Question

If $A=\{A_1,A_2,A_3,A_4,A_5,A_6\}$ be the set of six unit circles with centres at $C_1,C_2,C_3...C_6$ as shown in the diagram. The relation $R$ on $A$ is defined by
$(A_i,A_j)\in R⟺C_i{C}_j\le 2\sqrt{2},i,j\in \{1,\mathrm{2...},6\}$ where $C_i{C}_j$ represents distance between the centres $C_i\text{and}{C}_j$ , then

• A. $R$ is symmetric and transitive but not reflexive relation.
• B. $R$ is transitive relation only.
• C. $R$ is reflexive and symmetric but not transitive relation.
• D. $R$ is neither reflexive nor transitive but it is symmetric relation

Answer 1

The correct option is C $R$ is reflexive and symmetric but not transitive relation.

As distance between centres of same circle $=0$
$\Rightarrow C_i{C}_i=0\le 2\sqrt{2}$
$\Rightarrow (A_i,A_i)\in R\forall i=\{1,2,3,4,5,6\}$
So, $R$ is reflexive relation.

Let, $(A_i,A_j)\in R$
$\Rightarrow C_i{C}_j\le 2\sqrt{2}$
$\Rightarrow C_j{C}_i\le 2\sqrt{2}$
$\Rightarrow (A_j,A_i)\in R$
If $(A_i,A_j)\in R\Rightarrow (A_j,A_i)\in R$
So, $R$ is symmetric relation.

From diagram $(A_1,A_2)\in R$ and $(A_2,A_3)\in R$ but $(A_1,A_3)\notin R$
as, $C_1{C}_3=4>2\sqrt{2}$
So, $R$ is not transitive relation.