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.
How to convert radicals to decimals without a calculator
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0Take a look at this: http://www.wikihow.com/Calculate-a-Square-Root-by-Hand – 2012-08-27
4 Answers
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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0@Sasha: Of course, the better your initial guess, the quicker you converge to the desired precision. – 2012-08-28
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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0There'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 $\sqrt{37} = \sqrt{\frac{73^2-1}{12^2}} = \sqrt{\frac{73^2}{12^2}\cdot\frac{73^2-1}{73^2}} = \sqrt{\frac{73^2}{12^2}}\sqrt{\frac{73^2-1}{73^2}} = \frac{73}{12}\sqrt{1 - \frac{1}{73^2}} \approx \frac{73}{12}\left({1 - \frac{1}{2\cdot73^2}}\right)$. – 2012-08-27
I think the Newton–Raphson method is one of the simplest methods.
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1Explicitly; to compute $\sqrt[n]{y}$, take x_0>0 and then compute $x_{k+1} = \frac{(n-1)x_k^n+y}{n x_k^{n-1}}$. – 2012-08-27
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.