A Discrete Transition to Advanced Mathematics [3 ed.] 9780821847893, 2008047393, 0821847899

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Pure, ajuC/\M£le4j

UNDERGRADUATE JJTEXTS • 3

A Discrete Transition to Advanced Mathematics Bettina Richmond Thomas Richmond

American M a t h e m a t i c a l Society Providence, Rhode Island

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2000 Mathematics

Subject

Classification.

Primary 00-01.

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Library of C o n g r e s s C a t a l o g i n g - i n - P u b l i c a t i o n D a t a Richmond, Bettina. A discrete transition to advanced mathematics / Bettina Richmond, Thomas Richmond. p. cm. — (Pure and applied undergraduate texts ; v. 3) Originally published: Belmont, CA : Thomson/Brooks/Cole, c2004. Includes bibliographical references and index. ISBN 978-0-8218-4789-3 (alk. paper) 1. Mathematics—Textbooks. I. Richmond, Thomas. II. Title. QA39.3.R53 510—dc22

2009

2008047393

C o p y i n g and reprinting. Individual readers of this publication, and nonprofit libraries acting for them, are permitted to make fair use of the material, such as to copy a chapter for use in teaching or research. Permission is granted to quote brief passages from this publication in reviews, provided the customary acknowledgment of the source is given. Republication, systematic copying, or multiple reproduction of any material in this publication is permitted only under license from the American Mathematical Society. Requests for such permission should be addressed to the Acquisitions Department, American Mathematical Society, 201 Charles Street, Providence, Rhode Island 02904-2294, USA. Requests can also be made by e-mail to reprint-permissionQams.org. © 2004 held by the American Mathematical Society. All rights reserved. The American Mathematical Society retains all rights except those granted to the United States Government. Printed in the United States of America.

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Preface A Discrete Transition to Advanced Mathematics is designed to bridge the gap between more-computational lower level courses and more-theoretical upper level courses in mathematics. While the focus is on building understanding, sharpening critical thinking skills, and developing mathematical maturity, topics from discrete mathematics provide the means. The text contains more material than can be covered in one semester. There are several reasons for this. First, this makes the book appropriate for Discrete Mathematics courses for second- or third-year mathematics majors, as well as for Introduction to Proofs courses. Second, this will allow each instructor some flexibility in the selection of topics. Perhaps the best reason for the inclusion of so much material, however, is that the book is designed for students who should be learning to read mathematics on their own, and the extra sections should provide enjoyable reading at an appropriate level for these students. Besides more standard topics, the topics mentioned below will distinguish this text from others and, if not presented in class, would provide excellent material for independent projects. • Divisibility tests, long familiar to many students, are explained and proved in Section 3.4. • The surprising elementary number patterns in Section 3.5 emphasize the importance of pattern recognition. • The binomial coefficients are introduced and applied geometrically in Section 4.1 before the formula for them is presented in Section 4.3. • Modular arithmetic is introduced in Section 5.4 as a quotient construction and quotient spaces are used to investigate partial order relations on the blocks of a partition of a set A (i.e., quasiorders on A). • The study of sequences in Chapter 8 provides a discrete version of analysis. Finite differences and their relation to sequences generated by polynomials are investigated. Limits are treated formally, providing an introduction to epsilon-N proofs for those who may have missed epsilon proofs in the calculus sequence. • Infinite series, infinite products, and nested radicals in Section 8.5 provide an introduction to some forms of infinite arithmetic. • Fibonacci numbers and Pascal's triangle in Chapter 9 provide a delightful array of surprising results that provide a unifying synthesis of topics from the previous chapters. • Continued fractions and their applications are discussed in Chapter 10. iii Licensed to AMS. License or copyright restrictions may apply to redistribution; see http://www.ams.org/publications/ebooks/terms

The remarkable connections between the Fibonacci numbers, Pascal's triangle, and the golden ratio in Chapter 9 were the original impetus for our writing this text. We considered a course based on these connections and patterns, many of which are very easily grasped. As we debated the appropriate level of presentation, we concluded that these ideas would serve as an excellent capstone to A Discrete Transition to Advanced Mathematics course. We have taught courses based on Chapters 1-6 with Sections 3.4, 3.5, 4.5, and 6.4 optional, with additional topics and projects selected from the later chapters. Chapters 1-3 are required for all subsequent chapters and should be presented in order. The subsequent chapters need not be covered in order, but Chapter 5 is required for Chapter 6 and Sections 6.1 and 6.2 are needed for Section 7.2 and ' Chapter 8. The material presented here should be accessible to students with the mathematical maturity provided by two or three semesters of calculus or an introductory linear algebra class. No calculus or linear algebra is used, but on a few occasions, connections to these subjects are noted. Besides many classic results, we also include many elegant or surprising results which are not as widely known. We have tried to attain an engaging writing style that emphasizes precision through an intuitive understanding of the underlying concepts. However, simply reading the text will not be enough: Every student should work lots of exercises! There are over 650 exercises of varying difficulty designed to reinforce and extend the material presented. We hope that the selection of topics, examples, and exercises will provide each reader with some of the marvel and amazement we still enjoy.

Ancillaries The following ancillaries are available: Student Solutions Manual The Student Solutions Manual provides worked out solutions to selected problems in the text. Complete Solutions Manual The Complete Solutions Manual provides complete worked out solutions to all of the problems in the text and is available only to instructors.

