Question

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

Answer 1

Given: Relation $R$ in the set $R$ of real numbers, defined as $R=\left\{\right(a,b):a\le b^2\}$

Checking reflexivity of a relation:
If the relation is reflexive, then $(a,a)\in R$
$\Rightarrow a\le a^2$
Taking $a=\frac{1}{2}$
$\Rightarrow \frac{1}{2}\le \frac{1}{4}$ which is incorrect.
$\therefore R$ is not reflexive.

Checking given relation is symmetric or not:
If $(a,b)\in R,$ then $(b,a)\in R$
$\Rightarrow a\le b^2$, then $b\le a^2$
Taking $a=2,b=5$
$\Rightarrow 2\le 5^2$ which is correct but $\Rightarrow 5\le 2^2$ which is incorrect.
$\therefore R$ is not symmetric.

Checking transitivity of a relation:
If $(a,b)\in R\&(b,c)\in R$, then $(a,c)\in R$
$\Rightarrow a\le b^2\&b\le c^2$ then $a\le c^2$
Taking $a=2,b=-2$ and $c=-1$
$2\le (-2{)}^2\Rightarrow 2\le 4$ which is correct
And $-2\le (-1{)}^2$ which is also correct but $2\le (-1{)}^2\Rightarrow 2\le 1$ which is incorrect
Since if $a\le b^2\&b\le c^2$ then $a\le c^2$ is not true for all values of $(a,b,c)\in R$
$\therefore R$ is not transitive.