There is a relationship on $\Bbb R$ defined aRb if a-b is a rational number. I already proved its an equivalence relation in $\Bbb R$. My question is how to describe the equivalence classes? Here is my attempt at the answer:
[0]=$\{x\in \Bbb R : xR0\}$ = $\{x\in \Bbb R : x-0 $ is rational $\}$
[a]=$\{x\in \Bbb R : xRa\}$ = $\{x\in \Bbb R : x-a $ is rational$\}$
Let $x \in \mathbb{R}$, then
$[x] = \{x + q : q \in \mathbb{Q}\}$
Thus $[0] = \{0 + q : q \in \mathbb{Q}\} = \{q : q \in \mathbb{Q}\} = \mathbb{Q}$.
To see this, note that by your definition, $[x]$ is the set of all $a$ such that $a - x = q$ for some $q \in \mathbb{Q}$. Hence $a = x + q$ for $q \in \mathbb{Q}$.
If you know some group theory, an alternatively (but essentially equivalent) way of thinking of the equivalent using coset. $(\mathbb{R}, +)$ is an abelian group. $(\mathbb{Q}, +)$ is a normal subgroup. Hence the equivalence classes are the elements of the group $\mathbb{R} / \mathbb{Q}$.