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Question Number 101473 by 175 last updated on 02/Jul/20

Answered by 1549442205 last updated on 03/Jul/20

We prove by induction method that ∃m∈N such that A_n =(2+(√2))^n =(m_n +(√(m_n ^2 −2^n )))and q_n =(√((m_n ^2 −2^n )/2))∈N A_n =m_n +q_n (√2) +For n=1 we have 2+(√2)=2+(√(2^2 −2)) and (√((2^2 −2)/2))=1∈N⇒State true +Suppose State was true for ∀n≤k means that we had A_k =(2+(√2))^k =(m_k +(√(m_k ^2 −2^k ))) and q_k =(√((m_k ^2 −2^k )/2))∈N,A_k =m_k +q_k (√2) +Then A_(k+1) =(2+(√2))^(k+1) =(2+(√2))^k (2+(√2)) =A_k (2+(√2))=(m_k +(√(m_k ^2 −2^k )))(2+(√2)) =2m_k +(√(2m_k ^2 −2^(k+1) ))+(√2)m_k +2(√(m_k ^2 −2^k )) =(2m_k +2(√((m_k ^2 −2^k )/2)))+(√2)(m_k +(√(2m_k ^2 −2^(k+1) ))) =2m_k +2q_k +(m_k +2q_k )(√2)=2(m_k +q_k )+(√(2(m_k +2q_k )^2 )) =2(m_k +q_k )+(√(2m_k ^2 +8m_k q_k +8q_k ^2 )) =2(m_k +q_k )+(√(2m_k ^2 +8m_k q_k +4q_k ^2 +4.((m_k ^2 −2^k )/2))) =2(m_k +q_k )+(√((4m_k ^2 +8m_k q_k +4q_k ^2 )−2^(k+1) )) =2(m_k +q_k )+(√(4(m_k +q_k )^2 −2^(k+1) .)) Putting m_(k+1) =2(m_k +q_k ),q_(k+1) =m_k +2q_k then A_(k+1) =m_(k+1) +(√(m_(k+1) ^2 −2^(k+1) ))and q_(k+1=) (√(((m_(k+1) ^2 −2^(k+1) )/2) )) ∈N which shows that State is true for n=k+1 By indution principle State is true ∀n∈N .Thus,there is always exists a natural number m so that (2+(√2))^n =(m+(√(m^2 −2^n )) ∀n∈N (q.e.d)

