Homework 10 Solutions - MATH 74: HOMEWORK 10 COMMENTS Book...

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MATH 74: HOMEWORK 10 COMMENTS Book Problem # 14, p. 227 Let a and b be positive integers. Let A = { am + bn | m,n Z and am + bn > 0 } . Note that A 6 = because a,b A . By the Well-ordering principle A has a smallest element. Call this element c . We claim that c | a and c | b . Suppose for a contradiction that it doesn’t. Then either c - a or c - b . First we’ll treat the case where c - a . In this case the division theorem tells us there exist unique q and r such that a = cq + r 0 < r < c, (note r cannot equal zero by our assumption that c - a . Since c A there exist integers m 0 and n 0 such that c = am 0 + bn 0 . Hence r = a - cq = a - ( am 0 + bn 0 ) q = a (1 - m 0 q ) + b ( - n 0 q ) . Since in addition r > 0 we see that r A . But r < c , and c is the smallest element of A , and this is a contradiction. Hence c | a . A similar argument shows that c | b . We claim that c is in fact the greatest common divisor of a and b . We must verify two things: (i) c | a and c | b and (ii) if c 0 | a
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Homework 10 Solutions - MATH 74: HOMEWORK 10 COMMENTS Book...

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