It's a paper version of doing it in binary on a computer.
Wherever the divide-by-two results in a bit being lost, you add on the steadily doubling number on the other side.
( , Wed 5 Feb 2020, 16:58, Reply)
Wherever the divide-by-two results in a bit being lost, you add on the steadily doubling number on the other side.
( , Wed 5 Feb 2020, 16:58, Reply)
it's more to do with nature of binary itself, which goes up by two with each full column
so 7 (binary: 111) x 5 is the same as multiplying by (2,2,2,1)(adding 2x5 + 2x5 + 2x5 and itself together). The column format and rounding down give you a nice way of rationalising this. So 2,2,2,1 becomes 4,2,1. 6 Binary (110) becomes 4,2,0. 10 (binary 1010) 8,0,2,0 instead of 2,2,2,2,2
( , Thu 6 Feb 2020, 12:13, Reply)
so 7 (binary: 111) x 5 is the same as multiplying by (2,2,2,1)(adding 2x5 + 2x5 + 2x5 and itself together). The column format and rounding down give you a nice way of rationalising this. So 2,2,2,1 becomes 4,2,1. 6 Binary (110) becomes 4,2,0. 10 (binary 1010) 8,0,2,0 instead of 2,2,2,2,2
( , Thu 6 Feb 2020, 12:13, Reply)
7 is 111 in binary. Not 1111. That's 15.
6 is 110 and 10 is 1010.
So. Err.
( , Thu 6 Feb 2020, 18:12, Reply)
6 is 110 and 10 is 1010.
So. Err.
( , Thu 6 Feb 2020, 18:12, Reply)
fucksocks. i realised I'd made an error but pasted over the wrong bit
still, the proof is solid. multiplying something by a binary value means adding double the value with every column, as the binary column represents 2. So his method is breaking one of the factors up into twos (converting to binary), with a handy method of rationalising this.
( , Thu 6 Feb 2020, 23:05, Reply)
still, the proof is solid. multiplying something by a binary value means adding double the value with every column, as the binary column represents 2. So his method is breaking one of the factors up into twos (converting to binary), with a handy method of rationalising this.
( , Thu 6 Feb 2020, 23:05, Reply)