"If the R is not symmetric, then R^-1 is not symmetric even R=R^-1."

so can you give any example to support your claim?

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3 votes

Let $R$ be a relation from a set $A$ to a set $B.$ The inverse relation from $B$ to $A,$ denoted by $R^{-1}$ , is the set of ordered pairs $\{(b,a) \mid (a,b) \in R\}$ .

- $S1: R$ is reflexive relation iff $R^{-1} = R$
- $S2: R$ is a symmetric relation iff $R^{-1} = R$

Which one of the following statements is true?

- Only $S1$
- Only $S2$
- Both $S1$ and $S2$
- None

4 votes

**Suppose A = {1, 2, 3}.**

**For S1 :**

Suppose R = {(1,1), (2,2), (3,3), (1,2)}.

This relation R is reflexive.

Now, $R^{-1}$ = {(1,1), (2,2), (3,3), (2,1)}.

$R^{-1}$ $\neq$ R even when R is reflexive.

**Thus, S1 is False.**

**For S2 : **

Assume, R is symmetric ie $\forall$a,b aRb $\implies$ bRa.

aRb $\implies$ b$R^{-1}$a and bRa $\implies$ a$R^{-1}$b.

Therefore, (in simpler terms) all elements of R will form pair like (a,b) and (b,a) due to symmetry and all elements of $R^{-1}$ will have pairs like (b,a) and (a,b) for ever pair in R.

**This makes R = $R^{-1}$.**

Now, assume R = $R^{-1}$.

Suppose, (a,b) $\in$ R $\implies$ (b,a) $\in$ $R^{-1}$.

But R = $R^{-1}$ is given $\implies$ (b,a) $\in$ R and (a,b) $\in$ $R^{-1}$.

Now, we have (a,b) $\in$ R $\implies$ (b,a) $\in$ R.

**This makes R symmetric.**

**Thus, S2 is True.**

**Answer :- B.**