#### Need solution for RD Sharma maths Class 12 Chapter Relation Exercise 1.1 Question 7 maths textbook solution.

$R$ is neither reflexive nor symmetric nor transitive.

Hint:

A relation R on set A is

Reflexive relation:

If  $(a,a)\in \; R$ for every $a\in \; R$

Symmetric relation:

If $(a,b)$ is true then $(b,a)$  is also true for every $a, b \in A$

Transitive relation:

If $(a,b)$ and $(b, c) \in R$, then $(a, c) \in R$  for every $\mathrm{a}, \mathrm{b}, \mathrm{c} \in A$

Solution :

Reflexive :

It is observed that $(1 / 2,1 / 2) \text { is } R \text { as } 1 / 2>(1 / 2)^{3}=\frac{1}{8}$

$\therefore R \text { is not reflexive. }$

Symmetric :

Now $(1,2) \in R\left(\text { as } 1<2^{3}=8\right)$

But $(2,1) \notin R\left(\text { as } 2>1^{3}=1\right)$

$R$ is not symmetric

Transitive:

We have $\left(3, \frac{3}{2}\right),\left(\frac{3}{2}, \frac{6}{5}\right) \operatorname{in} R \text { as } 3<\left(\frac{3}{2}\right)^{3}$

and $\left(\frac{3}{2}\right)<\left(\frac{6}{5}\right)^{3}$

but $\left(3, \frac{6}{5}\right) \notin R \text { as } 3>\left(\frac{6}{5}\right)^{3}$

$R$ is not transitive

Hence, $R$ is neither reflexive, nor symmetric nor transitive