Check whether the relation $R$ on

Solution:

Reflexivity:

Since $\frac{1}{2}>\left(\frac{1}{2}\right)^{3}$

$\left(\frac{1}{2}, \frac{1}{2}\right) \notin R$

So, $R$ is not reflexive.

Symmetry:

Since $\left(\frac{1}{2}, 2\right) \in R$,

$\frac{1}{2}<2^{3}$

But $2>\left(\frac{1}{2}\right)^{3}$

$\Rightarrow\left(2, \frac{1}{2}\right) \in R$

So, $R$ is not symmetric.

Transitivity:

Since $(7,3) \in R$ and $\left(3,3^{\frac{1}{3}}\right) \in R$,

$7<3^{3}$ and $3=\left(3^{\frac{1}{3}}\right)^{3}$

But $7>\left(3^{\frac{1}{3}}\right)^{3}$

$\Rightarrow\left(7,3^{\frac{1}{3}}\right) \notin R$

So, $R$ is not transitive.