Complete set of equivalence class representative Let  $\sim$ be a relation on $\mathbb{R}$ and $x\sim y$ if and only if $x-y\in \mathbb{Z}$.
(a) Show that $\sim$ is an equivalence relation 
(b) Give a complete set of equivalence class representatives.

(a) is easy to show, but I really don’t understand (b).
I know that: 
$$[a]_{\sim}:=\{y\in \mathbb{R} \mid a\sim y\}$$
so
$$[0]_{\sim}=\{\dots,-2,-1,0,1,2,\dots\}$$
and that 
$$\dots=[-2]_{\sim}=[-1]_{\sim}=[0]_{\sim}=[1]_{\sim}=[2]_{\sim}=\,\dots$$
because if $x\sim y$ then $[x]=[y]$.
But how can I find a complete set of equivalence class representatives?
Thanks in advance!
 A: Hint: two numbers are equivalent under this relation iff they have the same fractional part.
A: By a complete set of equivalence class representatives, I think the question wants a set $X \subset \mathbb{R}$ such that for each equivalence class, there is exactly one representative in the set - this is the sense in which it is complete (there is a $1-1$ correspondence between $X$ and $\mathbb{R}/\sim$, the set of equivalence classes).
So what you need to do is for each equivalence class, you need to pick one representative and let $X$ be the set of all your choices. As you point out, there are lots of different representatives you can choose for any given equivalence class. To ensure that your set $X$ contains only one representative for each equivalence class and that each equivalence class does have a representative in $X$, it would be good if you were able to make some sort of consistent choice of representative. For example, there is a reason I would choose $\frac{1}{2}$ to be the element in $X$ which represents $[\frac{5}{2}]_{\sim}$, and I would choose all the other representatives of equivalence classes for the same reason.
