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This is in reference to the same proof given in the post

Is greatest common divisor of two numbers really their smallest linear combination?

I couldn't add a comment there so I'm asking it here. I am trying to understanding the same proof but can't digest a fact.

The part that d <= gcd(a,b) is understandable coz g in gcd implies greatest, so d can only be less than/equal to gcd(a,b). ( Fact -1)

The other part says that as gcd(a,b) | sa + tb then gcd(a,b) | d. This is the part where I am getting confused.

Ques: Fact 2 -> m | sa + tb is true and gcd(a,b) | sa + tb is also true.

Now we could write gcd(a,b) | m or m | gcd(a,b). I can well reason that as gcd(a,b) >= m , so m |gcd(a,b) is possible but I can't reason why we can write gcd(a,b) | m knowing until as of now that gcd(a,b) can be > = m. if gcd is greater than m then the result will not be an integer ?

Can anyone explain with some logic ? Would be really helpful.

Thanks in Advance

Ankit

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  • The primary point is that both $d\le m$ and $d\ge m$ are true, and this can only happen when $d=m$. – abiessu Oct 10 '14 at 02:05
  • yeah but sounds like I have half the proof. And coz the Proof has to be proved I take the second fact to be true. – Ankit Oct 10 '14 at 02:27
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    The other significant part of the referenced proof is the "reductio ad absurdum" or whatever the Latin term is... The assumption is that the claim is false, and this leads to an impossible conclusion. In particular it is assumed that $m=ax+by$ (guessing that is the variable you mean to be in place of $e$) and that $m$ is the smallest linear combination of $a,b$, but that $m\mid a$ is false. – abiessu Oct 10 '14 at 03:18
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    This leads to the creation of $0\lt r\lt m$ such that $a=qm +r$, but this also leads to the result that $r$ is a linear combination of $a,b$. $r$ is positive and smaller than $m$, which contradicts our assumption that $m$ was the smallest such number. – abiessu Oct 10 '14 at 03:19

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