Find the equivalence classes for $a T b \iff \frac a b \in \Bbb Q$

Question

Given the set $S = \{ x − \sqrt 5 y : x,y \in \Bbb Q, \ x − \sqrt 5 y \ne 0 \}$, assume the relation $T$ is defined on $S$ by $a T b \iff \frac a b \in \Bbb Q$.

How can I find the distinct equivalence classes of $T$?

Answer 1

One equivalence class is $\Bbb Q$ itself. Each other equivalence class contains exactly one element of the form $x+\sqrt 5$, and all other elements are rational multiples thereof, i.e., $tx+t\sqrt 5$ with $0\ne t\in\Bbb Q$.

Answer 2

Fix $a,b\in \Bbb Q$ (not both zero). Consider any $x,y\in \Bbb Q$ (not both zero). Note that $$ (x+y\sqrt5)/(a+b\sqrt 5) \in \Bbb Q \iff \\ (x+y\sqrt5)(a-b\sqrt5) \in \Bbb Q \iff \\ (ay-bx)=0 \iff \\ x/y=a/b $$

Answer 3

Let $a = u - \sqrt 5 v$ and $b = s - \sqrt 5 t$. Then

$$\frac a b = \frac {u - \sqrt 5 v} {s - \sqrt 5 t} = \frac {(u - \sqrt 5 v) (s + \sqrt 5 t)} {(s - \sqrt 5 t) (s + \sqrt 5 t)} = \frac {us - 5vt} {s^2 - 5 t^2} + \frac {ut - vs} {s^2 - 5 t^2} \sqrt 5 .$$

Clearly, the first fraction is a rational number. If you want $\frac a b \in \Bbb Q$, then the second part must be $0$, since it is irrational. This forces $ut - vs = 0$.

If $v \ne 0$, then this means $s = \dfrac u v t$; if $t = 0$ then $s = 0$, but this is not possible, because $0 \notin S$, so $t \ne 0$ and the above becomes $\frac s t = \frac u v$, so $a$ and $b$ are proportional.

If $v = 0$, then $ut = 0$ and since $u \ne 0$ (for the same reason as above), necessarily $t = 0$, so $a = u$ and $b = s$, and again $a$ and $b$ are proportional (any given two rational numbers are proportional).

Thus, we have discovered that $a T b \iff a = \lambda b, \ \lambda \in \Bbb Q ^*$.

This means that one equivalence class is formed by the elements of $S$ of the form $v \sqrt 5$, a representative of which is $\sqrt 5$ itself. No element of the form $u - \sqrt 5$ with $u \ne 0$ is equivalent to it, because none can be proportional to it.

Next, note that if $u \ne 0$, then $(u - \sqrt 5 v) T (1 - \sqrt 5 \dfrac v u)$, so for each $q \in \Bbb Q$ we have another equivalence class containing all the elements of the type $u - \sqrt 5 uq$, a representative of which is $1 - q \sqrt 5$ itself.

To summarize, the set of equivalence classes (the "quotient set") is in bijection with $\Bbb Q$, and the representatives are $\sqrt 5$ and $1 - \sqrt 5 q, \ q \in \Bbb Q$.