Congruences on Rational Numbers

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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 $\times$
\(\displaystyle \leadsto \ \ \) \(\displaystyle x\) \(\RR\) \(\displaystyle 0\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \eqclass 0 \RR\) \(=\) \(\displaystyle \Q\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \RR\) \(=\) \(\displaystyle \Q \times \Q\)

$\blacksquare$


Sources