1. Proofs

Introduction Writing Mathematics Proofs More Proofs Even More Proofs Conclusion

1. Informal Introduction to Methods of Proof

Stephane Bressan

August 7, 2010


Introduction

1.1. Introduction

“This sentence contradicts itself - noactually it doesn’t.”Godel, Escher, Bach: An EternalGolden Braid,by Douglas Hofstadter


The MU System

The MU System

Sentences

Sentences are sequences of the letters M, U and I .

M

UIUI

MIU

MU


The MU System

Axioms

The only axiom is MI .

Inference Rules

If an axiom or theorem ends with I then U can be added atthe end to form a theorem.

If an axiom or theorem starts with M then the remainder canbe duplicated to form a theorem.

III can be replaced by U anywhere in an axiom or theorem toform a theorem.

UU can be removed anywhere from an axiom or theorem toform a theorem.


The MU System

From the axiom MI , we can derive MIU by the first rule. Wesay that MIU is a theorem.

If MIII is a theorem then we can derive MU by the third rule.


The Formal MU System

The Formal MU System

Sentences

Sentences are sequences of the letters M, U and I .

Axioms

MI

Inference Rules

xI ` xIU

Mx ` Mxx

xIIIy ` xUy

xUUy ` xy


Theorems in the MU System


Is MUII a theorem?

1 MI

2 MI ` MII by the second rule where x = I .

3 MII ` MIIII by the second rule where x = II .

4 MIIII ` MIIIIIIII by the second rule where x = IIII .

5 MIIIIIIII ` MIIIIIIIIU by the first rule where x = MIIIIIII .

6 MIIIIIIIIU ` MIIIIIUU by the third rule where x = MIIIII andy = U.

7 MIIIIIUU ` MIIIII by the fourth rule.

8 MIIIII ` MUII by the third rule where x = M and y = II .



Lemma 1.1.1

The number of I in a theorem is never a multiple of 3.

Theorem 1.1.2

MU is not a theorem of the MU system.


1.2. Writing in the Language of Mathematics

“Everything is vague to a degree youdo not realize till you have tried tomake it precise.”The Philosophy of Logical Atomism,by Bertrand Russell


Definitions, Propositions and Proofs


The structure of a mathematical text usually consists of:

Definitions.

Propositions.

Proofs.



definitions, propositions and proofs involve one or a severalstatements expressed in a rigorous language. The language can benatural language or a formal language such as first order logic.While the former may be ambiguous, the latter is not.

Unfortunately, the former may appear more intuitive and the latterunnecessarily convoluted.


Definitions

Definitions

Definitions associate a name, the name of the concept beingdefined, to a necessary and sufficient condition expressed as astatement. It often contains the English language structure “if andonly if” written in shorthand as “iff”.


Definitions

Definition 1.2.1

The set of natural numbers, N, is {0, 1, 2, 3 etc .}This is not a proper definition!


Definitions

Definition 1.2.2

A natural number n is even iff there exists a natural number msuch that n = 2×m.

Definition 1.2.3

A natural number n is odd iff there exists a natural number m suchthat n = (2×m) + 1.


Definitions

We sometimes use phrases like “n is said to be even” or “n iscalled even” to make clear that the definition introduces the termfor the first time.

The fact that the condition is necessary is often implicit (byclosure). We only indicate that it is sufficient and use “if” or “suchthat”


Definitions

Definition from [Kranz, 2009]

A natural number is said to be even if, when it is divided by 2,there is an integer quotient and no remainder.

Definition from [Rosen, 2003]

The integer n is even if there exists an integer k such that n = 2k .

Definition from [Epp, 2004]

An integer n is even if, and only if, n equals twice some integer.

n is even ⇔ ∃ an integer k such that n = 2k.


Definitions

Definition from [Gosset, 2003]

An integer n, is even if there exists an integer, k , such that n = 2k .

Definition from [Ensley and Crawley, 2006]

An integer n is even if it can be written in the form n = 2.K forsome integer K.


Definitions

A definition is often given by a list of propositions called axioms orlaws.

Definition 1.2.4

Equality on a set S , noted =, a the binary relation on a set S suchthat:

(Reflexivity) ∀x ∈ S (x = x).

