TITLE: How can a ratio containing only real numbers have a complex value?
QUESTION [26 upvotes]: Solving this equation
$$x = 1-\cfrac{2}{1-\cfrac{2}{1-\cfrac{2}{\ddots}}}$$
Sub in $x$
$$x=1-\frac{2}{x}\implies x^2=x-2 \implies x^2-x=-2$$
Solve through completing the square
\begin{align}
x^2-x+\frac{1}{4}&=-2+\frac{1}{4}\\
\left(x-\frac{1}{2}\right)^2&=-\frac74\\
x-\frac{1}{2}&=\pm\sqrt{\frac{7}{4}}i\\
x&=\frac{1}{2}\pm\sqrt{\frac{7}{4}}i\\
\end{align}
When you substitute this value back in for $x$ it works.
But i don't understand why a equation like this can equal a complex value
Maybe I am missing something important here?
EDIT: So this equation diverges right. Does that make my working invalid, or just explain the non-real part.
Can anything useful be done by defining this recursively
$$f(x)=1-\frac{2}{f(x)}$$
and then using a seed value
I also wonder whether we would end up dividing by zero at some point. Can you prove we do or don't?

REPLY [11 votes]: The simple answer is that this repeated fraction isn't convergent for any initial value. Here's a plot of the first thousand iterations of $f_n(x) = 1 - \frac 2 {f_{n-1}(x)}$. 
The animation shows how the plot changes for different values of $f_0$. As you can tell, it bounces around a lot. Here's what the first 200 values look like with a line drawn between them (using $f_0=0.001$):

The only stable values of the function are in the complex plane. And even in the complex plane, the function isn't exactly convergent (here the initial value is $f_0=1+i$:

Instead of converging to a single value, the output of the function "orbits" around one of the stable points.