 # 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. ### Discrete Mathematics With Applicat...

5th Edition
EPP + 1 other
Publisher: Cengage Learning,
ISBN: 9781337694193 ### Discrete Mathematics With Applicat...

5th Edition
EPP + 1 other
Publisher: Cengage Learning,
ISBN: 9781337694193

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Chapter 8.4, Problem 41ES
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