# On the set S of all real numbers, define a relation

Question:

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.

Solution:

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.