Mathematical Encounters and Pedagogical Detours: Stories of Disturbance and Learning Opportunities in Teacher Education 303058433X, 9783030584337

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Boris Koichu Rina Zazkis

Mathematical Encounters and Pedagogical Detours Stories of Disturbance and Learning Opportunities in Teacher Education

Mathematical Encounters and Pedagogical Detours

Boris Koichu • Rina Zazkis

Mathematical Encounters and Pedagogical Detours Stories of Disturbance and Learning Opportunities in Teacher Education

Boris Koichu Department of Science Teaching Weizmann Institute of Science Rehovot, Israel

Rina Zazkis Faculty of Education Simon Fraser University Vancouver, BC, Canada

ISBN 978-3-030-58433-7    ISBN 978-3-030-58434-4 (eBook) https://doi.org/10.1007/978-3-030-58434-4 © Springer Nature Switzerland AG 2021 This work is subject to copyright. All rights are reserved by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors, and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This Springer imprint is published by the registered company Springer Nature Switzerland AG The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland

Dedicated to Olga Koichu and Michael Zazkis, our respective soulmates and spouses

Preface

As mathematics educators and mathematics education researchers, we live in quite an idiosyncratic mathematical world. It is rather remote from the mathematics that occupies contemporary research mathematicians, though we are eager to catch up as much as we can with mathematical practices and ideas, if not with the content. Our mathematical world borders with the advanced-level mathematics that we have studied decades ago when completing our formal degrees, and it occasionally touches mathematics employed in mathematics competitions. Roughly speaking, our mathematical world is located somewhere between elementary school mathematics and mathematics encompassed by undergraduate-level mathematics courses. This world is essentially shaped by mathematical concepts and topics included in school curricula, by tasks that appear in textbooks and problem books, by mathematics present in books on the history and philosophy of mathematics, and by non-conventional school-level questions that circulate within the mathematics education communities we belong to. Our mathematical world is confined but not small. It is cozy, fairly inclusive, and to us it is an endless source of mathematical encounters, that is, stimulating and often surprising engagements with mathematical concepts and problems situated within our world or in proximity to its boundaries. This book is about some of these encounters. It is also about our effort to make our mathematical encounters accessible and interesting to mathematics students and teachers we work with. Our students, mainly prospective and in-service mathematics secondary-school teachers, live  – either happily or not  – in their own mathematical worlds. Bridging these worlds with ours requires some pedagogical sophistication. Therefore, the book includes pedagogical detours that intersperse discussions of mathematical encounters. These detours comprise of accounts of our learning about teaching, learning via teaching, and learning for teaching mathematics. This book started as a long conversation between the authors (hereafter, B. and R.), including sharing the tasks that each of us has designed in his/her work in mathematics teacher education. Reflecting on how the task design was initiated, a common theme has emerged, that of disturbance and of how we learn from disturbance. This brings us to the words of John Mason: vii

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Preface What triggers a new phase of personal development? Most frequently there is some form of disturbance which starts things off. It may be a surprise remark in a lesson, a particularly poor showing on a test, something said by a colleague, something asserted in a journal or book, or a moment of insight (Mason 2002, p.10).

We resonate with this quote as problem solvers, as teacher educators, and as mathematics education researchers. As such, we set to elaborate upon the notion of disturbance as a driving force in the design of original tasks in teacher education and as a common feature of various stories of disturbance and learning opportunities included in this book. The forms and sources of disturbance vary and so vary our responses. We revisit a few in Chap. 1 and make explicit a theoretical framework that explains our choices when writing the book. In Chap. 2, we present and illustrate a particular format of narrative inquiry, which we use in most of the chapters. An essential characteristic of the format is that it includes a structured dialogue between the narrators, in this case, between B. and R. Of note is that B. and R. cannot be fully equated with the authors, though all (fictional) dialogues between R. and B., as well as the dialogues among student-characters, are based on actual discussions between the authors as well as on the data collected from and with our students. Chapters 3, 4, 5, 6, and 7 consist of a collection of mathematical encounters and pedagogical detours in various parts of our mathematical world, from arithmetic to undergraduate calculus and 3-D geometry. The chapters are self-contained, that is, the readers are invited to engage with the book by means of any reading tactics that suit them. Have a nice encounter. Rehovot, Israel  Boris Koichu Vancouver, BC, Canada  Rina Zazkis

Contents

1 Disturbance as a Driving Force��������������������������������������������������������������    1 1.1 Form and Sources of Disturbance����������������������������������������������������    1 1.2 A Framework of Professional Responding to Disturbing Encounters����������������������������������������������������������������������������������������    2 1.2.1 Responding to Disturbance as a Problem Solver������������������    3 1.2.2 Responding to Disturbance as a Teacher in Search for a Good Explanation��������������������������������������������������������    5 1.2.3 Responding to Disturbance as a Teacher in Search of Ways of Teaching Difficult Problems While Preserving Student Autonomous Learning ��������������������������    8 1.2.4 Responding to Disturbance as a Teacher-Educator Teaching Mathematics����������������������������������������������������������   11 1.2.5 Responding to Disturbance as a Teacher-Educator Teaching How to Teach��������������������������������������������������������   11 1.2.6 Coda: Response to Disturbance as a TeacherEducator-­Researcher ������������������������������������������������������������   14 References��������������������������������������������������������������������������������������������������   15 2 A Fictional Dialogue on Infinitude of Primes����������������������������������������   17 2.1 Introduction��������������������������������������������������������������������������������������   18 2.2 Theoretical Underpinnings����������������������������������������������������������������   19 2.2.1 Duoethnography��������������������������������������������������������������������   19 2.2.2 Virtual Monologue����������������������������������������������������������������   20 2.3 Introducing and Exemplifying Virtual Duoethnography������������������   21 2.3.1 What is Virtual Duoethnography?����������������������������������������   21 2.3.2 Lakatos’ Dialogue as an Example of Virtual Duoethnography��������������������������������������������������������������������   21 2.4 Virtual Duoethnography: Situating a Fictional Dialogue ����������������   22 2.4.1 Learning and Teaching by Scripting Instructional Interactions����������������������������������������������������������������������������   22 2.4.2 Background and Task������������������������������������������������������������   23 ix

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2.5 Fictional dialogue on Infinity of Primes ������������������������������������������   25 2.5.1 Same Theorem?��������������������������������������������������������������������   26 2.5.2 On Number and Unit������������������������������������������������������������   26 2.5.3 On “Measured” ��������������������������������������������������������������������   27 2.5.4 Prime or Not Prime ��������������������������������������������������������������   28 2.5.5 Absurd? ��������������������������������������������������������������������������������   30 2.5.6 The essence and the difference ��������������������������������������������   31 2.5.7 The Gap��������������������������������������������������������������������������������   33 2.6 Discussion ����������������������������������������������������������������������������������������   34 References��������������������������������������������������������������������������������������������������   36 3 Encounters with Euclidean Propositions ����������������������������������������������   39 3.1 Introduction��������������������������������������������������������������������������������������   39 3.2 How It Started ����������������������������������������������������������������������������������   39 3.3 Leaving the Choice to Students��������������������������������������������������������   41 3.4 On Continued Proportions����������������������������������������������������������������   42 3.4.1 Warm-Up Tasks��������������������������������������������������������������������   42 3.4.2 Proposition 4 from Book IX, Its Proof, and Students’ Script��������������������������������������������������������������   43 3.4.3 Solid Numbers and Similar Solid Numbers��������������������������   45 3.4.4 On Mean Proportional Numbers������������������������������������������   46 3.4.5 Interpreting Euclid’s Proof of Proposition IX.4��������������������   48 3.4.6 Geometric Sequences and Geometry������������������������������������   51 3.5 New Proposition VIII.6 ��������������������������������������������������������������������   52 3.6 On the Euclidean Algorithm ������������������������������������������������������������   57 3.6.1 Proposition X.2 and Mike’s Script����������������������������������������   58 3.6.2 Which Proposition Is Euclid’s Algorithm? ��������������������������   61 3.6.3 Detour on Applicability��������������������������������������������������������   65 3.6.4 On Rational and Irrational����������������������������������������������������   66 3.7 Concluding Remarks������������������������������������������������������������������������   70 References��������������������������������������������������������������������������������������������������   71 4 Stories on Problem-Solving Instruction������������������������������������������������   73 4.1 Introduction��������������������������������������������������������������������������������������   73 4.2 Explore – Launch – Re-explore – Relate������������������������������������������   75 4.2.1 Explore����������������������������������������������������������������������������������   75 4.2.2 Launch����������������������������������������������������������������������������������   76 4.2.3 Explore����������������������������������������������������������������������������������   76 4.2.4 Relate������������������������������������������������������������������������������������   80 4.3 Launch – Explore – Re-launch – Re-explore – Relate���������������������   80 4.3.1 Launch����������������������������������������������������������������������������������   80 4.3.2 Explore����������������������������������������������������������������������������������   81 4.3.3 Re-launch������������������������������������������������������������������������������   81 4.3.4 Re-explore����������������������������������������������������������������������������   84 4.3.5 Relate������������������������������������������������������������������������������������   85

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4.4 Launch – Get Stuck – Intervene – Relate ����������������������������������������   86 4.4.1 Background ��������������������������������������������������������������������������   86 4.4.2 Launch����������������������������������������������������������������������������������   87 4.4.3 Get Stuck������������������������������������������������������������������������������   88 4.4.4 Intervene ������������������������������������������������������������������������������   89 4.4.5 Relate������������������������������������������������������������������������������������   91 4.4.6 Follow-Up ����������������������������������������������������������������������������   92 4.5 Launch – Intervene – Re-launch – Explore – Relate������������������������   93 4.5.1 Background ��������������������������������������������������������������������������   93 4.5.2 Launch����������������������������������������������������������������������������������   94 4.5.3 Intervene (Formalize)������������������������������������������������������������   95 4.5.4 Detour: Types and Effects of Teacher Interventions ������������   96 4.5.5 Re-launch������������������������������������������������������������������������������   98 4.5.6 Explore����������������������������������������������������������������������������������   99 4.5.7 Relate������������������������������������������������������������������������������������  100 4.6 Concluding Reflections��������������������������������������������������������������������  102 References��������������������������������������������������������������������������������������������������  103 5 Encounters with Cardano’s Method������������������������������������������������������  105 5.1 Introduction��������������������������������������������������������������������������������������  105 5.2 Background ��������������������������������������������������������������������������������������  106 5.3 Collaboratively Re-telling the Story ������������������������������������������������  107 5.4 The Initial Engagement and Structuring the Method������������������������  109 5.5 Detour: Reverse Heuristic Engineering��������������������������������������������  111 5.6 How Could Such a Clever Solution Be Found?��������������������������������  112 5.7 Can the Law of Transitivity Fail?�����������������������������������������������������  115 5.7.1 Reversible or Not and Does It Matter? ��������������������������������  117 5.8 Reverse Heuristic Engineering and Problem Posing������������������������  122 5.8.1 Detour: The Variation Theory of Learning as a Task-­Design Heuristics��������������������������������������������������  124 5.9 Explorations Based on the Examples Generated by Students����������  125 5.9.1 “Nice” Roots – Difficult Solutions ��������������������������������������  128 5.9.2 “Nice” Roots – “Nice” Solutions������������������������������������������  129 5.9.3 Detour: “Niceness” and Artificiality ������������������������������������  136 5.9.4 Appearance of “Nice Monsters” ������������������������������������������  139 5.10 Concluding Remarks������������������������������������������������������������������������  143 References��������������������������������������������������������������������������������������������������  144 6 Synthesizing Exception-Barring and What-If-Not: If Not, What Yes? ������������������������������������������������������������������������������������  145 6.1 Introduction��������������������������������������������������������������������������������������  145 6.2 Background ��������������������������������������������������������������������������������������  146 6.3 “If Not, What Yes?”: An Instructional Aspect����������������������������������  147 6.3.1 The Account��������������������������������������������������������������������������  147 6.3.2 Illustration in the Context of Calculus����������������������������������  148

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6.3.3 Detour: A Happy End and (Somewhat Awkward) Follow-Up of the Story ��������������������������������������������������������  155 6.4 “If Not, What Yes?” A Design Aspect����������������������������������������������  160 6.4.1 Student-Generated Example in the Context of Calculus����������������������������������������������������������������������������  160 6.4.2 Student-Generated Example in the Context of Number Theory����������������������������������������������������������������  161 References��������������������������������������������������������������������������������������������������  164 7 From “Obviously Wrong” Methods to Surprisingly Correct Answers��������������������������������������������������������������������������������������  165 7.1 Introduction��������������������������������������������������������������������������������������  165 7.2 Tax or Discount? ������������������������������������������������������������������������������  166 7.3 Surprises with Order of Operations: A Story in Three Parts������������  167 7.3.1 Part 1: Mathematical Conventions Task��������������������������������  168 7.3.2 Part 2: Convention: Order of Operations������������������������������  169 7.3.3 Part 3: Order of Operation Tasks������������������������������������������  174 7.3.4 Order of Operations: Arbitrary Choice? ������������������������������  179 7.4 Modeling and Conversion of Square Units��������������������������������������  180 7.4.1 Part 1: Unexpected Solution ������������������������������������������������  180 7.4.2 Part 2: Scripting on Unexpected Solution����������������������������  182 7.4.3 Towards a Conventional Approach ��������������������������������������  183 7.4.4 Conversion Motivated ����������������������������������������������������������  184 7.4.5 Attending to a Variety of Shapes������������������������������������������  185 7.4.6 Computation “Works”����������������������������������������������������������  187 7.4.7 Towards a Geometric Explanation����������������������������������������  189 7.4.8 Modeling as Scaling: Mathematical Notes ��������������������������  190 7.5 Familiar Formula in Unfamiliar Situation����������������������������������������  192 7.5.1 Affine Coordinates: An Introduction������������������������������������  192 7.5.2 Exploring Medians����������������������������������������������������������������  195 7.5.3 Eureka! Mathematical Resolution����������������������������������������  199 7.5.4 Further Applicability������������������������������������������������������������  201 7.6 Concluding Remarks������������������������������������������������������������������������  202 References��������������������������������������������������������������������������������������������������  203 Epilogue������������������������������������������������������������������������������������������������������������  205 Index������������������������������������������������������������������������������������������������������������������  207

Chapter 1

Disturbance as a Driving Force

1.1  Form and Sources of Disturbance Disturbance is a multifaceted word whose use in the language overlaps but does not coincide with the use of such words as interference and perturbation. A Google search for “disturbance vs. perturbation” first led us to the WikiDiff website, where we read that disturbance refers to the act of disturbing, whereas perturbation refers to the state of being perturbed. Our next Google stop was at the ResearchGate forum, to a thread initiated in 2014 by the question “Is there any difference between perturbation and disturbance?” The answers were more or less consistent with the act/state distinction. Synthesizing two of our favorite answers, we infer that disturbance can be treated as an external input to the system affecting its output whereas perturbation concerns a change or uncertainty in the system itself. Mason and Johnston-Wilder (2004) approach the term of disturbance from several perspectives. In line with Wertheimer (1961), they interpret the notion of disturbance “as underlying or producing learner actions” (Mason, 2004, p. 118). Following Bruner (1996), they allude to disturbance as the basis for the construction of any good narrative (ibid, p. 68). Extending Piagetian vocabulary, Mason and Johnston-­ Wilder assert that equilibration, including assimilation and accommodation, is the process of resolving disturbance (ibid, pp. 55, 119). The closest to the concerns of this book discussion of the roles of disturbance in learning and professional development takes place in Mason (2002). After pointing out that the term disturbance is intended to signal that something surprising or irritating happens, Mason interprets this term as rather neutral with respect to response. Depending on situation, response to disturbance may occur through rejection, adjustment, assimilation, or accommodation and also be particular to the current disturbance or address a particular disturbance as a case of a more general disturbance. He also reminds us that disturbance can be small and as such can be encompassed, or it can be large and thus disruptive. Accordingly, the urge to respond to disturbance can be seen as an opportunity or as a pressure. Of course, we, as ­anybody © Springer Nature Switzerland AG 2021 B. Koichu, R. Zazkis, Mathematical Encounters and Pedagogical Detours, https://doi.org/10.1007/978-3-030-58434-4_1

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1  Disturbance as a Driving Force

else, encounter both kinds of disturbance, but this book embraces only those accounts in which disturbance has turned to be an opportunity. Some occurrences of responding to disturbance Mason (2002) characterizes as professional. These include locating and acting upon disturbances in watching colleagues at work, in following up research articles and in collecting and exchanging striking accounts. To the last category we add disturbance stipulated by interaction with students’ mathematical worlds. In Mason’s (2002) words, “the most useful accounts are ones which are striking in some way, which set up at least a frisson of disturbance. If a collection of accounts seems merely to confirm what you ‘have already thought’, then there will be no stimulus to action.” (p. 140). This brief and vivid account of good accounts suggests that professional responding to disturbance is a desirable skill for a teacher or a teacher-educator. However, our focus on disturbance as one of the central themes of this book should not be misinterpreted as a call to turn our professional lives into a series of disturbing events. Nobody likes to be disturbed too often. As a matter of fact, we like imagining our professional lives as a journey in our mathematical world, which embraces large plateaus of robust knowledge and routines, but also a few mountain ridges in which disturbing mathematical encounters happen. Each chapter of this book presents one or more of our hikes.

1.2  A  Framework of Professional Responding to Disturbing Encounters We now present a framework that serves for us as an umbrella for different stories presented in this book (see Fig. 1.1). It can also be seen as part of an emergent theoretical framework for making sense of professional response to disturbing mathematical encounters in mathematics teacher education. The framework is based on Mason’s (2002, 2004) premise that disturbing events of different characteristics can lead to different types of response. In addition, the framework makes explicit the fact that one can respond to the same disturbing event in different ways, depending on the situation, and also on one’s personal agenda for action stipulated by the prevailing (at the time of coping with disturbance) facet of our multifaceted professional beings, in the capacities of problem solvers, teachers, teacher-educators, and mathematics education researchers. These roles are not mutually exclusive and so neither are the responses. In particular, some of our roles embrace the others; this fact is schematically represented in Fig. 1.1 by arrows of different lengths. Needless to say, the framework does not aim at embracing all possible types of disturbance, all facets of our professional beings, and all possible response pathways. It concerns only those pathways which are presented in this book. We now illustrate parts of the framework by a few snapshots and relate each part to the chapters of the book, in which various disturbing encounters are discussed in depth.

1.2  A Framework of Professional Responding to Disturbing Encounters

3

Fig. 1.1 An emergent framework of professional responding to disturbing mathematical encounters

1.2.1  Responding to Disturbance as a Problem Solver One common source of disturbance is inability to solve a problem, especially when a problem appears accessible. A problem appears accessible when, for example, it is presented by a colleague who declares that he or she found the solution.

Rina: The following problem was shared at one of the lunch tables at the CMESG (Canadian Mathematics Education Study Group) over 20 years ago. This problem reappears at many CMESG meetings and always attracts new solvers.

4

1  Disturbance as a Driving Force

The Spinning Table Problem There is a spinning round table with four indistinguishable pockets, similar to those of a billiard table, equally distributed around the table. Four cups are placed in the four pockets, each either face up or face down. You cannot see what is in the pockets. But you can reach at any two of the four pockets simultaneously, determine the position of the cups, and possibly switch it (that is, switch from face up to face down or vice versa). Then the table spins again (so you do not know what pockets you checked previously). You can again reach into two pockets of choice and change the position of the cups, if you decide to. Then the table spins again. When all the cups are in the same position (either face up or face down), a bell rings. What is the minimal number of steps that this can be assured from any initial position of the cups?

Theoretically speaking, the reader can find himself or herself in one of the following positions: (i) you are familiar with the solution, and the Spinning Table Problem1 does not disturb you; (ii) you are not familiar with the problem or have forgotten its solution, but the problem does not disturb you anyway; (iii) you are not familiar with this problem and it disturbs you. B: Well, when a colleague introduced this problem to me, my immediate response was “Let me save it for later”. However, I remember a story in which I felt urged to respond quickly by solving the problem. Many years ago, while being a PhD student, I had a meeting with my advisor. At the meeting, he mentioned in passing the 100 Coins Problem as an interesting one. He did not ask me to solve it, but it was so important for me to impress the advisor by my problem-solving skills. The best scenario would be of course to solve the problem on the spot. Unfortunately, I couldn’t. It took about an hour after the meeting, and until today I vividly remember my ahaexperience with this problem. I shared the solution with the advisor next morning.

The 100 Coins Problem In a completely dark room, 100 identical coins are spread out on a table. Exactly 40 coins lie heads up. You can feel the coins but cannot determine by touching them if any individual coin rests head up or down. Divide 100 coins into two groups so that one group of coins contains the same number of heads­up coins as the other group contains.

1  Indeed, the acute reader may have noticed that the Spinning Table Problem is a variation of the Four Glasses Puzzle, also known as the Blind Bartender’s Problem, which has been published in Martin Gardner’s “Mathematical Games” column in 1979.

1.2  A Framework of Professional Responding to Disturbing Encounters

5

R: Was he impressed? B: I would not say so. After all, the problem has a two-step solution, which is accessible to middle-school students. I myself was impressed when my 13-year-old son solved the problem quicker that I did2. In the Book: Mathematical problems that have been disturbing to us can be found in all chapters of the book. These are encounters in the contexts of arithmetic, algebra, plane geometry, 3-D geometry, number theory, and calculus.

1.2.2  R  esponding to Disturbance as a Teacher in Search for a Good Explanation A disturbance for teachers has to do with seeking good explanations to counterintuitive mathematical phenomena. One of the famous examples of this type can be traced to Papert (1980, p. 146). The String Around the Earth Problem Imagine a string around the circumference of the Earth, which for this purpose we shall consider to be a perfectly smooth sphere, four thousand miles in radius. Someone makes a proposal to place a string on six-foot-high poles. Obviously this implies that the string will have to be longer. A discussion arises about how much longer it would have to be.

A common intuition is that the string should be much longer. The intuition-based estimates may reach hundreds or even thousands of kilometers3. A surprising result is that only about 38 more feet (about 12 meters) of string are needed, and this is invariant of the radius R of the sphere. The question that may disturb a teacher is how to make this counterintuitive result intuitively appealing, given that the algebraic manipulation that solves the problem (2(R + 6)π − 2πR = 12π ≈ 38 feet) is not considered as a convincing explanation by many. Let us briefly recall the ingenious way of resolving this disturbance which has been offered by Papert. Suppose our planet is a cube so that its circumference is a square. We are lifting the string h feet from the square. In this case, the 2  A variation of this problem appears in Burger (2007). The problem has been used in the job interviews for Apple (e.g., see www.glassdoor.co.uk/Interview/You-have-a-100-coins for the forum at which different attempted solutions to the problem are presented). 3  Of note is that this problem also belongs to the repertoire of questions asked at the job interviews in high-tech companies.

6

1  Disturbance as a Driving Force

additional string is needed only in four corners of the square, and each corner would require a quarter of the circle with the radius h. Let us then consider an octagon, 100-gon, and 1000-gon, each time coming closer to the circle-shape. We observe that for 8, 100, or 1000 corners, all that is needed is to add 8, 100, or 1000 pieces of the string, respectively, which lengths together sum up into a circle of the radius h, namely, 2πh. Arcavi (2003) brings up this example to argue that explanation by smart visualization can add meaning and conviction to a symbolic proof. There are many other problems, the formal solutions to which, even when presented in full, remain counterintuitive and thus call for a good explanation by means of a smart visualization (see Ejersbo, Leron & Arcavi, 2014). Consider, for example, the Two Kings Problem. The Two Kings Problem Consider the probability of choosing a king from a deck of 52 cards. Now, once a card is chosen, not looked at, and not returned to the deck, what is the probability that the second chosen card is a king?

A possible analytic-symbolic approach is to look at different cases, where the first pulled card was either a king or not a king. This brings us to the following computation: P ( second king ) = P ( second king after first king ) + P ( second king after not-king )

=

4 3 48 4 4 × 3 + 48 × 4 4 × ( 3 + 48 ) 4 × 51 4 × + × = = = = 52 51 52 51 52 × 51 52 × 51 52 × 51 52



The result is surprising, as the probability of the “second king” is the same as the probability of the first pulled card being a king. The surprise can be intensified, if in a similar way a probability of the “third king” is calculated. P ( third king ) = 4 3 2 4 48 3 48 4 3 48 47 4 = × × + × × + × × + × × 52 51 50 52 51 50 52 51 50 52 51 50 ( 4 × 3) ( 2 + 48 ) + ( 48 × 4 ) ( 3 + 47 ) 4 × ( 3 + 48 ) × 50 4 = = = 52 × 51 × 50 52 × 51 × 50 52

=



While we trust the computation, we are disturbed by it. We feel that the result should have a simpler explanation. What can tame our disturbance is the following isomorphic view of the problem. Suppose (in fact, visualize) that instead of pulling cards from the deck, the cards are spread on the table and you are simply pointing to a card. Then it does not matter what card is pointed to, first, second or third. Sometimes disturbance comes from consideration of what students perceive as a convincing explanation, especially when their explanations are not mathematically

1.2  A Framework of Professional Responding to Disturbing Encounters

7

valid. Zazkis and Mamolo (2009) reported an experience of disturbance when a secondary school teacher, Sean, claimed that he found a way to create a one-to-one correspondence between natural numbers and real numbers. Since this was after Cantor’s diagonal method was presented in class, the story naturally attained the name “Sean vs. Cantor.” Sean’s correspondence considered a sequence of decimal numbers with one digit after the decimal point, followed by numbers with two digits after the decimal point, then 3, etc., and this sequence was mapped to the ordered sequence of natural numbers. What convinced Sean of inappropriateness of his suggestion was not Cantor’s method but a question of what corresponds in his mapping to 1 3 . Furthermore, sometimes disturbance comes from dissatisfaction with normative explanations. For example, it is a “known” fact that the graph of parabola y = (x − 3)2 is located 3 units to the right (the positive direction) of the parabola y = x2. “Why not to the left?” ask many students while associating “minus” with translation to the left and “plus” with translation to the right (Fig. 1.2). The normally available explanations in terms of locating the vertex first, plotting the points or considering relocation of the axes, appear unconvincing to many students and thus disturb many teachers. As a result of this disturbance, alternative

Fig. 1.2  Horizontal translation of parabola

8

1  Disturbance as a Driving Force

explanations have been presented. One of them situates transformations of ­functions in a general conversation on transformations (see Zazkis, Liljedahl & Gadowsky, 2003, for detail of this research). We first consider the horizontal transformation T(a, b) = (a + 3, b), then apply it to all the points of the parabola y = x2, and only then seek the equation of the resulting object. In the Book: Elaborated examples of ways of coping with disturbance caused by the need to invent explanations of mathematical phenomena appear in chapter 2 (in the context of proofs of infinity of primes), chapter 3 (in the context of proportional numbers and the Euclidean algorithm for finding the greatest common divisor), chapter 4 (in the context of algebra), chapter 5 (in the context of the method of Cardano for solving cubic equations), and chapter 7 (in the context of ordering operations in arithmetic).

1.2.3  R  esponding to Disturbance as a Teacher in Search of Ways of Teaching Difficult Problems While Preserving Student Autonomous Learning An example presented below depicts the teacher disturbance stipulated by difficulty to find ways of how to include in teaching a difficult problem while preserving an autonomy of students as problem solvers. Inevitably, this sort of disturbance embraces the previously discussed ones: a teacher first may encounter disturbance related to his or her inability to solve a seemingly accessible problem, then additional disturbance based on the difficulty of explaining a solution to students in a way that would appeal to them, and then a third form of disturbance about how to use the problem in a lesson with another group of students, in a way that enables them to solve it with minimum assistance possible, that is, how to tell without telling. B: Once upon a time a student-teacher came to the university in moderate panic. After his regular geometry lesson on the use of triangle congruence theorems in proving, he gave the 14-year-old pupils an innocent-looking problem as part of homework. A little after that, when reviewing the materials for his next lesson, he realized that he did not know how to solve the problem. Here it is.

The Equal Angle Bisectors Problem Prove that if two angle bisectors of a triangle are equal, then the triangle is isosceles.

1.2  A Framework of Professional Responding to Disturbing Encounters

9

R: It does look innocent. I recall similar-looking problems in several geometry textbooks. However, these problems were about the equal medians or equal heights of a triangle.

Indeed, two similar-looking problems, namely, “Prove that if two medians (or two heights) are equal, then the triangle is isosceles” are within the reach of a good middle-school student who has used the theorems on triangle congruence as tools for proving. The Equal Angle Bisector Problem belongs to another league. In fact, it is known as the Steiner-Lemus Theorem, which appears in the works of many renowned mathematicians. Over 60 proofs were found during the nineteenth century. The theorem attained particular importance when it triggered a general discussion on direct vs. indirect proofs in mathematics (see Lewin, 1974, for the history of the Steiner-Lemus Theorem). American Mathematics Magazine conducted a competition for the best proof of the Steiner-Lemus Theorem in 1963. Many proofs were offered, including the previously unknown ones. Among various proofs, an indirect proof published in Coxeter (1969) is praised as one of the simplest ones. In fact, it is a proof within the reach of a school-level geometry course. Let us consider the following synopsis of the Coxeter proof of the Steiner-Lehmus Theorem. Assume that though CE = BD, ∠B  A), find Y, Z, and B that satisfy A X Y Z = = = X Y Z B 2. Given two natural numbers A and B, find X, Y,and Z, such that A X Y Z = = = X Y Z B 3. (Can serve as a clue to Tasks 1 and 2) Find five natural numbers such that A X Y Z = = = X Y Z B 4. Find four natural numbers such that A X Y = = X Y B

3.4 On Continued Proportions

43

5. (Can serve as a clue for previous tasks) Given natural numbers A and B, find X and Y such that A X Y = = X Y B For what values of A and B are the numbers X and Y natural?

Can you generalize? Notice your problem-solving strategies. We return to these tasks in a few pages.

3.4.2  P  roposition 4 from Book IX, Its Proof, and Students’ Script When attending to student’s work, several particular propositions attracted our attention. One of these was Proposition 4 from Book IX, chosen for the script-­ writing task by Donna and Allan2. The proposition and its proof are in Fig. 3.13. “So much effort for the simple a3  ×  b3  =  (ab)3” was our immediate reaction. However, looking at students’ work and later exploring Euclid’s proof ourselves resulted in unforeseen interest. Let us look at the dialogue created by Donna and Allan. The separation into sections is from the student’s work. We added the section numbers for reference. 1

2

3

4

G: What is a cubic/cube number? E: A cube number is number which is multiplied by the same number and then again by the same number G: OK, you mean a number to the power of 3 G: What do you mean by “is to B as D is to C”? E: I mean that there is a number of times that we need to take A in order to fit it into B. This is the same number that we need to take C and fit it into D G: So you mean that the ratio between A and B is the same as the ratio between C and D. I got it G: So, what is a solid number? E: A solid number is a number, which is produced by multiplying three numbers by one another. Any cubic number is a solid number G: OK, so any 3 multiplication will do it G: What is “two mean proportional numbers”?

 All names are pseudonyms.  All the propositions in this chapter are from Heath’s “canonic” English translation of Euclid’s Elements. (See Euclides, I. (2002). In D. Densmore (Ed.), Euclid’s elements: All thirteen books complete in one volume, the Thomas L. Heath translation. Santa Fe, NM: Green Lion Press.) 2 3

44

3  Encounters with Euclidean Propositions Book IX, Proposition 4

If a cube number by multiplying a cube number make some number, the product will be cube. For let the cube number A by multiplying the cube number B make C. I say that C is cube. A B C D For let A multiplying itself make D; therefore D is a cube. [IX.3] And, since A by multiplying itself has made D, and by multiplying B has made C, therefore, as A is to B, so is D is to C. [VII.17] And, since A, B are cube numbers, A, B are similar solid numbers. Therefore, two mean proportional numbers fall between A, B; [VIII.19] so that two mean proportional numbers fall between D, C also. [VIII.8] And D is cube, therefore C is also cube.

[VIII.23] Q.E.D.

Fig. 3.1  Euclid’s proof of Proposition 4, Book IX (Euclides 2002, p. 213)

4.1 E: This is tricky, it is based on a previous proposition of mine (Book VIII, Proposition 12). But let’s first understand what a mean proportional number is. Let A, B be two numbers. A mean proportional number is to B as A is to the mean proportional number 4.2 In case of two mean proportional numbers (C and D), let’s assume there are two numbers B. Then A is to C as D is to B. And there is a previous proposition which states that if there are two cubic numbers, then there are two mean proportional numbers between them 4.3 G: OK, so you are saying that between any two cubic numbers (A, B) there are two numbers (X, Y), which form the following: A/X = Y/B 4.4 G: Now I got it. If we multiply everything by A will get (A × A)/(X × A) = (Y × A)/(B × A) => D/( X × A) = (Y × A)/C And we got again two mean proportional numbers between D and C, which makes them cubic numbers. ☺

We find in sections #2 and #4  in an explicit claim “I got it,” which signals Goldboom’s understanding of the discussed ideas. We explored the issue of “sense of understanding” explicitly in Koichu and Zazkis (2018). We suggested that scriptwriters’ sense of understanding is denoted by attributing to the fictional characters’ claims such as “I get it” or “OK, I see it now” at the end of the dialogue’s section devoted to a particular topic or implied by transition to the new topic. Actually, in the script above, Goldboom’s understanding is marked by “OK,” which is followed by interpreting Euclidean claim or a concept in contemporary terms.

3.4 On Continued Proportions

45

In particular, in section #1 “cube number” is interpreted as “a number to the power of 3.” In section #2 the relationship between the four numbers is interpreted as a common ratio. Sections #3 and #4 attend to “solid numbers” and to “mean proportional numbers,” ideas that we explore in detail in the next sections.

3.4.3  Solid Numbers and Similar Solid Numbers According to Euclidio in Donna and Allan’s dialogue, “A solid number is a number, which is produced by multiplying three numbers by one another,” which Goldboom interprets as “any 3 multiplication will do it.” This invites a question: is 6, for example, a solid number as 6 = 1 × 2 × 3? Allowing for 1 be one of those numbers in multiplication will imply that any number is a solid number. Was this Euclid’s intention? We recall that for Euclid 1 is not a number, it was a unit4, and a number5 is a multitude composed of units. So we searched for Euclid’s definition of a solid: • When three numbers having multiplied one another make some number, the number so produced is solid, and its sides are the numbers which have multiplied one another [VII. Def.17, p. 157]. It clearly invites a geometric view of a number, where a solid could be considered as a volume of a rectangular prism and the numbers multiplied are the measures of its sides. Another interesting observation, while searching for a definition of Euclidean solid, was found in the previous two definitions: • When two numbers having multiplied one another make some number, the number so produced is called plane, and its sides are the numbers which have multiplied one another [VII. Def.16, p.157]. • A number is said to multiply a number when that which is multiplied is added to itself as many times as there are units in the other, and this some number is produced [VII.Def.15, p.157]. So Euclid’s plane number reminds of an area of a rectangle. We note that the possibility for a unit to be a side of a plane or a solid is excluded by the definitions. Furthermore, Euclid’s definition of multiplication presents a strong additive disposition. We note that contemporary research suggests that repeated addition as the main model for multiplication is limiting, and other models such as scaling should be considered (e.g., Luke 1988; Park and Nunes 2001). Still, the repeated addition model is prevalent in school (e.g., Van de Walle et  al. 2011). Should Euclid be blamed?

 A unit is that by virtue of which each of the things that exist is called one Book VII, Definition 1  Book VII, Definition 2

4 5

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3  Encounters with Euclidean Propositions

A related concept that appears in Euclid’s proof, but not in the students’ script, is that of similar solid numbers. To confirm what the term stands for, we considered several definitions from Book VII: • Numbers are proportional when the first is the same multiple, or the same part, or the same parts, of the second that the third is of the fourth [VII.Def.20, p.157]. • Similar plane and solid numbers are those which have their sides proportional [VII.Def.21, p.157]. That is to say, solid numbers a, b, c and ar, br, cr are similar, as their sides have the same ratio. Considering proportional sides leads to the conclusion that any two cubic numbers are similar solid numbers. This conclusion will be used later, when we return to the warm-up tasks.

3.4.4  On Mean Proportional Numbers We return now to Proposition 4 from Book IX and the students’ work. Section #4 in the students’ work and exploration of “two mean proportional numbers” deserves our extended attention. We agree with the script-writers about the need to explain first the meaning of a (one) mean proportional number. As Euclidio in Donna and Allan’s script explains: Let A, B be two numbers. A mean proportional number is to B as A is to the mean proportional number. [4.1]

This explanation is fine, but we see here a missed opportunity (a) to express the idea in contemporary formalism, that is, if M is a mean proportional number for A M numbers A and B, then = , and (b) to connect the claim to a contemporary M B terminology, namely, that M is a geometric mean of A and B. In fact, we became familiar with the term “mean proportional number” only recently, when exploring “Dialogues concerning two new sciences” by Galileo Galilei (1638/1991)6. Recall that these dialogues feature conversations among three characters, Simplicio, Sagredo, and Salviati. There, at some point in the dialogue, Salvati claims: Let A and B be two similar polygons of which A circumscribes the given circle and B is isoperimetric with it. The area of the circle will then be a mean proportional between the areas of the polygons.

We mention this claim here to explain our previous familiarity with the “mean proportional” term.

 Our thoughts about the potential of Galilei’s dialogues and some other historical dialogues in mathematics teacher education are presented in Zazkis and Koichu (2018). 6

3.4 On Continued Proportions

47

In the script, a brief explanation of the meaning of a mean proportional number is followed in Donna and Allan’s script with the explanation of two mean proportional numbers: In case of two mean proportional numbers (C & D), let’s assume there are two numbers A, B. Then A is to C as D is to B. And there is a previous proposition which states that if there are two cubic numbers, then there are two mean proportional numbers between them. [4.2]

It is not explicitly noted what “previous proposition” that is referenced here. Goldboom interprets the claim but decodes the proportion using X and Y: A/X = Y/B: So you are saying that between any two cubic numbers (A, B) there are two numbers (X, Y), which form the following: A/X = Y/B. [4.3]

This follows by some algebraic manipulation to find two mean proportional numbers between D and C [4.4], which concludes the proof with a smiley. The incomplete definition of “two mean proportional numbers” in students’ work led us to explore the issue in further detail. In fact, two numbers, say X and Y, are mean proportional numbers for A and B if the following proportion holds:



A X Y = = X Y B

So skipping the middle ratio  – as has been done by the students  – creates a proportion



A Y = X B

which can be satisfied with various values of X and Y and does not imply that these are two mean proportional numbers for A and B. For example, if A = 6 and B = 20, the possible pars for X and Y are (8, 15), (12, 10), (10, 12), etc. That is,



6 15 6 10 6 12 = = ; = ; , 8 20 12 20 10 20

but neither pair (X, Y) satisfies the continued proportion



6 X Y = = X Y 20

How can we find two mean proportional numbers between 6 and 20? How can we find two mean proportional numbers between any two natural numbers? We attend to these questions in the next section.

