1
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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 $\mathbf Q[\sqrt5]$.

$\endgroup$
  • $\begingroup$ when writing $Q$, should we exclude zero? $\endgroup$ – user2694307 Dec 20 '15 at 14:25
  • 1
    $\begingroup$ 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. $\endgroup$ – Bernard Dec 20 '15 at 14:26
  • $\begingroup$ For the projective line, yes. When we speak of the vector space, no. I've added a detail, I hope it makes it clearer. $\endgroup$ – Bernard Dec 20 '15 at 14:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.