# What's the partition associated with the equivalence relation?

This is a question from Pinter p.124. It asks me to describe the partition of the set of nonzero real numbers, $\mathbb{R}^*$ where the associated equivalence relation is $a$ ~ $b$ iff $\frac{a}{b}$ $\in \mathbb{Q}$.

I'm not sure if I can describe it as either:

{$A_r : r \in \mathbb{Q} \times \mathbb{R}^*$} where $A_r = \{x\in\mathbb{R}^* : \frac{x}{b} = a, r = (a,b)\}$

or

{$[y]:y\in \mathbb{R}^*$} where $[y] = \{ x \in\mathbb{R}^*:\frac{x}{y} \in \mathbb{Q}\}$

assuming that each disjoint set can have repeating labels? Thanks a lot.

• Your descriptions, while not wrong, do not really shed light on the structure of the partition. You are just using the definition of the equivalence class, and I believe the question is asking for more than that. – theyaoster May 19 '18 at 19:46

Hint 1: Given a rational number $q$, for every non-zero rational $p$, $p/q$ is also rational. Furthermore, for an irrational $x$, $q/x$ is irrational. This argument classifies the equivalence classes whose elements are all rational.
Hint 2: For the equivalence classes containing irrational numbers, take an irrational $x$. Then, if $x \sim y$ for $y \in \mathbb{R}^*$, then $y$ must be irrational (why?). Recall that $[x]$ contains all $y$ such that $x \sim y$, so determining the elements of $[x]$ comes down to determining when the ratio $x/y$ of two irrational numbers is rational.
• Hi Brian, thanks for your hints. I have thought those before but got stuck on the condition when the ratio of 2 irrational numbers is rational. It is clear that the ratio of $ax$ and $bx$ where $a$ and $b$ are rational and $x$ is irrational will yield a rational number. However, I'm not sure if 2 distinct irrationals that cannot be written as a multiple of a common irrational, their rational can still be rational. And if that's not the case, I suppose there'll still be infinitely many classes because there are infinitely many irrational numbers. – mackbox May 19 '18 at 20:03
• Good reasoning. Given an irrational $x$, assume $x/y$ is rational. Then, consider two cases: $y$ is either a rational multiple of $x$, or an irrational multiple of $x$. You already said what happens in the first case. What happens in the second? In fact, is it even possible? – theyaoster May 19 '18 at 20:13
• Thanks again Brian. That's the problem, mate. If $a$ and $b$ are distinct, irrational and cannot be written as a multiple of any irrational number besides themselves, I'm not sure if you divide $a$ by $b$, whether it can give you a rational number. Intuitively, that's not true but I don't know how to prove it. (say if I start by saying $a/b = c$, then $a = b \times c$, then what?). Could you give me a few more pointers so I can see why? – mackbox May 19 '18 at 20:40
• It could be easier to think about the contrapositive: show that if $a/b$ is rational, then $a$ and $b$ can be written as rational multiples of one another. Start by assuming $a/b = c$ is rational. Then, $a = bc$ shows that $a$ is a rational multiple of $b$ since $c$ is rational. Does this clear it up? – theyaoster May 19 '18 at 21:11
• That's very clear thanks a lot! I didn't realize your previous comment was actually based on contrapositive! So now the equivalent classes for the irrational numbers in $\mathbb{R}^*$ can be stated as $[x] = \{ y \in \mathbb{R}^* : y$ is a rational multiple of $x \}$, is that right? – mackbox May 19 '18 at 21:37