$\sqrt{3}$ represented as continued fraction

Question

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

Answer 1

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

Answer 2

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.

Answer 3

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.