Proving $\;x-y \in2\pi \mathbb Z\;$ defines an Equivalence Relation

Question

Prove that $\;x-y \in2\pi \mathbb Z\;$ defines an equivalence relation on $\;\mathbb R.$

Answer 1

I'm assuming that "$x-y \in 2\pi Z$ defines an equivalence relation on $R$" $\;$ means that for $x, y \in \mathbb R$, the equivalence relation in question is the relation $R = \{(x, y) \in \mathbb R \times \mathbb R \mid x - y \in 2\pi \mathbb Z\}\;$ where $\mathbb R$ is the set of real numbers, $\mathbb R \times \mathbb R$ is the set of all ordered pairs of reals, and $2\pi \mathbb Z$ is the set of all integer multiples of $2\pi$.

Put differently, $x R y$ means $x, y$ are real numbers such that $x - y \in 2\pi \mathbb Z \iff 2\pi\mid (x - y)$: that is, $2 \pi$ divides $x - y$, or $x \equiv y \pmod {2\pi}$

Now, you need to establish whether $R$ is an equivalence relation:

Is $R$ reflexive? Is it the case that for $x \in \mathbb R$, $x - x = 2\pi k$, where $n$ is some integer? Note, for any $x \in \mathbb R$, put $k = 0$...and your have reflexivity.

Is $R$ symmetric?

Is it the case that for all $x, y \in \mathbb R$, that if $x - y = 2\pi k$, then $y - x = 2\pi j$, where k, j are some integers? Hint: consider $j = -k \in \mathbb Z$, then $x - y = 2\pi k = -(y - x)$

Is $R$ transitive? Is it the case that for all $x, y, z \in \mathbb R$, that IF $x - y = 2\pi k$ AND IF $y - z = 2\pi j$, then it must follow that $x - z = 2\pi m$ for some integers j, k, m? If so, and this CAN be easily shown to be true, then the relation is transitive.

If you can answer yes to each of the above, then you will have proven that $R$ is an equivalence relation.

If you have questions about any particular property, and/or specify what you've figured out, and where you're stuck, I'll be happy to give you some more nudges to clear things up.