We Need To Go Deeper - 3

$\text {Let} y = 1 + \cfrac1{1 + \cfrac1{1 + \ddots} } \Rightarrow 1+ \dfrac {1}{y} = y \\ \rightarrow y^2 = y+1 . \\ \therefore y^2 - y - 1 = 0$

$\large 1 + \cfrac1{1 + \cfrac1{1 + \cfrac1x}} = 1 - \cfrac1{1 - \cfrac1{1 -\cfrac1x}} \\ \rightarrow 1+ \dfrac {1}{1+ \dfrac {x}{x+1}} = 1- \dfrac {1}{1- \dfrac {x}{x-1}}$

Simplifying fractions further, we get:

$1+ \dfrac {x+1}{2x+1} = 1 - \dfrac {x-1}{-1}$

$\dfrac {3x+2}{2x+1} = x$

$2x^2 + x = 3x+2 \\ \therefore x^2 - x- 1 = 0$

$\therefore x=y \\ \therefore \text { y is the positive solution of the equation above. }$

$y$ is indeed the solution because $1 + \cfrac1{1 + \cfrac1{1 + \ddots} }$ is positive, and thus y is positive. Also, $\large 1 + \cfrac1{1 + \cfrac1{1 + \cfrac1x}}$ has to be positive. Thus x is positive.