Nice trick. It works in any even base b: xy = b((x - b/2) + (y - b/2)) + (b - x)(b - y) For b = 2 in particular, x - b/2 and x - b/2 are the negative 1's

Nice trick. It works in any even base b: xy = b((x - b/2) + (y - b/2)) + (b - x)(b - y) For b = 2 in particular, x - b/2 and x - b/2 are the negative 1's complements of x and y, and b - x and b - y are the 2's complements. More generally, they are the same kinds of complements generalized to base b. Because of that symmetric relationship with complementation, I suspect there is a conceptual proof waiting to be found. By conceptual I mean something that isn't just blind formula crunching. That would get us much closer to "why".