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How can one convert radicals to decimals(approximate value) when the number is not perfect such as $\sqrt2$, $\sqrt3$, $\sqrt5$, etc. Without the use of calculators.

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+1 This article has an explanation. Google search gives many more. There is also a wiki page devoted to methods of computing of square roots. – Sasha Aug 27 '12 at 20:25
@Sasha - Thanks – James Fair Aug 27 '12 at 20:27
Wow, guys these are some excellent answers, I going to look at each method. Thanks – James Fair Aug 27 '12 at 20:35
Take a look at this: – Thomas Aug 27 '12 at 20:41
up vote 7 down vote accepted

You can use Heron's algorithm to estimate $\sqrt{a}$ by computing some iterations of:

$$ x_{n+1} = \frac12 \left( x_n + \frac{a}{x_n} \right) $$

You should start with some $x_0 \approx \sqrt{a}$.

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+1: I find this method even better by replacing $x_n$ by $\frac {N_k}{D_k}$ getting $N_{k+1}=N_k^2+a\, D_k^2$ and $D_{k+1}=2\,N_k\,D_k$. Example $\sqrt{5}\approx \frac 21\approx \frac {4+5}{4}\cdots$ – Raymond Manzoni Aug 27 '12 at 20:41
Actually, for $a>1$, starting with $x_0 = \frac{1+a}{2}$ gets the job done, since $1<x_n<a$ implies $1<x_{n+1}<a$. – Sasha Aug 27 '12 at 21:26
@Sasha: Of course, the better your initial guess, the quicker you converge to the desired precision. – Hurkyl Aug 28 '12 at 2:37

You could see Wikipedia. The digit-by-digit method used to be taught in school. The various root-finding approaches are often easier. When you are close, you can also use things like $\sqrt {37}=\sqrt {36} \sqrt {\frac {37}{36}}=6 \sqrt {1+\frac 1{36}}\approx 6(1+\frac 1{72})\approx 6.083$

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There's a nice variation on this that comes from Pell equations: Find integers $x$ and $y$ such that $37x^2 - y^2 = \pm 1$; there are guaranteed to be arbitrarily large solutions. In this case, we have $37\cdot12^2 - 73^2 = -1$. Then write $\sqrt{37} = \sqrt{\frac{73^2-1}{12^2}} = \sqrt{\frac{73^2}{12^2}\cdot\frac{73^2-1}{73^2}} = \sqrt{\frac{73^2}{12^2}}\sqrt{\frac{73^2-1}{73^2}} = \frac{73}{12}\sqrt{1 - \frac{1}{73^2}} \approx \frac{73}{12}\left({1 - \frac{1}{2\cdot73^2}}\right)$. – MJD Aug 27 '12 at 20:46

I think the Newton–Raphson method is one of the simplest methods.


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Explicitly; to compute $\sqrt[n]{y}$, take $x_0>0$ and then compute $x_{k+1} = \frac{(n-1)x_k^n+y}{n x_k^{n-1}}$. – copper.hat Aug 27 '12 at 21:07

Mental/longhand arithmetic is a time-space tradeoff between how long it takes you to get the answer versus how many rules you need to memorize. While other people have given faster ways to compute the answer, since roots are a problem I rarely need to do without a calculator I've always taken a simpler if less efficient approach.

My method of finding $\sqrt{n}$ is by bracketing it between the two adjacent integer roots ${x^2 < n < y^2}$ and then using a modified binary search (eg if I'm looking for $\sqrt{84}$, I'll probably start with a candidate of 9.1 or 9.2 instead of 9.5) and then using my toolset for mental multiplication to square my candidate until I've reached a sufficiently precise approximation.

If I found myself needing to compute roots on a daily basis instead of a few times a year learning a faster method might be worthwhile; but until then this meets the good enough criteria.

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