Acknowledgments We would like to thank Bob Pirtle and his excellent team at Brooks/Cole. Laura Horowitz has been invaluable in ushering the project through production. For their diligent work, we would like to thank the reviewers of this text: Yuanqian Chen, Central Connecticut State University Garry Lee Johns, Saginaw Valley State University Mary Y. Kutter, Florida State University Douglas David Mooney, Battelle

and we would like to thank our many students and colleagues who have inspired and encouraged us to complete the text before you. Bettina Richmond Tom Richmond

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Contents 1

Sets and Logic 1.1 1.2 1.3 1.4 1.5 1.6

1

Set s 1 Se t Operation s 8 Partition s 20 Logi c an d Trut h Table s Quantifier s 34 Implication s 39

25

2 Proofs 49 2.1 Proo f Technique s 49 2.2 Mathematica l Inductio n 2.3 Th e Pigeonhol e Principl e

3

Number Theory

3.1 3.2 3.3 3.4 3.5

4

11

Divisibilit y 77 Th e Euclidea n Algorith m 87 Th e Fundamenta l Theore m o f Arithmeti c Divisibilit y Test s 102 Numbe r Pattern s 11 1

Combinatorics

4.1 4.2 4.3 4.4 4.5

60 70

95

123

Gettin g fro m Poin t A to Point s 123 Th e Fundamenta l Principl e o f Countin g 132 A Formul a fo r th e Binomia l Coefficient s 14 1 Combinatoric s wit h Indistinguishabl e Object s 14 6 Probabilit y 155

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vi

Content s

5

Relations

5.1 5.2 5.3 5.4

163

Relations 163 Equivalence Relations 170 Partial Orders 178 Quotient Spaces 186

6

Functions and Cardinality

6.1 6.2 6.3 6.4

7

Graph Theory

7.1 7.2 7.3 7.4

8

207 223

231

Graphs 231 Matrices, Digraphs, and Relations Shortest Paths in Weighted Graphs Trees 263

Sequences

8.1 8.2 8.3 8.4 8.5 8.6

197

Functions 197 Inverse Relations and Inverse Functions Cardinality of Infinite Sets 215 An Order Relation for Cardinal Numbers

242 253

273

Sequences 273 Finite Differences 280 Limits of Sequences of Real Numbers Some Convergence Properties 297 Infinite Arithmetic 303 Recurrence Relations 316

289

9

Fibonacci Numbers and Pascal's Triangle

9.1 9.2 9.3 9.4 9.5

331

Pascal's Triangle 331 The Fibonacci Numbers 344 The Golden Ratio 355 Fibonacci Numbers and the Golden Ratio 362 Pascal's Triangle and the Fibonacci Numbers 370

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10

Continued Fractions 10. 1 10.2 10.3 10. 4

377

Finit e Continue d Fraction s 37 7 Convergent s o f a Continue d Fractio n Infinit e Continue d Fraction s 39 2 Application s o f Continue d Fraction s

Answers or Hints for Selected Exercises Bibliography Index

38 5 39 8 40 9

41 9

42 1

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Sets and Logic

[The universe] cannot be read until we have learnt the language and become familiar with the characters in which it is written. It is written in mathematical language... —Galileo Galilei (1564-1642)

For the things of this world cannot be made known without a knowledge of mathematics. —Roger Bacon (1214-1294)

Sets A set is a collection of objects. The objects of the set are called the elements of the set. One way to specify a set is to list all the elements inside set brackets "{" and " } " . For example, {Alabama, Alaska, Arizona, Arkansas} is a set with four elements. We may also specify a set in words. The set given above could be specified by stating "the set of all U.S. states that start with the letter A." It is convenient to give sets names, and conventionally, sets are named by capital letters. Thus, we may write A = {Alabama, Alaska, Arizona, Arkansas}. Alabama is an element of A. Birmingham, Atlanta, and 1 Licensed to AMS. License or copyright restrictions may apply to redistribution; see http://www.ams.org/publications/ebooks/terms

2

CHAPTE R 1

Set s an d Logi c

Wyoming are not elements of A. The symbol for "is an element of" is e. Putting a slash through this symbol gives the symbol for "is not an element of." Thus, we may write Alabama e A Alaska e A Birmingham £ A Atlanta £ A Wyoming ^ A. Let us consider the set consisting of the natural numbers less than 6, and let us call this set B. The previous cumbersome sentence may be shortened to this: Let B = {1, 2, 3,4, 5}. Here we are listing the elements of B in rosterform rather than giving a verbal description of the elements. Counting down from 6, you may determine that the set of natural numbers less than 6 should be {5, 4, 3, 2, 1}. This is also correct. The elements of a set may be listed in any order. Thus, B = {1, 2, 3,4, 5} = {5, 4, 3, 2, 1} = {3, 5, 2, 1, 4}, and there are many more correct representations of the set B. Any set U must be well-defined; that is, for every object x, there must be an unequivocal answer to the question "Is x e UT We may not always know the answer to this question, but we must know that an unequivocal answer exists. Consider the set F of all living people who have an ancestor with the name Fletcher. Are you a member of this set? Though you may not know the answer, you should recognize that there is an indisputable answer—either yes or no. The set of good books, however, is not a well-defined set. The answer to the question "Is War and Peace a good book?" may be subject to dispute. The usage of the word good is subjective, and this makes the word an improper choice to use in specifying well-defined sets. Two sets are equal if they contain exactly the same elements. The set B = {4, 3, 1, 5, 2} and the set {|, V i , \/9, 2 2 , 5} are equal since they contain exactly the same elements, namely 1 = | , 2 = \[\, 3 = \/9, 4 = 2 2 , and 5. The set of kangaroos on the moon is a well-defined set that contains no elements. The set { } containing no elements is called the empty set or null set and is denoted 0 or { }. A set is finite if there is a whole number that tells the number of elements in the set. The set B = {1, 2, 3, 4, 5} is finite, and the number of elements in B is five. 1.1.1

