Mathematical Problem Solving

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Solving

Darij Grinberg

^*

December 25, 2021 (unfinished!)

Preface

This is the text accompanying my Math 235 (Mathematical Problem Solving) class at Drexel University in Fall 2020. The website of this class can be found at

http://www.cip.ifi.lmu.de/~grinberg/t/20f .

This document is a work in progress. It might become a textbook one day, but for now is a construction zone.

Please report any errors you find todarijgrinberg@gmail.com.

What is this?

This course is an introduction to the art of mathematical problem solving. This means solving the type of problems that usually are posed in competitions (like the Putnam Competition or the International Mathematical Olympiad) or in jour- nals (like the American Mathematical Monthly or Crux Mathematicorum): self- contained mathematical questions that do not require any specialized knowledge (beyond the basic undergraduate curriculum) to understand, but often a significant amount of ingenuity, creativity and effort to solve, although the resulting solutions again are readable without specialized knowledge.

There is no panacea for solving such problems; the hardest ones constitute serious challenges for even the most well-trained solvers. Yet there is a variety of ideas, techniques and tools that can help attacking these problems; while they offer no

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guarantee of success, they are sufficiently useful that they nowadays get systemat- ically taught to olympiad participants in many countries and universities. Most of these techniques are also helpful in mathematical research – even though the typical research question poses a rather different kind of challenge than a competition problem.¹

Some of these techniques are very basic. Mathematical induction, for example, is fundamental to all of mathematics. The Pigeonhole Principle says (roughly speak- ing) that if you put 6 or more pigeons into 5 pigeonholes, then at least two pigeons will be roommates. The Extremal Principle says that (at least) one of the pigeons has the longest beak. These principles are neither surprising nor deep; but they can often be applied in unexpected and surprising ways. We will study them mainly by exploring their applications, as there is little to say about these principles them- selves.

Another part of the problem solving toolkit is a selection of “little” theories (“little” in the sense of not going deep; typically, the main results and their proofs fit on 20 pages at most). The most well-known such theory is elementary number theory, due to a large extent to Euclid (or to whoever discovered the results in his Elements) and expanded somewhat by Fermat, Euler and Gauss. The lore of polynomials (commonly over the rational or complex numbers) can be regarded as another such theory, with results like Viete’s theorems. The art of elementary inequalities (like Cauchy–Schwarz, Hölder, rearrangement), too, can be counted as a theory.

The tools and theories we will see are thus a cross-section of elementary mathematics. This text will omit most of analysis (which will only appear in some exercises; see [GelAnd17] for a lot more) and all of elementary geometry (which is its own subject, worthy of its own books; see [Honsbe12] for a selection of neat results and [Hadama09] for a systematic introduction). We will also not use abstract algebra beyond a few basic ideas that can be explained from scratch. As to linear algebra, we will occasionally make use of it, but we will not introduce it, as there are enough good texts for that.

This is not the first introductory text on mathematical problem solving, and has no ambitions to be the last. The classics in this genre are Polya’s [Polya81]

and Engel’s [Engel98]²; many other sources have been published since, including

1For starters, a researcher attacking a research question does not know whether it has an answer and, if it has one, how difficult it will be to find; meanwhile, questions posed in competitions have been “pre-solved” by their proposers and/or the selection committees, and usually have a somewhat predictable level of difficulty. Also, one of the frequent challenges in research is asking the right question; but competitions preempt this challenge (although it tends to resurface in the solving process, as one looks for useful auxiliary results).

However, competition problems can (and often do) have a lot in common with research questions; in particular, several have crystallized out of the research of their proposers. For the solver, mathematical competitions can provide inspiration and training for proper research; when de- signed well, they are research “writ small”.

2The even earlier book [Polya73] by Polya is, in some ways, a harbinger of this kind of book, although in itself it is too short (just a few dozen solved problems), too philosophical (replete

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[GelAnd17], [Zeitz17], [ZawHit09] and [Chen20, OTIS Excerpts]³. There is also a plenitude of problem books – i.e., collections of (solved) problems either taken from a specific contest (or journal) over the years (such as [DJMP11] or [Strasz65], or the

“Hungarian Problem Books” published by the MAA), or chosen from specific subjects (the classics in the latter genre are [DynUsp06], [ShChYa62], [YagYag64] and [YagYag67]). The present text intends to differ from these both in the choice of subjects and in the choice of examples and problems (based on my personal taste);

that said, there is likely to be more overlap than difference, since both the most im- portant techniques and their most prominent examples are generally agreed upon.

Prerequisites

I expect you, the reader, to be familiar with mathematical proofs. In particular, you should be able to

• apply (and verify) arguments by mathematical induction, by contradiction, by case distinction and by similar fundamental tactics,

• write up your proofs at a reasonable level of mathematical rigor, and read similarly rigorous proofs written by others (as frequently seen in textbooks written for mathematics majors),

• understand and use basic mathematical language such as the summation sign⁴ or set builder notation⁵.

