# Integers Divided by GCD are Coprime/Proof 2

## Theorem

Any pair of integers, not both zero, can be reduced to a pair of coprime ones by dividing them by their GCD:

$\gcd \set {a, b} = d \iff \dfrac a d, \dfrac b d \in \Z \land \gcd \set {\dfrac a d, \dfrac b d} = 1$

That is:

$\dfrac a {\gcd \set {a, b} } \perp \dfrac b {\gcd \set {a, b} }$

## 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:

$\gcd \set {s, t} = 1$

Aiming for a contradiction, suppose $\gcd \set {s, t} \ne 1$.

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:

 $\ds$  $\ds k \in \N \land k \ne 1$ $\ds \leadsto \ \$ $\ds$  $\ds k > 1$ $\ds \leadsto \ \$ $\ds$  $\ds d k > d$

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:

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

$\blacksquare$