Arrange-in-descending-order-5-2-7-5-13-11-19-17-

Question Number 209234 by Tawa11 last updated on 04/Jul/24

Arrange in descending order :

\sqrt{5} - \sqrt{2}, \sqrt{7} - \sqrt{5}, \sqrt{13} - \sqrt{11}, \sqrt{19} - \sqrt{17}

Answered by A5T last updated on 04/Jul/24

S a m e a r r a n g e m e n t :

\sqrt{5} - \sqrt{2} (> \sqrt{5} - \sqrt{3}) > \sqrt{7} - \sqrt{5} > \sqrt{13} - \sqrt{11} > \sqrt{19} - \sqrt{17}

\sqrt{x + 2} - \sqrt{x} i s d e c r e a s i n g a s x^{+} i n c r e a s e s

\sqrt{x + 2} - \sqrt{x} \overset{?}{>} \sqrt{x + 3} - \sqrt{x + 1}

\Leftrightarrow x + 2 - x - 2 \sqrt{x (x + 2)} > x + 3 - x - 1 - 2 \sqrt{(x + 1) (x + 3)}

\Leftrightarrow x (x + 2) < (x + 1) (x + 3) w h i c h i s t r u e .

Commented by Tawa11 last updated on 05/Jul/24

Thanks sir . I appreciate .

Answered by BaliramKumar last updated on 07/Jul/24

\sqrt{5} - \sqrt{2} = \frac{(\sqrt{5} - \sqrt{2}) (\sqrt{5} + \sqrt{2})}{(\sqrt{5} + \sqrt{2})} = \frac{2}{(\sqrt{5} + \sqrt{2})} \dots \dots (i)

\sqrt{7} - \sqrt{5} = \frac{(\sqrt{7} - \sqrt{5}) (\sqrt{7} + \sqrt{5})}{(\sqrt{7} + \sqrt{5})} = \frac{2}{(\sqrt{7} + \sqrt{5})} \dots \dots (ii)

\sqrt{13} - \sqrt{11} = \frac{(\sqrt{13} - \sqrt{11}) (\sqrt{13} + \sqrt{11})}{(\sqrt{13} + \sqrt{11})} = \frac{2}{(\sqrt{13} + \sqrt{11})} \dots (iii)

\sqrt{19} - \sqrt{17} = \frac{(\sqrt{19} - \sqrt{17}) (\sqrt{19} + \sqrt{17})}{(\sqrt{19} + \sqrt{17})} = \frac{2}{(\sqrt{19} + \sqrt{17})} \dots . (iv)

(i) > (ii) > (iii) > (iv)