# 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 $\RR \ne \Delta_\Q$.

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

Then:

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

$\blacksquare$