Can we work out square roots by hand? The last page finished with a suggestion to work out the square root of the date's largest factor, 3365119 which is done like this:

First find the size of the answer; this is a seven digit number so the root, R, will be 4 figures long. Th e1st digit is 3 so the root starts with a 1 (but less than 2). Thus we establish a first attempt at bounds bounds, 1000<R<2000. Using numbers you know, we can do better: 18²=324 and 19²=361 so 1800<R<1900.

Since 336.5 - 324 = 12.5 and 361-324 = 37, we might estimate R as being (18 + 12.5/37) x 100 = 1833.8. So 1834 is expected to be the nearest integer solution.

To find √230 we recognise the root lies between 15 and 16. 230 = 15² remainder 5. This might be good enough. The next approximation would say 15+5/31 = 15.161, which is pretty good, as √230=15.166. If we looked at √23000, two more zeroes multiplies the root by ten, so our first stab at √23000 is 150 (remainder 500) and we can soon see that the next approximation puts us between 151 and 152, where 151² = 22801 and 152²=23104 using either the Next Square principles or the Other Squares technique. 23000-22801= 199 and 23104-22801= 303, so our next best guess at √23000 is 151+199/303 = 151.6568. To 4 d.p. the answer is 151.6575, so we are already correct to 2 d.p. This shows you a technique to produce steadily better approximations. It helps to know that each linear approximation (the fancy term for what we are doing) will fall short of the true answer. This is because we are using straight line approximations to a quadratic curve, working from below the solution. You might draw a graph to help you see this, as I would do in a lesson.

Can we do this in an even simpler way? Start by pairing off digits from the right, so 23456 is seen as 02 34 56. The root of this starts with a 1. In general, a 2n-digit number has an n-digit root, and so a 2n+1 digit number has an n-digit root.

Investigation using 5 and 6 digit numbers: √23456 is approximately 150 (between 22500 and 25600, so perhaps 153). Check: 153²=23409 (by hand), so 153r47.

A smart way to find square roos is to start by doing prime factorisation of the number. Examples: √200 = 10√2, √20 = 2√5, √60 = 2√3√5 - so one could reduce many problems to only that of finding or remembering the roots of primes. For example I know √2=1.414. √3=1.732, √5=2.236, √10=√2√5 = 3.162 — and I haven't used that information for at least ten years. Knowing √10 is a big help when you've an answer of the wrong size.

I vaguely remember a hand-technique from my teens using a hand calculator—by which I mean mechanical, not electrical—that involved subtracting successive odd numbers in each column until some criterion was reached. I have tried to re-discover it, but decided that no-one would be interested any more, given that electronic solutions are now so prevalent.

A new puzzle launched in the Times recently is **Square Routes**, a small grid to be filled in with letters such that tracking adjacent letters (no diagonal moves) spells out a group of related nouns, usually names. poor link. Surprisingly difficult, for something taking up so little page space.

There's a relatively simple way to find square roots by considering (10a+b)²=10a)²+2.10.ab+b². For example, to work towards the square root of 7184.26; start with 80something, i.e a=8. Subtract the 6400 from 7184.26 and you have 7184.26, which is, mostly, (2(80b).b, 784 is then 160b => b is a little under 5. 85²=7225 is too high (and is found very quickly from previous techniques), so we can cycle again, this time saying a=74 and look for a new value for b, the next correction. 84²= 7225-85-84=7056, which is 128.26 too low. But 128.26 is *mostly* 2.84.b => b = 0.76, suggesting that our next good guess is 84.75. 84.75² = 7182.5625. Repeating that cycle one more time, gap of 1.6972 is mostly 2.84.75.b = 169.5 b => b = .0100, and our third iteration says √7184.26 = 84.76001. My calculator says this is correct to 5 s.f.

we can write this as one would in numerical methods work; let √S be the problem and let x_{n }be the latest estimate. Then xₙ₊₁ - xₙ = f(xₙ) / f'(xₙ), which you might see as one of Newton's ideas. But f(x) = x² and f'(x) = 2x....

In effect, then, xₙ₊₁= xₙ - (xₙ² - S) / 2xₙ = ½ (xₙ + S/xₙ )

For S=7184.26, x₁=80, x₂=84.901625, x₃ = 84.7601322568, x₄ = 84.760 014 1577, x₅ = 84.760 014 1576. so these answers are correct to significant figures of 1, 2, 5, 10, effectively doubling the precision with each cycle. One of the benefits of this technique is that you do not *need* to take the next xₙ as provided by the algorithm, only something suitably close, as I did in the written paragraph.

Note that subscripts may lose themselves on download. I have elsewhere preferred the xⁿ superscript UnicodeU=207F rather than try to promote the n-character to an upper baseline. Subscript-n is unicode 2099, U+2099, which prints on the screen as the 'unknown' character, . I want subscript-n, as in x_{n}. This is a pity, as the letters ₐₑₒₓₔ are available. in large, that is ₐₑₒₓₔ Some time later I changed some of these, so that we have xₙ using the unicode character and proof against switches awway from rich text and font swapping. I may noy have succeeded in doing this to all the subscript characters on the page, such as xn+1, which ought to turn into xₙ₊₁