DEFINITION The cardinality of a finite set S is the number of elements in the set S and is denoted |5|. Counting the number of elements in a set may not be as easy as it sounds, especially if the set is described instead of listed. How many elements does the set of letters in the word throughout have? Stated another way, find the cardinality \C\ of the set C = {t, h, r, o, w, g, h, o, u, t}. To the question "Is r e CT we should answer "Yes." To the question "Is t e C?" we should answer "Yes, yes." Though it is more emphatic, the affirmative outcome "Yes, yes" is not different from the affirmative outcome "Yes," so the element t e C only counts as one element, despite the fact that we listed it twice. If we let D be the set of letters in the word trough, then D = {t,r,o,u,g,h}. The sets C and D have exactly the same elements, so C = Z>, and thus \C\ = \D\ = 6. Repeated

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1.1 Set s

3

elements in a set should only be counted once. Recognizing the duplication is frequently more difficult than in this example. Sometimes we may not be able to count the elements of a finite set. The set F of living people with an ancestor named Fletcher is a finite set, and though we do not know the exact number of elements in F, we know \F\ cannot exceed the current world population, and thus must be finite. If a set is not finite, it is infinite. The set of natural numbers, for example, is infinite. We will not be able to list all the elements of an infinite set, but we may indicate an infinite set by a verbal description or by listing several of the elements in a clear pattern followed by an ellipsis ("..."). Some standard notation for some standard sets will illustrate this. The set of natural numbers = N =

{1, 2, 3, 4 , . . . }

The set of whole numbers = W = {0, 1, 2, 3 , . . . } The set of integers = Z =

{0, 1, - 1 , 2, - 2 , 3, - 3 , . . . }

= { . . . , - 3 , - 2 , - 1 , 0 , 1,2,3,...}

Another convenient way to specify a set symbolically is by set-builder notation, which we illustrate here. The notation {x \ x e N and x < 6} is read "the set of all x such that I G N and x < 6." In general, { j c | * * * * * * * } i s read "the set of all x such that x satisfies the properties * * * * * * * stated." Unknown elements of a set are conventionally denoted by lowercase letters, such as the x above. If the elements x are to come from some specified set, we may include this information before the "pipe" symbol "|". The set {x \ x e N and x < 6} could be written as {x e N\x < 6} and read "the set of natural numbers x such that x < 6." This is the set B = {1, 2, 3, 4, 5} we have seen earlier. We may now introduce the notation for two other frequently used infinite sets. The set of real numbers = E = {x \ x is a real number} The set of rational numbers = Q = { | | < z , £ e Z , b ^= 0} The set of rational numbers Q consists of Quotients of integers, with the usual restriction that division by 0 is not allowed. If we take an arbitrary set S and remove some, none, or all of its elements, the set T of remaining elements is called a subset of S, and we write T C S. Formally, a set T is a subset of S if and only if every element of T is also an element of S. If T is a subset of S, then S is a superset of T and we may write S 3 T. The notation T C S may also be read "T is contained in S." We will illustrate this notation with some examples. {red, white, blue} C {red, white, blue, green} {1,3,5} C { 1 , 2 , 3 , 4 , 5 } {2} C { 1 , 2 , 3 , 4 , 5 } 0 = { } c {1,2,3,4,5} {1,2,3,4,5} C {1,2,3,4,5} {5,6,7} g { 1 , 2 , 3 , 4 , 5 } Licensed to AMS. License or copyright restrictions may apply to redistribution; see http://www.ams.org/publications/ebooks/terms

4

CHAPTE R 1

Set s an d Logi c

The symbol % used in the last example above means "is not a subset of." It is critical to use the correct terminology and symbols for subsets and elements of a set. Observe that 3 G {1, 3, 5} but 3 £ {1, 3, 5}. Since 3 is not a set, it cannot be a subset of anything. Similarly, {3} c {1, 3, 5} but {3} j^ {1,3, 5}. If T C S but T ^ S, we say T is a proper subset of S and write T C S. (Compare this notation to < and (S). Thus, &>(S) = {A | A c S}. Example 1.1.3 shows that the power set of {1, 2, 3} is ^ ( { 1 , 2, 3}) = { 0, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3} }. We may consider subcollections of ^ ( { 1 , 2, 3}) such as the collection & of all subsets of {1, 2, 3} that contain the element 2: & =

{A e ^ ( { 1 , 2 , 3 } ) | 2 G A} =

{ {2}, {1, 2}, {2, 3}, {1, 2, 3} }.

