I have to write $x= \sqrt{n^2+2}$ as a continued fraction, where $n \in N^*$.

I tried something like this: $$n< \sqrt{n^2+2}<n+1 \text{ so } [a_{0}]=n\\ x_1= \frac{1}{x-a_0}=\frac{1}{2}(\sqrt{n^2+2}+n), [a_1]=n $$

but for $x_2$ I obtained $\frac{2}{\sqrt{n^2+2}}$. From here I don't know how to proceed.

  • 1
    $\begingroup$ Welcome to Mathematics Stack Exchange. Do you mean continued fraction? $\endgroup$ – J. W. Tanner May 31 at 15:41
  • $\begingroup$ I don't really know English terms, thanks for the correction $\endgroup$ – Maria May 31 at 15:42

Consider instead $y=x-n$ then we have \begin{align} y &=\sqrt{n^2+2}-n\\ &=\frac{\left(\sqrt{n^2+2}-n\right)\left(\sqrt{n^2+2}+n\right)}{\sqrt{n^2+2}+n}\\ &=\frac2{\sqrt{n^2+2}+n}\\ &=\frac1{n+\frac{y}2}\\ &=\frac1{n+\frac{1/(n+y/2)}2}\\ &=\frac1{n+\frac1{2n+y}}\\ \end{align} Applying this final recurrence should give a continued fraction representation of $$y=[0;\overline{n,2n}]$$ and so we have $$x=[n;\overline{n,2n}]$$ $x$ is then irrational because any rational number has a finite continued fraction representation.

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