For a set $E $ of real numbers, define two points in $E $to be rationally equivalent if their difference belongs to $Q $. Prove that this defines an equivalence relation.

(i) is trivial as 0 is a rational number

(ii) Suppose $r - q $ is rational. Then as $s \in Q \implies -s \in Q $, $q-r= -(r-q) \in Q $

How can I do (iii)?

Thanks in advance!


I think you mean $E$ a set of real numbers. $r-q\in \Bbb Q$ and $q-s\in \Bbb Q$, so $r-s=(r-q)+(q-s)\in \Bbb Q$.

  • $\begingroup$ Yes $E $ is a set of real numbers. $\endgroup$ – Alexander Feb 24 '14 at 13:21

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