MCQ
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
  • B
    Both $R_1$ and $R_2$ are transitive
  • C
    Both $R_1$ and $R_2$ are not symmetric
  • D
    $R_1$ is not symmetric but it is transitive

Answer

Correct option: A.
$R_2$ is symmetric but it is not transitive
a
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.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free