# Proving $k$ is a common divisor of $a$ and $b$ iff $k\,|\gcd(a,b)$ — converse

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!

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?
• 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 $k|b$ – Mathmo123 Jun 24 '14 at 23:48