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 Post subject: Equivalence RelationsPosted: Wed, 27 Feb 2008 04:01:44 UTC
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Im not sure if this is the correct section to put this in or not, but I was wondering if I was on the right track for this proof.

Q: Let R be an equivalence relation on A = {a, b, c, d, e, f, g} such that aRc, cRd, dRg, and bRf. If there are three-distinct equivalence classes resulting from R, then determine these equivalence classes and determine all elements of R.

I know how to get the equivalence classes, but I am unsure how to write it appropriately. I just have something like...

Equivalence classes

First equivalence class:
blah, blah, blah (these are the equivalence classes)

Second equivalence class:
same idea here

Third equivalence class:
same idea here

And then I have that

R = {(a,a) ...etc.}

I know there is a more mathematical way to write the 3 equivalence classes, but I wasnt sure what that would look like.

Thanks so much,

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 Post subject: Posted: Wed, 27 Feb 2008 04:04:40 UTC
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You just write the members of the classes.

Obviously you have the first class has at least a,c,d, and g, the second has at least b and f, but since you're told you have three, e must be the sole inhabitant of the last class.

You don't need to write out the entire relation, simply putting the members of the same class into sets is enough to completely determine the whole relation.

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 Post subject: Posted: Wed, 27 Feb 2008 04:07:45 UTC
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Okay, thank you.
But how would you write that?
That is where I'm having trouble.
Just seeing what it looks like.

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 Post subject: Posted: Wed, 27 Feb 2008 04:11:39 UTC
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 Post subject: Posted: Wed, 27 Feb 2008 04:14:13 UTC
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Why do you not need to include all possible relations of R? Like I was showing before.

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 Post subject: Posted: Wed, 27 Feb 2008 04:25:46 UTC
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Because of the properties of an equivalence relation, naming all the equivalence classes is equivalent to naming the relation. For example, say for some equivalence class of the relation Then the equivalence relation on any set is merely being in the same equivalence class. The two notions of an equivalence relation, listing all the pairs and listing the equivalence class, are equivalent.

Proof: (i)Since a is in its own equivalence class, (a,a) would be in the relation, which shows that being in an equivalence class is reflexive.
(ii)if a and b are in the same equivalence class, both of the pairs (a,b) and (b,a) are in the relation, so its symmetric
(iii) if all of a, b, and c are in the same class then (a,b),(b,c) and (a,c) are all in the same equivalence class, so transitivity is also shown.

And if any two relations yield the same equivalence classes, then they're the same relation, so listing the classes is just as good (and much more efficient) than listing all the pairs of the relation.

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 Post subject: Posted: Wed, 27 Feb 2008 04:31:36 UTC
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So it would be correct to answer like this:

Asub(1) = {a,c,d,g}

Asub(2) = {b,f}

Asub(3) = {e}

R = (Asub(1), Asub(2), Asub(3)}

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 Post subject: Posted: Wed, 27 Feb 2008 04:33:11 UTC
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Yes, that's perfectly valid. And you can name the classes whatever you like. Maybe even use instead of for it to make more sense.

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 Post subject: Posted: Wed, 27 Feb 2008 04:36:55 UTC
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Thanks so much!! You were very helpful!

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 Post subject: Posted: Wed, 27 Feb 2008 07:39:50 UTC
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^_^

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