 Let's try some larger numbers this time. I start with a binary number like, and I subtract. We should get some interesting results out. So 1 minus 0 gives me 1. 0 minus 1 means I need to borrow something. So I'll have 10 minus 1, which leaves me with 1. Now I don't have anything left here, so 0 minus 0 gives me 0. 1 minus 1 gives me 0. 1 minus 0, this block ends up being the same as this block. So 1 minus 0 gives me 1. 0 minus 1, well, need to borrow something again. So 10 minus 1 gives me 1. 0 minus 0 is 0. And 1 minus 1 is 0. So two binary numbers, do the subtraction, and you get your result. So there's a couple more binary numbers. Again, 0 minus 1, can't do anything, so I need to borrow something. So 10 minus 1 gives me with 1. Now I've got a 0 here, so 0 minus 0 is 0. This block will pretty much be the same. So 10 minus 1 gives me with 1, and 0 minus 0 is 0. Now I've got 0 minus 1. Again, I need to borrow something. So this will become a 0. Now I've got a 1 there. 10 minus 1 gives me 1. 0 minus 1, can't do that, have to borrow something again. So 10 minus 1 leaves me with 1, and then 0 minus 0 is 0. 1 minus 1 is 0. So there's another example for binary. If I try doing something in octal, might have 7, 6. So 4 minus 4 is 0. 3 minus 6, well, I'll need to borrow something. So 6 minus 1 is 5. Now I've got a 13 here. So 13 minus 6. Well, again, remember that we're working in octal. So I've got an 8. 8 minus 6 would be 2. 2 plus 3 would be 5. Or I could say that this is really 11. 11 minus 6 is 5. Either way, I'm getting 5 for my result. Now I have 5 minus 5, which is 0, and 7 minus 6, which is 1. So I do these two numbers in octal. Now I've got 4 minus 7. So I'll need to borrow something. It becomes a 6. Now I've got a 14. So this 14 would actually be 12 in decimal. 12 minus 7 would leave me with 5. Or I can say, well, 10 minus 7 is 1. 1 plus 4 is 5. Either way, I get 5 for my result. Then 6 minus 6 is 0. 3 minus 7. I'll need to borrow something again. 5. Now I have a 13. Again, this should leave me with 4. I can get there either by saying that this 13 in octal is 11 in decimal, and 11 minus 7 is 4. Or I can say 8 minus 7 is 1. 1 plus 3 is 4. And then 5 minus 5 is 0. If I do some in hexadecimal, I might subtract b from dead. So d is less than f, so I'll need to go borrow something. a minus 1 is 9. Now I have 1d. So I can say this is a 16 plus a 13, which is 29. 29 minus 15 is 14. And then 14 is e. Or I can do 1 minus f is 1. 1 plus d is e. 9 is less than e, so I can't do that subtraction. I will have to borrow something from this e. It will become a d. And now I have 19. So now I have 19 minus e. Well, I think I'll do what I did before. I'll say 10 minus e is 2. 9 plus 2 is b. Again, d is less than e, so can't do that. This one will be easier, though. This one becomes a c. So d is 1 less than e. So I know that anytime I try this 1d subtract, 1d minus e subtraction, I'm going to get something that's actually 1 less than my 10. You can kind of think of this like doing, say, 16 minus 7 will be 9. 17 minus 8 gives you 9. But in this case, 1 less than r10 is f. So we will get f as a result. And then c minus b is 1. So there's my answer for hexadecimal. I do cab minus a. So e is less than b. So I will need to borrow something from this a. And a minus 1 is 9. So now I have 1b minus e. So 10 minus e leaves me with 2. If I add that 2 to the b, then I'll get a d. The 9 is less than c, so I'll need to borrow something. C becomes a b. Now I have a 19, 19 minus c. So if I add 4 to c, I would get the 10. So 10 minus c leaves me with 4. 4 plus 9, that gets me to 13, which would be d, d over there. And then b minus a is 1.