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I'm currently trying to solve the converse of this statement is true after proving the normal version is true. If $k$ is a common divisor of $a$ and $b$, then $k \,| \gcd(a, b)$

So far I know the converse states that if $k \,| \gcd(a, b)$ then $k$ is a common divisor of $a$ and $b$.

Can anyone help me out and thanks!

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I'll start you off. Let $d = \text{gcd}(a,b)$. Then by definition $d|a$ and $d|b$. Now suppose $k|d$. Can you use what you know about division to prove the result?

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  • $\begingroup$ Hold on... which way are you trying to prove? $\endgroup$ – Mathmo123 Jun 24 '14 at 23:42
  • $\begingroup$ if k|d and ax+by=d then k|ax+by. Is that in the right direction? $\endgroup$ – James Jun 24 '14 at 23:47
  • $\begingroup$ Sorry - I should clarify. Both directions are true. Which was the direction that you wanted help with? I was assuming that you were stuck trying to prove that if $k|\text{gcd}(a,b)$, then $k|a$ and $k|b$ $\endgroup$ – Mathmo123 Jun 24 '14 at 23:48
  • $\begingroup$ yes your assumption was correct $\endgroup$ – James Jun 24 '14 at 23:49
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    $\begingroup$ by transitivity of division you can say that k|a and k|b? $\endgroup$ – James Jun 24 '14 at 23:55

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