Proofs that Really Count

Proofs that Really Count: The Art of Combinatorial Proof

Arthur T. Benjamin
Jennifer J. Quinn
Volume: 27
Copyright Date: 2003
Edition: 1
Pages: 209
https://www.jstor.org/stable/10.4169/j.ctt6wpwjh
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  • Book Info
    Proofs that Really Count
    Book Description:

    Mathematics is the science of patterns, and mathematicians attempt to understand these patterns and discover new ones using a variety of tools. In Proofs That Really Count, award-winning math professors Arthur Benjamin and Jennifer Quinn demonstrate that many number patterns, even very complex ones, can be understood by simple counting arguments. The arguments primarily take one of two forms: - A counting question is posed and answered in two different ways. Since both answers solve the same question they must be equal. -Two different sets are described, counted, and a correspondence found between them. One-to-one correspondences guarantee sets of the same size. Almost one-to-one correspondences take error terms into account. Even many-to-one correspondences are utilized. The book explores more than 200 identities throughout the text and exercises, frequently emphasizing numbers not often thought of as numbers that count: Fibonacci Numbers, Lucas Numbers, Continued Fractions, and Harmonic Numbers, to name a few. Numerous hints and references are given for all chapter exercises and many chapters end with a list of identities in need of combinatorial proof. The extensive appendix of identities will be a valuable resource. This book should appeal to readers of all levels, from high school math students to professional mathematicians.

    eISBN: 978-0-88385-333-7
    Subjects: Mathematics

Table of Contents

  1. Front Matter
    (pp. i-viii)
  2. Foreword
    (pp. ix-xii)

    Every proof in this book is ultimately reduced to a counting problem| typically enumerated in two different ways. Counting leads to beautiful, often elementary, and very concrete proofs. While not necessarily the simplest approach, it offers another method to gain understanding of mathematical truths. To a combinatorialist, this kind of proof is the only right one. We offerProofs That Really Countas the counting equivalent of the visual approach taken by Roger Nelsen inProofs Without Words I & II[37, 38].

    As human beings we learn to count from a very early age. A typical 2 year old...

  3. Table of Contents
    (pp. xiii-xiv)
  4. CHAPTER 1 Fibonacci Identities
    (pp. 1-16)

    Definition TheFibonacci numbersare defined by F0= 0, F1= 1, and for$n \ge 2$, Fn= Fn−1+ Fn−2.

    The first few numbers in the sequence of Fibonacci numbers are 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, . . ...

    How many sequences of 1s and 2s sum to n? Let’s call the answer to this counting question fn. For example, f4=5 since 4 can be created in the following 5 ways:

    1 + 1 + 1 + 1, 1 + 1 + 2, 1 + 2 + 1, 2 + 1 + 1, 2 + 2....

  5. CHAPTER 2 Gibonacci and Lucas Identities
    (pp. 17-34)

    Definition TheGibonacci numbers${G_n}$are defined by nonnegative integers${G_0},{G_1}$and for$n \ge 2,{G_n} - {G_{n - 1}} + {G_{n - 2}}$.

    Definition TheLucas numbers${L_n}$are defined by${L_0} = 2,{L_1} = 1$and for$n \ge 2,{L_n} = {L_{n - 1}} + {L_{n - 2}}$.

    The first few numbers in the sequence of Lucas numbers are 2, 1, 3, 4, 7, 11, 18, 29, 47, 76, 123, 199,....

    In this chapter, we pursue identities involvingGibonacci numbers,which is shorthand for generalized Fibonacci numbers. There are many ways to generalize the Fibonacci numbers, and we shall pursue many of these generalizations in the next chapter, but for our purposes, we say a sequence of nonnegative integers${G_0},{G_1},{G_2},...$is a...

  6. CHAPTER 3 Linear Recurrences
    (pp. 35-48)

    Definition Given integers${c_1},...,{c_k}$, akth order linear recurrenceis defined by${a_0},{a_1},...,{a_{k - 1}}$, and for$n > k,$,${a_n} = {c_1}{a_{n - 1}} + {c_2}{a_{n - 2}} + \cdot \cdot \cdot + {c_k}{a_{n - k}}$.

    Definition Given integerssandt,theLucas sequence of the first kindis defined by${U_0} = 0,{U_1} = 1$and for$n > 2$,${U_n} = s{U_{n - 1}} + t{U_{n - 2}}$. For combinatorial convenience, we also define for$n \ge - 1$,${u_n} = {U_{n + 1}}$. Whens=t= 1, these are the Fibonacci numbers:${U_n} = {F_n}$, and${U_n} = {F_n}$.

    Definition Given integerssandt,theLucas sequence of the second kindis defined by${V_0} = 2,{V_1} = s$and for${V_n} = s{V_{n - 1}} + t{V_{n - 1}} + t{V_{n - 2}}$. When$s = t = 1$, these are the

    Lucas numbers:${V_n} = {V_n}$.

