Congruence (Number Theory) is an Equivalence

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Theorem

For all $z \in \R$, congruence modulo $z$ is an equivalence relation.


Proof

Checking in turn each of the critera for equivalence:


Reflexive

We have that Equal Numbers are Congruent:

$\forall x, y, z \in \R: x = y \implies x \equiv y \bmod z$

so it follows that:

$\forall x \in \R: x \equiv x \bmod z$

and so congruence modulo $z$ is reflexive.


Symmetric

\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle x\) \(\equiv\) \(\displaystyle y \bmod z\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \implies\) \(\displaystyle \) \(\displaystyle x - y\) \(=\) \(\displaystyle k z\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \implies\) \(\displaystyle \) \(\displaystyle y - x\) \(=\) \(\displaystyle \left({-k}\right) z\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle y\) \(\equiv\) \(\displaystyle x \bmod z\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)                    

So congruence modulo $z$ is symmetric.


Transitive

\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle x_1\) \(\equiv\) \(\displaystyle x_2 \bmod z\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \land\) \(\displaystyle \) \(\displaystyle x_2\) \(\equiv\) \(\displaystyle x_3 \bmod z\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \implies\) \(\displaystyle \) \(\displaystyle \left({x_1 - x_2}\right)\) \(=\) \(\displaystyle k_1 z\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \land\) \(\displaystyle \) \(\displaystyle \left({x_2 - x_3}\right)\) \(=\) \(\displaystyle k_2 z\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \implies\) \(\displaystyle \) \(\displaystyle \left({x_1 - x_2}\right) + \left({x_2 - x_3}\right)\) \(=\) \(\displaystyle \left({k_1 + k_2}\right) z\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \implies\) \(\displaystyle \) \(\displaystyle \left({x_1 - x_3}\right)\) \(=\) \(\displaystyle \left({k_1 + k_2}\right) z\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle x_1\) \(\equiv\) \(\displaystyle x_3 \bmod z\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)                    

So congruence modulo $z$ is transitive.


So we are justified in supposing that congruence, as we have defined it, is an equivalence.


$\blacksquare$


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