Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

Thanks in advance.

share|improve this question
    
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

2 Answers 2

up vote 1 down vote accepted

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.)

share|improve this answer
1  
Using x = 2 and y = 4, gave me 4 and (1/4) those that prove they are anti-symmetric? –  Adegoke A Oct 30 '12 at 17:42

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?

share|improve this answer
1  
OK. I did: 1^2/ 2^2 = 1/4 and 2^2 / 1^2 = 4 They are both rational numbers but x those not equal to y –  Adegoke A Oct 30 '12 at 17:38

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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