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Show that the relation $R$ in the set $R$ of real numbers, defined as
$\mathbf{R}=\left\{(\mathbf{a}, \mathbf{b}): \mathbf{a} \leq \mathbf{b}^2\right\}$ is neither reflexive nor symmetric nor transitive.
$\mathbf{R}=\left\{(\mathbf{a}, \mathbf{b}): \mathbf{a} \leq \mathbf{b}^2\right\}$ is neither reflexive nor symmetric nor transitive.
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(i) $\mathrm{R}$ is not reflexive, $\because$ a is not less than or equal to $\mathrm{a}^2$ for all $\mathrm{a} \in \mathrm{R}$, e.g., $\frac{1}{2}$ is not less than $\frac{1}{4}$.
(ii) $\mathrm{R}$ is not symmetric since if $\mathrm{a} \leq \mathrm{b}^2$ then $\mathrm{b}$ is not less than or equal to $\mathrm{a}^2$ e.g. $2 < 5^2$ but 5 is not less than $2^2$.
(iii) $\mathrm{R}$ is not transitive : If $\mathrm{a} \leq \mathrm{b}^2, \mathrm{~b} \leq \mathrm{c}^2$, then $\mathrm{a}$ is not less than $c^2$, e.g. $2 < (-2)^2,-2 < (-1)^2$, but 2 is not less than $(-1)^2$
(ii) $\mathrm{R}$ is not symmetric since if $\mathrm{a} \leq \mathrm{b}^2$ then $\mathrm{b}$ is not less than or equal to $\mathrm{a}^2$ e.g. $2 < 5^2$ but 5 is not less than $2^2$.
(iii) $\mathrm{R}$ is not transitive : If $\mathrm{a} \leq \mathrm{b}^2, \mathrm{~b} \leq \mathrm{c}^2$, then $\mathrm{a}$ is not less than $c^2$, e.g. $2 < (-2)^2,-2 < (-1)^2$, but 2 is not less than $(-1)^2$
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