Proofs From the Book

Martin Aigner Giinter M. Ziegler

Proofs from THE BOOK Third Edition With 250 Figures Including Illustrations by Karl H. Hofmann

Springer

Preface

Paul Erd˝os

VIII

Table of Contents

Cornbinatorics

137

22 . Pigeon-hole and double counting . . . . . . . . . . . . . . . . . . . . . . . . . . . . .139 23 . Thre 63Lamousg

..........................

Six proofs of the infinity of primes

5

8

Bertrand's postulate

Leg ndre's The number

theorem

10

Bertrand's

12

Bertrand's postu40.036

Binomial coefficients are (almost) never powers

There is an epilogue to Bertrand's postulate which leads to a beautiful result on binomial coefficients.

Binomial coeflicients are (almost)never powers So for each i the number of multiples of pi among n, . . . , n-k+1, and hence among the aj9s,is bounded by 1sJ

15

<

16

Binomial coefficientsare (almost)never uowers Now since mi

<

18

Representing numbers as sums of two squares 0

x

Representing numbers as sums of two squares

20 Proof. We study the set

This set is 208sinite.This Indeed,oof.

24

Every finite division ring

EveryJinite

26

Every finite division ring is afield we find

k

By assumption, all ao, . . .

29

Some irrational numbers

and this is approximately (-I)"+'

E. More precisely: for 29

30

Some irrational numbers where for r

> 0 thTm(where )Tj01.6207 0 Td(denominator )Tj/T1_0 1 Tf0.

31


F ( x ) may also be written as an infinite sum F(x)

=

s2" f ( x ) - sZn-' f l ( x )

since the higher derivatives

+s

~ f U ~( x )7- . . .~,

n)

32


integral of a function that is positive (except on the boundary). However, < 1, then from part (ii) of the lemma we if we choose n so large that obtain

a contradiction. Here comes our final irrationality result.

Theorem 3. For every odd integer n 2 3, the number 1


is rational (with integers k , 4 > 0). Then

33

Three times 7r2/6

36 This evaluation

Three times n 2 16

37

This proof extracted the value of Euler's series from an integral via a rather simple coordinate transformation. An ingenious proof of this type - with an entirely non-trivial coordinate transformation - was later discovered by Beukers, Calabi and Kolk. The point of departure for that proof is to split $ int4(p18(the )11(even )-4(terms )3(and )6(the )]TJ0.03951 Tc 11.7473 0 Td(odd )Tj0.0132 Tc 1.8695 the sum

38

Three times .ir2 16 Beautiful

Form = l , 2 , 3 this yields cot2 =

5 ; =2 cot2 ;+ cot2 cot2 $ + cot2 + cot2

=5

-

Three times 7r2 16

Three times r216

40

This is very easy with the technique of "comparing with an integral" that we have reviewed already in the appendix to Chapter 2 (page 10). It yields

for an upper bound and

for a lower bound on the "remaining summands"

if you are willing to do a slightly more

-

or even

Three times n2/6

4

42

Three times n216

References [I] K . BALL& T. RIVOAL:

Hilbert 's third problem:

Hilbert's third problem: decomposinn polyhedra

48

Proof. The

Lines in the plane and decompositions of graphs

Perhaps the best-known problem on configurations of lines was raised by Sylvester in 1893 in a column of mathematical problems. QUESTION8 FOR SOLUTION.

11851. (Professor SYLFEBTEH.)-Provethat it is not poaaible to

a m n g c any finite number of real point8 EO that a right line. through every two of them shall pas8 through third, unless they

54

Lines in the plane, and decompositions of graphs

and x' are contained in precisely one set

55

56

A graph G with 7 vertices and 11 edges. It has one loop, one double edge and one triple edge.

The complete graphs K , on n vertices and (); edges

The

Lines in the plane, and decompositions o f graphs

Lines in the plane, and decompositions o f graphs

57

u'

The paths

The slope problem

Try for yourself -

Chapter 10

The slope uroblem

61

62

The slope problem

is crossing of order d

A touching move

An ordinary move

Three applications of Euler's formula

Chapter 11

.4 graph isplonar if it can be drawn in th515956n7llane

gW2

66

Three applications of Euler's formula Euler's formula thus produces a strong numerical conclusion from a geometric-topological the

The five platonic solids

Here the degree is written next to each vertex. Counting the vertices of given 1, degree yields n2 = 3, n s = 0, n4 n s = 2.

The number of sides is written into each region. Counting the faces with a given 3, number of sides yields fl = 1, f2

andf,=0otherwis.

