Multiplicative Inverse in Ring of Integers Modulo m

Theorem
Let $$\left({\Z_m, +_m, \times_m}\right)$$ be the Ring of Integers Modulo m.

Then $$\left[\!\left[{k}\right]\!\right]_m \in \Z_m$$ has an inverse in $$\left({\Z_m, \times_m}\right)$$ iff $$k \perp m$$.

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

Then $$u k \equiv 1 \left({\bmod\, m}\right)$$ and $$\exists v \in \Z: u k + v n = 1$$.

Thus from Bézout's Identity, $$k \perp m$$.