3 ms·
Ha, what a coincidence. It won't work for decimal 10s complement numbers: For x=10 we get 1 + 10 + 100 + 1000 + 10000 + ... = -1/9 And -1 in a decimal com
by jonsen 13y ago
Ha, what a coincidence.
It won't work for decimal 10s complement numbers:
For x=10 we get
1 + 10 + 100 + 1000 + 10000 + ... = -1/9
And -1 in a decimal computer is
9 + 90 + 900 + 9000 + 90000 + …
so … OMG!
9*(1 + 10 + 100 + 1000 + 10000 + ...) = 9*(-1/9) = -1
- thaumasiotes 13y ago> Ha, what a coincidence. > It won't work for decimal 10s complement numbers: I don't get why you say this -- you go on to demonstrate the opposite? This is just the other-side-of-the-decimal-point inverse of the well known 0.9999999... = 1 equality. One of my favorite proofs for 0.9999... being equal to 1 goes like so: imagine subtracting it from 1. You'll get a number which has a 0 at every decimal place: 0.0000000.... Obviously, the number with a 0 in every decimal place is 0 itself. Similarly, if you have your integer represented decimally as a bunch of coefficients (0 <= c < 10) of powers of 10, and all of the coefficients are 9, it's fairly straightforward to see that adding 1 will get you a new number for which all of the coefficients are 0. Since adding 1 to the original number gave us 0, we can treat the original number as -1.
- jonsen 13y agoI don't get why you say this Just a narrative of my own experience thinking this through :-) I thought "OMG" would tell you that, sorry.