Find the equivalence class of: [(1,2)],[(2,3)] and [3,-1]. In $ \mathbb{R^2}$ we consider the relation $(x,y)R(a,b)$ if and only if exists $n \in  \mathbb{Z} $ such that $n-1<y\leq n \ $ and $ n-1<b\leq n \  $. The relation is an equivalence relation.
Find the equivalence class of $(1,2)$, $(2,3)$, and $(3,-1)$.
[(1,2)]
$ [(1,2)]=\{(x,y)\in \mathbb{R^2} :(x,y)R(1,2) \}=\{(x,y)\in \mathbb{R^2} : n-1<y\leq n \ $ and $n-1<2\leq n \}  $
[(2,3)]
$ [(2,3)]=\{(x,y)\in \mathbb{R^2} :(x,y)R(2,3) \}=\{(x,y)\in \mathbb{R^2} : n-1<y\leq n \ $ and $n-1<3\leq n \}  $
[(3,-1)]
$ [(3,-1)]=\{(x,y)\in \mathbb{R^2} :(x,y)R(3,-1) \}=\{(x,y)\in \mathbb{R^2} : n-1<y\leq n \ $ and $n-1<-1\leq n \}  $
Is correct my answer?
 A: Technically, you've correctly applied the definition, but the intended answer was probably a simple description of the equivalence classes.
A note on terminology before deriving a simpler solution: $n$ is not quantified in your answers. Correctly, you would have to add $\exists n\in\mathbb Z$ to the set comprehensions. For example:
$$[(1,2)] = \{(x,y)\in\mathbb R^2: \color{red}{\exists n\in\mathbb Z \text{ s.t. }} n-1<y\le n \text{ and } n-1 < 2 \le n\}$$
Firstly, note that the relation is independet of $x$ and $a$, i.e. the first component is free.
This means, if $(x, y) R (a,b)$, so does $(x', y) R(a', b)$ for any $x', a' \in\mathbb R$.
Equivalence classes are therefor always of the form $\mathbb R \times A$ where $A$ is some subset of $\mathbb R$.
To further examine the definition, you will find that these sets $A$ are described by a whole number $n$:

If $A_n := (n-1, n]$, then $\mathbb R \times A_n$ is an equivalence class of $R$.

The intended answers therefor are likely the following:

 $[(1,2)] = \mathbb R \times (1, 2]$
 $[(2,3)] = \mathbb R \times (2, 3]$
 $[(3,-1)] = \mathbb R \times (-2, -1]$

All of these are, of course, equivalent to your answer.

A slightly more general way to tackle such problems is finding a function $f$, such that $xRy \Leftrightarrow f(x) = f(y)$. In this case its $f=\lceil \cdot \rceil \circ \pi_2$ (first project on the second coordinate, then take the ceiling ($\lceil x\rceil = n \in \mathbb Z \Leftrightarrow n-1 < x \le n$).
The equivalence class of a point $x$ is then $ [x]_R = f^{-1} \circ f(x)$ (i.e. the preimage of the image under $f$)
A: You seem to be having the same issue as before. You should start with something along the lines of
\begin{align*}
[(1,2)]
&= \{(x,y)\in\mathbb{R}^2 : (x,y)R(1,2)\} \\
&= \{(x,y)\in\mathbb{R}^2
    : \exists n\in\mathbb{Z}\ \text{satisfying}\ n-1 <y \le n\ \text{and}\ 
      n-1 < 2 \le n\}.
\end{align*}
Note the part "$\exists n\in\mathbb{Z}$ satisfying...". It's very important that you describe the variables you are using. In your case, the "$n$" in each of your sets was undefined.
Now from here you should notice that there is only one $n\in\mathbb{Z}$ which satisfies $n-1 < 2 \le n$, namely $n=2$.
This gives us the condition $1=2-1 < y \le 2$, or $y\in(1,2]$ whenever $(x,y)R(1,2)$.
Therefore $(x,y)R(1,2)$ if and only if $y\in(1,2]$, so that we can write
$$
[(1,2)] = \{(x,y)\in\mathbb{R}^2 : y\in(1,2]\} = \mathbb{R}\times (1,2].
$$
Doing the same thing you can find the equivalence classes $[(2,3)]$ and $[(3,-1)]$.
