Notes for Math 290 using Introduction to Mathematical Proofs byCharles E. Roberts, Jr.

Chapter 1: Logic

Topics:

1.1 Statements, Negation, and Compound Statements1.2 Truth Tables and Logical Equivalences1.3 Conditional and Biconditional Statements1.4 Logical Arguments1.5 Open Statements and Quantifiers

Notes:

Statement: a sentence that can be assigned a truth value. Example: 5 is anodd number. Example: 6 is an odd number.

Negation: If P represents a statement, ∼ P represents the statement withthe opposite truth value.

Compound statements can be formed with the logical connectives and, or,implies, if and only if.

The truth value of a compound statement is determined by truth tables.These are arbitrary rules which everyone agrees on.

In P =⇒ Q, P is called the hypothesis and Q is called the conclusion.

Equivalences: for example, DeMorgan’s Laws.

Statements derived from P =⇒ Q: Converse, inverse, contrapositive.

Necessary versus sufficient. Avoid using vague terms subject to misinterpre-tation.

Logical argument: Given premises P1 through Pn, deduce conclusion C. Theargument is valid if and only if the statement P1 ∧ · · · ∧ Pn =⇒ C is truefor all truth values assigned to the variables.

Illustrate by proving that the sum of two odd numbers is an even number.Premises: m and n are odd. Conclusion: m+ n is even.

A quantified statement P (x, y, z, . . . ) whose truth depends on the value ofthe variables x, y, z, . . . .

Associated ideas: universe of variable values, universal and existential state-ments and their negation.

1

Chapter 2: Deductive Mathematical Systems and Proofs

Topics:

2.1 Deductive Mathematical Systems2.2 Mathematical Proofs2.2.1 Techniques for proving P =⇒ Q2.2.2 Additional Proof Techniques2.2.3 Conjectures, Proofs, and Disproofs2.2.4 The System of Rational Numbers and the System of Real Numbers

Notes:

Many mathematical theorems are stated in the following form:

Theorem: ∀n : P (n) =⇒ Q(n).

Proof: Let n be an arbitrary element of the universe. Assume P (n) is true.Then argue that Q(n) is true.

Define even numbers as natural numbers ending in a digit in {0, 2, 4, 6, 8}.Define odd numbers as natural numbers ending in a digit in {1, 3, 5, 7, 9}.Every number is either even or odd, and cannot be both.

Here are some typical theorems about natural numbers:

Theorem: Every even natural number can be written in the form 2k forsome k ∈ N.

Logical Form: ∀n : n is even =⇒ ∃k ∈ N : n = 2k.

Proof: Let n be given. If n is odd, the implication is true. If n is even, thenby definition n = 10a + b where b ∈ {0, 2, 4, 6, 8}, and in every case n = 2kfor some k ∈ N.

Theorem: Every odd natural number can be written in the form 2k+ 1 forsome k ∈ N.

Logical Form: ∀n : n is odd =⇒ ∃k ∈ N : n = 2k + 1.

Proof: Similar.

Theorem: The square of an even number is even.

Logical form: ∀n ∈ Z : n is even =⇒ n2 is even.

Proof: Let n be given. If n is odd, the implication is true. If n is even, wemust show that n2 is also even. It is: n2 = (2k)2 = 2(2k2).

2

This is a direct proof.

Theorem: If the square of a number is even, the number must be even.

Logical Form: ∀n : n2 even =⇒ n even.

Proof: Let n be given. Assume n2 is even. If n is odd then n2 = (2k+1)2 =4k2 + 4k + 1 is odd, a contradiction, therefore n is even.

The idea behind proof by contradiction is that a statement must be true orfalse. If a choice of truth value forces another statement to have both truthvalues, the choice is wrong.

Another way to prove this theorem is to note that it is logically equivalentto its contrapositive, then to prove the contrapositive statement.

Rational numbers are quotients of integers. Real numbers are numbers withdecimal expansions.

Theorem:√

2 is not a rational number.

Logical Form: ∀a : ∀b 6= 0 :√

2 6= ab.

