GCD of two numbers divided by their greatest common divisor is 1 Im trying to prove that, given $a,b$ with at least one of $a,b \neq 0$,
$$
\gcd\left(\frac{a}{\gcd(a,b)},\frac{b}{\gcd(a,b)}\right)=1
$$
I have tried to prove the identity 
$$
\gcd(c\cdot a, c\cdot b) = c\cdot \gcd(a,b)
$$
with $c =  \dfrac{1}{\gcd(a,b)}$

However I'm having trouble understanding the proof.

Thank you for your time.
Let $c = \gcd\left( \dfrac a {\gcd(a,b)}, \dfrac b {\gcd(a,b)} \right).$

Then $c$ divides $\dfrac a {\gcd(a,b)}$ and $c$ divides $\dfrac b {\gcd(a,b)}.$

Thus $c\cdot\gcd(a,b)$ divides $a$ and $c\cdot\gcd(a,b)$ divides $b.$

So $c\cdot\gcd(a,b)$ is a common divisor of $a$ and $b$.

If $c&gt;1$ then $c\cdot\gcd(a,b)$ is a greater common divisor than the greatest common divisor.
Let g be a common divisor of both a and b, and let g' be a common divisor of both a/g and b/g. Then (a / g) / g' = a / (gg') is an integer, and so is (b / g) / g' = b / (gg'). Therefore g*g' is a common divisor of a and b. 

If g is the greatest common divisor of a and b, then g*g', which is also a common divisor, is not greater than g, and therefore g' = 1.