3 ms·
I didn't get where this comes from until I saw the second answer from the StackOverflow question another commenter shared. https://math.stackexchange.com/a/226
by oersted 1y ago
I didn't get where this comes from until I saw the second answer from the StackOverflow question another commenter shared.
https://math.stackexchange.com/a/2268896 https://math.stackexchange.com/a/2268896
Apparently 1/9^2 is well known to be 0.12345679(012345679)...
EDIT: Yes it's missing the 8 (I wrote it wrong intially): https://math.stackexchange.com/questions/994203/why-do-we-miss-8-in-the-decimal-expansion-of-1-81-and-98-in-the-decimal-expansi https://math.stackexchange.com/questions/994203/why-do-we-mi...
Interesting how it works out but I don't think it is anywhere close to as intuitive as the parent comment implies. The way its phrased made me feel a bit dumb because I didn't get it right away, but in retrospect I don't think anyone would reasonably get it without context.
- iso1631 1y ago9^2 is 81 1/81 is 0.012345679012345679.... no 8 in sight
- madcaptenor 1y agoThe 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]
- alyxya 1y agoIt actually skips the 8 in its repeating decimal. It’s better to think of 1/9^2 as the infinite sum of k * 10^-k for all positive integers k. The 8 gets skipped because you have something like ...789(10)(11)... where the 1 from the “10” and “11” digits carry over, increment the 9 digit causing another carry, so the 8 becomes a 9.
- dpacmittal 1y agoAlso 12345679*x*9 = xxxxxxxxx Eg 12345679*6*9 = 666666666
- madcaptenor 1y agoI think your formatting is off.