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Equivalence Relations

  1. Nov 1, 2008 #1
    1. The problem statement, all variables and given/known data
    Find relations that satisfying
    just Reflexive
    just Symmrtic
    just Transitive

    (R) & (S), but not (T)
    (R) & (T), but not (S)
    (S) & (T), but not (R)




    2. Relevant equations
    S=Z

    (a,b) [tex]\in[/tex]R if <=> a>b (T) but, not (S) & (R).

    the ex is given in the class, but nothing else was explained. I am confused. Please help if you can.



    3. The attempt at a solution

    How about some cartisian product.
    (1,1) (1,2) (1,3)
    (2,1) (2,2) (2,3)
    (3,1) (3,2) (3,3)

    for (1,1) we have equiv. rel (R, S & T)
     
  2. jcsd
  3. Nov 1, 2008 #2
    The problem above is first course of Abstarct Algebra. We use textbook (Dan Saracino, second edition).
     
  4. Nov 1, 2008 #3

    Office_Shredder

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    (a,b) is in R iff a>b. Then (a,a) is not in R as a>a is false. (a,b) in R means (b,a) is not in R as if a>b, b>a is false. And if (a,b) and (b,c) are in R, a>b>c implies a>c so (a,c) is in R. Note in this example (1,1) is NOT in R as 1>1 isn't true.

    I think a more intuitive notation is aRb to denote 'a is related to b' where R is a relation. Then aRb iff a>b it's clear that if aRb is true, bRa is false since a>b implies b>a is false. Similarly, aRa is false since a>a is false. But aRb, bRc implies a>b, b>c so a>b>c implies a>b and aRc holds. hence > is a relation which is transitive but not symmetric or reflexive.

    Try to think up other relations that satisfy, say, aRa but not aRb -> bRa or aRb, bRc -> aRc
     
  5. Nov 1, 2008 #4
    how about
    aRb iff a-b[tex]\geq[/tex]0
    not (S), but is (T) and (R)
     
  6. Nov 1, 2008 #5
    I can't think of example that is only (S) , and only (T). I don't think I understand the differences og being only (R,) only (S), only (T), and when we have (T), but not (R) & (S) and so on.
     
  7. Nov 1, 2008 #6
    Thank you so much. I found similar questions with answers and I get it now.
     
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