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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.

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

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

  2. The transitive property is a consequence of the fact that the sum of two rational numbers is rational.

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  • $\begingroup$ How would you show transitivity for the irrationals? it's easy to show that if a-b is rational, and b-c is rational, then a-c is also rational- But how do you show the converse? $\endgroup$ – Dr. John A Zoidberg Feb 29 '16 at 5:21
  • $\begingroup$ @user3677952: You don't need to show the converse. Transitivity is a one-way implication. $\endgroup$ – Eric Wofsey Feb 29 '16 at 5:39

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