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Discrete Mathematics With Applicat...

5th Edition
EPP + 1 other
ISBN: 9781337694193

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Discrete Mathematics With Applicat...

5th Edition
EPP + 1 other
ISBN: 9781337694193
Textbook Problem
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a. Use mathematical induction and Euclid’s lemma to prove that for every positive integer s. if p and q 1 , q 2 , , q s are prime numbers and p | q 1 q 2 q s , then p = q i for some i with 1 i s .

b. The uniqueness part of the unique factorization theorem for the integers says that given any integer n if n = p 1 p 2 p r q 1 q 2 q s for some positive integers r and s and prime numers p 1 p 2 p r and q 1 q 2 q s then r = s and p i = q i for every integer i with 1 i r .

Use the result of part (a) to fill in the details of the following sketch of a proof: Suppose than n is an integer with two different prime faclorizations: n = p 1 p 2 p t = q 1 q 2 q u . All the prime factors that appear on both sides can be cancelled (as many times as they appear on both sides) to arrive at the situation where p 1 p 2 p r = q 1 q 2 q s , p 1 p 2 p r , q 1 q 2 q s , and p i q j for any integers i and j. Then use part (a) to deduce a contradiction. and conclude that the prime factorization of n is unique except, possibly, for the order in which the prime factors are written.

To determine

(a)

To prove:

For all positive integers s, if p and q1,q2,...,qs are prime numbers and p|q1q2....qs then p=qi for some i with 1is, using mathematical induction and with Euclid’s lemma

Explanation

Given information:

Mathematical induction and Euclid’s lemma.

Concept used:

Mathematical Induction: This method has two steps to be followed:

  1. Basis Step
  2. Inductive Step.

Proof:

Basis Step:

Prove the statement for the first possible value s=1.

For s=1

The statement becomes, “if p and q1 are prime numbers and p|q1 then p=q1

By definition of a prime number, it has only two factors itself and 1.

Since, q1 is a prime number so it is divisible by 1 and q1 only.

But it is divisible by p. So, p has to be either 1 or q1.

It is given that p is a prime number so, it can’t be 1. So, p=q1.

Therefore, the statement holds true for s=1.

Inductive Step:

Assume that the statement holds true for s=m.

Therefore, it is assumed that if p and q1,q2,....,qm are prime numbers and p|q1,q2,q3,......,qm then p=qi for some i with 1<i<m is true ........(1).

The next case in the series is m+1.

The following statement has to be proved as true.

If p and q1,q2,.....,qm,qm+1 are prime numbers and p|q1,q2,.....,qm,qm+1 then p=qi for some iwith 1im+1

To determine

(b)

To prove:

The uniqueness part of the unique factorization theorem for integers.

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