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:


The table representation is $[1; 1,2]$.

This is what I have so far:


Which seems to lead me to $[1;2,2,...]$ any help would be appreciated, thanks

Martin Sleziak
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    Your continued fraction works. If you want a simple continued fraction, you should make all the partial numerators equal to 1, which is as simple as dividing both numerators and denominators by two and carefully moving deeper and deeper. You will obtain the simple CF that way – Yuriy S Oct 22 '16 at 21:15
  • I agree with the remark of @Yuriy (numerators = 1). Besides, a little Latex hint: there is a special Latex primitive \cfrac for continuous fractions, that I use here: $\begin{equation*} a_0+\cfrac{1}{a_1+\cfrac{1}{a_2+\cfrac{1}{a_3+\cdots}}} \end{equation*}$ – Jean Marie Oct 22 '16 at 21:17
  • Related: (http://math.stackexchange.com/q/265690) – Jean Marie Oct 22 '16 at 21:22
  • The continued fraction of $\sqrt{3}$ with numerators all equal to 1 can be found in (https://en.wikipedia.org/wiki/Square_root_of_3). It is periodical like all real numbers that are roots of a quadratic. – Jean Marie Oct 22 '16 at 21:26

3 Answers3


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

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You seem to have worked out something that looks correct. Now prove it. Put


$$x^2+2x-2=0\implies x_{1,2}=\frac{-2\pm\sqrt{12}}{2}=-1\pm\sqrt3$$

as all above is positive we must have $\;x=-1+\sqrt3\;$ , and then $\;1+x=\sqrt3\;$ , as expected.

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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


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.

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