Prove that gcd(a,b)=dgcd(a/d,b/d).

Question

I think the strategy for this problem would be i assume some arbitrary element is in gcd(a,b) and chase it into the rhs and then do it the other way around , but i am not sure how to even start this one.

Answer 1

We can rephrase the conditions as $a=p\cdot d$ and $b=q\cdot d$ with $p,q\in\mathbb{N}$.

Then $$\gcd(a,b)=\gcd(pd,qd)=d\cdot\gcd(p,q)=d\cdot\gcd(\frac{a}{d},\frac{b}{d})$$