(Symmetry) ∀x ∈ S ∀y ∈ S (x = y ⇔ y = x).

(Transitivity)∀x ∈ S ∀y ∈ S ∀z ∈ S ((x = z ∧ z = y)⇒ x = y).

(Closure) ∀x ∈ S ∀y (x = y ⇒ y ∈ S).


Propositions

Propositions

The main type of proposition is the theorem.Ancillary propositions are rather called lemma if they are enouncedand proved in preparation of the proof of a theorem. They arecalled corollary if they follow more or less directly from a theorem.Statements that have not yet been proved but are believed to betrue by their inventors are called conjectures.


Propositions

The difference between lemmas, theorems and corollaries (andeven conjectures) is subjective. Proposition are named by traditionand convention.Fermat’s Last Theorem remained a conjecture for almost 450 years.Other names are occasionally used for propositions depending onthe context and the inspiration of their inventors and readers.Propositions can be called principle, postulate, hypothesis, law,algorithma or else.

aPreferably when they are operational and constructive and can betranslated into computer programs)


Propositions

Famous propositions are given a name: the Fifth Postulate, theContinuum Hypothesis, Napoleon’s Theorem, The PoincareConjecture, the Hand-shaking Theorema.

athe list is made of propositions discussed in the book [Crilly, 2007].


1.3. Methods of Proof

“If it is a miracle, any sort ofevidence will answer, but if it is afact, proof is necessary.”by Mark Twain


Some Methods of Proof


Some Methods of Proof:

Direct Proof,

Equivalence Proof,

Proof by Exhaustion,

Proof by Counter Example,

Proof by Construction,

Pure Existence Proof.



More Methods of Proof:

Contrapositive Proof,

Proof by Contradiction,

Proof by Induction,

Pigeon Hole Principle,

Combinatorial Proof,

Inclusion-Exclusion Principle.


Direct Proof

Direct Proof

Proposition 1.3.1

The square of an even natural number is even.


Direct Proof

Proof Sketch

This is a direct proof. We show that if n is a natural number and nis even then n × n is even by factorizing 2.

Proof.

1 Let n be a an even natural number.

2 Since n is even then there exist a natural number m such thatn = 2×m.

3 Therefore n × n = (2×m)× (2×m).

4 Therefore n × n = 2× (2×m ×m) by associativity andcommutativity of multiplication.

5 Therefore n × n is even.

Q.E.D.


Equivalence Proof

Equivalence Proof

Proposition 1.3.2

The square of a natural number is even if and only if the naturalnumber is even.

Proof Sketch

First, we prove that if a natural number is even then its square iseven. Second, in order to prove that if the square of naturalnumber is even then the number is even, we prove thecontrapositive (see Subsection 4.5): if a natural number is noteven then its square is not even.


Equivalence Proof

Proof.

1 Let n be a natural number.2 If: Let us that prove that if n is even then its square is even.

1 We know that if n is even then its square is even byProposition 1.3.1.

3 Only if: let us prove that if n is not even then its square is noteven.

The proof is omitted.

Q.E.D.


Exhaustion

Exhaustion

Proposition 1.3.3

Natural numbers of the form n7 − n are divisible by 7.


Exhaustion

Proof Sketch

We factorize n7 − n. If we try and divide n by 7 then we may beleft with a remainder. In other words n = 7× q + r . We show thatfor each of the possible values of r , namely 0, 1, 2, 3, 4, 5 and 6one of the factors is divisible by 7.

Proof.

1 Let n be a natural number.2 Let us prove that

n7 − n = n× (n− 1)× (n2 + n + 1)× (n + 1)× (n2 − n + 1).1 n7 − n = n × (n6 − 1).2 Therefore n7 − n = n × (n3 − 1)× (n3 + 1).3 Therefore

n7 − n = n × (n − 1)× (n2 + n + 1)× (n + 1)× (n2 − n + 1).

...


Exhaustion

Proof Cont.

3 Let us now consider the seven cases for n = 7q + r where requals 0, 1, 2, 3, 4, 5 and 6, respectively.

1 Let us assume that r = 0.

1 Therefore n = 7× q.2 Therefore n7 − n has the factor n, which is divisible by 7.


1 Therefore n = 7× q + 1.2 Therefore n7 − n has the factor n − 1 = 7× q, which is

divisible by 7.