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3  Encounters with Euclidean Propositions

3.4.5  Interpreting Euclid’s Proof of Proposition IX.4 Several mathematical inaccuracies and incomplete or skipped explanations in the students’ script led us to consider the proof of Proposition 4 in Book IX ourselves. In contemporary notation, to demonstrate that the product of two cube numbers is a cube, one can use the following manipulation:

a 3 × b3 = ( a × a × a ) × ( b × b × b ) = ( a × b ) × ( a × b ) × ( a × b ) = ( a × b )

3



Note that the main part of this apparently trivial proof is based on the properties of associativity and commutativity of multiplication. In Euclid’s proof, each claim is justified by referring to a prior proposition. In the table below, we revisit Euclid’s proof, elaborate on these prior propositions, and also interpret the claims of the proofs with the contemporary symbols. Interpretation with contemporary symbols Claim in the proof For let the cube number A Given: A cubic by multiplying the cube B cubic number B make C Prove: I say that C is cubic A × B = C For let A multiplying itself A × A = D = > D cubic make D; therefore D is a cube A × A = D And, since A by A × B = C multiplying itself has Therefore made D, and by multiplying B has made C, A AA D therefore, as A is to B, so = = B AB C is D is to C

References and comments The statement is if A and B are cubic numbers and C is their product, then C is cubic

[IX.3] If a cube number multiplying itself makes some number, the product will be cube [VII.17] If a number by multiplying two numbers makes certain numbers, the numbers so produced will have the same ratio as the numbers multiplied

A D = B C A cubic, And, since A and B are cube numbers, A and B are B cubic => A~B similar solid numbers

This can be derived from the definition of similar solids, but we do not find this conclusion explicitly justified in Elements

3.4 On Continued Proportions

Claim in the proof Therefore two mean proportional numbers fall between A and B; so that two mean proportional numbers fall between D and C also

49

Interpretation with contemporary symbols There exist X and Y such that

References and comments

[VIII.19] Between two similar solid numbers, there fall two mean proportional numbers, [and the A X Y solid number has to the similar solid number = = X Y B the ratio triplicate of that which the and therefore X ' and corresponding side has to the corresponding Y' side] NOTE: the second part of this proposition is D X′ Y ′ not needed for the proof = = X′ Y ′ C [VIII.8] If between two numbers there fall numbers in To elaborate continued proportion with them, then, AA XA YA however many numbers fall between them in = = XA YA BA continued proportion, so many also fall in continued proportion between the numbers D XA YA which have the same ratios with the original = = numbers XA YA C

= > two mean proportional numbers between D and C are XA and YA And D is cube; therefore C D cubic [VIII.23] is also cube => C cubic If four numbers be in continued proportion, and the first be cube, then fourth will also be cube

We explore further the last claim in the proof. Euclid concludes that D is a cube which implies that C is a cube. This is because two mean proportional numbers are found between D and C. In fact, the claim “D is a cube, therefore C is also a cube” is justified by the following propositions: • If four numbers be in continued proportion, and the first be cube, then fourth will also be cube (VIII. 23). • If two mean proportional numbers fall between two numbers, the numbers are similar solid numbers (VIII. 21). Actually, cubic numbers are a special case of similar solid numbers. Note that existence of two mean proportional numbers, which we denoted X and Y, does not tell us what these numbers are. A numerical example can help here. Let us start with two cubic numbers, 8 and 27, and look for mean proportional X and Y such that



8 X Y = = X Y 27

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3  Encounters with Euclidean Propositions

By trial and error, we almost immediately get



8 12 18 = = 12 18 27 However, we can extrapolate, acknowledging that A and B are cubic numbers:



A X Y = = X Y B



aaa aab abb = = aab abb bbb What satisfied the proportion is X = aab and Y = abb. Similarly



D XA YA = = XA YA C



aaaaaa Xaaa Yaaa = = Xaaa Yaaa aaabbb



aaaaaa aabaaa aabaaa = = aabaaa abbaaa aaabbb (Note that the claim that C is a cube does not tell us what the side of this cube is.) So, considering



8 12 18 = = 12 18 27

We are looking at 8, 12, 18, and 27, which are four numbers in continued proportion, mentioned in Book VIII, Proposition 23. The case is more apparent if we list these numbers as

23 , 22 × 3, 2 × 32 , 33

AHA! These numbers form a geometric sequence with ratio 2/3. In order to explain the proposition, starting with the cube a3 and ratio of a geok metric sequence , we get a

a 3 , a 2 k, ak 2 , k 3 In general, if the ratio is r, we get



a 3 , a 3r , a 3r 2 , ( ar )

3



3.4 On Continued Proportions

51

This clearly shows that if in a geometric sequence an element in the n-th place is a cube, then the element in place (n + 3) is also a cube – a claim that can be seen as a contemporary generalization of Euclid’s VIII.23. Furthermore, a generalization of VIII.21 can be stated as follows: If in a geometric sequence, with ratio r, in the n-th place there is a solid number abc, then in place (n + 3) there is an element abcr3 = (ar)(br)(cr), which is a similar solid number to abc We are ready now to find two mean proportional numbers between 6 and 20, a problem posed at the end of the previous section. 20 = 6 × r 3





r3 =

20 6

r=

20 6

3

So X and Y that satisfy 6 X Y = = X Y 20

are

2

6× 3

 20  20 and 6 ×  3  6  6 

Of course these X and Y are not “numbers” in the Euclidean sense, if the unit is chosen to be 1. Had we started with cubic numbers, the mean proportional numbers between them would have been more aesthetically appealing and would have been accepted by Euclid as “numbers.”

3.4.6  Geometric Sequences and Geometry We are thankful to Donna and Allan, whose choice of a proposition and whose script prompted us to explore the proof and the related terminology. For us, the main insight from this work is the connection between Euclidean notion of continued proportion and a modern notion of a geometric sequence. We can return now to the warm-up tasks. Consider, for example, Task 2: Given two natural numbers A and B, find X and Y, such that



A X Y Z = = = X Y B B

52

3  Encounters with Euclidean Propositions

We have tried this task with several colleagues, and most begin by setting several equations with several unknowns. However, interpreting the tasks as a consideration of a geometric sequence A, X, Y, Z, B paves the way for faster and more elegant solutions. We also gain a new appreciation for the notion of a geometric mean. The definition that is most often given in school is that m = ab is the geometric mean of a and b. However, looking at m as a mean proportional number, that is, satisfying a m brings to mind two similar rectangles, a × m and m × b, which are referred = m b in the Elements as proportional planes. Having made the connection to continued proportions, we have extended our view of a geometric sequence, which is usually defined as a sequence of numbers with the same ratio between consecutive elements. Also, we gained a new appreciation for the term “geometric” as related to geometric sequences. We now consider a geometric sequence as a sequence of similar objects, and a ratio of the geometric sequence as a factor of dilation. Similarly, we have developed an extended concept image associated with Euclid’s notion of “similar solid numbers.” Any cubic numbers are similar as any cubes are. Moreover, dimensions of similar prisms are represented by similar solid numbers.

3.5  New Proposition VIII.6 Recall that we asked students to choose a proposition from Elements and write a dialogue in which Euclidio and Goldboom discuss this proposition. In previous sections we attended to IX.4 We now turn to VIII.6, which was chosen by Lily for this assignment. • If there be as many numbers as we please in continued proportion, and the first does not measure the second, neither will any other measure any other [VIII.6, p. 191]. Before turning to the proof of this proposition, we sought a set of numbers in continued proportion. The first continued proportion that comes to mind,



1 2 4 8 = = = …, 2 4 8 16

does not satisfy the requirement of “first does not measure the second.” Finding a set of numbers in continued proportion, where the first does not divide (measure) the second, brings us back to the warm-up Task #1. In fact, we sought

3.5 New Proposition VIII.6

53

numerical examples to make sense of Proposition VII.6 and then turned our experience into a task for the readers and for our students. Below is an example of five numbers that satisfy the requirements for the warm­up Task #1:

16, 24, 36, 54, 81 These numbers do not look too strange if expressed as



2 4 , 23 × 3, 2 2 × 32 , 2 × 33 , 34 They form a geometric sequence with ratio 32 . In fact, n + 1 such numbers can be generated as follows:



a n , a n −1b, a n − 2 b 2 ,…, ab n −1 , b n This is a geometric sequence with ratio ba .

Our experience in generating such a sequence was informed by our previous work with the Proposition IX.4 and connecting the notion mean proportional numbers and continued proportions to the idea of a geometric sequence. We believe that without such experience that task is more demanding. We turn now to the Proposition VIII.6 (Fig. 3.2). Below is the dialogue composed by Lily. We numbered the sections for the ease of reference. As mentioned previously, we consider the students’ choices and their scripts as an opportunity to examine and extend our personal understanding. 1 G: Good day, Euclidio. I read Proposition 6 in Book VIII, and there are a few things that I didn’t get E: Book VIII is one of my favorite ones. How can I help you? G: The proposition states that if there are numbers in continued proportion, and the first does not measure the second, then no number measures another one. Do I understand this correctly? E: And if so, then what? G: First, what is the meaning of a number measures a number? E: Let’s say there are two segments A and B; these are two lengths. Then we say that A measured B if A fits into B a number of times such that there is no part of A outside of B. G: AAAA! WE say that A divides B without a remainder. Now I get the idea of “measures”

54

3  Encounters with Euclidean Propositions Book VIII, Proposition 6

If there are as many numbers as we please in continued proportion, and the first do not measure the second, neither will any other measure any other. Let there be as many numbers as we please, A, B, C, D, and E, in continued proportion, and let A not measure B. I say that neither will any other measure any other. Now it is manifest that A, B, C, D, and E do not measure one another in order, for A does not even measure B. I say, then, that neither will any other measure any other. A F B

G C

H D E

For, if possible, let A measure C. And, however many A, B, C are, let as many numbers F, G, H, the least of those which have the same ratio with A, B, C, be taken. [VII.33] Now, since F, G, H are in the same ratio with A, B, C, and the multitude of the numbers A, B, C is equal to the multitude of the numbers F, G, H, therefore, ex aequale as A is to C, so is F to H. [VII.14] And since A is to B, so is F to G, while A does not measure B, therefore neither does F measure G; [VII.Def.20] therefore, F is not a unit, for the unit measures any number. Now F, H are prime to one another. [VIII.3] And F is to H, so is A to C, therefore, neither does A measure C. Similarly we can prove that neither does any other measure any other.

QED

Fig. 3.2  Euclid’s proof of Proposition 6 Book VII (Euclides 2002, p. 191) 2 G: Will it be OK to take the following sequence of numbers, 32, 16, 8, and 4 and say that 32 does not measure 16? E: This is correct. G: But according to the proposition, no number in the sequence should measure another number, but can we say that 4 measures 16? E: True G: Isn’t this a contradiction to the proposition? E: No, because 16 does not measure 4, and 8 does not measure 4, and 16 does not measure 8, 32 does not measure 4… G: But this is not exactly the claim. The claim is that no number measures another number in a sequence. Here only those numbers that come after the number do not measure it. E: OK, I see what you are saying. There is a need for a small amendment that no number measures numbers that come later in the sequence

3.5 New Proposition VIII.6

55

3 G: So, why does this hold true? E: Assume there is a sequence a, b, …, c and a does not measure b. Let us take the same sequence, but reduced. (That is, we divide each number in the sequence by the greatest common divisor.) So we get a sequence f, g, …, h. The ratio remains after we reduce the numbers, so a : b = f : g. So f does not measure g G: Why is there the same ratio in the original sequence and in the reduced one? E: It is a good question. Let there be two numbers A and B and also number K that measures A and B. We can see A and B by K. So looking at the ratio A to B, there is no need to consider K, because it does not distinguish between A and B. Only the number of times K fits into A and B distinguishes between lengths A and B. This is Proposition 3; I recommend you check it yourself. ☺ G: If I translate to my language, you are saying the following: A = Kg, B = Kf. The ratio A : B equals f : Kg = f : g, reducing by the common factor K. And this was from the beginning f : g; that is actually the new sequence created by reducing E: I do not understand exactly, but I trust that you translate me correctly 4 E: The next step in the argument is that because f does not measure g, then f is not a unit, as unit measures any number. The unit is the basic smallest segment with which all the other segments are measured G: OK, we simply call this number 1 5 E: Additionally, f and h are relatively prime. This is from another proposition by Euclid (Book VIII, Proposition 3). And if so, f does not measure h. But f is to h as a is to c. Therefore a does not measure c G: And why is this true that f and h are relatively prime? E: This can be a topic for another lesson, maybe a lesson on Book VII Proposition 3

In section #1, the characters revisit the notion of “measures,” of the meaning of one number measures another. We noted previously (in line with Artmann 1999) that in Elements there is no separate definition for the concept of “measures” (Zazkis and Koichu 2015). While we translate the Euclidean “A measures B” as “A divides B” or “B is a multiple of A,” the description in Lily’s script (“A fits into B a number of times such that there is no part of A outside of B) corresponds with the understanding of “no remainder.” This echoes the findings of prior research (e.g., Zazkis 2002), where students find the notion of “divides” not sufficiently self-explanatory and often supplement the description of divisibility with an added emphasis “divides without remainder.” In section #2, there is an attempt to provide a counterexample to the proposition. The considered counterexample is a sequence 32, 16, 8, 4, where the first two elements satisfy the requirement that A (A = 32) does not measure B (B = 16), but later 4 measures 8 and 16, etc. This example leads Euclidio to clarify the proposition, claiming that no number measures (divides) another number that appears later in the sequence. This explains the insertion of “later” – then neither does any other measure any [later] other – that appears, for example, in some Internet resources7, but not in Heath’s prevalent translation.

7

 E.g., http://aleph0.clarku.edu/~djoyce/elements/bookVIII/propVIII6.html

56

3  Encounters with Euclidean Propositions

In sections #3 and #4, we note the conflation between the Euclidean terminology and the modern one. For example, nowhere in Euclid’s text, there is a notion of a sequence or of reducing. Section #4 attracts our attention further as in the dialogue composed by Lily what appears is the restatement of the claim that f is not a unit, followed by an explanation of what is meant by a unit. This explanation is interesting as it implies a temporal order in the choice of the unit (first segments, then unit). It also conflates the contemporary usage of measurement and Euclidean notion of “measures” as divides. However, while Euclidio restated the claim that f cannot be a unit, we wondered why this is necessarily so. After all, isn’t it possible that in a “reduced” geometric sequence, the first element is 1? Obviously, such geometric sequence can be created. But it cannot be obtained by reduction of a sequence in which the second element is not a multiple of the first. In section #5, Euclidio claims that f and h are relatively prime; this claim is not explained but referenced to “another proposition.” The claim appeared not self-­ explanatory, so we looked up the related proposition: • If as many numbers as we please in continued proportion be the least of those which have the same ratio with them, the extremes of them are relatively prime [VIII.3, p.187]. What could have added an explanation to Euclid’s proof (as well as to Lily’s script) is the reference to extremes, that is, end numbers in a sequence. As often, a numerical example can help. Let us consider five numbers in a continued proportion, where the first element does not divide the second.

80, 120, 180, 270, 405 Then consider the sequence with the same ratio reduced to lowest terms,



16, 24, 36, 54, 81

It could be more convenient to consider the elements in their prime decomposition:

2 4 × 5, 23 × 3 × 5, 2 2 × 32 × 5, 2 × 33 × 5, 34 × 5



2 4 , 23 × 3, 2 2 × 32 , 2 × 33 , 34

The extremes, here 24 and 34, are indeed relatively prime numbers. Of course, we wonder why this geometric sequence stops at the fifth element. After all, in considering geometric sequences, we are mostly used to infinite sequences. We can definitely continue the sequence; can’t we?



2 4 , 23 × 3, 2 2 × 32 , 2 × 33 , 34 ,

35 36 , ,… 2 22

3.6 On the Euclidean Algorithm

57

The issue is that the new elements in the sequence, while preserving the common ratio, are not “numbers” in the Euclidean definition of a number (assigning 1 to a unit); these are ratios. For us, the consideration of geometric sequences provided an easily accessible solution for the warm-up tasks. Recall the Columbus egg?

3.6  On the Euclidean Algorithm In mathematics, the Euclidean algorithm is known as an efficient method for computing the greatest common divisor (GSD) of two numbers. When introduced to the Euclidean algorithm in school, many students do not appreciate its utility. After all, for “school numbers” there is almost always an easily accessible prime decomposition. Moreover, for today’s students, there is a GCD function on a graphing calculator. So we wondered about the “usefulness” of the algorithm: R: Suppose you need to find the greatest common divisor of 270 000 and 1620 B: I would immediately use the Euclidean Algorithm: 270,000 – 1620 × 166 = 270,000 – 268,920 = 1080 1620 – 1080 = 540 And, 540 divides 1080, 1080 = GCD(270000, 1620) R: Well, but you used a calculator to find 166. I can build factor trees, and so 270,000 = 53 × 33 × 24 and 620 = 51 × 34 × 22 So considering the exponents of common factors, GCD (270000, 1620) = 51 × 33 × 22 = 540 B: But using the Euclidean algorithm is so much faster! R: For now, I do not care much about the speed; I rather care about which method is more explanatory. The Euclidean algorithm looks like a magic trick. Of course we can prove that it works, but when working with teachers I like emphasizing the idea of the greatest common divisor. So we need to look for divisors, then for common divisors, and then for the greatest one among the common ones. This sequence leads itself naturally to considering the prime decomposition. B: And what if prime decomposition is not easily available? R: Your concern sounds as if we were talking about RSA algorithm8 for ciphering and deciphering. We are talking about teacher education. So the preferences might differ …

 See https://en.wikipedia.org/wiki/RSA_(cryptosystem) for a brief discussion of RSA algorithm.

8

58

3  Encounters with Euclidean Propositions

Our conversation about Euclidean algorithm versus other methods continued and resulted in the following Algorithm Comparison Task. We suggest that interested readers consider the following task before reading the rest of this section. Algorithm Comparison Task Consider two methods of finding the greatest common divisor (GCD) of two numbers: prime factorization and the Euclidean algorithm. Compare the methods by pointing to advantages and disadvantages of each.

3.6.1  Proposition X.2 and Mike’s Script Our interest in the Euclidean algorithm sparked when considering a script of Mike related to Proposition 2 from Book X. To remind, Mike, as well as Donna, Alan, and Lily, wrote his script in response to the scripting task on a proposition of his choice. As required in the task, Mike wrote a dialogue between Euclidio and Goldboom, attempting to explain and understand the proof. As before, below we provide the proposition and its proof (Fig. 3.3), followed by the script composed by Mike. We then comment on the script and describe our mathematical engagement that followed. Book X, Proposition  If, when the less of two unequal magnitudes is continually subtracted in turn from the greater, that which is left never measures the one before it, the magnitudes will be incommensurable.

E C

F

A

G

B D

For, there being two unequal magnitudes AB, CD, and AB being the less, when the less is continually subtracted in turn from the greater, let that which is left over never measure the one before it; I say that the magnitudes AB, CD are incommensurable. For, if they are commensurable, some magnitude will measure them. Let a magnitude measure them, if possible , and let it be ; let AB, measuring FD, leave CF less than itself, let CF measuring BG, leave AG less than itself, and let this process be repeated continually, until there is left some magnitude which is less than E. Suppose this done, and let there be left AG less than E. Then, since E measures AB, while AB measures FD, therefore E also measures FD. But it measures the whole CD also; therefore it will also measure the remainder CF. But CF measures BG, therefore E also measures BG. But it measures the whole AB also, therefore it also measures the remainder AG, the greater the less; which is impossible. Therefore no magnitude will measure the magnitudes AB, CD; therefore the magnitudes AB, CD are incommensurable. [X.Def.1] Therefore, etc. QED

Fig. 3.3  Book X, Proposition 2 and its proof (Euclides 2002, p. 238–239)

3.6 On the Euclidean Algorithm

59

We present below the script written by Mike. We numbered the sections for the ease of reference. 1

2.1

2.2

2.3

2.4 3.1

3.2

4.1

4.2 4.3 4.4

G: Clearly I cannot understand the proof without understanding the proposition itself. So can you explain to me the meaning of “incommensurable.” E: I’m always ready to explain Those magnitudes are said to be commensurable which are measured by the same measure, and those incommensurable cannot have any common measure. This is Definition 1 in Book X G: Two magnitudes not having a common measure? Not sure if I understand E: In my world it is said that two magnitudes A and B have a common measure C if they both are multiples of C. And if there is no C like that, then they do not have a common measure G: I suppose you forgot to mention that magnitude C is not a unit, as any two numbers are multiples of a unit, so any two numbers have a common measure, so this makes the definition meaningless E: You are correct, my friend. Any two numbers have a common measure, but pay attention that the definition does not say two numbers, but two magnitudes. A number is a magnitude composed of units, but not every magnitude is a number. A, B, and C are some magnitudes, not necessarily numbers G: OK, I understand now. I simply did not pay attention to the word “magnitude.” Now I also see a clearer difference between the definition of magnitudes that do not have a common measure and the definition of relatively prime numbers, which is Definition 12 in book VII“Numbers relatively prime are those which are measured by a unit alone as a common measure.” E: Obviously these are different definitions G: I suppose that by referring to segments without a common measure, you mean numbers that the ratio between them is not rational E: Ratio that is not rational? This appears so strange, as if the concept is not rational G: This is just a term; I will explain later. By the way, how many magnitudes without a common measure are known? E: Not many Strange voice: many, very many, more than you know G: This is Cantorio, drinking again G: But let us return to the proposition and its proof. I note that the proof relies on the same algorithm that was used in proving Proposition 2 in Book VII. But there a different conclusion is reached. Is it because “magnitude” instead of “number”? E: Not exactly. That proposition assumes that the process of continuous subtraction of pairs of segments brings to a remainder of 1. Here we do not have such an assumption G: Indeed. Since every “remainder” in the process does not measure the previous one, the process does not end; it is actually infinite E: Infinite? This sounds irrational

Before commenting on Mike’s script, we remind the reader that our purpose is not to critique or downplay the student work. Such scripts, as the above one, were commented and graded generously. Our purpose is to systematically scrutinize the script for discrepancies between Mike’s interpretation of Euclid with ours, while realizing that our interpretation is not a golden standard. From our experience, such

60

3  Encounters with Euclidean Propositions

discrepancies can always be found when looked for. We then treat them as disturbing encounters and use as a springboard for our own learning and inspiration for future task development. In #1 we agree with Mike’s character Goldboom that it is impossible to understand the proof without understanding the related terminology. The notion “incommensurable” is explained by quoting the definition, which in turn introduces a new term, “common measure,” for which an explanation is sought. The term is explained as “two magnitudes A and B have a common measure C if they both are multiples of C.” This explanation makes perfect sense taking the contemporary definition of a multiple, as well as the Euclidean one (“The greater number is a multiple of the less when it is measured by the less” [VII, Def. 5]). However, we note that Euclidio refers to magnitudes having a common measure, where the Euclidean definitions attend to numbers. We note that every Euclidean number is a “magnitude” (as every natural number is also a real number). This was likely obvious for Mike and served as “first principles” for Euclidio. In fact, the converse that “not every magnitude is a number” is stated explicitly in [2.2] of the script, drawing attention to the difference between numbers and magnitudes. We appreciate the attention to the Euclidean concept of a “unit,” which is used to amplify the distinction between numbers and magnitudes. However, Goldboom’s statement in [2.3], “Now I also see a clearer difference between the definition of magnitudes that do not have a common measure and the definition of relatively prime numbers, which is Definition 12 in book VII,” attracted our interest. Obviously – as Euclidio claims – these are different definitions [2.4]. We decided to explore the analogy that was brought forward and considered the notes provided by Mike that accompanied his script. There Mike explicitly notes his initial confusion between “incommensurable” and “relatively prime” and then notes that the first term is applicable for any numbers, while the second is considered only for whole numbers, “which are ‘numbers’ for ancient people.” Upon further conversation with Mike, we believe that his temporary confusion originated with a common reference to relatively prime numbers. Note that Euclid defined relatively prime numbers as “those which are measured by a unit alone as a common measure” [VII, Def. 12], and contemporary definition refers to relatively prime numbers as those with greatest common factor of 1. However, a common reference used in school is that numbers are relatively prime if they do not have a common factor, other than 1. So “not having a common measure,” and “not having a common factor,” creates a confusing analogy; when for natural/Euclidean numbers, a common factor/measure always exists, it is 1/unit. In section #3 we note two humorous vignettes. In [3.1], the phrase “ratio not rational” appeared strange to Euclidio and also to us. It appears from the script that Goldboom interprets the concept of incommensurability with contemporary terminology, related to irrational numbers, which is foreign to Euclidio, and pun is intended. In [3.2] we note an introduction of Cantorio (by humoristic extrapolation, a student of Cantor), through which Mike demonstrates his knowledge of the existence of infinitely many irrational numbers (magnitudes without a common measure),

3.6 On the Euclidean Algorithm

61

which is not known to Euclidio. While humor is always appreciated, it is also sometimes used by students as a “shield.” That is, when a humorous intermezzo directs attention away from the main theme, it could sometimes point to the writer’s inability or unwillingness to delve into the core. It is not of course the case with Mike’s script, who used a humoristic introduction of Cantorio in order to deepen the discussion. In section #4 of Mike’s script, it is noted that the proof method relies on the same method as in Book VII Proposition 2, “Given two numbers not prime to one another, to find their greatest common measure.” Euclidio notes correctly that the described processes have one important difference: the process of continuing subtraction in VII.2 actually ends. “That proposition assumes that the process of continuous subtraction of pairs of segments brings to a remainder of 1” [4.3]. We note that here Euclidio, and likely Mike, relies on the contemporary interpretation of what is known as “Euclidean algorithm” for finding the greatest common factor. However, Euclid distinguishes between two cases: common measure is a unit, and common measure is a number. That is why he needs two separate propositions, which basically describe the same method of continued subtraction. • Two unequal numbers are set out, and the less is being continually subtracted in turn from the greater; if the number which is left never measures the one before it until a unit is left, then the original numbers will be prime to one another [VII.1, p.158]. • Given two numbers not prime to one another, to find their greatest common measure [VII.2, p.159]. Indeed, had Euclid considered a unit to be a number, two separate propositions would be unnecessary. In [4.3] and [4.4], we find another humorous use of the concept of infinity, which is unfamiliar to Euclidio and is referred to as irrational. This ends the dialogue. We did not suspect at the time that “sounds irrational” phrase would lead us to another investigation, which follows the notions of rational and irrational and which is presented below.

3.6.2  Which Proposition Is Euclid’s Algorithm? Having immediately noted the resemblance of the method in X.2 with the familiar Euclidean algorithm, we wondered how it is described in Elements. We thought we found it in Book X, Proposition 3. • Given two commensurable magnitudes, to find their greatest common measure [X.3, p.239]. But, Mike in his script referred to Book VII, Proposition 2. • Given two numbers not prime to one another, to find their greatest common measure [VII.2, p.158].

62

3  Encounters with Euclidean Propositions

Which one is the Euclidean algorithm? In the table below, we compare line by line the proofs of VII.2 and X.3. Both proofs follow the same method of continued subtraction, till the “remainder” measures what is subtracted. Organized side by side in Fig. 3.4, it is easy to see that the proofs follow the same steps and are supported by similar diagrams (though, as in many other cases, we do not find the diagrams very helpful). The only claim in VII.2 that is not “matched” with the corresponding claim from X.3 is the one that assures that the “remainder” is number and not a unit. In fact, a similar algorithm, where the remainder is a unit, is found in VII.1, attending to relatively prime numbers. So why did Euclid need two separate propositions? One possible answer is that there are many redundancies in Elements, if judged by contemporary standards. However, the main difference is that one proposition attends to numbers, where the other attends to magnitudes. And as it has already been mentioned, Euclidean numbers are natural numbers, while magnitudes can be of any length. As we know, there exists a common measure for any two rational numbers. Consider, for example, 2 1 and 1 . When looking for a common measure, we are 3 2 looking for a number that “fits” in the given two numbers. Given these “easy” numbers, we detect that a common measure is 61 , and in fact this is the greatest common measure. It “measures” 2 1 15 times; it “measures” 1 twice: 2 3

2 1 = 1 × 15 and 1 = 1 × 2 2 6 3 6

It can also be found by applying Euclidean algorithm: continued subtraction of 1 from 2 1 until what is left “measures” what is subtracted: 3 2

2 1 – 7× 1 = 1 2 3 6

However, this example is too brief – it contains only one solution step – so it does not give a proper sense for the iterative nature of the Euclidean algorithm. Let us consider another example: Find common measure of 6 23 and 5 15 .

6 2 − 5 1 = 22 3 5 15



5 1 − 3 × 22 = 12 5 15 15



22 12 10 − = 15 15 15

3.6 On the Euclidean Algorithm Book VII, Proposition 2 Given two numbers not prime to one another, to find their greatest common measure. A C F

63 Book X, Proposition 3 Given two commensurable magnitudes, to find their greatest common measure.

G

E

G C

A

F

E

B D

D B

Let AB, CD be the two given numbers not prime to one another. This it is required to find the greatest common measure of AB, CD.

Let the two given commensurable magnitudes be AB, CD of which AB the less; thus it is required to find the greatest common measure of AB, CD.

If now CD measures AB – and it also measures itself – CD is a common measure of CD, AB.

Now the magnitude AB either measures CD or it does not. If then it measures it – and it measures itself also – AB is a common measure of AB, CD. And it is manifest that it is also the greatest; for a greater magnitude than the magnitude AB will not measure AB.

And it is manifest that it is also the greatest, for no greater number than CD will measure CD. But, if CD does not measure AB, then, when the less of the numbers AB, CD being continually subtracted from the greater, some number will be left which will measure the one before it.

Next, let AB not measure CD. Then, if the less is continually subtracted in turn from the greater, that which is left over will sometime measure the one before it, because AB and CD are not incommensurable. [cf. X.2]

For a unit will not be left, otherwise AB and CD will be prime to one another, which is contrary to the hypothesis. [VII.1] Therefore some number will be left which will measure the one before it. Now let CD, measuring BE, leave EA less than itself, let EA, measuring DF, leave FC less than itself, and let CF measure AE.

Let AB, measuring ED, leave EC less than itself, let EC, measuring FB, leave AF less than itself, and let AF measure CE.

Fig. 3.4  Proofs of VII.2 (pp. 158–159) and X.3 (pp. 239–240) matched

64

3  Encounters with Euclidean Propositions Since then, CF measures AE, and AE measures DF, therefore CF will also measure DF. But it also measures itself; therefore it will also measure the whole CD. But CD measures BE; therefore CF also measures BE. But it also measures EA; therefore it will also measure the whole BA.

Since, then, AF measures CE, while CE measures FB, therefore AF will also measure FB. But it measures itself also; therefore AF will also measures the whole AB. But AB measures DE; therefore AF will also measure ED.

But it also measures CD, therefore CF measures AB, CD. Therefore CF is a common measure of AB, CD.

But it measures CE also, therefore it also measures the whole CD. Therefore AF is a common measure of AB, CD.

I say next that it is also the greatest.

I say next that it is also the greatest.

For, if CF is not the greatest common measure of AB, CD, some number which is greater than CF will measure the numbers AB, CD. Let such a number measure them, and let it be G.

For if not, there will be some magnitude greater than AF which will measure AB, CD. Let it be G. Since then G measures AB, while AB measures ED, therefore G also measures ED.

Now, since G measures CD, and CD measures BE, G also measures BE. But it also measures the whole BA, therefore it will also measure the remainder AE.

But it measures the whole CD also, therefore G will also measure the remainder CE.

But AE measures DF, therefore G will also measure DF.

But CE measures FB, therefore G will also measure FB.

But it also measures the whole DC; therefore it will also measure the remainder CF, that is, the greater will measure the less: which is impossible.

But it measures the whole AB also, and it will therefore measure the remainder AF, the greater the less: which is impossible.

Therefore no number which is greater than CF will measure the numbers AB, CD; therefore CF is the greatest common measure of AB, CD. Q.E.D.

Therefore no magnitude greater than AF will measure AB, CD; therefore AF is the greatest common measure of AB, CD. Therefore the greatest common measure of the two given commensurable magnitudes AB, CD has been found. Q.E.D.

Fig. 3.4 (continued)

3.6 On the Euclidean Algorithm

65

12 10 2 − = 15 15 15 2 2 1 and we stop here as “measures” both 6 and 5 15 3 5

6

2 2 = × 50 3 15

and

5 1 = 2 × 39 5 15

This brings us back to the warm-up task from the beginning of this section, in which we asked to compare two methods of finding the greatest common divisor (GCD). While finding the GCD of two numbers can appear more explanatory when looking at prime decomposition, prime decomposition of large numbers may not be easy to find. As such, the Euclidean algorithm presents an efficient computational method, which is faster than prime decomposition when applied for large numbers. However, efficiency is not the main difference. We learned that the Euclidean algorithm, that is, the method of continued subtraction, provides a way for finding the greatest common measure of any commensurable magnitudes, such as rational numbers. We note that a common reference to the Euclidean algorithm as a method for finding the greatest common divisor of two natural numbers limits the scope of applicability of the remarkable idea of continued subtraction.

3.6.3  Detour on Applicability The need to find a common measure for rational numbers manifests itself, for example, when one is required to find a period of certain trigonometric functions, in particular of sums of sinusoids. Zazkis and Truman (2015) extended the definition of GCD (greatest common divisor) and LCM (least common multiple) to GCD-X and LCM-X (X for extended). Consider, for example, (x) +  sin (2x). As the period of sin(x) is 2π and the period of sin(2x) is π, it is easy to see that the period of the sum will also be π. But what a c about sin ( sin x ) and ( sin x ) , for natural numbers a, b, c, and d? b d In Zazkis and Truman (2015), it was shown that the sought period is a c LCM-X  .  . b d

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 a c  LCM ( a,c ) It was further proved that LCM-X  .  = .  b d  GCD ( b,d )  a c  GCD ( a,c ) Similarly, GCD-X  .  = .  b d  LCM ( b,d ) 8 For example, to find the period of h(x), which is the sum of f ( x ) = sin x and 3 3 9 16 g ( x ) = sin x , we note that the period of f(x) is π and the period of g(x) is π. 4 8 9 3 9 9 The common measure, LCM-X ( , ) = , so the period of the sum of these 4 8 4 9 functions h(x), is π. It is illustrated in Fig. 3.5 below. 4

3.6.4  On Rational and Irrational We return now to the last lines of Mike’s script. It could be the case that Mike believed, as did we at some point, that the concepts of rational or irrational were not part of Euclid’s repertoire. That is why, we believe, when “rational ratio” was mentioned, Mike’s character Euclidio claimed that “Ratio that is not rational? This appears so strange, as if the concept is not rational.” However, we found that both terms, rational and irrational, appear in Euclid’s definition: • With these hypotheses, it is proved that there exist straight lines infinite in multitude which are commensurable and incommensurable, respectively, some in length only and others in square also, with an assigned straight line. Let then the assigned straight line be called rational, and those straight lines

Fig. 3.5  Illustrating the period of the sum of trigonometric functions

3.6 On the Euclidean Algorithm

67

which are commensurable with it, whether in length and in square or in square only, rational, but those that are incommensurable with it irrational [X, Def.3, p.237]. “Hypotheses” are likely the references to prior definitions. But what does it mean “commensurable with it, whether in length and in square”? We were confused by this expression. But, luckily, the confusion was resolved when attending to the definitions 1 and 2 from Book X. • Those magnitudes are said to be commensurable which are measured by the same measure and incommensurable which cannot have any common measure [X, Def.1, p.237]. • Straight lines are commensurable in square when the squares on them are measured by the same area and incommensurable in square when the squares on them cannot possibly have any area as a common measure [X, Def.2, p.237]. We make several observations considering these definitions: °° Euclid’s “rational” refers to (A) commensurable in lengths or (B) commensurable in square. While (A) implies (B), (B) does not imply (A). For example, 1 and 2 are commensurable in length and in square, while 1 and 2 are commensurable in square only. °° Euclid’s notions of rational and irrational are defined in reference to a particular unit of choice. Actually, in trigonometry we use a unit different from 1 as our measuring unit: in considering trigonometric functions, the x-axis is labeled with multiples of π. °° Euclid’s “rational” include natural numbers and their square roots. In fact, we plan to engage our future students with the task of comparing what Euclid’s refers to as “rational” or “irrational” with the contemporary terms in the mathematical register. Definition Comparison Task Consider Euclid’s definitions of rational and irrational and compare Euclid’s notions (use E-rational and E-irrational to refer to Euclid’s notions) with the contemporary meanings assigned to (or definitions of) rational and irrational numbers. Provide examples of E-rational and E-irrational. Consider four sets: E-rational, E-irrational, rational, and irrational. What can you say about set inclusion? Set cardinality?

Here another disturbance catches us. We concluded (in haste) that E-rational include natural numbers and their square roots. We did so by extrapolating from our understanding that 1 and 2 are commensurable in square, as those can be measured by the same area.

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However, noting Euclid’s geometrical worldview, we looked for an area unit that will measure both a square with side 1 and a square with side 2. This is not hard to find, and such a unit is demonstrated in Fig. 3.6. The Definition Comparison task became challenging when we tried, unsuccessfully, to subdivide the square with area of 3 (and side 3 ) into three areas of ­measure 1. This was our “literal” interpretation of “measured by the same area” and the search for the measuring area unit, in some analogy to Fig. 3.6. The geometric solution for s is clear in two cases: for square numbers and for powers of 2. But what about other square roots? Having failed to demonstrate geometrically the commensurability in square of 3 and 1, and suspecting that this may not be possible, we wondered whether a direct geometric demonstration was needed. Obviously in contemporary terms, 3 square units are three times larger than 1 square unit, but how was this conclusion accessible in Euclidean terms? Artmann noted that “Euclid takes measuring to be an undefined notion” (Artmann 1999, p. 224). So we sought a construction that transforms given areas to “easily measurable” areas. Proposition 14 from Book II helped us find such a transformation: • To construct a square given to a given rectilineal figure [II.14, p.49]. In fact, “Rectilineal figure” is a (convex) polygon, and the proposition shows how to construct a square of the same area as the given polygon. In particular, it shows how a square can be constructed whose area equals that of the given rectangle. As such, our dilemma with respect to the square of area 3 is solved by reversing the procedure described in II.14. That is, starting with a square of area 3, we can construct a rectangle with sides 1 and 3. Then it is clear that a rectangle with sides 1 and 3, having the same area as the square with a side of 3 , is “measured” by the square of area 1 geometrically (Fig. 3.7). This construction can be extrapolated for any square root, which supports our initial assumption that square roots are E-rational as commensurable in square with natural numbers (if the chosen unit is commensurable in length with natural numbers).

Fig. 3.6 Illustrating 2 and 1 are commensurable in square

1

3.6  On the Euclidean Algorithm

69

2

2

3 1 1

Fig. 3.7  Illustrating that have area of 3

3 and 1 are commensurable in square: the square and the rectangle

In Fig. 3.8, ABCD is a given square with side n , and AK = AN = 1. Point O is constructed as equidistant from B and K, and R = OB = OK is the radius of the circle with center O. Based on the construction, the area of the rectangle is 2R − 1 as the sides of the rectangle are of length 2R − 1 and 1; this is also the area of the square: From triangle ABO:

( n)

2

+ ( R − 1) = R 2 2



n = R 2 − ( R − 1) = 2 R − 1 2





As we engage in distinguishing rational from E-rational based on Heath’s canonic translation, we find that Fowler (1998) prefers to avoid the use of “rational” and “irrational” in translating Elements while noting that the Greek words rhetos and alogos are “almost always rendered into English as ‘rational’ and ‘irrational’ respectively” (p.  162). He wrote, “I have rendered rhetos as ‘expressible’ but have left alogos in transliterated form” (ibid). This is because “the usual translation can lead to mathematical confusions and misunderstandings” (ibid). While we agree that the usual translation can lead to confusion, we also believe that it can lead to a worthwhile task of making distinctions and seeking geometrical justifications – a task that let us to disturbance and enjoyment, which we plan to experience further with our students.

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Fig. 3.8  Illustrating that square side rectangle have the same area

3  Encounters with Euclidean Propositions

n and 1 are commensurable in square: the square and the

3.7  Concluding Remarks This chapter is an example of “learning through teaching” (Leikin and Zazkis 2010) as well as “learning for teaching.” For the former, we showed how our reflections on students’ work resulted in our personal exploration of Euclidean propositions, proofs, and definitions. For the latter, we showed how our mathematical explorations resulted in new tasks for students, which in turn are designed to expand students’ mathematical experience.

References

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In particular, °° We explored interesting connections between continued proportions, mean proportional numbers, and geometric sequences, which highlighted the connection of geometric sequences to geometry. °° We showed that the Euclidean algorithm, as described by Euclid, is more general than it looks in some widespread contemporary interpretations. °° We focused on the difference between Euclidean notion of “commensurable” and “rational” and the contemporary conventions. °° We hope to have inspired interest in further personal mathematical and pedagogical encounters with Elements.

References Artmann, B. (1999). Euclid: The creation of mathematics. New York: Springer. Boyer, C. (1991). A history of mathematics. Hoboken: Wiley. Dewey, J. (1938/1963). Experience and education (reprint). New York: Collier (Original work published 1938). Euclides, I. (2002). In D. Densmore (Eds.). Euclid’s elements: All thirteen books complete in one volume, the Thomas L. Heath translation. Santa Fe: Green Lion Press. Fowler, D.  H. (1998). The mathematics of Plato’s academy: A new reconstruction. Oxford: Clarendon Press. Galilei, G. (1638/1991). Dialogues concerning two new sciences (trans: Crew, H., & de Salvio, A.). Amherst: Prometheus Books. Hazzan, O., & Leron, U. (1996). Students’ use and misuse of mathematical theorems: The case of Lagrange’s theorem. For the Learning of Mathematics, 16(1), 23–26. Koichu, B., & Zazkis, R. (2013). Decoding a proof of Fermat’s little theorem via script writing. Journal of Mathematical Behavior, 32, 364–376. Koichu, B., & Zazkis, R. (2018). “I understand” talk in script writing: A case from Euclid’s elements. In R.  Zazkis & P.  Herbst (Eds.), Scripting approaches in mathematics education: Mathematical dialogues in research and practice. Cham: Springer. Leikin, R., & Zazkis, R. (Eds.). (2010). Learning through teaching mathematics: Developing teachers’ knowledge and expertise in practice. Dordrecht: Springer. Leron, U. (1985). A direct approach to indirect proofs. Educational Studies in Mathematics, 16(3), 321–325. Luke, C. (1988). The repeated addition model of multiplication and children’s performance on mathematical word problems. The Journal of Mathematical Behavior, 7(3), 217–226. Park, J.  H., & Nunes, P. (2001). The development of the concept of multiplication. Cognitive Development, 16(3), 763–773. Ribenboim, P. (1996). The new book of prime number records. New York: Springer. Van de Walle, J.A., Folk, S., Karp, K.S. Bay-Williams, J.M. (2011). Elementary and middle school mathematics: Teaching Developmentally. 3rd Canadian Edition. Toronto: Pearson Canada. Zazkis, R. (2002). Language of number theory: Metaphor and rigor. In S.  R. Campbell & R. Zazkis (Eds.), Learning and teaching number theory: Research in cognition and instruction (pp. 83–96). Westport: Ablex Publishing.

