NumberTheory by Frances Odumodu NAU

8/16/2019 NumberTheory by Frances Odumodu NAU

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Nnamdi Azikiwe University

2013/2014 Second Semester

MAT 241: Number Theory

Author: Frances Odumodu


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Contents

1 Basic Set Theory 41.1 Relations between sets . . . . . . . . . . . . . . . . . . . . . . 61.2 Set Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.3 Quantifiers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.4 Indexed sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

2 Symbolic Logic 112.1 Basic Logical Operations . . . . . . . . . . . . . . . . . . . . . 122.2 Important equivalences . . . . . . . . . . . . . . . . . . . . . . 152.3 Quantifiers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152.4 Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

3 Methods of Mathematical Proofs 173.1 Direct Method . . . . . . . . . . . . . . . . . . . . . . . . . . . 173.2 Indirect Method . . . . . . . . . . . . . . . . . . . . . . . . . . 18

3.2.1 Proof by Contraposition . . . . . . . . . . . . . . . . . 193.2.2 Proof by Contradiction . . . . . . . . . . . . . . . . . . 19

3.3 Proof by Cases (Exhaustion) . . . . . . . . . . . . . . . . . . . 203.4 Existence Proofs . . . . . . . . . . . . . . . . . . . . . . . . . 213.5 Uniqueness proofs . . . . . . . . . . . . . . . . . . . . . . . . . 22

3.6 Proof of Universal statements . . . . . . . . . . . . . . . . . . 223.7 Proof of Statements of the form p ⇒ (q ∨ r) . . . . . . . . . . 223.8 Principle of Mathematical Induction . . . . . . . . . . . . . . 233.9 Pigeon Hole Principle . . . . . . . . . . . . . . . . . . . . . . . 253.10 Counter Examples and General Comments . . . . . . . . . . . 263.11 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

4 Relations 274.1 Ordered pairs and cartesian products . . . . . . . . . . . . . . 274.2 Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

4.3 Equivalence Relation . . . . . . . . . . . . . . . . . . . . . . . 314.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

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Outline

Mathematics is the queen of the sciences and number theory is the queen of Mathematics. – Johann Carl Friedrich Gauss

Basic set theory. Symbolic logic. Methods of mathematical proof. Relations-partial ordering, equivalence, upper and lower bounds, maximal, minimal,maximum and minimum elements of sets of real numbers. Elementary treat-ment of the well ordering principle, Zorn’s lemma and axiom of choice.Prime numbers - infinitude of divisibility and modulo systems. GCDs andLCMs. Euclid’s division algorithm. The fundamental theorem of arithmetic

or unique factorization theorem. Congruences and residue classes. Euler’stotient and the chinese remainder theorem. Continued fractions and the solv-ability of linear congruence. Transversals and the solvability of polynomialcongruence (elementary treatment only).

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Part I: Set Theory and Logic

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1 Basic Set Theory

Set theory is at the foundation of Mathematics, and nearly every Mathemat-ical object of interest is a set of some kind. – Larry Gerstein

A set is a well-defined collection of objects. That is, it is defined in such away that we can determine for a given object x whether or not x belongs tothe set.

Example

• The set of all sentences of a language.

• The set of living human beings.• The set of vowels of the English language.• The set of students absent from this class.• The even integers 2, 4, 6, . . ..

If S is a set and s belongs to S , we write s ∈ S and say that s is an elementof S . If s does not belong to S , we write s ∈ S . Example. If S = {s,t,u}then s ∈ S and 4 ∈ S .

Sets are determined by their elements. Thus, we may define a set by listing itsmembers. The set is then said to be in tabular form. Example. A = {1, 2, 3}.The order in which the elements are listed is irrelevant. Thus, {1, 2, 3} and{3, 1, 2} represent the same set. Also, repetition of elements does not affectthe set being represented. For example, {1, 2, 2, 2, 3, 3} is the same set as{1, 2, 3}.Many sets that occur frequently are given standard names:

• N = {1, 2, 3, 4, . . .}, the natural numbers or positive integers.

• Z =

{. . . ,

−3,

−2,

−1, 0, 1, 2, 3, . . .

} the set of integers or whole num-

bers.

• Q = {x : x = ab

for some a, b ∈ Z, b = 0} the set of rational numbersor the rationals.

• R the set of real numbers or the reals.

A set may also be defined by stating a property that completely characterisesthe elements of a set. That is, it determines whether or not an object x

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belongs to a set. The set is then said to be in set builder form. Examples. The

set of all planets in the solar system. The set of Mathematical Instruments.

Given a property P and an element s of a set S . We wrte P (s) to indicatethat s has property P . For a given element s, the statement P (s) shouldeither be True or False. Thus, we can write a set A = {s ∈ S : P (s)}, theset of all elements of S having the property P .

1. E = {x : x is an even integer and x > 0}= {x ∈ Z : x = 2y for some y ∈ Z}.

2. B =

{x : x

∈Z and 0

≤ x

≤ 1000

} =

{0, 1, 2, . . . , 1000

}.

Note that if we are given a set of the form S = {π, Goodluck , −√ 53, spaghetti , 4},then it would be difficult to find a property that characterises the elements.

Exercise.

1. Describe the following sets in the form {x : p(x)}.(a) A = {0, 2, 4, 6, 8, 10, 12, 14, . . .}.(b) B = {1, 2, 5, 10, 17, 26, 37, 50, . . .}.(c) C =

{1, 5, 9, 13, 17, 21, . . .

}.

(d) D = {1, 12 , 13 , 14 , 14 , . . .}.(e) E = {lemon , 1, 2, 3, 4, . . .}.

2. List the elements in the set {x ∈Q : x ∈ N and 1 ≤ x ≤ 3}.3. True or False.

(a) {1, 2, 3} = {3, 1, 2}.(b) {a,c,b} = {b,b,c,c,a,c}.(c) t

∈ {2, 5

}.

(d) ∅ ∈ {1, 2}.(e) {1, 2} = {1, 2}.(f) {x ∈ R : x2 = −2} = ∅.

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1.1 Relations between sets

Subset relation Given two sets, we may speak of whether or not one set

is contained in the other. A set A is said to be a subset of a set B if everyelement of A is also an element of B. Notation: A ⊆ B .In symbols: A ⊆ B if and only if , for all x, if x ∈ A, then x ∈ B. That is.

A ⊆ B ⇔ ∀x, x ∈ A =⇒ x ∈ B

If A is not a subset of B, we write A B. To get the characterisation, wenegate the statement for A

⊆ B . That is

A B ⇔ ¬(∀x, x ∈ A =⇒ x ∈ B)⇔ ∃x such that x ∈ A x ∈ B⇔ ∃x such that x ∈ A and x ∈ B

Thus, to show that A B, it suffices to show that A contains an elementwhich is not in B.

