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Can we say that $\sqrt{2}= \cfrac{2}{\cfrac{2}{\cfrac{2}{\cfrac{2}{\ldots}}}}$?

We have $$\phi=1+\cfrac{1}{\phi}=1+\cfrac{1}{1+\cfrac{1}{\phi}}=1+\cfrac{1}{1+\cfrac{1}{1+\frac{1}{\phi}}}=\cdots$$

(with $\phi$ being the Golden Ratio)

Which gives us the confirmed infinite fraction $$\phi=1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{\ldots}}}$$

We also have $$\sqrt{2}=\cfrac{2}{\sqrt{2}}=\cfrac{2}{\cfrac{2}{\sqrt{2}}}=\cfrac{2}{\cfrac{2}{\cfrac{2}{\sqrt{2}}}}=\cdots$$

So by analogy we can deduce that $$\sqrt{2}=\cfrac{2}{\cfrac{2}{\cfrac{2}{\cfrac{2}{\ldots}}}}$$

The sequence $(a_n)_{n\in\mathbb Z^+}$ such that $a_1=\sqrt{2}, a_{n+1}=\frac{2}{a_n}$ gives $\lim_{n\to +\infty}a_n=\sqrt{2}$, so indeed the representation should be correct.

Everywhere on the Internet that I see a continued fraction of $\sqrt{2}$, it is $$\sqrt{2}=1+(\sqrt{2}-1)=1+\cfrac{1}{1+\sqrt{2}}=1+\cfrac{1}{2+\cfrac{1}{1+\sqrt{2}}}=1+\cfrac{1}{2+\cfrac{1}{2+\cfrac{1}{1+\sqrt{2}}}}=\cdots$$

This gives $$\sqrt{2}=1+\cfrac{1}{1+\cfrac{1}{2+\cfrac{1}{2+\cfrac{1}{2+\cfrac{1}{\ldots}}}}}$$

Why haven't I seen the representation $\sqrt{2}=2/(2/(2/(2/\ldots)))$ mentioned, and see the above used instead? Is something wrong about my representation?

I would say it isn't mentioned because it is not useful: you cannot approximate $\sqrt{2}$ using the representation, unlike you can using the above one (which you can do by removing $\frac{1}{1+\sqrt{2}}$ in any member).

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    $\begingroup$ Is there any possible advantage to using such representation (if indeed it is correct)? By the way, if $a_1 = \sqrt{2}$ and $a_{n+1} = \frac{2}{a_n}$, then $a_n = \sqrt{2}$ for every positive integer $n$, i.e. your sequence is constant. $\endgroup$ – A.P. Apr 23 '15 at 15:41
  • $\begingroup$ One of the points behind the continued fraction representation is that it converges no matter where you 'start from'. As long as $a_0\gt 0$, if you take $a_{n+1}=1+\frac1{a_n}$ then $\lim_{n\to\infty}a_n=\phi$. Your formula only works if you start with $a_1=\sqrt{2}$; it doesn't converge otherwise. $\endgroup$ – Steven Stadnicki Apr 23 '15 at 15:44
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    $\begingroup$ A continued fraction expansion of a real number defines a sequence of rational numbers converging to that real number. $2/(2/(2/(2/\ldots)))$ doesn't define a sequence in any reasonable way except as the sequence alternating between $2$ and $1$ so you are unlikely to make any sense of it equaling $\sqrt 2$. $\endgroup$ – PVAL-inactive Apr 23 '15 at 15:49
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The problem is that if you take $a_0$ non-zero and you define $a_n = 2/a_{n-1}$, you have the relation $a_n = a_{n-2}$ (check that) so that the sequence converges if and only if $a_n = a_{n-1}$, which implies $a_0 = \pm \sqrt{2}$. So we cannot "deduce by analogy" because there is no analogy to make. The difference with the continued fractions is that in this case, the recurrence relation building the pattern with the $\cdots$ dots gives rise to a sequence which converges.

Hope that helps,

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You have: $\sqrt 2 := \dfrac{2}{\frac{2}{\frac{2}{\frac{2}{\vdots\over\sqrt 2}}}} = \dfrac{2}{2}\cdot\frac{2}{2}\cdots\sqrt 2$

Which is not a very helpful definition.

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Point of view : Cauchy sequences constructing $\Bbb R$

The limitation of the sequence $\{x_i\}$ where $x_1 = 2$, $x_n = \frac{2}{x_{n-1}}$ does not exist. Thus $\{x_i\}$ is not a cauchy sequence.

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