સાબિત કરો કે વાસ્તવિક સંખ્યાઓના ગણ R પર R =le | vedclass

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Question

$( i)$   $R = \left\{ {(a,b):a \leqslant {b^2}} ight\}$

It can be observed that $\left(\frac{1}{2}, \frac{1}{2}ight) 
otin R,$

si

Vedclass · Accepted Answer

Vedclass · Answer

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

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.

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