Integers Divided by GCD are Coprime/Proof 1

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

By definition of divisor:

$d \divides a \iff \exists s \in \Z: a = d s$
$d \divides b \iff \exists t \in \Z: b = d t$

So:

\(\ds \exists m, n \in \Z: \, \) \(\ds d\) \(=\) \(\ds m a + n b\) Bézout's Identity
\(\ds \leadstoandfrom \ \ \) \(\ds d\) \(=\) \(\ds m d s + n d t\) Definition of $s$ and $t$
\(\ds \leadstoandfrom \ \ \) \(\ds 1\) \(=\) \(\ds m s + n t\) dividing through by $d$
\(\ds \leadstoandfrom \ \ \) \(\ds \gcd \set {s, t}\) \(=\) \(\ds 1\) Bézout's Identity
\(\ds \leadstoandfrom \ \ \) \(\ds \gcd \set {\frac a d, \frac b d}\) \(=\) \(\ds 1\) Definition of $s$ and $t$

$\blacksquare$


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