# 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:

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 $\times$ | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds x\) | \(\RR\) | \(\ds 0\) | |||||||||||

\(\ds \leadsto \ \ \) | \(\ds \eqclass 0 \RR\) | \(=\) | \(\ds \Q\) | |||||||||||

\(\ds \leadsto \ \ \) | \(\ds \RR\) | \(=\) | \(\ds \Q \times \Q\) |

$\blacksquare$

## Sources

- 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): $\S 11$: Example $11.3$