# What is the sum that the square root button on calculator does so I can do it without the calculator button [duplicate]

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.

• Just to clarify: you are asking for methods of calculating the square root without a calculator? – angryavian Jul 9 '14 at 19:15
• Start here, perhaps. – Namaste Jul 9 '14 at 19:15
• Newtons method is probably the fasted with a good initial guess – ClassicStyle Jul 9 '14 at 19:22
• Do a search (google) for "square-root algorithms". There will be lots of hits. Explore hits to search for what you're looking for. – 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

$\frac{65}{46}=1.4130434$

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.

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

• 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. – Maths Fail Jul 9 '14 at 22:38