Proof: Let a and b be given. Suppose√

2 = ab. Let k be the largest common

factor of a and b, so that a = kA and b = kB where A and B have no commonfactor. Then

√2 = A

B, and after cross-multiplying and examining the results,

we can see that A and B have common factor 2, a contradiction. Therefore√2 6= a

b.

3

Chapter 3: Set Theory

Topics:

3.1 Sets and Subsets3.2 Set Operations3.3 Additional Set Operations3.4 Generalized Set Union and Intersection

Terminology: set, element. List elements or give rule for belonging. When xis a variable and P (x) is a statement about x, then by {x : P (x)} we meanthe set of all x for which P (x) is true. Empty set.

Russell’s Paradox: The set A = {1} satisfies A 6∈ A. Now consider theexpression S = {A|A 6∈ A}. Then either S ∈ S or S 6∈ S, both of which leadto a contradiction. So our idea of what constitutes a set is not well-defined.

Interval notation.

Union, intersection, set difference, complement.

Set Inclusion: Quantified definitions of A ⊆ B and A = B. Logical proofsof set statements, for example DeMorgan’s Laws and others stated in reviewsection.

Indexed sets and their use in forming unions and intersections.

Ordered pairs and the cartesian product of two sets. Ordered pair definitionis (a, b) = {{a}, {a, b}}.

Power set.

4

Chapter 4: Relations

Topics:

4.1 Relations4.2 The Order Relations <, ≤, >, ≥4.3 Reflexive, Symmetric, Transitive, and Equivalence RelationsSupplementary: the ring Zn.

Notes:

We will skip most of Section 4.1 and Section 4.2, and introduce some sup-plementary material.

Let A and B be sets. A relation from A to B is any subset R of A × B.Domain, codomain, range.

Let A and B be sets. A function from A to B is relation f with the additionalproperty that if (a, b) ∈ f and (a, c) ∈ f then b = c. In this case, we can usethe notation f(a) to denote the second coordinate. We will study functionsin more detail in Chapter 5.

Relation on a set S: a subset R of S × S. aRb if and only (a, b) ∈ R. Oftenwritten a ∼ b. Reflexivity, symmetry, transitivity. Examples of relations: <,⊆, a|b, etc.

Equivalence relation: reflexive, symmetric, transitive. How to rememberthese properties: (1) Alphabetical order R,S,T. (2) Number of elements in-volved 1,2,3.

An equivelence relation: a ∼ b iff 3|(a− b).

Not an equivalence relation: A ∼ B iff A ⊆ B.

When ∼ is an equivalence relation, we can define equivalence classes.

Theorem: a ∼ b ⇐⇒ [a] = [b].

Corollary: The set of distinct equivalence classes partitions the set.

Example: Congruence modulo n partitions all integers into n equivalenceclasses. To decide which class a number falls in, find the remainder afterdivision by n. In the case of a negative number, add n.

The Ring Zn: Given [a] and [b] where 0 ≤ a, b ∈ {0, 1, . . . , n − 1}, define[a] + [b] = [a+ b] and [a][b] = [ab], where 0 ≤ a, b < n.

5

Theorem: The associative properties hold for addition and multiplication.

Proof: (1) We have [a] + ([b] + [c]) = [a] + [r] = [a+ r] and ([a] + [b]) + [c] =[s]+[c] = [s+c] where r, s ∈ {0, 1, . . . , n−1}. We must show [a+r] = [s+c].Write b + c − r = nj and a + b − s = nk. Then (a + r) − (s + c) =(a+ b+ c− nj)− (a+ b− nk + c) = nk − nj = n(k − j), as desired.

(2) We have [a]([b][c]) = [a][bc] = [a][r] = [ar] and ([a][b])[c] = [s][c] = [sc]where r, s ∈ {0, 1, . . . , n−1}. We must show [ar] = [sc]. Write bc−r = nj andab−s = nk. Then ar−sc = a(bc−nj)−(ab−nk)c = anj−nkc = n(aj−kc),as desired.

Theorem: The distributive property holds.

Proof: We have [a]([b]+[c]) = [a][b+c] = [a(b+c)] = [ab+ac] = [ab]+[ac] =[a][b] + [a][c].

Additive and multiplicative identities: [0] and [1].

