TITLE: Continued fraction $[0;0,0,0,\ldots] = \pm 1$
QUESTION [1 upvotes]: I would like to represent each $x\in\mathbb{R}_{>0}$ as a unique continued fraction. We represent each $x$ with $[a_{0};a_{1},a_{2},\ldots]$ as usual for the continued fraction. for irrational numbers, we know the sequence is infinite. For rational numbers we have a finite sequence. To have a unique representation in this notation I would like to extend the sequence with zero's. For example, $\frac{415}{93} = [4;2,6,7]$ would be extended to $[4;2,6,6,0,0,\ldots]$. We compute $$0+\frac{1}{0+\frac{1}{0+\ddots}},$$
$x_{n+1}=0+\frac{1}{x_{n}}\Rightarrow x^{2}=1\Rightarrow x=\pm 1.$
Would this be a unique representation for $x\in\mathbb{R}_{>0}$, or would this give a problem since the continued fraction of $[0;0,0,\ldots]$ is also $-1$. Possible way to work around this problem?

REPLY [0 votes]: The value of the continued fraction is, by definition, $\lim r_n$ where the $r_n$ are the convergents
$$
r_0 = 0,\\
r_1 = 0 + \frac{1}{0},\\
r_2 = 0 + \frac{1}{0+\frac{1}{0}},\\
\vdots\\
r_{n+1} = 0+\frac{1}{r_n}\\
\vdots
$$
Because of the zeros in the denominator, we cannot do this in the real numbers. So let's do it in the Riemann sphere (a.k.a. projective line).  Then
$$
r_0 = 0\\
r_1 = 0+\frac{1}{0} = \infty\\
r_2 = 0+\frac{1}{\infty} = 0\\
r_3 = 0+\frac{1}{0} = \infty
$$
alternating between $0$ and $\infty$.  So, in the Riemann sphere, this sequence does not converge.

In the conventional theory of continued fractions with real entries, there is a theorem that if $b_k \ge c$ for some $c>0$, then
$$
b_0 + \frac{1}{b_1+\frac{1}{\displaystyle b_2+\ddots}}
\tag1$$
converges.  Even if the $b_k$ are positive but converge to $0$ we could get divergence for the continued fraction $(1)$.
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