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

\(\ds \)

\(\)

\(\ds \forall x \in \R: x - x = 0 \in \Q\)

\(\ds \)

\(\leadsto\)

\(\ds x \sim x\)

So $\sim$ is reflexive.

$\Box$

Symmetric

\(\ds x\)

\(\sim\)

\(\ds y\)

\(\ds \)

\(\leadsto\)

\(\ds x - y \in \Q\)

\(\ds \)

\(\leadsto\)

\(\ds x - y = \frac p q\)

for some integers $p, q$

\(\ds \)

\(\leadsto\)

\(\ds -\paren {x - y} = \frac {-p} q\)

Definition of Diagonal Relation

\(\ds \)

\(\leadsto\)

\(\ds y - x \in \Q\)

as $-p$ is an integer

\(\ds \)

\(\leadsto\)

\(\ds y \sim x\)

So $\sim$ is symmetric.

$\Box$

Transitive

\(\ds \)

\(\)

\(\ds \forall x, y, z \in \R: x \sim y \land y \sim z\)

\(\ds \)

\(\implies\)

\(\ds x - y = \frac p q \land y - z = \frac r s\)

for some integers $p, q, r, s$

\(\ds \)

\(\leadsto\)

\(\ds x - z = \frac p q - \frac r s\)

adding the second equality to the first

\(\ds \)

\(=\)

\(\ds \frac {s p - q r} {q s} \in \Q\)

as $s p - q r$ and $q s$ are integers

\(\ds \)

\(\leadsto\)

\(\ds x \sim z\)

So $\sim$ is transitive.

$\Box$

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

Hence by definition it is an equivalence relation.

$\blacksquare$