Zero property: [0]x = [0] for all x ∈ Zn.

An equation: solve [2]x+ [3] = [4] when n = 5.

Solution: Adding [2] to both sides, [2]x = [6]. Multiplying both sides by [3],[6]x = [18]. Therefore [1]x = [18] therefore x = [18] = [3].

An equation with no solution: [2]x = [3] when n = 6. Reason: suppose thereis a solution. Multiplying through by [3], [0]x = [9], therefore [0] = [9], acontradiction.

6

Chapter 5: Functions

Topics:

5.1 Functions5.2 Onto Functions, One-to-One Functions, and One-to-One Correspondences5.3 Inverse of a Function5.4 Images and Inverse Images

Notes:

Function was defined in Chapter 4. Review the definition.

Domain, codomain, range, image, inverse image.

The quantified definitions of injective, surjective, and bijective function.

Composition of functions. Associative property. A composition of injectionsis injective. A composition of surjections is surjective. A composition ofbijections is bijective. (This last is useful for proving that two sets are inthe same cardinality class.) Cancellation laws. The inverse of a bijectivefunction.

7

Chapter 6: Mathematical Induction

Topics:

6.1 Mathematical Induction6.2 The Well-Ordering Principle and the Fundamental Theorem of Arith-metic

Notes:

Mathematical Induction: Let the universe be U = {n ∈ Z : n ≥ a}. Thestatement

∀n : P (n)

is equivalent to the statement

P (a) ∧ (∀n : P (n) =⇒ P (n+ 1)).

Base case is P (a), induction hypothesis is P (n).

Examples: Sum of consecutive integers. Sum of consecutive powers. Numberof subsets of a set.

Example: For x ≥ 0 and n ∈ N, (1 + x)n ≥ 1 + nx.

Strong Mathematical Induction: Let the universe be U = {n ∈ Z : n ≥ a}.The statement

∀n : P (n)

is equivalent to the statement

P (a) ∧ (∀n : P (1) ∧ · · · ∧ P (n) =⇒ P (n+ 1)).

Base case is P (a), induction hypothesis is P (1) ∧ · · · ∧ P (n).

Applications of Mathematical Induction: Some Number Theory

Review the definitions of natural number and integer.

Define greatest common divisor of two natural numbers.

Theorem: When a > b, gcd(a, b) = gcd(a− b, b).

Proof: Any common divisor of a and b is a common divisor of a− b and b,and vice-versa.

Theorem: Given a, b ∈ N there exist x, y ∈ Z such that gcd(a, b) = ax+ by.

8

Proof: By strong induction on a + b. Base case is trivial using a = b = 1,x = 1, y = 0. Assume true whenever a + b < n. Now suppose a + b = n. Ifa = b we can use x = 1, y = 0. If a > b, we have

gcd(a, b) = gcd(a− b, b) = (a− b)x+ by = ax+ b(y − x).

Use a similar argument if a < b.

Prime number: A natural number with exactly two divisors.

Corollary: When a, p ∈ N and p is prime and p 6 |a, there is a solution toax+ pb = 1.

Theorem: If p|ab and p is prime then p|a or p|b.Proof: If p|a, we’re done. If p 6 |a, write ab = pk. Then

px+ ay = 1

pxb+ aby = b

pxb+ pky = b

p|b.

Corollary: If p|a1 · · · an then p|ai for some i.

Proof: Induction on n.

Theorem: Every n ≥ 2 is prime or a product of primes.

Proof: By induction on n. Base case n = 2 is true. Assume for all integers≤ n. If n + 1 is prime, fine. If it is not prime, factor as two integers in{2, . . . , n}. Each factor prime or a product of primes, hence so is n+ 1.

Theorem: There are infinitely many primes.

Proof: If not, let the primes be p1, . . . , pn. Then x = p1 · · · pn+1 is divisibleby one of these. Contradiction.

Theorem: If p1, . . . , pm and q1, . . . , qn are primes and p1 · · · pm = q1 · · · qnthen both lists contain the same distribution of primes.

Proof: By induction on m. When m = 1, p1|qi for some i, which forcesp1 = qi. Dividing through by p1 we see that there are no other qj since theirproduct is greater than 1.

