Symmetric relation

  • #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:
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
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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".
 
  • #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?
 
  • #4
help? please :frown:
 

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