Integers Divided by GCD are Coprime/Proof 2

Proof
Let $d = \gcd \set {a, b}$.

We have:


 * $(1): d \divides a \iff \exists s \in \Z: a = d s$


 * $(2): d \divides b \iff \exists t \in \Z: b = d t$

We have to prove:
 * $\gcd \set {s, t} = 1$

$\gcd \set {s, t} \ne 1$.

So:


 * $(3): \exists k \in \N \setminus \set 1$ such that $k \divides s \land k \divides t$

So:


 * $(4): \exists m, n \in \N: s = k m, t = k n$

Substituting from $(4)$ in $(1)$ and $(2)$:


 * $a = d k m$, $b = d k n$

Therefore:
 * $ d k \divides a \land d k \divides b$

From $(3)$ we have:

As $d k$ is a common divisor of $a$ and $b$ greater than $d$, this contradicts $d = \gcd \set {a, b}$.

So our initial assumption that $\gcd \set {s, t} \ne 1$ is false.

Therefore, from Proof by Contradiction, we have:


 * $\gcd \set {s, t} = 1 \implies \gcd \set {\dfrac a d, \dfrac b d} = 1$