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Now assume true for a given m. Suppose p1 · · · pm+1 = q1 · · · qn. Then p1 = qifor some i. Dividing through by p1, we can use the induction hypothesis tosay that the remaining prime distributions are the same.

Theorem:√

2 is irrational.

Proof: Suppose√

2 = p1···pmq1···qn . Then 2q21 · · · q2n = p21 · · · p2m. This is impossible

because 2 appears an odd number of times on the left and an even numberof time on the right.

Theorem: (75)13 is irrational.

Proof: At some point we get to 3 · 52 · q31 · · · q3n = p31 · · · p3m. Impossiblebecause the left product contains 1 + 3j factors of 3 and the right productcontains 3k factors of 3, therefore 1 + 3j = 3k, a contradiction. Or: the leftproduct contains 2+3j factors of 5 and the right product contains 3k factorsof 5, therefore 2 + 3j = 3k, a contradiction.

10

Chapter 7: Cardinalities of Sets

Topics:

7.1 Finite Sets7.2 Denumerable and Countable Sets7.3 Uncountable Sets

Notes:

This chapter does not contain a chapter review.

Definition: A is a finite set if and only if there is a bijection f : A →{1, 2, . . . , n} for some n. Notation: |A| = n. Counting the elements of Agives rise to f .

Definition: A is an infinite set if and only if it is not a finite set.

Definition: A is a countably infinite set (denumerable set) if and only ifthere there is a bijection f : A→ N. Notation: |A| = ℵ0.Note that |A| = ℵ0 if and only if it is possible to list the elements of Asequentially.

Examples of countably infinite sets: N, Z, kN, N× N (one way to do this isto organize the pairs (a, b) by the size of a+ b).

Theorem: P (N) 6= ℵ0.Proof: Let S1, S2, S3, . . . be any infinite sequence of distinct subsets of N.Then the set X = {n ∈ N : n 6∈ Sn} is not any of these sets. So you cannotlist all the subsets of N sequentially.

Theorem: If a1, a2, a3, . . . is an infinite sequence, and if A = {ai : i ∈ N} isinfinite, then A is countably infinite.

Proof: The list of distinct elements in A is an1 , an2 , an3 , . . . , where n1 <n2 < n3 < · · · .Corollary: Q is countably infinite.

Proof: List the elements of N× N sequentially as

(a1, b1), (a2, b2), (a3, b3), . . . .

Then the sequencea1b1,a2b2,a3b3, . . .

11

contains all positive rational numbers, therefore Q+ = {aibi

: i ∈ N} is count-able, therefore Q is contable.

Theorem: If A is countably infinite and B ⊆ A is infinite, then B iscountably infinite.

Proof: If the elements of A are a1, a2, a3, . . . then the elements of B arean1 , an2 , an3 , . . . where n1 < n2 < n3 < · · · .Theorem: c 6= ℵ0.

Proof: If suffices to show that R contains a subset X that is not countablyinfinite.

Let X be the set of real numbers of the form 0.a1a2a3 · · · , where ai ∈ {0, 1}for each i. There is a one-to-one corerspondence between X and P (N). SinceP (N) is not countably infinite, neither is X.

Theorem: If |A| = ℵ0 and |B| = ℵ0 then |A ∪B| = ℵ0.Proof: The elements of A ∪B can be listed a1, b1, a2, b2, . . . .

Corollary: The set of irrational numbers is not countably infinite.

Proof: Since k√

2 is irrational for all k ∈ N, there are infinitely manyirrational numbers. Let I be the set of irrational numbers. If |I| = ℵ0 thensince R = I ∪Q, c = ℵ0, a contradiction. Therefore |I| 6= ℵ0.Theorem: If |A| = ℵ0 and |B| = ℵ0 then |A×B| = ℵ0.Proof:

Q×Q = {(i1, j1), (i2, j2), (i3, j3), . . . }A×B = {(ai1 , bj1), (ai2 , bj2), (ai3 , bj3), . . . }.

Equivalence class on sets: A ∼ B if and only if there is a bijection f : A→ B.

