# Difference is Rational is Equivalence Relation

## Theorem

Define $\sim$ as the relation on real numbers given by:

$x \sim y \iff x - y \in \Q$

That is, that the difference between $x$ and $y$ is rational.

Then $\sim$ is an equivalence relation.

## Proof

Checking in turn each of the criteria for equivalence:

### Reflexive

 $\displaystyle$  $\displaystyle \forall x \in \R: x - x = 0 \in \Q$ $\displaystyle$ $\leadsto$ $\displaystyle x \sim x$

So $\sim$ is reflexive.

$\Box$

### Symmetric

 $\displaystyle x$ $\sim$ $\displaystyle y$ $\displaystyle$ $\leadsto$ $\displaystyle x - y \in \Q$ $\displaystyle$ $\leadsto$ $\displaystyle x - y = \frac p q$ for some integers $p, q$ $\displaystyle$ $\leadsto$ $\displaystyle -\paren {x - y} = \frac {-p} q$ Definition of Diagonal Relation $\displaystyle$ $\leadsto$ $\displaystyle y - x \in \Q$ as $-p$ is an integer $\displaystyle$ $\leadsto$ $\displaystyle y \sim x$

So $\sim$ is symmetric.

$\Box$

### Transitive

 $\displaystyle$  $\displaystyle \forall x, y, z \in \R: x \sim y \land y \sim z$ $\displaystyle$ $\implies$ $\displaystyle x - y = \frac p q \land y - z = \frac r s$ for some integers $p, q, r, s$ $\displaystyle$ $\leadsto$ $\displaystyle x - z = \frac p q - \frac r s$ adding the second equality to the first $\displaystyle$ $=$ $\displaystyle \frac {s p - q r} {q s} \in \Q$ as $s p - q r$ and $q s$ are integers $\displaystyle$ $\leadsto$ $\displaystyle x \sim z$

So $\sim$ is transitive.

$\Box$

Thus $\sim$ has been shown to be reflexive, symmetric and transitive.

Hence by definition it is an equivalence relation.

$\blacksquare$