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1.Relation and Function
medium
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.
Option A
Option B
Option C
Option D
Solution
$( i)$ $R = \left\{ {(a,b):a \leqslant {b^2}} \right\}$
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.
Standard 12
Mathematics