    The recurrence for Fibonacci numbers can be extended in many different...

  7. CHAPTER 4 Continued Fractions
    (pp. 49-62)

    Definition Given integers${a_0} \ge 0,{a_1} \ge 1,{a_2} \ge 1,...,{a_n} \ge 1$, define$[{a_0},{a_1},...{a_n}]$to be the fraction in lowest terms for

    $a{}_0 - \frac{1}{{{a_1} + \frac{1}{{{a_2} + \frac{1}{{ \ddots - \frac{1}{{{a_n}}}}}}}}}$.

    For example,$[2,3,4] = \frac{{30}}{{13}}$

    You might be surprised to learn that the finite continued fraction

    $3 + \frac{1}{{7 + \frac{1}{{15 + \frac{1}{{1 + \frac{1}{{292}}}}}}}}$and its reversal$292 + \frac{1}{{1 + \frac{1}{{15 + \frac{1}{{7 + \frac{1}{3}}}}}}}$

    have the same numerator. These fractions simplify to$\frac{{103993}}{{33102}}$and$\frac{{103993}}{{355}}$respectively. In this chapter, we provide a combinatorial interpretation for the numerators and denominators of continued fractions which makes this reversal phenomenon easy to see. Our interpretation also allows us to visualize many important identities involving continued fractions.

    First, we define some basic terminology. Given an infinite sequence of integers${a_0} \ge 0,{a_1} \ge 1,{a_2} \ge 1$, . . . let...

  8. CHAPTER 5 Binomial Identities
    (pp. 63-80)

    Definition Thebinomial coefficient$\left( \begin{array}{l} n \\ k \\ \end{array} \right)$is the number ofk-element subsets of {1, . . . , n}.

    Definition Themultichoose coefficient($\left( \begin{array}{l} n \\ k \\ \end{array} \right)$) is the number ofk-element subsets of {1, . . . , n}.

    Examples of binomial coefficients are$\left( {\begin{array}{*{20}{c}} 4 \\ 0 \\ \end{array}} \right) = 1$,$\left( {\begin{array}{*{20}{c}} 4 \\ 1 \\ \end{array}} \right) = 4$,$\left( {\begin{array}{*{20}{c}} 4 \\ 2 \\ \end{array}} \right) = 6$,$\left( {\begin{array}{*{20}{c}} 4 \\ 3 \\ \end{array}} \right) = 4$, and$\left( {\begin{array}{*{20}{c}} 4 \\ 4 \\ \end{array}} \right) = 1$.

    Examples of multichoose coefficients are ($\left( {\begin{array}{*{20}{c}} 4 \\ 0 \\ \end{array}} \right) = 1$,$\left( {\begin{array}{*{20}{c}} 4 \\ 1 \\ \end{array}} \right) = 4$,$\left( {\begin{array}{*{20}{c}} 4 \\ 2 \\ \end{array}} \right) = 6$,$\left( {\begin{array}{*{20}{c}} 4 \\ 3 \\ \end{array}} \right) = 4$, and$\left( {\begin{array}{*{20}{c}} 4 \\ 4 \\ \end{array}} \right) = 35$.

    Binomial coefficients were born to count! Unlike most of the quantities we have discussed in this book, binomial coefficients are almost always defined as the answer to a counting problem. Specifically, we define$\left( \begin{array}{l} n \\ k \\ \end{array} \right)$to be the number of...

  9. CHAPTER 6 Alternating Sign Binomial Identities
    (pp. 81-90)

    In the last chapter, we proved, for$n > 0$,$\sum\nolimits_{k > 0} {\left( {\begin{array}{*{20}{c}} n \\ {2k} \\ \end{array}} \right) = {2^{n - 1}}} $. Since$\sum\nolimits_{k \ge 0} {\left( {\begin{array}{*{20}{c}} n \\ {2k} \\ \end{array}} \right) = {2^n}} $, this implies that half of all subsets of$\left\{ {1, \ldots ,n} \right\}$are even. Consequently

    $\sum\limits_{k \ge 0} {\left( {\begin{array}{*{20}{c}} n \\ {2k} \\ \end{array}} \right)} = \sum\limits_{k \ge 0} {\left( {\begin{array}{*{20}{c}} n \\ {2k + 1} \\ \end{array}} \right)} $.

    This suggests that there should be a simple one-to-one correspondence between the even subsets of$\left\{ {1,2, \ldots ,n} \right\}$and the odd ones. We begin with a bijective proof of this fact.

    Identity 167 For$n > 0$,

    $\sum\limits_{k = 0}^n {\left( {\begin{array}{*{20}{c}} n \\ k \\ \end{array}} \right)} {( - 1)^k} = 0$.

    Set 1: Let$\varepsilon $denote the set of even subsets$\left\{ {{a_1}, \ldots ,{a_k}} \right\}$of$\left\{ {1, \ldots ,n} \right\}$wherekis an even number. This set has size${\sum _k}$even$\left( \begin{array}{l} n \\ k \\ \end{array} \right)$.