1,f9=1,

f4=

Three a~nlicationso f Euler's formula

Let us deduce from this - together with Euler's formula- quickly that the complete graph K g and the complete bipartite graph K3,3are not planar. 5 For a hypothetical plane drawing of K5 we calculate

67

Sylveste475

70

Three applications of Euler's formula

Lattice bases A basis

72

Cauchy's rigidity theorem Now we mark by

+ the angles of Q for which the corresponding angle

73

Cauchv's ripidin theorem

>

Now let n 4. If for any i E ( 2 , . . . , n corresponding vertex can be cut off by c

-

1) we have ai = a : , then the u t

Touching simplices

Chapter 13

How many d-dimensional simplices can be positioned in Rdso that they touch pairwise, that is, so that thatpaim(so )TjETEMC BT/T1_1 1 Tf0.0128 Tc 9.75 0 0 10.25 281.27969 476

t1

Touching simplices

Proof. For d = 2 the family of four triangles that we had considered does have such a transversal line. Now consider any d-dimensional configuration of touching simplices that has a transversal line k'. Any nearby parallel line e' is a transversal line as well. and

Touching simplices In contrast to this exponential lower bound, tight upper bounds are harder to get. A naive inductive argument (considering all the facet hyperplanes in

77

78

The first row of the C-matrix represents the shaded triangle, while the second row corresponds to an empty intersection of

Touchina simulices

80 Every large point set has

Every large point set has an obtuse angle

This correspondence is illustrated by the sketch in the margin. Furthermore, from

81

83

Evew laree voint set has an obtuse anale We have sete aboe

thata

n

l

l

84 84

Borsuk's conjecture

Karol Borsuk's paper "Three theorems on the n-dimensional euclidean sphere" from on

Borsuk's conjecture

87

(3) From R, we obtain the set of points in R(;) whose coordinates are the subdiagonal entries of the corresponding matrices:

S

:= { ( z x ~ ) ~ , ~

Borsuk's coniecture

Claim 4. There is

89

90

Borsuk's conjecture To obtain a general bound for large d, we use monotonicity and unimodality of the binomial coefficients and the estimates n! > e(:)" and n! < en(:)" (see the appendix to Chapter 2) and derive

95

Sets, functions, and the continuum hypothesis

How do we obtain this sequence, and hence the Calkin-Wilf listing of the positive fractions? Consider the infinite binary tree in the margin. We immediately note its recursive rule: is on top of the tree, and every node 4 has two sons: the left son is 3

& and the right son is y. 1 -


96 To

Sets, jitnctions, and the continuum hypothesis

So how does one get from one rational to the next? To answer this, we first

97

98

Sets. functions. and the continuum hv~othesis

Let us move on to the real numbers R. Are they still countable? No,44(98(th

Sets, functions, and the continuum hypothesis Next we find that any two intervals (of finite length > 0) have equal size by considering the central projection as in the figure. Even more is true: Every interval (of length

99

100


Now we are faced with a basic problem. We would certainly like to have that the usual laws concerning inequalities also hold for cardinal numbers. n, But is this true for infinite cardinals? In particular, is it true that m n 5 m imply m = n? This

<


(Case 2), it may stop in an element of A l that does not lie

101

102


As a corollary to N o 5 m for any infinite cardinal m, we can immediately prove "Hilbert's hotel" for any infinite cardinal number m, that is, we have With this we have also proved a result

1M

U

Sets, functions, and the continuum hypothesis Very shortly afterwards Erdos showed that, surprisingly, the answer depends on the continuum hypothesis.

Theorem 5. Ifc > N1, then evVery

103

104

"A legend talks about St. Augustin who, walking along the seashore and contemplating injinitj, saw a child trying to empty the


Sets, ,functions, and the continuum hypothesis

-

p :N subset U

P ( A f ) is a bijection of N

C N of all

bl onto P ( M ) . Consider the

105

106

Sets, functions, and the continuum hypothesis number a. Indeed, if the infinite well-ordered set M has ordinal 6 ber

set element


What we finally need is a considerable strenghthesning

107

In praise of inequalities

Analysis abounds with inequalities, as witnessed for example by the famous book "Inequalities" by Hardy, Littlewood and

110

In praise of ine~ualities

For n = 2, we have ala2 5 (%jQ)2

(a1 - ~

2

)

~


where the left-hand side equals

while the right-hand side is

6

We conclude integrals in ( I )

-

1

2 0, which is A 2 G. In the case of equality, all

111


Second proof. The following proof of Theorem 3, using the inequality of the arithmetic and the geometric mean, is a

