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Unformatted text preview: Homework 1 Solutions 1. Case 2 of the Division Algorithm follows almost identically to Case 1. The big change is replacing q + 1 in the existence proof with q 1 since a is now negative. 2. Problem 1.51: Let I = { k Z : k = 1 } Z . We claim that I meets the hypotheses of Corollary 1.37. (1) follows clearly since = 1. (2): Suppose a,b I then a b = a b = 1 1 = 1 which implies a b I . (3): Suppose a I and q Z . Then qa = ( a ) q = 1 q = 1 which implies qa I . Since I meets the hypotheses of Corollary 1.37 we know that there exists a nonnegative integer d such that I consists precisely of all the multiples of d . Furthermore, d 6 = 0 since is a root of unity and thus k 6 = 0 such that k = 1. Hence we have a positive d as desired. 3. Problem 1.56: Let a,b Z such that as + bt = 1 for some s,t Z . From the proof of Theorem 1.35 we know that ( a,b ) is the smallest positive linear combination of a and b . Since as + bt = 1, 1 must be the smallest positive linear combination (there are no smaller positive integers). Thereforethe smallest positive linear combination (there are no smaller positive integers)....
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 Spring '11
 Johnson
 Algebra, Division

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