 This time we're going to try a couple of large examples. Work through the entire problem and just see how it works, even in random cases. So I'll start with a number and I will divide this binary number by this binary number. So I have a five-bit binary number here, which means I need at least five bits before I can start doing any subtraction. It turns out this binary number matches up nicely with this one. So I will write that one down, do my subtraction, pull down a zero, one. And I'll continue pulling down bits until I get at least five bits down in the bottom. So this five-bit binary number is still larger than this one, so I need to bring down at least one more bit. Now I have a five-bit binary number and a six-bit binary number. The six-bit binary number is obviously larger than the five-bit binary number, so now I can do the subtraction again. I'll pull down my next bit, and again I've got a six-bit binary number, so I know I can do my subtraction, and 100 minus 11 is one. Now I have a four-bit binary number, I'll pull down one zero. This five-bit binary number is still larger than this five-bit binary number, so I'll need to pull down another bit, and then I can do my subtraction again. This time I need to do lots of borrowing, and then I've got one more one that I can pull down. If I do this subtraction, I'll end up with something small, but I'll have some remainder. I could continue doing this, just adding in a binary point, and continuing to pull down zeroes as long as I'd like. In this case, I've run out of space, so I think I'll stop here, but this number would have several floating-point bits behind it. I could try a slightly smaller example. There's another example that will probably not work out precisely. Here I've got a four-bit binary number that will go into this four-bit binary number, so I can do the subtraction starting at this point. Now I need to pull down bits until I've got at least four, but this four-bit binary number is smaller than this four-bit binary number, so I'll pull down another bit. Now I have a five-bit binary number, which is certainly larger than my four-bit binary number. Since I did this subtraction, I'll write down a one again, pull down another one. Now I can do my subtraction again, and here again I'll have lots of borrowing to do. Now I can pull down this zero. I've got a four-bit binary number, which is larger than this four-bit binary number, so I'll do my subtraction again, and write down a one. I'll pull down a zero. This four-bit binary number is smaller than this one, so I'll write down a zero and pull down another bit. And I'd have one more one I can pull down. I can certainly do the subtraction. That would leave me with one zero one afterwards, and then I could pull down another one bit. That would actually divide out cleanly, and I'd get a point one there with no remainder. So in this case, the result ends up kind of nice, ran out of space at the end, but the last little bit wasn't too hard to do. So this would be the result for this division problem.