    Set 2: Let$o$denote the set of odd subsets$\left\{ {{a_1}, \ldots ,{a_k}} \right\}$of$\left\{ {1, \ldots ,n} \right\}$wherekis an odd number. This set has size${\sum _k}$even$\left( \begin{array}{l} n \\ k \\ \end{array} \right)$.

    Correspondence: For...

  10. CHAPTER 7 Harmonic and Stirling Number Identities
    (pp. 91-108)

    Definition Thenth harmonic numberis${H_n} = 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}$. The first few harmonic numbers are${H_1} = 1$,${H_2} = \frac{3}{2}$,${H_3} = \frac{11}{6}$,${H_4} = \frac{25}{12}$.

    Definition TheStirling number of the first kind$\left[ {\begin{array}{*{20}{c}} n \\ k \\ \end{array}} \right]$counts the number of permutations ofnelements with exactlykcycles. Some examples when$k = 2:\left[ {\begin{array}{*{20}{c}} 4 \\ 0 \\ \end{array}} \right] = 0$,$\left[ {\begin{array}{*{20}{c}} 2 \\ 2 \\ \end{array}} \right] = 1$,$\left[ {\begin{array}{*{20}{c}} 3 \\ 2 \\ \end{array}} \right] = 3$,$\left[ {\begin{array}{*{20}{c}} 4 \\ 2 \\ \end{array}} \right] = 11$,$\left[ {\begin{array}{*{20}{c}} 5 \\ 2 \\ \end{array}} \right] = 50$.

    Definition TheStirling number of the second kind$\left[ {\begin{array}{*{20}{c}} n \\ k \\ \end{array}} \right]$counts the number of partitions ofnelements with exactlyksubsets. Some examples when$k = 2:\left[ {\begin{array}{*{20}{c}} 4 \\ 0 \\ \end{array}} \right] = 0$,$\left[ {\begin{array}{*{20}{c}} 2 \\ 2 \\ \end{array}} \right] = 1$,$\left[ {\begin{array}{*{20}{c}} 3 \\ 2 \\ \end{array}} \right] = 3$,$\left[ {\begin{array}{*{20}{c}} 4 \\ 2 \\ \end{array}} \right] = 7$,$\left[ {\begin{array}{*{20}{c}} 5 \\ 2 \\ \end{array}} \right] = 15$.

    Harmonic numbers are defined to be partial sums of the harmonic series. That is, for...

  11. CHAPTER 8 Number Theory
    (pp. 109-124)

    In this chapter, we have collected identities from arithmetic, algebra and number theory.

    What could be simpler than the sum of the firstnnumbers? You probably already know this first identity. In fact, it’s a special case of Identity 135 in Chapter 5. You may recall that

    $\sum\limits_{k = 1}^n k = \frac{{n(n + 1)}}{2}$.

    Combinatorially, this can be rewritten and explained in two different ways. The subsequent identities and their counting proofs hinge on the interpretation of$\frac{{n(n + 1)}}{2}$as$\left( {\frac{{n + 1}}{2}} \right)$, a selection without repetition, or ($\left( {\frac{{n }}{2}} \right)$), a selection with repetition.

    Identity 211 For$n \ge 0$,

    $\sum\limits_{k = 1}^n k = \left( {\frac{{n + 1}}{2}} \right)$.

    Question: How many ways can two different numbers be selected from the set {0, 1, . . . , n}?...

  12. CHAPTER 9 Advanced Fibonacci & Lucas Identities
    (pp. 125-146)

    We end this book as we began it, by exploring more Fibonacci and Lucas identities (Lucas might say that we’ve come “full circle”!) We include some of the proofs that we found particularly challenging. As a climax, we add a dose of probability to obtain combinatorial proofs of theBinet formulas

    ${F_n} = \frac{1}{{\sqrt 5 }}\left[ {{{\left( {\frac{{1 + \sqrt 5 }}{2}} \right)}^n} - {{\left( {\frac{{1 - \sqrt 5 }}{2}} \right)}^n}} \right]$

    and

    ${L_n} = {\left( {\frac{{1 + \sqrt 5 }}{2}} \right)^n} + {\left( {\frac{{1 - \sqrt 5 }}{2}} \right)^n}$

    We conclude with some known identities that, as far as we know, have not yet succumbed to combinatorial interpretation.

    Recall by Combinatorial Theorem 1 that fnis the number of square and domino tilings of a lengthnboard. The first identity is a warm-up to...

  13. Some Hints and Solutions for Chapter Exercises
    (pp. 147-170)
  14. Appendix of Combinatorial Theorems
    (pp. 171-172)
  15. Appendix of Identities
    (pp. 173-186)
  16. Bibliography
    (pp. 187-190)
  17. Index
    (pp. 191-193)
  18. About the Authors
    (pp. 194-194)