#### Provide solution for RD Sharma maths Class 12 Chapter Relations Exercise 1.1 Question 3 Subquestion (i) maths textbook solution.

Answer: $R_{1}$ is symmetric but neither reflexive nor transitive.

Hint :

A relation R on set A is

Reflexive relation:

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

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 $a, b, c \in A$

Given :

$R_{1} \text { on } \mathrm{Q}_{0}$ defined by $(a, b) \in R_{1} \Leftrightarrow a=\frac{1}{b}$

Solution :

Reflexivity:

Let  $a$ be an arbitrary element of $R_{1}$

Then, $a \in R_{1}$

$a \neq \frac{1}{a} \text { for all } a \in Q_{0}$

So, $R_{1}$ is not reflexive

Symmetry:

Let $(a, b) \in R_{1}$

Therefore, we can write 'a' as $a=\frac{1}{b}$

$b=\frac{1}{a}$

Then $(b, a) \in R_{1}$

So, $R_{1}$ is symmetric.

For Transitive:

Let   $(a, b) \in R_{1} \text { and }(b, c) \in R_{1}$

\begin{aligned} &a=\frac{1}{b} \text { and } b=\frac{1}{c} \\ &a=\frac{1}{\left(\frac{1}{c}\right)} \Rightarrow c \\ &a \neq \frac{1}{c} \\ &(a, c) \notin R_{1} \end{aligned}

So, $R_{1}$ is not transitive.