Prove: if $d=\text{gcd}(m,n)$ so $\text{gcd}\left(\frac{m}{d},\frac{n}{d}\right)=1$

Question

$$\newcommand{\gcd}{\text{gcd}}$$

Prove: if $d=\gcd(m,n)$ so $\gcd\left(\frac{m}{d},\frac{n}{d}\right)=1$

Intuitively it is obvious, but I am having a hard time to formalize the proof, what I have came to this:

$d=\gcd(m,n)$ so $d|m$ and $d|n$ therefore $m=dx$ and $n=dy$ now if $\gcd\left(\frac{m}{d},\frac{n}{d}\right)\neq 1$ that mean that $m$ and $n$ have a common factor, after the division in $d$ which is the greatest common divisor is contradiction.

Answer 1

Your proof seems fine although you could show the contradiction more evidently.

We want to prove that if $d = \gcd(m,n)$ then $\gcd(m/d, n/d) = 1$.

Assume for the sake of contradiction that this is not the case i.e. $d = \gcd(m,n)$ but $\gcd(m/d, n/d) = k$ where $k > 1$. Then we have:

$$\frac{m}{d} = kx$$

and

$$\frac{n}{d} = ky$$

for some $x$ and $y$, so $m = kdx$ and $n = kdy$ giving:

$$ \gcd(m, n) \geq kd > d $$

which is a contradiction since we assumed $\gcd(m,n) = d$.

Answer 2

If $d=\gcd(m,n)$, then $\exists \, x,y \in \mathbb{Z}$ such that $d=mx+ny$. From this we can also write $$1=\frac{m}{d}x+\frac{n}{d}y.$$ Thus $\gcd(m/d,n/d)=1$.

Answer 3

Suppose that $gcd(\frac{m}{d},\frac{n}{d})=d'>1.$ Then $d' \mid m$ and $ d' \mid n$. It implies that $d'd \mid m$ and $ d'd \mid n$ and we find a common divisor $d d'$ which is greater then $d.$ Contradiction.