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Question Number 108453 by gospelkenny last updated on 17/Aug/20

$$\mathrm{If}R={(7+4\sqrt{3})}^{2n}=I+f,\mathrm{where}I\in N$$ $$\mathrm{and}0<f<1,\mathrm{then}R(1-f)\mathrm{equals}$$

Answered by 1549442205PVT last updated on 17/Aug/20

$$\mathrm{Set}\mathrm{Q}={(7-4\sqrt{3})}^{2\mathrm{n}}\Rightarrow \sqrt{\mathrm{RQ}}=[{(7+4\sqrt{3})}^{\mathrm{n}}{(7-4\sqrt{3})}^{\mathrm{n}}=1(\ast )$$ $$\mathrm{We}\mathrm{prove}\mathrm{that}\mathrm{S}=\sqrt{\mathrm{R}}+\sqrt{\mathrm{Q}}\in {\mathbb{N}}^{\ast }.\mathrm{Indeed},$$ $$\mathrm{i})\mathrm{For}\mathrm{n}=1\mathrm{we}\mathrm{get}\mathrm{S}=(7+4\sqrt{3})+(7-4\sqrt{3})$$ $$=14\Rightarrow \mathrm{State}\mathrm{is}\mathrm{true}\mathrm{for}\mathrm{n}=1$$ $$\mathrm{ii})\mathrm{Suppose}\mathrm{State}\mathrm{is}\mathrm{true}\mathrm{\forall }\mathrm{n}=\overline{1...\mathrm{k}}$$ $$\Rightarrow {\mathrm{S}}_{\mathrm{k}}={(7+4\sqrt{3})}^{\mathrm{k}}+{(7-4\sqrt{3})}^{\mathrm{k}}={\mathrm{m}}_{\mathrm{k}}\in \mathbb{N}$$ $$\mathrm{iii})\mathrm{Consider}{\mathrm{S}}_{\mathrm{k}+1}={(7+4\sqrt{3})}^{\mathrm{k}+1}+{(7-4\sqrt{3})}^{\mathrm{k}+1}$$ $$=[{(7+4\sqrt{3})}^{\mathrm{k}}+{(7-4\sqrt{3})}^{\mathrm{k}}][(7+4\sqrt{3})+(7-4\sqrt{3})]$$ $$-\left[(7+4\sqrt{3})\times (7-4\sqrt{3})\right][{(7+4\sqrt{3})}^{\mathrm{k}-1}+{(7-4\sqrt{3})}^{\mathrm{k}-1}]$$ $$={14\mathrm{S}}_{\mathrm{k}}-{\mathrm{S}}_{\mathrm{k}-1}={14\mathrm{m}}_{\mathrm{k}}-{\mathrm{m}}_{\mathrm{k}-1}\in {\mathbb{N}}^{\ast }$$ $$(\mathrm{by}\mathrm{the}\mathrm{introduction}\mathrm{hypothesis}{\mathrm{m}}_{\mathrm{k}},{\mathrm{m}}_{\mathrm{k}-1}\in {\mathbb{N}}^{\ast })$$ $$\mathrm{This}\mathrm{shows}\mathrm{that}\mathrm{the}\mathrm{state}\mathrm{is}\mathrm{also}\mathrm{true}\mathrm{for}\mathrm{n}=\mathrm{k}+1$$ $$\mathrm{Hence}\mathrm{by}\mathrm{the}\mathrm{imtroduction}\mathrm{principle}$$ $$\mathrm{the}\mathrm{state}\mathrm{is}\mathrm{true}\mathrm{for}\mathrm{\forall }\mathrm{n}\in {\mathbb{N}}^{\ast }\mathrm{which}$$ $$\mathrm{means}(\sqrt{\mathrm{R}}+\sqrt{\mathrm{Q}})\in {\mathbb{N}}^{\ast }\Rightarrow {(\sqrt{\mathrm{R}}+\sqrt{\mathrm{Q}})}^2\in {\mathbb{N}}^{\ast }$$ $$\iff \mathrm{R}+\mathrm{Q}={(\sqrt{\mathrm{R}}+\sqrt{\mathrm{Q}})}^2-2\in {\mathbb{N}}^{\ast }$$ $$\Rightarrow {(7+4\sqrt{3})}^{2\mathit{n}}+{(7-4\sqrt{3})}^{2\mathit{n}}=\mathbf{q}\in {\mathbb{N}}^{\ast }(\ast \ast )$$ $$\mathrm{On}\mathrm{the}\mathrm{other}\mathrm{hands},\mathrm{by}\mathrm{above}\mathrm{we}\mathrm{have}$$ $$\left[{(7+4\sqrt{3})}^{2n}\times {(7-4\sqrt{3})}^{2n}\right]=1$$ $$\Rightarrow 0<{(7-4\sqrt{3})}^{2n}=\frac{1}{{(7+4\sqrt{3})}^{2n}}<1$$ $$\mathrm{Therefore},\mathrm{if}\mathrm{R}={(7+4\sqrt{3})}^{2n}=\mathrm{I}+\mathrm{f}\mathrm{and}$$ $$\mathrm{I}\in \mathbb{N}\mathrm{and}0<\mathrm{f}<1\mathrm{then}\mathrm{by}(\ast \ast )$$ $$\mathbf{I}=\mathbf{q}\mathbf{and}\mathbf{f}={(7-4\sqrt{3})}^{2\mathit{n}}.\mathrm{It}\mathrm{follows}\mathrm{that}$$ $$\mathrm{R}(1-\mathrm{f})=\mathrm{R}-\mathrm{Rf}={(7+4\sqrt{3})}^{2n}-\left[{(7+4\sqrt{3})}^{2n}\times {(7-4\sqrt{3})}^{2n}\right]$$ $$={(7+4\sqrt{3})}^{2\mathit{n}}-1$$ $$\mathrm{Thus},\mathbf{R}(1-\mathbf{f})={(7+4\sqrt{3})}^{2\mathit{n}}-1$$

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