#### Need solution for RD Sharma maths Class 12 Chapter Relation Exercise 1.1 Question 5 Subquestion (ii) maths textbook solution.

Reflexive and symmetric but not 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 $\mathrm{a}, \mathrm{b}, \mathrm{c} \in A$

Given :

$a R b \text { if } 1+a b>0$

Solution :

Reflexivity:

Let $a$ be an arbitrary element of $R$

Then, $a\in \; R$

\begin{aligned} &1+a \times a>0 \\ &1+a^{2}>0 \end{aligned}
Since, Square of any number is positive.

So, the given relation is reflexive.

Symmetry:

Let  $(a, b) \in R$

\begin{aligned} &1+a b>0 \\ &1+b a>0 \\ &(b, a) \in R \end{aligned}

So, the given relation is symmetric.

Transitivity:

\begin{aligned} &\text { Let }(a, b) \in R \text { and }(b, c) \in R \\ &\text { Let } \mathrm{a}=-8, \mathrm{~b}=-2, \mathrm{c}=\frac{1}{4} \\ &\text { Then } 1+a b>0 \text { i.e; } 1+(-8)(-2)=17>0 \\ &\text { and } 1+b c>0 \text { i.e; } 1+(-2) \frac{1}{4}=\frac{1}{2}>0 \text { But } 1+a c \neq 0 \text { i.e; } 1+(-8) \frac{1}{4}=-1 \ngtr 0 \\ &(a, c) \epsilon R \end{aligned}

So, the given relation is not transitive.