Binary divison with zero-filling, rather than bringing down a bit

Question

I'm having trouble trying to figure out a method of binary division. This method does not do the "bring down a bit" that wikihow suggests, instead one always fills in all of the available digits at all times.

Example 1

   0011
11|1011
  -0110
   ----
   0101 # How do I know what to put in the quotient here?
  -0011
   ----
   0010

Example 2

     0001010
1010|1101011
     1010000
    -0010000
     -------
     0011011 # and here?
    -0010100
     -------
     0000111

Where I get confused is filling in the quotient. I know that I need to look through the dividend until the divisor fits, and then it put a 1. And proceed to multiply the divisor by that 1, adding in the zeros depending on how far left it sits in the quotient.

But once I have done that the first time, I get stuck with where the rest of the values in the quotient can be found. The wikihow method suggests comparing the result of the subtraction to the divisor, and seeing if the divisor fits, but this does not help here. In both cases the first subtraction gives a number that is larger than the divisor, but one has a 0 and the other has 1 as the next value in the quotient.

I'm also confused by the fact that example #2 has only two steps filled out, but four bits to its answer.

I have been trying to figure this out for some time, these examples come off some of my old notes but I can't remember what my method was for finding what goes in the quotient. Any help would be greatly appreciated.

Answer 1

When you are working with binary representations of numbers, tell yourself:

"If a computer can arithmetically 'slice and dice' this stuff, so can I."

Finding answers is more mechanical than $\text{Base-}10$ arithmetic, and long division can actually be less mentally taxing (uhh, $1 \times 1 = 1$, $0 \times 0 = 0$ and $1 \times 0 = 0$).

To avoid errors and make it easier to follow/check, I will work on your problems using a less compact method than long division notation, but the steps are the same and you can port the work over and rework your notes.

Example 1

Divide 1011 by 11.
Mechanically align 11 under 1011 and subtract.
   1011
    110
    ---
    101
Math Result: 1011 = 10 × 11 + 101

Divide 101 by 11.
Mechanically align 11 under 101 and subtract.
    101
     11
    ---
     10
Math Result: 101 = 1 × 11 + 10

WORK: 1011 = 10 × 11 + (101) = 10 × 11 + (1 × 11 + 10) = 11 × 11 + 10

ANS: 11, R = 10

Example 2

Divide 1101011 by 1010.
Mechanically align 1010 under 1101011 and subtract.
   1101011
   1010000
   -------
     11011
Math Result: 1101011 = 1000 × 1010 + 11011

Divide 11011 by 1010.
Mechanically align 1010 under 11011 and subtract.
     11011
     10100
   -------
       111
Math Result: 101011 = 10 × 1010 + 111

WORK: 1101011 = 1010 × 1010 + 111

ANS: 1010, R = 111

Notice that in Example 2 we did not show much work. We simply 'collected' the quotient $1^{'}$s, putting them in the correct positions. Once you believe in this pattern, you should have no trouble using the standard (compact) long division format.