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Zazkis, R., & Koichu, B. (2015). A fictional dialogue on infinitude of primes: Introducing virtual duoethnography. Educational Studies in Mathematics, 88(2), 163–181. Zazkis, R., & Koichu, B. (2018). Dialogues on dialogues: The use of classical dialogues in mathematics teacher education. In R. Zazkis & P. Herbst (Eds.), Mathematical dialogue: Scripting approaches in mathematics education research and practice (pp. 365–387). Cham: Springer. Zazkis, R., & Truman, J. (2015). From trigonometry to number theory and back: Extending LCM to rational numbers. Digital Experiences in Mathematics Education, 1, 79–86. Zazkis, D., & Zazkis, R. (2016). Prospective teachers’ conceptions of proof comprehension: Revisiting a proof of the Pythagorean theorem. International Journal of Mathematics and Science Education, 14, 777–803. Zazkis, R., Liljedahl, P., & Sinclair, N. (2009). Lesson plays: Planning teaching vs. teaching planning. For the Learning of Mathematics, 29(1), 40–47. Zazkis, R., Sinclair, N., & Liljedahl, P. (2013). Lesson play in mathematics education: A tool for research and professional development. Dordrecht: Springer.

Chapter 4

Stories on Problem-Solving Instruction

4.1  Introduction Like many of our colleagues, we are fond of problem-solving instruction. As such, we adhere to both teaching for problem solving and teaching via problem solving in our courses for prospective and in-service teachers. This distinction – for problem solving vs. via problem solving  – is referred to in many literature sources (e.g., Schroeder and Lester 1989; NCTM 1980, 1989; Schoenfeld 1992; Lester 2013). According to Schroeder and Lester (1989), in teaching via problem solving, developing a new solution method by which a certain class of non-routine problems can be transformed into routine ones is the goal of the intended learning. In teaching for problem solving, the main goal of the intended learning is applying the previously introduced knowledge and transferring it from one problem context to another. Of note is that a theoretical border between the two approaches is blurred. This is simply so because the border between developing and applying is blurred. Indeed, both developing and applying knowledge can require a big deal of creativity. In particular, developing new knowledge can occur in the process of relating pieces of the existing knowledge to each other (e.g., Bruner 1960), which in fact is based on applying knowledge in a different way. The process of relating is pivotal also when one needs to develop a way to apply knowledge taught in one problem context to another problem context, especially when the contexts are perceived by the problem solver as remote. At the same time, the border between teaching via problem solving and teaching for problem solving seems to be very clear in terms of traditional instructional practices that these two approaches entail. Roughly speaking, teaching via problem solving frequently corresponds to the launch-and-explore teaching strategy, and teaching for problem solving to the demonstrate-and-apply teaching strategy. The former strategy consists of a stage of launching, by the teacher, of a non-routine mathematical question having the potential to engage students with important mathematical ideas. This stage is followed by a stage of exploring the launched problem. © Springer Nature Switzerland AG 2021 B. Koichu, R. Zazkis, Mathematical Encounters and Pedagogical Detours, https://doi.org/10.1007/978-3-030-58434-4_4

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The latter strategy consists of demonstrating, by a teacher, of a problem-solving method, which is followed by applying it by students, who are given more or less sophisticated problems to be solved by means of the demonstrated method. Both launch-and-explore and demonstrate-and-apply sequences can be complemented by summarizing, at which the students’ actual and attempted solutions are considered and conclusions are made. Stein et  al. (2008) noted that the launch-explore-­ summarize strategy can be observed in a typical reform-oriented lesson, as contrasted to a “traditional” show-and-tell lesson. Schoenfeld (2018) made a similar point when using the terms demonstrate-and-practice pedagogy and transmission pedagogy as synonyms and arguing that both of them are not particularly productive. R: I value both strategies, but prefer using the launch-and-explore strategy in my lessons. Just consider a difference between the following two lessons. One begins from introducing a formula for solving quadratic equations and continues by applying it to various cases (akin to the demonstrate-and-apply). The other begins from the question “how can we find roots of an equation x2 − 2x − 3 = 0? ”, continues by a guided invention of the completing-the-square method, and concludes by formalizing this method to all equations of the form x2 + bx + c = 0 (akin to the launch-and-explore). B: The difference is huge indeed. Let’s say, focusing on the second option, that the students already know the abridged multiplication formulas (a ± b)2 = a2 ± 2ab + b2 and a2 − b2 = (a + b) × (a − b). Then the guided invention may be supported by the Pairwise Comparison Task. R: Nice task! This is one option of course. We can imagine many other exploration lines towards the invention.

Pairwise Comparison Task Compare solutions to the following pairs of equations. How can a way of solving one of the equations in the pair be used for solving another equation? Pair 1: (x − 1)2 − 4 = 0 and x2 − 2x − 3 = 0. Pair 2: (x − 2)2 − 9 = 0 and x2 − 4x − 5 = 0. 2 5 1  Pair 3:  x −  − = 0 and x2 − 5x + 6 = 0. 2 4  Pair 4: (x − m)2 − n2 = 0 and x2 − 2mx + c = 0

Needless to say, we are far from assuming that there exists one preferable instructional strategy. The devil is in the details, as usual. For example, Stein et al. (2008) showed that there are more productive and less productive ways of enacting the launch-explore-summarize strategy. In particular, they analyzed a lesson that had been designed and conducted in accordance with the launch-explore-summarize strategy but eventually had characteristics of a launch-and-demonstrate lesson.

4.2  Explore – Launch – Re-explore – Relate

75

As a prelude to this chapter, we would like to say that for many mathematics educators, us included, teaching for problem solving as well as via problem solving is an instant source of mathematical and pedagogical disturbances and surprises. This chapter is devoted to some of them and to our attempts to experiment with and reflect on different scenarios of problem-solving instruction. Specifically, we present four stories, two of which are elaborations of the teaching for problem-solving instructional approach (but with an eye to teaching via), and two other stories are elaborations of the teaching via problem-solving approach (but with an eye to teaching for). We refer to these stories as follows: 1. 2. 3. 4.

Explore – launch – re-explore – relate Launch – explore – re-launch – re-explore – relate Launch – get stuck – intervene – relate Launch – intervene – re-launch – re-explore – relate

4.2  Explore – Launch – Re-explore – Relate 4.2.1  Explore Our first story stems from the fact that one of us turns quite frequently to graphing software, and the other one does so as the last resort when solving algebraic problems. Accordingly, we have a lot of fun when translating each other’s problem-­ solving ideas from graphical to algebraic language and backward, or when teasing each other by saying that a particular approach to a specific problem, either graphical or algebraic, is the only “natural” one. The story began during one of our conversations, in which we playfully used a graphical software without having any particular problem in mind. Availability and easy access to a graphing software invite a free exploration of graphs of functions, initially without any specific agenda. Focusing or quadratic functions, one can be invited simply to choose any values of a, b, and c in y = ax2 + bx + c and see what a related graph looks like. Of note is that this is very different from our own school memories of plotting and connecting points, as part of the function learning activity. In the school setting employing conventional approaches, the exploration can be funneled, for example, by focusing on how the change of one of the coefficients (while keeping others constant) influences the resulting graph (e.g., Edwards 1996). Staring at parabolas, one can see that some intersect the axes at points that have integer coordinates, while others have vertices at points with integer coordinates. For brevity, we refer to a point (a, b) where a and b are integers as an “integer point.” Both cases are frequently generated by teachers: by focusing on the roots, that is, keeping the coefficient a = 1 for simplicity, expanding y = (x − a)(x − b), with integer values a and b results in a graph that intersects the x axis in integer points (a, 0)

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4  Stories on Problem-Solving Instruction

and (b, 0); focusing on the general form f(x)  =  (x  −  h)2  +  k, where h and k have integer values assures that the vertex of the parabola (h, k) is at the integer point. Again, this is an application of conventional curricular knowledge. Extending these observations while looking at about 10 graphs of parabolas having integer roots or having integer vertices, we wondered, what if we had BOTH?

4.2.2  Launch The “what if both?” question brought us to the Integer Parabola Task. Integer Parabola Task What choice of integer coefficients b and c of a monic1 quadratic function y = x2 + bx + c assures that the resulting graph has integer intersections with both axes and an integer vertex?

4.2.3  Explore Engaging in a more focused exploration, we admit that with immediate access to graphing technology, there is sometimes an urge to graph before engaging in any analysis of the task. As such, we decided to start with a case where one of the roots is zero. The following graphs were generated among others (Figs. 4.1 and 4.2). It is only after enjoying the graphical display, we noticed the connection.

x 2 − 8 x = ( x − 4 ) − 16



x2 − 8x + 7 = ( x − 4) − 9





x 2 − 8 x + 12 = ( x − 4 ) − 4





x 2 − 8 x + 15 = ( x − 4 ) − 1



2



2

2

2

And similarly x 2 − 10 x = ( x − 5 ) − 25 2





 To recall, a polynomial is called monic if its highest-degree coefficient equals one.

1

4.2  Explore – Launch – Re-explore – Relate

77

Fig. 4.1  Integer parabolas with vertices at x = 4



x 2 − 10 x + 9 = ( x − 5 ) − 16





x 2 − 10 x + 16 = ( x − 5 ) − 9





x 2 − 10 x + 21 = ( x − 5 ) − 4





x 2 − 10 x + 24 = ( x − 5 ) − 1



2

2

2

2

78

Fig. 4.2  Integer parabolas with vertices at x = 5

4  Stories on Problem-Solving Instruction

4.2  Explore – Launch – Re-explore – Relate

79

Fig. 4.3  Integer parabolas with vertices on x = 4 and roots of different signs

Which brings us to the general form for the solution. y = ( x − h ) − w2 2





or

(

y = x 2 − 2hx + h 2 − w 2

)

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That is, when generating parabola equations in the form y = x2 + bx + c, we need to assure that the coefficient of x is even and that c is a specific difference of squares. Note that this also assures integer intersection with the y-axis (Fig. 4.3). Of interest is that extending the sequence of square numbers in the equations generates integer roots, where one of the roots is negative:

( x − 4)



( x − 4)

2

2

− 25 = x 2 − 8 x − 9



− 36 = x 2 − 8 x − 20



etc.

4.2.4  Relate Looking back, the task could have been approached systematically in a more straightforward fashion. Starting with y = x2, we get an integer vertex and integer intersection at (0,0). Then y = x2 − k, results in a vertical translation (down, to ensure existence of real roots). To assure integer roots (integer intersections with x-axis), we need to choose k = w2, that is, k has to be a square number. Then the parabola intersects the x-axis in integer points (w, 0) and (−w, 0). From here all we need is to add horizontal translation by an integer number of units, which brings us to the form y = (x − h)2 − w2 – a parabola with an integer vertex and integer intersections with the x-axis in points (h + w, 0) and (h − w, 0). Note that this also assures intersection with the y-axis at the integer point (0, h2 − w2). In closing this story, which in a way is a mathematical prologue to the next one, let us recall that Chazzan (1999) has argued that relating special properties of roots of a quadratic polynomial and special properties of its vertex can be a source of remarkable student misconceptions and thought-provoking explorations. He brought up a story of one of his students, who formulated an interesting (though wrong) conjecture: the distance between intersections of a parabola with the x-axis and between the vertex and the x-axis is the same. Well, it is true, for instance, for y = x2 − 4 or y = x2 − 6x + 5. We leave it to the readers to find additional examples of parabolas with this property (if any) and also invite them to reflect on which approach to this question seems them the most “natural” one, graphical or algebraic.

4.3  Launch – Explore – Re-launch – Re-explore – Relate 4.3.1  Launch The story below has been launched for us by the Miscopying Task (appears in Lehoczky and Rusczyk 2004, p. 87).

4.3  Launch – Explore – Re-launch – Re-explore – Relate

81

Miscopying Task Two students attempted to solve a quadratic equation, x2 + bx + c = 0. Although both students did the work correctly, the first miscopied the middle term and obtained the solution {−6, 1}. The second student miscopied the constant term and obtained the solution set {2, 3}. What is the right solution?

4.3.2  Explore The solution to the Miscopying Task is based on a modest exploration of the from-­ the-­end-to-the-beginning nature. Indeed, the knowledge of the solutions enables us to restore the miscopied equations as the first step. Namely, the first student solved the equation

( x + 6 ) ( x − 1) = x 2 + 5 x − 6 = 0,



and the second student solved the equation

( x − 2 ) ( x − 3) = x 2 − 5 x + 6 = 0.



Next, it is given that the first student miscopied the middle term (i.e., 5),but the constant (i.e., −6) was not miscopied; therefore it belongs to the right equation. Similarly, the middle term of the second equation (i.e., −5) belongs to the right equation. Therefore the right equation is x2 − 5x − 6 = 0, and its roots are {-1, 6}, possibly obtained by the formula for solving quadratic equations: x1,2 =





x1,2 =

− ( −5 ) ±

−b ± b 2 − 4c : 2

( −5) 2 ⋅1

2

+ 4 ⋅1 ⋅ 6

=

5±7 → {−1,6} . 2

4.3.3  Re-launch The re-launch, possibly stimulated for us by the story presented in the previous section, began from the following observation. The student-characters in the above task miscopied the right equation in a very specific way: they changed the signs of the middle term and the constant term of a given equation and in this way obtained two additional equations. What is interesting is that all three equations have integer roots. Is it by chance? It did not feel so, and therefore a question arose: Are there

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more sets of quadratic equations that all have integer roots when alternating the signs of the middle-term coefficient and of the constant? And if yes, how to find them or at least some of them? Our immediate search for such pairs of equations failed (try it!). Thus we decided to attack the question in a systematic way and inquired about the relationships among the roots of four equations with integer non-zero coefficients:

x 2 + bx + c = 0



x 2 − bx + c = 0



x 2 + bx − c = 0.



x 2 − bx − c = 0.

Specifically, we asked under which conditions equations (a)-(d) can all have (or do not have) integer roots. B: I remember that when in school I was wondering why changing the sign of the middle term is never a troublemaker, but changing the sign of the constant frequently is, as the following pair of equations shows. x2 − 3x + 2 = 0 has roots {1, 2},  3 ± 17  x2 − 3x − 2 = 0 has roots  .  2  R: Well, when integer solutions are expected, miscopying often leads to roots that are not integers, or to equations with no solutions in real numbers. However, the pairs of equations (a)-(b) and (c)-(d) have or do not have the integer roots simultaneously. This is because the roots of these equations are pairwise symmetrical: if equation (a) has roots x1 and x2, the equation (b) has the roots −x1 and −x2. This can easily be seen when sketching graphs of the functions y = x2 + bx + c and y = x2 − bx + c, as we have done in the first story for this chapter. The question remains about (a)-(c) and (b)-(d) pairs. Clearly, our question required formalization to become approachable. To reiterate, we were seeking equation pairs of the form (a)-(c), with non-zero integer coefficients b and c, that both have integer roots. At this point, we recalled the Rational Zero Theorem2 and its special case for monic polynomials (i.e., for polynomials with the highest-degree coefficient 1). If P is a monic polynomial with integer coefficients having non-zero integer roots, they all are factors of the last (constant) coefficient of P.

 See, for example, http://mathworld.wolfram.com/RationalZeroTheorem.html

2

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In the hope to somehow use this theorem, we formalized our question so that it would be expressed by means of one monic polynomial with integer coefficients, as follows: Consider the polynomial P(x) = (x2 − bx + c)(x2 − bx − c) = x4 − 2bx3 + b2x2 − c2, b ≠ 0, c ≠ 0. It is given that two of its roots, x1 and x2 are integers. Find all (or at least some of) the pairs (b, c) so that P(x) would have four integer roots. Well, the Rational Zero Theorem did not appear useful for attacking the above question. We did not know how to proceed from here (the readers are invited to try!) and moved on. We then recalled the Viet Theorem, often used when solutions of quadratic equations are sought by factoring the standard form. That is, if {x1, x2} is a set of roots of x2 + bx + c = 0, then c is the product of the roots and b is the opposite of the sum. So, seeking four integer roots of two equations, we developed another formalization of our question, as follows: For which pairs of non-zero integers {b, c} there exist quadruples of integers {x1, x2, x3, x4} such that x1 + x2 = x3 + x4 = b and x1x2 =  −(x3x4) = c Unfortunately, the Viet Theorem also failed us, and after a while, we decided to change the direction. Also, we felt that the above two formalizations would take us far beyond mathematics that can reasonably be used in our courses with teachers, and this discouraged us from continuing to try. We then came back to the formula for solving quadratic equations (simplified for the monic polynomial we are using here), −b ± b 2 − 4c and to its troublemaking part, b 2 − 4c . Whether or not an 2 equation x2 + bx + c = 0 with integer coefficients b and c has integer roots depends on whether or not b 2 − 4c is an integer. Given that the discriminant of the equation x1,2 =



x 2 + bx + c = 0 is b 2 − 4c,

and the discriminant of the equation

x 2 + bx − c = 0 is b 2 + 4c,

we decided to re-launch our quest as the Double Perfect Squares Task. Double Perfect Squares Task Find all (or at least some) of the pairs of non-zero integers {b, c} so that b2 − 4c and b2 + 4c are perfect squares.

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4.3.4  Re-explore Here the main part of the story begins. We began looking for pairs {b, c} that fulfill the following system of conditions for some integers p and s: b 2 − 4 ⋅ c = p 2  2 2 b + 4 ⋅c = s



Naturally, we turned to exploring the only example known to us at this stage, the pair of equations from the Miscopying Task. Note that the equation x2 + 5x + 6 = 0 has the discriminant 25 − 24 = 1, and the equation x2 + 5x − 6 = 0 has the discriminant 25 + 24 = 72. Looking for some time at the numbers that showed up, 7, 24, 25, we noticed they form a Pythagorean triple. Indeed: 72 + 24 2 = 49 + 576 = 625 = 252.



It was a surprise. And it looked encouraging, mainly because Pythagorean triples have many beautiful properties and therefore can be a good resource. But how to use this observation? After many attempts, we decided to simplify the task. Inspired by our initial equation, x2 + 5x + 6 = 0 having the discriminant 1, we decided to look only at the case p2 = 1. Accordingly, our system turned to be  b2 − 4 ⋅ c = 1  2 2 b + 4 ⋅ c = s





As we were interested in b and c, we transformed the system into the form  2 1 + s2 b = 2 , in which the second equation looked to us especially promising, since  2  c = s −1  8 it has an integer on its left side and division by 8 on its right side. This structure immediately lead us to the conclusion that s should be an odd number. So, let s = 2k + 1, and with this, let us come back to the first equation. 1 + s 2 1 + ( 2 k + 1) 2 = = 2 k 2 + 2 k + 1 = k 2 + ( k + 1) 2 2 2

b2 =

Eureka!

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4.3.5  Relate Why eureka? Because the equation b2 = k2 + (k + 1)2 suggests that b is a member of special Pythagorean triples, namely, triples in which the two smaller members are consecutive numbers. The most famous triple of this type is 3, 4, 5. Let us check. 72 − 1 If k = 3 then b =  ± 5, s = 2 ∙ 3 + 1 = 7 and c = =6. 8 That is, {b, c} = {±5, 6}, which gives us our initial two pairs of equations: x 2 + 5 x + 6 = 0 ( roots: − 3 and − 2 ) , x 2 + 5 x − 6 = 0 ( roots: − 6 and 1)





and

x 2 − 5 x + 6 = 0 ( roots: 3 and 2 ) , x 2 − 5 x − 6 = 0 ( roots: 6 and − 1) .



More importantly, we now have a method of how to look for additional pairs of equations: we just need to use Pythagorean triples in which the two smaller members are consecutive numbers. What a nice relation between quadratic equations with special properties and Pythagorean triples with special properties! Fortunately, there is an indefinite number of such triples, though the way of generating them is beyond the scope of this chapter3. The next Pythagorean triple having two consecutive numbers as its smaller members is 20, 21, 29. So, let us apply the method. In this case, k = 20, then



b = ± 20 2 + 212 = ±29, s = 2 ⋅ 20 + 1 = 41 and c =

412 − 1 = 210 8

Thus, our next pair is as follows:

x 2 + 29 x + 210 = 0 ( roots: − 14 and − 15 ) and



x 2 + 29 x − 210 = 0 ( roots: − 35 and 6 ) .





And the next appropriate triple is 119, 120, 169, and the next one is 696, 697, 985, and so on. R: Wait a minute. From the triples 3, 4, 5 and 20, 21, 29, we obtained pairs of equations so that the roots in one equations in the pair differ by 1. Would the property be preserved for the next triples?

 The way of generation of Pythagorean triples in which two smaller members are consecutive integers is described, for example, here: http://www.math.wichita.edu/~richardson/mathematics/ pythagoras/pythagoreantriples.html 3

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B: It should be, because we consider a special case, in which the discriminant of one equation is 1, or, in our earlier notation, p2 = 1. It is tempting to continue as there are so many additional questions to explore. Anyway, let’s stop here. I believe that we’ve fulfilled our goal for this story, which was to illustrate the launch – explore – relaunch – re-explore – relate scenario.

4.4  Launch – Get Stuck – Intervene – Relate 4.4.1  Background The two stories presented above stem from our mathematical disturbances in the context of algebra and from our hypothetical thinking of how these disturbances can be used as learning opportunities for our students. On the contrary, stories presented in this and in the next section are based on our actual teaching experiences with real students, in the context of geometry. You can easily imagine the students we refer to in this section. They are pre-­service secondary-school teachers with diverse mathematical backgrounds, who, even when excelling in algebra, frequently struggle with geometry. Some of them periodically complain that “they cannot see,” not in a perceptual sense, but in that mysterious sense that enable humans, as Arcavi (2003) puts it, to see the unseen when doing mathematics. In other words, we refer to students for whom visualization is difficult. R: Doesn’t this description apply to all students, and also to us? B: To us, and to many of our students – for sure. Arcavi (2003) proposes the following conceptualization of visualization: Visualization is the ability, the process and the product of creation, interpretation, use of and reflection upon pictures, images, diagrams, in our minds, on paper or with technological tools, with the purpose of depicting and communicating information, thinking about and developing previously unknown ideas and advancing understanding. (p. 217)

In the story below, we will use the collocation “visualization of a drawing” in Arcavi’s sense. In addition, we draw on the distinction between drawing and figure developed by Laborde (1993, 1995). Briefly speaking, a drawing refers to what actually is drawn, while a figure is a theoretical referent attached to the drawing not only from what is perceptually seen but also in a discursive way, by which the intended geometrical properties are explicitly stated. For reading this section, imagine students who can readily perform a computational part of a geometry problem, say, by using a formula S = 12 b ⋅ h for finding the area S of a triangle with base b and corresponding height h, but for whom it is difficult “to see” where a and h are on a non-conventional drawing. The story below has repeated itself many times in different classes.

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4.4.2  Launch The launching problem is chosen from a large collection of problems based on our own struggle “to see,” as well as by our wish to help our students see what we have seen, by means of some non-particularly intrusive interventions. Here it is.

Five Triangles Problem In a triangle ABC, AB = 60, CB = 48. Place points X and Y on AB and points Z, W on CB such that 5 “small” triangles (AXC, CZX, XYZ, ZWY and YWB, see Fig. 4.4) have the same area.

Keeping suspense is not in our plans this time. We invite the readers to solve the problem or to look at the intended solution. Solution to the Five Triangles Problem  Area of AXC is one-fifth of the area of ABC. 1 Therefore, AX = ⋅ 60 = 48. 5 Area of CXZ is one-fourth of the area of CXB. 1 ⋅ 48 = 12. 4 Area of XZY is one-third of the area of XZB. Therefore, CZ =

1 4 Therefore, XY = ⋅ ⋅ 60 = 16 and YB = AB − AX − XY = 32. 3 5 Area of ZWY is one-half of the area of ZYB. Therefore, ZW = WB =

Fig. 4.4  Five triangles

1 1 ZB = ( CB − CZ ) = 18. 2 2

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4.4.3  Get Stuck Some students can solve this problem quickly but most of the students can’t. As a rule, the latter students make the following move: they construct the heights of all the triangles mentioned in the problem formulation (see example in Fig. 4.5) Once the heights are drawn, the students realize that they are not particularly helpful. After all, as van Essen and Hamaker (1990) have pointed out, one of the roles of a drawing is to relieve working memory, not overload it. Apparently, visualizing the drawing presented in Fig. 4.5 is a cognitively loaded enterprise. This is especially true in light of the fact that the added heights are not among the givens of the problem so it is not immediately evident how to use them. As a result, the students get stuck. Of note is that, as ineffective as it is, the described problem-solving start is sensible. Selden et al. (2000) have described similar situations in their study on undergraduate students’ problem solving. The scholars found that students frequently employ unnecessarily complicated methods that lead nowhere in problem-solving situations having simple and within-the-student’s-reach solution methods (of course, from the expert problem-solver viewpoint). The explanation of this phenomenon was given in the following terms: any problem evokes a problem situation image containing tentative solution starts, i.e., tentative general ideas for beginning the process of finding a solution. The students do not retain the knowledge most appropriate to the problem situation since they have particular tentative solution starts in their possession. So, which situational image of the Five Triangles Problem can evoke such a start? One can speculate (but in a good sense of this verb!) that when reading the problem formulation, the students immediately identify it as a problem about areas of triangles and thus connect it with the piece of knowledge most robust for them on this topic, the formula S = 12 base × height. Accordingly, the initial visualization of the Five Triangles Problem consists of a figure in which all elements needed in order to use the formula are explicit. Hence the above drawing with heights. Looking from the position of a person who knows the solution, visualizations with heights are not helpful, in particular, because they are insufficiently goal-directed: the students just do what they can do in the hope that their move will somehow progress the solution (cf. Koichu et  al. 2007 for finding-what-is-easy-to-find heuristics.). Importantly, in order to continue attacking the problem, the students actually need to monitor their idea, realize that it is as an unhelpful one and then take a step back.

Fig. 4.5  Five Triangles Problem with student-­ added heights

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As scholars noted (e.g., Schoenfeld 1985), this course of action is quite unrealistic to expect. Thus, the question to us as mathematics educators is, how to get the students unstuck without depriving them of the pleasure of being independent problem solvers, and with an eye to introducing important ideas that they can learn on the way?

4.4.4  Intervene In light of the above considerations, one way to get the students unstuck can be as follows: the students should experience a situation, in which the known formula S = 12 b ⋅ h can be used not as a computational tool but as an exploratory tool. The intervention invites students to leave the problem and think of another one, focusing in medians. In response, the students do exactly what they have done before – they draw the height of the triangle in order to use the formula, and this time this works. The conclusion is, the areas of BAD and CAD are equal.

B: And then they go back to the Five Triangles Problem and… R: Well, not that quickly. Some of them indeed go back and observe that for areas of triangles ZYW and YWB be equal, YW must be a median of ZYB (Fig. 4.4). However, as a rule, this is all the progress that some of them can make at this stage. B: Indeed, it is so easy to underestimate the difficulty of the problem when you know its solution. I now think that even this modest progress is not trivial. Actually, in order to make it, the students should use the converse

Medians and Area In a triangle ABC, AD is the median. (see Fig. 4.6). What can you say about the areas of triangles BAD and D ?

Fig. 4.6  Medians in a triangle

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Fig. 4.7  What if not median?

of the theorem “If AD is a median of a triangle BAC, then the areas of triangles BAD and CAD are equal” that they’ve just proved. R: Well, as a matter of fact they do, but I doubt that they notice that they use the converse at this stage. As we know, students tend to confuse if-then theorems with if-and-only-if theorems when translating drawings into figures, as many scholars have shown (e.g., Connor et  al. 2007; Edwards 1997; Marrades and Gutiérrez 2000; Lachmy and Koichu 2014). More importantly, the students’ attention at this stage is still drawn to applying the formula for area to each triangle taken separately, whereas what is needed in order to progress the Five Triangles Problem is to make comparative conclusions about their areas, and do so without visualizing the heights. Therefore, my next step is to vary the problem and generalise. What if not medians? What if the base is cut not-in-half, but into three equal segments? Four equal segments? Five equal segments, etc.? (see Fig. 4.7)

As a rule, the students answer this question quickly, and, importantly, without drawing the height. B: And then they go back to the Five Triangles Problem and… R: Well, not that quickly. As you can see, the above question invites thinking of equal areas of small triangles in a particular partitioning, but the Five Triangles Problem requires attending to areas of small triangles as compared with the areas of bigger triangles, in different configurations. Therefore, if no progress is made, I propose at this stage the next question, which I refer to as the One-Fifth Problem.

One-Fifth Problem In a triangle ABC, place point X on side CB, such that the area of AXC is 1 of the area of ABC (Fig. 4.8). 5

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Fig. 4.8 One-Fifth Problem

4.4.5  Relate

B: And then they go back to the Five Triangles Problem and… R: Well, not that quickly. Some of them indeed do, but the others need additional support in order to construct a proper visualization of the initial drawing.

Indeed, one more piece of the puzzle is needed in order to relate the above sequence of auxiliary questions with the target problem. As one can see (in the most straightforward meaning of this word), all intervention questions have employed only drawings of triangles that Clement (2003) called prototypical, that is, triangles with a base flat to the ground and one top vertex. The target problem however requires one to apply the results of the auxiliary questions not only to prototypical triangles having their bases on the side CB but also to non-prototypical triangles having their bases on the side AB (Fig. 4.4). Those students who still “cannot see,” may benefit from one of two following intervention options, a weaker one and a stronger one, as follows: • The teacher can advice the students to rotate the drawing for the Five Triangles Problem upside down so that AB would be “the base.” • Without straightforwardly appealing to the target problem, the teacher can ask the students to re-solve the One-Fifth Problem for the “upside-down drawing” (Fig. 4.9).

B: And then they go back to the Five Triangles Problem and… R: Well, and solve it! While still having the feeling that the solution belongs to them, and while learning how the known (but occasionally forgotten) fact – that a median divides triangle into two equal-area triangles – is generalised to other division of a “base” and implemented in a solution. Eventually, it is a lesson, in which both teaching for and teaching via problem solving takes place.

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Fig. 4.9  “An upside-­ down” drawing for the One-Fifth Problem

4.4.6  Follow-Up In closing this section and as an apology for breaking suspense (if any) at the beginning of this section, we invite the readers to consider The Tetrahedron Problem (adapted from Pr. 46.1 in Pólya and Kilpatrick 1974). The Tetrahedron Problem In a tetrahedron (which is not necessarily regular), two opposite edges have the same length of 60 cm, and they are perpendicular to each other. Moreover, they are each perpendicular to a line of length 48  cm which joins their midpoints. Find the volume of the tetrahedron, and prove your answer. The reason for which we bring this problem is that it has a brief but difficult to see solution based on an upside-down visualization, just as in the Five Triangles Problem. Specifically, we invite interested readers to visualize, with the help of Fig. 4.10, the following solution idea. Image a plane through one edge of length 60 cm and the perpendicular of length 48 cm. This plane divides the tetrahedron into two congruent tetrahedra having the common base and equal heights of 30 cm each.

R: Well, I’ve tried. It is difficult. What about you? B: I did it, but also with some difficulty. The formula for the volume of a pyramid, V = 13 S ⋅ H (where S is area of the base and H is height) entered my reasoning all the time, in the same way as the formula S = 12 b ⋅ h entered the above lesson. And I intuitively considered ABC as the base of the pyramid (Fig.  4.10). However, I was familiar with a particularly bright student, she appears as Marilyn in Chap. 6, who was able to visualize (actually, invent!) the above idea almost effortlessly in front of my eyes. R: Well, let’s not be jealous.

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Fig. 4.10 Tetrahedron Problem

4.5  Launch – Intervene – Re-launch – Explore – Relate 4.5.1  Background The story4 presented in this section is based on events that occurred during two consecutive 90-min lessons of the Technion course “Selected mathematical problems.” Eight pre-service secondary mathematics teachers participated. All of them had experience in deductive proving, and none of them had systematically studied 3-D geometry. One of us (Boris) was the lecturer of the course, and one of us (Rina) was a participating observer. The lesson design has been rooted in our disturbances of both a mathematical and a pedagogical nature. As problem solvers and learners of mathematics, we were curious for some time of how to prove that the Penrose tribar (see Fig. 4.11), one of the most famous impossible objects5, is indeed impossible. As mathematics educators, we were keen to experiment with formats of problem-solving instruction so that some of our mathematical disturbances would also become disturbing to our students. As mathematics education researchers, we were interested in finding ways to intellectually necessitate for students, via problem solving, such central mathematical activities as defining, formalizing, and proving. Having all these considerations in mind, it should not be surprising that the lesson design turned to be sophisticated. It should also not be surprising that the reality turned to be even more sophisticated. In the exposition below, we contrast our plans and actual events and attempt to analyze the differences.

 The story is a re-worked version of a case study presented in Koichu (2012).  Wikipedia refers to an impossible object as a type of optical illusion consisting of a 2-D figure which is instantly and subconsciously interpreted by the visual system as representing a projection of a 3-D object, although it is not actually possible for such an object to exist, at least not in the form interpreted by the visual system. 4 5

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Let us consider two pictures. The picture on the left-side represents the tribar, created by Roger Penrose in 1954, under the influence of Escher’s drawings. The right-side picture is a photo of a sculpture of the Penrose tribar standing in East Perth, Western Australia, which was created by Bryan McKay in 1999.

Using a 1-10 scale, express your opinion as to whether the Penrose tribar can indeed be made as a 3-D sculpture or there is some “trick” involved. “1” means “I am certain that the Penrose tribar can be made”, and “10” means the opposite. Fig. 4.11  Two-Picture task

4.5.2  Launch The story has been launched by the Two-Picture Task presented in Fig. 4.11. In response, one student chose “1,” two students “10,” one student “9,” and four students chose intermediate responses, like “5,” “6,” or “7.” The students who chose “1” or “10” were given an opportunity to persuade the rest of the class in their ­rightness. Below are ideas expressed by Ahmed, who chose “10,” and Ronnie, who chose “1.” Ahmed: The sculpture misses what we clearly see at the picture, namely, interior and exterior parts. If the sculpture would be painted in three colors, as the picture, we would see that it is impossible. Ronnie: I think that it is possible to make, but not in the way we think about it… The class found these explanations too vague and no one changed his or her opinion. Moreover, three students, who had been confident in their decisions at the beginning of the discussion, now looked unsatisfied by their own explanations. Overall, the launching stage developed exactly as it had been planned: consideration of two pictures converged to disturbance, where some students felt the need for ascertaining, and others felt the need for persuading the rest. At this point, Ahmed made a prominent suggestion: Ahmed: We should define the problem.

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4.5.3  Intervene (Formalize) B: I was so happy to hear Ahmed! His suggestion was just in time and in line with what I wanted to achieve, that is, evoke in students the need to define the problem by mathematical means, out of necessity to ascertain or persuade. R: It was a great moment indeed, but I recall that the students perceived YOU as someone who must do the defining work. B: Indeed. I was forced to heavily intervene. And I should confess, when I tried the same lesson with another group of students, no one entered the Ahmed’s shoes, so I had no choice but to initiate the defining activity myself.

This is what happened next. Echoing the Ahmed’s suggestion, the students were asked to formulate and agree upon a list of properties of the Penrose tribar that could serve as a definition of the object. The request surprised some of the students, who expected that the definition should come from an authority. Ralph asked: Ralph: How can we define it? Should we vote? How can something be defined in mathematics just by an agreement among us? Boris: How do you think a definition of, say, a square, was formulated? This question triggered a discussion, which led the class to the conclusion that all definitions in mathematics are, in words of Ralph, “a sort of agreements among people.” We introduced notation (see Fig.  4.12), and then the students were split into small groups in order to discuss properties of the tribar that can serve as a basis for its mathematical definition. When in 15 min the class reconvened, it became clear that the students visualized the drawing of the Penrose tribar in very different ways. Or, in Laborde’s (1993, 1995) terms, the same drawing of the Penrose tribar was associated with different figures for the students. For example, Ronnie opposed Ahmed’s suggestion to incorporate into the definition a condition a ∥ b ∥ c ∥ d, that is, that the segments labeled as a, b, c, and d are parallel (see Fig. 4.12).

Fig. 4.12  A drawing of the Penrose tribar with notation

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She argued that though line d looked parallel to a, b,and c, it was probably not parallel to them. Ahmed saw the opposite, but how they could make each other see what each of them saw? There was also no agreement among the students regarding whether the segments looking as equal on the picture (e.g., d, n, and h) were “really” equal. Another disagreement concerned the angles on the picture, namely, whether ∠(e, b) = 60° or ∠(e, b) = 90°. The agreement among all eight students was eventually achieved about the following statements only: 1. The Penrose tribar looks like a 3-D object. 2. The close parts of the tribar of different shades are plane figures, and 3. a ∥ b ∥ c; e ∥ f ∥ g; k ∥ l ∥ m. It was planned (or rather naively hoped) that the students would eventually arrive at the “right” definition, that is, the definition that would later work well in proving that the Penrose tribar cannot be made as a 3-D object. The intended definition, with reference to Fig. 4.12, is as follows: 1. Penrose tribar is shaped by three sets of parallel segments, namely, a ∥ b ∥ c ∥ d; e ∥ f ∥ g ∥ h, and k ∥ l ∥ m ∥ n. 2. The parts of the tribar of different shades are plane figures, which belong to three different planes (denoted as α, β, and γ). The students came quite close to this definition. However, they lacked shared criteria for deciding whose vision of the Penrose triangle should be accepted by all and why. Notably, a similar phenomenon was described in Zaslavsky and Shir (2005), who studied how a small group of students decided whether particular statements can be accepted as definitions of mathematical objects. That situation was devoid of the actual use of the definitions, so the agreements were usually reached based on communicative and social reasons. In our case, a definition of the tribar should have been constructed for further mathematical use. The discussion was stuck, and the next intervention was in order. Boris: It looks like we cannot fully agree upon the definition. But is it necessary at this stage? Let us do not forget that we wish to agree upon the definition in order to examine whether an object satisfying it exists. Which definition is really needed for this purpose? We don’t know yet. This may become clearer when you start proving. So, please start proving based on the properties agreed so far, and modify the definition if you feel you need to.

4.5.4  Detour: Types and Effects of Teacher Interventions As the reader might have noted from the story so far, the teacher attempted to intervene only out of necessity (i.e., when the class was stuck and the discussion would die without a new input), avoided comments that could be interpreted by the students as endorsing a particular idea and attempted to get the students unstuck with-

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out straightforwardly providing them with relevant information. The rationale for this intervention strategy was our wish not to deprive the students from the fun of autonomous learning. The same can be said also about the intervention during the solution of the Five Triangles Problem in the previous section. More generally speaking, our instructional strategy has been inspired by Brousseau’s (1997) idea of teaching by means of problematic situations that permit the students to construct mathematical knowledge autonomously, where “the knowledge will appear as the optimal and discoverable solution to the problems posed” (p. 22). Furthermore, Brousseau and Gibel (2005) argued that “[ideally], the teacher must propose problems (whose solutions require knowledge that has not been institutionalized in class yet) to be solved completely autonomously by the students” (p. 23). In practice, asserts Brousseau (1997), classroom problem-solving situations that allow students to construct new knowledge without the teacher's mediating actions barely exist. The teacher’s interventions are often inevitable, for instance, when students get stuck when coping with a given problem or are uncertain about their ability to solve it. Brousseau and Gibel (2005) pointed out three types of teacher interventions in a situation where a problem is open for students (but not for the teacher, who knows its solution): Type 1: to bring into play the relevant knowledge students had been taught Type 2: to point out information as given in the text of the problem Type 3: to refer to conditions which are not included in the presumed students’ knowledge and which cannot be logically deduced from the text of the problem Furthermore, Brousseau and Gibel (2005) pointed out that autonomous learning through problem solving is highly sensitive to the types of teacher interventions. In particular, they showed that Type 3 interventions make the students’ autonomous learning impossible because “[i]n the third option, the student can accept the solution only upon his trust in the teacher’s authority” (p. 20). In these terms, the teacher in the above story (1) avoided Type 3 interventions, (2) used one Type 1 intervention (when asking the students about defining a square), and (3) used one Type 2 intervention (when reminding the students about the overall purpose of defining the tribar). As it will be evident shortly, the story will arrive at a point where a Type 4 intervention should have been implemented. We refer to Type 4 intervention as a teacher’s attempt to get the solution process unstuck by offering students an opportunity to stop thinking for a while on a target problem, consider an easier problem having an analogical connection with the target one, and after solving the auxiliary problem, to come back to the target problem while being equipped with the autonomously developed new knowledge. The reader may notice that this intervention is aligned with one of Pólya’s (1945/1973) problem-solving heuristics: when stuck, recall a simpler but similar problem. The obvious difference, of course, is that Pólya addresses problem solvers in the middle of the struggle, and a Type 4 intervention comes from a person (a teacher) who already knows the solution and thus knows in advance how the recalled problem can help. Elaborated examples follow. For now, the reader can think of the first intervention in the story on solving the Five Triangles Problem as Type 4 intervention.

