# Discrete Maths - Sets & Relations

I'm trying to get my assignment done and I'm finding it hard to understand Relations. The question says:

Let $Q$ be the relation on the set $R$ of non-zero real numbers, where non-zero real numbers $x$ and $y$ satisfy $xQy$ if and only if $x^2/y^2$ is a rational number. Determine:

(i) whether or not the relation $Q$ is reflexive,

(ii) whether or not the relation $Q$ is symmetric,

(iii) whether or not the relation $Q$ is anti-symmetric,

(iv) whether or not the relation $Q$ is transitive,

(v) whether or not the relation $Q$ is a equivalence relation,

(vi) whether or not the relation $Q$ is a partial order.

So far I'm on the 3rd part. I understand that anti-symmetric means when $xQy$ and $yQx$ then $x=y$. This, to me looks a bit like the reflexive relation or maybe I'm wrong.

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Reflexive means that $x\sim x$. – user123123 Oct 30 '12 at 17:30
@Peter I understand that, thank you. I'm just unsure of part 3. – Adegoke A Oct 30 '12 at 17:31
Well, if $xQy$ and $yQx$, then we know that $x^2/y^2$ is rational and $y^2/x^2$ is rational. Do those two facts taken together imply that $x=y$ ? This should be easy to answer. Also, once you know the answer to (iv), you should be able to immediately answer both (v) and (vi). – Bey Oct 30 '12 at 17:32
I try to solve these type of question by using numbers. Like, x = 2 and y = 4. Is that a good way of solving them? – Adegoke A Oct 30 '12 at 17:35
This is a good way to do it. Using those values for $x$ and $y$, what can you conclude about the relation $Q$ regarding anti-symmetry? – Bey Oct 30 '12 at 17:36

The anti-symmetric relation property, as you have defined it, is the following: Whenever both $x^2/y^2$ and $y^2/x^2$ are both rational, $x=y$. So your goal is either to prove that this is the case, or find a counterexample. Can you come up with two different numbers $x$ and $y$ so that $x^2/y^2$ and $y^2/x^2$ are both rational, but $x\neq y$? (Hint: In this particular problem, you can consider nice numbers, like positive integers.)
To prove that $Q$ is antisymmetric, you must prove that if $xQy$ and $yQx$, then $x=y$. Break that down: you must prove that if $\frac{x^2}{y^2}$ is rational and $\frac{y^2}{x^2}$ is rational, then $x=y$. Does that seem likely? What if $x=1$ and $y=2$, say?