Question

If $ R $ is an equivalence relation on a set $ A $ , then $ {{R}^{-1}} $ is
(a) Reflexive but not transitive
(b) Symmetric but nor transitive
(c) Reflexive and transitive but not symmetric
(d) An equivalence relation

Answer 1

Hint: We will look at the definition of an equivalence relation. Then we will see the meaning of the terms reflexive, symmetric, and transitive. Then we will look at the inverse relation $ {{R}^{-1}} $ in terms of the equivalence relation $ R $ . We will check whether the inverse relation is reflexive, symmetric, and transitive.

Complete step by step answer:
We say that a relation is an equivalence relation when it is reflexive, symmetric and transitive. Let us see the definitions of reflexive, symmetric and transitive. Suppose we have three elements $ x $ , $ y $ and $ z $ belonging to a set with a relation $ r $ , then the relation is
(i) reflexive, if $ xrx $ that is, $ x $ is related to itself
(ii) symmetric, $ xry $ if and only if $ yrx $
(iii) transitive, if $ xry $ and $ yrz $ then $ xrz $ .
We are given that $ R $ is an equivalence relation on a set $ A $ . Let us assume that $ a,b,c\in A $ . So, we have the following,
(i) reflexive, if $ aRa $
(ii) symmetric, $ aRb $ if and only if $ bRa $
(iii) transitive, if $ aRb $ and $ bRc $ then $ aRc $ .
Let us denote a relation between two elements as $ R\left( a,b \right) $ . So, the inverse relation can be defined as $ {{R}^{-1}}\left( a,b \right)=R\left( b,a \right) $ .
Now, we can see that the inverse relation is reflexive since $ {{R}^{-1}}\left( a,a \right)=R\left( a,a \right) $ and $ R $ is reflexive. Next, we have to check if the inverse relation is symmetric. Consider the following,
 $ {{R}^{-1}}\left( a,b \right)=R\left( b,a \right) $
But, $ R $ is symmetric. hence we can write the above expression as the following,
 $ \begin{align}
  & {{R}^{-1}}\left( a,b \right)=R\left( a,b \right) \\
 & \therefore {{R}^{-1}}\left( a,b \right)={{R}^{-1}}\left( b,a \right) \\
\end{align} $
Hence, the inverse relation is symmetric.
Similarly, for transitivity, we have that $ {{R}^{-1}}\left( a,b \right)=R\left( b,a \right) $ and $ {{R}^{-1}}\left( b,c \right)=R\left( c,b \right) $ . Since $ R $ is transitive, and we have $ R\left( c,b \right) $ and $ R\left( b,a \right) $ , we get $ R\left( c,a \right) $ which is equal to $ {{R}^{-1}}\left( a,c \right) $ . Hence, the inverse relation is also transitive. This implies that the inverse relation is also an equivalence relation. Hence, the correct option is (d).

Note:
 It is important that we are familiar with the concept of relations on sets. The functions that we define on sets are relations such that we get only one output. This leads us to defining one-one functions and onto functions. It is necessary to verify all the definitions of reflexivity, symmetry, and transitivity for every relation to seeing whether it is an equivalence relation or not.