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Comment by alyxya | Hacker News Reader

I like to think of 0.987654... and 0.123456... as infinite series which simplify to 80/81 and 10/81, hence the ~8 ratio.

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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.

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?

This illustrates it nicely: https://math.stackexchange.com/a/994214

The 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.

Ah, that's cool. Thanks!

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.

Care to elaborate? Why does 0.987654 simplify to 80/81 and 0.123456 to 10/81?

If you set x = 0.123456..., then multiplying it by (10 - 1) gives 9x = 1.111111..., and multiplying it by (10 - 1) again gives 81x = 10, or x = 10/81. I’m not writing things formally here but that’s the rough idea, and you can do the same procedure with 0.987654... to get 80/81.

.123456... = x + 2 x^2 + 3 x^3 + ... with x = 1/10.

Then you have (x + 2 x^2 + 3 x^3 + ...) = (x + x^2 + x^3 + x^4 + ...) + (x^2 + x^3 + x^4 + x^5 + ...) + (x^3 + x^4 + x^5 + x^6 + ...) (count the number of occurrences of each power of x^n on the right-hand side)

and from the sum of a geometric series the RHS is x/(1-x) + x^2/(1-x) + x^3/(1-x) + ..., which itself is a geometric series and works out to x/(1-x)^2. Then put in x = 1/10 to get 10/81.

Now 0.987654... = 1 - 0.012345... = 1 - (1/10) (10/81) = 1 - 1/81 = 80/81.

I don't know who downvoted this, but it's correct.

The use of series is a little "sloppy", but x + 2 x^2 + 3 x^3 + ... has absolute uniform convergence when |x|<r<1, even more importantly that it's true even for complex numbers |z|<r<1.

The super nice property of complex analysis is that you can be almost ridiculously "sloppy" inside that open circle and the Conway book will tell you everything is ok.

[I'll post a similar proof, but mine use -1/10 and rounding, so mine is probably worse.]

Don't need the clutter of infinite series and polynomials:

    1/9 = 0.1111...

    1/81 = 1/9 * 1/9 = 0.111... * 0.111... =

    Sum of:
       0.0111...
       0.00111...
       0.000111...
       ...
    
    =  0.012345...

This is better than my answer, at least if you can get your brain to interpret it in base b. In that case the first two lines would become

  1/(b-1) = 0.1111...
  1/((b-1)^2) = 1/b * 1/b = 0.111... * 0.111... =

Isn't it essentially the same thing, but less formal

0.1111... is just a notation for (x + x^2 + x^3 + x^4 + ...) with x = 1/10

1/9 = 0.1111... is a direct application of the x/(1-x) formula

The sum of 0.0111... + 0.00111... ... = 0.012345... part is the same as the "(x + 2 x^2 + 3 x^3 + ...) = (x + x^2 + x^3 + x^4 + ...) + (x^2 + x^3 + x^4 + x^5 + ...)" part (but divided by 10)

And 1/81 = 1/9 * 1/9 ... part is the x/(1-x)^2 result

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