If not, there is a whole industry of texts from which these skills can be learned (such as [LeLeMe16, Chapters 1–5], [Day16], [Hammac15], [Newste20, Part I and

with didactical comments and Socratic dialogues) and too basic (typical strategies suggested are

“do you know a related problem?” and “introduce suitable notations”). Even in the much more content-rich and concrete [Polya81], Polya spends quite a lot of space attempting to construct a general theory of mathematical problem solving, with input from philosophy and psychol- ogy (two chapters are called “Rules of Discovery?” and “On Learning, Teaching, and Learning Teaching”, although they are more hands-on than the names may suggest).

Engel’s main innovation in [Engel98] was to scrap the philosophy and fill as much of his book as possible with actual problems and solutions. The problem solving strategies he presents are mathematical ones, and they are taught by doing (worked examples, then problems). The occasional personal story or heuristical remark gives some variety (but the mathematics itself is already pretty varied, as one and the same strategy often has applications in very different fields).

In the notes you are now reading, I intend to stick to Engel’s paradigm (but aim at a higher level of detail).

3also, various books in foreign languages: e.g., [Grinbe08] and [Carl17] (in German) and [Jarvin20]

(in Finnish) and [DLPS16] (in Dutch)

4for example, ∑^b

k=a

1 k = ¹

a+ ¹

a+1+· · ·+¹ b

5for example,{i∈_Z|iis even}for the set of all even integers

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Appendices A–B] and [Swanso20, Chapters 1–3], just to mention a few freely avail- able ones⁶).

Acknowledgments

I thank Natalia Grinberg, Sam Hopkins, Pasha Pylyavskyy, Tom Roby and David Speyer for helpful conversations on the problem solving literature.

6I recommend the MIT text [LeLeMe16] in particular for its lively writing and its enjoyable exercises. It is long and goes far beyond the basics; it has significant overlap with what I will be doing in this class.

Newstead’s [Newste20] (work in progress) is also far more than an introduction to proofs. It covers a lot of the “no man’s land” between such an introduction and advanced courses.

Swanson’s [Swanso20] is focussed on setting up the prerequisites for a rigorous treatment of analysis.

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1. Introduction

1.1. Homework set #0: Diagnostic

This is a special problem set: Its main purpose is to give you an idea of what is to come in this course, and to give me an idea of your level of familiarity with certain things (including proof writing). Do not expect to solve all these problems.

This problem set (and only this one!) will be graded on completion – each problem will be graded out of 2; any serious work on a problem (e.g., a partial solution, a particular case solved, or even an attempt that shows you have thought about it) will get 2 points. (All future problem sets will be gradedon correctness, with a correct and well-written solution worth 10 points.)

Tips:

• Ways to make partial progress can include

– trying out small values of n (in problems that depend on an integer n – even if it is not explicitly mentioned⁷),

– replacing some numbers by smaller numbers (to make the problem eas- ier to check on examples),

– ruling some cases out, or

– rewriting some of the expressions involved in the problem.

• You don’t need to be too detailed in your solutions (nor will I always be in mine). If you solve (say) Exercise 1.1.2 by induction onn, I trust you that you can do the base case without you going into the details. Likewise, you don’t need to show any intermediate steps when claiming algebraic identities like

a²+b²+c²

−(bc+ca+ab) = ¹ 2

(b−c)²+ (c−a)²+ (a−b)²; it is clear enough that the reader can just expand both sides and compare. (Thus, your solution to Exercise 1.1.7 can consist of a single equation.) Details are needed when a reader (who has been following the course, but doesn’t already know the solution of the specific problem) might realistically stumble or get stuck.

• I will proofread your solutions, so this is a good chance to get feedback on your writing before we get to homework worth more serious points.

Exercise 1.1.1. Let n be an even positive integer. Find a q ∈ {1, 2, . . . , 2n} such that 1!·2!· · · · ·(2n)!

q! is a perfect square.

7E.g., a problem about a polynomial always has an “implicitn”, namely the degree of the polynomial involved. Thus one can start by considering polynomials of degrees 1, 2, 3.

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Exercise 1.1.2. Define a sequence (t0,t₁,t2, . . .) of positive rational numbers recursively by setting

t₀=1, t₁ =1, t₂=1, and tn = ¹+t_n−1t_n−2

tn−3 for eachn ≥_3.

(For example,t3= ¹+t2t1

t₀ = ¹+1·1

1 =2 andt₄ = ¹+t3t2

t₁ = ¹+2·1 1 =3.) (a)Prove that tn+2 =4tn−tn−2 for eachn ≥2.

(b)Prove that tn is a positive integer for each integern ≥0.

The next exercise uses thefloor function, which is defined as follows:

Definition 1.1.1. For every real x, we let bxc denote the largest integer that is

≤x. This integer bxc is called thefloor(or theinteger part) of x.

