• 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.

    • 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)

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

        I’d just say that not all fractions can be broken down into a proper decimal for a whole number, just like pie never actually ends. We just stop and say it’s close enough to not be important. Need to know about a circle on your whiteboard? 3.14 is accurate enough. Need the entire observable universe measured to within a single atoms worth of accuracy? It only takes 39 digits after the 3.

        • sp3ctr4l@lemmy.zipEnglish
          0·
          6 months ago

          There are a lot of concepts in mathematics which do not have good real world analogues.

          i, the _imaginary number_for figuring out roots, as one example.

          I am fairly certain you cannot actually do the mathematics to predict or approximate the size of an atom or subatomic particle without using complex algebra involving i.

          It’s been a while since I watched the entire series Leonard Susskind has up on youtube explaining the basics of the actual math for quantum mechanics, but yeah I am fairly sure it involves complex numbers.

          • myslsl@lemmy.worldEnglish
            01·
            6 months ago

            i has nice real world analogues in the form of rotations by pi/2 about the origin (though this depends a little bit on what you mean by “real world analogue”).

            Since i=exp(ipi/2), if you take any complex number z and write it in polar form z=rexp(it), then multiplication by i yields a rotation of z by pi/2 about the origin because zi=rexp(it)exp(ipi/2)=rexp(i(t+pi/2)) by using rules of exponents for complex numbers.

            More generally since any pair of complex numbers z, w can be written in polar form z=rexp(it), w=uexp(iv) we have wz=(ru)exp(i(t+v)). This shows multiplication of a complex number z by any other complex number w can be thought of in terms of rotating z by the angle that w makes with the x axis (i.e. the angle v) and then scaling the resulting number by the magnitude of w (i.e. the number u)

            Alternatively you can get similar conclusions by Demoivre’s theorem if you do not like complex exponentials.

      • DeanFogg@lemm.eeEnglish
        01·
        6 months ago

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

    • 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.

    • 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.

  • rustyfish@lemmy.worldEnglish
    0·
    6 months ago

    Remember when US politicians argued about declaring Pi to 3?

    Would have been funny seeing the world go boink in about a week.

    • myslsl@lemmy.worldEnglish
      01·
      6 months ago

      Some software can be pretty resilient. I ended up watching this video here recently about running doom using different values for the constant pi that was pretty nifty.