[msuvakov]

Why the b > 2 condition? In the b=2 case, all three formulas also work perfectly, providing a ratio of 1. And this is interesting case where the error term is integer and the only case where that error term (1) is dominant (b-2=0), while the b-2 part dominates for larger bases.

[listeria]

in the b=2 case, you get:

  1 / 1 = 1 = b - 1
  1 % 1 = 0 = b - 2

they are the other way around, see for example the b=3 case:

  21 (base 3) = 7
  12 (base 3) = 5
  7 / 5 = 1 = b - 2
  7 % 5 = 2 = b - 1

[gowld]

In the b=2 case, 1/1 = 1 = (b-2) + (b-1)/denom(b) = (b-2) + (b-1)/1 = 2b - 3 = (b-1)*b^1 -1 (b-1)

In base 2 (and only base 2), denom(b) >= b-1, so the "fractional part" (b-1)/denom(b) carries into the 1's (units) place, which then carries into the 2's (b's) place, flipping both bits.