$\sqrt{3}$ represented as continued fraction Am I incorrect in the assumption that I deal with $\sqrt{3}$ in the same way I would approach $\sqrt{2}$, by adding and subtracting $1$ such that:
$\sqrt{3}=1+\sqrt{3}-1=1+\cfrac{2}{1+\sqrt{3}}$  
The table representation is $[1; 1,2]$.  
This is what I have so far:  
$\sqrt{3}=1+\cfrac{2}{2+\cfrac{2}{1+\sqrt{3}}}$   
Which seems to lead me to $[1;2,2,...]$ any help would be appreciated, thanks
 A: The numerators should be all $1$. The procedure is to write $x>0$ as
$$
x=x_0+y_0
$$
with $0\le y_0<1$. If $y_0=0$, we're done. Otherwise we set
$$
\frac{1}{y_0}=x_1+y_1
$$
in the same fashion and go on with the same rule.
In your case, $y_0=\sqrt{3}-1$, so
$$
\frac{1}{y_0}=\frac{1}{\sqrt{3}-1}=\frac{\sqrt{3}+1}{2}=
1+\frac{\sqrt{3}-1}{2}
$$
Therefore $x_1=1$ and $y_1=\frac{\sqrt{3}-1}{2}$. Then
$$
\frac{1}{y_1}=\frac{2}{\sqrt{3}-1}=\sqrt{3}+1=2+(\sqrt{3}-1)
$$
Hence $x_2=2$ and $y_2=\sqrt{3}-1$.
OK, now we'll go on forever with the same numbers. Hence $x_n=1$ for odd $n$ and $x_n=2$ for even $n>0$:
$$
\sqrt{3}=[1;1,2,1,2,\dots]=1+
\cfrac{1}{
  1+\cfrac{1}{
    2+\cfrac{1}{
      1+\cfrac{1}{
        2+\ddots
      }
    }
  }
}
$$
See http://planetmath.org/tableofcontinuedfractionsofsqrtnfor1n102
A: You seem to have worked out something that looks correct. Now prove it. Put
$$x=\frac2{2+\frac2{2...}}\implies1+x=1+\frac2{2+x}\implies2x+x^2=2\implies$$
$$x^2+2x-2=0\implies x_{1,2}=\frac{-2\pm\sqrt{12}}{2}=-1\pm\sqrt3$$
as all above is positive we must have $\;x=-1+\sqrt3\;$ , and then $\;1+x=\sqrt3\;$ , as expected.
A: If you obtain a rational approximation of $\sqrt{3}$, for example $\frac{362}{209}$ you'll find its continued fraction is $$1+\frac{1}{1+\frac{1}{2+\frac{1}{1+\frac{1}{2+\frac{1}{1+\frac{1}{2+\frac{1}{3}}}}}}}$$ That 3 looks extraneous, so we might form the conjecture that $$\sqrt{3}=1+\frac{1}{1+\frac{1}{2+\frac{1}{1+\frac{1}{2+\frac{1}{1+\frac{1}{2+\ddots}}}}}}$$
Then we can prove that by calling the RHS x
$$x=1+\frac{1}{1+\frac{1}{1+\color{red}{1+\frac{1}{1+\frac{1}{2+\frac{1}{1+\frac{1}{2+\ddots}}}}}}}=1+\frac{1}{1+\frac{1}{1+x}}$$
Then by simple algebra we can verify that $x=\sqrt{3}$. 
On a small digression, these rational approximations might seem hard to obtain but I found that with the chakravala method in a few minutes.
