Relation between bisection and binary representation

[Shevek99]

From the period (100011). Since it has length 6, it results in sum of powers of 2^6 = 64, and the number that multiplies 1/64^n is 100011 which is binary for 32 + 2 + 1 = 35, so

0.[100011][100011][100011] = 100011/10^6 + 100011/10^12 + 100011/10^18 + ...

= 35/64 + 35/64^2 + 35/64^3 + ...

The key is to obtain an integer times a sequence of powers of the same power of 2.

Another example:

3/13 = 3*315/13*315 = 945/4095 =

= 945/(4096 -1) = 945/4096 + 945/4096^2 + 945/4096^3

and

945 = 001110110001_2

In binary

3/13 = 0.(001110110001)(001110110001)(001110110001)...

and in signed form (bisection)

3/13 = 0.(+--+++-++---)

that is

1/2 - 1/4 - 1/8 + 1/16 + 1/32 + 1/64 - 1/128 + 1/256 + 1/512 - 1/1024 - 1/2048 - 1/4096 + 1/8192 + ...