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For $x,y\in\mathbb{R}$ define x ~ y to mean that $x-y\in\mathbb{Z}$. Prove that ~ is an equivalence relation on $\mathbb{R}$. Describe its equivalence classes.

I've successfully proved x ~ y relation is reflexive, symmetric and transitive.

What I'm not able to do is to describe the equivalence classes. In my view it can be

  • $\mathbb{Z}$ as every integer $x,y$ is $x~R~y$ as demonstrated above

  • or more specifically even $\mathbb{R}$ given that $x-y\in\mathbb{Z}$.

Am I missing something?

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    Given a single $x\in \Bbb R$, its equivalence class under $\sim$ is given by those $y$ such that $y = x+n$ where $n\in \Bbb Z$. How would you write this in set-builder notation? Can you pick a nice interval to work in for a simple description..? – Cameron Williams Dec 20 '19 at 15:30
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    Try some examples to get a feel for it. For instance, if $y = 1.27$ then $x \sim 1.27 \iff x - 1.27 \in \mathbb Z \iff \exists n \in \mathbb Z$ such that $x-1.27 = n \iff \exists n \in \mathbb Z$ such that $x = 1.27+n$. Now substitute some values of $n$ into $x=1.27+n$, like $n=0,1,2,3$ and $n=-1,-2,-3$. See if you observe a pattern, and see if you can generalize that pattern. – Lee Mosher Dec 20 '19 at 15:43
  • I think I can see it now. It's pretty much what has been given by in the answer section, so I guess it's ${x+n\in\mathbb{Z}|x\in[0,1)}$. So it's not complete $\mathbb{R}$, but something close enough. –  Dec 20 '19 at 15:48

1 Answers1

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Each equivalence class is a subset of $\mathbb R$ and therefore the set of all equivalence classes must be a set of subsets of $\mathbb R$.

In your case, for each $x\in[0,1)$, consider the set $x+\mathbb Z$. There you have it. The set of all equivalence classes is$$\{x+\mathbb Z\mid x\in[0,1)\}.\tag1$$Note that if $x\in\mathbb R$, then $x-\lfloor x\rfloor\in[0,1)$, that $x\sim x-\lfloor x\rfloor$ and that $x-\lfloor x\rfloor$ is the only $y\in[0,1)$ such that $x\sim y$. Therefore, $(1)$ describes all equivalence classes and each equivalence class appears there only once.

José Carlos Santos
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