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

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Solution

## $\mathrm{We}\mathrm{have},\phantom{\rule{0ex}{0ex}}R=\left\{\left(x,y\right):x-y+\sqrt{2}\mathrm{is}\mathrm{an}\mathrm{irrational}\mathrm{number};x,y\in \mathbf{R}\right\}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{As},x-x+\sqrt{2}=\sqrt{2},\mathrm{which}\mathrm{is}\mathrm{an}\mathrm{irrational}\mathrm{number}\phantom{\rule{0ex}{0ex}}⇒\left(x,x\right)\in R\phantom{\rule{0ex}{0ex}}\mathrm{So},R\mathrm{is}\mathrm{reflexive}\mathrm{relation}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Since},\left(\sqrt{2},2\right)\in R\phantom{\rule{0ex}{0ex}}\mathrm{i}.\mathrm{e}.\sqrt{2}-2+\sqrt{2}=2\sqrt{2}-2,\mathrm{which}\mathrm{is}\mathrm{an}\mathrm{irrational}\mathrm{number}\phantom{\rule{0ex}{0ex}}\mathrm{but}2-\sqrt{2}+\sqrt{2}=2,\mathrm{which}\mathrm{is}\mathrm{a}\mathrm{rational}\mathrm{number}\phantom{\rule{0ex}{0ex}}⇒\left(2,\sqrt{2}\right)\notin R\phantom{\rule{0ex}{0ex}}\mathrm{So},R\mathrm{is}\mathrm{not}\mathrm{symmetric}\mathrm{relation}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Also},\left(\sqrt{2},2\right)\in R\mathrm{and}\left(2,2\sqrt{2}\right)\in R\phantom{\rule{0ex}{0ex}}\mathrm{i}.\mathrm{e}.\sqrt{2}-2+\sqrt{2}=2\sqrt{2}-2,\mathrm{which}\mathrm{is}\mathrm{an}\mathrm{irrational}\mathrm{number}\mathrm{and}2-2\sqrt{2}+\sqrt{2}=2-\sqrt{2},\mathrm{which}\mathrm{is}\mathrm{also}\mathrm{an}\mathrm{irrational}\mathrm{number}\phantom{\rule{0ex}{0ex}}\mathrm{But}\sqrt{2}-2\sqrt{2}+\sqrt{2}=0,\mathrm{which}\mathrm{is}\mathrm{a}\mathrm{rational}\mathrm{number}\phantom{\rule{0ex}{0ex}}⇒\left(\sqrt{2},2\sqrt{2}\right)\notin R\phantom{\rule{0ex}{0ex}}\mathrm{So},R\mathrm{is}\mathrm{not}\mathrm{transitive}\mathrm{relation}$ Hence, the correct alternative is option (a).

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