤ θ + φ ≤ 360 in equation (4.1) is necessary because we have not yet defined rotations of degrees that are < −360 or > 360. The simple proof of (4.1) is best left as an exercise (see Exercise 6 on page 216).37 The examples in the last paragraph would seem to suggest that the composition of transformations is commutative, in the sense that if F and G are transformations, then it is always the case that F ◦ G = G ◦ F . It is instructive, as well as essential, to look at a simple example to see that this is false in most cases.38 Let A and B be two distinct points on a line and let ρA , ρB be rotations of 90 degrees around A and B, respectively. Now consider (ρA ◦ ρB )(A),

(ρB ◦ ρA )(A)

(ρA ◦ ρB )(B),

(ρB ◦ ρA )(B).

and

We want to show that the two points are different in each case. To this end, we have to look at the following picture, where lines perpendicular to LAB through A and B have been drawn. Let points C, N , P , and D be chosen on these lines, as shown, so that LCN and LP D are perpendicular to LAB at A and B, respectively, and |AB| = |AC| = |AN | = |BD| = |BP |. In addition, let E be a point on LAB so that A ∗ B ∗ E and |AB| = |BE|. 37 Equation (4.1) should be counterbalanced by the fact that the composition of two rotations with distinct centers is in general not a rotation. See Exercise 11(iii) and (iv) on page 372. 38 The following discussion of this example will assume some geometric facts that we have not proved. There is no harm in doing this because this example is a side remark rather than an integral part of our logical development.

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4. BASIC ISOMETRIES AND CONGRUENCE

C q

rP

q q A @ B @ @ q @r D N

q E

We want to find out what (ρA ◦ρB )(A) is. By definition, this is the point ρA ( ρB (A)). So we have to first find out what the point ρB (A) is. This is the point obtained by rotating A 90 degrees counterclockwise around B. First of all, ρB (A) rotates the ray RBA to the ray RBD . So ρB (A) must lie on the ray RBD . But we are assuming that |AB| = |DB|, so by (L5)(ii) on page 184, ρB moves A to D, and therefore ρB (A) = D. Hence in order to find out what (ρA ◦ ρB )(A) is, we now must find out what ρA (D) is. Notice that we are looking strictly at the effect of the transformation ρA on the point D, and we ignore what ρB is or the fact that D = ρB (A). In other words, in finding out about the effect of a composition of two transformations, say ϕ ◦ , we first observe what the first transformation  does to a point P , say (P ) = Q; once that is done, we forget about  and concentrate entirely on Q to find out what ϕ does to the point Q. Please keep this in mind. To return to our task at hand, we have to find out what ρA (D) is. So what does ρA do? It turns every point 90 degrees counterclockwise around A. For example, ρA (B) lies on the ray RAC , but since |AB| = |AC| by construction and ρ(AB) has the same length as AB, ρA (B) = C by (L6)(iv) on page 188. Similarly, ρA (N ) = B. By the same reasoning, ρA (D) lies on the ray perpendicular to LAD and lying in the same closed half-plane of LAD as P . Now, by elementary geometry (see, e.g., Section 6.2 of [Wu2020b]), we know that |∠P AB| = |∠BAD| = 45◦ , so that |∠P AD| = |∠P AB| + |∠BAD| = 90◦ (see (L6)(iv) on page 188), and that |AD| = |AP | because AD and AP are the diagonals of the squares AN DB and ABP C, respectively. (For our need here, it suffices to verify both facts experimentally.) Therefore, ρA (D) = P for reasons similar to the above. Consequently, (4.2)

(ρA ◦ ρB )(A) = ρA (D) = P.

Now similar considerations lead to the conclusion that (4.3)

(ρB ◦ ρA )(A) = ρB (A) = D.

So we see from equations (4.2) and (4.3) that (ρA ◦ ρB )(A) = (ρB ◦ ρA )(A) and the composition of transformation is in general not commutative. The proofs that (ρA ◦ ρB )(B) = C, (ρB ◦ ρA )(B) = N

4.3. TRANSFORMATIONS OF THE PLANE

211

are similar and will be left as an exercise (Exercise 8 on page 216). In any case, we also have (4.4)

(ρA ◦ ρB )(B) = (ρB ◦ ρA )(B).

Conclusion: Given two transformations F and G of the plane, it is in general false that F ◦ G = G ◦ F . With these preliminaries out of the way, we now come to the main point. Given a transformation F , suppose there is a transformation G so that both F ◦ G and G ◦ F are equal to the identity transformation I on the plane. Then we say that G is an inverse transformation of F (and of course also that F is an inverse transformation of G). Often, we simply say F is an inverse of G. Again, referring to rotations, let ρ be the rotation of degree θ around O, where −360 ≤ θ ≤ 360, and let ρ be the rotation of degree −θ around the same point O, where −360 ≤ θ  ≤ 360; then it can be immediately verified by using the definition of a rotation that (4.5)

ρ ◦ ρ = ρ ◦ ρ = I

so that ρ is an inverse transformation of ρ. The following theorem characterizes transformations which have an inverse transformation. Theorem 4.15. (i) If a transformation of the plane has an inverse transformation, then it is a bijection. (ii) If a transformation is a bijection, then it has an inverse transformation. Proof. In an exercise (Exercise 5 on page 216), you will prove (ii). We can prove (i) very simply by use of a standard argument, one that deserves to be learned. Let G be an inverse of a given transformation F . Then F is injective because if F (P1 ) = F (P2 ) for two points P1 and P2 , then also G(F (P1 )) = G(F (P2 )) and therefore P1 = P2 because G◦F = I. Thus if P1 = P2 , then F (P1 ) = F (P2 ). Also, F is surjective because given Q ∈ Π, if we let P = G(Q), then F (P ) = F (G(Q)) = Q, because F ◦ G = I. The proof of the theorem is complete. In short, a transformation F being a bijection is equivalent to its having an inverse transformation. Observe that equation (4.5) and Theorem 4.15 together give an abstract proof that every rotation is a bijection, something for which we have already given a direct proof on page 211. You will also show in an exercise (Exercise 5 on page 216) that the inverse of a transformation (if it has one) is unique, i.e., if there are transformations G and G relative to a given F , so that F ◦ G = I, G ◦ F = I and F ◦ G = I, G ◦ F = I, then G = G . From now on we can speak of the inverse of a transformation. The inverse of a bijection F is traditionally denoted by F −1 (read: "F inverse"). If F and G are bijections, then the inverse of F ◦ G is G−1 ◦ F −1 (note the order!) (see Exercise 1 on page 215). Mathematical Aside: Because the composition of bijections is a bijection (Exercise 1 on page 215) and every bijection has an inverse bijection (Theorem 4.15), the set of all bijections forms a group whose binary operation is ordinary composition of transformations. This group of bijections of the plane has many interesting

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4. BASIC ISOMETRIES AND CONGRUENCE

subgroups, as we will point out in due course, e.g., pp. 235, 240, and 286. (Note that while it is clear that we are here talking about bijections of the plane, we have purposely omitted any reference to the plane because this discussion is valid for the bijections of any set.) Appendix We will outline a definition of counterclockwise rotation. It will be clear that once that is done, clockwise rotation can be similarly defined. The overall strategy is to fix a point O to define counterclockwise and clockwise rotations around O, and then we use translations (page 234) to propagate these concepts to other points in the plane. First, we have an informal discussion. Let a point O be fixed, and let x be a number satisfying 0 < x < 180. Consider the problem of defining the counterclockwise rotation  of x◦ around O. By definition, (O) = O. If P is a point not equal to O, then according to (L6)(ii) (page 188), there are two unique rays LOQ and LOQ — residing in the two closed half-planes of LOP —so that |∠QOP | = |∠Q OP | = x◦ . Without loss of generality, we may also assume that |OP | = |OQ| = |OQ |, as shown (see (L5)(ii) on page 184).

Q P

H

+

H

_

O

xo xo

Q

Let us denote the half-plane of LOP in which Q lies by H + and denote the halfplane of LOP in which Q lies by H − . According to Lemma 4.10 on page 190, the point Q in H + is unique, and the point Q in H − is also unique. If it is counterclockwise rotation that we want, then intuitively, we would choose Q in H + and define (P ) = Q, but if we want instead the x◦ clockwise rotation of P , then we would take Q in H − . Thus for the purpose of defining (P ), we simply choose the half-plane H + of LOP and define (P ) to be the unique point Q in H + so that Q ∈ H + , |OP | = |OQ|, and |∠QOP | = x◦ . The definition of counterclockwise rotation therefore boils down to the "consistent" choice of a half-plane of a given line passing through O. Perhaps we should point out that, just as we have used "left" and "right" regarding the number line without any formal definition, we will also use "up" and "down" in the rest of this appendix without a formal definition. The fact is that we could define all these concepts if we must, but given that this volume is already overloaded with (uninteresting) technicalities, we have chosen not to given these formal definitions. We can now give the formal definition. Fix a point O in the plane. Let a point P be given, P distinct from O. First, we are going to single out a specific half-plane HP of the line LOP , in the following way. Let ∠A1 OA2 be a right angle with vertex

4.3. TRANSFORMATIONS OF THE PLANE

213

at O so that the side ROA1 is horizontal and right-pointing and the side ROA2 is vertical and upward-pointing. By (L5)(ii), we may assume without loss of generality that |A1 O| = |A2 O|. Similarly, on the line LOA1 , let a point A3 be chosen so that A1 ∗ O ∗ A3 and |OA1 | = |OA3 |, and on the line LOA2 let a point A4 be chosen so that A2 ∗ O ∗ A4 and |OA2 | = |OA4 |, as shown:39 A2

A3

O

q P

A1

A4 q P For the definition of HP , we first dispose of four special cases:

(1) If P lies on the that contains A2 . (2) If P lies on the that contains A3 . (3) If P lies on the that contains A4 . (4) If P lies on the that contains A1 .

ray ROA1 , then HP is the half-plane of LOP ray ROA2 , then HP is the half-plane of LOP ray ROA3 , then HP is the half-plane of LOP ray ROA4 , then HP is the half-plane of LOP

The HP in each of these four cases is represented as the shaded region in the following:

Next, assume that P lies on neither the horizontal line LA1 A3 nor the vertical line LA2 A4 . If P lies in ∠A3 OA4 (let us say), then the two points A3 and A4 will lie on different sides of the line LOP by the crossbar axiom on page 250.40 The following definition of HP makes use of this fact: (i) If P lies in ∠A1 OA2 , then HP is the half-plane of LOP that contains A2 . (ii) If P lies in ∠A2 OA3 , then HP is the half-plane of LOP that contains A3 . (iii) If P lies in ∠A3 OA4 , then HP is the half-plane of LOP that contains A4 . 39 Anticipating the use of coordinates in Chapter 6, we may think of the ray R OA1 as the nonnegative x-axis and ROA2 as the nonnegative y-axis. 40 There is no logical difficulty here because the crossbar axiom (L8) could be stated right after assumption (L6) on page 188.

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(iv) If P lies in ∠A4 OA1 , then HP is the half-plane of LOP that contains A1 . The HP in each of these four cases is represented as the shaded region in the following: HP

A2 P

A3

O

A1

A2

P

A2

O

A3

A3

A1

O

A1

A3

A2

HP

O

A1

A4

P

P A4

A4

HP

A4

HP

We are now in a position to define the x-degree counterclockwise rotation  of P around O where 0 ≤ x ≤ 180. First of all, (O) = O. Moreover, the case of a 0-degree or a 180-degree counterclockwise rotation is easy to dispose of: the 0-degree rotation (clockwise or counterclockwise) of a point P = O is just P itself, while the 180-degree rotation of P (clockwise or counterclockwise) is the point P so that O is the midpoint of the segment P P ; i.e., |OP | = |OP | (see the picture on p. 213; the possibility of choosing such a P is guaranteed by (L5)(ii)). Thus it suffices for us to define the x-degree counterclockwise rotation  of P around O where 0 < x < 180 and P = O. In this case, (P ) is the unique point Q lying in the specified half-plane HP so that |∠P OQ| = x◦ and so that |OP | = |OQ| (see (L6)(ii) on page 188 and (L5)(ii) on page 184). The following picture is for the case of a P lying in ∠A2 OA3 :

A2

P

xo

A3

HP

Q

O

A1

A4

It is now easy to define the x-degree counterclockwise rotation  around O when 0 ≤ x ≤ 360. First, we do it intuitively. If we have to rotate a point P = O counterclockwise through (let us say) 235 degrees, we can stop the rotation after 180 degrees and then resume the counterclockwise rotation for another 55 degrees (235 = 180 + 55). The advantage of doing this is that after 180 degrees of rotation, we know exactly where P is, namely, the point P which lies on the line LOP so that O is the midpoint of the segment P P (see the picture on p. 213). Therefore to define the 235-degree counterclockwise rotation of P , all we need to do is carry out the 55-degree counterclockwise rotation of P . But the latter is something we already know how to do. Formally, suppose a number x is given so that 180 < x ≤ 360. Write x as x = 180 + x, where 0 < x ≤ 180. Then the x-degree counterclockwise rotation

4.3. TRANSFORMATIONS OF THE PLANE

215

of a given point P around O, where 180 < x ≤ 360, is by definition the x-degree counterclockwise rotation of the point P which is the 180-degree rotation of P . In summary, for a number x so that 0 ≤ x ≤ 360, the x-degree counterclockwise rotation  around O is a well-defined transformation of the plane. Observe that the 0-degree and the 360-degree counterclockwise rotations are just the identity transformation of the plane. It is now clear how one should go about defining the x-degree clockwise rotation around O for x satisfying 0 ≤ x ≤ 180: for each point P = O, we specify the preferred half-plane of LOP to be the opposite half-plane of HP . Let us denote this half-plane by HP− . Then for each x satisfying 0 ≤ x ≤ 180 and for each P = O, the x-degree clockwise rotation of P is by definition the point P  so that |OP | = |OP  |, |∠P OP  | = x◦ , and P  lies in the specified half-plane HP− . The x-degree clockwise rotation around O, where 180 < x ≤ 360, is then defined as the x-degree clockwise rotation of P (the 180-degree rotated image of P as above), where x = 180 + x and 0 < x ≤ 180. Altogether, we have defined the x-degree clockwise rotation of any point P in the plane and for any x satisfying 0 ≤ x ≤ 360. It remains to define the x-degree counterclockwise rotations around an arbitrary point O  of the plane, where 0 ≤ x ≤ 360. Let T be the translation −−→ along the vector OO  (page 234). For each point P  , let P be the point in the plane so that T (P ) = P  , and let Q be the x-degree counterclockwise rotation of P around O. Then, by definition, the x-degree counterclockwise rotation of P  around O  is the point T (Q). The x-degree clockwise rotation is defined similarly. Exercises 4.3. (1) (a) Prove that the composition of two isometries is an isometry. (b) Prove that the composition of two surjections is a surjection and the composition of two injections is an injection. (Hence the composition of bijections is a bijection.) (c) If F , G are bijections, then prove that the inverse of F ◦ G is G−1 ◦ F −1 . (2) (This exercise makes use of coordinates; see the warning on pp. 207 and 208.) (a) Let F and G be transformations of the plane defined by F (x, y) = (x, y + 1) and G(x, y) = (xy, y). Are the transformations F ◦ G and G ◦ F equal? (b) Is F ◦G injective? Surjective? (c) Is G◦G injective? Surjective? (3) (This exercise makes use of coordinates; see the warning on pp. 207 and 208.) (a) Let F be the transformation of the plane defined by F (x, y) = (x + 1, y + 1), and let C denote the unit circle, i.e., the circle of radius 1 around the origin (0, 0). Give a rough description of F (C), but be as precise as you can. (b) Let G be the transformation of the plane defined by G(x, y) = (2x, y), and let C be the unit circle as before. Give a rough description of G(C), but be as precise as you can. (c) Let H be the transformation of the plane defined by H = G ◦ F , so that H(x, y) = (2x + 1, y + 1), and let C be the unit circle as before. Give a rough description of H(C), but be as precise as you can. (4) (This exercise makes use of coordinates; see the warning on pp. 207 and 208.) (a) Consider the transformation G of the plane defined by G(x, y) = (x2 , y). Is it injective? Is it surjective? (b) Consider the transformation F

216

4. BASIC ISOMETRIES AND CONGRUENCE

(5)

(6) (7)

(8)

(9)

(10)

(11)

(12)

of the plane defined by F (x, y) = (x, y 3 ). Is it injective? Is it surjective? (c) With F and G as in (a) and (b), what is (F ◦ G)(x, y) for any point (x, y)? Is the composite injective? Surjective? (a) Prove that a bijection F of the plane must have an inverse G. (b) Prove that if the inverse of a transformation F exists, then it must be unique (in other words, if G and G are both inverse transformations of F , then G = G ). Make use of (a) on page 201 to prove equation (4.1) on page 209. For each of the following assertions about transformations F and G of the plane, if it is true, prove it. If always false, prove it. If sometimes true and some times false, give examples of each kind. (a) If F ◦ G is injective, then G is injective. (b) If F ◦ G is injective, then F is injective. (c) If F ◦ G is surjective, then G is surjective. (d) If F ◦ G is surjective, then F is surjective. (a) Prove equation (4.4) on page 211 that (ρA ◦ ρB )(B) = (ρB ◦ ρA )(B). (b) Exhibit two rotations F and G in the plane so that F ◦ G = G ◦ F and so that these are not the same as the ρA and ρB on page 209. (i) Make use of the appendix (pp. 212ff.) to prove that if P1 on a circle is turned φ degrees (0 < φ ≤ 180) counterclockwise to P2 and P2 is then turned θ degrees (0 < θ ≤ 180) counterclockwise to P3 , then P2 lies in ∠P1 OP3 (here ∠P1 OP3 is taken to be the union of the convex angles ∠P1 OP2 and ∠P2 OP3 ). (ii) Now prove (a) on page 201. (This exercise makes use of coordinates; see the warning on pp. 207 and 208.) Show that the transformations of the plane defined by H(x, y) = (ax3 + bx + c, y) for constants a, b, and c are bijective if ab > 0. (You may assume that any cubic polynomial has a (real) root.) (This exercise makes use of coordinates; see the warning on pp. 207 and 208.) Consider the transformation F of the plane defined as follows. Let P = (x, y). If x ≤ 1, then we define F (P ) = P . If 1 ≤ x ≤ 2, then we define F (P ) = (2 − x, y). If 2 ≤ x, then we define F (P ) = (x − 2, y). Is F injective? Is it surjective? Can you roughly describe what F does to the plane? Let F be a transformation of the plane. (a) Show that if F is either the identity transformation I or a constant transformation, then F ◦ F = F . (b) Exhibit a transformation F which is neither the identity transformation nor a constant transformation and yet F ◦ F = F . (c) Show that if F ◦ F = F and F is surjective, then F is the identity transformation. (d) Show that if F ◦ F = F and F is injective, then F is the identity transformation.41 4.4. The basic isometries: Rotations

This and the next section will be devoted to a discussion of the three transformations of the plane that we call the basic isometries: rotations, reflections, and translations. The assumptions we make about them will be summarized in assumption (L7) on page 237. We will draw a few consequences from the definition of a rotation, which was defined in the last section, and make a few tentative steps 41 (c)

and (d) are due to N. Ackerman.

4.4. THE BASIC ISOMETRIES: ROTATIONS

217

toward proving theorems in geometry in the process. These theorems are needed for the definitions of reflections and translations in the next section, and among them, the most important is Theorem G1 on page 220. Theorem G1 will have many applications. Assumptions about rotations and first consequences (p. 217) Theorem G1 and its proof (p. 220) Theorems G2–G4 (p. 223) Assumptions about rotations and first consequences At this point, we will assume that the first six assumptions, (L1)–(L6), have been committed to memory and we will freely make use of them to prove some simple geometric theorems. Recall that (L2) is the parallel postulate. Our attitude toward geometric proofs at this point is strictly utilitarian: we prove the minimum number of theorems that are needed for the discussion of linear equations in beginning algebra. A more systematic presentation of the proofs of the basic theorems in plane geometry will be given in Chapter 6 of [Wu2020b]. Moreover, Chapter 8 in [Wu2020b] will discuss the nature of proofs in geometry from a broader perspective. We have already defined rotations on page 202. It remains to make explicit our assumptions about rotations: (1) Any rotation maps a line to a line, a segment to a segment, a ray to a ray, and an angle to an angle. (2) Any rotation preserves lengths of segments and degrees of angles. Thus every rotation is by assumption an isometry (see (2)). Note that, by (2), a rotation preserves not only lengths but also degrees of angles (see page 203 for the definition of degree-preserving). Rotation is the first of three isometries to be studied in detail that will be referred to as the basic isometries of the plane,42 the other two being reflection (page 229) and translation (page 234). These three are the basic building blocks of the concept of congruence (page 240). Note that the rotation of zero degrees around a point is just the identity transformation I of the plane. Rotations of 180 degrees play a major role in the logical development of plane geometry; see, for example, Theorem G1 on page 220 and Theorem G12 on page 259. Let θ and σ be numbers so that −360 ≤ θ, σ ≤ 360 and so that −360 ≤ θ + σ ≤ 360. Let θ and σ be rotations of θ and σ degrees, respectively, around the same center. Then according to equation (4.1) on page 209, the composition of θ and σ satisfies (4.6)

θ ◦ σ = θ+σ .

Recall that the restriction −360 ≤ θ + σ ≤ 360 has to be imposed on equation (4.6) because otherwise the right side of (4.6) will not make sense (rotations are thus far defined only for angles in the range [−360, 360]). Equation (4.6) has to be supplemented by three remarks. First, the composition of two rotations with distinct centers may not be a rotation in general; see Exercise 42 But

see page 237 for further comments on the terminology of "basic isometries".

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4. BASIC ISOMETRIES AND CONGRUENCE

10(b) on page 238 and Exercise 11 on page 372. Second, if σ = −θ, then (4.6) reduces to equation (4.5) on page 211 (because a rotation of 0 degrees is the identity transformation I): θ ◦ −θ = I

and −θ ◦ θ = I.

As noted on page 211, this implies that, by virtue of Theorem 4.15 on page 211, each rotation θ is a bijection. A third remark is that when rotations of any degree have been defined, equation (4.6) will be seen to hold for any two number θ and σ with no restrictions (see equation (1.94) in Section 1.6 of [Wu2020c]). We now point out that there are "plenty of" rotations as a result of our assumption about the existence of angles with a prescribed degree in (L6)(ii) on page 188. Lemma 4.16. Given a point O and a number t so that −360 ≤ t ≤ 360, there exists a t-degree rotation around O. Proof. We have to show that, given a number t so that 0 ≤ t ≤ 360, we can define a t-degree rotation  around a given point O using only the given assumptions. We do so as follows. Since such a rotation maps O to O, we only have to define the rotated image of P for a point P distinct from O. According to (L6)(ii), there are two angles ∠P OQ1 and ∠P OQ2 (where Q1 and Q2 lie in different half-planes of LOP ) sharing the side ROP so that |∠P OQ1 | = |∠P OQ2 | = t◦ . The following picture shows the case 0 < t < 180: q P H HH qQ1 HH t◦ HH H t◦ O A A A A A Aq Q2 A Without loss of generality, we may assume that Q1 and Q2 are the points so that |OQ1 | = |OQ2 | = |OP |. Then by the definition of rotation (page 202), the t-degree counterclockwise rotation of P has to be one of Q1 and Q2 ; according to the picture above, it is Q1 . Thus (P ) = Q1 . We have now defined how  moves an arbitrary point P = O, so  is well-defined. If t satisfies, instead, −360 ≤ t ≤ 0, then the t-degree rotation of P will be defined in a similar way, except that we now look for the clockwise rotation of P of degree |t|. The proof of Lemma 4.16 is complete. It may not be entirely obvious that the assumptions (1) and (2) about rotations on page 217, when coupled with (L1)–(L6), already allow us to prove very interesting geometric theorems. See Theorems G1–G4 in the remainder of this section. In addition to their intrinsic interest, these four theorems are needed for a meaningful discussion of the definitions of reflection and translation in the next section. To get a preview of some of the mathematical issues involved in these definitions, we give an intuitive discussion of reflections and translations.

4.4. THE BASIC ISOMETRIES: ROTATIONS

219

Intuitively, the reflection Λ (capital Lambda) across a given line L assigns to each point on L the point itself, and it assigns to any point P not on L the point Λ(P ) which is symmetric to P with respect to L, in the sense that L is the perpendicular bisector (page 192) of the segment joining P to Λ(P ). L Λ(P r)

Pr

A r

Λ(A) r

For simplicity let us denote by P  the point Λ(P ). Implicit in this definition is the fact that (a) there is such a point P  so that L is the perpendicular bisector of the segment P P  and (b) there is only one such point P  . Neither is obvious at the moment. The need for (a) is clear, but the need for (b) may be less so. The fact is that if there is another point Q distinct from P  so that L is also the perpendicular bisector of P Q, then the definition of a reflection implies that we can also define Λ(P ) = Q. This raises the question: which point does R assign to P , P  or Q?

L

P

O.

.

P

Q If we cannot verify that both (a) and (b) are valid, then the concept of a reflection would not be well-defined (see page 45) on two levels. Given a line L and a point P in the plane, either the putative reflection Λ across L cannot assign a point to P (this would be the case if (a) fails) or there is more than one candidate for such a P  so that the assignment of Λ to P becomes ambiguous (this would be the case if (b) fails). To go forward, what we need is a confirmation of the following. Wish List #1. Given a line L and a point P , there is one and only one line passing through P and perpendicular to L. We must also keep in mind that any confirmation of this claim must be based only on the properties of rotations such as (1) and (2) on page 217, together with assumptions (L1)–(L6). Next, consider the concept of a translation along a vector. First, a vector is by definition a segment AB with one of its two endpoints designated as the starting point and the other as the endpoint. (We put an arrowhead at the endpoint.) Given a segment AB, there are two ways to make it into a vector: if the starting −→ point is A, then we denote the resulting vector by AB, and if the starting point is

220

4. BASIC ISOMETRIES AND CONGRUENCE

−→ B, then the resulting vector is denoted by BA. Therefore, while AB and BA are −− → −− → the same segment, AB and BA are different vectors because they have different −− → starting points and different endpoints. The length of a vector AB is by definition the length of the segment AB. −− → Intuitively, the translation T along a given vector AB is the transformation that "moves a point P in the plane the same distance and in the same direction as −−→ AB". For example, if P does not lie on line LAB , we can describe T (P ) as follows. Draw the line  passing through P and parallel to line LAB ; then Q = T (P ) is the intersection of  and the line passing through B and parallel to the line LAP . B MBB −− → B AB B AB

BBQ BMB BB BB BB P B B

−−→ Pictorially, this description of Q = T (P ) is believable as far as "P Q pointing in −− → −−→ the same direction as AB" is concerned. But now how do we know that, with P Q −− → −−→ so defined, AB and P Q indeed have the same length? In other words, noting that ABQP is by definition a parallelogram (page 193), we need the following to be true. Wish List #2. Opposite sides of a parallelogram have the same length. Theorem G1 and its proof We now set out to prove both wish list items. The key ingredient in both proofs is the following basic theorem about rotations of 180 degrees: For the sequence of geometric theorems to follow, we will adopt a special convention for their enumeration. Henceforth, all the theorems in plane geometry will be numbered consecutively by G1, G2, G3, etc. This is because, in Chapter 6 of [Wu2020b], we will bring all these theorems together to give a coherent account of plane geometry.

Theorem G1. Let O be a point not contained in a line L, and let  be the rotation of 180◦ around O. Then the image of L by  is a line parallel to itself; i.e., (L)  L. We note first of all that Lemma 4.16 on page 218 guarantees that there is such a rotation of 180◦ around O. That said, let us begin by describing explicitly this rotation . Given a point P distinct from O, let P  denote the image point (P ) of P by . Then, by (L6)(iii) on page 188, P  is the point on the ray RP O so that |P  O| = |P O|. This is because ∠P OP  is a straight angle ( is a 180degree rotation) and a rotation is distance-preserving (assumption (2) on page 217). Similarly, for any other point Q, the image Q of Q by  is the point on RQO so that |Q O| = |QO|.

4.4. THE BASIC ISOMETRIES: ROTATIONS

q Q    Or  q  P     Q q

221

qP 

One should also get an intuitive feeling for why Theorem G1 is true by some hands-on activities. Activity 1. Draw a line L on a piece of paper. Also draw a point P on L and a point O not belonging to L. Copy this picture exactly on a transparency using a different color. Now pin the transparency to the paper at the point O and rotate the transparency by 180 degrees. Does the rotated image of L look like a line parallel to L itself? If we denote the rotated image of P by P  , what do you observe to be the relationship between the three points P , O, and P  ? Try this experiment with different choices of O. Proof of Theorem G1. We give two proofs of this simple but important theorem. The first proof argues by contradiction. The second proof does not use a contradiction argument and is one that can be used directly in a school classroom.  (L)     qO    

 Q

q P

L

By the assumptions about rotations (page 217), we know (L) is a line. Suppose (L) is not parallel to L. Then they intersect at a point Q. The fact that Q ∈ (L) means that there is a point P ∈ L so that (P ) = Q. Since  is a rotation of 180◦ around O, the three points P , O, and (P ) are collinear; i.e., P , O, and Q are collinear (see (L6)(iii) on page 188). As usual, call this line LP Q . Now, not only is P on L, but Q is also on L because Q = L ∩ (L). Thus L and LP Q have two points P and Q in common and therefore they coincide: L = LP Q (see (L1) on page 165). But O also lies on LP Q , so O lies on L, and this directly contradicts the hypothesis that O is not contained in L. Therefore (L) has to be parallel to L. Theorem G1 is proved. Now a second proof. Let Q be a point so that Q ∈ (L). We have to prove that Q does not belong to L. By the definition of (L), Q being in (L) means that there is a point P on L so that (P ) = Q.   O q      P 

q   Q (= (P ))

L

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4. BASIC ISOMETRIES AND CONGRUENCE

Since  is a 180-degree rotation around O, the three points P , O, Q are collinear and therefore Q lies on LOP . Observe that since P is on L and O does not lie on L by hypothesis, O and P are distinct so that P and Q are also distinct points (because P and Q lie on opposite half-lines with respect to O by the definition of a 180◦ angle). Now O lies on LOP and O is not on L, so that LOP and L are distinct lines. By Lemma 4.2 on page 165, LOP and L can have at most one point in common. Since P is already known to be on both L and LOP , no other point on LOP can be on L. In particular, since Q is distinct from P , Q does not lie on L. The proof is complete. As mentioned on page 166, we can now prove an existence result that complements the parallel postulate (page 165). Corollary to Theorem G1. If a point P and a line L are given so that P does not lie on L, then there exists a line passing through P and parallel to L. Proof of corollary. Let Q be any point on L and let O be the midpoint of segment P Q. If  is the 180-degree rotation around O, then of course (Q) = P so that (L) passes through P . Moreover, Theorem G1 says (L)  L, and the corollary is proved. We can now fulfill the promise made on page 166 by giving a direct proof of Lemma 4.3. We are given three lines L1 , L2 , and L3 , and they satisfy L1  L2 and L2  L3 . We have to prove that if L1 and L3 are distinct, then L1  L3 . Suppose L1 and L3 are distinct. By Lemma 4.1 on page 165, there is a point P on L3 so that P is not on L1 . If we can prove that any line  passing through P distinct from L3 must intersect L1 , then this would imply that any line passing through P that is not L3 is not parallel to L1 . But since the preceding corollary implies that there is a line passing through P that is parallel to L1 , we conclude that L3 must be that line. This then shows L3  L1 . Thus let  be a line passing through P and distinct from L3 . We are going to prove that  intersects L1 .

 L3

P

L2 L1

P P 

We cannot prove in one step that  intersects L1 . Instead, we first prove that  intersects L2 . Indeed, since P ∈ L3 and L3  L2 by hypothesis, P does not lie on L2 . By the parallel postulate, through P passes at most one line parallel to L2 . Since, by hypothesis, L3 is that line, then L3 is the only line passing through P that is parallel to L2 . Therefore  is not parallel to L2 and  intersects L2 at some point P  . As usual, P  ∈ L2 and L2  L1 (by hypothesis) imply that P  does not lie on L1 . By the parallel postulate again, through P  passes only one line

4.4. THE BASIC ISOMETRIES: ROTATIONS

223

parallel to L1 , and since L2 is that line by hypothesis,  is not parallel to L1 . Thus  must intersect L1 at some point P  . By the remark above, this proves Lemma 4.3. Theorems G2–G4 The next two theorems are both intuitively obvious, and both are simple consequences of Theorem G1. Our immediate concern is whether given a line L and a point P not on L, there can be two distinct lines passing through P and both perpendicular to L (see Wish List #1 on page 219). Your instinct tells you "of course not", because if this happens, the lines would create a triangle whose angles (have degrees that) add up to > 180 degrees.



 

 

P C  C  C

C C

C C

C

L

However, that theorem about the sum of angles of a triangle is not known at this point and therefore cannot be invoked. We must find another way to show that this is impossible. This is the content of the following theorem. Theorem G2. Two lines perpendicular to the same line are either identical or parallel to each other. Proof. Let L1 and L2 be two lines perpendicular to a line  at A1 and A2 , respectively. L1

L2

 

 (L1 )

  M q    A2 A1  We have noted in Lemma 4.11 on page 192 that the line passing through a given point of a line and perpendicular to that line is unique. Thus if A1 = A2 , L1 and L2 are identical. So suppose A1 = A2 . We need to prove that L1  L2 . Let  be the rotation of 180 degrees around the midpoint M of A1 A2 . If we can show that the image of L1 by  is L2 , then we know L2  L1 by virtue of Theorem G1. To this end, note that (L1 ) is a line, by assumption (1) on page 217. Furthermore, (L1 ) contains A2 because (A1 ) = A2 . Since  is the rotation of 180 degrees around M , it is clear that also (A2 ) = A1 . Thus () is a line that passes through A1 and A2 , and since  is the unique line passing through these two points (by (L1)), we have () = . Now, L1 ⊥ . By assumption (2) on page 217, rotations map perpendicular lines to perpendicular lines. Thus we have (L1 ) ⊥ (); i.e., (L1 ) ⊥ . Therefore each of (L1 ) and L2 is a line that passes through A2 and is perpendicular to . By the preceding observation about the

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4. BASIC ISOMETRIES AND CONGRUENCE

uniqueness of the line perpendicular to a line  at a given point of , we see that, indeed, (L1 ) = L2 and therefore, by Theorem G1, L1  L2 . Theorem G2 is proved. We will make a digression. Recall that we have introduced the concept of a rectangle as a quadrilateral whose adjacent sides are all perpendicular to each other (see p. 193). As a result of Theorem G2, we now have Corollary. A rectangle is a parallelogram. Returning now to the main line of our discussion, we see from Theorem G2 that given a point outside a line , there can be at most one line passing through the point and ⊥ . Now comes the question of existence: is there such a line? We show that such must be the case by a clever argument: so far we only have one nontrivial existence theorem, namely, the Corollary to Theorem G1 on page 222. We will use that to produce a line perpendicular to a given line. First, we give a definition. Given two lines L1 and L2 , a transversal of L1 and L2 is a line  that is distinct from L1 and L2 and intersects both. We will prove

Theorem G3. A transversal of two parallel lines that is perpendicular to one of them is also perpendicular to the other.

Proof. Let L1  L2 and let the transversal  meet L1 and L2 at A1 and A2 , respectively. Assuming L1 ⊥ , we will prove that L2 ⊥ . Again we consider the rotation  of 180◦ around the midpoint M of the segment A1 A2 . L1



L2

M q A1

A2

As before, (A1 ) = A2 so that (L1 ) is a line passing through A2 . By Theorem G1, we also know that (L1 )  L1 . Since L2 is likewise a line passing through A2 and parallel to L1 , the parallel postulate (page 165) implies that (L1 ) = L2 . Now L1 ⊥ , and  preserves degrees of angles. Therefore (L1 ) ⊥ (); i.e., L2 ⊥ (). Since  passes through M and  is a 180◦ rotation around M , we see that () = . Thus from L2 ⊥ (), we conclude L2 ⊥ , as desired. Theorem G3 has two interesting corollaries. The first proves Wish List #1 on page 219, and the second one justifies our definition of a rectangle: rectangles do exist in the plane. Corollary 1 of Theorem G3. Given a point P not lying on a line , there exists one and only one line L passing through P and perpendicular to .

4.4. THE BASIC ISOMETRIES: ROTATIONS

Λ

225

L qP



A

Proof. Let A be a point on  and let Λ be the line passing through A and perpendicular to . If Λ passes through P , we have proved the existence part. If not, then by the Corollary to Theorem G1 on page 222, there exists a line L passing through P and parallel to Λ. We shall prove presently that L intersects . Thus  intersects both Λ and L and is therefore a transversal of Λ and L. By Theorem G3, we have L ⊥ . This then completes the proof of the existence of such an L provided we can show that L intersects the line . Suppose L does not intersect ; then L  . But we already know that L  Λ. Therefore, by Lemma 4.3 on page 166, Λ  , and this contradicts the fact that Λ meets  at A. Thus L intersects  after all. To prove the uniqueness of L, suppose another line L passes through P and is also perpendicular to . By Theorem G2, since L and L are not parallel (they have P in common), they have to be identical. Thus L = L . Corollary 1 to Theorem G3 is proved. Corollary 2 of Theorem G3. There exist rectangles in the plane. Proof. Indeed, let two lines L1 and L2 be perpendicular to a third line L at A and D, respectively. Let a line L be perpendicular to L1 at a point B on L1 , and let L meet L2 at a point C. L2

L1 B

A

L

C

D

L

Then ABCD is a rectangle because L  L by Theorem G2, and therefore, since L2 ⊥ L, we get L2 ⊥ L , by Theorem G3. Therefore all four angles of ABCD are right angles and ABCD is a rectangle. Corollary 2 to Theorem G3 is proved. Corollary 2 to Theorem G3 justifies our definition of a rectangle, because we now know that there is such a geometric figure as a rectangle. Let us put this statement in perspective. Suppose we define a slantangle to be a quadrilateral each of whose angles is 80 degrees. If you think this is a silly definition, you should ask yourself why you think it is silly. In fact, a rectangle is as likely to exist as a "slantangle" until we can prove that the sum of the angles of a quadrilateral is 360 degrees. Since the latter theorem will not be available for a while (see Section 6.5

226

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of [Wu2020b]), the virtue of Corollary 2 to Theorem G3 is to give assurance in the meantime that rectangles do exist and have to be taken seriously.43 The next theorem proves Wish List #2 on page 220. Theorem G4. Opposite sides of a parallelogram have the same length. Theorem G4, together with the fact that a rectangle is a parallelogram (Corollary to Theorem G2 on page 224) implies that the opposite sides of a rectangle have the same length. This reconciles the usual definition in school mathematics of a rectangle (a quadrilateral with four right angles and opposite sides of the same length) with our definition of a rectangle (a quadrilateral with four right angles). The proof of Theorem G4 requires the following lemma. Lemma 4.17. Let F be a bijection of the plane that maps lines to lines, and let L1 and L2 be two distinct lines. Then the image lines F (L1 ) and F (L2 ) are also distinct. Furthermore, if L1 and L2 intersect at a point P , then F (L1 ) and F (L2 ) also intersect, and their point of intersection is F (P ).

L2 @ L1

F (L2 ) @

r

F (P )

@P r

@

@

F (L1 )

Proof of Lemma 4.17. Because F is a bijection, the proof of the distinctness of F (L1 ) and F (L2 ) is routine and may be left as an exercise (Exercise 2 on p. 228). Now let the distinct lines L1 and L2 intersect at P . Since P lies on the line L1 , F (P ) lies on the line F (L1 ). But P also lies on L2 , so F (P ) lies on the line F (L2 ) as well. Therefore F (P ) lies in the intersection of the distinct lines F (L1 ) and F (L2 ). By Lemma 4.2 on page 165, two distinct lines intersect at exactly one point. Hence F (P ) is the point of intersection of F (L1 ) and F (L2 ). The proof is complete. Proof of Theorem G4. Given parallelogram ABCD, we must show |DA| = |BC| and |AB| = |CD|. It suffices to prove the former. A @

D



@



@q M



@



@

@ C B 43 To further firm up this discussion, note that "slantangles" exist in the hyperbolic plane while rectangles do not. The latter fact is proved in Chapter 6 of [Greenberg].

4.4. THE BASIC ISOMETRIES: ROTATIONS

227

We have so few tools at our disposal that our first thoughts have to be: how can we make use of Theorem G1? If we look at the picture of a parallelogram, sooner or later the idea will surface that we should do a 180-degree rotation around the midpoint of a diagonal, e.g., around the midpoint M of the diagonal AC. (The diagonal AC is not in the original picture of ABCD, but putting it there helps us see the situation better.) Let  be the rotation of 180 degrees around M . Then (C) = A so that (LBC ) is a line passing through A and (by Theorem G1) parallel to LBC . Since the line LAD has exactly the same two properties by assumption, the parallel postulate (page 165) implies that (LBC ) = LAD . Similarly, (LAB ) = LCD . Thus, the intersection of (LBC ) and (LAB ) = the intersection of LAD and LCD = D. On the other hand, the intersection of LBC and LAB is B. By Lemma 4.17, we have (4.7)

(B) = D.

Recall we also have (C) = A. Since  maps segments to segments (by assumption (1) on page 217), we have (BC) = DA. Since  is an isometry (by assumption (2) on page 217), we have |BC| = |DA|, and Theorem G4 is proved. Pedagogical Comments. In a school classroom, the preceding proof is plenty good enough. However, if we want to insist on 100% mathematical clarity, then there are two steps in the preceding proof that may appear obvious but should be proved in detail if students need them. The first is why (C) = A. Here is the reason. The rotation  interchanges the two rays RM C and RM A ; i.e., (RM C ) = RM A , so that (C) ∈ RM A . But  is also an isometry, so |(M C)| = |M C|. Now if we let C  = (C), then C  ∈ RM A . We claim (M C) = M C  . To see this, observe that (M ) = M , and since  maps segments to segments (by assumption (L7)(i) on page 237), (M C) is a segment joining M to C  . However, by assumption (L1) on page 165, the only segment joining M to C  is the segment M C  on the line LM C  , which is LM A . Thus (M C) = M C  , as desired. It follows that the equality |(M C)| = |M C| becomes |M C  | = |M C|. Since M is the midpoint of AC, |M A| = |M C| and therefore |M C  | = |M A|. Now both C  and A are in the ray RM A , so we can conclude that C  = A, by (L5)(ii) on page 184; i.e., (C) = A. A second step that may need more details is similar: why |BC| = |DA|. We start with the fact that (B) = D (as in (4.7)) and (C) = A. Since  maps segments to segments, by assumption (L7)(i),  maps the segment BC to a segment joining D and A. But there is only one segment joining D and A, namely, the segment DA in the line LDA , by assumption (L1) on page 165. Hence we have (BC) = DA. Since  is an isometry, we have |BC| = |DA|. Our recommendation is that such details should be presented only if students press for them. Generally speaking, they are too much of a good thing in a beginning class on geometry. End of Pedagogical Comments. Corollary to Theorem G4. The angles of a parallelogram at opposite vertices (i.e., vertices in a quadrilateral that are not adjacent vertices) have the same degree.

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4. BASIC ISOMETRIES AND CONGRUENCE

The proof is already implicit in the proof of Theorem G4 and will therefore be left as an exercise (Exercise 3 on p. 228). Theorem G4 allows us to introduce a useful concept. Given two parallel lines, we can now define the distance between them. First, let P be a point not lying on a line . The distance of P to the line  is by definition the length |P Q|, where Q is the point of intersection of the line  and the line passing through P and perpendicular to . See the following picture on the left:

P

Q



P

P

Q

Q

 

(

Now suppose we have parallel lines  and  and P ∈  (see picture on the right). If P  is another point on  , then we claim that the distance of P to  is the same as the distance of P  to .44 Indeed, let the line passing through P  and perpendicular to  intersect  at Q . By Theorem G2 (page 223), LP Q  LP  Q . Therefore P QQ P  is a parallelogram. Consequently, |P Q| = |P  Q |, by Theorem G4. This proves the claim. The common distance from points on one of two parallel lines to the other is called the distance between the parallel lines. Exercises 4.4. (1) Prove that a parallelogram with one right angle is a rectangle. (2) Let F be a bijection of the plane that maps lines to lines. Prove that F maps distinct lines to distinct lines. (3) Prove the Corollary to Theorem G4 on page 227. (4) (This exercise is a further refinement of the proof of Theorem G4.) Recall from the proof of Theorem G4 that if M is the midpoint of the diagonal AC and  is the 180◦ rotation around M , then (B) = D. Prove that (i) B, M , and D are collinear and that (ii) the diagonal BD and the diagonal AC bisect each other, in the sense that the point of intersection of AC and BD is the midpoint of both AC and BD. (5) Fix two points P and Q in the plane. Let 1 be the counterclockwise rotation of 45◦ around P , and let 2 be the clockwise rotation of 90◦ around Q. Also write L for LP Q in the interest of notational simplicity. Now describe as precisely as you can the two lines 1 2 (L) and 2 1 (L). In particular, does 1 2 (L) equal 2 1 (L)? (6) Prove the following slight generalization of Lemma 4.17 on page 226: let F be a bijection of the plane and let U and V be two subsets of the plane. Then F (U ∩ V) = F (U) ∩ F (V). (7) Given a line L, prove that all the points of a fixed distance k from L form two lines each parallel to L. 44 This

length.

explains why the sleepers (cross ties) across rail tracks can afford to be all of the same

4.5. THE BASIC ISOMETRIES: REFLECTIONS AND TRANSLATIONS

229

(8) Given positive numbers a and b, prove that there exists a rectangle whose sides have lengths a and b. (Do not skip any steps!) (9) Let L1 and L2 be parallel lines and let O be a point equidistant from L1 and L2 . Let two lines passing through O intersect L1 and L2 at A, B and C, D, respectively, as shown. Prove that |AB| = |CD|. D J J

C L2    J JO  J  J  L1 A B (10) (This exercise makes use of a coordinate system; see the warning about the use of coordinates on page 207.) Let A = (1, 0) and B = (0, 1) and let A be the 90◦ counterclockwise rotation around A and let B be the 90◦ clockwise rotation around B. Let  = B ◦ A . What is (A) and what is (B)? 4.5. The basic isometries: Reflections and translations This section gives the definitions of the remaining two basic isometries, reflection and translation, by making use of the tools derived from the assumption we made about rotations in the last section, specifically, Corollary 1 of Theorem G3 on page 224 and Theorem G4 on 226. It concludes by summarizing the assumptions we make about the basic isometries. Reflections (p. 229) Translations (p. 231) Assumption (L7) about basic isometries (p. 236) Reflections Without further ado, we proceed to define reflections. Definition. Given a line L, the reflection across L (or with respect to L) is by definition the transformation ΛL of Π, so that: (1) If P ∈ L, then ΛL (P ) = P . (2) If P is not in L, then ΛL (P ) is the point Q so that L is the perpendicular bisector of the segment P Q. q Q (= ΛL (P )) @ @S @ @ @@ q @ P @ L@ We hasten to show that a reflection is well-defined, in the sense that for a given point P not on L, there do not exist two distinct points Q and Q so that L is the

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4. BASIC ISOMETRIES AND CONGRUENCE

perpendicular bisector of both segments P Q and P Q . If this were to happen, then each time we would reflect a point, we wouldn’t know what we are getting as there would be more than one reflection of the point. So suppose there were such Q and Q , and we will deduce a contradiction. Observe that both lines LP Q and LP Q would be perpendicular to L. By Corollary 1 to Theorem G3 (page 224), there is only one line passing through P and perpendicular to L. Therefore LP Q = LP Q ; let us say this line intersects L at S. Then on the line LP S , the points Q and Q lie in the half-plane of L opposite to that of P , so Q and Q are in the same half-line of LP S relative to S. Since |SQ| = |SQ | (as both are equal to |P S|), Q = Q by (L5)(ii) on page 184. This is the desired contradiction. Thus a reflection is well-defined. As in the case of rotations, it would be helpful to do an activity to gain some intuitive understanding of reflections. Activity 2. On a piece of paper, draw a line, to be called  for the sake of discussion. Draw some figure on the paper. Then use a piece of overheadprojector transparency to carefully copy (i.e., trace over) the figure on the paper, using a different color, say red. In particular, make sure the line  is also on the transparency. Flip over the transparency along  and superimpose it on the paper, making sure that the red line  on the transparency matches point for point the line  on the paper. Now a comparison between the figure on the paper and the corresponding red figure on the transparency gives a clear idea of how the reflection across  moves the figure around. In the following picture, the original figure is a collection of black dots as indicated, and the red figure is represented by the white dots. Notice that the original figure is spread over both sides of , and so is the reflected figure. c

c c

 c

c ss s s

s

s

s

c c cc

s s

A subset S of the plane is said to be symmetric with respect to a line L if the reflection Λ across L maps S onto itself, i.e., if Λ(S) = S. It is also common to say that the set S has a line symmetry or has bilateral symmetry if it is symmetric with respect to some line. Each of the following capital letters of the alphabet, for example, has bilateral symmetry with respect to a vertical line:

A, H, I, M, O, T, U, V, W, X, Y. The same is true of the following capital Greek letters:

Δ, Θ, Λ, Ξ, Π, Υ, Φ, Ψ, Ω. By contrast, it is relatively easy to convince oneself that letters such as J, F, L, and P have no bilateral symmetry with respect to any line. Reflections enjoy a remarkable property. Fix a line L, and let Λ be the reflection with respect to L. Then it is straightforward to check that Λ ◦ Λ = I, where I denotes the identity transformation of the plane as usual. But this means Λ is its

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231

own inverse. It follows from Theorem 4.15 on page 211 that every reflection is a bijection. As in the case of rotations, we make the following entirely plausible assumptions about reflections: (Λ1) Any reflection maps a line to a line, a segment to a segment, a ray to a ray, and an angle to an angle. (Λ2) Any reflection preserves lengths of segments and degrees of angles. By assumption, a reflection is an isometry (see (Λ2)). Here is a simple application of these assumptions. Lemma 4.18. Every point on the perpendicular bisector of a segment is equidistant from the endpoints of the segment. qC

@ @S @ @ @ @ q q @A B @ @ Proof. Indeed, let  be the perpendicular bisector of BC and let A ∈ . We have to prove |AB| = |AC|. Let Λ be the reflection with respect to . By the definition of reflection, we see that Λ(B) = C and Λ(A) = A, and therefore Λ(AB) = AC. By assumption (Λ2), we have |AB| = |AC|. The proof is complete. As in the case of rotations, we point out that there are "plenty of" reflections: every line has a unique reflection that leaves each of its points fixed. Lemma 4.19. Given a line in the plane, there is a reflection across that line. The lemma follows immediately from the definition of reflection and Corollary 1 of Theorem G3 on page 224. Translations The last basic isometry to be introduced is translation. We have to introduce −− → a new concept before we can give the definition of a translation. Let AB and −−→ −−→ −−→ P Q be two vectors. We will need to know what it means for AB and P Q to be pointing in the same direction. By definition, this means (i) either LAB = LP Q or LAB  LP Q and (ii) there is a line L0 , distinct from LAB and LP Q and parallel to neither, so that one of the closed half-planes of L0 contains both rays, RAB and RP Q . (Recall: "closed half-plane" of L0 means the union of a half-plane of L0 with L0 itself; see page 176.) We hasten to add that if the line L0 in (ii) is not parallel to either LAB or LP Q , then by virtue of Lemma 4.3 on page 166, it is also not parallel to the other.

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4. BASIC ISOMETRIES AND CONGRUENCE

−−→ −−→ If AB and P Q are pointing in the same direction, then typically they look something like the following, where the "right" closed half-plane of L0 contains RAB and RP Q :

L0

If LAB this:

r P

Q

r A B −− → −−→ = LP Q , then AB and P Q pointing in the same direction would look like



L0

Pr Ar B -Q



Still assuming that LAB = LP Q , if we make LAB into a number line (as (L3) on −− → −−→ page 167 says we could), then the preceding picture suggests that AB and P Q point in the same direction if and only if either P < Q and A < B, or P > Q and A > B. The simple proof may be left to Exercise 5 on page 237. An alternate formulation − − → −−→ is this: suppose LAB = LP Q . Then AB and P Q point in the same direction if and only if RAB ⊂ RP Q or RP Q ⊂ RAB . One can also prove this easily by making LAB into a number line or making use of Lemma 4.6 on page 174 (see Exercise 12 on page 239). −−→ We usually abuse the language and say that "AB and the line L are parallel" to mean that LAB and L are parallel. With this understood, it is essential that, in the −− → −−→ preceding definition, the line L0 be not parallel to either AB or P Q. Otherwise, we could have the following situation where L0  LAB  LP Q and the "lower" closed −− → half-plane of L0 does contain both rays, RAB and RP Q , and yet the vectors AB −−→ and P Q are in no way "pointing in the same direction". L0

r P

Q  B

r A

The following lemma gives some substance to the preceding definition. −− → −−→ Lemma 4.20. Given a vector AB and a point P , there is a unique vector P Q −− → pointing in the same direction as AB so that |AB| = |P Q|. Proof. [Since the lemma is intuitively obvious (try drawing lots of pictures) and the proof is tedious, it is suggested that this proof not be given in a school classroom.] First assume P ∈ LAB . If P = A, then we simply let Q = B. From now on, we may assume P = A. Either P lies on the ray RAB or P does not.

