Question

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

Answer 1

$R=\left\{\right(a,b):a\le b^2\}$
It can be observed that $\left(\frac{1}{2},\frac{1}{2}\right)\notin R$, since $\frac{1}{2}>{\left(\frac{1}{2}\right)}^2=\frac{1}{4}$.
$\therefore R$ is not reflexive.
Now, $(1,4)\in R$ as $1<4^2$
But, $4$ is not less than $1^2$.
$\therefore (4,1)\notin R$
$\therefore R$ is not symmetric.
Now, $(3,2),(2,1.5)\in R$
$(\text{as}3<2^2=4\text{and}2<(1.5{)}^2=2.25)$
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.