Finding the equivalence classes of a relation Define a relation on the set of all real numbers $x, y \in \mathbb{R}$:
$x ≃  y$  if  and only if $x − y \in \mathbb{Z}$
Prove that this is an equivalence relation, and find the equivalence class of the number $1/3$.
I proved that the relation is:
reflexive
$$x-x = 0 \in \mathbb{Z}$$
symmetric
$$x-y = y-x \in \mathbb{Z}$$
transitive
$$x-y \in \mathbb{Z}$$
$$y-z \in \mathbb{Z}$$
$$x-z \in \mathbb{Z}$$
Are these proofs enough?
I'm stuck on the step where I need to find the equivalence class.

An equivalence class of an element $x$ is the set of all elements which are in relation $\sim$ with $x$. In your case $x=1/3$, so the equivalence class is the set of all elements $y$ such that $\frac{1}{3}-y = k \in \mathbb Z$. It follows that $y=\frac{1}{3}-k = l+\frac{1}{3}$, $l \in \mathbb Z$. Therefore, the equivalence class of $\frac{1}{3}$ is
$$\left[\frac{1}{3}\right]_\sim = \left\{k+\frac{1}{3} : k\in \mathbb Z\right\}$$
Described in common language, this is the set of all fractions with $3$ in the denominator and their numerator divided by $3$ gives the remainder $1$. Some of these numbers are $\frac{1}{3}, \frac{4}{3}, \frac{7}{3}, \cdots$
