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    show that...

    Prove that if a=bq+r, then gcd(a,b)=gcd(b,r)
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by mandy123 View Post
    Prove that if a=bq+r, then gcd(a,b)=gcd(b,r)
    Assuming b \ne 0, note that a is a multiple of (b,r), since it can be written as a linear combination of b and r. So (b,r) \mid (a,b) since b is a multiple of (b,r) as well.

    Now use r = a - bq to show that (a,b) \mid (b,r). thus it will follow that (b,r) \mid (a,b) and (a,b) \mid (b,r) so that (a,b) = (b,r)
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