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If I want to find the continued fraction of $\sqrt{n}$ how do I know which number to use for $a_0$? Is there a way to do it without using a calculator or anything like that? What's the general algorithm for computing it? I tried to read the wiki article but was overwhelmed and lost. I tried Googling but couldn't find a website that actually explained this question.

If anyone has a good site that answers these questions either, please let me know. Thanks!

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Let's just do an example. Let's find the continued fraction for $\def\sf{\sqrt 5}\sf$. $\sf\approx 2.23$ or something, and $a_0$ is the integer part of this, which is 2.

Then we subtract $a_0$ from $\sf$ and take the reciprocal. That is, we calculate ${1\over \sf-2}$. If you're using a calculator, this comes out to 4.23 or so. Then $a_1$ is the integer part of this, which is 4. So: $$\sf=2+\cfrac{1}{4+\cfrac1{\vdots}}$$

Where we haven't figured out the $\vdots$ part yet. To get that, we take our $4.23$, subtract $a_1$, and take the reciprocal; that is, we calculate ${1\over 4.23 - 4} \approx 4.23$. This is just the same as we had before, so $a_2$ is 4 again, and continuing in the same way, $a_3 = a_4 = \ldots = 4$: $$\sf=2+\cfrac{1}{4+\cfrac1{4+\cfrac1{4+\cfrac1\vdots}}}$$


This procedure will work for any number whatever, but for $\sf$ we can use a little algebraic cleverness to see that the fours really do repeat. When we get to the ${1\over \sf-2}$ stage, we apply algebra to convert this to ${1\over \sf-2}\cdot{\sf+2\over\sf+2} = \sf+2$. So we could say that: $$\begin{align} \sf & = 2 + \cfrac 1{2+\sf}\\ 2 + \sf & = 4 + \cfrac 1{2+\sf}. \end{align}$$

If we substitute the right-hand side of the last equation expression into itself in place of $ 2+ \sf$, we get:

$$ \begin{align} 2+ \sf & = 4 + \cfrac 1{4 + \cfrac 1{2+\sf}} \\ & = 4 + \cfrac 1{4 + \cfrac 1{4 + \cfrac 1{2+\sf}}} \\ & = 4 + \cfrac 1{4 + \cfrac 1{4 + \cfrac 1{4 + \cfrac 1{2+\sf}}}} \\ & = \cdots \end{align} $$

and it's evident that the fours will repeat forever.

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  • $\begingroup$ I always thought they were in the form a0 + 1/(a1 + 1/ (a2 + 1/( ... $\endgroup$ – user51819 Dec 27 '12 at 2:01
  • $\begingroup$ @user51819 Quite so (although $a_0$ could be 0.) I had the typesetting wrong, and have corrected it. $\endgroup$ – MJD Dec 27 '12 at 2:03
  • $\begingroup$ Sorry, one last question; how do we know, in the general sense, when we've hit the point where it's periodic? When you encounter an $a_k$ you've seen before? Or when a reciprocal equals a reciprocal you've seen before? I am guessing the latter? $\endgroup$ – user51819 Dec 27 '12 at 2:09
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    $\begingroup$ When the reciprocal is one you have seen before. It couldn't be when you reach an $a_k$ you've seen before or else you couldn't have a repeating sequence like $1,1,2,1,1,2,1,1,2,\ldots$. $\endgroup$ – MJD Dec 27 '12 at 2:15
  • $\begingroup$ Incredible, really. $\endgroup$ – Meitar Jun 13 '15 at 12:19
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Confirm the algebraic identity: $$\sqrt n=a+\frac{n-a^2}{a+\sqrt n}$$ Then chose whatever value of 'a' you want, and just keep on pluging in $\sqrt n$