Exercise. Show that A ⊆ A. Show also that if A ⊆ B and B ⊆ C , thenA ⊆ C .Example. N ⊆ N ⊆ Q ⊆ R ⊆ C.A set A is said to be a proper subset of a set B if A is contained in B butB is not contained in A. That is, there is an element in B which is not anelement of A. In symbols, A

⊂ B

⇔ A

⊆ B and A

= B. Notation: A

⊂ B

or A B or A B.Example. {1, 7} ⊂ {1, 7} ⊂ {1, 2, 7} and ∅ ⊂ A for any nonempty set A.Let A = {1, 2}, B = {1, 2, 3}, C = {3, {1, 2}} and D = {1, 2, {1, 2}}. Then,A ⊂ B, A /∈ B, A ∈ C , A C , A ∈ D and A ⊂ D.Set equality. Two sets A and B are equal when they consist of the sameelements. That is, A = B if and only if, for all x, x ∈ A if and only if x ∈ B.If A and B are not equal, we write A

= B.

Exercise. Prove that A = B if and only if A ⊆ B and B ⊆ A.Exercise. Let A = {x ∈ Z : x = 2(y − 2) for some y ∈ Z} and B = {x ∈ Z :x = 2z for some z ∈ Z}. Are A and B equal? Justify.Set operation We would like to define operations on sets that are some-


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Union. The union of two sets A and B is the set of elements that belong to

either A or B or both.

A ∪ B = {x : x ∈ A or x ∈ B}

Intersection. The intersection of two sets A and B is the set consisting of elements of both A and B.

A ∩ B = {x : x ∈ A and x ∈ B}

Set difference. The difference of two sets A and B is the set consisting of those elements of A that do not belong to B. The operation is called set

complement or complement of B in A.

A − b = {x : x ∈ A and x /∈ B}Example. Let A = {2, 3, 5, 7, 11, 13} and B = {A, 2, 11, 18} then A ∪ B ={A, 2, 3, 5, 7, 11, 13, 18}, A ∩ B = {2, 11} and A − B = {3, 5, 7, 13}.Universal set. Usually, the sets we work with are susbet of some biggerset. For example, the even numbers and the odd numbers are both subsets of the set of integers. We call such a set a universal set or the set of discourse,denoted by U . Thus, a universal set is the underlying set that all the setsunder examination are subsets of.Thus, we may speak of the set difference U − A, which is the set of thoseelements of U that do not belong to A. That is,

Ac = U − A = {x ∈ U : x /∈ A}.

Empty set. Set operations may lead to a set containing no elements. If Ais any set and E is a set containing no elements, then E ⊆ A. Thus, there isa unique set containing no elements.

We call the set containing no elements the empty set or nullset. Notation:

{} or ∅. Thus, ∅ = {x ∈ Z : x = x}.Exercise. Verify that (Ac)c = A, ∅c = U and U c = ∅.Disjoint sets. Two sets A and B are said to be disjoint if A ∩ B = ∅.Singleton sets. A set with only one element is called a singleton set. Ex-ample. {x}.Cardinality. A set is finite if it contains exactly n distinct elements wheren = 0 or n is a natural number. Otherwise, the set is infinite. For example,∅ and { the English alphabets}.

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Power Set Let A be a set. The power set of A is the class of all subsets of

A, denoted by P (A). If A is a finite set, then so is its power set. If n(A) isthe number of elements in A, then n(P (A)) = 2n(A).

1.2 Set Identities

There are a number of set identites that the set operations of union, inter-section and set difference must satisfy. They are very useful in calculationswith sets. Let U be the universal set and A, B, C be any sets. Then

1. A

∪B = B

∪A; A

∩B = B

∩A. Commutativity.

2. (A ∪ B) ∪ C = A ∪ (B ∪ C ); (A ∩ B) ∩ C = A ∩ (B ∩ C ). Associativity.3. A ∩ (B ∪ C ) = (A ∩ B) ∪ (A ∩ C ); A ∪ (B ∩ C ) = (A ∪ B) ∩ (A ∪ C ).

Distributivity.

4. A ∪ A = A; A ∩ A = A. Idempotent.5. A ∩ (A ∪ B) = A; A ∪ (A ∩ B) = A. Absorption laws.6. A ∪ ∅ = A; A ∩ ∅ = ∅. Identity laws.

7. (A ∪ B)c = Ac ∩ Bc; (A ∩ B)c = Ac ∪ Bc. De Morgan’s laws.8. A ∪ Ac = U ; A ∩ Ac = ∅; U c = ∅; ∅c = U ; (Ac)c = A. Complement

laws.

9. A − B = A ∩ Bc.

Exercise. Prove all the laws.Prove that A ∩ (((B ∪ C c) ∪ (D ∩ E c)) ∩ ((B ∪ Bc) ∩ Ac)) = ∅.Exercise.Let A =

{1, 2, 3, 4, 5

}, B =

{3, 4, 5, 6

} and C =

{x ∈Z : x is even

}. Then

1. Find A − B; B − A; B ∩ A; C − B; (A ∩ B) ∩ C ; (A ∩ B) ∪ C .2. If Z is the universal set. Find Ac, C c, (A ∩ C )c.

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1.3 Quantifiers

Let x be a variable and P a property. P (x) asserts that : x has property P .The sentence P (x) is not a proposition since we cannot tell if P is true ornot. If we however replace x by a specific element a, then P (a) becomes aproposition.

If we want to say that P (x) is true for at least one substitution of an elementx, we use the notation

∃ x, P (x)which reads, ‘there exists an x, such that P holds’ or ‘P holds for some x’or ‘There is at least one x for which P holds’.

∃ is called the existential

quantifier .

If we want to say that every element in a certain set has property P , we write

∀ x, P (x)

which reads, ‘P holds for all x’ or ‘P holds for every x’. ∀ is called theuniversal quantifier .

Example. ∀ x, x2 ≥ 0∃ x such that x + 5 = 0.

If by ¬P , we mean that ‘P does not hold’, then ¬(∀ x, P ) is the same as(∃ x, ¬P ) and ¬(∃ x, P ) is the same as (∀ x, ¬P ).

1.4 Indexed sets

Suppose you attend four Number Theory classes. We can denote the set of classes by C = {c1, c2, c3, c4} where c1 = first class, c2 = second class, etc.Here I = {1, 2, 3, 4} forms our labels or indexes . The set I is called the index set . Having chosen an index set, we can write C = {ci : i ∈ I }.Example. If we have the terms of a sequence

{x1, x2, x3, . . .

}, we can instead

define our index set as N = {n : n ∈ N} so that our sequence becomes{xn}n∈N.Example. If we have a set of circles with different radii which are positivereal numbers then

C r = {(x, y) :

x2 + y2 = r}.If we define R+ to be the set of positive real numbers, then we have a familyof circles with center (0, 0) for each r: {C r}r∈R+.

