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.
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.