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
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-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.
It 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.
Also 12345679*x*9 = xxxxxxxxx
Eg 12345679*6*9 = 666666666
I think your formatting is off.
9^2 is 81
1/81 is 0.012345679012345679....
no 8 in sight
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.
Shouldn't wee see two zeros then?
The reason you don't see two zeroes is as follows: you have
then add 10 on the end, as the tenth digit after the decimal point, to get 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 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 or after carrying and now there is only one zero.Ah, that's cool. Thanks!
This illustrates it nicely: https://math.stackexchange.com/a/994214