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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.

  • 0
    Yes: $x\neq 4$, and yet both $x^2/y^2=1/4$ and $y^2/x^2=4$ are rational. So $Q$ is not anti-symmetric.2012-10-30

2 Answers 2

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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.)

  • 1
    Using x = 2 and y = 4, gave me 4 and (1/4) those that prove they are anti-symmetric?2012-10-30
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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?

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    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 y2012-10-30