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Math Help - invertible relation

  1. #1
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    invertible relation

    hi
    my teacher wants us to think a conjecture for the invertible relation and then prove it, but I have absolutely no idea how to find that conjecture. Any help is appreciated.

     R,S  \subset X \times  X  , we define  R \bullet S \text{ to be the set:} (x,y) \in X \times X , \text{s.t there exsits one and only one}   z \in X : (x,z) \in S,(z,y) \in R

    like a relation  \mathcal{R} \subset X \times  X  is invertible with respect to  \bullet if and only if (what condition does R has to satisfy?)
    any idea?

    cheers.
    Last edited by xixihaha; March 15th 2010 at 03:25 AM.
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  2. #2
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    any idea?
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  3. #3
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    , we define
    This is not a regular composition of relations. In the standard definition of composition there is no restriction that there is at most one z. Is this a point of the exercise or a typo?
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  4. #4
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    Also, by R being invertible you mean that there exists an S such that S\bullet R=Id or ( R\bullet S=Id) where Id is the diagonal relation?

    Probably this has something to do with surjection and injection. It is well-known that a function is an injection iff it has a left inverse, and is a surjection iff it has a right inverse (I hope I did not mix it up). Maybe something similar is expected here.
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  5. #5
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    Quote Originally Posted by emakarov View Post
    This is not a regular composition of relations. In the standard definition of composition there is no restriction that there is at most one z. Is this a point of the exercise or a typo?
    yea this is the exericese that our lecturer gives us.
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  6. #6
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    Quote Originally Posted by emakarov View Post
    Also, by R being invertible you mean that there exists an S such that S\bullet R=Id or ( R\bullet S=Id) where Id is the diagonal relation?

    Probably this has something to do with surjection and injection. It is well-known that a function is an injection iff it has a left inverse, and is a surjection iff it has a right inverse (I hope I did not mix it up). Maybe something similar is expected here.
    S\bullet R= R\bullet S=Id) where Id is the diagonal relation. yes.
    any idea what the conjecture is??

    my thought is  \forall y \in x there exists one and only one  x \in X, (x,y) \in R
    is this correct?
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  7. #7
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    my thought is there exists one and only one
    I think this not correct. Let X = {1, 2} and let R = {(1, 1), (1, 2), (2, 2)}. Then R^2 = Id.

    I don't know what the answer is, it requires some thinking and considering examples.
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  8. #8
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    Quote Originally Posted by emakarov View Post
    I think this not correct. Let X = {1, 2} and let R = {(1, 1), (1, 2), (2, 2)}. Then R^2 = Id.

    I don't know what the answer is, it requires some thinking and considering examples.

    i dont know either, but anyway thanks for your attention.
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