Question

Mark the correct alternative in the following question:

For real numbers x and y, define xRy iff $x-y+\sqrt{2}$ is an irrational number. Then the relation R is

(a) reflexive (b) symmetric (c) transitive (d) none of these

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ R=\left\{\left(x,y\right):x-y+\sqrt{2}\mathrm{is}\mathrm{an}\mathrm{irrational}\mathrm{number};x,y\in \mathbf{R}\right\} \\ \mathrm{As},x-x+\sqrt{2}=\sqrt{2},\mathrm{which}\mathrm{is}\mathrm{an}\mathrm{irrational}\mathrm{number} \\ \Rightarrow \left(x,x\right)\in R \\ \mathrm{So},R\mathrm{is}\mathrm{reflexive}\mathrm{relation} \\ \mathrm{Since},\left(\sqrt{2},2\right)\in R \\ \mathrm{i}.\mathrm{e}.\sqrt{2}-2+\sqrt{2}=2\sqrt{2}-2,\mathrm{which}\mathrm{is}\mathrm{an}\mathrm{irrational}\mathrm{number} \\ \mathrm{but}2-\sqrt{2}+\sqrt{2}=2,\mathrm{which}\mathrm{is}\mathrm{a}\mathrm{rational}\mathrm{number} \\ \Rightarrow \left(2,\sqrt{2}\right)\notin R \\ \mathrm{So},R\mathrm{is}\mathrm{not}\mathrm{symmetric}\mathrm{relation} \\ \mathrm{Also},\left(\sqrt{2},2\right)\in R\mathrm{and}\left(2,2\sqrt{2}\right)\in R \\ \mathrm{i}.\mathrm{e}.\sqrt{2}-2+\sqrt{2}=2\sqrt{2}-2,\mathrm{which}\mathrm{is}\mathrm{an}\mathrm{irrational}\mathrm{number}\mathrm{and}2-2\sqrt{2}+\sqrt{2}=2-\sqrt{2},\mathrm{which}\mathrm{is}\mathrm{also}\mathrm{an}\mathrm{irrational}\mathrm{number} \\ \mathrm{But}\sqrt{2}-2\sqrt{2}+\sqrt{2}=0,\mathrm{which}\mathrm{is}\mathrm{a}\mathrm{rational}\mathrm{number} \\ \Rightarrow \left(\sqrt{2},2\sqrt{2}\right)\notin R \\ \mathrm{So},R\mathrm{is}\mathrm{not}\mathrm{transitive}\mathrm{relation} \end{gathered}$$

Hence, the correct alternative is option (a).