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4.5.5  Re-launch Following the above prompt, the students returned to the work in small groups but showed no progress. In retrospect, the difficulty of this stage for the students could have been attributed to three kinds of uncertainty that they experienced at once. The first one was regarding the existence of the Penrose tribar, the second one – regarding the choice of the properties of the Penrose tribar that could be taken for granted, that is, to serve as a shared visualization, and the third one – regarding how to get started proving either that the tribar can exist or not. Note that the intended proof included elements that had not yet appeared in the discussion: (1) the decision to invest effort in proving that the Penrose tribar is indeed an impossible object, (2) the idea to look for a proof by contradiction, and (3) the idea of how to construct a contradiction. For the sake of coherency of the story, let us break the suspense and state that the contradiction can be constructed based on an axiom of 3-D geometry stating that the intersection of two planes is a straight line. As mentioned, the students have never systematically studied 3-D geometry, and for this reason the teacher encountered a dilemma that could be formulated as a paraphrase of learning paradox (e.g., Brousseau 1997; Sfard 1998): How could the students autonomously understand that an axiom about the intersection of two planes could help them if they did not know about its existence? For the reasons discussed in the detour above, the teacher did not want to use a Type 3 intervention, that is, to straightforwardly introduce the axiom. He offered instead a Type 4 intervention, as follows: Boris: Let us leave for a while the Penrose tribar and consider another object. The drawing on the screen (Fig. 4.13) consists of a tetrahedron ABCD and its plane section MNPK. In your opinion, does this drawing represent a possible or an impossible object?

Fig. 4.13  An auxiliary impossible

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All the students quickly observed that “something is wrong in the picture.” Ronnie was the most specific among all and uttered that point N was problematic. She explained: Ronnie: PNM belongs to a face of a tetrahedron and to the cutting plane. But if two planes intersect, it should be a straight line! The instructor re-voiced her explanation as follows: the object is impossible because it violates the fact that two planes intersect by a straight line. He then informed the students that this fact has a status of a solid geometry axiom. Thus, the axiom of the intersection of two planes entered the stage as a tool for solving an auxiliary problem:

R: So, eventually you informed the students. Isn’t it a Type 3 intervention? B: Perhaps, but somehow it felt differently. I informed the students only about the status of the fact that they put into play themselves, in response to the auxiliary problem. I think that it was Type 4 intervention.

The lesson time was over, and the teacher asked the students to keep thinking about possible proofs of the impossibility of the Penrose tribar until the lesson next week. This was the main re-launch, namely, changing the open question (exists or does not exist?) into a proving question (how to prove that the tribar does not exist by working from the definition of the tribar that suits you).

4.5.6  Explore Two students, Iris and Ronnie, announced their success at the beginning of the next lesson. Iris and Ronnie first presented their proofs in small groups working in parallel, and then the proofs were brought to a whole-class discussion. The proofs appeared to be similar in two ways. First, they both relied on definitions of the Penrose tribar which were fully compatible with the intended definition (see above). This meant, in particular, that both students decided to include the property “all lines that look parallel on the Penrose tribar’s drawing are parallel” into the definition. To recall, this property was not agreed upon a week ago, but Iris and Ronnie managed to convince the rest of the students by simply pointing out that without this property their proofs would not work. Thus, Iris and Ronnie succeeded to construct a proof-generated definition (Lakatos 1976), which by itself was a prominent accomplishment (cf. Ouvrier-­ Buffet 2006, for elaboration on the difficulties in creating situations with the potential to lead students to constructing proof-generated definitions). Our point here is that the constructed definition became for the students a piece of knowledge, whose emergence was stipulated not only by communicative or social reasons (cf. Zaslavsky and Shir 2005) but also by a reason of a mathematical nature.

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Fig. 4.14  Drawings supporting Iris’ and Ronnie’s proofs. (a) Iris. (b) Ronnie

Second, both proofs included a final step of constructing a contradiction with an axiom of the intersection of two planes. The difference between the proofs resided in different auxiliary constructions (see Fig. 4.14). Specifically, Iris constructed line AB, when A = b ∩ l and B = c ∩ k (Fig. 4.14a). In this case, the contradiction with the axiom was that planes α and γ intersect by f and AB. Ronnie constructed point A (Fig. 4.14b) and, importantly, justified its existence. In her proof, the contradiction with the axiom was that planes α and γ intersect by f and A. In a slightly polished form, her proof was as follows: Let A be an intersection of the straight lines b and l: A = b ∩ l (see Fig. 4.14b). Point A exists because straight lines a, b and l belong to the same plane (condition 2 of the definition), a ∥ b (condition 1) and lines a and l intersect. In addition, b ⊂ α, l ⊂ γ, and then A ∈ α and A ∈ γ. Consider now planes α and γ. They intersect by line f and point A, which does not belong to f. Thus, the following solid geometry axiom is violated: “An intersection of two planes is a straight line.” We obtained a contradiction because we assumed that α, β, and γ are different planes (conditions 2). Consequently, the Penrose triangle cannot be a 3-D object, QED. In fact, Ronnie’s proof was fully compatible with the intended proof that we (Boris and Rina) had in mind. In the spirit of preserving autonomous learning, the teacher did not reveal this fact but asked the rest of the students to express their opinions.

4.5.7  Relate Several students noted that Ronnie’s auxiliary construction was the most parsimonious among the two. In addition, Ronnie observed that Iris’ auxiliary construction (Fig. 4.14a) involves an implicit assumption about the existence of point B, which

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belonged to the “invisible” part of the drawing. She wondered whether this assumption must be incorporated in the definition. Ronnie’s question was difficult to understand for some students, and the teacher decided to unstack the discussion by an additional Type 4 intervention. Namely, he introduced an additional auxiliary problem. He drew on the blackboard a sketch presented on Fig. 4.15 and asked the class: Boris: Please leave the Iris drawing for a while and think, what may point M represent? Students who perceived the drawing as a plane figure suggested that point M may represent intersection of NP and KL, and those who perceived the drawing as a drawing of a 3-D object, said that point M “does not exist” or may belong to either NP or KL. The students then concluded that it was impossible to answer the question without additional (verbal) information regarding the drawing. More importantly, the students realized the connection between the auxiliary problem and the question asked by Ronnie. As a result, the class decided that the proper use of Iris’s auxiliary construction indeed required incorporating into the definition of the Penrose tribar a condition that would ensure that point B exists (Fig.  4.14a). Our point is that Type 4 intervention allowed the students to get unstuck, quite autonomously, by relating different pieces of knowledge and different though analogous problems. By the end of the lesson, the teacher came back to the question about whether the Penrose tribar can be made as a 3-D object. All the students except Ronnie confidently stated that the Penrose tribar cannot be made “because we've just proved it.” Ronnie (definitely our favorite character for this chapter!) said: Ronnie: It is still difficult for me to decide. Perhaps, it can be made, but not by our definition… Though it is beyond the scope of this chapter, it is appropriate to mention here that Ronnie’s intuition can be mathematically backed up (e.g., Sugihara 2011).

Fig. 4.15  What may point M represent?

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4.6  Concluding Reflections We began this chapter by recalling the theoretically blurred but practically clear distinction between teaching for problem solving vs. teaching via problem solving. In our view, the presented scenarios demonstrate to what extent the two approaches can be interlaced. Every time when a scenario included a well-formulated problem and a realistic pathway toward its successful solution, one can argue that teaching (and probably learning) for problem solving has taken place. Every time when a scenario included a re-launch of the initial question or Type 4 intervention, one can argue that teaching (and probably learning) via problem solving has taken place. The common aspect in all the scenarios, other than our initial disturbances as the departure points, is the presence of the relate stage, which, apparently, can be thought of as a locus of learning either for or via problem solving. Also, three out of four scenarios included the re-launch stage. In the above stories, re-launch has been referred to in two ways. First, re-launch is a re-­consideration of the initial question after formalizing it, as in the story of the Penrose tribar. Second, it is a consideration of a new question which is intrinsically related to the initial one. The related question can emerge as a follow up to the initial question (e.g., the Double Perfect Square Task is related to the Miscopying Task in this way). Alternatively, the related question can inform the solution of the initial question (e.g., the problem on equal areas of “small” triangles formed by a median is related to the Five Triangles Problem in this way). The launch-relaunch dyad reminded us about another dyad which is broadly considered in the mathematics education community, namely, the distinction between problem solving and problem posing. Some scholars (e.g., Kilpatrick 1987) discuss these two activities as complementary to each other, whereas some other scholars (e.g., Silver et al. 1996) focus on their inherent connections. Our re-launch in the first meaning (a more difficult question follows a less difficult question) can be seen as sequencing problem-solving and problem-posing activities as counterparts. Our re-launch in the second meaning (a less difficult question enters the solution pathway to a more difficult question) is closer to problem posing as problem reformulation in due course of problem solving, as Silver (1994) has pointed out. In both cases, we feel that presence of the re-launch stage makes the presented scenarios vivid and to some extent mathematically genuine. In closing, we would like to invite the readers to join us in constructing, trying in practice and sharing fairly sophisticated scenarios of problem-solving instruction, moving beyond the aforementioned launch-and-explore and demonstrate-and-­ apply distinction. These scenarios can tentatively be thought of as different ways of combining the stages illustrated above, namely, launching, exploring, getting stuck, intervening with care to preserving students’ autonomous learning, re-launching in different meanings, re-exploring, and relating. So far, we believe that this list is rich enough in order to embrace a considerable portion of what can be done with problem-­solving instruction in school reality.

References

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References Arcavi, A. (2003). The role of visual representations in the learning of mathematics. Educational Studies in Mathematics, 52(3), 215–241. Brousseau, G., & Gibel, P. (2005). Didactical handling of students’ reasoning processes in problem solving situations. Educational Studies in Mathematics, 59(1/3), 13–58. Brousseau, G. (1997). Theory of didactical situations in mathematics (N. Balacheff, M. Cooper, R. Sutherland, & V. Warfield: trans. and eds.). Dordrecht: Kluwer. Bruner, J. S. (1960). The process of education. Cambridge, MA: Harvard University Press. Chazzan, D. (1999). On teachers’ mathematical knowledge and student exploration: A personal story about teaching a technologically supported approach to school algebra. International Journal of Computers for Mathematical Learning, 4(2-3), 121–149. Clements, D.  H. (2003). Teaching and learning geometry. In J.  Kilpatrick, W.  G. Martin, & D. Schifter (Eds.), A research companion to principles and standards for school mathematics (pp. 151–178). Reston: NCTM. Connor, J., Moss, L., & Grover, B. (2007). Student evaluation of mathematical statements using dynamic geometry software. International Journal of Mathematical Education in Science and Technology, 38(1), 55–63. Edwards, L.  D. (1997). Exploring the territory before proof: Students’ generalizations in a computer microworld for transformation geometry. International Journal of Computers for Mathematical Learning, 2(3), 187–215. Edwards, T. G. (1996). Exploring quadratic functions. The Mathematics Teacher, 89(2), 144–146. Kilpatrick, J. (1987). Problem formulating: Where do good problems come from? In A.  H. Schoenfeld (Ed.), Cognitive science and mathematics education (pp.  123–147). Hillsdale: Routledge. Koichu, B. (2012). Enhancing an intellectual need for defining and proving: A case of impossible objects. For the Learning of Mathematics, 32(1), 2–7. Koichu, B., Berman, A., & Moore, M. (2007). Heuristic literacy development and its relation to mathematical achievements of middle school students. Instructional Science, 35(2), 99–139. Laborde, C. (1993). The computer as part of the learning environment: The case of geometry. In C. Keitel & K. Ruthven (Eds.), Learning from computers: Mathematics education and technology (pp. 48–67). Berlin: Springer. Laborde, C. (1995). Designing tasks for learning geometry in a computer-based environment. In L. Burton & B. Jaworski (Eds.), Technology in mathematics teaching – a bridge between teaching and learning (pp. 35–68). London: Chartwell-Bratt. Lachmy, R., & Koichu, B. (2014). The interplay of empirical and deductive reasoning in proving "if" and "only if" statements in a Dynamic Geometry environment. Journal of Mathematical Behavior, 36, 150–165. Lakatos, I. (1976). Proof and refutation: The logic of mathematical discovery. Cambridge: Cambridge University Press. Lehoczky, S., & Rusczyk, R. (2004). The art of problem solving. Volume 1: The basics. Alpine: AoPS Incorporated. Lester, F. (2013). Thoughts about research on mathematical problem-solving instruction. The Mathematics Enthusiast, 10(1-2), 245–278. Marrades, R., & Gutiérrez, Á. (2000). Proofs produced by secondary school students learning geometry in a dynamic computer environment. Educational Studies in Mathematics, 44(1-2), 87–125. National Council of Teachers of Mathematics (NCTM). (1980). An agenda for action: Recommendations for school mathematics of the 1980s. Reston: Author. National Council of Teachers of Mathematics (NCTM). (1989). Curriculum and evaluation standards for school mathematics. Reston: Author. Ouvrier-Buffet, C. (2006). Exploring mathematical definition construction processes. Educational Studies in Mathematics, 63(3), 259–282.

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Pólya, G. (1945/1973). How to solve it. Princeton: Princeton University Press. Pólya, G., & Kilpatrick, J. (1974). The Stanford mathematics problem book with hints and solutions. New York: Teachers College Press. Schoenfeld, A. (1985). Mathematical problem solving. New York: Academic Press. Schoenfeld, A. (1992). Learning to think mathematically: Problem solving, metacognition, and sense making in mathematics. In D. A. Grouws (Ed.), Handbook of research on mathe- matics teaching and learning (pp. 334–370). New York: Macmillan. Schoenfeld, A. (2018). Learning from Malcolm. Mathematics in School, 47(2), 15–26. Schroeder, T. L., & Lester, F. K. (1989). Developing understanding in mathematics via problem solving. In P. R. Trafton (Ed.), New directions for elementary school mathematics (pp. 31–56). Reston: NCTM. Selden, A., Selden, J., Hauk, S., & Mason, A. (2000). Why can’t calculus students access their knowledge to solve non-routine problems? In A. Schoenfeld, J. Kaput, & E. Dubinsky (Eds.), Research in Collegiate Mathematics Education, 4, 128–153. Sfard, A. (1998). On two metaphors for learning and the danger of choosing just one. Educational Researcher, 27(2), 4–13. Silver, E. A. (1994). On mathematical problem posing. For the Learning of Mathematics, 14(1), 19–28. Silver, E. A., Mamona–Downs, J., Leung, S., & Kenny, P. A. (1996). Posing mathematical problems: an exploratory study. Journal for Research in Mathematics Education, 27(3), 293–309. Stein, M.  K., Engle, R.  A., Smith, M.  S., & Hughes, E.  K. (2008). Orchestrating productive mathematical discussions: Five practices for helping teachers move beyond show and tell. Mathematical Thinking and Learning, 10(4), 313–340. Sugihara, K. (2011). Spatial realization of Escher’s impossible world. Asia Pacific Mathematics Newsletter, 1, 1–5. Van Essen, G., & Hamaker, C. (1990). Using self-generated drawings to solve arithmetic word problems. Journal of Educational Research, 83(6), 301–312. Zaslavsky, O., & Shir, K. (2005). Students’ conceptions of a mathematical definition. Journal for Research in Mathematics Education, 36(4), 317–346.

Chapter 5

Encounters with Cardano’s Method

5.1  Introduction Many people, including ourselves, have been intrigued when first introduced to a story of inventing a formula for solving cubic equations. We invite the reader to recall a sixteenth-century story, in which characters possessing sonorous Italian names – del Ferro, Tartaglia, Cardano, and Bombelli – were on the stage. Different authors have depicted this story in relation to inventing complex numbers and legitimizing negative numbers (e.g., Guilbeaur 1930; Kenney 1989; Feldmann 1961). The story is an important milestone in the history of mathematics, and many teachers and mathematics educators like bringing it to a secondary school classroom or a course for mathematics teachers. B: I heard the story for the first time as a schoolboy, at one of the mathematics lessons taught by my teacher Boris G. Orach. Mr. Orach was at his seventh decade at that time. He was a master storyteller and told the story so vividly that we, 14-year-­olds, felt as if he had witnessed the mathematics duel between Antonio Fiore and Niccolo Tartaglia. R: I see that this lesson is still in your memory. B: Part of it. I guess that I was more impressed by a human drama and the idea of a duel than by the mathematical part of the story. As I understand now, Mr. Orach used the version in which Cardano was an undisputable negative character and Tartaglia was an undisputable hero. Tartaglia won the mathematical duel with Fiore by the virtue of pure talent. Then Cardano convinced Tartaglia to reveal to him the solution to cubic equations under the promise to keep it in secret until Tartaglia would publish his book. And then Cardano broke the oath and published the solution in his own book, Ars Magna. This nearly destroyed Tartaglia’s life.

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R: As we know, this is one version. Another version is that Cardano waited for Tartaglia to publish his book for years and decided to include the method of solving cubic equations in Ars Magna only when he discovered that the priority of discovery should have been credited to another mathematician, Scipione del Ferro. ­Anyway, Cardano gave in his book an equal credit to del Ferro and Tartaglia for independently discovering the method. B: Well, let’s leave this debate to historians. When I had become a teacher myself, I began wondering why the mathematics of the story almost completely disappeared from my memory. I only remembered that there existed a formula known as “Cardano formula” and that it was so complicated and technical that no one wanted to use it. This disturbed me, and I decided that one day I would devise a different experience with the story for my students. I got such a chance when teaching a course “Educational aspects in history of mathematics.” The story below is partly based on a series of real classroom situations and tasks at that course and partly on our hypothetical reasoning about what might have occurred if we could devote more time to cubic equations. The goal of this chapter is to present various learning opportunities, tasks, and explorations that stem from our encounters with the Cardano formula and with the great story of its invention.

5.2  Background The course “Educational aspects in history of mathematics” is a one-semester elective course for mathematics education graduate and advanced undergraduate students at the Technion. Usually 15–25 students are enrolled in this course. The course focuses on ways of teaching high school mathematics based on historical materials. Each 90-min lesson of the course is problem-solving-based. The students are required to read materials devoted to the history of a particular mathematical topic toward each lesson. In particular, historical methods of solving quadratic equations were discussed at the beginning of the course, including the method that stems from Euclid’s Propositions 5 and 6 in Book II.1  In a canonic translation by Heath (Euclides 2002), Proposition 5 Book II states:

1

If a straight line be cut into equal and unequal segments, the rectangle contained by the unequal segments of the whole together with the square on the straight line between the points of section is equal to the square of the half. Proposition 6 Book II states: If a straight line be bisected and a straight line be added to it in a straight line, the rectangle contained by the whole with the added straight line and the added straight line together with the square on the half is equal to the square on the straight line made up of the half and the added straight line. These two propositions underlie what is known nowadays as a method of solving quadratic equation by completing the square.

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Prior to the two-lesson module on cubic equations, the students were assigned to read two papers: the paper by Merino (2006) that overviews the history of inventing complex numbers and the paper by Ekert (2008) that focuses on Cardano’s life and achievements. Two additional papers were recommended as complementary reading: the paper by Feldmann (1961), which was devoted to the Cardano-Tartaglia dispute, and Kenney’s (1989) paper, which included a geometric demonstration for a quadratic equation having imaginary solutions, as based on Cardano’s Ars Magna. Overall, these four papers served as an introduction to different versions and aspects of the story.

5.3  Collaboratively Re-telling the Story During the lesson, it quickly became evident that some of the students got lost in the details and different points of view. To clarify the confusion, some shared understanding of the main events of the story should have been constructed. This is what we did in order to achieve this goal. The students had access to the course website from their laptops and smartphones. Thus, it was technologically possible to offer the students the following task (Task 1). The task was given with 15 min allotted for work in groups of four or five. Task 1: Collaboratively Re-telling the Story Collaboratively re-create the story of discovering the solution to cubic equations in as much detail as possible. A shared document is opened for your group at the course website. Every group member should add in his or her turn a written sentence to the document. Begin from the following sentences: 1. Scipione del Ferro found a method of solving equations of the type x3 + px = q but kept his method a secret. 2. Del Ferro revealed the secret to his students, Antonio Fiore and Annibale della Nave, at his death. Continue from here.

Four versions of the story were created and became available at the course website. Well, the versions were far from being comprehensive or consistent. We worked hard on comparing and merging the versions. Below is our best attempt to produce an account of the story based on the mentioned above two papers and the students’ contributions (see Fig. 5.1).

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Fig. 5.1  An account of the story of cubic equations, from students’ re-telling

5.4 The Initial Engagement and Structuring the Method

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5.4  The Initial Engagement and Structuring the Method The story set a good frame for delving into mathematics. The students were ready to learn what all the fuss was about. Some even tried to work out the Cardano method at home. R: Really? B: Well, this was a course for graduate and advanced undergraduate students. Not all, but some really did. We decided that it would be too complicated to follow the solution to cubics while adhering to the original medieval notation. So, we stopped talking about the historical papers for a while and used a file named “The method of Cardano in modern terms.” The file was prepared by one of the course participants on the basis of several Internet resources. He wrote the file voluntarily at home in order to help himself while reading the papers (see Fig. 5.2). A quick look at the file can just reinforce the impression “the story is great, but the method is too messy.” Furthermore, reading the file in a line-by-line mode can lead to the same impression. Being aware of this, we approached the file by means of Task 2. The task was inspired by Leron’s (1983) approach to structuring proofs. Leron suggested that long proofs written in a linear line-by-­line form become more comprehensible when they are described or structured in levels. The top-structure level gives a general outline of the main line of the proof but without detail, and the second level elaborates on the details omitted at the top level. Task 2: Structuring the Proof Work in groups of three or four for about 15 min. Your task is to identify a small number of the pivotal solution moves, that is, the moves that appear to you to be the bearers of the main line of the method of Cardano.

When the fog of long algebraic transformations disappeared, the main line of the proof was structured by means of only two pivotal moves. The first one consists of a (line 1) and leads to a reduced cubic equation a substitution of y for x − 3 x3 − px + q = 0 (line 6). The second one looks like a clever trick: an unknown x is substituted by a sum of two unknowns, x = u + v, and not by any u and v, but by those that fulfill an additional condition, 3uv =  − p (line 7). (Wow! What a genius one should be in order to invent such a move!) The move opens a way for ­representing a reduced cubic equation as a system of two equations, in which the sum and the product of u3 and v3 are given (lines 11–13). The resulting system of equations is not difficult to solve. The solution can be obtained either geometrically (as Euclid did or Cardano demonstrated; see Kenney 1989) or algebraically, as we do nowadays in

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Fig. 5.2  The method of Cardano in modern terms

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middle school (that is, by transforming the system into a quadratic equation with one unknown, w = u3v3 in our case; see line 15 in Fig. 5.2). Voilà! Voilà?

5.5  Detour: Reverse Heuristic Engineering To us, the two pivotal moves of the method looked mysterious and called for explanation. We share below our thoughts about how these moves could have been invented. Do we intend to be speculative? We definitely do! In defense, let us recall that we do not engage with the solution of cubics as historians of mathematics, but as mathematics educators and learners who are trying to make sense of a difficult topic. As a matter of fact, we invite the reader to join us in an activity that we call reverse heuristic engineering. We mean by this term an observer’s effort to make sense of a given solution to a mathematical problem by building a plausible heuristic model of how the solution idea – which initially appears a magic trick – could have been constructed by its inventor. In traditional reverse engineering, it is of little importance whether the process of manufacturing a particular product suggested by a reverse engineering specialist is exactly the same as the (unknown or lost) original process. What is of importance is whether the specialist can devise a ready-to-use technology of producing a like product after closely examining the original one. Stretching the analogy, there is little chance that reverse heuristic engineering applied to a given solution would result in a reliable account of the first solver’s thinking. Nonetheless, we find this activity useful as an effort to demystify the emergence of non-intuitive solutions and make sense of them. Capitalizing upon Hewitt’s (1999) terminology, having a plausible model of how a non-intuitive idea could have being invented helps us to transfer the idea, from a pool of arbitrary ideas to be memorized to a pool of ideas that are necessitated by some reasoning and thus are more comprehensible. The rationale for using reverse heuristic engineering can be constructed based on several precedents in mathematics education research and practice. One precedent is known as virtual monologue. It is offered by Uri Leron (Leron and Hazzan 1997; Ejersbo and Leron 2005) as a useful tool in mathematics education in general and in mathematics teacher professional development in particular. To recall, virtual monologue is a tool that engages mathematics education researchers or mathematics teachers with someone’s solution to a mathematical problem by creating a text of a monologic format, giving a solver’s voice in first-person narration. The monologue is about what might be going on in the solver’s mind during problem solving. Leron and Hazzan (1997) argue that the strength of the tool is in enhancing communication and reflection. Our own past work on script writing and virtual duoethnography (Koichu and Zazkis 2013; Zazkis and Koichu 2015) can be indicated as another relevant precedent. As we have argued in Chapters 2 and 3, creating scripts of fictional interactions on proving provides mathematics teachers an avenue for exposing their own struggles and expressing empathy with students by introducing their potential struggles.

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In addition, Abraham Arcavi (oral communication at a workshop, 2011) observed teacher-student interactions in a Japanese mathematics class, in which students were given a chance to act as reverse heuristic engineers. In one episode, a master teacher responded to one student’s (mistaken) solution by asking another student to guess how the first student might have been thinking in order to produce that solution. Arcavi featured that teacher’s response as particularly meaningful and productive. We now come back to the story of the cubics. In particular, we focus on pivotal ideas in developing the formula and use reverse heuristic engineering to describe how these ideas could have emerged.

5.6  How Could Such a Clever Solution Be Found? As mentioned, the pivotal solution moves in the above file (Fig. 5.2) looked mysterious to us and called for explanation. The next several minutes of the lesson were shaped by the following request: Task 3: Put Yourself in Tartaglia’s Shoes Imagine that you are Tartaglia getting prepared for the duel with Fiore. Suggest how the two pivotal solution moves could have come to your mind. For the sake of simplicity, assume that Tartaglia, who thought on algebraic matters geometrically, could also use negative numbers and modern algebraic notation.

Below is the result of our attempt to contemplate the solution to cubics as reverse heuristic engineers. Our attempt to reason toward the first pivotal move (elimination of a quadratic term of an equation y3 + ay2 + by + c =0 by means of the substitution a y = x − ; see line 1 Fig. 5.2) was not particularly successful. It was easy to suggest 3 that the underlying reasoning was geometric (what else could it be?) and related to the “completing the cube” (by analogy with “completing the square”) process. However, our algebra-poisoned minds did not find the idea of reducing cubics by geometric manipulations with a solid cube to be intuitively appealing2. Then we recalled that Tartaglia might just have known the trick. According to Merino (2006), a way of reducing cubics was first described in two anonymous Italian manuscripts written at the end of the fourteenth century. Some other authors3 suggested that the substitution was due to Cardano who might have discovered it under the influence 2  The process of reducing cubic equations represented on solid cubes is demonstrated in Henning (1972). 3  See, for example, http://www.ms.uky.edu/~sohum/ma330/files/eqns_2.pdf

5.6 How Could Such a Clever Solution Be Found?

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of al-Khayyam and al-Tusi. We also recalled that Tartaglia and Fiore exchanged only the reduced equations at the duel. That is, whether they knew the trick or not, they did not need to use it. In recap, the historical allusions were interesting, but we did not find a way to use them in order to address the question “how the trick could have been invented” in an intuitively appealing to us way. That is, the heuristic considerations underlying the process of reducing cubics, whoever invented it and whenever it was invented, remained a mystery to us. We come back to this issue later in this chapter when equipped with more conceptual tools. As to the second move (“Let us look for x in the form x = u + v when 3uv = − p”; see line 7 Fig. 5.2), our guess was as follows. The idea to look for an unknown as a sum of two unknowns might be counterintuitive to us4 but could have been quite a cliché for sixteenth-century mathematicians. This is because the idea is perfectly compatible with Euclid’s method for solving quadratic equations by “completing the square” (see Euclid’s Propositions 5 and 6 Book II, Footnote 1). The guess is further supported by the fact that Cardano’s Ars Magna contains several Euclid-like demonstrations of solutions to quadratic equations (Kenney 1989). Accordingly, we boldly attributed to the ancient and medieval solvers of quadratic and cubic equations the thought “let’s divide an unknown segment into two unequal segments and see what can be done from here.” The remaining question is where the condition 3uv = − p could have come from. Again, after reading Henning (1972), we find that the historically plausible geometric interpretations are too complicated in order to be intuitively appealing to us. However a close look at lines 7–12 of the above file (Fig. 5.2) led us to a simple conjecture: the condition was introduced for the sole purpose of eliminating the term (3uv + p)x from the equation u3 + v3 + (3uv + p)x + q = 0 (line 10). Don’t you believe this explanation? Let’s see if a virtual monologue in the voice of our homemade Tartaglia will convince you. Tartaglia So, I need to solve the equation x3 + px + q = 0. How? Hmm… What would I do if these were not cubics but quadratics? I’d probably substitute x for u + v like Euclid and many other wise men have done. Well, my equation is not quadratic but let me try anyway. x 3= u + v, 3 x = ( u + v ) = u3 + v 3 + 3u 2 v + 3uv 2 = u3 + v 3 + 3uv ( u + v. ) Now I combine these two rows into x3 = u3 + v3 + 3uvx and come back to the reduced equation, x3 + px + q = 0. What do I get? OK, the equation transforms into u3 + v3 + (3uv + p)x + q = 0. What now? Hmm…One equation with three unknowns, u, v and x … Not helpful, there are too many unknowns... How can I get rid of some of them? If by any chance (3uv + p) were equal to 0, the situation would be better. Does it make sense to introduce a condition 3uv + p = 0? Actually, why not!

 By the way, this idea is widely used for solving certain types of differential equations.

4

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5  Encounters with Cardano’s Method

An analogy comes to my mind: in order to solve an equation of the type x2 + y2 = 0, one should just assert that the left side can be equal to 0 if and only if x = y = 0. This is how I would solve, say, an equation (x3 − 1)4 + (x − 1)2020 = 0.  x3 − 1 = 0 I would say that it is equivalent to the system  , and this would be it.  x −1 = 0 Well, this each-part-­equals-zero trick only works when we know that each part should be non-negative, as in my example. But what if I needed to solve another equation, say, (x3 − 1)5 + (x − 1)2020 = 0? The second addend is non-negative, but the first one can either be negative or non-­negative. There is no reason for both addends to be simultaneously equal to 0. Nevertheless, even though they are not forced to be 0, they can be 0! If by any chance there exists an x for which (x3 − 1)5 = (x − 1)2020 = 0 (and it does exist, x = 1), it would be a solution. A solution, not the solution, but it is better than nothing. So, let me try. My equation is u3 + v 3 + ( 3uv + p ) x + q = 0, and I can consider it as a two-part equation, u3 + v 3 + q  + ( 3uv + p ) x  = 0.  Each part of the above are not forced to be 0, but they can be 0, why not? In other words, Let’s say that 3uv + p = 0 and u3 + v3 + q = 0. Now what? The unknowns in the second equation are u3 and v3. And the first equation, 3uv + p = 0, includes a product of u and v … I am getting close, I am close… Actually, I have got two equations and need two unknowns… Bingo! I have uv in the second equation (3uv + p = 0) and can turn it into u3v3 . Now I have a system of two equations with two unknowns: u3 + v 3 = −q  m + n = −q u3 + v 3 = −q    3  p or  3 3 p3 p or  u v = −  mn = −  uv = − 3 27   27  3 3 for u = m and v = n. Even ancient Greeks would know how to approach it! But does it really have a solution? Let’s see… [Tartaglia moves on, completes the solution and enters the story and history]. You still don’t believe that this is what happened with Tartaglia the night before the duel with Fiore? Well, in this case we would not insist. We just note that the above exercise in reverse heuristic engineering helped us to somewhat demystify the solution.

5.7 Can the Law of Transitivity Fail?

115

5.7  Can the Law of Transitivity Fail? The Cardano method began making sense, and this pleasant feeling should have been reinforced. Obviously, we needed examples. It was natural to consider in some depth those cubics that were repeatedly mentioned in the historical papers that we had read. We decided to stop on an equation from Bombelli’s L’Algebra, x3 = 15x + 4. This equation had an obvious-by-inspection integer solution of 4 and a relatively nice solution by Cardano’s formula:

x = 3 2 + −121 + 3 2 − −121 ( check it!) .



We also knew that the breakthrough in the history of complex numbers occurred when Bombelli showed that 4 = 3 2 + −121 + 3 2 − −121 , so that “casus irreducibilis” attained meaning. We identified an opportunity here for an interesting exploration and formulated Task 4.1, which, as will be evident shortly, became a prelude to a serious conversation on several interesting issues. Task 4.1 Check if 4 = 3 2 + −121 + 3 2 − −121 .

At the first glance, the task can be approached in the same manner as many textbook identities in algebra and trigonometry are approached. Namely, begin from the identity to be proved, and transform it, by means of allowed transformations, into an identity known to be true5. The realization of this idea in our case is based on two formulas of abridged multiplication, (a  +  b)3  =  a3  +  3a2b  +  3b2a  +  b3  =  a3  +  b3  +  3ab(a  +  b) and a2 − b2 = (a − b)(a + b)(see Fig. 5.3): Looks right, doesn’t it? And perhaps it is right, though it is not entirely clear whether all the transformations are reversible… Anyway, no one suspected anything wrong at this point and we moved on. The next step was to find two additional roots of the equation. The knowledge of one root (4 in our case) opens an opportunity to find all the roots of a cubic by long division of polynomials. Dictum-factum, as Cardano might have said, and two additional solutions were quickly found:

(x

3

)

− 15 x − 4 ÷ ( x − 4 ) = x 2 + 4 x + 1



Thus, we look for solutions for the quadratic equation and find x2,3 = −2 ± 3 . 5  See Koichu and Leron (2015) for a discussion of mathematical and psychological quandaries associated with this way of proving.

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5  Encounters with Cardano’s Method

Fig. 5.3  Solution to Task 4.1

What now? On the one hand, we have three solutions for our cubic equation x3 = 15x + 4, obtained by inspection (x = 4) and long division of polynomials and solving quadratic equation ( x2,3 = −2 ± 3 ). On the other hand, we have a solution obtained by Cardano’s formula:



(x =

3

)

2 + −121 + 3 2 − −121 .

Hmm… How can we be sure that Cardano’s formula gives us the solution x = 4 and not one of the other solutions? Bombelli apparently was not aware of the existence of additional roots besides 4, so the query was worth checking. The students were given Task 4.2. Task 4.2 Check if 2 ± 3 = 3 2 + −121 + 3 2 − −121 .

The manipulations that worked so well for the root 4 were repeated for the root −2 + 3 and then for the root −2 − 3 , and – surprise, surprise – it worked again! Below is a computation for the root −2 + 3 (Fig. 5.4): Taking together the results of Tasks 4.1 and 4.2, and using the law of transitivity (if a = b and a = c, then b = c; in our case, a = 3 2 + −121 + 3 2 − −121 , b = 4, and c = −2 ± 3 ), we obtain an obviously wrong conclusion, 4 = −2 ± 3 . Task 4.3 immediately followed: Task 4.3 What (the hell) is going on here? Your task is to refute the obviously wrong conclusion 4 = −2 ± 3 .

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5.7 Can the Law of Transitivity Fail?

Fig. 5.4  Solution to Task 4.2

5.7.1  Reversible or Not and Does It Matter? The students began from double- to triple-checking the above calculations but did not find any computational mistake. Henceforth we were compelled to come back to the previously postponed question “are all the transformations reversible?” Recall that verifying the solution used two main steps: raising both sides to the third power (cubing) and “self-substitution.” Now both solution steps looked suspicious to us, and we decided to closely inspect them. Below we summarize the results of our inspection in a fictional (though inspired by real interactions) dialogue. 5.7.1.1  Exploration of the First Step: Cubing Albert: If we were squaring the initial equality, the issue would be clear. Obviously, if a  =  b, then a2  =  b2, but if a2  =  b2, then a  =  b  or a =  − b. For example, x = 1 implies that x2 = 1, but x2 = 1 only implies that x =  ± 1. Barbara: But we were cubing, not squaring! So how is this relevant? Albert: I don’t know… I just try thinking by analogy: x = 1 implies that x3 = 1, but does x3 = 1 implies that x = 1? Carol: It does for real numbers… Barbara: Indeed! The cubing transformation is reversible for real numbers, but we are dealing with complex numbers like −121 and also with cubic roots of complex numbers. Does it matter? Carol:  Let’s see… If we switch to complex numbers, the equation x3 = 1 has three roots, and 1 is just one of them… Albert: Can we find these roots? Carol: Of course! x3 = 1, x3 − 1 = 0, (x − 1)(x2 + x + 1) = 0, and then x1 = 1 or x2,3 =

−1 ± −3 . 2

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5  Encounters with Cardano’s Method

Barbara: So what? Carol: That’s it! We’ve just shown that cubing both sides of an equation is not reversible if complex numbers are involved, because this operation creates extraneous roots. Coming back to your [Albert’s] analogy with quadratics, all three numbers (i.e., 1 and −1 ± −3 ) are hidden in the equation x3 = 1, just as ±1 are hidden 2 in the equation x2 = 1. Barbara: But we had a numerical equality to check, not an equation to solve… Carol: It’s all related! Let’s say that some numerical expression A = 1, and some other expression B =  −1. After squaring, both expressions become equal, A2 = B2 = 1, but it does not mean that A = B and −1 = 1. Albert: I see… Do you mean, by analogy, that in our case, A  =  4, B = −2 + 3 , and A3 = B3, but it does not mean that 4 = −2 + 3 ? Carol: Exactly! Barbara: Just a second. In our case, not only A3 = B3, but also A = B as they both are equal to 3 2 + −121 + 3 2 − −121 . Simultaneously, A = 4 and = −2 + 3 ? How is this possible? Albert: Indeed, something strange is going on. It seems that the analogy with quadratics stops working here… Barbara: It looks like the expression 3 2 + −121 + 3 2 − −121 is multivalued, that is, it denotes not one number but three numbers… Because only in this case it is possible that , 4 = 3 2 + −121 + 3 2 − −121 2 ± 3 = 3 2 + −121 + 3 2 − −121 , but 4 ≠ −2 + 3 … Teacher: Great insight! You’ve just discovered that our computations were correct, but the final conclusion (i.e., the use of the law of transitivity) was not. In fact, the expression 3 2 + −121 + 3 2 − −121 denotes all three roots of the equation x3 − 15x − 4 = 0. It took a long time after Cardano and Bombelli to realize this fact. Dana: Oh, now I get how we got the contradiction. Let’s say that instead of checking if 3 2 + −121 + 3 2 − −121 equals this or that number, we would be asked to find its value. In this case, we could just denote it as an unknown x: x = 3 2 + −121 + 3 2 − −121 , x 3 = 2 + −121 + 2 − −121 + 3 3 (2 + −121) ⋅ 3 2 − −121 × ×

(

3

)

2 + −121 + 3 2 − −121 ,

5.7 Can the Law of Transitivity Fail?

119

x 3 = 4 + 3 3 4 − ( −121) ⋅ x ), x3 = 15x + 4 But we already know the solutions to the equation, x3 = 15x + 4, they are 4, −2 ± 3. Consequently, x attains all three values, and it is all right that the values are different. Teacher: Exactly! This is as if we concluded that −1  =  1  because both numbers are solutions to the equation x2 = 1. 5.7.1.2  Exploration of the Second Step: “Cubing and Self-substituting” Barbara: Just a second! [the acute reader may have noticed that Barbara is our favorite character: she is the one who exclaims “just a second!” when the rest of the students feel relieved]: I understand now how the mystery is resolved  – cubic roots of complex numbers are three-valued – but I still don’t understand whether cubing the both sides of the equation and using what should have been proved in the middle of the proof, let’s call this method “cubing and self-substituting,” had any role in creating the trouble. Teacher: Let’s explore this question. To begin with, let’s solve the following two equations: (1)

3

x + 25 − 3 x + 6 = 1 ,

(2) 3 x − 3 − 3 x −1 = 1 . Well, after coping with Cardano’s method, solving these equations was technically easy (see Fig. 5.5). Teacher: Anything strange? Barbara: Not really…Perhaps it is strange that all the roots appeared to be nice numbers, but you know most of the equations in the textbooks are of this kind… Teacher:  Check the numerical answers please. Do they all fit the equations? The students performed the checking and found (1): If x = 2, then 3 2 + 25 − 3 2 + 6 = 3 27 − 3 8 = 3 − 2 = 1 . Thus, x = 2 is a root of (1). Similarly, x =  −33 is a root of (1), because 3 −33 + 25 − 3 −33 + 6 = 3 −8 − 3 −27 = −2 + 3 = 1. (2): For x = 2, 3 2 − 3 − 3 2 − 1 = 3 −1 − 1 = −2 ≠ 0. Thus, the solution 2 is not a root of Equation (2).

