64 Relatively Prime Integers You already know that if d is a

6.4 Relatively Prime Integers You already know that if d is a common divisor of a and b. then d lax b more is and (Proposition 6.2.13). If d is the greatest common divisor y fra natural true. It is possible to find integers x and y hat d of a b. ax by and this 6.4.1 THEOREM Section Leta and bbe integers, not both zero. Let d gcd(a, b). Then there exist integers a y so that is so im Yo unique diligem the the (Hint: Let k be the least element of the set are inv be wri Use proof by contradiction and the division algorithm to show that k is a common divisor of a and b. Then show that d is in S, also. it is pr can b can br this th is, This fact is surprisingly useful especially when a and b are relatively prime, that when goda, Every time allowed to assume that a pair of numbersis 6.5. relatively prime, keep your eye out for a trick that uses this fact!) 6.4.2 COROLLARY Ever a and b be et integers. The following are equivalent prim 1. a and b are relatively prime, 2. There room you

Solution

First, let a and b be integers, not both zero, and suppose a and b are relatively prime.

Then 1 = gcd(a, b)

so,

there exist integers m and n such that 1 = ma + nb.

In the other direction, let a and b be integers, not both zero, and suppose there exist integers m and n such that 1 = ma + nb.

If S = { c Z |there exist integers m and n such that c = ma + nb } then we are assuming that 1 S.

Let d = gcd(a, b).

By Theorem Let a and b be integers, not both zero. Let d = gcd(a, b) and let S = { c Z |there exist integers m and n such that c = ma + nb }. Then d is the smallest positive integer in S.

d is the smallest positive integer in S.

Clearly 1 is the smallest positive integer there is. Since 1 S and d is the smallest positive integer in S, it follows that d = 1.