For example, b3c = b_πc = 3 and b−_πc = −4. It is easy to see⁸ that for every x ∈ _{R, we have}

bxc ≤ x<bxc+1. (1) (Note that the chain of inequalities (1) determines bxc uniquely, and thus can be used as an alternative definition of bxc. This is how bxc is defined in [Grinbe16, Definition 1.1.1], for example.) Older sources use the notation [x] for bxc; this notation, however, can mean many other things.

Exercise 1.1.3. Let x ∈_Rand let nbe a positive integer. Prove that

n−1 k

∑

=₀

x+ ^k

n

=bnxc.

Exercise 1.1.4. Let a,b,c,n be positive integers such that a | bⁿ and b | cⁿ and c| aⁿ. Prove that abc |(a+b+c)ⁿ²⁺ⁿ⁺¹.

Exercise 1.1.5. A mountain ridge has the form of a (finite) line segment, bordered on each end by a cliff. Several (finitely many) lemmings are walking along the ridge, with equal speeds (but not necessarily in the same direction). Whenever two lemmings meet, they “bounce off” one another, keeping their respective speeds but reversing their directions. Whenever a lemming arrives at an end- point of the ridge, it falls off the cliff. Prove that eventually, all lemmings will fall off the cliff.

8and well-known enough to be used without proof on any contest

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[Example: Here is a possible lemming configuration:

−→ 1 ←−

2 ←−

3 −→

4 ←−

5

(with 1, 2, 3, 4, 5 being the lemmings, and the arrows signifying their walking directions). The first two lemmings to meet here will be 1 and 2, after which they both change their directions:

←− 1 −→

2 ←−

3 −→

4 ←−

5

Now lemming 1 is on its way to the cliff, which it will reach without interference from other lemmings.]

Exercise 1.1.6. You have 20 white socks and 20 black socks hanging on a clothesline, in some order.

(a)Prove that you can pick 10 consecutive socks from the clothesline such that 5 of them are black and the other 5 white. (You can call such a pickcolor-balanced.) (b) Prove the same if the number 20 is replaced by 13 (so you have 13 white and 13 black socks).

[Example: If the number 20 is replaced by 7, then the claim does not hold.

For example, the configuration B³W⁷B⁴ (standing for “3 black socks, followed by 7 white socks, followed by 4 white socks”) has no color-balanced pick of 10 consecutive socks.]

Exercise 1.1.7. Factor the polynomial

bc(b−c) (b+c) +ca(c−a) (c+a) +ab(a−b) (a+b) into a product of four linear (i.e., degree-1) polynomials.

Exercise 1.1.8. (a)Let a,b,c be three nonnegative reals. Prove that

|ca−ab|+|ab−bc|+|bc−ca| ≤b²−c² +

c²−a² +

a²−b² . (b)Is this still true if the word “nonnegative” is omitted?

Exercise 1.1.9. Let n ≥ 1. Let a₁,a₂, . . . ,an be any n integers. Prove that there exist some p,q ∈ {1, 2, . . . ,n} with p≤q and n| ap+ap+1+· · ·+aq.

Exercise 1.1.10. Briefly review the problems above: Which ones did you like?

Which ones did you not like? Why? How long did they take you? Which parts did you get stuck on? Did you learn anything from solving (or trying to solve) them? If you knew the solution already (nothing wrong with that!), where did you learn it? (No need to rate every exercise; just say some words about some 4 of them.)

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2. Induction I

Mathematical induction is essentially the main strategy for proving theorems in mathematics: Any statement that involves a natural number (even implicitly, such as by talking of a finite set) potentially lends itself to be proved by induction. This does not mean induction is always the best way to go, or even a viable one, but it is a method you should certainly not be surprised to see working. Only recently, two long books [AndCri17] and [Gunder10] (with 400 and 850 pages, respectively) entirely devoted to applications of induction have appeared⁹. We will not go into this depth or detail; but we will show examples for the most frequent ways in which induction can be applied.

2.1. Standard induction

We assume you are familiar with standard mathematical induction. Just as a re- minder, it is a way to prove that a certain statement A(n) holds for every integer n≥ g, where g is a fixed integer. Stated as a theorem itself, here is what it says:

Theorem 2.1.1. Let g ∈ Z. For each integer n ≥ g, let A(n) be a logical statement. Assume the following:

• Assumption 1: The statementA(g) holds.

• Assumption 2: Ifmis an integer such thatm≥gand such thatA(m)holds, then A(m+1) also holds.

Then, A(n) holds for all integersn ≥g.

Theorem 2.1.1 is one of the facts known as the principle of mathematical induction.

Some consider it to be an axiom; others use essentially equivalent axioms that it can easily be derived from.¹⁰ We will not dwell on its logical status, but rather explore its use.