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5  Encounters with Cardano’s Method

Fig. 5.5  Solution to equations (1) and (2)

Carol: You know, now when we’ve checked it, I see that Equation (2) could not have real solutions. Its structure is “a smaller number minus a greater number equals a positive number.” But it is strange anyway! Both solutions are based on the same method, “cubing and self-substituting.” So, it looks like the “cubing and self-­substituting” method SOMETIMES leads to extraneous roots and sometimes it does not… How so? Carol’s “How so?” question might trigger6 an exploration of a mechanism by which the “cubing and self-substituting” method can create extraneous roots. The exploration might begin from a close look at squaring. Indeed, how does the squaring of both parts of an equation create extraneous roots? The answer might be, in three steps, as follows: Step 1: If a + b = c, then (a + b)2 = c2. Step 2: The squared equation is factorized: (a + b − c)(a + b + c) = 0. Step 3: The extraneous solutions appear if and only if the equation a + b + c = 0 has a solution that differs from the solutions of the equation a + b − c = 0. Let us now adapt these three steps to the “cubing and self-substituting” method. Step 1: If a + b = c, then

6  This question was not discussed with the students due to time constraints. The presented scenario of dealing with the question is based on several lengthy discussions between us, B. and R.

5.7 Can the Law of Transitivity Fail?



(a + b)

3

121

= a 3 + b3 + 3a 2 b + 3ab 2 = a 3 + b3 + 3ab ( a + b ) .



By substituting c for a + b, we obtain a 3 + b3 + 3abc = c 3 or a 3 + b3 − c 3 + 3abc = 0



Step 2: Factorizing the equation a3 + b3 − c3 + 3abc = 0 It is reasonable to assume that one of the factors should be a + b − c. If so, the second factor can be found straightforwardly, as follows. Consider an expression a3 + b3 − c3 + 3abc as a cubic polynomial of the variable a: a3 + 0 ∙ a2 + 3bc ∙ a + (b3 − c3). Similarly, consider the expression a + b − c to be a linear polynomial of a + (b − c). By long division of the polynomials, we obtain (check it!)

(

a 3 + b3 − c 3 + 3abc = ( a + b − c ) a 2 + b 2 + c 2 − ab + ac + bc

)

(An alternative way of proving this would be to somehow guess the factorization and prove it by transforming the right side into the left side.) Step 3: Solving the equation a2 + b2 + c2 − ab + ac + bc = 0. Here we use our virtual Tartaglia’s idea for solving the equations of the type x2 + y2 = 0. Recall that he solved such equations by observing that if the left side of an equation consists of a sum of non-negative addends and the right side equals 0, then each addend must be 0. But does our equation belong to this type? Let’s see:

a 2 + b 2 + c 2 − ab + ac + bc = 0



2 a 2 + 2b 2 + 2c 2 − 2 ab + 2 ac + 2bc = 0



(a

2

) (

(a − b) + (a + c) + (b + c) 2



) (

)

− 2 ab + b 2 + a 2 + 2 ac + c + a 2 + 2bc + c 2 = 0 2

2

=0





Thus, the left side of the equation transforms into a sum of three non-negative addends. The equality is possible if and only if each addend is 0. Thus, a = b =  − c. The above three-step logical chain proves the following claim, which henceforth is named the Extraneous Solution Criterion7.

7   A variation of this claim, in Russian, can be found here: http://festival.1september.ru/ articles/532757/

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5  Encounters with Cardano’s Method

Extraneous Solution Criterion: Solving an equation of the form a + b = c by the “cubing and self-substituting” method leads to the appearance of extraneous solutions if and only if a = b = − c. 5.7.1.3  Intermediate Closure The above proof does not require that a, b, or c be real-valued expressions, and the proof also holds in the case when a, b, or c attains complex values. Consequently, the Extraneous Solution Criterion helps to answer the question that was a motivation for our exploration: are all the moves in proving the statements 4 = 3 2 + −121 + 3 2 − −121 (Fig. 5.3) and 2 ± 3 = 3 2 + −121 + 3 2 − −121 (Fig. 5.4) reversible? The explicit answer to this question is counterintuitive: generally speaking, the transformations are not reversible because additional roots may emerge when both sides of the equation a + b = c are cubed. However, in the considered case, they are reversible! To verify this answer with the help of the Extraneous Solution Criterion, we observe that the equality 3 2 + −121 + 3 2 − −121 = 4 is of the “a + b = c” type but it does not fit the condition a = b = − c. The implied transitivity leading to the erroneous claim “ 4 = −2 ± 3 ” has been clarified in the above dialogue between Alfred, Barbara, Carol, and Dana: the cubic root of a complex number is three-valued, so the law of transitivity is not applicable.

5.8  Reverse Heuristic Engineering and Problem Posing In addition to the above, the Extraneous Solution Criterion provides us with some insight into the mathematical mechanism underlying the creation of such “special” equations as the equation discussed in the dialogue between Teacher, Barbara, and Carol. Indeed, it is nice and even efficient for a mathematics teacher to have such equations in their pocket, but how one can know in advance that a given equation with cubic radicals would have extraneous solutions when solved by the “cubing and self-substituting” method? A reverse heuristic engineer would suggest that the equation 3 x − 3 − 3 x − 1 = 1 could have been created to meet the conditions of the above claim for a particular x. Indeed, in the equation 3 x − 3 − 3 x − 1 = 1

a = 3 x − 3 , b = 3 1 − x , c = 1, so for x = 2, a = b = −c.

Just a second! (Here we use again Barbara’s favorite rhetoric move). Another reverse heuristic engineer can raise another suggestion. She might point out that the appearance of the extraneous solution for the equation 3 x − 3 − 3 x − 1 = 1 was a result of an incorrect assumption, namely, the assumption that the equation had a

5.8 Reverse Heuristic Engineering and Problem Posing

123

real-valued solution when in fact it did not. In other words, the self-substitution in the middle of the solution assumes the existence of a real number x for which 3 x − 3 − 3 x − 1 = 1 (see Movshovitz-Hadar and Webb 1998, pp.  20–21, for the elaboration of this reasoning line). However, since this assumption is wrong (i.e., there is no such a number), an extraneous root may appear. R: May appear or does appear? B: Movshovitz-Hadar and Webb (1998) use the modal verb “may,” but it seems that we can be more specific with the help of the Extraneous Solution Criterion. After a lengthy discussion, we suggested that both explanations were correct but of a different extent of specificity and generality. Namely, we observed that the explanation “if the equation does not have solutions then the self-substitution may add them” was a particular case of the explanation “solving a cubic equation by the ‘cubing and self-substituting’ method leads to the appearance of extraneous roots if and only if the equations conform to the Extraneous Solution Criterion.” This is because the case “there are no solutions to the initial equation but there are solutions to the intermediate polynomial equation” is a particular case of a more general situation, namely, “there are some solutions to the initial equation but there are more solutions to the intermediate polynomial equation.” Note that the Extraneous Solution Criterion is applicable in both cases. To further explore this suggestion, we needed more examples of the equations that would have either true or extraneous solutions or both when solved by the “cubing and self-substituting” method. Consequently, we engaged ourselves (and plan to engage our future students) into the following problem-posing task. Task 5: Creating Equations with a Pre-defined Property Compose a pair of similar-looking equations with the following property: when solved by the “cubing and self-substituting” method, the solution to one equation returns an extraneous root, and the solution to other equation does not return an extraneous root.

To make the task more specific, we decided upon the following conditions: (1) we wished both equations to be based on the same simple numerical equality, for example, 3 1 + 3 0 = 1 or 3 27 − 3 8 = 1 ; (2) we wished both new equations to have roots among real numbers; and (3) we wished that x = 2 would be a true root to one equation and an extraneous root to another. Dictum-factum, as Cardano might say (again). The first pair of the equations was created based on the equality 3 1 + 3 0 = 1 . To be even more specific, we made a decision that one of the equations would have roots x1 = 2 and x2 = 3 and worked out the possible expressions for the radicals. Practically, we found the coefficients of the generic cubic equation

124

5  Encounters with Cardano’s Method

3



 A⋅2 + B = 1 C ⋅ 2 + D = 0  Ax + B + 3 Cx + D = 1 out of the following conditions :   A⋅3 + B = 0  C ⋅ 3 + D = 1



As a result, the equation 3 x − 2 + 3 3 − x = 1 was developed. For the second equation, we wished that x1 = 2 would become an extraneous root after cubing and self-substituting, so we worked out the expressions under the radicals from the condition provided by the Extraneous Solution Criterion. The equation 3 x − 3 + 3 2 x − 5 = 1 was composed in this way ( a = 3 x − 3 , b = 3 2 x − 5 , c = 1, then for x = 2 a = b = − c). The reader is invited to double-check the above two equations or, even better, to act as a reverse heuristic engineer and re-create them. After all, the experiences of creating and re-creating are interesting, enlightening, and pleasant, especially if an online calculator for solving cubic equations8 is available as a tool for escaping from the computational routine. We can testify that knowing a mechanism of creating a non-trivial mathematical problem evokes a wonderful feeling of being in control over the situation. And once you have the feeling of being in control, you just cannot stop. As a result, we created many additional equations having or not having the extraneous root when solved by the “cubing and self-substituting” method. Some of them are presented in Fig. 5.6. Needless to say, the readers are invited to double-check the equations, re-create them, and then to create their own “special” equations. Note that the process is not completely algorithmic as it leaves room for making decisions about the roots and the coefficients of the expressions under the radicals, as described above.

5.8.1  D  etour: The Variation Theory of Learning as a Task-­Design Heuristics This detour is needed in order to make explicit a theoretical framework that informed the process of creating “special” equations presented in the previous section. Namely, we would like to acknowledge the influence of the variation theory of learning on problem-posing choices. The variation theory (Marton and Booth 1997; Runesson 2005) postulates that what is learned is what remains invariant in a given space of variations. In practice, mathematical tasks constructed with the aid of this theory usually consist of sequences of slightly different and carefully ordered ­mathematical situations. The sequences are aimed at inviting the learners to contemplate the common structure underlying the situations.

 Such a calculator can be found here: http://www.mathforyou.net/en/online/equation/cubic/

8

5.9 Explorations Based on the Examples Generated by Students

125

Various instances of such mathematical sequences are discussed in Watson and Mason (2006). The scholars argue for the importance of visual similarity among the situations in a sequence as a means to foster the search for “looking the same” and “looking different” (p. 108) in learners. In turn, these processes are commended as those that underlie the search for regularity, structure and generalization. For example, Watson and Mason discuss the following sequence of multiplication exercises: 1 7 1 7 1 7 3 7 3 14 3 21 9 14 × = , × = , × = ,…, × = , × = , × = ,…, × =. 7 2 7 3 7 4 7 8 7 15 7 22 21 22 They note that “7” is used in the sequence as a generic placeholder, which role is kept constant for a while and then gradually changes. As a result, the learners’ attention is hopefully moved from the details of calculations to the multiplicative ­relationships between numerators and denominators and to the understanding of how the sequence was constructed. We hope that the introduced specifications for creating “special” equations in the previous section may play the same role as “7” in Watson and Mason’s example. To review, Task 5 asked for a pair of similar-looking equations for which the same solution method would lead to a particular variation related to one of the solutions. The first pair of constructed equations and the first equation in Fig. 5.6 share the same underlying numerical equality ( 3 1 + 8 0 = 1 ) and revolve around “x = 2” as a possible solution, depending on whether or not the conditions of the Extraneous Root Criterion are fulfilled. The next three equations in Fig.  5.6 continue revolving around “x = 2” but are based on another numerical equality ( 3 27 − 8 8 = 1) . Finally, the last equation in Fig.  5.6, which is also based on the equality 3 27 − 8 8 = 1 , involves a new pair of solutions, 4 and −3. To recall, the chosen learning goal of the sequence (that is, the invariant around which the variations revolve) is to contemplate the structure presented in the Extraneous Solution Criterion. So, what equations might come next in the sequence, in accordance with the introduced logic of variations? We invite interested readers to answer this question while posing their own equations in continuation of the sequence presented in Fig. 5.6. We now come back to the story.

5.9  E  xplorations Based on the Examples Generated by Students Part of the tasks and explorations presented so far occurred during and around two meetings of the course. A homework task (see below) was aimed at practicing the Cardano method and at creating an opportunity for repeated reasoning about the nature of complex numbers. As will be evident shortly, the homework turned out to be a source of mathematical surprises to us and some of the students, but also a source of boredom for some other students.

126

5  Encounters with Cardano’s Method

Fig. 5.6  Additional examples of equations with pre-defined properties

5.9 Explorations Based on the Examples Generated by Students

127

Fig. 5.6 (continued)

Task 6: Homework Your homework assignment is as follows: 1. Construct a cubic equation with “nice” answers. 2. Solve the constructed equations by the method of Cardano. Note: don’t use the final formula, but develop the solution in accordance with the steps of the method. 3. Show that the answers from the method of Cardano are indeed those “nice” answers that you’ve chosen when constructing the equation. 4. Did you receive all three solutions by the method of Cardano? If not, explain when and how you missed the solution(s).

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5  Encounters with Cardano’s Method

5.9.1  “Nice” Roots – Difficult Solutions Michel, one of 23 students who responded to Task 6, chose to experiment with randomly chosen9 coefficients of the cubic equation. Not surprisingly, the roots appeared “not nice,” and the solution process terribly difficult. Probably more surprisingly, the same can be said about many of the submitted assignments, in which the students constructed their equations beginning from some “nice roots.” The work of Ivan represents this cluster. It is presented below with some abridgments.

( x − 4 ) ( x + 2 ) ( x + 5) = 0



Then x3 + 3x2 − 18x − 40 = 0. The task is to solve this equation by the method of Cardano. Let us reduce the equation by using the first substitution of the method: x = y − 1. … The reduced cubic is y3 − 21y − 20 = 0. Let’s use the second substitution y = u + v so that uv = 7. … The equation y3 − 21y − 20 = 0 transforms into the following system of equations:



u3 = 10 + 9 3 ⋅ i u3 + v 3 = 20 , …,   3 3 3  u v = 243  v = 10 + 9 3 ⋅ i

And here the real challenge began! To compute u, Ivan, as well as many other students, made a mathematically efficient though historically unfair step: he represented the complex numbers in polar form and used de Moivre’s theorem. To recall: De Moivre’s Theorem: If z is a complex number, written in polar form as z = r(cosx + i sin x), then the n roots of z are given by x + 2π k  x + 2π k  r n  cos + i sin , n  n  1

where k varies over the integer values from 0 to n − 1. By means of this theorem and by using approximations, Ivan obtained the fol1.0004 + 2π k 1.0004 + 2π k   lowing: u = 2.64575  cos + i sin  , k = 0,1, 2. 3 3   He then computed the three values of u, the three values of v, the three values of y, and the three values of x. The work was free from computational mistakes, and  “Randomly chosen” means here that the student’s choice of the coefficients was apparently detached from her thinking about how to solve the equation. 9

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the resulting solutions were good approximations of the pre-defined answers, namely, x1 = 4, x2 = −2, and x3 = −5. Needless to say, we commended Ivan’s precision, insistence, and smart use of the prior knowledge. The main lesson learned however was that, generally speaking, cubic equations with three integer roots are about as difficult to solve by the Cardano method as randomly chosen equations.

5.9.2  “Nice” Roots – “Nice” Solutions Several interesting learning opportunities stemmed from the works, in which the students looked for the shortcuts. Some of the students succeeded to construct their equations so that solving them by Cardano’s method did not involve difficult computations. In a way, they utilized what is known as the principle of intellectual parsimony in problem solving. This principle presumes: When solving a problem, one intends not to make more intellectual effort than the minimum needed. In other words, one makes more effort only when forced to do so by the evidence that the problem cannot be solved with less effort. (Koichu 2010, p. 271)

For example, Nadin rightly decided that the simplest cubic to solve by the Cardano method is x3 = 0. Nadin however felt that the equation was too simple and changed it to (x + 2)3 = 0 or x3 + 6x2 − 12x + 8 = 0. The last equation looks “serious” at the first glance, but it reduces to y3 = 0 by means of the first substitution of the method of Cardano. a (Namely, x = y − ; a = 6 in Nadin’s case; see Fig. 5.2). 3 Two students created their cubics so that they had two equal integer roots and an additional integer root. Consider the work of Alex as an example. Alex composed equation x3 + 9x2 + 24x + 20 = 0, having the solutions −5, −2, and −2 in mind. The corresponding reduced equation was

y 3 − 3 y + 2 = 0.

What is nice about this equation is that Cardano’s method transforms it into the system



u3 + v 3 = 2 u3 = 1 , …,  3 .  3 3  u v =1 v = 1



One more shortcutting idea was to consider three different integer solutions having symmetric configurations. For example, the triple “−3, −1, and 1” has a symmetric configuration in the following meaning: the differences between the numbers are the same:

−1 − ( −3 ) = 1 − ( −1) = 2.



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5  Encounters with Cardano’s Method

More generally speaking, A configuration of real-number roots x1, x2, x3  to a cubic equation is referred to as symmetric with the difference k if for x1 ≤ x2≤ x3, x2 − x1 = x3 − x2 = k. The (unspoken) idea of symmetric configurations worked well for several students. For example, Lena composed the equation x 3 + 6 x 2 − 12 x + 8 = 0

out of the equation

( x + 3) ( x + 1) ( x − 1) = 0.



She obtained, by means of the substitution x = y − 1, the reduced equation y 3 − 4 y = 0.



This one is not difficult to solve by Cardano’s method as it transforms (see Fig. 5.2) into the following system: 8  3  u3 + v 3 = 0 i u = ±  3 3  3  3 3  4  ,…,  v3 =  8 i u v =   3   3 3



Another student, Ahmed, constructed the equation x 3 − 12 x 2 + 44 x + 8 = 0

by extending the equation

( x − 2 ) ( x − 4 ) ( x − 6 ) = 0. The reduced equation in Ahmed’s case appeared to be the same as in Lena’s one:



y 3 − 4 y = 0.

Was it by coincidence? We would probably leave this question without attention  – after all, not every disturbance becomes an impetus for time-consuming explorations – but it so happened that the question was reinforced when Dana shared with us her difficulty with the homework. Dana: Something is wrong with Cardano’s method! All the equations that I’ve tried at home have the same reduced equation. Boris: Which equations did you try?

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Dana (shows her drafts): x3 + 6x2 + 11x + 6 = 0 x3 − 6x2 + 11x − 6 = 0 x3 + 9x2 + 26x + 24 = 0 x3 + 12x2 + 47x + 60 = 0 Dana: You name it! The reduced equation is always y3 − y = 0. We quickly realized that Dana’s examples were not as random as she thought. It so happened that she created equations from expanding three binomials having the same structure: (x + r)(x + r + 1)(x + r + 2). That is, all Dana’s equations have the same symmetric configuration of roots with difference 1. (Observe that the roots of the first Dana’s equation are −3, −2, and −1,of the second one 1, 2, and 3, etc.) This observation calmed down Dana, but we became curious about the nature of the relationship between the roots to canonic cubics and the ease of the solutions of their corresponding reduced equations. We began the search by constructing an Excel worksheet for computing the coefficients of canonic equations having the real-number solutions of our choice and the corresponding reduced equations. The first three columns (x1, x2, and x3; see Table 5.1) are independent variables in the worksheet. The next three columns (a, b, c) represent coefficients of an equation x3 + ax2 + bx + c = 0 having the roots x1, x2, and x3. These coefficients are computed by Vieta’s formulas: a =  − (x1 + x2 + x3); b = x1 x2 + x1x3 + x2x3; and c =  − x1 x2 x3. The last two columns are for coefficients of the corresponding reduced cubics x3  +  px  +  q  =  0 computed by the formulas a2 2 a 3 − 9ab p = b− and q = +c. 3 27 Creating a sequence from Nadin’s, Dina’s, Lena’s, Ahmed’s, and Alex’s equations in the Excel worksheet (rows 1–10 in Table 5.1) was a good exercise in the use of the variation theory for constructing a pattern-sniffing activity10. Note that the first two rows of the table (Nadin’s equations) concern the situations where the cubics have one triple root or, in other words, they have a symmetric configuration of roots with the difference 0. The next four rows (Dina) represent cubics having the symmetric configuration with the difference 1 and rows 8 and 9 (Lena and Ahmed) with the difference 2. Naturally, we continued by experimenting with non-symmetrical configurations. The configuration of solutions in Alex’s equation is characterized by the differences 0 and 3 (row 10). We added two more equations with the same configuration (Experiment 1). Then we considered the equations with the differences 0 and 6 (Experiment 2). Experiment 3 concerns the configuration in which the first difference between the ordered solutions is 2 and the second is 1. In Experiment 4, the differences are 8 and 9, and in Experiment 5 the differences are generic (so we intended), 13.15 and 1.55. Note in addition that in all the experiments, at least one of the equations has 0 as a root.  Activities that enable students to act as “pattern-sniffers” are recognized, by Cuoco et al. (1996), as those having the potential for developing mathematical habits of mind.

10

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5  Encounters with Cardano’s Method

Table 5.1  Excel-based search for patterns 1 Nadin 2 3 Dana 4 5 6 8 Lena 9 Ahmed 10 Alex 11 Experiment 1 12 13 Experiment 2 14 15 Experiment 3 16 17 18 19 Experiment 4 20 21 22 23 Experiment 5 24 25 26

x1 0 −2 −3 1 −4 −5 −3 2 −5 2 0 0 −2 −3 0 −1 −4 −7 −2 15 0 −13 −9 −23 0

x2 0 −2 −2 2 −3 −4 −1 4 −2 2 0 0 −2 −1 2 1 −2 1 6 23 8 0.15 4.15 −9.85 13.15

x3 0 −2 −1 3 −2 −3 1 6 −2 5 3 6 4 0 3 2 −1 10 15 32 17 1.7 5.7 −8.3 14.7

a 0 6 6 −6 9 12 3 −12 9 −9 −3 −6 0 4 −5 −2 7 −4 −19 −70 −25 11.15 −0.85 41.15 −27.9

b 0 12 11 11 26 47 −1 44 24 24 0 0 −12 3 6 −1 14 −67 48 1561 136 −23.795 −64.995 499.205 193.305

c 0 8 6 −6 24 60 −3 −48 20 −20 0 0 −16 0 0 2 8 70 180 −11040 0 3.315 212.895 1880.365 0

p 0 0 −1 −1 −1 −1 −4 −4 −3 −3 −3 −12 −12 −2.3333 −2.3333 −2.3333 −2.3333 −72.3333 −72.3333 −72.3333 −72.3333 −65.2358 −65.2358 −65.2358 −65.2358

q 0 0 0 0 0 0 0 0 2 −2 −2 −16 −16 0.7407 0.7407 0.7407 0.7407 −24.0741 −24.0741 −24.0741 −24.0741 194.4343 194.4343 194.4343 194.4343

As a result of the experimentation presented in Table 5.1, several claims/suggestions were formulated. Claim 5.1  All cubic equations x3 + ax2 + bx + c = 0 with the same configuration of roots can be transformed, by means of the first Cardano’s substitution a x = y − , to the same reduced equation. 3 An outline of the algebraic proof of Claim 5.1 follows. Let x3 + ax2 + bx + c = 0 be a cubics with roots x1, x2, and x3. By Vieta’s formulas

a = − ( x1 + x 2 + x 3 ) , b = x1 x 2 + x1 x 3 + x 2 x 3 , c = −x1 x 2 x 3



a2 The coefficients of the corresponding reduced equation are p = b − ; 3 3 2 a − 9ab q= + c (see Fig. 5.2) 27

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133

Now consider another cubic equation x3 + Ax2 + Bx + C = 0 having the same configuration of roots. These solutions can be represented as x1  +  m, x2  +  m, and x3 + m for some m ≠ 0. By Vieta’s formulas and in terms of a, b, and c, the coefficients of the second equation are A = − ( x1 + m + x2 + m + x3 + m ) = a − 3m B = ( x1 + m) ( x2 + m ) + ( x1 + m ) ( x3 + m ) + ( x2 + m ) ( x3 + m ) = … = b − 2 am + 3m 2

C = − ( x1 + m) ( x2 + m ) ( x3 + m ) = … = am 2 − bm − m 3 + c



Now we can find the coefficients of the corresponding reduced equation,



P = B−

2 A3 − 9 AB A2 ,Q= + C, 3 27

and show (show it!) that P = p and Q = q, QED. Note that the proof works either for real number of complex roots as well as for the equations with complex coefficients. R: Claim 5.1 is a result of Dina’s surprise and our experimentation. What disturbs me is that Claim 5.1 is totally unsurprising when considered from graphical-­functional perspective. Just recall that the substitution a x → x − translates a graph of a function along the x-axis, and every3 thing becomes almost trivial… B: Totally unsurprising? Hmm…Let’s make a sketch [constructs Fig. 5.7a that corresponds to row 18 of Table 5.1]. OK, y = x3 + 7x2 + 14x + 8 is a 7 20 function corresponding to the original cubics, and y = x 3 − x + is a 3 27 function corresponding to the reduced cubics. The translation is in left7 to-right direction, and the size of the translation is … So, what exactly 3 does the translation do? Hmm… It takes a graph and translates it so that the center of symmetry of the original cubic polynomial (point B) moves, along the x-axis, until it meets the y-axis (point A). R: And this translation works for any cubic polynomial with the same configuration of roots. In other words, wherever the graph of a cubic polynomial is located in the coordinate system, the graph of its corresponding reduced polynomial has a center of symmetry on y-axis [adds to Fig. 5.7a two more graphs corresponding to rows 16 and 17 of Table  5.1; see Fig. 5.7b]. This is the graphical-functional meaning of the algebraic idea “let the term of the equation with x2 disappear.” It is so good to visualize!

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5  Encounters with Cardano’s Method

Fig. 5.7  Graphical-functional interpretations of the first Cardano substitution

5.9 Explorations Based on the Examples Generated by Students

135

B: Interestingly, we can now do what we couldn’t do at the beginning of our journey. I mean that we now can make sense of the first Cardano’s a substitution, x → x − (see Fig.  5.2), as reverse heuristic engineers. a 3 Let’s reverse our reasoning and suggest that the size of the translation 3 is being chosen in order to move the center of symmetry of the original polynomial (points B, C, and D on Fig. 5.7b) to y-axis (point A). R: This suggestion is easy to check. The center of symmetry of a cubic polynomial is also its inflection point. Accordingly, its coordinates can easily be found by calculation of the second derivative. Let y = x3 + ax2 + bx + c, then y′ = 3x2 + 2ax + b; and then y′′ = 6x + 2a. a Then we have an equation, 6x + 2a = 0 and then x = − . 3 B: Just a second. It means that Cardano’s substitution re-emerges as the first coordinate of the inflection point! R: It looks so. The second coordinate is 3

2

a 3 a 2 ba 2 a 3 − 9ab  a  a  a  a y − = − + a  −  + b  −  + c == − + − +c= +c .  3   3  27 9 3 27  3  3

B: Just a second! I recognize the last expression. It is a coefficient q of the reduced cubics x3 + px + q = 0 2 a 3 − 9ab q= + c (see Fig. 5.2). 27 Well, this expression does not look weird anymore. R: Note that the reasons for which the reduced equation looks as it is became intuitively appealing to us only when we employed reasoning in terms of functions. This sort of reasoning did not yet exist in the sixteenth century. B: Hmm… Let me sum up: some things were difficult to see, also to us, when we adhered to algebraic reasoning of the sixteenth century, even when we used the modern notation. And then the same things became almost trivial when reasoning in terms of functions and their graphs was employed. R: Indeed. There is probably no mathematical news in our exploration, but what a wonderful motivation for a concept of function we now have, either for us or our students! To reiterate, we connected Cardano’s method to transformation of functions, and we noted that reducing a canonic cubic equation x3 + ax2 + bx + c = 0 to the form x3 + px + q = 0 corresponds to translating a graph of a cubic function y = x3 + ax2 + bx + c to the graph of the function y = x3 + px + q so that this graph has its symmetry center on y-axis.

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5  Encounters with Cardano’s Method

Two additional claims concerning particular configurations of roots were deduced from Table 5.1. Claim 5.2  If a cubic equation x3 + ax2 + bx + c = 0 has roots r, r, and t, then the 2

3

r −t  r −t  corresponding reduced equation is y 3 − 3   y + 2 3  = 0 . 3     Claim 5.3  If a cubic equation x3 + ax2 + bx + c = 0 has a symmetric configuration of roots with the difference k (that is, the roots are −k, r, r + k), then it has a corresponding reduced equation y3 − k2y = 0. We produced the proofs of the claims with some technical effort but quite straightforwardly. The following plan was realized. Given the solutions, find out the expression for a, b, and c by Vieta’s formulas, and then find p and q given that a2 2 a 3 − 9ab p = b− and q = + c. It is of note that the (quite substantial!) techni3 27 cal work is essentially reduced by using Claim 5.1. By Claims 5.1 and 5.2 can be proved without loss of generality for the equation having the roots 0, 0, and (r − t) instead of r, r, and t. Similarly, it is enough to prove Claim 5.3 for the equation having the roots −k, 0, and k instead of r − k, r, and r + k.

5.9.3  Detour: “Niceness” and Artificiality As mentioned previously, our motivation for exploring cubic equations with “nice” roots was of problem-posing nature: we inquired how to construct cubic equations so that their solutions by the method of Cardano would neither be trivial nor exhaustingly difficult. As we have seen in the work of Ivan, choosing three “nice” integer solutions does not guarantee the reasonable “niceness” of the calculations. In fact, Cardano’s method constitutes a serious technical challenge especially when an iniu3 + v 3 = −q  tial equation transforms into the system  3 3 p3 (see Fig. 5.2) with solutions u v = − 27  for u3 and v3 that are too awkward. Claims 5.2 and 5.3 provided us with some problem-posing ideas as to this issue. This is because the claims show that particular configurations of roots transform into relatively easy-to-solve systems of equations. Specifically, the equations having roots r, r, and t (Claim 5.2) transform into a 3  3 3 r −t  u + v = −2  3    3  r −t  3 3 system  that has a “nice” solution u v = = −  3  6    r −t  3 3 = u v  3      (check it!).

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137

Next, the equations having symmetric configurations of solutions (Claim 5.3) u3 + v 3 = 0 k3  transform into the system  3 3 k 6 , which has the solutions u3 = −v 3 = ± i 3 3 u v = 27  (check it!). With respect to the last case, we note that the notions of “niceness” and “simplicity” are relative. The solution to the last system is not that simple as in the previous case, but it is much “nicer” (that is, much less technically demanding) than the solution to Ivan’s case. R: Let me play the devil’s advocate. These claims definitely show some effective strategies of dealing with the task “create an equation from some ‘nice’ roots and solve it by the method of Cardano.” I think that this task was justified in the context of the course, but in many other contexts, it would be artificial. After all, “nice solutions” are usually equated with “integer solutions.” We know that if a cubic equation has an integer or even rational root, then it can be found by means of the Rational Zero Theorem,11 that is, by a cleverly organized inspection, then long division, and that’s it… B: Would you extend this comment also to the formula for the roots of a quadratic equation? R: Only to some extent. We know that in school practice most quadratic equations that students are asked to solve indeed have “nice” roots, either integer or rational. Those indeed can be found without the formula, by factorization, or by the Rational Zero Theorem or just by inspection. We also know that extensive use of “nice numbers” is a problematic pedagogical practice. Harel (1995), for example, has argued against it. Nevertheless, good teachers and good textbooks care to provide students with a certain amount of quadratic equations having “not nice” roots. In some cases, the teachers even require their students to solve “random” quadratic equations. By so doing they demonstrate the need for the formula and its generality. B: Here I see the difference. Whether or not the roots of a quadratic equation are “nice,” the amount of the technical work involved in using the general formula is about the same. But when it comes to cubics… Well, we’ve seen the gap between the efforts needed in order to solve a random cubic equation vs. a specially constructed one. We’ve also seen the gap between Ivan’s equation having three ­randomly chosen “nice” roots and the equations made up by Alex or Ahmed. Those equations also had “nice” roots, but chosen with some care.  See http://mathworld.wolfram.com/RationalZeroTheorem.html

11

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5  Encounters with Cardano’s Method

R: The gap, in terms of Cardano’s method, is huge indeed, though all these equations can also be solved by the Rational Zero Theorem. I am not sure of course that the Rational Zero Theorem would be a good strategy for solving such equations as No. 21 in Table 5.1 (x3 − 70x2 + 1561x − 11040 = 0). B: I worry that the excessive technical effort associated with the use of Cardano’s method may cause anxiety and contribute to the bad reputation that the formula has. Remember? I was exposed to the formula and its story as a schoolboy, but in time I only remembered that “the story was great, but the formula was ugly.” This is why I wanted to create a different experience for my students. As a mathematics educator, I do want my students, prospective teachers, to appreciate Cardano’s method for its deepness, elegance, and surprise (just look at the “proof” of the numerical paradox 4 = −2 ± 3 !), and I don’t want them to hate the method for its technical difficulty. Practically, I want to have some control over the level of difficulty of the use of Cardano’s method. R: As a matter of fact, you argue that if the Cardano method were associated for learners with problems of reasonable difficulty, then the pedagogical feasibility of the formula would improve. Perhaps. And from this argument you deduce that all these claims (Claims 5.1, 5.2 and 5.3) are needed… B: Right. This is not mentioning the fun that we had when formulating the claims and our increased mathematical understanding of some of the phenomena that we’ve observed in student-generated examples. It is exciting to see something new, and I believe that it is good to share this excitement with the students. I also believe that having repeated opportunities to apply Cardano’s method to reasonably difficult exercises could change the aesthetic value of the formula in the eyes of the learners. R: Well, I have an argument in support of the point about aesthetics. Do you remember our conversation about which historical equation to choose in order to demonstrate the mentioned above numerical paradox? A famous example from Cardano’s Ars Magna was x3  =  8x  +  3, and a famous example from Bombelli’s L’Algebra was x3 = 15x + 64. The equations looked similar to us, and they both were appropriate for constructing the paradox. But the solution to x3 = 8x + 3 was 3

3 1805 3 3 1805 + − + − − , and the solution to x3 = 15x + 64 was 2 108 2 108

2 + −121 + 3 2 − −121 . Of course, Bombelli’s equation was chosen! B: Indeed, this is a good example in support of the idea “aesthetic considerations matter.” 3

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139

R: I wish we would also have some mathematical argument, say, some useful implications for Claims 5.1, 5.2 and 5.3 in problem solving. B: Don’t we have useful implications for them? We’ve shown how they can be used in making up “special” cubic equations. The claims are needed because it is definitely not enough to make up “good” equations from randomly chosen integer roots, as Ivan and many others have tried. And this is in clear contrast with the case of making up quadratic equations. In this case any pair of “nice” roots guarantees the “niceness” of the solution by the formula. After all, there is no “casus irreducibilis” in the case of quadratics. R: Indeed, but here we come back to the issue of artificiality of our problem-posing exercises. I wish the claims would help us to make up some “nice” cubic equation without rational roots. This would demonstrate our students the need for Cardano’s formula in the same vein as quadratic equations with irrational roots demonstrate the need for the general formula for the roots of a quadratic equation. B: That is, you wish we had in our arsenals some cubic equations for which the Rational Zero Theorem would be inapplicable, but solving them by Cardano’s method would not be excessively difficult…Why not? The conversation continued. As a result, we made up several cubic equations. Two similar-looking equations are chosen to be presented below: x 3 − 3 2 x 2 + 5 x − 2 = 0 (roots: 2 − 1, 2 , 2 + 1) and

x 3 − 4 2 x 2 + 10 x − 4 2 = 0 (roots: 2 , 2 , 2 2 )



We plan to offer these equations to our future students. Though these equations do not have rational roots, it is our hope that the reasonable “niceness” of solving them by Cardano’s method would be encouraging for the students, either mathematically or aesthetically.

5.9.4  Appearance of “Nice Monsters” 5.9.4.1  One Real-Number Root – “Too Nice” Solutions Rachel constructed her cubic equation based on another shortcutting idea: she considered the factorization (x  +  1)(x2  +  3). The corresponding cubic equation to solve was

x 3 + x 2 + 3 x + 3 = 0. 1 The first substitution of Cardano’s method ( x = y − ) transformed it into 3

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5  Encounters with Cardano’s Method

y3 +



8 y 56 + = 0. 3 27

To our surprise, solving this reduced equation by means of the second Cardano’s 8 method substitution (y = u + v when 3uv = − p = − ) was quite a pleasant experi3 ence and not only in comparison with Ivan’s equation – anything is “nice” in comparison with Ivan’s equation – but also in comparison with almost all the equations from our previous Excel-based experiments11 (see Table 5.1). Specifically, the second Cardano’s method substitution led Rachel to a system of equations for u3 and v3 with real-number solutions in the form of cubic numbers:



56   3 3 64 8  3  u + v = − 27  u13 = u1 = − 27   27 . orr  3 ,  u3 v 3 =  − 8  v 3 = − 64  v 3 = 8 1 1    27 27   9  

Having encouraged by the “niceness” of the system and still being in a mode of looking for “special” configurations, we tried two more equations of the same structure:

( x + 1) ( x 2 + 4 ) = x 3 + x 2 + 4 x + 4 = 0 ( x + 2 ) ( x 2 + 4 ) = x 3 + 2 x 2 + 4 x + 4 = 0,

and then generalized:

( x + a ) ( x 2 + b2 ) = x 3 + ax 2 + b2 x + ab2 = 0.

 2 a 3 + 18ab 2 a2  The generalized equation was reduced ( x 3 +  b 2 −  x + = 0 ) and 3  27  then checked for the “niceness” of its solution. We found that a trouble-making part 2

3

q  p  2  +  3  (it is frequently     referred to as a discriminant of a reduced cubic), was of the following form: of Cardano’s method, namely, the expression

(



2 3 b2 b2 + a 2 q  p  2 + 3  =…= 27    

)

2

=

1 3 3

(

b b2 + a 2

) ( check it!) .



 Note that we could not explore Rachel’s equation by means of the Excel worksheet. This is because Excel works only with real numbers and Rachel’s equation has one real-number root and two complex-number roots.

11

5.9 Explorations Based on the Examples Generated by Students

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Accordingly, we ascertained that the “niceness” of Rachel’s equation was not accidental. We felt however that the considered configuration was a bit “too nice” in order to formulate a claim about its properties. Our feelings stem from the above discussion about “niceness” and artificiality. Namely, it would be too artificial to insist that Cardano’s method should be used for solving cubics of the form x3 + ax2 + b2x + ab2 = 0. Indeed, a factorization of such equations is transparent:

(

) (

)

(

)

x 3 + ax 2 + b 2 x + ab 2 = x 3 + ax 2 + b 2 x + ab 2 = ( x + a ) x 2 + b 2 .

5.9.4.2  Creating “Nice Monsters” Being a bit tired after thinking about “niceness,” we switched the direction and constructed several equations of a slightly more complicated structure than Rachel’s equation, namely, (x + a)(x2 + bx + c) = 0 where the factor x2 + bx + c has complex-­ number roots. For example, (x − 1)(x2 + x + 4) = 0, and then x3 + 3x − 4 = 0. The discriminant of this equation is 2

3

2

3

q  p  −4   3   2  +  3  =  2  +  3  = 5.        