4.5. THE BASIC ISOMETRIES: REFLECTIONS AND TRANSLATIONS

233

First, assume P ∈ RAB , as shown: L0 rA

-Br

Pr

r -Q

We make LAB into a number line with A = 0 and B > 0. Then the ray RAB consists of nonnegative numbers. Since P = A and P ∈ RAB , P > 0. Define Q so that Q > P and so that |P Q| = |AB|. Then RP Q consists of numbers ≥ P and therefore RP Q consists of positive numbers. Hence, if L0 is the line perpendicular to LAB at A, then the closed half-plane of L0 that contains B contains all the nonnegative numbers in LAB (by Lemma 4.7 on page 176) and therefore contains −− → −−→ both RAB and RP Q . This shows AB and P Q point in the same direction. Next, suppose P does not lie in the ray RAB . Then we make LAB into a number line so that P = 0 and A > 0. If B < A, then the ray RAB would consist of all the numbers ≤ A and, since P = 0 < A, we would have P ∈ RAB . Contradiction. Thus A < B, and RAB consists of all the positive numbers ≥ A. We now choose Q to be a positive number on LAB so that |P Q| = |AB|. Then RP Q consists of all nonnegative numbers on LAB . Let L0 be the line perpendicular to LP Q at P . L0 rP

r -Q

Ar

-Br

Then the same reasoning shows that the closed half-plane of L0 that contains A contains all the nonnegative numbers on LAB and, therefore, contains both RAB −− → −−→ and RP Q . This again shows AB and P Q point in the same direction and the existence part of the lemma is proved if P lies in LAB . −−→ This vector P Q is unique because, by the definition of "pointing in the same direction", we know that the point Q must lie in LAB and therefore Q must lie in one of the two rays issuing from P on LAB determined by P . The requirement that |P Q| = |AB| implies that once a ray issuing from P is chosen, there can be only one Q in that ray so that |P Q| = |AB| (by (L5)(ii) on page 184). Since the preceding existence proof specifies the ray issuing from P in which Q resides, it is −−→ clear that this Q is unique and therefore P Q is unique. Next, suppose P does not lie in LAB . Let L0 be the line LAP . Let L1 be the line passing through P and parallel to LAB . Let L2 be the line passing through −− → the endpoint B of AB and parallel to the line L0 . The point Q is the intersection of L1 and L2 , as shown: L0 L2





Q q

P

L1







A

B



Now observe that LP Q  LAB . Furthermore, because ABQP is a parallelogram by construction, |AB| = |P Q| by Theorem G4 on page 226. We claim that both rays RP Q and RAB lie in the closed half-plane of L0 containing B. This is because B

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4. BASIC ISOMETRIES AND CONGRUENCE

and Q lie on a line L2 parallel to L0 , so the segment BQ contains no point of L0 and therefore B and Q lie in the same half-plane of L0 . By Lemma 4.7 on page 176, both RP Q and RAB lie in this closed half-plane, thereby proving the claim. −−→ It remains to prove the uniqueness of P Q. Since RP Q and RAB point in the same direction, Q has to lie on the unique line L1 passing through P and parallel to LAB (the uniqueness of L1 comes from the parallel postulate). The requirement that |P Q| = |AB| implies there are only two possibilities for Q, namely, the two points on L1 of distance equal to |AB| from P (see (L5)(ii) on page 184). The fact that Q must lie in the half-plane of L0 containing B then uniquely determines Q. The proof of Lemma 4.20 is complete. Remark. Why must the lines L1 and L2 in the preceding picture intersect? This is probably not a question one wants to address in a proof during a lesson in a school classroom, but it is something a teacher should be ready to explain if the question is raised. So suppose not, then L1  L2 . Since we also have L0  L2 by construction, we have L0  L1 or L0 = L1 by Lemma 4.3 on page 166. But L0 = L1 because A ∈ L0 and A does not lie in L1 ; therefore, L0  L1 . This is absurd because L0 intersects L1 at P . Hence, L1 must intersect L2 after all. With the availability of Lemma 4.20, we can now define a translation. −− → −→ Definition. Given a vector AB, the translation along AB is the transformation TAB of the plane so that, for a point P in the plane, TAB (P ) = Q, where −−→ −− → Q is the endpoint of the vector P Q which points in the same direction as AB and so that |P Q| = |AB|. It follows immediately from the definition of the translation TAB that if a point P lies in LAB , then TAB (P ) lies in LAB (see the definition of pointing in the same direction on page 231). In particular, TAB (A) = B. We would also like to make explicit the following property of TAB that is basically contained in the proof of Lemma 4.20. Lemma 4.21. Suppose a point P does not lie on LAB , and suppose TAB (P ) = Q. Then Q is the unique point so that ABQP is a parallelogram.

A

P



- q Q = TAB (P )



- B

Proof. The only thing that is not already in the proof of Lemma 4.20 is the uniqueness of such a Q. With A, B, P given, ABQP being a parallelogram means Q has to be the intersection of the line passing through P and parallel to LAB , and the line passing through B and parallel to LAP . Since these two lines are uniquely determined once A, B, and P are given (parallel postulate), Q is also uniquely determined. This completes the proof. We observe that every translation has an inverse transformation (see page 211) that is also a translation. To see this, let us keep the same notation as Lemma 4.21

4.5. THE BASIC ISOMETRIES: REFLECTIONS AND TRANSLATIONS

235

−− → so that TAB is the translation along AB. We will prove that the translation TBA −−→ is inverse to TAB . If Q lies on LBA , then one can easily prove that, since P Q −− → −−→ −−→ and AB point in the same direction, QP and BA also point in the same direction. Therefore, since |QP | = |BA|, we have TBA (Q) = P in this case (Exercise 2 on 237). If Q does not lie on LBA , then since ABQP is a parallelogram, BAP Q is also a parallelogram. The uniqueness part of Lemma 4.21 now shows that TBA (Q) = P . Therefore for any point P , we have TBA (TAB (P )) = P , or, TBA ◦ TAB = I. By switching the points A and B, we obtain TAB ◦ TBA = I. −− → This means that for any vector AB, the translation TAB has an inverse transformation TBA . By Theorem 4.15 on page 211, every translation is a bijection of the plane. Mathematical Aside: Because the composition of translations is a translation (Exercise 10 on page 238), the fact that the inverse of a translation is a translation means that the set of all translations in the plane is a group whose binary operation is the composition of transformations. This group of translations is a subgroup of the group of bijections of the plane defined on page 211. As in the case of reflections, the following hands-on activity is highly recommended as a way to enhance one’s intuitive understanding of translations. Activity 3. We use a piece of paper as a model for the plane. On the paper, −− → draw a vector AB, and also extend the segment AB to a line, denoted as usual by LAB . Draw some figures on the paper in black. Then use a sheet of overheadprojector transparency to copy (i.e., trace over) everything on the paper, using (let −−→ us say) a red pen. In particular, make sure that both the vector AB and the line LAB are on the transparency. Holding the paper in place, slide the transparency along the black line LAB on the paper until the red point A on the transparency is on top of the (black) point B on the paper. The new positions of all the red figures on the transparency will then display how the translation from A to B moves the figures on the paper. Here is an example (the starting point of the red vector, a red A, is not shown in the picture).

B B A

236

4. BASIC ISOMETRIES AND CONGRUENCE

We proceed to make the same assumptions about translations as those on rotations and reflections: (T 1) Any translation maps a line to a line, a segment to a segment, a ray to a ray, and an angle to an angle. (T 2) Any translation preserves lengths of segments and degrees of angles. Again, we point out that, by assumption, a translation is an isometry and that there are "plenty of" translations. The following lemma follows immediately from the definition of translation and Lemma 4.20 on page 232. −− → Lemma 4.22. Given any vector AB, there exists a unique translation along − −→ AB. Translations have a noteworthy property. Theorem G5. Given a line LAB that joins two distinct points A and B, let  −− → be a line in the plane, and let T be the translation TAB along AB. (i) If  is equal to LAB or parallel to LAB , then T () = . (ii) If  is neither equal to LAB nor parallel to LAB , then the translation T () is a line parallel to  itself. Proof. Part (i) follows immediately from the definition of a translation and the parallel postulate, so we may leave its proof as an exercise (Exercise 1 on page 237). We will give the proof of part (ii). So let  be a line neither parallel to LAB nor equal to LAB . Suppose T () is not parallel to . We will deduce a contradiction. Since T () is not parallel to , either T () =  or T () intersects  at a point Q. In either case, we have a point Q on T () ∩ . Since Q ∈ T (), there is a point P ∈  so that Q = T (P ).  r T ()   Q = T (P )   r  P 



Either P does lie in LAB or it does not. Suppose P ∈ LAB ; then T (P ) ∈ LAB (as was pointed out right below the definition of a translation on page 234), so that Q ∈ LAB . Therefore both P and Q lie on LAB . But P = Q because |P Q| = |AB| by the definition of a translation and A and B are distinct. So |P Q| = |AB| > 0. Thus  and TAB are two lines passing through the distinct points P and Q; by (L1),  = LAB . This contradicts the hypothesis of (ii) that  is not equal to LAB . Next, suppose P does not lie on LAB . By Lemma 4.21 on page 234, P Q  LAB ; i.e.,   LAB . This again contradicts the working hypothesis that  is not parallel to LAB . Hence T ()   and the theorem is proved. Assumption (L7) about the basic isometries We have just finished the definitions of the basic isometries (i.e., rotations, reflections, and translations), and it remains to make some concluding remarks. We have made assumptions about rotations (see (1) and (2) on page 217), reflections

4.5. THE BASIC ISOMETRIES: REFLECTIONS AND TRANSLATIONS

237

(see (Λ1) and (Λ2) on page 231), and translations (see (T 1) and (T 2) on page 236). We can now summarize these assumptions in one all-embracing statement, as follows. (L7) The basic isometries (rotations, reflections, and translations) have the following properties: (i) A basic isometry maps a line to a line, a segment to a segment, a ray to a ray, and an angle to an angle. (ii) A basic isometry preserves lengths of segments and degrees of angles. Once again, we note that rotations, reflections, and translations are, by assumption, isometries and that there are "plenty of" basic isometries in the sense of Lemmas 4.16, 4.19, and 4.22 (page 218, page 231, and page 236, respectively). In one sense, the terminology of "basic isometries" is unfortunate because, at least for now, the basic isometries possess properties that seem not to be shared by isometries in general. Indeed, whereas by definition, isometries only preserve distance, the basic isometries are assumed to preserve not only distance but also degrees of angles; they also map lines to lines and angles to angles (see (L7) above). We will have to wait until Section 6.6 of [Wu2020b] before we can resolve this apparent discrepancy; every isometry will turn out to satisfy properties (i) and (ii) of the basic isometries in assumption (L7). Needless to say, this fact is far from obvious. The next section gives the most basic applications of the basic isometries, but a lot more will be said about them in Chapter 6 of [Wu2020b]. The reason we do not pursue this discussion further in this volume is that we are trying to present just enough of the geometry needed to begin the study of linear equations. This constraint comes from the practical reality of the middle school mathematics curriculum: to the extent that the study of linear equations is taken up in middle school, we must try to do enough with similar triangles to support this study.

Exercises 4.5. (1) Prove part (i) of Theorem G5 on page 236. (2) Let a translation TAB be given, and let TAB (P ) = Q. If Q lies on LBA , prove that TBA (Q) = P . (3) Prove that if the diagonals of a parallelogram are perpendicular to each other, then the parallelogram is a rhombus; i.e., all four sides of the parallelogram are equal. (See Exercise 4 on page 228.) (4) (i) Let ABCD be a parallelogram. Suppose F is a point on AD and E is a point on BC such that |AF | = |BE|. Prove that ABEF is a parallelogram. (ii) Suppose in part (i) the parallelogram ABCD is a rectangle. Prove that ABEF is also a rectangle. (5) Suppose the four points A, B, P , and Q lie on a number line. Prove that −− → −−→ AB and P Q point in the same direction if and only if A < B and P < Q, or A > B and P > Q.

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−−→ −−→ (6) Suppose the two vectors AB and P Q point in the same direction and the −−→ −−→ two vectors P Q and U V also point in the same direction. Does it follow −− → −−→ that AB and U V point in the same direction? (7) (This exercise makes use of a coordinate system; see the warning about the use of coordinates on page 207.) In the picture below, L is the horizontal −−→ line {y = 1}, and CD is the vector that starts at C = (3, 0) and ends at D = (2, 1). The points A and B are as shown. Let R = the reflection −−→ across L, T = the translation along CD, and  = the 90◦ rotation around B (according to the definition of rotations on page 202, this is a counterclockwise rotation). What is ( ◦ T ◦ R)(A) and what is ( ◦ R ◦ T )(A)? Are the two points the same? y

O

r A = (1, 2)

r B = (1, 0)

D = (2, 1) L I @ @ @ @ @ 45◦ @

x

C = (3, 0)

(8) In terms of the concept of a translation, we saw in the proof of Lemma 4.20 on page 232 that if TAB (P ) = Q and P does not lie on LAB , we can draw a picture to show where the point Q is. Using exactly the same notation, draw a picture to show where Q is if TBA (P ) = Q . (9) Prove that every point on the angle bisector of a convex angle is equidistant from both sides of the angle. (Hint: Recall the concept of distance of a point from a line on page 228.) (10) (a) Prove that the composition of two translations is a translation. (b) Is the composition of two rotations (with possibly distinct centers of rotation) a rotation?45 (Hint: Consider the example in Exercise 10 on page 229, and also prove that if  is a rotation, then for a point P distinct from the center O of the rotation, O must lie on the perpendicular bisector of the segment joining P to (P ).) (11) Let L1 , L2 , and L3 be parallel lines and let  and L be transversals so that they intersect L1 , L2 , and L3 at A, B, and C, and D, E, and F , respectively. Suppose |DE| = |EF |. Then prove that |AB| = |BC|. (This implies that equidistant parallel lines intercept equal segments on any transversal.)

45 Part

(b) has a continuation in Exercise 11 on page 372.

4.6. CONGRUENCE, SAS, AND ASA

 A

L1

L2

B

L C DC C C C EC C

239

P

Q

C C C F C L3 CC (Hint. There are many ways to do this, and here is one that uses translations. Let the line parallel to  and passing through E intersect L1 at P , and let the line parallel to  and passing through F intersect L2 at Q. −−→ Let T be the translation along the vector DE. Then of course T (D) = E. Prove T (E) = F , and then prove T (P ) = Q.) −− → −−→ (12) Prove that if LAB = LP Q , then AB and P Q point in the same direction if and only if RAB ⊂ RP Q or RP Q ⊂ RAB . (Hint: Make use of Lemma 4.6 on page 174 and imitate its proof by making LAB into a number line −−→ −−→ and argue with < rather than with ∗. Suppose AB and P Q point in the same direction. Let L0 be a line whose closed half-plane contains both RAB and RP Q , and let LAB intersect L0 at O. If O ∗ P ∗ A, then prove RAB ⊂ RP Q , but if O ∗ A ∗ P , then prove RP Q ⊂ RAB . Conversely, if RAB ⊂ RP Q , let L0 be the line ⊥ LP Q at P . Then the closed half-plane of L0 that contains Q contains both RP Q and RAB .) (13) (a) Prove that every translation is equal to the composition of two reflections. (b) Prove that every rotation is also equal to the composition of two reflections. Comments: The net effect of this exercise seems to be that we can forget rotations and translations because we only need reflections. This is an algebraic afterthought on the geometry of basic isometries, and it must be said that, in advanced mathematics, this algebraic point of view has paid immense dividends. On the other hand, this algebraic fact is only something to keep in mind from the point of view of algebra, but no more than that. Geometers continue to think in terms of translations and rotations directly.

4.6. Congruence, SAS, and ASA This section gives the definition of a congruence and proves two of the three most basic criteria for triangle congruence: SAS and ASA. It concludes with a statement of the last assumption we need for doing geometry in these volumes: the crossbar axiom. This axiom has not made its appearance until now, but it will be seen to enter a few proofs at some of the most critical junctures. The definition and basic properties of congruence (p. 240) Two congruence criteria for triangles (p. 244) The crossbar axiom (p. 250)

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The definition and basic properties of congruence We begin with a key definition. Definition. A congruence is a transformation of the plane that is the composition of a finite number of basic isometries. The concept of congruence is one of the cornerstones in the school geometry curriculum. Here are its most basic properties: Theorem G6. (a) Every congruence is an isometry; it preserves lines, rays, segments, and the degrees of angles, and it is also a bijection. (b) The inverse of a congruence is a congruence. (c) Congruences are closed under composition in the following sense: if F and G are congruences, so is F ◦ G. Proof. It has been pointed out that every one of the basic isometries has the following three properties: it is a bijection (see pp. 205, 231, and 235), it is an isometry, and it maps lines to lines as well as preserves the degrees of angles (see (L7) on page 237 for each of these claims). Because these properties persist under composition, the proof of part (a) of the theorem is straightforward. To prove part (b), i.e., the inverse of a congruence is a congruence, let a congruence ϕ be the composition of three basic isometries F ◦ G ◦ H; then it is simple to directly verify that if ψ = H −1 ◦ G−1 ◦ F −1 , then ψ ◦ ϕ = I = ϕ ◦ ψ. So ψ is the inverse of ϕ. But the inverse of a basic isometry is a basic isometry, because the inverse of a rotation is a rotation (see page 211), the inverse of a reflection is itself (see page 231), and the inverse of a translation is a translation (see page 235), so ψ is also a congruence. The proof is similar if ϕ is the composition of any number of basic isometries. Part (c) follows immediately from the definition of a congruence as a composition of basic isometries. The proof of Theorem G6 is complete. Mathematical Aside: The fact that the composition of congruences is a congruence and the inverse of a congruence is also a congruence means that the set of all congruences in the plane form a group, with the binary operation being the composition of transformations. We call this the group of congruences of the plane. Clearly, recalling the group of bijections (page 211) and the group of translations (page 235), we have group of translations ⊂ group of congruences ⊂ group of bijections. In this chain of inclusions, each group is a subgroup of the next. This is one way to observe group theory in action in a "real world" setting. A subset of the plane S is said to be congruent to another subset S  of the plane if there is a congruence ϕ so that ϕ(S) = S  . In symbols, S ∼ = S  . Since the inverse of a congruence is a congruence (Theorem G6(b)) and since ϕ(S) = S  implies ϕ−1 (S  ) = S, we see that S ∼ = S  implies S  ∼ = S. This means "S being   congruent to S " is equivalent to "S being congruent to S", so that it does not matter whether S or S  comes first. Thus we can speak unambiguously about two sets S and S  being congruent. Incidentally, the usual terminology for the fact that S ∼ = S for every S and S  is that the relation of congruence = S  implies S  ∼ between two sets is a symmetric relation or that congruence is symmetric.

4.6. CONGRUENCE, SAS, AND ASA

241

We leave it as an exercise to show that if S1 is congruent to S2 and S2 is congruent to S3 , then S1 is congruent to S3 . This fact is usually expressed by saying that congruence is a transitive relation or more simply congruence is transitive. Finally, congruence is also reflexive, in the sense that S ∼ = S for any S (one also says congruence is a reflexive relation). In general, if there is a relation among a class of geometric figures that is reflexive, symmetric, and transitive, then the figures that are so related are intuitively "almost equal". Such a relation is called an equivalence relation. Therefore congruence is an equivalence relation. With the availability of the concept of congruence, we can now give new meaning to segments with the same length and angles with the same degree by proving the following lemma. The proof of the lemma is very instructive. Lemma 4.23. (i) Two segments have the same length if and only if they are congruent to each other. (ii) Two angles have the same degree if and only if they are congruent to each other. Proof. We first prove (i). Let AB and A B  be congruent segments. Since every congruence is an isometry (Theorem G6), the segments have the same length. Conversely, suppose |AB| = |A B  |, then we will prove that there is a congruence φ so that φ(AB) = A B  . B

A

ZZ Z Z

Z Z A

B

−−→ Let T be the translation along the vector AA . Then T maps A to A and—by (L7)(i) on page 237—maps AB to a segment whose one endpoint is at A and the other endpoint we call B0 ; i.e., B0 = T (B). (In the following picture, since we are trying to show the position of A B0 (= T (AB)), we use a dashed line to represent the segment AB in its original position. Of course T also moves A B  away from its original position, but since we are not concerned with T (A B  ), we omit T (A B  ) from the picture altogether. However, we choose to retain A B  in its original position in the picture for the benefit of this discussion.) B

ZZ Z Z

Z Z :   A      (= T (A))     B A

B0 (= T (B))

Let ∠B  A B0 denote the convex angle with vertex at A , and let |∠B  A B0 | = t◦ . Let  denote the t-degree rotation around A . (In this picture,  is clearly the

242

4. BASIC ISOMETRIES AND CONGRUENCE

counterclockwise rotation of t degrees, but if the endpoint B  of A B  were in the lower half-plane of the line LA B0 , then it would be the clockwise rotation.) B

ZZ Z Z

t Z Z :   A          A B

B0

We claim that the composition  ◦ T is the desired congruence. Let φ =  ◦ T . We already know that φ(A) = A , so we have to prove that φ(B) = B  . The main weight of proving this claim lies in the proof that (B0 ) = B  .

(4.8)

Denote (B0 ) by B1 ; then by the definition of , |∠B1 A B0 | = t◦ . By Lemma 4.10 on page 190, RA B1 coincides with RA B  . Therefore B1 lies on the ray RA B  . To prove (4.8), observe that |A B0 | = |AB|, because A B0 = T (AB) and T preserves lengths of segments (see (L7)(ii) on page 237). In addition, |A B0 | = |A B1 |, again because  preserves lengths of segments on account of (L7)(ii). Putting these two equalities together, we obtain |A B1 | = |AB|. But by hypothesis, |AB| = |A B  |, so we obtain |A B1 | = |A B  |. Recall that B1 (= (B0 )) and B  are two points on the ray RA B  , so (L5)(ii) on page 184 implies that B1 = B  ; i.e., (B0 ) = B  , and we have proved (4.8). We can now easily prove φ(B) = B  . Since B0 = T (B), we appeal to (4.8) to get B  = (B0 ) = (T (B)) = ( ◦ T )(B) = φ(B), as claimed. By (L7)(i) and the fact that φ(A ) = A , we see that φ(AB) is the segment joining A (= φ(A)) to B  (= φ(B)) and, therefore, φ(AB) = A B  by (L1) on page 165. The proof of part (i) is complete. Next, we prove part (ii). If two angles

B

are congruent, then because congruence

O

preserves the degrees of angles (Theorem G6), the angles have the same degree. Conversely, given two angles ∠AOB and 





∠A O B with the same degree, we have to produce a congruence that maps one angle to the other.

O BHH B

B

HH

H

B

B

BB

HA

A

4.6. CONGRUENCE, SAS, AND ASA

243

−−→ As before, we use the translation T along the vector OO  to map O to O  . T then moves ∠AOB to a new angle ∠A0 O  B0 (see (L7)(i) on page 237) with vertex at O  , where A0 = T (A) and B0 = T (B). See the left picture below. Let the degree of the convex angle ∠A0 O  A be t. Let  be the t-degree (clockwise or counterclockwise) rotation around O  (in the picture, it is counterclockwise) so that  maps the ray RO A0 to the ray RO A . Let  map B0 to a point B1 . Thus  maps ∠A0 O  B0 to ∠A O  B1 , as shown in the right picture below.

B O O

B O

A

t

A0 = T (A)

O

A B1

A B0 = T (B)

A B

B

Notice that the congruence  ◦ T has moved ∠AOB to ∠A O  B1 so that the latter now has one side in common with ∠A O  B  . In the right picture above, B1 and B  are in opposite half-planes of the line LO A . Now we use the reflection Λ across LO A to map ∠A O  B1 to an angle in the closed half-plane of LO A that contains ∠A O  B  . We claim that Λ maps the ray RO B1 to the ray RO B  . To prove the claim, let B 1 = Λ(B1 ). Then the claim becomes: the two rays RO B  and RO B 1 coincide.

O %

B1 % B % % % A

To this end, we will prove that |∠A O  B  | = |∠A O  B 1 | and appeal to Lemma 4.10 on page 190 to draw the desired conclusion. To prove |∠A O  B  | = |∠A O  B 1 |, observe that ∠A O  B 1 = Λ(∠A O  B1 ), and we also have ∠A O  B1 = (∠A0 O  B0 ) = (T (∠AOB)). Therefore, ∠A O  B 1 = Λ(∠A O  B1 ) = Λ((T (∠AOB))) = (Λ ◦  ◦ T )(∠AOB). Let φ denote the congruence Λ ◦  ◦ T . Then we have (4.9)

φ(∠AOB) = ∠A O  B 1 .

By assumption (L7) on page 237, φ preserves degrees of angles. Thus ∠AOB and ∠A O  B 1 have the same degree. By hypothesis, ∠AOB and ∠A O  B  also have the

244

4. BASIC ISOMETRIES AND CONGRUENCE

same degree. We conclude therefore that |∠A O  B  | = |∠A O  B 1 |, as desired. Since these two angles lie in the same closed half-plane of their common side which is the ray RO A , Lemma 4.10 implies that their other sides must coincide; i.e., the two rays RO B  and RO B 1 coincide, and the claim is proved. Equation (4.9) now reads: φ(∠AOB) = ∠A O  B  . Therefore, φ is the congruence that maps ∠AOB to ∠A O  B  . Finally, we observe that if the ray RO B1 is already in the same closed halfplane of LO A as the ray RO B  , then no reflection across LO A would be necessary because the preceding argument would have proved directly that RO B1 is equal to RO B  . Then the congruence  ◦ T maps ∠AOB to ∠A O  B  as before. The proof of Lemma 4.23 is complete. Pedagogical Comments. The simplicity of the reasoning in the preceding proof is marred by the unavoidably messy notation. In a school classroom, the proof can be made much clearer by using two plastic angles to represent ∠AOB and ∠A O  B  so that one can realize the translation T , the rotation , and the reflection Λ by actual movements of one of the plastic angles. End of Pedagogical Comments. The lemma shows the equivalence of "two segments have the same length" with "two segments are congruent". Following the tradition started by Euclid ([Euclid1]), it is customary to say that two segments are equal when what is meant is that they are congruent. The same goes for equal angles when what is meant is that the angles are congruent. When there is no fear of confusion, we will also abuse the language in this manner for the rest of these volumes, but it is important to note that this terminology is, strictly speaking, incorrect, because, for example, "equal segment" means literally that the segments are equal geometric figures in the sense of equal subsets of the plane (see page 141). Two congruence criteria for triangles The rest of the geometric discussion in this volume will focus on triangles and, at times, polygons. It is however worth pointing out that the concept of congruence applies not only to polygons, but to any geometric figures, including "curved" ones such as parabolas and ellipses. See Chapter 2 of [Wu2020b] and Chapters 1 and 4 of [Wu2020c]. At this point, a little reflection will reveal that the crude definition of congruence in TSM as "same size and same shape"—regardless of its intuitive appeal—can in no way be used as a definition of congruence. In mathematics, one cannot replace a precise concept with a vague intuitive one, no matter how appealing. For example, we can say that the following two strange looking figures are congruent, not because they "look the same", but because the left figure can be mapped to the right figure by a translation (such as from P to Q) followed by a (counterclockwise) rotation of 90◦ .

4.6. CONGRUENCE, SAS, AND ASA

245

P

Q

Congruent triangles occupy a special position in elementary geometry and have their own special conventions. Denote a triangle ABC by ABC. Then the congruence notation ABC ∼ = A B  C  will be understood to mean not only that there is a congruence ϕ so that ϕ(ABC) = A B  C  , but also that ϕ(A) = A , ϕ(B) = B  , and ϕ(C) = C  . Therefore, if ABC ∼ = A B  C  , then it is understood that ϕ(AB) = A B  ,   ϕ(AC) = A C , and ϕ(BC) = B  C  and also that ϕ(∠A) = ∠A , ϕ(∠B) = ∠B  , and ϕ(∠C) = ∠C  . Therefore, by Theorem G6(a) on page 240, ABC ∼ = A B  C  implies that |∠A| = |∠A |, |∠B| = |∠B  |, |∠C| = |∠C  | and |AB| = |A B  |, |AC| = |A C  |, |BC| = |B  C  |. We now prove the converse: Theorem G7. If for two triangles ABC and A B  C  , |∠A| = |∠A |,

|∠B| = |∠B  |,

|∠C| = |∠C  |

|AB| = |A B  |, then ABC ∼ = A B  C  .

|AC| = |A C  |,

|BC| = |B  C  |,

and

Conceptually, Theorem G7 is the correct theorem. However, in a practical sense, this theorem is an overkill, in that it is not necessary to require the equalities of all the angles and all the sides of two triangles before we can prove the congruence of the triangles. Typically, it suffices to impose three suitably chosen conditions46 to guarantee congruence, and the best known among these criteria are SAS, ASA, and SSS. The proof of SSS is a bit of an interruption at this juncture and will therefore be postponed to Section 6.2 in [Wu2020b], but we can prove the other two here: Theorem G8 (SAS). Assume two triangles ABC and A B  C  so that |∠A| = |∠A |, |AB| = |A B  |, and |AC| = |A C  |. Then the triangles are congruent. 

Theorem G9 (ASA). Assume two triangles ABC and A B  C  so that |AB| = |A B  |, |∠A| = |∠A |, and |∠B| = |∠B  |. Then the triangles are congruent. 

46 The

emphasis is on "suitably chosen". See, for example, Exercise 5 on page 251.

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4. BASIC ISOMETRIES AND CONGRUENCE

We will prove Theorem G9 below. The proof of Theorem G8 (SAS) is very similar and will be left as an exercise (Exercise 3 on page 251). However, we want to call attention to an animation (due to Larry Francis) that shows, under the hypothesis of SAS, how one triangle can be moved to the other: https://youtu.be/30dOn3QARVU. This video sheds light on the following proof of Theorem G9 as well. The proofs of SAS and ASA depend on the following simple lemmas. We note that for the proof of Theorem G9 (ASA), we only need Lemma 4.24 (whose proof is actually implicit in the proof of Lemma 4.23 on page 241). For the proof of Theorem G8 (SAS), however, Lemma 4.25 will be needed. Lemma 4.24. Assume two convex angles ∠M AB and ∠N AB so that |∠M AB| = |∠N AB|. Suppose they have one side AB in common and M and N are on opposite sides of the line LAB . Then the reflection across the line LAB maps ∠N AB to ∠M AB (and also maps ∠M AB to ∠N AB). M       AH B HH HH HH N Lemma 4.25. Suppose two convex angles ∠M AB and ∠N AB have the same degree and they have one side AB in common. Assume further that the segments AM and AN have the same length. Then either M = N (if M and N are on the same side of LAB ) or the reflection across LAB maps N to M (if M and N are on opposite sides of LAB ). r M      AH B HH HH HH Hr N For the proof of Lemma 4.24, observe that the reflection R across LAB maps ∠N AB to ∠N0 AB, where N0 = R(N ), so that ∠N0 AB and ∠M AB are now convex angles with the same degree, in the same half-plane of LAB , with one side RAB in common. So ∠N0 AB = ∠M AB, by Lemma 4.10 on page 190. This proves Lemma 4.24. As to Lemma 4.25, suppose M and N are on the same side of LAB . By Lemma 4.10 on page 190 again, we know that the rays AM and AN coincide. But since |AM | = |AN |, necessarily M = N . Now if M and N are on opposite sides of LAB , then Lemma 4.24 shows that the reflection across LAB maps the ray

4.6. CONGRUENCE, SAS, AND ASA

247

RAN to the ray RAM . Since a reflection preserves length (see (L7)(ii) on page 237), the reflection maps the segment AN to a segment of length equal to |AM |, and therefore M = N by virtue of (L5)(ii) on page 184. This proves Lemma 4.25. Proof of Theorem G9. We will prove that if the triangles ABC and A B  C  satisfy |AB| = |A B  |, |∠A| = |∠A |, and |∠B| = |∠B  |, then there is a congruence ϕ so that ϕ(ABC) = A B  C  . To this end, we break up the proof into three steps, going from a special case to the most general case: Case I. The triangles satisfy in addition, A = A , B = B  . Case II. The triangles satisfy in addition, A = A . Case III. The general case.

Case I. In this case, either C, C  are already in the same half-plane of LAB or they are in opposite half-planes of LAB . If the former, then we claim that C = C  , so that in this situation, we need only let ϕ be I, the identity transformation. To prove the claim, observe that because |∠CAB| = |∠C  AB| by hypothesis, the fact that C and C  are in the same half-plane of LAB implies that we have the equality of rays RAC = RAC  (Lemma 4.10 on page 190). Thus the two rays RAC and RAC  in the following picture in fact coincide:

A %

% % %

C % C %

B

In like manner, because |∠CBA| = |∠C  BA|, we have RBC = RBC  . Therefore the following intersections are equal: RAC ∩ RBC = RAC  ∩ RBC  which means of course that C = C  . So in this situation, ABC = A B  C  , or, more formally,

(4.10)

I(ABC) = A B  C  ,

where I is the identity transformation.

248

4. BASIC ISOMETRIES AND CONGRUENCE

Next, suppose A = A , B = B  but C, C  are in opposite half-planes of LAB : C @   @  @   @ B = B A = A  HH HH HH H H C By assumption, ∠CAB = ∠C  AB. Therefore if Λ is the reflection with respect to LAB , then Λ maps the ray RAC to the ray RAC  , by Lemma 4.24. For the same reason, Λ also maps RBC to RBC  . Thus, by Lemma 4.17 on page 226, Λ(RAC ∩ RBC ) = RAC  ∩ RBC  . Since RAC ∩ RBC is just C and RAC  ∩ RBC  is just C  , we get Λ(C) = C  . Since also Λ(A) = A = A and Λ(B) = B = B  , we have, (4.11)

Λ(ABC) = A B  C  .

Now let ϕ be either I or the reflection Λ across LAB , depending on whether C and C  lie in the same half-plane or different half-planes of LAB , respectively. Then (4.10) and (4.11) together imply that ϕ(ABC) = A B  C  . Case I is thus proved with this choice of the congruence ϕ. Case II. We now let the triangles satisfy the less restrictive condition that A = A , but B and B  may now be distinct. Let the degree of the convex angle between the rays RAB and RAB  be θ, as shown in the left picture below. C @  A = A r  @ 
A @B
A r  A = A 
A

A
θ
A
A C∗
C
@
@ C
@  @ B = B∗ @ @ B Then there is a θ-degree (clockwise or counterclockwise) rotation θ around the point A so that θ (RAB ) = RAB  (compare the proof of Lemma 4.23 on page 241). In the left picture above, it is a clockwise rotation, but a different configuration may require a counterclockwise rotation from RAB to RAB  . Let B ∗ = θ (B). Now B  and B ∗ are two points on the ray RA B  so that   |A B | = |AB| (by hypothesis) and |A B ∗ | = |θ (AB)| = |AB| (because of (L7)(ii) on page 237 and the fact that θ is a basic isometry). Therefore, B  and B ∗ are two points on the same ray RA B  equidistant from A ; we conclude that B  = B ∗ (by (L5)(ii) on page 184); i.e., B  = θ (B). Letting C ∗ = θ (C), we get (4.12)

θ (ABC) = A B  C ∗ .

4.6. CONGRUENCE, SAS, AND ASA

249

Therefore, the two triangles A B  C  and A B  C ∗ satisfy the condition of Case I; i.e., they share two vertices A and B  , and the angles at these vertices are pairwise congruent. Consequently, there is a congruence ω so that ω(A B  C ∗ ) = A B  C  .

(4.13)

If we apply the transformation ω to both sides of (4.12) and make use of (4.13), we get ω(θ (ABC)) = ω(ABC ∗ ) = A B  C  . By Theorem G6(c) on p. 240, ω ◦ θ is a congruence. Letting ϕ = ω ◦ θ , we get ϕ(ABC) = A B  C  . Thus the theorem is also proved for Case II. Case III. Finally, we deal with the general case where no restriction is placed on the triangles ABC and A B  C  . Assuming that the vertices A and A are distinct, −−→ let T be the translation along the vector AA ; note that T (A) = A . If we define B ∗ = T (B) and C ∗ = T (C), then T (ABC) = A B ∗ C ∗ .

(4.14)

C∗

A

B∗ C

A

C

B

B 

Now the triangles A B C and A B ∗ C ∗ have the vertex A in common. Furthermore, because T is a basic isometry and therefore preserves lengths of segments and degrees of angles, these two triangles satisfy |A B ∗ | = |A B  |, |∠A | = |∠B ∗ A C ∗ |, and |∠B  | = |∠A B ∗ C ∗ |. Thus Case II applies and, for a suitable congruence ω, we have (4.15)





ω(A B ∗ C ∗ ) = A B  C  .

Applying ω to (4.14) and making use of (4.15), we get ω(T (ABC)) = ω(A B ∗ C ∗ ) = A B  C  . Letting ϕ = ω ◦ T , we see that ϕ is a congruence (Theorem G6(c) on p. 240) and ϕ(ABC) = A B  C  . This completes the proof of Theorem G9. The preceding proof furnishes a classic example of a proof that "progresses from the simple to the complex", in the sense that it starts by proving a relatively simple case (Case I), then proceeds to a slightly more complex case (Case II), and then finally arrives at the most general case (Case III). It may remind you of the proof

250

4. BASIC ISOMETRIES AND CONGRUENCE

of Theorem 1.7 on page 49, in which we first proved that the area of a rectangle with sides of length 1 and n1 is 1 × n1 and, on the basis of this fact, we proceeded to prove the general case, namely, that the area of a rectangle with sides of length k m k m  and n is  × n . One should not hold onto the simplistic belief that all proofs yield to such a direct attack, but this direct approach is something we must keep in mind anytime we want to prove a theorem. We will prove in Section 6.6 of [Wu2020b] that every isometry of the plane is a congruence. In other words, every isometry is nothing but the composition of a finite number of basic isometries. This underscores the importance of the basic isometries: they are the basic building blocks of all the isometries of the plane. Once we know this, then we know that if a transformation preserves distance, it must be a congruence and therefore it is automatically surjective and it automatically preserves lines, segments, rays, angles, and degrees of angles. However, until we can prove this fact about isometries, we cannot assume that an isometry has all these desirable properties. So be careful. The crossbar axiom Before we leave the mathematical discussion of this chapter, we state the last assumption we need for the development of plane geometry. (L8) (Crossbar axiom) Given a convex angle AOB, for any point C not equal to O in ∠AOB, the ray ROC intersects the segment AB (indicated as point D in the following figure).  A    @    @D ` O `` C ``` ``` @ ``@ ` @``` B You may regard the crossbar axiom as frivolous, because "what else can the ray ROC do if it does not intersect AB"? First of all, so long as you consider this statement to be obvious, then our objective of agreeing on a common starting point has been met: we do want to assume only believable facts. As to whether the crossbar axiom is frivolous, we should point out that up to this point, none of the assumptions (L1)–(L7) explicitly guarantees that the ray ROC must intersect AB.47 The purpose of the crossbar axiom is therefore to firm up the intuitive idea that a ray is indeed "straight and infinite in one direction" and therefore cannot stay inside the bounded triangular region OAB. For example, (L8) guarantees that the angle bisector (see page 192) of an angle in a triangle must intersect the opposite side, any two medians (see page 252 for the definition) must meet, and the two diagonals of a parallelogram must intersect each other (page 260). It will also make a rather dramatic appearance in unexpected places, e.g., the proof of Theorem G14 (hidden in the proof of (♣) on page 262), the proof of Theorem G16 on page 270, the proof of Ceva’s theorem in Section 6.7 of [Wu2020b], etc.

47 See

the Mathematical Aside immediately following.

4.6. CONGRUENCE, SAS, AND ASA

251

Mathematical Aside: Since we are assuming that every line can be made into a number line (see (L3) on page 167), we get easy access to the definition of the subtle concept of "betweenness" and its basic properties (pp. 167ff.; also see the more elaborate discussion in Chapter 8 of [Wu2020b]). Such being the case, it is known that the crossbar axiom can be deduced from the plane separation property (L4) (page 176) if our goal is to pursue a strictly axiomatic development of plane (Euclidean) geometry. See page 116 of [Greenberg]. However the proof is too technical to be of real educational value for the purpose of teaching in K–12. Exercises 4.6. (1) Prove that congruence is a transitive relation (see page 241). (2) Prove that if ϕ is a congruence and S is a convex set in the plane, then ϕ(S) is also convex. (3) Prove Theorem G8 (SAS) on page 245. (4) Prove that any two circles with equal radii are congruent. (Caution: This is a slippery proof. Be very precise.) (5) Explain why two triangles with two pairs of congruent sides and one pair of congruent angles need not be congruent. (6) (This exercise makes use of a coordinate system; see the warning about the use of coordinates on page 207.) In the picture below, let C denote the lower left corner of the black figure. Suppose |∠CAB| = 45◦ , |AB| = |BC|, and line L makes an angle of 45 degrees with line LAB . Let F be the counterclockwise rotation of 45◦ with center at the point B, let G be the clockwise rotation of 90◦ with center at the point A, let H be the reflection across the line L, and let J be the translation along −−→ AB. Furthermore, let S denote the black figure at the point C.

C

L

A

B

Using a separate sketch for each of the following items, indicate the positions of (a) G(S), (b) F (G(S)), (c) G(H(S)) and H(G(S)) (are they equal?), (d) J(S), (e) J(F (S)) and F (J(S)), (f) H(J(S)) and J(H(S)), (g) G(H(J(S))), and (h) J(H(F (S))). (7) Recall that the opposite sides of a rectangle have the same length (see page 226), so that knowing the lengths of a pair of adjacent sides is equivalent to knowing the lengths of all four sides of a rectangle. It is common to refer to a rectangle with a pair of adjacent sides of lengths a and b as a rectangle with side lengths a and b. Now let R1 and R2 be two rectangles with side lengths a1 , b1 and a2 , b2 , respectively. Prove that R1 ∼ = R2 if and only if either a1 = a2 and b1 = b2 , or a1 = b2 and a2 = b1 .

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4. BASIC ISOMETRIES AND CONGRUENCE

(8) Let ABCD be a parallelogram. If a diagonal is an angle bisector (e.g., BD bisects ∠ABC), then prove ABCD is a rhombus; i.e., all four sides of ABCD are equal (see the definition on page 193). (9) Prove that the angle bisector from a vertex of a triangle is perpendicular to the opposite side if and only if the two sides of the triangle issuing from this vertex are equal. (Note that by the crossbar axiom, there is no question that the angle bisector must intersect the opposite side.) (10) A median of a triangle is a segment joining a vertex to the midpoint of the opposite side. Prove that the median from vertex A of ABC is the angle bisector of ∠A if and only if the median ⊥ BC. (Hint: Let D be the midpoint of BC. Suppose AD bisects ∠A. To show AD ⊥ BC, let E be the point on RAD so that |AD| = |DE|. Show ABEC is a parallelogram and consider the reflection across LAE .) 4.7. A brief pedagogical discussion of proofs We discuss a few pedagogical issues regarding the realities of teaching proofs of theorems in a high school classroom. We mentioned in the overview (pp. 157ff.) that the content of this chapter should also be taught in middle school, but in a more intuitive and informal manner. The Common Core Standards are in agreement ([CCSSM], page 55): the eighth-grade geometry standards call for an understanding of "congruence and similarity using physical models, transparencies, or geometry software" and the use of "informal arguments to establish facts about the angle sum and exterior angle of triangles, about the angles created when parallel lines are cut by a transversal, and the angle-angle criterion for similarity of triangles". It will take a delicate touch to achieve a balance between the nurturing of geometric intuition and the promotion of reasoning. Because this middle school issue has been taken up in Chapters 4 and 5 of [Wu2016a] (also see [Wu2010a]), we will instead concentrate on the corresponding problem in high school: how to introduce geometry in the high school curriculum. Transformations of the plane and concepts of surjectivity and injectivity are taxing topics even for college students, and it would not do to subject the average high school student to a treatment with the same degree of precision as in this chapter and the next.48 A teacher will have to judiciously simplify the content of these chapters in order to convey to students their main message, namely, that congruence and similarity are precise mathematical concepts. One suggestion is to confine the discussion of transformations only to bijections (one-to-one correspondences) in the plane and mention general transformations only in a few exercises or as activities for enrichment (cf. the examples on pp. 207ff.). Another suggestion is to make liberal use of transparencies at every turn when discussing basic isometries and dilations. See, for example, the Activities of Sections 4.4 and 4.5 (pp. 221, 230, and 235) and the Activity of Section 5.1 on page 258. One can even assign homework 48 One of the considerations that entered into the design of this geometry curriculum is precisely the awareness that students may initially experience difficulty with these concepts. Therefore we want these concepts to be taught, first intuitively in middle school and then more formally in high school.

4.7. A BRIEF PEDAGOGICAL DISCUSSION OF PROOFS

253

problems on such activities and ask students to report their findings on the effects of various transformations. With enough such hands-on experiences, students will build up their geometric intuition about the basic isometries and therefore about isometries in general. Using transparencies in a similar manner to illustrate the composition of transformations is also highly recommended. Such hands-on activities are meant, of course, to supplement the definitions and the accompanying mathematical discussions, not to replace them. At the same time, a high school presentation of some of the definitions, lemmas, and theorems can probably afford to specify that they be "skipped on first reading" so that they can be revisited later (and even then, perhaps soft-pedaled). For example, Theorem G1 (page 220) is so basic that ample time should be devoted to its proof, but the proofs of Theorems G2–G3 (pages 223–224) do not quite occupy the same exalted status. Recall that the purpose of these two theorems is to show that, from a point outside a given line, one can drop one and only one perpendicular to the given line. This fact is needed to make the definition of a reflection well-defined. However, since this fact is so intuitively obvious, it may be pedagogically legitimate to simply state these two theorems but postpone their proofs so as to get to the definition of reflection as quickly as possible. Pedagogical decisions of this type are always the prerogative of the teacher. Having said that, we would suggest explicitly that Theorem G4 be carefully proved because the method of proof is powerful. In fact, the proof of Theorem G18 (page 277) can be given right after Theorem G4 if so desired. As another illustration of what can be done to smooth geometric instruction in high school, it is not usually realized that the definitions of the alternate interior angles and corresponding angles of a transversal are by no means simple (see page 276). While the precise definitions should be given (they justify why the concepts of a half-line in Lemma 4.5 and a half-plane in (L4) are indispensable), one should not make a big deal of the precision; it is boring and cumbersome. Unless absolutely necessary, one should simply use a picture to identify alternate interior angles. There is a similar phenomenon with certain proofs. Take Theorem G14, for example. In this case, there is in fact an explicit recommendation on pp. 262ff. to suppress some of the subtleties inherent in the proof. Please note that this recommendation is based on two considerations: (a) if the subtle point is not explicitly brought out by the teacher, an overwhelming majority of the students will not be aware of it, and (b) in the context of mathematics learning, learning about the proof of such a subtle point at this stage of students’ mathematical development is of secondary importance. What we are suggesting is that good pedagogy always involves some subjective judgment: there is no ironclad rule that dictates in a given situation what should be emphasized and what could be soft-pedaled. Some compromise is essential when the conflict between what is ideal and what is achievable becomes extreme. Overall, we wish to advocate a certain flexibility in teaching proofs in geometry. There should be no doubt about the importance of proofs all through the school curriculum. But the main message we are trying to convey—as we did in the discussion of axiomatic developments of plane geometry on page 160—is that a slavish adherence to the mindset of "proving absolutely everything" is counterproductive when it comes to the teaching of high school geometry. Compare the discussion on pp. 160ff. As an example of the kind of flexibility we have in mind, it would be

254

4. BASIC ISOMETRIES AND CONGRUENCE

at times worthwhile to skip a less interesting proof and use the time for a detailed discussion of the evolution of another proof. A potential candidate for the latter would be the proof of Theorem G15 on page 263. The preceding discussion is a reminder of the realities about the mathematical education of teachers: what we teach teachers is not always what we can use, unchanged, to teach school students. Pedagogical considerations will necessarily modify pure mathematical knowledge, or, in the terminology of [Wu2006], the "mathematical engineering" aspect of school mathematics education cannot be ignored. In particular, these volumes contain more proofs than is optimal for students’ mathematics learning. Experience in the actual classroom will suggest the proper give-and-take between what ideally should be taught and what could actually be taught. What such pedagogical considerations cannot do, however, is lighten your mathematical load as a prospective teacher. If you hope to teach certain concepts or certain proofs effectively by making the correct mathematical decisions, then you must know the whole mathematical story first before you can decide what message must be conveyed and which details can be harmlessly left out. One cannot write a faithful twenty-page plot outline of War and Peace without first carefully reading through the thousand and more pages of the uncondensed version. Likewise, without a complete knowledge of the relevant mathematics, you will not know what to keep and what to leave out in your lessons because you won’t be able to distinguish between what is truly essential and what is expendable. Besides, if by chance you get a precocious youngster who wants the whole truth and nothing but the truth, then you will have to supply the whole truth and nothing but the truth. This too is part of your basic duty as a teacher, and these volumes are designed to get you ready for such contingencies.

CHAPTER 5

Dilation and Similarity This chapter introduces the other basic concept in school geometry: similarity. Like congruence, similarity has not fared well in TSM.1 Middle school students are taught that two sets are similar if they are the same shape but not the same size. Intuitively, this is a useful description of similarity, but as in the case of congruence, TSM has the habit of confusing nice-sounding intuitive statements with precise mathematical definitions. When students get to high school under the illusion that "same shape but not necessarily the same size" is all they need to know about similarity, they are shocked to be confronted with the fact that similarity henceforth will only mean equal angles and proportional sides for triangles but no further mention is made about the similarity of curved figures. Consequently, students’ understanding of similarity upon graduation from high school consists of two disconnected sound bites: a definition of similar triangles in terms of proportional sides and equal angles and a vague conception of "same shape but not the same size" for anything other than triangles. Thus TSM even fails to give students a correct understanding of a concept as basic as similarity. Fortunately, a correct definition of similarity, one that is discussed below, can be easily introduced as early as middle school through ample hands-on experiments plus a judicious amount of reasoning. A more detailed description of how this can be accomplished in middle school has been given in Chapter 4 of [Wu2016a]. We are happy to point out that this curricular advocacy has been adopted by the CCSSM ([CCSSM]), and if the CCSSM is rigorously implemented, at least one of the egregious errors of TSM will be rectified in the near future. We will not pretend that a successful implementation in school classrooms of this new point of view will be easy or straightforward. It will require some hard work by knowledgeable teachers to bring about a true understanding of similarity. The main goal of this chapter is to provide teachers with the content knowledge they will need for such a successful implementation. In this context, what was said in Section 4.7 (pp. 252ff.) is just as relevant, if not more so, to the material of this chapter. 5.1. The fundamental theorem of similarity The discussion of the concept of similarity will rest on a single theorem that we call the fundamental theorem of similarity. At this point, we have not yet said what "similarity" means (the definition will be given in Section 5.3), much less why this theorem is fundamental. However, it will become all too clear that this theorem dominates the whole discussion of similarity. In this section, we will only prove a very special case of the theorem, Theorem G15 on page 263, but this special 1 See

page xiv of the preface. 255

256

5. DILATION AND SIMILARITY

case—on account of its simplicity—is of independent interest, not the least because its proof requires two new characterizations of a parallelogram. Statement of the theorem (p. 256) Two characterizations of a parallelogram (p. 259) Proof of FTS when r = 12 (p. 263) Statement of the theorem Theorem G 10 (Fundamental theorem of similarity). Let ABC be given, and let D, E be points on the rays RAB and RAC , respectively, so that |AD|

|AE|

D is not equal to A or B. If |AB| = |AC| and their common value is denoted by r, then |DE| = r. DE  BC and |BC| A
@
@
@
@
@
@ E D
@ @C B


The fundamental theorem of similarity is usually referred to by its acronym, FTS. We only prove a special case of this theorem in this volume and will leave the complete proof to Section 6.4 of [Wu2020b] (for the case where r is rational) and Section 2.6 of [Wu2020c] (for the general case where r is any positive number). It should be mentioned that the long and intricate proof of FTS can in fact be given at this point; it uses either ideas we have already developed or those that are independent of what we have been doing (such as the least upper bound axiom). There is thus no fear of circular reasoning in the rest of this chapter when we make extensive use of the FTS to prove other results about similarity. The decision not to give the proof at this juncture is due to the pressing need to get to the AA criterion for similarity (Theorem G22 on page 288) as quickly as possible; the latter is needed for the study of linear equations in two variables. The number r in the theorem is generally referred to as the scale factor. The statement above on DE BC is a standard abuse of notation for LDE  LBC ; i.e., the line containing the segment DE is parallel to the line containing the segment BC. We will continue to use this abuse of notation for the rest of these volumes. In applications, it is sometimes more convenient to assume, instead of (5.1)

|AE| |AD| = , |AB| |AC|

5.1. THE FUNDAMENTAL THEOREM OF SIMILARITY

257

the equivalent condition when D = B and E = C, that (5.2)

|AE| |AD| = . |DB| |EC|

Because the proof of the equivalence of (5.1) with (5.2) is entirely elementary and straightforward, we will leave it as an exercise (Exercise 1 on page 267). We will give a proof of the special case of FTS where r = 12 at the end of this section (page 263), i.e., for the case that D and E are midpoints of AB and AC, respectively. To this end, we first give an equivalent formulation of FTS that is sometimes more convenient to use: Theorem G11 (FTS*). Let ABC be given, and let D be a point on the ray RAB not equal to A or B. Let  be the line parallel to BC and passing through D. Then  intersects the ray RAC at a point E, and |AE| |DE| |AD| = = . |AB| |AC| |BC| A
QQ
Q Q
Q
Q E D Q
Q
QC B

Let us explain in greater detail what we mean when we say Theorems G10 and G11 are equivalent. It means that, if we assume the validity of (L1)–(L8) in Chapter 4 and Theorems G1–G9, then Theorem G10 implies Theorem G11 and, conversely, Theorem G11 implies Theorem G10. In other words, if you know Theorem G10 is true, then you also know Theorem G11 is true and, conversely, if you know Theorem G11 is true, then you also know Theorem G10 is true. These two theorems are therefore interchangeable in the precise sense above. Leaving the details of the converse to an exercise (Exercise 9 on page 268), we will prove that, assuming Theorem G10, we can prove Theorem G11. Proof of Theorem G11. Let the line passing through D and parallel to LBC be  as in the theorem. We will first prove that  intersects not just the line LAC , but the ray RAC ; it will be an indirect proof. Let E be chosen on the ray RAC so that |AE| =

|AD| · |AC|. |AB|

E could be between A and C or C could be between A and E, depending on whether |AD|/|AB| is < 1 or > 1, which in turn depends on whether D is between A and B or B is between A and D. Both cases are shown in the picture below.

