If $\gcd(a,b)=d$ then prove that $\gcd(\frac{a}{c},\frac{b}{c})=\frac{d}{c}$ My attempt:
$ax+by=d$(as gcd(a,b)=d).
Dividing by c
$\frac{a}{c}x+\frac{b}{c}y=\frac{d}{c}$
Let $k$ be the gcd of$ (\frac{a}{c},\frac{b}{c})$ so
$k|\frac{a}{c}$ and $ k|\frac{b}{c}$ then $k|\frac{a}{c} x + \frac{b}{c}y$  and so $k|\frac{d}{c}$.
As $d|a $ implies that $\frac{d}{c}|\frac{a}{c} $ similarly it divides $\frac{b}{c}$ . Hence $\frac{d}{c}|k$(as $k$ is the greatest common factor). Thus $k= \frac{d}{c}$

Clearly $\frac {d}{c} \mid \frac {a}{c}$ and $\frac {b}{c}$ as $\gcd(a,b)=d$. Let $m$ be a common divisor of $\frac {a}{c}$ and $\frac{b}{c}$. Then $m$ divides the left side which implies that $m \mid \frac{d}{c}$ and $\gcd(\frac{a}{c},\frac{b}{c}) = \frac{d}{c}$.
