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# Relations Watch

1. So when there's a question like find the equivalence classes, its usually just writing out some of the classes and trying to find a relationship between them? Because I would've never got {mn^2: n in N}.
2. (Original post by JBKProductions)
So when there's a question like find the equivalence classes, its usually just writing out some of the classes and trying to find a relationship between them? Because I would've never got {mn^2: n in N}.
The equivalence classes aren't {mn² : n in N}. Well, they are, but that can be deduced later; you wouldn't immediately jump to the conclusion that those were the equivalence classes. In general, if R is an equivalence relation, then the equivalence class of an element (a, say) is [a] = {b : aRb}. So in this case, "aRb" means "ab is a perfect square", and so [a] = {b : ab is a perfect square}.
3. (Original post by nuodai)
The equivalence classes aren't {mn² : n in N}. Well, they are, but that can be deduced later; you wouldn't immediately jump to the conclusion that those were the equivalence classes. In general, if R is an equivalence relation, then the equivalence class of an element (a, say) is [a] = {b : aRb}. So in this case, "aRb" means "ab is a perfect square", and so [a] = {b : ab is a perfect square}.
Oh ok. Thank you!
4. Ok, I'm trying another question similar to this one. It has equivalence relation a^2+a=b^2+b. Again I'm trying to find the equivalence classes for it so factorised it to (a-b)(a+b+1)=0, but I'm not sure how to find its equivalence class. This is on the set S = R where R is the real numbers. Can someone give me some hints please?
5. (Original post by JBKProductions)
Ok, I'm trying another question similar to this one. It has equivalence relation a^2+a=b^2+b. Again I'm trying to find the equivalence classes for it so factorised it to (a-b)(a+b+1)=0, but I'm not sure how to find its equivalence class. This is on the set S = R where R is the real numbers. Can someone give me some hints please?
Well if (a-b)(a+b+1)=0 then what is b in terms of a? Then use the fact that [a] = {b : (a-b)(a+b+1)=0} to find [a].
6. (Original post by nuodai)
Well if (a-b)(a+b+1)=0 then what is b in terms of a? Then use the fact that [a] = {b : (a-b)(a+b+1)=0} to find [a].
so if b=a and b=-a-1 then [a]={b: (2a+1)(0)=0) a=-1/2 so [-1/2]?
7. (Original post by JBKProductions)
so if b=a and b=-a-1 then [a]={b: (2a+1)(0)=0) a=-1/2 so [-1/2]?
Uh... your notation's managed to get majorly screwed up there.

If (a-b)(a+b+1)=0 then b=a or b=-a-1, that's correct. So the set of values of b for which (a-b)(a+b+1)=0 is the set {a, -a-1}, and so [a] = {a, -a-1}. So for example [0] = {0, -1}, and [4] = {4, -5}.

I don't know where you were going with "{b: (2a+1)(0)=0)" <-- that. You seem to have set b=0 in both brackets and a=0 in the second bracket and then set it equal to 0 and solved for a... why? Think about what you're doing!
8. (Original post by nuodai)
Uh... your notation's managed to get majorly screwed up there.

If (a-b)(a+b+1)=0 then b=a or b=-a-1, that's correct. So the set of values of b for which (a-b)(a+b+1)=0 is the set {a, -a-1}, and so [a] = {a, -a-1}. So for example [0] = {0, -1}, and [4] = {4, -5}.

I don't know where you were going with "{b: (2a+1)(0)=0)" <-- that. You seem to have set b=0 in both brackets and a=0 in the second bracket and then set it equal to 0 and solved for a... why? Think about what you're doing!
Oh ok I see thanks! So hard to get my head around this stuff!
9. (Original post by JBKProductions)
Oh ok I see thanks! So hard to get my head around this stuff!
I found that when I was first learning about relations, but then it clicked... be patient, it'll happen Just don't go crazy and start setting things equal to zero and triple-exponentiating things if you don't know why.
10. (Original post by nuodai)
I found that when I was first learning about relations, but then it clicked... be patient, it'll happen Just don't go crazy and start setting things equal to zero and triple-exponentiating things if you don't know why.
Lol . Thanks.

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Updated: December 29, 2010
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