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Unformatted text preview: 39 + 112 n, 13267 n ) is a solution for all n ∈ Z . Question 8 This could be proven several di²erent ways. Below is one method. 1 Let a and b be natural numbers, and let gcd( a, b ) = d . By defnition, we always have d ≥ 1. Let S = { ar + bs ≥ 1  r, s ∈ Z } . Then by Bezout’s thereom, there exists an r , s ∈ Z such that ar + bs = d . ThereFore, d ∈ S . Let e = ar + bs be any element in S . So r and s are integers, and hence by prop. 4, d  e . This means e = dn For some n ∈ Z . But e, d ≥ 1, so we must also have n ≥ 1. ThereFore the smallest possible value For e is d · 1 = d . Since d ∈ S , gcd ( a, b ) is indeed the least element in the set. 2...
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 Spring '08
 BILLERA
 Math, Algebra, Inductive Reasoning, #, Natural number, Bezout, 3 2 k, Partial Solutions Question

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