4 ms·

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.

by alyxya 1y ago

Stolpe 1y ago
Care to elaborate? Why does 0.987654 simplify to 80/81 and 0.123456 to 10/81?
- alyxya 1y ago
  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.
- madcaptenor 1y ago
  .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.
  - gus_massa 1y ago
    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.]
  - gowld 1y ago
    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...
    - madcaptenor 1y ago
      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... =
    - GuB-42 1y ago
      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
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 ago
  9^2 is 81 1/81 is 0.012345679012345679.... no 8 in sight
  - 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 ago
      Shouldn't wee see two zeros then?
      - oersted 1y ago
        This illustrates it nicely: https://math.stackexchange.com/a/994214 https://math.stackexchange.com/a/994214
      - madcaptenor 1y ago
        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.

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.

by alyxya 1y ago

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.

Stolpe 1y ago

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

alyxya 1y ago
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.
madcaptenor 1y ago
.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.
- gus_massa 1y ago
  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.]
- gowld 1y ago
  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...
  - madcaptenor 1y ago
    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... =
  - GuB-42 1y ago
    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

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 ago
9^2 is 81 1/81 is 0.012345679012345679.... no 8 in sight
- 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 ago
    Shouldn't wee see two zeros then?
    - oersted 1y ago
      This illustrates it nicely: https://math.stackexchange.com/a/994214 https://math.stackexchange.com/a/994214
    - madcaptenor 1y ago
      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.