 So this time we'll try some slightly harder numbers. For example, I might try 47 minus 22. We should know that'll leave us with 25, but that'll be a little more interesting in different bases. So 47 in binary is, and then we go back, find 22, which is 1, 0, 1, 1, 0. So now if I do this subtraction, 0 minus 1 is 1, 1 minus 1 is 0, 1 minus 0 is 1, and now I have 0 minus 1. So I'll borrow 1 from this position. So I have 10 minus 1, which leaves me with 1. So if I look next to 25, I do get 11, 0, 0, 1 for my binary representation. So this worked out again. If I try this again in octal, I'd start with 57 for my decimal 47, and I'll subtract off 26. So this one won't be terribly interesting. 7 minus 6 is 1, 5 minus 2 is 3, and next to 25, I see 31. I can do this again for hexadecimal where I start with 2f and subtract 16. So f minus 6, well, f is 15, 15 minus 6 is 9. Then 2 minus 1 is 1. That gives me 19, and I can again see that 19 and 25 are the same number. So all three of those do the same thing. We're just working in slightly different bases.