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

\(\displaystyle \RR\) | \(\ne\) | \(\displaystyle \Delta_\Q\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \exists r, s \in \Q: r\) | \(\ne\) | \(\displaystyle s \land r \RR s\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \exists h \in \Q, h \ne 0: h\) | \(=\) | \(\displaystyle r - s\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle h\) | \(\RR\) | \(\displaystyle 0\) | 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

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