The collection & of subsets of {1, 2, 3} that have cardinality 2 is & = {A € ^ ( { 1 , 2, 3}) | \A\ = 2} = { {1, 2}, {1, 3}, {2, 3} }. We have 9 C ^ ( { 1 , 2, 3}) and & C ^ ( { 1 , 2, 3}). Observe the distinction between "containing two elements" and "containing the element 2." As with any set, we may consider the cardinality of a set of sets—that is, of a collection. Here we have \(3\ = 4, \&\ = 3 a n d | ^ ( { l , 2 , 3 } ) | = 8 . The following example will reinforce the importance of distinguishing between an element of a set and a subset of a set. Licensed to AMS. License or copyright restrictions may apply to redistribution; see http://www.ams.org/publications/ebooks/terms

6

CHAPTE R 1 Set s an d Logi c

1.1.5

EXAMPLE Let A = {Alabama, Alaska, Arizona, Arkansas} E= ® I = {Illinois, Indiana, Iowa} O = {Ohio, Oklahoma, Oregon}, and U = {Utah}. Now, if we let y = {A, E, I, O, U}, then V is a collection of five sets and we have Alaska e A {Alaska, Arizona} C A Alaska £ y {Alaska, Arizona} £ V {Alaska, Arizona} g y I = {Illinois, Indiana, Iowa} e y

/ gr

{O,E,U}

c

y

{£,£/} = { { } , { U t a h } } c y

{c/j c r .

Note that Utah € £/ and {Utah} C U (in fact, {Utah} = U), but Utah g £/. Furthermore, C / e r = {A,£, /, O, U] and {{Utah}} = {U} C r , but {Utah} = £/ g y , Utah g y , and Utah ^ ^ . There are also some subtleties involving the empty set in this example. From the definition of Y, we see that £ = 0 e f. The empty set, however, is a subset of any set, and in particular, the empty collection is a subcollection of any collection. For our collection y, we have 0 c f Now we have shown that E = 0 e V and E = 0 C f. This is a rare occurrence. Only in extraordinary circumstances will an element of a set also be a subset of that set. Note that E = 0 c Y and also {E} = {0} C r , b u t 0 ^ {0}. Generally, x ^ {x}, and there is no exception for x = 0. While 0 has no elements, {0} has one element, namely 0. Before leaving this example, we should note that the collection y of the five sets A, E, I, O, and U is not the same as the set {Alabama, Alaska, Arizona, Arkansas, Illinois, Indiana, Iowa, Ohio, Oklahoma, Oregon, Utah} of the 11 states that start with a vowel. In the next section we will see that this latter set is the union of the collection y. Large collections of sets are often expressed using an "index" for each set. For example, suppose a certain class meets for 36 days. Let S\ be the set of students present on the first day, S2 be the set of students present on the second day, and in general, let Sk be the set of students present on the &-th day (k e {1, 2, 3, . . . , 36}). The subscript k is called the index (plural: indices). The collection 5? of all the sets [S\9 S2, . . . , S36} Licensed to AMS. License or copyright restrictions may apply to redistribution; see http://www.ams.org/publications/ebooks/terms

1.1 Set s

7

may be represented as an indexed collection using various notations: y =

{Sk\k=

1,2,3, . . . , 3 6 }

= { S * | * €{ 1 , 2 , 3 , . . . , 3 6 } } = {Sk\ke

/} w h e r e / = {1, 2, 3 , . . . , 36}

= {&> * i = {Skhei where / = {1, 2, 3 , . . . , 36}. The index k is a "dummy variable"—we could just as well use /, j , A, or any other symbol. The set of values that the index may assume is called the index set. In the example at hand, the index set is / = {1, 2, 3 , . . . , 36}.

EXERCISES 1. (a) True or false? {Red, White, Blue} = {White, Blue, Red}. (b) What is wrong with this statement: Red is the first element of the set {Red, White, Blue}? 2. Which has the larger cardinality? The set of letters in the word MISSISSIPPI the set of letters in the word FLORIDAl 3. Fill (a) (b) (c)

in the blank with the appropriate {1,2,3} {1,2,3,4} 3 {1, 2, 3, 4} {3} {1, 2, 3, 4}

or

symbol, e or C. (d) {a} {{a}, {b}, {a, b}} (e) 0 {{«}, {M, {a, b}} (f) {{«}, {b}} {{a}, {b}, {a, b}}

4. Draw a Venn diagram showing the proper relationship between these sets: N, Q, R, W, and Z. 5. (a) How many subsets does the empty set have? (b) How many subsets does the set {1} have? (c) Noting the number of subsets of a two-element set and of a three-element set from Examples 1.1.2 and 1.1.3, how many subsets do you think a four-element set {1, 2, 3, 4} would have? (d) List all the subsets of the four-element set {1, 2, 3, 4}. (e) How many subsets do you think a five-element set would have? A six-element set? An n -element set? 6. Determine whether the sets below are well-defined or not. For each well-defined set, state whether it is finite or infinite. (a) The set of women pregnant with twins at some time during this year. (b) The set of kangaroos in Australia. (c) The set of tall buildings. (d) The set of grains of sand on the earth. (e) The set of even integers. (f) The set of hairs on your head. (g) The collection of all subsets of the set of hairs on your head. (h) The set of people who shook hands with George Washington. Licensed to AMS. License or copyright restrictions may apply to redistribution; see http://www.ams.org/publications/ebooks/terms

8

CHAPTE R 1 Set s an d Logi c

7. (a) Are there well-defined sets in Exercise 6 for which we may not know the answer to the question "Is x an element of this set?" for every object xl (b) Are there finite sets in Exercise 6 for which we do not know the cardinality? 8. Let (a) (b) (c) (d)

S\ = {o, n, e}, 5 2 = {t, w, o], S3 = {t, h, r, e, e}, and so on. Find all k e {1, 2 , . . . , 10} with 15*1 = 4. Find distinct indices j,k eN with Sj = Sk. Find the smallest value of k G N with a e Sk. Let y = {Sk)fLx. Determine whether the following statements are true or false. i. S13 = {n, e,i,t, h, e, r} ix. Si c 52i ii. [n, e, t] c 52o x. Si C 5 21 iii. S\ e y xi. {n, i, e} G y iv. S3 c y xii. {{f,o,u,r}}^y v. vi. vii. viii.

