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)

Commented by175 last updated on 03/Jul/20

thank, please answer the quation 2

Commented by175 last updated on 03/Jul/20

khd

Commented by1549442205 last updated on 04/Jul/20

You are welcome sir.