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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
$R_2$ 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 $\left( {x,y} \right) \in {R_1}$, we have
$\left( {y,x} \right) \in {R_1}$ and similarly for ${R_2}$
Now, for ${R_2},\left( {b,a} \right) \in {R_2},\left( {a,c} \right) \in {R_2}$ but $\left( {b,a} \right) \notin {R_2}$.
Similarly, for ${R_1},\left( {b,c} \right) \in {R_1},\left( {c,a} \right) \in {R_1}$ but $\left( {b,c} \right) \notin {R_1}$.
Therefore, neither ${R_1}$ nor ${R_2}$ is transitive.
