Prove that if $d$ is a common divisor of $a$ & $b$, then $d=\gcd(a,b)$ if and only if $\gcd(a/d,b/d)=1$ 
Prove that if $d$ is a common divisor of $a$ & $b$, then $d=\gcd(a,b)$ if and only if $\gcd(\frac{a}{d},\frac{b}{d})=1$

I know I already posted this question, but I want to know if my proof is valid: 
So for my preliminary definition work I have: 
$\frac{a}{d}=k, a=\frac{dk
b}{d}=l,b=ld $
so then I wrote a linear combination of the $\gcd(a,b)$, 
$$ax+by=d$$ and
substituted:
$$dk(x)+dl(y)=d
             d(kx+ly)=d
             kx+ly=1
             a/d(x)+b/d(y)=1$$
Is this proof correct? If not, where did I go wrong? Thanks!
 A: It seems fine however this proof can be reduced massively. To prove '$\Leftarrow$',we know $gcd(\frac{a}{d},\frac{b}{d})=1$ therefore we can write this as a linear combination. $$\frac{a}{d}x+\frac{b}{d}y=1$$
Now multiply through by $d$: $$ax+by=d$$.Therefore $gcd(a,b)=d$
To prove'$\Rightarrow$' is mostly the same logic, try reducing yours.
A: Let us look at the logic of your proof. 
It uses the fact that if $d$ is the greatest common divisor of $a$ and $b$, then there exist integers $x$ and $y$ such that $ax+by=d$. Then you conclude correctly that $\frac{a}{d}x+\frac{b}{d}y=1$, and therefore $\frac{a}{d}$ and $\frac{b}{d}$ are relatively prime. 
But you are asked to prove an "if and only if" statement, and you have not shown that if $d$ is a common divisor of $a$ and $b$, and $\frac{a}{d}$ and $\frac{b}{d}$ are relatively prime, then $d$ is the gcd of $a$ and $b$. You can prove this result in various ways. One way close to how you handled the first part is to use the fact that there exist integers $s$ and $t$ such that $\frac{a}{d}\cdot s+\frac{b][d}\cdot t=1$. 
