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Source: Discrete Mathematics with Applications by Susanna Epp

Is $[a]_R$ the same as [a]=a/R? Then, $x \in ([a]_R$=a/R) is the same as $x \in ([a]=a/R)$, right?

buzzee
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    As the book says, $[a]_R$ is a more precise version of $[a]$ when there are several equivalence relations floating around, but I have no idea what you mean by $a/R$. – Ted Jan 26 '16 at 06:27
  • @Ted a/R is another notation for [a] in that definition. I learn chapters of set theory first before going throug each chapter of discrete mathematics. FYI Definition 6. in Set Theory by You-Feng Lin. http://math.stackexchange.com/questions/1626126/isnt-x-e-y-e-%e2%87%94-x-e-y-deduced-not-just-x-e-y-e-%e2%87%92-x-e-y-from-a-x-e%e2%89%a0-%c3%98 – buzzee Jan 26 '16 at 07:14

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$a/R$ is an alternate notation for the equivalence class of $a$, so yes, $[a]_R = a/R$. It's not a bad notation, as it's consistent with "$X/R$" denoting the set of all $R$ equivalence classes. You'll probably never see them both in the same document, however -- or if you do, one of them will have another meaning (let's hope). Well, except for documents like this one, which discuss whether they mean the same thing and conclude that they do :)

BrianO
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