Show that the relation R in the set R of real numbers, defined as


Show that the relation R in the set of real numbers, defined as

$R=\left\{(a, b): a \leq b^{2}\right\}$ is neither reflexive nor symmetric nor transitive.



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

It can be observed that $\left(\frac{1}{2}, \frac{1}{2}\right) \notin \mathbf{R}$, since $\frac{1}{2}>\left(\frac{1}{2}\right)^{2}=\frac{1}{4}$.

∴R is not reflexive.

Now, $(1,4) \in R$ as $1<4^{2}$

But, 4 is not less than $1^{2}$.

$\therefore(4,1) \notin \mathrm{R}$

∴R is not symmetric.


$(3,2),(2,1.5) \in \mathrm{R}$

(as $3<2^{2}=4$ and $2<(1.5)^{2}=2.25$ )

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

$\therefore(3,1.5) \notin \mathrm{R}$

∴ R is not transitive.

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




Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now