Continued fraction $[0;0,0,0,\ldots] = \pm 1$ 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?
 A: My gut feeling is that you should not proceed in this direction at all, unless this new notation buys you something really big (some result that is impossible or very hard to get otherwise) - which is very unlikely, as it would be a result, essentially, about rational numbers.

I will elaborate.
This is because, with ordinary continued fractions, even if they are infinite, the resulting irrational number is the limit (in the ordinary calculus sense) of the sequence of the truncated (finite) continued fractions. However, once you start the tail of zeros, truncation is impossible. You will not be able to obtain the value of $0+\frac{1}{0+\frac{1}{0+\ldots}}$ using standard analytic limit. Your technique ("$x=0+\frac{1}{x}$") is useful to calculate limits that already exist, but here you don't even have a sequence that should converge to anything, because any finite truncation results in division by zero!
Of course, you may neglect all that and say "let me introduce some new notation, in which $0+\frac{1}{0+\frac{1}{0+\ldots}}$ has the value (say, $1$), and let me develop the mathematics that proves some rules about that new notation. It might be possible (although my guess is that it is very unlikely you will get to something that you cannot prove without using this notation).
A few last remarks:

*

*Introducing zeros will make the continued fractions ambiguous. Say, $[1;1,1,1,1,1,\ldots]=\varphi$ but if you introduce zeros, $[1;0,0,1,1,0,0,1,1,0,0,1,1,\ldots]$ would be $\varphi$ too.

*Also, what is $[1;0,1,0,1,0,1,\ldots]$? (It's got to be a number $x$ such that $x=x+1$.)

Altogether, I think by adding the possibility that some numbers in the continued fraction expansion are zeros, you lose some important properties of continued fractions - do you gain enough that would justify that loss?
A: 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 $(1)$.
