Prove R^n is Symmetric for All Positive Integers n

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Homework Help Overview

The discussion revolves around proving that the n-th power of a symmetric relation R on a set A is also symmetric for all positive integers n. The original poster presents an initial approach to the problem, focusing on the definition of symmetry and the properties of relations.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to establish a base case for n=1 and proposes an inductive step for n=k to n=k+1, but expresses uncertainty about the next steps. Some participants question the accuracy of the definition of a symmetric relation provided by the original poster, leading to a discussion about the correct interpretation of symmetry.

Discussion Status

The discussion is ongoing, with participants exploring the definition of symmetry and its implications. There is no explicit consensus yet, as some participants are clarifying definitions while others are trying to build on the original poster's approach.

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physicsuser
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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)<br /> [/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:
 
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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".
 
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?
 
help? please :frown:
 

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