1 Therefore n = 7× q + 2.2 Therefore n7 − n has the factor

n2+n+1 = (7×q+2)2+(7×q+2)+1 = 49×q2+35×q+7,which is divisible by 7.

...


Exhaustion

Proof Cont.

3 3 and 4.4 Let us assume that r = 3.


n2−n+1 = (7×q+3)2−(7×q+3)+1 = 49×q2+35×q+7,which is divisible by 7.



n2+n+1 = (7×q+4)2+(7×q+4)+1 = 49×q2+63×q+21,which is divisible by 7.

...


Exhaustion

Proof Cont.

3 5 and 6.6 Let us assume that r = 5.


n2−n+1 = (7×q+5)2−(7×q+5)+1 = 49×q2+63×q+21, which is divisible by 7.


1 Therefore n = 7× q + 6.2 Therefore n7 − n has the factor n + 1 = 7× q + 7, which is

divisible by 7.

Q.E.D.


Proof by Construction


Proposition 1.4.1

There exists an irrational number whose power to an irrationalpower is rational.



Proof Sketch

We exhibit two irrational numbers p and q such that pq is rational.



Proof.

1 Let p and q be real numbers and p =√

2 and q = log2(25).

2 We know that√

2 is irrational by Proposition 1.4.3.3 Let us prove that log2(25) is irrational.

The proof is omitted.

4 Let us prove that pq is rational.

1 log2(pq) = log2(√

2log2(25)

) by (1)2 Therefore log2(pq) = log2(

√2)× log2(25).

3 Therefore log2(pq) = log2(2)× 12 × log2(25).

4 Therefore log2(pq) = log2(√

25).5 Therefore pq =

√25 = 5, which is rational.

Q.E.D.


Proof by Counter Example


The following proposition was thought to be true but it is false.

Proposition 1.4.2

Natural numbers of the form 2(2n) + 1, where n is a positiveinteger, are prime.



Proof Sketch

We disprove the proposition by exhibiting a number n such that2(2n) + 1 is not prime.

Proof.

1 Let n be a natural number and n = 5.

2 Therefore 2(25) + 1 = 4294967297.

3 Therefore 4294967297 = 641× 6700417a

Q.E.D.

a641 and 6700417 are prime numbers


Pure Existence Proof


Proof Sketch

We exhibit two irrational numbers p and q such that if pq isrational then we can construct a rational number of the form r s

where r and s are irrational and if pq is irrational then we canconstruct a rational number of the form uv where u and v areirrational. Whatever the rationality of pq there exists an irrationalnumber whose power to an irrational power is rational.

Alternative Proof of Proposition 1.4.1.

1 Let p and q be irrational numbers.

2 We know that√

2 is irrational by Proposition 1.4.3.

...



Proof Cont.

3 Let us assume that√

2√

2is irrational.

1 Let p =√

2√

2and q =

√2 both irrational.

2 Therefore pq = (√

2√

2)

√2

.

3 Therefore pq =√

2√

2×√

2.

4 Therefore pq =√

22.

5 We know that pq = 2 is rational.

4 Let us assume that√

2√

2is rational.

1 Let p =√

2 and q =√

2 both irrational.

2 We know that pq =√

2√

2is rational by hypothesis.

Q.E.D.


Proof by Contradiction


Proposition 1.4.3√

2 is irrational.

Proposition 1.4.3

There is no rational number whose square is 2.



Proof Sketch

We prove by contradiction that if x is rational then x2 6= 2. Weassume such an x exists and derive a contradiction.

Proof.

1 Let x be a rational number.2 Let us preliminarily prove that there exists two natural

numbers a and b such that x = ab and one of a or b is not

even.1 a and b such that x = a

b exist by definition of rationalnumbers.

2 If both a and b are even we simplify the fraction by dividingthem both by 2 until one of them is odd. We replace a and bby the new nominator and denominator, respectively.

...



Proof Cont.

3 Let us assume that x2 = 2.1 We know that there exists a and b such that x = a

b and one ofthem is not even by (2).

2 Therefore x2 = a2

b2 = 2 .3 Therefore a2 = 2× b2.