258

5. DILATION AND SIMILARITY

A D B

A E B

C

D

C E

Then by the cross-multiplication algorithm on page 22 (applied to real numbers by appealing to FASM—see page 133), we have |AD| |AE| = . |AB| |AC| By FTS (Theorem G10), we have LDE  LBC . Since   LBC by hypothesis,  and LDE are two lines both passing through D and parallel to LBC . By the parallel postulate (page 165), the lines  and LDE coincide, so that  does intersect the ray RAC at E. Moreover, by FTS once again, |AD| |AE| |DE| = = . |AB| |AC| |BC| The proof of Theorem G11 is complete (when Theorem G10 is assumed). To a student, the proof of Theorem G11 (FTS*) is probably not as convincing as some direct experimental verifications of the theorem. To this end, it is worthwhile to point out that the lined papers in ordinary notebooks provide fertile ground for experimentations related to FTS*, as the following Activity explains. Activity. We will assume that the lines of lined papers are equidistant parallel lines, in the sense that the distances between adjacent parallel lines (see page 228 for the definition) are equal. Therefore on a given transversal LAB , the segments intercepted on LAB by adjacent parallel lines will be all of the same length (see Exercise 11 on page 238). It follows that if a segment on LAB has its endpoints on two of the lines on a piece of lined paper (such as AB below), then any one of the parallel lines in between will divide the segment into two parts whose lengths can be instantly read off by counting the number of parallel lines. To be explicit, consider the following situation: A \ \ \ P q \qQ \ D q \qE \ \ \ \ B C If the length of the segment on AB trapped between two adjacent parallel lines is s, then |AD| = 3s and |AB| = 5s. Because the lines LBC and LDE are parallel,

5.1. THE FUNDAMENTAL THEOREM OF SIMILARITY

259

Theorem G11 predicts that (5.3)

|AE| |DE| |AD| = = . |AB| |AC| |BC|

The first equality is not in doubt because if the length of the segment on AC trapped between two adjacent parallel lines is t, then we have |AE| = 3t and |AC| = 5t for the same reason. Therefore, |AE| 3 |AD| = = . |AB| |AC| 5 What remains uncertain is the prediction in the second equality in (5.3): |DE| 3 = . |BC| 5

(5.4)

Of course Theorem G11 has already been proved in general so that, in theory, (5.4) has to be true. But here we are talking about the visceral feeling of conviction. Some can probably get it from the proof itself, but others may need something extra, beyond the proof itself. This is where this Activity comes in: never mind the theory, but direct measurements of |DE| and |BC| will actually confirm the validity of (5.4). This reinforces one’s faith in Theorem G11. Similar considerations apply to the segments AP and AQ in the picture above so that |AP | |AQ| 2 = = . |AB| |AC| 5 Theorem G11 now predicts that (5.5)

|P Q| 2 = . |BC| 5

Again, it should be a satisfying experience for students to verify the prediction (5.5) by directly measuring |BC| and |P Q|. In a school classroom, such an experiment, when repeated for many different variations of this configuration, should provide an excellent opportunity for students to build up their intuition about Theorem G11 which, as we noted, is equivalent to FTS. Two characterizations of a parallelogram The proof of FTS for the special case of r = 12 requires the ability to recognize a parallelogram when we see one. This subsection takes a step in that direction. First, a general observation. Theorem G12. Let O be a point on a line L, and let  be the rotation of 180◦ around O. Then  maps each half-plane of L to its opposite half-plane. Proof. Let the half-planes of L be H+ and H− . The theorem says (H− ) = H+

and (H+ ) = H− .

It suffices to prove the first assertion; i.e., (H− ) = H+ . As usual, this means proving both (H− ) ⊂ H+ and H+ ⊂ (H− ). Let us first prove (H− ) ⊂ H+ . So

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5. DILATION AND SIMILARITY

let P be a point in H− , and let Q = (P ). By the definition of a 180◦ rotation, P , O, and Q are collinear and O, being the midpoint of the segment P Q, lies in P Q. q  Q      O     H− P q H+

L

Now the segment P Q contains a point of L, namely, O, so P and Q are in opposite half-planes of L (see (L4) on page 176). Since P ∈ H− , we have Q ∈ H+ ; i.e., (P ) ∈ H+ , as claimed. Next we need to prove that H+ ⊂ (H− ). Thus given Q ∈ H+ , we must show that there is a point P ∈ H− so that (P ) = Q. We simply let P = (Q). Since obviously  ◦  is the identity transformation, we see, by applying  to both sides of the equation P = (Q), that (P ) = ( ◦ )(Q) = Q. Since O lies in P Q (for the same reason as before) and O ∈ L, we see that P and Q have to be in opposite half-planes of L. Thus P ∈ H− , and the proof is complete. The next two theorems give characterizations of a parallelogram that will prove to be useful. Two segments AB and CD are said to bisect each other if they intersect at a point which is the midpoint of both segments. With this understood, the first theorem says that parallelograms are the quadrilaterals whose diagonals bisect each other (recall that a diagonal of a polygon is a segment; see page 171). Theorem G13. A quadrilateral is a parallelogram ⇐⇒ its diagonals bisect each other. A "D @ "  "   @ "   " @M "   " @  "" @   " @ B C Proof. We will give a proof similar to the one outlined in Exercise 4 on page 228. It is well to point out that, at the outset, the two diagonal segments AC and BD are not known to intersect each other (see the Pedagogical Comments after the proof), much less bisect each other. Let M be the midpoint of the diagonal AC, and let  be the rotation of 180◦ around M . In the proof of Theorem G4 (see (4.7) on page 227), we proved that (B) = D. Since  is a 180◦ rotation, the points B, M , and D are collinear, and since  is an isometry and (M B) = M D, M is also the midpoint of the diagonal BD. Thus AC and BD bisect each other. Next, we look at the converse. Suppose a quadrilateral ABCD (the definition of a quadrilateral is given on p. 171) has the property that its diagonals AC and BD meet at M and M is the midpoint of both AC and BD. We will prove that ABCD is a parallelogram. As usual, let  be the rotation of 180◦ around M . Then

5.1. THE FUNDAMENTAL THEOREM OF SIMILARITY

261

(B) = D and (C) = A. Thus (LBC ) = LAD because, by (L7)(i),  maps lines to lines (page 237) and there is only one line passing through A and D (by (L1) on page 165). By Theorem G1 (page 220), LBC  LAD . In the same way, we can prove LAB  LCD . This proves that ABCD is a parallelogram, and the proof of Theorem G13 is complete. Pedagogical Comments. The preceding proof shows indirectly that the diagonals AC and BD of a parallelogram meet at a point M . The first comment is that a direct proof of this fact can be given using the crossbar axiom (page 250), but it is tedious. A second comment is that few high school students will ever conceive of the need for such a proof. In a beginning geometry class, one should probably take something this obvious for granted. Once the formal proof is over, however, a teacher may want to point out that this visually obvious fact actually requires a proof because a quadrilateral ABCD could look like this: D @ @ H @  B HH@  H@ H@  HC A  and the two diagonal segments AC and BD do not intersect. The question then becomes why this cannot happen to the diagonals of a parallelogram: which property exactly about a parallelogram makes its diagonals intersect? If we can get students to be curious about the answer to this question, then (and perhaps only then) would they find such a proof to be meaningful. End of Pedagogical Comments. Theorem G14. A quadrilateral is a parallelogram ⇐⇒ it has one pair of sides which are equal and parallel.

Proof. The fact that a parallelogram has a pair of sides which are equal and parallel is implied by Theorem G4, page 226. We prove the converse. Let ABCD be a quadrilateral so that |AD| = |BC| and AD  BC. We have to prove that ABCD is a parallelogram. It suffices to prove that AB  CD. Let  be the rotation of 180 degrees around the midpoint M of the diagonal AC.

B





A B

B

Bq

M B

B

B C







D

As usual, (A) = C by the definition of , and we also have (LAD )  LAD , by Theorem G1, page 220. Therefore (LAD ) is a line passing through C and parallel to LAD itself. Since LBC is also a line passing through C and parallel to LAD , the

262

5. DILATION AND SIMILARITY

parallel postulate (page 165) implies that (LAD ) = LBC . In particular, (D) lies in LBC . Denote (D) by D , so D ∈ LBC . We are going to show that D = B. To this end, observe that on the line LBC , D either lies in the ray RCB or in the opposite ray so that B ∗ C ∗ D , as shown: B

C (A)

D

We want to show that the second alternative is impossible. Indeed, if D lies on the opposite ray of RCB as in the drawing above, then the segment BD contains a point C of LAC and therefore B and D lie in opposite half-planes of LAC . Since B and D also lie in opposite half-planes of LAC , D and D must lie in the same half-plane of LAC . In other words, D and (D) lie in the same half-plane of LAC . This contradicts Theorem G12 on page 259. Therefore D lies in the ray RCB . Since  is an isometry, |(AD)| = |AD|. Since (AD) = CD , we have |CD | = |AD|. But |AD| = |CB| by hypothesis, so |CD | = |CB|. Hence B and D , being two points at the same distance from C and lying on the same ray RCB , must coincide (see (L5)(ii) on page 184). Thus D = B; i.e., (D) = B. Coupled with the fact that (C) = A, this shows (CD) = AB, and therefore, (LCD ) = LAB . But according to Theorem G1, (LCD )  LCD . Hence LAB  LCD , as desired. Remark. In the preceding proof, the fact that B and D lie in opposite halfplanes of the diagonal line LAC was taken for granted and this fact allowed us to conclude that B = (D). But as we have seen from the picture on page 261, the vertices B and D of a quadrilateral ABCD may very well lie on the same side of the diagonal line LAC . In that case, (D) and B would be in opposite half-planes and the preceding proof of Theorem G14 would fall apart. Therefore this proof of Theorem G14 tacitly assumes that the following assertion holds: (♣) If two sides AD and BC of a quadrilateral ABCD are parallel, then the vertices B and D lie on opposite sides of the diagonal line LAC . We now supply a proof of (♣). Suppose D @  @ of LAC , as shown. We will deduce  HH@@  a contradiction by showing that this B HH@  H@  implies the segments AB and CD must HC A  intersect. B and D lie on the same side

To this end, we will prove the following two steps. Step I. The ray RAB intersects the segment CD. Step II. A and B lie on opposite sides of the line LCD .

5.1. THE FUNDAMENTAL THEOREM OF SIMILARITY

263

To prove Step I, we claim that B lies in the convex angle ∠DAC (see page 182 for the definition of a convex angle). This means we have to show both of the following: (i) B and D lie on the same side of LAC and (ii) B and C lie on the same side of LAD . Now (i) is true because this is our working hypothesis. To prove (ii), observe that since LAD  LBC , the segment BC cannot intersect LAD . By assumption (L4)(ii) on page 176, B and C lie on the same side of LAD , as desired. Thus B lies in ∠DAC. The crossbar axiom (page 250) now implies that RAB intersects CD, and Step I is proved. Next, to prove Step II, assume that A and B lie on the same side of LCD and we will show that this is impossible. We claim that B lies in the convex angle ∠DCA. As usual, this means we have to prove both of the following: (a) B and D lie on the same side of LAC and (b) B and A lie on the same side of LDC . There is no need to prove (a) because it is our overall working hypothesis (see (i) above), and (b) is true because we are assuming A and B lie on the same side of LCD . Thus the claim is true that B lies in ∠DCA. By the crossbar axiom, RCB intersects AD, and this contradicts the fact that LBC  LAD . The proof of Step II is complete. We can now deduce the contradiction we are after, namely, that if B and D lie on the same side of LAC , then the segments AB and CD must intersect. Indeed, Step I shows that the line LAB intersects the segment CD at a point X, and Step II shows that the line LCD intersects the segment AB at a point Y . Now X and Y both lie on LAB and LCD , and since two distinct lines intersect at exactly one point, X = Y . But X lies on CD and Y lies on AB, therefore the segments CD and AB intersect at X (= Y ). This contradicts the fact that, ABCD being a quadrilateral, only its adjacent sides can intersect (see the definition of a polygon on page 171). So B and D lie on opposite sides of LAC after all, and we have proved (♣). Pedagogical Comments. The preceding proof of (♣)—which one must remember is part of the proof of the simple and intuitive Theorem G14—together with the Pedagogical Comments on page 261, reveals one of the less attractive features of plane geometry, namely, the fact that there are many intuitively obvious geometric details that are quite subtle and whose proofs are tedious. For the case at hand, it is most unlikely that an average high school student would be interested in seeing a proof of something as obvious as B and D being in opposite half-planes of LAC when AD  BC. A reasonable way to proceed would be for a teacher to announce at the outset that the pictorially plausible fact that B and D lie in opposite half-planes of the diagonal AC will be assumed without proof. Then one can proceed as in the above proof of Theorem G14 by concentrating on the idea of using the rotation  to complete the proof by contradiction. However, in order to win your students’ trust, you will have to be ready to produce at least an outline of the proof of (♣) when some of them ask for it. This is part of the reason that, for your education as a teacher, we will continue to supply complete proofs. End of Pedagogical Comments. Proof of FTS when r =

1 2

We can now prove the special case of FTS (page 256) when r = 12 : Theorem G15. Let ABC be given, and let D and E be midpoints of AB and AC, respectively. Then DE  BC and |BC| = 2|DE|.

264

5. DILATION AND SIMILARITY

A
C
C
C CE D

C
C
C CC B
Analysis. Let us see how we can prove such a theorem. The situation is this: we know (L1)–(L8) and Theorems G1–G11, and we are confronted with the statement of Theorem G15. The question is how we can prove (among other things) |BC| = 2|DE|. This is awkward, because we know how to prove two segments have the same length—find a congruence that carries one segment to the other—but not one segment being equal to twice another. One way out of this predicament is to look for a segment that has twice the length of DE and then we can try to prove that this segment has the same length as BC. In this light, extending the segment DE to F so that DF has twice the length of DE (i.e., |DF | = 2|DE|) would be a very natural move, as shown: A
C
C
C CE D
F

C
C

C
C
C B
It would be equally natural at this point to connect C to F by a line segment and, once this is done, we see that if we can prove the quadrilateral DBCF is a parallelogram, then Theorem G4 on page 226 would immediately yield the desired conclusion that |BC| = |DF |. So once we get to the "augmented figure" with the additional line segments EF and CF added to the original figure, we see a clear path to our goal, the proof of the theorem. The line segments EF and CF that are added to the original picture of ABC together with the segment DE are called auxiliary lines in the school education literature. TSM makes a big deal out of "adding auxiliary lines" as a kind of magical tool for learning how to prove theorems, but there is in fact nothing "magical" about these "add-ons". Think of a theorem as an edifice; then the analog of proving a theorem is finding ways to build a given edifice. Of course when one shows off an edifice, one first takes down all the scaffolding and removes all traces of the construction process. If we are serious about building the edifice, however, we must first mentally remove the pristine picture of the edifice and put back the scaffolding and begin thinking about the messy construction process itself. Likewise, when a textbook presents a theorem, the textbook will only give the finished product—the geometric figure that goes with the theorem—without including the messy details of the thinking process that may have gone into the proof of the theorem. If we want to learn to prove the theorem ourselves, we cannot be limited by the pristine figure attached to the theorem but must put back some of the lines or circles (the

5.1. THE FUNDAMENTAL THEOREM OF SIMILARITY

265

"scaffolding") that are integral to the proof itself. So the "add-ons", such as the segments EF and CF , are neither random nor magical, but are things that come up naturally when we try to look for ways to better understand the "construction process". In the above picture, we have chosen to extend DE along the ray RDE , but the proof does not change if we extend instead along the opposite ray RED . (See Exercise 5 on page 267.) Let us continue with our attempt at arriving at a proof of Theorem G15. Referring specifically to the preceding picture now, we wish to prove |DF | = |BC|. But how do we prove DBCF is a parallelogram? At this point, we realize that our repertoire in this regard is extremely limited: we only have Theorems G13 and G14 for this purpose. The latter prompts us to try to show that BD  CF and |BD| = |CF |. By hypothesis, |AD| = |DB|, so our focus shifts to proving AD  CF and |AD| = |CF |. Such would be the case if DCF A is a parallelogram. Since AC and DF bisect each other, Theorem G13 gives us exactly what we need. Now, the whole proof comes together. One observation of the above analysis is worth mentioning. We see that the reasoning process is built on a solid knowledge base: students who do not have Theorems G1, G4, G13, and G14 at their fingertips will be handicapped in trying to prove this theorem. (We are not saying that memorizing Theorems G1, G4, G13, and G14 will be all it takes to prove Theorem G15, but, rather, that having easy access to these theorems is a sine qua non for the task.) What we have is therefore a simple illustration of the fact that doing mathematics requires a solid memory bank of basic facts. Do not listen to anyone telling you that "conceptual understanding"—but no memorization—is all it takes to do mathematics.

Proof of Theorem G15. On the ray RDE , we take a point F so that |DE| = |EF |. Join CF . Now recall that |AE| = |EC| by hypothesis. Therefore ADCF is a parallelogram, by Theorem G13, and we see that CF  AD, or in other words, CF  BD. Moreover, |CF | = |DA| by Theorem G4, and hence |CF | = |BD|

A
@ C
C @
C @ CE @ F D

@
C

@ C
@ C
@C
C B


because |BD| = |DA| by hypothesis.

Theorem G14 now implies that the quadrilateral DBCF , having a pair of sides which are equal and parallel, is a parallelogram. Thus DF  BC, which is the same as DE  BC. Furthermore, |BC| = |DF | (Theorem G4 on page 226), and since |DE| = |EF |, we have |BC| = 2|DE|. The proof is complete. Theorem G15 has a surprising consequence: if ABCD is any quadrilateral, then the quadrilateral obtained by joining the midpoints of all four pairs of adjacent sides in ABCD is always a parallelogram. (See Exercise 6 on page 267).

266

5. DILATION AND SIMILARITY

Because of the importance of Theorem G15 in our work, we will give it a second proof using translations. The strategy is to first prove the following Theorem G15* directly and then use it to prove Theorem G15. Theorem G15*. Let ABC be given and let D be the midpoint of AB. Suppose a line parallel to BC passing through D intersects AC at E. Then E is the midpoint of AC and 2|DE| = |BC|. A
e
e D
eE
e e
e e e eC B
F Proof. Observe first of all that the pictorially obvious fact is correct, namely, that the point E lies on the segment AC; this is because of Lemma 4.8 on page 178. −−→ Now let T denote the translation along the vector AD. Because |AD| = |DB| by hypothesis, the definition of T implies that T (D) = B. Since T maps any line  not equal to or parallel to LAD to a line parallel to  itself (Theorem G5 on page 236), T (LDE ) is a line passing through B and parallel to LDE . Since LBC  LDE by hypothesis, the parallel postulate implies that T (LDE ) = LBC so that T (E) is a point on line LBC . Let T (E) = F . Note that F is a point on the segment BC on account of, once again, Lemma 4.8 on page 178 (because T (E) = F implies LEF  LAB , by Lemma 4.21 on page 234). In any case, we have T (DE) = BF , and since T is an isometry (page 236), we get (5.6)

|DE| = |BF |.

Next consider T (LAC ). Because T (A) = D and T (E) = F , it follows that T (AE) = DF and therefore T (LAC ) = LDF . Using Theorem G5 once more and the fact that a translation is an isometry, we have DF  AC

and

|AE| = |DF |.

Since DE  BC by hypothesis, DF CE is a parallelogram and therefore, (5.7)

|DE| = |F C| and |DF | = |EC|,

by Theorem G4. Coupled with equation (5.6), the first equality of (5.7) implies 2|DE| = |BC|. Finally, the equality |AE| = |DF |, together with the second equality of equation (5.7), implies that E is the midpoint of AC. The proof of Theorem G15* is complete. Proof of Theorem G15 using Theorem G15*. Using the notation and picture of Theorem G15, we draw a line L through D parallel to BC. By Theorem G15*, L passes through the midpoint E of AC and therefore DE  BC. Since Theorem G15* also says 2|DE| = |BC|, we are done.

5.1. THE FUNDAMENTAL THEOREM OF SIMILARITY

267

Exercises 5.1. (1) Let D and E be points on sides AB and AC, respectively, of ABC, so that D = A, B.. Make use of |AB| = |AD|+|DB| and |AC| = |AE|+|EC| to prove |AE| |AD| = |AB| |AC|

⇐⇒

|AD| |AE| = |DB| |EC|

⇐⇒

|DB| |EC| = . |AB| |AC|

(2) In ABC, let D, F ∈ AB and E, G ∈ AC, as shown: A
@
@ D
@E
@ F
@G
@
@
@C B Suppose DE  BC and F G  BC. Prove: (a) If |AD| = |F B|, then |AD|

|AE|

|AE| = |GC|. (b) More generally, |F B| = |GC| . (3) Assume ABC so that |AB| = |AC|. Let the angle bisector of ∠A meet BC at F . Prove that AF is the perpendicular bisector of BC. (4) Let D, E, F be the midpoints of sides AB, AC, and BC, respectively, of ABC. Prove that the four triangles ADE, DBF , DEF , EF C are congruent. A
@
@
@ D
@E
@
@
@
@
@
@
@
@C B F (5) Give a proof of Theorem G15 by following the proof in the text, but this time, extend DE along the ray RED (rather than along the opposite ray RDE ) to a point F , so that |F D| = |DE|. (6) If ABCD is any quadrilateral. Prove that the quadrilateral obtained by joining midpoints of the adjacent sides of ABCD is always a parallelogram. (7) In ABC, let D be a point on AB. Let the line passing through D and parallel to BC intersect AC at F , and let the line passing through D and parallel to AC intersect BC at E. Prove that if |DF |/|BC| = |DE|/|AC|, then D is the midpoint of AB.

268

5. DILATION AND SIMILARITY

(8) Use the idea in the proof of Theorem G15, but do not assume FTS, to prove that if in triangle ABC, D and E are points on AB and AC, respectively, so that |AB| = 3|AD| and |AC| = 3|AE|, then DE  BC and |BC| = 3|DE|. (9) Prove that FTS* (Theorem G11) implies FTS (Theorem G10). More precisely, this means: assume everything we have proved up to and including Theorem G9 plus Theorem G11, and prove Theorem G10. (10) Let a segment AC lie in a half-plane of a given line , and let B be the midpoint of AC. Let LAD , LBE , and LCF be three parallel lines which meet  at D, E, F , respectively. Prove that 2|BE| = |AD| + |CF |. 5.2. Dilation We will give the definition and prove the basic properties of dilation, the first transformation of the plane worthy of our serious attention that is not an isometry. It is also the new ingredient we need to define similarity. It will be clear from the discussion in this section why the FTS is the fundamental theorem of similarity. Definition of a dilation (p. 268) Basic properties of dilations (p. 269) Effects of dilations on lengths and degrees (p. 275) Definition of a dilation We have been considering isometries almost exclusively thus far. Now we have to look seriously into an important class of transformations that are not isometries. Definition. A transformation D of the plane is a dilation with center O and scale factor r (r > 0) if (1) D(O) = O. (2) If P = O, the point D(P ), to be denoted by P  , is the point on the ray ROP so that |OP  | = r|OP |. Oq



Pq





Pq 

r|OP | Remark. We call attention to the fact that this definition of a dilation requires the scale factor to be positive. Some authors allow the scale factor to be any real number so that a negative scale factor means a dilation in the opposite direction of O. There are pros and cons to either convention. Observe that if r = 1 in this definition, then D is the identity map and there is nothing to discuss. In the following, we will tacitly assume that r = 1. The definition of a dilation is starkly simple: a dilation with center at O maps each point by "pushing out" or "pulling in" the point along the ray from O to that point, depending on whether the scale factor r is bigger than 1 or smaller than 1, respectively. Roughly, it is a kind of "central projection from the point O". In particular, each ray issuing from O is mapped to the same ray. (Caution: All this

5.2. DILATION

269

says is that the ray is mapped onto itself, but each point on the ray will in general be mapped to another point on the same ray.) Here is an example of how a dilation with r = 2 maps four different points (for any point P , we let the corresponding letter with a prime, P  , denote the image D(P ) of P ): c U cr c

 c

r Q c c Ur

c

r c

Q c

c cs

O r V r  V

r P

r P

Basic properties of dilations A fundamental property of dilations, one that makes possible the simple drawings of the dilation of rectilinear figures (i.e., figures composed of line segments), is given in the following theorem. It will be clear from its proof and other proofs related to dilation that the FTS and the parallel postulate lie at the heart of the matter. Theorem G16. Dilations map segments to segments. More precisely, a dilation D maps a segment P Q to the segment joining D(P ) to D(Q). Moreover, if the line LP Q does not pass through the center of the dilation D, then the line LP Q is parallel to the line containing D(P ) and D(Q). Proof. Let D be a dilation with center O and scale factor r. If LP Q passes through O, then either P and Q lie on the same side of O or they lie on opposite sides of O. In either case, the fact that D maps P Q to the segment in LP Q from D(P ) to D(Q) follows immediately from the definition of a dilation. We may therefore assume that LP Q does not pass through O. Let P  = D(P ) and Q = D(Q). We will show that D(P Q) is the segment P  Q joining P  and Q P

C C

C C





U

  U



 Q C

 C  C

 O As usual, there are two steps. Step I. D(P Q) ⊂ P  Q . Step II. P  Q ⊂ D(P Q). C P C



Q    

270

5. DILATION AND SIMILARITY

To prove Step I, let LP Q be  and let LP  Q be  . We will first show D() ⊂  . Let U be any point of , and we have to show that D(U ) is on  . Let U  = D(U ); then by the definition of a dilation, U  lies on the ray ROU . Consider OP  U  . Because D maps P and U to P  and U  , respectively, |OP  | |OU  | = = r. |OP | |OU | By FTS, LP  U   LP U ; i.e., LP  U   . If we apply the same reasoning to OP  Q , then the fact that |OP  | |OQ | = =r |OP | |OQ| implies (by FTS) that LP  Q  LP Q ; i.e.,   . Thus both  and LP  U  are lines passing through P  and parallel to . By the parallel postulate (page 165), LP  U  =  , and, in particular, D(U ) (= U  ) lies on  , thereby proving D() ⊂  . It remains to prove that D(P Q) ⊂ P  Q ; i.e., if U ∈ P Q, then we must prove U  ∈ P  Q . Since U ∈ P Q, clearly U ∈ ∠P OQ. By the crossbar axiom (page 250), the ray ROU must intersect the segment P  Q at some point; let us say V . But U  is the point of intersection of LOU and LP  Q (=  ), so V coincides with U  because the two distinct lines LOU and  can intersect only at one point. Since V  ∈ P  Q , we have U  ∈ P  Q , and the proof of Step I is complete. We pause to observe that we have also proved in the process that, assuming  does not contain O, then   , or equivalently, if the line LP Q does not contain O, it is parallel to the line containing D(P ) and D(Q) (as claimed in the theorem). Next, we prove Step II; i.e., we show that P  Q ⊂ D(P Q). Let U  ∈ P  Q , and we have to show that there exists a point U ∈ P Q so that U  = D(U ). Because of the crossbar axiom, the ray ROU  intersects P Q at a point and we claim that if we denote this point by U , then in fact D(U ) = U  . To this end, recall that  (= LP Q ) is parallel to  (= LP  Q ), by FTS. We now apply FTS* (Theorem G11 on page 257) to OP  U  to obtain |OP  | |OU  | = = r. |OU | |OP | Therefore |OU  | = r|OU |. Since U  ∈ ROU , by the definition of the dilation D, we get U  = D(U ) and Step II is proved. As mentioned above, Steps I and II prove that D(P Q) = P  Q , and the proof of Theorem G16 is complete. There are two useful corollaries of this theorem. Corollary 1. A dilation maps lines to lines and rays to rays. Proof. Let the dilation be D. Given a line , we must prove that D() is also a line. Let P , Q be points on . If  passes through the center O of D, then it is easy to see that D() = . So we assume  does not pass though O. The theorem says D(P Q) is the segment P  Q , where P  = D(P ) and Q = D(Q). Let  be the line containing P  Q . We will prove D() =  . First, we prove that D() ⊂  . Thus, if U ∈ , we have to prove that D(U ) ∈  . For the proof itself, it does not matter where U is located in , and the picture below shows the case of P ∗ Q ∗ U , as shown:

5.2. DILATION

P B B B B P B

Q







 Q

U B

 B

 B  B

 O

271

U     





Let U  = D(U ), and let 0 = LP  U  . Then P  ∈ 0 and Theorem G16 implies that 0 is parallel to . Likewise, P  ∈  and   . Since both 0 and  pass through P  and are parallel to , the parallel postulate implies that  = 0 . Consequently, the point U  on 0 is now a point on  ; i.e., D(U ) ∈  , as claimed. We have proved that D() ⊂  . It remains to prove that  ⊂ D(). Let U  ∈  , and let U be the point of intersection of the ray ROU  with . Then we prove exactly as in the proof of Theorem G16 that U  = D(U ) and therefore U  ∈ D(). Thus  ⊂ D(), and D() =  . The fact that D maps rays to rays is proved in the same manner. Corollary 2. Let triangles ABC and A B  C  be given. If a dilation D maps vertices to vertices, then it also maps one triangle to the other. Precisely, if D(A) = A , D(B) = B  , and D(C) = C  , then D(ABC) = A B  C  . Proof. Recall that ABC is the union of the segments AB, BC, and AC. Thus we must prove D(AB) = A B  , D(BC) = B  C  , and D(AC) = A C  . By Theorem G16, D(LAB ) is a line joining A and B  because D(A) = A , D(B) = B  . Since assumption (L1) implies that there is a unique line joining two points, we see that D(LAB ) = LA B  , from which D(AB) = A B  immediately follows. The same is true for D(BC) = B  C  and D(AC) = A C  . Corollary 2 is proved. Armed with Theorem G16, we see that it is very simple to draw the image of a segment by a dilation. Indeed, to draw the image of a given segment AB by a dilation D, we simply find the image points A and B  of the endpoints A and B under D and then draw the segment joining A to B  . Here are two examples. In the first, the original figure is a triangle ABC, and the scale factor is 2.5. A H HH   HHC  





 A 

HHC

 





 



O B B

272

5. DILATION AND SIMILARITY

Here A = D(A), B  = D(B), and C  = D(C). In the next example, we specify a scale factor of 2.1; the original figure S is a quadrilateral ABCE, and D(S) is the enlarged quadrilateral to the right:

Q D(S)  

Q Q Q J

J 

J  E A

 J Q 

J Q J C

J J

 J J O B You are encouraged to make many such drawings of dilated rectilinear figures. You may also have noticed that the dilation of a rectilinear figure "has the same shape" as the original figure. But what about the dilation of curved figures? There is no simple replacement of Theorem G16 in that case, but in practical terms (in a sense to be made precise below), the procedure of getting a dilated figure is only slightly more complicated. Consider for example the following curve:

Let us shrink it by a scale factor of 12 . Here is what we do: we pick an arbitrary point O as the center of dilation, and then pick 10 judiciously chosen points on the curve, as shown below. For convenience, we shall refer to the chosen points on the original geometric figure as data points. r r r

O

r

r

r r

r

r

r

q

Now we draw the rays from O to each of the data points on the curve and dilate the latter by a scale factor of 12 (i.e., "shrink it by half" in everyday language) along these rays and ignore the curve itself. We thereby obtain a collection of points, and these will be points on the dilated curve. It should not be difficult to discern, just from these 10 dilated points, the general shape of the dilated curve.

5.2. DILATION

O

273

, r ,  " , ,  "  ,r""  ,r"  !! " ! r   , "r !!  , r r r " !!   " !  r ,   r  r " !   r( , r ! "  r r ! ((((  ( ,  (  r " ! (    r  !  r r ((( " r ,   !     " ! ((((  (   , ( !    (  "  ((   ( " ,    q(!   !

In the preceding picture, we used 10 data points to demonstrate how to carry out the approximate dilation of the given curved figure; namely, we simply dilate these points one by one. The dilated points then give a suggestion of what the dilated curve would look like. If we delete the rays, we may see better the data points and their dilated images. r r r rr rr

O

r

r

rr r r r

r

r r

r

r

r

q

It is obvious that the more data points we choose on the original curve, the better we will be able to approximate the dilated curve. To drive home this idea, let us use 40 data points instead of 10 on the original curve and dilate them from O to get the following picture. (We omit the rays issuing from O in the interest of visual clarity.)

O

q

274

5. DILATION AND SIMILARITY

Next we double the number of data points and use 80 instead of 40. The approximation of this finite collection of points to the curve itself is already remarkably good.

O

q

If we use 400 points, then the images can almost pass for the real thing, except that if we look very carefully we can see that they are not entirely "smooth".

O

q

What we have described is a basic principle of constructing the dilated image of any object: to dilate a given object by a scale factor of r, replace the object by a finite collection of judiciously chosen points on the object, still to be called data points, and then simply dilate these data points one by one by a scale factor of r. By increasing the number of data points, their dilated images yield a closer and closer approximation to the true dilated object.2 A few experiments with this kind of drawing (see Exercises 5 and 6 on page 281) should suffice to convey the idea that the dilated object so obtained "has the same shape" as the original but is magnified or shrunk by a scale factor of r (depending on whether r > 1 or r < 1). This is how we can enlarge or shrink arbitrary figures regardless of how curved they may be. It would be very instructive in the school classroom for students to magnify or shrink many simple curved figures by such hands-on activities. We will elaborate on these ideas in the next section. Incidentally, what we have described is also the basic operating principle of digital photography: approximate any real object by a large number of data points on the object, and then magnify or shrink these data points by dilation.

2 Mathematical Aside: In computer graphics, one decides on a finite number of data points to use for the purpose of dilation and then uses spline interpolation to complete the dilated image.

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Effect of dilations on lengths and degrees The following theorem summarizes what a dilation does to lengths and degrees.

Theorem G17. Let D be a dilation with center O and scale factor r. Then: (a) D is a bijection. In fact, its inverse is the dilation with the same center O but with a scale factor 1/r. (b) For any segment AB, |D(AB)| = r|AB|. (c) D maps angles to angles and preserves degrees of angles. Remark. Observe the delicate point that the statements of part (b) and part (c) depend on the validity of Theorem G16 and its Corollary 1. Indeed, without knowing that D(AB) is a segment, the notation |D(AB)| would not even make sense (because the notation |σ| only makes sense when σ is a segment or an angle), and without knowing that D maps rays to rays, it would not make sense to say that D maps angles to angles. Proofs of parts (a) and (b). (a) Let D be the dilation with center O and scale factor 1r . From the definition of a dilation, it is easy to check that D ◦ D = I = D ◦ D. Thus D is a bijection (Theorem 4.15 on page 211). Part (b) has been implicitly proved in the proof of Theorem G16. Indeed, in the notation above, if P  = D(P ) and Q = D(Q), then we have shown that D(P Q) = P  Q . If LP Q contains O, then the fact that |D(P Q)| = r|P Q| follows trivially from the definition of a dilation. So suppose LP Q does not pass through O; then LP  Q also does not pass through O as the definition of D clearly shows. We may therefore look at OP  Q , and FTS implies that |P  Q |/|P Q| = r; i.e., |D(P Q)| = r|P Q|. Since P and Q are arbitrary points, (b) is proved. For the proof of part (c) of Theorem G17, we first observe that a dilation maps convex angles to convex angles. Indeed, a convex angle is the intersection of closed half-planes, so it suffices to prove that (i) a dilation maps closed half-planes to closed half-planes, (ii) if H+ and H− are closed half-planes of a line L, then D(H+ ) and D(H− ) are the closed half-planes of the line D(L), and (iii) if H1 and H2 are closed half-planes (of possibly different lines), then D(H1 ∩ H2 ) = D(H1 ) ∩ D(H2 ). Because the proofs of (i)–(iii) are as tedious as they are straightforward, we will leave their verifications to the reader. Now once we know a dilation maps convex angles to convex angles, then it also maps nonconvex angles to nonconvex angles because every nonconvex angle is the union of the complement (in the plane) of a convex angle together with the two sides of the convex angle. To complete the proof of part (c), we have to make a long digression to discuss the angles associated with parallel lines. The proof of part (c) itself will be given on page 280. We will need some definitions. Here is the first one. Let L and L be two lines meeting at a point O. On L (respectively, L ), let P , Q (respectively, P  , Q ) be

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points so that P ∗ O ∗ Q (respectively, P  ∗ O ∗ Q ), as shown in the following figure:  L X Qq  XXXPq   XXX O   X X  XXXX  XqXX q  X   P Q L Then the angles ∠P OP  and ∠QOQ are called opposite angles at the point O.3 There is a simple observation about opposite angles. Lemma 5.1. Opposite angles at a point are equal. Proof of Lemma 5.1. We make use of the preceding figure. Each of the two numbers, |∠P OP  | and |∠QOQ |, when added to |∠P  OQ| is 180 because ∠P OQ and ∠P  OQ are both straight angles. So |∠P OP  | = |∠QOQ |. We want to give a different proof for the purpose of demonstrating how to make use of basic isometries to prove theorems. In this case, we argue as follows. Consider the rotation  of 180◦ around O. Clearly (ROP ) = ROQ and (ROP  ) = ROQ . Therefore (∠P OP  ) = ∠QOQ . Since  preserves angles (by assumption (2) of rotations on p. 217), we have |∠P OP  | = |∠QOQ |. The proof of the lemma is complete. Let two distinct lines L1 , L2 be given. Recall that a transversal of L1 and L2 is any line  distinct from L1 and L2 that intersects both. Suppose  meets L1 and L2 at P1 and P2 , respectively. Let Q1 , Q2 be points on L1 and L2 , respectively, so that they lie in opposite half-planes of . Then ∠Q1 P1 P2 and ∠Q2 P2 P1 are said to be alternate interior angles4 of the transversal  with respect to L1 and L2 .



EE E qS E R ( E ((((2q(( L2 Q2 (((( EP ((q( E 2 E E E E q E L1 P1 E Q1 E An angle which is the opposite angle of one of a pair of alternate interior angles is said to be the corresponding angle of the other angle. For example, let S be a point on the transversal  so that P1 ∗ P2 ∗ S, and let R2 be a point on L2 so that Q2 ∗ P2 ∗ R2 . Then, because ∠Q1 P1 P2 and ∠Q2 P2 P1 are alternate interior angles 3 Thus also ∠P OQ and ∠P  OQ are opposite angles at O. Most school textbooks in the U.S. call these vertical angles. 4 My colleague Ole Hald correctly suggests that opposite interior angles might be a better terminology because they lie in opposite half-planes.

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and because ∠R2 P2 S is the opposite angle of ∠Q2 P2 P1 , ∠R2 P2 S is the corresponding angle of ∠Q1 P1 P2 . Observe that, by the definition of opposite angles, ∠Q2 P2 P1 and ∠R2 P2 S lie on opposite closed half-planes of . Therefore the corresponding angles ∠Q1 P1 P2 and ∠R2 P2 S lie on the same closed half-plane of the transversal . Pedagogical Comments. In the school classroom, we suggest that alternate interior angles be defined simply by drawing a picture as above and pointing to ∠Q1 P1 P2 and ∠P1 P2 Q2 . The correct definition (the one just given), using the precise concept of the half-planes of a line, deserves to be pointed out to open students’ minds to the potential of complete logical precision but perhaps should not be emphasized at this point of the curriculum. Most school students do not take kindly to the need for such precision in the proofs of theorems at the beginning of their excursions in geometry (as in the proof of Theorem G18); they would likely consider the investment of so much effort in something so visibly obvious to be ridiculous. So some compromise in the school classroom may be necessary. See the Pedagogical Comments on page 263. Nevertheless, we are obliged to give such a precise presentation of geometry because mathematics demands no less. There will be a real need for such precision later on, such as in the proofs of several theorems, including the one on the angle sum of a triangle (see Theorem G32, Section 6.5 of [Wu2020b]). End of Pedagogical Comments. The basic theorem about parallel lines and angles is the following: Theorem G18. Alternate interior angles of a transversal with respect to a pair of parallel lines are equal. The same is true of corresponding angles.

 Q2 q

E E E P2 E E Er M E E P1 E E E

L2

q Q1

L1

Proof of Theorem G18. We will give two proofs of this theorem. First, let L1  L2 , and let  be the transversal that meets L1 at P1 and L2 at P2 . Let Q1 be a point on L1 distinct from P1 , and let Q2 be a point of L2 so that Q1 and Q2 lie in opposite half-planes of . We will prove that the alternate interior angles ∠Q1 P1 P2 and ∠Q2 P2 P1 are equal. Let M be the midpoint of P1 P2 , and let  be the rotation of 180 degrees around M . Because (P1 ) = P2 , (L1 ) is a line passing through P2 . By Theorem G1 (page 220), (L1 )  L1 , and the hypothesis says L2 is also a line passing through P2 and parallel to L1 . Hence the parallel postulate (page 165) says (L1 ) = L2 . In particular, (Q1 ) lies on L2 and by Theorem G12 on page 259, (Q1 ) and Q1 lie on opposite half-planes of the transversal . Without loss of generality, we may let Q2 = (Q1 ) so that (RP1 Q1 ) = RP2 Q2 . Now, we also have (RP1 P2 ) = RP2 P1 . Hence, (∠Q1 P1 P2 ) = ∠Q2 P2 P1 . Since  preserves degrees of angles, |∠Q1 P1 P2 | = |∠Q2 P2 P1 |, as desired. The last assertion of the

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theorem about corresponding angles follows from the equality of opposite angles at a point (Lemma 5.1). Now we give a second proof. Let L1  L2 with transversal  intersecting L1 and L2 at P1 and P2 , respectively, as before. Also let Q1 , Q2 be as before. Choose a point R2 on L2 so that Q2 ∗ P2 ∗ R2 , and choose a point S on  so that P1 ∗ P2 ∗ S, as shown:

E E

Q2

qS E E E E E P2 E E E E E P1 E E

R2 EE E

L2 E E Q1

L1

E

Since Q2 ∗ P2 ∗ R2 is equivalent to Q2 and R2 being two points on L2 lying on opposite sides of  and since Q2 and Q1 were chosen to be on opposite sides of , we see that R2 may be characterized as a point on L2 that lies on the same side of  as Q1 . By the definition of opposite angles at a point (see page 276), ∠R2 P2 S and ∠Q2 P2 P1 are opposite angles at P2 . It follows that ∠Q1 P1 P2 and ∠R2 P2 S are corresponding angles of the transversal , and we will prove that they are equal. −−−→ Let T be the translation along the vector P1 P2 . Then T (P1 ) = P2 . By Theorem G5 (page 236), the translation T maps L1 to a line parallel to itself. So T (L1 ) is a line passing through P2 and parallel to L1 . But by hypothesis, L2 is also a line with the same properties. The parallel postulate therefore implies that T (L1 ) = L2 . Now let W denote T (Q1 ); then W ∈ L2 . We claim that W is on the same side of  as −−−→ Q1 . To see this, observe that Lemma 4.21 on page 234 implies that Q1 W  . In particular, the segment Q1 W is disjoint from  and therefore W and Q1 lie on the same side of  (see (L4)(ii) on page 176), thereby proving the claim. It follows that W is a point on L2 that lies on the same side of  as Q1 . By the characterization of R2 in the preceding paragraph, there is no loss of generality if we let R2 = W . Thus we have T (Q1 ) = R2 . Next, let us turn to S, and we want to show that we may likewise let S = T (P2 ). To this end, let U = T (P2 ) and we claim that P1 ∗ P2 ∗ U . Indeed, U has to lie in either half-line of  determined by P2 . Suppose U lies in the ray RP2 P1 . Then clearly RP2 U = RP2 P1 . Since U = T (P2 ), the definition of a translation (page 234) −−→ −−−→ implies that P2 U points in the same direction as P1 P2 , and this means that there  is a line  not parallel to  so that one of its closed half-planes contains both rays, RP1 P2 and RP2 P1 (= RP2 U ). Since the union of RP1 P2 and RP2 P1 is the line , there is no such  . Therefore U must lie in the opposite ray of RP2 P1 on . Consequently, P1 ∗ P2 ∗ U by Lemma 4.6(i) on page 174, and the claim is proved. It follows that both U and S are points on  lying on the opposite side of P1 on  with respect to P2 . Thus we may let S = U without any loss of generality; i.e., we may let

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S = T (P2 ). Now, we have T (∠Q1 P1 P2 ) = ∠R2 P2 S. Since T preserves degrees of angles, |∠Q1 P1 P2 | = |∠R2 P2 S|, and Theorem G18 is proved for the case of corresponding angles. The case of alternate interior angles follows from Lemma 5.1 on page 276. Pedagogical Comments. Because the second proof includes all the technical details, it masks the very intuitive underlying argument. In a high school classroom, we can present the same proof in the following way, with an explicit caution to the class that we will rely on the picture itself to justify whether or not two points lie on the same side or opposite sides of  or L2 : Let L1  L2 , and let  intersect L1 and L2 at P1 and P2 , respectively, as before. Also let Q1 , Q2 be as before. Let T be −−−→ the translation along the vector P1 P2 . Then T (P1 ) = P2 . By Theorem G5 (page 236), the translation T maps L1 to a line parallel to itself. So T (L1 ) is a line passing through P2 and parallel to L1 . But by hypothesis, L2 is also a line with the same properties. The parallel postulate therefore implies that T (L1 ) = L2 . So T maps Q1 to a point R2 on L2 , and it maps P2 to a point S on , as shown in the preceding picture. Now, we have T (∠Q1 P1 P2 ) = ∠R2 P2 S. Since T preserves degrees of angles, |∠Q1 P1 P2 | = |∠R2 P2 S|, and Theorem G18 is proved for the case of corresponding angles. The case of alternate interior angles follows from Lemma 5.1 on page 276. Such a proof would be far more instructive to a beginning student in geometry than the (completely correct) formal proof above. End of Pedagogical Comments. Remark. Readers who are familiar with some high school geometry may be tempted at this point to immediately use Theorem G18 to prove the well-known fact that the sum of (the degrees of) angles in a triangle is 180◦ . The argument goes as follows. Given ABC, extend the ray RBA to a point D; through A draw a ray RAE that is parallel to BC so that E lies in ∠CAD, as shown: D A  JJ J B

E

J J

C

By Theorem G18, |∠DAE| = |∠B| and |∠CAE| = |∠C|, so that |∠BAC| + |∠B| + |∠C| = |∠BAC| + |∠DAE| + |∠CAE| = 180◦ . This would seem to finish the proof. Let us affirm that this intuitive argument is indeed how a high school student should remember why the angle sum of a triangle is 180. For a teacher to really come to grips with the delicate points about Euclidean geometry, however, it is necessary to point out that for Theorem G18 to be applicable, we must first prove that ∠C and ∠EAC are alternate interior angles and ∠B and ∠DAE are corresponding angles. See the Pedagogical Comments both

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before and after Theorem G18. In Section 6.5 of [Wu2020b], we will present a proof of the angle sum theorem with such details filled in. We can finally give the Proof of part (c) of Theorem G17. Let O be the center of the dilation D. Since D maps rays to rays, it maps angles to angles. Given ∠P QR, let D map the ray RQP to the ray RQ P  and let it map the ray RQR to the ray RQ R , so that D(∠P QR) = ∠P  Q R . We have to prove that |∠P QR| = |∠P  Q R |. If one of P , Q, and R is equal to O, the argument is simpler (this will be evident below). So we may assume that none of P , Q, and R is equal to O. Furthermore, suppose (let us say) O, P , Q are collinear. Then by the definition of a dilation, P  and Q will also lie on the line containing O, P , Q and Q R  QR (Theorem G16 on page 269). Then the fact that |∠P QR| = |∠P  Q R | follows directly from Theorem G18 (the case of corresponding angles).  R       R      r  r r r Q Q O P

r P

In general, we have a situation depicted by the following picture:

Q

 P      P         Q R B     

R

Without loss of generality, we may assume that neither angle is the zero angle (see (L6) part (ii), page 188). We claim that LQ P  must intersect LQR . If not, then LQ P   LQR . But we already know from part (b) that LQ R  LQR . Thus we have two distinct lines LQ P  and LQ R passing through Q and parallel to LQR , and this contradicts the parallel postulate (page 165). Thus LQ P  intersects LQR at a point B, as shown. By Theorem G16 (page 269) and its Corollary 1 (page 270), LQR  LQ R and LQP  LQ P  . Therefore, according to Theorem G18 about corresponding angles, (notation as in the preceding figure) |∠P QR| = |∠P  BR| = |∠P  Q R |, as desired. The proof of Theorem G17 is complete. In the preceding proof, it is simply asserted that certain angles are corresponding angles without any proof; this is because the details are similar to those in the second proof of Theorem G18 (see especially the Pedagogical Comments on page 279). Going through such uninspiring arguments once is quite enough and we will

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continue to skip such arguments in the future. The following converse of Theorem G18 will also be useful; the proof is sufficiently straightforward to be left to Exercise 1 below. Theorem G19. If the alternate interior angles of a transversal with respect to a pair of distinct lines are equal, then the lines are parallel. The same is true of corresponding angles. To conclude this discussion of dilation, it would be pleasant to be able to report that a composite of two dilations (whose centers may be different) is also a dilation (with respect to some other center), but unfortunately such is not the case. An example will be given in Exercise 7 on page 282.

Exercises 5.2. (1) Prove Theorem G19 on page 281. (Hint: Use Theorem G18 and Lemma 4.10 on page 190.) (2) Prove: (a) The dilation of a convex set is a convex set. (b) The dilation of a polygon is a polygon. (c) The dilation of a regular polygon is a regular polygon. (3) Let ABCD and A B  C  D be two quadrilaterals. Suppose there is a point K so that the rays RKA , RKB , RKC , RKD contain A , B  , C  , D , respectively. Assume also |KB| |KC| |KD| |KA| = = = . |KA | |KB  | |KC  | |KD | Prove that if ABCD is a square, then so is A B  C  D . (Caution: Be careful about what you say and how you say it.) (4) Let O be a point not on a given circle C with center K. Let D be the dilation with center O and scale factor r. Prove that the image D(C) is a circle and that the center of D(C) is the image under D of the center of C. (Caution: This is a slippery proof. Follow the precise definitions of a circle and a dilation.) (5) Let D and E be the midpoints of AB and AC, respectively, of ABC, and let K be the midpoint of DE (see picture below). Let D be the dilation with center A and scale factor 12 . (a) If  is the rotation of 180◦ around K, describe precisely the figure D((ABC)). (b) If T is the translation along AD, describe precisely the figure T (D(ABC)). (c) How are the figures in (a) and (b) related? A
@
@
@ D
r @E
@ K
@
@ @C B


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(6) Assume a point O and the following curve in the plane:

O r

(7)

(8)

(9)

(10)

(11)

Trace both onto a piece of paper, and choose enough points on the curve so that, by dilating these points with center O and scale factor 2, the dilated points give a reasonable picture of the dilated curve with scale factor 2. (i) Let P and Q be two distinct points in the plane and let DP , DQ be two dilations with center at P , Q, respectively, and with scale factor 12 and 2, respectively. Prove that DP ◦ DQ is a translation along the vector −−→ QM , where M is the midpoint of P Q. (ii) Generalize part (i). Let DP , DQ be two dilations with centers at two distinct points P and Q and with scale factors r and s, respectively. If rs = 1, prove that there is a point X so that DP ◦ DQ is a dilation with center X. (This exercise refers to a coordinate system. See the warning on page 207.) (a) Let D be the dilation with center O and scale factor 2, and let ϕ be the congruence which is the reflection across the horizontal line corresponding to {y = 1}. Are ϕ ◦ D and D ◦ ϕ equal? (b) Repeat part (a) if ϕ is now the congruence which is the rotation of 90 degrees around the point (1, 0). (c) Repeat part (a) if ϕ is now the congruence which is the translation that sends a point (x, y) to (x + 2, y). Let ϕ be a congruence and let D be a dilation with center O and scale factor r. Prove that ϕ−1 ◦ D ◦ ϕ is a dilation; be sure to state what its center of dilation is and what its scale factor is. Let ROA , ROB , and ROC be three rays issuing from O. Let A ∈ ROA , B  ∈ ROB , and C  ∈ ROC . Suppose AB  A B  and BC  B  C  . Prove that AC  A C  . A H HH   HHC  

 

A 

HHC











O B B

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5.3. Similarity The goals of this section are to introduce a correct definition of similarity, prove two basic criteria for triangle similarity, and, as an application, prove the most famous theorem of elementary mathematics: the Pythagorean theorem. We will also prove the converse of the Pythagorean theorem. Definition of similarity, its symmetry, and its transitivity (p. 283) Two criteria for triangle similarity (p. 287) The Pythagorean theorem and its proof (p. 290)

Definition of similarity, its symmetry, and its transitivity Let S and S be two sets in the plane. How do we say correctly that they are "similar"? First and foremost, the phrase "having the same shape" lacks precision and cannot be used as a definition of similarity, contrary to what TSM tells you. Moreover, the only precise definition of similar figures that TSM offers is that of "similar polygons", and the problem with such a narrow definition is that it leaves out the consideration of the similarity of geometric figures like parabolas. We cannot afford any ambiguity about the similarity of parabolas because it will limit our understanding of the graphs of quadratic functions and conic sections (see, e.g., Sections 2.2 and 2.3 of [Wu2020b], respectively). We need a definition of similar figures that not only applies to all plane figures but also coincides with the TSM definition in the case of polygons. Now in Section 5.2, we saw that if one figure is a dilation of another, then they do appear to have the same shape. Why not just say a figure is similar to another if one is the dilation of the other? To answer this question, consider the following figures:

S

S

One can convince oneself that S is obtained from S by a dilation of scale factor 1   2 . Now rotate S clockwise by 90 degrees around the center of the circle in S to obtain S*, as shown:

S

S*

Now S* is of course congruent to S and therefore must have "the same shape" as S, but can S* be a dilation of S? Not according to Theorem G16 (page 269) because if it were, then the horizontal segment of S* would have to be parallel to the vertical segment of S.

