On the set $S$ of all real numbers, define a relation $R=\{(a, b): a \leq b\}$.
Show that $R$ is
(i) reflexive
(ii) transitive
(iii) not symmetric.
Let $R=\{(a, b): a \leq b\}$ be a relation defined on $S .$
Now,
We observe that any element $x \in S$ is less than or equal to itself.
$\Rightarrow(x, x) \in R \forall x \in S$
$\Rightarrow \mathrm{R}$ is reflexive.
Let $(x, y) \in R \forall x, y \in S$
$\Rightarrow \mathrm{x}$ is less than or equal to $\mathrm{y}$
But $y$ cannot be less than or equal to $x$ if $x$ is less than or equal to $y$.
$\Rightarrow(\mathrm{y}, \mathrm{x}) \notin \mathrm{R}$
For e.g., we observe that $(2,5) \in R$ i.e. $2<5$ but 5 is not less than or equal to $2 \Rightarrow(5,2) \notin R$
$\Rightarrow \mathrm{R}$ is not symmetric
Let $(x, y) \in R$ and $(y, z) \in R \forall x, y, z \in S$
$\Rightarrow x \leq y$ and $y \leq z$
$\Rightarrow x \leq z$
$\Rightarrow(x, z) \in R$
For e.g., we observe that
$(4,5) \in R \Rightarrow 4 \leq 5$ and $(5,6) \in R \Rightarrow 5 \leq 6$
And we know that $4 \leq 6 \therefore(4,6) \in \mathrm{R}$
$\Rightarrow \mathrm{R}$ is transitive.
Thus, $R$ is reflexive and transitive but not symmetric.