Can anything interesting be said about this fake proof? The Facebook account called BestTheorems has posted the following. Can anything of interest be said about it that a casual reader might miss?

Note that \begin{align}
\small 2 & = \frac 2{3-2} = \cfrac 2 {3-\cfrac2 {3-2}} = \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3-2}}} = \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3-2}}}} = \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {\ddots}}}}} \\[10pt]
\text{and } \\
\small 1 & = \frac 2 {3-1} = \cfrac 2 {3 - \cfrac 2 {3-1}} = \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3-1}}} = \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3-1}}}} = \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {\ddots}}}}} \\
\text{So } & 2=1.
\end{align}

 A: Simple, the problem is the assumption the three dots aka the ellipsis, in the first equation equal the ellipsis in the second.  They are not. 
A: $$x = \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {\ddots}}}}}$$
$$x = \frac 2 {3 - x}\\
x^2 - 3x + 2 = 0\\
(x-1)(x-2) = 0$$
$1,2$ are both solutions.  However, if we consider this recurrence relation:
$$x_n = \frac 2 {3 - x_{n-1}}$$
when $x_{n-1} <1 \implies x_{n-1}<x_n<1$
And the squence converges to 1.
And
$$1<x_{n-1} < 2 \implies 1<x_n  <x_{n-1}$$
and the sequence again converges.
but,
$$2<x_{n-1} < 3 \implies x_{n-1}<x_n$$
and
$$x_{n-1} >3 \implies x_n< 0$$
The sequence isn't stable in a neighborhood of $2.$
and it converges to $1$ for nearly all starting conditions.
A: For any $a$ and $b$, finding $n$ and $k$ such that
$a=\dfrac{n}{k-a}$
$b=\dfrac{n}{k-b}$
"proves" that $a=b\;\;\forall a,b\in\mathbb{N}$.
A: Suppose
$$
x = 2 + \cfrac 1 {3 + \cfrac 1 {2 + \cfrac 1 {3 + \cfrac 1 {2 + \cfrac 1 {3 + \cfrac 1 {\ddots}}}}}}
$$
Then
$$
x = 2 + \cfrac 1 {3 + \cfrac 1 x}.
$$
Solving this by the usual method, one gets
$$
x = \frac {3\pm\sqrt{15}} 3.
$$
At this point I'd be inclined to say we obviously want the positive solution. And that that's what the fraction converges to.  However, the answer from Doug M. suggests this point of view: The function
$$
x \mapsto 2 + \cfrac 1 {3 + \cfrac 1 x}
$$
has one attractive fixed point and one repulsive fixed point.