Accordingly, Cardano’s formula gives the following root: 2



3

2

3

q q q  p q  p x = 3 − +   +   + 3 − −   +   = 3 2 + 5 + 3 2 − 5. 2 2 2 3 2 3 Since we already knew that x = 1 is the only real root of the equation, we obtained

the proof of the following fact: 3 2 + 5 + 3 2 − 5 = 1. We immediately recognized in the last equality one of those numerical “monsters” that appear in many problem books and teaching units in the field of advanced elementary algebra. Examples are presented in Fig. 5.8.

Fig. 5.8  Examples of numerical equalities with cubic radicals

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The solutions to these exercises are not particularly difficult for algebraically skilled individuals, though they (the solutions) contain certain mathematical subtleness. For example, the discussion presented in the “Reversible or Not and Does It Matter?” section of this chapter implies that the straightforward use of the “cubing and self-substituting” method is problematic because this method contains potentially non-reversible steps. One way for safely solving the above exercises (see Harel 2013, for alternatives) is to denote the left part as an unknown, reveal the cubic equation behind the equality, and solve it. Consider exercise (a) as an example: Let

3

9 + 80 + 3 9 − 80 = x . By cubing both sides, we obtain

9 + 80 + 9 − 80 + 3 3 9 + 80 ⋅ 3 9 − 80



(

3

)

9 + 80 + 3 9 − 80 = x 3

18 + 3 3 (9 + 80)(9 − 80) ⋅ x = x 3







x 3 − 3 x − 18 = 0.



By inspection, x = 3 is a root. Long division of polynomials results in the following factorization: x3 − 3x − 18 = (x − 3)(x2 + 3x + 6). The equation x2  +  3x  +  6  =  0 does not have solutions in the real numbers. Accordingly, the question of which of the roots to equate with 3 9 + 80 + 3 9 − 80 does not arise. Since both x = 3 and x = 3 9 + 80 + 3 9 − 80 are solutions to the above cubic equation and since the equation has only one real-number solution, 9 + 80 + 3 9 − 80 = 3, QED. As previously noted, solving such exercises is not a particularly illuminating experience once you solve one of them. But what about making up such exercises? As mathematics teachers who are familiar with many algebra textbooks, we have always been curious about how such “nice monsters” could have been created and created wholesale!12 Our curiosity was satisfied when we showed that 3 2 + 5 + 3 2 − 5 = 1 while experimenting with equations of the structure (x + a)(x2 + bx + c) = 0. Generally speaking, a method for creating “nice monsters” is to begin from a cubic equation having exactly one real root and solve it by Cardano’s formula. For example, begin from the equation (x − 4)(x2 + 4x + 19) = x3 + 3x − 76 = 0 and arrive at…Can you guess? We arrive at exercise (b) from Skanavi (1992); it was mentioned at the beginning of this section: 3



3

38 + 1445 + 3 38 − 1445 = 4

You need an even more monstrous equality? Well, just produce it.  For example, a problem book by Skanavi (1992), which is a highly influential mathematical resource in the countries comprising the former USSR, contains tens of such “nice monsters.”

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5.10  Concluding Remarks

143

As to us, we were glad to find a straightforward implication of Cardano’s formula in elementary algebra. “Nice numerical monsters” are usually offered to secondary school students, in the context of studying the abridged multiplication formulas and polynomials, that is, a long time before they have a chance to get familiar with complex numbers. As a result, the question “where do these ‘monsters’ come from?” remains out of reach for the students and frequently also out of reach for their teachers. What a nice surprise it would be for them (as it has been for us) to find out that the mechanism underlying the construction of “nice real-number monsters” for the textbooks is rooted in the history of cubic equations and complex numbers!

5.10  Concluding Remarks So, which of our manifold mathematical disturbances were addressed while exploring the story of Cardano’s formula with our students? Our answer to this question in the form of a bullet list is as follows: • The method of Cardano now belongs, for us, to the realm of knowledge which is necessitated by some reasoning rather than to the realm of arbitrary knowledge that can only be memorized13. This point alludes to our attempts at demystifying the pivotal moves of the method as reverse heuristic engineers. • We better understand consequences of the fact that roots of complex numbers are multi-valued. Here we refer to the “ 4 = −2 ± 3 ” paradox. • Exploration of the “cubing and self-substituting” method for solving cubic equations helped us to distinguish between situations where the use of the method may lead to the appearance of extraneous roots and does lead to the appearance of extraneous roots. Specifically, the discussions of reversibility of a “cubing” step and of a “self-substituting” step of the method allowed us to formulate the Extraneous Solution Criterion. In turn, the criterion opened for us a way of making up “special” irrational equations. • We better understand now the relationships between the roots of canonic cubic equations, on one hand, and the roots of their corresponding reduced equations on the other hand. In other words, we developed some approaches to constructing cubic equations so that the level of technical difficulty of solving them would be reasonable (see Claims 5.1 and 5.2). This understanding decreases our dependence on the existing problem books and increases our qualifications as designers of new instructional sequences on cubics, especially in cases where the variation theory of learning serves as a powerful task-design heuristic. • We observed a surprisingly simple connection between algebraic and functional-­ graphical representations of the process of reducing canonic cubics. • We also satisfied our curiosity as to how “monstrous” numerical equalities with cubic radicals that appear in many textbooks and teaching units could be made up. This point refers to a straightforward implication of Cardano’s formula in school algebra. 13

 Here we resort once more to Hewitt’s (1999) terminology and ideas.

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References Cuoco, A., Goldenberg, E. P., & Mark, J. (1996). Habits of mind: An organizing principle for mathematics curricula. The Journal of Mathematical Behavior, 15(4), 375–402. Ejersbo, L. R., & Leron, U. (2005). The didactical transposition of didactical ideas: The case of the virtual monologue. In M. Bosch (Ed.), Proceedings of the 4rd conference of the European Society for Research in Mathematics Education (pp. 1379–1388). Sant Fceliu de Guíxols. Ekert, A. (2008). Complex and unpredictable Cardano. International Journal of Theoretical Physics, 47(8), 2101–2119. Euclides. (2002). D. Densmore (Ed.), Euclid’s Elements: All thirteen books complete in one volume, the Thomas L. Heath translation. Santa Fe: Green Lion Press. Feldmann, R. (1961). The Cardano-Tartaglia dispute. The Mathematics Teacher, 54(3), 160–163. Guilbeaur, L. (1930). The history of the solution of the cubic equation. Mathematics News Letter, 5(4), 8–12. Harel, G. (1995). From naive interpretist to operation conserver. In J. Sowder & B. Schappelle (Eds.), Providing a foundation for teaching mathematics in the middle grades (pp. 143–165). New York: SUNY Press. Harel, G. (2013). DNR-based curricula: The case of complex numbers. Journal of Humanistic Mathematics, 3(2), 2–61. Henning, H. (1972). Geometric solutions to quadratic and cubic equations. The Mathematics Teacher, 65(2), 113–119. Hewitt, D. (1999). Arbitrary and necessary part 1: A way of viewing the mathematics curriculum. For the Learning of Mathematics, 19(3), 1–9. Kenney, E. (1989). Cardano: Arithmetic subtlety and impossible solutions. Philosophia Matemática, II, 4(2), 195–216. Koichu, B. (2010). On the relationships between (relatively) advanced mathematical knowledge and (relatively) advanced problem solving behaviours. International Journal of Mathematical Education in Science and Technology, 41(2), 257–275. Koichu, B., & Zazkis, R. (2013). Decoding a proof of Fermat’s little theorem via script writing. Journal of Mathematical Behavior, 32, 364–376. Koichu, B., & Leron, U. (2015). Proving as problem solving: The role of cognitive decoupling. Journal of Mathematical Behavior, 40, 233–244. Leron, U. (1983). Structuring mathematical proofs. The American Mathematical Monthly, 90(3), 174–185. Leron, U., & Hazzan, O. (1997). The world according to Johnny; a coping perspective in mathematics education. Educational Studies in Mathematics, 32(3), 265–292. Marton, F., & Booth, S. (1997). Learning and awareness. Mahwah: Erlbaum. Merino, O. (2006) A short history of complex numbers. Retrieved from http://www.math.uri. edu/~merino/spring06/mth562/ShortHistoryComplexNumbers2006.pdf. Movshovitz-Hadar, N., & Webb, J. (1998). One equals zero and other mathematical surprises: Paradoxes, fallacies, and mind bogglers. Berkeley: Key Curriculum Press. Runesson, U. (2005). Beyond discourse and interaction. Variation: A critical aspect for teaching and learning mathematics. The Cambridge Journal of Education, 35(1), 69–87. Skanavi, M. I. (Ed.). (1992). Sbornik zadach po matematike dlja postupajushix v VUZU (Problem book in mathematics for college applicants) (6th ed.). Moscow: ONIKS Mir i Obrazovanie. Watson, A., & Mason, J. (2006). Seeing an exercise as a single mathematical object: Using variation to structure sense-making. Mathematical Thinking and Learning, 8(2), 91–111. Zazkis, R., & Koichu, B. (2015). A fictional dialogue on infinitude of primes: Introducing virtual duoethnography. Educational Studies in Mathematics, 88, 163–181.

Chapter 6

Synthesizing Exception-Barring and What-If-Not: If Not, What Yes?

6.1  Introduction Boris: Can we imagine a thrill experienced by a mathematician who sees an unexpected structure in a string of numbers (consider 43, 47, 53, 61, 71, 83, 97, 113, 131…) or by a mathematics learner who has a sudden illumination that leads her to formulating a statement about an interesting mathematical property, which she could not find in her textbook (consider “Any convex quadrilateral with two opposite equal angles and two opposite equal sides is a parallelogram.)"? Rina: Well, I think we can… But let me look at your examples. So, what is special about this string of numbers? They are all odd, they all seem to be prime, yes, they are primes… Let me Google them. Oh, these are numbers generated by the Euler’s polynomial F(x) = x2 + x + 41. The values of the polynomial are prime numbers for 1 ≤ x ≤ 40, but x = 41 breaks the pattern and the polynomial stops working as a generator of primes. B: Indeed. I learned when in school that a formula for all prime numbers does not exist and felt a lot of empathy for Leonard Euler imagining how he had checked one integer after another, in the hope that so-desired formula was found... R: And I did not believe my grade-5 teacher, who mentioned that there was no formula for generating primes. So I spent my grade-5 searching for such a formula… and this is before the era of calculators. Fortunately, in grade-6 I had other interests. Anyway, your second example, isn’t it just a school theorem?

© Springer Nature Switzerland AG 2021 B. Koichu, R. Zazkis, Mathematical Encounters and Pedagogical Detours, https://doi.org/10.1007/978-3-030-58434-4_6

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B: Not really. The statement I mentioned is: Any convex quadrilateral with two opposite equal angles and two opposite equal sides is a parallelogram. The closest to the above statement school theorems I can recall are: Any quadrilateral having two pairs of opposite equal angles is a parallelogram. Any quadrilateral having two pairs of opposite equal sides is a parallelogram.

As plausible as the above statement looks, a counterexample to the statement can be constructed (we invite the readers to find it!). The point of the above examples is that inventing and scrutinizing conjectures that refer to plausible but not necessarily correct statements is an instant source of thrill when experiencing mathematics either as a discovery-based science or as a subject of study inviting exploration. Many people have argued for this point before us. In particular, Lakatos (1976) does in his Proof and Refutation, which is probably the most influential book on historical methods of mathematical discovery. Also, Brown and Walter (1993) do in The Art of Problem Posing, which is probably the most influential book on making systematic conjecturing feasible in school setting. In this chapter, we depict our mathematical and pedagogical disturbances related to the use of insights from these two books in our work with mathematics teachers and students.

6.2  Background The question “If not, what yes?” put in the title of this chapter is the name of an approach to constructing and conducting discovery-oriented activities in a mathematics classroom; the approach is influenced by Lakatos’ (1976) exception-barring method and by Brown and Walter’s (1993) “What if not?” method. Lakatos (1976) introduced the exception-barring method as “Piecemeal exclusions. Strategic withdrawal or playing for safety” (p. 120). The method is part of a convoluted way from an initial primitive conjecture into a theorem. The method consists of restricting a theorem domain only to those cases for which the theorem works, and its proof is not endangered. This is done by considering certain cases, for which the suspect theorem is uncertain, as exceptions. Lakatos (footnote 4, p. 124) attributes the idea of exceptions to Abel, who, out of considerations of keeping proofs rigorous, restricted the domain of suspect theorems about functions to functions that can be represented as power series. The exceptions, according to Lakatos, should not be confused with monstrous counterexamples invented especially in order to fail the theorem. The difference is in that the exceptions are “interesting examples in their own right, worthy of a separate investigation” (p. 120) and the monstrous counterexamples are not. The exception-barring method is supposed to create “a stronghold for a conjecture” (p. 125) but can itself be a danger for the theo-

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rem, as “playing for safety” can be too radical and leave outside the walls some interesting and legitimate cases. A Brown and Walter’s (1993) “What if not?” method is a systematic method of conjecturing and problem posing based on the existing theorems. The method consists of 5 levels. At level 0, a starting point, a well-known theorem is chosen. At level 1, all apparent attributes of the theorem’s formulation are listed. The authors point out that even obvious and seemingly irrelevant attributes can originate deep mathematical explorations and lead to new and interesting results. At level 2, alternatives to the listed attributes are pointed out. In a detailed example concerning the Pythagorean Theorem, the authors list 10 attributes of the theorem and ask the question “What if not?” about each of them. As a result, they construct at least two alternatives for each attribute. For example, an alternative to the attribute “The Pythagorean Theorem concerns right-angled triangles” can be “Consider acute-­angled triangles or obtuseangled triangles.” Level 3 consists of posing new questions inspired by these alternatives, for instance, “What would an analogue of the Pythagorean Theorem for acute-angled triangles look like?” At level 4, the students analyze the posed questions and try to address the most promising and interesting ones. For instance, they can discover that the cosine theorem is a generalization of the Pythagorean Theorem that works for right-angled, acute-angled, and obtuse-­angled triangles. Although the levels of the “What if not?” method seem to be highly formalized and linear, Brown and Walter stress that their scheme is nonlinear and, when properly implemented, “provides a touchstone for a spirit of investigation and free inquiry” (p. 62). The “If not, what yes?” method presented in this chapter consists of two aspects: an instructional aspect, which is a version of exception-barring, and a design aspect, which is a version of “What if not?” Let us discuss and illustrate these aspects one by one.

6.3  “If Not, What Yes?”: An Instructional Aspect 6.3.1  The Account In accordance with the “If not, what yes?” method, the students are asked to prove or refute a mathematical statement at the beginning of a lesson. The statement is chosen by the teacher to induce some counterexample as an immediate response and, in turn, the conclusion “The statement is incorrect.” That is, the chosen statement is a school-level analogue of Lakatos’ primitive conjecture. The teacher acknowledges that the statement is wrong and asks: “If that statement is wrong, which one would be correct?” or, in brief, “If not, what yes?” Then the teacher modifies the initial statement in such a way that the provided counterexample (akin to exception) would be neutralized (akin to excluded). The modification must be minimal. The students have a choice: to look for a new exception or try proving the modified statement. If some new exception is found, the question “If not, what yes?” arises again, and a new minimal modification is suggested, this time preferably by the students or with the students, in order to exclude the exception. The multi-step process of modifying the initial statement continues until the last modified state-

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ment is proved or otherwise found by the class not worth further effort. Generally speaking, with every new exception excluded, the conjecture at stake becomes more plausible, the search for new exceptions becomes more challenging, and the requested argumentation more sophisticated. The role of the teacher in such a lesson is in providing the students with the initial wrong but somewhat plausible mathematical statement and in flexibly mediating the follow-up small-group and whole-class discussions in the spirit of the “If not, what yes?” question. The expected role of the students is, at the beginning, in refuting the initial statement and then in gradually constructing and proving or refuting the refined statements. In a successful “If not, what yes?” lesson, students have a feeling that they work on constructing a mathematical result that would also be new to their teacher.

6.3.2  Illustration in the Context of Calculus1 A scenario presented in this section has been tried in more than 10 90-min workshops with in-service and pre-service mathematics teachers. The scenario is remarkably typical in a sense that its essential events were observed with almost no changes in all the workshops. Moreover, the mathematical part of the scenario is remarkably similar to our own attempts to generalize and formalize a particular property of even functions. These attempts had resulted from our own mathematical disturbance some time before we decided to transpose them into the “If not, what yes?” activity for our students. So, let the story begin. Each workshop has been opened with the following monologue: Boris: Recently I looked at many examples of even functions, and noticed an interesting property: Every even function has a local strict extremum at x=0. To me, the statement looks promising, but it does not appear in any algebra or calculus textbook I’ve looked at. Is there somebody familiar with this property of even functions? What do you think about it? A Small Detour The opening monologue is carefully designed. It is aimed at allocating the teacher as a person that shares with the students his personal struggle. In addition, the monologue takes into account that mathematics textbooks serve for many students as a depersonalized mathematics authority. The information that the statement offered by the teacher – the initial statement – does not appear in textbooks is given to raise uncertainty about its correctness. It is important in order to develop the lesson towards a situation, in which the mathematical authority of the teacher is challenged and any reference to the professional literature is impossible. Last but not

 This illustration is based on a story presented in Koichu (2008).

1

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least, all the information provided in the monologue is true, which is important for ethical reasons. Now back to the story. For many students, the initial statement Every even function has a local strict extremum at x=0 looks plausible as it works for the most frequently mentioned examples of even functions. Indeed, it is reasonable to assume that those who keep in mind y = x2, y = cosx, or y =  ∣ x∣ as representative examples of even functions feel that the above statement can be true. Note that there are many additional examples that support that feeling (y =  − x10, y = sin2x, y = x4 + 3x2 −  cos x, etc., see Fig. 6.1). In terms of Tall and Vinner (1981), many available concept images of the even function support the initial statement. However, the information that the statement does not appear in algebra and calculus textbooks makes some participants rather skeptical, and they start from looking for a counterexample. This is not difficult. In about 3–5 min of working individually or in small groups, the participants suggest two ways of refutation of the above statement, always the same ones. Those who apparently pay special attention to the word “strict” in the initial formulation respond as follows: Response 1:  The statement is wrong as it does not work for constant functions, for example, for y = 2. Those who apparently are used to connecting the notion of extremum with derivatives and, in turn, with a condition of continuity, suggest: Response 2:  The statement is wrong as it does not work, for example, for the 1 1 functions y = x and y = 2 . x

Fig. 6.1  Supporting examples for the initial statement

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At this point, the teacher sketches on the class board the suggested functions and, after a quick check of the initial statement, admits that those are proper counterexamples and that his conjecture is wrong. Then he tries to find a way to deal with the issue. He thinks out loud how to fix the situation in the spirit of the “If not, what yes?” question. Boris: Well, it is pity, but my conjecture is wrong as it does not account for your examples. Apparently, it is not that easy to discover a new theorem in mathematics. However, let me try. First, to neutralize the first counterexample, I should assure that the formulation only concerns non-constant 1 1 functions. As to y = and y = 2 , these functions are not defined at x x x = 0, so the formulation should include a condition restricting it to functions determined at x = 0. So, what do you think about the following statement (referred hereafter as Formulation 1): Formulation 1:  Every even and nonconstant function which is defined for all real numbers has a local strict extremum at x = 0. At this point, the participants are usually split into those who try to independently prove Formulation 1 and those who feel that the formulation is not “clean enough” and try to construct a new counterexample. With no exceptions, the latter participants arrive first at ideas for a whole group discussion. Two ideas are usually suggested: Response 3:  Formulation 1 is wrong as it does not work, for example, for the functions sketched on Fig. 6.2: Response 4:  Formulation 1 is wrong as it does not take into account, for exam1  ,x ≠ 0 ple, the function y =  x 2 (see Fig. 6.3).  1, x = 0 Discussion of Response 3 usually starts from attempts to represent the functions sketched on Fig. 6.2 analytically as some participants feel that graphical representations are not convincing enough. For instance, appropriate formulas for the functions

Fig. 6.2  Counterexamples to Formulation 1 (Response 3)

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Fig. 6.3  “Counterexample” to Formulation 1 (Response 4)

on Fig. 6.2 are y = |x − 1| + |x + 1| and y = (−1)[|x|], respectively. Then the question “If not, what yes?” is asked again, but this time the instructor encourages the participants to help him in modifying Formulation 1. One way to neutralize Response 3 is as follows: Formulation 2:  Every even function defined for all real arguments that is nonconstant in a neighborhood of x = 0 has a local strict extremum at x = 0. Whole class discussion of Response 4 turns to be very different from the previ1  ,x ≠ 0 ous discussions. This is because the function y =  x 2 is not a proper coun 1, x = 0 terexample. Indeed, the function satisfies all conditions of Formulation 1 and Formulation 2, yet x = 0 is a local minimum of the function. Indeed, y(0) = 1 and y(x) > 1 for x ∈ (−1, 0) ∪ (0, 1). This conclusion is not obvious for many participants, so the natural need for recalling the formal definitions of an even function and a local extremum arises. This adds a new dimension to the discussion and supports the feeling that Formulation 2 is correct. The following remarkable dialogue occurs at this point in one of the workshops with one of the students, Peter. Peter: Well, the formulation [Formulation 2] looks right, but it cannot be right as you said that it does not appear in any algebra or calculus textbook.

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Boris: No, I said so about the initial formulation, and you’ve seen that the last formulation was created on the spot. It is completely up to us – to prove or refute Formulation 2. From this moment, the feeling that some new, unknown also to the teacher, result can be reached becomes dominant and creates a very special atmosphere in the classroom. In Brousseau’s (1997) terms, the responsibility for obtaining valuable mathematical outcomes is devolved in the participants. Typically, most of the students work on proving Formulation 2 using the previously recalled definition of even functions, and some participants try to change the function suggested in Response 4 to make it a proper counterexample. After a while, both groups of participants arrive at an idea to incorporate a condition of differentiability or at least continuity into the formulation. The former ones do since they wish to utilize the mathematical apparatus associated with the extremum concept, while the latter ones do since they feel that a natural way to improve the last ­counterexample is to make the functions under consideration continuous. The teacher suggests: Boris: I see that there is no progress in proving or refuting Formulation 2 and that several groups consider continuity. Well, it happens that when a mathematician does not know how to cope with some conjecture, he or she works for a while on a weaker statement, that is, the statement with more restrictive conditions. Have you heard about the Abel prize, one of the most prestigious prizes for research mathematicians? It is named after Niels Henrik Abel, a Norwegian mathematician who proved that it is impossible to solve polynomial equations of 5th degree in radicals. So, Abel used this strategy a lot. Therefore, I think it would be all right if we leave for a while Formulation 2 and think of Formulation 3. Formulation 3:  Every continuous even function defined for all real arguments which is nonconstant in a neighborhood of x = 0 has a local strict extremum at x = 0. Surprisingly for most of the participants, Formulation 3, and in turn Formulation 2, appear wrong. The counterexample to both formulations is as follows: 1   x sin , x ≠ 0 (see Fig. 6.4). y= x 0, x=0 This function is frequently mentioned in university-level courses of calculus. It is continuous, even, defined for all real numbers, nonconstant in a neighborhood of x  =  0, and does not have a local strict extremum at x  =  0. Indeed, by definition y(0) = 0 and in any neighborhood of x = 0 it takes either positive or negative values. In some cases, the participants discovered this counterexample independently. In other workshops, the teacher prompted them by saying “Think about ‘pathological’ examples of continuous functions you’ve learned in an undergraduate calculus course.”

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Fig. 6.4  A counterexample to Formulation 2 and Formulation 3

R: Coming back to Lakatos, can a pathological example be a legitimate exception? B: It is called “pathological” in calculus courses, but it is not a monstrous example invented especially in order to fail the conjecture. In my view, it is legit.

After the whole group discussion of the above function, the next natural modification, namely, incorporation of the condition of differentiability is considered. Formulation 4:  Every differentiable even function defined for all real arguments which is non-constant in a neighborhood of x = 0 has a local strict extremum at x = 0. At this point, most of the participants feel that the venue of differentiability has no potential in constructing a correct statement about extrema of even functions at x = 0. However, a counterexample should be constructed to justify that feeling. In some workshops, it is built by the participants independently, in two steps. First, the following qualitative analysis of the expression x sin 1x can be considered. In this expression, the factors x and sin 1x interact so that the component sin 1x oscillates between the functions y =  − x and y = x, and the function x sin 1x is not ­differentiable at x = 0 for the same reasons as y =  ∣ x∣. To build a counterexample to Formulation 4, one can substitute the boundary function y = x by y = x2 in order to make the

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component sin 1x oscillate between y =  − x2 and y = x2, which are differentiable at  x 2 sin 1x , x ≠ 0 x = 0. Thus, the function y =  appears. x=0 0, Second, the student note that the latter function is still not a proper counterexample as it is odd, not even. However, it is not difficult to fix, for instance, by one of the following modifications:



 x 2 sin 1x , x ≠ 0  x 2 sin 12 , x ≠ 0  x 2 cos 1x , x ≠ 0 x or y =  or y =  y= x=0 x=0 x=0 0, 0, 0,



The last function is sketched in Fig. 6.5.  x 2 cos 1x , x ≠ 0 The function y =  satisfies all conditions of Formulation 4 but x=0 0, does not have an extremum at x  =  0. This counterexample helps the participants appreciate the fact that discovery of a new theorem in calculus is a challenging task with no assured success. In turn, this stimulates the most persistent participants to look for alternative ideas. Typically, the described “If not, what yes?” 90-min lesson includes 5–7 formulation-counterexample-reformulation cycles and ends up with the following conjecture: Formulation 5:  Every even function that is monotonic and not constant in a right-­side neighborhood of x = 0 has a local strict extremum at x = 0.

Fig. 6.5  A counterexample to Formulation 4

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In many cases, the participants, who are prospective or in-service teachers with considerable mathematics background, are able to prove Formulation 5 independently or with minimal assistance. Indeed, its proof straightforwardly relies on three definitions – those of an even function, of a monotonous function, and of a local extremum. Of note is that many students do not like Formulation 5 even after completion of its proof, on the grounds that it is not particularly surprising or elegant. In spite of some disappointment at the end of the lesson, the students usually appreciate their “If not, what yes?” experience. In the words of one of them, “It was something real…that’s how mathematicians work.” The feeling of “working on something real” is sometimes so strong that the students are very surprised when the teacher distributes by the end of the lesson the previously prepared account of the lesson, including the presumed sequence of formulations and counterexamples in relation to the initial statement. As a rule, the presumed sequence is remarkably close to the actual sequence. On the one hand, it means for the participants that their mathematical thinking and behaviors could have been predicted, at least partially. Some of them even feel that their free will was somehow predetermined during the lesson. On the other hand, the account serves as evidence for that a teacher conducting a “If not, what yes?” activity can be responsive to the students’ in-the-moment ideas but still well prepared. By the end of the described lesson, the teacher invites the participants to keep thinking about a more elegant formulation of the looked-for theorem on extrema of even functions as well as about additional topics for “If not, what yes?” activities.

6.3.3  D  etour: A Happy End and (Somewhat Awkward) Follow-Up of the Story2 R: Actually, I agree with the students: the process is interesting, but its mathematical result, Formulation 5, is not so much. B: I also agree. This was disturbing for me for several years. But let me tell you a story that resolved this disturbance, though it created a new one. The story occurred when I taught course “Selected problems in mathematics” in a class, in which a particularly bright student, Marilyn, took part. Doing “business as usual,” the teacher taught the above “If not, what yes?” lesson and by the end of the lesson encouraged the students to keep looking for some correct and interesting statement about extremum of even functions. A night before the next lesson, Marilyn sent the teacher an e-mail. The e-mail contained Formulation 6, as follows, and its two-page proof.

 This story is presented on the basis of a section in Koichu (2010).

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Formulation 6:  Every even, nonconstant twice differentiable in a neighborhood of x = 0 function f(x), such that f  ′′(0) ≠ 0, attains a local strict extremum at x = 0. B: I was happy, though a small cloud remained at the horizon: Marilyn’s proof contained a mistake, but a fixable one (see below). Without disclosing my opinion about the correctness of the statement, I suggested to Marilyn to present it at the lesson.

Technically, Marilyn's proof was shown to the class on a large screen as a Word document, and the teacher assisted her to shape the presentation by modifying the document in a real-time mode. During the presentation, the proof was divided into steps, which were given appropriate titles. This instructional approach is known as structuring proofs (Leron 1983). It helped Marilyn and the rest of the students to understand the proof better. At the beginning, Marilyn mentioned two lemmas that were well-known to the students from their first-year calculus course. The lemmas can be found, for example, in Salas and Hill (1974). Lemma 1:  Let g(x)  be twice differentiable at the point x0 and g′(x0)  =  0. If g′′(x0) > 0 (or g′′(x0)  0 (or g′′(x)  0. By Lemma 1, x = 0 is a local minimum point (not necessarily strict!), so there is a neighborhood of x = 0 in which for every x, f(x) ≥ f(0). Step 2: proving strict convexity of f Any neighborhood of x = 0 includes a closed interval; therefore, by Lemma 2, the function f is strictly convex on that interval. Step 3: reductio ad absurdum If every x in the closed interval satisfies f(x) > f(0), then we are done. Otherwise, there exists a point x1 in the interval such that f(x1) = f(0). The function f is nonconstant in the neighborhood of x = 0; hence, there exists a point x2 such that 0  0 so that f(x1) = f(0) in the neighborhood of x = 0. As x = 0 is a local minimum point, there exists a neighborhood for which f(x) > f(0). Hence, f has a local strict minimum at x = 0.

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It was a little difficult to follow the proof; in addition, the class knew that it had just been created. Therefore, it was decided to double-check the proof. The students moved slowly through the steps trying to be extra critical. With a little help, Marilyn eventually realized that Step 2 was problematic. Namely, the use of Lemma 2 was not sufficiently justified. Indeed, Formulation 6 requires f   ′′(0) ≠ 0, while Lemma 2 requires f   ′′(x) > 0 in some closed interval containing x = 0. In other words, Lemma 2 can be used only when f   ′′(x) is continuous in a neighborhood of x = 0. This understanding immediately led us to Formulation 7. Formulation 7:  Every even, nonconstant function f, twice continuously differentiable in a neighborhood of x  =  0 such that  f   ′′(0)  ≠  0, attains a local strict extremum at x = 0. In fact, this statement (Formulation 7), not Formulation 6, is proved above. B: We were all happy that the mission seemed to be accomplished as we had found a nice, and to our knowledge new, property of even functions. R: Happy for you! I can imagine your thrill at this moment. But I also hear a reservation touch… B: Indeed. As will be evident shortly, our accomplishment, as nice as it was, already bore seeds of a danger pointed out by Lakatos for the exceptionbarring method. Back to the story.

The question about the correctness of Formulation 6 remained open as we neither proved it nor found a counterexample. Marilyn decided to keep looking at two directions: to look for a counterexample and also search the literature for an idea that could help. After several days of searching, Marilyn found the following lemma in a textbook Advanced Calculus by Buck (1987). Lemma:  Let f : Rn → R be a real-valued function, whose second partial derivatives are continuous. Let H( f) be the Hessian matrix of f. If ∇f (x0) = 0 at some point x0, and H(f) is positive (or negative) definite at x0, then f attains a local strict minimum (or maximum) at x0. This lemma strengthened the feeling that Formulation 6 can be refuted as the above lemma does require continuity of the second partial derivatives for assuring existence of a local strict extremum. For this reason, Marilyn decided to concentrate on looking for a counterexample. In a way, her reasoning at this stage was similar to the reasoning of a student, Peter, quoted above. Namely, she felt that Formulation 6 is most likely incorrect since if it were right, it or its equivalent would appear in a textbook. Marilyn looked for a counterexample to Formulation 6 for several weeks. The main idea was to somehow modify the counterexamples produced during the lesson, such as the one presented in Fig. 6.5. The best attempt was building a function with the following properties: it is even, nonconstant, differentiable in a neighborhood of x = 0; its second derivative exists but is not continuous at x = 0; the function

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Fig. 6.6  Attempted counterexample to Formulation 6

attains a minimum at x = 0, but the minimum is not strict. The function follows:  x 4 sin 2 1x , x ≠ 0 f ( x) =  (see Fig. 6.6). x=0 0, This function is not a counter example to Formulation 6 for one reason: its second derivative at x = 0 equals zero. Many modifications of this function were considered while trying to make its second derivative non-zero, but finally none of the modifications worked. Marilyn also tried to think of alternative ideas, but unsuccessfully. At this stage she started to heavily use MATLAB for sketching functions. Logically speaking, there was little chance that a function whose graph does not look appropriate for our needs on the interval (−10−3; 10−3) would behave better on the interval (−10−30; 10−30). In spite of this understanding, Marilyn spent many hours trying to sketch functions in narrower and narrower neighborhoods of zero. Finally, she gave up. Only several months later the story got to a mathematical closure: a counterexample to Formulation 6 cannot be constructed! This is because Formulation 6 and Formulation 7 are both true. As a matter of fact, they are both covered by a more general statement, which was suggested by Kenneth Shilling as a solution to a problem published in the American Mathematical Monthly in 2000 (AMM, vol. 107, p. 958).

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Final Formulation:  If f(x) is an even function and f  ʺ(0) exists and is non-zero, then f(x) has a strict local extremum at x = 0. Proof:  Without loss of generality, suppose that f  ʺ(0) > 0. Since f(x) is even, f ′(0)  =  0  (see Step 1  in the “proof” of Formulation 6). Therefore, by definition, f ′( x) f ′′ ( 0 ) = lim > 0 , so there exists h > 0 such that f ′(x)   0 for x in the interval (0, h). Therefore, f(x) is decreasing on (−h, 0) and increasing on (0, h), which implies that f(0) is a strict local minimum at x = 0, QED. R: So brief, so elegant! And it is nice that the condition of continuity of the second derivative, all this heavy stuff, is not needed at all. B: Indeed! However, I must admit that when Kenneth Shilling sent me a reference to this proof, my spontaneous reaction was "How could I miss such a simple formulation?" Marilyn's reaction was similar. We both felt that the solution was within our reach. Interestingly, according to Shilling, the simple solution was missed also by many other people3. I analyze this phenomenon in Koichu (2010). While being engaged in exception-barring as a way of inventing a correct statement in accordance with “If not, what yes?” instructional strategy, the students were engaged, in terms of Lakatos (1976) in “strategic retreat hoped to be a stronghold of the conjecture” (p. 125). Namely, they repeatedly practiced inventing modifications of the primitive conjecture, in response to the apparent counterexamples, towards determining the domain of safety of the conjecture. On the one hand, a valid formulation (Formulation 7) was proved, and this is definitely a remarkable achievement for a mathematics education course. On the other hand, a more general and more elegant correct formulation within the students’ reach was overlooked. This overlook would definitely not count for success in the mathematics community. Therefore, as much as the “If not, what yes?” method can engage students in what mathematicians sometimes do, it is probably not a proper approximation of their most praised practices. We choose to conclude this section by presenting the voice of Lakatos’ Teacher, who argues for pros and cons of the exception-barring method in the context of Euclidean polyhedra. Teacher [talks to Beta]: You are playing for safety. But are you as safe as you claim to be? You still have no guarantee that there will not be any exceptions inside your stronghold. Besides, there is the opposite danger. Could you have withdrawn too radically, leaving lots of Euclidean polyhedra outside the walls? (Lakatos, 1976, p. 125).

 See the editorial comments in AMM vol. 107(5), pp. 463-464, and in vol. 107(10), p. 958.

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6.4  “If Not, What Yes?” A Design Aspect As mentioned, the above scenario has arisen from one of our own mathematical disturbances. Fortunately, the mathematical content of this disturbance was more or less aligned with the mathematical background of our students, in particular their proficiency in calculus, so it could be transposed to the lesson. However, the above story brings forth another everlasting disturbance of ours, scalability. If we accept the potential of the presented instructional method for engaging students in mathematician-­like experiences (but see the reservations above!), how can we, or our student-teachers, design “good” primitive conjectures for “If not, what yes?” lessons on various topics and in ways that would be aligned with various mathematical backgrounds of students? At a first glance, the design strategy looks quite straightforward. Similarly, the “What if not?” method can look straightforward, but only at a first glance. For instance, one can think of any nontrivial theorem (analogue of Level 0 in what-if-­ not, see the Background section), list systematically all apparent conditions included in the theorem (analogue of Level 1), then ask “What if not?” about each condition (analogue of Level 2), and then, instead of alternating the conditions as prescribed by the “What if not?” method, to look for counterexamples that enter the stronghold of the theorem when its conditions are removed one by one (analogue of Levels 3 and 4). Of course, the process is not as linear as it might look. After presenting and illustrating their “What if not?” method as an emerging scheme for posing new problems, Brown and Walter (1993) warned their readers that they should not adopt the scheme in a mechanical manner in their classes, but rather modify it to fit their own needs. The same holds for “If not, what yes?” Consider two examples proposed by Marilyn’s classmates after the “If not, what yes?” lesson. It is perfectly understandable that undergraduate university students attended to what they had recently learned in their mathematics courses.

6.4.1  Student-Generated Example in the Context of Calculus Rolle’s Theorem, usually studied in proof-oriented calculus courses for first-year university students, states: If a function f is continuous on the closed interval [a, b] and differentiable on the open interval (a, b) such that f(a) = f(b), then f  ′(x) = 0 for some x with a ≤ x ≤ b. Sarah choose this theorem as a basis for designing an “If not, what yes?” activity, as follows. Formulation 1  For any function defined on the closed interval [a, b] so that f(a) = f(b), there exists x0 ∈ [a, b] so that f  ′(x0) = 0.

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 x 2 , x ∈ [ −1, 0 ) ∪ ( 0, 1] Counterexample: ( x ) =  . The counterexample exploits a  1, x ≠ 0 noncontinues function, hence, Formulation 2:  For any continuous function defined on the closed interval [a, b] so that f(a) = f(b), there exists x0 ∈ [a, b] so that f  ′(x0) = 0. Counterexample: f(x) = |x| for x ∈ [−1, 1]. This time the function serving as a counterexample is not differentiable. Adding the condition of differentiability leads exactly to the Rolle’s Theorem formulation, which is correct and thus is a final formulation.

R: Formally speaking, Sarah designed a three-step “If not, what yes?” activity, which can illuminate for learners the role of continuity and differentiability in the Rolle’s Theorem. And it looks like she designed the activity in accordance with the aforementioned levels 0–4. However, some artificiality shows. This is probably because Formulations 1 and 2 include a derivative of a function without explicitly requiring that the function is differentiable. B: Indeed. As mentioned, following the scheme does not guarantee a good-­ quality result. Let us look at one more example.

6.4.2  S  tudent-Generated Example in the Context of Number Theory Lena chose the Wilson theorem that she had studied at her number theory course4: For any integer p > 1, (p − 1)!  + 1 is divisible by p if and only if p is prime. More precisely, Lena chose the “only if” direction as a basis for her design: Any prime number p divides (p − 1)!  + 1. Though it looks like this theorem does not have many conditions to play with, Lena nicely decomposed the fact that for the theorem be true, p must be a prime number. She treated this as a strong restriction in relation to the entire set of integers, and considered the following “weakened” cases: What if p is not prime but any odd number? The following primitive conjecture stems from this question: Formulation 1:  Any odd number p divides (p − 1)!  + 1 Supporting examples: 4   The proofs can be Wilson%27s_Theorem

found

here:

https://artofproblemsolving.com/wiki/index.php/

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p = 3, ( 3 − 1)! + 1 = 3; p = 5, ( 5 − 1)! + 1 = 25; p = 7, ( 7 − 1)! + 1 = 721, 721 ÷ 7 = 103. Counterexample: p = 9, (9 − 1) !  + 1 = 40,321,40,321 ÷ 9 = 4,480.111… Then she observed that “9” is a square number. The following modification excludes it. Formulation 2:  Any odd and non-square number p divides (p − 1)!  + 1 Supporting examples: p = 3, 5, 7 see above. In addition : p = 11, (11 − 1)! + 1 = 3, 628, 801, 3, 628, 801 ÷ 11 = 329, 891 p = 13, (13 − 1)! + 1 = 479, 001, 601, 479, 001, 601 ÷ 13 = 36, 846, 277 Counterexample: p = 15, (15 − 1)! + 1 = 87, 178, 291, 201, 87, 178, 291, 201 ÷ 15 = 5, 811, 886, 080.066 … And then, having the supporting examples for p = 3, 5, 7, 11, 13, Lena proposed the Wilson theorem as the final formulation. R: It looks nicer than Sarah’s example. Perhaps because Lena thought not only about counterexamples but also about supporting examples. But the numbers in the above computations increase so quickly… Is there a shortcut? B: As a matter of fact, there is. For p = 15, Sarah observed that 14! includes 3 and 5 among the factors, and thus 15 obviously divides 14 !  = 1 · 2 · 3 · 4 · 5 · … · 12 · 13 · 14. As a result, 15 obviously cannot divide 14! +1. The same could have been done for p = 9. R: Nice! Lena’s choice of the Wilson theorem makes me rethink an example of the Euler’s polynomial, which we discussed in the Introduction section. I wonder how Leonard Euler did the checking of the values of the polynomial F(x) = x2 + x + 41 for the first 40 integers. B: Of course, we cannot know for sure, but his letter to Bernoulli in 1792 implies that Euler did a lot of computational work when checking his hypotheses. By the way, this example convincingly shows that even 40 supporting examples in a row cannot guarantee that there is room for a generalization that would work. R: Anyway, the Euler’s polynomial can hardly be a topic for a school-level “If not, what yes?” activity, though, as we know, it inspired several firstclass mathematicians, including Legendre and Goldbach, to look for other polynomials that generate many prime numbers in a row. B: Why do you think that Euler’s polynomial cannot be turned into “If not, what yes”?