0 ey 0c y 0 c y S\ c Sn

xiii. xiv. xv. xvi.

U G 540

&&) C ^ ( 5 1 9 ) { J , i } €^ ( 5 6 ) u; G ^ ( 5 2 )

9. For t € {1, 2 , . . . , 20}, let Dk = {x \ x is a prime number that divides /:} and let # = {0*1* €{1,2,...,20}}. (a) Find Du D2, D10, and D 20 . (b) True or false: vii. {5} G 3f i. D2 C Dio viii. {4, 5} G 9 11. D7 C Dio ix. {{3}} c ® iii. £>io C Z)20 iv. 0 G ® x. &(D9) c ^ ( D 6 ) v. xi. ^ ( { 3 , 4}) c ® vi. 5 e ® xii. {2, 3} G ^ ( D 1 2 ) (c) Find|D 10 and |Z>i9|. (d) Find \@\. 10. Give an example of an indexed collection y =

{Sk}5k= with \y\ = 3.

Set Operations There are some standard set operations used to derive new sets from given sets.

Intersection and Union Given sets 5 and T, the intersection of 5 and 7\ denoted 5 H T, is the set of elements which are in both 5 and T. The union of sets 5 and T, denoted 5 U 7\ is the set of all elements which are in either 5 or T or both. SOT =

{x\x e Sandx

SUT =

{x\x e Sorx

eT} eT}

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1.2 Se t Operation s

9

It is clear from the definitions that for any sets S and T, S DT = T Pi S and S U T = T U S. The Venn diagrams in Figure 1.2 depict the sets S and T, the intersection S C\T, and the union S U T.

QD SnT

C

j \

Intersections and unions.

su r

3rd Ave.

Main Street

Figure 1.2

QD

/

Intersection of 3rd Ave. and Main Street

European Union

For example, suppose A = {1, 3, 5, 7}, B = {3, 4, 5, 6}, and C = {2, 4}. Then we have AH B =

AUB = ADC

{3,5}

{1,3,4,5,6,7}

= 0

A U C = {1,2,3,4,5,7} B n C = {4} BUC =

{2,3,4,5,6}

Two sets with no "overlap," such as A and C above, are said to be disjoint. Formally, sets S and T are disjoint if S n T = 0. Let us consider another example. Let £/ be the set of students enrolled at Ottawa University. In this example, we will only consider subsets of this set U. Such a set U containing all the objects to be considered is called the universal set for the problem in question. Let H = {x e U \x has black hair}. Let E = {x e U \x has green eyes}. Licensed to AMS. License or copyright restrictions may apply to redistribution; see http://www.ams.org/publications/ebooks/terms

10

CHAPTE R 1

Set s an d Logi c

Now H is the set of Ottawa University students with black hair and E is the set of Ottawa University students with green eyes. H n E = {x e U | x e H and x e E} = the set of Ottawa University students with black hair and green eyes. HUE =

[x € U\x e H ovx e E) = the set of Ottawa University students with black hair or green eyes.

Observe that those Ottawa University students with black hair and green eyes qualify to be an element of H U E in two ways. If a waitress asks you if you would like french fries or a baked potato, she is using an "exclusive or": She means you may select one or the other but not both. In mathematics, the word or is interpreted as an "inclusive or," so saying "P or Q" means "P or Q or both." If a student is admitted to the Ottawa University black-haired green-eyed student union HUE because she has black hair, she will not be thrown out if she also happens to have green eyes. Recognizing that some students may qualify as a member of H U E by two criteria is important in counting the number of elements in the union HUE. Suppose we know that Ottawa University has 127 blackhaired students and 73 green-eyed students; that is, suppose \H\ = 127 and \E\ = 73. Without further information, we cannot determine the number | H U E | of students with black hair and green eyes. If we simply add \H\ and \E\, the dually qualified students— those with black hair and green eyes; that is, the elements of H Pi E—are counted twice (Fig. 1.3). Tofindthe correct answer, we must add |H| and \E\, then subtract the number \H C\ E\ that were counted twice. Thus we have \HUE\ =

\H\ +

\E\-\HHE\.

Figure 1.3

\H\ + \E\ counts the elements ofHDE twice. To compensate, subtract \H D E\ once.

Without further information on the size of the overlap H C\ E, we may only find a possible range of values for | H U E |. At one extreme, if H and E do not overlap at all—that is, if H and E are disjoint (Fig. 1.4)— then there are no students with both black hair and green eyes, so no one is counted twice and \HUE\ =

\H\ + \E\ -\HDE\ = \H\ + \E\-\0\ = 127 + 7 3 - 0 = 200.

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1.2 Se t Operation s

11

Figure 1.4

If H and E are disjoint, then|#| + |£| = \HUE\.

