1. Induction question..

Hello,

Q - Ler R be a symmetric relation defined on a set C. Prove or disprove thatr R^5 is a symmetric relation.

My question is.. do I have to prove that by Induction ??? or is there any other way to solve it ??

Thanks.

2. If R is symmetric then the standard proof that so is $\displaystyle R \circ R$ follows:
$\displaystyle \begin{gathered} (a,b) \in R \circ R \Rightarrow \;(a,c) \in R \wedge (c,b) \in R \hfill \\ \Rightarrow \;(c,a) \in R \wedge (b,c) \in R \hfill \\ \Rightarrow (b,a) \in R \circ R \hfill \\ \end{gathered}$.

You should be able to extend that

3. Originally Posted by Plato
If R is symmetric then the standard proof that so is $\displaystyle R \circ R$ follows:
$\displaystyle \begin{gathered} (a,b) \in R \circ R \Rightarrow \;(a,c) \in R \wedge (c,b) \in R \hfill \\ \Rightarrow \;(c,a) \in R \wedge (b,c) \in R \hfill \\ \Rightarrow (b,a) \in R \circ R \hfill \\ \end{gathered}$.

You should be able to extend that
Yeah.. I do get what you mean.
because we have 5 as a power.. 5 is an odd number. So we will be missing a pair making the relation not symmetric..
So the answer should be No.. I guess..??

Anyway, thanks for the help man. Really appreciate it.

4. Originally Posted by MMM88
Yeah.. I do get what you mean.
because we have 5 as a power.. 5 is an odd number. So we will be missing a pair making the relation not symmetric..
So the answer should be No.. I guess..??
You have been mislead by my posting!
We have shown that $\displaystyle R \circ R$ is symmetric.
Now suppose that $\displaystyle (x,y) \in R \circ R \circ R$ then for some z, $\displaystyle (x,z) \in R \circ R\quad \wedge \quad (z,y) \in R$.
But we already have shown that $\displaystyle (x,z) \in R \circ R\quad \Rightarrow \quad (z,x) \in R \circ R$.

This is the reason that you may have been lead to think that it is a proof by induction.
In fact, $\displaystyle R^5 = R \circ R \circ R \circ R \circ R$ is symmetric.