Question Number 60955 by Tawa1 last updated on 27/May/19 Commented by alphaprime last updated on 27/May/19 hello sir , I want you to join my community to resolve the question no. 60723 , so please provide me your email Commented by Tawa1 last updated on 27/May/19 $$\mathrm{How}\:?…
Question Number 191997 by Mastermind last updated on 05/May/23 $$\mathrm{Show}\:\mathrm{that}\:\mathbb{C}=\left\{−\mathrm{1},\mathrm{1},−\imath,\imath\right\}\:\mathrm{where} \\ $$$$\imath=\sqrt{−\mathrm{1}}\:\mathrm{with}\:\mathrm{addition}\:\mathrm{operation}\:\mathrm{is}\:\mathrm{a} \\ $$$$\mathrm{group}. \\ $$$$ \\ $$$$\mathrm{Help}! \\ $$ Commented by Rasheed.Sindhi last updated…
Question Number 191986 by Mastermind last updated on 04/May/23 $$\mathrm{Ques}.\:\mathrm{1} \\ $$$$\mathrm{Let}\:\left(\mathrm{G},\ast\right)\:\mathrm{be}\:\mathrm{a}\:\mathrm{group},\:\mathrm{then}\:\mathrm{show} \\ $$$$\mathrm{that}\:\mathrm{for}\:\mathrm{each}\:\mathrm{a}\in\mathrm{G},\:\exists\:\mathrm{a}\:\mathrm{unique}\: \\ $$$$\mathrm{element}\:\mathrm{e}\in\mathrm{G}\:\mid\:\mathrm{a}\ast\mathrm{e}=\mathrm{e}\ast\mathrm{a}=\mathrm{a} \\ $$$$ \\ $$$$\mathrm{Ques}.\:\mathrm{2} \\ $$$$\mathrm{If}\:\mathrm{a}\in\mathrm{G}\:\Rightarrow\:\mathrm{x}\in\mathrm{G}\:\mathrm{and}\:\mathrm{x}\:\mathrm{is}\:\mathrm{unique} \\ $$$$\mathrm{show}\:\mathrm{that}\:\mathrm{if}\:\mathrm{x}\ast\mathrm{a}=\mathrm{e},\:\mathrm{then}\:\mathrm{a}\ast\mathrm{x}=\mathrm{e}. \\…
Question Number 126400 by F_Nongue last updated on 20/Dec/20 $${If}\:{a}_{{n}} =\mathrm{6}^{{n}} +\mathrm{8}^{{n}} \:{find}\:\frac{{a}_{\mathrm{1991}} }{\mathrm{49}}. \\ $$ Commented by talminator2856791 last updated on 20/Dec/20 $$\:\mathrm{are}\:\mathrm{you}\:\mathrm{wanting}\:\mathrm{remainder}? \\…
Question Number 191937 by Mastermind last updated on 04/May/23 $$\mathrm{Check}\:\mathrm{whether}\:\left(\mathrm{Q},\:\centerdot\right)\:\mathrm{is}\:\mathrm{a}\:\mathrm{group}\:\mathrm{or} \\ $$$$\mathrm{not} \\ $$$$ \\ $$$$\mathrm{Hello}\:\mathrm{bosses}! \\ $$ Answered by AST last updated on 04/May/23…
Question Number 191926 by Rupesh123 last updated on 04/May/23 Answered by AST last updated on 04/May/23 $$\left[{x}\right]+\left\{{x}\right\}={x} \\ $$$$\left[{x}\right].\left[{y}\right]=\left[{x}\right]+\left[{y}\right]+\left\{{x}\right\}+\left\{{y}\right\}<\left[{x}\right]+\left[{y}\right]+\mathrm{2} \\ $$$$\Rightarrow{Case}\:{I}:\left[{x}\right].\left[{y}\right]=\left[{x}\right]+\left[{y}\right]\:{or}\:{II}:\left[{x}\right].\left[{y}\right]=\left[{x}\right]+\left[{y}\right]+\mathrm{1} \\ $$$${I}\Rightarrow\left[{x}\right]\left(\left[{y}\right]−\mathrm{1}\right)=\left[{y}\right]\Rightarrow\left[{x}\right]=\frac{\left[{y}\right]}{\left[{y}\right]−\mathrm{1}}=\mathrm{1}+\frac{\mathrm{1}}{\left[{y}\right]−\mathrm{1}} \\ $$$$\Rightarrow\left(\left[{y}\right]−\mathrm{1}\right)\mid\mathrm{1}\Rightarrow\left[{y}\right]−\mathrm{1}=\mathrm{1}\:{or}\:\left[{y}\right]−\mathrm{1}=−\mathrm{1}…
