Showing $ x-y\in\mathbb{Q}$ is an equivalence relation? How would one show the the following is an equivalence relation.
The relation R on the real numbers given by xRy iff number
$ x-y\in\mathbb{Q}$.
This is what I did.
Reflexive 
Let $x \in \mathbb{R}$ and $y\in \mathbb{R}$ then $x-y \in \mathbb{Q}$ is the the same as $x-y \in \mathbb{Q}$
Thus xRx yRy
Symmetric 
Let xRy then $x-y \in \mathbb{Q}$ is the same as 
$y-x \in \mathbb{Q}$ thus yRx.
Transitive
x,y,z are real numbers/
Let xRy be $x-y \in \mathbb{Q}$, then let yRz be $y-z \in \mathbb{Q}$. Thus 
$x-z \in \mathbb{Q}$ and in conclusion xRz showing transitive. 
Now I have to find the equivalence class of the following
$0=\{1/2-1/2,2-2,3-3\}$  any real minus itself
$1=\{2-1,3-2,4-3\}$
$\sqrt{2}$= empty set b/c it cannot be written as a rational.
 A: Your proof of reflexivity is incorrect.  To show that $xRx$, show that $x-x$ is rational (and it is because $0 \in \mathbb Q$).  Your proof of symmetry and transitivity should make some reference to the fact that $\mathbb Q$ is closed under negatives and addition, respectively.
Now, your discussion of the equivalence classes of $0$, $1$, and $\sqrt{2}$ are incorrect.  The equivalence class of $x$ is the set of all $y$ such that $x R y$.  So we are looking for all $y$ such that $y - x = q$ is rational.  Solving this equation, we're looking for all numbers of the form $x + q$, for $q$ rational.  If you know some abstract algebra, the equivalence class in this case is the coset $x + \mathbb Q$.
Therefore, the equivalence class of $\sqrt{2}$ is not empty ($\sqrt{2}$ should be in that set, at the very least), but rather should include all numbers of the form $\sqrt{2} + q$ for $q \in \mathbb Q$.  This thought process should be similar for the other two cases.
A: Your "reflexive" proof is wrong. It should read $$\forall x\in\mathbb{R}, \quad x-x=0\in\mathbb{Q}\implies xRx$$
I'm not sure if this was implied, but for transitivity the correct argument is $$\begin{align}xRy, \;yRz&\implies x-y,y-z\in\mathbb{Q} \\ &\implies (x-y)-(y-z)\in\mathbb{Q} \\ &\implies x-z\in \mathbb{Q} \\ &\implies xRz.\end{align}$$ 
The equivalence class of a number is the set of all numbers at a rational distance away from it. In other words, $$\bar{0}=\mathbb{Q}, \\ \bar{1}=1+\mathbb{Q}=\mathbb{Q}=\bar{0}, \\ \overline{\sqrt 2}=\sqrt 2 +\mathbb{Q}.$$ I have used "coset" notation. That $1+\mathbb{Q}=\mathbb{Q}$ follows from $1\in\mathbb{Q}$.
A: Your phrasing of things is a little odd; I think you might be misunderstanding something.
For example, to prove the transitive property, you need to assume that you start with $x,y,z$ such that $xRy$ and $yRz$ and make an argument that eventually concludes $xRz$. e.g. you might write

Let $xRy$ and $yRz$. Therefore ... and thus $xRz$.

or better yet,

Let $x,y,z$ satisfy $xRy$ and $yRz$. Therefore ... and thus $xRz$.

Your actual statements, like

Let $xRy$ be $x - y \in \mathbf{Q}$

sound more like you're stating a definition of $R$ rather than, for example, stating the hypothesis that you have variables $x$ and $y$ that satisfy $xRy$, and then inferring that $x - y \in \mathbf{Q}$.