At the other extreme, the largest overlap H Pi E occurs if all 73 green-eyed students also have black hair. In this case, E C. H, so H C\ E = E and

H

Figure 1.5

I f £ C tf,then \HUE\ = \H\.

1.2.1

\HUE\ = \H\ + \E\ - \HDE\ = \H\ + \E\-\E\ = \H\ = 127. Equivalent^, if E C H (Fig. 1.5), then H U E = H and therefore \H U E\ = \H\ = 127. Thus, if we only know |//| = 127 and \E\ = 73, we may conclude that the intersection HOE has n elements for some n satisfying 0 ^ n ^ 7 3 . The formula \H U E\ = \H\ + \E\-\HnE\ now tells us that the union H U E has from 127 to 200 elements. The exact answer can be determined only if we know \H C\ E\ exactly. If we determine that there are exactly 32 black-haired, green-eyed students at Ottawa University, then we would know that | HUE | = \H\ + \E\ - \H D E\ = 127 + 7 3 - 3 2 = 168. In the discussion above on \H U E\, we implicitly encountered the fact that the number of elements of a subset T of set S cannot exceed the number of elements of S. That is, if T C S, then \T\ < \S\. In particular, since H n E c H and H n ^ c £, we h a v e | / / n £ | ^ | H | a n d | / / n £ | ^ |£|. Similarly, since// c // U EandE c i/ U £, we have | # | ^ |/f U E\ and | £ | < \H U £|. We summarize our results here. THEOREM

Suppose S and r are finite sets. Then

(a) isuri = |5| + | r | -

\SHT\. (b) If S and T are disjoint, then \SUT\ (c) If S c T,then|5| < | 7 | .

isi + m.

We may define the intersection or union of more than two sets. If ff = {St•] i e 1} is a collection of sets S,- indexed by the set /, then we define the intersection of the collection ^ to be the intersection of all the sets in the collection: p | tf = p | Si: = {x |x e St for every i e I}. iel

The intersection of a collection is thus the set of elements common to all of the sets in the collection. Licensed to AMS. License or copyright restrictions may apply to redistribution; see http://www.ams.org/publications/ebooks/terms

Set s an d Logi c

The union of the collection ^ — {St\i e 1} is the union of all the sets in the collection: |^J ^ = |^J Sti = {x | x e Si for at least one i €/ } . iel

The union of a collection consists of all the elements that belong to at least one of the sets in the collection. If the index set / is finite, say / = {1, 2, 3 , . . . , n}, then we may write n

p | s( = f] St = Si n 52 n 53 n . . • n sn iel

=i l

and a similar notation holds for unions. We say that the sets of a collection {S(\i e 1} are mutually disjoint (or pairwise disjoint) if 5/ # Sj implies St 0 Sj = 0, or equivalently, if 5/ H Sj• ^ 0 implies 5/ = Sj. Thus, {5,-11 G /} is a collection of mutually disjoint sets if every pair of distinct sets in the collection is disjoint. A collection of sets {5,-1 i e 1} is called a nested collection if for any i, y e / , either St c S; orS, c St. 1.2.2

EXAMPLE Given below are three collections of intervals on the real line. For each collection, determine whether the sets of the collection are mutually disjoint. Determine whether the collection is nested. Find the union of the collection and the intersection of the collection. ^ =

{[n,n + l)\n e Z}

@ = {(n,n + 2)\n e Z]

tf = {Hr.i + ir)l"€N} Solution The sets of srf are mutually disjoint, for if m and n are distinct integers, then the intervals [m, m + 1) and [n, n + 1) are disjoint. The sets of SS are not mutually disjoint, for given any integer n, the number n + 1.5 is an element of two distinct sets (n, n + 2) and (n + 1, n + 3) of 38. The sets of ^ are not mutually disjoint, for 0 is contained in the intersection of any pair of sets of ^ . Neither srf nor ^ is a nested collection, for each contains pairs of nonempty disjoint sets. ^ is a nested collection, for given any m, n e N, we have either m ^ n or n ^ m. Without loss of generality, assume m ^ n. Then (—-, 1 + -) c (—-, 1 -f - ) . to

j->

\

\

n->

n

/

—

v

m>

m

/

As every real number x is contained in the interval [n,n + 1) where n is the greatest integer less than or equal to x, we see that E c ( J ^ . On the other hand, since [n,/i + l ) c R for every integer n, we have (J«g^ = Unezt^' w + 1) £ R. and thus (J ^ = R. Similarly, U 38 = R. As the intervals of the collection ^ were nested with the intervals getting smaller as the index increases, we find that the largest interval in ^ corresponds to the smallest index. That is, every interval of ^ is nested inside the first interval (-•f, 1 + \) = (-1,2). Thus, \J^ = (-1,2). Since both collections srf and 38 contain pairs of disjoint intervals, we have (~) srf — 0 = p | 38. Since ^ is a nested collection, f] ^ would equal the smallest interval of Licensed to AMS. License or copyright restrictions may apply to redistribution; see http://www.ams.org/publications/ebooks/terms

1.2 Se t Operation s

13

^ if ^ had a smallest interval. However, ^ has no smallest interval. It is easy to see that every element of [0, 1] is contained in every interval (—-, 1 + -) of ^ and that every number outside of the interval [0, 1] is excluded from some interval of ^ . Thus,

n ^ = [o,i].

