irrational-base number system

3021_pi = 3pi³ + 0 + 2pi + 1

You might be asking yourself: what's the use of that? Wouldn't any useful number have an infinite pi-cimal representation? Well, yes and no. 0 and 1 are still 0 and 1 in any base. And if you're working with circles, spheres, and "spheres" in more than 3 dimensions, base pi is fairly useful. But it's true that most rational numbers would have an infinite number of digits in base pi.

Base sqrt(2) is even more useful, since it becomes base 2 (binary) if you set every other digit to 0. Think about it:

(a_i x 2ⁱ) + (a_i-1 x 2^i-1) + ... + (a₁ x 2¹) + (a₀ x 2⁰) ...

is really the same as

(a_i x sqrt(2)²ⁱ) + (a_i-1 x sqrt(2)^2(i-1)) + ... + (a₁ x sqrt(2)^2x1) + (a₀ x sqrt(2)^2x0) ...

and so you have a system that can comfortably be used for rational numbers, by skipping every other digit, and can also be used to express numbers that have more to do with fractional powers of 2.

For example, I could represent decimal 12 as 1010000_sqrt(2), because 4+8=12. But a base conversion algorithm would give the result: 100000000.01001000000001..._sqrt(2) because sqrt(2)^7, or 11.313712, is closer to 12 than sqrt(2)^6, or 8. Thus, in base sqrt(2), the number 12 is rational and irrational at the same time.

A writeup in base pi points out that this problem occurs in any non-integral base.

Actually, FordPrefect, it isn't the fact that it's less than 2 that causes the problem.

Consider base sqrt(5), which is greater than 2. (sqrt(5) ~ 2.236). Take decimal 280, which is 2010100_sqrt(5). But the usual conversion routines will give 10000000.1001..._sqrt(5), since sqrt(5)⁷ is 279.51, which is closer to 280 than 2*sqrt(5)⁶, or 250.

Same for base sqrt(10), which in some ways is easier to work in, and is greater than 3. 322 decimal is "obviously" 30202_sqrt(10), but it's also 100012.12..._sqrt(10), by similar reasonings.

I'm thinking it may also have to do with the way the base⁰ place works. Ponder: the equations 10-1=(b-1) (where b-1 is the digit for one less than the base b) works for all integer bases, but does not work for irrational bases. The problem is that we're using integer multiples of integer powers of the base, and when the base isn't an integer that isn't so well-behaved anymore.