Multiplicative Inverse in Ring of Integers Modulo m/Proof 1

Proof
First, suppose $k \perp m$.

That is:
 * $\gcd \set {k, m} = 1$

Then, by Bézout's Identity:
 * $\exists u, v \in \Z: u k + v m = 1$

Thus:
 * $\eqclass {u k + v m} m = \eqclass {u k} m = \eqclass u m \eqclass k m = \eqclass 1 m$

Thus $\eqclass u m$ is an inverse of $\eqclass k m$.

Suppose that:
 * $\exists u \in \Z: \eqclass u m \eqclass k m = \eqclass {u k} m = 1$.

Then:
 * $u k \equiv 1 \pmod m$

and:
 * $\exists v \in \Z: u k + v m = 1$

Thus from Bézout's Identity:
 * $k \perp m$