# Binary relation R is symmetric and transitive iff?

This is a homework question and I am stuck.

Binary relation R is transitive and symmetric if and only if $R=R^{-1}∘R$

The "only if" way is trivial.

On the "if" way, I worked out that given $R=R^{-1}∘R$ we have $R\subseteq R^{-1}$ , but this doesn't gives symmetry since symmetry requires $R^{-1}\subseteq R$.

• Showing symmetry isn't very hard if you know what's written here: math.stackexchange.com/questions/234891/… (Let $S = R^{-1} \circ R$ then compute $S^{-1}$.) – Myself Mar 5 '14 at 12:36
• @Myself You're right. I should try work out transitivity given symmetry now. – sjtufs Mar 5 '14 at 12:44
• Hint Note that transitivity means $R\circ R\subseteq R$. Spoiler But you have just shown that $R = R^{-1}$. Thus $R\circ R = R\circ R^{-1} = R$. – Myself Mar 5 '14 at 12:50
• @Myself Yes, Thank you – sjtufs Mar 5 '14 at 12:54

As mentioned in the comments, we can use the hypothesis $R = R^{-1} \circ R$ to derive: $$R^{-1} = (R^{-1} \circ R)^{-1} = R^{-1} \circ (R^{-1})^{-1} = R^{-1}\circ R = R$$ Subsequently, $R \circ R = R^{-1} \circ R = R$ and it follows that $R$ is both symmetric ($R= R^{-1}$) and transitive ($R \circ R \subseteq R$).