# Inequality related to the continued fraction expansion of sqrt(3)

I am working on a problem related to the continued fraction expansion of $\sqrt3$. If $p_k$ and $q_k$ denote the numerator and denominator, respectively, of the $k$th convergent, I should show that $$\left|\sqrt{3}-\frac{p_{2n+1}}{q_{2n+1}}\right| < \frac{1}{2\sqrt{3}q_{2n+1}^2}\;.$$

I have determined that the continued fraction expansion of $\sqrt3$ is $[1;\overline{1,2}]$ and I am able to show the equality apart from the factor $2\sqrt3$ in the denominator.

Any suggestions?

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Note that $p_{2n+1}^2 - 3 q_{2n+1}^2 = 1$. – Daniel Fischer Aug 19 '13 at 19:37
Use the standard fact that the absolute value of the difference is less than $\dfrac{1}{q_{2k+1}q_{2k+2}}$. – André Nicolas Aug 19 '13 at 19:48
And then what? From the recursive relation $q_{2k+2}=2q_{2k+1}+q_{2k}$, it seems like I want to show that $2q_{2k+1}+q_{2k}>2\sqrt3 q_{2k+1}$. This, however, implies $q_{2k}>2(\sqrt{3}-1)q_{2k+1}>q_{2k+1}$, a contradiction. – Alexandre Vandermonde Aug 19 '13 at 19:59

From the period length of $2$, you obtain that for all $n$, you have

$$p_{2n+1}^2 - 3q_{2n+1}^2 = 1,$$

and from that you can deduce

\begin{align} \frac{p_{2n+1}}{q_{2n+1}} - \sqrt{3} &= \frac{p_{2n+1} - \sqrt{3}q_{2n+1}}{q_{2n+1}}\\ &= \frac{(p_{2n+1} - \sqrt{3}q_{2n+1})(p_{2n+1} + \sqrt{3}q_{2n+1})}{q_{2n+1}(p_{2n+1} + \sqrt{3}q_{2n+1})}\\ &= \frac{1}{\left(\frac{p_{2n+1}}{q_{2n+1}}+\sqrt{3}\right)q_{2n+1}^2}\\ &< \frac{1}{2\sqrt{3}q_{2n+1}^2} \end{align}

since $\frac{p_{2n+1}}{q_{2n+1}}>\sqrt{3}$ (which you can either deduce from the general fact that odd convergents are larger than [or equal to] the value of the continued fraction, or from the fact mentioned above).

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Thank you your reply! If you have the time, would you consider leaving a hint as to why $p_{2n+1}^2-3q_{2n+1}^2=1$? I can't seem to get it right. – Alexandre Vandermonde Aug 19 '13 at 20:28
Uh, sorry, didn't think of the fact that you're probably not yet at that part of the theory. I guess the simplest way would then be an induction showing that $p_{2n}^2 - 3q_{2n}^2 = -2$, $p_{2n+1}^2-3q_{2n+1}^2 = 1$ and $p_{n+1}p_p - 3q_{n+1}q_n = (-1)^{n+1}$, using $p_{2n+2} = 2p_{2n+1} + p_{2n}$, $p_{2n+3} = p_{2n+2}+p_{2n+1}$ and analogously for $q_k$. – Daniel Fischer Aug 19 '13 at 21:03
Thanks, that solved it! – Alexandre Vandermonde Aug 20 '13 at 13:04