Quirino27's question at Yahoo! Answers (R symmetric implies R^2 symmetric)

[Fernando Revilla]

Here is the question:

Can someone please help me with this problem? I can't seem to figure it out by myself... Any help would be appreciated!

Here is a link to the question:

Prove that if a relation R on a set A is symmetric, then the relation R² is also symmetric? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

 

[Fernando Revilla]

Hello Quirino27,

If $U$ is a relation from $A$ to $B$ and $V$ a relation from $B$ to $C$, i.e. $U\subset A\times B$ and $V\subset B\times C$, then the relation $V\circ U$ from $A$ to $C$ is defined in the following way: $$(a,c)\in V\circ U\Leftrightarrow \exists b\in Ba,b)\in U\mbox{ and } (b,c)\in V$$ In our case, suppose $(x,y)\in R^2=R\circ R$, then exists $y\in A$ such that $(x,y)\in R$ and $(y,z)\in R$. But $R$ is symmetric, so $(y,x)\in R$ and $(z,y)\in R$ and by definition of composition of relations, $(z,y)\in R^2$. That is, $R^2$ is symmetric. $\qquad \square$