Ronn Blankenship wrote: > >> Then, of course, there's my secondary question, which is how can you >> have a counting system with an irrational number at the base in the first >> place? > ><number> = d(n)*pi^n + d(n-1)*pi^(n-1) + . . . + d(2)*pi^2 + d(1)*pi + d(0) >+ d(-1)*pi^-1 + d(-2)* pi^(-2) + . . . > >where n is the larger of zero or the integer such that pi^n .le. <number> >.lt. pi^(n-1), and the series formally does not terminate on the right, >though it would be possible that all the d(i) = 0 after a certain point. > The problem is that, unlike the integer-base case, this representation using a non-integer base is not unique. A trivial example, using base 2.5:
(2.5)^3 = 2 * (2.5)^2 + 1 * (2.5)^1 + 0.625, so the number 15.625 has, at least, two representations in "base" 2.5, which contradicts the whole idea of a _base_. >The d(i) are what we call in base ten "digits," but since that term is >sometimes considered only applicable in base ten, > No! bit = binary digits. Digits come from the latin word for fingers Alberto Monteiro