•

Complements Suppose A is the set of month names that contain an a, R is the set of month names that contain an r, and Y is the set of month names that contain a y. The universal set in this setting would be the set U of all twelve month names, and we have A = {January, February, March, April, May, August} R = {January, February, March, April, September, October, November, December} Y = {January, February, May, July} The intersection A fl R Pi Y is the set of month names that contain an a, an r, and a v; and thus, AD ROY = {January, February}. The union of A, R, and Y is A U R U Y = {January, February, March, April, May, July, August, September, October, November, December}. Observe the placement of the 12 months in the eight regions of the Venn diagram in Figure 1.6.

fl • August

Figure 1.6

All ROY = {January, February); A U RUY = {January, February, March, April, May, July, August, September, October, November, December}.

V.

May

I

Y

July

Vr ^ March April

f f

L

January February

> ,

j

j

September October November December !

June

In the previous example, the 11-element subset A U R U Y of the 12-element universal set U could be described by specifying which element of U was left out. The set of elements left out of a set S is called the complement of the set S. 1.2.3

DEFINITION If S is any subset of the universal set U, then the complement of S in U (Fig. 1.7), denoted Sc or U \ S, is the set Sc = U \ S = {x e U \ x # S}.

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14

CHAPTE R 1 Set s an d Logi c

Figure 1.7

Sc = U \ 5, the complement of S, is shaded. From the definition or the Venn diagram, it should be clear that (Sc)c = S. The notation U \ S has the advantage that it explicitly names the universal set, which we may want to alter or restrict at times. Suppose Al moves to the gulf coast so he can catch and eat fresh oysters every month, except December and January, which are too cold for Al. The set of months that Al eats fresh oysters may be described as U \ {December, January}, where U is the universal set of the 12 months. However, once Al learns from the locals that oysters should only be eaten in months with an r, the set of months that Al eats fresh oysters may be described as R \ {December, January}, where R is the set of months that contain an r. 1.2.4

DEFINITION The set R \ S = {x e R | x £ 5} is called the complement of S in R. To emphasize that the set R may not be the entire universal set, sometimes R \ S is called the relative complement of S in R. To find R \ S, we start with all the elements of R and throw out any that are also in S (Fig. 1.8). For this reason, R \ S is typically called a set difference. For example, if R is the set of right-handed people and S is the set of people under 5 feet tall, then R \ S is the set of right-handed people who are 5 feet tall or taller.

Figure 1.8

The set difference R \ S is shaded. It is easy to show that if R and S are subsets of the universal set £/, then R \ S = R n Sc. The proof is left as an exercise. The set R \ Q = {x € R \ x £ Q} is the set of real numbers that are not rational. This set is called the set of irrational numbers. Licensed to AMS. License or copyright restrictions may apply to redistribution; see http://www.ams.org/publications/ebooks/terms

1.2 Se t Operation s

15

Two important set-theoretic properties describing interactions of intersections, unions, and complements are given below. A more complete discussion and proofs of these statements will appear in Sections 1.4 and 1.5. De Morgan's Laws: If {S,-},-€/ is a collection of sets, then

u*y=n

and

i€/

/

iel

(oy==uIto\iel

)

iel

These may be stated as the complement of a union of sets is the intersection of the complements and the complement of an intersection of sets is the union of the complements. Distributive Laws: For sets A, B, and C,

(AU B)nc = and

(Anc)u(Bnc)

(A n B) u c = (A u c) n (B U C).

Cartesian Products Suppose you have three vehicles: a pickup, a Buick, and a Jaguar. You have two options for Saturday evening entertainment: the opera or bowling. To decide what to drive and where to go, you must choose one element from the set V = {pickup, Buick, Jaguar} of vehicles and one element from the set E = {opera, bowling} of excursion options. If you decide to drive the pickup, there remain two options for the excursion. We will list these two possible outcomes as ordered pairs (pickup, opera) and (pickup, bowling). Similarly, the other possible outcomes are (Buick, opera), (Buick, bowling), (Jaguar, opera), and (Jaguar, bowling). We can visualize the decision process and the possible outcomes by a tree diagram as shown in Figure 1.9. pickup Buick Figure 1.9 A tree diagram.

\

opera bowling opera bowling

Jaguar < ^ ° p e m ^ ^ bowling

The set of outcomes is the six-element set {(pickup, opera), (pickup, bowling), (Buick, opera), (Buick, bowling), (Jaguar, opera), (Jaguar, bowling)} consisting of every possible ordered pair whose first coordinate is an element of V and whose second Licensed to AMS. License or copyright restrictions may apply to redistribution; see http://www.ams.org/publications/ebooks/terms

Set s an d Logi c

coordinate is an element of E. This set is called the Cartesian product of the sets V and E, denoted V x E. 1.2.5

DEFINITION is the set

The Cartesian product of two nonempty sets A and B, denoted A x B,

Ax B = {(a,b)\a

eA,beB}

of all ordered pairs whosefirstcoordinate is an element of A and whose second coordinate is an element of B. The most familiar Cartesian product is the product R x R = R2 = {(JC, y) \x, y e R}. By considering "solution sets" of equations, we have a link between algebraic equations like y = 3x + 1 and the geometric subset {(x, y) e R2 \ y = 3x + 1} of the plane. This link is the foundation of analytic geometry and was utilized by Rene Descartes (1596-1650), Pierre de Fermat (1601-1665), and others. Cartesian products are named in honor of Rene Descartes. In the example above, |V\ = 3, \E\ = 2 and \V x E\ = 3 • 2 = 6. If A and B are any finite sets, we may count the elements of A x B = {(a, b) \a e A,b e B} by considering the tree diagram for A x B. At the first branching, there is one limb for each element of A. At the second branching, we append | B | limbs onto the end of each of the \A\ existing limbs. Thus, the total number of limb tips on the right of the tree diagram is | B | • | A | = | A | • | B |. This gives us the following result. 1.2.6

THEOREM

If A and B are nonempty finite sets, then \A x B\ = \A\ • \B\.

