Symmetric relation

1. Mar 13, 2005

physicsuser

Question: Let R be a symmetric relation on set A. Show that $$R^n$$ 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,
$$R^1=R$$.
Next, assume that $$(a,b) and (b,a) \in R^k$$, for k a possitive integer. So $$R^{k+1}=R^k \circ R$$.

Then what?

2. Mar 13, 2005

HallsofIvy

Staff Emeritus
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. Mar 13, 2005

physicsuser

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. Mar 14, 2005