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If we have an indexed family of sets:

{Ai

}i∈I . Then, we can define the union

i∈I

Ai

as the set of elements that belong to at least one Ai and the intersection

i∈I

Ai

as the set of elements that belong to all Ai. More compactly,

i∈I

Ai = {x : x ∈ Ai for some i ∈ I } = {x : ∃ i ∈ I such that x ∈ Ai}

and i∈I

Ai = {x : x ∈ Ai for all i ∈ I } = {x : x ∈ Ai ∀ i ∈ I }.

Note that if x /∈ i∈I Ai then ∃ i ∈ I, x /∈ Ai. That is, if x is not simul-taneously in Ai for all i ∈ I , then there is an Ai that does not contain x.

Note: Extension of DeMorgan’s laws:i∈I Bi

c =

i∈I Bci and

i∈I Bi

c =

i∈I Bci .

Exercise. If A1 = {1, 2, 3, 4}, A2 = {0, 1, 2}, A3 = {−1, 0, 1} and I ={1, 2, 3}. Then, determine ∪i∈I Ai and ∩i∈I Ai. If U = Z, determine ∪i∈I Aciand ∩i∈I Aci .Exercise. Prove the extension of DeMorgan’s law.

An indexed family of sets {Ai}i∈I is said to be disjoint if

i∈I Ai = ∅ andpairwise disjoint if Ai ∩ A j = ∅.Pictorially:

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Exercise. A1 =

{1, 2, 3, 4

}, A2 =

{3, 4, 5, 6

} and A3 =

{1, 6, 7

}. Then, show

that {A1, A2, A3} is disjoint but not pairwise disjoint.Recall: [−α, α] = {x ∈ R : −α ≤ x ≤ α}. Determine the following sets:

α>0

[−α, α],α>0

[−α, α] andα≥5

[−α, α].

2 Symbolic Logic

Mathematics, like ice cream and politics, is discussed in sentences. – Larry J. Gerstein

The development of symbolic logic can be traced back to the work of GeorgeBoole. Boolean algebra, an important branch of Symbolic Logic, is namedafter him.

Definition 1. A statement or a proposition is a declarative sentence that is either true or false, but not both true and false. The designation True (T)or False (F) is called the truth value of a statement. One can assign one and only one of them to a statement.

Examples. sin2 x + cos2 x = 1 (T) and 6 is a prime number (F).

The following are not statements.

1. Is (eπ)2 = e2π? (This is a question.)

2. If only every day could be like this one! (Exclamation.)

3. This proposition is false. (This contradicts itself)

4. 2 + 3i is less than 5 + 3i. (there is no less than in the set of complexnumbers).

5. x > 5. (This is an open sentence. Neither true nor false because itcontains a variable. It becomes True or False when we quantify it orsubstitute a specific object for its variable.)

Simple statements are not composed of any other statements. We can formcompound statements by joining two or more components statements usingone or more logical connectives: ∨, ∧, →, ⇔, ¬.

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2.1 Basic Logical Operations

1. Negation ∼ or ¬Let P be a statement. Then, we can form the negation of P by writing‘It is not the case that P holds’ or ‘It is false that P holds’ or ‘NOTP ’. In symbol, ∼ P or ¬P . The truth values of ¬P can be representedin a truth table:

P ¬P T FF T

The truth value of the negation of P is always the opposite of the truthvalue of P . Example.P : Paris is in France.¬P : Paris is NOT in France.P : 6 > 9.¬P : 6 > 9 i.e. 6 ≤ 9.

2. Conjunction p ∧ q “AND”Two statements p and q may be combined using “and” to form a com-pound statement p ∧ q read p and q . Its truth table is given by

p q p ∧ q T T TT F FF T FF F F

p ∧ q is only true when p and q are both true.Example. If p: Paris is in France and q : 2+2 = 5. Then p ∧ q is NOTtrue since q is false.

3. Disjunction p ∨ q ‘OR’In natural languages, there are two distinct uses of “OR”:Inclusive: meaning p or q or bothExclusive: meaning p or q but NOT both.We will always use ‘OR’ in the inclusive sense. The truth table is givenby

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p q p

∨q

T T TT F TF T TF F F

So, p ∨ q is false only when p and q are both false.4. Conditional p ⇒ q ‘If p then q

This is read as ‘ p implies q ’ or ‘If p then q ’. This has truth table:

p q p

⇒ q

T T TT F FF T TF F T

Thus, p ⇒ q is false only when p is true and q is false.5. Biconditional p ⇔ q

This is read as “ p if and only if q ”. That is, p ⇒ q and q ⇒ p. Insymbols, ( p ⇒ q ) ∧ (q ⇒ p). The truth table is given by:

p q p ⇔ q T T TT F FF T FF F T

The biconditional is true whenever p and q have the same truth valueand false otherwise.

If p and q are statements, so are (¬P ), ( p ∧ q ), ( p ∨ q ) ( p ⇒ q ) and ( p ⇔ q ).

Let P be a proposition. P is a tautology if it contains only T in the lastcolumn of its truth table. That is, it is true for any truth values of itsvariables. Example.

P ¬P P ∨ ¬P T F TF T T

A proposition P is a contradiction if it contains only F in the last column of its truth table or if it is false for any truth values of its variables. Example.

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P

¬P P

∧ ¬P

T F FF T F

A proposition P is a contingency if it is true under some conditions and falseunder others.

Two propositions p and q are said to be (logically) equivalent if they haveidentical truth tables. Example. p ⇒ q and ¬q ⇒ ¬P are logically equiva-lent.

p q p ⇒ q ¬q ¬ p (¬q ) ⇒ (¬ p)T T T F F T

T F F T F FF T T F T TF F T T T T

Definition 2. Given the proposition p ⇒ q , we call the (logically) equivalent proposition (¬q ) ⇒ (¬ p) its contrapositive and vice versa.Examples.

1. p XOR q and ¬( p ⇔ q ) are equivalent

p q p XOR q p q p ⇔ q ¬( p ⇔ q )T T F T T T FT F T T F F TF T T F T F TF F F F F T F

2. The propositions p ⇒ q , ¬q ⇒ ¬ p and (¬ p) ∨ q are equivalent. p q p ⇒ q ¬q ¬ p ¬q ⇒ ¬ p ¬ p q (¬ p) ∨ q T T T F F T F T T

T F F T F F F F FF T T F T T T T TF F T T T T T F T

3. Let x ∈ R, p : x = 10 and q : x2 = 100. Then p ⇒ q : If x = 10 then x2 = 100.= q ⇒ ¬ p: If x2 = 100 then x = 10.(¬ p) ∨ q : either x = 10 or x2 = 100.

