Proof.
Let c = (a/d, b/d). Then c | a/d, and so cd | a. Also c | b/d, and so cd | b.
My question is, did this happen because of this simple algebra? ie. c | a/d So cx = a/d for some integer x then cx/d = a
Next step of proof.. Thus, cd is a common divisor of a and b. Therefore, cd <= d, which implies c = 1.
My question for this part: how did I conclude cd <= d?
If $cd$ is a divisor of $a$ and $b$, it is a divisor of the greatest common divisor, which is $d$. Hence $cd|d$.
Remark:
If you are familiar with Bezout's identity.
We have $$ax+by=d, x, y \in \mathbb{Z}$$
$$x\left( \frac{a}d\right)+ y\left( \frac{b}d\right)=1$$
and we can conclude.