Definition: |A| = |B| if and only if A ∼ B. We say that A and B have thesame cardinality iff |A| = |B|.Definition: |A| ≤ |B| if and only if there is an injective function f : A→ B.So for example |N| < |P (N)| and |N| < c.

Some facts about cardinality:

(1) |A| ≤ |B| and |B| ≤ |C| implies |A| ≤ |C|. Reason: composition ofinjective functions.

12

(2) |A| = |B| and |B| = |C| implies |A| = |C|. Reason: composition ofbijective functions.

(3) |A| = |A′| and |B| = |B′| implies |A× B| = |A′ × B′|. Reason: it’s easyto construct a bijection.

(3) |A| ≤ |B| and |B| ≤ |A| implies |A| = |B|. This is called the Schroeder-Bernstein Theorem.

Proof of Schroeder-Bernstein Theorem: Let f : A→ B and g : B → Abe injective. Let A1 be the elements in A not in the range of g. Then setA2 = g(f(A1)), A3 = g(f(A2), A4 = g(f(A3)), etc. Define h : A→ B by

h(a) =

{f(a) a ∈ Ai for some i

g−1(a) a 6∈ Ai for all i.

h is injective: Suppose h(a1) = h(a2).

Case 1: f(a1) = f(a2). Then a1 = a2.

Case 2: f(a1) = g−1(a2). Then g(f(a1)) = a2. So we have a1 ∈ Ai for somei, which places a2 ∈ Ai+1, which is not possible. So this case does not occur.

Case 3: g−1(a1) = f(a2). Not possible either.

Case 4: g−1(a1) = g−1(a2). Then a1 = a2.

h is surjective: Let b ∈ B be given. Then g(b) 6∈ A1.

Case 1: g(b) 6∈ Ai for all i. Then h(g(b)) = g−1(g(b)) = b.

Case 2: g(b) ∈ Ai for some i ≥ 2. Then g(b) = g(f(a)) for some a ∈ Ai−1,therefore b = f(a).

Theorem: |[0, 1)| = |P (N)|.

Proof: P (N) is equinumerous with binary sequences, which can be mappedinjectively into [0, 1) using a decimal expansions. Conversely, each x ∈ [0, 1)has a decimal expansion, and we can uniquely encode 0.b1b2b3 · · · by a stringof 0s and 1s (first 10 digits encodes b1, second 10 digits encodes b2, etc), hence[0, 1) can be injectively mapped into binary sequences, hence into P (N).

Notation: |R| = c.

Theorem: c = |[0,1)|.

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Proof: [0, 1) can be bijectively mapped into (0, 1] via x 7→ 1 − x, and(0, 1] can be bijectively mapped into [0,∞) via x 7→ 1

x− 1. So we have

[0, 1) ∼ [0,∞). We also have (0,∞) ∼ R via x 7→ lnx. To complete theproof, we just need to show [0,∞) ∼ (0,∞). We use the following lemma.

Lemma: If X is infinite and x0 ∈ X then X ∼ X − {x0}.

Proof: Choose a sequence of elements x1, x2, x3, . . . distinct from x0. Definef : X → X − {x0} via

f(x) =

{x x 6= xi for all x

xi+1 x = xi.

One can check that f is a bijection.

Corollary: |P (N)| = c.

Cantor’s Theorem: For all sets A, |A| < |P (A)|.

Proof: The injection that establishes |A| ≤ |P (A)| is a 7→ {a}. Now suppose|A| = |P (A)|. Then we have P (A) = {f(a) : a ∈ A} for some bijectionf : A → P (A). Let X = {a ∈ A : a 6∈ f(a)} ∈ P (A). If X = f(b) thenb ∈ X ⇐⇒ b 6∈ f(b), so X 6= f(b), a contradiction. Hence |A| 6= |P (A)|.

It is not known whether or not there exists a set X such that ℵ0 < |X| < c.

Continuum Hypothesis: No such X exists.

Generalized Continuum Hypothesis: For all infinite sets A, there is noset X satisfying |A| < |X| < |P (X)|.

