Difference is Rational is Equivalence Relation

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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.

$\blacksquare$