# Proving that rational equivalence is an equivalence relation on any set.

I seek to prove that the rational equivalence relation is an equivalence relation, in that it is reflexive, symmetric, and transitive.

The rational equivalence relation is as follows "Two numbers in a set are rationally equivalent provided their difference is rational".

I know that it is reflexive, since for two points in a set E, a and b, if |a-b| is rational then |b-a| is also rational, because the two are equivalent, and similarly for the irrational case.

I'm not sure how to prove symmetry and transitivity.

• What you proved is the symmetry. For reflexiveness note that $|a-a|=0$ and $0$ is a rational number. – Sak Feb 29 '16 at 5:10
• That would explain my confusion! – Dr. John A Zoidberg Feb 29 '16 at 5:11

1. The absolute value signs are a needless distraction: $|a|$ is rational if and only if $a$ is rational.