Exercise: R× R ∼ R

Solution: It suffices to prove B × B ∼ B, where B is the set of binarysequences. A bijection is

(a, b) 7→ (a1, b1, a2, b2, . . . ).

14

Chapter 8: Proofs from Real Analysis

Topics:

8.1 Sequences8.2 Limit Theorems for Sequences8.3 Monotone Sequences and Subsequences8.4 Cauchy Sequences

Notes:

There is no chapter review for this chapter. Also, we will skip the materialon Cauchy Sequences since it is not put to any use.

Trichotomy Law: for all a, b ∈ R, either a > b or a = b or a < b. These aremutually exclusive conditions.

The absolute value of a rational number:

|r| =

{r r ≥ 0

−r r < 0.

Distance between two rational numbers: |r − s|.Triangle Inequalities for Rational Numbers: |r + s| ≤ |r| + |s| and||r| − |s|| ≤ |r − s|.Proof: |r + s| = (r + s)θ = rθ + sθ ≤ |r|+ |s| where θ ∈ {−1, 1}. To provethe second inequality, for any u and v we have

|u+ v| − |v| ≤ |u|.

Setting u = r − s, v = s, we obtain

|r| − |s| ≤ |r − s|.

Reversing r and s we obtain

|s| − |r| ≤ |s− r| = |r − s|.

So||r| − |s|| ≤ |r − s|.

15

Limit of a sequence

Let xn ∈ R for all n, and let x ∈ R. Then we define

limn→∞

xn = x

if and only if for all ε ∈ R+ there exists N ∈ N such that

|xN − x| < ε, |xN+1 − x| < ε, |xN+2 − x| < ε, . . . .

In other words, |xn − x| < ε for all n beyond a certain point. Notation:xn → x.

Example: limn→∞1n

= 0. Let ε > 0 be given. We want to find N such thatn ≥ N implies 1

n< ε. We can use any natural number N ≥ 1

ε.

Example: If a ∈ R and 0 < |a| < 1 then limn→∞ an = 0.

Proof: Let ε > 0 be given. We wish to find N so that n > N implies |an| < ε,

or equivalently(

1|a|

)n> 1

ε. Write 1

|a| = 1 + b where b > 0. We proved earlier

that(

1|a|

)n= (1 + b)n ≥ 1 + nb. We wish to require 1 + nb > 1

ε. We just

need any natural N satisfying N ≥1ε−1b

=1ε−1

1|a|−1

.

Example: a = 99100

and ε = 1100

. Our formula yields N(an, ε) = 9801, a grossoverestimate according to the graph below:

100 200 300 400 500 600

0.05

0.10

0.15

0.20

0.25

0.30

a[n] =99

100

n

16

Notation: When (an) is a sequence which converges to a limit a we willdenote by N(an, ε) a number such that n ≥ N(an, ε) =⇒ |an − a| < ε. Wecan summarize our last two examples by saying

N(1

n, ε) =

1

ε

and

N(an, ε) =1ε− 1

1|a| − 1

.

Example of a sequence that does not have a limit: xn = (−1)n. Forany x ∈ R, every other value of |xn − x| is ≥ 1.

Properties of Convergent Sequences

Lemma: A convergent sequence is bounded from above and from below.

Proof: Suppose an → a. Then for n ≥ N , |an − a| < 1, which impliesa− 1 < an < a+ 1. Therefore for all n we have

min(a1, . . . , aN , a− 1) ≤ an ≤ max(a1, . . . , aN , a+ 1).

Sum rule: ak → a and bk → b implies ak + bk → a+ b. Details: |ak + bk −a− b| ≤ |ak − a|+ |bk − b| → 0. Say that n > N(an, ε) =⇒ |an− a| < ε andn > N(bn, ε) =⇒ |bn − b| < ε. Then we can use

N(an + bn, ε) = max(N(an,

ε

2

), N(bn,

ε

2

)).

Product rule: ak → a and bk → b implies akbk → ab. Details: |akbk−ab| =|akbk − abk + abk − ab| ≤ |ak − a||bk|+ |a||bk − b| ≤ |ak − a|B +A|bk − b| → 0where A and B are positive upper bounds for (|an|) and (|bn|). We can use

N(anbn, ε) = max(N(an,

ε

2B

), N(bn,

ε

2A

)).

