Let $a,b,c \in\mathbb Z$ where $a \neq0$ or $b \neq 0$. Suppose that $c \neq 0$ and is a common divisor of $a$ and $b$.
Prove that:
${gcd(a,b)\over |c|}= gcd ({a \over c}, {b \over c})$
What I have so far is
Let $a,b \in \mathbb Z$ , both not zero. If $d = gcd(a,b)$, then $gcd({a \over d},{b \over d})=1$.
But I am not sure how I can use this in the equation we have to prove. Can anyone help please?
It is of course not restrictive to assume $a$, $b$ and $c$ non negative.
Suppose $d$ is a common divisor of $a/c$ and $b/c$; then $cd$ is a common divisor of $a$ and $b$; therefore $cd$ is a divisor of $\gcd(a,b)$, so $d$ is a divisor of $\gcd(a,b)/c$.
Suppose $d$ is a divisor of $\gcd(a,b)/c$. Then $cd$ is a divisor of $\gcd(a,b)$, hence a common divisor of $a$ and $b$. Therefore $d$ is a common divisor of $a/c$ and $b/c$.
