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x = 0.999... x = 9/10 + 9/10^2 + 9/10^3 ... 9/10^inf x = 9/10 + (9/10 + 9/10^2 + 9/10^3 ... 9/10^(inf-1))/10 x = 9/10 + (x - 9/10^inf)/10 x - 9/

by drran 7y ago

x = 0.999... x = 9/10 + 9/10^2 + 9/10^3 ... 9/10^inf x = 9/10 + (9/10 + 9/10^2 + 9/10^3 ... 9/10^(inf-1))/10 x = 9/10 + (x - 9/10^inf)/10 x - 9/10 = (x - 9/10^inf)/10 10x - 9 = x - 9/10^inf 9x - 9 + 9/10^inf = 0 9x = 9 - 9/10^inf x = 1 - 1/10^inf x = 1 - 0.000...1

nybble41 6y ago
x = 9/10 + 9/10^2 + 9/10^3 ... 9/10^inf x = 9/10 + (9/10 + 9/10^2 + 9/10^3 ... 9/10^(inf-1))/10 This is exactly the issue I was referring to. You're assuming the sequence stops "at infinity" but infinity is not a concrete number of steps, it's the absence of any end condition. Subtracting one step from "no end condition" is nonsense. The sequence (0.999… - 0.9)×10 does not end earlier than 0.999…; these are exactly the same sequence, repeating 9s without end. The difference between them is zero in every digit, with no trailing 1.
- drran 6y ago
  The sequence doesn't end earlier, but it does start later by one element, so they are not exactly same sequences. One infinite sequence has one more element than another infinite sequence, so difference is 1/inf.
  - nybble41 6y ago
    No, they start at the same element (they're both 0.999… and thus start with 9/10) and have the same (infinite) "number" of elements. If you lined them up digit by digit there is never a point where one digit is a 9 and the other is a 0.
    - nybble41 6y ago
      Incidentally, this confusion is part of the reason why actual infinite sequences are written with a trailing ellipsis (9/10 + 9/10² + 9/10³ + …) and not a final term involving infinity. There is no final term—not even 9÷10^∞. So the correct way to write the separation of the leading term in your formula is: x = 9÷10 + 9÷10² + 9÷10³ + … x = 9÷10 + (9÷10 + 9÷10² + …) ÷ 10 Note that the number of elements in the sequence is the same no matter how many leading terms you write, so long as the pattern doesn't change. The notation { 1, 3, 5, 7, 9, … } and { 1, 3, 5, … } both refer to exactly the same set; the first notation is merely a bit more verbose. Similarly, the parenthesized portion of the second formula above is exactly equal to x, despite being written with two explicit leading terms rather than three.

by drran 7y ago

nybble41 6y ago

x = 9/10 + 9/10^2 + 9/10^3 ... 9/10^inf x = 9/10 + (9/10 + 9/10^2 + 9/10^3 ... 9/10^(inf-1))/10 This is exactly the issue I was referring to. You're assuming the sequence stops "at infinity" but infinity is not a concrete number of steps, it's the absence of any end condition. Subtracting one step from "no end condition" is nonsense. The sequence (0.999… - 0.9)×10 does not end earlier than 0.999…; these are exactly the same sequence, repeating 9s without end. The difference between them is zero in every digit, with no trailing 1.

drran 6y ago
The sequence doesn't end earlier, but it does start later by one element, so they are not exactly same sequences. One infinite sequence has one more element than another infinite sequence, so difference is 1/inf.
- nybble41 6y ago
  No, they start at the same element (they're both 0.999… and thus start with 9/10) and have the same (infinite) "number" of elements. If you lined them up digit by digit there is never a point where one digit is a 9 and the other is a 0.
  - nybble41 6y ago
    Incidentally, this confusion is part of the reason why actual infinite sequences are written with a trailing ellipsis (9/10 + 9/10² + 9/10³ + …) and not a final term involving infinity. There is no final term—not even 9÷10^∞. So the correct way to write the separation of the leading term in your formula is: x = 9÷10 + 9÷10² + 9÷10³ + … x = 9÷10 + (9÷10 + 9÷10² + …) ÷ 10 Note that the number of elements in the sequence is the same no matter how many leading terms you write, so long as the pattern doesn't change. The notation { 1, 3, 5, 7, 9, … } and { 1, 3, 5, … } both refer to exactly the same set; the first notation is merely a bit more verbose. Similarly, the parenthesized portion of the second formula above is exactly equal to x, despite being written with two explicit leading terms rather than three.