Theorem 2.1.1 provides a way to prove any result that can be stated in the form

“some statement A(n) holds for each integer n ≥ g” (commonly, g is taken to be 0 or 1, but any integer is fine). All one needs to do is to prove Assumption 1 and Assumption 2. The proof of Assumption 1 is commonly called the induction base (or base case), while the proof of Assumption 2 is commonly called the induction step. In the induction step, the assumption thatA(m) holds is called the induction hypothesis (or induction assumption). The whole proof is called an “induction on n”

(or “induction over n”). Here is a simple example:

9This is not to say that the method has not been appreciated before; it is just that the idea of gathering mathematical results by their method of proof (as opposed to their objects of concern) is fairly new.

10You may have seen it stated forg =0 only (or forg =1 only). But the general case of Theorem 2.1.1 easily follows from these particular cases (see, e.g., [Grinbe15, proof of Theorem 2.53]).

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Exercise 2.1.1. Let b be a real number. Prove that (b−1)b⁰+b¹+· · ·+bⁿ⁻¹

=bⁿ−1 (2) for each integer n≥0. ¹¹

Solution to Exercise 2.1.1. We will prove (2) by induction onn(that is, we shall apply Theorem 2.1.1 to g=0 and to A(n) being the statement (2)).

Induction base: We have(b−1)b⁰+b¹+· · ·+b⁰⁻¹

| {z }

=(empty sum)=0

=0 =b⁰−1 (sinceb⁰ =1).

In other words, (2) holds forn =0. This completes the induction base.

Induction step: Let m ≥ 0 be an integer. Assume (as the induction hypothesis) that (2) holds for n = m. We must prove that (2) holds for n = m+1. In other words, we must prove that(b−1)b⁰+b¹+· · ·+b⁽^m⁺¹⁾⁻¹

=b^m⁺¹−1.

Our induction hypothesis says that (2) holds forn=m. In other words, (b−1)b⁰+b¹+· · ·+b^m⁻¹

=b^m−1. (3) Now,(m+1)−1=m, so that

(b−₁)b⁰+b¹+· · ·+b⁽^m⁺¹⁾⁻¹

= (b−₁)b⁰+b¹+· · ·+b^m

| {z }

=(^b⁰⁺^b¹^+···+^b^m−1)⁺^b^m

= (b−1)b⁰+b¹+· · ·+b^m⁻¹

+b^m

= (b−1)b⁰+b¹+· · ·+b^m⁻¹

| {z }

=_b^m−1 (by (3))

+ (b−1)b^m

| {z }

=bb^m−b^m

=b^m−1+bb^m−b^m = bb^m

|{z}

=b^m+1

−1=b^m⁺¹−1.

But this is precisely what we need to prove. Thus, the induction step is complete.

Hence, (2) is proved. Thus, Exercise 2.1.1 is solved.

We note that if bis a real number distinct from 1, then the claim of (2) entails b⁰+b¹+· · ·+bⁿ⁻¹ = ^b

n −1

b−1 = ¹−bⁿ

1−b (4)

for each integer n ≥ 0. This is the classical formula for the sum of a (finite) geo- metric progression. There are other proofs of this formula (in particular, we shall

11Note that if n = 0, then the sum b⁰+b¹+· · ·+bⁿ⁻¹ is an empty sum and thus equals 0 by definition.

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prove something slightly more general in a later chapter), but the induction proof that we gave above has the advantage of being entirely straightforward.

A particularly simple particular case of Exercise 2.1.1 is the following: Each integern≥0 satisfies

2⁰+2¹+· · ·+2ⁿ⁻¹=2ⁿ−1. (5) Indeed, this follows by applying Exercise 2.1.1 tob =2 (since 2−1=1).

Here is another basic example for a proof by induction (which appears, e.g., in [Grinbe15, Proposition 2.164], in [Gunder10, Exercise 74] and in [AndCri17, Prob- lem 2.2]):

Exercise 2.1.2. Prove that every integer n≥0 satisfies 1

1−¹ 2 +¹

3−¹

4 +· · ·+ ¹

2n−1− ¹

2n = ¹

n+1+ ¹

n+2 +· · ·+ ¹

2n. (6) Solution to Exercise 2.1.2. Let us denote the left hand side of (6) by L(n), and let us denote the right hand side by R(n). That is, we set

L(n) = ¹ 1 −¹

2+¹ 3 −¹

4+· · ·+ ¹

2n−₁ − ¹

2n and

R(n) = ¹

n+1+ ¹

n+2+· · ·+ ¹ 2n

for each n ≥ 0. Thus, we need to prove that L(n) = R(n) for each n ≥ 0. We shall do this by induction onn (that is, we shall apply Theorem 2.1.1 to g =0 and A(n) = (“L(n) = R(n)”)).

Induction base: The equality L(0) = R(0) holds, since both L(0) and R(0) are empty sums (and thus 0). In other words, L(n) = R(n) holds for n = 0. This completes the induction base.

Induction step: Let m ≥ 0 be an integer. Assume (as the induction hypothesis) that L(m) = R(m). We must prove that L(m+1) = R(m+1).