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2.2 Important equivalences

1. Double negation:¬(¬ p) = p.2. Idempotent laws: p ∨ p = p and p ∧ p = p.3. Commutative laws: p ∨ q = q ∨ p, p ∧ q = q ∧ p and p ⇔ q = q ⇔ p.4. Associative laws: ( p ∨ q ) ∨ r = p ∨ (q ∨ r) and ( p ∧ q ) ∧ r = p ∧ (q ∧ r).5. Distributive laws: p ∧ (q ∨ r) = ( p ∧ q ) ∨ ( p ∧ r) and

p ∨ (q ∧ r) = ( p ∨ q ) ∧ ( p ∨ r).

6. De Morgan’s laws: ¬( p ∨ q ) = ¬ p ∧ ¬q and ¬( p ∧ q ) = ¬ p ∨ ¬q .7. Contrapositive: p ⇒ q = ¬q ⇒ ¬ p.8. Implication: p ⇒ q = (¬ p) ∨ q 9. p ⇔ q = ( p ⇒ q ) ∧ (q ⇒ p) and p ⇔ q = ((¬ p) ∧ q ) ∧ ( p ∨ (¬q ))

10. p ∨ q ≡ (¬ p) ⇒ q .

Exercise. Prove these using truth tables.

2.3 Quantifiers

Recall quantifiers: the universal quantifier ∀ and the existential quantifier ∃.The following are propositions: ∀x, P (x) and ∃x, P (x).The proposition ∀x, P (x) is true precisely when the statement P (a) is true for every substitution of objects from the domain of discussion while theproposition ∃x, P (x) is true precisely when P (a) is true for at least one substitution of an object in the domain of discussion.

Example. U = R.1. ∃x such that x2 = 4 is True. and ∀x, x2 = 4 is False.2. ∃x, x2 ≥ 0 and ∀x, x2 ≥ 0 are both True.

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2.4 Exercise

The only way to learn Mathematics is to do Mathematics.– Paul Halmos

1. Construct truth tables for

(a) ¬( p ∧ q )(b) ( p ⇒ q ) ∧ (q ⇒ p).(c) ( p ⇒ q ) ∨ (¬ p)

2. Show that p ⇒ q and q ⇒ p are not equivalent.

3. Use truth tables to prove the following equivalences

(a) ¬( p ⇒ q ) ≡ p ∧ (¬q ).(b) ¬( p ⇒ q ) ≡ ( p ∧ (¬q )) ∨ (q ∧ (¬ p))(c) p ⇒ (q ∧ r) ≡ ( p ⇒ q ) ∧ ( p ⇒ r)(d) p ⇒ (q ∨ r) ≡ ( p ⇒ q ) ∨ ( p ⇒ r)

4. Write as statements using only letters.

(a) Either Sam will come to the party and Max will not, or Sam will

not come to the party and Max will enjoy himself.(b) A sufficient condition for x to be odd is that x is prime.

(c) A necessary condition for a sequence s to converge is that it isbounded.

(d) Fiorello goes to the movies only if a comedy is playing.

(e) If x is positive, x2 is positive.

(f) The bribe will be paid if and only if the goods are delivered.

5. Given the following propositions

A = Tomorrow is SaturdayB= I will visit my friend PeterC = There are more than 200 elephants in Nigeria.Write down the new propositions:

(a) A ⇒ B .(b) A ∧ (B ⇔ C ).(c) ((¬B) ∨ A) ⇒ C .

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3 Methods of Mathematical Proofs

If it is a miracle, any sort of evidence will answer, but if it is a fact, proof is necessary. – Mark Twain

Proofs are the heart of Mathematics. You must be able to read, understandand write them. A good proof should be easy to follow.

A Theorem is a statement which has been proved to be True.

A proof of a theorem is a finite sequence of claims, each of which is derivedlogically from the previous claims as well as theorems whose truth has beenalready established.–(reference missing). In otherwords, a proof is a series of

statements each of which is either

• an assumption (axiom), or• a conclusion, clearly following from an assumption or some previously

proved result.

and each step should be clear or at least clearly justified.

To prove any thing, we must know some previous truths. Logic supplies waysby which we can deduce a statement from other statement, but we need somestatements to begin with. These initial statements are what we call axioms .

Thus, an axiom is a statement that is assumed to be true.

A Lemma is smaller theorem that is used as a step in larger proofs of moreinteresting results.

A Corollary Corollaries are interesting in their own right, but follow withvery few extra steps after the underlying theorem is proved.

3.1 Direct Method

Many theorems are of the form if p then q . This is the standard form of

theorems, though it can be disguised. To prove p ⇒ q , we assume p as wellas other established truths and prove q .Most proofs are and should be direct proofs. Always try direct proofs first,unless you have a good reason not to.

Examples.

Theorem 1. The sum of two odd integers is even.

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Proof. An integer x is odd if we can write x = 2k + 1 for some k

∈Z and x

is even if x = 2k for some k ∈ Z.Let x and y be odd integers. Then x = 2k + 1 and y = 2l + 1 for some k, l∈ Z. Now,

x + y = 2k + 1 + 2l + 1

= 2k + 2l + 2

= 2(k + l + 1)

= 2z where z = k + l + 1

Theorem 2. If r1 and r2 are distinct roots of the polynomial p(x) = x2+bx+c

then r1 + r2 = −b and r1r2 = c.Proof. From our assumption, p(x) factors into linear equations. Now,

p(x) = x2 + bx + c (1)

andP (x) = (x

−r1)(x

−r2) = x

2

−(r1 + r2)x + r1r2 (2)

Comparing corresponding equation in (1) and (2) we have that r1 + r2 = −band r1r2 = c.

Theorem 3. Let A and B be sets. If A ∪ B = A ∩ B then A ⊆ B.Proof. Assume A ∪ B = A ∩ B. We show that x ∈ A ⇒ x ∈ B .Let x ∈ A. Since A ⊆ A ∪ B, we have that x ∈ A ∪ B. But A ∪ B = A ∩ B.So x ∈ A ∩ B. That is, x ∈ A and x ∈ B . This implies that x ∈ B . Hence,A

⊆ B .

3.2 Indirect Method

When a theorem cannot be proved by direct methods, we try an indirectmethod.

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3.2.1 Proof by Contraposition

If the dog is dead, he smells If he doesn’t smell, he is not dead.

Sometimes the direct route is just too difficult to deduce. In that case we doan indirect proof. The contrapositive of p ⇒ q is ¬q ⇒ ¬ p. We do a directproof on the contrapositive.

Example.

Theorem 4. If 3x + 7 is even, then x is odd.

Proof. We prove a contrapositive statement: If x is even then 3x + 7 is odd.

Suppose x is even, then x = 2k for some k ∈ Z. Then 3x + 7 = 3(2k) + 7 =6k + 7 = 6k + 6 + 1 = 2(3k + 3) + 1. Since 3k + 3 is an integer, we have that3x + 7 is odd.

Theorem 5. If x2 is even, then x is even.

Proof. We prove the contrapositive statement: If x is odd, then x2 is odd.

Suppose x is odd, then x = 2k + 1 for some k ∈ Z. Thus, x2 = (2k + 1)2 =4k2 + 4k + 1 = 2(2k2 + 2k) + 1 which is odd.