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What this simple example shows is that it is too restrictive to define "similarity" in terms of dilations alone. One must allow for compositions with congruences as well since, intuitively, each congruence preserves both shape and size. In the preceding example, for instance, a dilation of S by a scale factor of 12 , followed by a clockwise rotation of 90 degrees yields the figure S* which still "has the same shape" as S. With this in mind, we now give the formal definition of "similarity". Definition. A figure S is said to be similar to another figure S*, in symbols, S ∼ S*, if there is a finite sequence of dilations and congruences whose composition maps S to S*. A composition of a finite sequence of congruences and dilations is called a similarity. Thus a congruence is a similarity. In this terminology, S is similar to S* if there is a similarity F so that F (S) = S*. Let F be a similarity. We claim that there is a positive number r so that if A and B are any two points in the plane and A∗ = F (A) and B ∗ = F (B), then (5.8)

|A∗ B ∗ | = r|AB|.

This is because if ϕ is a congruence, then |ϕ(A)ϕ(B)| = |AB| for any two points A and B, and if D1 is a dilation with scale factor r1 , then |D1 (A)D1 (B)| = r1 |AB| for any two points A and B. Therefore, if F is the composition of p congruences and q dilations (presumably with different centers) D1 , . . . , Dq and if the scale factor of each Di is ri , then it is straightforward to see that (recall A∗ = F (A) and B ∗ = F (B)) |A∗ B ∗ | = (r1 r2 · · · rq )|AB| for any two points A and B. Therefore (5.8) holds with r = r1 r2 · · · rq . The positive number r in (5.8) is, by definition, the scale factor of the similarity F . It follows that a similarity with scale factor 1 is an isometry and, as soon as we can show that every isometry is a congruence (Theorem G39 in Section 6.6 of [Wu2020b]), then we will know that a similarity with scale factor 1 is a congruence. At the moment, we must be careful about one aspect of the definition of S being similar to S*; namely, if we know that S is similar to S*, do we also know that S* is similar to S? This question is probably confusing at first, so let us try to make sense of it. We are used to saying "the two figures S and S* are similar" because we tend to think of S and S* as interchangeable parts so that if S is similar to S*, then, "of course", S* would also be similar to S. But the fact that S ∼ S* means there is a similarity F so that F (S) = S*. If we want to say S* ∼ S, then by definition, we must produce a similarity G so that G(S*) = S. The question then becomes whether the existence of such an F will always imply the existence of such a G. The next lemma answers this question affirmatively. (Recall that symmetric relation and transitive relation are defined on page 241.) Lemma 5.2. (i) The composition of a finite number of similarities is a similarity. (ii) Each similarity has an inverse transformation that is also a similarity.

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(iii) The similarity of two figures S ∼ S* is a symmetric relation. (iv) The similarity of two figures S ∼ S* is also a transitive relation. Proof. (i) follows immediately from the definition of similarity. For (ii), recall that congruences and dilations are bijections (Theorem G6 on page 240 and Theorem G17 on page 275) and, since a composition of bijections is a bijection, each similarity—being a composition of bijections—has an inverse transformation (see page 211). It remains to prove that this inverse transformation is also a similarity. To this end, we will look at a special case to avoid notational excesses, and it will be seen that the reasoning behind the proof of this special case is perfectly general. Suppose a similarity F is equal to a composition F = ϕ1 ◦ D1 ◦ D2 ◦ ϕ2 , where ϕ1 and ϕ2 are congruences and D1 and D2 are dilations (possibly with different centers). Let G be the composition of the following congruences and dilations: −1 −1 −1 G = ϕ−1 2 ◦ D2 ◦ D1 ◦ ϕ1 , where ϕ−1 denotes the inverse transformation of ϕ, etc. It is straightforward to check that both F ◦ G = I and G ◦ F = I, where I is the identity transformation of the plane, so that G is the inverse transformation of F . But G is a similarity because the inverse of a congruence is a congruence and the inverse of a dilation is also a dilation (Theorem G6 and Theorem G17 again). Thus G, being a composition of congruences and dilations, is a similarity, as claimed. The case of a similarity F being a composition of an arbitrary number of congruences and dilations can be proved by an essentially identical argument. We can now prove (iii); i.e., the similarity of two figures S ∼ S* is a symmetric relation with respect to S and S*. Here is the reason: by definition, S ∼ S* means there is a similarity F so that F (S) = S*. This implies F −1 (S*) = S, where as usual F −1 denotes the inverse transformation of F . But we have just seen that F −1 is a similarity, so S* ∼ S, by definition. Finally, suppose S1 ∼ S2 and S2 ∼ S3 ; then we will prove S1 ∼ S3 . This is because S1 ∼ S2 implies that there is a similarity F1 so that F1 (S1 ) = S2 , and S2 ∼ S3 implies that there is a similarity F2 so that F2 (S2 ) = S3 . Thus (F2 ◦ F1 )(S1 ) = S3 . Since F2 ◦ F1 is a similarity by part (i), we conclude that S1 ∼ S3 . This proves part (iv) and hence the lemma. Obviously, every figure is congruent to itself and hence similar to itself. Thus similarity is also a reflexive relation (see page 241). Since similarity is a reflexive, symmetric, and transitive relation, this says that similarity is an equivalence relation (see page 241). The fact that the similarity relation ∼ is both symmetric and transitive is not an abstraction for its own sake. It has substantive intuitive content. As we pointed out above, the symmetry of the relation allows us to say that "two figures S and S* are similar" without having to worry about whether it is S ∼ S* or S* ∼ S, because they are equivalent. We will freely avail ourselves of this terminology from now on. In addition, the transitivity of similarity leads to the following intuitive conclusion: Lemma 5.3. If two geometric figures are each similar to a third, then they are similar to each other.

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Proof. Suppose S1 ∼ S3 and S2 ∼ S3 . Then we have to prove that S1 ∼ S2 . By the symmetry of the ∼ relation, S2 ∼ S3 implies S3 ∼ S2 . Therefore, we now have S1 ∼ S3 and S3 ∼ S2 , so that the transitivity of the ∼ relation implies immediately that S1 ∼ S2 , thereby proving the lemma. We note explicitly that, although most of our attention will be lavished on triangles, this definition of similarity makes it possible to prove the similarity of geometric figures which do not consist of segments, or, as we say, which are not rectilinear (see page 269). For example, it follows directly from the definition that all circles are similar to each other (see Exercise 5 on page 294). More importantly, a correct definition of similarity is fundamental to the study of graphs of quadratic functions which will be taken up in Chapter 2 of [Wu2020b]. In Section 2.2 of [Wu2020b], we will prove that the graphs of all quadratic functions are similar to each other and, in fact, all parabolas are similar to each other. Mathematical Aside: (a) This concept of similarity is meaningful not only for any geometric figure in the plane but also for figures in Euclidean spaces of higher dimensions, as soon as we extend the definitions of rotations, reflections, translations, and dilations to higher dimensions. (b) Parts (i) and (ii) of Lemma 5.2 imply that the set of all similarities of the plane form a group, the group of similarities. On pp. 211, 235, and 240, we have introduced certain groups and we can now bring them together: group of translations ⊂ group of congruences ⊂ group of similarities ⊂ group of bijections. In this chain of inclusions, each group is a subgroup of the next. As soon as we prove that every isometry of the plane is a congruence (Section 6.6 of [Wu2020b]), then we will be able to replace the group of congruences above by the isometry group of the plane. It remains to point out that the definition of a similarity as a finite composition of congruences and dilations raises the specter that some similarities may require the composition of "many" congruences and dilations for their definitions. Such turns out not to be the case, as the following theorem shows. Theorem. For a transformation F of the plane, the following three conditions are equivalent: (i) F is a similarity. (ii) F is the composition of a dilation followed by a congruence. (iii) F is the composition of a congruence followed by a dilation.

This theorem implies that we could have defined a similarity to be, for example, the composition of a dilation followed by a congruence. However, the disadvantage of such a definition is that it is actually clumsy to work with; e.g., this definition makes it difficult to prove that ∼ is a symmetric and transitive relation (see remark (b) in the preceding Mathematical Aside). Since the proof of this theorem is quite

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abstract and technical, and decidedly not simple, we will relegate it to a file, A Theorem about Similarity, to be posted on the author’s homepage, https://math. berkeley.edu/~wu/. Two criteria for triangle similarity As in the case of congruence, the notation in the similarity of triangles, by tradition, is made to carry more information. We say ABC ∼ A B  C  if there is a similarity F so that F (A) = A ,

F (B) = B  ,

F (C) = C  .

In other words, ABC ∼ A B  C  means not only that there is a similarity F so that the sets F (ABC) and A B  C  are equal, but that F specifically maps A to A , B to B  , and C to C  . Theorem G 20. Given two triangles ABC and A B  C  , then ABC ∼ A B  C  , if and only if 

|∠A| = |∠A |, and

|∠B| = |∠B  |,

|∠C| = |∠C  |

|AB| |AC| |BC| =   =   . |A B  | |A C | |B C |

It is common to express the second set of equalities, i.e., the equality of the ratios of corresponding sides of the two triangles, by saying that the corresponding sides are proportional. Remark. It is in the proof of this theorem that we get to see why a similarity is defined as the composition of dilations and congruences (rather than just isometries). The reason is that we need a similarity to preserve the degrees of angles whereas an isometry is, at this point, not yet known to do that (compare Theorems G6 on page 240 and Theorem G17 on page 275). It is the property of a congruence to also preserve degrees of angles that accounts for the validity of Theorem G20. Proof. If we have ABC ∼ A B  C  , then the assertions about angles and sides follow from Theorems G6 (page 240) and G17 (page 275). For the converse, we prove something stronger: Theorem G21 (SAS for similarity). Given two triangles ABC and A B  C  , if |∠A| = |∠A | and |AC| |AB| =   ,   |A B | |A C | then ABC ∼ A B  C  . Proof. The idea of the proof is to use a congruence to move A B  C  into a position so that a dilation with center at A will map it to ABC.

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If |AB| = |A B  |, then the hypothesis would imply |AC| = |A C  | and we are reduced to the SAS criterion for congruence. Thus we may assume that |AB| and |A B  | are not equal. Without loss of generality, suppose |A B  | < |AB|. Then the hypothesis that |AB|/|A B  | = |AC|/|A C  | implies |A C  | < |AC|. On the segment AB, let B0 be the point so that |AB0 | = |A B  |. Similarly, on AC, let C0 be the point satisfying |AC0 | = |A C  |. 

 

Because |∠A| = |∠A | by hypothesis, the SAS criterion for congruence (Theorem G8, page 245) implies that A B  C  ∼ = AB0 C0 . Let ϕ be the congruence that maps A B  C  to AB0 C0 . Moreover, if r denotes the common value of |AB|/|A B  | and |AC|/|A C  |, then the dilation D with center A and scale factor r maps A to A (of course), but also B0 to B because by the definition of dilation, D(B0 ) is the point on the ray RAB so that the distance of D(B0 ) from the center A is r|AB0 | =

|AB| |AB| |AB0 | =   |A B  | = |AB|.   |A B | |A B |

Thus D(B0 ) = B. Similarly, D(C0 ) = C. Therefore, D maps AB0 C0 to ABC, thanks to Corollary 2 of Theorem G16 (see page 271). Consequently, we have (D ◦ ϕ)(A B  C  ) = D(ϕ(A B  C  )) = D(AB0 C0 ) = ABC. Since D ◦ ϕ is a similarity, ABC and A B  C  are similar and the proof of the theorem is complete. We next give the proof of the most easily applied criterion of similarity: the AA criterion (angle-angle criterion) for similarity.

Theorem G22 (AA for similarity). Two triangles with two pairs of equal angles are similar.

Remark. Of course as soon as we prove that the sum of angles in a triangle is 180◦ (Theorem G32 in Section 6.5 of [Wu2020b]), then knowing the equality of two pairs of angles is seen to be equivalent to knowing that all three pairs of angles are equal. This is why this criterion is sometimes cited as the AAA criterion.

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289

Proof. Let two triangles ABC and A B  C  be given. We may assume |∠A| = |∠A | and |∠B| = |∠B  |.

We have to prove that ABC ∼ A B  C  . If |AB| = |A B  |, then the hypothesis would imply ABC ∼ = A B  C  because of the ASA criterion for congruence (Theorem G9, page 245). Thus we may assume that |AB| and |A B  | are not equal. Suppose |A B  | < |AB|. On AB, choose a point B0 so that |AB0 | = |A B  |, and let the line parallel to BC and passing through B0 intersect the line LAC at C0 . By Theorem G11 (FTS*) (page 257), C0 lies on the ray RAC and

(5.9)

|AC| |AB| = . |AB0 | |AC0 |

On the other hand, we have |AB0 | = |A B  | by construction and |∠A| = |∠A | by hypothesis. In addition, |∠AB0 C0 | = |∠B  | because |∠AB0 C0 | = |∠B| by Theorem G18 on page 277 concerning corresponding angles with respect to parallel lines, and because |∠B| = |∠B  | by hypothesis. Thus ASA implies that AB0 C0 ∼ = A B  C  .     Hence |AB0 | = |A B | and |AC0 | = |A C |. Therefore equation (5.9) becomes

|AB| |AC| =   . |A B  | |A C |

Now recall that ∠A and ∠A are assumed to be equal. Therefore, triangles ABC and A B  C  are similar because they satisfy the conditions of SAS for similarity (Theorem G21). Theorem G22 is proved. We emphasize once again that inasmuch as the validity of Theorem G21 depends on Theorem G16 which depends on the parallel postulate, and the proof of Theorem G22 depends on FTS* which also depends on the parallel postulate, both theorems ultimately depend on the parallel postulate. The fact that all conclusions about similar figures depend crucially on the parallel postulate will be underscored once more in the last section of Chapter 8 in [Wu2020b].

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To round off the picture, let us also mention the fact that, in analogy with the case of congruence, there is also an SSS criterion for similarity. This will be proved in Section 6.4 of [Wu2020b]. The Pythagorean theorem and its proof For the purpose of learning about linear equations, students have to learn how to apply Theorems G21 and G22 in specific situations. There is probably no better illustration of such applications than the following proof of the Pythagorean theorem.5 Note that there will be a second proof of this theorem in Section 4.4 of [Wu2020c] using the concept of area. Let us fix the terminology. Given a right triangle ABC with C being the vertex of the right angle. Then the sides AC and BC are called the legs of ABC, and AB is called the hypotenuse of ABC. A HH HH c HH HH b HH HH C B a

Theorem G23 (Pythagorean theorem). If the lengths of the legs of a right triangle are a and b and the length of the hypotenuse is c, then a2 + b2 = c2 .

The basic idea of the proof is very simple. Referring to the same picture, we draw a perpendicular from C to line LAB . The perpendicular meets the segment AB at a point D (see the definition of segment on page 169), as the middle figure in the following picture shows:

AH b

H HD
H





C


AH b

H D H
HH c HH
H

HH

H H

C


a

D

HB

C








H
HH H
H



H

a

HH

H H

H

B

5 Pythagoras of Samos is a pivotal figure in the development of mathematics, yet little is known about him or his work with certainty. He was a Greek philosopher-mathematician who lived around 500 BC. He founded a school devoted to mathematics and philosophy, but it was also a school shrouded in secrecy and infused with a large dose of mysticism. The recognition of the mathematical relationship between musical notes and the existence of numbers which are not rational are attributed to this school. The so-called Pythagorean theorem was actually known independently, and earlier, to the Babylonians, Hindus, and Chinese ([Katz, Chapters 1 and 2]). The Babylonians made extensive computations with this theorem around 1800 BC; see Exercise 9 on page 384.

5.3. SIMILARITY

291

Then a simple use of Theorem G22 reveals that both right triangles CBD and ACD are similar to ABC and therefore their corresponding sides are proportional. This immediately leads, via the cross-multiplication algorithm, to several equalities between the products of (lengths of) the sides of these triangles. If you already know what to prove, then by trial and error, you cannot help but arrive at a combination of these equalities that will give you what you want. On the other hand, if you do not know what to prove, then these identities are not likely to do you much good. There are many proofs of the Pythagorean theorem, but regardless of the proof, one is always aware that guessing a correct statement of the Pythagorean theorem is a higher order of achievement than finding a proof of the theorem. The first person to discover this theorem must have been an extraordinary mathematician. Proof. We will prove that ABC ∼ CBD and also ABC ∼ ACD. H H A Hβ HH c HH D HH
HH
H H

HH HαH HH b

HH H
HH C
B a It suffices to prove ABC ∼ CBD as the other similarity can be proved in the same way. The two triangles ABC and CBD have two pairs of equal angles: |∠CDB| = |∠ACB| = 90◦ and |∠CBD| = |∠ABC|. By the AA criterion for |BA|

|BC|

similarity (Theorem G22 on p. 288), the triangles are similar. Hence |BC| = |BD| . Letting |AC| = b,

|AB| = c,

|BC| = a,

(see the preceding picture), we get algorithm,

c a

=

a α,

|AD| = β,

|BD| = α

so that by the cross-multiplication

a2 = αc.

(5.10)

By considering the similar right triangles ABC and ACD, we conclude in the same |AC|

|AD|

way that |AB| = |AC| , so that cb = βb . Therefore, (5.11)

b2 = βc.

Adding (5.10) and (5.11) and making use of α + β = c, we finally obtain a2 + b2 = αc + βc = (α + β)c = c2 . The proof of the Pythagorean theorem is complete. There is an animation of the preceding proof by Larry Francis that also makes the striking observation that the algebraic manipulations above actually have a geometric interpretation in terms of area: https://youtu.be/QCyvxYLFSfU.

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5. DILATION AND SIMILARITY

Pedagogical Comments. The fact that the perpendicular from C to the line LAB meets the line at a point D between A and B is of critical importance in the preceding proof. Without knowing this fact, we would not have been able to conclude in the last step of the proof that |BD| + |DA| = |AB|. From a mathematical standpoint, a proof of this fact is essential. Unfortunately, this proof is too intricate and too abstract for the school classroom, as we shall see presently. Thus, once again, we are confronted with a common dilemma in the teaching of school geometry: what should be done in terms of mathematics is incompatible with what can be done in terms of pedagogy. Our recommendation is that, because this fact is so pictorially obvious, the preceding proof therefore should be allowed to stand in a school classroom.6 That said, we will proceed to give this proof because our goal is to help you as a teacher to know the whole truth about things you are supposed to teach. Note that the proof would be considerably simpler if we had at our disposal the theorem that the sum of (the degrees of) the angles of a triangle is 180◦ . However, in the way we are developing plane geometry, this theorem about the angle sum of a triangle—whose full proof involves technical details that are equally intricate—will not be taken up until Section 6.5 of [Wu2020b]. We will now prove that the perpendicular from C to line LAB must meet LAB at a point D between A and B; i.e., A ∗ D ∗ B. We will argue by contradiction. Suppose not. Then either D is equal to A or B, or D lies outside the segment AB. If D = A (let us say), then CA ⊥ AB and LCB and LAB are both perpendicular to LAC and therefore LCB  LAB (Theorem G2 on page 223). This is impossible because these two lines intersect at B. Therefore we may assume D lies outside AB; i.e., either D ∗ A ∗ B or A ∗ B ∗ D (Lemma 4.4 on page 169). Without loss of generality, we may assume the former; i.e., A is between D and B, as shown: HH D
H A
H
HH HH

b HcH HH HH

HB HH a C HH HH HH HE Let line LCE be parallel to LAB (use the corollary to Theorem G1 on page 222). We may assume the point E to be so chosen that E and B lie in the same half-plane of LCD . The strategy is to show that (5.12)

|∠ACB| < |∠DCE|.

Assuming this for the moment, we will show how to conclude the proof. Recall that, by hypothesis, ∠ACB is the right angle of the right triangle ABC. Therefore |∠ACB| = 90◦ . But LCE  LAB and BD ⊥ CD; therefore by Theorem G3 on p. 224, EC ⊥ CD. It follows that also |∠DCE| = 90◦ , and (5.12) now leads to the absurd statement that 90◦ < 90◦ . Thus D has to be between A and B to begin with. 6 Although

students should be informed that it is possible to prove A ∗ D ∗ B.

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293

It remains to prove (5.12). To this end, we will prove a more detailed statement: (5.13)

|∠ACB| < |∠DCB| < |∠DCE|.

We begin by proving the second inequality in (5.13); i.e., |∠DCB| < |∠DCE|. We claim that B lies in the convex angle ∠DCE. Thus we must prove (i) E and B lie in the same half-plane of LCD and (ii) D and B lie in the same half-plane of LCE (see the definition of angle on p. 182). Now (i) is true because this was how we chose E. To see that (ii) is true, observe that since LCE  LAB , the line LAB does not contain any point of LCE and therefore neither does the segment DB; by assumption (L4)(ii) on p. 176, D and B lie in the same half-plane of LCE . Thus (ii) is also true and B lies in ∠DCE, thereby proving the claim. It follows that ∠DCB and ∠BCE are adjacent angles with respect to ∠DCE (see page 186 for the definition of adjacent angles). Therefore by assumption (L6)(iv) on p. 188, |∠DCB| + |∠BCE| = |∠DCE|. Since LCE  LAB , B does not lie on LCE so that, in particular, B, C, and E are not collinear. Thus |∠BCE| > 0. Therefore |∠DCB| < |∠DCE|, and the second inequality in (5.13) holds. The proof of the first inequality in (5.13), i.e., |∠ACB| < |∠DCB|, is entirely similar, but simpler. We want to show that A lies in ∠DCB, and this requires the proof that A and B lie in the same half-plane of LCD and that A and D lie in the same half-plane of LCB . Both follow immediately from the hypothesis that A is between D and B, so that AB does not contain any point of LCD and DA does not contain any point of LCB (see (L4)(ii) again). Therefore ∠DCA and ∠ACB are adjacent angles with respect to ∠DCB, and assumption (L6)(iv) implies that |∠DCA| + |∠ACB| = |∠DCB|. Since |∠DCA| > 0, we have |∠ACB| < |∠DCB|. We have therefore completely proved (5.13) and, therewith, also (5.12). As explained right after (5.12), this means that the point D on LAB has to be between A and B. End of Pedagogical Comments. The converse of the Pythagorean theorem is also true. It is intriguing that the proof of the converse makes use of the Pythagorean theorem itself. Since it is sufficiently simple, it will be left as an exercise with an ample supply of hints (Exercise 7 on page 294). Theorem G 24 (Converse of Pythagorean theorem). If triangle ABC satisfies |CA|2 + |CB|2 = |AB|2 , then |∠C| = 90◦ . An immediate consequence of the Pythagorean theorem is the following extension of the SAS criterion for triangle congruence in the case of right triangles: Theorem G25 (HL). Two right triangles with equal hypotenuses and a pair of equal legs are congruent. Here HL stands for "hypotenuse-leg". By the Pythagorean theorem, the other pair of legs of these two right triangles must be equal. The SAS criterion then yields the desired congruence. The details are left to Exercise 7 on page 294. Before leaving the Pythagorean theorem, we wish to bring out the fact that this theorem is a consequence of the parallel postulate, in the following sense. The proof

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5. DILATION AND SIMILARITY

of the theorem on page 291 depends on the concept of similar triangles, which in turn depends on the concept of dilation. It is manifest that almost every property of dilation rests on the parallel postulate, e.g., the fact that a dilation maps a line to a line (see the proof of Theorem G16 on page 269). It is therefore clear that the parallel postulate plays a critical role in validating the truth of the Pythagorean theorem. Of course, it is possible that there is another proof of the Pythagorean theorem that does not make use of the parallel postulate, but what we want to emphasize is that such a proof does not exist. Without the parallel postulate, the Pythagorean theorem will cease to hold. In fact, in hyperbolic geometry (see Section 8.4 of [Wu2020b]), where it is assumed that through a point not lying on a line  pass two distinct lines parallel to  (the opposite of the parallel postulate), the Pythagorean theorem fails. There, a2 + b2 < c2 . The Pythagorean theorem is therefore a characteristic theorem of Euclidean geometry. Exercises 5.3. (1) Let D, E, F be the midpoints of the sides BC, AC, AB, respectively, of a triangle ABC. Prove that DEF ∼ ABC with a scale factor of 2. (2) Let ABC be a right triangle with AC ⊥ CB. Let the perpendicular line |AC|·|BC| from C to AB meet AB at D. Prove that |CD| = |AB| . (3) Let ABC be a right triangle so that |AC| = 3, |BC| = 4, and |AB| = 5. Let the perpendicular line from C to AB meet AB at D, and let the perpendicular line from D to AC meet AC at E. Find |CE|. (4) (This exercise generalizes Exercise 11 on page 238.) Assume FTS. Let L1 , L2 , and L3 be three mutually parallel lines, and let  and  be two distinct transversals which intersect the three parallel lines at A1 , A2 , A3 and B1 , B2 , B3 , respectively. Prove that |B1 B2 | |A1 A2 | = . |A2 A3 | |B2 B3 | (5) Prove that all circles are similar to each other. (Caution: This is a slippery proof. Given two circles C1 and C2 , suppose you have found a dilation D and congruence ϕ so that (ϕ ◦ D)(C1 ) = C2 . Then you will have to prove that the two sets (ϕ ◦ D)(C1 ) and C2 are equal. This means you will have to prove that each is contained in the other (see page 141). Do not skip steps.) (6) Prove that two rectangles are similar to each other if and only if either the ratios of (the lengths of) their sides are equal or the product of these ratios is 1. Precisely, let the lengths of the sides of one rectangle be a and b and those of the other be a and b ; then the rectangles are similar if and   only if either ab = ab or ab · ab = 1. (7) (a) Write a detailed proof of Theorem G25. (b) Prove Theorem G24 (Converse of the Pythagorean theorem). (Hint for (b): Suppose in ABC that |AB| = c, |AC| = b, |BC| = a, a2 + b2 = c2 , and yet |∠C| = 90◦ . Deduce a contradiction as follows: let D be the point on LBC so that AD ⊥ BC. There are two cases: D lies in BC and D lies outside BC. The two cases are similar, so consider the former case where B∗D∗C. Compare the hypothesis that a2 + b2 = c2 with the results obtained by applying the

5.3. SIMILARITY

295

Pythagorean theorem to the right triangles ABD and ACD to arrive at a contradiction.) (8) Let L and L be two lines intersecting at a point O. Take any point P on L, and let the line passing through P and perpendicular to L meet L at |P P  | a point P  . Prove that the ratio |OP  | is independent of the position of P on L; i.e., if Q is another point on L and if the line passing through Q and perpendicular to L meets L at a point Q , then |QQ | |P P  | = . |OP  | |OQ | (9) (a) Let |∠B| = |∠C| in ABC. Prove that |AB| = |AC|. (b) Prove that every point on the angle bisector of an angle is equidistant from (the lines containing) the two sides of the angle, in the sense of page 228. (Note: In some sense, these two assertions should be proved in the setting of congruence, not similarity; any theorem related to similar triangles requires the FTS, which is a more sophisticated theorem than anything about congruent triangles. Indeed, we will revisit these assertions again in Chapter 6 of [Wu2020b] (to be precise, Theorem G29 in Section 6.2 and Exercise 1 in Exercises 6.7), and you will prove them using only theorems about congruence. That said, the virtue of this exercise is that you get to see another approach to these standard facts.) (10) Suppose we have two parallel lines L and L and a point O not lying on either line. Let three lines passing through O intersect L and L at points A, B, C and A , B  , C  , respectively, as shown:









(This picture puts O between L and L , but O could be anywhere.) Prove that |AB| |BC| |AC| =   =   .   |A B | |B C | |A C | (11) Suppose in ABC, AB is longer than AC. Let a point D on the segment BC be such that AD ⊥ BC. A J  J  J   J  J  J  B C D (a) Prove that |BD| > |DC|. (b) Prove that |AB| + |AC| > |BC|. (c) Prove that |BD| − |DC| > |AB| − |AC|.

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(12) (This exercises assumes a familiarity with symbolic computations; see, e.g., Section 6.1 on pp. 298ff.) Given ABC, let a point D on the segment BC be such that AD ⊥ BC. Let also |AB| = c,

|AC| = b,

|AD| = h,

|DC| = ,

|BC| = a.

A J  J  c Jb h  J  J  J  B C    D  a (a) Prove that  =

a2 + b2 − c2 . (b) Prove that 2a

1 (a + b + c)(a + b − c)(−a + b + c)(a − b + c). 2a (This exercise essentially proves Heron’s formula for the area of a triangle in terms of its sides; see Section 4.5 in [Wu2020c].) (13) Suppose you are a teacher in middle school and you are handed a textbook series that takes up similarity in grade 7 and congruence in grade 8. (Such a series did exist in 2013.) (a) Do you believe such a curricular decision is defensible? Explain. (b) If you are a seventh-grade teacher, what would you do? (Obviously there will be no unique answer to part (b), but the idea is that you had better start thinking about such real-world situations because your ability to adjust is, alas, part of your responsibility.). h =

CHAPTER 6

Symbolic Notation and Linear Equations In this chapter, we begin the study of algebra. The main topics of this chapter are the use of symbols, linear equations in one or two variables, and systems of two linear equations in two variables. In the context of school mathematics, the most urgent task in helping students to achieve success in algebra may very well be getting them to be fluent in the correct use of symbols. There is at present an unhealthy preoccupation in TSM1 with the concept of a "variable" in the teaching of algebra, to the point of elevating it to a formal mathematical concept. The truth is that "variable" is not a mathematical concept. A main goal of this chapter is to explain why, if students know the basic etiquette in the use of symbols,2 there would never be any need for them to understand what a "variable" is. Clearly the word "variable" is suggestive, and it is often used in mathematical discussions as a shorthand; for example, we have just used it to talk about "linear equations in one or two variables". However, we did so only because the meaning of this phrase is universally understood, and there is no need to find out what "variables" means in this context.3 So when all is said and done, students should just concentrate on learning the basic etiquette in the use of symbols and learn it well. A major stumbling block in students’ learning about linear equations in two variables is the concept of slope (cf. [PG]). The fairly voluminous literature in education research on slope indicates an awareness of students’ difficulty on this topic. One of the many symptoms of this difficulty is articulated in [BeckmannIzsák]: They might not see slope as a number, but instead think of it as a pair of numbers separated by a slash, basically "rise slash run." It is surprising that [Beckmann-Izsák]—in discussing why slope is hard to teach— did not mention the glaring absence of a correct definition for slope in TSM. Just as in the case of fractions, students, teachers, and educators have been forced to learn about slope without the benefit of knowing precisely what it is. Under the circumstances, the nonlearning of slope—like the nonlearning of fractions—becomes all but inevitable. It is a mathematical and pedagogical imperative that students understand why one single number can be attached to a line to supposedly describe its "slant" (whether it is this way \ or that way /) and its "steepness". To this end, we devote all of Section 6.4 to a detailed discussion of a correct definition of slope: what 1 See

page xiv of the preface for the definition of TSM. page 299. 3 In the same way we understand "Faustian bargain" or "Catch 22" without having to find out who Faust is or what "22" is all about. 2 See

297

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6. SYMBOLIC NOTATION AND LINEAR EQUATIONS

it is supposed to measure, why the definition—relying on the concept of triangle similarity4 —has to be so elaborate, and how the usual formula of "rise-over-run" follows logically from the definition. It is a complex concept, and it deserves a mathematical treatment that recognizes its complexity.5 Then we prove that a line is horizontal if and only if it has 0 slope and that two lines passing through the same point with the same slope must coincide. These theorems should begin to convince students that the hard work of learning the definition is worth the effort. The precise definition of slope enables us to dispel the mystery behind the interplay between the geometry and the algebra of a linear equation in two variables. Indeed, it is this mystery that has bedeviled students, teachers, and educators. The availability of a precise definition of slope also enables us to prove that the graph of a linear equation is a line and that each line is the graph of a linear equation (see Theorem 6.11 on page 354). We recommend, strenuously, that all students learn this proof, because the reasoning imbedded in the proof renders all assessment items related to equations of lines to be nothing more than routine exercises. It also goes without saying that the discussion of systems of two linear equations in two variables gains immeasurably in transparency as a consequence. We hope that, building on such a correct mathematical foundation, education research on student learning—or nonlearning—of slope will acquire greater validity. This chapter makes a great effort to combat the negative impact on student learning by both the abuse of the "concept of a variable" and the nondefinition of slope in TSM. The extended pedagogical comments on pp. 318ff., 327ff., and 361ff. will likely give you an even better idea about what goes into this chapter, and why. 6.1. Symbolic expressions This section has the modest goal of introducing readers to the correct use of symbols. Such a discussion would seem to have little mathematical substance, but we will strenuously argue that it may very well be the most important section of this chapter because it asks you to shed any bad habits you may have acquired in your encounters with TSM concerning the use of symbols. You have been told that mastering the concept of a "variable" is the gateway to algebra (cf. the pedagogical comments in the subsection on pp. 318ff.). You were also told how to "manipulate symbolic expressions" in a symbol x without giving any thought to what x may be (cf. the Pedagogical Comments on page 327). These are not valid mathematical practices, and the dual purpose of this section is to explain why not and, more importantly, make suggestions on how to do better. The basic etiquette in the use of symbols (p. 299) Expressions and identities (p. 302) An important identity (p. 306) Mersenne primes (p. 307) The finite geometric series (p. 309) Polynomials and "order of operations" (p. 310) Rational expressions (p. 316) Pedagogical comments on the teaching of "variables" (p. 318) 4 This

is the reason we take up slope after Chapters 4 and 5. than just heuristic argument after heuristic argument or lots of manipulatives and storytelling without mathematical substance. 5 Rather

6.1. SYMBOLIC EXPRESSIONS

299

The basic etiquette in the use of symbols In mathematics, we use symbols to expedite the expression of ideas. The beginning of algebra, as we understand this term, is the introduction of generality and abstraction by using symbols6 to represent numbers. In order to convince students with only a background in arithmetic that the use of symbols is something well worth learning, we have to demonstrate the benefits of so doing. Consider the problem of asking students to interpret a string of symbols, such √ as y = 3x − 7. There are education researchers who believe that a problem of this type can be used to assess mature ways of understanding mathematics and mature ways of thinking about mathematics. This view is, however, erroneous. In mathematics, such a string of symbols has no meaning , because they are the exact analog of the question, "Is he someone with 225 pounds on a six-foot-five frame?" Without knowing who "he" is, this statement may be true√or it may be false. By the same token, without knowing what y and x are in y = 3x − 7, there is no interpretation to give and no conclusion to draw. Let us do better. In mathematics, the correct use of symbols dictates that each symbol must be quantified, i.e., clearly described as to what it stands for each time it is used. This may be called the basic √ etiquette in the use of symbols. For example, we can make sense of "y = 3x − 7" by specifying what x and y are and by providing a context. Here are four variations on this theme: For √ all real numbers x, we can find a real number y so that y = 3x − 7. For some real numbers x, we can find a real number y so that √ y = 3x − 7. There are an infinite number of fractions x and y so that y = √ 3x − 7. There √ are an infinite number of positive integers x and y so that y = 3x − 7. The importance of quantification can be seen by noting that, despite the similarity between the first two statements, the first is false (e.g., x = 0) and the second is √ true (e.g., x = 3 and y = 2). Similarly, despite the similarity between the last two statements, the first is true whereas the second is false (see Exercise 1 on page 320). A pertinent remark in this connection is that many school students7 commit the elementary error of √ writing down symbolic expressions without quantifying the symbols, such as "y = 3x − 7" above. Very likely, the only way to combat this widespread abuse is to not allow TSM to take root in students’ thinking right from the beginning. Let us teach them to always quantify their symbols.

6 Usually using letters of the English alphabet, but often using letters from the Greek alphabet as well because it is easy to run out of appropriate symbols for a particular task. 7 And a good number of college students too.

300

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To make sure you see why it is important to always quantify your symbols, we take up another example that has more mathematical substance. Consider the following three statements: (C1) n ≥ 3 and an + bn = cn . (C2) For any positive integer n ≥ 3, there are no positive numbers a, b, and c so that an + bn = cn . (C3) For any positive integer n ≥ 3, there are no positive integers a, b, and c so that an + bn = cn . The statement (C1) has no meaning, because we do not know what the symbols a, b, c, and n stand for. If a and b in (C1) are 2 × 2 matrices and c is a 3 × 3 matrix, √ then (C1) is false, but of course (C1) is true if n = 3 and a = 1, b = 2, c = 3 9 (the cube root of 9). (C2) is totally false because no matter what n may be and no matter what the positive numbers a and b may be, letting c be the positive n-th root of an + bn (see Theorem 4.2 in Section 4.2 of [Wu2020b]) will always yield the desired equality of numbers, an + bn = cn . Finally, one may recognize statement (C3) as the famous Fermat’s Last Theorem, first conjectured by Pierre Fermat in 1637 but not proved until Andrew Wiles did so in 1995 (see [WikiFermat]; we will have more to say about Fermat on page 308). Not to harp on the obvious, but the statements (C2) and (C3) differ by just one word in the quantifications of a, b, and c. Moral: Precise quantification of symbols is important. Once the need for quantification of symbols is understood, we now clarify the use of the word "variable". First we give an example. Consider the problem of finding all the numbers x which satisfy 3x + 7 = 5. In the usual jargon, this is known as solving the linear equation 3x + 7 = 5. We will take a serious look at "what an equation means and how to solve an equation" in Section 6.2 on pp. 322ff., but we will proceed informally at this juncture to get our point across. With this understood, the usual procedure for solving such equations yields 3x = 5 − 7, and therefore 5−7 . x= 3 There is a reason why we do not carry out the computation in the numerator to 1 write the solution as −2 3 , and it is because if we consider 3x + 2 = 13 instead, then we get 13 − 12 . x= 3 Or, consider 3x − 25 = 4.6 and by rewriting it as 3x + (−25) = 4.6, we get x=

4.6 − (−25) . 3

Or, consider 5x − 25 = 4.6 and get 4.6 − (−25) , 5 and so on. There is an unmistakable abstract pattern here: one can easily verify that, with a, b, and c (a = 0) understood to be three fixed numbers throughout the following discussion, the solution of the linear equation ax + b = c is x=

x=

c−b . a

6.1. SYMBOLIC EXPRESSIONS

301

We have now witnessed the fact that in some symbolic expressions, the symbols stand for elements in an infinite set of numbers,8 e.g., the statement that mn = nm for all real numbers m and n, while in others, the symbols stand for the element in a set consisting of exactly one element (in other words, they stand for a fixed value throughout the discussion), e.g., the numbers a, b, and c in the preceding linear equation ax + b = c. In the former case, the symbols m and n are called variables, and in the latter case, a, b, and c are called constants. Notice that such terminology is no more than an afterthought when we have carefully quantified the symbols in each situation. There is in fact no need for the words variables and constants when such information is already contained in the quantification. However, we will continue to use them not only because they have been in use for over three centuries and are everywhere in the mathematics literature, but also because they are at times an indispensable shorthand. There are compelling reasons for singling out the terminology of "variable" and "constant" for such an extended discussion. See the pedagogical comments on pp. 318ff. and 327ff., respectively. In a situation where we try to locate any numbers x that satisfies a given equation (such as 2x2 + x − 6 = 0 or 2x = x), the value of the number x is unknown to us, of course. For this reason, we will conveniently refer to the symbol x as an unknown, just to save verbiage. To the extent that we will never make logical deductions based on the properties of an "unknown", it is not necessary to make this terminology more precise.9 At the risk of pointing out the obvious, note that we have been making use of symbols from the very beginning of this volume out of necessity. One example is the addition formula for fractions (equation (1.12) on page 33): for any two fractions k m  and n , where k, , m, n are whole numbers (the product n = 0), k m kn + m + = .  n n If we do not use symbols, we would be forced to express the formula as follows: The sum of two fractions is the fraction whose numerator is the sum of the product of the numerator of the first fraction with the denominator of the second, and the product of the numerator of the second with the denominator of the first, and whose denominator is the product of the denominators of the given fractions.10 Even if you are inordinately fond of the English language, you will have to admit that the symbolic statement is far more clear, and this is not even taking into account the difficulty of trying to provide a mathematical derivation of this addition formula without the benefit of symbols. 8 Strictly speaking, all that matters is that the symbols stand for elements in a set consisting of more than one element. But for school algebra, "infinite" suffices for the purpose at hand. 9 This saves us from the need to discuss the relationship between an unknown and a variable. 10 This was the way mankind had to express formulas from al-Khwarizmi (c. 780 to c. 850)— the person whose name gave birth to the word "algorithm"—all through the Middle Ages to the time of François Viète (1540–1603). The codification of the symbolic notation is generally attributed to R. Descartes (1596–1650). See [Bashmakova-Smirnova].

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This example may serve the purpose of explaining to students why the use of symbols is a necessity. Of course, there are innumerable other examples as well. Expressions and identities We now begin the mathematical discussion. By a number expression, or more simply an expression, in a given collection of numbers x, y, . . . , w, we mean a number obtained from these x, y, . . . , w and from a collection of specific real numbers by the use of a combination of the four arithmetic operations (i.e., +, −, ×, ÷). For example, if x, y, z are numbers, then

4 21 xy + x3 (16z − y 2 ) − z xyz − 7 13 is an example of a number expression in the numbers x, y, z (we have to assume xyz = 7). More precisely, it is the number obtained by applying +, −, ×, and ÷ 4 . to the numbers x, y, z and to the specific numbers 7, 16, and 13 We note explicitly that since the definition of an expression requires that we compute with numbers x, y, etc., that may not be rational, FASM (page 133) has been implicitly invoked for this definition to make sense. The meaning of number expression will be enlarged, in due course, to include the use of specific functions of the numbers x, y, . . . , w (see the end of Section 1.1 in [Wu2020b]) and the use of the operation of "taking the n-th root" (see Section 4.2 of [Wu2020b]). Because all the symbols we use are numbers, we can apply all we know about numbers (including FASM) to number expressions without having to learn anything new, including the fact that the associative, commutative, and distributive laws are automatically valid for computations with number expressions. The importance of the latter fact for teaching and learning cannot be overstated, because in TSM, a "variable" is considered to be a different animal from a number and therefore the arithmetic operations on expressions (involving "variables") can only be justified by an arbitrary decree—a prime example of teaching by rote. In a number expression such as 1 x , x4 − 5x3 y 2 + x2 y 2 − xy 3 + 2y 4 + 2 1 + y2 which involves the numbers x and y, we may regard it as nothing more than a sum of products, namely, 1 (x4 ) + (−5x3 y 2 ) + (x2 y 2 ) + (−xy 3 ) + (2y 4 ) + (x · (1 + y 2 )−1 ). 2 (You may wish to review at this point the definition of subtraction in terms of addition on page 96 and the interpretation of division as multiplication by a multiplicative inverse in equation (2.29) on page 115). Any of the expressions x4 , −5x3 y 2 , 1 2 2 x 3 4 2 x y , −xy , 2y , and 1+y 2 , which are separated by two consecutive +’s (except x for the first one x4 and the last one 1+y 2 ), is called a term of the expression. As x is the custom, the writing of the expression x4 − 5x3 y 2 + 12 x2 y 2 − xy 3 + 2y 4 + 1+y 2 has made implicit use of three notational conventions: Retiring the multiplication symbol ×: The multiplication sign × is omitted in expressions except that if emphasis on a particular multiplication is needed, a dot "·" is used, as in

6.1. SYMBOLIC EXPRESSIONS

303

(x · (1 + y 2 )−1 ). As is well known, the reason for retiring × is that it is too easily confused with the letter x when written by hand. The way of writing specific numbers in an expression: The numbers −5 and 12 in the expression x4 − 5x3 y 2 + 12 x2 y 2 are called coefficients or, more precisely, the coefficients of x3 y 2 and x2 y 2 , respectively. By convention, the coefficients are always placed in front of the symbols, e.g., never x3 (−5)y 2 or even x3 y 2 (−5) unless there is a compelling reason for doing so. The term x4 also has a coefficient because x4 is the abbreviated form of 1x4 , but the number 1 as a coefficient is always suppressed. The order of arithmetic operations among symbols in an expression: It is understood that (i) we do the multiplications of each letter symbol (in this case, x and y) indicated by x the exponent first (e.g., x3 and y 2 in −5x3 y 2 , or y 2 in 1+y 2 ), 2 −1 then (ii) the multiplications within each term (e.g., x · (1 + y ) x in 1+y 2 ), and finally (iii) the addition of the various terms. We will have more to say about the third convention presently. Two (number) expressions in numbers x, y, . . . , w are said to be equal expressions if the two numbers are equal for all values of x, y, . . . , w. A well-known example is the two expressions in x and y, (x + y)(x − y) and x2 − y 2 . We will prove presently (see (6.3) below) that, indeed, these two expressions are equal no matter what x and y may be. Thus they are equal in the sense just defined and we can write (x + y)(x − y) = x2 − y 2 . By the definition of equal expressions, the equal sign between the expressions automatically means the two sides are equal as numbers for all numbers x and y. Such an equality between two expressions is then called an identity. In TSM, equal expressions are said to be "equivalent expressions". Suffice it to say that this terminology is unnecessary and is, in any case, not used in mathematics. In general, it can happen that the equality of two expressions in a collection of numbers x, y, z, . . . is valid, not for all values of x, y, z, . . . , but for "many values" of x, y, z, . . . . Here, the meaning of "many" will have to be understood in context. It could mean all numbers with a small number of exceptions, as in11 1 + tan2 x = sec2 x

for all numbers x = an odd integer multiple of

π . 2

This equality makes no sense when x is equal to an odd integer multiple of π2 because tan x and sec x are not defined at those values of x . Or, "many" could 11 We continue to make use of some mathematics that we have not yet discussed—but which you most likely know—to illustrate a point. For the issue at hand, see the appendix in Section 1.4 of [Wu2020c].

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mean all nonzero whole numbers only, such as 1 + 2 + 3 + · · · + (n − 1) + n =

n(n + 1) 2

for all whole numbers n ≥ 1.

By tradition, both of these equalities, thus carefully quantified, are also called identities. You recognize that we have not offered a precise definition of what an identity is, other than that an identity is a figure of speech that alerts you to the fact (already made explicit above) that it is an equality between two expressions which is valid for "many" values of the numbers in question. By a common abuse of language, one simply writes, for example, 1 + tan2 x = sec2 x as an identity without any qualifications and leaves it to the readers to figure out that the equality is claimed only for x not equal to an odd integer multiple of π2 . Therefore, you have to be careful about the range of the values of x for which the equality is supposed to hold in each case. This is but one example among many where we sometimes use mathematical terms out of respect for tradition but not as precise mathematical concepts. Usually, these terms do manage to suggest a pleasant mental image, and that should count for something. "Variable" is another example of such usage, as we already discussed. Fortunately, a majority of the well-known identities are valid for all numbers, with no exceptions. We will focus on these in this section. Now with the terminology of an identity understood, let us list the three most common identities: (6.1)

(x + y)2

= x2 + 2xy + y 2 ,

(6.2)

(x − y)2

= x2 − 2xy + y 2 ,

(6.3)

(x + y)(x − y)

= x2 − y 2 .

At least three comments should be made. The first is that when x and y are rational numbers, these three identities follow from routine number computations using the associative, commutative, and distributive laws. For example, here is a proof of (6.3): for any numbers x and y, (x + y)(x − y) =

(x + y)x − (x + y)y

(dist. law)

=

x + yx − xy − y

2

(dist. law, Theorem 1 on p. 87)

=

x + xy − xy − y

2

(comm. law of mult.)

=

x −y

2 2 2

2

(Theorem 1 on p. 87).

Because FASM (page 133) assures us that the associative, commutative, and distributive laws (for both + and ×) continue to hold for real numbers, these identities are valid for all real numbers x and y. A second comment is that once we have the first identity (6.1), the second one (6.2) becomes a trivial consequence of the first because (6.1), being valid for all numbers x and y, is also valid for x and −y. Therefore, for any numbers x and y, (x + (−y))2 = x2 + 2x(−y) + (−y)2 .

6.1. SYMBOLIC EXPRESSIONS

305

But by equation (2.23) on page 109, this is the same as (x − y)2 = x2 − 2xy + y 2

for any x and y,

which is the same equation as (6.2). Thus we have proved that (6.1) implies (6.2). In a similar manner, we can prove that, conversely, (6.2) implies (6.1). In the terminology of page 22, the first identity (6.1) is equivalent to the second one (6.2). You may be wondering why we bother with the preceding proof of the second identity since a simple direct computation already proves (x − y)2 = (x − y)(x − y). Our point is that, at the beginning stage of algebra, your students are put in touch with the idea of generality for the first time; namely, the first identity is not just the equality of the expressions (x + y)2 and x2 + 2xy + y 2 for certain numbers x and y, but that it is valid for all numbers x and y. If you can teach them to take the latter statement seriously and make them realize that the validity of the equality when y is replaced by −y immediately implies the validity of the second identity, then you have taught them something valuable. One may paraphrase by saying that, because of the generality of the first identity, the first identity already contains the second identity as a special case. An important part of learning algebra is to become alert to the potential implications of a general statement. From this vantage point, the derivation of the second identity from the first now becomes a noteworthy demonstration of the power of generality. A third comment is that, in practice, the usefulness of these identities is, more often than not, derived from one’s ability to read these identities also from right to left, i.e., the ability to recognize in a given situation that, for any two numbers x and y, x2 + 2xy + y 2

is equal to (x + y)2 ,

x2 − 2xy + y 2

is equal to (x − y)2 ,

x2 − y 2

is equal to (x + y)(x − y).

For example, 25x2 + 49y 2 − 70xy is equal to (5x − 7y)2 , because 25x2 + 49y 2 − 70xy = (5x)2 − 2(5x)(7y) + (7y)2 . We pause to make another remark about identities. We may rewrite, for example, the identity (x − y)(x + y) = x2 − y 2 as x2 − y 2 = x + y. x−y Remembering that we cannot divide by 0, we see that this equality holds for all numbers x and y except when x = y. This equality is of course also considered to be an identity—keeping in mind the exceptions. We have thus far discussed the concept of an equation informally and the concepts of an expression and an identity in some depth. On pp. 322ff. below, we will elaborate on the concept of an equation. It is to be noted that our view on these fundamental concepts in beginning algebra deviates from those typically found in education research; see, e.g., [McCrory et al.].

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An important identity There is more to be said about the identity (x − y)(x + y) = x2 − y 2 ! We can ask if there is an analogous identity that has x3 − y 3 on the right side. There is, because a straightforward computation making repeated use of the distributive law (and of course also Theorems 1 and 2 of the appendix in Chapter 1, page 87) shows that (x − y)(x2 + xy + y 2 )

= x(x2 + xy + y 2 ) − y(x2 + xy + y 2 ) = x3 + x2 y + xy 2 − yx2 − yxy − y 3 = x3 + x2 y + xy 2 − x2 y − xy 2 − y 3 = x3 − y 3

for all numbers x and y. In other words, (x − y)(x2 + xy + y 2 ) = x3 − y 3 . Similarly, we have for the 4-th and 5-th powers: (x − y)(x3 + x2 y + xy 2 + y 3 ) =

x4 − y 4 ,

(x − y)(x + x y + x y + xy + y ) =

x5 − y 5 .

4

3

2 2

3

4

The pattern is now clear: for any positive integer n and for all numbers x and y, (6.4)

(x − y)(xn + xn−1 y + xn−2 y 2 + · · · + xy n−1 + y n ) = xn+1 − y n+1 .

Let us rewrite this identity in the following way: for all numbers x and y, (6.5)

xn+1 − y n+1 = (x − y)(xn + xn−1 y + xn−2 y 2 + · · · + xy n−1 + y n ).

Thus, the difference of two numbers x and y raised to the same power can always be expressed as a product of x − y and xn + xn−1 y + · · · + y n . By any measure, this is a nice-looking identity. We will examine its consequences in some detail in two different settings, and the two groups of results—which would seem to be unrelated to each other—will nicely illustrate the concept of generality. Since this identity is valid for all numbers x and y, it is certainly valid when x and y are positive integers. In that case, observe in particular that the right side of (6.5) is a product of two integers. If it happens that x > y > 0, then also xn+1 > y n+1 > 0 (see Exercise 11 on page 132) and the identity says that the positive integer xn+1 − y n+1 has a factorization (in the sense of page 138) as the product of two positive integers: x − y and xn + xn−1 y + · · · + xy n−1 + y n . For the sake of clarity, we restate it as follows: for all positive integers a, b, and n, so that a > b > 0, we have the following factorization of the positive integer an+1 − bn+1 , which is a special case of identity (6.5): (6.6)

(an+1 − bn+1 ) = (a − b)(an + an−1 b + an−2 b2 + · · · + abn−1 + bn ).

If a−b > 1, then in particular a > 1 so that an +an−1 b+an−2 b2 +· · ·+abn−1 +bn > 1. Therefore the positive integer an+1 − bn+1 , being the product of two integers each bigger than 1, is not a prime (page 148). We have thus proved the following. Lemma 6.1. If a, b are positive integers and a − b > 1, then for all positive integers n, an+1 − bn+1 is not a prime.

6.1. SYMBOLIC EXPRESSIONS

307

For example, 1273 −663 = 1,760,887, and this lemma guarantees that 1,760,887 is not a prime. By no means is this fact obvious because its smallest divisor is 61. In fact, the prime decomposition (see Theorem 3.6 on page 149) of 1,760,887 is 61 × 28867. In a similar vein, you can show off to your friends by challenging them to check whether 13,997,513 is a prime. You of course know that it is not a prime because 13,997,513 = 2413 − 23 . What makes the testing of the primality of this number difficult is that the prime decomposition of 13,997,513 is 239 × 58567. In other words, its smallest divisor is 239, so that guess-and-check will not work efficiently in this case. (Again, the identity 2413 − 23 = (241 − 2)(2412 + 241 · 2 + 22 ) also happens to exhibit the prime decomposition of 13,997,513.) Activity. Let a and b be integers. Does (a5 − b5 ) divide a15 − b15 in the sense of page 138? Does (a2 + ab + b2 ) divide a15 − b15 ? Suppose b = 1 in identity (6.6). Then we get, for all positive integers a and n, (6.7)

an+1 − 1 = (a − 1)(an + an−1 + an−2 + · · · + a2 + a + 1).