The definition ofL(n)shows that L(m+1)is the same sum as L(m)except with the two extra addends 1

2(m+1)−1 and − ¹

2(m+1) added to it. Thus, L(m+1) = L(m) + ¹

2(m+1)−1 − ¹

2(m+1)^. ⁽⁷⁾

The definition ofR(n)shows thatR(m+1)is the same sum as R(m) except without the addend 1

m+1 but with the two new addends 1

2m+1 and 1

2m+2 added to it. Thus,

R(m+1) = R(m)− ¹

m+1 + ¹

2m+1 + ¹

2m+2. (8)

Now it is clear what to do: We need to show that the left hand sides of the two equalities (7) and (8) are equal, but we know that the L(m) and R(m) terms on

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their right hand sides are equal (since L(m) = R(m) by assumption). So we need to show that the remaining terms on the right hand sides are also equal – i.e., that we have

1

2(m+1)−1− ¹

2(m+1) =− ¹

m+1+ ¹

2m+1 + ¹ 2m+2.

But this can be checked by a straightforward computation (which is simplified by the facts that 2(m+1)−1 =2m+1 and 2m+2=2(m+1)). Thus, the right hand sides of (7) and (8) are equal (since L(m) = R(m)), and therefore so are the left hand sides. In other words, L(m+1) = R(m+1). This completes the induction step. Exercise 2.1.2 is thus solved.

Equalities between finite sums (such as in Exercise 2.1.2) provide lots of exercise in using induction; for example, you can use induction to prove the equalities

1+2+· · ·+n= ⁿ(n+1)

2 ; (9)

1−2+3−4± · · ·+ (−1)ⁿ(n+1) =

(n/2+1, if nis even;

−(n+1)/2, if nis odd;

1²+2²+· · ·+n²= ⁿ(n+1) (2n+1)

6 ;

1³+2³+· · ·+n³=

n(n+1) 2

2

;

1⁴+2⁴+· · ·+n⁴= ⁿ(2n+₁) (n+₁) 3n+3n²−₁ 30

for each integer n ≥ 0 (see [Grinbe15, §2.4] and [Grinbe15, Exercise 2.9], respectively, for detailed proofs of the first two equalities).

We shall now come to a more combinatorial application. Recall that a bit is defined to be an element of the 2-element set {0, 1}.

Exercise 2.1.3. Fix a positive integer n. An n-bitstring shall mean an n-tuple (a1,a2, . . . ,an) ∈ {0, 1}ⁿ of bits. Prove that there exists a list(b1,b2, . . . ,b2ⁿ) containing all n-bitstrings (each exactly once) such that for every i ∈ {1, 2, . . . , 2ⁿ}, the two n-bitstrings b_i and b_i−1 differ in exactly one entry. Here, we understand b0to mean b2ⁿ.

[Example: Forn =3, one such list is(b₁,b₂, . . . ,b₈), where

b1= (0, 0, 0), b2 = (0, 0, 1), b3 = (0, 1, 1), b4 = (0, 1, 0), b5= (1, 1, 0), b6 = (1, 1, 1), b7 = (1, 0, 1), b8 = (1, 0, 0). It is far from the only such list.]

Before we solve this exercise, a few comments on its significance are in order.

A list (b₁,b₂, . . . ,b₂ⁿ) satisfying the requirement of Exercise 2.1.3 is known as a (circular) Gray code. Quoting the (extensive) Wikipedia article on these:

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“Gray codes are used in linear and rotary position encoders (absolute encoders and quadrature encoders) in preference to weighted binary encoding. This avoids the possibility that, when multiple bits change in the binary representation of a position, a misread will result from some of the bits changing before others.”

That is, the property thatb_i andb_i−1 differ in exactly one entry ensures thatb_i−1

can be turned into b_i by only changing a single bit, which removes the need for changing several bits simultaneously to prevent unintended intermediate states.

Another way to think of Exercise 2.1.3 is as follows: Imagine a combination lock with n dials, where each dial has exactly 2 disks labeled 0 and 1. Then, a Gray code provides a way to cycle through all 2ⁿ possible combinations in a way that getting from each combination to the other (and from the last back to the first) only requires rotating a single dial. Variants of Gray codes exist for dials with more than 2 disks, and for various other similar situations; a comprehensive discussion of various such contraptions can be found in [TAoCP4A, §7.2.1.1].¹²

Solution to Exercise 2.1.3 (sketched). We define ann-Gray codeto be a list(b₁,b₂, . . . ,b₂ⁿ) containing all n-bitstrings (each exactly once) such that for everyi ∈ {1, 2, . . . , 2ⁿ}, the two n-bitstrings bi and bi−1 differ in exactly one entry, where b0 means b2ⁿ. Thus, the exercise wants us to prove that there exists an n-Gray code.

We shall prove this by induction on n, so we forget that we fixed n. Note that we have to start our induction at n = 1 (that is, we must apply Theorem 2.1.1 to g = 1), since the exercise is stated for all positive integers n (rather than for all integers n≥0, as the two previous exercises were).

