Prove that $\sim$ defines an equivalence relation on $\mathbb{Z}$.

Question

For $a,b \in \mathbb{R}$ define $a \sim b$ if $a - b \in \mathbb{Z}$

I don't understand how I'm suppose to prove this:

Prove that $\sim$ defines an equivalence relation on $\mathbb{Z}$

Also can you help me with finding the equivalence class of 5. In other words what I'm trying to describe is the set $[5]$ = {$y : 5 \sim y$}. And $[5]$ is just the name of the set.

Edit: Please read my comment below (an attempt to solve this problem) does it make any sense? I'm sorry if I sound really dumb I'm new to this stuff.

Answer 1

Every relation on a set $X$ that is defined as $x \sim y$ iff $f(x)=f(y)$ for some function $f:X\to A$, where $A$ is another set, must be an equivalence relation just because equality is an equivalence relation.

In your example, $X=A=\mathbb R$ and $f(x)=$ the fractional part of $x$.

Answer 2

You need to check 3 things :

1. Reflexivity : $a\sim a$ because $a-a = 0 \in \mathbb{Z}$

2. Symmetry : $a\sim b$ implies that $a-b\in \mathbb{Z}$, and so $b-a\in \mathbb{Z}$ and hence $b\sim a$

3. Transitivity : If $a\sim b$ and $b\sim c$, then $a-b, b-c\in \mathbb{Z}$, and hence $$ a-c = a-b + b-c \in \mathbb{Z} $$ So $\sim$ is an equivalence relation.

Now, $[5] = \{a \in \mathbb{Z} : 5-a\in \mathbb{Z}\} = ?$