Congruence Modulo Integer is Equivalence Relation

Theorem

For all $z \in \Z$, congruence modulo $z$ is an equivalence relation.

Proof

Checking in turn each of the criteria for equivalence:

Reflexive

We have that Equal Numbers are Congruent:

$\forall x, y, z \in \Z: x = y \implies x \equiv y \pmod z$

so it follows that:

$\forall x \in \Z: x \equiv x \pmod z$

and so congruence modulo $z$ is reflexive.

$\Box$

Symmetric

$\ds x$

$\equiv$

$\ds y \bmod z$

$\ds \leadsto \ \ $

$\ds x - y$

$=$

$\ds k z$

$\ds \leadsto \ \ $

$\ds y - x$

$=$

$\ds \paren {-k} z$

$\ds y$

$\equiv$

$\ds x \bmod z$

So congruence modulo $z$ is symmetric.

$\Box$

Transitive

$\ds x_1$

$\equiv$

$\ds x_2 \pmod z$

$\, \ds \land \, $

$\ds x_2$

$\equiv$

$\ds x_3 \pmod z$

$\ds \leadsto \ \ $

$\ds \paren {x_1 - x_2}$

$=$

$\ds k_1 z$

$\, \ds \land \, $

$\ds \paren {x_2 - x_3}$

$=$

$\ds k_2 z$

$\ds \leadsto \ \ $

$\ds \paren {x_1 - x_2} + \paren {x_2 - x_3}$

$=$

$\ds \paren {k_1 + k_2} z$

$\ds \leadsto \ \ $

$\ds \paren {x_1 - x_3}$

$=$

$\ds \paren {k_1 + k_2} z$

$\ds x_1$

$\equiv$

$\ds x_3 \pmod z$

So congruence modulo $z$ is transitive.

$\Box$

So we are justified in supposing that congruence, as we have defined it, is an equivalence.

$\blacksquare$

Sources

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