Induction base: There exists a 1-Gray code – namely, the list ((0),(1)). Thus, Exercise 2.1.3 is proved¹³ forn=1. This completes the induction base.

Induction step: Let m ≥ 1 be an integer. Assume (as the induction hypothesis) that there exists an m-Gray code. We must show that there exists an(m+1)-Gray code.

We introduce a notation: For anym-bitstringa = (a1,a2, . . . ,am), we let 0adenote the (m+1)-bitstring (0,a1,a2, . . . ,am), and we let 1a denote the (m+1)-bitstring (1,a₁,a₂, . . . ,am). Thus, 0a and 1a are the two possible (m+1)-bitstrings that can be obtained from a by inserting a new bit at the front. For example, ifa = (1, 1, 0) (so m=3), then 0a = (0, 1, 1, 0)and 1a = (1, 1, 1, 0).

We have assumed that there exists an m-Gray code; let (b₁,b₂, . . . ,b₂^m−1,b₂^m) be thism-Gray code. Then,

(0b1, 0b2, . . . , 0b2^m−1, 0b2^m, 1b2^m, 1b2^m−1, . . . , 1b2, 1b1)

12We note that Exercise 2.1.3 is also a popular contest problem. For instance, Problem A3 on the Putnam Mathematical Competition 1968 is a thinly-veiled restatement of Exercise 2.1.3 (more precisely, of a weaker version of this exercise, which requires bi and bi−1 differ in exactly one entry only fori∈ {1, 2, . . . , 2ⁿ−1}rather than for alli∈ {1, 2, . . . , 2ⁿ}).

13Here, we are being slightly sloppy with our language: When we say “Exercise 2.1.3”, we mean

“the claim of Exercise 2.1.3”. Thus, “Exercise 2.1.3 is proved” means “the claim of Exercise 2.1.3 is proved”. We shall use this sloppy language occasionally, as it saves us space.

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(note that the subscripts first go up from 1 to 2^m, then go back down from 2^m to 1) is an (m+1)-Gray code (check this!). Thus, there exists an (m+1)-Gray code.

This completes the induction step. Thus, Exercise 2.1.3 is solved by induction.

The verb “induct” is often used to mean “perform an induction”. Thus, for instance, in the above proof, we could have said “We induct on n” instead of “We shall prove this by induction on n”. A proof by induction is also known as an inductive proof.¹⁴

As mentioned above, when one inducts on n to prove a statement, the n needs not explicitly appear in the statement; it suffices that the statement can be restated in a way that contains an n. As an example, let us show the following (seemingly evident) fact:

Proposition 2.1.2. Let S be a nonempty finite set of integers. Then, S has a maximum.

Recall that a “maximum” (also known as a “largest element”) of a setSof integers¹⁵ is defined to be an elements ∈ S such that we haves ≥tfor all t∈ S. It is easy to see that any set of integers has at most one maximum (because ifs1 and s2are two maxima of the same set, thens₁≥s₂ ands₂ ≥s₁, whences₁ =s₂); but some sets of integers have none. (For example, the sets ∅ and Z have none.) Proposition 2.1.2 says that for a nonempty finite set, there always is a maximum (and thus, according to what we just said, a unique maximum).

Proof of Proposition 2.1.2 (sketched). (See [Grinbe15, first proof of Theorem 2.35] for details¹⁶.) Forget that we fixed S. We thus must prove the following:

Claim 1: Every nonempty finite setSof integers has a maximum.

We can rewrite this claim as follows:

Claim 2: Letnbe a positive integer. Then, everyn-element set of integers has a maximum.

Claim 2 is equivalent to Claim 1, because a setS is a nonempty and finite if and only if its size|S|is a positive integer. But Claim 2 has the advantage of depending on a positive integern, which gives us an opportunity to use induction. And that’s what we will do now:

[Proof of Claim 2: We induct onn:

14This is not to be confused with the epistemological concept of “inductive reasoning”, which is not a proof method.

15You can replace “integers” by “rational numbers” or “real numbers” here, and nothing will change.

16This said, [Grinbe15, first proof of Theorem 2.35] is organized somewhat differently (in particular, most of the argument is relegated into a lemma), and uses the nonnegative integer|S| −1 instead of the positive integer|S|to do the induction.

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Induction base:Every 1-element set of integers has a maximum (namely, its unique element). In other words, Claim 2 holds for n = 1. This completes the induction base.

Induction step: Let m be a positive integer. Assume (as the induction hypothesis) that Claim 2 holds for n = m. We must prove that Claim 2 holds for n = m+1.

In other words, we must prove that every (m+1)-element set of integers has a maximum.

So let S be an (m+1)-element set of integers. Then, |S| = m+1 ≥ 1 > 0, so that S is nonempty. Hence, there exists some s ∈ S. Consider such an s. The set S\ {s} then must be anm-element set (sinceSis an (m+1)-element set), and thus has a maximum (because our induction hypothesis says that every m-element set of integers has a maximum). Lett be this maximum.

