Equivalence relation and Distinct equivalence classes 
Given the set $S = \{x-y \sqrt5: x, y$ are rational numbers and $x-y \sqrt5 \neq 0\}$. Assume the relation $T$ is defined on the set $S$ by $a T b$ if $a/b$ is a rational number.

Question has two parts;
a. Prove that $T$ is an equivalence relation.
b. Find the distinct equivalence classes of $T$.
I know if $T$ is equivalence relation, it should be reflexive, symmetric and transitive. I should prove them in order I guess.However, for this question for set $S$ there is $x$ and $y$. After that it says $a$ and $b$. I don't understand their relation. Also, how should I use $x-y \sqrt5$, what is the relation of $x-y \sqrt5$ and $a/b$?
 A: $a$ and $b$ are just elements of the set $S$ having the form of $x-y\sqrt{5}$
Say $a= x_1-y_1\sqrt5$ and $b=x_2-y_2\sqrt5$ where $x_1,x_2,y_1,y_2\in \mathbb{Q}$.
A: Hint
You have to consider two elements $a,b \in S$ and show that : $aTa$, if $aTb$, then $bTa$, and so on.
To say that $a,b \in S$ means that :

$a=x_1−y_1 \sqrt 5 \ne 0$ and $b=x_2−y_2 \sqrt 5 \ne 0$

with : $x_1, y_1, x_2, y_2 \in \mathbb Q$.

Thus, for reflexivity : $aTa$ iff $a/a$ is a rational number. So, we have only to "calculate" $a/a$...
For symmetry, we have to show that : if $a/b$ is a rational number, then also $b/a$ is. Say : $a/b=q$; then, what is $b/a$ ?
A: For showing that $T$ is an equivalence relation :-
1) Reflexive :- $\frac{x-y\sqrt{5}}{x-y\sqrt{5}}=1$
2) Symmetric :-If $a=x-y\sqrt{5}$ and $b=p-q\sqrt{5}$ and $aTb$ then clearly $bTa$ as if $x\in \mathbb{Q}$then so is $\frac 1x$.
3) Transitive :- Check yourself (if you have understood it).
For equivalence classes :-
$[x-y\sqrt5]=\{p-q\sqrt5\}\mid \frac {x-y\sqrt5}{p-q\sqrt5}\in \mathbb{Q}\}$ and hence $(x-y\sqrt5).(p+q\sqrt5)\in \mathbb{Q}$ (why ? ), thus giving the relation that $xq-yp=0$.
