Initial Segment of Natural Numbers forms Complete Residue System

Theorem
Let $m \in \Z_{\ne 0}$ be a non-zero integer.

Let $\N_m = \set {0, 1, 2, \ldots, m - 1}$ denote the initial segment of $\N$

Then be a complete residue system modulo $m$.

Then $s = m$.

Proof
Let $n \in \Z$.

From the Division Theorem there exist unique integers $q$ and $u$ such that:
 * $n = m q + u$

such that $0 \le u < m$.

That is:
 * $n \equiv u \pmod m$

and $u$ is one of $0, 1, \ldots, m - 1$.

Also, since $\forall i, j \in \N_m: \size {i - j} < m$, no two elements of $\N_m$ are congruent.

Thus $\N_m = \set {0, 1, 2, \ldots, m - 1}$ is a complete residue system modulo $m$.