Menu Close

Author: Tinku Tara

coukd-you-prove-please-that-11-is-not-racional-number-

Question Number 128462 by Ari last updated on 07/Jan/21 $$\mathrm{coukd}\:\mathrm{you}\:\mathrm{prove}\:\mathrm{please}\:\mathrm{that}\:\sqrt{\mathrm{11}\:}\mathrm{is}\:\mathrm{not}\:\mathrm{racional}\:\mathrm{number} \\ $$ Answered by Dwaipayan Shikari last updated on 07/Jan/21 $$\sqrt{\mathrm{11}}=\mathrm{3}+\sqrt{\mathrm{11}}−\mathrm{3}=\mathrm{3}+\frac{\mathrm{2}}{\:\sqrt{\mathrm{11}}+\mathrm{3}}=\mathrm{3}+\frac{\mathrm{2}}{\mathrm{3}+\mathrm{3}+\frac{\mathrm{2}}{\mathrm{3}+\mathrm{3}+\frac{\mathrm{2}}{\mathrm{3}+\mathrm{3}+\frac{\mathrm{2}}{\mathrm{3}+\mathrm{3}…}}}} \\ $$$$=\mathrm{3}+\frac{\mathrm{2}}{\mathrm{6}+\frac{\mathrm{2}}{\mathrm{6}+\frac{\mathrm{2}}{\mathrm{6}+\frac{\mathrm{2}}{\mathrm{6}+\frac{\mathrm{2}}{\mathrm{6}+…}}}}}\:\:\:\:\: \\ $$$${It}\:{is}\:{an}\:{Infinte}\:{continued}\:{fraction}\:.{So}\:{it}\:{can}\:{never}\:{be}\:{Rational}…

Three-friends-Boakye-kwame-and-kojo-thinking-that-they-are-decieving-their-parents-decided-to-take-turns-to-run-away-from-their-parent-Boakye-run-away-on-monday-of-the-first-week-After-how-m

Question Number 62925 by otchereabdullai@gmail.com last updated on 26/Jun/19 $$\mathrm{Three}\:\mathrm{friends}\:,\:\mathrm{Boakye}\:,\:\mathrm{kwame}\:\mathrm{and}\: \\ $$$$\mathrm{kojo}\:\mathrm{thinking}\:\mathrm{that}\:\mathrm{they}\:\mathrm{are}\:\mathrm{decieving}\: \\ $$$$\mathrm{their}\:\mathrm{parents}\:\mathrm{decided}\:\mathrm{to}\:\mathrm{take}\:\mathrm{turns}\:\mathrm{to}\: \\ $$$$\mathrm{run}\:\mathrm{away}\:\mathrm{from}\:\mathrm{their}\:\mathrm{parent}\:. \\ $$$$\mathrm{Boakye}\:\mathrm{run}\:\mathrm{away}\:\mathrm{on}\:\mathrm{monday}\:\mathrm{of}\:\mathrm{the}\:\mathrm{first} \\ $$$$\mathrm{week}.\:\mathrm{After}\:\mathrm{how}\:\mathrm{many}\:\:\mathrm{weeks}\:\mathrm{will}\: \\ $$$$\mathrm{he}\:\mathrm{run}\:\mathrm{again}\:\mathrm{on}\:\mathrm{monday}? \\ $$ Answered…

Question-128458

Question Number 128458 by SLVR last updated on 07/Jan/21 Commented by BHOOPENDRA last updated on 07/Jan/21 $${Givenf}\left({f}\left(\mathrm{1}\right)\right)=\mathrm{0},{f}\left({f}\left(\mathrm{2}\right)\right)=\mathrm{0} \\ $$$${i}.{e}\:{equation}\:{f}\left({x}\right)=\mathrm{0} \\ $$$${has}\:{two}\:{root}\:{f}\left(\mathrm{1}\right){andf}\left(\mathrm{2}\right) \\ $$$${f}\left(\mathrm{1}\right)+{f}\left(\mathrm{2}\right)=−\alpha\:{and}\:{f}\left(\mathrm{1}\right).{f}\left(\mathrm{2}\right)=\beta \\ $$$${so}\:\mathrm{5}+\mathrm{3}\alpha+\mathrm{2}\beta=−\alpha…