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

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.

 

Solution:

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

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.

Now,

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

(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.

 

 

 

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