Question:
Consider the following equivalence relation on the real numbers:
$(x,y) \in R \iff x-y \in \mathbb{Z}$
In the following list of equivalence classes find two equivalence classes that are equal.
$[-5],\ [\frac{8}{3}], \ [\frac{-2}{6}], [\frac{4}{3}], [\frac{1}{6}] $
My attempt:
Notice that $ \frac{-1}{3} \in [\frac{8}{3}], [\frac{-2}{6}] $.
Can we conclude that $[\frac{8}{3}], \ [\frac{-2}{6}]$ are equal? In general if I can find one common element that is in two equivalence classes would that imply they are equal? I think there is a theorem that states that two equivalence classes have nothing in common or are equal.
Note that any $r\in\Bbb R$ ,$r\in$ $[ fractional$ $part$ $(x)]$.i.e.$r_1$ and $r_2$ are in same equivalence class iff they have the same fractional part. Now compute fractional parts of the given numbers and the conclusion follows