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I was solving this question:

On a set S , $a\;R\;b$ if $a=b$ , prove it's an equivalence relation

Edited proof:

  1. $R = \{(x,x) \;|\; x \in S \} $

    since every element $x=x$ under the relation $R$ for all $x \in S$ , hence it's reflexive

  2. $ \forall\; (x,y) \in R$ , since $x=y$ then $y=x$ also, hence it's symmetric

  3. $ \forall\; (x,y),(y,z) \in R$ since $x=y$ and $y=z$ $=>$ $x=z$, hence it's transitive

I want to ask whether my proof is right? And if right did I do mathematically or more in layman's term?

Thanks

Edit: Now it seems R contains only one element $(a,a)$

2 Answers 2

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1) What you wrote here is wrong: you must show

$\forall\,x\in S\,\,,\,\,xRx\,\,\text{(or}\,\,(x,x)\in R)$

and not $\,R=\{(x,x)\;:\;x\in S\}$

About the whole thing: you say $\,x=y\Longrightarrow y=x\,$ "by logic"...whose logic??

This, and the next transitivity point depend generally on what you can rely on: if you've studied and/or have been said the equality relation is symmetric/transitive then...well, there's nothing to prove in fact, is there? If you haven't then you'll have to use other tools that most probably were given to you in that course.

The point is: you didn't prove anything, you just wrote down the definitions of every characteristic of an equivalence relations.

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    I honestly can't @Mr.Anubis as if you're not given *anything* else I'd say the equality relation is an equivalence one *per definition*2012-09-23
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I totally agree with DonAntonio. You didn't prove anything here. In fact, information is insufficient to prove that $R$ is an equivalence relation. You need to specify some properties on set $S$.

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    Please give a look at my edited proof2012-09-23