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 :
$$x=\frac{1+\sqrt{5}}{2}$$
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.