Show that the relation $R$ in the set $R$ of real numbers, defined as $R =\left\{(a, b): a \leq b^{2}\right\}$ is neither reflexive nor symmetric nor transitive.

Vedclass pdf generator app on play store
Vedclass iOS app on app store

$( i)$   $R = \left\{ {(a,b):a \leqslant {b^2}} \right\}$

It can be observed that $\left(\frac{1}{2}, \frac{1}{2}\right) \notin R,$

since, $\frac{1}{2}>\left(\frac{1}{2}\right)^{2}$

$R$ is not reflexive.

Now, $(1,4)\in R$ as $1<42$ But, $4$ is not less than $1^{2}$.

$\therefore $ $(4,1) \notin R$

$\therefore R$ is not symmetric.

Now,

$(3,2),\,(2,1.5) \in R$              $[$ as $3<2^{2}=4 $ and $2<(1.5)^{2}=2.25]$

But, $3 >(1.5)^{2}=2.25$

$\therefore $ $(3,1.5) \notin R$

$\therefore $ $R$ is not transitive.

Hence, $R$ is neither reflexive, nor symmetric, nor transitive.

Similar Questions

A relation from $P$ to $Q$ is

The probability that a relation $R$ from $\{ x , y \}$ to $\{ x , y \}$ is both symmetric and transitive, is equal to

  • [JEE MAIN 2022]

Let $R$ be a relation over the set $N × N$ and it is defined by $(a,\,b)R(c,\,d) \Rightarrow a + d = b + c.$ Then $R$ is

Let a relation $R$ on $\mathbb{N} \times \mathbb{N}$ be defined as : $\left(\mathrm{x}_1, \mathrm{y}_1\right) \mathrm{R}\left(\mathrm{x}_2, \mathrm{y}_2\right)$ if and only if $\mathrm{x}_1 \leq \mathrm{x}_2$ or $\mathrm{y}_1 \leq \mathrm{y}_2$

Consider the two statements :

($I$) $\mathrm{R}$ is reflexive but not symmetric.

($II$) $\mathrm{R}$ is transitive

Then which one of the following is true?

  • [JEE MAIN 2024]

Let $r$ be a relation from $R$ (Set of real number) to $R$ defined by $r$ = $\left\{ {\left( {x,y} \right)\,|\,x,\,y\, \in \,R} \right.$ and $xy$ is an irrational number $\}$ , then relation $r$ is