A more formal proof of the theorem above follows on page 22. Using " x " to represent multiplication of real numbers as well as the Cartesian product of sets, the result can be restated as \A x B\ = \A\ x \B\. 1.2.7

EXAMPLE Jimmy and Stacy want to go to the movies on Wednesday, Thursday, Friday, or Saturday of next week. There are three movies playing at the cinema: Amadeus, Beethoven, and Critter from the Brown Lake. How many possible outcomes do they have for the selection of a day and a movie? Draw a tree diagram to illustrate the possible outcomes. Solution Jimmy and Stacy must choose a day from the set D = {W, T, F, S] and a movie from the set M = {A, B, C}. The set of all their possible outcomes is {(d, m) \ d e D,m e M) = D x M. Thus, they have \D x M\ = |D| • \M\ = 4 • 3 = 12 possible outcomes, as illustrated in the tree diagram of Figure 1.10. • Extending the problem above, suppose that each of the movies has two showtimes each night—7 PM and 9 PM. If Jimmy and Stacy wish to select a day, movie, and time, they must select one element from D as above, one element from M as above, and one element from the set T = {7,9} of possible showtimes. Their selection will be an ordered triple (d, m, t), where d e D,m € M, and t e T. The set of all such ordered triples is the Cartesian product of D, M, and T: D x M xT =

{(d, m,t)\d

e D,m e M,t

eT}.

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1.2 Se t Operation s

17

Figure 1.10

Tree diagram for Example 1.2.7.

The elements of D x M x T can be illustrated by a tree diagram with three branching stages (Fig. 1.11). The first two branchings correspond to the selection of day and movie as shown in Example 1.2.7 above. Having selected a particular d e D and a particular m e M, there are two time options, 7 or 9; thus, each existing branch forks into two branches at the third branching stage. The total number of (day, movie, time) options is 4 • 3 • 2 = 24. That is, \D x M

xT\

\D\.\M\

Our motivation for defining Cartesian products of three sets and the method for counting the elements in a Cartesian product of three sets extend to Cartesian products of any finite number of sets. 1.2.8

DEFINITION If S\, S2, S3, . . . , Sn are nonempty sets, then the Cartesian product S\ x S2 x S3 x • • • x Sn is the set {(s\, S2,..., sn) \s\ G Si, s2 e S2, . . . , sn G Sn} of all ordered n-tuples whose A;-th coordinate s^ is an element of the A>th set Sk for each k = 1,2, . . . , n . We may denote the Cartesian product S\ x S2 x S3 x • • • x Sn by YYi=\ Si. The symbol Y\ is the Greek letter pi, corresponding to the first word product. This notation is comparable to the "sigma notation" X^= i ai a\ + CL2 + • - - + an. Observe that the Greek sigma YL corresponds to the the word sum.

the notation letter of the f ° r the sum initial letter of

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18

CHAPTE R 1

Set s an d Logi c

Figure 1.11

Tree diagram. 1.2.9

THEOREM

If S\, 52, 5 3 , . . . , Sn are nonempty finite sets, then

| 5 1 x 5 2 x 5 3 x . . . x 5 n | = |51|.|52|.|53|

\Sn\.

A formal proof of this theorem will be presented in Section 2.2. For now, it is enough to keep in mind our arguments based on what the tree diagram for the Cartesian product would look like. Observe that ifn = 2, the theorem reduces to Theorem 1.2.6. If S\ = 5 2 = 5 3 = • • • = S„ = 5, then the Cartesian product S\ x 5 2 x 5 3 X - - - x 5 „ =

V

5x5x---x5x5 v

n factors

'

is denoted 5 n . The familiar examples of R2 = R x R = {(JC, v) \x, y e R] and R3 = R X R X R = { ( J C , V , Z ) | J C , V , Z G R } follow this convention. If 5 is a finite set, the theorem above implies \Sn\ = \S\n. EXERCISES 1. Let 5 = {1, 3, 5, 7, 9}, T = {1, 2, 3, 4, 5}, and V = {3, 6, 9}. List the elements of the specified sets. (a) 5 n T (d) SUV (g) V x T (b) 5 U T (e) (T n V) U 5 ( h ) V x ( m 5)

(c) 5 n v

(f) r n ( v u 5 )

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1.2 Se t Operation s

19

2. Let U be the set of 52 cards in a standard deck. (For a description of a standard deck, see the paragraph following Example 4.5.2 on page 158.) Let S be the set of spades, D the set of diamonds, A the set of aces, and K the set of kings. Tell which cards belong to each set below, and find the cardinality of each set. (a) A f l D (f) Dc (k) A x Kc c (b) SOD (g) Kns (l) s x s c c (c) A n (5 U D) (h) KD(SU D) (m) S\K (d) (A U /O fl (5 U £>) (i) (AUKYDS (n) /