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Math Help - When R is antisymmetric, R inverse is antisymmetric

  1. #1
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    When R is antisymmetric, R inverse is antisymmetric

    Hello,

    I need to prove: when R is antisymmetric, then R inverse is antisymmetric. Can someone tell me whether I am correct or wrong?

    Let R be antisymmetric, therefore (a,b), (b,a) \in R where a = b. This gives us an inverse function which is (b,a), (a,b) \in R^{-1} but because it is impossible to get a \neq b when a = b (in R), therefore R^{-1} is antisymmetric.

    Is this correct? I have the feeling that this is wrong... (If this is wrong can someone give me a counterexample?

    Thanks for everything!
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  2. #2
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    I would expect students to do the following.
    If (a,b)\in R^{-1}~\&~(b,a)\in R^{-1} then a=b.
    So much of what you have done is correct, but not the starting place.
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  3. #3
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    I'm sorry I do not understand why do you start with the inverse of R? Because normally I would assume R antisymmetric then try to prove R inverse antisymmetric.
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  4. #4
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    Quote Originally Posted by umbrella View Post
    prove: when R is antisymmetric, then R inverse is antisymmetric.
    Here the given is: “R is antisymmetric”.
    You are to prove that: “R inverse is antisymmetric”.

    Quote Originally Posted by umbrella View Post
    I'm sorry I do not understand why do you start with the inverse of R?
    Because that is the object of the proof; you are proving that R^{-1} is antisymmetric.
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  5. #5
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    I disagree a bit, I see it as follow: R antisymmetric -> R inverse is antisymmetric. But starting from the conclusion, doesn't seem right to me, I think you should start from the assumption that R is antisymmetric and not from the assumption that R inverse is already antisymmetric.
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  6. #6
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    Quote Originally Posted by umbrella View Post
    I disagree a bit, I see it as follow: R antisymmetric -> R inverse is antisymmetric. But starting from the conclusion, doesn't seem right to me, I think you should start from the assumption that R is antisymmetric and not from the assumption that R inverse is already antisymmetric.
    Well of course you do. I never said otherwise.
    Here is the proof.
    Suppose that (a,b)\in R^{-1}~\&~(b,a)\in R^{-1}.
    Then by definition (b,a)\in R~\&~(a,b)\in R.
    But we are given that R is antisymmetric.
    That means a=b. Thus R^{-1} is also antisymmetric.
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