Suppose $R$ is a relation on $A$ and $R^{-1}$ is the inverse. Give a proof or counterexample for each of the following statements.
(a) If $R$ is reflexive, then $R^{-1}$ is reflexive.
Let $x \in A$; since $R$ is reflexive, $(x,x) \in R$. By definition of inverse $(x,x)\in R^{-1}$; therefore, $R^{-1}$ is reflexive.
(b) If $R$ is transitive, then $R^{-1}$ is transitive.
Let $x,y,z \in A$ be such that $x R^{-1} y$ and $y R^{-1} z$. Then by definition of the inverse, $y R x$ and $z R y$. By the transitivity of R we then know that $z R x$. Applying the definition of the inverse again gives that $x R^{-1} z$, and therefore $R^{-1}$ is transitive.
(c) If $R$ is symmetric, then $R^{-1}$ is symmetric.
Let $x,y,z \in A$ be such that $x R^{-1} y$; then by definition of the inverse, $y R x$. $R$ is symmetric, so we know that $x R y$. By definition of the inverse again, we have $y R^{-1} x$ and therefore that $R^{-1}$ is symmetric.
(d) If $R$ is antisymmetric, then $R^{-1}$ is antisymmetric.
Don't know about (d).
What about: Let $a,b \in A$ be such that $aR^{-1}b$ and $bR^{-1}a$. Then by definition of the inverse $bRa$ and $aRb$. But $R$ is antisymmetric, so $a = b$. Thus $R^{-1}$ is also antisymmetric.