As before, if a > 2, then a − 1 > 1 and an + an−1 + · · · + a2 + a + 1 > 1 so that an+1 − 1 is never a prime. This is a special case of Lemma 6.1. It turns out that the case of a = 2 provides a different kind of intrigue. Mersenne primes If a = 2 in (6.7), then a − 1 = 1, and (6.7) no longer provides a nontrivial factorization (see page 138 for the definition) of 2n+1 − 1 and, therefore, no longer exhibits the whole number 2n+1 − 1 as a composite. More to the point, 2n+1 − 1 is actually a prime for certain values of n, as we can see from the first five cases of n = 1, 2, 3, 4, 5: If n = 1, then 2n+1 − 1 = 3. If n = 2, then 2n+1 − 1 = 7. If n = 3, then 2n+1 − 1 = 15. If n = 4, then 2n+1 − 1 = 31. If n = 5, then 2n+1 − 1 = 63. So among the first five values of 2n+1 − 1, three of them are primes but two are not. The question naturally arises: among all possible values of 2n+1 − 1 as n runs through all the positive integers, which of them are primes? There is an easy reduction of this question, as the following proposition tells us that there is no need to look for primes among the numbers 2n+1 − 1 where n + 1 is a composite: Lemma 6.2. If m is a composite positive integer, then 2m −1 is also a composite.

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Proof. Indeed, if m = pq for positive integers p and q so that p, q > 1, then 2m = (2p )q . Thus,

 2m − 1 = (2p )q − 1q = (2p − 1) (2p )q−1 + (2p )q−2 + · · · + (2p ) + 1 . (Notice that, once more, we rely on identity (6.6) for the second equality.) But p > 1 implies (2p − 1) > 1 because 2p − 1 ≥ 3. Moreover, q > 1 implies 2p − 1 < (2p )q − 1 = 2m − 1. Therefore 1 < (2p − 1) < 2m − 1, and 2p − 1 is a proper divisor of 2m − 1. The lemma is proved. Lemma 6.2 explains why 24 − 1 (= 15) and 26 − 1 (= 63) in the above list are not primes. In view of Lemma 6.2, our original question about which of 2n+1 − 1 are primes can now be simplified to the following: Which of 2p − 1 are primes, as p runs through all the primes? In 1644, Father Mersenne12 claimed that, among all the primes p < 258, 2p − 1 is a prime exactly when p = 2, 3, 5, 7, 13, 17, 19, 31, 67, 127, 257. Keep in mind that the computer is a creation of the latter part of the twentieth century, so it was a nontrivial matter in the days of Mersenne to check the primality of a (whole) number with, say, ten digits such as 231 −1 = 2, 147, 483, 647. Mersenne probably did not test the primality of all the numbers he wrote down, and his statement was likely a mixture of guessing and wishful thinking. It is not even clear whether he actually proved that his list was correct for p ≤ 31 because, if he did, he would have to verify the following two assertions: (A) The number 2p − 1 is composite for p equal to 11, 23, and 29. (B) The number 2p − 1 is prime for p equal to 13, 17, 19, 31 (the cases of p = 2, 3, 5, 7 are obvious). For (A), the fact that 211 − 1 = 23 · 89 was known back in 1536, and the fact that 223 − 1 is composite was found by Fermat13 in 1640 as a refutation of the opposite claim made by Pietro Cataldi (1548–1626) in 1603. The case of 229 − 1 was not known at the time Mersenne made his conjecture, and it would stay that way until Euler,14 almost a century after Mersenne made his conjecture, managed to find the 12 Marin

Mersenne (1588–1648) was a French theologian and amateur mathematician. He was a friend of all the French mathematical luminaries of his time, including Descartes, Fermat, Pascal, and Desargues, and performed the critical service of disseminating mathematical information among them at a time when mathematical publications were basically nonexistent. 13 Pierre Fermat (1607–1665) was a French lawyer; he was also an amateur mathematician but one of the greatest mathematicians of all time nonetheless. The terminology of "Cartesian coordinates" masks the fact that Fermat was a codiscoverer of analytic geometry with Descartes; in fact Fermat made the discovery a few years earlier and seemed to have a better understanding of the potential of his discovery. Fermat was a cofounder of the theory of probability (with Blaise Pascal), and the modern theory of numbers also owes its existence to Fermat. We already had occasion to mention Fermat’s Last Theorem on page 300. His definition of the tangent to a curve at a point inspired Newton’s definition of the derivative (see Theorem 6.19 in [Wu2020c]). 14 Leonhard Euler (1707–1783), the most productive mathematician of all time, was the dominant mathematician of the eighteenth century and rightfully ranks among the greatest. He made important contributions in every part of mathematics, physics, and astronomy as they were known in his time.

6.1. SYMBOLIC EXPRESSIONS

309

prime decomposition 229 − 1 = 233 · 1103 · 2089 in 1738. As for (B), the primality of 213 − 1 = 8191 only takes a little patience and was known in any case as far back as the fifteenth century. The primality of 217 − 1 and 219 − 1 was verified by the same Cataldi in 1588. But the primality of 231 − 1 was finally proved only in 1772 by Euler, some 130 years after Mersenne made his conjecture. With the help of an electronic computer, we can easily see that Mersenne’s list contains five errors: 61, 89, 107 should have been on his list but they were not, and the numbers 67 and 257 which were on his list shouldn’t have been there (i.e., 267 − 1 and 2257 − 1 are composites). As a result of Mersenne’s list, a number of the form 2p − 1 which is a prime is called a Mersenne prime. You can get more information about Mersenne primes from http://mersenne.org/. So what are the Mersenne primes? Unfortunately, we know very little about this question. We do not even know if there are an infinite number of Mersenne primes. A search for bigger and bigger Mersenne primes is an ongoing enterprise (see the website above), and one of the side benefits of this search is that each time a larger Mersenne prime was found, it also turned out to be the largest prime number known to mankind. As of January 2020, 51 Mersenne primes are known, and the largest is a number with more than 24 million digits corresponding to the prime p = 82,589,933 (discovered on December 7, 2018). From a mathematical perspective, this search would acquire greater significance if we knew that the number of Mersenne primes is finite. The finite geometric series Now, let us take a second look at identity (6.5), xn+1 − y n+1 = (x − y)(xn + xn−1 y + xn−2 y 2 + · · · + xy n−1 + y n ) for all numbers x and y and all positive integers n. (We emphasize that, at this juncture, x and y are no longer restricted to positive integers but are arbitrary numbers.15 ) Letting y = 1, we get (6.8)

xn+1 − 1 = (x − 1)(xn + xn−1 + xn−2 + · · · + x2 + x + 1)

for all numbers x and for all positive integers n. We now explore the implications 1 of identity (6.8) from another angle. If x = 1, multiplying both sides by x−1 and switching the left and the right sides gives (6.9)

1 + x + x2 + · · · + xn =

xn+1 − 1 x−1

for all x = 1

for any number x = 1 and for any positive integer n. The sum 1 + x + x2 + · · · + xn is called a finite geometric series in x, and identity (6.9) is usually called the summation formula for the finite geometric series. A slight variant of (6.9) is the following identity obtained from (6.9) by applying the distributive law: for any a = 0, (6.10)

a + ax + ax2 + · · · + axn = a ·

xn+1 − 1 x−1

for all x = 1.

15 Once we have introduced complex numbers (see Section 5.2 in [Wu2020b]), this identity will be seen to be valid for complex numbers x and y as well.

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6. SYMBOLIC NOTATION AND LINEAR EQUATIONS

We may assume that in both (6.9) and (6.10), x = 0 because the case of x = 0 is not interesting. Recalling16 that 1 = x0 , we may consider this identity as the expression of the sum of all the whole number powers of x up to and including xn as a quotient (xn+1 − 1)/(x − 1). For example, if x = −3 and n = 12, then 1 − 3 + 32 − 33 + 34 − · · · − 311 + 312 = But if x =

3 4

1594324 (−3)13 − 1 = = 398581. −3 − 1 4

and n = 15, then we have 1+

3 4

+ ( 43 )2 + ( 43 )3 + · · · + ( 43 )15 = {( 34 )16 − 1}/( 43 − 1)

which is equal to approximately 0.9899774 = 3.9599096, 0.25 or more crudely, 3.96. The summation of the finite geometric series is usually tucked away at the end of the second year of school algebra, with the result that it is often not taught, or taught hastily, for lack of time. This is unfortunate because this summation formula is one of the most basic pieces of mathematics students should know for advanced mathematical or scientific work. As we have just seen, it is also one of the most elementary and therefore should be taught at the beginning of school algebra, not at the end of it. Please be aware of this fact when you teach algebra. Let us cast a backward glance at the last three subsections which are devoted to a discussion of the identity (6.5) for all positive integers n: xn+1 − y n+1 = (x − y)(xn + xn−1 y + · · · + xy n−1 + y n ). On account of its generality—the fact that it is valid for all numbers x and y— this identity lends itself to two divergent trains of thought. One is to ask whether certain positive integers are primes, and the other is to obtain a formula for the sum of a finite geometric series. This is a trivial illustration of why mathematics pursues generality, because general theorems always raise the possibility that they will have many interesting potential applications. Mathematical Aside: It is worthwhile to point out that the identity (6.5) also gives a short proof of the calculus fact that the derivative of xn+1 (where n is a positive integer) is equal to (n + 1)xn . Briefly, the proof goes as follows (see the proof of Theorem 6.16 in [Wu2020c]). The derivative of xn+1 at a is the limit of the difference quotient (xn+1 − an+1 )/(x − a) as x goes to a. Because of (6.5), the numerator of the difference quotient is equal to the product of (x − a) and xn + xn−1 a + · · · + xan−1 + an . Thus the difference quotient itself becomes xn + xn−1 a + · · · + xan−1 + an after (x − a) has been canceled from the numerator and denominator. So as x converges to a, we get an + an + · · · + an (n + 1 times), which is (n + 1)an . Polynomials and "order of operations" We next introduce polynomials, but first a remark before the formal definition. Underlying the whole discussion of polynomials in school algebra is a basic technique 16 For

a fuller discussion of the 0-th power of x, see Section 4.2 in [Wu2020b].

6.1. SYMBOLIC EXPRESSIONS

311

known as collecting like terms. It is nothing more than a simple observation based on the distributive law, and we deal with this first. Suppose we have a sum (18 × 53 ) + (53 × 23) + (69 × 53 ). One can compute this sum by first multiplying out each term 18 × 53 , 53 × 23, and 69 × 53 and then adding the resulting numbers to get (18 × 53 ) + (53 × 23) + (69 × 53 ) = 2250 + 2875 + 8625 = 13750. Now if we reflect for a moment, we would realize that we wasted precious time doing three multiplications before adding. If we apply the distributive law, then the computation simplifies: (18 × 53 ) + (53 × 23) + (69 × 53 ) = =

(18 + 23 + 69) × 53 110 × 125 = 13750.

Notice that we have made use of the commutative law of multiplication to change 53 × 23 to 23 × 53 in the process. You may think that, with the advent of high speed computers, it does not matter if we get the answer by multiplying three times and then adding once or (as in the second case) adding three times and multiplying once. This is true, but the difference in conceptual and visual clarity between (18 × 53 ) + (53 × 23) + (69 × 53 ) and (18 + 23 + 69) × 53 is substantial. This is because multiplication is a far more complicated concept than addition: 234 + 677 is simply the addition of two three-digit numbers, but 234 × 677 means adding 677 to itself 234 times. It is therefore far simpler conceptually to add three times and multiply once than to multiply three times and add once. Because both conceptual and visual clarity are important in the learning and doing of mathematics, we will collect together terms involving the same numbers raised to a fixed power (such as 53 in (18 × 53 ) + (53 × 23) + (69 × 53 )) by using the distributive law. For example, we will always rewrite (181 × 25 ) + (67 × 25 ) + (25 × 96) − (257 × 25 ) as 87 × 25

(= (181 + 67 + 96 − 257) × 25 ).

Similarly, we will write the sum       8   8 8



14  14  3 3 3 14 24×59 − × 89 + 59 ×73 + 59 ×66 + 25 × × 11 + 5 5 5 as

 163 × 5914 −



 8  3 53 × , 5

where 163 = 24 + 73 + 66 and −53 = −89 + 25 + 11. Recall once more that we refer to both of the above expressions as a "sum" because subtraction is just addition in disguise (see page 96). In an entirely similar manner, suppose we are given a sum of multiples of nonnegative integer powers of a fixed number x, where multiple here means simply

312

6. SYMBOLIC NOTATION AND LINEAR EQUATIONS

multiplication by any number and not necessarily by a whole number and "nonnegative integers" refers to the whole numbers 0, 1, 2, . . . . Then we would automatically collect together the terms involving the same power of x as before. For example, we will write 1 1 3 x + 16 − 8x2 + x3 − x5 − 6x2 + 75x + 2x3 2 3 as 17 (6.11) −x5 + x3 − 14x2 + 75x + 16. 6 A sum of multiples of nonnegative integer powers of x is called a polynomial in x.17 We also agree to call any expression in a number x a polynomial in x if it is equal to a sum of multiples of nonnegative integer powers of x after applications of the associative, commutative, and distributive laws to the expression. Thus the expression in (6.11) is a polynomial in x, as is (x − 1)(x + 2) because it is equal to x2 + x − 2 after expansion. We have followed three notational conventions that are generally followed in discussions about polynomials: (A) Parentheses are usually suppressed, with the understanding that exponents are computed first, multiplications second, and additions third. (B) The earlier convention is that the power of x is placed last in each term, so that we write −14x2 instead of x2 (−14). (C) The terms are written in decreasing powers of the number x in question. (The term 16 is the term 16x0 , where, by definition, x0 = 1; incidentally, this is where we need the concept of the zeroth power of a number18 .) In this connection, a caveat about (C) should be mentioned: a polynomial is sometimes written in increasing powers of the variable for a reason.19 As on page 303, the number in front of a power of x is called the coefficient of that particular power of x. For the polynomial in (6.11), it is in reality equal to 17 3 x + (−14)x2 + 75x + 16x0 6 when it is written strictly as a sum of multiples of decreasing powers of x. Therefore, the coefficient of x5 in (6.11) is −1, the coefficient of x4 is 0 (remember, x is just 2 a number, so that 0 · x4 = 0), the coefficient of x3 is 17 6 , the coefficient of x is −14, the coefficient of x is 75, and the so-called constant term 16 is actually the coefficient of x0 . A multiple of a single nonnegative power of x, such as 58x12 , is called a monomial. Thus, a monomial is a polynomial with only one term. The highest power of x with a nonzero coefficient in a polynomial is called the degree of the polynomial. The terminology about "nonzero coefficient" refers to the fact 3 2 that the preceding polynomial −x5 + 17 6 x − 14x + 75x + 16 could be written as 17 0·x37 −x5 + 6 x3 −14x2 +75x+16, but the 37-th power of x clearly does not count. (−1)x5 + 0 · x4 +

17 Mathematical Aside: This definition of a polynomial with real coefficients is the appropriate one for school mathematics and is based on the fact that the polynomial ring over R is ringisomorphic to the ring of R-valued polynomial functions. 18 Note that in this case, we have to make an ad hoc definition by agreeing to write x0 = 1 regardless of whether x = 0 or not. 19 Mathematical Aside: The Taylor polynomials of a differentiable function are usually written in increasing powers.

6.1. SYMBOLIC EXPRESSIONS

313

This polynomial has degree 5, and not 37 (and not any whole number different from 5, for that matter). This is the place to make a comment on the notational convention (A) above. As we said earlier, visual clarity in the notation we use is important. We find the polynomial 17 −x5 + x3 − 14x2 + 75x + 16 6 easy to work with because it is visually simple. As written, though, this symbolic expression is a priori ambiguous because it could mean, among other things, the following:   3 17 (−x)5 + x − {(14x)2 + 75}x + 16. 6 But of course what we have in mind is   17 3 {−(x5 )} + (x ) + {−14(x2 )} + {75x} + 16. 6 (We need not specify the order of doing the additions at this point because of the general associative law; cf. Theorem 1 in the appendix of Chapter 1, page 87.) So the net effect of notational convention (A) is to eliminate the need to write the 3 last cumbersome-looking expression by declaring that the expression −x5 + 17 6 x − 2 14x + 75x + 16 already means the same thing. This is all there is to the notational convention (A). It is a convention, and one should not invest more mathematical significance in a convention than what it truly is. As is well known, school mathematics is sometimes led astray by misplaced emphases. Convention (A) has somehow become enshrined in the middle school curriculum under the name of order of operations. In TSM, mnemonic devices were created to help students memorize it (PEMDAS, "Please Excuse My Dear Aunt Sally"), and standard assessments likewise contribute to promote the importance of this convention. A mathematics classroom has to deal with conventions, of course, but a convention should be put in its proper place and not be magnified out of proportion. A more moderate approach would be to explain the genesis of the convention (A) above, quiz students at the beginning to make sure they get it, and go on to more important things. For a fuller discussion, see [Wu2004b] on the so-called "order of operations". A polynomial of degree 1 is called a linear polynomial, and a polynomial of degree 2 is called a quadratic polynomial. Because a general quadratic polynomial has only three terms, ax2 + bx + c, it is also called a trinomial in school mathematics. However, the term "trinomial" is used only in school mathematics, not in higher mathematics; it should therefore be avoided in general discussions. We will discuss quadratic polynomials in some detail in Chapter 2 of [Wu2020b]. A polynomial of degree 3 is called a cubic polynomial. The most familiar polynomials are the so-called expanded forms of whole numbers; these are polynomials in the number 10. For example, the expanded form of 75,018 is (7 × 104 ) + (5 × 103 ) + (0 × 102 ) + (1 × 101 ) + (8 × 100 ) which is a fourth-degree polynomial in 10. Of course the expanded form of any k-digit whole number is a polynomial of degree (k − 1) in 10. On the other hand,

314

6. SYMBOLIC NOTATION AND LINEAR EQUATIONS

the so-called complete expanded form of a decimal such as 32.58, (3 × 101 ) + (2 × 100 ) + (5 × 10−1 ) + (8 × 10−2 ), is not a polynomial in 10, for the reason that it contains negative powers of 10. It should also be pointed out that as a polynomial in 10, the expanded form of a whole number is not a typical polynomial because each coefficient is a single-digit whole number. By contrast, the coefficients of a general polynomial in 10 can be any number. Because polynomials are just numbers, we can add, subtract, multiply, and divide them as numbers. Therefore, with the exception of division, the other three arithmetic operations produce another polynomial in a routine manner. (Division of polynomials does not generally produce a polynomial and will be looked at separately.) Consider, for example, the product of two linear polynomials ax + b and cx + d: (ax + b)(cx + d)

=

(ax + b)(cx) + (ax + b)d 2

(dist. law)

= acx + bcx + adx + bd

(dist. and com. laws)

= acx2 + (ad + bc)x + bd

(dist. law).

Because of the definition of a polynomial, we had to collect terms of the same degree in x using the distributive law and rearrange the terms so that they are in descending powers of x. Other than that, this shows that the multiplication of polynomials is no different from the usual operations with numbers. If the arithmetic of numbers (whole numbers and rational numbers) were taught correctly, such operations with polynomials would be just routine rather than a significant problem to reckon with. As is well known, TSM has fostered the uncivilized practice of "FOILING" to teach the multiplication of two linear polynomials in beginning algebra classrooms. This does serious damage to mathematics learning for at least two reasons. First, the mnemonic device of FOIL is only applicable to the product of two linear polynomials, but what students of algebra must learn is how to apply the distributive law in general (cf. (6.4) on page 306). Second, teaching FOIL does nothing so much as promote the TSM paradigm of replacing mathematical reasoning by the memorization of rote skills such as PEMDAS, "the butterfly method", etc. Such practices are antithetical to the goal of a good mathematics education: teaching students how to reason. We have mentioned the need to sometimes look at an equality backwards, and we will repeat this message once again. What we obtained above, (ax + b)(cx + d) = acx2 + (ad + bc)x + bd, is nothing but routine applications of the distributive law. However, when this equality is read from right to left, it becomes (6.12)

acx2 + (ad + bc)x + bd = (ax + b)(cx + d).

This equality is no longer routine by any stretch of the imagination! In general, if the polynomials p(x), q(x), r(x) in x satisfy p(x) = q(x)r(x), then we say q(x)r(x) is a factorization of p(x) if the degrees of both q(x) and r(x) are positive. (Thus 5 3 2 1 5 3 2 2 3 2 2 3 x −2x + 3 = ( 3 )(5x −6x +2) is not a factorization of 3 x −2x + 3 because the 1 degree of 3 is zero.) In this terminology, the identity (6.12) furnishes a factorization of acx2 + (ad + bc)x + bd as a product (ax + b)(cx + d), provided a = 0 and c = 0.

6.1. SYMBOLIC EXPRESSIONS

For example, we get 1 2 5 x + x − 3 = (2x − 3) 2 4



315

 1 x+1 4

by letting a = 2, b = −3, c = 14 , and d = 1. However, this is by no means an invitation for students to memorize the identity (6.12)! There are more reasonable ways to learn how to obtain the factorization of 12 x2 + 54 x − 3. One way is the following. Since it is much easier to deal with integers rather than rational numbers, we rewrite the quadratic polynomials by using the distributive law to take out the denominators of all the coefficients, as follows: 1 1 2 5 x + x − 3 = (2x2 + 5x − 12). 2 4 4 Then we recognize that (2x2 + 5x − 12) = (2x − 3)(x + 4) because the 0-degree term (i.e., 12) of 2x2 + 5x − 12 has to be the product of the 0-degree terms −3 and 4 of 2x − 3 and x + 4, and the coefficient 2 of 2x2 + 5x − 12 has to be the product of the coefficients 2 and 1 of 2x − 3 and x + 4, respectively. So a few trials and errors would get it done. Hence, we obtain as before, 1 2 5 1 1 x + x − 3 = (2x2 + 5x − 12) = (2x − 3)(x + 4). 2 4 4 4 At present, the teaching of factoring quadratic polynomials with integer coefficients figures prominently, not to say obsessively, in school courses in algebra. For this reason, some perspective on this subject is called for. One should keep in mind that all it does is factor two integers A and C into products of integers so that a given quadratic polynomial Ax2 + Bx + C can be written as acx2 + (ad + bc)x + bd (which then equals (ax + b)(cx + d)). There is no denying that beginning students ought to acquire some facility with decomposing integers into products. It is also important that they are able to effortlessly factor a simple quadratic polynomial such as x2 + 2x − 35 into the product (x + 7)(x − 5). But as sometimes happens, although a little bit of something is good for you, a lot of it may actually be harmful. This would seem to be the case here where a minor skill gets blown up to a major topic, with the consequence that other topics that are more central and more substantial (such as learning about why the graphs of linear equations in two variables are lines or how to solve rate problems correctly) get slighted in terms of time and emphasis. There is a memorable passage on this issue in an introductory textbook on abstract algebra: Very early in our mathematical education—in fact in junior high school or early in high school itself—we are introduced to polynomials. For a seemingly endless amount of time, we are drilled, to the point of utter boredom, in factoring them, multiplying them, dividing them, simplifying them. Facility in factoring a quadratic becomes confused with genuine mathematical talent. ([Herstein, p. 153]) Teachers of algebra should avoid this pitfall. Please also keep in mind the fact that once the quadratic formula becomes available (see Theorem 2.8 in Section 2.1 of [Wu2020b]), there will be a two-step algorithm to accomplish this factorization no matter what the coefficients of the quadratic polynomials may be.

316

6. SYMBOLIC NOTATION AND LINEAR EQUATIONS

We give one more illustration of the factorization of polynomials. We begin with a multiplication of polynomials where each step except the last makes use of the distributive law:         1 1 1 1 3 2 3 2 3 3 5x − 5x − (x + 2x + 8) = x + 5x − 2x + 5x − 8 2 2 2 2   1 = 5x5 − x2 + (10x4 − x) + (40x3 − 4) 2 1 = 5x5 + 10x4 + 40x3 − x2 − x − 4. 2 Reading this equality from right to left results in a factorization that is not so trivial:   1 2 1 5 4 3 3 5x + 10x + 40x − x − x − 4 = 5x − (x2 + 2x + 8). 2 2 Note the fact that if p(x) and q(x) are polynomials of degree m and n, respectively, then the degree of the product p(x)q(x) is (m + n). In other words, the degree of a product is the sum of the degrees of the individual polynomial factors. For example, the preceding calculation which multiplies a degree 3 polynomial with a degree 2 polynomial yields a polynomial of degree 5 (= 3 + 2). Rational expressions Finally, a quotient (i.e., division) of two polynomials in a number x is called a rational expression in x. Here is an example: 3x5 + 16x4 − 25x2 − 7 . x2 − 1 We note that in the case of rational expressions, we need to exercise some care in not allowing division by 0 to take place. For example, in the preceding rational expression, x can be any number except ±1 because if x = ±1, then x2 − 1 = 0 and the denominator would be 0. In writing rational expressions, it is understood (unless stated to the contrary) that only those values for which the denominator is nonzero are considered. In middle school, we are mainly interested in rational numbers and, as a consequence, all computations with numbers tacitly assume that the numbers involved are rational numbers. (But keep in mind that, by FASM (page 133), we can carry out the same computations with arbitrary real numbers as well.) With this understood, since x is a (rational) number, a rational expression is just a rational quotient (in the sense defined on page 117) and can therefore be added, subtracted, multiplied, and divided like any fraction (see items (a)–(d) on page 118). For example, in case x = 12 in the foregoing rational expression, we would be looking at the rational quotient 1 1 3( 32 ) + 16( 16 ) − 25( 41 ) − 7 , ( 41 ) − 1 5 . In general, no matter what x is, we can compute with which is equal to 16 24 rational expressions in x in the usual way (see (c) and (d) on page 118 and (2.33)

6.1. SYMBOLIC EXPRESSIONS

317

on page 119): 0.5x3 + 1 2x7 + 3 3 = 8 x +x−2 x +7 and

3 2 2x + 1 2 x + 4x −

7

·

(0.5x3 + 1)(x3 + 37 ) + (2x7 )(x8 + x − 2) (x8 + x − 2)(x3 + 37 )

( 32 x2 + 1)(6) 6 = 3x4 − 5 (x2 + 4x − 7)(3x4 − 5)

and 2x+1 x2 −0.3 3 4x −x+11 2x

=

(2x + 1)(2x) . (x2 − 0.3)(4x3 − x + 11)

These are exactly the same as any computation with rational numbers. This realization is important in the teaching of algebra because, when you teach rational expressions, you should remind students that if they know how to handle rational numbers, then they already know all there is to know about this topic. There is so much in introductory algebra that is just a revisit of arithmetic. Because the cancellation law is valid for rational quotients (item (a) on page 118, B which says AB AC = C for all rational numbers A, B, C, with A = 0, C = 0), some rational expressions can be simplified. Sometimes the cancellation presents itself, as in (5x4 − x3 + 2)(2x − 15) , (14x2 + 3x − 28)(5x4 − x3 + 2) Here, the number (5x4 − x3 + 2) in both the numerator and denominator can be canceled, resulting in 2x − 15 (5x4 − x3 + 2)(2x − 15) = . (14x2 + 3x − 28)(5x4 − x3 + 2) 14x2 + 3x − 28 Sometimes, the cancellation can be less obvious. For example, the rational expression 1 3 8x − 1 x2 + 2x + 4 can be simplified to 18 (x − 2) because, by an identity (6.5) on page 306, 1 3 1 1 1 x − 1 = (x3 − 8) = (x3 − 23 ) = (x − 2)(x2 + 2x + 4) 8 8 8 8 and we can cancel the number (x2 + 2x + 4) from the numerator and denominator. (As we know from the theory of quadratic equations, e.g., Section 2.1 in [Wu2020b], this particular rational expression in x actually makes sense for all (real) numbers x because x2 + 2x + 4 has a negative discriminant and is therefore never equal to 3 −8 = x − 2 is in fact an identity for all (real) 0. Consequently, the identity x2x+2x+4 numbers x.) In beginning algebra, often there is too much emphasis on simplifying rational expressions. This is a leftover from the ill-informed practice of teaching fractions by insisting on the reduction of all fractions to lowest terms. It remains to round off this discussion by mentioning that one can easily define polynomials in several numbers x, y, z, etc., and therefore one can likewise define rational expressions in x, y, z, etc.

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SUMMARY: All computations involving expressions in a number x are the same as ordinary computations in rational numbers. A thorough knowledge of rational numbers is therefore the foundation for learning algebra (see the recommendations on pp. 17–18 of [NMAP1]; see also pp. 3-40 to 3-41 of [NMAP2]). Any effort to teach algebra to students who do not know rational numbers is probably doomed from the start. Mathematical Aside: Rational functions are to polynomials as the rational numbers, Q, are to the integers, Z. Precisely, rational functions are elements of the quotient field of the integral domain of polynomial, in the same way that Q is the quotient field of the integral domain of Z. This is the advantage of the abstract viewpoint in advanced mathematics: it allows us to better perceive the coherence of mathematics. Pedagogical comments on the teaching of "variable" Although getting every school student to learn the fundamentals of algebra is a national goal (e.g., page xviii of [NMAP1]), the education establishment does not seem to have tried very hard to make algebra learnable. One reason is that TSM20 erects an artificial road block in students’ learning path by making the understanding of the "concept of a variable" a sine qua non of algebraic proficiency. A typical pronouncement on the importance of the "concept of a variable" goes as follows: Understanding the concept of variable is crucial to the study of algebra; a major problem in students’ efforts to understand and do algebra results from their narrow interpretation of the term. . . . Two particularly important ways in grades 5–8 are using a variable as a placeholder for a specific unknown, as in n + 5 = 12, and as a representative of a range of values, as in 3t + 6. ([NCTM1989, p. 102]) One way for students to acquire more than a "narrow interpretation" of the "concept of variable" would be to teach them the precise meaning of the "concept of variable" in a forthright and understandable manner. What then does TSM have to offer? Nothing that would serve this purpose, as can be seen from the following random sample from school textbooks: A symbol is a variable. A variable is a letter used to represent one or more numbers. A variable is a quantity that changes or varies. Variables represent quantities whose values vary. A variable is a letter or other symbol that can be replaced by any number (or other object) from some set. . . . A sentence with a variable is called an open sentence, and it is called open because its truth cannot be determined until the variable is replaced by values. There is also a video on the internet ([KhanAcad]) that illustrates very well why students are confused over the TSM "concept of variable". Incidentally, the reference 20 See

page xiv of the preface for the definition of TSM.

6.1. SYMBOLIC EXPRESSIONS

319

to "open sentences" in the last item of the preceding list is part of the erroneous thinking that√led to the problem of asking students to interpret a string of symbols such as y = 3x − 7 on page 299. You may have gotten the idea by now that, in order to make mathematics learnable, we must express mathematical messages clearly and precisely. But if clarity and precision are what you are after, then TSM would be the wrong place to look because, for instance, none of the preceding so-called definitions of a "variable" has the necessary clarity and precision that we normally demand of a mathematical definition. A "variable" is certainly not "a quantity that changes or varies", because nothing in mathematics ever changes or varies. Let us get down to the details of this assertion. Anticipating a later discussion about functions, suppose we want to say one real-valued function f is less than another function g; i.e., in symbols, f < g. In the spirit of "a quantity that changes or varies", this is a formidable statement about one quantity f (x) that varies being smaller than another quantity g(x) that also varies. This suggests that, even when two quantities wiggle all over the place, one can discern in some mysterious fashion that one is "smaller" than the other. This is heady stuff, but for the purpose of understanding and doing mathematics, the correct definition of f < g is much simpler and much more mundane, and it is this: for every point x0 in their common domain of definition, the two numbers f (x0 ) and g(x0 ) satisfy f (x0 ) < g(x0 ). In other words, just check one point at a time, so that at each x0 , f (x0 ) < g(x0 ). Nothing varies. The truth is that a variable is not a mathematical concept. It is rooted in tradition and was used as suggestive language back in the days when the very concept of a function had not been clearly formulated and mathematics did not possess the transparency and precision that it does today. In the year 2020, the "concept of variable" is kept alive as an integral part of mathematics only in TSM. It is time to teach students how to use symbols correctly and not allow this bogus "concept" to wreak havoc on student learning. It is possible that you consider this insistence on the proper use of symbols— the basic etiquette in the use of symbols (page 299)—to be nothing more than a piece of pedantry flaunted only by professional mathematicians. To dispel this misconception, consider a typical passage from TSM:21 y = −3 implies y sin x = −3 sin x, for all real values of x but y sin x = −3 sin x does not imply y = −3 because the principle that ac = bc a = b [sic] does not apply when c = 0 (i.e., a(0) = b(0) does not imply that a = b). ([MUST], page 395)22 We will take for granted that x and y are numbers. Consider the second phrase, but y sin x = −3 sin x does not imply y = −3. Without knowing what x is, this phrase can have no meaning; it may be true or it may be false, as the ensuing discussion shows. Let us try to properly quantify the symbol x. In context, the beginning of the passage ("y = −3 implies y sin x = −3 sin x, for all real values of x") suggests that, perhaps, what is intended is the 21 As noted in the footnote on page 303, we continue to make use of some mathematics that we have not yet discussed, but which you most likely know, to illustrate a point. 22 There is an obvious typo in the original, which states "x = −3 implies x sin x = −3 sin x but x sin x = −3 sin x does not imply x = −3". We have done our best to make sense of it.

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statement that "but y sin x = −3 sin x for all real values of x does not imply y = −3". If so, then this statement is incorrect since by letting x = π2 , we get y·1 = −3·1 so that we get y = −3, contrary to the claim. Now it is possible that what is intended is, instead, the following: but if x = π, then y sin x = −3 sin x does not imply y = −3. Indeed, we then get y · 0 = −3 · 0, so that 0 = 0 and one cannot reach the conclusion that y = −3. This would be consistent with the last part of the passage, although if this is the case, it would be far simpler to replace the whole passage by a direct statement: "if a, b, and c are real numbers, then unless we know c = 0, the equality ac = bc does not imply a = b." In summary, we see that the above passage is meaningless because symbols are used without proper quantification. Moreover, when the symbols are properly quantified, the passage becomes either wrong or trivial. But we should not take the easy way out by ascribing such errors to the carelessness of the authors of the preceding passage. This must be recognized for what it is, a systemic error in TSM. Those of us who allow TSM to be taught to generations of students and ensure that TSM (rather than mathematics) remains the enduring content component of mathematics education research are the main culprits. Let us do better and teach the basic etiquette in the use of symbols in mathematics education. Exercises 6.1. (1) (a) Show that every positive integer can be expressed as 3k, 3k+1, or 3k+2 for a whole number k. (Hint: Use division-with-remainder.) (b)√Use (a) to show that there are no positive integers x and y so that y = 3x − 7. (c) Show that there are an infinite number of fractions x and y so that √ y = 3x − 7. (2) Suppose you know (x − y)2 = x2 − 2xy + y 2 for all numbers x and y. Using this fact alone, prove that (x + y)2 = x2 + 2xy + y 2 for all numbers x and y. (3) If x is a nonzero number and n is a positive integer, what is 1 1 1 1 1 1 − 6 + 9 − 12 + · · · + 6n−3 − 6n ? 3 x x x x x x (4) Find the sum of −1 +

57 58 59 510 532 − + − + · · · − . 68 69 610 611 633 (5) If x is a number that makes all the denominators nonzero in the following, simplify (2x3 − 9x2 − 5x)/(x − 2)2 . (x2 − 3x − 10)/(x4 − 16) y 1 + 3 =? (6) If x and y are numbers and x = y, what does 2 x − y2 x + y3 Simplify your answer as much as possible. (7) It was known to Archimedes that for any positive integer k > 1 and for any positive integer n, 1 1 1 1 k 1 + + 2 + ··· + n + . = n k k k (k − 1)k k−1

6.1. SYMBOLIC EXPRESSIONS

(8)

(9)

(10)

(11)

(12)

321

(a) Prove that Archimedes’ identity is correct even when k is any number different from 0 and 1. (b) Conversely, prove that the generalized Archimedes identity of part (a) implies the summation formula for the finite geometric series. We have seen that identity (6.8) on page 309 is a special case of identity (6.5) on page 306. Now prove the converse: identity (6.8) implies identity (6.5). a−b 2 2 (a) Prove the following identity: ab = ( a+b 2 ) − ( 2 ) for all numbers a and b. (b) Use this identity to give a different proof of part (b) of Exercise 18 in Exercises 2.6, page 132; i.e., among all rectangles with a fixed perimeter, the square has the biggest area. Let a, b be positive integers, not both equal to 1, and let n be an odd positive integer > 1. Prove that an + bn is not a prime. Would this hold if n is even? (a) Factor 4x2 − 12x + 9 and 30x2 + 16x + 2 for any number x. (b) Factor (30x2 − 16x + 2)2 − 5(30x2 − 16x + 2) − 14 for any number x. (c) Factor s4 + s2 t2 + t4 into polynomials in s and t with integer coefficients for any numbers s and t. (d) Factor s4k + s2k t2k + t4k for any positive integer k. Simplify (assuming the denominators below are never zero for the numbers x and y):

4x4 − 9y 4 ; (i) 4x4 + 12x2 y 2 + 9y 4

15x3 y 4 − x4 y 3 (ii) ; 60x5 y 2 − 4x4 y 3

(iii)

2x2 +7x+3 x3 −7x+6 x+2 x−1

.

(13) How much money is in an account at the beginning of the sixteenth year if the initial deposit is $500, the annual interest rate is 5%, and at the beginning of each year starting with the second, $10 is added to the account? Write down the formula, and then use a scientific calculator to get a numerical answer. In each of the following exercises, you are asked only to write the equations that fully capture the verbal information. No solution is required. (14) Two women started at sunrise and each walked at constant speed. One went straight from City A to City B while the other went straight from B to A. They met at noon and, continuing with no stop, arrived respectively at B at 4 pm and at A at 9 pm. If the sunrise was x hours before noon and if L is the speed of the woman going from A to B and R is the speed of the woman going from B to A, transcribe the information above into equations using the symbols L, R, and x. (15) The sum of the squares of three consecutive integers exceeds three times the square of the middle integer by 2. If the middle integer is x, express this fact in terms of x. If the smallest of the three integers is y, express the same fact in terms of y. (16) I have $4.60 worth of nickels, dimes, and quarters. There are 40 coins in all, and the number of nickels and dimes is three times the number of quarters. If N , D, and Q denote the number of nickels, dimes, and quarters, respectively, write equations in terms of these symbols to capture the given information.

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(17) We are given two whole numbers so that, when the larger number is divided by the smaller number, the quotient is 9 and the remainder is 15, and so that the larger number is 97.5% of ten times the smaller number. If x is the larger number and y is the smaller number, express the given information in equations in terms of x and y. (18) A video game manufacturer sells out every game he brings to a game show. He has two games, an A Game and a B Game. He can bring 50 of A Games and B Games in total to the show. Each A Game costs $75 to manufacture and brings in a profit of $ 125. Each B Game costs $165 to manufacture and brings in a profit of $185. However, he only has $6,000 to spend on manufacturing. If he brings x A Games and y B Games, describe in terms of x and y how he can maximize his profit. 6.2. Solving linear equations in one variable This section explains what an equation in one variable is and what it means to solve it. Because "solving equations" is the most basic part of school algebra, it is difficult to imagine that TSM could get it wrong—completely wrong—for so long, but this is the reality of school mathematics education in the grips of TSM. We will present two ways to make sense of the usual TSM procedure of solving an equation and, in the process, lay bare the fact that, without the ability to compute fluently with rational numbers, it is impossible to learn algebra. What is an equation? (p. 322) How to solve an equation (p. 324) A second way to solve an equation (p. 328) What is an equation? We have already discussed "equations" informally on pp. 300ff., but it is time to define this concept precisely. An equation in one variable x is a question that asks, when two expressions f (x) and g(x) in a number x are given, whether there is a number k so that f (x) is equal to g(x) when x = k, i.e., so that f (k) = g(k). Such a number k, if it exists, is called a solution of the equation. Sometimes we also say k satisfies the equation. By tradition, this question (i.e., the equation) is symbolically written in the form f (x) = g(x). We can understand this better by looking at an example. Let the two expressions be 2x2 − x − 1 and 0 (= 0 · x). Then the question of whether there is a number k so that 2k2 − k − 1 = 0 is known as a quadratic equation in one variable and, of course, this question usually appears as 2x2 − x − 1 = 0. In this case, a solution k is a number so that 2k2 − k − 1 = 0. For example, it is easy to check that 3 is not a solution of the equation 2x2 − x − 1 = 0, because 2 · 32 − 3 − 1 = 14 = 0, whereas 1 and − 12 are solutions of 2x2 − x − 1 = 0. Because we do not know ahead of time what the solutions of an equation are going to be or in fact if there is any solution at all, the letter x is also called an unknown (see page 301).

6.2. SOLVING LINEAR EQUATIONS IN ONE VARIABLE

323

To solve an equation is to obtain all the solutions of the equation. In the preceding example of 2x2 − x − 1 = 0, we already know that 1 and − 12 are solutions, but we have not yet solved this equation because there may be other solutions. It turns out that 1 and − 12 are all the solutions of 2x2 − x − 1 = 0 (see Section 2.1 of [Wu2020b]). Assuming this fact, we have solved the equation 2x2 − x − 1 = 0, and its solutions are 1 and − 12 . In general, we have no prior guarantee of how many solutions an equation has. There may be exactly one (e.g., 3x = 2), there may be an infinite number (e.g., x2 + 2x + 1 = (x + 1)2 ), or there may be none (e.g., x2 + 1 = x2 − 3). To the extent that there may be many solutions to a given equation in x, the symbol x is sometimes called the variable of the equation to indicate that there may be more than one number x to make the equation valid (see page 301 again). Activity. Prove the assertions about the number of solutions of the given equations in the preceding paragraph. When the two expressions are linear polynomials, let us say ax + b and cx + d in x, where a, b, c, d are given numbers, then the question of whether there is a number k so that ak + b = ck + d is called a linear equation in one variable, the variable being x. The given numbers a, b, c, d are the constants (see page 301) of the equation. The usual representation of this linear equation is ax + b = cx + d. Be that as it may, one should not lose sight of the fact that "ax + b = cx + d" is nothing more than a compact representation of the preceding question. More generally, if two number expressions in a number x have the property that each expression becomes an expression of the form ax + b after applications of the commutative, associative, and distributive laws, then the question about whether they are equal or not will also be called a linear equation in one variable. Thus the equation in x, 5x − 75 + (2 − 4x) = 6x − (68 + 3x) − 7, is a linear equation in one variable because it is easily seen to be the linear equation x − 73 = 3x − 75 (5x − 75 + (2 − 4x) = x − 73 and 6x − (68 + 3x) − 7 = 3x − 75). What has been said about equations in one variable extends to equations in any number of variables. For our need, we single out the case of two variables. Thus, an equation in two variables x and y, normally written as f (x, y) = g(x, y), is a question that asks, when two expressions f (x, y) and g(x, y) in the numbers x and y are given, whether there is a pair of numbers k and  so that f (k, ) = g(k, ). Such a pair k and  is called a solution of f (x, y) = g(x, y). To solve the equation f (x, y) = g(x, y) is to find all its solutions. Pedagogical Comments. Some readers may be aghast that we actually take the trouble to give a definition of an equation. Isn’t an equation something that students solve, and don’t they all recognize this fact with ease? Perhaps, but there are at least two reasons for offering such a definition. The first is education researchers’ finding that many elementary and middle school students interpret the equal sign as an operational command to perform a calculation rather than as a statement about two numbers, two sets, etc., being "equal" and that this misconception hampers their learning of algebra (see, e.g., [Carpenter et al.] and [McNeil et al.]). In other words, deep down, many students do not know what an

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equation is! A second reason is more subtle. We have emphasized all along the need for precision and transparency to make school mathematics more learnable. Now, an equation such as x2 = 7x − 10 has a "variable" x on each side, and a "variable" is something that "varies", according to TSM. So if both of the expressions x2 and 7x − 10 are "varying" all the time, what does it mean to say that they are "equal"? It certainly does not mean that no matter how x varies, the equality x2 = 7x − 10 holds for all values of x (try x = 0, for instance). It is a fact that TSM does not even explain what the equality x2 = 7x − 10 involving a variable x is all about. Yet, by routinely asking students to solve such an "equation", TSM is as good as telling them not to ask what an "equation" is23 but just go ahead and "do something with it to get an answer". We can all agree that it is imperative to remove this mindset from school classrooms. To this end, a first step will be to offer students a precise definition of an equation and, strictly on the basis of this definition, show them how to solve the equation. We will do just that in the next subsection. We note that this definition of an equation in a symbol is not the traditional one in education research (cf. [McCrory et al.] and further references therein). End of Pedagogical Comments. How to solve an equation In this subsection, we will limit ourselves to solving linear equations of the form ax + b = cx + d. The main theorem about solutions to a linear equation is Theorem 6.3 on page 326. Note, however, that the reason we spend the effort to explain how to solve linear equations is that the same principle will be broadly applicable to the solution of any equation. Therefore it will be effort well spent. In TSM, a linear equation in the number x such as 27 x+1 (6.13) 3x − 4 = 5 is supposed to be solved via the following easily recognizable steps. 27 27 (a) − 27 5 x + (3x − 4) = − 5 x + ( 5 x + 1). (b) − 12 5 x − 4 = 1. (c) (− 12 5 x − 4) + 4 = 1 + 4. (d) − 12 5 x = 5. 5 5 (e) (− 12 )(− 12 5 x) = (− 12 ) · 5.

(f) x = − 25 12 . 27 The conclusion is that − 25 12 is the solution of the equation 3x − 4 = 5 x + 1. Perhaps you are so accustomed to this routine method of solution that you can no longer see its many flaws or imagine how it can stultify a beginner’s mathematics learning. Let us consider a few questions that a beginner might raise: (A) How do we know that the "solution" in step (f), i.e., − 25 12 , is a solution of equation (6.13)? Nothing in the computations given in (a)–(f) shows that − 25 12 satisfies equation (6.13) or that it is the only solution. (B) Recall that, in TSM, the x is a "variable". The six steps (a)–(f) make the assumption that we can do arithmetic with the "variable" x as if it were a number. If one interprets x as a "quantity that varies", then this makes no sense because, 23 Once

again, "Ours is not to reason why."

6.2. SOLVING LINEAR EQUATIONS IN ONE VARIABLE

325

thus far, there is no theorem that says "quantities that vary" satisfy the associative, commutative, and distributive laws. Yet, (a)–(f) freely make use of these laws. For example, in going from (a) to (b), the distributive law is used to conclude that 12 − 27 5 x + 3x = − 5 x. But why? (C) Now suppose we interpret x as just a number (cf. the second item in the TSM definitions of a "variable" on page 318); then the equations in (a)–(f) become equalities between numbers. The question is whether the number x has to be some special number or just any number for (a)–(f) to hold. For example, if x = 0, then (b) would read "−4 = 1", which is absurd, and (d) would read 0 = 5, equally absurd, and so on. If x has to be a special number, then what is it, and why wasn’t this announced from the beginning? It is possible that beginners do not know enough to articulate these doubts, but very likely they have them in the back of their minds. Yet, after years of not getting satisfactory answers—because TSM does not provide answers to basic questions— most students have learned to suppress their natural curiosity by the time they get to middle school for the sake of "getting the right answer". So they stop asking why and just follow instructions. The end result is that students coming out of TSM know how to get answers by memorizing facts and following template solutions, but apparently not much else. Unfortunately, in the year 2020, robots are beginning to excel exactly at carrying out instructions to perform preassigned tasks (see, e.g., [Paquette] and [WikiAlphaGo]). TSM therefore threatens to produce students who, upon graduating from high school, possess no technical skills that can help them outperform a robot. In other words, until we improve school mathematics education, our public education system will run the risk of producing students who become instantly obsolete upon graduation. We have to avert this catastrophe by making a real effort to teach students how to reason, which seems—for the time being, at least—to be beyond the capabilities of robots. Let us forsake TSM and make an effort to shed light on every murky corner of school mathematics to make it transparent and learnable so that we can answer students’ questions, and let us encourage them to never cease asking why. For the case at hand, let us explain how to solve a linear equation correctly. We begin by confronting the fact that, as it stands, the solution method described in (a)–(f) makes no sense.24 However, because of its simplicity and the fact that it works, this solution method is not going away anytime soon. For this reason, we will reinterpret (a)–(f) in two different ways to help students make sense of them (see pp. 327 and 329). But first, we will explain a correct method of solving equation (6.13), i.e., 3x − 4 = 27 5 x + 1. To begin with, we do not claim that there will be any solution to this equation. Rather, we take the position that if there is such a solution, let us say x0 , then we can find out what x0 must be. Let us therefore assume that there is such a solution x0 to equation (6.13). Then, by the definition of a solution, we have 27 3x0 − 4 = x0 + 1. 5 24 Mathematical Aside: From an advanced standpoint, one may be tempted to argue that x + 1 is taking place in the polynomial ring R[x]. This still makes no the equation 3x − 4 = 27 5 sense, however, because two polynomials in R[x] are equal if and only if the coefficients of each monomial in the two polynomials are pairwise equal. In this case, the coefficient of x on the left . Therefore the two sides can never be equal in is 3, whereas the coefficient of x on the right is 27 5 R[x] either.

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We may therefore apply the ordinary arithmetic operations to the numbers on both 27 sides of this equality. By adding − 27 5 x0 to each side of this equality, we get − 5 x0 + 3x0 − 4 = 1, which is (6.14)

−

12 x0 − 4 = 1. 5

In standard terminology, what we have just done is transpose the term 27 5 x0 to the other side. Using the same terminology, let us transpose −4 to the right side by adding 4 to both sides, thereby getting − 12 5 x0 = 1 + 4, which is (6.15)

−

12 x0 = 5. 5

5 5 Now multiply both sides of this equality by − 12 and we get x0 = (− 12 ) · 5, which is

(6.16)

x0 = −

25 . 12

We pause to observe that—if we are willing to conflate x0 with x—then equations (6.14)–(6.16) are identical with the equations in (b), (d), and (f) on page 324.25 To resume our discussion, let us summarize what we have accomplished: we have proved that if there is a solution x0 to equation (6.13), i.e., 3x − 4 = 27 5 x + 1, then this x0 must be equal to − 25 . In other words, we have just proved that the 12 solution to equation (6.13) is unique: if it exists at all, it will have to be − 25 . Note 12 that this says nothing about − 25 12 being a solution to the original equation (6.13). However, the verification that, indeed, − 25 12 is a solution to (6.13) is straightforward: simply check that     25 27 25 3 − −4= − + 1. 12 5 12 This is routine, provided one is fluent in computations with rational numbers: both 25 27 sides are equal to − 123 12 . So − 12 is a solution of 3x − 4 = 5 x + 1. 25 Altogether, we have just proved that − 12 is the unique solution of the equation 27 25 3x − 4 = 27 5 x + 1. Thus we have solved 3x − 4 = 5 x + 1, and its solution is − 12 (remember that solving an equation means obtaining all its solutions). The preceding reasoning can be seen to be independent of the specific equation used (3x − 4 = 27 5 x + 1) and is valid in a general context. It leads to the following theorem: Theorem 6.3. The equation ax + b = cx + d where a, b, c, d are constants and d−b . a − c = 0 has the unique solution a−c

25 The reader cannot help but notice that we have just referred to (6.14)–(6.16) and (b), (d), (f) as "equations", whereas these are not equations in the sense of page 322. Here then is another instance where we have to face up to a common linguistic abuse and learn to tiptoe through a linguistic minefield: it is a common practice in mathematics to refer to any displayed collection of symbols involving the equal sign as an "equation".

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327

Proof. The proof is broken into two steps: d−b Step 1. If x0 is a solution of ax + b = cx + d, then x0 = a−c . d−b Step 2. a−c is a solution of ax + b = cx + d. (Compare Exercise 9 on page 120.)

Because the proofs of these two steps are so similar to the solution of the equation 3x − 4 = 27 5 x + 1, we can safely leave them as Exercise 2 on page 330. The theorem can therefore be considered to be proved. Pedagogical Comments. We have explained why the TSM method of solution given in (a)–(f) on page 324 makes no sense. By contrast, since the solution of the equation 3x − 4 = 27 5 x + 1 given above involves nothing more than simple computations with rational numbers (and nothing about "variables"), it should be understandable by one and all. The key point is not to compute with a "variable" but to work with an assumed solution x0 from the beginning and, through direct computations, deduce what x0 has to be (i.e., Step 1 in the preceding proof) and then turn around to check (i.e., prove) that this candidate for a solution is in fact a solution (i.e., Step 2 in the preceding proof). In this light, we come to an understanding of what (a)–(f) on page 324 are all about: although (a)–(f), as is, are fatally flawed because they purport to compute with a "variable" and the equalities cannot be justified, they are seen to be procedurally identical to our computations with the assumed solution x0 (compare the remarks about equations (6.14)–(6.16) and (b), (d), and (f) on page 326). This is where Theorem 6.3 comes in: it guarantees that the computations in (a)–(f), when x is replaced by x0 , do lead to the correct solution of the equation (as confirmed by Step 2 in the proof of the theorem). Therefore, as far as getting the correct solution is concerned, (a)–(f) are good for that purpose. This explains why, procedurally, school textbooks can get away with (a)–(f) as the "method of solution" even though it is fundamentally nonsensical. Needless to say, for the purpose of solving equations, we do not want school students to repeat this long-winded ritual, every time, of first getting a candidate for the solution by assuming that there is one, finding out what it is, and then verifying that it is a solution by substituting it back into the original equation. Our new understanding of (a)–(f) suggests a more reasonable way to teach how to solve linear equations (or any equation) in the school classroom: explain to students at least the essence of the proof of Theorem 6.3, especially the idea that one begins by assuming there is a solution in order to find out what it is and that the computations in (a)–(f) are not computations with a "variable" but those with a number, namely, an assumed solution of the equation. Make sure they know that solving an equation is an exercise in number computations rather than some magical incantation in the land of the great beyond about something called a "variable". Once they can demonstrate an understanding of the overall structure of solving an equation encoded in Step 1 and Step 2 of the preceding proof, students should be allowed to use the mechanical procedure of (a)–(f) above as a shorthand method for finding a solution. A pleasant byproduct of such an explanation is that students will better understand why they are required to check their work by verifying the solution so obtained (because Step 2 above is an integral part of the solution).