...



Proof Cont.

3 Cont.4 Let us assume that a is odd.

1 This is a contradiction since the square of an odd number isan odd number.

5 Let us assume that a is even.

1 Therefore b is odd by (3.1)2 Therefore there exist a number c such that a = 2× c and b is

odd.3 Therefore a2 = 2× 2× c2 = 2× b2.4 Therefore 2× c2 = b2.5 This is a contradiction since the the square of an odd number

is an odd number.

4 Therefore x2 6= 2.

Q.E.D.


Contrapositive Proof


Proposition 1.4.4

If the sum of two natural numbers is even then the two numbershave the same parity.



Proof Sketch

We prove the contrapositive, namely that if two numbers havedifferent parity then their sum is odd.

Proof.

1 Let n and m be natural numbers. Let n be even and m beodd without loss of generality (because of commutativity ofaddition).

2 Therefore there exist two natural numbers k and p such thatn = 2× k and m = 2× p + 1 by definition of even and oddnatural numbers.

...



Proof Cont.

3 Therefore we obtain the following equation.

(1) n + m = (2× k) + (2× p + 1)

4 Therefore we obtain the following equation from equation 1 bycommutativity of addition and associativity of multiplication.

(2) n + m = (2× (k + p)) + 1

5 Therefore n + m is odd.

Q.E.D.


Proof by Induction

Proof by Induction

Proposition 1.5.1

The number of I in a theorem of the MU system is never amultiple of 3.


Proof by Induction

Proof Sketch

We prove by induction on the length of the derivation that everytheorem cannot have three I.

Proof.

1 Base case: the axiom has one I .

2 Induction hypothesis: Let us assume that all theorems derivedby n inference rules do not have a number of I which is amultiple of 3.

...


Proof by Induction

Proof Cont.

3 Induction step:1 Rule 1 does not change the number of I .2 Rule 2 doubles the amount of I .3 Rule 3 replaces III by U.4 Rule 4 does not change the number of I

4 Therefore the number of I in a theorem of the MU system isnever a multiple of 3.

Q.E.D.


Pigeon Hole, Dirichlet’s Box principle, and Ramsey’s Theory

Pigeon Hole, Dirichlet’s box principle and Ramsey’s Theory

Proposition 1.5.2 (Pigeon Hole or Dirichlet’s box principle)

If n + 1 objects are put into n boxes then at least one box containstwo or more objects.



Proof Sketch

This is a proof by contradiction. If no box contains two objects ormore then we are missing objects.

Proof.

1 Let us assume that each box contain 1 or 0 object1 Therefore we have less than n objects in total.2 This is a contradiction.

2 Therefore at least one box contains two or more objects.

Q.E.D.



Proposition 1.5.3 (Generalized Pigeon Hole Principle)

If n objects are put into k boxes then at least one box contains⌈nk

⌉objects.

Proposition 1.5.4 (Infinite Pigeon Hole Principle)

If an infinite number of objects are put into k boxes then at leastone box contains an infinite number of objects.



Ramsey’s Theory

There is local order in sufficient global chaos.

Proposition 1.5.5 (Ramsey’s Theorem for six vertices)

A complete graph with six vertices whose edges are colored withtwo colors contains at least one monochrome triangle.



Six Vertices

A

B C

D E

F



Proof Sketch

Using twice the Generalized Pigeon Hole Principle, we prove thatthere are three edges of the same color leaving any node and thatthere is a triangle between the node and two of the verticesconnected by these edges or betwen the three vertices connectedby these edges.



Proof.

1 Let G be a complete simple graph whose edges are coloredwith two colors.

2 Let v be any vertex of G .

3 We know that v has five incident edges because we know thatthe graph is simple and complete by (1).

4 Therefore three of these vertices at least have the same colorby the Generalized Pigeon Hole Principle (Proposition 1.5.3).

5 Therefore there are three different vertices connected to v bythese edges haveing the same color because we know that thegraph is simple by hypothesis.

6 Let w , x and y be these three different vertices.

...



Proof Cont.

