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.
Best Answer
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 $1\in\mathbb{Q}$.