3.2.2 Proof by Contradiction

The basic structure of a proof by contradiction is this: Assume that theconclusion of your statement is false. Find a contradiction.

Suppose you want to prove a statement of the form p ⇒ q , Assume that p ⇒ q is false. This only happens when p is true and q is false. So we assumethat p is true and q is false and then we proceed to find a contradiction.

Examples.

Theorem 6. The square root of √ 2 is irrational.Proof. We prove this by contradiction. Suppose

√ 2 is rational. Then

√ 2 =

a

b; b = 0.

Assume that a and b have no common divisors (factors). That is, ab

is in itslowest term. Then.

√ 2 =

a

b ⇒ 2 = a

2

b2 ⇒ a2 = 2b2.

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Thus, a2 is even. This implies that a is even. Since a is even, we can write

a = 2k for some k ∈ Z. So,a2 = 2b2 ⇒ 4k2 = 2b2 ⇒ b2 = 2k2.

Thus, b2 is even implying that b is even.

Thus a and b are both even, that is, they have a common factor of 2. Thisleads to a contradiction. Hence,

√ 2 is irrational.

Example. A diophantine equations: equations for which you seek only integersolutions.

Theorem 7. There are no positive integer solutions to x2 − y2 = 1.Proof. Suppose for contradiction that there are positive integers x, y suchthat x2 − y2 = 1. Then, x2 − y2 = (x − y)(x + y) = 1.Since x, y ∈ Z, then EITHER x − y = 1 and x + y = 1 OR x − y = −1 andx + y = −1. If the first condition holds then x = 1 and y = 0 and we get acontradiction. If the second condition holds, then x = −1 and y = 0 and wealso get a contradiction. Thus, the original statement must be true.

3.3 Proof by Cases (Exhaustion)We use this method of proof when a hypothesis naturally divides itself intodifferent cases. We divide what we want to prove into cases and prove eachseparately. We need to make sure that all the cases are exhausted.

Theorem 8. For a, b ∈ R, then |ab| = |a||b|.Proof. Four cases arise:

1. a > 0, b > 0 then ab > 0, |a| = a, |b| = b. Then |ab| = |a||b|.2. a > 0, b < 0 then ab < 0, |a| = a, |b| = −b. Then |ab| = −ab =

a · −b = |a||b|.3. a < 0, b > 0 then ab < 0, |a| = −a, |b| = b. Then |ab| = −ab =

−a · b = |a||b|.4. a < 0, b < 0 then ab > 0, |a| = −a, |b| = −b. Then |ab| = ab =

(−a)(−b) = |a||b|.

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Theorem 9. Let x

∈Z, then x2 + x is even.

Proof. Two cases arise

1. If x is even, then x = 2n for some n ∈ Z. Thus,

x2 + x = 4n2 + 2n = 2(2n2 + n)

which is even.

2. If x is odd, then x = 2n + 1 for some n ∈ Z. Thus,

x2 + x = (2n + 1)2 + 2n + 1 = 4n2 + 6n + 2 = 2(2n2 + 3n + 1)

which is even.

Since every integer must be either even or odd but not both, we have ex-hausted all the cases.

3.4 Existence Proofs

Existence proofs are proofs of the form ∃x, P (x). Here we aim to find a csuch that P (c) holds. There are three ways to show the existence of such c.

1. Constructive Here we construct such a c. In otherwords, we demon-strate how to actually find or construct a specific element c for whichP (c) is true.

Example. A square exists that is a sum of two squares.

Proof. Pythagorean triple. We find a square 25 and we show that it isa sum of two squares 9 and 16. That is, 32 + 42 = 52.

2. Nonconstructive Here we show that such a c exists but we do notactually find it.

3. Contradiction Here, we assume that no such c exists and then wearrive at a contradiction.

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3.5 Uniqueness proofs

A theorem may state that there is only one value c such that P (c) holds. Weprove such a theorem in two parts. Existence: We first show that such a valuedoes indeed exist using any of the methods of the last section. Uniqueness:We show that there is only one such value.

Example.

Theorem 10. If 5x + 3 = 0 has a solution in R, then it is unique.

Proof. We first find a solution. Solving 5x+3 = a yields the solution x = a−35

.Question: Is this a constructive or nonconstructive proof?

Next, we show that if there are two solutions x and y to this equation, thenthey must be equal. So we equate, 5x+3 = a = 5y +3. Thus, 5x+3 = 5y +3and this yields that x = y. And thus, we have proved uniqueness of thesolution.

3.6 Proof of Universal statements

Universal statements are of the form: ∀x ∈ D, P (x). We prove such state-ment using

1. Direct proof: Take an arbitrary element x in D. Then show that P (x)

is true.

2. Contradiction: Assume that there is some c ∈ D such that P (c) is false.Show that a contradiction results.

3.7 Proof of Statements of the form p ⇒ (q ∨ r)To prove statemetnts of this form, instead of attempting to prove it directly,we replace this problem by the problem of showing that r follows from pand ¬q : that is p ∧ (¬q ) ⇒ r OR that q follows from p and (¬r): that is, p

∧(

¬r)

⇒ q . Either approach would do.

Example

Theorem 11. For all x, y ∈ R, if xy = 0 then x = 0 or y = 0.Proof. Let x, y ∈ R and x = 0, we prove that y = 0. Since x = 0, thereciprocal 1

x exists and we have

xy = 0 ⇒ 1x · xy = 1

x · 0 ⇒ y = 0.

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3.8 Principle of Mathematical Induction

Mathematical induction is a powerful technique used for proving statementsabout large sets, such as the set of natural numbers. If we can prove that ev-ery natural number n has a certain property P , then we have proved infinitelymany theorems P (1), P (2), P (3), . . ., etc. The principle of mathematical in-duction allows us to do this in finitely many steps. Let P (n) denote “theinteger n has property P ”.

The principle of mathematical induction (1st version) Suppose P isa property of the natural numbers for which the following statements hold:

1. P (1) is true.

2. for each k ∈ N, the following implication is true: P (k) =⇒ P (k + 1).Then, P (n) holds for every n ∈ N.In otherwords, it is enough to show that 1 has the property, and to show thatwhenever some number k has the property then k + 1 also has the property.P (1) is called the base step and P (k), the induction step.

Example. Prove that 1 + 2 + . . . + n = n(n+1)2

.

Proof. put P (n) : 1 + 2 + . . . + n = n(n+1)

2 .Base step: P (1) : 1 = 1(1+1)

2 = 1

Induction step: Suppose that P (k + 1) holds. That is suppose

1 + 2 + . . . + k = k(k+1)2

.We show that P (k + 1) holds:

P (k + 1) : 1 + 2 + . . . + k + k + 1 = k(k + 1)

2 + k + 1

= 1

2(k + 1)(k + 2)

=

1

2 (k + 1)((k + 1) + 1)

So P (k + 1) holds. Hence, P (n) holds for all n ∈ N.

