1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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,
    \exists a,b \in A ((a,b) \in R \wedge (b,a) \in R)


    For n=1,
    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


    User Avatar
    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:
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook