Let \(gcd(a, b) = c\) and \(gcd(ma, mb) = d\), so:
\[ \displaylines{ c|a \text{ and } c|b \\ \therefore mc|ma \text{ and } mc|mb \\ \therefore mc|d } \]
This means:
$$ d = mcx $$
For some integer \(x\), so:
\[ \displaylines{ d|ma \text{ and } d|mb \implies \\ mcx|ma \text{ and } mcx|mb \\ \therefore cx|a \text{ and } cx|b } \]
Since \(c\) is the greatest common divisor, then \(x=1\), so \(d = mc\).