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Unformatted text preview: ditional comment on the negative case for those who are interested: If m > 0 and n < 0, then we know from above that
n = qm + r so n = (
q)m + (– r). This is the right form but now r is negative or zero. If r = 0, then 0 ≤ r < m and we are done. But if
m <
r < 0, we add m to r to get 0 < m – r < m. So if we rewrite the equation as n = (
q
1)m + (m– r). So this has the form n = q'm + r, with q' =
q
1 and r' = m
r. For negative m, it is simpler. If m is negative and so –m is postive, then for any n, n = q(
m) +r and so n = (
q)m + r. is of the right form. If these equations seem obscure, write out some simple examples. Problem 9
3: Prove: If m, n, q, and r and integers, then the set of common divisors of m and n is the same as the set of common...
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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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