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Equivalance Relations Could some one check my work please?

  1. Apr 13, 2004 #1
    Question, List the members of the equivalance relation on {1,2,3,4,5} by the given partition. Identify the equivalance classes
    A) {(1,2,3),(4,5)}
    B) {(1),(2,4),(5,3)}

    My solution is;
    A) {(1,1),(1,2),(1,3),(2,2),(2,1),(2,3),(3,1),(3,2),(3,3),(4,4),(4,5),(5,4),(5,5)}

    B) {(1),(2,2),(2,4),(4,2),(4,4),(3,3),(3,5),(5,5),(5,3)}

    Then the next qusetion for which I dont know where to begin is;
    For the above (A and B) find the matrix of the relation from X to X. Show the ordering that you are using :confused:

    Any help would be greatly appreciated.
     
  2. jcsd
  3. Apr 13, 2004 #2

    HallsofIvy

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    Science Advisor

    What you have is correct. For the "matrix" question, do this:
    List all members of X across the top and vertically on the left on your paper. For each intersection, IF the member of X on the left IS equivalent to the member on the top, write "1", otherwise write "0".

    For B (the easier of the two) this is

    Code (Text):

        1   2   3   4   5
    1   1   0   0   0   0
    2   0   1   0   1   0
    3   0   0   1   0   1
    4   0   1   0   1   0
    5   0   0   1   0   1
    The matrix is
    [1 0 0 0 0]
    [0 1 0 1 0]
    [0 0 1 0 1]
    [0 1 0 1 0]
    [0 0 1 0 1]

    The diagonal is all 1s because an equivalence relation is reflexive and the matrix is symmetric because an equivalence relation is symmetric. The number of 1s in each row and column is the number of elements equivalent to that member.
     
    Last edited by a moderator: Apr 13, 2004
  4. Apr 13, 2004 #3
    Thank you but if its not too much trouble why is it that 1 and 2 as in the first "1" is equivalent but not 1 and 3?
     
  5. Apr 13, 2004 #4
    First of all his horizontal row is just a space off, 1 is not related to 2 in example B. Just like 1 is not related to 3 in example B. He didn't do example A.
     
  6. Apr 14, 2004 #5
    Thank you now I see it.
     
  7. Apr 14, 2004 #6
    I long the same lines could some one check this?

    R={(x,y)|x<y};ordering of X:1,2,3,4

    My solution is;
    ? 1 2 3 4
    1 0 1 1 1
    2 0 0 1 1
    3 0 0 0 1
    4 0 0 0 0

    [0 1 1 1]
    [0 0 1 1]
    [0 0 0 1]
    [0 0 0 0]

    OK?
     
  8. Apr 15, 2004 #7

    HallsofIvy

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    Staff Emeritus
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    Yes (It wont let me post less than 10 letters!)
     
  9. Apr 15, 2004 #8
    Kool I think I get it. Thank you
     
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