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.
 A: 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 $c>1$ then $c\cdot\gcd(a,b)$ is a greater common divisor than the greatest common divisor.
A: Let $d = (a,b)$.
If $\big ( \frac{a}{d}, \frac{b}{d} \big ) = p$, not 1.
Then, $pd \mid a$ and $pd \mid b$ and so the gcd of $a$ and $b$ will become $pd$.
A: If you know the fact that whenever $\gcd(a,b) = d $, we have integers $x$ and $y$ such that $ax+by = d$. This can be shown by the division algorithm.  Suppose $ax+by =1$ for some integers $x$ and $y$, then  $d \mid (ax+by)$ as $d\mid a$ and $d\mid b$, therefore $d\mid 1$. As $d$ is integer $d=1$. So we proved that $ax +by=1$ iff $a$ and $b$ are co-prime. Then  as $1=(a/d)x+(b/d)y$ where $a/d, b/d$ are integers as $d$ is $\gcd$,  $1 = \gcd(a/d, b/d).$
A: All variable-symbols stand for non-zero integers here.
We have $a=a'd$ and $b=b'd$ where $d=\gcd (a,b).$ 
Suppose $n$ divides both $a'$ and $b'.$ Then $a'=na''$ and $b'=nb'',$ so $a=a''(nd)$ and $b=b''(nd).$ So $nd$ is a common divisor of $a$ and $b,$ and by the def'n of $\gcd$ we have  $d\geq |nd|.$ And we also have $nd\ne 0.$ So $|n|=1.$ 
Therefore $1=\gcd (a',b')=\gcd (a/d,b/d)=\gcd (a/\gcd (a,b), b/\gcd (a,b)). $
A: 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. 
A: since d is a common divisor of a and b, we can have integers r and s such that  a=dr and b=ds. now gcd(a/d,b/d)=1 which implies gcd(r,s)=1, multipliying d both sides leads to d.gcd(r,s)=d------(1)
  we know by theorem, that gcd(ka,kb)=k.gcd(a,b).
thus d.gcd(r,s)=gcd(dr,ds)=gcd(a,b)-----(2)
from (1)and (2)   gcd(a,b)=d
