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Math Help - relations proofs

  1. #1
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    relations proofs

    suppose p is an equivalence relaiton.
    i) (x,y) E p --> [x]p = [y]p
    ii) (x,y) not E p --> [x]p intersection [y] = null

    sorry but hw to use mathematical symbols here...
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  2. #2
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    i guess the second one has to be proved by contradicton?
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by doresa View Post
    sorry but hw to use mathematical symbols here...
    See this thread

    CB
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  4. #4
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    thanks
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  5. #5
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by doresa View Post
    suppose p is an equivalence relaiton.
    i) (x,y) \in \rho \Rightarrow [x]\rho = [y]\rho
    ii) (x,y) \notin \rho \Rightarrow [x]\rho \cap [y]\rho = \{\phi\}

    sorry but how to use mathematical symbols here...
    I presume that's what you mean (see my quote)...I can't quite rememeber all the formalitites surrounding the notation of equivalence relations and sets...anyway, for the code, simply click on the LaTeX image. There are two of them, one for i) and one for ii). If you ever get stuck by LaTeX, just click on someone elses code that contains the characters you want to use.

    For i), use transitivity to show that  [x]\rho \subseteq [y]\rho, and that [x]\rho \supseteq [y]\rho.

    For ii) you are quite right, use contradiction. How far have you got with your proof?
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  6. #6
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    by Swlabr;

    For i), use transitivity to show that
    [x]\rho \subseteq [y]\rho, and that [x]\rho \supseteq [y]\rho.

    For ii) you are quite right, use contradiction. How far have you got with your proof?
    thanks im already done both.
    for the first one its the same way as you told. first assume (x,y) E p and let z be an arbitary elment belongs to [x]p then using first assumption and the transitivity we can show that z E [y]p and the other way round.


    for the second one its quite simple when use p --> q = ~q --> ~p

    thats all....
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  7. #7
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    Thumbs up

    (x,y) E p --> [x]p = [y]p

    use transitivity to show that  [x]\rho \subseteq [y]\rho, and that [x]\rho \supseteq [y]\rho.
    i was lately thinking of this....
    why cant i do it in one proof? like this

    suppose (x,y) E p
    let z be arbitrary and suppose
    z E [x]p <--> (x,z) E p
    <--> .........
    like wise by using iff then u dont have to split it into 2 parts,
    because i dont see any wrong in it.

    thanks in advance..
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