7 We know that there are three edges between the vertices w , xand y because the graph is complete by hypothesis.

8 Therefore either the three vertices have the same color or oneof the vertices is of the same color as the vertices from v tow , x and ya

1 Let us assume that the three vertices have the same color.

1 Therefore they form a monochrome triangle (w , x , y).

2 Let us assume that one of the vertices is of the same color asthe vertices from v , say (x , y) without loss of generality.

1 Therefore they form a monochrome triangle (v , x , y).

9 Therefore there exists at least one monochrome triangle.

Q.E.D.

aNotice that the two cases are not exclusive but cover all possibilities.


Combinatorial Proof

Combinatorial Proof

Proposition 1.5.6 (Fermat’s Little Theorem)

If p is a prime number and n natural number, then np − n isdivisible by p.


Combinatorial Proof

Proof from “Combinatorial Proof of Fermat’s ‘Little’ Theorem” byS.W. Golomb.

Suppose that we have beads in n different colors, and we wish tomake necklaces using exactly p beads.First we put p beads on a string. Since each of the beads can bechosen in n ways, there are np possible strings.For each of the n colors, there is one string entirely of that color.We throw these away, leaving np − n strings.We will join the two ends of each of these strings to formnecklaces. But we observe that if two strings differ only by a cyclicpermutation of the beads, the resulting necklaces will beindistinguishable.Since there are p cyclic permutations of p beads on a string, thenumber of distinguishable necklaces is (np − n)/p which musttherefore be an integer.Q.E.D.


Inclusion-Exclusion Principle


Proposition 1.5.7

Let S, A and B be a finite sets such that S = A ∪ B. Thefollowing equation holds.

(1) | S |=| A | + | B | − | A ∩ B |



Problem

Let n be a strictly positive natural number. Let p and q be twoprime positive natural numbers.How many multiples of p or q are there between 1 and n × p × q?

Proposition 1.5.8

Let n be a strictly positive natural number. Let p and q be twoprime positive natural numbers.There are n × p + n × q − n multiples of p or q between 1 andn × p × q.



Proof Sketch

We use the Inclusion-Exclusion Principle to compute the numberof multiple of p or q from the number of multiple of p, the numberof multiple of q and the number of multiples of both p and q.

Proof.

1 Let n be a strictly positive natural number.

2 Let p and q be two prime positive natural numbers.

3 Let P be the set of multiples of p between 1 and n × p × q.

4 We know that there are n × q multiples of p between 1 andn × p × q by construction and because q is prime byhypothesis.

...



Proof Cont.

5 Let Q be the set of multiples of q between 1 and n × p × q.

6 We know that there are n × p multiples of q between 1 andn × p × q by construction and because p is prime byhypothesis.

7 We know that P ∩ Q is the set of multiples of p and qbetween 1 and n × p × q by construction.

8 We know that there are n multiples of p and q between 1 andn × p × q by construction and because p and q are prime byhypothesis.

...



Proof Cont.

9 We know that P ∪ Q is the set of multiples of p or q between1 and n × p × q by construction.

10 Therefore| P ∪Q |=| P | + | Q | − | P ∩Q |= n× p + n× q − n by theIncludion-Exclusion Principle (Proposition 1.5.7).

11 Therefore there are n × p + n × q − n multiples of p or qbetween 1 and n × p × q.

Q.E.D


Conclusion

1.6. Conclusion

Quod Erat Facienduma.

aWhat needed to be done (that is all for today)


Conclusion

Further Readings I

T. Crilly,50 mathematical ideas you really need to know,Quercus Publishing Plc. (2007).

B.A Dumas and J.E. McCarthy,Transition to Higher Mathematics, Structure and Proofs,Fifth Edition, Mcgraw-Hill (2007).

D.E. Ensley and J.W. Crawley,Discrete Mathematics, Mathematical Reasoning and Proofswith Puzzles, Patterns, and Games,John Wiley & Sons Inc. (2006).

S.S. Epp,Discrete Mathematics with Applications,Third Edition, Thomson Brooks/Cole (2004).


Conclusion

Further Readings II

E. Gosset,Discrete Mathematics with Proof,Pearson Education (2003).

S.G. Kranz,Discrete Mathematics Demystified,Mcgraw-Hill (2009).

K.H. Rosen,Discrete Mathematics and its Applications,Fifth Edition, Mcgraw-Hill (2003).

1. Proofs

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