# Finding distinct equivalence classes

Q: Given the set $S = \{x - y\sqrt 5 : \text{x, y are rational numbers and }x - y \sqrt5 \neq 0 \}$. Assume the relation defined on the set $S$ by $a\ T\ b$ if $a/b$ is a rational number. Find the distinct equivalence classes of T.

The question actually first asks a proof that $T$ is an equivalence relation. It is easy to prove that. But I'm not sure how to show distinct equivalence classes of $T$. The following is what I know about the equivalence classes

• No two distinct equivalence class should have an element in common.
• If $a \sim b$, they belong to the same equivalence class.
• The union of all the equivalence classes should give you the set $S$.

So, when you think about the elements of $S$, for example,

$1 - \sqrt 5,\ \dfrac{1}{2}(1 - \sqrt 5), \ \dfrac{1}{3}(1 - \sqrt 5)$, ...

$2/3 - \sqrt 5,\ \dfrac{1}{2}(2/3 - \sqrt 5), \ \dfrac{1}{3}(2/3 - \sqrt 5)$, ...

$\sqrt 5,\ \dfrac{1}{2}(-\sqrt 5), \ \dfrac{1}{3}(\sqrt 5)$, ...

$1,\, \ 2, \ 3, \ \dfrac{1}{2}, \ - \dfrac{1}{2}$, ...

are all distinct equivalence classes of T.

But I know that there are more lines to add since $x,y \in Q$ in $x - y\sqrt5$, and of course more elements to add to each line since any rational multiple of $x - y\sqrt5$ is in the same equivalence class as $x - y\sqrt5$.

So, in what form should I give the equivalence classes?

Translate the equivalence in equations: $$\frac{x-y\sqrt5}{ u-v\sqrt5}\in\mathbf Q\iff(xu-5vy)+(xv-yu)\sqrt5\in\mathbf Q\iff (x,y)\enspace\text{and}\enspace(u,v)\enspace\text{are collinear}.$$ Hence the equivalence class of $x-y\sqrt5$ is simply $\;\mathbf Q^*\cdot(x-y\sqrt5)$. In other words, you obtain the projective line over $\mathbf Q$ associated to the vector space $\mathbf Q[\sqrt5]$.
• when writing $Q$, should we exclude zero? – user2694307 Dec 20 '15 at 14:25
• It's the same set. Or am I misunderstanding ? Btw, you for the $y$ in your comment, and I forgot a dot in my answer to avoid any ambiguity. – Bernard Dec 20 '15 at 14:26