Based on the fact that you asked this three times and got the same answer three times, I suspect the interpretation made by the users that posted those answers was incorrect, and that you meant to ask about dividing in base 2.
We have
111001₂ = 1×2⁵ + 1×2⁴ + 1×2³ + 1×2⁰ = 57
1101₂ = 1×2³ + 1×2² + 1×2⁰ = 13
and 57/13 = (4×13 + 5)/13 = 4 + 5/13.
4 = 2² is already a power of 2, so we have
111001₂/1101₂ = 1×2² + 5/13
we just need to convert 5/13. To do this, we look for consecutive negative powers of 2 that 5/13 falls between, then expand 5/13 as the sum of the smaller power of 2 and some remainder term. For instance,
• 1/4 < 5/13 < 1/2, and
5/13 - 1/4 = (20 - 13)/52= 7/52
so that
5/13 = 1/4 + 7/52
or
5/13 = 1×2 ⁻² + 7/52
Then a partial conversion into base 2 gives us
111001₂/1101₂ = 1×2² + 1×2 ⁻² + 7/52
111001₂/1101₂ = 100.01₂ + 7/52
Continuing in this fashion, we find
• 1/8 < 7/52 < 1/4, and
7/52 = 1/8 + 1/104
==> 111001₂/1101₂ = 100.011₂ + 1/104
• 1/128 < 1/104 < 1/64, and
1/104 = 1/128 + 3/1664
==> 111001₂/1101₂ = 100.0110001₂ + 3/1664
• 1/1024 < 3/1664 < 1/512, and
3/1664 = 1/1024 + 11/13312
==> 111001₂/1101₂ = 100.0110001001₂ + 11/13312
• 1/2048 < 11/13312 < 1/1024, and
11/13312 = 1/2048 + 9/26624
==> 111001₂/1101₂ = 100.01100010011₂ + 9/26624
• 1/4096 < 9/26624 < 1/2048, and
9/26624 = 1/4096 + 5/53248
==> 111001₂/1101₂ = 100.011000100111₂ + 5/53248
and so on.
It turns out that this pattern repeats, so that
[tex]\displaystyle \frac{111001_2}{1101_2} = 100.\overline{011000100111}_2[/tex]
