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FootnoteWhat I have to say here is kind of a double footnote to what I said about base 60 and to what I said about denominators of the form b^{k}1.The thing about base 60 is, it ought to be a good base for computation, but to use it, you either have to learn sixty symbols or mess around with alternating bases. However, while I was writing all this, I realized you could get almost the same computational benefits by using a different base, base b let's say, and looking at fractions with repeating sequences of length k. Such fractions can be written with denominator b^{k}1; and if we choose b and k well, that can be a nice number. Sequences of length 2 are especially nice, due to a lovely identity that I already mentioned in another context.
x^{2}1 = (x+1)(x1) So, for example, just as sequences of length 2 in decimal have denominator 11×9 = 99, so sequences of length 2 in base 11 have denominator 12×10 = 120. Here are all the fractions 1/n that can be written that way.
How is that good for computation? Well, the idea is, if you use only fractions with those denominators, you can ignore the fact that they repeat. If, for example, you want to add 1/3 and 1/8, you can just think to yourself, 37 + 14 = 50. Honestly, I'd rather work with alternating bases, in which case I'd think, 20 + 07.30 = 27.30. But, for now, let's keep pretending that repeating sequences are good. Next, here's a table of some other likely bases.
The first two don't have any factors of 5, and the last two require learning lots of symbols, so base 11 is pretty clearly the best, which of course is why I used it as the main example. The real problem with repeating sequences is that they can't handle smaller numbers. If I want to look at 1/360 in base 60, no problem, I can just add another “decimal” place, but if I want to look at it in base 11, I'm sunk. The denominator for sequences of length 4 is not the useful 120^{2} but rather the nearlyuseless 14640. So, repeating sequences are fine for, say, measuring fractions of an inch, but not for much else. Speaking of length 4, though, there is one case where it works pretty well.
This can be thought of as a result of applying the identity above two times.
x^{4}1 = (x^{2}+1)(x^{2}1) = (x^{2}+1)(x+1)(x1)

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