I am not very good when it comes to Maths but the current work I am doing means I need to get better and quick.

I have been teaching myself about areas, diagonals and square roots. However I am struggling to understand how the calculator works out the square root. For example I know that a diagonal in a square from corner to corner is approx. $1.414$ of the side distance of the square. so if I have a square that is $a=1 \times b=1$ I can do $a \times \sqrt{2}(1.414)$. I know that the distance of the diagonal is $d=1.414$.

I want to know how to work out the square root without a calculator?

Or am I just getting this whole thing wrong?

Sorry I did say I am not very good at Maths.

  • $\begingroup$ Just to clarify: you are asking for methods of calculating the square root without a calculator? $\endgroup$ – angryavian Jul 9 '14 at 19:15
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    $\begingroup$ Start here, perhaps. $\endgroup$ – Namaste Jul 9 '14 at 19:15
  • $\begingroup$ Newtons method is probably the fasted with a good initial guess $\endgroup$ – ClassicStyle Jul 9 '14 at 19:22
  • $\begingroup$ Hi, Thanks for reply, yes I am, Its just i am really terrible at maths. I am looking at the Wiki page you sent but I have a couple of issues with it. 1 Its states "This article has multiple issues. Please help improve it or discuss these issues on the talk page. This article possibly contains original research. (January 2012) This article may be too technical for most readers to understand. (September 2012)" 2. I am a long way off from understanding that kind of algebra i have only been learning the basics for the last few days, It is algebra in that wiki link right? $\endgroup$ – Maths Fail Jul 9 '14 at 19:24
  • $\begingroup$ Do a search (google) for "square-root algorithms". There will be lots of hits. Explore hits to search for what you're looking for. $\endgroup$ – Namaste Jul 9 '14 at 19:31

Here is an extremely simple idea of

$$ \frac{a}{b}\to\frac{a+2b}{a+b}$$

For instance to get the square root of 2 you can start with any 2 counting integers like a = 1 and b = 2.

$\frac{1}{2}\to\frac{1+2 (2)}{1+2}=\frac{5}{3}\to \frac{5+2(3)}{5+3}=\frac{11}{8}\to \frac{11+2(8)}{11+8}=\frac{27}{19}\to \frac{27+2(19)}{27+19}=\frac{65}{46}...$

continue as long as you need and notice that


is already close. This method was derived by Terry Goodman and John Bernard. For the $ \sqrt{n} $ you would use $ \frac{a}{b}\to\frac{a+nb}{a+b}$

The advantages are you can keep the intermediate answers in fraction form which means you only have to multiply and add whole numbers. Closer initial estimates of a and b will yield better answers more quickly.

  • $\begingroup$ Can this technique be generalized to higher order roots? $\endgroup$ – frogeyedpeas Jul 9 '14 at 19:56
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    $\begingroup$ Yes it can according to them, I will edit the post if you like. $\endgroup$ – bobbym Jul 9 '14 at 19:59
  • $\begingroup$ A link would suffice, unless you want to put that information out for anyone else that comes across this question and may be curious $\endgroup$ – frogeyedpeas Jul 9 '14 at 20:05
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    $\begingroup$ I can not find it online but it is from the magazine, Mathematics Teacher 73 May 1979 p344-345. $\endgroup$ – bobbym Jul 9 '14 at 20:18

What could $\sqrt 2$ be? You know that $1$ is too small because $1^2=1<2$. And you know that $2$ is too big becasue $2^2=4>2$. So the truth must be somewehre inbetween. The average of $1$ and $2$, that is $1.5$ suggests itself as a better guess. In general, if we have two different positive numbers $a,b$ such that $a\cdot b=2$, then one of them is too big and the other is too small, and wie might try their average $\frac{a+b}2$ as a better guess. Of course, if we only have one guess, $a$ say, we obtain the suitable $b$ by division: $b=\frac 2a$.

So let's start again:

  • From the ridiculously bad $a=1$, we obtain $b=2$ and the improved guess $\frac{1+2}2=1.5$.
  • Taking $a=1.5$ as guess, we obtain $b=1.333\ldots$ and the improved guess $\frac{1.5+1.333\ldots}{2}=1.41666\ldots$
  • Taking $a=1.4166$ as guess, we obtain $b=1.4118$ (I'm using rounded numbers throughout beacause higher accuracy is not needed as long as we do not require high accuracy for the final result) and the improved guess $\frac{1.4166+1.4118}{2}=1.4142$

This is already quite good as approximation to $\sqrt 2$ and you may notice that it took only few steps even thouihg our starting value was really awful.

Remark: This is called Heron's or the Babylonian method.

  • $\begingroup$ Hi, this is all really interesting and already more is starting to make sense, I had to chuckle at Hagen von Eitzen comment "From the ridiculously bad a=1" I am sorry I left school at 14 because I couldn't grasp any of the work. Thanks I am going to need some more understanding of algebra etc before I full y get this and I am sure I will. I will keep referring back to this to aid my learning. Perhaps i should stick to learning the layers of the pyramid in the correct order before trying to jump up levels. Its hard teaching yourself. Thanks again. $\endgroup$ – Maths Fail Jul 9 '14 at 22:38

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