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 May 5th, 2010, 03:41 AM #1 Newbie   Joined: Oct 2009 Posts: 14 Thanks: 0 Equivalence Classes Hi, I have a task here that ask me to find the equilvalence classes. The task is written as following: Let X = {2, ..... , 12} ? N and define a relation R on X with nRm <=> m/n ? N Now let S be the least (or smallest) equivalence relation that expand R so that we have a relation 2S3. Find S by describing all the equivalence classes to S. The problem for me here is that I can't see how we have a relation 2S3 at all. And I can't describe the equivalence classes without knowing what the relation S is here. I do know that S is an equivalence relation since it is said in the task's text, which also mean that S is a relation that is reflexive, symmetric and transitive. But still, I just can't see what the S relation is, especially when we have a relation 2S3 somehow. So I would appreciate any kind of help or suggestions here.
 May 5th, 2010, 07:57 AM #2 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: Equivalence Classes What does "expand R" mean? I suspect that if I understood that I could answer the question.
 May 5th, 2010, 08:14 AM #3 Newbie   Joined: Oct 2009 Posts: 14 Thanks: 0 Re: Equivalence Classes Thanks for reply, If I understand the text correctly, "expand R" might mean that S is an equvalence relation on X so that nRm <=> nSm, or something like that. There are no clear definitions of the word "expand" in the text so that's what I guess is the meaning of "expand". Maybe you have a different opinion on what S expanding R might be?
May 5th, 2010, 08:36 AM   #4
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Re: Equivalence Classes

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 Originally Posted by Zhai Thanks for reply, If I understand the text correctly, "expand R" might mean that S is an equvalence relation on X so that nRm <=> nSm, or something like that. There are no clear definitions of the word "expand" in the text so that's what I guess is the meaning of "expand". Maybe you have a different opinion on what S expanding R might be?
There is only one relation S such that, for all n and m, nRm <=> nSm, and in your case it's not an equivalence relation.

Perhaps it means that S is a superset of R, viewed as a collection of ordered pairs (m, n)?

 May 5th, 2010, 08:44 AM #5 Newbie   Joined: Oct 2009 Posts: 14 Thanks: 0 Re: Equivalence Classes Can you give me a concrete example on that please, i don't quiet get it yet. Because you might be right, and if so, that can be what I have missed to solve this task.
May 5th, 2010, 09:08 AM   #6
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Re: Equivalence Classes

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 Originally Posted by Zhai Can you give me a concrete example on that please, i don't quiet get it yet. Because you might be right, and if so, that can be what I have missed to solve this task.
In other terms, for all (m, n), mRn => mSn (but maybe mSn even if not mRn).

If that was the case, then since 1Rn for n = 1,2,...,12 then 1Sn for those n, and the same if we change variables: 1Rm and 1Sm. Then since S is an equivalence relation, nS1 and mS1. By transitivity, mS1 and 1Sn give mSn. So if that interpretation is right, then S is the trivial relation where everything is related to everything else.

 May 5th, 2010, 09:58 AM #7 Newbie   Joined: Oct 2009 Posts: 14 Thanks: 0 Re: Equivalence Classes I see... But the relation is based on X, which was X = {2, ..... , 12} ? N You explained it with 1, and 1 is not in X, so would that give another meaning if we only consider the numbers in X? And if what you are saying is correct, wouldn't it be existing a lot of equivalence classes which is what we are looking for to begin with?
May 5th, 2010, 10:26 AM   #8
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Re: Equivalence Classes

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 Originally Posted by Zhai I see... But the relation is based on X, which was X = {2, ..... , 12} ? N You explained it with 1, and 1 is not in X, so would that give another meaning if we only consider the numbers in X?
Oh good, I missed that -- I must have misread the "2" as a "1". Let's see what we get.

2Rn for n even, 2Rm for m even, thus 2Sn, 2Sm, nS2, and mS2. By transitivity on mS2 and 2Sn, all even numbers are related to each other by S. We can then glob all the evens together; I'll call them [2].

Similarly, all multiples of three are S-related: mSn for m,n divisible by 3 (call such elements [3]). Since 6 is divisible by both 2 and 3, we have [2] = [3].

So right now we have [2], 5, 7, and 11.

I'll let you finish this off. So far we know that 5S5, 7S7, 11S11, and mSn for any m,n in X other than 5, 7, and 11. But do any other pairs hold?

 May 5th, 2010, 12:21 PM #9 Newbie   Joined: Oct 2009 Posts: 14 Thanks: 0 Re: Equivalence Classes Ok, think things are starting to get more clear for me, I think. So if I am gonna write the equivalence classes, we would have a class for 2, which you called [2]? If we take that as example, an equivalence class where we pair numbers with 2 would be something like this? E(2) = {2, 4, 6, 8 ,10, 12} ? Since the pairings with 2 are the even numbers in X?
May 5th, 2010, 01:18 PM   #10
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Re: Equivalence Classes

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 Originally Posted by Zhai So if I am gonna write the equivalence classes, we would have a class for 2, which you called [2]? If we take that as example, an equivalence class where we pair numbers with 2 would be something like this? E(2) = {2, 4, 6, 8 ,10, 12} ? Since the pairings with 2 are the even numbers in X?
Well, you knew that [2] contained at least 2,4,6,8,10,12, so you can write $\{2,4,6,8,10,12\}\subseteq[2]$. You don't know that they're equal, and in fact they're not since it also contains 3 and 9 and I showed above.

Edit: Of course you don't need to use this notation, but I imagine you'll cover it soon and it's good to recognize it.

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