115

118

A theorem of Po'lya on polynomials

120 which means that It

122

A theorem of Pdlya on polynomials Equating the real parts in (4) and ( 5 ) we obtain by

125

On a lemma of Littlewood and Offord

and an easy calculation shows that the first sum adds the k and the second sum the largest k binomial coefficients

+ 1 largest

126

On a lemma ofLittlewood and

128

Cotangent and the Herglotz

Addition theorems: sin(x t y) = sin cos(x y) = cos

+

sn(x+)= X ); =

+

y

+ cos -

sin

- sinx

= 2sin$cosI

5.

y y

Cotangent and the

130

Cotangent and the Herglotz trick

So to finish our story let us see how Euler

Cotangent and the Herglotz trick

From

we obtain by comparing coefficients for z n :

We may compute the Bernoulli numbers recursively

131

Buffon's needle problem

Buffon's needle problem

135

The circle can be approximated by polygons. Just imagine that together with the circular needle C we are dropping an inscribed polygon as well a46 25(a6 22(circumscribed )-62ed )-118(polyg/T1_0 1 Tf0.0667 Tc 11.211.75 0 01211149.5195 556.32019 Tm(Pn

Pigeon-hole and double counting

Some mathematical principles, such as the two in the title of this chapter, are so obvious that you might think they would only produce equally obvious results. To convince you that "It ain't necessarily so" we illustrate them with examples that were by Paul

Chapter 22

142


For the proof we set N = { 0 , 1 , . . . , n} and R = { 0 , 1 , . . . , n sider the map f : N +

-

1). Con-


144

Pigeon-hole and double counting ois n o this section the following beautiful application to an extremal problem on graphs. Here is the problem: length G = (V,E ) has n vertices and contains no cycle of Suppose 4 (denoted by C4), that


Theorem. Ifthe graph G on n vertices contains no 4-cycles, then

For n = 5 this gives JEl 5 6, and the

145

146


i


From linear algebra we know that the trace equals the sum of the e5values. And here comes the trick: While

147

148


We discuss Sperner's lemma, and Brouwer's theorem as a consequence, for the first interesting case, that of dimension n = 2. The reader

152

Threefamous theorems onJinite sets Tw

Three famous theorems on finite sets

As we mentioned, Hall's theorem was the beginning of the now vast field of matching theory [6]. Of the many variants and ramifications let us state one particularly appealing result which the reader is invited to prove for himself: Suppose the sets A l . . . . , A,

155

158

Shufflingcards

than which is smaller than for n = 23 (this is the paradox"!), less 9 percent for n = 42, and exactly 0 for n > 365 (the "pigeon-hole principle," see Chapter 22). The formula is easy to see - if we take the persons in some fixed order: If in

where at the end we sum a geometric series (see page 28).

Shuffling cards

Indeed, if Ai denotes the event that the ball i is not drawn in the first m drawings, then rob [v,>

159

160

Shufling cards

Two

Shufflingcards

Let T be the number 3161

161

162

Shufflingcards

and we have seen that Prob[Ti - T,-1 for

= j] =

Prob[V,-i+l

-

Vn-i

= j]

163

Shuffling cards

>

+

0 and k := [nlog n cnl . Then after peiforming k top-in-at-random shuffles on a deck qf n cards, the variation distance,from the uniform distribution satisjies

Theorem 1. Let c

One can also verify that the variation distance d ( k ) stays large if we do significantly fewer than n log n top-in-at-random shuffles. The reason is that a smaller number of shuffles will not suffice to destroy the relative ordering on the lowest few cards in the deck. Of course, top-in-at-random shuffles are extremely ineffective - with the

164

Shuflinn cards Bell Labs "Mathematics of Communication" department at the time), has several virtues: 0 it is elegant, simple, and ca3ms natural, 0 it models quite well the way an amateur would perform riffle shuffles, a and we have a chance to analyze it. Here are three descriptions - all of them describe the

Lattice paths

168

Lattice paths and determinants and the weight of the path system 'P,

Lattice paths and determinants

Proof. A typical summand of det(A1) is sign a r n ~ , ( ~. .j . rn,,(,), which can be written as

Summing over a

169

170


Theorem. If P is an ( r x

171


Now look at the figure to the right, where A, is placed at the point (0, - a i ) and B, at ( b j , - b j ) . The number of paths from Ai to B, in this onlygrid steps thatto the use ( b 3 + (b, " ' p b ~north )) = and other east In is, by what we just proved, words, the matrix of binomials Af is precisely the path frommatrix A to B in thegraph directed for which lattice all edges have weight A

(z;).