$$ \\ $$$$\mathrm{We}\:\mathrm{prove}\:\mathrm{by}\:\mathrm{induction}\:\mathrm{method}\:\mathrm{that} \\ $$$$\exists\mathrm{m}\in\boldsymbol{\mathrm{N}}\:\mathrm{such}\:\mathrm{that}\: \\ $$$$\mathrm{A}_{\mathrm{n}} =\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)^{\mathrm{n}} =\left(\mathrm{m}_{\mathrm{n}} +\sqrt{\mathrm{m}_{\mathrm{n}} ^{\mathrm{2}} −\mathrm{2}^{\mathrm{n}} }\right)\mathrm{and}\:\mathrm{q}_{\mathrm{n}} =\sqrt{\frac{\mathrm{m}_{\mathrm{n}} ^{\mathrm{2}} −\mathrm{2}^{\mathrm{n}} }{\mathrm{2}}}\in\mathbb{N} \\ $$$$\mathrm{A}_{\mathrm{n}} =\mathrm{m}_{\mathrm{n}} +\mathrm{q}_{\mathrm{n}} \sqrt{\mathrm{2}} \\ $$$$+\mathrm{For}\:\mathrm{n}=\mathrm{1}\:\mathrm{we}\:\mathrm{have}\:\mathrm{2}+\sqrt{\mathrm{2}}=\mathrm{2}+\sqrt{\mathrm{2}^{\mathrm{2}} −\mathrm{2}} \\ $$$$\mathrm{and}\:\sqrt{\frac{\mathrm{2}^{\mathrm{2}} −\mathrm{2}}{\mathrm{2}}}=\mathrm{1}\in\boldsymbol{\mathrm{N}}\Rightarrow\mathrm{State}\:\mathrm{true} \\ $$$$+\mathrm{Suppose}\:\mathrm{State}\:\mathrm{was}\:\mathrm{true}\:\mathrm{for}\:\forall\mathrm{n}\leqslant\mathrm{k} \\ $$$$\mathrm{means}\:\mathrm{that}\:\mathrm{we}\:\mathrm{had}\:\mathrm{A}_{\mathrm{k}} =\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)^{\mathrm{k}} =\left(\mathrm{m}_{\mathrm{k}} +\sqrt{\mathrm{m}_{\mathrm{k}} ^{\mathrm{2}} −\mathrm{2}^{\mathrm{k}} }\right) \\ $$$$\mathrm{and}\:\mathrm{q}_{\mathrm{k}} =\sqrt{\frac{\mathrm{m}_{\mathrm{k}} ^{\mathrm{2}} −\mathrm{2}^{\mathrm{k}} }{\mathrm{2}}}\in\boldsymbol{\mathrm{N}},\mathrm{A}_{\mathrm{k}} =\mathrm{m}_{\mathrm{k}} +\mathrm{q}_{\mathrm{k}} \sqrt{\mathrm{2}} \\ $$$$+\mathrm{Then}\:\mathrm{A}_{\mathrm{k}+\mathrm{1}} =\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)^{\mathrm{k}+\mathrm{1}} =\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)^{\mathrm{k}} \left(\mathrm{2}+\sqrt{\mathrm{2}}\right) \\ $$$$=\mathrm{A}_{\mathrm{k}} \left(\mathrm{2}+\sqrt{\mathrm{2}}\right)=\left(\mathrm{m}_{\mathrm{k}} +\sqrt{\mathrm{m}_{\mathrm{k}} ^{\mathrm{2}} −\mathrm{2}^{\mathrm{k}} }\right)\left(\mathrm{2}+\sqrt{\mathrm{2}}\right) \\ $$$$=\mathrm{2m}_{\mathrm{k}} +\sqrt{\mathrm{2m}_{\mathrm{k}} ^{\mathrm{2}} −\mathrm{2}^{\mathrm{k}+\mathrm{1}} }+\sqrt{\mathrm{2}}\mathrm{m}_{\mathrm{k}} +\mathrm{2}\sqrt{\mathrm{m}_{\mathrm{k}} ^{\mathrm{2}} −\mathrm{2}^{\mathrm{k}} } \\ $$$$=\left(\mathrm{2m}_{\mathrm{k}} +\mathrm{2}\sqrt{\frac{\mathrm{m}_{\mathrm{k}} ^{\mathrm{2}} −\mathrm{2}^{\mathrm{k}} }{\mathrm{2}}}\right)+\sqrt{\mathrm{2}}\left(\mathrm{m}_{\mathrm{k}} +\sqrt{\mathrm{2m}_{\mathrm{k}} ^{\mathrm{2}} −\mathrm{2}^{\mathrm{k}+\mathrm{1}} }\right) \\ $$$$=\mathrm{2m}_{\mathrm{k}} +\mathrm{2q}_{\mathrm{k}} +\left(\mathrm{m}_{\mathrm{k}} +\mathrm{2q}_{\mathrm{k}} \right)\sqrt{\mathrm{2}}=\mathrm{2}\left(\mathrm{m}_{\mathrm{k}} +\mathrm{q}_{\mathrm{k}} \right)+\sqrt{\mathrm{2}\left(\mathrm{m}_{\mathrm{k}} +\mathrm{2q}_{\mathrm{k}} \right)^{\mathrm{2}} } \\ $$$$=\mathrm{2}\left(\mathrm{m}_{\mathrm{k}} +\mathrm{q}_{\mathrm{k}} \right)+\sqrt{\mathrm{2m}_{\mathrm{k}} ^{\mathrm{2}} +\mathrm{8m}_{\mathrm{k}} \mathrm{q}_{\mathrm{k}} +\mathrm{8q}_{\mathrm{k}} ^{\mathrm{2}} } \\ $$$$=\mathrm{2}\left(\mathrm{m}_{\mathrm{k}} +\mathrm{q}_{\mathrm{k}} \right)+\sqrt{\mathrm{2m}_{\mathrm{k}} ^{\mathrm{2}} +\mathrm{8m}_{\mathrm{k}} \mathrm{q}_{\mathrm{k}} +\mathrm{4q}_{\mathrm{k}} ^{\mathrm{2}} +\mathrm{4}.\frac{\mathrm{m}_{\mathrm{k}} ^{\mathrm{2}} −\mathrm{2}^{\mathrm{k}} }{\mathrm{2}}} \\ $$$$=\mathrm{2}\left(\mathrm{m}_{\mathrm{k}} +\mathrm{q}_{\mathrm{k}} \right)+\sqrt{\left(\mathrm{4m}_{\mathrm{k}} ^{\mathrm{2}} +\mathrm{8m}_{\mathrm{k}} \mathrm{q}_{\mathrm{k}} +\mathrm{4q}_{\mathrm{k}} ^{\mathrm{2}} \right)−\mathrm{2}^{\mathrm{k}+\mathrm{1}} } \\ $$$$=\mathrm{2}\left(\mathrm{m}_{\mathrm{k}} +\mathrm{q}_{\mathrm{k}} \right)+\sqrt{\mathrm{4}\left(\mathrm{m}_{\mathrm{k}} +\mathrm{q}_{\mathrm{k}} \right)^{\mathrm{2}} −\mathrm{2}^{\mathrm{k}+\mathrm{1}} .} \\ $$$$\mathrm{Putting}\:\mathrm{m}_{\mathrm{k}+\mathrm{1}} =\mathrm{2}\left(\mathrm{m}_{\mathrm{k}} +\mathrm{q}_{\mathrm{k}} \right),\mathrm{q}_{\mathrm{k}+\mathrm{1}} =\mathrm{m}_{\mathrm{k}} +\mathrm{2q}_{\mathrm{k}} \\ $$$$\:\mathrm{then}\:\mathrm{A}_{\mathrm{k}+\mathrm{1}} =\mathrm{m}_{\mathrm{k}+\mathrm{1}} +\sqrt{\mathrm{m}_{\mathrm{k}+\mathrm{1}} ^{\mathrm{2}} −\mathrm{2}^{\mathrm{k}+\mathrm{1}} }\mathrm{and}\: \\ $$$$\mathrm{q}_{\mathrm{k}+\mathrm{1}=} \sqrt{\frac{\mathrm{m}_{\mathrm{k}+\mathrm{1}} ^{\mathrm{2}} −\mathrm{2}^{\mathrm{k}+\mathrm{1}} }{\mathrm{2}}\:}\:\in\boldsymbol{\mathrm{N}}\:\mathrm{which}\:\mathrm{shows} \\ $$$$\mathrm{that}\:\mathrm{State}\:\mathrm{is}\:\mathrm{true}\:\mathrm{for}\:\mathrm{n}=\mathrm{k}+\mathrm{1} \\ $$$$\mathrm{By}\:\mathrm{indution}\:\mathrm{principle}\:\mathrm{State}\:\mathrm{is}\:\mathrm{true}\: \\ $$$$\forall\mathrm{n}\in\boldsymbol{\mathrm{N}}\:.\mathrm{Thus},\mathrm{there}\:\mathrm{is}\:\mathrm{always}\:\mathrm{exists}\:\mathrm{a} \\ $$$$\mathrm{natural}\:\mathrm{number}\:\mathrm{m}\:\mathrm{so}\:\mathrm{that} \\ $$$$\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)^{\boldsymbol{\mathrm{n}}} =\left(\boldsymbol{\mathrm{m}}+\sqrt{\boldsymbol{\mathrm{m}}^{\mathrm{2}} −\mathrm{2}^{\boldsymbol{\mathrm{n}}} }\:\forall\boldsymbol{\mathrm{n}}\in\boldsymbol{\mathrm{N}}\:\left(\boldsymbol{\mathrm{q}}.\boldsymbol{\mathrm{e}}.\boldsymbol{\mathrm{d}}\right)\right. \\ $$

Commented by 175 last updated on 03/Jul/20

thank, please answer the quation 2

Commented by 175 last updated on 03/Jul/20

khd

Commented by 1549442205 last updated on 04/Jul/20

You are welcome sir.

$$\mathrm{You}\:\mathrm{are}\:\mathrm{welcome}\:\mathrm{sir}. \\ $$

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