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?
Best Answer
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 $\mathbf Q[\sqrt5]$.