Consider the following two binary relations on the set $A = {a, b, c}$:
$R_1 = {(c, a), (b, b), (a, c), (c, c), (b, c), (a, a)}$ and
$R_2 = {(a, b), (b, a), (c, c), (c, a), (a, a), (b, b), (a, c)}$.
Then:
(A)
both $R_1$ and $R_2$ are not symmetric.
(B)
$R_1$ is not symmetric but it is transitive.
(C)
$R_2$ is symmetric but it is not transitive.
(D)
both $R_1$ and $R_2$ are transitive.
MEDIUM
Correct Answer:C
Explanation
Here, both $R_1$ and $R_2$ are symmetric, as for any $(x, y) \in R_1$, we have $(y, x) \in R_1$ and similarly for any $(x, y) \in R_2$, we have $(y, x) \in R_2$.
In $R_1$, $(b, c) \in R_1$, $(c, a) \in R_1$ but $(b, a) \notin R_1$.
Similarly, in $R_2$, $(b, a) \in R_2$, $(a, c) \in R_2$ but $(b, c) \notin R_2$.
Therefore, $R_1$ and $R_2$ are not transitive. Thus, $R_2$ is symmetric but not transitive.