S.O.S. Mathematics CyberBoard

Your Resource for mathematics help on the web!
It is currently Fri, 22 Aug 2014 12:46:10 UTC

All times are UTC [ DST ]




Post new topic Reply to topic  [ 10 posts ] 
Author Message
 Post subject: Equivalence Relations
PostPosted: Wed, 27 Feb 2008 04:01:44 UTC 
Offline
Member

Joined: Tue, 12 Feb 2008 04:23:09 UTC
Posts: 26
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,

_________________
*~*KaysiE*~*


Top
 Profile  
 
 Post subject:
PostPosted: Wed, 27 Feb 2008 04:04:40 UTC 
Offline
Moderator
User avatar

Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 13993
Location: Austin, TX
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.

_________________
(\ /)
(O.o)
(> <)
This is Bunny. Copy Bunny into your signature to help him on his way to world domination


Top
 Profile  
 
 Post subject:
PostPosted: Wed, 27 Feb 2008 04:07:45 UTC 
Offline
Member

Joined: Tue, 12 Feb 2008 04:23:09 UTC
Posts: 26
Okay, thank you.
But how would you write that?
That is where I'm having trouble.
Just seeing what it looks like.
:)

_________________
*~*KaysiE*~*


Top
 Profile  
 
 Post subject:
PostPosted: Wed, 27 Feb 2008 04:11:39 UTC 
Offline
Moderator
User avatar

Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 13993
Location: Austin, TX
\{a,c,d,g\} = A_1
\{b,f\}=A_2
\{e\}=A_3

_________________
(\ /)
(O.o)
(> <)
This is Bunny. Copy Bunny into your signature to help him on his way to world domination


Top
 Profile  
 
 Post subject:
PostPosted: Wed, 27 Feb 2008 04:14:13 UTC 
Offline
Member

Joined: Tue, 12 Feb 2008 04:23:09 UTC
Posts: 26
Why do you not need to include all possible relations of R? Like I was showing before.

_________________
*~*KaysiE*~*


Top
 Profile  
 
 Post subject:
PostPosted: Wed, 27 Feb 2008 04:25:46 UTC 
Offline
Moderator
User avatar

Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 13993
Location: Austin, TX
Because of the properties of an equivalence relation, naming all the equivalence classes is equivalent to naming the relation. For example, say a,b,c\in R_k for some equivalence class of the relation R 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.

_________________
(\ /)
(O.o)
(> <)
This is Bunny. Copy Bunny into your signature to help him on his way to world domination


Top
 Profile  
 
 Post subject:
PostPosted: Wed, 27 Feb 2008 04:31:36 UTC 
Offline
Member

Joined: Tue, 12 Feb 2008 04:23:09 UTC
Posts: 26
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)}

_________________
*~*KaysiE*~*


Top
 Profile  
 
 Post subject:
PostPosted: Wed, 27 Feb 2008 04:33:11 UTC 
Offline
Moderator
User avatar

Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 13993
Location: Austin, TX
Yes, that's perfectly valid. And you can name the classes whatever you like. Maybe even use R_i instead of A_i for it to make more sense.

_________________
(\ /)
(O.o)
(> <)
This is Bunny. Copy Bunny into your signature to help him on his way to world domination


Top
 Profile  
 
 Post subject:
PostPosted: Wed, 27 Feb 2008 04:36:55 UTC 
Offline
Member

Joined: Tue, 12 Feb 2008 04:23:09 UTC
Posts: 26
Thanks so much!! You were very helpful! :D

_________________
*~*KaysiE*~*


Top
 Profile  
 
 Post subject:
PostPosted: Wed, 27 Feb 2008 07:39:50 UTC 
Offline
Moderator
User avatar

Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 13993
Location: Austin, TX
Glad to help!

^_^

_________________
(\ /)
(O.o)
(> <)
This is Bunny. Copy Bunny into your signature to help him on his way to world domination


Top
 Profile  
 
Display posts from previous:  Sort by  
Post new topic Reply to topic  [ 10 posts ] 

All times are UTC [ DST ]


Who is online

Users browsing this forum: No registered users


You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum

Search for:
Jump to:  
Contact Us | S.O.S. Mathematics Homepage
Privacy Statement | Search the "old" CyberBoard

users online during the last hour
Powered by phpBB © 2001, 2005-2011 phpBB Group.
Copyright © 1999-2013 MathMedics, LLC. All rights reserved.
Math Medics, LLC. - P.O. Box 12395 - El Paso TX 79913 - USA