Question and Answers Forum

All Questions      Topic List

Number Theory Questions

Previous in All Question      Next in All Question      

Previous in Number Theory      Next in Number Theory      

Question Number 108516 by Rasheed.Sindhi last updated on 17/Aug/20

$$\mathrm{If}\mathrm{x}+\frac{1}{\mathrm{x}}=2(\mathrm{x}\ne 0),\mathrm{prove}\mathrm{that}$$$${\mathrm{x}}^{\mathrm{n}}+\frac{1}{{\mathrm{x}}^{\mathrm{n}}}=2\mathrm{\forall }n\in \mathbb{Z}$$

Commented by mr W last updated on 17/Aug/20

$$ifx>0:$$$$x+\frac{1}{x}⩾2\sqrt{x\times \frac{1}{x}}=2$$$$=onlyforx=\frac{1}{x}=1$$$$$$$$ifx<0:$$$$x+\frac{1}{x}=-(-x+\frac{1}{-x})⩽-2$$$$\Rightarrow nosolutionforx+\frac{1}{x}=2$$$$$$$$\Rightarrow onlysolutionforx+\frac{1}{x}=2is$$$$x=1$$$$$$$$\Rightarrow x^n+\frac{1}{x^n}=1+1=2$$

Commented by Rasheed.Sindhi last updated on 17/Aug/20

$${\mathbf{e}}^{\mathbf{x}}\mathrm{cellent}!$$

Commented by udaythool last updated on 17/Aug/20

$$\mathrm{By}\mathrm{induction}\mathrm{it}\mathrm{follows}...$$

Commented by Rasheed.Sindhi last updated on 17/Aug/20

$$Thankssirwaitingforproofby$$$$induction....$$

Commented by udaythool last updated on 17/Aug/20

$$\mathrm{Let}n=0,1,...$$$$\mathrm{Induction}\mathrm{statment}:$$$$P_{\mathrm{n}}:x^n+\frac{1}{x^n}=2$$$$\mathrm{For}n=0\mathrm{and}1,P_n\mathrm{is}\mathrm{true}$$$$\mathrm{Induction}\mathrm{hypothesis}:$$$$\mathrm{Let}{P}_n\mathrm{is}\mathrm{true}\mathrm{for}\mathrm{all}n=0,...,k$$$$\therefore (x^{k+1}+\frac{1}{x^{k+1}})=(x+\frac{1}{x})(x^k+\frac{1}{x^k})-2=2\mathrm{x}2-2=2$$$$\mathrm{i}.\mathrm{e}.P_{\mathrm{k}}\Rightarrow P_{k+1}$$$$\mathrm{Thus}\mathrm{by}\mathrm{PI}{P}_n\mathrm{is}\mathrm{true}\mathrm{for}\mathrm{all}n\in {\mathbb{Z}}^{+}$$$$\mathrm{If}n\in {\mathbb{Z}}^{-}\Rightarrow n=-m\mathrm{for}m=\mid n\mid \in {\mathbb{Z}}^{+}$$$$\therefore x^n+\frac{1}{x^n}=x^{-m}+\frac{1}{x^{-m}}=x^m+\frac{1}{x^m}=2$$$$\mathrm{And}\mathrm{hence}{P}_n\mathrm{is}\mathrm{true}\mathrm{\forall }n\in \mathbb{Z}$$$$\mathrm{etc}...$$$$$$

Commented by Rasheed.Sindhi last updated on 18/Aug/20

$$Nicesir!$$$$Thelinebelowneedssome$$$$detail:\mathcal{T}hanks.$$$$\therefore (x^{k+1}+\frac{1}{x^{k+1}})=(x+\frac{1}{x})(x^k+\frac{1}{x^k})-2=2\mathrm{x}2-2=2$$$$$$

Commented by udaythool last updated on 13/Sep/20

$$\mathrm{Actually};$$$$(x+\frac{1}{x})(x^k+\frac{1}{x^k})=(x^{k+1}+\frac{1}{x^{k+1}})+(x^{k-1}+\frac{1}{x^{k-1}})$$$$\Rightarrow (x^{k+1}+\frac{1}{x^{k+1}})=(x+\frac{1}{x})(x^k+\frac{1}{x^k})-(x^{k-1}+\frac{1}{x^{k-1}})$$$$\mathrm{we}\mathrm{have}\mathrm{assumed}(\mathrm{induction}$$$$\mathrm{hypothesis}){\mathrm{p}}_{\mathrm{n}}\mathrm{is}\mathrm{true}\mathrm{for}\mathrm{n}=\mathrm{k}$$$$\mathrm{i}.\mathrm{e}.{\mathrm{p}}_{\mathrm{n}}=2\mathrm{for}\mathrm{n}=1,2,\dots ,\mathrm{k}$$$$\mathrm{Thus}(x^{k+1}+\frac{1}{x^{k+1}})=2\mathrm{x}2-2=2$$

Answered by floor(10²Eta[1]) last updated on 17/Aug/20

$${\mathrm{x}}^2-2\mathrm{x}+1={(\mathrm{x}-1)}^2=0\Rightarrow \mathrm{x}=1$$$$\Rightarrow {\mathrm{x}}^{\mathrm{n}}=1\therefore {\mathrm{x}}^{\mathrm{n}}+\frac{1}{{\mathrm{x}}^{\mathrm{n}}}=2=1+\frac{1}{1}$$

Commented by Rasheed.Sindhi last updated on 17/Aug/20

$$\theta \alpha n\mathcal{X}sir!$$

Answered by mathmax by abdo last updated on 17/Aug/20

$$\mathrm{x}+\frac{1}{\mathrm{x}}=2\Rightarrow {\mathrm{x}}^2+1=2\mathrm{x}\Rightarrow {\mathrm{x}}^2-2\mathrm{x}+1=0\Rightarrow {(\mathrm{x}-1)}^2=0\Rightarrow \mathrm{x}=1\Rightarrow$$$$\mathrm{\forall }\mathrm{n}\mathrm{we}\mathrm{hsve}{\mathrm{x}}^{\mathrm{n}}+\frac{1}{{\mathrm{x}}^{\mathrm{n}}}=2$$

Commented by Rasheed.Sindhi last updated on 18/Aug/20

$$Thanxsir$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com