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R: Well, just because the “yes” part is still murky, and in school you normally wish to get to some closure of an exploration. B: Sounds as a design principle for “If not, what yes?” activities. Another one is that the primitive conjecture should rely on many supporting examples within the reach to the students.

Our collection of “If not, what yes?” primitive statements also includes examples proposed by in-service teachers. For example, at one of the workshops the teachers suggested quite a long list of possible primitive statements after just 10 min of work in small groups. These are some of them, in the context of functions. • For a differentiable function to have an inflection point at x0, its second derivative must be zero at x0. Supporting example: y = x3, counterexample: y = x ·  ∣ x∣. • If f′(x) = 0 on the domain of the function, then f(x) is a constant.  1, x > 0 Supporting example: y  =  1, counterexample: =  . Note the trick: −1, x < 0 domain for the latter function is (−∞ ; 0) ∪ (0, +∞) and not all real numbers. It can be a matter of debate to which extent this counterexample is “monstrous.” • If a function has a horizontal asymptote for x →  +∞, then the function must have a horizontal asymptote also for x →  −∞ . Supporting example: any rational function; counterexample: y = 2−x. • If a function y = f(x) has an horizontal asymptote at x →  +∞, then it has the same asymptote at x →  −∞. Supportive example: any rational function; Counterexample: y =  arctan x. All the above primitive statements seem to invite rather brief two-step activities of the format “primitive statement – counterexample – final statement.” However, they are all sourced in the teachers’ knowledge on widespread student misconceptions and in nuanced understanding of school curricula, instead of anchoring the statements to known theorems. For this reason, the in-service teachers’ two-step “If not, what yes?” activities look to us more natural than the above three-step activities proposed by pre-service teachers. The teachers appreciated the fact that instead of warning the students against making a particular mistake  – as we know, this approach is problematic (e.g., Smith, diSessa & Roschelle, 1993) – they can play with them in “If not, what yes?” After consideration of the above examples, we are now in better position in order to discuss similarities and differences between the “If not, what yes?” method and the “What if not?” method. As argued, both methods are based on systematic modifications of an initial statement, posing new problems and exploring their mathematical correctness. However, the mechanism of producing and justifying conjectures in these two methods are different. First, a “What if not?” activity begins

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from a correct statement, which is opened for exploration by alternating its conditions. Designing an “If not, what yes?” activity can also begin from consideration of a correct statement as a final formulation, but then the final statement is worked backwards in order to offer the students a plausible (on the virtue of supporting examples) but incorrect initial statement to be refuted by a counterexample. Second, the “What if not?” method leads to numerous conjectures, each of which is usually a one-step modification of the initial statement. The “If not, what yes?” method involves a gradual process of refining the initial statement, in which every new conjecture is a step further from the initial statement than the previous one. Third, conjectures produced in “If not, what yes?” process are intrinsically necessitated – they appear as neutralizers of particular counterexamples. Also, analysis of the suggested conjectures in an “If not, what yes?” activity is being conducted for many times and not only at the end of the activity, as in the “What if not?” activity. In fact, problem posing and problem solving are interlaced in an “If not, what yes?” activity; the same holds for conjecturing and proving. In closing, it looks like designing a good initial statement, and conducting a meaningful “If not, what yes?” lesson, involves various design considerations. Besides those mentioned so far, it requires in-depth thinking on possible responses of learners and detailed knowledge of their mathematical background and experiences. It requires the envisioning of several multi-step trajectories of the lesson and, consequently, solid knowledge of students' epistemology. In addition, it requires sensitivity to the students' ideas, readiness to be flexible, and, last but not least, a high level of self-confidence.

References Brousseau, G. (1997). Theory of didactical situations in mathematics (N. Balacheff, M. Cooper, R. Sutherland, & V. Warfield: trans. and eds.). Dordrecht: Kluwer. Brown, S., & Walter, M. (1993). The art of problem posing. Mahwah: Lawrence Erlbaum. Buck, R. C. (1987). Advanced calculus. New York: McGraw-Hill. Koichu, B. (2008). If not, what yes? International Journal of Mathematical Education in Science and Technology, 39(4), 443–454. Koichu, B. (2010). On the relationships between (relatively) advanced mathematical knowledge and (relatively) advanced problem solving behaviours. International Journal of Mathematical Education in Science and Technology, 41(2), 257–275. Lakatos, I. (1976). Proofs and refutations: The logic of mathematical discovery. Cambridge: Cambridge University Press. Leron, U. (1983). Structuring mathematical proofs. The American Mathematical Monthly, 90, 174–185. Salas, L., & Hill, E. (1974). Calculus. Lexington: Xerox. Smith, J.  P., diSessa, A., & Roschelle, J. (1993). Misconceptions reconceived: A constructivist analysis of knowledge in transition. Journal of the Learning Sciences, 3(2), 115–163. Tall, D., & Vinner, S. (1981). Concept image and concept definition in mathematics, with special reference to limits and continuity. Educational Studies in Mathematics, 12, 151–169.

Chapter 7

From “Obviously Wrong” Methods to Surprisingly Correct Answers

7.1  Introduction We stated previously that the tasks described in this book are triggered by our previous interactions with our students, many of whom are prospective teachers. There is a repeated “experience of disturbance” when students draw erroneous conclusions or offer suggestions which lead to erroneous conclusions. However, at times, ideas that appear at first as “obviously wrong” lead to a correct answer. Exploration of these ideas leads to new insights and some surprises. It further also leads to new insights into mathematics and deeper appreciation of mathematical relationships. The idea of “capitalizing on errors” (Borasi 1994) is not new. In fact, any experienced and thoughtful teacher attempts not to reject immediately a student’s (erroneous) thinking but engage the students in an activity that would lead to correcting the error. We follow (or at least attempt to follow) the strategy of “capitalizing on errors” in teaching. Moreover, there are some student approaches that were initially presumed as “errors” that have attracted an extended attention. These approaches become “tasks” for the next generation of students. We refer to such tasks as “capitalizing on surprises.” We begin with an example of a solution that students considered as wrong but was “obviously correct” for the teacher. We then turn to three examples of student solutions that led to correct results, but in which the implemented methods were thought of as wrong by a teacher. We describe the resulting explorations of these examples that led to mathematical insights, insights that we share with new groups of students.

© Springer Nature Switzerland AG 2021 B. Koichu, R. Zazkis, Mathematical Encounters and Pedagogical Detours, https://doi.org/10.1007/978-3-030-58434-4_7

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7.2  Tax or Discount? What should be calculated first, tax or discount? For example, if there is 30% discount on a suitcase and the tax is 14%, how does one calculate the price to be paid? Note that in some countries the tax is included in the price presented to customers, while in Canada the price-before-tax is displayed to the buyer and the tax is added to the displayed price when a payment is made. When the question was presented for discussion in a class of prospective elementary school teaches, the students argued vigorously that discount should be applied first, otherwise the customer is overcharged. Most assertive were several students with experience in retail, who claimed that “that’s how it is done.” One student even pulled out a receipt from a recent purchase as irrefutable evidence that applying discount after tax was “obviously wrong.” Following some deliberation, students were asked to calculate “both ways.” That is, to calculate the final price to be paid when discount is applied before tax and when tax is applied before discount. To a surprise for most, both ways led to the same result. However, many students were not confident in the results and repeated the calculations for different choices of numbers for the initial price, for the discount, and for the applied tax. Numerical variation confirmed the initial conclusion that the order of applying tax and discount does not matter for the amount to be paid by a customer. Actually, in addition to the students’ surprise with the calculation results, what was found noteworthy was the popular approach to this calculation. To exemplify such an approach, consider that the price of the suitcase was $100, 30% discount on sale, 14% tax. 30 % of 100 = 30 100 – 30 = 70 14 % of 70 = 9.8 70 + 9.8 = 79.8

calculating the amount of discount finding the price after discount is applied calculating the amount of tax calculating the price to be paid

OR 14 % of 100 = 14 calculating the amount of tax 100 + 14 = 114 finding the price after tax is applied 30 114 × = 34.2 100 ⇒ 30 % of 114   = 34.2 calculating the discount 114 – 34.2 = 79.8

calculating the price by subtracting the discount

Of course, the calculation makes sense if one is interested in the amount of discount or the amount of tax, which is indeed the case for bookkeeping. But if we are interested only in the total price for the customer, these calculations look unnecessarily complicated. It took a while in a class of prospective elementary school teachers to establish that the amount after 14% tax can be calculated with one operation, multiplication by 1.14. Similarly, amount after 30% discount is found by multiplying the initial amount by 0.7.

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Even for students familiar with this method, its justification was not obvious. It took time to establish that the price after the tax of 14% is added equals to 114% of the price before the tax. Similarly, the price after applying 30% discount equals to 70% of the price before the discount. These steps, that may appear obvious to a mathematically inclined reader, were essential to proceed towards a generalization. For the initial price of P, applying tax after discount results in 1.14 × ( 0.7 × P ) and applying discount after price results in

0.7 × (1.14 × P )



Luckily, by associativity and commutativity, both expressions are equivalent regardless of the price P. As a result of this exercise, the “obviously wrong” method was not considered by students as wrong any longer. The consideration of a “real world” situation related to tax and discount resulted not only in learning a faster method of calculating percentages but also in revisiting properties of arithmetic operations with (integer, rational, real) numbers: distributive, commutative, and associative. As a follow up exercise, we like to offer the following task: Estimate, what is bigger, 27% of 63 or 63% of 27? This could be a tricky question, as the “correct” answer is not given as one of the possible choices in the question’s formulation. The choice of numbers is deliberate to avoid easy mental calculations. Also the request to estimate guides away from thinking about calculations. However, the solution presents a wonderful opportunity to revisit properties of operations, as no calculation is needed to conclude that 0.27 × 63 = 0.63 × 27 This story sets the stage for the next one, which is also concerned with arithmetic operations and their properties.

7.3  S  urprises with Order of Operations: A Story in Three Parts Before turning to our story, we invite readers to consider the following claim: “According to the established order of operations, division should be performed before multiplication.”

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This claim, suggested by a teacher, initially appeared as “obviously wrong” to ourselves, as well as to many students who learned the conventional order of operations in arithmetic. That is, while multiplication is given priority over addition and subtraction, multiplication and division are to be attended to in order of their appearance, left to right. However, the above claim surprisingly at first, leads to correct results when only division and multiplication are involved. (To exemplify, consider solving 10  ×  20  ÷  5  in order presented, vs. performing division first, that is, 10 × (20 ÷ 5)). In what follows we share with the reader how, capitalizing on our surprise, the surprise for students was initiated, how it was explored with different groups of teachers, and how it enriched our repertoire of tasks used in teacher education.

7.3.1  Part 1: Mathematical Conventions Task One of the assignments for secondary mathematics teachers enrolled in a problem-­ solving course was to consider mathematical conventions. This assignment followed discussion of the choice of a particular mathematical convention: superscript (−1) that appears in different contexts: fractions, functions, or matrices. In prior research, Zazkis and Kontorovich (2016) analyzed prospective secondary teachers’ explanations of the “curious appearance” of superscript (−1) in two contexts  – inverse of a function and reciprocal of a fraction. Consider, for example, f−1 and 3−1. The superscript (−1) points to an inverse of a function f in one case and to a recipro−1 cal of a fraction in the other case. While 3−1 = 13 , f ≠ 1f . It was found that the majority of participants do not attend to the inverse notion with respect to different operations, that is, they do not view “reciprocal” as a multiplicative inverse. Rather, the differences between the contexts were emphasized, that is, the meaning for the symbol (−1) that appears in superscript should be determined differently in the context of functions vs. the calculations with numbers and fractions with negative exponents. The participants also referred to analogy with other words and symbols, whose meaning is context dependent. In the conversation with a different group of teachers about the superscript (−1), similar ideas have initially been voiced, but later an agreement converged towards group-theoretic perceptions of inverse, as exemplified in two different contexts. This provoked interest in the choice of other mathematical conventions, conventions that are often introduced and perceived as givens, without any particular explanation. Mathematical Conventions Task was designed to address this interest.

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Mathematical Conventions Task Choose a mathematical convention and consider possible explanations for the particular choice. In your submission: 1. Reflect on the process of choosing the particular mathematical convention for this task. Share other conventions that you considered for this task and explain why they were not chosen. (1–2 paragraphs) 2. Write a script for a dialogue in which interlocutors consider possible explanations for the convention you explored. The dialogue should reflect possible doubts, uncertainties, and arguments regarding the suggested explanations. The dialogue should end either with an explanation that interlocutors accept or a summary of the disagreement between the characters. (3–5 pages) The dialogue can begin in the following way: Sam: Dina: Sam: Dina: Sam: Dina:

Hey Dina, have you ever noticed that (the chosen convention)? Well, everybody knows that. Yes, but did you ever think about why it is so? Why should I think about it? It’s a convention. But, still… Can you propose an explanation? Maybe, this is because…

Feel free to modify the proposed beginning of the dialogue. 3. What have you learned, if anything, from completing this task? (1–2 paragraphs)

7.3.2  Part 2: Convention: Order of Operations One of the repeated examples for a convention (chosen by 4 out of 16 students) was order of operations when performing arithmetic calculations. We present the following excerpt from the script written by Andy, who describes an imaginary conversation occurring in Grade 8 class and a follow-up class activity. An abbreviated account of these is analyzed in Zazkis (2017) utilizing the notions of local and nonlocal mathematical landscape, developed by Wasserman (2016a, b). Sam: Hey Mr. X, a couple of us can’t decide on the answer to the following question: 25 + 5 × 7 − 2 × 10 ÷ 5 Mr. X: What do you mean? Mary: I bet them a dollar that they couldn’t get the correct answer to this exercise. Sam: Well I got 40. Jane says it’s 56. Cam believes it’s 436, and no one can agree on a solution. Cam: Mine is correct! I know it.

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Mr. X: Cam: Mr. X: Cam:

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Cam, why do you say that? I had a process of how I did mine. How so? I just did one operation after another: 26 plus 5 times 7 and so on. See: 25 + 5 × 7 − 2 × 10 ÷ 5 30 × 7 − 2 × 10 ÷ 5 210 − 2 × 10 ÷ 5 218 × 10 ÷ 5 2180 ÷ 5 436

Sam:

I did something similar, but I started on the right side of the problem:

25 + 5 × 7 − 2 × 10 ÷ 5 25 + 5 × 7 − 2 × 2 25 + 5 × 7 − 4 25 + 5 × 3 25 + 15 40 Mary: You guys did the operations in the wrong order. Jane: I agree with you Mary. Mr. X: What order would you suggest? Jane: Well I did the division first, followed by multiplication, addition, and subtraction. 25 + 5 × 7 − 2 × 10 ÷ 5 25 + 5 × 7 − 2 × 2 25 + 35 − 4 60 − 4 56 Cam: I don’t understand why you started with division. Why would you start there? Jane: Everybody knows that’s the proper order to do operations. Sam: Mr. X, is that correct? Mr. X: Jane is correct. That is the correct order to do those operations. Sam: But why? Mr. X: A long time ago, a group of people has had a very similar situation that we have now. They were confused and couldn’t figure out who had the correct solution to a problem that involved the very operations you are having problems with. It happened around the early 15th century in a small kingdom, which was called the Kingdom of Math. The King of Math, as it

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were, was a very intelligent leader and believed that his people should always come together to solve their problems. Jane: Really, Mr. X, a Kingdom of Math? Mr. X: Oh yeah, they were a very progressive country. Several of the King’s subjects had come to him to settle a problem that they were having. They couldn’t decide on an order of the operations that needed to be used. Sounds familiar? Mary: Very funny, Mr. X. Cam: So what did the King do? Mr. X: The king commanded his most trusted advisors, members of the Order of Knowledge, to look into the problem. It took several months before the Order had a response for the King. They proposed that the only way to solve this problem was for the King to proclaim an order to the operations so that everyone would know the correct way to solve the mathematical problem. Sam: That makes sense. Then everyone would follow the same order and no one would be confused about what steps to do first. Andy offered the following comment at the end of the assignment: I felt that there was only one reason that I could [offer] students: “We need to have an order that everyone follows so we can be consistent.” “This is the way we all do it.” “We” being us, the math community. Whether you’re in France, New Zealand, or Canada, it’s the same. This is because we’ve all agreed to use the same order so as to have the same understanding of the operations. I tried to find an actual history of the order of operations, but couldn’t find anything concrete. So I decided to make up a story that would hopefully give them some connection to the problem and some entertainment along the way. Andy’s decision to make his teacher-character Mr. X agree with the student-­ character statement, “division first followed by multiplication, addition, and subtraction” could have been overlooked, as the result was correct. It is further unclear from the script whether the listed order refers to the general convention or to the particular case. Nevertheless, both the claim of “division first,” and the order in which the operations were performed in Jane’s example, attracted our attention. For simplicity, let us consider only the last short computation that involves multiplication and division only. 2 × 10 ÷ 5 Performing “division first” means interpreting this calculation “as if” there are parenthesis around the operation of division: 2 × (10 ÷ 5 )





But actually,

2 × 10 ÷ 5 = 20 ÷ 5 = 4

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and also

2 × (10 ÷ 5 ) = 2 × 2 = 4



Is this a coincidence? In other words, is it a “general case” that

a × b ÷ c = a × (b ÷ c)?

The idea for giving priority to division appeared peculiar to us, and at first seemed “obviously wrong,” as it contradicted our experience in learning, teaching, and performing computations with respect to the order of operations. In our (shared) knowledge, in cases involving the four arithmetic operations and no brackets, multiplication, and division have priority over addition and subtraction; however, other than acknowledging this priority, operations are calculated in the left-to-right order. The reason for prioritizing division over multiplication was puzzling to us, and the first intuitive idea was to show that the approach was wrong by means of a counterexample. (The explanation for the correctness of the approach came moments later.) However, the reason for suggesting priority of division over multiplication was “obvious” for students schooled in Canada: BEDMAS! In Canadian schools, BEDMAS is used as a mnemonic abbreviation, which is supposed to assist students is remembering the order of operations: Brackets, Exponents, Division, Multiplication, Addition, Subtraction. Of note, in American and British schools, the prevailing mnemonic is PEMDAS1, where “P” denotes parentheses, and it further assists memory with phrase “Please Excuse My Dear Aunt Sally.” Note that while “parentheses” and “brackets” are synonyms, the order of division and multiplication is reversed in PEMDAS vs. BEDMAS. Following “division first” suggestion in Andy’s script, the following assertion was presented to class discussion: According to the established order of operations, division should be performed before multiplication. Four students (out a class of 16) agreed with the claim, while others insisted on the “left to right” order when only division and multiplication appear in a computation. A majority claimed that “division first” was wrong and attempted to find a counterexample. For example, for 100 × 4 ÷ 2 try calculating (100 × 4) ÷ 2 vs. 100 × (4 ÷ 2). When “simple” computations did not lead to a counterexample, students turned to more complicated examples. These examples included a longer chain of computations, fractions, and negative numbers. Additional conjecture was voiced that “division first” works only in case there is divisibility between the chosen numbers. This resulted in more complicated examples, but the conjecture was also refuted after several tests. (The ideas of “multiplication first out of left-to-right-order” or

 Other mnemonics related to the order of operations are PIMDAS (I for “indices”) or POMDAS (O for “of”). 1

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“order does not matter” were suggested, but rejected by a counterexample, such as (100 ÷ 2) × 5 vs. 100 ÷ (2 × 5). In a class session, a search for a counterexample lasted for about 25 min. There were occasional exclamations of “Eureka!” which eventually resulted in double checking and facing a computational error. A failure to come up with a ­counterexample resulted in a conjecture that this (that is, prioritizing division over multiplication) will always work. A natural follow up for this conjecture was an invitation to explain why this was indeed the case. In what follows, we try to recreate a conversation that occurred. Dina: So, you see, it works! It always works. Division comes first. That is, before multiplication. Lora: But why this works is still unclear. Dina:  BEDMAS, BEDMAS, BEDMAS.  Haven’t you learned BEDMAS? This makes everything clear. Lora: Actually, I heard about it from students and other teachers, when I started teaching in Canada, but in my schooling [in Romania] this was never mentioned. We learned to do multiplication and division before addition and subtraction. But other than that – follow the order. Dina: So now you have BEDMAS. Lora: This still doesn’t explain why BOTH ways – do it in order and do division first – give the same result. Dina: I see what is unclear to you now. Lora: Maybe this has something to do with the relationship between multiplications and division. Every division can be rewritten as multiplication. Dina: OK, that is why. Division first works because division can be rewritten as multiplication. Teacher: I agree with your observation that division can be rewritten as multiplication. But this does not explain the phenomenon we observed with order of operations. After all, any multiplication can be rewritten as division. Lora: But this is not the same thing. Teacher: What do you mean? Lora: We can write, say, 1 2 × 10 ÷ 5 as 2 × 10 × 5 1 Teacher: Sure, but you can also rewrite this as 2 ÷ ÷ 5 10 Lora: You cannot turn it like that. Teacher:

Have I done something wrong?

At the end of this conversation, the teacher played a “devil’s advocate” in an attempt to direct students’ attention to the properties of operations. Dina: In multiplication you can change the order. There is commutativity. There is no such thing in division.

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Teacher: Indeed. Multiplication is commutative. But when you describe commutativity as changing the order, what do you actually change? Lora: You change the order on numbers, of factors, like a × b = b × a Teacher: So we have not changed that order, the order of numbers. We only varied what operation is done first. The conversation continued for several minutes, until another student brought forward the idea of associativity. Indeed, multiplication is associative, while division is not.

a × b × c = (a × b) × c = a × (b × c)



a ÷ b ÷ c = (a ÷ b) ÷ c ≠ a ÷ (b ÷ c)



As such, giving priority to division can be described as replacing division with multiplication and applying the associative property of multiplication. Symbolically,



(a × b) ÷ c = (a × b) ×

1  1 = a ×  b ×  = a × (b ÷ c) c  c

There were welcoming exclamations of AHA! when this conclusion was voiced, which was followed by “I could have/should have thought of this myself.” There were claims of denouncing BEDMAS, as well as altering it and offering alternative D A recording of the learned acronym, such as BE(DM)(AS) or BE . M S The observation of associativity resolves the question of “why” performing division first, out of order, results in a correct answer. However, it raised several additional questions and, as our pedagogical convention, resulted in developing tasks for the new groups of students.

7.3.3  Part 3: Order of Operation Tasks Recall that the conversation around the order of operations originated from the assignment for secondary school teachers to examine mathematical conventions, in the form of a scripted dialogue. However, the topic of the order of operations can be addressed in elementary school. As such, we wondered, how do prospective elementary school teachers explain and implement the order of operations in working out simple computations. We prepared two tasks for prospective elementary teachers. In Task 1, students were asked to perform simple calculations. They responded to Task 1 before the notion of order of operations was mentioned in class. In Task 2, the students were presented with four arguments related to the order of performing operations and were asked to agree or disagree with the characters and explain their views.

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Task 1: Simple Calculations Solve the following without the use of brackets or fractions: 600 ÷ 5 × 2 = ____ 10 × 4 ÷ 2 = ____ 100 ÷ 10 ÷ 2 = ____ 100 ÷ 5 × 4 = ____ 200 ÷ 2 × 50 = ____

Task 2: Four Arguments on Order of Operations Mary, David, Laurie, and Ned considered numerical expressions that have only multiplication and division, and disagreed on how the calculations should be carried out Mary says that you must always do the multiplication operation first. Is she right? Please circle: YES / NO If yes, please explain your reasoning. If no, please explain your reasoning and provide a counterexample. David says that you must always do the division operation first. Is he right? Please circle: YES / NO If yes, please explain your reasoning. If no, please explain your reasoning and provide a counterexample. Laurie says that you must always do the operations from left to right in the order they appear. Is she right? Please circle: YES / NO If yes, please explain your reasoning. If no, please explain your reasoning and provide a counterexample. Ned says that the order does not matter when there is only multiplication and division in the calculation. Is he right? Please circle: YES / NO If yes, please explain your reasoning. If no, please explain your reasoning and provide a counterexample. Note that the numbers in Task 1 are chosen carefully to assure that the calculations can be performed with ease without raising the need for the use of a calculator. We added “without the use of brackets or fractions” to the instructions in Task 1, as we noted this tendency among several students in our informal pilot observations. Furthermore, we witnessed students’ claims that “order does not matter, we chose the order for convenience.” While such flexibility in choosing the order is applicable

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for computations such as 376 × 25 × 40 (noting that 25 × 40 = 1000), there is no privilege of choosing the order when different operations are involved. We wondered where the attention to 5 × 2 in the first task and 2 × 50 in the last task will influence the solution. In Task 2, we presented four different ways the order of operations can be approached. The names Mary (Multiplication), David (Division), Laurie (left-to-­ right,) and Ned (no difference) were chosen to help us follow the discussion and not get confused about the characters. After each claim there was about half a page of space to provide a counterexample or explain reasoning. The tasks were administered in order, so there was no option for students to correct their responses to Task 1 after considering the claims in Task 2. However, we asked students to use the same pseudonym for both tasks, to allow us to connect their responses to Tasks 1 and 2. The detailed analysis of students’ responses is presented elsewhere (Zazkis and Rouleau 2018), connecting this analysis to the notion of relearning and the robustness of students’ “met-befores” (Tall, 2013). Here we share a few comments on the classroom discussion that emerged after the questionnaires with both tasks were completed. To initiate the discussion, the four character names were presented on the board, and students were invited to indicate with which character they agreed. No one agreed with Mary, and two students who initially agreed with Ned changed their choice when presented with counterexamples. Eventually, out of 21 students in the class, 16 agreed with David and 5 agreed with Laurie. That is, 21 students agreed with the claim that division should be performed before multiplication and only 6 claimed that the operations should follow the order in which those are listed. And the reason, yet again, was in the “magic” acronym BEDMAS. As this experience followed the conversation with the secondary school teachers, we were prepared for the “BEDMAS” argument. However, a strong preference of agreement with David (“division first”) was first surprising. Following the tally of preferences, we attempted to create a disturbance by inviting students to perform computations both ways, in order of listing and giving priority to division. This time, unlike in the group of secondary teachers, the disturbance was minimal. The students did confirm that computations result in the same answer both ways, but there was no evident curiosity to understand the phenomenon. Still, following the planned activity, the students were offered the explanation, which involved associativity of multiplication versus lack of associativity of division. A conversation during the break evolved around teaching the order of operations. Sara: So how would you teach order of operations? Would you totally avoid BEDMAS? Teacher: This indeed would be my preference, but… Sara: So not mention this at all. How will students remember? Teacher: I have never heard of it before coming to Canada. So it is indeed possible to learn without mnemonic. Sara: So do not mention it in class.

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Teacher: Maybe totally avoiding BEDMAS is impossible today. Even if you do not mention this in class, students will hear it from siblings or parents. Kelly: And yes, if they come from the States, like my mom, they will tell you about Please Excuse My Dear Aunt Sally and PEMDAS. Teacher: So you see how confusing it can be to follow acronyms without understanding. If a student brings up the acronym, there is a need to explain its drawbacks, and that division and multiplication have the same priority. We have just seen that “division first” gives a correct answer, but if a student thinks of PEMDAS as “multiplication before division,” it will lead to mistakes. Anna: So now we not only know that BEDMAS works, you explained why it works. Anna’s last declaration was a totally unintended conclusion. But, disturbingly to us, we have not seen any opportunity to continue or elaborate, and point to the problematic order implied by A-S in BEDMAS, as students were leaving for the next class. As previously, personal disturbances lead to developing tasks for students. The same group of students was invited to write a script for a dialogue between a teacher and students, the beginning of which is presented as Task 3. Task 3: Focus on Addition/Subtraction Order Suppose that you have discussed the order of operations with your students, and assigned a variety of calculations to carry out. As students engage in the task, focusing first on the calculations that involve only addition and subtraction, you attended to the following argument: Adel: In these ones (pointing to a few tasks), we have only addition and subtraction. So we have to do addition first. Sue: Why addition first? I did subtraction first and got everything right! Just try this example… Continue from here Adel: … Sue: … … Teacher: … You "eavesdrop" on the students for a few minutes and then decide to join their conversation. Note for the script: consider for example a + b − c + d − e “addition first” is interpreted as (a + b) − (c + d) − e “subtraction first” is interpreted as a + (b − c) + (d − e)

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Students could follow several different prompts to write a script for a play, and five have chosen to extend the dialogue related to the order of operations. In fact, doing “addition first” fails, and doing “subtraction first,” out of left-to-right order, works, while the infamous BEDMAS (and PEMDAS) suggests otherwise. In Sue’s suggestion in the prompt, we attempted to create a conflict with the expectation. However, we did not find in the plays a conflict or the expected analogy to the division vs. multiplication case. That is, “subtraction first” works because subtraction can be replaced with addition, and addition is associative. We found mostly reiteration by the teacher-character of the correct way to perform the order of operations, and not explicitly addressing the issue of “addition first” vs. “multiplication first,” when performed out of order. With the next iteration of the task we plan to make a conflict explicit, as in Task 4. Task 4: Revised Prompt, with a Focus on Addition-Subtraction Order Students attend to the order of operations, focusing first on the calculations that involve only addition and subtraction. You attended to the following argument: Adel: In these ones (pointing to a few tasks), we have only addition and subtraction. So we have to do addition first. Remember BEDMAS? Addition comes before subtraction. Sue: Why addition first? I did subtraction first and got everything right! Just try this example, 50 + 40 − 5 … Adel: You likely got it wrong, you should go 50  +  40  =  90 and then 90 − 5 = 85 Sue: But I did 40 − 5 = 35 and then 50 + 35 = 85 Adel: The same result? Strange… Sue: But look, if subtraction comes first in the exercise, and you do addition first, it will be a disaster. Look at 50 − 40 + 5. If we follow your BEDMAS and do addition first, you get 45, and then 50 − 45 = 5, but this in NOT the answer on our key… Adel: You are confusing me! … Teacher: … You "eavesdrop" on the students for a few minutes and then decide to join their conversation.

This remains to be seen how the explicit examples and explicit reference to the mnemonic will influence students’ thinking about addition-subtraction order and how this will be featured in their scripts.

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7.3.4  Order of Operations: Arbitrary Choice? From our conversations with both elementary and secondary teachers, and in particular as exemplified in Andy’s comments, we learn that most teachers believe that the convention is just an arbitrary choice, an agreement for consistency without a particular reason behind it. However, a possible reason can be found when repeated addition is replaced by multiplication. Consider, for example, the following calculation:

20 + 3 + 3 + 3 + 3 + 3

However, since 3 + 3 + 3 + 3 + 3 = 5 × 3 it is convenient to record the operations above as

20 + 5 × 3 Of course, one can use parentheses: 20 + (5 × 3) However, in order to assure the equality



20 + 3 + 3 + 3 + 3 + 3 = 20 + 5 × 3

there is a need for an agreement about the order of operations: multiplication before addition. A similar – and for some students a more convincing – example is the case where repeated addition appears more than once:

3+3+3+3+3+ 4+ 4+ 4+ 4+ 4+ 4+ 4

This example can be accompanied with a story: “Candy worked at a chocolate factory for a year. For the first 5 months, Candy was earning 3 goldickles a month. Then she was promoted and her earnings increased for 4 goldickles a month. How many goldickles did she earn for the year?” Repeated addition then is replaced by multiplication2, resulting in a sum of two products:

3× 5 + 4 × 7

and in order to assure equality with the sum below:

3 + 3 + 3 + 3 + 3 + 4 + 4 + 4 + 4 + 4 + 4 + 4 = 3× 5 + 4 × 7

multiplication should be given priority over addition. Once the relationship between addition and multiplication is established, we can move to considering division and multiplication. These two have the same priority as division is multiplication by the inverse, while similarly subtraction and addition have the same priority as subtraction is addition of the opposite (additive inverse). 2  Note that we are NOT claiming that multiplication IS repeated addition (conceptually), we only claim that computationally repeated addition can be replaced by multiplication and vice versa.

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Note that a similar explanation can be extrapolated and used to explain the priority of exponentiation over multiplication. In summary, what was presumed as an arbitrary choice can be explained as assuring consistency when arithmetic operations are either expanded or compressed, the use of brackets is avoided, and the relationships of multiplication-division and addition-subtraction are taken into account.

7.4  Modeling and Conversion of Square Units Our next story starts in a class of prospective elementary school teachers who were working on conversion of square units in their mathematics (content) course. It then continues in with a group of prospective secondary school teachers enrolled in a problem-solving course.

7.4.1  Part 1: Unexpected Solution Despite the common knowledge that 1 m = 100 cm, it often comes to prospective elementary school teachers as a surprise that 1 square meter does not equal 100 cm2. A visual representation is helpful here, when a square meter is represented as a square with a side of 1 m (see Fig. 7.1). Then, the side of the same square can be labeled as 100 cm, and its area in square centimeters was calculated. Area (square)  =  100  cm  ×  100  cm  =  10,000  cm2, which supported the conclusion that 1 m2 = 10,000 cm2. When practicing the idea of relating linear units to square units, the students were tasked with solving a variety of associated problems, such as: If the size of the sides of a square is doubled, how does this affect the area? If the size of the sides of a square is tripled, how does this affect the area? If the size of the sides of a square is increased (or decreased) by 20%, how does this affect the area?

Fig. 7.1  Visual representation of 1 m2 = 10, 000 cm2

1m

1m2

1m

Þ

100 cm

10000cm2

100 cm

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In one of the homework assignments, the following task was included: An architect is building a model of a park, in which every 10 m are represented by 3 cm. There is a lake in the park. The area of the lake is 7200 m2. What is the area of the lake in the model? Similar problems were solved in class using proportions as well as visual representation that helps in conversion of square units. For the particular problem, since 10 m is represented by 3 cm, then 100 m2 is represented by 9 cm2. Possible proportions useful for solving the problem include



7200 100 7200 x = = or x 9 100 9

which yield as a final answer x = 648 cm2 9 An alternative approach (not an alternative calculation) was to consider as 100 a conversion factor and obtain the result by



7200 ×

9 = 648 100

When submitting a homework assignment, one student presented the following solution: 7200 = 80 × 90 If 10 meters = 3 cm, then 80 meters = 24 cm 90 meters = 27 cm Then, 24 × 27 = 648, so the area of the lake on the model is 648 cm2. When the homework was marked by a teaching assistant, the student did not get a full credit for this solution. So he approached the course instructor with a complaint. Below is a recollection of the conversation that occurred: Teacher: Please explain how you got your answer. Student: If the lake had length of 90 m and width of 80… Teacher: Did you know the shape of the lake? Student: So if it were a rectangle… Teacher: Have you ever seen a rectangular lake? Student: I pretended it was and I got the correct answer. Teacher: And what if it were a different rectangle, would you get the same answer? Student: I have not checked but this one worked. The solution method initially appeared as “obviously wrong,” but the answer was indeed correct (see Zazkis & Mamolo, 2018, for an extended focus on a mathematics educator and her knowledge on the mathematical horizon). The student could not explain why the “pretended” solution worked. However, for us, this was an

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opportunity to expand our mathematical knowledge as well as our repertoire of tasks presented to teachers.

7.4.2  Part 2: Scripting on Unexpected Solution As a result of the unexpected solution for the modeling problem presented above, the Building a Model task for a group of prospective secondary school mathematics teachers was developed. This scripting task invited teachers to imagine a conversation with a student and present it in the form of a dialogue. The particular instructions for the task asked students to respond to three parts: • Part 1 (1–2 pages): Extend the imaginary interaction in the form of a dialogue between a teacher and a student (or several students). • Part 2 (1–2 paragraphs): Explain your choice of action. • Part 3 (optional, 1–2 paragraphs): Explain how you would clarify the issue for yourself or a “mathematically mature colleague,” acknowledging that this could be different from your approach with a student.

Scripting Task: Building a Model Your class worked on conversion of square units. Then your class was given the following task: An architect is building a model of a park, in which every 10 m are represented by 3 cm. There is a lake in a park. The area of the lake is 7200 m2. What is the area of the lake in the model? For the solution, you expected that students will note that 100 m2 are represented by 9 cm2. However, Alex suggested a different solution. Here is what he wrote: 7200 = 80 × 90 If 10 meters = 3 cm, then 80 meters = 24 cm 90 meters = 27 cm 24 × 27 = 648, so the area of the lake on the model is 648 cm2. Do you accept Alex’s approach? Explain in detail. Write a script that extends the following conversation: Teacher: Alex: Teacher: Alex:

Alex, please explain how you got your answer. If the lake had length of 90 m and width of 80… Did you know the shape of the lake? …

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The scripts written by prospective teachers can be clustered as utilizing three main approaches, each of which is exemplified below: 1. Guiding a student towards a conventional conversion of square units while ignoring the discussion of Alex’s solution 2. Discussing Alex’s approach by considering other possible shapes of the lake with particular units of measure 3. Exemplifying algebraically why Alex’s approach “works” No script attended to the geometric idea of scaling/dilation or provided a complete general justification as to why Alex’s approach resulted in a correct answer. Neither did any script attend to the “abuse of the equal sign” (in claiming “10 m = 3 cm”) in the student’s solution. In the next sections, we exemplify recurring approaches in students’ scripts.

7.4.3  Towards a Conventional Approach The following excerpt demonstrates an approach, variations of which were presented by script writers. As in prior studies that used scripting tasks (e.g., Zazkis et al. 2013), the moves of the teacher-character represent the chosen approach of a script writer. Teacher: Alex, please explain how you got your answer. Alex: If the lake had length of 90 m and width of 80… Teacher: Did you know the shape of the lake? A: Ummm… I assumed that the shape of the lake is a rectangle. T: Why? A: Because we don’t know how to find the area of nonpolygons. T: You don’t have to know it. You will only use the ratio equation to solve this problem. A: Okay. How? T: First, we know that 10  m is represented by 3  cm. What is the number that represents 100 m2? A: I don’t know. The units are different. How can I find it out? T: Assume that you have a 10 m by 10 m2. What is the area of it? A: That is 100 m2. T: Then what about 3 cm by 3 cm2? A: That is 9 cm2.

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T: Look at the diagram. We know that 10m is represented by 3cm. And 100 m2 is represented by 9 cm2. A: Oh I understand. T: So we are supposed to find the number that represents 7200cm2. How are you going to find it now? A: Hmmm…. Using the ratio equation? T: Okay, then make the equation. 7200 x A:  = , and if we do the cross multiplication, 100 9 100x = 7200 × 9, so x = 648. So the area is still 648 cm2. I was right! T: Your original “answer” was right, but your original method only covered the certain shape of the lake. So you should be able to find the answer in general. A: Okay, I see. As we see in this excerpt, the teacher makes no attempt to explore the correctness of Alex’ answer or the correctness of the chosen approach. The teacher-character suggests that it is not necessary to know how to find the area of a nonpolygon region. The teacher leads the student toward a solution that utilizes a proportion, which is referred to as “ratio equation.” From the final comments of the teacher – “Your original ‘answer’ was right, but your original method only covered the certain shape of the lake” – we see that the “method” is rejected for its lack of generality. There is no curiosity invoked around the correct result.

7.4.4  Conversion Motivated The following excerpt includes an additional student-character, who joins the conversation between the teacher and Alex.