Example. Find the sum of the first n odd natural numbers. Prove your resultby mathematical induction.

Solution:The odd natural numbers are 1, 3, 5, 7, . . . where the nth term is given by

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U n = 1 + (n

−1)2 = 2n

−1. Their sum to n terms is

S n = n2 (2a + (n − 1)d) = n2 (2 + (n − 1)2) = n2 · 2n = n2.Set P (n) : 1 + 3 + 5 + 7 + . . . + (2n − 1) = n2. Then,P (1) : 1 = 12

Suppose P (k) holds, that is 1 + 3 + 5 + 7 + . . . + (2k − 1) = k2. Then,

P (k + 1) : 1 + 3 + 5 + 7 + . . . + (2k − 1) + (2(k + 1) − 1) = k2 + 2k + 2 − 1= k2 + 2k + 1

= (k + 1)2

So P (k + 1) holds. Hence, P (n) holds for all n ∈ NExample. Prove that 1 + r + r2 + . . . + rn = 1−r

n+1

1−r ; r = 1 for all n ∈ N.

Proof. P (1) : 1 = 1−r1−r

= 1

P (k) : Suppose 1 + r + r2 + . . . + rk = 1−rk+1

1−r P(k+1):

1 + r + r2 + . . . + rk + rk+1 = 1 − rk+1

1 − r + rk+1

= 1 − rk+1 + (1 − r)rk+1

1

−r

= 1 − rk+2

1 − r=

1 − r(k+1)+11 − r

So, P (n) holds for all n ∈ N and r = 1 by induction.

Example. Prove by induction: for all n ≥ 1, 13 + 23 + . . . + n3 =n(n+1)

2

2.

Proof. P (1) : 1 = 1·222 = 1.

P (k). Suppose 13 + 23 + . . . + k3 =k(k+1)

2

2.

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Then, P (k + 1):

13 + 23 + . . . + k3 + (k + 1)3 =

k(k + 1)

2

2+ (k + 1)3

= k2(k + 1)2

4 + (k + 1)3

= (k + 1)2

k2

4 + (k + 1)

= (k + 1)2

k2 + 4k + 4

4

= (k + 1)2(k + 2)2

4

=

(k + 1)(k + 2

2

2

=

(k + 1)((k + 1) + 1)

2

2

So P (k + 1). Hence P (n) holds for all n ≥ 1.

Exercises. Prove:

1. 12 + 22 + . . . + n2 = n(n+1)(2n+16 for all n ∈ N.2. for every integer n ≥ 2,

1

1 · 2 + 1

2 · 3 + . . . + 1

(n − 1)n = 1 − 1

n.

3.

3.9 Pigeon Hole Principle

THis is often used in proofs of existence.

Suppose you have n pigeon holes, but more than n pigeons. If you are toput all your pigeons in the holes then the pigeon hole principle says that inat least one hole you will have to put more than one pigeon.

Mathematically,Let A be a finite set with order |A| = n. Pigeon hole principle: If A andB are finite sets with |A| > |B| and f : A → B is a function, then f is notone-to-one. That is, there exists x, y ∈ A, x = y such that f (x) = f (y).

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3.10 Counter Examples and General Comments

Given a universally quantified statement, find a single example which is nottrue. This is disproving a universal statement by a counter example.

Every integer is the square of another integer. FalseCounter example: the square root of 5 is 2.236 which is not an integer.

Note:

1. You CANNOT prove a statement by an example.

2. You can only prove an existential by an example.

3. You can only disprove a universal by an example.

3.11 Exercises

Where there are problems, there is life.– Zinoviev (1980)

1. Prove by direct method

(a) For any two positive numbers x and y, √

xy ≤ x+y2

.

(b) If a, b ∈ R then 2ab ≤ z 2 + b2.(c) The sum of two rational numbers is rational.(d) If r1, r2, r3 are three distinct roots of a polynomial p(x) = x

3 +bx2 + cx + d then r1r2 + r1r3 + r2r3 = c.

(e) Every odd integer is the difference of two squares.

2. Prove by contraposition

(a) If 3n5 is an odd integer, then n is an odd integer.

(b) If x, y ∈ Z are such that xy is odd then both must be odd.(c) Let a, b

∈R, if ab is an irrational number, then either a or b must

be an irrational number.

(d) For any integers a and b, we have that a + b ≥ 15 implies thata ≥ 8 or b ≥ 8.

3. Prove by contradiction.

(a) The number 3√

2 is irrational.

(b) There are no positive integer solutions to the diophantine equationx2 − y2 = 10.

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(c) If a

∈Q and b /

∈Q then a + b /

∈Q.

(d) There do not exist integers x and y such that 9 = 4x + 2y.

4. Proof by cases: Prove that for all n ∈ Z, we have n2 − n is even.5. Prove by contraposition: If f is increasing on an interval I , then f is

one-to-one on I .6. Prove: If x,a, b, ∈ R such that xa = xb, then either x = 0 or a = b.7. Given A and B sets such that A ⊆ B and A C . Prove that B C .8. Prove or disprove: If x, y

∈R are irrational then x + y is irrational? xy

is irrational?

9. Use direct method and contraposition to prove: Let A, B and C be setswith A ⊆ B and B ⊆ C . Show that A ⊆ C . Compare the effectivenessof each approach.

10. Describe the general approach you would take to prove these theorems.

(a) If S ⊆ R is compact, then S is closed and bounded.(b) If A and B are sets such that A ∪ B is finite, then A is finite and

B is finite.

(c) THe subset 2Z of R consisting of all even integers is closed underaddition and multiplication.

Note that it is not necessary to know the meaning of technical termssuch as “bounded” in order to solve this question.

4 Relations

4.1 Ordered pairs and cartesian products

An ordered pair (a, b) is such that a is the first element and b is the secondelement. Order is important in the sense that (a, b) = (b, a). Ordered pairsare also called 2-tuples. Note also that (a1, b2) = (a2, b2) if and only if a1 = a2and b1 = b2.

Let A and B be sets. The Cartesian product A × B consists of all orderedpairs (a, b) with a ∈ A and b ∈ B . That is,

A × B = {(a, b) : a ∈ A, b ∈ B}.

Examples.

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1. Let A =

{3, 4, 5

}and B =

{b, c

}. Then A

×B =

{(3, b), (3, c), (4, b), (4, c), (5, b), (5, c)

}and B × A = {(b, 3), (b, 4), (b, 5), (c, 3), (c, 4), (c, 5)}.2. R2 = R×R = {(x, y) : x, y ∈ R}.

Cartesian product is not commutative: A × B = B × A except when1. A = B. Example A = B = {a, b} Then, A × B = B × A =

{(a, a), (a, b), (b, a), (b, b)}2. OR when A or B is empty. Example: A = {1, 2} and B = ∅. Then,

A × B = B × A = ∅.

