# Congruences on Rational Numbers

## Theorem

There are only two congruence relations on the field of rational numbers $\left({\Q, +, \times}\right)$:

$(1): \quad$ The diagonal relation $\Delta_\Q$
$(2): \quad$ The trivial relation $\Q \times \Q$.

## Proof

From:

Diagonal Relation is Universally Compatible and
Trivial Relation is Universally Congruent

we know that both these relations are compatible with both addition and multiplication on $\Q$.

Now we need to show that these are the only such relations.

Let $\RR$ be a congruence on $\Q$, such that $\RR \ne \Delta_\Q$.

 $\ds \RR$ $\ne$ $\ds \Delta_\Q$ $\ds \leadsto \ \$ $\ds \exists r, s \in \Q: r$ $\ne$ $\ds s \land r \RR s$ $\ds \leadsto \ \$ $\ds \exists h \in \Q, h \ne 0: h$ $=$ $\ds r - s$ $\ds \leadsto \ \$ $\ds h$ $\RR$ $\ds 0$ as $\RR$ is a congruence relation compatible with $+$

Then:

 $\ds \forall x \in \Q: \paren {x / h}$ $\RR$ $\ds \paren {x / h}$ $\ds \leadsto \ \$ $\ds \paren {\paren {x / h} \times h}$ $\RR$ $\ds \paren {\paren {x / h} \times 0}$ as $\RR$ is a congruence relation compatible with $\times$ $\ds \leadsto \ \$ $\ds x$ $\RR$ $\ds 0$ $\ds \leadsto \ \$ $\ds \eqclass 0 \RR$ $=$ $\ds \Q$ $\ds \leadsto \ \$ $\ds \RR$ $=$ $\ds \Q \times \Q$

$\blacksquare$