Define a relation on the set of all real numbers $x,y \in \mathbb{R} $ as follows: Define a relation on the set of all real numbers $x,y \in \Bbb{R} $  as follows:
$x \sim y$ if and only if $x - y \in \Bbb{Z}$
Prove this is an equivalence relation and find the equivalence class of number $\dfrac {1}{3}$.
ATTEMPT:
Reflexive: For any $x\in\Bbb Z$, ${x} - {x}=0$ and $0 \in \mathbb{Z}$.
Symmetric: For any $x,y \in \Bbb Z$, if  $x - y \in \Bbb Z$ then $y -  x \in \Bbb Z$.
Transitive: For any $x,y,z \in \Bbb Z$, if $ {x} - {y} = 2k$ and ${y} - {z}= 2l$ for some $k,l \in \Bbb Z$, then ${x} - {z} = 2(k+l)$
Is the proof correct? And how do I find equivalence class?
 A: Your ideas are good, but your proof contains some errors. First of all we are talking about $x,y\in\mathbb{R}$ not $x,y\in\mathbb{Z}$. Fixing your errors, the proof would look something like this:

Reflexive: $\forall x\in\mathbb{R}$, $x-x=0$ so $x\sim x$. Therefore $\sim$ is reflexive.
Symmetric: For $x,y\in\mathbb{R}$, if $x\sim y$ then $x-y\in\mathbb{Z}$ so $x-y=k$ for some $k\in\mathbb{Z}$. Since $-k\in\mathbb{Z}$ it follows that $y-x\in\mathbb{Z}$ and so $y\sim x$. Therefore $\sim$ is symmetric.
Transitive: For $x,y,z\in\mathbb{R}$, if $x\sim y$ and $y\sim z$ then $x-y\in\mathbb{Z}$ and $y-z\in\mathbb{Z}$. Since addition is closed in the integers $(x-y)+(y-z)=x-z\in\mathbb{Z}$ and so $x\sim z$. Therefore $\sim$ is transitive.

To find the equivalence class of $\frac13$ you need to find all $x\in\mathbb{R}$ such that $\frac13-x\in\mathbb{Z}$ i.e. 
$$\frac13-x=k,k\in\mathbb{Z}$$
Therefore $x=\frac13-k$ and so the equivalnce class of $\frac13$ is just 
$$\frac13+\mathbb{Z}=\{\dots,\frac13-1,\frac13,\frac13+1,\dots\}$$
A: Here is another way to proceed - one that works quite generally.  Notice that $\rm\,x\sim y\,$ iff $\rm\,x\,$ and $\rm\,y\,$ have equal fractional parts, i.e. $\rm\:x\sim y \iff f(x) = f(y)\ $ for $\rm\:f(x) = $ fractional part of $\rm\,x.$
Now it is straightforward to prove that any relation of the above form is an equivalence relation.
More generally, suppose $\rm\ u\sim v\ \smash[t]{\overset{\ def}{\iff}}\, f(u) \approx f(v)\ $ for a function $\rm\,f\,$ and equivalence relation $\,\approx.\, \ $ Then the equivalence relation properties of $\,\approx\,$ transport (pullback) to $\,\sim\,$ along $\rm\,f\,$ as follows:


*

*reflexive  $\rm\quad\ f(v) \approx f(v)\:\Rightarrow\:v\sim v$

*symmetric  $\rm\,\ u\sim v\:\Rightarrow\ f(u) \approx f(v)\:\Rightarrow\:f(v)\approx f(u)\:\Rightarrow\:v\sim u$

*transitive $\rm\ \ \ u\sim v,\, v\sim w\:\Rightarrow\: f(u)\approx f(v),\,f(v)\approx f(w)\:\Rightarrow\:f(u)\approx f(w)\:\Rightarrow u\sim w$
Such relations are called (equivalence) kernels. One calls $\, \sim\,$ the $\,(\approx)\,$ kernel of $\rm\,f.$ 
A: Just to add that the collection of equivalence classes can be represented as a circle $S^1$ , since , as said by others here, $x$~$y$ iff $x,y$ have the same decimal expansion; if $x-y$ is in $\mathbb Z$ , then $x,y$ must have the same decimal representation. Then all the equivalence classes are represented in the segment $[0,1)$ . BUT: note that $0$~$1$ here, since $1-0=1$ is in $\mathbb Z$. After identifying $0,1$ to a point, we get a copy of $S^1$.
