How do you add, subtract, multiply, and divide Binary numbers?

Interesting question! I would have thought column methods would work. The conventional arithmetic method can also be used for division.

You just have to remember to carry the remainder from eight, sixteen, two or whatever base you are calculating in. For example, the sum of the hexadecimal numbers 3C and DD (remember A is ten (10 in base ten) and F is fifteen) can be determined by C+D=19 -counting back three digits from C in the units column because D is three short of 10 (sixteen) and adding one to the sixteens column- 3+D=10 -counting on from D: E, F, 10. With column addition, you would just carry the one from the units calculation over into the sixteens calculation.

Here, I can note that 30+D0=100 and add 19 onto this to get 119 (all in base sixteen). For multiplications, a grid might well be useful, but note that you could do a grid for addition as well. Let's now divide DD by 3C and see what we get.

Looking at the 3C times table we have (counting back four in the units column because C is four short of 10 (sixteen) and adding an extra one to the sixteens column in addition to the three most of the time) 3C, 78, B4, F0, 12C and so on. Okay, so we have 3 remainder DD minus B4, which is 29. If we were dividing DD.0 this remainder would be carried over to the sixteenths column and we we divide 290 by 3C.

3C*4=F0, so 3C*8=2*F0=1E0 and (3C times sixteen) 3C*10=3C0, which we knew anyway. Subtracting F0 from 3C0 gives us 2D0 and subtracting another two lots of 3C (78) gives 258. So we have 3C*(10-4-2)=3C*A=258 and the remainder is 38.

We could carry this into to the two-hundred-and-fifty-sixths column to get an answer to two hexadecimal places. Anyway, we have DD/3C is 3. B to one hexadecimal place and exactly 3+29/3C.

Regarding whether 29/3C can be simplified using common factors of the two numbers, 3C is 40-4 and therefore clearly a multiple of four such that 4*F=3C. F (fifteen) we know has factors of 3 and 5 (three and five). So 3C=2*2*3*5 as a product of prime numbers.

Does 29 have any of these prime factors? Well, it is 30-7, so not 3. It is 20+9, so not 2 either.

5 in base sixteen is a little bit like three in base ten, because 5*3=10-1. So also, 5*9=30-3=2D. But D-5=8, so 5 is not a factor of 29 either.

I hope this is all error-free, but more importantly it gives you some more ideas about how the methods you will be familiar with from working in base ten can be applied to other number bases. In base two, work will be far more robotic, relying almost exclusively on the formalities of the column or quotient methods.

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