172


References [ I ] I. M. GESSEL&

Cayley's formula for the number of trees

Chapter 26

One of the most beautiful formulas in enumerative combinatorics concerns the number of labeled trees. Consider the set N = {1,2, . . . ,n). How many different trees can we form on this vertex set? Let us denote this number by Enumeration "by hand" yields TI = 1, = 1, = 3, T4 = 16, with the trees shown in the following table:

T,.

T2

T3

174

T2 1 1 L

The four trees of

Cavlev's formula for the number o f trees

Cayley's

176

Cayley's formula for the number of trees

1 Third proof (Recursion). Another classical method in enumerative combinatorics is to establish a recurrence relation and to solve it by induction. The following idea is essentially due to Riordan and RCnyi. To find the proper recursion, we consider a

Cayley'sformula for the number of trees

177

178

Cavlev's formula for the number o f trees

Completing Latin squares

Comnletinp Latin sauares 4

181

184

Completing Latin squares

In our example the "exchange case" happens for k = 5: the element x5 = 3 does already occur in the last column, so that entry has to be moved back to column k = 5. But the exchange element xk = 6 is not new either, it is exchanged by x:

186

The

The Dinitz ~ r o b l e m

188

The Dinitz uroblem

Now we come to an important concept, "stable matchings," with a downto-earth interpretation. A matching M

190

The Dinitz problem

of a new graph, joining two such vertices if and only if as edges in K,,, they have a common endvertex, then we clearly obtain the square graph S,. Let us say that

Construction of a line graph

Identities versus bijections

+

+

+

Chapter 29

+

Consider the infinite product ( I x ) ( 1 x 2 )( 1 x 3 )(1 x 4 ) . . . and expand it in the usual way into a series Enroanxn by grouping together those products that yield the same power xn. By inspection we find for the first terms

So we have e. g. a6

=

4, a7

=

5,


Problem. Let P,(n) and Pd(n)be the partitions of n into odd and into

193

196


References [I] G. E. ANDREWS: The Theory of

Five-coloring plane graphs

Plane graphs and their colorings have been the subject of intensive research since the beginnings of graph theory because of their connection to the fourcolor problem.

Chapter 30

200

Five-coloring plane graphs From part (A) of the proposition on page 67 we know that G has a vertex v of degree at most 5. Delete 11 and all edges incident

202

Five-coloring plane graphs is still plane and 3-colorable. We assign { 5 , 6 , 7 , 8 ) to the top vertex and {9,10, 11,

204

A museum with n = 12 walls

A triangulation of the museum

The interior dihedral angles po-0.039 Tnc 8.875 0.026 Tc - 170674368rd

How to guard a museum


convex vertex A. In fact, there must be at least three of them: In essence this is an application of the pigeonhole principle! Or you may consider the convex hull of the polygon, and note that all its vertices are convex also for the original polygon. Now look at the two neighboring vertices B and C of A.If the segment BC lies entirely in then this is our diagonal. If

205

206

"Museum guards" (A 3-dimensional art-gallery problem)


208

Turdn's graph theorem

Let us turn to the general case. The first two proofs due to Turin and to ErdBs, respecttivly.

induction and are

First proof. We -6e induction on n. One easily computes tha5 0 0 9.8123 Tc

Turan's graph theorem

all vertices adjacent to u,, and define s, similarly for uj, where we may

209

Tura'n's

214

Communicating without errors

set Vl x V2 =

Cornmunicatin~without errors

cliques of size 1, the edges are the cliques of size 2, the triangles are cliques of size 3, and so on. Let C be the set of cliques in G. Consider an arbitrary probability distribution

215

216

The


Communicatin~without errors Let us see how LovAsz proceeded

217


Of friends and politicians

It is not known who first raised the following problem or who gave it its

Suppose in a group of people we have the situation that any pair of persons have precisely one common friend. Then there is always a

Chapter 34

224

Of

O f friends and voliticians Now if the square root

225

Probability makes countinn (sometimes) easy

Theorem 2, 2. For all numbers: R(k:k)

>

229

the following lower bound holdsfor the Ramsey

2

2s.

. .

. * ______.....,_______ L _ _ _ _ _ _ _ _ _ _ _

H 2) Proof. = 2. o We mhave (2) we R (know 2 3

3)

Probability makes counting (sometimes) easy

23 1

Probability makes counting (sometimes) easy

232

Since the right-hand side goes to 0 with n going to infinity, we infer that ):

>

p(X

234

Probability makes counting (sometimes) easy By linearity of expectation we thus find

Set

which is

: p = Here comes the punch line:

Proofs From the Book

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