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    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}.
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    (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}.
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    (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!
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    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?
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    (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].
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    (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]?
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    (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? :p: Think about what you're doing!
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    (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? :p: Think about what you're doing!
    Oh ok I see thanks! So hard to get my head around this stuff!
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    (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.
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    (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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