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Math 251W: Foundations of Advanced Mathematics, Spring 2011Portfolio Assignment 11: 6.1-4

Name: Sean Fogerty

Problem 6.1.6

proposition: Let A and B be finite sets such that |A| = |B|. Let f : A B. Show that f isbijective iff it is injective iff it is surjective.

proof (Direct) Cases

Surjective Injective.Assume f, defined as f : A B, is surjective. Let A and B be finite sets such that |A| = |B|.As f is surjective, it follows that for each b B, there exists some a A such that f(a)=b.Because the cardinality ofA matches B, it must be the case that each a A map to exactly

1 b B. Therefore, f is injective.

Injective SurjectiveAssume f, defined as f : A B, is injective. Let A and B be finite sets such that |A| = |B|.As f is injective, then the image ofA must have the same cardinality as A, or |f(A)| = |A|.Because the cardinality ofA matches B, it must be the case that the image ofA must havethe same cardinality as B, or |f(A)| = |B|. Therefore, because B is a finite set, f must besurjective.

Bijective InjectiveAssume f, defined as f : A B, is bijective. Let A and B be finite sets such that |A| = |B|.By the definition of bijectivity, f must also be injective.

Injective Bijective

Assume f, defined as f : A B, is injective. Let A and B be finite sets such that |A| = |B|.As we already proved that when fwas injective, then it was necessarily surjective. Therefore,because f is injective and surjective, then it must be bijective.

Bijective SurjectiveAssume f, defined as f : A B, is bijective. Let A and B be finite sets such that |A| = |B|.By the definition of bijectivity, f must also be surjective.

Surjective BijectiveAssume f, defined as f : A B, is surjective. Let A and B be finite sets such that |A| = |B|.As we already proved that when fwas surjective, then it was necessarily injective. Therefore,because f is injective and surjective, then it must be bijective.

Problem 6.1.13

proposition: Let A and B be countable sets. Show that AB is countable.

proof (Direct)

As A and B are countable sets, there must exist a bijection between them and the naturalnumbers. Therefore, there must exist a surjection. Let g be defined as g : N A and let f bedefined as f : N B. Consider BAi for some i I, where I is a non-empty set. The union of the

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cartesian products of all the BA AB must be countable, as A is itself countable. Therefore,AB must also be countable.

Problem 6.3.11

proposition: For all n N, ifp(x) is a reductible polynomial of degree n, then it can be writtenas the product irreductible polynomials.

proof (Mathematical Induction)

Base Case We will prove that the case holds for n = 2.

Let p(x) be an arbitrary reductible polynomial of degree 2. By the definition of reductible,p(x) can be written as the product of polynomials of a lesser degree. The only degree that isless than 2 is 1, which is an irreductible degree. Therefore, reductible polynomials of degree2 can be written as the product of irreductible polynomials.

Induction Hypothesis Suppose p(x) is an arbitrary reductible polynomial of degree i such

that i N and 2 i n for some n N. Consider g(x), a polynomial of reductible degreen + 1. By the definition of reductible, it can be rewritten as the product ofj(x) and k(x),where j(x) and k(x) are polynomials with degrees less than n + 1. If its degree is lessthan n + 1, then it is n or less. Therefore the degrees of j(x) and k(x) are i. By inductivehypothesis we know that j(x) and k(x) can be rewritten as products of lesser irreductiblepolynomials. Therefore g(x) can be written as the product of irreductible polynomials.

Problem 6.3.11

proposition: Show thatn

i=2

(1 1

i2) =

n + 1

2n

for all n N such that n 2.

proof (Mathematical Induction)

Base Case We will prove that the case holds for n = 2.

Substituting, we have: 1 122 =2+12(2) . Simplifying, we have

34 =

34 . Therefore, the principle

holds for n = 2.

Induction Hypothesis Supposen

i=2(11i2

) = n+12(n) . Consider the casen+1

i=2 (11i2

). This

can be expanded as (1 122 ) (11n

) (1 1n+1). By substitution, we have

n+12n (1

1n+1).

This can be simplified to n+12n n(n+2)n2+2n+1 . Factoring , we have

(n+2)(n+1)2(n+1)(n+1) . Reducing, we

have n+22(n+1) . This can be rewritten asn+1+12(n+1) . Therefore, by the principle of mathematical

induction, we have

n+1i=2 (1

1i2

) = n+1+12(n+1) for all n N such that n 2.

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Problem 6.4.14.1.i

proposition: Prove that the following equation holds for all n N.

Pn = 3Tn 2Ln

proof (Mathematical Induction)

Base Case We will prove that the case holds for n = 1.Substituting, we have 1+(3(1)+1) = 3(1+(1+1))2(1+ 1). Simplifying, we have 5 = 94.Therefore the principle holds for n = 1.

Induction Hypothesis Suppose the case Pn = 3Tn2Ln. Consider the case 3Tn+12Ln+1.

By the definition of triangular and linear numbers, we can rewrite it as 3(Tn+(n+1))2(Ln+1). By distribution, we have 3Tn + 3n+ 3 2Ln + 3n 2 and, reducing, 3Tn 2Ln + 3n+ 1.Substituting, we have Pn + 3n + 1. By the definition of pentagonal numbers, we have Pn+1.Therefore, by the principle of mathematical induction Pn = 3Tn 2Ln for all n N.

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