Consider the following two binary relations on the set $A=\{a, b, c\}: R_1=\{(\mathrm{c}, a)(b, b)…
Consider the following two binary relations on the set $A=\{a, b, c\}: R_1=\{(\mathrm{c}, a)(b, b),(\mathrm{a}, c),(c$, $c),(b, c),(a, a)\}$ and $\mathrm{R}_2=\{(\mathrm{a}, \mathrm{b}),(\mathrm{b}, \mathrm{a}),(\mathrm{c}, \mathrm{c})$, (c, a), (a, a), (b, b), (a, c). Then
$R_2$ is symmetric but it is not transitive
Both $R_1$ and $R_2$ are transitive
Both $R_1$ and $R_2$ are not symmetric
$R_1$ is not symmetric but it is transitive
Solution
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 $R_2$ Now, for $R_2,(b, a) \in R_2,(a, c) \in R_2$ but $(b, c) \notin R_2$. Similarly, for $R_1,(b, c) \in R_1,(c, a) \in R_1$ but $(b, a) \notin R_1$.
Therefore, neither $R_1$ nor $R_2$ is transitive.