Cartesian product is not associative. Exercise: Prove that (A × B) × C =A × (B × C ). (Hint: find a counter example).If n(A) is the number of elements in A. Then, n(A × B) =n(A)n(B). Also,n(A × B) is infinite if either n(A) is infinte or n(B) is infinite and the otherset is not empty.

4.2 Relations

In Mathematics, there are lots of ways in which an entity may be related to

another. Consider the following statements 5 < 10, 5|80, x = y , 6 ∈ Z. Thesymbols < . |, =, are relations. They express some relationship betweentwo entities.

Motivation

Consider the set A = {1, 2, 3, 4, 5}. There is nothing special about this set.It just serves as an example. Compare elements in A by “


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• (a, b) /

∈ R that is a

Rb; a is not related to b.

Thus, R defines uniquely a subset of A × B by R = {(a, b) : aRb}. On the other hand, any subset R of A × B defines a relation by: aRb if and only if (a, b) ∈ R. Thus, there is a correspondence between subsets of A × B and relations from A to B.

Definition 4. A relation R from A to B is a subset of A × B.

The domain of R is the set of all first elements of the ordered pairs belonging to R. Domain of R = {a : (a, b) ∈ R} ⊂ A}. The Range of R is the set of all second elements of the ordered pairs belonging to R. Range of R = {b : (a, b) ∈ R} ⊂ B}.

A relation R from a set A to itself is called a relation on A and is a subsetof A × A.Example. Let A = {1, 2, 3, 4}. LetR = {(1, 1), (2, 1), (2, 2), (3, 3), (3, 2), (3, 1), (4, 4), (4, 3), (4, 2), (4, 1)} ⊆ A×A.Then R is a relation on A. Note that (2, 2) ∈ R, so 2R2.

Let A = {1, 2, 3, 4, 5, 6}. Let R be the relation “divides” on A. That is,(a, b) ∈ R if and only if a|b. Then, we have the setR = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 4), (2, 6), (2, 2), (3, 3), (3, 6), (4, 4), (5, 5), (6, 6)}Pictorially

Example. Let S be a set. Then S × S is the universal relation on S and ∅ isthe empty relation on S . These relations are not interesting.

Example. An important relation on the set A is that of equality. That is

R = {

(a, a) : a ∈

A}

This is called the identity or diagonal relation on A.

Let R be a relation from A to B. The inverse of R, denoted by R−1 is therelation from B to A which is given by R−1 = {(b, a) : (a, b) ∈ R}. Note thatthe domain of R is the range of F −1 while the range of R is the domain of R−1.

Example. R = {(a, b), (c, d), (a, a), (c, c)}. Then, R−1 = {(b, a), (d, c), (a, a), (c, c)}.Properties of Relations

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A relation xRy is a statement. It is either True or False. Since relational

expressions have true or false values, we can combine them with logical op-erators. Example. aRb ⇒ bRa is a statement whose truth value depends onthe truth value of a and b.

Definition 5. Let R be a relation on a set S . Then,

1. R is reflexive if (x, x) ∈ R ∀ x ∈ S that is, xRx, ∀ x ∈ S .2. R is symmetric if (x, y) ∈ R ⇒ (y, x) ∈ R ∀ x, y ∈ S that is,

xRy ⇒ yRx.3. R is transitive if (x, y)

∈ R and (y, z )

∈ R

⇒ (x, z )

∈ R

∀ x, y,z

∈ S

that is, xRy and yRz ⇒ xRz .4. R is antisymmetric if (x, y) ∈ R and (y, x) ∈ R ⇒ x = y ∀ x, y ∈ S

that is, xRy and yRx ⇒ x = y.

Example. A = Z.

1. Reflexive: ≤, =, |.2. Not Reflexive: >, ,, =, .3. Symmetric:

=, =.

4. Not symmetric: ≤.5. Transitive ≤, , =.

Let A = {b,c,d,e}. Then the relation R on A defined by(b, b), (b, c), (c, b), (c, c), (d, d), (b, d), (d, b), (c, d), (d, c)

is not reflexive, is symmetric and is transitive.

Partitions

A partition of a set A is a set of nonempty subsets of A, such that the unionof all subsets equals A and the intersection of any two different subsets is ∅.In otherwords, A partition

of a set A is a family

= {Ai : i ∈ I } of

nonempty subsets of A such that

1. every element of A belongs to some Ai. So,i∈I

Ai = A

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2. Ai

∩A j =

∅ for i

= j That is, they are pairwise disjoint or nonoverlap-

ping.

The Ais are called the blocks of partition.

Example. A = {a,b,c,d}. One partition of A is {{a, b}, {c}, {d}} whereA1 = {a, b}, A2 = {c} and A3 = {d}. Note that, A1 ∪ A2 ∪ A3 = A andA1 ∩ A2 = ∅, a1 ∩ A2 = ∅ and A2 ∩ A3 = ∅.Other partitions of A are

1. {a, b}, {c, d}}

2. {{a, c}, {b}, {d}}3. {{a}, {b}, {c}, {d}}4. {{a,b,c,d}}.

Example. The oddness or evenness of an integer is called its parity . Therelation “having the same parity as” leads to a partition of Z into two blocks

1. A1 = set of odd integers

2. A2 = set of even integers.

every integer is either odd or even and no element is both odd and even.

4.3 Equivalence Relation

A relation R on a set A is called an equivalence relation if and only if it isreflexive, symmetric and transitive. One frequently denotes an equivalencerelation by ∼.Example.

1. “is equal to” on the set of real numbers R.2. “has the same parity as” on the set of integers Z.

3. R = {(−1, −1), (1, 1), (2, 2), (3, 3), (4, 4), (1, 3), (3, 1), (2, 4), (4, 2)}4. “is congruent to” on the triangles in the plane.

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Example. Similarity of matrices.

Let A and B be two 2 by 2 matrices with entries in R. Define an equivalencerelation by A ∼ B if there is an invertible matrix P such that P AP −1 = B .Example. With

A =

1 2−1 1

, B =

−18 33−11 20

and P =

2 51 3

We see that A ∼ B since P AP −1 = B. Now let I be the 2 by 2 identitymatrix.

1. The relation is reflexive:IAI −1 = I −1AI = A, thus A

∼ A.

2. The relation is symmetric: A ∼ B implies that there exists an invertiblematrix P such that P AP −1 = B. So, this gives, A = P −1BP =P −1B(P −1)−1. Thus, B ∼ A.

3. Transitivity: A ∼ B and B ∼ C implies that there exists invertiblematrices P and Q such that P AP −1 = B and QBQ−1 = C . Thus,C = QBQ−1 = QPAP −1Q−1 = (QP )A(QP )−1. Hence, A ∼ C .

Two matrices that are equivalent in this manner are said to be similar.

