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Math Help - a haunting problem in relations

  1. #1
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    a haunting problem in relations

    supose p is an equivalent relation
    (x,y) E p --> [x]p = [y]p

    can anyone pls tell me whether the below approach is correct..

    suppose (x,y) E p
    let z be arbitrary and suppose
    z E [x]p <--> (x,z) E p
    <--> .........

    like wise by using biconditional arrow, then u dont have to split it into 2 parts,
    because i dont see any wrong in it.( since p is equivalent)

    thanks in advance..
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  2. #2
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    If \mathcal{P} is an equivalent relation then it is true that
    z\in [{x}]_\mathcal{P}\Leftrightarrow (x,z)\in \mathcal{P}.
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  3. #3
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    so then the proof of above will be very easy. isnt it? but i havnt seen people using this for proofs in relations.(what they do is prove it in two parts) but in functions relations they use this kind of bidirectional proofs. why is this?
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