Integer is Congruent to Integer less than Modulus

Theorem

Let $m \in \Z$.

Then each integer is congruent (modulo $m$) to precisely one of the integers $0, 1, \ldots, m - 1$.


Proof

Proof of Existence

Let $a \in \Z$.

Then from the Division Theorem: $\exists r \in \set {0, 1, \ldots, m - 1}: a \equiv r \pmod m$.


Proof of Uniqueness

Suppose that:

$\exists r_1, r_2 \in \set {0, 1, \ldots, m - 1}: a \equiv r_1 \pmod m \land a \equiv r_2 \pmod m$

Then:

$\exists r_1, r_2 \in \Z: a = q_1 m + r_1 = q_2 m + r_2$


This contradicts the uniqueness clause in the Division Theorem.

$\blacksquare$


Sources

  • 1964: Walter Ledermann: Introduction to the Theory of Finite Groups (5th ed.) ... (previous) ... (next): Chapter $\text {I}$: The Group Concept: $\S 6$: Examples of Finite Groups: $\text{(iii)}$
  • 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 14.2 \ \text{(ii)}$: Congruence modulo $m$ ($m \in \N$)
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