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anonymous
 one year ago
Is this a true statement? Given two nonzero integers a,b. If there exists integer x,y such that ax + by = 1, then gcd(a,b) = 1?
anonymous
 one year ago
Is this a true statement? Given two nonzero integers a,b. If there exists integer x,y such that ax + by = 1, then gcd(a,b) = 1?

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I believe it's true. But not sure. So here is my attempt. Suppose not, that gcd(a,b) = c, for some positive integer c > 1. Then by definition, c  a and cb. I.e cm = a and cn = b for some integer m,n. Substitute we have: cmx + cny = 1 c (mx + ny) = 1. This implies c  1. But c > 1. Hence this is a contradiction. The only value that divide 1 is 1. So c = 1 Right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, I think this is used for Euclid algorithm

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I believe that is the converse (with more assumptions) of Bézout's identity

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the converse of Bézout's identity is not true in general. But if we restrict a,b to be nonzero, then it's true. (provided that the proof above is correct)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I never had any confidence in the validity of my proof. Which is why I asked this question to see if any one agrees.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it's true, this is a special case of Bezout's identity

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0any integer combination of \(x,y\) must be a multiple of \(\gcd(x,y)\), so if you have a combination \(ax+by=1\) it follows that \(\gcd(x,y)\,\, 1\) so \(\gcd(x,y)=1\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@oldrin.bataku awesome! thank you :D

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0consider that if \(\gcd(x,y)=z\) then \(x=mz,y=nz\) so it follows that \(amz+bnz=z(am+bn)=1\) so \(z\,\, 1\implies z=1\) since we restrict \(z\gt0\)
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