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Unformatted text preview: ote that max of r and r1 < m
1. So we continue with r = q2 r1 + r2. Again the divisors of m and n are the same as the divisors of r1 + r2. And now the max of r1 + r2 is less than m – 2. So at each stage, the remainder, which began as a number < m decreases by at least 1. So in no more than m steps, the remainder will be 0. At this point we have rk = qk rk
1 + 0. And the divisors of m and n will be the same as the divisors of rk
1 and 0. But this is the same as the set of divisors of rk
1 since any integer divides 0. So we conclude that rk
1 itself (a divisor of rk
1) divides m and n, and any common divisor of m and n is a divisor of rk
1. So rk
1 must be the gcd of m and n. Example 1: Let...
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 Winter '11
 JamesKing
 Math, Integers

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