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If you as a teacher can make sense to students about what they are learning, they will repay your efforts by making sense to you. End of Pedagogical Comments. A second way to solve an equation We will now revisit the proof of Theorem 6.3 and recast it in a more abstract setting. First we define two equations to be equivalent if they have the same solutions; i.e., every solution of one is also a solution of the other. We also introduce the following two operations on a linear equation in a number x. (E1) Add the same number or same monomial to the expressions on both sides of an equation. (For example, adding 21 to both sides of 15x−21 = 28x+3 results in the expressions 15x−21+21 and 28x + 3 + 21.26 ) (E2) Multiply the expressions on both sides of an equation by the same nonzero number. (For example, multiplying both sides 1 1 leads to the expressions 15 (15x − 21) of 15x − 21 = 28x + 3 by 15 1 and 15 (28x + 3).) We can see the relevance of these operations by noting that we used (E1) to obtain equation (b) on page 324 from equation (6.13) by adding − 27 5 x to both sides of the 27 original equation 3x − 4 = 5 x + 1, and we obtained (f) from (d) on the same page 5 by multiplying both sides of the equation in (d) by (− 12 ). We should also point out a key feature of both (E1) and (E2): each operation is reversible in the sense that if, in (E1), we add a number or monomial A to both sides, then if we follow it by adding −A to both sides, we get back the original equation. Similarly, in (E2), if we multiply both sides of an equation by a nonzero number B, then if we follow it by multiplying both sides by B1 , again we get back to the original equation. Incidentally, the second comment shows why in (E2) we stipulate that the number B be nonzero, as otherwise B1 would make no sense. With these preparations out of the way, we now come to the main point: Lemma 6.4. Applying either of the operations (E1) and (E2) to a given linear equation of one variable results in an equivalent equation. Proof. Instead of a general proof of Lemma 6.4, we will offer a proof of the lemma for equation (6.13), which is 27 x + 1. 3x − 4 = 5 It will be seen that the reasoning in this special case is in fact perfectly general. First, let us prove the part of the lemma about (E1). So let us add, let us say, − 27 5 x 12 to both sides of (6.13) to get the expressions − 5 x − 4 and 1, and the resulting equation is 12 (6.17) − x − 4 = 1. 5 We will show that equations (6.13) and (6.17) are equivalent. Suppose x0 is a solution of equation (6.13); then we have an equality of numbers: 3x0 − 4 = 27 5 x0 + 1. If 26 Notice that we are making use of Theorem 1 of the appendix in Chapter 1 (page 87) so that there is no need to use any parentheses on either side.

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329

12 we add − 27 5 x0 to both sides of this equality, we get − 5 x0 − 4 = 1, which is exactly the statement that x0 is also a solution of equation (6.17). Conversely, suppose x1 is a solution of equation (6.17) and we must show that it is also a solution of 27 equation (6.13). Thus by hypothesis, − 12 5 x1 − 4 = 1. Now we add the number 5 x1 to both sides of the last equality to get 3x1 − 4 = 27 5 x1 + 1. But this says x1 is a solution of equation (6.13), and the proof of the part of Lemma 6.4 about (E1) is complete. We note that this reasoning depends on the fact that (E1) is reversible. The reasoning with the part of Lemma 6.4 about (E2) is similar. We may therefore consider the proof of Lemma 6.4 to be complete.

We are now in a position to give a Second Proof of Theorem 6.3 (see p. 326). Thus let the equation ax + b = cx + d be given so that a − c = 0. By repeated applications of Lemma 6.4, we have the following: (A) The equation ax + b = cx + d is equivalent to (−cx) + ax + b = (−cx) + cx + d, (B) which is equivalent to (a − c)x + b = d, (C) which is equivalent to (a − c)x + b + (−b) = d + (−b), (D) which is equivalent to (a − c)x = d − b, 1 1 (a − c)x = a−c (d − b), (E) which is equivalent to a−c d−b . (F) which is equivalent to x = a−c d−b , we see that the equation Since the last equation clearly has the unique solution a−c d−b . The proof of Theorem 6.3 is complete. ax + b = cx + d has the unique solution a−c

This new proof of Theorem 6.3 should be complemented by two comments. The first is that it gives us another way to salvage the TSM method of solving an equation given in (a)–(f) on page 324. Indeed, if we specialize the proof of Theorem 6.3 to equation (6.13), then what we get is the following: (A) The equation 3x − 4 = 27 5 x + 1 is equivalent to 27 27 − 27 x + (3x − 4) = − 5 5 + ( 5 x + 1), 12 (B) which is equivalent to − 5 x − 4 = 1, (C) which is equivalent to (− 12 5 x − 4) + 4 = 1 + 4, (D) which is equivalent to − 12 5 x = 5, 5 5 (E) which is equivalent to (− 12 )(− 12 5 x) = (− 12 ) · 5, 25 (F) which is equivalent to x = − 12 . 25 Therefore the solutions of 3x−4 = 27 5 x+1 are exactly the same as those of x = − 12 , 25 which consist of one number, − 12 . Now, the foregoing steps (A), (B), . . . , (F) are the correct mathematical counterparts of the steps (a), (b), . . . , (f) on page 324, respectively. In greater detail, what (a)–(f) really try to say—in view of (A)–(F)—is not that the solutions of 3x − 4 = 27 5 x + 1 can be obtained by the computations in (a)–(f), but that (a)–(f) present a succession of equations, each being equivalent to 3x − 4 = 27 5 x + 1, so that at the end the equation so obtained is so simple (i.e., x = − 25 ) that its solutions can be read off. 12 A second comment is that the proof of Theorem 6.3 shows that solving a linear equation is conceptually very simple: use (E1) to show that the given equation is equivalent to an equation of the form Ax = B, where A and B are constants (this is called isolating the variable; see step (D) above), and then use (E2) to conclude

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that the equation Ax = B has the unique solution unique solution of the original equation.

B A.

Therefore, the latter is the

Pedagogical Comments. School textbooks traditionally teach the solving of linear equations by breaking it up into the solving of one-step equations, two-step equations, and multistep equations. When something as simple as the solution of a linear equation is broken up in this incomprehensible manner and especially if at the end of the discussion no overview about what has taken place is given, distortion of the meaning of solving an equation is bound to take place in students’ minds (see the comment about students’ misconception of the equal sign on page 323). Please do not do this to your students when you teach. Rather, teach students to use (E1) to isolate the variable in a linear equation, thereby getting an equivalent equation Ax = B for some constants A and B, from which one obtains the desired solution B A. It is worth noting that either proof of Theorem 6.3 makes it obvious why fluency with rational number computations is a pre-condition for learning algebra. If one has the correct understanding of what it means to solve an equation, then one would realize that solving a linear equation is nothing more than a careful application of computations with rational numbers. Your job now is to bring this message to all your students. End of Pedagogical Comments. Finally, we caution about a possible misunderstanding about Theorem 6.3: it does not say that every linear equation of one variable, ax+b = cx+d, has a unique solution. What it says is that this is true if a − c = 0. We now give some examples to round off the picture. The equation 2x − 1 = 2x + 1 clearly has no solution, and the equation 2x − 3 = 7x − (5x + 3) has an infinite number of solutions as every number is a solution. We leave the general case to an exercise below (Exercise 5). Exercises 6.2. (1) Prove that ab is a solution of the equation ax = b, where a = 0, and that any solution of the equation is equal to ab . (2) Give the details of the proofs of Step 1 and Step 2 in the proof of Theorem 6.3 on page 326. (3) Write out a proof of Lemma 6.4 on page 328 for a general linear equation ax + b = cx + d. (4) Imagine that you are teaching eighth-grade algebra and you have to explain for the first time how to solve 2x + 11 = 2 − x. How would you explain (a) what an equation is, (b) what it means to solve this equation, and (c) how to actually solve it, step by step? (5) Formulate a precise statement about a linear equation in x, ax+b = cx+d, where a, b, c, d are the constants, so that the equation has a unique solution, has no solution, and has an infinite number of solutions. (Of course you will have to prove it.) 5 of a number N exceeds a third of N by 8, what is N ? (This is (6) If 13 Exercise 15 on page 68.) (7) (In this exercise, you are assumed to know the formula for the circumference of a circle.) Aaron and Warren start walking around a circle from the same spot at the same time. They walk in the same direction,

6.3. SETTING UP COORDINATE SYSTEMS

331

Aaron at a constant speed of F feet per minute and Warren at 2F feet per minute. After T minutes, they are at the greatest distance apart for the first time. (i) What is the radius of the circle in terms of F and T ? (ii) Without any computations, can you explain how many laps around the circle Warren has walked after T minutes? (8) In a pen containing chickens and rabbits, the ratio of chickens to rabbits is 1 : 3. If we count the total number of feet of the animals in the cage, there are 294. How many chickens and how many rabbits are there? (9) Given three consecutive whole numbers, if the sum of the smallest plus twice the next number plus three times the largest number is 110, what are these numbers?

6.3. Setting up coordinate systems To prepare for the discussion of linear equations in two variables, this section introduces the basic idea of associating each point of the plane with a unique ordered pair of numbers (x, y), called the coordinates of the point. This is the familiar process of fixing a pair of perpendicular lines as coordinate axes and using them as the reference for the introduction of coordinates. Having coordinates for points makes it possible to translate algebraic objects into geometric objects, and vice versa; the rest of this volume and [Wu2020b] and [Wu2020c] may be regarded as nothing more than a sequence of exercises to demonstrate how this is done. One striking example is the rendering of the distance between two points (a geometric concept) into an algebraic expression in terms of their coordinates (see the distance formula on page 336). We are going to introduce coordinates in the plane, in the sense that we will associate to each point of the plane an ordered pair of numbers, and vice versa. Because you already have a procedural familiarity with these concepts, we will be brief. Incidentally, we will get to see in this process another reason why we took the trouble to prove Theorem G4 on page 226, which asserts that opposite sides of a parallelogram have equal length. Choose two perpendicular lines in the plane so that one of them is horizontal;27 let them intersect at a point to be called O. The horizontal line is traditionally designated as the x-axis, and the vertical one the y-axis. Together, the x-axis and the y-axis are called the coordinate axes and the point O is called the origin. The totality of the coordinate axes and the origin is said to form a coordinate system. We proceed to specify the choices of 0 and 1 on each coordinate axis to make it into a number line (see assumption (L3) on page 167). The point O will be the 0 on the x-axis, and the point A on the x-axis so that A is to the right of O and so that dist(A, O) = 1 will be the 1 on the x-axis. Now let ϕ be the 90-degree counterclockwise rotation around O, then ϕ(A) lies on the y-axis above O. On the

27 Mathematical Aside: As is well known, it is irrelevant whether either line is horizontal, but it is customary to keep things simple at the beginning by making this requirement.

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6. SYMBOLIC NOTATION AND LINEAR EQUATIONS

y-axis, let O be the 0 and ϕ(A) be the 1, as shown: Y ϕ(A) = 1 s

@

ϕ

O

s A=1

X

Of course the 1 on the y-axis could also have been specified as the point B on the y-axis above O so that dist(B, O) = 1. This is because ϕ, being distance-preserving, maps A to B (see (L7)(ii) on page 237). By our choices, the positive numbers on the x-axis are to the right of O and the positive numbers on the y-axis are above O on the y-axis. Naturally, the half-line (see Lemma 4.5 on page 173) on the x-axis to the right of O is called the positive x-axis; the opposite half-line on the xaxis is then the negative x-axis. The half-line on the y-axis above O is similarly called the positive y-axis, and the opposite half-line on the y-axis is the negative y-axis. Making a particular choice of the origin and the x- and y-axes is referred to as setting up a coordinate system. Clearly, there are an infinite number of ways to set up a coordinate system; i.e., there are an infinite number of choices of a pair of perpendicular lines (so that one of them is horizontal) as the x-axis and the y-axis. In the following discussion, it will be understood that we have made a fixed choice of a coordinate system. Let us pause to observe what we have achieved. Let R2 denote the set of all ordered pairs of real numbers,28 which by convention are written as (x, y) (where x, y ∈ R), so that the pair (a, b) and (b, a) are considered to be distinct if a = b. Therefore in R2 , (3, 4) = (4, 3), and in general, (a, b) = (c, d)

⇐⇒

a = c, b = d.

Again, note that there is no ambiguity as to what the equality between two ordered pairs of numbers means. Now, relative to a fixed coordinate system, we will assign an ordered pair of numbers to each point Q in the plane in the following way. We call any line that is either the given x-axis or parallel to the x-axis a horizontal line, and also any line that is either the given y-axis or parallel to the y-axis a vertical line. Then through Q draw two lines, one vertical and one horizontal, so that they intersect the x-axis at a point A and the y-axis at a point B, respectively. Let the point A correspond to the number a on the x-axis and the point B correspond to the number b on the y-axis. Then the ordered pair of numbers (a, b) are said to be the coordinates of Q (relative to the chosen coordinate axes); we write Q = (a, b), and a is called the x-coordinate and b the y-coordinate of Q (relative to the chosen coordinate axes). For clarity, we denote the x-axis and the y-axis by the letters X and Y (sometimes x and y), respectively.

28 The

superscript "2" in R2 reminds us that two real numbers are involved.

6.3. SETTING UP COORDINATE SYSTEMS

333

Y Qr

r A = (a, 0)

rB = (0, b)

O

X

Observe that, for this particular Q, a = −|OA| and b = |OB|. Also observe that A = (a, 0) and B = (0, b). It is customary to abuse the notation and replace the points (a, 0) on the x-axis by a and the points (0, b) on the y-axis by b. Therefore the preceding picture is normally labeled as follows: Y Q r

rb

r a

O

X

Activity. Prove that a horizontal line is always perpendicular to a vertical line. The assignment Q → (a, b) then defines a transformation Φ (upper case phi) from the plane to R2 once a coordinate system is fixed.29 Thus, we have Φ : the plane → R2 so that Φ(Q) = (a, b) as above. Conversely, with respect to the same pair of coordinate axes, we now show how to define a transformation Ψ : R2 → the plane (Ψ = upper case psi). Thus, given an ordered pair of numbers (x, y) in R2 , we define Ψ(x, y) to be the point of intersection of the vertical line passing through (x, 0) and the horizontal line passing through (0, y).30 These two lines being unique, by virtue of the parallel postulate, the point of intersection is also unique. Thus the point Ψ(x, y) is well-defined. It is easy to see that the composition Ψ ◦ Φ is the identity map of the plane and the composition Φ ◦ Ψ is the identity map of R2 . There is thus a bijection between all the points in the plane and R2 , the set of all the ordered pairs of numbers with respect to a fixed coordinate system in the plane (see page 205 for the definition of bijection). In view of the bijection Φ : the plane → R2 , we will often denote the plane by R2 . But be aware that, in so doing, it is always understood that the choice of 29 Strictly speaking, we have only defined transformations from the plane to the plane, so Φ is not a "transformation". However, (1) we will define a function in Section 1.1 of [Wu2020b], and this Φ will be seen to be a function, and (2) we can broaden the definition of a transformation at this point to include such a Φ if we wish. 30 Notice that strictly speaking, we should write Ψ((x, y)) rather than Ψ(x, y). But tradition dictates that we use the latter and not the former, and it must be said that, as a matter of convenience, tradition got it right this time!

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6. SYMBOLIC NOTATION AND LINEAR EQUATIONS

a coordinate system has been made. Still with the bijection Φ : the plane → R2 understood, we usually identify a point with its corresponding ordered pair of coordinates. The x-axis, and more generally a horizontal line H that intersects the y-axis at a number c, separates the plane into two parts. Denote by U p (respectively, Lo) all the points P so that the horizontal line passing through P intersects the y-axis at a number > c (respectively, < c). Then it is clear that the plane is the disjoint union (see page 175) of the nonempty sets U p, Lo, and H. U p and Lo are called the upper half-plane and lower half-plane of H, respectively. A point in U p is said to be above H; likewise, a point in Lo is said to be below H. Y Up

c

H Lo X

O

Some clarification of the terminology is in order. In assumption (L4) on page 176, we introduced the concept of the half-planes of a line, and it may be thought that the terminology of upper and lower half-planes for U p and Lo could lead to confusion. Actually this is not the case, because we will show in Section 1.4 of [Wu2020b] that U p and Lo are indeed "half-planes" in the sense of (L4) (compare also Exercise 3 on page 337). However, since the only thing that matters to us in the present discussion is to be able to refer to the sets U p and Lo but not so much whether U p and Lo satisfy the properties listed in (L4), we will defer the discussion of their relationship with (L4) to Section 1.4 of [Wu2020b]. There is clearly an analogous discussion for vertical lines. Thus, suppose a vertical line V intersects the x-axis at a number d; then we can define the left half-plane L of V (respectively, the right half-plane R of V ) to be all the points so that the vertical line passing through each of them intersects the x-axis at a number < d (respectively, > d). Then the plane is the disjoint union of the nonempty sets L, R, and V . A point in L is said to be to the left of V , and a point in R is said to be to the right of V . Y

V L

R

d

O

X

6.3. SETTING UP COORDINATE SYSTEMS

335

We will now use the new terminology of upper half-plane, etc., to give another interpretation of the coordinates of a point P . First we recall the picture: Y rB = (0, b)

P r

r A = (a, 0)

O

X

By construction, P AOB is a parallelogram. By Theorem G4 on page 226, the length of the segment P B is just |a|. Likewise, the length of the segment P A is just |b|. Since the line LP A is parallel to the y-axis and the y-axis is perpendicular to the x-axis, we see that LP A is perpendicular to the x-axis (Theorem G3 on page 224). For the same reason, LP B is perpendicular to the y-axis. Thus, |a| is in fact the distance of P from the y-axis in the sense of the definition on page 228, and |b| is the distance of P from the x-axis. We have therefore obtained a new interpretation of the coordinates of P : The x-coordinate of P is the distance of P from the y-axis if P is in the right half-plane of the y-axis, and it is minus this distance of P from the y-axis if P is in the left half-plane of the y-axis. The y-coordinate of P is likewise the distance of P from the x-axis if P is in the upper half-plane of the x-axis, and it is minus this distance of P from the x-axis if P is in the lower half-plane of the x-axis. The intersections of the four half-planes of the x-axis and the y-axis of a given coordinate system form four regions in the plane R2 . These are called the four quadrants of the coordinate system and are labeled I, II, III, and IV, as shown. Notice that the four quadrants do not include any points on the coordinate axes.

II

I

q O

III

IV

More formally, the four quadrants are defined as follows: Quadrant I: all the points (x, y) so that x > 0 and y > 0. Quadrant II: all the points (x, y) so that x < 0 and y > 0. Quadrant III: all the points (x, y) so that x < 0 and y < 0. Quadrant IV: all the points (x, y) so that x > 0 and y < 0. By introducing coordinates in the plane, we establish a bridge between geometry (points in the plane) and algebra (an ordered pair of numbers). This was the

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6. SYMBOLIC NOTATION AND LINEAR EQUATIONS

great insight of Pierre Fermat and René Descartes.31 The study of geometry using coordinates is usually called analytic geometry. An example of the immediate impact of the introduction of coordinates is the following distance formula between two points in the plane. Suppose two points (a, b) and (c, d) in the coordinate plane are given. We want to compute the distance between (a, b) and (c, d) in terms of the four coordinates. By (L5) (page 184), this distance is the length of the hypotenuse of the right triangle whose legs are the horizontal and vertical segments, as shown: V

Y

(c, d) rH HH H H

b (c, b)

HH HHr (a, b)

X c O Now, (a, b) and (c, b) are points on the horizontal line H passing through the point b on the y-axis. Hence, (6.18)

distance between (a, b) and (c, b) = |a − c|

(see (2.37) on page 126). Likewise, (c, b) and (c, d) are points on the vertical line V passing through the point c on the x-axis. Hence, (6.19)

distance between (c, b) and (c, d) = |b − d|.

It therefore follows from the Pythagorean theorem that the distance between (a, b)  and (c, d) is |a − c|2 + |b − d|2 . Since |x|2 = x2 for any number x, we have |a − c|2 = (a − c)2 and |b − d|2 = (b − d)2 . Hence,  (6.20) distance between (a, b) and (c, d) = (a − c)2 + (b − d)2 . This is the distance formula we are after. Before we leave the topic of setting up a coordinate system, we point out a practical issue for future reference about the pictorial representation of coordinate axes in the plane. If a point P is a point on the positive x-axis representing the number t, then dist(P, O) = t (see (L5)(iii) on page 184) and the number on the positive y-axis represented by ϕ(P ) (where ϕ is the 90◦ -counterclockwise rotation around O) would also be t because ϕ is distance-preserving. Then, in the usual pictorial presentations of the coordinate plane in textbooks, it is not practical to insist at all costs that the 1’s on the x- and y-axes be visibly equidistant from O because if t is large, then it would be difficult to represent a point such as (t, 50t) on the ordinary page of a book. We will have further comments on this issue on page 360. 31 The French mathematician René Descartes (1596–1650) is probably more famous as a philosopher ("I think, therefore I am"). In addition to discovering analytic geometry with Fermat, he was instrumental in codifying the modern symbolic notation. We will come across him again in Section 3.2 of [Wu2020b].

6.4. LINES IN THE PLANE AND THEIR SLOPES

337

Exercises 6.3. (1) Let ρ be the rotation of 180◦ with respect to the origin O. Prove that ρ(x, y) = (−x, −y) for all (x, y). (2) (a) Let L be the vertical line x = c. Prove that the reflection with respect to L is given by the transformation Λ so that Λ(x, y) = (2c − x, y) for all (x, y). (b) Formulate the corresponding statement for reflection with respect to a horizontal line y = d. (3) (a) Prove that the set U p defined on page 334 is convex. (Caution: This is not as straightforward as you may think. Use Lemma 4.8 on page 178.) (b) Prove that the set Lo defined on page 334 is also convex. (4) Given two distinct points P = (a, b) and Q = (c, d), if P1 , P2 , . . . , Pn are points on the segment P Q which divide P Q into n + 1 segments of equal lengths, what are the coordinates of each Pi , for i = 1, . . . , n? (Hint: Make use of Exercise 11 on page 238.) (5) (a) Write down the set of all the points which are equidistant from (1, 0) and (4, 0) and describe the relationship of this set with (1, 0) and (4, 0). (b) Write down the set of all the points which are equidistant from P = (p1 , p2 ) and Q = (q1 , q2 ) and describe the relationship of this set with P and Q. (6) (a) Given two points P = (p1 , p2 ) and Q = (q1 , q2 ), show that the midpoint of the segment P Q has coordinates ( 12 (p1 + q1 ), 12 (p2 + q2 )). (Hint: Compare Exercise 4 on page 294.) (b) Given P = (1, 2) and Q = (3, −1), |P A| find the point A on the segment P Q so that |AQ| = 32 . (c) Given two points P = (p1 , p2 ) and Q = (q1 , q2 ), find the point B on the segment P Q |P B|

so that |BQ| = m n , where m and n are given positive integers. (7) If S is a geometric figure in the plane, define S1 to be the collection of all the points P in the plane so that there is some point Q ∈ S of distance ≤ 1 from P , i.e., so that |P Q| ≤ 1 for some Q ∈ S. (i) If S is a point O, then show that S1 is the closed disk of radius 1 centered at O. (ii) If S is the unit segment from O to (1, 0), what is S1 ? Give reasons. 6.4. Lines in the plane and their slopes The main goal of this section is to give a detailed exposition of the key concept— the slope of a line—that underlies the proof of the theorem in the next section that the graph of a linear equation in two variables is a line. Students’ confusion about the concept of slope is well known. The root cause of this confusion is TSM’s failure to give slope a correct definition. The main purpose of this section is to set the record straight regarding slope: first, by defining it correctly and, second, by showing that slope is, above all, a number attached to the line itself that describes the "slant" and the "steepness" of the line (assumed to be nonvertical). There are some subtleties in the definition of slope that should be left out of the typical middle or high school classroom but which every teacher should be aware of nonetheless, and these are duly pointed out (see (‡) on page 344). The local slopes of lines at a point (p. 338) The slope of a line (p. 342) A formula for slope (p. 347)

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The local slopes of lines at a point We will begin by explaining what purpose the concept of the slope of a line is supposed to serve and then show how to define it correctly. To put the last statement in context, think of the corresponding situation regarding the concept of a "fraction". As far as TSM is concerned, it is sufficient that students know a fraction is a piece of pizza or, more generally, a part-of-a-whole. By contrast, we have seen the need to define a fraction as a certain point on the number line before we can develop the subject of fractions in a logical and coherent manner. We are going to do the same with the slope of a line.32 Our starting point is the consideration of all the nonvertical lines passing through a fixed point P . To further simplify matters, let P be the origin O. In what way can we distinguish the following lines from one another? Y

L1

L2 X

O

L3 L4 Intuitively, these lines differ in their "steepness": L1 is more "steep" than L2 , and L4 is more steep than L3 , though L1 and L2 differ from L3 and L4 because the two groups slant differently. Part of the challenge will be to figure out how to separate the "steepness" of these two groups of lines. From this point of view, we eliminate the vertical line—the y-axis—from our consideration at the outset because, having the ultimate "steepness", the vertical line has no need for any discussion. The next step is to convert these appealing but vague ideas into precise mathematics: can we use a number to measure the "steepness" of the nonvertical lines passing through O? Let us denote the vertical line passing through the point (1, 0) on the x-axis by {x = 1}; this notation will be explained in the first subsection of Section 6.5 on pp. 351ff. Observe that, by the definition of the coordinates of a point, all the points on {x = 1} have coordinates (1, y), where y ∈ R. Now a nonvertical line L passing through O is not parallel to {x = 1} and must intersect the latter at a point (1, s) for some s ∈ R. We assign the number s to the line L as a measure of its "steepness". See the following picture:

32 The following exposition on slope has been used in [EngageNY] and [Eureka] with the author’s consent.

6.4. LINES IN THE PLANE AND THEIR SLOPES

Y

L1

339

L (1, s)

L2

(1, s 2)

O

(1,0)

X

(1, s 3)

L 4 { x= 1} L 3 We now explain intuitively how this number s serves the purpose of revealing "the steepness of L", on two different levels. First, on the qualitative level, if {s > 0}, then L intersects the vertical line {x = 1} at a point above the x-axis (in the sense of page 334) and therefore L will slant this way /. However, if s < 0, then L intersects the vertical line x = 1 at a point below the x-axis (in the sense of page 334) and therefore L will slant this way \. See L3 in the preceding picture, for example. Thus the sign of s (i.e., whether s is positive or negative) already tells us the way the line L slants, whether it is / or \. One can go further, however. It is visibly obvious that the closer L gets to being vertical, the larger the absolute value |s| of s is going to be. For example, L1 will intersect the line {x = 1} at a point (1, s1 ) very high above the x-axis so that s1 is going to be very large, whereas L4 will intersect the line {x = 1} at a point (1, s4 ) very far below the x-axis and therefore s4 will be "large negative" or, more correctly, s4 < 0 and |s4 | is very large. This number s is therefore a good measure of the "steepness" of L. We will call this s the local slope of L at O. We emphasize the need to refer to the whole phrase, "the local slope of L at O", because, up to this point, this concept refers only to the behavior of L at one point: at the origin O. If two nonvertical lines passing through O have the same local slope at O, then these lines join O to the same point on the vertical line {x = 1} and therefore are the same line (assumption (L1) on page 165). Conversely, if the two lines passing through O coincide, then of course they have the same slope. We may therefore summarize this short discussion in the following lemma. Lemma 6.5. Two lines passing through O have the same local slope at O if and only if they are the same line. There is another way to look at the use of the vertical line {x = 1} to measure the local slope of a nonvertical line at O. When we say we assign s to be the local slope of L at O if L intersects the vertical line {x = 1} at the point (1, s), it is equivalent to saying that we are looking at the line {x = 1} as a number line so that its 0 is the point (1, 0) and its 1 is the point (1, 1). Equivalently, we are identifying the line {x = 1} with the y-axis by using the translation along the horizontal vector from O to (1, 0). This point of view will be useful below. Pedagogical Comments. In the school classroom, the preceding definition of the local slope of L at O is likely to raise at least two questions. The first is:

340

6. SYMBOLIC NOTATION AND LINEAR EQUATIONS

why use the vertical line {x = 1} instead of the vertical line {x = 2} that passes through (2, 0), for example? The answer is that the choice of the line {x = 1} to measure the local slope of L at O is just a convention, as we now explain. Suppose L intersects the line {x = 1} at the point (1, s), so that the local slope of L at O is s. If we had chosen to use the line {x = 2} instead, then the same L would intersect the line {x = 2} at (2, 2s) (use Theorem G15* on page 266; see picture).

It follows that if the line x = 2 is used instead of x = 1, then the local slope of L at O would change from s to 2s for every s. In terms of our understanding of what the "local slope of L at O" means, this hardly matters. More generally, the following Activity gives the complete picture. Activity. Let the local slope of a line L at O be s. Show that L intersects the vertical line x = k at the point (k, ks) for every k, positive or negative. (Use Theorem G11 (FTS*) on page 257 instead of Theorem G15*.) This Activity shows that if we use the vertical line x = k for a number k > 0, instead of the line x = 1, to measure the local slope of L at O, where L is any line passing through the origin O, then the only effect that this change will have is that the local slope of L at O will be multiplied by a factor of k. Therefore we may as well keep things simple by using the line x = 1. A second question that is likely to be asked is why not use vertical lines to the left of O to measure the local slope of L at O? Notice that the lines x = k for k > 0 are to the right of O, so this question is tantamount to asking why not use the vertical lines x = k for k < 0? In fact, why not use the vertical line x = −1? The reason for not using x = −1 is again a matter of convention. So suppose we consider a line L passing through O which is slanted this way /. In our present definition, the local slope of L at O is a positive number, let us say, s. Now suppose we switch over to the line x = −1 for measuring the local slope of L at O. Then the same L will intersect the vertical line x = −1 at the point (−1, −s), which is easy to see (e.g., use congruent triangles or the 180-degree rotation around O).

6.4. LINES IN THE PLANE AND THEIR SLOPES

341

Y L

s s −1

1

O

X

(−1,−s )

x= 1

x= −1

In particular, the local slope of L at O is now −s, and −s is negative. However, a longstanding tradition in mathematics is that lines that are slanted this way / should have positive local slope at O. This tradition then rules out using a vertical line to the left of O to measure the local slope of a line at O. End of Pedagogical Comments. Now let P be an arbitrary point, and consider all the nonvertical lines passing through P . We ask the same question of how to distinguish these lines one from the other.

Y

L1 L2 P

L4 O

L3

X

There is an obvious answer. We can pretend that P is just the origin O and do to it what we did to O to get the local slope of each line at P and then use the local slope (steepness) to distinguish these lines. More precisely, let X  and Y  be the horizontal and vertical lines passing through P , respectively. These correspond to the x-axis and y-axis, respectively.

342

6. SYMBOLIC NOTATION AND LINEAR EQUATIONS

Y

LY

Y

L1

L

2 1

s 1

P

1

X

Q 0.75 L2

O

X

Let Q be the point on X  of distance 1 to the right of P and let LY be the vertical line through Q. Then LY corresponds to the vertical number line {x = 1}. The distance between the parallel lines Y  and LY is thus 1 (see page 228 for the concept of distance between parallel lines). If the coordinates of P are (x0 , y0 ), then the coordinates of Q are (x0 + 1, y0 ). Thus P = (x0 , y0 ) and Q = (x0 + 1, y0 ). All the points on LY have coordinates of the form (x0 + 1, y0 + y), where y is a real number. Clearly, y > 0 if and only if (x0 + 1, y0 + y) is above Q, and y < 0 if and only if (x0 + 1, y0 + y) is below Q. We can now show how to define the local slope of a line L at P by making use of the horizontal line X  and the vertical line LY . So let L be a line passing through P and let L intersect the line LY at the point (x0 + 1, y0 + s). Then the local slope of L at P is by definition s. For example, still referring to the preceding picture, the line L1 intersects the line LY at (x0 + 1, y0 + 1), and therefore its local slope at P is 1. The line L2 intersects the number line LY at the number (x0 + 1, y0 − 0.75), so its local slope at P is −0.75. The same reasoning that proves Lemma 6.5 on page 339 also proves the following lemma. Lemma 6.6. Two lines passing through a point P have the same local slope at P if and only if they are the same line.

The slope of a line So far we have only looked at all the lines passing through a fixed point P and we learned how to assign a local slope at P to each of these lines. Now we change our vantage point by focussing on a single line instead. If we fix a line L in the coordinate plane and if P1 and P2 are two points on L, then a critical question naturally arises: is the local slope of L at P1 equal to the local slope of L at P2 ? The following lemma answers this question affirmatively. Lemma 6.7. The local slope of a nonvertical line L at P for a point P ∈ L does not depend on P .

6.4. LINES IN THE PLANE AND THEIR SLOPES

343

Proof. By relabeling the points P1 and P2 if necessary, we may assume that P1 is to the left of P2 , in the sense that if P1 = (p1 , p1 ) and P2 = (p2 , p2 ), then p1 < p2 (compare page 334).

Y P2 1

M1

Q2

P1 1

L

M2

Q1 X

O

At P1 , we do the usual construction to get the local slope of L at P1 ; namely, on the horizontal line passing through P1 , pick the point Q1 that is 1 unit to the right of P1 , and let the vertical line passing through Q1 intersect L at a point M1 . Similarly, repeat this construction at P2 to obtain the points Q2 and M2 .33 First consider the case that M1 is above the horizontal line LP1 Q1 (see the definition of "above" on page 334). Then M2 is also above the horizontal line LP2 Q2 , as shown in the preceding picture. Therefore the local slope of L at P1 is positive and is equal to the length |M1 Q1 | and the local slope of L at P2 is also positive and is equal to the length |M2 Q2 |. We have to show |M1 Q1 | = |M2 Q2 |. If we can prove the congruence of P1 Q1 M1 and P2 Q2 M2 , then we would be done. Now the congruence is a consequence of ASA, because |P1 Q1 | = |P2 Q2 | = 1, |∠P1 Q1 M1 | = |∠P2 Q2 M2 | = 90◦ , and |∠M1 P1 Q1 | = |∠M2 P2 Q2 | because they are corresponding angles of the transversal L with respect to the parallel lines LP1 Q1 and LP2 Q2 (Theorem G18 on page 277). Therefore the local slopes of L at P1 and P2 are equal. The other possibility is that M1 is below the horizontal line LP1 Q1 , as shown in the picture below. Then M2 is also below the horizontal line LP2 Q2 .

Y

P1

1

Q1 M1

P2

1

Q2 M2

O

X L

33 Note that in this picture, M is put between P and P on the line L for the sake of visual 1 1 2 clarity. In general, P2 could very well be between P1 and M , but the validity of the reasoning to follow does not depend on the relative positions of P1 , P2 , and M .

344

6. SYMBOLIC NOTATION AND LINEAR EQUATIONS

The local slopes of L at P1 and P2 are therefore equal to −|Q1 M1 | and −|Q2 M2 |. The same reasoning as above then shows that |Q1 M1 | = |Q2 M2 | so that, once again, the local slopes of L at P1 and P2 are equal. The proof of the lemma is complete. Remark. There is a subtle point in the preceding proof that we have intentionally glossed over. How did we know that if M1 is above (respectively, below) LP1 Q1 , then M2 would also be above (respectively, below) the horizontal lines LP2 Q2 ? This is a critical part of the proof because this allows us to conclude that, regardless of whether the local slopes of L at P1 and P2 are equal or not, they are at least both positive or both negative. It is only because of this fact that the equality of the two local slopes reduces to the equality of the lengths of Q1 M1 and Q2 M2 . In general, imagine that L is not a line but a curve; then we can have the following situation where M1 is above LP1 Q1 but M2 is below LP2 Q2 :

Y

P2

Q2

1

M1

P1

M2

L

Q1 1

O

X

One’s instinctive reaction is that this anomalous situation cannot occur when L is actually a straight line. (Recall that line is synonymous with straight line in this volume and in [Wu2020b] and [Wu2020c].) In other words, our intuition tells us that the "straightness" of a line will eliminate this anomalous situation. The following assertion then gives mathematical substance to our intuition: (‡) Let P1 and P2 be two distinct points on a line L and let Q1 (respectively, Q2 ) be the point on the horizontal line passing through P1 (respectively, P2 ) that is 1 unit to the right of P1 (respectively, P2 ). Let the vertical line passing through Q1 (respectively, Q2 ) intersect L at M1 (respectively, M2 ). Then M1 being above LP1 Q1 implies M2 is above LP2 Q2 , and M1 being below LP1 Q1 implies M2 is below LP2 Q2 . The key to this proof is Lemma 4.8 on page 178. Since the points P1 and P2 are interchangeable in (‡), we may assume that P1 is to the left of P2 . Thus we are assuming that (6.21)

for P1 = (p1 , p1 ) and P2 = (p2 , p2 ), p1 < p2 .

Let the coordinates of M1 be (m1 , m1 ). Because M1 and Q1 are on the same vertical line, they have the same x-coordinate. But the x-coordinate of Q1 is (p1 + 1), by definition; therefore, m1 = p1 + 1. Thus M1 = (p1 + 1, m1 ). Similarly, M2 = (p2 + 1, m2 ) for some number m2 , as shown:

6.4. LINES IN THE PLANE AND THEIR SLOPES

345

   

We summarize these facts for a later reference: (6.22)

M1 = (p1 + 1, m1 ),

M2 = (p2 + 1, m2 ).

Let vertical lines be drawn from P1 , M1 , P2 , and M2 . Then the points of intersection of these lines with the x-axis are the x-coordinates of the aforementioned points, and these are the numbers p1 , p1 + 1, p2 , and p2 + 1, according to (6.21) and (6.22). Since p1 < p2 (by (6.21)), we have p1 < p1 + 1 < p2 + 1,

p1 < p2 < p2 + 1.

In terms of the concept of betweenness (page 167), this implies (6.23)

p1 ∗ (p1 + 1) ∗ (p2 + 1),

p1 ∗ p2 ∗ (p2 + 1).

Since vertical lines are parallel to each other, by applying Lemma 4.8 on page 178 to the line L and the x-axis, we obtain by virtue of (6.21) and (6.23) (6.24)

P1 ∗ M1 ∗ M2 ,

P1 ∗ P2 ∗ M2 .

Now draw horizontal lines through P1 , M1 , P2 , and M2 ; then their points of intersection with the y-axis are the y-coordinates of these same points, which are, by virtue of (6.21) and (6.22), p1 , m1 , p2 , and m2 . Because of (6.24) and because horizontal lines are parallel, by applying Lemma 4.8 on page 178 to the line L and the y-axis, we obtain (6.25)

p1 ∗ m1 ∗ m2 ,

p1 ∗ p2 ∗ m2 .

Now the y-axis is a number line, and we know from the discussion on pp. 167ff. that (6.25) means either (6.26)

p1 < m1 < m2 ,

p1 < p2 < m2

p1 > m1 > m2 ,

p1 > p2 > m2 .

or (6.27)

Suppose M1 is above the horizontal line LP1 Q1 ; then p1 < m1 , so that only (6.26) is possible. Such being the case, the second part of (6.26) implies p2 < m2 , which means that M2 is above the horizontal line LP2 Q2 , as desired.

346

6. SYMBOLIC NOTATION AND LINEAR EQUATIONS

On the other hand, if M1 is below the horizontal line LP1 Q1 , then m1 < p1 and only (6.27) is possible. See this figure:

Y p1 P

1

Q1

1

M1

m1 p2

P2

1

Q2

m2 O

M2 p1

p +1 p2 1

p +1 2

X L

Such being the case, the second part of (6.27) implies p2 > m2 , which means that M2 is below the horizontal line LP2 Q2 . The proof of (‡) is complete. Pedagogical Comments. We do not recommend that the proof of (‡) be presented in a school classroom except in special situations that call for it. First of all, the reasoning is too sophisticated, and when it is applied to the proof of something as pictorially obvious as (‡), it may very well turn students off. In addition, what is asserted in (‡) has always been taken for granted in the school classroom. At this point of students’ learning trajectory, they have to learn the need for a correct definition of "slope" and must be made aware of the need to use congruent triangles to justify the correctness of the definition. Their plate is already full. Good sense in pedagogy would suggest that something like the proof of (‡) be put away for the moment, though it might be mentioned in passing after the proof of Lemma 6.7 has been given. By now we have come across too many similar cases of proofs that, while germane to the school curriculum, really do not belong in a school classroom (e.g., the Pedagogical Comments on page 204 and (♣) on page 262). This is a fact of life that we should acknowledge. Nevertheless, we repeat that such proofs are important for teachers: these proofs give them the reassurance that, indeed, everything in school mathematics can be supported by reasoning. End of Pedagogical Comments. Now to return to the discussion of slope, Lemma 6.7 tells us that it is no longer necessary to refer to the local slope of a line L at a particular point P of L, because it does not matter which point P on L is used. This then allows us to finally introduce the following definition. Definition. Let L be a nonvertical line in the plane. The slope of L is the local slope of L at P for any point P on L. The following simple lemma is now an immediate consequence of the definition.

Lemma 6.8. A nonvertical line has 0 slope if and only if it is horizontal.

6.4. LINES IN THE PLANE AND THEIR SLOPES

347

Remark. We emphasize that this definition of the slope of a (nonvertical) line brings out the fact that slope is, above all else, a single number. Moreover, we have explained clearly the purpose this number (the slope) is supposed to serve (see the discussion around page 339). Contrast this definition with the definition of slope in TSM34 as two quantities, "rise" and "run", as in "rise over run". The concept of slope also yields a simple criterion for deciding whether two lines are the same. This criterion will be a key ingredient for the proof of Theorem 6.11 on page 354. Theorem 6.9. If two lines pass through the same point and have the same slope, then they are the same line. Proof. This theorem is nothing more than a rewrite of Lemma 6.6 on page 342. Suppose two lines L and L have the same slope and both pass through the same point P . By the definition of slope, the local slopes of L and L at P are equal. Therefore by Lemma 6.6, L = L . The proof is complete. A formula for slope For all the virtues of the preceding definition of the slope of a line, one should not be blind to the fact that it is too clumsy for computations. Thus, to compute the slope of a line L, the best one can do so far is to pick a point P on L; draw a horizontal line through P and let Q be the point 1 unit to the right of P on this horizontal line. Let Q = (x0 , y0 ). Through Q draw a vertical line LY and let L intersect LY at the point with coordinates (x0 , y). Then the slope of L is equal to y − y0 . (An equivalent way of looking at this is to regard LY as a number line whose 0 is at Q and whose 1 is 1 unit above Q; the slope of L is the number on LY at which L intersects LY (see page 342)). If this were the only (excruciating) way to get the slope of a line, the concept of slope might have been abandoned long ago. Fortunately, the following theorem shows a way to compute slope that removes the clumsiness. In fact, the ratio in (6.28) below is usually taken to be the definition of slope. Theorem 6.10. On a given nonvertical line L, let any two distinct points P1 = (x1 , y1 ) and P2 = (x2 , y2 ) be chosen. Then the slope of L is equal to the ratio y2 − y1 (6.28) . x2 − x1 We will preface the proof of the theorem with a little discussion. The ratio in (6.28) is called the difference quotient of P1 and P2 . Remarks. (1) The ratio (6.28) is a rational quotient if the coordinates x1 , x2 , etc., are rational numbers. (See page 118.) This will be the last reminder of why a detailed discussion of the arithmetic of complex fractions and rational quotients is essential. Notice that insofar as we will allow x1 , x2 , etc., to be arbitrary real numbers, we have to rely on FASM (page 133) to affirm that this definition makes sense even when x1 , x2 , etc., are not rational numbers. 34 See

page xiv for the definition of TSM.

348

6. SYMBOLIC NOTATION AND LINEAR EQUATIONS

(2) Because for any numbers a and b, ab = −a −b (Theorem 2.10 on page 116 when a and b are rational; then use FASM), we see that the difference quotient (6.28) of P1 = (x1 , y1 ) and P2 = (x2 , y2 ) enjoys an important symmetry property with respect to P1 and P2 : y1 − y2 y2 − y1 = . x2 − x1 x1 − x2

(6.29)

Therefore, in writing the difference quotient, the order of P1 and P2 does not matter. We will henceforth assume that P1 is to the left of P2 ; i.e., x1 < x2 . (3) Observe that the denominator of the difference quotient (6.28) is never 0 because, if it were, then x2 − x1 = 0 and x1 = x2 . The distinct points P1 and P2 on L now would have the same x-coordinate and L, being a line that joins two such points, would have to be a vertical line, contradicting the hypothesis that L is nonvertical. Thus the denominator of this ratio is never zero, and the ratio makes sense. Before giving the proof of Theorem 6.10, we have to discuss a geometric interpretation of the difference quotient (6.28) that is of independent interest. This interpretation will play a crucial role in all the considerations of slope. If the slope of L is 0, then L is horizontal (by Lemma 6.8 on page 346) so that the difference quotient (6.28) is identically zero for all P1 and P2 . From now on we may assume that the slope of L is not 0. With this understood and with P1 = (x1 , y1 ) and P2 = (x2 , y2 ), x1 < x2 , as always, let the horizontal line passing through P1 and the vertical line passing through P2 intersect at R, so that P1 P2 R is a right triangle with the right angle at R. Then we claim ⎧ |P2 R| ⎪ ⎪ if the slope of L > 0, ⎪ ⎪ ⎨ |P1 R| y2 − y1 = (6.30) ⎪ x2 − x1 ⎪ |P2 R| ⎪ ⎪ if the slope of L < 0. ⎩ − |P1 R| The two cases of positive and negative slopes are illustrated by the left figure and right figure below, respectively.

Y

P2

L

1

P1

O

Q

Q R

P1

M 1

LY

Y

LY

M R X

O

P2 X L

For the proof of (6.30), we begin with two general comments that are valid for both cases in (6.30). We first claim that the coordinates of R are (x2 , y1 ). This is because P1 and R, being on the same horizontal line, must have the same ycoordinate, which is y1 (the y-coordinate of P1 ). Similarly, P2 and R, being on the same vertical line, must have the same x-coordinate, which is x2 (the x-coordinate

6.4. LINES IN THE PLANE AND THEIR SLOPES

349

of P2 ). Thus, indeed, R = (x2 , y1 ). Next, since P1 and R lie on the same horizontal line, |P1 R| = |x1 − x2 | (see (6.18) on page 336). By assumption, x2 > x1 , so we get (6.31)

x2 − x1 = |P1 R|.

To prove (6.30), first suppose the slope of L is positive. Since P2 and R lie on the same vertical line x = x2 , |P2 R| = |y2 − y1 | (see (6.19) on page 336). We claim that P2 lies above R so that y2 > y1 and, therefore (still with the assumption that the slope of L is positive), (6.32)

y2 − y1 = |P2 R|.

Together with (6.31), this proves the first half of (6.30). It remains to prove (6.32). For this, we will need the assumption of the positivity of the slope of L. On the horizontal line LP1 R , let Q be the point to the right of P1 so that |P1 Q| = 1, and let the vertical line LY passing through Q meet L at M . From the definition of local slope (page 339), the positivity of the local slope of L at P means that M is above the horizontal line LP1 R (see the left picture above; note that the picture shows the case where Q is to the left of R, but the reasoning is the same even if Q is to the right of R). It is obvious from the picture that since R is also to the right of P1 , the intersection P2 of the vertical line passing through R with L is also above LP1 R ;35 i.e., P2 is above R as claimed. We have therefore proved (6.30) in case the slope of L is positive. If the slope of L is negative, the local slope of L at P1 is negative. Now the definition of local slope shows that if Q is the point on the horizontal line LP1 R which is 1 unit to the right of P1 and if the vertical line LY passing through Q meets L at M , then M is below the horizontal line LP1 Q (see the right picture above). Clearly P2 , being the point of intersection of L with the vertical line passing through a point R to the right of P1 , is also below LP1 R .36 Thus P2 lies below R. Consequently, a similar reasoning shows that y2 < y1 and −(y2 − y1 ) = |P2 R|. Together with (6.31) and Theorem 2.10 on page 116, this proves the second half of (6.30). The proof of (6.30) is complete. We are now ready for the proof proper of Theorem 6.10. Proof of Theorem 6.10. The claim (6.30) transforms the difference quotient (6.28) into a geometric quantity, namely, ± the ratio of the lengths of two sides of a right triangle P1 P2 R. This then invites the consideration of similar triangles (Section 5.3 on page 283). More precisely, let Q be the point on the horizontal line P1 R so that Q is to the right of P1 and |P1 Q| = 1. Let the vertical line LY passing through Q intersect L at M as usual. See the pictures below for both cases of a positive and a negative local slope of L. 35 In fact, the reasoning that proves (‡) on page 344 also suffices to prove that M being above LP1 Q implies P2 is above LP1 Q . But see the Pedagogical Comments on page 346. 36 But see the preceding footnote.

350

6. SYMBOLIC NOTATION AND LINEAR EQUATIONS

Y

P2

L 1

P1

O

Q

Q R

P1

M 1

LY

Y

LY

M

R X

O

P2 X L

We have drawn the pictures so that P2 is to the right of M purely for the sake of clarity; the subsequent reasoning will be independent of this fact. We will prove (6.33)

P1 P2 R ∼ P1 M Q.

Assuming (6.33) for the moment, we will finish the proof of Theorem 6.10. By the proportionality of corresponding sides of similar triangles, we have the equality |P1 R| |P2 R| = |M Q| |P1 Q| which, by virtue of the cross-multiplication algorithm ((b) on page 70) and FASM, is equivalent to |M Q| |P2 R| = . |P1 R| |P1 Q| Suppose the local slope of L at P1 is positive. Then by (6.30), this implies y2 − y1 |M Q| = |M Q|. = x2 − x1 1 Since |M Q| is by definition the local slope of L at P1 , the preceding equality proves Theorem 6.10 in the case of positive local slope. If the local slope of L at P1 is negative, then an entirely analogous argument shows that |M Q| y2 − y1 = −|M Q|. =− x2 − x1 1 Since the local slope of L at P1 is, by definition, equal to −|M Q| in this case, Theorem 6.10 is now proved also in the case of a negative local slope. It remains to prove (6.33). Observe that P1 P2 R and P1 M Q obviously satisfy the AA criterion for similarity (Theorem 22 on page 288) because |∠P2 P1 R| = |∠M P1 Q| and |∠P2 RP1 | = |∠M QP1 | = 90◦ . The desired similarity in (6.33) follows. The proof of Theorem 6.10 is complete. Activity. Let P = (x0 , y0 ) be a point in Quadrant III of a coordinate system with origin O. What is the slope of the line joining P to O? If Q is a point in Quadrant II, what is the slope of the line joining Q to O? The great significance of Theorem 6.10 is that, given a line, we can find its slope by computing the difference quotient of any two points on the line that suit our purpose. This is not only a useful idea to keep in mind in general, but also one

6.5. THE GRAPHS OF LINEAR EQUATIONS IN TWO VARIABLES

351

that is critical for the proof that the graph of a linear equation of two variables is a line (Theorem 6.11 on page 354). Exercises 6.4. (1) Let  be the line joining (1, 2) and (−3, 4). If (x, y) is a point on , what y−4 is the value of x+3 ? (2) (a) Let  be the line with slope m passing through ( 12 , 34 ). For which value of m would  pass through ( 53 , 13 )? (b) Let  be the line joining (− 32 , 4) and ( 45 , q), where q is some number. For what value of q would  pass through (2, −3)? (3) Does the line joining (3, −2) and (6, 2) contain the point (9, 6)? Explain your answer two different ways. (4) Let P be a point in quadrant II and let Q be a point in quadrant IV. If L is a line joining P to Q, what is the sign (page 339) of the slope of L? Explain. (5) Let L be the graph of a linear equation 3x + 4y = c for some constant c. (a) What is the slope of L? (b) Suppose L passes through the point (−1, 5). What is c? (6) Let a, b be positive numbers. Can the three points (a, b), (2a, b + 2), (−a3 , b − 1) be collinear? Explain. 6.5. The graphs of linear equations in two variables The main theorem of this section affirms that the graph of a linear equation of two variables is a line (recall that in these volumes, a line is synonymous with a straight line, as on page 165) and that, conversely, every line is the graph of a linear equation in two variables. This theorem is absolutely fundamental for the understanding of all aspects of the graphs of linear equations in two variables, but it is never stated in TSM, much less proved. One reason for such a spectacular failing in TSM is that it never defines the concept of the slope of a line correctly. We will demonstrate how a clear understanding of what slope is renders the standard problems about finding the equation of a line completely routine. Generalities about graphs of equations (p. 351) The main theorem (p. 354) Some applications of the main theorem (p. 357) Generalities about graphs of equations The concept of the graph of an equation is basic in mathematics, but it gets short shrift in TSM. We will begin by defining it precisely. Because we will be mainly interested in equations in two variables x and y, the following discussion will focus on equations of two variables. The basic idea is the same, however, for equations in any number of variables. Recall that an equation in two variables, x and y, is a question asking whether two given expressions in the numbers x and y are equal (page 323). We identify such a pair of numbers x and y with the point (x, y) in the coordinate plane. Then such an equation is equivalent to asking whether there are points (x0 , y0 ) in the coordinate plane so that their coordinates make the two given expressions equal. We will call such points (x0 , y0 ) solutions of

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the given equation. The collection of all the solutions of a given equation is called the graph of the equation. The graph of an equation is therefore a subset of R2 . For example, the equation x4 + y 2 + 1 = 0 has no solutions (because x4 + y 2 ≥ 0 for all numbers x and y), so that the graph of x4 + y 2 + 1 = 0 is the empty set, i.e., the set with no elements. On the other hand, the graph of x2 + y 2 = 25 is the circle of radius 5 around the origin (0, 0). Activity. Use the distance formula (see equation (6.20) on page 336) to prove the claim about the graph of x2 + y 2 = 25. Do not skip steps. A linear equation in two variables x and y is a question about whether there are numbers x and y so that ax + by = c, where a, b, and c are fixed constants and at least one of a and b is nonzero. Sometimes one gains more flexibility by defining such an equation to be an equation of two expressions in x and y which, after applications of the two operations (E1) and (E2) on page 328, can be put in the form of ax + by = c. For example, −2x + y = 25 y + 7 and 3x2 + 6 + 2y = 19 − 5x − 3y + 3x2 are examples of linear equations in two variables because, after the use of (E1) and (E2), they can be brought to −2x + 35 y = 7 and 5x + 5y = 13, respectively. The point (− 72 , 0) is a solution of the first equation, and ( 35 , 2) is a solution of the second. In general, the graph of ax+by = c is the collection of all the points (x0 , y0 ) in R2 so that ax0 + by0 = c. Your experience with TSM concerning linear equations in two variables immediately tells you that this graph is a line and, in fact, TSM leaves you with no doubt of this fact—even as it never tells you why. The main theorem of this chapter (Theorem 6.11 on page 354) affirms this fact, and everything we have done in this chapter up to this point is nothing but a preparation for its proof. Consider the example of a linear equation in two variables, x − 2y = −2. We observe that in this situation, it is easy to find all the solutions of this equation with a prescribed first number x0 or a prescribed second number y0 . For example, with the first number prescribed as 3, we solve the linear equation in y, 3 − 2y = −2, and get y = 52 . Therefore (3, 52 ) is the sought-for solution. Or, if the second number is prescribed to be y0 , then we solve the linear equation in x, x − 2(y0 ) = −2, to get x = 2y0 − 2. The solution is now (2y0 − 2, y0 ). Thus we see and will continue to bear witness that the study of linear equations in two variables is grounded in the study of linear equations in one variable. Using the above method of getting all the solutions of the equation x−2y = −2, we can plot as many points of the graph of x − 2y = −2 as we please to get a good idea of the graph. For example, the following points are on the graph of x−2y = −2: (−5, −1.5), (2.5, 2.25),

(−4, −1), (4, 3),

(−2, 0), (6, 4),

(0, 1), (7, 4.5).

