Question Number 209234 by Tawa11 last updated on 04/Jul/24
$$\mathrm{Arrange}\:\mathrm{in}\:\mathrm{descending}\:\mathrm{order}: \\ $$$$\:\:\:\:\sqrt{\mathrm{5}}\:\:−\:\:\sqrt{\mathrm{2}},\:\:\:\:\:\sqrt{\mathrm{7}}\:\:−\:\:\sqrt{\mathrm{5}}\:,\:\:\:\sqrt{\mathrm{13}}\:\:−\:\:\sqrt{\mathrm{11}}\:,\:\:\:\sqrt{\mathrm{19}}\:\:−\:\:\sqrt{\mathrm{17}} \\ $$
Answered by A5T last updated on 04/Jul/24
$${Same}\:{arrangement}: \\ $$$$\sqrt{\mathrm{5}}−\sqrt{\mathrm{2}}\left(>\sqrt{\mathrm{5}}−\sqrt{\mathrm{3}}\right)>\sqrt{\mathrm{7}}−\sqrt{\mathrm{5}}>\sqrt{\mathrm{13}}−\sqrt{\mathrm{11}}>\sqrt{\mathrm{19}}−\sqrt{\mathrm{17}} \\ $$$$\sqrt{{x}+\mathrm{2}}−\sqrt{{x}}\:{is}\:{decreasing}\:{as}\:{x}^{+} \:{increases} \\ $$$$\sqrt{{x}+\mathrm{2}}−\sqrt{{x}}\overset{?} {>}\sqrt{{x}+\mathrm{3}}−\sqrt{{x}+\mathrm{1}} \\ $$$$\Leftrightarrow{x}+\mathrm{2}−{x}−\mathrm{2}\sqrt{{x}\left({x}+\mathrm{2}\right)}>{x}+\mathrm{3}−{x}−\mathrm{1}−\mathrm{2}\sqrt{\left({x}+\mathrm{1}\right)\left({x}+\mathrm{3}\right)} \\ $$$$\Leftrightarrow{x}\left({x}+\mathrm{2}\right)<\left({x}+\mathrm{1}\right)\left({x}+\mathrm{3}\right)\:{which}\:{is}\:{true}. \\ $$
Commented by Tawa11 last updated on 05/Jul/24
$$\mathrm{Thanks}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate}. \\ $$
Answered by BaliramKumar last updated on 07/Jul/24
$$\sqrt{\mathrm{5}}−\sqrt{\mathrm{2}}\:=\:\frac{\left(\sqrt{\mathrm{5}}−\sqrt{\mathrm{2}}\right)\left(\sqrt{\mathrm{5}}+\sqrt{\mathrm{2}}\right)}{\left(\sqrt{\mathrm{5}}+\sqrt{\mathrm{2}}\right)}\:\:=\:\frac{\mathrm{2}}{\left(\sqrt{\mathrm{5}}+\sqrt{\mathrm{2}}\right)}\:\:……\:\left(\mathrm{i}\right) \\ $$$$\sqrt{\mathrm{7}}−\sqrt{\mathrm{5}}\:=\:\frac{\left(\sqrt{\mathrm{7}}−\sqrt{\mathrm{5}}\right)\left(\sqrt{\mathrm{7}}+\sqrt{\mathrm{5}}\right)}{\left(\sqrt{\mathrm{7}}+\sqrt{\mathrm{5}}\right)}\:\:=\:\frac{\mathrm{2}}{\left(\sqrt{\mathrm{7}}+\sqrt{\mathrm{5}}\right)}\:\:……\left(\mathrm{ii}\right) \\ $$$$\sqrt{\mathrm{13}}−\sqrt{\mathrm{11}}\:=\:\frac{\left(\sqrt{\mathrm{13}}−\sqrt{\mathrm{11}}\right)\left(\sqrt{\mathrm{13}}+\sqrt{\mathrm{11}}\right)}{\left(\sqrt{\mathrm{13}}+\sqrt{\mathrm{11}}\right)}\:\:=\:\frac{\mathrm{2}}{\left(\sqrt{\mathrm{13}}+\sqrt{\mathrm{11}}\right)}\:…\left(\mathrm{iii}\right) \\ $$$$\sqrt{\mathrm{19}}−\sqrt{\mathrm{17}}\:=\:\frac{\left(\sqrt{\mathrm{19}}−\sqrt{\mathrm{17}}\right)\left(\sqrt{\mathrm{19}}+\sqrt{\mathrm{17}}\right)}{\left(\sqrt{\mathrm{19}}+\sqrt{\mathrm{17}}\right)}\:\:=\:\frac{\mathrm{2}}{\left(\sqrt{\mathrm{19}}+\sqrt{\mathrm{17}}\right)}\:….\left(\mathrm{iv}\right) \\ $$$$\left(\mathrm{i}\right)\:>\:\left(\mathrm{ii}\right)\:>\:\left(\mathrm{iii}\right)\:>\:\left(\mathrm{iv}\right) \\ $$$$ \\ $$