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0.000...1 can be written as 1/inf, which has sense in Surreal Numbers math.

by drran 7y ago

nybble41 7y ago
You could just as easily say that 0.000…54234 can be written as 1/∞. Surreal Numbers is a bit of a detour in this case. The premise of the idea that 1 - 0.999… could be written 0.000…1 is the mistaken concept that there is some point "after an infinite number of steps" where the expansion of 0.999… stops and you can leave the remaining 1. The expansion never stops and there is no final remainder. The result is 0.000…, which is more typically written as 0. Plain zero, not an infinitesimal. Personally I find the following algebraic proof to be the most approachable: x = 0.999… x = 0.9 + 0.0999… x = 0.9 + (0.999… ÷ 10) x = 0.9 + (x ÷ 10) x - 0.9 = x ÷ 10 10x - 9 = x 9x - 9 = 0 9x = 9 x = 1 0.999… = 1
- 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.

0.000...1 can be written as 1/inf, which has sense in Surreal Numbers math.

by drran 7y ago

0.000...1 can be written as 1/inf, which has sense in Surreal Numbers math.

You could just as easily say that 0.000…54234 can be written as 1/∞. Surreal Numbers is a bit of a detour in this case. The premise of the idea that 1 - 0.999… could be written 0.000…1 is the mistaken concept that there is some point "after an infinite number of steps" where the expansion of 0.999… stops and you can leave the remaining 1. The expansion never stops and there is no final remainder. The result is 0.000…, which is more typically written as 0. Plain zero, not an infinitesimal. Personally I find the following algebraic proof to be the most approachable: x = 0.999… x = 0.9 + 0.0999… x = 0.9 + (0.999… ÷ 10) x = 0.9 + (x ÷ 10) x - 0.9 = x ÷ 10 10x - 9 = x 9x - 9 = 0 9x = 9 x = 1 0.999… = 1

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.