TITLE: $\sqrt{3}$ represented as continued fraction
QUESTION [10 upvotes]: 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

REPLY [6 votes]: 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