Now, we can see that S has a maximum: Indeed, if t ≥ s, then t is a maximum ofS, whereas otherwise, sis a maximum ofS. (Check this!)

Forget that we fixed S. We thus have proved that every (m+1)-element set S of integers has a maximum. In other words, Claim 2 holds for n = m+1. This completes the induction step. Claim 2 is thus proved.]

Since Claim 2 is equivalent to Claim 1, we thus conclude that Claim 1 also holds, i.e., Proposition 2.1.2 is proved.

In advanced mathematics, most inductions are of the kind we just showed (although usually more complicated) – i.e., the n that is being inducted upon does not explicitly appear in the claim that is being proved, but rather is a “derived quantity” (like the size|S| of the setSin the above proof). Commonly, this is done quickly and tacitly – that is, instead of restating the claim in terms of n as we did above, one simply says that one is doing an induction on the derived quantity (i.e., in the above example, an induction on |S|). See [Grinbe15, §2.5.3] for the exact convention that is being used here; let me just show how the above proof could be rewritten using this convention:

Proof of Proposition 2.1.2 (short version). (See [Grinbe15, second proof of Theorem 2.35] for details.) Forget that we fixed S. Notice that |S| is a positive integer wheneverS is a nonempty finite set. Hence, we induct on|S|:

Induction base: If S is a nonempty finite set of integers satisfying |S| = 1, then S has a maximum (namely, its unique element). In other words, Proposition 2.1.2 holds for|S| =1. This completes the induction base.

Induction step: Let m be a positive integer. Assume (as the induction hypothesis) that Proposition 2.1.2 holds for|S| =m. We must prove that Proposition 2.1.2 holds for|S| =m+1. In other words, we must prove that every nonempty finite set Sof integers satisfying |S|=m+1 has a maximum.

So let S be a nonempty finite set of integers satisfying |S| = m+1. Then, |S| = m+1 ≥ 1 > 0, so that S is nonempty. Hence, there exists some s ∈ S. Consider such ans. The set S\ {s} then satisfies|S\ {s}|=|S| −1=m (since|S| =m+1), and thus is nonempty (since m is a positive integer); hence, this set S\ {_s} has a

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maximum (because our induction hypothesis says that every nonempty finite set of integers that has size mhas a maximum). Let tbe this maximum.

Now, we can see that S has a maximum: Indeed, if t ≥ _{s, then} _t is a maximum ofS, whereas otherwise, sis a maximum ofS. (Check this!)

Forget that we fixedS. We thus have proved that every nonempty finite setS of integers satisfying |S| = m+1 has a maximum. In other words, Proposition 2.1.2 holds for |S| = m+1. This completes the induction step. Proposition 2.1.2 is thus proved.

We shall see more complicated examples of induction on a derived quantity soon.

2.2. Fibonacci numbers I

The Fibonacci sequence, with its recursive definition and multiple properties, is a veritable induction playground. But it is also an object of serious research; a whole book about it has been written [Vorobi02], and the Fibonacci Association publishes theFibonacci Quarterlysince 1963 and organizes the two-yearlyFibonacci Conference.

Arguably these are only partly concerned with the Fibonacci sequence, but its role in them is substantial if not leading.

Let us recall the definition of this venerable sequence and show a few of its properties; we’ll see more of it at times later on.

Definition 2.2.1. The Fibonacci sequence is the sequence (f0, f1, f2, . . .) of integers which is defined recursively by

f0 =0, f₁ =1, and fn = fn−1+ fn−2 for all n≥2.

The entries of this sequence are called theFibonacci numbers.

The first Fibonacci numbers are

f0=0, f1 =1, f2 =1, f3=2, f4=3, f5 =5, f6=8, f7 =13, f8 =21, f9 =34, f₁₀ =55,

f₁₁ =89, f₁₂ =144, f₁₃ =233.

Some authors¹⁷ prefer to start the sequence at f1 rather than f0; of course, the recursive definition then needs to be modified to require f2 =1 instead of f0 =0.

The first property of Fibonacci numbers that we prove is the following:

Exercise 2.2.1. Let (f0, f1, f2, . . .)be the Fibonacci sequence. Prove that each integern ≥0 satisfies

f₁+f2+· · ·+ fn = fn+2−1. (10)

17such as Vorobiev in his book [Vorobi02]

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Solution to Exercise 2.2.1. We prove (10) by induction on n:

Induction base: We have f1+ f2+· · ·+ f0 = (empty sum) = 0 = f2−1 (since f2 =1). In other words, (10) holds for n=0. This completes the induction base.¹⁸

Induction step: Let m ≥ 0. Assume (as the induction hypothesis) that (10) holds forn =m. We must show that (10) holds forn =m+1.

Our assumption says that (10) holds for n = m; in other words, f1+ f2+· · ·+ fm = fm+₂−1.

