Congruences on Rational Numbers

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Theorem

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

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


Proof

From:

Diagonal Relation is Universally Congruent and
Trivial Relation is Universally Congruent

we know that both these relations are congruent 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 $\RR \ne \Delta_\Q$.


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


Then:

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

$\blacksquare$


Sources