If $x$ and $y$ are any two positive real numbers, then $x > y$ implies

- x > - y

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- x < - y

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\frac{1}{x} > \frac{1}{y}

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- \frac{1}{x} > \frac{1}{y}

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Solution

The correct option is B $- x < - y$
$Giventhatxandyarepositivenumberandx>y.Wemaycalculate(a)onthepositivesideofnumberlinexwillbetotherightofyand(b)onthenegativesideofthenumberlinexwillbetotheleftofy.StatementA⟶−x>−yi.e−xtotherightsideof−ywhichisnotcomplyingwith(b).Soitisnottrue.StatementB⟶−x<−yi.e−xliestotheleftof−y.Thiscomplieswith(b).Sothestatementistrue.StatementC⟶1x>1y.Herethenumeratorsofthetwofractionsareequal.Butthedenominatorof1xisgreaterthanthatof1y.Sothisstatementisnottrue.StatementD⟶−1x>1y,−1xisanegativenumbersoitwillalwaysbesmallerthanapositivenumber.∴−1x≯1y.Sothisstatementisnottrue.Answer−StatementBiscorrect.$

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