• Valthorn@feddit.nuEnglish
    2·
    6 months ago

    x=.9999…

    10x=9.9999…

    Subtract x from both sides

    9x=9

    x=1

    There it is, folks.

    • yetAnotherUser@discuss.tchncs.deEnglish
      1·
      6 months ago

      Unfortunately not an ideal proof.

      It makes certain assumptions:

      1. That a number 0.999… exists and is well-defined
      2. That multiplication and subtraction for this number work as expected

      Similarly, I could prove that the number which consists of infinite 9’s to the left of the decimal separator is equal to -1:

      ...999.0 = x
...990.0 = 10x

Calculate x - 10x:

x - 10x = ...999.0 - ...990.0
-9x = 9
x = -1

      And while this is true for 10-adic numbers, it is certainly not true for the real numbers.

    • barsoap@lemm.eeEnglish
      1·
      6 months ago

      Somehow I have the feeling that this is not going to convince people who think that 0.9999… /= 1, but only make them madder.

      Personally I like to point to the difference, or rather non-difference, between 0.333… and ⅓, then ask them what multiplying each by 3 is.

      • Buglefingers@lemmy.worldEnglish
        1·
        3 months ago

        The thing is 0.333… And 1/3 represent the same thing. Base 10 struggles to represent the thirds in decimal form. You get other decimal issues like this in other base formats too

        (I think, if I remember correctly. Lol)

      • DeanFogg@lemm.eeEnglish
        01·
        6 months ago

        Cut a banana into thirds and you lose material from cutting it hence .9999

    • ColeSloth@discuss.tchncs.deEnglish
      0·
      6 months ago

      X=.5555…

      10x=5.5555…

      Subtract x from both sides.

      9x=5

      X=1 .5555 must equal 1.

      There it isn’t. Because that math is bullshit.

      • blue@ttrpg.networkEnglish
        1·
        6 months ago

        x = 5/9 is not 9/9. 5/9 = .55555…

        You’re proving that 0.555… equals 5/9 (which it does), not that it equals 1 (which it doesn’t).

        It’s absolutely not the same result as x = 0.999… as you claim.

  • Tomorrow_Farewell [any, they/them]@hexbear.netEnglish
    1·
    6 months ago

    The decimals ‘0.999…’ and ‘1’ refer to the real numbers that are equivalence classes of Cauchy sequences of rational numbers (0.9, 0.99, 0.999,…) and (1, 1, 1,…) with respect to the relation R: (aRb) <=> (lim(a_n-b_n) as n->inf, where a_n and b_n are the nth elements of sequences a and b, respectively).

    For a = (1, 1, 1,…) and b = (0.9, 0.99, 0.999,…) we have lim(a_n-b_n) as n->inf = lim(1-sum(9/10^k) for k from 1 to n) as n->inf = lim(1/10^n) as n->inf = 0. That means that (1, 1, 1,…)R(0.9, 0.99, 0.999,…), i.e. that these sequences belong to the same equivalence class of Cauchy sequences of rational numbers with respect to R. In other words, the decimals ‘0.999…’ and ‘1’ refer to the same real number. QED.