Page 5 of 5 FirstFirst 12345
Results 61 to 66 of 66

Thread: Hi explain why 0.999.... != 1 please.

  1. #61
    One Hundred Chimneys Recognized Member Tavrobel's Avatar
    Join Date
    Jan 2006
    Location
    Conjunction Junction
    Posts
    10,455
    Articles
    102
    Contributions
    • Former Site Staff

    Default

    Have you people completely forgotten that .999... is an approximation?
    Intergers, radicals, and fractions are exact values.
    .999... vs. 1 is comparing an infinte concept (an approximation, at that) to an exact value. That is like asking someone to divide by 0.

    Ok, so 1/3 == .333..., but .333... is still a approximation. You cannot have an exact value for 1/3 in decimal notation because it does not end. So stop saying that 1/9 * 9 == .999... == 1 or 1/3 * 3 == .999... == 1 is true. 1/7 * 7 is also an infinite set of numbers, yes, even though it does not equal .999..., but rather, a smaller number, it still would equal one.
    Any number divided by itself is 1 (except 0/0). This the FIRST grand assumption that people miss/omit.

    However, the "pick out a number between the two and you define them as different numbers", that one, I couldn't attack even if I were smart enough to explain it.

  2. #62
    Shlup's Retired Pimp Recognized Member Raistlin's Avatar
    Join Date
    Dec 1999
    Location
    Spying on Unne and BUO
    Posts
    20,583
    Articles
    101
    Blog Entries
    45
    Contributions
    • Former Cid's Knight
    • Former Editor

    Default

    It's not an approximation at all. 1/9 (or 9/9, in the case of .999999...) is not an approximation. Any real number is not an approximation.

  3. #63
    ORANGE Dr Unne's Avatar
    Join Date
    Dec 1999
    Posts
    7,394
    Articles
    1
    Contributions
    • Former Administrator
    • Former Developer
    • Former Tech Admin

    Default

    Quote Originally Posted by Tavrobel
    Have you people completely forgotten that .999... is an approximation?
    Intergers, radicals, and fractions are exact values.
    .999... vs. 1 is comparing an infinte concept (an approximation, at that) to an exact value. That is like asking someone to divide by 0.

    Ok, so 1/3 == .333..., but .333... is still a approximation. You cannot have an exact value for 1/3 in decimal notation because it does not end. So stop saying that 1/9 * 9 == .999... == 1 or 1/3 * 3 == .999... == 1 is true. 1/7 * 7 is also an infinite set of numbers, yes, even though it does not equal .999..., but rather, a smaller number, it still would equal one.
    Any number divided by itself is 1 (except 0/0). This the FIRST grand assumption that people miss/omit.

    However, the "pick out a number between the two and you define them as different numbers", that one, I couldn't attack even if I were smart enough to explain it.
    It's not an approximation. It's a method of describing a number as an infinite geometric series. So let's rewrite this without decimals.

    Been a while since I've had calc, so forgive me if I mess this up. A decimal number

    0.x1x2x3x4x5x6...xn

    (those numbers are meant to be subscripts) is actually a way of writing

    0 + x1/10 + x2/10^2 + x3/10^3 + x4/10^4 + x5/10^5 + ... + xn/(10^n)

    It so happens in the case of .99999999 (however many 9's you'd like) that all of x1...xn all are 9.

    9/10 + 9/100 + 9/1000 + ... + 9/10^n

    Easier to write it like this:

    SUM(k = 1 to n) of [9/(10^k)]

    Or you could rewrite it equivalently as this:

    SUM(k = 0 to n) of [(9/10) / (10^k)]

    or

    SUM(k = 0 to n) of [(9/10) * (1/10)^k]

    This so happens to be a geometric series.

    Now, imagine that we let n approach infinity. In the case of .999.... n does approach infinity. In this case we have an infinite geometric series.

    Some infinite series add up to (converge to) a finite sum. It seems non-intuitive, but that's calculus for you. If it wasn't possible to add up an infinite number of numbers and (sometimes) get a finite result, we'd be screwed. See <a href="http://en.wikipedia.org/wiki/Zeno%27s_paradoxes">Zeno's paradoxes</a>.

    It so happens that this infinite geometric series is one of the types of infinite series that converges to a finite sum. But you have to use limits.

    lim(n -> infinity) SUM(k = 0 to n) of [a * r^k] = a / (1 - r)

    The proof of this is simple and can be seen <a href="http://mathworld.wolfram.com/GeometricSeries.html">here</a> for example. This is true when |r| < 1, and clearly|1/10| < 1.

    Substituting from above, a = 9/10 and r = 1/10, we have

    .999...
    = lim(n -> infinity) SUM(k = 0 to n) of [(9/10) * (1/10)^k]
    = (9/10) / (1 - (1/10))
    = (9/10) / (9/10)
    = 1

    So no approximations or decimals needed.

  4. #64
    Yuffie ate my avatar Sefie1999AD's Avatar
    Join Date
    Mar 2001
    Location
    Finland
    Posts
    3,151

    Default

    Unne: Yeah, that seems right, considering I did pretty much the same thing on the previous page, except I took 9 as a factor outside the sum and kept 10^(-n) inside the sum (where n goes from 1 to infinity).
    People dislike FFIX because they're horrible idiots. - Kawaii Ryűkishi
    "One-Winged Angel" is far and away the best final boss song ever
    composed.
    - Kawaii Ryűkishi



  5. #65
    Banned Avathar's Avatar
    Join Date
    Mar 2006
    Location
    Antwerp, greatest city of Belgium
    Posts
    854

    Default

    http://pi.ytmnd.com/
    It's so...........hypnotic:kaodizzy:

    Edit: http://whatismath.ytmnd.com/

  6. #66

    Default

    mai gawd PG its cause the world is round! therefore we round! 9.5=10 .: .9999999...=1.

    meh...
    Making a slow come back since 2008

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •