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Provide solution for RD Sharma maths Class 12 Chapter Relations Exercise 1.1 Question 4 maths textbook solution.

Answers (1)

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

$R_{2}$ is symmetric but neither reflexive nor transitive

$R_{3}$ is transitive but neither reflexive nor symmetric

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 $\mathrm{a}, \mathrm{b}, \mathrm{c} \in \mathrm{A}$

Given :

$\begin{aligned} &R_{1}=\{(1,1),(1,3),(3,1),(2,2),(2,1),(3,3)\} \\ &R_{2}=\{(2,2),(3,1),(1,3)\} \\ &R_{3}=\{(1,3),(3,3)\} \end{aligned}$

Solution :

Consider $R_{1}$

$R_{1}=\{(1,1),(1,3),(3,1),(2,2),(2,1),(3,3)\}$

Reflexivity:

Here $(1,1),(2,2),(3,3) \in R$

So, $R_{1}$ is Reflexive

Symmetric :

Here $(2,1) \in R_{1}$

But, $(1,2) \notin R_{1}$

So, $R_{1}$ is not symmetric

Transitivity:

$\begin{aligned} &\text { Here }(2,1) \in R_{1} \text { and }(1,3) \in R_{1} \\ &\text { But }(2,3) \notin R_{1} \end{aligned}$

So, $R_{1}$ is not transitive.

Now consider $R_{2}$

$R_{2}=\{(2,2),(3,1),(1,3)\}$

Reflexivity :

Clearly, $(1,1) \text { and }(3,3) \notin R_{2}$

So, $R_{2}$ is not reflexive.

Symmetric :

$\text { Here }(1,3) \in R_{2} \text { and }(3,1) \in R_{2}$

So, $R_{2}$ is symmetric.

Transitivity:

$\begin{aligned} &\text { Here }\\ &(1,3) \in R_{2} \text { and }(3,1) \in R_{2}\\ &\text { But }(3,3) \notin R_{2} \end{aligned}$

So, $R_{2}$ is not transitive.

Now consider $R_{3}$

$R_{3}=\{(1,3),(3,3)\}$

Reflexivity:

Clearly, $(1,1) \notin R_{3}$

So, $R_{3}$ is not reflexive.

Symmetry:

Here $(1,3) \in R_{3} \text { but }(3,1) \notin R_{3}$

So, $R_{3}$ is not symmetric.

Transitivity:

Here $(1,3) \in R_{3} \text { and }(3,3) \in R_{3}$

But $(1,3) \in R_{3}$

So, $R_{3}$ is transitive.

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