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$A$ is the "absolute value" relation on $\Bbb R$: For all real numbers $x$ and $y$, $x A y\Leftarrow\Rightarrow |x| = |y|$.

Determine whether the given relation is reflexive, symmetric, or transitive, or none of these. Justify your answer.

I am struggling with the proofs for this question. I believe it is all three, but I have confused myself overthinking it.

AlienRem
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bacon
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  • Welcome to MSE! What've you tried exactly? – Arpit Kansal Sep 25 '15 at 12:10
  • I know that reflexive means that for all x, (x,x) is in the relation. Of course it is, but I don't know how to formally write the proof. I know that for symmetry, it means that if (x,y) is in the relation, then (y,x) is also in the relation - also true. I think for transitivity, it would be vacuously true. – bacon Sep 25 '15 at 12:26
  • I've added the details in my answer. – Arpit Kansal Sep 25 '15 at 12:33

1 Answers1

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To show reflexivity, note that for every $x\in R$, we have $xAx$ (just by definition)

To show symmetry, note that for every $x,y\in R$, we have $xAy$ implies $yAx$(again just by definition)

To show transitivity, note that for every $x,y,z\in R$, we have $xAy$ and $yAz$ implies $xAz$ because $xAy$ means $\vert x \vert =\vert y \vert$ and $yAz$ means $\vert y \vert =\vert z \vert$ therefore $\vert x \vert =\vert z \vert$,hence $xAz$.

Arpit Kansal
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  • thank you - so it seems there isn't anything 'complex' in the proof, as I thought i needed to show. It just is reflexive and symmetric by definition. i appreciate it! – bacon Sep 25 '15 at 12:34
  • yeah,It just follows from definition.if you are okay with my solution you can accept/upvote my answer :) – Arpit Kansal Sep 25 '15 at 12:36