Multiplicative Inverse in Ring of Integers Modulo m/Proof 1

Proof
First, suppose $k \perp m$. That is, $\gcd \left\{{k, m}\right\} = 1$.

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

Thus:
 * $\left[\!\left[{u k + v m}\right]\!\right]_m = \left[\!\left[{u k}\right]\!\right]_m = \left[\!\left[{u}\right]\!\right]_m \left[\!\left[{k}\right]\!\right]_m = \left[\!\left[{1}\right]\!\right]_m$

Thus $\left[\!\left[{u}\right]\!\right]_m$ is an inverse of $\left[\!\left[{k}\right]\!\right]_m$.

Suppose that:
 * $\exists u \in \Z: \left[\!\left[{u}\right]\!\right]_m \left[\!\left[{k}\right]\!\right]_m = \left[\!\left[{u k}\right]\!\right]_m = 1$.

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

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

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