17

Reciprocal rule: ak → a and ak, a 6= 0 implies 1ak→ 1

a. Details: | 1

ak− 1

a| =

|ak−a||a||ak|

≤ |ak−a|(1/2)|a|2 → 0, assuming |a− ak| ≤ |a|

2. We can use

N

(1

an, ε

)= max

(N

(an,|a|2

), N

(an,|a|2

2ε

)).

Quotient rule: ak → a and bk → b and bk, b 6= 0 implies akbk→ a

b. Details:

this follows from product rule and reciprocal rule. Assuming A is a positiveupper bound for (|ak|) and B is a positive upper bound for ( 1

|bk|), we can use

N

(anbn, ε

)= max

(N(an,

ε

2B

), N

(1

bn,ε

2A

)).

Example: Let xn = 1n2−5n+1

. To prove limn→∞ xn = 0 using an ε-N argu-ment, apply limit properties to

xn =1

n2

1

1− 5n

+ 1n2

.

We have N( 1n, ε) = 1

ε, and an upper bound for ( 1

n) is 1, hence using the

product rule we have

N(1

n2, ε) = N

(1

n,ε

2

)=

2

ε.

Note that for n ≥ 6, 1− 5n

+ 1n2 >

16, hence |xn − 0| < 6

n2 . So a solution is

N(xn, ε) = max(6,1

3ε).

Example: Let an = (2 + 1n)2. To prove that limn→∞ (2 + 1

n)2 = 4 using an

ε-N argument, observe that 2 + 1n≤ 3 for all n. We have

N(an, ε) = N((2 + 1/n)(2 + 1/n), ε) =

N(2 +1

n,ε

6) = N(

1

n,ε

6) =

6

ε.

Exercise: Compute N(

2n3+100n3−3n2−3 , ε

).

18

Exercise: Assume that (an) converges to a and that (|an|) has positiveupper bound A ≥ 1. Prove that for k ∈ N,

N(akn, ε) = N(an,ε

2k−1Ak−1)

(hint: use induction). Using this, find N so that

n > N =⇒ (2 +1

n)100 − 2100 <

1

10.

Theorem: Suppose that (xn) converges to a and (xn) converges to b. Thena = b.

Proof: If a > b then beyond a certain point |xn−a| < a−b2

and |xn−b| < a−b2

,which implies that beyond a certain point

a−b = |a−b| = |a−xn+xn−b| ≤ |a−xn|+ |xn−b| <a− b

2+a− b

2= a−b,

a contradiction. Therefore a ≤ b. Similarly, b ≤ a, therefore a = b.

Theorem: If an ≤ bn ≤ cn for all n and limn→∞ an = limn→∞ cn = L, thenlimn→∞ bn = L.

Solution: Let ε > 0 be given. Beyond a certain point an > L − ε andcn < L− ε, hence L− ε < an ≤ bn ≤ cn < L+ ε, hence |bn − L| < ε.

Bounded Monotonic Sequences

Definition.

Example: Let sn = 112

+ 122

+ · · · + 1n2 . The sequence (sn) is bounded and

monotonic increasing.

Reason:

s2k = 1+k−1∑i=0

1

(2i + 1)2+

1

(2i + 2)2+ · · ·+ 1

(2i+1)2≤ 1+

k−1∑i=0

2i

22i= 1+

1− 2−k

1− 2−1.

Hence the subsequence (s2k) is bounded above 1 + 11−2−1 , hence the entire

sequence is bounded above by this number since the sequence is increasing.The sequence is plotted below through s256.

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50 100 150 200 250

1.60

1.61

1.62

1.63

1.64

Example: Let sn = 11

+ 12

+ · · ·+ 1n. The sequence (sk) is unbounded.

Reason: We have

s2k = 1 +k−1∑i=0

1

2i + 1+

1

2i + 2+ · · ·+ 1

2i+1≥ 1 +

k−1∑i=0

2i

2i+1= 1 +

k

2.

Since (s2k) is unbounded, (sn) is unbounded and does not converge. Thesequence is plotted below through s256.

