#### Provide solution for RD Sharma maths Class 12 Chapter Relations Exercise 1.1 Question 2 maths textbook solution.

$R_{1}$ is reflexive and transitive but not symmetric

$R_{2}$ is reflexive, symmetric, and transitive.

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

$R_{4}$ is neither reflexive nor symmetric nor transitive

Hint :

A relation R on set A is

Reflexive relation:

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

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 $a, b \in A$

Given :

Set  $A=\{a, b, c\}$

\begin{aligned} &R_{1}=\{(a, a),(a, b),(a, c),(b, b),(b, c),(c, a),(c, b),(c, c)\} \\ &R_{2}=\{(a, a)\} \\ &R_{3}=\{(b, c)\} \\ &R_{4}=\{(a, b),(b, c),(c, a)\} \end{aligned}

Consider   $R_{1}=\{(a, a),(a, b),(a, c),(b, b),(b, c),(c, a),(c, b),(c, c)\}$

Reflexive :

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

So, $R_{1}$ is reflexive.

For Symmetric:

We see that the ordered pairs obtained by interchanging the components of $R_{1}$ are not in $R_{1}$.

For ex : $(a, b) \in R_{1} \text { but }(b, a) \notin R_{1}$

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

For Transitive:

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

So, $R_{1}$ is transitive

(ii) Consider $R_{2}$

$R_{2}=\{(a, a)\}$

Reflexive:

clearly $(a, a) \in R_{2}$

So, $R_{2}$ is reflexive.

Symmetric:

Clearly $(a,a)\in \; R_{2}$

So, $R_{2}$  is symmetric.

Transitive:

$R_{2}$ is a transitive relation, since there is only one element in it.

(iii)  Consider $R_{3}$

$R_{3}=\{(b, c)\}$

Reflexive:

Here neither $(b, b) \notin R_{3}$ nor $(c, c) \notin R_{3}$

So, $R_{3}$  is not reflexive

Symmetric:

Here neither $(b, c) \in R_{3}$ nor $(c, b) \notin R_{3}$

So, $R_{3}$ is not symmetric.

Transitive:

$R_{3}$ has only one element

Hence $R_{3}$ is transitive.

(iv)   Consider  $R_{4}=\{(a, b),(b, c),(c, a)\}$

Reflexive:

Here  $(a, b) \in R_{4} \text { but }(b, a) \notin R_{4}$

So, $R_{4}$ is not reflexive

Symmetric:

Here $(a, b) \in R_{4} \text { but }(b, a) \notin R_{4}$

So,$R_{4}$ is not symmetric

Transitive:

Here $(a, b) \in R_{4},(b, c) \in R_{4}$ but  $(a, c) \notin R_{4}$

Hence $R_{4}$ is not transitive.