(2, 2),

Students need the experience of plotting points on the graph of an equation by hand, and they should form this good habit right from the outset for the case of a linear equation in two variables. In the age of the graphing calculator, there is all the more reason to emphasize this need. Consider next the graph of the linear equation in two variables y = 3 which, as an equation in two variables, is in reality the abbreviated form of the equation

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353

(0 · x) + (1 · y) = 3. We want to prove that the graph of y = 3 is the horizontal line passing through the point (0, 3).37 Here is the simple proof. As always, in order to show the equality of sets, we must show that the two sets have the same collection of elements. For the case at hand, let the horizontal line be denoted by  and the graph of y = 3 be denoted by G. We have to show that G and  have the same collection of points in the plane. Recall that there is a standard way to do this (page 141), which is to show that each set is contained in the other set, or in standard set-theoretic notation: ⊂G

and

G ⊂ .

Observe that G consists of all the points (s, t) which are solutions of y = 3 and therefore consists of all the points (s, t) so that t = 3, i.e., all the points whose ycoordinate is 3. Let us now prove  ⊂ G. Let P be a point of ; we must prove that P ∈ G; i.e., the y-coordinate of P is 3. But the horizontal line  passing through P intersects the y-axis at the number 3 (by assumption on ), so the y-coordinate of P is 3, by definition. Hence P ∈ G. Conversely, we prove G ⊂ . Let P ∈ G, and we must prove that P ∈ . Since P ∈ G, the y-coordinate of P is 3. Thus the horizontal line passing through P intersects the y-axis at the number 3 (definition of the y-coordinate of a point). This horizontal line is precisely , so P ∈ . Y

 3 O

X

The same reasoning shows that the graph of the equation y = b in the plane for any number b is exactly the horizontal line L0 passing through the point (0, b) on the y-axis. We can do the same to vertical lines. In summary, we have the following: the graph of x = c for any number c is the vertical line passing through the point (c, 0) on the x-axis, and the graph of y = b for any number b is the horizontal line passing through the point (0, b) on the y-axis. Since there is only one horizontal (respectively, vertical) line passing through a given point of the plane (why?), it follows that every vertical line is the graph of the equation x = c, where (c, 0) is the point of intersection of the line and the x-axis, and every horizontal line is the graph of an equation y = b, where (0, b) is the point of intersection of the line with the y-axis. 37 This obvious fact is usually taught by rote, but we want to establish the fact that if the definition of the graph of an equation means anything at all, the fact that the graph of y = 3 is a line should follow logically from the definition. For this reason, we want to prove this fact clearly and correctly. If there are any doubts about the need to take the definition of the graph of an equation or function seriously, it suffices to consult the blog of [Meyer].

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In the next subsection, we generalize these assertions to graphs of arbitrary linear equations. The main theorem It is well known that the graph of a linear equation in two variables is a line, and vice versa. What is not well known is the reason why this is true. The main purpose of this section is to prove this fact. Precisely, we have the following theorem. Theorem 6.11. The graph of a linear equation in two variables is a line. Conversely, every line is the graph of a linear equation in two variables. We would like to immediately address a question that is implicitly raised by the second part of the theorem. If the graph of ax + by = c is L, it is common to say that L is the line defined by ax + by = c. By the second part of the theorem, every line L is defined by a linear equation in two variables ax + by = c for some constants a, b, c. Now suppose L is also defined by a x + b y = c for some constants a , b , c . Are the two equations related? The answer is given by the following lemma. Lemma 6.12. Suppose a line is the graph of both ax + by = c and a x + b y = c . Then there is a nonzero number k so that a = ka, b = kb, and c = kc. Conversely, the graphs of the two equations ax + by = c and kax + kby = kc, where k = 0, are the same line. The proof of the first part of Lemma 6.12 is quite straightforward once you consider the three cases separately: the line is vertical, horizontal, or neither. Of course the proof of the second part is trivial. We will leave the details to an exercise (see Exercise 4 on page 362). Lemma 6.12 shows that if L is the graph of ax + by = c, then every equation defining L is necessarily of the form kax + kby = kc for some k = 0. We naturally regard all such equations (for any value of k) as the same equation. With this understood, we are in the habit of saying in this case that the equation of L is ax + by = c. Proof of Theorem 6.11. We first prove that the graph of a linear equation of two variables, ax + by = c (where a, b, c are constants), is a line. If b = 0 in ax + by = c, then the equation becomes ax = c, whose graph is clearly the same as the graph of x = c/a. The fact that the graph of x = c/a is the vertical line passing through (c/a, 0) has been proved on page 353. We may therefore assume from now on that in the equation ax + by = c, b = 0. The graph of ax + by = c is clearly the same as the graph of y = −(a/b)x + (c/a). Therefore, to complete the proof of Theorem 6.11, it suffices to prove the following: If G is the graph of an equation y = mx + k, where m and k are constants, then G is a line.38 38 In many textbooks, the standard notation for this equation is y = mx + b, but the "b" will not work for us because our notation for a general linear equation is ax + by = c.

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355

Let P1 = (x1 , y1 ) and P2 = (x2 , y2 ) be two points on G. Let the line joining P1 and P2 be denoted simply by L. We are going to prove G = L, thereby proving that the graph of y = mx + k is a line. Recall that to prove G = L, we have to first prove G ⊂ L and then prove L ⊂ G. What is needed for both proofs is the fact that the slope of L is equal to m. To see this, observe that since P1 and P2 are points on L, y1 = mx1 + k

and y2 = mx2 + k.

Therefore the difference quotient of P1 and P2 is y2 − y1 (mx2 + k) − (mx1 + k) m(x2 − x1 ) = = = m. x2 − x1 x2 − x1 x2 − x1 By Theorem 6.10 on page 347, the slope of L is m, as desired. First: G ⊂ L. Given P  = (x , y  ) ∈ G, we must prove P  ∈ L. We do so by proving that the line L that joins P1 and P  coincides with L. To this end, Theorem 6.9 on page 347 implies that all we need to do is prove that L and L have the same slope and pass through the same point. They clearly pass through the same point P1 . As to their slopes, we have just seen that the slope of L is m. As to the slope of L , note that since P  is on the graph G of y = mx + k, we have y  = mx + k. Therefore the difference quotient of P1 and P  is y  − y1 (mx + k) − (mx1 + k) m(x − x1 ) = = = m.   x − x1 x − x1 x − x1 By Theorem 6.10, the slope of L is also m. Thus L = L. Therefore P  ∈ L now means P  ∈ L. This proves G ⊂ L. Next: L ⊂ G. This time, let P  = (x , y  ) ∈ L and we have to prove that  P ∈ G, i.e., y  = mx + k. Since P1 is already a point of G, we may assume that P  = P1 . So by Theorem 6.10, we may compute the slope of L (which is m) using the points P1 and P  to obtain m=

y  − y1 . x − x1

Simplifying, we get y  = mx + (y1 − mx1 ). However, recalling that P1 ∈ L, we also have y1 = mx1 + k, which implies (y1 − mx1 ) = k. Altogether, we get y  = mx + k. Thus P  ∈ G after all, and L ⊂ G. The proof that the graph of y = mx + k is a line is complete. Now that we know the graph of every linear equation in two variables is a line, we can prove the second part of Theorem 6.11 that every line is the graph of a linear equation in two variables. If the line is vertical, then we have already proved on page 353 that it is the graph of an equation x = c. Henceforth, we may assume the line is nonvertical. Thus let L be a nonvertical line and let P1 = (x1 , y1 ) and P2 = (x2 , y2 ) be two distinct points on L. Let m be the slope of L. Then by Theorem 6.10, y2 − y1 (6.34) m= . x2 − x1 Rewriting (6.34) as y2 − y1 = m(x2 − x1 ), we are led to consider the equation (6.35)

y − y1 = m(x − x1 ).

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Observe that (6.35) is a linear equation in two variables because it is equivalent to (6.36)

y = mx + (y1 − mx1 ).

Furthermore, what (6.34) affirms is that (x2 , y2 ) is a solution of (6.35). Obviously, (x1 , y1 ) is also a solution of (6.35). Thus, if G denotes the graph of the linear equation (6.35), then both P1 = (x1 , y1 ) and P2 = (x2 , y2 ) lie in G. By the first part of this theorem, G is a line. Therefore G is a line passing through P1 and P2 . Recall that L is also a line passing through P1 and P2 . By assumption (L1) (page 165), G = L, which implies L is the graph of equation (6.35). The proof of Theorem 6.11 is complete. Students who master this proof will never have to memorize the different forms of the equation of a line. They will be able to derive the needed equation easily at a moment’s notice or, at worst, they can memorize the different forms with much greater ease because they now possess a mental framework in which each of those forms can take its proper place. We will illustrate with some simple examples. To this end, let us extract two useful facts from the preceding proof. For the first lemma, define the y-coordinate of the point at which a line L intersects the y-axis to be the y-intercept of the line; sometimes the point of intersection is itself called the y-intercept of L. Lemma 6.13. A nonvertical line L is the graph of the linear equation y = mx + k, where m is the slope of L and k is the y-intercept of L. Remark. This lemma explains why the equation y = mx + k is called the slope-intercept form of the equation of L. Proof. By Theorem 6.11, L is the graph of a linear equation ax + by = c for some constants a, b, and c. Since L is nonvertical, b = 0. Thus this equation is equivalent to (i.e., has the same set of solutions as) y = mx + k, where m = −a/b and k = c/b. Since (0, k) is a solution of y = mx + k, k is the y-intercept of L. To see that m is the slope of L, we use Theorem 6.10: let P1 = (x1 , y1 ) and P2 = (x2 , y2 ) be any two distinct points on L. Then the slope of L is y2 − y1 . x2 − x1 Since P1 and P2 are points on the graph of y = mx + k, we also have y1 = mx1 + k and y2 = mx2 + k. Therefore the slope of L is y2 − y1 (mx2 + k) − (mx1 + k) m(x2 − x1 ) = = = m. x2 − x1 x2 − x1 x2 − x1 The proof of the lemma is complete. Lemma 6.14. Let L be the nonvertical line passing through two distinct points P1 and P2 , where P1 = (x1 , y1 ) and P2 = (x2 , y2 ). Then the equation of L is (6.37)

y − y1 = m(x − x1 ),

where (6.38)

m=

y2 − y1 . x2 − x1

6.5. THE GRAPHS OF LINEAR EQUATIONS IN TWO VARIABLES

357

Proof. Let m denote the slope of L. By Theorem 6.10 on page 347, when m is computed using P1 and P2 , it is exactly as in (6.38). Let (x, y) be an arbitrary point on L different from P1 . Then the slope of L computed using (x, y) and P1 is y − y1 (6.39) m= . x − x1 This is equivalent to y − y1 = m(x − x1 ), which is precisely equation (6.37). The advantage of (6.37) is that every point (x, y) on L now satisfies this equation. Let G be the graph of (6.37). Note that G is a line, by Theorem 6.11. The point P2 lies in G because of (6.38), and P1 lies in G for trivial reasons. Thus G is also a line passing through P1 and P2 . By assumption (L1) (page 165), G = L, which implies L is the graph of equation (6.37). This proves Lemma 6.14. Using (6.38), we can write equation (6.37) as   y2 − y1 (x − x1 ). (6.40) y − y1 = x2 − x1 The equation (6.40) is called the two-point form of the equation of a line. The virtue of equation (6.40) is that it is essentially equation (6.39), and the latter is entirely memorable because it says the slope of L computed by using P1 and P2 (= left side) is equal to the slope of L computed by using P1 and another point (x, y) on L (= right side). Some applications of the main theorem We begin with some examples about finding the equation of a line satisfying certain geometric conditions; the preceding lemmas will come in handy. For the first example, we introduce another definition: the x-coordinate of the point at which a line L intersects the x-axis is called the x-intercept of L; sometimes the point of intersection is itself called the x-intercept of L. Example 1. Given a line which is the graph of y = mx + k, we have seen from Lemma 6.13 that k is the y-intercept of the line. We can also determine its xk , 0) satisfies the equation y = mx+k, the x-intercept intercept: since the point (− m of the line is −k/m. Example 2. What is the equation of the line which passes through (−1, 3) and ( 12 , 4) ? Call this line . The slope of  computed using the points (−1, 3) and ( 21 , 4) is 1 2

4−3 2 = . 3 − (−1)

By equation (6.37), the equation of  is (y − 3) = 23 (x − (−1)), which is y = 23 x + 11 3 . Alternately, we can use Lemma 6.13. Then the equation of  has the form y = 23 x + k. Since the point (−1, 3) lies on , we have 3 = 23 (−1) + k. Thus k = 11 3 and the equation of  is y = 23 x + 11 . (We can equally well use the other point 3 ( 21 , 4) to evaluate k in the equation y = 23 x + k. Then we get 4 = 23 · 12 + k, so that k = 4 − 13 = 11 3 as before.)

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Suppose you do not remember the exact statement of Lemma 6.14; then you start from scratch. You would begin by computing the slope of the line as before, obtaining 23 . Now you have to remember Theorem 6.10,39 which says the slope can be computed using any two points. So you use (−1, 3) and an arbitrary point (x, y) on  not equal to (−1, 3) to get 2 y−3 = . x − (−1) 3 Therefore y − 3 = 23 (x + 1) for all (x, y) on  not equal to (−1, 3). Notice that in the latter form, the equality continues to hold even when (x, y) = (−1, 3). Therefore rewriting y − 3 = 23 (x + 1) as (6.41)

y=

2 11 x+ , 3 3

what you have then proved is that every point (x, y) on  satisfies (6.41). Thus  lies in the graph of (6.41). Since the latter is itself a line (Theorem 6.11),  is the graph of (6.41), by (L1) on page 165. So, once again, the answer is y = 23 x + 11 3 . Example 3. What is the equation of the line with slope the point (3, 4)?

1 2

and passing through

By Lemma 6.13, the equation is of the form y = 12 x + k for some constant k. Since (3, 4) has to be a solution of this equation, we have 4 = 12 · 3 + k, which implies that k = 2 12 . Thus the equation is y = 12 x + 2 12 . An alternative solution is the following. Let L be the line in question, and let (x , y  ) ∈ L. Then by Theorem 6.10 on page 347, we have y − 4 1 = , x − 3 2 1  1  or, equivalently, y = 2 x + 2 2 . This shows that (x , y  ) lies on the graph of y = 12 x + 2 12 . Since this graph is also a line (Theorem 6.11 on page 354) which passes through (3, 4) and whose slope is 12 , L is equal to this graph, by Theorem 6.9 on page 347. Thus the equation of L is y = 12 x + 2 12 . We conclude this section with two applications. First, we will give an explicit description of the segment joining two points. Lemma 6.15. Let two distinct points P = (p, p ) and Q = (q, q  ) be given. (i) If p = q but p < q  , then the segment P Q consists of all the points {(p, t)}, where p ≤ t ≤ q  . (ii) If p < q, let P and Q lie on the line whose equation is y = mx + k for some constants m and k. Then P Q consists of all the points {(s, ms + k)}, where p ≤ s ≤ q. Proof. Let L be the line joining P to Q. (i) If p = q, then L is the vertical line x = p. If T is the translation along the vector from (0, 0) to (p, 0), then T maps the y-axis to L (Theorem G5 on page 236), 39 You

cannot afford to forget something this basic.

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359

so that T (0, y) = (p, y) for all real numbers y. Thus if P0 and Q0 are the points (0, p ) and (0, q  ) on the y-axis, respectively, then T (P0 ) = P

and T (Q0 ) = Q

so that T maps the segment P0 Q0 to the segment P Q; i.e., T (P0 Q0 ) = P Q. Since the segment P0 Q0 consists of all the points {(0, t)} for all t satisfying p ≤ t ≤ q  , it follows that P Q = T (P0 Q0 ) = {(p, t)}, where p ≤ t ≤ q  . This proves (i). Y

L Q = (p, q  )

Q0

- P = (p, p )

T

P0

X

p

O

(ii) If p < q, then L is not vertical and is therefore the graph of an equation y = mx + k, where m is the slope of L; i.e., m=

q  − p . q−p

(See Lemma 6.13.) Every point of L is therefore of the form (x, mx + k) for a real number x. Let S = (s, ms+k) be a point on the segment P Q, where P = (p, mp+k) and Q = (q, mq+k). Thus S is between P and Q. The vertical lines passing through P , S, and Q then intersect the x-axis at the numbers p, s, and q, respectively.

Y S

Q

L

P

O

p

s

q

X

These vertical lines being parallel, Lemma 4.8 on page 178 implies that s is between p and q; i.e., p < s < q. Since the segment P Q consists of P , Q, and all the points between P and Q, this shows that P Q consists of all the points {(s, ms + k)} where p ≤ s ≤ q. The proof of the lemma is complete. A second application is to give an explicit description of the half-lines of a given line L (in the sense of Lemma 4.5 on page 173) in the coordinate plane. Lemma 6.16. (i) Let a nonvertical line L be the graph of y = ax + b, and let P be a point on L with coordinates (p, ap+b). Then the two half-lines of L determined by P are the following two subsets of L:

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6. SYMBOLIC NOTATION AND LINEAR EQUATIONS

P − = all the points (x, ax + b) so that x < p. P + = all the points (x, ax + b) so that p < x.

Y

P +

 

 P  − r P 

p

X

(ii) For a vertical line L defined by x = c, if P = (c, p) is a point of L, then the half-lines of L determined by P are the subsets P − = all the points (c, y) so that y < p. P + = all the points (c, y) so that p < y. Proof. We first prove (i). Fix a point Q ∈ P + . Since L is the disjoint union of {P } and the two half-lines L+ , L− with respect to P (see Lemma 4.5 on page 173), Q lies in either L+ or L− . By changing the notation if necessary, we may assume Q ∈ L+ . Thus, Q is a point in both L+ and P + . We claim that P + = L+ . To show P + ⊂ L+ , let U ∈ P + and we will prove U ∈ L+ . Let the coordinates of Q and U be (q, aq + b) and (u, au + b), respectively. Because Q and U are both in P + , we have p < q and p < u. For definiteness, let q < u. Then by the preceding lemma, the segment QU consists of all the points (x, ax + b) so that q ≤ x ≤ u. Thus P cannot lie in QU because P = (p, ap + b) and p < q. So QU does not contain P and, by Lemma 4.5(ii), Q and U lie in the same half-line of P . Since Q lies in L+ , so does U . Thus, P + ⊂ L+ . Next, we prove L+ ⊂ P + . Let V ∈ L+ , and we must show V lies in P + . Suppose V lies in P − , and we will deduce a contradiction. By the definition of P − , V = (v, av + b), where v < p. Consider the segment V Q. By the preceding lemma, V Q consists of all the points (x, ax + b) so that v ≤ x ≤ q. But v < p < q and P = (p, ap + b), so P lies in V Q. Now both V and Q are points of L+ , so the convexity of L+ implies that V Q does not contain P . This contradiction shows that V cannot lie in P − . Since L is obviously also a disjoint union of P + , P − , and {P }, V has to lie in P + after all. We have proved that P + = L+ . It remains to show that P − = L− . We first prove P − ⊂ L− . Let U ∈ P − . U cannot lie in L+ because if it did, U would lie in P + (since L+ = P + ), contradicting the disjointness of P + and P − . So U lies in L− . Since U is an arbitrary point of P − , we have P − ⊂ L− . Conversely, we show that L− ⊂ P − . Suppose V is a point of L− . Then V cannot lie in P + because if it did, V would lie in L+ (since P + = L+ ), contradicting the disjointness of L− and L+ . So V lies in P − and therefore L− ⊂ P − . Together, we have proved P − = L− . (i) is proved. Having gone through the proof of (i) in such detail, we see that the proof of (ii) is similar but simpler: replace (x, ax + b) by (c, y) and reason with y instead of x, and there will be no need to invoke Lemma 6.15 because we will be looking at a number line consisting of all the (c, y), where y ∈ R. The proof of the lemma is complete. Finally, the issue of graphing a linear equation of two variables brings out the confrontation of theory with practice in graphic presentations. Consider the graph L of an equation such as y = 25x + 50. Clearly, the two points (0, 50) and (−2, 0) are on L, and by Theorem 6.11 on page 354, L is the line passing through these two points. Now, how do we present this graph on the blackboard or on the pages of

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361

a book? By choice, the two points (1, 0) and (0, 1) are equidistant from the origin O of the coordinate system; see the discussion on pp. 336ff. Such being the case, once the two points (0, 0) and (1, 0) have been chosen on the horizontal x-axis, the point (0, 50) on the y-axis would have to be, vertically, 50 times the distance between (0, 0) and (0, 1) above (0, 0). This is simply not practical. A reasonable compromise is to introduce a scaled coordinate system, i.e., rescale the y-axis, so that the old distance between (0, 0) and (0, 1) is now interpreted to be 10 instead of 1. Now the graph L of y = 25x + 50 can be presented as follows: Y 50

40 30 L 20 10

X −5 −4 −3 −2 O 1 −1 What must be borne in mind is the fact that, in this graphic representation, the 90-degree rotation around O is no longer length-preserving, among other geometric anomalies. The reason is clear: the 90-degree counterclockwise rotation now maps (1, 0) to (0, 10), so that it maps the unit segment [1, 0] on the x-axis to a segment of length 10 (namely, the segment from (0, 0) to (0, 10)). Pedagogical Comments. The teaching of slope (or, more to the point, the widespread nonteaching of slope) in school mathematics furnishes an excellent example of how TSM40 causes massive nonlearning in school mathematics. In TSM and standard professional development materials, there appear to be two ways to approach the definition of the slope of a line. One way is to choose two points on the line, compute the "rise-over-run" using these two fixed points, and declare that this "rise-over-run" is the slope of the line. There is no hint of the fact that the "rise-over-run" computed with any two points on the line would also be equal to the same number. The standard excuse is that since similar triangles are not taught until high school geometry—after slope has been taught and used—this independence cannot be explained. A second way is to inform students that what we call nonvertical lines are the graphs of equations of the form y = mx + b and then define the slope of such a line to be the number m (in other words, Theorem 6.11 on page 354 is implicitly assumed). Such a definition of a line is totally inappropriate for K–12 as it likely shatters students’ confidence about whether they even know what a line is anymore. Either of these approaches inevitably leads to rote-teaching and rote-learning of linear equations and their graphs. A recent study by Postelnicu and Greenes ([PG]) of students’ understanding of straight lines in algebra reveals that the most difficult problems for students are those requiring the identification of slope of a 40 See

page xiv of the preface for the definition of TSM.

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line from its graph. Let us pause for a moment to absorb the absurdity of the situation: how could a routine computation (see (6.28) on page 347) become "the most difficult problem"? One can well imagine that—with respect to the first approach—if students do not know they can use any two points on a line to compute its slope, they would naturally be confused about "how to measure rise-and-run" (see page 15 of [PG], left column). It therefore stands to reason that, since teaching slope strictly by analogies and metaphors for lack of a correct definition of slope has led to widespread, abject failures in student learning, we should try returning to first principles by beginning with a correct definition of it for a change. Let us teach similar triangles and explain the reasoning surrounding the proof of Theorem 6.11 (pp. 354ff.). One can appreciate the stranglehold TSM has exerted on school mathematics education from the following three facts. In the mathematics education reform of the 1990s, the fact that the concept of the slope of a line has almost never been presented to students correctly did not seem to merit any discussion (compare [NCTM1989] and [PSSM]). Moreover, for at least two decades since 1990, many states pushed for "algebra in grade 8" without paying any attention to this glaring defect in the teaching of slope and the attendant massive nonlearning of the geometry of linear equations in two variables. Finally, even after the publication of [CCSSM], which took over the recommendation of (a draft of) [Wu2016b] and made a major change in the curriculum so that the concept of similar triangles is taught in grade 8 to make possible a correct definition of slope, there still seems to be little awareness in the mathematics education literature that slope has to be correctly defined. For example, the lack of a correct definition of slope in TSM is not mentioned in a 2014 article on the teaching of slope ([Nagle-Moore-Russo]) or a 2015 volume on Mathematical Understanding for Secondary Teaching ([MUST]). By now, it should be obvious that, in large part, these three volumes (this volume, [Wu2020b], and [Wu2020c]) have been written in response to just such abuses in TSM. End of Pedagogical Comments.

Exercises 6.5. (1) Solve for x: (a) 4bx + 13 = 2x + 26b, where b is a number not equal to 1 2 1 15 2 . Simplify your answer. (b) 5 ax − 17 = 3 ax − 2 , where a is a nonzero number. (2) Find the equation of the line in each of the following; understand that, in this situation, getting the right answer is only half the work because your solution must be supported by reasoning at each step. (a) The line  passing through (− 21 , 3) with slope −5. (b) The line passing through the two points (−7, 2) and (3, −4) (write your answer in the form of ax + by = c). (c) The line L with slope −2 15 and x-intercept −4. (3) (a) What is the equation of the line joining the two points (X, Y ) and (Z, W )? (Write your answer in the form of ax + by = c.) (b) What is the equation of the line whose slope is A and whose y-intercept is B? (4) Prove Lemma 6.12 on page 354. (5) For large x, e.g., x ≥ 106 , which of the graphs of the following two equations is above the other: y = 10x − 5,000,000 and y = x + 1,500,000?

6.6. PARALLELISM AND PERPENDICULARITY

363

(6) Explain directly as if to an eighth grader why the slope of the line defined by 2x − 5y = 7 is 25 by making use of only the definition of slope and without invoking Theorem 6.10. (7) Find the equation of the line passing through (c, c3 ) and (d, d3 ), where c and d are numbers so that c = 1, d = 1, and c = d. Simplify your answer. 6.6. Parallelism and perpendicularity In this section, we show how the parallelism and perpendicularity of lines can be characterized in terms of slope. In TSM, these characterizations are either stated without proof or, worse, stated as "new" definitions of parallel and perpendicular lines. As an application of the characterization of parallelism in terms of slope, we give a coordinate description of a translation along a vector. Characterization of parallelism (p. 363) Characterization of perpendicularity (p. 366) A coordinate description of translation (p. 368) Characterization of parallelism Theorem 6.17. Two distinct nonvertical lines have the same slope ⇐⇒ they are parallel. Remark. We have been talking informally about lines that slant this way / or that way \. It is time to point out that, with the availability of Theorem 6.17, we can give precision to these expressions: we say a line slants this way / if the line passing through the origin and parallel to it lies in quadrants I and III,41 and similarly, we say a line slants this way \ if the line passing through the origin and parallel to it lies in quadrants II and IV. It follows from Theorem 6.17 that a nonvertical and nonhorizontal line slanting this way / has positive slope, and one slanting this way \ has negative slope. Proof. We give two proofs. First proof. This uses algebra by way of Theorem 6.11 on page 354. Suppose lines L and L are nonvertical and are parallel and we want to prove that they have the same slope. Suppose not, and we will deduce a contradiction. By Theorem 6.11, we may assume that L and L are the graphs of y = mx + k and y = m x + k , respectively. By Lemma 6.13 on page 356, m and m are the slopes of L and L and, therefore, m = m under the present assumption. It is easy to verify that the point        k −k k −k ,m +k P = m − m m − m is a solution of both equations y = mx + k and y = m x + k .42 By the definition of the graphs of linear equations (page 352,) P is the point of intersection of L and L . This contradicts L  L . 41 By

a common abuse of language, we ignore the point of the line at the origin O. P is obtained by solving the given simultaneous linear equations, a subject we will take up in earnest in Section 6.7 below. However, the verification that this P is a solution of both equations is independent of any theory about simultaneous equations. 42 Obviously,

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6. SYMBOLIC NOTATION AND LINEAR EQUATIONS

Next, we look at the converse. Suppose L and L are distinct lines with the same slope and we have to prove that they are parallel. As before, let L and L be the graphs of y = mx + k and y = mx + k , respectively. Note that the same m appears as the coefficient of x in both equations because the lines have the same slope (Lemma 6.13 again) and k = k because L and L are distinct lines, by hypothesis. Again we use a contradiction argument. If they are not parallel, let them intersect at a point P = (p1 , p2 ). By the definition of the graphs of linear equations, P is a solution of both y = mx + k and y = mx + k . Thus, p2 = mp1 + k

and

p2 = mp1 + k .

It follows that k = p2 − mp1 = k , contradicting k = k . The proof is complete. Second proof. This uses geometry. If one of the lines 1 and 2 is horizontal, this theorem is easily disposed of on account of the characterization of horizontal lines as those with zero slope (Lemma 6.8 on page 346). Assume then that neither is horizontal. First, suppose 1  2 , and we will prove they have the same slope. For clarity, we will assume that 1 and 2 have positive slope, but the case of negative slope can be handled in exactly the same way. Take a point P on 1 and let a vertical line through P intersect 2 at Q. (This vertical line must intersect 2 because the latter is not vertical.) Since the lines are distinct, P = Q. Go along the ray RP Q from P to Q and stop at some point R so that |P Q| = |QR|. From Q, draw a horizontal line which meets 1 at S, and from R also draw a horizontal line which meets 2 at T . The following gives one representation of this situation.  1 P  2           Q   S         T R  O Y

X

Because P SQ and QT R are right triangles with legs parallel to the coordinate axes (Theorem G3 on page 224), the slopes of 1 and 2 are |P Q| |SQ|

and

|QR| , |T R|

respectively. We have to show that these two numbers are equal. Since we already know |P Q| = |QR|, it suffices to prove |SQ| = |T R|. We do this by showing that P SQ and QT R are congruent, which then immediately yields the desired equality |SQ| = |T R|, because the corresponding sides of congruent triangles have the same length (Theorem G7 on page 245). To prove the congruence, we will make use of the ASA criterion of congruence (Theorem G9 on page 245). We have |P Q| = |QR| by construction. Also |∠P QS| = |∠QRT | = 90◦ , and ∠SP Q and ∠T QR have the same degrees because they are corresponding angles of the parallel lines 1 and 2 (Theorem G18 on page 277).

6.6. PARALLELISM AND PERPENDICULARITY

365

Hence triangles SP Q and T QR are congruent, from which we conclude that |SQ| = |T R|. This then completes the proof that 1 and 2 have the same slope. Before proving the converse, we should ask what strategy we might use. As always, what we can do depends on what tools are available. Up to this point, what tools (theorems) are at our disposal that would guarantee that two lines are parallel? Basically there is only one: Theorem G19 on page 281 in Chapter 5, which says that if the corresponding angles (or alternate interior angles) of a transversal with respect to a pair of lines are equal, then the lines are parallel. With the same construction as above, we see that proving the equality of ∠SP Q and ∠T QR would be our best bet. It then follows that we would try to prove the congruence of triangles P SQ and QT R to achieve our goal. Y  1  2 P            Q  S        T R  X O Conversely, suppose two distinct, nonvertical lines 1 and 2 have the same slope and we have to show that they are parallel. We are assuming they are not horizontal, so we may perform the same construction as before to get |P Q| = |QR| and right triangles P SQ and QT R. We are going to prove that the triangles are congruent by using the SAS criterion for congruence (Theorem G8 on page 245). We already have right angles ∠P QS and ∠QRT . We also have the equality of one pair of sides, |P Q| = |QR|, by construction. To get the equality of the other pair of sides, we use the hypothesis on the equality of the slopes of 1 and 2 : |QR| |P Q| = . |SQ| |T R| Since |P Q| = |QR|, we have |SQ| = |T R|. Hence the triangles P SQ and QT R are congruent, and consequently, the corresponding angles43 ∠SP Q and ∠T QR are equal (Theorem G6 on page 240). This implies 1  2 because their corresponding angles relative to the transversal P R are equal (Theorem G19 on page 281). The proof is complete. Pedagogical Comment. We could have rephrased the second proof so that the concept of congruence is replaced by the concept of similarity. Then, we could have let U be any point on the ray RP Q and not finesse it so that |P Q| = |QR|, and the whole argument would still be valid, provided we replace the SAS criterion for congruence by the SAS criterion for similarity (Theorem G21 on page 287). Such a proof would be a trifle more natural. But in terms of teaching in 43 Having given explicit details in the second proof of Theorem G18 (see page 278) on how to prove that a pair of angles are corresponding angles with respect to a transversal of parallel lines, we will henceforth omit such arguments (see especially the Pedagogical Comments on page 279).

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6. SYMBOLIC NOTATION AND LINEAR EQUATIONS

the school classroom, especially at the beginning of teaching geometry, congruence is the more intuitive and the more elementary concept, and students find it more accessible than similarity. To the extent possible, we would therefore exploit congruence. This accounts for the above proof. End of Pedagogical Comment. Characterization of perpendicularity Theorem 6.18. Two distinct, nonvertical lines are perpendicular ⇐⇒ the product of their slopes is −1. We begin with a general observation about why the slopes of perpendicular lines that pass through the origin of a coordinate system must have opposite signs (i.e., one is negative and the other is positive; see page 125). Restricting the discussion to nonvertical and nonhorizontal lines passing through the origin, we see that (excepting the origin, as always) they must lie completely inside either quadrants I and III, or quadrants II and IV (see page 335), as shown:



O





T

T T

T

OT

T

T

T

We are now going to explain why the lines in the left picture, those that lie completely inside quadrants I and III, are the ones with positive slope, and those in the right picture, those that lie completely inside quadrants II and IV, are the ones with negative slope. Take a point (a, b) on such a line ((a, b) = (0, 0)); then using (a, b) and (0, 0) to compute its slope, we see that the slope is ab . Since b and a have the same sign in quadrants I and III (see page 122 for the definition) and opposite signs in quadrants II and IV, it follows that the slope is positive for lines lying in quadrants I and III, and negative for lines lying in quadrants II and IV (recall Theorem 2.10 on page 116). As a consequence, if we have two rays issuing from O with positive slopes (i.e., the lines containing them have positive slopes), then they lie in either quadrant I or III and therefore the degree of the angle between these rays is either greater than 90◦ or less than 90◦ . Similarly for two lines with negative slopes. It follows that two lines whose slopes have the same sign can never be perpendicular to each other. Hence we have proved the following lemma. Lemma 6.19. If two nonvertical lines passing through O are perpendicular, their slopes have opposite signs. We can now give the Proof of Theorem 6.18: first suppose 1 and 2 are perpendicular lines. Let lines L1 and L2 be lines passing through the origin so that 1  L1 and 2  L2 . (If 1 or 2 passes through the origin, we will let L1 or L2 be 1 or 2 , as the case may be.) By Theorem 6.17, 1 and L1 have the same slope. The same is true for 2 and L2 . Therefore it suffices to prove that the product of the slopes of L1 and L2 is equal to −1. By hypothesis, neither line is vertical and

6.6. PARALLELISM AND PERPENDICULARITY

367

they are perpendicular to each other; thus neither line is horizontal either. So both L1 and L2 are nonvertical and nonhorizontal. By Lemma 6.19, we already know that the product is a negative number. Hence, it suffices to prove that the product of the absolute values of the slopes of L1 and L2 is equal to 1. Observe that, because 1 ⊥ 2 , L1 ⊥ L2 (Exercise 3 on page 372). It follows from the preceding discussion that we may assume that L2 lies in quadrants I and III and L1 lies in quadrants II and IV. Let P2 be some point on the line L2 and in quadrant I, and let  be the rotation of 90 degrees around the origin O. Then (L2 ) = L1 and therefore if P1 = (P2 ), we have P1 ∈ L1 . Furthermore, let the vertical line from P2 meet the x-axis at Q2 ; then (Q2 ), to be denoted by Q1 , lies on the y-axis. As  is a congruence, we have (6.42)

|P1 Q1 | = |P2 Q2 |

and

|OQ1 | = |OQ2 |.

Y

L1JJ J

P1 J

Q1

J

J J

   J  J O

 P2  

L2

X

Q2

By (6.30) on page 348, we see that the absolute value of the slope of L1 is |OQ1 |/|P1 Q1 | and the absolute value of the slope of L2 is |P2 Q2 |/|OQ2 |. Thus, taking into account the equalities in (6.42), the product of the absolute values of the slopes of L1 and L2 is |OQ1 | |P2 Q2 | |OQ2 | |P2 Q2 | · = · = 1. |P1 Q1 | |OQ2 | |P2 Q2 | |OQ2 | This completes the first part of the proof of Theorem 6.18. How shall we approach the proof of the converse? We want to prove that if the product of the slopes of L1 and L2 is −1, then L1 ⊥ L2 . In other words, if P2 and P1 are two random points on L2 and L1 , respectively, we want to prove |∠P1 OP2 | = 90◦ . Y L1 J J

 L2

J

P1 J

J

Q1

P2 

  J   J  J O



Q2

X

Since we are already given that |∠Q1 OQ2 | = 90◦ and clearly |∠Q1 OQ2 | = |∠P2 OQ2 | + |∠Q1 OP2 |, it means that if we can show |∠P1 OQ1 | = |∠P2 OQ2 |,

368

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then we would have |∠P1 OP2 | = |∠P1 OQ1 | + |∠Q1 OP2 | = |∠P2 OQ2 | + |∠Q1 OP2 | = 90◦ . So how can we show |∠P1 OQ1 | = |∠P2 OQ2 |? We resort to the standard reasoning of identifying these angles as corresponding parts of similar or congruent triangles. Now we proceed to the formal proof of the converse statement in Theorem 6.18. We have a choice of using congruence or similarity, but since we already used congruence in the second proof of Theorem 6.17, we will phrase the present proof in terms of similarity (see the Pedagogical Comment on page 365). Suppose we have two lines 1 and 2 so that the product of their slopes is −1. We must prove that 1 ⊥ 2 . Again we let L1 and L2 be lines passing through the origin and parallel to 1 and 2 , respectively (or equal to 1 , 2 as the case may be if 1 or 2 already passes through the origin), and it suffices to prove that L1 ⊥ L2 (Exercise 3 on page 372). By Theorem 6.17, the product of the slopes of L1 and L2 is also −1. In particular, since the slopes of L1 and L2 have opposite signs, we may assume that L2 lies in quadrants I and III and L1 lies in quadrants II and IV. Let P2 be some point on L2 lying in quadrant I. Drop a vertical line from P2 so that it meets the x-axis at Q2 (see preceding picture). Let P1 be some point on L1 lying in quadrant II, and let a horizontal line from P1 meet the y-axis at Q1 . If we can prove that P1 OQ1 ∼ P2 OQ2 , then we would have |∠P1 OQ1 | = |∠P2 OQ2 | (Theorem G20 on page 287) so that |∠P1 OP2 | = |∠P1 OQ1 | + |∠Q1 OP2 | = |∠P2 OQ2 | + |∠Q1 OP2 | = 90◦ . In other words, L1 ⊥ L2 , and therefore 1 ⊥ 2 . It remains to prove that P1 OQ1 ∼ P2 OQ2 . Since the product of the slopes of L1 and L2 is −1, the product of the absolute values of slopes of L1 and L2 is equal to 1. By a reasoning that is familiar to us by now, this means |P2 Q2 | |OQ1 | · = 1. |OQ2 | |P1 Q1 | Multiplying both sides of the equality by |OQ2 |/|OQ1 |, we get |OQ2 | |P2 Q2 | = . |P1 Q1 | |OQ1 | Since ∠P2 Q2 O and ∠P1 Q1 O are right angles, the SAS criterion for similarity (Theorem G21 on page 287) implies that P1 OQ1 ∼ P2 OQ2 , as desired. The proof of Theorem 6.18 is complete. A coordinate description of translation We are now in a position to give the precise coordinate descriptions of a few basic isometries, the most important of these being the following. −−→ Lemma 6.20. Let T be the translation along the vector BC, where B = (b1 , b2 ) 2 and C = (c1 , c2 ). Then for all (x, y) in R , T (x, y) = (x + a1 , y + a2 ), where (a1 , a2 ) = (c1 − b1 , c2 − b2 ).

6.6. PARALLELISM AND PERPENDICULARITY

369

Remark. When B is the origin O = (0, 0), the lemma simplifies to the follow−−→ ing: let T be the translation along the vector OC, where C = (c1 , c2 ). Then for any (x, y) in R2 , T (x, y) = (x + c1 , y + c2 ). Proof. Let the line passing through B and C be denoted by L. Let P be a point with coordinates (p1 , p2 ). We will prove that T (P ) has coordinates (p1 +a1 , p2 +a2 ). According to the definition of translation on page 234, the proof is broken into two cases. Case 1. P ∈ L. [The theorem in this case is pictorially obvious but its proof is tedious. The suggestion is to skip this proof in a school classroom and concentrate on proving Case 2 only.] First assume that L is not vertical; i.e., b1 = c1 . Let Q denote the point (p1 + a1 , p2 + a2 ), where, as in the lemma, (a1 , a2 ) = (c1 − b1 , c2 − b2 ). We want to prove that Q = T (P ). The following picture is for the case a1 > 0, a2 > 0, and p1 < b1 : Y

C  L : Br Q  : Pr   X

O

According to the definition of translation on page 234, we have to prove that −−→ −−→ Q lies on L and |P Q| = |BC| and P Q and BC point in the same direction. Let us first observe that |P Q| = |BC| because the distance formula (equation (6.20) on page 336) says  ((p1 + a1 ) − p1 )2 + ((p2 + a2 ) − p2 )2 |P Q| =   = a21 + a22 = (c1 − b1 )2 + (c2 − b2 )2 = |BC|. Next, we prove that Q lies on L. Let LP Q denote the line containing P and Q as usual. Now L and LP Q are two lines that contain the point P , and they also have the same slope because the slope of LP Q is (p2 + a2 ) − p2 a2 c2 − b2 = = (p1 + a1 ) − p1 a1 c1 − b1 and the latter is the slope of L. Therefore, by Theorem 6.9 on page 347, the lines L and LP Q coincide. Therefore Q lies on L = LP Q . −−→ −−→ Finally, we prove that P Q and BC point in the same direction. There are two cases: a1 > 0 and a1 < 0. It suffices to take up the case of a1 > 0 as the second case is similar. Recall that P = (p1 , p2 ) and B = (b1 , b2 ). Suppose p1 < b1 , as in the preceding picture. Then we claim that the ray RBC is contained in the ray RP Q . Since a1 > 0, we have p1 < q1 (= p1 + a1 ). Therefore with Q = (q1 , q2 ), the ray RP Q consists of all the points (x, y) on L so that p1 ≤ x, according to Lemma 6.16 on page 359. By the same token, since c1 = b1 + a1 > b1 , the ray RBC consists of all the points (x, y) on L so that b1 ≤ x. Thus for any point (x , y  ) lying in RBC , we have p1 < b1 ≤ x . It follows that RBC ⊂ RP Q . Now if we let L0 be the vertical line passing through P , then the closed right half-plane of L0 (consisting of all the

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points (x, y) in the plane so that p1 ≤ x) contains both RP Q and RBC . This shows −−→ −−→ that P Q and BC point in the same direction. Suppose b1 < p1 instead. Y

Q  L : Pr C :   B r  X

O

Then if we go through the preceding argument by interchanging the pair B, C by the pair P , Q, we will be able to conclude that RP Q ⊂ RBC . Now if we let L0 be the vertical line passing through B, then the closed right half-plane of L0 will −−→ −−→ contain both RP Q and RBC . Once again, this shows that P Q and AB point in the same direction. Thus T (P ) = Q if P ∈ L and L is not vertical. If L is vertical, then b1 = c1 and a1 = 0. It is straightforward to see that, in this case, the preceding argument simplifies; e.g., we prove that if B is above (respectively, below) C, then P is also above (respectively, below) Q. The proof of Case 1 is complete. Case 2. P is not on L. Again let P = (p1 , p2 ) and let T (P ) = Q. According to Lemma 4.21 on page 234, Q is the intersection of the line L which passes through P and is parallel to L, and the line  passing through C and is parallel to LBP . We have to prove that Q = (p1 + a1 , p2 + a2 ). Y

 L C  :   B  r     L       Q    P



O

X

Our strategy to prove Q = (p1 +a1 , p2 +a2 ) is to let Q be the point (p1 +a1 , p2 +a2 ) and then show that Q coincides with Q. To this end, we will prove below that L  LP Q and LCQ  LBP . Assuming this for the moment, we see that, by the parallel postulate, LP Q has to be the line passing through P and parallel to L, and therefore L = LP Q . Similarly, LCQ has to be the line passing through C and parallel to LBP , and therefore  = LCQ . Since two distinct lines can intersect at only one point (Lemma 4.2 on page 165), Q = Q and we are done in this case. Let us now prove L  LP Q . First assume that L is not vertical. Then b1 = c1 , so that a1 = c1 − b1 = 0, We will compute the slope of LP Q using (of course) the two points P = (p1 , p2 ) and Q = (p1 + a1 , p2 + a2 ):

slope of LP Q =

(p2 + a2 ) − p2 a2 = . (p1 + a1 ) − p1 a1

6.6. PARALLELISM AND PERPENDICULARITY

371

Next, we compute the slope of L using the two points B = (b1 , b2 ) and C = (c1 , c2 ). Because a1 = c1 − b1 and a2 = c2 − b2 , we have C = (a1 + b1 , a2 + b2 ). Hence, c2 − b2 a2 = . c1 − b1 a1 These two slopes being equal, we see that L  LP Q (Theorem 6.17 on page 363) when L is not vertical. Now if L is vertical, then b1 = c1 and a1 = c1 − b1 = 0. Since Q = (p1 + a1 , p2 + a2 ) = (p1 , p2 + a2 ), Q and P = (p1 , p2 ) have the same first coordinates and LP Q is also vertical. Thus L  LP Q as before. Next, we will prove that LCQ  LBP . First assume that LBP is not vertical, so that b1 = p1 . Then using Theorem 6.17 again, we will prove that the slopes of LCQ and LBP are equal. Using the fact that C = (a1 + b1 , a2 + b2 ), we compute the slope LCQ using C and Q : slope of L =

slope of LCQ =

(a2 + b2 ) − (p2 + a2 ) b2 − p2 = slope of LBP . = (a1 + b1 ) − (p1 + a1 ) b1 − p1

Thus LCQ  LBP in case LBP is not vertical. It remains to consider the case that LBP is vertical. Then b1 = p1 so that a1 + b1 = a1 + p1 . Since a1 = c1 − b1 , this implies c1 = p1 + a1 , and the first coordinates of C and Q are equal. Therefore LCQ is also vertical and again LCQ  LBP . The proof of the lemma is complete. Lemma 6.20 can be used to give the shortest proof that the composition of two translations is a translation (see Exercise 10 on page 238). The same lemma also shows that the composition of translations is commutative in the sense that if T −−−→ −− → is the translation along a vector AB and T  is the translation along a vector A B  , then for all (x, y) in the plane, (T ◦ T  )(x, y) = (T  ◦ T )(x, y). Mathematical Aside: Recall that we introduced the group of translations in the plane in the Mathematical Aside on page 235. The preceding remark about the composition of translations being commutative means that the group of translations in the plane is in fact an abelian group. It turns out that, with the help of trigonometric functions and complex numbers, we will be able to express all the basic isometries in terms of coordinates, at least in principle (see Section 1.6 in [Wu2020c]). In the meantime, here are some simple rotations and reflections in terms of coordinates. (We leave their proofs to Exercises 8, 9, and 11 on page 372). (i) Let  denote the counterclockwise rotation of 90 degrees around the origin O of R2 . Then for every (x, y) ∈ R2 , (x, y) = (−y, x).44 (ii) Let 0 be the 180-degree rotation around the origin O. Then for every (x, y) ∈ R2 , we have 0 (x, y) = (−x, −y). (iii) If Λ1 denotes the reflection with respect to the x-axis, then for every (x, y) in R2 , Λ1 (x, y) = (x, −y). (iv) If Λ2 denotes the reflection with respect to the y-axis, then for every (x, y) ∈ R2 , Λ2 (x, y) = (−x, y). 44 If

(x, y) lies in the first quadrant, this is implicit in the proof of Theorem 6.18.

372

6. SYMBOLIC NOTATION AND LINEAR EQUATIONS

Exercises 6.6. (1) What is the equation of the line which is perpendicular to the graph of ax + by = c, where a, b, c are constants, a = 0, and b = 0, and which passes through the point (− 12 , 3)? (2) Assume a triangle with vertices at (1, 1), (5, 1), and (7 21 , 4). Does the point (2, − 61 ) lie on the line passing through the vertex (1, 1) and perpendicular to the opposite side of the triangle? Explain. (3) Prove the following assertion which was used to prove Theorem 6.18: let L and V be two perpendicular lines. If L and V  are lines so that L  L and V  V  , then also L ⊥ V  . (4) Let O be the origin of a coordinate system in R2 , and let r be a positive number. Prove that the transformation T of R2 which assigns to each point (a, b) the point (ra, rb) is exactly the dilation with center O and scale factor r. (Caution: This is a slippery proof.) (5) (a) Let L be the line joining O (the origin (0, 0)) to the point (a, b), and let L be the line joining O to the point (a , b ). Prove: L ⊥ L if and only if aa + bb = 0. (You may recognize the latter as the dot product of the vectors (a, b) and (a , b ) that you came across in calculus.) (b) Let L and L be the lines defined by the equations ax + by = c and a x + b y = c , respectively. Prove that L ⊥ L if and only if aa + bb = 0. (6) Write out a self-contained proof, using only congruent triangles and without using similar triangles, of the second half of Theorem 6.18; i.e., if the product of the slopes of two lines is −1, then the lines are perpendicular. (7) Let T be the translation along the vector from O to a fixed point (a1 , a2 ) in R2 . (a) If L is the vertical line defined by x = 27 , what is the equation of T (L)? (b) If L is the horizontal line defined by y = 51, what is the equation of T (L)? (c) If L is the line defined by 2x − 3y = 1, what is the equation of T (L)? (8) (i) Let 0 be the 180-degree rotation around the origin O. Prove that for every (x, y) ∈ R2 , we have 0 (x, y) = (−x, −y). (ii) Let φ be the 180degree rotation around the point (a, b). Prove that for every (x, y) ∈ R2 , we have φ(x, y) = (2a − x, 2b − y). (9) (i) If Λ1 denotes the reflection with respect to the x-axis, prove that for every (x, y) ∈ R2 , Λ1 (x, y) = (x, −y). (ii) If Λ2 denotes the reflection with respect to the y-axis, prove that for every (x, y) ∈ R2 , Λ2 (x, y) = (−x, y). (10) Let L be the line defined by 3x − 4y = 3, and let Λ denote the reflection across L. Compute Λ(7, 13 ). (11) (i) Let ρ be the 90◦ counterclockwise rotation around the origin O and let φ be the 90◦ clockwise rotation around O. Prove that for any point (x, y), ρ(x, y) = (−y, x) and φ(x, y) = (y, −x). (ii) Let (a, b) be a fixed point in the coordinate plane and let  be the 90◦ counterclockwise rotation around (a, b) and let ϕ be the 90◦ clockwise rotation around (a, b). Prove that for any point (x, y) (x, y) = (−y + a + b, x − a + b)

and

ϕ(x, y) = (y + a − b, −x + a + b).

(iii) Let P and Q be two distinct points and let P be the 90◦ counterclockwise rotation around P and ϕQ be the 90◦ clockwise rotations around

6.7. SIMULTANEOUS LINEAR EQUATIONS

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Q. Prove that both ϕQ ◦ P and P ◦ ϕQ are nontrivial translations; i.e., they are not identity transformations of the plane. (iv) Prove that the compositions of rotations in part (iii) are not rotations by proving that a nontrivial translation is never equal to a rotation. (This puts in perspective Exercise 10 on page 238.) 6.7. Simultaneous linear equations In this section, we give a complete analysis of the solution set of a linear system of two equations in two unknowns, first geometrically in terms of the two lines defined by the pair of equations and then algebraically in terms of the so-called determinant of the linear system. In the last subsection, we also give an explanation of why the standard method of solution of such linear systems by substitution is correct. The solution set of a linear system: Geometry (p. 373) The solution set of a linear system: Algebra (p. 377) Solution by substitution (p. 380) The solution set of a linear system: Geometry Suppose we are given constants a, b, . . . , f and we want to know if there are numbers x and y so that they are solutions of both of the following linear equations:  ax + by = e, cx + dy = f. Such a pair of linear equations is variously called a linear system, or a system of linear equations, or sometimes, simultaneous (linear) equations in the numbers x and y. To be precise, one would have to refer to such a pair of equations as a linear system of two equations in two unknowns or in two variables, where the "unknowns" or "variables" refer to the symbols x and y. An ordered pair (x0 , y0 ) is called a solution of the system if it is a solution of both equations. To solve the system is to find all the solutions of the system. Sometimes we also call the collection of all these (x0 , y0 )’s the solution set of the system. Consider, for example,  x + y = 2, (6.43) 3x + 3y = 6. This linear system has as solution all ordered pairs of the form (t, 2 − t), where t is an arbitrary number. In this case, the solution set is an infinite collection of points, being all the points on the graph of x + y = 2 (the second equation has the same graph as the first). On the other hand, the system  x + y = 2, 3x + 3y = 5 clearly has no solution, because if (x0 , y0 ) is a solution, the fact that it is a solution of the first equation means x0 + y0 = 2. By multiplying both sides of this equality by 3, we get 3x0 + 3y0 = 6. But since (x0 , y0 ) is (supposedly) also a solution of the second equation, we also have 3x0 + 3y0 = 5, and therefore 5 = 6, a contradiction. Therefore, the solution set in this case is the empty set, i.e., no solutions.

