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How can I find the square root of a number without using a calculator?

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    $\begingroup$ Here is an algorithm that can be done by hand: en.wikipedia.org/wiki/… $\endgroup$ Commented Jun 16, 2013 at 13:25
  • $\begingroup$ @AymanHourieh: there is a similar method for computing cube roots (and higher). I used these algorithms to implement square and cube roots for a numerics package when I worked at Apple. $\endgroup$
    – robjohn
    Commented Jun 16, 2013 at 15:34

2 Answers 2

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Newton's method says that $$ x_{k+1}=\frac{n+x_k^2}{2x_k} $$ converges to $\sqrt{n}$.

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One approach: find a perfect square close to your number. E.g. for $23$, we could take $25$. Now write (for this example) $\sqrt{23} = 5 \sqrt{23/25} = 5 \sqrt{1 - 2/25}.$ Now if we want to compute $\sqrt{1 - x}$ where $|x| < 1$, there is the binomial series:

$$\sqrt{1 - x} = 1 - \dfrac{1}{2} x - \dfrac{1}{8} x^2 + \cdots.$$

In our examples this gives $\sqrt{1 - 2/25} = 1 - \dfrac{1}{25} - \dfrac{1}{1250} + \cdots,$ giving $\sqrt{23} = 5 - \dfrac{1}{5} - \dfrac{1}{250} + \cdots \sim 4.796,$ which isn't a bad approximation.


Another approach is to define the sequence $x_{n+1} = (x_n + A/x_n)/2$, where $A$ is the number whose square root you are trying to compute. (Here you start with $x_0$ some reasonable ball-park estimate to $\sqrt{A}$.) Then $x_n$ will converge to $\sqrt{A}$. (This is a special case of Newton's method, applied here to find a root of $x^2 - A = 0$.)

In the case of $A = 23$, we might take $x_0 = 5$, then $x_1 = 4.8$, and $x_2 \sim 4.796$ again.

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  • $\begingroup$ Typo in your last line: "$250/+$" should be "$1/250-$". $\endgroup$ Commented Jun 16, 2013 at 15:02

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