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Unformatted text preview: e integer n, there are integers q and r, with 0 ≤ r < m so that n = qm + r. So m is fixed throughout the proof. We do an inducation on n. Base case: n = 0. In this case, n = 0m +0, so q = r = 0. Inductive step: Assume n = qm + r, with 0 ≤ r < m (inductive hypothesis). Prove: n+1 = q'm + r', with 0 ≤ r' < m. (Note: we need a new q and a new r for this case.) Now we start with n + 1 = qm + r + 1 from the inductive hypothesis. If 0 ≤ r +1 < m, then we can take r' = r+1 and q' = q. If r < m
1, this is the case and we are done. In the remaining case of r = m
1, then r+1 = m. So in this case n + 1 = qm + r + 1 = qm + m = (q+1)m + 0. So if we set q' = q+1 and r' = 0, we have proved what was needed in each case. QED. A very optional ad...
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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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