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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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