Now, our goal is to prove that (10) holds forn=m+1; in other words, our goal is to prove that f1+ f2+· · ·+ fm+1 = fm+3−1 (since this is what (10) says for n=m+1). But this follows by comparing

f1+ f2+· · ·+ fm+1 = (f1+ f2+· · ·+ fm)

| {z }

=f_m+2−1

+fm+1 = fm+2−1+ fm+1

with

f_m+3

| {z }

=f_m+2+f_m+1 (by the recursive definition

of the Fibonacci sequence)

−1= f_m+2+f_m+1−1 = f_m+2−1+f_m+1.

So we have shown that (10) holds forn=m+1. This completes the induction step, and thus proves (10). Hence, Exercise 2.2.1 is solved.

Let us briefly discuss two notational aspects of induction. So far, we have always been using the letter m for the integer that is introduced in the induction step, because it was called min the statement of Theorem 2.1.1. Clearly, we can use any other letter (for example, k) instead, as long as that letter does not already have a different meaning. It is also perfectly fine to use k−1 for it – i.e., instead of assuming that the claim holds for n = m and proving that it holds for n = m+1, we can just as well assume that the claim holds for n = k−1 and prove that it holds forn =k. (Of course, if we do this, then we need to assumek >ginstead of k ≥ g.) This just boils down to substituting k−1 for m in Theorem 2.1.1. Here is how Theorem 2.1.1 looks like after this substitution:

Theorem 2.2.2. Let g ∈ Z. For each integer n ≥ g, let A(n) be a logical statement. Assume the following:

• Assumption 1: The statementA(g) holds.

• Assumption 2: If k is an integer such that k > g and such that A(k−1) holds, then A(k) also holds.

Then, A(n) holds for all integersn ≥_g.

18We will no longer write this sentence at the end of an induction base, since the “Induction step:”

that follows it should make it clear enough that the induction base ends here.

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As an example for an induction proof written in this way, let us prove another property of the Fibonacci sequence (the so-calledCassini identity):

Exercise 2.2.2. Let (f0, f₁, f2, . . .) be the Fibonacci sequence. Prove that fn+1fn−1− f_n²= (−1)ⁿ

for any positive integer n.

Solution to Exercise 2.2.2. We induct on n(using Theorem 2.2.2):

Induction base: We have f₀ =0, f₁ =1 and f₂ =1. Thus, f₂f₀−f₁² =1·0−1² =

−1= (−1)¹. In other words, Exercise 2.2.2 holds forn =1.

Induction step: Let k be an integer such that k > 1. Assume (as the induction hypothesis) that Exercise 2.2.2 holds for n = k−1. We must show that Exercise 2.2.2 holds forn =k. In other words, we must show that f_k+1f_k−1− f_k²= (−1)^k.

Our induction hypothesis says that Exercise 2.2.2 holds for n = k−1. In other words, it says that f_kf_k−2− f_k²₋₁ = (−1)^k⁻¹. But the recursive definition of the Fibonacci sequence yields f_k+1 = f_k+ f_k−1 (sincek >_{1). Hence,}

fk+₁

|{z}

=f_k+fk−1

fk−1 = (fk+ fk−1) fk−1 = fkfk−1+ f_k²₋₁

|{z}

=f_kf_k−2−(−1)^k−1 (since f_kfk−2−f_k−1² =(−1)^k−1)

= f_kf_k−1+ f_kf_k−2−(−1)^k⁻¹= f_k (f_k−1+f_k−2)

| {z }

=f_k

(since the recursive definition of the Fibonacci sequence

yieldsf_k=f_k−1+f_k−2)

−(−1)^k⁻¹

| {z }

=−(−1)^k

= f_kf_k−−(−1)^k= f_k²+ (−1)^k,

so that fk+1fk−1− _f_k² = (−₁)^k. This is exactly what we wanted to prove. This completes the induction step, and with it the solution to Exercise 2.2.2.

Some writers shorten their induction proofs even further by reusing the letter n itself (instead ofmor kas we did above) in the induction step. That is, instead of fixing ak > g and assuming that the claim holds for n= _k−1 and proving that it holds for n= _{k, they} fix ann> g and assume that the claim holds “forn−1 instead ofn” (i.e., the claim holds ifn is replaced byn−1 in it) and prove that it holds forn as well. This saves a letter and a bit of writing, at the cost of being potentially more confusing and slippery (as it requires you to check that you have properly replacednbyn−1 in the induction hypothesis); thus I do not recommend it. But you should be aware that it is a commonly used “figure of speech”.¹⁹

The next exercise (a particular case of [Grinbe15, Theorem 2.26(a)]) illustrates a point about choosing the right claim when doing induction.

19For example, here is how the above solution to Exercise 2.2.2 could be rewritten using this con-

Mathematical Problem Solving

Solving

Darij Grinberg

December 25, 2021 (unfinished!)

Contents

Preface

What is this?

Prerequisites

Acknowledgments

1. Introduction

1.1. Homework set #0: Diagnostic

∑

2. Induction I

2.1. Standard induction

2.2. Fibonacci numbers I