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**dwsmith** R is symmetric implies that $\displaystyle R^2$ is symmetric.

Let $\displaystyle aR^2b$ in R. Therefore, $\displaystyle (a,x)\in R \ \mbox{and} \ (x,b)\in R$ for some x in A. By the symmetric hypothesis, $\displaystyle (b,x)\in R$ and $\displaystyle (x,a)\in R$. We now have a path of length 2 from b to x to a in R and therefore $\displaystyle R^2$ is symmetric.

If R is asymmetric, then $\displaystyle R\cap R^{-1}=\phi$

By contraction: Suppose R is asymmetric and $\displaystyle R\cap R^{-1}\neq\phi$

Let $\displaystyle (a,b)\in R\cap R^{-1}\rightarrow (a,b)\in R \ \mbox{and} \ (a,b)\in R^{-1}\leftrightarrow (b,a)\in R$. Now, we have $\displaystyle (a,b)\in R \ \mbox{and} \ (b,a)\in R$ but R is asymmetric. Therefore, we have reached a contradiction and $\displaystyle R\cap R^{-1}=\phi$.