Let $S$ be the set of all points in a plane and let $R$ be a relation in $S$ defined by $R=\{(A, B): d(A, B)<2$ units $\}$, where $d(A, B)$ is the distance between the points $A$ and $B$.
Show that $\mathrm{R}$ is reflexive and symmetric but not transitive.
Given that, $\forall A, B \in S, R=\{(A, B): d(A, B)<2$ units $\} .$
Now,
$\underline{R}$ is Reflexive if $(A, A) \in \underline{R} \forall \underline{A} \in \underline{S}$
For any $A \in S$, we have
$\mathrm{d}(\mathrm{A}, \mathrm{A})=0$, which is less than 2 units
$\Rightarrow(A, A) \in R$
Thus, $R$ is reflexive.
$\underline{R}$ is Symmetric if $(A, B) \in \underline{R} \Rightarrow \underline{(B, A)} \in \underline{R} \underline{\forall} \underline{A, B} \in \underline{S}$
$(A, B) \in R$
$\Rightarrow d(A, B)<2$ units
$\Rightarrow \mathrm{d}(\mathrm{B}, \mathrm{A})<2$ units
$\Rightarrow(B, A) \in R$
Thus, $R$ is symmetric.
$\underline{\mathrm{R}}$ is Transitive if $(\mathrm{A}, \mathrm{B}) \in \underline{\mathrm{R}}$ and $(\mathrm{B}, \mathrm{C}) \in \underline{\mathrm{R}} \Rightarrow \underline{(\mathrm{A}, \mathrm{C})} \in \underline{\mathrm{R}} \underline{\forall} \underline{\mathrm{A}, \mathrm{B}, \mathrm{C}} \in \underline{\mathrm{S}}$
Consider points $A(0,0), B(1.5,0)$ and $C(3.2,0)$.
$\mathrm{d}(\mathrm{A}, \mathrm{B})=1.5$ units $<2$ units and $\mathrm{d}(\mathrm{B}, \mathrm{C})=1.7$ units $<2$ units
$\mathrm{d}(\mathrm{A}, \mathrm{C})=3.2 \nless 2$
$\Rightarrow(A, B) \in R$ and $(B, C) \in R \Rightarrow(A, C) \notin R$
Thus, $R$ is not transitive.
Thus, $R$ is reflexive, symmetric but not transitive.