50 100 150 200 250

1

2

3

4

5

6

Theorem: Every increasing and bounded sequence in R converges to a limit.

Proof: Proved in Math 417.

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Therefore (∑n

k=11k2

) converges to a limit, which we call∑∞

k=11k2

. It can be

shown using methods of complex analysis (Math 511) that∑∞

k=11k2

= π2

6≈

1.64493406684822643647241516665.

Remark: More generally, every subset of R that has an upper bound hasa least upper bound in R. Hence R forms a complete ordered field. This isnot true of the rational numbers. For example, the decimal expansion of

√2

yields an increasing sequence of rational numbers bounded above by√

2 butdoes not have a rational least upper bound. In fact, the least upper boundis√

2, which is irrational.

Bolzano-Weierstrass Theorem: Every bounded sequence in R has a con-vergent subsequence.

Proof: If suffices to find a monotonic subsequence.

If (xi) has a a strictly increasing subsequence, we’re done. Now suppose(xi) does not have a strictly increasing subsequence. Then for any k thesequence (xk, xk+1, . . . ) must have a maximum element, otherwise it yields astrictly increasing subsequence of (xi). Let xk1 be the maximum element of(x1, x2, . . . , ), let xk2 be the maximum element of (xk1+1, xk1+2, . . . ), let xk3 bethe maximum element of (xk2+1, xk2+2, . . . ). We have xk1 ≥ xk2 ≥ xk3 ≥ · · · ,therefore (xki) is a decreasing subsequence of (xi).

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Chapter 9: Proofs from Group Theory

Topics:

9.1 Binary Operations and Algebraic Structures9.2 GroupsSubgroups and Cyclic Groups

Notes:

Examples of groups: Z7 under addition, Z∗7 under multiplication. Give theoperation tables.

Group in general: a set and a binary operation on the set. The binaryoperation must be associative, there must be an identity element, and everyelement should have an inverse.

Some examples of sets with binary opertions are given in Exercises 9.1, pp.297–298.

Some examples of groups are given in Exercises 9.2, pp. 302–304.

Zn is always a group under addition.

Z∗6 is not a group under multiplication. Reason: identity is [1] but [2]−1 doesnot exist.

Z∗p is always a group under multiplication. Reason: For 1 ≤ a ≤ p− 1, p 6 |a,so there is a solution to px + ay = 1. Therefore [a][y] = [1], and moreover[y] ∈ Zp since [y] 6= [0].

Groups also arise as automorphisms of a geometric figure. Example: au-tomorphisms of a square. The elements of the group are functions from{1, 2, 3, 4} to {1, 2, 3, 4} that preserve the edges 12, 23, 34, 41.

The operation table for D4 shows that the group is not Abelian. For example,σ = (1, 2, 3, 4) ∈ D4 and τ = (1, 2)(3, 4) ∈ D4, yet στ = (3, 1)(2, 4) andτσ = (1)(2, 4)(3).

A subgroup of a group is simply a subset that forms its own group using thesame group operation.

Cyclic subgroup generated by g: 〈g〉 = {gn : n ∈ Z}.Cyclic subgroups are always abelian.

〈(1, 2, 3, 4)〉 is an abelian subgroup of S4. D4 is a non-abelian subgroup ofS4.

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Some examples of subgroups are given in Exercises 9.3, pp. 311–312.

We can derive theorems about numbers by thinking in terms of groups. Forexample, given any g in a finite group G, the list g, g2, g3, . . . must eventuallyrepeat itself, so that gi = gj where i < j, which implies gj−i = e. The leastpositive value of k such that gk = e is called the order of g.

It is a theorem that o(g)|o(G) when G is finite. To prove this, form thefunction σ : G → G via σ(a) = ga. Then σ is a permutation. If we write itin cycle structure, we will see that all the cycles have length equal to o(g).Since all the elements of G are present in the cycles, O(G) = k · o(g) wherek is the number of cycles of σ.

Example: Let G = Z∗7 and g = [2]. Then σ = ([1], [2], [4])([3], [6], [5]). Wecan see that o(g) = 3 divides o(G) = 6.

There are some order problems in Exercises 9.3.

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