# If you know how to add, you know (almost) everything about mathematics

joão pestana

Computers are essentially glorified calculators and all they know is how to add numbers in binary. From there, it's easy to do multiplication — or glorified adding —, subtraction and division — or glorified subtraction. To do a multiplication, you just add the same number over and over the times that you want. To do a division, you do the same procedure, but you remove a certain number over and over until you reach the point when, if you remove it one more time, your result becomes negative. You count the number of times that you did it and the remainder is... well, the remainder. But can we really subtract by adding? Yes! All the mechanical calculators did it — something called the method of complements.

How does it work? It's very simple. Imagine you have two numbers X and Y and you want to subtract Y from X. Each number has a certain number of digits, lets say N and M respectively. So our numbers are X = x_1 x_2 x_3 ... x_N and Y = y_1 y_2 y_3 ... y_M where x_i,y_i in {0,1,2,3,4,5,6,7,8,9}.

For example, if X=1954 it means x_1=1, x_2=9, x_3=5 and x_4=4. I want to subtract from that the number Y=1756 which has the digits y_1=1, y_2=7, y_3=5 and y_4=6. Usually, we'd do something like X-Y=1954-1756=198. We already know our final answer so now we just need to reach it using only the operation of addition.

To do this we need to create Z as the complement of Y. What this means is that for every y_i ranging from i=1 to i=M-1, we write down z_i as the distance from that digit to 9 and for the last y_M we do just the same thing for z_M but this time it's the distance to 10. In our example, the complement of y_1=1 is z_1=8, y_2=7 is z_2=2, y_3=5 is z_3=4 and finally for y_4=6 is z_4=4. Now we have our Z=8244 as the complement of Y.

I added this paragraph some time after writing the article to include a suggestion made by a friend of mine. He pointed out to me that this complement was a subtraction and I agree, but I refuse to call it as such. What I told him is that it's a fixed correspondence. A 1 will always be replaced by an 8 at any digit except the least significant one and will be replaced by a 9 at the later. We can work out the correspondence table 1 for future use.

 Original digit y_i Most significant digits z_1 ... z_(M-1) Least significant digit z_M 0 9 0 1 8 9 2 7 8 3 6 7 4 5 6 5 4 5 6 3 4 7 2 3 8 1 2 9 0 1
Table 1 — Correspondence table for the most and least significant digits of Z.

As a final step, to conclude our subtraction by addition, we need to add Z to X and ignore the first digit — which will be a 1. That is, X+Z=1954+8244=10198 and by ignoring the first 1 we get 198. If you recall, it was our answer!

Why does this work? This exploits a weakness in the machines that we build. In this sense, we can think like the machine and achieve the same result. Imagine that we have a calculator which can only display 4 digits. Our calculator can display exactly 10000 different numbers ranging from 0000 to 9999. What would happen if we were to add 1 to 9999? The answer is 10000, but the display can only show the last 4 digits so it discards the first and we'd see the result as 0000. The calculator has a limit and we cannot use it to perfom calculations that will result in a number that's larger than 9999, because an overflow will occur and we'll only be able to see the last 4 digits of the result. If we were to perform 9999+9999=19998 in our special 4-digit calculator, we'd see 9998 as a result and be surprised at it — by adding two large numbers we got a smaller one!

What we're actually doing is forcing the calculation to overflow exactly up to the point of the difference between X and Y. You can think of the complement Z as what it would take to make the calculator overflow — or reset — starting at Y. So whatever you add to it will yield the difference, because in our calculator Y+Z=1756+8244=0000 (remember we disregard the first 1).

I can give you another visual example of how this is supposed to work. Imagine that we have 3 identical glasses over some large container. The first is filled with some water and the second is also filled with water, but less than the first one.  X Y
You want to find out the amount of water that the first glass has more than the second one. How would you do that? That's when the third glass enters the picture. Z
If it would so happen that it contained the exact amount of water that would fill up the second glass to the top and you were to pour ir over the first glass, you'd see the exact difference between the first and the second spill out into the larger container.  Full glass X-Y (spillage)
From there, you'd just have to collect and measure the spilled water. It's simple! Now... Could you actually do it using only glasses of water? Here are some rules. You can somehow precisely collect the spillage. You can use how many glasses you need. You have no way of measuring water so you can only rely on overflows.