Define a relation $R$ on $\mathbb R$ by $xRy$ if and only if $y − x \in \mathbb Z$.
(a) Show that $R$ is an equivalence relation on $\mathbb R$.
(b) Describe the set of all $x \in \mathbb R$ which are $R$ equivalent to the number $1$. What are the elements of the equivalence class $[\sqrt 2]$?
I believe that I have part (a) correctly but I am not sure about $xRx$. For reflexive I have if $x\in \mathbb R$, then $x-x=0$ and therefore $xRx$. I'm not sure if this is supposed to be $y\in \mathbb R$ and $y-y=0$.
I do not know where to start in order to find the equivalence class
To show that $R$ is an equivalence relation, we have to show three properties: reflexivity, symmetry, and transitivity. (I assume that when you say a relation "$R$ on $R$," you mean that $R$ is a subset of $\mathbb{R} \times \mathbb{R}$. So this answer will assume that your second $R$ is in fact the real numbers. If not, tell me and I will edit.)
Reflexivity: For any $a \in \mathbb{R}$, we have $a - a = 0 \in \mathbb{Z}$, so $aRa$ is satisfied.
Symmetry: For $a, b \in \mathbb{R}$, suppose that $aRb$, so $a - b \in \mathbb{Z}$. But $-(a-b) = b - a \in \mathbb{Z}$ (the integers are closed under multiplication), so $bRa$ holds.
Transitivity: Suppose $a, b, c \in \mathbb{R}$, and that $aRb$ and $bRc$ are satisfied. Hence, $a - b \in \mathbb{Z}$ and $c - b \in \mathbb{Z}$. Define $x := a - b$ and $y := c - b$. We have to show that $a - c \in \mathbb{Z}$, but we have $$a - c = (x+b) - (y+b) = (x+y) + (b - b) = x + y,$$ but the sum of two integers is another integer, as the integers are closed under addition. Hence, $aRc$ holds.
(b) The set of all numbers $R$ equivalent to $1$ is $$[1] = \{x \in \mathbb{R} \mid xR1\} = \{x \in \mathbb{R} \mid x - 1 \in \mathbb{Z}\}.$$
The equivalence class $\sqrt{2}$ contains the set of all elements $R$ equivalent to $\sqrt{2}$. That is, $$[\sqrt{2}] = \{x \in \mathbb{R} \mid x - \sqrt{2} \in \mathbb{Z}\}.$$