• Valthorn@feddit.nu
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    5 months ago

    x=.9999…

    10x=9.9999…

    Subtract x from both sides

    9x=9

    x=1

    There it is, folks.

    • barsoap@lemm.ee
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      5 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.

      • ColeSloth@discuss.tchncs.de
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        5 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.

        • Wandering_Uncertainty@lemmy.world
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          5 months ago

          The problem is, that’s exactly what the … is for. It is a little weird to our heads, granted, but it does allow the conversion. 0.33 is not the same thing as 0.333… The first is close to one third. The second one is one third. It’s how we express things as a decimal that don’t cleanly map to base ten. It may look funky, but it works.

        • rockerface@lemm.ee
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          5 months ago

          pi isn’t even a fraction. like, it’s actually an important thing that it isn’t

        • force@lemmy.world
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          5 months ago

          Pi isn’t a fraction – it’s an irrational number, i.e. a number with no fractional form in integer bases; as opposed to rational numbers, which are defined as being able to be expressed as a ratio of two integers (a fraction). Furthermore, π a transcendental number, meaning it’s never a solution to f(x) = 0, where f(x) is a non-zero finite-degree polynomial expression with rational coefficients. That’s like, literally part of the definition. They cannot be compared to rational numbers like fractions.

          Since |r|<1 => ∑[n=1, ∞] arⁿ = ar/(1-r), and 0.999... is that sum with a = 9 and r = 1/10 (visually, 0.999... = 9(0.1) + 9(0.01) + 9(0.001) + ...), it’s easy to see after plugging in, 0.999... = 9(1/10) / (1 - 1/10) = 0.9/0.9 = 1). This was a proof present in Euler’s Elements of Algebra.

    • Blum0108@lemmy.world
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      5 months ago

      I was taught that if 0.9999… didn’t equal 1 there would have to be a number that exists between the two. Since there isn’t, then 0.9999…=1

      • wieson@feddit.org
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        5 months ago

        Not even a number between, but there is no distance between the two. There is no value X for 1-x = 0.9~

        We can’t notate 0.0~ …01 in any way.

    • Shampiss@sh.itjust.works
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      5 months ago

      Divide 1 by 3: 1÷3=0.3333…

      Multiply the result by 3 reverting the operation: 0.3333… x 3 = 0.9999… or just 1

      0.9999… = 1

      • ColeSloth@discuss.tchncs.de
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        5 months ago

        You’re just rounding up an irrational number. You have a non terminating, non repeating number, that will go on forever, because it can never actually get up to its whole value.

        • WldFyre@lemm.ee
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          5 months ago

          1/3 is a rational number, because it can be depicted by a ratio of two integers. You clearly don’t know what you’re talking about, you’re getting basic algebra level facts wrong. Maybe take a hint and read some real math instead of relying on your bad intuition.

          • ColeSloth@discuss.tchncs.de
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            5 months ago

            1/3 is rational.

            .3333… is not. You can’t treat fractions the same as our base 10 number system. They don’t all have direct conversions. Hence, why you can have a perfect fraction of a third, but not a perfect 1/3 written out in base 10.

            • WldFyre@lemm.ee
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              5 months ago

              0.333… exactly equals 1/3 in base 10. What you are saying is factually incorrect and literally nonsense. You learn this in high school level math classes. Link literally any source that supports your position.

            • pyre@lemmy.world
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              5 months ago

              .333… is rational.

              at least we finally found your problem: you don’t know what rational and irrational mean. the clue is in the name.

              • Klear@sh.itjust.works
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                5 months ago

                TBH the name is a bit misleading. Same for “real” numbers. And oh so much more so for “normal numbers”.

                • pyre@lemmy.world
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                  5 months ago

                  not really. i get it because we use rational to mean logical, but that’s not what it means here. yeah, real and normal are stupid names but rational numbers are numbers that can be represented as a ratio of two numbers. i think it’s pretty good.

                  • Klear@sh.itjust.works
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                    5 months ago

                    I know all of that, but it’s still misleading. It’s not a dumb name by any means, but it still causes confusion often (as evidenced by many comments here)

      • ArchAengelus@lemmy.dbzer0.com
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        5 months ago

        In this context, yes, because of the cancellation on the fractions when you recover.

        1/3 x 3 = 1

        I would say without the context, there is an infinitesimal difference. The approximation solution above essentially ignores the problem which is more of a functional flaw in base 10 than a real number theory issue

        • Shampiss@sh.itjust.works
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          5 months ago

          The context doesn’t make a difference

          In base 10 --> 1/3 is 0.333…

          In base 12 --> 1/3 is 0.4

          But they’re both the same number.

          Base 10 simply is not capable of displaying it in a concise format. We could say that this is a notation issue. No notation is perfect. Base 10 has some confusing implications

          • ColeSloth@discuss.tchncs.de
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            5 months ago

            They’re different numbers. Base 10 isn’t perfect and can’t do everything just right, so you end up with irrational numbers that go on forever, sometimes.

        • chaonaut@lemmy.world
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          5 months ago

          This seems to be conflating 0.333...3 with 0.333... One is infinitesimally close to 1/3, the other is a decimal representation of 1/3. Indeed, if 1-0.999... resulted in anything other than 0, that would necessarily be a number with more significant digits than 0.999... which would mean that the ... failed to be an infinite repetition.

    • yetAnotherUser@discuss.tchncs.de
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      5 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 is equal to -1:

      ...999 = x
      ...990 = 10x
      
      Calculate x - 10x:
      
      x - 10x = ...999 - ...990
      -9x = 9
      x = -1
      

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

      • Valthorn@feddit.nu
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        5 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.

        • yetAnotherUser@discuss.tchncs.de
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          5 months ago

          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.

    • ColeSloth@discuss.tchncs.de
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      5 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.

      • force@lemmy.world
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        5 months ago

        Lol what? How did you conclude that if 9x = 5 then x = 1? Surely you didn’t pass algebra in high school, otherwise you could see that getting x from 9x = 5 requires dividing both sides by 9, which yields x = 5/9, i.e. 0.555... = 5/9 since x = 0.555....

        Also, you shouldn’t just use uppercase X in place of lowercase x or vice versa. Case is usually significant for variable names.

      • blue@ttrpg.network
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        5 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.

      • Redex@lemmy.world
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        5 months ago

        ?
        Where did you get 9x=5 -> x=1
        and 5/9 is 0.555… so it checks out.