Question

For real numbers $xandy,$we write$xRy\Rightarrow x–y+\surd 2$ is an irrational number. Then the relation $R$ is

• A. Reflexive
• B. Symmetric
• C. Transitive
• D. None of these

Answer 1

The correct option is A

Reflexive

Explanation for the correct option:

Prove that relation is reflexive:

For a relation to be reflexive $xRx$

$$\begin{gathered} \mathrm{For}\mathrm{real}\mathrm{x} \\ \mathrm{xRx}\Rightarrow \mathrm{x}-\mathrm{x}+\sqrt{2}=\sqrt{2} \\ \sqrt{2}\mathrm{is}\mathrm{an}\mathrm{irrational}\mathrm{number}. \\ \therefore \mathrm{xRx}\mathrm{is}\mathrm{reflexive}. \end{gathered}$$

Hence, Option(A) is the correct answer.

Explanation for the incorrect answer:

Prove that relation is not symmetric.

For a relation to be symmetric $xRy=yRx$

$$\begin{gathered} \mathrm{For}\mathrm{real}\mathrm{number}\mathrm{x}\mathrm{and}\mathrm{y} \\ \mathrm{xRy}\Rightarrow \mathrm{x}-\mathrm{y}+\sqrt{2} \\ \mathrm{yRx}\Rightarrow \mathrm{y}-\mathrm{x}+\sqrt{2} \\ \Rightarrow \mathrm{xRy}\ne \mathrm{yRx} \end{gathered}$$

$\therefore$The relation is not symmetric.

Hence, Option(B) is not the correct answer.

Prove that relation is not transitive.

For a relation to be transitive $xRy=yRz\Rightarrow xRz$

For real numbers $\mathrm{x},\mathrm{y}\mathrm{and}\mathrm{z}$

Let

$\begin{array}{rcl}\mathrm{x}& =& -\sqrt{2}\\ \mathrm{y}& =& 3\sqrt{2}\\ \mathrm{z}& =& 2\left[\mathbf{Substitute}\mathbf{}\mathbf{the}\mathbf{}\mathbf{values}\mathbf{}\mathbf{of}\mathbf{}\mathbf{x}\mathbf{,}\mathbf{}\mathbf{y}\mathbf{,}\mathbf{}\mathbf{and}\mathbf{}\mathbf{z}\mathbf{}\mathbf{in}\mathbf{}\mathbf{realtion}\right]\\ \mathrm{xRy}& \Rightarrow & \mathrm{x}-\mathrm{y}+\sqrt{2}=-\sqrt{2}-3\sqrt{2}+\sqrt{2}\\ & =& -3\sqrt{2}\mathrm{is}\mathrm{an}\mathrm{irrational}\mathrm{number}.\\ \mathrm{yRz}& \Rightarrow & \mathrm{y}-\mathrm{z}+\sqrt{2}=3\sqrt{2}-2+\sqrt{2}\\ & =& 4\sqrt{2}-2\mathrm{is}\mathrm{an}\mathrm{irrational}\mathrm{number}\\ \mathrm{xRz}& \Rightarrow & \mathrm{x}-\mathrm{z}+\sqrt{2}=-\sqrt{2}-2+\sqrt{2}\\ & =& -2\mathrm{is}\mathrm{not}\mathrm{an}\mathrm{irrational}\mathrm{number}\end{array}$

$\therefore \mathrm{xRy},\mathrm{yRx}\mathrm{then}\mathrm{x}\mathrm{is}\mathrm{not}\mathrm{related}\mathrm{to}\mathrm{z}$

The relation is not transitive.

Hence, Option(C) is not the correct answer.