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$$ x=1+\cfrac{1}{1+\cfrac{1}{1+...}}\implies x=1+\frac{1}{x}\implies x=\frac{1\pm \sqrt{5}}{2} $$ Can the negative solution be considered as a solution? If yes, how is it possible to have a negative solution for a positive continued fraction? If no, how do we prove that it can't be a solution?

Edit 1: I want to understand the assumption we are considering while forming the equation which results in the "extraneous solution".

arnabanimesh
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    The right side of the equation is a positive number. So $x>0$ – Roman83 Apr 11 '16 at 11:40
  • the right side of x=1+2+3+4+...=>x=-1/12. What's your reasoning for this? I don't think your reasoning holds ground when it comes to infinite sequences. Moreover, here you put the negative solution and it solves just fine. Why is it not possible? – arnabanimesh Apr 11 '16 at 11:43
  • I know it's different. But I just want a solid proof because it is an infinite pattern. Something substantial other than "since it's positive it must be positive" which holds good for finite sequences. – arnabanimesh Apr 11 '16 at 11:49
  • If each $a_n>0$ and if $a\equiv\lim_{n\to\infty}a_n$ exists, then of course $a\geq 0$ (otherwise, there would be a neighborhood of $a$ containing no $a_n$, contrary to the definition of $a$). So the "positive because it's positive" is essentially correct. – MPW Apr 11 '16 at 12:28
  • @elvarox: "the right side of x=1+2+3+4+...=>x=-1/12." That is simply not true. What it is true is that $1^{-s}+2^{-s}+3^{-s}+\dots=\zeta(s)$ when $\mathrm{Re}(s)\gt1$, and that $\zeta(s)$ can be analytically extended to $\mathbb{C}\setminus{0}$, and that $\zeta(-1)=-\frac12$. However, this does not say that $1+2+3+\dots=-\frac12$. The left side of this equation does not converge. – robjohn Apr 11 '16 at 13:16
  • @robjohn I know what you are talking about. Check out my Edit. – arnabanimesh Apr 11 '16 at 13:32

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No, the negative number is not a solution. You showed that if $x$ is equal to that fraction, then it is either $\frac{1+\sqrt 5}{2}$ or $\frac{1-\sqrt5}{2}$. You calculated possible candidates for solutions, not the solution itself.

You can prove that $x$ must be positive by simply arguing that $x$ is a limit of a sequence with only positive elements, so the limit (if it exists, which should also be proven) must be positive.

5xum
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  • But you didn't provide any proof. It's your statement. Check my comment in the question. – arnabanimesh Apr 11 '16 at 11:47
  • @elvarox $1+2+3+\cdots \neq -\frac1{12}$. But that's beside the point. I did provide proof, or at least I provided an argument: The point is that $x$ is defined as a limit of a sequence, and the sequence is a sequence of positive numbers, and a limit of such a sequence is positive. – 5xum Apr 11 '16 at 11:49
  • https://www.youtube.com/watch?v=E-d9mgo8FGk&nohtml5=False – arnabanimesh Apr 11 '16 at 11:50
  • @elvarox As I said, that's beside the point. The video is wrong in a mathematical sense. What they are talking about is not the sum of all natural numbers, because sums are only defined for finite sets. But as I said, that's not what this question is about and I will no longer comment on it. There was plenty said about the subject here: http://math.stackexchange.com/questions/816250/sum-of-all-the-positive-integers-problem – 5xum Apr 11 '16 at 11:53
  • @elvarox Can we go back on point? Do you understand my argument now? (the limit of a positive sequence is positive one?) – 5xum Apr 11 '16 at 11:55
  • But, as the comment in that thread says "But that result is used in string theory!". Your argument is completely understandable. But, infinite pattern is a tricky business. I can't simply consider this argument itself as a solid proof. – arnabanimesh Apr 11 '16 at 12:00
  • Let us continue this discussion in chat. – arnabanimesh Apr 11 '16 at 12:04
  • @elvarox As I said, I refuse to comment on the "sum of integers is -1/12" right now because that is not the question here. You are going off-topic, please don't do that. The question was "is the negative solution here valid", and my answer is "no", and the proof is that $x$ is a limit of a positive sequence, and a limit of a positive sequence is not negative. Which part of my argument are you not happy with? – 5xum Apr 11 '16 at 12:08
  • I'm not going off topic. I just want a proof. Just tell me one thing. I didn't see in any other problem, without any assumption, a candidate for solution was ruled out just like that. At least, it should not be able to solve when substituted. If we ruled out candidates just like that, then there would be no existence of complex numbers. – arnabanimesh Apr 11 '16 at 12:12
  • @elvarox I repeated my proof five times now, here is the sixth: Infinite fractions are defined as limits. In your particular example, $x$ is the limit of a sequence with positive elements. Sequences of positive elements have non-negative limits. Therefore, $x$ is positive. Can you please address my proof? – 5xum Apr 11 '16 at 12:14
  • @elvarox Also, bringing in complex numbers is further pushing this discussion off topic. I never said that the negative number is not the solution to the equation you got. But, the negative number is not equal to $x$ as you defined it, because you defined it as a limit of a positive sequence, so $x$ must surely be a positive real number. – 5xum Apr 11 '16 at 12:15
  • You don't have to comment anything on complex numbers or -1/12. Ok, I get your point. – arnabanimesh Apr 11 '16 at 12:22
  • @elvarox I am commenting on complex numbers and $-1/12$ because you refuse to address my comments and instead go off on tangents. I wrote my argument an hour ago and you only now commented on it. – 5xum Apr 11 '16 at 12:29
  • Can you explain my concern in "Edit 1"? – arnabanimesh Apr 11 '16 at 13:31
  • @elvarox It's not really any assumption in particular. I can start, for example, with the equation $x=1$, and after squaring this equation, I get the equation $x^2=1$, which has two solutions. Would you say I am "considering some assumption" here? – 5xum Apr 11 '16 at 13:37
  • But in this case, $x=-1$ doesn't satisfy when substituted, hence is an extraneous solution. While squaring you are assuming that x might also be negative and that's why those two equations are not the same. – arnabanimesh Apr 11 '16 at 13:53
  • "While squaring you are assuming that x might also be negative"... see, here's where I sign off. This is very hand-wavy non-mathematical debating here, and while I believe such a debate is somethimes useful, I don't think it's contributing to this question any more. Also, the negative number doesn't satisfy in your case as well, so I don't even see your point. – 5xum Apr 11 '16 at 13:55
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Look at what you've written (assuming the continued fraction converges): $$ x=1+\cfrac1{\color{#C00000}{1+\cfrac1{1+\cfrac1{1+\dots}}}}\tag{1} $$ implies that $$ x=1+\frac1{\color{#C00000}{x}}\tag{2} $$ which, in turn, implies that $$ x=\frac{1+\sqrt5}2\quad\text{or}\quad x=\frac{1-\sqrt5}2\tag{3} $$ By transitivity, $(1)\implies(3)$. Since $x$, as defined in $(1)$ is greater than $0$, we know that $x=\frac{1-\sqrt5}2$ is false. However, since $(3)$ is true, we must have $$ x=\frac{1+\sqrt5}2\tag{4} $$

robjohn
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