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Composition of Relations

  1. Apr 30, 2012 #1
    1. The problem statement, all variables and given/known data

    R = { (1,2), (3,5), (2,2), (2,5) }
    S = { (2,1), (5,3), (5,1), (5,5) }

    Explicitly find the relation R^-1 o S^-1

    2. Relevant equations



    3. The attempt at a solution

    This was on my test.

    First I just wrote down the inverses:

    R^-1 = { (2,1), (5,3), (2,2), (5,2) }
    S^-1 = { (1,2), (3,5), (1,5), (5,5) }

    I didn't know what to do because the definition we learned defines 3 other sets, and all of the exercises in my test book has those 3 other sets defined.

    For example, there are usually sets A, B, and C along with the sets R and S. So I have no idea how I can apply the definition to do this.
     
  2. jcsd
  3. May 1, 2012 #2

    HallsofIvy

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    So [itex]S^{-1}[/itex] "maps" 1 to 2 and [itex]R^{-1}[/itex] maps 2 to 1. Therefore [itex]R^{-1}oS^{-1}[/itex] maps 1 to 1 and contains the pair (1, 1).

    [itex]R^{-1}[/itex] also maps 2 to 2 so [itex]R^{-1}oS^{-1}[/itex] also maps 1 to 2 and contains the pair (1, 2).

    What 3 sets?

    fog contains the pair (a, b) if and only if there exist some c such that g contains (a, c) and f contains (c, b).
     
    Last edited: May 1, 2012
  4. May 1, 2012 #3
    So (3,3) is in the composition because we have (5,3) and (3,5)?
     
  5. May 1, 2012 #4

    SammyS

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    (3, 3) is in [itex]\displaystyle R^{-1}\circ S^{-1}[/itex] because, (3, 5) is in [itex]S^{-1}[/itex] and (5, 3) is in [itex]R^{-1}\ .[/itex]
     
  6. May 1, 2012 #5
    I think the other three sets in my definition are A, B, and C and are dupposed to be the domain of R, the Range of R/domain of S, and the range of S.

    Sound reasonable?
     
  7. May 1, 2012 #6

    SammyS

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    As Halls said earlier, "What 3 sets?"

    The domain of R is {1,2,3}.

    The domain of S is {2,5}.

    etc.
     
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