why $1+\frac{1}{1+\frac{1}{\frac{1}{1+\dots}}}$ cannot be equal to $\frac{1-\sqrt{5}}{2}$

I know that some of you will answer because it is positive.I know it but I don't know how to prove it.Maybe the second part is positive how do you know that?Thanks in advance.

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    $\begingroup$ Maybe because the LHS is a positive number? $\endgroup$ – Jack D'Aurizio Jul 16 '16 at 18:47
  • $\begingroup$ @Jack D'Aurizio .My question is exactly this why every body think it is positive? $\endgroup$ – Taha Akbari Jul 16 '16 at 18:49
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    $\begingroup$ You need to really understand what a continued fraction is. When you do the trick $1+1/x=x$ you aren't fully using the essence of what a continued fraction is. Look up how it is actually defined. $\endgroup$ – abnry Jul 16 '16 at 18:49
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    $\begingroup$ Well... how do you define $1+\frac{1}{1+\frac{1}{\frac{1}{1+\dots}}}$? I would think any reasonable definition would be the limit of the sequence: $1, 1+\frac{1}{1}, 1+\frac{1}{1+\frac{1}{1}}, 1+\frac{1}{1+\frac{1}{1+\frac{1}{1}}}, \dots$. All of which are positive and this will converge upon a specific limit. $\endgroup$ – JMoravitz Jul 16 '16 at 18:50
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    $\begingroup$ The infinite continued fraction is the limit of the finite continued fractions, all of which are clearly at least $1$. $\endgroup$ – user296602 Jul 16 '16 at 18:51

The following answer summarizes the comments. By definition, the value of a continued fraction $a_0 + \frac1{a_1 + \frac1{a_2 + \dots}}$ is the limit of its truncations to fractions. Assuming the $a_i$s are positive, the sequence $s_0 = a_0, s_1 = a_0 + \frac1{a_1},\dots$ of truncations is alternating in the sense that $s_n$ is always strictly between $s_{n-1}$ and $s_{n-2}$. The sequence therefore converges to a positive number.

The value $x$ of the continued fraction $1 + \frac1{1+\frac1{1+\dots}}$ solves $x = 1 + \frac1x$ since the operations $x \mapsto \frac1x$ and $y \mapsto 1+y$ are both continuous. Since $x = 1 + \frac1x$ has a unique positive solution, $1 + \frac1{1+\frac1{1+\dots}}$ must converge to that solution.

I remark that the question is not entirely trivial, the comments above notwithstanding. In particular, the formula "$\frac{1 + \sqrt 5}2$" can be interpreted either purely algebraically or in terms of the real numbers. The real numbers distinguish between $\sqrt 5$ and $-\sqrt 5$: the former has a square root in $\mathbb R$ whereas the latter does not. The rational numbers, if you do not consider any topology on them, do not distinguish between these two elements. The field $\mathbb Q[\sqrt 5]$ has an automorphism taking $\sqrt 5$ to $-\sqrt 5$.

Fractions are purely algebraic, but continued fractions, being infinitary, necessarily involve topology. One can study topologies on $\mathbb Q$ other than the one coming from $\mathbb R$ --- indeed, such study is very important throughout modern mathematics. One could imagine that in some other topology, the continued fraction at question would have a limit, and if in that topology you can distinguish $\pm \sqrt 5$, perhaps the limit might be $\frac{1-\sqrt 5}2$.

I know of no topology for which that works, however. The topologies on $\mathbb Q$ usually studied are: (0) the "real" topology, whose completion is $\mathbb R$, which ultimately can be derived from the idea that as you continue to add $1$ to an integer, you get further and further away from $0$; (p) the "$p$-adic topology", whose completion is $\mathbb Q_p$, which ultimately can be derived from the declaration that multiplying by $p$ (some chosen prime) moves you closer to $0$, and multiplying by numbers relatively prime to $p$ doesn't change your distance from $0$.

In $p$-adic topologies, continued fractions tend not to converge at all. For example, the truncations of $1 + \frac1{1+ \frac1{1+\dots}}$ are ratios $F_{n+1} / F_{n}$ of Fibonacci numbers. For any given prime $p$, occasionally the $n$th Fibonacci number $F_n$ will be divisible by a large power of $p$ (in which case $F_{n+1}$ will not be divisible by $p$ at all), making $F_n$ very small for the $p$-adic topology, making $F_{n+1} / F_{n}$ very large, thereby preventing the sequence of truncations from converging.

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    $\begingroup$ I think it is better not to comment on the courtesy of others as part of an answer. $\endgroup$ – quid Jul 16 '16 at 20:06
  • $\begingroup$ @quid I have removed the parenthetical remark, because I wasn't sure of whether to include it, and because I certainly think whether and how to call out courtesy of others is a complicated question. I've seen much worse behavior here and on MO, and I know I sometimes forget to be polite myself. (I should also mention that I spend much more time on MO than here, so I don't entirely know the culture.) In this case, I felt like there was something to complain about, but only barely. Likely tone-of-voice clues are all I was missing. $\endgroup$ – Theo Johnson-Freyd Jul 16 '16 at 23:46
  • $\begingroup$ I have no issue with you remarking on it, it should just not be part of the answer IMO (note the emphasis). Thanks for removing it from the answer. $\endgroup$ – quid Jul 16 '16 at 23:57

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