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Need solution for RD Sharma maths Class 12 Chapter Relation Exercise 1.1 Question 7 maths textbook solution.

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R is neither reflexive nor symmetric nor transitive.


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

Solution :

Reflexive :

It is observed that (1 / 2,1 / 2) \text { is } R \text { as } 1 / 2>(1 / 2)^{3}=\frac{1}{8}

\therefore R \text { is not reflexive. }

Symmetric :

Now (1,2) \in R\left(\text { as } 1<2^{3}=8\right)

But (2,1) \notin R\left(\text { as } 2>1^{3}=1\right)

R is not symmetric


We have \left(3, \frac{3}{2}\right),\left(\frac{3}{2}, \frac{6}{5}\right) \operatorname{in} R \text { as } 3<\left(\frac{3}{2}\right)^{3}

and \left(\frac{3}{2}\right)<\left(\frac{6}{5}\right)^{3}

but \left(3, \frac{6}{5}\right) \notin R \text { as } 3>\left(\frac{6}{5}\right)^{3}

R is not transitive

Hence, R is neither reflexive, nor symmetric nor transitive

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