Proving two numbers $q1, q2$ are relatively prime, related to the the $gcd(a,b)$ $a,b > 1$ and are integers, and $g: = gcd(a,b)$ is their greatest common divisor. Show that if $a= g * q1$ and $b = g * q2$, then $q1$ and $q2$ are relatively prime.
 A: What you want to show is that $$\left(\frac{a}{d},\frac{b}{d}\right)=1$$ if $d=(a,b)$. We can prove more, that is:

PROP Let $d>0$. Let $d$ be a common divisor of $a,d$. Then $\left(\dfrac{a}{d},\dfrac{b}{d}\right)=1$ if, and only if, $d=(a,b)$.

P First, suppose that $d=(a,b)$. Let $f$ be such that $f\; \left|\;\dfrac ad,\dfrac bd\right.$. We prove that $f=1$. But the above means that $fd\mid a,b$. Thus $fd\mid d$. This means that $f\mid 1$; so $f=1$. 
Converesely, let $d$ be such that $\left(\dfrac{a}{d},\dfrac{b}{d}\right)=1$. We prove $d=(a,b)$. It is clear $d$ is a common divisor, so if $d'=(a,b)$; $d \mid d'$.   We have that  $$\frac{d'}{d}\frac{a}{d'}=\frac{a}{d}$$  $$\frac{d'}{d}\frac{b}{d'}=\frac{b}{d}$$
Thus $\dfrac{d'}{d}\left|\; \dfrac{a}d,\dfrac bd\right.$ whence $\dfrac{d'}{d}\left|\;\right.\left( \dfrac{a}d,\dfrac bd\right)=1$, so $\dfrac{d'}{d}=1$, $d=d'$. $\blacktriangle$.
A: Hint: If $\gcd(q_1, q_2) = \alpha$, then
$$a = g q_1 \implies \alpha g | a$$
and likewise
$$b = g q_2 \implies \alpha g | b$$
So what  can you conclude?
A: We can find integers $m,n$ such that $am+bn=g$.  Then $\frac{a}{g}m+\frac{b}{g}n=1$.  So $q_1 m+q_2 n=1$.  So $\gcd(q_1,q_2)=1$.
