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Unformatted text preview: n, there is an integer q and an integer r, with 0 ≤ r < m so that n = qm + r. Hint: Induction on n. Correction and comment: The integer m cannot be 0. Also, if the case for positive m and nonnegative n is proved, the case for negative m or n can be quickly proved from the positive case (as will be noted at the end). So the revised instructions for the problem were to prove the result for positive m and nonnegative n. Answer: The idea of the proof is that we can fix m (in class there was an example of m = 5) and then do an induction on n. Also, it suffices to prove the theorem for nonnegative n, as will be explained at the end of the proof. So we can rephrase what we want to prove thus: To prove: Given a positive integer m, for any nonnegativ...
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This note was uploaded on 01/29/2014 for the course MATH 300A taught by Professor Jamesking during the Winter '11 term at University of Washington.
 Winter '11
 JamesKing
 Math, Integers

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