Where did the negative answer come from in the continued fraction $1+\frac{1}{1+1/(1+\dots)}$? In this question we increased solution domain by squaring both sides of equation but what about this one ? 
Here the question is to evaluate $1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{\ddots  }}}}}$
$$x=1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{\ddots}}}}}$$
$$x-1=\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{\ddots  }}}}}$$
$$\cfrac{1}{x-1}=1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{\ddots }}}}}$$
$$\cfrac{1}{x-1}=x$$
$$x^2-x-1 = 0$$
$$x=\frac{1\pm \sqrt{5}}{2}$$
Now, it is obvious that answer can't be negative so : 
$$x=\frac{1+\sqrt{5}}{2}$$
 A: Let $f(z)=1+\frac1z$. The function $f$ has two fixed points, which are $\frac{1\pm\sqrt{5}}{2}$.
At first, when you write
$$x=1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{\ddots}}}}}$$
you're defining $x$ to be the particular fixed point of $f$ that attracts the orbit of $1$, if it exists. It so happens that $x$ does exist, and it equals $\frac{1+\sqrt{5}}{2}$.
Later, when you write
$$x=1+\frac1x$$
you're asserting that $x$ is some fixed point of $f$, but this equation doesn't specify which fixed point that is.
A: If we set $x = \dfrac{1}{-1 +\underbrace{\dfrac{1}{-1+\dfrac{1}{-1+\dfrac{1}{-1+\cdots}}}}_{x}}$, then we have $x = \frac{1}{-1 + x}$, which gives also $x^2 - x - 1=0$
That's a similar presentation of the negative answer.
A: I think it's obviously a negative number of magnitude less than 1. After all, $1 / x$ is clearly a negative number of magnitude greater than $1$!
Okay not really. But the reason you think it's obviously positive is that there is a hidden condition in your question: you intend $x$ to be the limit of the sequence $$1, 1 + \frac{1}{1}, 1 + \frac{1}{1 + \frac{1}{1}}, \ldots $$ rather than, say, just a number that satisfies the self-referential relationship $x = 1 + 1/x$. It is this hidden condition that differentiates between the two solutions of this equation.
