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Let L be the set of all lines in a plane and $\mathrm{R}$ be the relation in $\mathrm{L}$ defined as $\mathrm{R}=\left\{\left(\mathrm{L}_{1}, \mathrm{L}_{2}\right): \mathrm{L}_{1}\right.$ is perpendicular to $\left. \mathrm{L} _{2}\right\}$. Show that $\mathrm{R}$ is symmetric but neither reflexive nor transitive.
Solution
$R$ is not reflexive, as a line $L_{1}$ can not be perpendicular to itself, i.e., $\left(L_{1}, \,L_{1}\right)$ $\notin R$. $R$ is symmetric as $\left(L_{1}, L_{2}\right) \in R$
$\Rightarrow $ $L_{1}$ is perpendicular to $L_{2}$
$\Rightarrow $ $L_{2}$ is perpendicular to $L_{1}$
$\Rightarrow $ $\left(L_{2},\, L_{1}\right) \in R$
$R$ is not transitive. Indeed, if $L_{1}$ is perpendicular to $L_{2}$ and $L _{2}$ is perpendicular to $L _{3},$ then $L _{1}$ can never be perpendicular to $L _{3} .$ In fact, $L _{1}$ is parallel to $L _{3},$ ie., $\left( L _{1},\, L _{2}\right) \in R ,\left( L _{2}, L _{3}\right) \in R$ but $\left( L _{1}, L _{3}\right) \notin R$.
