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Symmetric relation

  1. Mar 13, 2005 #1
    Question: Let R be a symmetric relation on set A. Show that [tex]R^n[/tex] is symetric for all positive integers n.

    My "solution":
    Suppose R is symmetric,
    [tex]
    \exists a,b \in A ((a,b) \in R \wedge (b,a) \in R)

    [/tex]

    For n=1,
    [tex]R^1=R[/tex].
    Next, assume that [tex] (a,b) and (b,a) \in R^k[/tex], for k a possitive integer. So [tex]R^{k+1}=R^k \circ R[/tex].

    Then what? :confused:
     
  2. jcsd
  3. Mar 13, 2005 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Your definition of "symmetric relation" is not correct.

    What you wrote is, in words, "there exist a and b such that (a,b) is in R and (b,a) is in R". The correct definition is "for any a, b IF (a,b) is in R, then (b,a) is also in R".
     
  4. Mar 13, 2005 #3
    Uhm, yes. That is the definition. So if R is symmetric doesn't that mean that there exist (a,b) and (b,a) in R?
     
  5. Mar 14, 2005 #4
    help? please :frown:
     
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