Proving $\gcd \left(\frac{a}{\gcd (a,b)},\frac{b}{\gcd (a,b)}\right)=1$ How would you go about proving that $$\gcd \left(\frac{a}{\gcd (a,b)},\frac{b}{\gcd (a,b)}\right)=1$$
for any two integers $a$ and $b$?
Intuitively it is true because when you divide $a$ and $b$ by $\gcd(a,b)$ you cancel out any common factors between them resulting in them becoming coprime. However, how would you prove this rigorously and mathematically?

Assume WLOG that $a, b \geq 1$. Let $m = \dfrac{a}{\gcd(a,b)}$, and $n = \dfrac{b}{\gcd(a,b)}$, and let $c = \gcd(m,n)$. Then $c \mid m$, and $c \mid n$. This means: $(c\cdot \gcd(a,b)) \mid a$, and $(c\cdot \gcd(a,b)) \mid b$. So $(c\cdot \gcd(a,b)) \mid \gcd(a,b)$. but $\gcd(a,b) \mid (c\cdot \gcd(a,b))$. Thus: $c\cdot \gcd(a,b) = \gcd(a,b)$, and this means $c = 1$.