Definition 6. Let R be an equivalence relation on a set A. Given any ele-ment a ∈ A, the equivalence class of a is the set containing all the elements of A that are related to a.

[a] = {x ∈ A : xRa} = {x ∈ A : (a, x) ∈ R}.If ∼ is an equivalence relation on A, then the set of representatives is a subset of A which contains exactly one element from each equivalence class. The set of representatives is denoted by S/R and called the quotient set.

Properties of equivalence classes:Theorem 12. Let [x] and [y] be two equivalence classes. Then,

1. x ∈ [x] i.e. an equivalence class contains its representatives.2. x ∼ y ⇔ [x] = [y] i.e. if two elements are related then their equiva-

lence classes are equal.

3. [x] ∩ [y] = ∅ or [x] = [y]. i.e. two equivalence classes are either disjoint or equal.

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Proof. (1) and (2) are left as exercises.

We prove the third property. Notice that this is a proof of the type p ⇒ (q ∨r),where

• p : [x] and [y] are equivalence classes.• q : [x] ∩ [y] = ∅.• r : [x] = [y].

Thus we prove instead, p and ¬q implies r.Now, Suppose [x] and [y] are equivalence classes and that [x]

∩[y]

=

∅. Then,

there exists z ∈ [x] ∩ [y]. This implies that z ∼ x and z ∼ y. Thus, x ∼ z and z ∼ y by symmetry. By transitivity, this yields, x ∼ y. Hence, by (2),we have that [x] = [y].

Example. Integers modulo nWe know that 2 divides 10, i.e. 2|10, because we can write 10 = 2 × 5. Whenwe write a|b, we mean that a divides b without any remainder. If there is aremainder, then a does not divide b. More formally, a|b if and only if thereis an integer k such that b = ka.

Fix a natural number n. Define a relation “congruence modulo n” denotedby ≡ on the set of integers Z by a is congruent to b modulo n if and only if n divides a − b. That is,

a ≡ b( mod n) if and only if n|(a − b).

Claim: The relation “is congruent to modulo n” is an equivalence relationon Z.

Proof. We show that the relation is reflexive, symmetric and transitive.

1. Reflexivity: Let x ∈ Z. Observe that n|0. So, n|(x − x). Thus x ≡ xmod n for every integer x.

2. Symmetry: Direct proof: Suppose x ≡ y mod n. Then by definition,n|(x − y). This implies that x − y = na for some integer a. But thisgives that y − x = n(−a) for some integer −a. Hence y ≡ x mod n.

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3. Transitivity: Suppose x

≡ y mod n and y

≡ z mod n. Then by

definition, n|(x − y) and n|(y − z ). That is, x − y = kn and y − z = lnfor some integers k and l. Thus,

x − z = x − y + y − z = (x − y) + (y − z ) = kn + ln = (k + l)n = tnwhere t = k + l ∈ Z. Hence, x ≡ z mod n.

Example. n = 3. We find the equivalence classes of the equivalence relation≡ mod 3.

1. Equivalence class of 0:

[0] = {x ∈ Z : x ≡ 0(mod 3)} = {x ∈ Z : 3|(x − 0)}= {x ∈ Z : 3|x} = {. . . , −3, 0, 3, 6, 9, . . .}

Thus the class of zero, [0] is the set of all multiples of 3. That is,all integers which leave a remainder of 0 on division by 3. Note that:x ∼ y ⇔ [x] = [y]. Thus, [0] = [3] = [6]

2. Equivalence class of 1:

[1] = {x ∈ Z : x ≡ 1(mod 3)}= {x ∈ Z : 3|(x − 1)}= {. . . , −5, −2, 1, 4, 7, 10, . . .}

Thus, the class of 1, [1] is the set of integers which leave a remainder 1on division by 3.

3. [2] = {. . . , −4, −1, 2, 5, 8, 11, . . .}. Thus, the class of 2, [2] is the set of integers which leave a remainder 2 on division by 3.

The equivalence classes are {[0], [1], [2]}. These form a partition of Z :{{. . . , −3, 0, 3, 6, 9, . . .}, {. . . , −5, −2, 1, 4, 7, 10, . . .}, {. . . , −4, −1, 2, 5, 8, 11, . . .}}.A set of representatives is {0, 1, 2}.Similarly, we can show that the equivalence relation

≡ mod n has n equiva-

lence classes: [0], [1], [2], . . . , [n − 1]. These form a partition of Z.Note: There are infinitely many different partitions of Z; one for each n ∈ N.

The next result gives a correspondence between equivalence relations andpartitions. We state the following theorem without proof.

Theorem 13. Given an equivalence relation ∼ on a set A, the equivalence classes of A form a partition of A. Conversely, if

= {Ai} is a partition of

A, then there is an equivalence relation on A with equivalence classes Ai.

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4.4 Exercises

1. Define a relation R on R2 by (a, b) ∼ (c, d) if and only if a2+b2 ≤ c2+d2.Show that ∼ is reflexive and transitive but not symmetric.

2. Define a relation on R2 by (x−1, y1) ∼ (x2, y2) if and only if x−12+y21 =x22 +y

22. Show that ∼ is an equivalence relation. Describe geometrically

the equivalence classes.

3. Determine whether or not the following are equivalence relations onthe given set. If it is an equivalence relation, describe the partition, if it is not, state why it fails to be one

(a) x ∼ y in R if x ≥ y.(b) m ∼ n in Z if mn > 0.(c) x ∼ y in R if |x − y| ≤ 4.(d) m ∼ n in Z if mn( mod 6).(e) m ∼ n in Z if m − n > 0.(f) m ∼ n in Z if m + n = 10.(g) m ∼ n in Z if m + n is even.(h) m

∼ n in

Z if gcd(a, b) = 1.

(i) R = {(0, 0), (1, 1), (1, 2), (2, 1), (2, 2), (3, 3)}(j) R = {(0, 0), (1, 1), (2, 2), (3, 3)}.

4. Create as many possible partitions of the integers Z as you can.

5. Let A = {1, 2, 3, 4}. Give a relation R on A that is(a) Reflexive and symmetric but not transitive.

(b) Reflexive and transitive but not symmetric.

(c) Symmetric and transitive but not reflexive.

6. Let A be the set of all 2 by 2 matrices with entries in R. Define arelation on A by

A ∼ B if there is an invertible matrix C such that B = C A.Show that ∼ is an equivalence relation on A.

7. Let R be the relation “is a cousin of”: xRy if x is a cousin of y. Is Rsymmetric? transitive?

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8. Let A =

{1, 2, 3, . . . , 20

}. Define a relation on A by aRb if a and b

have the same prime divisors. Give the set describing R. RepresentR pictorially. show that R is an equivalence relation. Describe theequivalence classes of the corresponding partition of A.

9. Find all the equivalence classes of the equivalence relation “≡ mod 5”.

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NumberTheory by Frances Odumodu NAU

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