I'm trying to learn how to express a square root as continued fraction, but I can't get one thing.

The following example of $\sqrt{14}$ is from this page (click the image to see it at full size):

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In the 2nd row of the table, can anyone please tell me where the 1 (in red) comes from? For instance, why can't it be 2, or 3, or 4?

This is my only doubt, which I've been trying hard to understand, but unable to.

  • $\begingroup$ @Zev, thank you for your nice editing! :) $\endgroup$ – xylon97 Feb 20 '13 at 16:39
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    $\begingroup$ Glad to help! In general, it's best to capture things from external websites here so that if the link ever dies the question will still make sense :) $\endgroup$ – Zev Chonoles Feb 20 '13 at 16:42

Because $1$ is the greatest integer that is $\le \dfrac{\sqrt{14}+3}{5}$.

Remark: Recall the general continued fraction procedure. At any stage, we are looking at a number $x_n$, which is $\ge 1$ except possibly at the beginning. We compute $a_n=\lfloor x_n\rfloor$. If $a_n\ne x_n$, we define $x_{n+1}$ by $x_{n+1}=\dfrac{1}{x_n-a_n}$. Then $$x_n=a_n +\frac{1}{x_{n+1}}.$$ (There is a "better" algorithm for square roots of integers, and related numbers.)

  • $\begingroup$ Ah, thanks a lot. I don't think this has mentioned anywhere. Thanks again. $\endgroup$ – xylon97 Feb 20 '13 at 16:43
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    $\begingroup$ @hkapur97: You are welcome. It helps to carry out the algorithm on a calculator a few times. This is particularly easy if the calculator has a "$1/x$" button. Do it a few times. In particular, you might want to do it with the value of $\pi$ stored in your calculator. I am doing it now. So I called for $\pi$, got $3.14\dots$ on the display. Write down the greatest integer $\le \pi$, which is $3$. Subtract $3$, press the $1/x$ button. We get something a bit over $7$. Write down the $7$, subtract $7$ from the calculator result, press $1/x$. We get something very close to $16$, a bit under. $\endgroup$ – André Nicolas Feb 20 '13 at 16:59
  • $\begingroup$ Write down $15$. Subtract $15$, flip. We get something a tiny bit over $1$. Write down a $1$. Continue. But soon it is unreliable, because calculator accuracy issues. However, look at the first approximation, namely $3$. The second is $3+\frac{1}{7}$, familiar! This way we get very good rational approximations to $\pi$ with smallish denominator. $\endgroup$ – André Nicolas Feb 20 '13 at 17:03

More generally, if $x$ is real and $n$ is a natural number, then $$\left\lfloor \frac{x}{n}\right\rfloor = \left\lfloor \frac{\lfloor x\rfloor}{n}\right\rfloor$$

So in computing the continued fraction expansion of a square root of $D$, all the $x_i$ are of the form $$\frac{\sqrt D + A}{B}$$ and to find the integer part of this, we only need to find the integer part of:

$$\frac{\lfloor\sqrt D \rfloor+ A}{B}$$

That greatly simplifies the process of computing the integer parts at each step - the only "approximation" we need for $\sqrt{D}$ to compute this step is $\lfloor \sqrt{D}\rfloor$. Consider how much harder it would be if the $x_i$ could be of the form:


This essentially requirs a "better" estimate for $\sqrt{D}$ to find the integer part of this expression.


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