374

6. SYMBOLIC NOTATION AND LINEAR EQUATIONS

In between the two preceding extreme cases, "most" linear systems have exactly one pair (x0 , y0 ) as a solution. We will explain what "most" means and why this is true by way of geometry. Let 1 , 2 be the lines which are the graphs of the equations ax + by = e and cx + dy = f , respectively. Suppose (x0 , y0 ) is a solution of the linear system  ax + by = e, cx + dy = f. In particular, this means we are assuming that there is a solution of the system. We wish to interpret this solution geometrically. Since (x0 , y0 ) is a solution of ax + by = e, the point (x0 , y0 ) lies on 1 , by the definition of the graph of an equation. For the same reason, (x0 , y0 ) lies on 2 as well. Therefore (x0 , y0 ) lies on both 1 and 2 , and therefore it lies in the intersection of 1 and 2 . (We have to be careful not to assume that the intersection of 1 and 2 is a point, because we cannot a priori exclude the possibility that 1 = 2 as in (6.43) above, in which case the intersection of 1 and 2 is the line itself.) Conversely, suppose (x0 , y0 ) lies in the intersection of 1 and 2 ; then it must be a solution of the system  ax + by = e, cx + dy = f because (x0 , y0 ) being on 1 means ax0 + by0 = e and (x0 , y0 ) being on 2 means cx0 + dy0 = f . We have therefore proved the following basic fact relating the solutions of a linear system to the graphs of the equations in the system. Theorem 6.21. A simultaneous system of linear equations has a solution (x0 , y0 ) ⇐⇒ the point (x0 , y0 ) lies in the intersection of the (linear) graphs of the equations in the linear system. Theorem 6.21 gives the precise reasoning for why the solution of a linear system of two linear equations in two unknowns corresponds to the intersection of the two lines defined by the linear equations of the system. This is the reason why one can get the solution of a system of simultaneous linear equations by graphing the equations. Such a correspondence is usually decreed by fiat in TSM without explanation, probably because the precise definition of the graph of an equation is rarely given or, if given, is not put to use. It is worth noting that Theorem 6.21 shares a common feature with a coordinate system: they both provide a dictionary that mediates two disparate sets of information: the algebraic information about solutions of a linear system and the geometric information about intersections of lines. In this particular case, we know all about the intersections of lines (see (L1) and (L2) on page 165) and will therefore use this knowledge to shed light on the solutions of linear systems. We know that there are exactly three mutually exclusive possibilities for two lines in the plane: the lines are either identical or parallel or distinct but not parallel.

6.7. SIMULTANEOUS LINEAR EQUATIONS

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Correspondingly, the lines intersect at an infinite number of points or do not intersect or intersect exactly at one point. From Theorem 6.21, we therefore conclude the following. Corollary. Given a linear system of two equations in two unknowns, let the graphs of the linear equations be 1 and 2 . Then the linear system either has an infinite number of solutions (corresponding to 1 = 2 ) or has no solution (corresponding to 1  2 ) or has a unique solution (corresponding to 1 = 2 but 1 is not parallel to 2 ). We can now explain what is meant by "most" linear systems have a unique solution. Given two lines, what are the chances that they are either identical or parallel? This is in fact a precise mathematical question that can be answered completely: zero. To obtain this answer, one must do some advanced mathematics. Nevertheless, one can provide an intuitive understanding of the situation by fixing one of the lines, say 1 , and ask what the chances are that the other line 2 either coincides with 1 or is parallel to 1 . Clearly we can ignore the possibility of 2 actually equaling 1 because this almost never happens. What about 2  1 ? Look at it this way: restrict 2 to be a line passing through a fixed point P not lying in 1 ; then according to the parallel postulate (page 165), there is at most "one chance" that 2  1 , whereas there are infinitely many possibilities for 2 not to be parallel to 1 . Since this is true for any point P not lying on 1 , it is intuitively clear that, "almost always", 2 will be a line distinct from 1 and not parallel to 1 . So by the Corollary, a linear system will "almost always" have a unique solution. We now take this corollary of Theorem 6.21 to the next level: what are the algebraic properties of the linear system that would lead to an infinite number of solutions, no solution, and a unique solution? We can literally follow the prescription of the preceding corollary and just doggedly investigate the algebraic properties of the linear equations that correspond to 1 = 2 , 1  2 , and 1 = 2 but 1  2 . This would lead to a depressing case-by-case argument with thickets of details that would ultimately not be particularly enlightening. Here is one way this argument could be carried out. Case 1. The graphs 1 and 2 of ax + by = e and cx + dy = f , respectively, coincide. If they are vertical, then b = d = 0 and the equations become x = e/a and x = f /c. Their graphs are identical ⇐⇒ ae = fc . Therefore this case is equivalent to f e = . a c If they are not vertical, then both b = 0 and d = 0 and we may rewrite the system as  y = mx + k, y = m x + k , b=d=0

and

where m = − ab , k = eb , m = − dc , and k = fd . Then 1 and 2 being identical means they have the same slope and therefore m = m . The equations of 1 and 2 become y = mx + k and y = mx + k ,

376

6. SYMBOLIC NOTATION AND LINEAR EQUATIONS

respectively. If k = k , then (0, k) would be a point lying on 1 but not 2 . Thus k = k . Therefore in this case, c e f a = and = . b d b d Case 2. The graphs 1 and 2 of ax + by = e and cx + dy = f , respectively, are parallel. Then they are distinct and either are both vertical or are both nonvertical and have the same slope (Theorem 6.17 on page 363). In the former case, being vertical means b = d = 0, and since they are distinct, the equations ax = e and cx = f have distinct graphs. Hence, ae = fc . To summarize, 1 and 2 being vertical and parallel is equivalent to f e = . b = d = 0 and a c In the latter case, since 1 and 2 are nonvertical, b = 0 and d = 0 and we can write the system as  y = mx + k, y = m x + k , where m = − ab , k = eb , m = − dc , and finally, k = fd . Since 1 and 2 have the same slope, m = m , so that ab = dc and therefore by the cross-multiplication algorithm, ad = bc. Moreover, the graphs of y = mx + k and y = mx + k (recall m = m ) are distinct ⇐⇒ k = k ; i.e., eb = fd . To summarize, 1 and 2 being nonvertical, parallel, and distinct is equivalent to f e = . ad − bc = 0 and b d Case 3. The graphs 1 and 2 of ax + by = e and cx + dy = f , respectively, are not parallel and not the same line. Assume first that both lines are not vertical. Then as in Case 1, b = 0 and d = 0, so that we may rewrite the system as  y = mx + k, y = m x + k , where m = − ab , k = eb , m = − dc , and k = fd . By Theorem 6.17 on page 363, 1 and 2 not being parallel is equivalent to m = m , and therefore ab = dc and therefore by the cross-multiplication algorithm, ad = bc. Hence this case is equivalent to ad = bc. Next, we consider the case where one of the lines is vertical; let us say 1 is vertical but 2 is not. Then 1 is defined by x = ae and 2 is defined by y = m x + k ,   c where m = − d and k = fd . Then a = 0, b = 0, and c, d = 0, and again ad = bc.

What we are going to do is to give a sophisticated algebraic analysis of the preceding corollary from the perspective of slope. This analysis is not something most school students can "discover" on their own. Instead, we are going to offer students an opportunity to learn from the wisdom of the past. Absorbing what others have

6.7. SIMULTANEOUS LINEAR EQUATIONS

377

to offer is one very good way for us to grow intellectually. It is pointless to try to "discover" everything yourself; it is impossible anyway. The solution set of a linear system: Algebra We now begin the algebraic analysis of the preceding corollary. Let 1 , 2 be the lines which are the graphs of the equations ax + by = e and cx + dy = f , respectively, in the linear system  ax + by = e, cx + dy = f. To motivate what is to come, suppose 1 and 2 are not vertical, so that their slopes are defined. We notice that there is one thing that distinguishes the first two cases (1 = 2 and 1  2 ) from the third (1 = 2 but 1  2 ); namely, in the first two cases, 1 and 2 have the same slope (Theorem 6.17 on page 363) whereas in the third case, 1 and 2 have different slopes. We now express this information about slope algebraically, as follows. Since 1 and 2 are both nonvertical, we have b = 0 and d = 0 so that we may rewrite the linear system as ⎧ ⎨ y = (− ab )x + eb , ⎩

y

=

(− dc )x +

f d.

Thus the slope of 1 is − ab , and that of 2 is − dc . The slopes are equal ⇐⇒ ab = dc , which is equivalent to ad = bc (the generalized cross-multiplication algorithm for rational quotients, item (b) on page 118 of Chapter 2), which in turn is equivalent to ad − bc = 0. For the same reason, the slopes of 1 and 2 being different is equivalent to ad − bc = 0. Thus whether or not the two lines of the linear system have the same slopes or different slopes is captured by the vanishing (i.e., equal to zero) or nonvanishing of the number ad − bc, respectively. Such considerations suggest that the number ad − bc is an important characteristic of the linear system. Indeed it is, and it is called the determinant of the linear system, usually denoted by Δ.45 The theorem we want to prove is then the following. Theorem 6.22. Assume a linear system  ax + by = e, cx + dy = f. Let Δ denote the determinant of the system, Δ = ad − bc. Then: (i) If Δ = 0, the linear system has a unique solution. (ii) If Δ = 0, the linear system has either an infinite number of solutions or no solution. Before giving the proof, we observe that since the definition of the determinant does not require that any of a, b, c, and d be nonzero, the lines 1 and 2 that correspond to the two equations in the system could therefore be vertical. We also wish to point out that the following proof could be presented slightly differently; see Exercise 3 on page 383. 45 Mathematical Aside: You may consider this discussion to be a review of the simplest case of linear algebra, namely, the case of dimension 2.

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Proof. We handle the two cases (i) and (ii) separately. Case (i). The determinant ad − bc is nonzero. In this case, clearly not both b and d are zero. Suppose b = 0. Then d = 0, and we see that 1 is vertical but 2 is not vertical. Therefore 1 intersects 2 at exactly one point, and the linear system (according to the corollary of Theorem 6.21) has a unique solution. Similarly, if d = 0, then b = 0 and again the linear system has a unique solution. If both b and d are nonzero, then the linear system can be rewritten as ⎧ ⎨ y = (− ab )x + eb , ⎩

y

=

(− dc )x +

f d.

The slope of 1 is therefore − ab and the slope of 2 is − dc . Since ad − bc = 0, ad = bc and ab = dc by the cross-multiplication algorithm. Thus 1 and 2 have different slopes and therefore (by Theorem 6.17 on page 363) are not parallel to each other. It follows that 1 intersects 2 at exactly one point and the linear system again has a unique solution. We have thus proved Case (i). Case (ii). The determinant ad − bc is zero. We claim that in this case, either both b and d are 0 or both b and d are not 0. To prove the claim, it suffices to show that if b = 0, then d must be 0, and if d = 0 then b = 0. Suppose b = 0. Then ad − bc = 0 implies that ad = 0. But by the definition46 of a linear equation of two variables, not both a and b can be 0 in ax + by = e. So a = 0. It follows that ad = 0 implies d = 0. Similarly, if d = 0, then also b = 0. The claim is proved. We now examine the first possibility: b = d = 0. Then both a and c are nonzero, by the definition of a linear equation of two variables. The linear system may therefore be rewritten as ⎧ ⎨ x = ae , ⎩

x

=

f c.

Clearly, these vertical lines are identical (and the system has an infinite number of solutions) if e/a = f /c, and they are parallel (and the system has no solution) if e/a = f /c. Next we examine the second possibility: b = 0 and d = 0. The linear system may therefore be rewritten as ⎧ ⎨ y = (− ab )x + eb , ⎩

= (− dc ) x + fd . Thus 1 and 2 have slopes equal to −a/b and −c/d, respectively. Now ad − bc = 0 by hypothesis, so ad = bc and the cross-multiplication algorithm implies that a/b = c/d. This implies 1 and 2 have the same slope. We have to decide if they are identical or parallel. If e/b = f /d, then the equations are identical and therefore so are their graphs (the linear system therefore has an infinite number of solutions), and if e/b = f /d, then the lines are distinct (because, for example, y

46 This is another reminder that we must take definitions seriously. Since we have defined a linear equation αx + βy = γ to be such that not both α and β are zero, it stands to reason that this part of the definition will play a critical role sooner or later.

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379

(0, e/b) is a point on 1 but not on 2 ) and are therefore parallel (the linear system therefore has no solution). This completes the proof of Case (ii) and, therewith, the proof of Theorem 6.22. Remark. From the proof itself, we see that the conclusion of Case (ii) can be made very precise; namely: (i) If the determinant is 0, then either b = d = 0 or both b = 0 and d = 0. (ii) In case b = d = 0, then the linear system has an infinite number of solutions if e/a = f /c, and it has no solution if e/a = f /c. (iii) In case b = 0 and d = 0, then the linear system has an infinite number of solutions if e/b = f /d and has no solution if e/b = f /d. However, it is imperative that you not try to memorize these conclusions. If you understand the reasoning, then you can use it in each situation to draw the right conclusion. We illustrate with some simple examples. If we are given a linear system with b = d = 0, e.g.,

4 3x = 5, −x

= − 15 ,

then common sense dictates that you multiply the second equation by −3 to change the system to ⎧ ⎨ 3x = 45 , ⎩

3x = 35 . Direct inspection now shows that the linear system has no solution. Suppose we are given

10.2x − 13.6y = 11.5, − 94 x

+

= − 21 4 .

3y

We note that 10.2 × 3 − (−13.6)(− 49 ) = 0 as both products are equal to 30.6, and so we are in the situation of ad − bc = 0 but bd = 0. We know from the preceding analysis that the linear system has either no solution or an infinite number of solutions, depending on whether the lines defined by the equations are distinct or identical, respectively. The simplest way to find out whether the lines defined by these equations are identical or distinct is to rewrite both equations in the form −1 and the of y = mx + b and compare. Thus multiplying the first equation by 13.6 1 second equation by 3 , we get

11.5 y = 10.2 13.6 x − 13.6 , y

=

3 4x

−

7 4.

Notice that in so doing, we do not need to bother with checking whether 10.2 13.6 and 3 are equal or not, as we already know that they must be equal (because the lines 4 7 have the same slope!). The only thing to compare is whether 11.5 and are equal. 13.6 4 Since the former is less than 1 and the latter is greater than 1, they are obviously

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6. SYMBOLIC NOTATION AND LINEAR EQUATIONS

not equal. Hence the two lines are distinct and this system has no solutions. Therefore the same is true of the original system as well. Solution by substitution Up to this point, we have not talked about the explicit algebraic method of solving a linear system taught in school classrooms. This is the well-known method of substitution (sometimes taught in the equivalent form of the method of elimination). This method is usually taught by rote in schools. What we will do next is to subject this method to a critical examination. In the remainder of this section, we will explain why this method produces the correct solution to any linear system with nonzero determinant and what it means geometrically. Let us first look at a simple case and then use the standard symbolic manipulations taught in schools to get a solution. Consider  2x − 3y = 1, (6.44) 3x + 2y = −1. Let us eliminate y. So from the second equation, we get 1 3 y =− x− . 2 2 Substituting this expression of y into the first equation gives   3 1 2x − 3 − x − = 1. 2 2 Simplifying, we get

13 2 x

= − 12 so that

1 . 13 Substituting this value of x into y = − 32 x − 12 then leads to x=−

5 . 13 1 5 TSM now says (− 13 , − 13 ) is a solution of (6.44). y=−

We pause to reflect on this method of solution. We have performed the whole computation without knowing what x and y are beyond the fact that they are some symbols. But in all the mathematics we have learned up to this point, we have never talked about computing with "symbols". All we know are numbers and geometric figures. So suppose x and y are numbers. But what numbers are they? Could x be 7 and y be −2? Not really, because it is not true that  2(7) − 3(−2) = 1, 3(7) + 2(−2) = −1. Therefore the above computation could not possibly be valid for any two numbers x and y and would make sense only if this pair (x, y) is a solution of the linear system (6.44). At this point, we must recall the basic etiquette in the use of symbols (see page 299): before we use any symbol, we must know what it stands for. Let us start over and do things in a way that makes sense. The following discussion is parallel

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381

to the discussion on the solution of a linear equation in one variable on pp. 324ff.; the reader may wish to review the latter before proceeding further. Suppose (x0 , y0 ) is a solution of the system (6.44); i.e., we assume that for an ordered pair of numbers (x0 , y0 ), we have  2x0 − 3y0 = 1, 3x0 + 2y0 = −1. Then these are two equalities of numbers, and we can proceed to compute with them in the usual way that we do arithmetic. From the second equation, we get 3 1 y0 = − x0 − . 2 2 Substituting this value of y0 into the first equation gives   3 1 2x0 − 3 − x0 − = 1. 2 2 Solving this linear equation in x0 (as in Section 6.2 on page 322), we get so that 1 x0 = − . 13 Substituting this value of x0 into y0 = − 32 x0 − 12 then leads to

13 2 x0

= − 12

5 . 13 Note that if we replace x0 by x and y0 by y, then this computation is formally identical to the previous method of solution taught in the school classroom. The only difference is that the second computation is the one that is mathematically valid, because it is nothing but an ordinary computation carried out with numbers. To summarize, what we have proved is this: (A) If (x0 , y0 ) is a solution of the given linear system (6.44), then 1 5 and y0 = − . x0 = − 13 13 1 5 Have we shown that (− 13 , − 13 ) is actually a solution of the given linear system? No. For that purpose, we need to prove the following assertion, which is in fact the converse statement of (A): 1 5 and y0 = − 13 , then (x0 , y0 ) is a solution of the (B) If x0 = − 13 linear system (6.44). A routine computation verifies that, indeed, ⎧ 1

5 − 3 − 13 = 1, ⎨ 2 − 13 y0 = −

⎩

5

1 + 2 − 13 = 3 − 13

1 5 (− 13 , − 13 )

−1.

So the ordered pair produced by the method of substitution is a solution of the system (6.44), and (B) is correct. Obviously, this pragmatic answer would be of little value if it were a singular occurrence that happens to furnish a solution for this linear system but for no others. Such is not the case. We now give a self-contained and coherent account that shows that the usual method of substitution is the procedural aspect of a mathematically valid method of solution. In other words, the rote procedure may seem to make

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6. SYMBOLIC NOTATION AND LINEAR EQUATIONS

no sense, but in fact it can be shown to make sense after all. Here is the general explanation. Assume a linear system  ax + by = e, (6.45) cx + dy = f. We assume that its determinant Δ = ad − bc = 0. We first assume that the system has a solution (x0 , y0 ). Then (6.46) (6.47)

ax0 + by0 cx0 + dy0

= e, = f.

We now follow the usual computations used in the method of substitution to determine the explicit value of (x0 , y0 ), as follows. If b = d = 0, then Δ = 0. Thus Δ = 0 implies that not both b and d are 0. Without loss of generality, we may assume d = 0. Then equation (6.47) implies y0 = − dc x0 + fd . Substitute this value of y0 into (6.46) to get   c f ax0 + b − x0 + = e. d d Multiplying both sides by d and simplifying, we obtain (ad − bc)x0 = de − bf , so that de − bf x0 = . Δ Substituting this value of x0 into the equation y0 = − dc x0 + fd , we obtain     1 −cde + bcf + adf − bcf c de − bf f y0 = − + = d Δ d d Δ and therefore

af − ce . Δ What we have just proved is that, assuming that the system (6.45) has a solution (x0 , y0 ) and that Δ = 0, then necessarily   de − bf af − ce , (6.48) (x0 , y0 ) = . Δ Δ y0 =

It is now routine to check that the (x0 , y0 ) in (6.48) is in fact a solution of the system (6.45). For example, let us check that with (x0 , y0 ) as in (6.48), (6.46) holds:     de − bf af − ce ax0 + by0 = a +b Δ Δ

=

ade − abf + abf − bce Δ

=

ade − bce =e Δ



ad − bc Δ

 .

But by definition Δ = ad−bc, so the last term is equal to e, and we have ax0 +by0 = e, showing that (6.46) is valid. Similarly, (6.47) also holds for the values of (x0 , y0 ) in (6.48) (see Exercise 1 on page 383).

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383

It remains to observe that the computations that led to (6.48) are exactly what we do when solving simultaneous equations by substitution. Pedagogical Comments. One may ask whether it is realistic to explain the solution of a general linear system (6.45) in an eighth-grade classroom. There is usually not one simple answer to a pedagogical question, but for an average eighthgrade class, the extensive symbolic computations in this case may be a bit much. However, it is possible to convey the same basic idea while minimizing the needed abstraction by using a system such as (6.44) on page 380. You are strongly encouraged to try. End of Pedagogical Comments.

Exercises 6.7. (1) Verify by direct computation that equation (6.47) holds for the (x0 , y0 ) in (6.48). (2) Discuss the solutions of each of the following systems without actually solving them.  Give reasons. 4x − 3y = 1, (a) 9x − 7y = 53 .  (b)

2.4x −0.264x

− 3.5y + 0.385y

= 43, = −4.63.



15x + 12y = −4, −35x − 28y = 28 3 . (3) (a) Give a direct proof of () below by using Lemma 6.13 on page 356 and Theorem 6.21 on page 374 without invoking Theorem 6.22 (on page 377). () Assume the linear system (where m, m , k, and k are constants)  mx + y = k, m x + y = k  . (c)

(i) If m = m , it has a unique solution, and (ii) if m = m , it has either an infinite number of solutions when k = k or no solution when k = k . (b) Prove that () implies Theorem 6.22. (4) If 3 is added to the numerator of a fraction and 7 subtracted from the denominator, its value is 67 . But if 1 is subtracted from the numerator and 7 added to the denominator, its value is 25 . Find the fraction. (5) If the digits of a two-digit number are reversed, the new number is 9 less than the original. The sum of the digits is also 9. (a) If x is the tens digit and y the ones digit, write down equations satisfied by x and y. (b) What is this number? (6) If the digits of a two-digit number are reversed, the new number is 1 less than twice the original number. Furthermore, if 10 times the tens digit is divided by the ones digit, the quotient is 4 and the remainder is 2. (a) If x is the tens digit and y the ones digit, write down the equations satisfied by x and y. (b) What is this number?

384

6. SYMBOLIC NOTATION AND LINEAR EQUATIONS

−5(x+1)

(7) Let x be a number not equal to −4 or 3. Express 3(x2 +x−12) as a sum of 1 1 and x−3 . (constant) multiples of x+4 (8) Suppose you are teaching a ninth grade algebra class and you want to show your students how to solve the following linear system by the method of substitution for the first time:  2x − y = −1, x + 2y = 7. Carefully explain how you would teach it. (9) Given positive integers s and t with s < t. (a) Solve for u and v in the linear system ⎧ ⎨ u + v = t, s

⎩ u − v

(10)

(11)

(12) (13)

(14)

=

s t.

(b) Show that u > 0 and v > 0. (c) If the solutions u and v in terms of s and t are written in the form of u = cb and v = ab , where a, b, c are positive integers expressed in terms of s and t, show that {a, b, c} forms a Pythagorean triple; i.e., a, b, and c are positive integers and a2 + b2 = c2 . (Note: See page 290. This way of generating Pythagorean triples dates back to the Babylonians circa 1800 BC. See [Robson] and also Chapter 1 of [Katz].) In each of the following, you are asked to solve the linear system in the preceding exercise with the given values of s and t to obtain Pythagorean triples. You may use a scientific calculator. (a) s = 1, t = 2. (b) s = 2, t = 3. (c) s = 2, t = 69. (d) s = 54, t = 125. (e) s = 8, t = 9907. A Pythagorean triple is said to be primitive if there is no (positive) common divisor among the triple of positive integers other than 1. Prove that the following are equivalent for a Pythagorean triple {a, b, c}: (i) {a, b, c} is primitive. (ii) {a, b} are relatively prime. (iii) {a, c} are relatively prime. (iv) {b, c} are relatively prime. Let {a, b, c} be a Pythagorean triple so that a2 + b2 = c2 . If {a, b, c} is primitive, prove that one of a and b is even and the other is odd. If s and t are relatively prime positive integers so that one is even and the other is odd, then prove that the Pythagorean triples produced in Exercise 9 are primitive. Let {a, b, c} be a Pythagorean triple. Prove that the following are equivalent: (i) {a, b, c} is a primitive Pythagorean triple with a odd and b even. (ii) There is a pair of relatively prime positive integers s and t, with s < t and one of them is even and the other odd, so that a = t2 − s2 , b = 2st, and c = t2 + s2 .

Glossary of Symbols N : the whole numbers, 6 Q : the rational numbers, 90 R : the real numbers, 6 Z : the integers, 91 ⇐⇒ : is equivalent to, 22 =⇒ : implies, 76 · · · a for a number a and a positive integer n, 14 an : the product aaa n

n!n: n factorial for a whole number n, 42n! k : binomial coefficient defined by k!(n−k)! , 42

n|m : n divides m for integers m and n (n = 0), 138 n  |m : n does not divide m for integers m and n (n = 0), 138 GCD(a, b) : the greatest common divisor of integers a and b, 138 B −1 : multiplicative inverse of a fraction B, 56 p∗ : mirror reflection of a number p, 90 → − x : the vector from the origin 0 of a number line to the fraction x on this number line, 101 x · y : product of the numbers x and y, 37 |x| : absolute value of a number x, 125 √ α : the positive square root of a positive number α, 148 [a, b] : the segment or the closed interval from a to b for numbers a < b, 5 (a, b) : the open interval from a to b for numbers a < b (it could also mean the point (a, b) in the coordinate plane), 126 < : less than, 12 ≤ : less than or equal to, 13 > : greater than, 13 ≥ : greater than or equal to, 13 AB : the segment joining A to B, 169 dist(A, B) : the distance between two points A and B in the plane, 184 A ∗ C ∗ B : the point C is between points A and B, 167 |P Q| : the length of segment P Q, 185 LP Q : the line joining P to Q, 166 H+ , H− : half-planes of a line L, 176 ROP : the ray from O to P , 174 −− → AB : the vector from A to B in the plane, 219  : is parallel to, 165 ⊥ : is perpendicular to, 192 ∠AOB : an angle with vertex O and sides ROA and ROB , 182 ABC : triangle ABC, 245 385

386

GLOSSARY OF SYMBOLS

∠A : the angle of a triangle or a polygon at a vertex A, 182 x◦ : x degrees (of an angle), where x is a positive real number, 188 ∼ = : is congruent to, 240 ∼ : is similar to, 284 ∈ : belongs to (as in a ∈ A), 9 A ⊂ B : A is contained in B, 91 ∪ : union (of sets), 195 ∩ : intersection (of sets), 165 R2 : the coordinate plane, 332 (x, y) : the coordinates of a point in the plane (it could also mean the open interval from the number x to the number y), 333 H+ , H− : half-planes of a line L, 176 

AB : one of two arcs joining the point A to the point B on a circle, 190

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47 This is referenced in [CSSM], page 92, as "Wu, H., Lecture Notes for the 2009 Pre-Algebra Institute".

Index

angles alternate interior, 276 corresponding, 276 opposite, 276 Arbaugh, Fran, xviii arc, 190 intercepted by an angle on a circle, 190 length of, 190, 197 major, 190 minor, 190 arctan function, 207 area, 17, 47 of rectangle, 48 arithmetic mean, 132 arithmetic mean-geometric mean inequality, 127, 132 ASA, 51, 245 associative law, 2, 44 for addition, 87 failure for subtraction, 96, 98 for addition of fractions, 33 for addition of rational numbers, 91–94 for multiplication, 87 for multiplication of fractions, 46 for multiplication of rational numbers, 105 for multiplication of real numbers, 133 for number expressions, 302, 304, 312, 323 makes no sense for variables, 325 assumptions about basic isometries, 237 about reflections, 231 about rotations, 217 about translations, 236

AA criterion, 288 above (a horizontal line), 334 absolute error, 130 absolute value, 121, 125, 125–126 rationale for, 130–131 Ackerman, Nate, 216 acute angle, 191 acute triangle, 191 addition algorithm for (finite) decimals, 34 addition of fractions, 32–36 addition of rational numbers, 92–96 using vectors, 102–104 addition of vectors (on number line), 101 additivity property of length, 98 adjacent angles (with respect to an angle), 186, 186–188 sides, 171 vertices, 171 al-Khwarizmi, 301 algorithm, 139 alternate interior angles, 276, 277, 281 analytic geometry, 336 angle, 182 acute, 191 convention about, 182 full, 183 interior (of a polygon), 197 obtuse, 191 right, 191 straight, 183 zero, 183 angle bisector, 192 angle-angle criterion (= AA criterion), 288 391

392

for geometry, 165–176, 184–188, 237, 250 on addition of rational numbers, 92 on multiplication of rational numbers, 105 auxiliary lines, 264 average rate, 81–82 lawn-mowing, 82 average speed (over an interval), 79 pitfall of the terminology, 79 relation with constant speed, 79 axiomatic system, xxxiv, 159–161 Babylonians, 290, 384 basic etiquette in the use of symbols, 299 basic isometries, 161–163, 199, 200, 216, 217, 229, 231, 237 assumption about, 237 below (a horizontal line), 334 between (a point between two points), 167 between (two points on a number line), 13 betweenness, 166, 167 bigger than among numbers, 13 bijection, 205, 333 bijective, 205 bilateral symmetry, 230 binomial coefficient, 42 bipolar mathematics education, xxxvii bisect, 192 bisect each other (two segments), 228, 260 bisector angle, 192 perpendicular, 192 boundary, 194 of disk, 198 boundary point, 194, 194 bounded, 194 cancellation law, 20 generalized, 118 cancellation rule for fractions, 47

INDEX

for rational quotients, 118 Cataldi, Pietro, 308 CCSSM, xvi–xxvii, 5–362 center of a rotation, 202 of a circle, 186 of a dilation, 268 circle, 186 center of, 186 exterior of, 186 inside, 186 of radius r, 186 unit, 186 clockwise rotation, 202 pedagogical issues with its definition, 204 clockwise rotation around O, 215 clockwise rotation around any point, 215 closed bounded interval, 5, 169 closed disk, 186 confused with circle, 193 closed half-plane, 176 closed interval, 126 closed set, 195 inside a circle, 195 closed under composition, 240 coefficient, 303, 312 coherence, xiii, 318 importance of, xxxii collecting like terms, 311 collinear (points), 169 Collins, David, 68 Common Core State Standards for Mathematics (= CCSSM), xvi common denominator, 23, 33 common divisor (of two integers), 138 commutative (in composition of transformations), 209, 371 commutative law, 2 for addition, 87 for addition of fractions, 33 for addition of rational numbers, 91–94 for multiplication, 87 for multiplication of fractions, 46

INDEX

for multiplication of rational numbers, 105 for multiplication of real numbers, 133 for number expressions, 302, 304, 312, 323 makes no sense for variables, 325 commutative ring, 133 comparing fractions, 38, 52 comparing rational numbers, 121–125 complement (of a geometric figure), 181 complete expanded form (of decimal), 35 complex fraction, 68 denominator of, 68 numerator of, 68 complex fractions, 72–74 basic formulas, 70–71 composite, 148 composite transformation, 208 composition of transformations, 208 concatenation, 12 congruence, 17, 240 closed under composition, 240 equivalence relation, 241 reflexive relation, 241 students’ confusion in TSM, 158–159 symmetric relation, 240 transitive relation, 241 congruent to, 240 symbol for, 240, 245 connected (region), 195 constant, 301 term, 312 constant rate, 76, 81–82 house-painting, 82, 86 lawn-mowing, 81 water flow, 82, 84 constant speed, 77, 78 importance of definition, 81 constant transformation, 200 converse, 22 conversion of fractions to decimals by long division, 62–65

393

convex (geometric figure), 172, 181–182 convex polygon, 197 coordinate axis, 331 coordinate system, 331 coordinate axes of, 331 four quadrants of, 335 origin of, 331 scaled, 361 setting up, 332 coordinates (of a point), 332 coordinates in the plane, 331 introduction of, 331 x-axis, 331 y-axis, 331 copies (of a fraction), 12 corresponding angles, 276, 277, 281 corresponding sides proportional, 287 counterclockwise rotation, 201 pedagogical issues with its definition, 204 properties of, 201 counterclockwise rotation around O, 212–215 counterclockwise rotation around any point, 215 cross-multiplication algorithm, 22, 134 mistaken for rote skill, 23 cross-multiplication inequality, 32, 36, 118 crossbar axiom, 250, 251 cubic polynomial, 313 data points, 272, 274 decagon, 171 decimal digits, 14 decimal fraction, 14 decimal point, 14 decimals, 4, 14 addition algorithm for, 34 finite, 14 infinite, 14 multiplication algorithm for, 51 subtraction algorithm for, 40 terminating, 14 definitions, xiii absence of, in TSM, xxviii

394

importance of, xxviii–xxix, 75 the role of, xxix–xxx degree (of a polynomial), 312, 316 degree (of a rotation), 202 negative, 203 degree (of an angle), 190, 191 preserve, 203, 231, 236, 237 denominator of a complex fraction, 68 of a fraction, 10 of a rational quotient, 118 derivative, 310 Desargues, Girard, 308 Descartes, René, 301, 308, 336 determinant, 377 and solutions of a linear system, 377 nonvanishing, 377 vanishing, 377 diagonal, 171 difference (of two fractions), 41 difference quotient (of two points), 347 different sides of a line in the plane, 176 of a point on a line, 174 dilation, 268 basic properties of, 269–271 center of, 268 effect on lengths and degrees, 275 scale factor of, 268 direction (of a vector on number line), 100 disjoint (sets), 173 disjoint union of sets in a line, 173 of sets in the plane, 175 disk, 186 closed, 186 open, 186 unit, 186 distance, 6–7, 125 between parallel lines, 228 between two points, 6, 184 function, 185 of a point to a line, 228 preserve, 200

INDEX

relation with length of a segment, 184 unit, 7 distance formula, 336 distinct lines, 165 distinct rays, 175 distributive law, 2, 88 for fractions, 46, 59 for number expressions, 302, 304, 312, 323 for rational numbers, 105 for real numbers, 133 makes no sense for variables, 325 divide (one integer by another), 138 division as multiplication, 47, 57, 115 division interpretation of a fraction, 28, 30 division of decimals, 60 division of fractions, 57 by a whole number, 46 relation to division of whole numbers, 55, 58 division of rational numbers, 112–115 division of whole numbers, 29, 58 division-with-remainder, 15, 137, 139 dividend of, 140 divisor of, 140 quotient of, 140 remainder of, 140 divisor (of an integer), 138 divisor of zero, 133 double inequality, 126 associated, 126 edge (of polygon), 170, 171 elimination, method of, 380 empty set, 352 endpoint, 6, 126, 169 left, 6 of vector, 100, 219 right, 6 equal angles, 244 equal expressions, 303 equal fractions, 12 equal ordered pairs of numbers, 332 equal parts (of a segment), 9 equal segments, 244 equal sets, 141 equal transformations, 208, 208

INDEX

equality of two sets, 141 equation, 322 in one variable, 322 in two variables, 323 linear, 323 quadratic, 322 solution of, 322 solve, 323 equation in two variables, 323 solve, 323 equation of a line, 354 slope-intercept form, 356 two-point form, 357 equidistant from 0, 90 points, 6 equidistant parallel lines, 258 equilateral triangle, 193 equivalence of two theorems, 257 equivalence relation, 241 equivalent equations, 328 equivalent expressions, 303 equivalent fractions, 12, 20–21, 25, 35, 47, 61 equivalent to, 22 Euclid, xlii, 137, 155, 160, 161, 244 proof of infinity of primes, 155 Euclidean algorithm, 140, 143–144 Euler, Leonhard, 308 existence, 56, 112, 114, 139, 150, 155, 166, 197, 216, 218, 222, 224, 225, 236, 284, 322 expression, 302 coefficients in, 303 notational conventions for, 302 number, 302 order of operations in, 303 expressions equal, 303 equivalent, 303 extension (of a concept), 29, 58, 92, 96 exterior of a circle, 186, 194 factor (of an integer), 138 factorial, 42 factorization, 314 potential harmful effects of overemphasis, 315

395

factorization (of an integer), 138 FASM, xix, 3, 76, 86, 107, 121, 127, 133, 258, 304, 316, 347–348, 350 Fermat, Pierre, 308, 336 FFFP, 23, 23–24, 33–34, 36, 39 field, 133 finite decimals, 14 division of, 60 five fundamental principles of mathematics, xiii, xxxiii FOIL, 314 folding transformation, 207 fraction, 10 complex, 68 convert to a decimal by long division, 62–65 convert to a finite decimal, 60 copies of, 12 denominator of, 10 division interpretation of, 28, 30 in lowest terms, 30, 138 numerator of, 10 proper, 36 reduced, 138 reduced form of, 138 TSM concept of, 4–5 unit, 10 fraction addition, 33 formula for, 33 fraction division, 57 formula for, 57 fraction multiplication, 45 formula for, 46 fraction subtraction, 38 formula for, 39 fractional multiple, 56 fractions, 10 addition of, 33 cancellation rule for, 47 common denominator of, 23, 33 division of, 57 equal, 12 equal to finite decimals, 152 equivalent, 12, 20–21 high school teachers’ need for, xxiv multiplication of, 26, 45 order among, 12 reducing, 20

396

research on the teaching of, xxi–xxii subtraction of, 38 sum of, 33 Francis, Larry, 246, 291 FTS, 51, 256, 263 FTS*, 257 full angle, 183 fundamental assumption of school mathematics (= FASM), 3 fundamental assumption of school mathematics (= F ASM ), 133 fundamental fact of fraction-pairs (= FFFP), 23 fundamental principles of mathematics, xiii, xxxix, 2, 75, 160, 178 fundamental theorem of arithmetic, 149 fundamental theorem of similarity (= FTS), 256 Gardiner, Tony, 198 GCD, 138 generality and abstraction, 299, 305–310 geometric figure, 47 bounded, 194 data points on, 272, 274 paved by other geometric figures, 47 rectilinear, 269 unbounded, 194 geometric mean, 132 geometric series, 309 finite, 309 geometry curriculum issues with, 157–164 Gödel’s incompleteness theorem, 169 Goldbach conjecture, 147 graph of a linear equation in two variables, 353–354 graph of an equation in two variables, 352 greater than among numbers, 13, 91 greatest common divisor (= GCD), 40, 138 group, 211, 235, 240, 286

INDEX

Hald, Ole, 67, 86, 156, 198, 276 half-line, 173, 253, 359 half-plane, 176, 253 closed, 176 left, 334 lower, 334 right, 334 upper, 334 half-planes opposite, 176 harmonic mean, 72 heptagon, 171 Heron’s formula, 296 hexagon, 170, 171 HL (criterion for triangle congruence), 293 horizontal line, 332 above, 334 below, 334 slope of, 347 house-painting, 82, 86 hypotenuse (of a right triangle), 290 hypotenuse-leg (= HL), 293 identity, 303–304 identity transformation, 200 if and only if, 22 image (of a transformation), 205 inequalities, 121 about absolute value, 127–130 about rational numbers, 121–125 inequality, 12 arithmetic and geometric means, 127, 132 double, 126 triangle, 129 infinite decimals, 14 infinity of primes, 155 injection, 205 injective, 205 inscribed (in a circle), 193 inside of a circle, 195 of a polygon, 196 inside a circle, 186 integers, 91 integral domain, 133 integral linear combination, 140 interior angle of a polygon, 197

INDEX

intermediate value theorem, 195 intersection (of sets), 165 interval closed, 126 closed bounded, 5, 169 length of, 126 open, 126 inverse, 56 multiplicative, 56, 113 inverse (of a transformation), 211, 211 inverse transformation, 211 of a congruence, 240 of a reflection, 231 of a rotation, 211 of a similarity, 284 of a translation, 234 invert and multiply rule, 1, 57, 59, 68, 71, 119 generalized form, 119 irrational number, 153 isolating the variable, 329 isometry, 200 relation with congruence, 237, 250 isosceles triangle, 193 controversy in TSM about its definition, 193 Jordan curve theorem, 196 key lemma, 144 Koswatta, Sunil, 101 (L1) (geometric assumption), 165 (L2) (geometric assumption), 165 (L3) (geometric assumption), 167 (L4) (geometric assumption), 176 (L5) (geometric assumption), 184 (L6) (geometric assumption), 188 (L7) (geometric assumption), 237 (L8) (geometric assumption), 250 lawn-mowing, 81 LCM, 40, 41, 156 least common denominator, xxii, 1, 32, 34, 40, 41, 134 least common multiple (=LCM), 34, 156 left (of a vertical line), 334 left endpoint, 6

397

left half-plane (of a vertical line), 334 left-pointing (vector on number line), 100 leg (of a right triangle), 290 length additivity property of, 98 of a segment, 7, 185 of a vector, 100, 220 of an interval, 126 preserve, 201 less than among numbers, 12 line defined by a linear equation in two variables, 354 joining two points, 166 segment (joining two points), 169, 358 slants / or \, 363 line separation, 173 line symmetry, 230 linear equation, 300 in one variable, 323 in two variables, 352 linear equations simultaneous, 373 system of, 373 linear polynomial, 313 linear system, 373 determinant of, 377 in two variables, 373 of two equations in two unknowns, 373 relation with geometry, 374–375 solution by elimination, 380 solution by substitution, 380 solution set of, 373 relation with determinant, 377 lines distinct, 165 intersecting, 165 parallel, 165 perpendicular, 191 locating a fraction on a number line, 15, 52 lower half-plane, 334 lowest terms (of a fraction), 138 lowest terms (of fraction), 30

398

m-th multiple (of a unit fraction), 10 major arc, 190 map (transformation), 200, 205 mathematical engineering, xii mathematical integrity, xiii, xv, xix, xx, xxiii, xxviii, 159 mathematics educator, xi, xxvii Mersenne prime, 309 Mersenne, Marin, 308 midpoint, 84, 192 minor arc, 190 minus sign, 97 mirror reflection, 90 relation with addition, 93–96 relation with multiplication, 105–106 mixed number, 36 mixed numbers, 15 subtraction of, 39–40 monomial, 312 move, 205 move (transformation), 200 mph, 77 multiple, 311 integral, 138 rational, 114 whole number, 10, 56 multiplication algorithm for decimals, 51 multiplication of fractions, 26, 45 motivation for the definition, 43–45 multiplication of rational numbers, 105–110 multiplicative inverse of a fraction, 56 of a rational number, 113, 114, 133, 134 n factorial, 42 n-gon, 171 n-sided polygon, 171 National Mathematics Advisory Panel, xxiv, 1 negative degree (of a rotation), 203 negative numbers, 91 negative sign, 91 negative times negative is positive, 107–110, 135

INDEX

nonagon, 171 nonconvex, 190 nonconvex angle, 182, 183, 186 nonempty (set), 173 nonvanishing, 377 notational conventions for expressions, 302 for polynomials, 312 number, 6 irrational, 153 negative, 91 positive, 91 real, 6 whole, 6 number expression, 302 number line, 5, 6 mirror reflection on, 90 numerator of a complex fraction, 68 of a fraction, 10 of a rational quotient, 118 obtuse angle, 191 obtuse triangle, 191 octagon, 171 of (as in fraction of a fraction), 20, 24, 24–28 one-to-one (= injective), 205 one-to-one correspondence, 205 (= bijective), 205 onto, 205 onto (= surjective), 205 open disk, 186 open interval, 126 opposite angles at a point, 276 opposite half-planes, 176 opposite rays, 174 opposite sides of a line in the plane, 176 of a point on a line, 174 of a quadrilateral, 193 opposite signs, 125 opposite vertices (of a quadrilateral), 227 order among fractions, 12 among numbers, 12, 91 order of operations, 313 ordered pair, 134, 332, 333

INDEX

origin of a coordinate system, 331 parallel lines, 165 distance between, 228 equidistant, 258 parallel postulate, 165 parallel segments as abuse of notation, 256 parallelogram, 193 characterizations of, 260, 261 equality of angles at opposite vertices, 227 equality of opposite sides, 226 opposite vertices of, 227 partitive interpretation of whole number division, 28 parts of a whole, 4 Pascal, Blaise, 308 pave, 47 PEMDAS, 313, 314 pentagon, 171 percent, 73 of a quantity, 73 perfect square, 153 perimeter of rectangle, 53 perpendicular bisector, 192 perpendicular lines, 191 plane separation assumption, 176 pointing in the same direction, 231 polygon, 171 confused with polygonal region, 197 convex, 197 diagonal of, 171 edge of, 171 inside of, 196 region enclosed by, 196 regular, 193, 197 side of, 171 polygonal region, 196 confused with polygon, 195 polygonal segment, 195 polynomials, 312 coefficients in, 312 cubic, 313 expanded forms of whole numbers as, 313 factorization of, 314 in several numbers, 317

399

linear, 313 notational conventions for, 312 order of operations, 313 quadratic, 313 positive numbers, 91 precision, xiii importance of, xxxi preserve degree (basic isometries), 203, 231, 236, 237 preserve distance (transformations), 200 preserve length (transformations), 201 prime, 148, 151 prime decomposition (of a whole number), 150, 151, 154 prime number, 148 primes infinity of, 155 problem solving, xxxiv, xxxvi–xxxvii, xxxviii, xxxviii product formula, 43, 46, 47–51, 53, 69, 71, 134 product of fractions, 45 progression from the simple to the complex, 51, 249 proper fraction, 36 purposefulness, xiii importance of, xiii, xxxii–xxxiii, 140 Pythagoras, 290 Pythagorean theorem, 290 converse of, 293 dependence on the parallel postulate, 293 Pythagorean triple, 384 quadrants (of a coordinate system), 335 quadratic equation in one variable, 322 quadratic polynomial, 313 quadrilateral, 171 quotient (of a division), 57, 115 quotient (of a division-with-remainder), 140 quotient field (of an integral domain), 133

400

radian, 191 radius, 186 of closed disk, 186 of open disk, 186 rate, 76, 81 ratio, 75, 75 use in everyday context, 75 rational expressions, 316 in several numbers, 317 rational multiple, 114 rational number addition, 92, 102 formulas for, 96, 97 rational number division, 115 formula for, 119 rational number multiplication, 105 formulas for, 107 rational number subtraction, 96 rational numbers, 90 as quotients of integers, 116 field of, 133 rationale for, 89 rational quotient, 117 ray, 174 from one point to another, 174 issuing from a point, 174 rays distinct, 175 opposite, 174 real number, 6, 76, 153 reasoning, xiii reciprocal (of a fraction), 57 rectangle, 193, 251 area of, 48 existence of, 225 perimeter of, 53 rectilinear figure, 269 reduced form (of a fraction), 138 reduced fraction, 138 reducing fractions, 20 reflection, 229 reflexive property, 121 reflexive relation, 241 region, 197 connected, 195 enclosed by a polygon, 196 polygonal, 196 triangular, 197, 198 regular polygon, 193, 197

INDEX

relation equivalence, 241 reflexive, 241 symmetric, 240 transitive, 241 relatively prime (integers), 138, 152 removing parentheses, 97, 111 research on the teaching of fractions, xxi–xxii rhombus, 193 right (of a vertical line), 334 right angle, 191 right half-plane (of a vertical line), 334 right triangle, 191 hypotenuse, 290 leg, 290 right-pointing (vector on number line), 100 rise-over-run, 298, 347 rotation, 202 center of, 202 counterclockwise, 201 of θ degrees around a point, 202 same direction, 231 same length, 6 same side of a line in the plane, 176 of a point on a line, 174 same sign (for two numbers), 122 SAS, 245 SAS for similarity, 287 satisfies an equation, 322 scale factor in FTS, 256 of a dilation, 268 of a similarity, 284 scaled coordinate system, 361 Schoenfeld, Alan, xv, xviii school geometry curriculum issues with, 157–164 school mathematics, xii segment, 5, 169 divided into n equal parts, 9 endpoints of, 169 length of, 7, 185 midpoint of, 84 short, 8

INDEX

unit, 6 sequence of n-ths, 9, 10 sequence of fifths, 9 sequence of thirds, 8, 9 setting up a coordinate system, 332 short segment, 8, 26 Shulman, Lee, xiv, xxiv side of a polygon, 170, 171 of an angle, 182 sign (of a number), 339 opposite, 125 same, 122 similar to, 284 similarity, 284 equivalence relation, 285 reflexive relation, 285 scale factor of, 284 students’ confusion in TSM, 158–159 symmetric relation, 285 transitive relation, 285 slant, 363 slantangle, 225 slope, 297, 346 formula for, 347, 348 local slope at O, 339 local slope at a point, 342 what it is for, 338–342 slope-intercept form (of the equation of a line), 356 slopes of parallel lines, 363 of perpendicular lines, 366 smaller than among numbers, 12, 37, 91 solution (of an equation), 322 solutions of an equation in two variables, 351 solving an equation, 300, 323 meaning of, 324–329 pedagogical comments on, 327 speed, 76–77, 78 square, 193 square root, 148 SSS, 51 standard representation (of a fraction), 8

401

starting point (of vector), 100, 219 straight angle, 183 straight line, 164 substitution, method of, 380 subtraction of rational numbers, 91 subtraction algorithm for decimals, 40 subtraction as addition, 39, 96 subtraction of fractions, 36, 38–41 subtraction of rational numbers, 96–98, 99 sum of fractions, 33 formula for, 33 sum vector (of two vectors on number line), 101 surjection, 205 surjective, 205 symbols basic etiquette in the use of, 299 need for, 301 need to quantify, 299 symmetric relation, 240 symmetric with respect to a line, 219, 230 Taylor polynomial, 312 terminating decimals, 14 Textbook School Mathematics (= TSM), xiv theorem on equivalent fractions, 20, 21–23, 24, 26, 37, 139, 153 transformation, 200 bijective, 205 composite, 208 constant, 200 folding, 207 identity, 200 image of, 205 image under, 205 injective, 205 inverse, 211 issues with using coordinates, 207 maps a point to a point, 200 moves a point to a point, 200 rationale for, 199–200 surjective, 205 transformations composition of, 208

402

equal, 208 transitive property, 121 transitive relation, 241 translation, 6 coordinate description, 368 translation (along a vector), 234 transversal (of lines), 224 trapezoid, 193 triangle, 171 acute, 191 equilateral, 193 isosceles, 193 obtuse, 191 right, 191 sum of angles, 279 triangle congruence, 245 special conventions of, 245 triangle inequality, 129 triangle similarity, 287 special conventions of, 287 triangular region, 197, 198 trichotomy law, 121 trinomial, 313 TSM, xiv, xxvii–xxxv, xxxvii, 1–5, 7, 13, 14, 29, 30, 32, 35–37, 40, 43, 48, 53–56, 62, 64, 67–70, 72–76, 81–82, 114–115, 119–120, 130, 139, 147, 154–155, 157–163, 166, 200, 244, 255–283, 297–298, 302, 313–314, 318–320, 322, 324–325, 327, 329, 337–338, 347, 351–352, 361–363, 374, 380 unbounded, 194 union (of sets), 9 uniqueness, 55–57, 71, 112, 113, 115, 139, 144, 147–151, 165, 184, 188, 192, 216, 232, 234, 236, 326, 331, 377 unit, 6 circle, 186 disk, 186 unit distance, 7 unit fraction, 10, 42 unit length, 7 unit segment, 6 unit square, 47

INDEX

unknown, 301, 322 upper half-plane, 334 vanishing, 377 variable, 297–298, 301, 323 defect of teaching in TSM, 302, 318–320 isolating the, 329 not a mathematical concept, 319 vector in the plane, 219 length of, 220 on a number line, 100 vertex, 170, 171 of a polygon, 171 of a ray, 174 of an angle, 182 vertical line, 332 left of, 334 right of, 334 Viète, François, 301 weak inequality, 13 well-defined, 45, 57, 138, 168–169, 185, 202, 215, 218, 219, 229, 253, 333 whole (as in parts of a whole), 4, 7 whole number multiple, 56 whole number multiple (of a unit fraction), 10 whole numbers, 6 division of, 29 x-axis, 331 negative, 332 positive, 332 x-coordinate of a point, 332 x-intercept, 357 y-axis, 331 negative, 332 positive, 332 y-coordinate of a point, 332 y-intercept, 356 zero zero zero zero

angle, 183 product property, 114 product rule, 114 vector, 100

This is the first of three volumes that, together, give an exposition of the mathematics of grades 9–12 that is simultaneously mathematically correct and grade-level appropriate. The volumes are consistent with CCSSM (Common Core State Standards for Mathematics) and aim at presenting the mathematics of K–12 as a totally transparent subject. The present volume begins with fractions, then rational numbers, then introductory geometry that can make sense of the slope of a line, then an explanation of the correct use of symbols that makes sense of “variables”, and finally a systematic treatment of linear equations that explains why the graph of a linear equation in two variables is a straight line and why the usual solution method for simultaneous linear equations “by substitutions” is correct. This book should be useful for current and future teachers of K–12 mathematics, as well as for some high school students and for education professionals.

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MBK/131