How can I solve the continued fraction of $\sqrt{3}$?
I know that a continued fraction is [1;1,2,1,2,1,2,....] -> this is the continued fraction of $\sqrt{3}$, but how can I solve? What are the steps?
I don't understand this proof.
Thank you so much.
We know $\sqrt{3}$ lies between 1 and 2 (because $1^2 < 3 < 2^2$). So the first digit of the continued fraction expansion is 1. So
$\sqrt{3} = 1 + (\sqrt{3}-1)$
and $\sqrt{3}-1$ must be between $0$ and $1$. So next we want to find the reciprocal of $\sqrt{3}-1$:
$\frac{1}{\sqrt{3}-1} = \frac{\sqrt{3}+1}{(\sqrt{3}-1)(\sqrt{3}+1)} = \frac{\sqrt{3}+1}{2}$
$\frac{\sqrt{3}+1}{2}$ lies between $1$ and $2$ so the next term in the continued fraction is 1. So we have $\sqrt{3} = [1;1,\dots]$.
The remaining amount is $\frac{\sqrt{3}+1}{2}-1 = \frac{\sqrt{3}-1}{2}$ so next we want to find the reciprocal of $\frac{\sqrt{3}-1}{2}$:
$\frac{2}{\sqrt{3}-1} = \frac{2(\sqrt{3}+1)}{2} = \sqrt{3}+1$
$\sqrt{3}+1$ is between 2 and 3, so the next term in the continued fraction is 2. So we have $\sqrt{3} = [1;1,2,\dots]$.
The remaining amount is $(\sqrt{3}+1)-2=\sqrt{3}-1$. We could continue, but we have already calculated the reciprocal of $\sqrt{3}-1$ above, so now the continued fraction just repeats and we have:
$\sqrt{3} = [1;1,2,1,2,1,2, \dots]$
An alternative way of seeing why this continued fraction repeats is to note that
$(\sqrt{3}+1)^2 = 2\sqrt{3} + 4 = 2(\sqrt{3}+1) + 2$
so $\sqrt{3}+1$ is a solution to $x^2=2x+2$, which we can re-write as:
$x = 2 + \dfrac{2}{x} = 2 + \dfrac{1}{\dfrac{x}{2}} = 2 + \dfrac{1}{1 + \dfrac{1}{x}}$
so $\sqrt{3} + 1 = [2;1,2,1,2,1,\dots]$.
Method described by Prof. Lubin at Continued fraction of $\sqrt{67} - 4$
$$ \sqrt { 3} = 1 + \frac{ \sqrt {3} - 1 }{ 1 } $$ $$ \frac{ 1 }{ \sqrt {3} - 1 } = \frac{ \sqrt {3} + 1 }{2 } = 1 + \frac{ \sqrt {3} - 1 }{2 } $$ $$ \frac{ 2 }{ \sqrt {3} - 1 } = \frac{ \sqrt {3} + 1 }{1 } = 2 + \frac{ \sqrt {3} - 1 }{1 } $$
Simple continued fraction tableau:
$$
\begin{array}{cccccccccc}
& & 1 & & 1 & & 2 & \\
\\
\frac{ 0 }{ 1 } & \frac{ 1 }{ 0 } & & \frac{ 1 }{ 1 } & & \frac{ 2 }{ 1 } \\
\\
& 1 & & -2 & & 1
\end{array}
$$
$$ \begin{array}{cccc} \frac{ 1 }{ 0 } & 1^2 - 3 \cdot 0^2 = 1 & \mbox{digit} & 1 \\ \frac{ 1 }{ 1 } & 1^2 - 3 \cdot 1^2 = -2 & \mbox{digit} & 1 \\ \frac{ 2 }{ 1 } & 2^2 - 3 \cdot 1^2 = 1 & \mbox{digit} & 2 \\ \end{array} $$
........................
We know that $\sqrt{3}$ is between $1$ and $2$, so $a_0 = 1$ and its continuous fraction will then have the following form: \begin{equation} \sqrt{3}=1+\cfrac{1}{x}\quad (1) \end{equation} Hence, $\sqrt{3}-1=\frac{1}{x}$, so $x=\frac{1}{\sqrt{3}-1}$. Multiplying up and down by $\sqrt{3}+1$ (to remove the root) we get that $x=\frac{\sqrt{3}+1}{2}.$
In this same expression we substitute $\sqrt{3}$ for its value of $(1)$, arriving at $x=1+\frac{1}{2x}$.
And now in this obtained expression we substitute $x$ for that value to which we have arrived indefinitely, obtaining the following partial results: $$x=1+\cfrac{1}{2x}=1+\cfrac{1}{2+\cfrac{1}{x}}=1+\cfrac{1}{2+\cfrac{1}{1+\cfrac{1}{2x}}}=1+\cfrac{1}{2+\cfrac{1}{1+\cfrac{1}{2+\cfrac{1}{x}}}}=\ldots$$ We see that the expressions are repeated. If we continue with the process indefinitelyu, the continous fraction will have that structure, with the same values repeating indefinitely.
Now we take the initial expression $(1)$ and substitute $x$ for what we have just obtained, finally arriving at the continuous fraction: $$\sqrt{3}=1+\cfrac{1}{1+\cfrac{1}{2+\cfrac{1}{1+\cfrac{1}{2+\cfrac{1}{1+\ddots}}}}}$$