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    $\begingroup$ With this you end up with generalized continuous fraction. I mean $n-a^2$ not necessarily $1$. $\endgroup$ – Yola Oct 24 '16 at 17:59
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    $\begingroup$ This is a sad case of changing the question to get a simpler answer (and that is assuming the question was how to find the continued fraction of $\sqrt n$; in fact it was rather uninterestingly just about its first term, the integer part of $\sqrt n$). $\endgroup$ – Marc van Leeuwen May 29 '18 at 10:50
  • $\begingroup$ -1 As @Yola points out, this doesn’t (necessarily) give you a continued fraction $\endgroup$ – HelloGoodbye Aug 30 '18 at 22:55
  • $\begingroup$ @HelloGoodbye OP changed his question, I answered what he originally asked. $\endgroup$ – Ethan Sep 1 '18 at 1:06
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    $\begingroup$ Okay. If so, shouldn't it be possible to see that here? $\endgroup$ – HelloGoodbye Sep 1 '18 at 23:39
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$a_0$ is simply the largest integer such that $a^2 \le n$ . You can determine the continued fraction for a square root by performing the $\frac1{\sqrt n - a_0}$ step and then using the conjugate to remove the square root from the denominator, and repeating.

I recommend Ron Knott's site: http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/cfINTRO.html . Good Luck.

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$a_0$ is the largest integer that is smaller than or equal to $\sqrt n$. Or put another way, you want $a_0^2$ to be smaller than or equal to $n$, and $(a_0+1)^2$ to be bigger than $n$.

If you really have no idea what integer to use, then you find it by guessing an integer $g$. Then you calculate $g^2$. If $g^2$ is bigger than $n$, your guess $g$ was too big, and you try a smaller guess. If $g^2$ is much smaller than $n$, your guess $g$ was too small, and you try a bigger guess. You keep doing this until you find a guess $g$ where $g^2 \le n$ and $(g+1)^2 > n$, and then $a_0 = g$.

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  • $\begingroup$ This is probably the most helpful answer to the original question. $\endgroup$ – LarsH Mar 14 at 13:17
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Here the easiest method to generate continued fraction for any square (or more) root. Lets take $\sqrt{5}$:

$$\sqrt{5} \approx 2,2360679775...$$ $$\sqrt{5} = 2 + 0,2360679775$$

So $2=\left \lfloor \sqrt(5) \right \rfloor$

$$\sqrt{5} =2 + x$$ with $x=0,2360679775$

$$\sqrt{5}^{2} = (x + 2)^{2} \Rightarrow 5= x^2 + 4x + 4$$ $$1= x^2 + 4x$$ $$\frac{1}{x}= x + 4 \Rightarrow x = \frac{1}{4+x}$$ So $x = \frac{1}{4+\frac{1}{4+\frac{1}{4+\frac{1}{4+\ddots}}}}$

Finally, we got: $\sqrt{5} = 2 + \frac{1}{4+\frac{1}{4+\frac{1}{4+\frac{1}{4+\ddots}}}}$.

So for any x, $\sqrt(x) = \left \lfloor \sqrt(x) \right \rfloor + \frac{x-\left \lfloor \sqrt(x) \right \rfloor^{2}}{2\left \lfloor \sqrt(x) \right \rfloor+\frac{x-\left \lfloor \sqrt(x) \right \rfloor^{2}}{2\left \lfloor \sqrt(x) \right \rfloor+\frac{x-\left \lfloor \sqrt(x) \right \rfloor^{2}}{2\left \lfloor \sqrt(x) \right \rfloor + \ddots}}}$

So to answer, $a_0 =\left \lfloor \sqrt(x) \right \rfloor$ (because any periodic continued fraction is smaller than 1).

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  • $\begingroup$ I am getting better at mentally doing square roots... generally to slide rule accuracy. But your answer mentions other roots, Could you amplify this a bit for cube roots? $\endgroup$ – richard1941 Apr 29 '18 at 0:57
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    $\begingroup$ How does this even work for say $\sqrt 7$? In a continued fraction the numerators have to be $1$, which you don't get by this method. $\endgroup$ – Marc van Leeuwen May 29 '18 at 10:48
  • $\begingroup$ @MarcvanLeeuwen see math.stackexchange.com/q/3075250/432081 $\endgroup$ – CopyPasteIt Jan 16 at 2:49

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