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185

T: So how did you use that fact when solving the problem; that you could think of the shape of the lake as a rectangle? A: Well I thought that if it were a rectangle of 7200 m2, that would let me think of its sides as multiples of 10 and that would simplify the problem. S2: But why multiples of 10? A: Because we know 10 m represent 3 cm in the model. So I said it should be 90 by 80. S2: Ok I think I understand now, because that will mean that 80m is represented by 24 cm and 90 by 27 cm. A: Exactly, so that will mean that we multiply 24 by 27, which is 648 cm2. T: Ok, you found a way to solve it, but what if you had other dimensions? Do you need multiples of 10? S2: I tried 72 by 100 and go the same result as Alex. T: But what if the lake had an area of 7248 m2. Do you think you will be able to use that method to find how big it will be in the model? A: Let me see. [The teacher gave Alex and her team a couple of minutes while she checked with the other teams.] S2: We can assume that the dimensions are 48 by 151. So we can just use multiples, not just multiples of 10! T: And what if… What if… instead of 7200 we had a prime number? Like 7507 A: It is pretty close to 7500… T: Pretty close is not good enough for our architect. Let me show you how to calculate this exactly, without knowing or assuming the shape of the lake. As the script continues, the teacher introduced conversion of square units. We see that Alex explains the choice of 90 by 80, focusing on multiples of 10, because “10 m represent 3 cm in the model.” While the issue is not explored further, we see this as a convenient choice of numbers for the ease of computation. The script writer uses the help of another student (S2), who suggests a representation of 7200 as 72 × 100, and then decomposes 7248 as 151 × 48, claiming “we can just use multiples, not just multiples of ten.” This suggestion leads the teacher to turn the focus to a prime number. The fact that this number cannot be represented as a product of two factors (when neither is 1) creates a motivation for focusing on conversion if square units.

7.4.5  Attending to a Variety of Shapes In the following excerpt, the teacher invites Alex to consider a different shape of the lake. Alex responds by choosing different dimensions for the rectangle. Teacher: Well, what would happen if you pick a different shape of lake and see if you get the same area?

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Alex: Teacher: Alex:

Why would it matter if I pick a differently shaped lake? Just try to figure out the area of the lake in the model. Ok well how about I pick the shape to look like: 7200 = 10 × 720 If 10 m = 3 cm, then 10 m = 3 cm & 720 m = 216 cm 3× 216 = 648, so the area of the lake on the model is still 648 cm2. Teacher: I am glad that it gives you the same response as your first attempt. But here is something to think about. What would happen if I gave you an area that you couldn’t figure out nice factors? For example, a number that is a decimal or irrational. This script leaves the reader in a suspense. The question “What would happen if I gave you an area that you couldn’t figure out nice factors?” is not answered in the script. It is not revealed what happened after Alex thought about the issue. The next excerpt moves away from “nice factors” and rectangles and attends to different shapes.

Teacher: Alex, please explain how you got your answer. Alex: If the lake had length of 90 m and width of 80… Teacher: Did you know the shape of the lake? Alex: No…but let’s say it was a rectangle. T: Is it fair to assume that? A: It makes it easier to solve for the area. T: Does it change the problem though, if the lake is circular, or some other shape? A: I’m not sure. I think the answer should still hold since it is a conversion factor. T: OK then let’s try it with a circular lake. What’s the formula for area of a circle? A: area = πr2 so 7200 = πr2 and r = 47.87 m Then I convert that radius to cm on the model:  47.87   10  × 3 = 14.36 cm   And finally the area on the model will be: area = π14.362 = 648 cm2 T: Very interesting finding! What if we used a square length, so the rectangle has 4 equal sides? A: Let us try this. OK, well in that case, if the area is still 7200, then each side length must be the square root of 7200, which is 84.85  m. Then I convert that to cm on the model, which is 84.85 × 3 = 25.46 cm . And finally I put that back into the area for 10 a rectangle, or square in this case, which is 25.462 = 648 cm2 T: Very good and interesting. Why do you think your procedure works, or when would it not work?

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A: It seems like it would always work. T: Well the important part is that you are using your conversion factor at the right time. The conversion gives you a path between meters and cm in real life to the model. You would run into trouble if you tried using 10 m = 3 cm to convert between area units, meters squared, or centimeters squared. As long as you are converting first simply to meters or centimeters and going between the two, your procedure should work out. In this excerpt, Alex confirms the calculation for a circular lake and for a square lake. Note that while working with different shapes, the conversion still attends to linear units, and only then area in the model is calculated. While it seems to Alex that “it would always work,” the teacher offers a warning: “You would run into trouble if you tried using 10 m = 3 cm to convert between area units… As long as you are converting first simply to meters or centimeters and going between the two, your procedure should work out.” Indeed, the procedure that worked in several checked cases “should work out” in the future. However, the why question is not raised and is not answered. In the comments that followed this script, Kate wrote: I was actually expecting that Alex’s procedure would not work with different shapes of lake. If I had run across this in the classroom, this is the process I would have gone through and had the student try out different cases to see if it worked. When I tutored students with these conversion factors, I found where they usually get stuck is trying to convert units of area with units of length (i.e., m2 to cm), but since Alex was still using his formula to find side lengths and then converting before solving for the final area, his procedure worked out. We find it interesting to note that students who included in their scripts checking for a variety of possible measures for a rectangle or possible shapes for a lake acknowledged their initial confusion and initial suspicion that Alex’ solution was wrong. However, we see in Kate’s note a confirmation of the procedure that Alex was using, rather than an explanation for its correctness.

7.4.6  Computation “Works” The next script excerpt we share looks at the situation by unpacking the computational procedure and in so doing connects conversion of linear units to conversion of square units. Teacher (T): Did you know the shape of the lake? S: Well, no. But I knew the area was 7200, so it doesn’t matter. T: Do we all agree? Can Alex do this and get the right answer? S2: I didn’t think so, but I got the same answer by doing it the way you taught us.

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T: Ah, by writing [7200/100 = x/9] and solving for x? S2: Yes, I got 648 this way too, but I don’t understand why Alex got the same answer as me. T: What if the lake was 72 x 100? 72 x 100 = 7200 m2 also. S: (Doing work). Then it’s 72 m to 21.6 cm and 100 m to 30 cm… T: And what’s 30 x 21.6? S: 648 cm2 T: Ok, so Alex here is getting a bit more confident. Are more people agreeing with Alex? How about… 150 x 45? (Students working) Yes? S2: Yes, that works too! T: What is going on? Alex, what do you think? S: Now I’m not sure why it works. T: Let’s compare your way to my way. I would have done it this way: 7200 100 7200 x = = or x 9 100 9 S2: This is what I did! The right one. T: Why did you set it up this way? S2: Because we were trying to make a comparison to 7200 m2, so we need to square the numbers from the scale as well so that we have (m2/m2 = cm2/cm2). I don’t understand why Alex’s way worked. T: You know, I was a little surprised myself but it makes sense. The proportion for the scale is 3:10 or = 0.3. But we can’t just apply this proportion to 7200 because it is a squared number. The proportion for a squared number then should be (0.3)2…, so 7200 m2 × 0.09 cm2/m2 = 648 cm2. S: Is that what I did? T: In a way, yes. What you did was equal to this because you had: (80 × 0.3) × (90 × 0.3) = (80 × 90) × (0.3 × 0.3) = 7200 × 0.09 So, since you multiplied by 0.3 twice, you ended up with the correct answer. Did you think about this relationship when you were solving the problem at first? S: No. I just tried it and it worked. T: OK, so the lesson here is that you need to be careful about making assumptions when you’re solving problems. Assuming that the lake was 80 x 90 m didn’t end up leading you astray here, but in a different problem or with different information it might! Ok, moving on… Ruti in her comments explained further: To solve this problem, we use 9/100 as a conversion factor. So the answer is given by 7200 x 9/100.

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189

But look what in fact is Alex’s solution is using 3/10 as a conversion factor twice. So the result is the same as when applying the square conversion. A more detailed general algebraic explanation was provided by Darien: At the start I decided to see if it mattered if Alex decided to pick his own dimensions of the lake. I discovered that it did not matter as long as you converted the dimensions to the models scale. This was hard for me to grasp at first. The way the conversion section at the beginning looked, I thought the student’s method was wrong, but in fact it was simply a different way to solve the problem. This is what found when combining the teacher’s solution with the student’s solution: Teacher’s solution: 7200 ×



9 100

Student’s solution:

l × w = 7200



3  3   l × 10   w × 10     But with some algebra we see that



3  3 9   3 3  l × 10   w × 10  = ( l × w )  10 × 10  = ( l × w ) × 100      So Alex’s method will work for any choice of rectangles’ length and width.

7.4.7  Towards a Geometric Explanation While algebra “works,” we also found some attention to the geometry of the situation. The is exemplified in the next excerpt. Teacher: Alex, please explain how you got your answer Alex: If the lake had length of 90 m and width of 80… Teacher: Did you know the shape of the lake? Alex: No, but I thought that did not matter. T: Well could you explain why the shape is irrelevant. A: I am going to try. So let's imagine we have a piece of dough. So we can give to that piece whatever shape we choose. S1: Like a cat?

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A: Or a lion if you prefer. However, the amount of dough we have will be the same, regardless of the shape. So I thought I could do something similar with an area. T: But how? Why did you say it had a length and a width? A: Well, I thought that imagining that the shape of the lake was a rectangle would make things easier. S2: So we can give the shape we wanted to the lake? A: No, this change of shape was just to simplify the work. If the shape of the lake is a star in real life, the model also has to be a star. Imagine that the lake was a rectangle was useful just to say how big it will be in the map; to make the multiplications easier. This excerpt is a good start leading towards the general geometric explanation. As Alex explains, “change of shape was just to simplify the work.” Many shapes can have the same area, and the shape in the model will be same as the shape in real life. What the script does not attend to is that objects that have the same area in reality will be modeled by objects that also have the same shape. This justifies the “change of shape,” while preserving area, for the ease of calculation.

7.4.8  Modeling as Scaling: Mathematical Notes What explains the general situation for us, in addition to the algebraic proof, is a perspective on modeling (that is, building a model) as a geometric transformation of scaling (dilation), which is a similarity transformation. With this perspective, a model can be seen as the image of reality under a similarity transformation. Scaling (dilation) preserves angles (shapes) and ratios of lengths. Consequently, shapes of the same area are transformed into shapes of the same area. We provide a visual demonstration of this idea in Fig. 7.2. For simplicity, and in analogy to the problem of building a model of a lake, the scale factor we use is 3/10. Under this transformation, the length of any segment on the image is 3/10 of the length of its 2 source. As such, the area of the image shape is ( 103 ) of area of the source shape. As such, Alex’s approach of assuming semi-random dimensions for a lake and finding the area of the transformed image of this shape, in fact, provided the area of any transformed shape with the same starting area. The “obviously wrong” method, which led to a correct answer, enriched our repertoire of tasks for teachers, which in turn enriched their appreciation of mathematical connections.

191

7.4 Modeling and Conversion of Square Units A

R

L Q

M

B AB = 10.00 cm BC = 7.20 cm Area R = 72.00 cm2

D

A' D'

R'

B' C'

C A'B' = 3.00 cm B'C' = 2.16 cm Area R' = 6.48 cm2

ML = 9.00 cm LK = 8.00 cm Area Q = 72.00 cm2

N

K M'

Q'

N'

L' M'L' = 2.70 cm L'K' = 2.40 cm

K' O

C K C'

J

JK = 4.79 cm Area C = 72.00 cm2

Fig. 7.2  Scaling figures of the same area

J' K' J'K' = 1.44 cm Area C' = 6.48 cm2

O

Area Q' = 6.48 cm2

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7  From “Obviously Wrong” Methods to Surprisingly Correct Answers

7.5  Familiar Formula in Unfamiliar Situation In this section, we share a story about an “obviously wrong” application of a formula that surprisingly leads to correct results. To set the context, we discuss briefly non-Cartesian coordinate systems. The students mentioned in this section are prospective secondary school teachers enrolled in a problem-solving course, in the last semester of their two-year-long teacher certification program.

7.5.1  Affine Coordinates: An Introduction The familiar Cartesian coordinate system is an orthogonal homogeneous system, that is, it is defined on a plane by a choice of two perpendicular lines, of a positive direction on each line and of a unit that is equal on both lines. This coordinate system induces one-to-one correspondence between ordered pairs of real numbers and points on the plane. But what if not? That is, what if the lines are not perpendicular? What if the units are not the same? In fact, the Cartesian system of coordinates is only one particular example of affine coordinates that can be neither orthogonal nor homogeneous. An affine coordinate system is defined on a plane by any three noncollinear points (O, I, J). Lines OI and OJ establish two axes intersecting at the point of origin O, and the directed segments OI and OJ determine the positive direction and a unit on each one of the axes (Fehr et al. 1973). Coordinates of a point on a plane are found by parallel projection, that is, by drawing parallel lines to the axes through this point, and noting points of intersection of these lines with the axes. Varying the familiar coordinate system offers a variety of interesting tasks and a possibility to reexamine what we are used to when working in the Cartesian system. An introduction of electronic graphing devices, such as graphing calculators and a variety of software packages, creates an essential need for such reexamination. For example, Zaslavsky et al. (2002) discuss the following task, a variation of which we usually offer to our students (see Fig. 7.3). According to Zaslavsky et al., the task was designed specifically to trigger reexamination of the notion of slope (e.g., is it 1 or 1/3?). Participants in their study, as well as ourselves and our students, experienced difficulty when their prior knowledge that the graph of y = x creates a 90 degree angle with the axes was not in accord with the visual image of the graph and the calculated slope was in discord with the slope perceived visually. We then move to nonorthogonal systems, considering general affine coordinates. Among the first tasks presented to teachers when introducing an affine coordinate system is to assign a correspondence between points on a plane and their coordinates (see Fig. 7.4 for examples). But even this simple task is challenging at first to individuals “stuck” in the Cartesian plane. For example, in describing how to find the coordinates of a point on a plane, many teachers report their attempt to “draw a perpendicular from the point to the axes.” Of course this description is valid only

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193

Fig. 7.3  Slope in orthogonal homogeneous and nonhomogeneous systems. (Adapted from Zaslavsky et al. 2002)

when the axes are orthogonal. Facing the limitations of this description, and striving for an appropriate one, can be a starting point for a proof that within an affine system of coordinates, every point on a plane has a unique representation as an ordered pair of numbers and vice versa.

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Fig. 7.4  Determining coordinates of a point in an affine plane

Fig. 7.5  Assigning functions to graphs of g(x) and h(x)

After becoming familiar with plotting points and assigning coordinates, teachers attend to graphs of polynomial functions of the first degree or linear functions. Possible tasks include graphing given functions and assigning functions to the given graphs; see Fig. 7.5 for examples. This may be a good opportunity to reconsider the knowledge of linear functions plotted with the Cartesian coordinates and to explore what holds for any affine coordinate system. Possible questions to be considered are: • How can you determine if three points, given by their affine coordinates, are collinear? • What is determined by parameters m and b in y=mx+b?

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195

• Do two lines have a unique point of intersection? • Does the graph of y=x form a 45-degree angle with the positive direction of the x-axis? If not, how can this angle be determined? A general idea of the task is to make a list of what is known about linear equations and their graphs and then examine which known facts and approaches hold in a general affine system.

7.5.2  Exploring Medians One basic notion that invites reexamination with the help of affine coordinates is that of length. Consider, for example, the well-known theorem about the medians of a triangle. The theorem states that (a) the three medians intersect at one point and (b) the ratio of the two segments of the median determined by this point is 2:1. The task for teachers was to prove this theorem by using the vertices of a triangle to assign the system of affine coordinates (see Fig. 7.6). Embedding the triangle in an affine coordinate system, as shown in Fig. 7.6, the equations of the three lines are given as follows:

AL : y = x



BM : y = −(1 / 2 ) x + 3



CK : y = −2 x + 6

Fig. 7.6  Medians in a triangle ABC

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The intersection point of each pair of these lines is then calculated to be (2,2). This completes the proof of (a). Considering (b), after calculating the coordinates of the intersection Q, the task for teachers was to prove that = AQ : QL BQ = : QM CQ : QK = 2 : 1



Surprisingly or not, most teachers used the “distance formula”: d=



( x1 − x2 ) + ( y1 − y2 ) 2

2



to determine the lengths of the segments. For example, CQ =



( 0 − 2 ) + (6 − 2 ) 2

QK =



2

= 4 + 16 = 20 = 2 5

( 2 − 3) + ( 2 − 0 ) 2

2

= 1+ 4 = 5





Therefore, CQ = : QK 2= 5 : 5 2 :1

or

BQ =

(6 − 2 ) + ( 0 − 2 )

QM =

2

2

= 16 + 4 = 20 = 2 5

( 2 − 0 ) + ( 2 − 3) 2

2

= 4 +1 = 5





BQ : QM = 2 : 1

The method appeared to us “obviously wrong”! The familiar “distance formula” is based on the Pythagorean Theorem, which is applicable only in case of a right-­ angled triangle, and is based on the assumption that the coordinates are Cartesian. However, why did the inappropriate application of a familiar formula lead to a correct result? In class, the calculations using the distance formula led to a conversation about the origins of the formula, its scope of applicability, and of the notion of length in general. That is, in the particular case of affine coordinates, the unit of length is determined only with reference to the particular axis and therefore there cannot be a numerical assignment to the length of a given segment. Nevertheless, comparison of lengths in terms of ratios is possible by attending to the difference in x- and y-coordinates, respectively. For example, consider the points A(0, 0), Q(2, 2), and L(3, 3). From A to Q there is a difference of 2 in both x- and y-coordinates. From Q to L there is a difference of 1  in both x- and y-coordinates. Therefore, AQ : QL = 2 : 1. Similarly, consider the points B(6,0), Q(2, 2), and M(0, 3). From B to Q there is a difference of 4  in the x-coordinates and a difference of 2  in the

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197

y-coordinates. From Q to M there is a difference of 2 in the x-coordinates and a difference of 1 in the y-coordinates. This proves the ratio BQ : QM = 2 : 1. As mentioned, the application of the “distance formula” in the case of affine coordinates was considered as an “obviously wrong” method. Still, it led to a correct result. In students’ solutions a correct statement was proven using an incorrect method. However, why did this work? We attend to this question in what follows. In fact, the proof presented above (that relied on the distance formula) attends only to a particular case of an isosceles right-angled triangle in the Cartesian coordinate system, in which the equal sides are located on the x and y axes (See Fig. 7.7). Why was this relationship of lengths preserved? Familiar formulas were applied outside of their scope of applicability in other cases as well. For example, considering medians in a triangle, another property is that a median divides a triangle into two triangles of equal area. Among the proofs for this relationship proposed by students, we found the following:

Fig. 7.7  The case of isosceles right triangle in the Cartesian coordinates

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Solution using the determinant of a matrix: For a triangle with vertices (x1, y1), (x2, y2), (x3, y3), the area is given by x1 y1 1 1 Area = x2 y2 1 2 x3 y3 1





Therefore, for a triangle with vertices (6, 0), (0, 6), (0, 0), Area ( ABC ) =



6 0 1 1 1 0 6 1 = × ( 6 × 6 ) = 18 2 2 0 0 1

3 0 1 1 1 Area ( AKC ) = 0 6 1 = × ( 3 × 6 ) = 9 2 2 0 0 1





Therefore, Area (AKC) = Area (KBC) Solution using Heron’s formula: For a triangle with sides of length a, b, and c, Area = s ( s − a ) ( s − b ) ( s − c )





a+b+c where s = 2 For triangle ABC,

Perimeter = 6 + 6 + 6 2



s = 6+3 2 Area ( ABC ) =

(6 + 3 2 )(3 2 )(3 2 ) (6 − 3 2 ) =

( 36 − 18 ) × 18

= 18 = 18 2





For triangle AKC,

Perimeter = 6 + 3 + 3 5



s = 4.5 + 1.5 5 Area ( AKC ) =



( 4.5 + 1.5 5 ) (1.5

)(

)(

5 − 1.5 1.5 5 = 1.5 4.5 − 1.5 5

=

( 4.5

=

( 20.25 − 11.25) (11.25 − 2.25) =

2

− 5 × 1.52

) ( 5 × 1.5

Therefore, Area (AKC) = Area (KBC)

2

− 1.52

)

) 92 = 9



7.5  Familiar Formula in Unfamiliar Situation

199

To reiterate, the results are correct, as indeed a median divides a triangle into two triangles of equal area. However, both methods appear to be “obviously wrong,” as they depend on assigning particular values to lengths and areas, which is meaningless in the affine coordinate system. Furthermore, the formula that uses the determinant relies on Cartesian coordinates, and Heron’s formula relies on the Pythagorean Theorem – both of these are inapplicable for affine coordinates. While students’ use of advanced tools is appreciated, as before, these solutions account for a particular triangle (see Fig. 7.7). Actually, by focusing on Fig. 7.7, the area of the triangle of interest is immediately identified using the familiar formula

1 2

bh.

However, we reiterate the question: why do solutions for the specific case (assuming Cartesian coordinates), using methods inappropriate for the general case (such as methods that rely on the Pythagorean Theorem), determine correct relationships for the general case (that is, the case of scalene triangle and affine coordinates)?

7.5.3  Eureka! Mathematical Resolution The surprise of seemingly incorrect approaches and reaching the correct solution is resolved by considering affine transformations (the following elaboration is adapted from Stein 1999). The general affine mapping is described as follows:

x ′ = ax + by + e



y′ = cx + dy + f

In this sense, the transformation of Cartesian coordinates to affine coordinates can be considered as an affine mapping. Though lengths are not preserved in an affine mapping, the mapping magnifies parallel line segments by the same factor and as such preserves the ratios of parallel line segments. Further, an affine mapping does not preserve areas, but it preserves ratios of areas. This property is best illustrated by transforming the grid of squares to the grid of parallelograms (see Fig. 7.8). The idea that each one of the squares, which can be made as small as necessary, corresponds to similar parallelograms is the intuitive basis for the preservation of the ratio of areas. More formally, adopting the proof from Stein (1999), the mapping given by T(x, y) = (ax + by + e, cx + dy + f) magnifies areas by |ad-bc|. This is shown by considering a triangle PQR, with Cartesian vertices P(0, 0), Q(1, 0), R (0, 1), which is transformed by T to triangle P′Q′R′ with vertices P′(e, f), Q′(a + e, c + f), R′(b + e, d + f). The area of the triangle P′Q′R′ is given by considering the difference between the area of a rectangle with sides a and d and the three triangles that complete P′Q′R′ to this rectangle, as shown in Fig. 7.9.

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7  From “Obviously Wrong” Methods to Surprisingly Correct Answers

Fig. 7.8  Intuitive view of an affine transformation

Fig. 7.9  Image of an affine transformation of a triangle

As such, Area ( P ′Q ′R ′ ) = ad – ½ ac + ½ bd + ½ ( a − b ) ( b − c )  =

= ad – [ ½ ad + ½ bc ] = ( ad − bc ) / 2



The area of PQR is ½. Therefore, comparing the areas of PQR and P′Q′R′, the magnifying factor of T for areas is (ad  −  bc). This explains why the “obviously wrong” method led to correct results. To reiterate, this explains why the statement that the ratios of segments at the intersection of medians are (2:1), as proved by students using the distance formula, holds in the general case, despite inappropriately invoking tools that assume Cartesian coordinates.

7.5  Familiar Formula in Unfamiliar Situation

201

7.5.4  Further Applicability The fact that some ratios hold under affine transformation has some nice implications in problem solving in geometry. If all the givens and the question of a problem are affine invariants, the problem can be solved for an affine transformation of an initial object, without loss of generality. That is, in some special cases, we can choose a particular triangle instead of an arbitrary triangle, a square instead of a parallelogram, etc. Such choices allow for the use of tools that are applicable in the transformed case but are not applicable in the general case. For example, consider the following proof for the ratios of medians at the point of intersection (2:1) in a triangle: For an equilateral triangle ABC, consider medians AL, BM, and CK, which intersect at point Q (assume that the common intersection point was already determined). It is clear (based on congruent triangles) that the three medians are of equal length, that is AL = BM = CK. Focusing on a triangle AQK, we note that its angles are 30, 60, and 90 degrees. As such, 2QK = AQ. However, AQ = CQ (for example, based on considering angles in triangle AQC. Therefore, CQ:QK = 2 : 1. From the consideration of symmetry, the same ratio holds for the other medians (Fig. 7.10). The proof may appear as “obviously wrong” as it relies on properties of equilateral triangle (equal sides and 60 degree angles) and based on these assumptions determines other properties used in the proof (medians are of equal lengths, median are heights, relationship in a 30 − 60 − 90 triangle). However, these considerations do prove the general case, as there exists an affine transformation that transforms an arbitrary triangle to an equilateral one. (Note that it is sufficient to claim the existence of such transformation, without determining the specific one.)

Fig. 7.10 Medians

202

7  From “Obviously Wrong” Methods to Surprisingly Correct Answers

Fig. 7.11  Ratio of squares

Having justified the above proof, we offer two comments of caution: One, consideration of special cases as proofs of general cases are applicable only to properties that are invariant under affine transformations. [These properties are linearity, that is, parallel lines are transformed to parallel lines, ratios of lengths (of segments on parallel lines), and ratios of areas, among others.] Two, rather often students use tools applicable in special cases without any awareness of whether the used tools are applicable in a general case. In the above discussion, such tools were formulas based on the Pythagorean theorem. However, the existence of an affine transformation in which an arbitrary triangle is transformed to a right-angle triangle explained the correctness of the solutions. With this in mind we invite readers to consider the following two problems: Problem 1:  Consider triangle ABC and points A′, B′, and C′ placed on the sides ′ = BA′ = CB′ = 1 BC, AC, and AB, respectively, such as AC AB BC CA 3 (Fig. 7.11). Find S ′ ′ ′ ABC , that is, the ratio of areas of triangles A′B′C′ and ABC. S ABC Problem 2:  Consider triangle ABC. On sides CA and CM points M and N are CN CM placed such that = n . CK is the median that intersects MN at = m and CB CA point E (Fig. 7.12). ME Find the ratio . MN

7.6  Concluding Remarks The first two stories – on the tax vs. discount and on the order of operations – rely on the notion of associativity as a solution for disturbance. They show an important application of associativity, a property that is often poorly understood and confused with commutativity.

References

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Fig. 7.12  Ratio of segment lengths

The last two stories – on the familiar distance formula and on modeling a lake – have many elements in common. They start with seemingly inappropriate use of prior knowledge: the use of a formula which is not applicable in the given situation and the assumption of a particular shape which ignores the expected generality. Nevertheless, in both cases the calculations lead to correct results. The mathematical explanation in both cases is based on the same idea: consideration of transformation and invariance under transformation. It is invariance of ratios under (affine and scaling, where the latter is a particular case of the former) transformations that justifies the assumptions (though our students did not provide such a justification) and serves as a tool of convenience for solving a variety of problems.

References Borasi, R. (1994). Capitalizing on errors as “springboards for inquiry”: A teaching experiment. Journal for Research in Mathematics Education, 25(2), 166–208. Fehr, H., Fey, J., & Hill, T. (1973). Unified mathematics. Menlo Park: AddisonWesley. Stein, S. (1999). Archimedes: What did he do besides cry Eureka? Washington, DC: MAA Press. Tall, D. (2013). How humans learn to think mathematically: Exploring the three worlds of Mathematics. New York: Cambridge University Press. Wasserman, N. (2016a). Abstract algebra for algebra teaching: influencing school mathematics instruction. Canadian Journal of Science Mathematics and Technology Education, 16(1), 28–47. Wasserman, N. (2016b). Nonlocal mathematical knowledge for teaching. In C. Csíkos, A. Rausch, & J.  Szitányi (Eds.), Proceedings of the 40th conference of the International Group for the Psychology of Mathematics Education (Vol. 4, pp. 379–386). Szeged: PME. Zaslavsky, O., Sela, H., & Leron, U. (2002). Being sloppy about slope: The effect of changing the scale. Educational Studies in Mathematics, 49(1), 119–140. Zazkis, R. (2017). Order of operations: On conventions, mnemonics and knowledge-in-use. For the Learning of Mathematics, 37(3), 18–20. Zazkis, R., & Kontorovich, I. (2016). A curious case of superscript (-1): Prospective secondary mathematics teachers explain. Journal of Mathematical Behavior, 43, 98–110. Zazkis, R., & Mamolo, A. (2018). From disturbance to task design, or a story of a rectangular lake. Journal of Mathematics Teacher Education, 21(5), 501–516. Zazkis, R., & Rouleau, A. (2018). Order of operations: on convention and met-before acronyms. Educational Studies in Mathematics, 97(2), 143–162. Zazkis, R., Sinclair, N., & Liljedahl, P. (2013). Lesson play in mathematics education: A tool for research and professional development. Dordrecht: Springer.

Epilogue

We are happy to have an opportunity to share our disturbances and some of our conversations. Traditional ways of reporting research in mathematics education do not allow for detailed accounts of personal experiences and personal growth. In the first chapter we described our different roles as problem solvers, teachers, teacher educators, and researchers. In addition, our interactions with students, in a classroom and beyond, are an incentive for continuous learning. Paraphrasing the distinction of teaching via vs. for problem solving (Schroeder and Lester 1989) we note that we learn via teaching, and we learn for teaching. For the former, teaching, taken broadly – that is, in and out of class interaction with students and reflecting on their work – is an opportunity to deepen personal mathematical knowledge. It involves making connections and attending to nuances. For the latter, teaching, taken broadly  – that is, including planning and task design – is an opportunity to shape both pedagogical and mathematical learning of our students, most of whom are prospective and practicing teachers. We titled this book Mathematical Encounters and Pedagogical Detours. As mathematics educators and researchers in mathematics education, we described mathematics-related experiences of disturbance, which resulted in different actions. We described how personal disturbances influenced our work with students. We described the design of tasks for our students, in which they engage in mathematics as problem solvers and in which they imagine themselves dealing with school students in a classroom. However, as the book was close to completion, we noted that or most of the described experiences of disturbance appeared as a result of communication with students and/or a reflection of this communication. As a result, we engaged in some mathematical explorations, which in turn were brought to our students, which potentially enriched their mathematics. As such, our work can be also viewed as Pedagogical Encounters and Mathematical Detours. This could be the title for the sequel, as a cycle of encounters – disturbances – detours saturates our work.

© Springer Nature Switzerland AG 2021 B. Koichu, R. Zazkis, Mathematical Encounters and Pedagogical Detours, https://doi.org/10.1007/978-3-030-58434-4

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Index

A Abridged multiplication formulas, 147 Addition/subtraction order, 209, 210, 212 Affine coordinates Cartesian coordinate system, 224 definition, 224 electronic graphing devices, 224 linear functions, 226 medians of triangle, 227, 229 nonorthogonal systems, 224 plotting points and assigning coordinates, 226 reexamination, 227 unit of length, 228 Affine mapping, 231 Affine transformations, 231–234 Algebra, 118 Algebraic and functional-graphical representations, 175 Algebraic problems, 107 Algorithm comparison task, 58 Aphorism, 14 Applicability, 65, 66, 228, 229 Arbitrary choice, 211 Artificiality, 173 Associative property, 206, 234 Associativity, 199, 206, 208, 234 Auxiliary constructions, 132 Auxiliary problem, 131 B Bombelli’s L’Algebra, 147, 170 Book VII, Proposition 2 [VII.2], 63 Book VII, Proposition 6 [VII.6], 54 Book VIII, Proposition 6 [VIII.6], 52–57

Book IX, Proposition 4 [IX.4], 53 Book X, Proposition 2 [X.2], 58–62 Book X, Proposition 3 [X.3], 61, 62 Building a Model task, 214 C Canonic cubics, 163 Canonic equations, 163 Cantor’s diagonal method, 7 Cardano’s Ars Magna, 139 Cardano’s method algebraic transformations, 141 cubic equations, 161, 168 expression, 172 formula, 173–175 gap, 170 graphical-functional interpretations, 166 modern terms, 141, 142 niceness, 168, 171 opportunities, 170 problem solving, 161 Rational Zero Theorem, 171 reasonable difficulty, 170 sense making, 147 substitution, 160, 161, 167, 171, 172 symmetric configurations, 161 task, 157, 159 technical effort, 170 transformations, 162, 167 Cartesian coordinate system, 224, 229 Casus irreducibilis, 147, 171 Columbus’ egg, 41 Commensurable, 59, 61, 65–68, 70 Commutativity, 199, 205, 234 Comparison task, 67

© Springer Nature Switzerland AG 2021 B. Koichu, R. Zazkis, Mathematical Encounters and Pedagogical Detours, https://doi.org/10.1007/978-3-030-58434-4

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Index

208 Complex coefficients, 165 Complex numbers, 137 Computation, 219, 221 Continued proportion, 50–53, 56, 71 Book IX, Proposition 4 [IX.4], 43–45 warm-up tasks, 42, 43 Continued subtraction, 65 Conversion of square units, 212 Cubic/cube number, 43, 45, 48 Cubic equations collaboratively re-create the story, 139, 140 configuration, 164, 165, 168 equality, 174 hypothetical reasoning, 138 solutions, 148 solving, 137 two-lesson module, 139 Cubic polynomial, 167 Cubing and self-substituting method, 151 complex numbers, 151 cubic equations solving, 175 cubic polynomial, 153 extraneous roots, 152, 156 extraneous solutions, 154 logical chain, 153 non-reversible steps, 174 Cubing method, 149, 150 Cubing transformation, 149 D De Moivre’s theorem, 160 Dialogic method, 17 Dictum-factum, 147 Discount/tax, 198, 199 Distance formula, 228, 229, 232, 235 Disturbance accessible problem, 8 Cantor’s method, 7 description, 1 the Equal Angle Bisector Problem, 8, 9 inability to solve problem, 3 learning and professional development, 1 as opportunity/pressure, 1 perspectives, 1 Steiner-Lemus Theorem, 9, 10 the Spinning Table Problem, 4 the String Around the Earth Problem, 5, 6 as teacher-educator teaching how to teach, 11, 12 as teachers’ mathematical experiences, 11 translation of parabola, 7 the Two Kings Problem, 6 vs. perturbation, 1

Double Perfect Squares Task, 115 Double-to triple-checking, 149 Duoethnography, 17 E Easy-to-solve systems, 168 Electronic graphing devices, 224 Elementary algebra, 175 Eminent mathematicians, 40 The Equal Angle Bisector Problem, 8, 9 E-rational and E-irrational, 67–69 Euclid’s elements, 39–42, 51, 52, 55–57, 61, 62, 69, 71 Euclid’s plane number, 45 Euclid’s proof, 17 Euclid’s Propositions, 138 Euclid’s text, 56 Euclidean algorithm applicability, 65, 66 Book VII, Proposition 2 [VII.2], 63 Book X, Proposition 2 [X.2], 58–62 Book X, Proposition 3 [X.3], 61, 62 comparison task, 58 GCD, 65 GSD, 57 iterative nature, 62 rational and irrational, 66–70 Euclidean propositions Book II, Proposition 14 [II.14], 68 Book VII, Proposition 2 [VII.2], 59, 61, 63 Book VII, Proposition 6 [VII.6], 53, 54 Book VII, Proposition 12 [VII.12], 59, 60 Book VIII, Proposition 3 [VIII.3], 55, 56 Book VIII, Proposition 6 [VIII.6], 52–57 Book VIII, Proposition 12 [VIII.12], 44 Book VIII, Proposition 21 [VIII.21], 49, 51 Book VIII, Proposition 23 [VIII.23], 49–51 Book IX, Proposition 4 [IX.4], 43–46, 48–53 Book IX, Proposition 20 [IX.20], 40 Book X, Proposition 2 [X.2], 58–61, 67 Book X, Proposition 3 [X.3], 61–63 continued (see Continued proportion) definitions, 39 geometric sequence, 51, 52 mean proportional numbers, 46–47 scripting task, 41 solid numbers, 45, 46 Euclidean solid definition, 45 Euclid-like demonstrations, 145 Excel-based search, 164 Exception-barring method, 178

Index

209

Explore-launch– re-explore-relate, 107, 108, 112 Extraneous Root Criterion, 157 Extraneous solutions, 152, 154 complex values, 154 cubing and self-substituting, 153, 155–157 irrational equations, 175 mathematical mechanism, 154 reversible, 154 Extremum local strict extremum, 181–186, 188, 189

Launch– Get Stuck– Intervene– Relate, 118–124 Launch–Intervene– Re-launch– Explore– Relate, 125, 127–133 Learning for teaching, 70 Learning through teaching, 70 Least common multiple (LCM), 65 Lesson plays, 39 Lily’s script, 55, 56 Line-by-line mode impressions, 141

F Factorization, 169 Fermat’s Little Theorem, 40 Fictional characters, 40 Final formulation, 191, 193, 196 Five Triangles Problem, 119–124, 129, 134 Four numbers, 49, 50

M Mathematical conventions task, 200, 201 Mathematics education, 40, 143 Mean proportional numbers, 46–47 Measures, 55 Medians, 121–123, 134 Mike’s script, 58–61 Miscopying task, 112, 113, 116, 134 Modeling, 222, 235 Monic polynomials, 114 Motivated conversion, 216 Multiplication, 45

G Geometric sequence, 50–53, 56, 57, 71 Geometry, 51, 52 Graphing calculators, 224 Graphing software, 107 Graphing technology, 108 Greatest common divisor (GCD), 57, 58, 65 H Heath’s canonic translation, 69 Heron’s formula, 230, 231 Heuristic scaffolding, 10 Homework assignment, 213 I Incommensurability, 60, 67 Infinitude of primes, 17 Infinity, 61 Integer parabola task, 108–111 Integer solutions, 169 Internet resources, 55 Invariance, 235 L Lagrange’s Theorem, 40 Lakatos, I. exception-barring method, 178, 189 mathematical discovery, 178 Launch– Explore– Re-launch– Re-explore– Relate, 112–117

N Nice monsters, 174 Nice numbers, 169 Nice real-number monsters, 175 Nice solutions, 169 Niceness, 169, 172 Non-Cartesian coordinate systems, 224 Nonconstant function, 182, 189 Non-prototypical triangles, 123 Non-symmetrical configurations, 163 Numerical monsters, 173 O Order of operations, 201, 203, 204, 206, 208–211 P Pairwise Comparison Task, 106 Pattern-sniffing activity, 163 Penrose tribar, 125, 127, 128, 130, 131, 133, 134 Perfect squares, 115 Plane, 45 Plausible geometric interpretations, 145 Prime decomposition, 56 Prime factorization, 58 Prime numbers, 56 Problem-posing, 134, 155, 156, 168, 171

Index

210 Problem-solving instruction demonstrate-and-apply teaching strategy, 105 developing vs. applying, 105 explore– launch– re-explore– relate, 107, 108, 112 launch– explore– re-launch– re-explore– relate, 112–117 launch– get stuck– intervene– relate, 118–124 Launch– Intervene– Re-launch– Explore– Relate, 125, 127–133 launch-and-explore teaching strategy, 105 launch-explore-summarize strategy, 106 Pairwise Comparison Task, 106 process of relating, 105 Proof-generated definition, 131 Pythagoras, 40 Pythagorean theorem, 40, 179, 228, 231, 234 Pythagorean triples, 116, 117 Q Quadratic equations Euclid’s method, 145 niceness, 171 polynomials, 148 random, 169 school practice, 169 solutions, 147 solving, 138 Quadratic functions, 107 R Rachel’s equation, 173 Ratio equation, 216 Rational and irrational, 66–70 E-rational and E-irrational, 67–69 Rational numbers, 62 Rational Zero Theorem, 115, 169, 171 Ready-to-use technology, 143 Real-number solution, 174 Rectilineal figure, 68 Reverse heuristic engineering, 143, 144 Rolle’s theorem, 192, 193 Ronnie’s approach, 12 Ronnie’s auxiliary construction, 132 Ronnie’s proof, 132 S Scaling (dilation), 222 Script writer, 217

Script writing, 143 Scripting task, 17, 39, 41, 58, 214, 215 proposition of choice, 41 Script-writers, 40, 41, 44 Script-writing task, 39, 43 Script-writing tasks, 40, 41 Secondary mathematics teachers, 41 Secondary school classroom, 137 Self-substitution, 155 Similar solid numbers, 45, 46 Simplicity, 169 Solid numbers, 43, 45, 51 and similar solid numbers, 45, 46 Solving quadratic equation, 138 Special irrational equations, 175 The Spinning Table Problem, 4 Steiner-Lemus Theorem, 9, 10 The String Around the Earth Problem, 5, 6 Structuring proofs, 188 Structuring proofs method, 10 Symmetric configurations, 169 T Task-design heuristic, 156, 157 Tax/discount, 199 Teacher interventions, 128, 129 Teacher-student interactions, 144 Technion, 138 Tetrahedron Problem, 124, 125 Traditional reverse engineering, 143 Translation of parabola, 7, 8 Trigonometry, 67 Two Kings Problem, 6 Two mean proportional numbers, 43, 44, 46, 47, 49, 51 Two-Pictures task, 126 V Variation theory, 163 Viet Theorem, 115 Vieta’s formulas, 164 Virtual dialogue, 40 Virtual duoethnography, 17, 40, 143 Virtual monologue, 143 Visualization, 118 W Warm-up task, 42, 43, 57, 65 Who-is-right task, 13 Wilson’s theorem, 193, 194