•  barsoap   ( @barsoap@lemm.ee ) 
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      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.

    • 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.

      •  Valthorn   ( @Valthorn@feddit.nu ) 
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        6 months ago

        While I agree that my proof is blunt, yours doesn’t prove that .999… is equal to -1. With your assumption, the infinite 9’s behave like they’re finite, adding the 0 to the end, and you forgot to move the decimal point in the beginning of the number when you multiplied by 10.

        x=0.999…999

        10x=9.999…990 assuming infinite decimals behave like finite ones.

        Now x - 10x = 0.999…999 - 9.999…990

        -9x = -9.000…009

        x = 1.000…001

        Thus, adding or subtracting the infinitesimal makes no difference, meaning it behaves like 0.

        Edit: Having written all this I realised that you probably meant the infinitely large number consisting of only 9’s, but with infinity you can’t really prove anything like this. You can’t have one infinite number being 10 times larger than another. It’s like assuming division by 0 is well defined.

        0a=0b, thus

        a=b, meaning of course your …999 can equal -1.

        Edit again: what my proof shows is that even if you assume that .000…001≠0, doing regular algebra makes it behave like 0 anyway. Your proof shows that you can’t to regular maths with infinite numbers, which wasn’t in question. Infinity exists, the infinitesimal does not.

        • Yes, but similar flaws exist for your proof.

          The algebraic proof that 0.999… = 1 must first prove why you can assign 0.999… to x.

          My “proof” abuses algebraic notation like this - you cannot assign infinity to a variable. After that, regular algebraic rules become meaningless.

          The proper proof would use the definition that the value of a limit approaching another value is exactly that value. For any epsilon > 0, 0.999… will be within the epsilon environment of 1 (= the interval 1 ± epsilon), therefore 0.999… is 1.

    •  sp3ctr4l   ( @sp3tr4l@lemmy.zip ) 
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      26 months ago

      The explanation I’ve seen is that … is notation for something that can be otherwise represented as sums of infinite series.

      In the case of 0.999…, it can be shown to converge toward 1 with the convergence rule for geometric series.

      If |r| < 1, then:

      ar + ar² + ar³ + … = ar / (1 - r)

      Thus:

      0.999… = 9(1/10) + 9(1/10)² + 9(1/10)³ + …

      = 9(1/10) / (1 - 1/10)

      = (9/10) / (9/10)

      = 1

      Just for fun, let’s try 0.424242…

      0.424242… = 42(1/100) + 42(1/100)² + 42(1/100)³

      = 42(1/100) / (1 - 1/100)

      = (42/100) / (99/100)

      = 42/99

      = 0.424242…

      So there you go, nothing gained from that other than seeing that 0.999… is distinct from other known patterns of repeating numbers after the decimal point.

      •  blue   ( @blue@ttrpg.network ) 
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        96 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.

  • Mathematics is built on axioms that have nothing to do with numbers yet. That means that things like decimal numbers need definitions. And in the definition of decimals is literally included that if you have only nines at a certain point behind the dot, it is the same as increasing the decimal in front of the first nine by one.

  • 0.999… / 3 = 0.333… 1 / 3 = 0.333… Ergo 1 = 0.999…

    (Or see algebraic proof by @Valthorn@feddit.nu)

    If the difference between two numbers is so infinitesimally small they are in essence mathematically equal, then I see no reason to not address then as such.

    If you tried to make a plank of wood 0.999…m long (and had the tools to do so), you’d soon find out the universe won’t let you arbitrarily go on to infinity. You’d find that when you got to the planck length, you’d have to either round up the previous digit, resolving to 1, or stop at the last 9.

  • The way I see it is the difference between equal numbers is zero.

    The difference between 0.999… and 1 is 0.000…, and since the nines don’t end, the zeros don’t end, so the difference is just zero.

    Meaning 0.999… = 1