3 ms·
The 8 is there but then it's followed by a 9 and a 10, and the carry from the 10 ends up bumping it up.
by madcaptenor 1y ago
The 8 is there but then it's followed by a 9 and a 10, and the carry from the 10 ends up bumping it up.
- zkmon 1y agoShouldn't wee see two zeros then?
- oersted 1y agoThis illustrates it nicely: https://math.stackexchange.com/a/994214 https://math.stackexchange.com/a/994214
- madcaptenor 1y agoThe reason you don't see two zeroes is as follows: you have .123456789 then add 10 on the end, as the tenth digit after the decimal point, to get .123456789(10) where the parentheses denote a "digit" that's 10 or larger, which we'll have to deal with by carrying to get a well-formed decimal. Then carry twice to get .12345678(10)0 .1234567900 So for a moment we have two zeroes, but now we need to add 11 to the 11th digit after the decimal point to get .1234567900(11) or after carrying .12345679011 and now there is only one zero.
- zkmon 1y agoAh, that's cool. Thanks!
- deleted 1y ago[deleted]