Question Number 60765 by readone97 last updated on 25/May/19 $${express}\:{in}\:{partial}\:{fraction}\:\mathrm{14}/\left({x}^{\mathrm{2}} +\mathrm{3}\right)\left({x}+\mathrm{2}\right) \\ $$ Commented by Prithwish sen last updated on 25/May/19 $$\frac{\mathrm{1}}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{3}\right)\left(\mathrm{x}+\mathrm{2}\right)}\:=\:\frac{\mathrm{A}}{\left(\mathrm{x}+\mathrm{2}\right)}\:+\frac{\mathrm{Bx}+\mathrm{C}}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{3}\right)} \\…
Question Number 60756 by readone97 last updated on 25/May/19 $${express}\:{in}\:{partial}\:{fraction}\:\mathrm{5}/\left({x}−\mathrm{2}\right)\left({x}+\mathrm{3}\right)^{\mathrm{2}} \\ $$ Commented by Prithwish sen last updated on 25/May/19 $$\frac{\mathrm{1}}{\left(\mathrm{x}−\mathrm{2}\right)\left(\mathrm{x}+\mathrm{3}\right)^{\mathrm{2}} }\:=\frac{\mathrm{A}}{\left(\mathrm{x}−\mathrm{2}\right)}\:+\frac{\mathrm{B}}{\left(\mathrm{x}+\mathrm{3}\right)}\:+\frac{\mathrm{C}}{\left(\mathrm{x}+\mathrm{3}\right)^{\mathrm{2}} } \\ $$$$\mathrm{and}\:\mathrm{then}\:\mathrm{proceed}.…
Question Number 60735 by readone97 last updated on 25/May/19 $${express}\:{in}\:{partial}\:{fraction}\:\mathrm{3}/\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} −\mathrm{4}\right) \\ $$ Answered by Forkum Michael Choungong last updated on 25/May/19 $$\frac{\mathrm{3}}{\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} −\mathrm{4}\right)\:}=\:\frac{{A}}{{x}+\mathrm{1}}\:+\:\frac{{B}}{{x}^{\mathrm{2}} −\mathrm{4}}…
Question Number 191787 by Mastermind last updated on 30/Apr/23 $$\mathrm{Ques}.\:\mathrm{2}\:\left(\mathrm{Metric}\:\mathrm{Space}\:\mathrm{Question}\right) \\ $$$$\:\:\:\:\:\:\mathrm{Let}\:\mathrm{d}\:\mathrm{be}\:\mathrm{a}\:\mathrm{metric}\:\mathrm{on}\:\mathrm{a}\:\mathrm{non}−\mathrm{empty} \\ $$$$\mathrm{set}\:\mathrm{X}.\:\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{function}\:\mathrm{U}\:\mathrm{is} \\ $$$$\mathrm{defined}\:\mathrm{by}\:\mathrm{U}\left(\mathrm{x},\mathrm{y}\right)=\frac{\mathrm{d}\left(\mathrm{x},\mathrm{y}\right)}{\mathrm{1}+\mathrm{d}\left(\mathrm{x},\mathrm{y}\right)},\:\mathrm{where} \\ $$$$\mathrm{x}\:\mathrm{and}\:\mathrm{y}\:\mathrm{are}\:\mathrm{arbitrary}\:\mathrm{element}\:\mathrm{X}\:\mathrm{is}\:\mathrm{also} \\ $$$$\mathrm{a}\:\mathrm{metric}\:\mathrm{on}\:\mathrm{X}. \\ $$ Terms of Service…