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I just cited myself.

⁨877⁩ ⁨likes⁩

Submitted ⁨⁨10⁩ ⁨months⁩ ago⁩ by ⁨fossilesque@mander.xyz⁩ to ⁨science_memes@mander.xyz⁩

https://mander.xyz/pictrs/image/08aa062f-89c2-4177-a017-d26a51990c4d.jpeg

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  • Valthorn@feddit.nu ⁨10⁩ ⁨months⁩ ago

    x=.9999… 10x=9.9999… Subtract x from both sides 9x=9 x=1

    There it is, folks.

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    • barsoap@lemm.ee ⁨10⁩ ⁨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 0.333… vs. ⅓, then ask them what multiplying each by 3 is.

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      • ColeSloth@discuss.tchncs.de ⁨10⁩ ⁨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.

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      • BugleFingers@lemmy.world ⁨7⁩ ⁨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)

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      • DeanFogg@lemm.ee ⁨10⁩ ⁨months⁩ ago

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

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      • SkyezOpen@lemmy.world ⁨10⁩ ⁨months⁩ ago

        Oh shit, don’t think I saw that before. That makes it intuitive as hell.

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    • Blum0108@lemmy.world ⁨10⁩ ⁨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

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      • wieson@feddit.org ⁨10⁩ ⁨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.

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    • Shampiss@sh.itjust.works ⁨10⁩ ⁨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

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      • ArchAengelus@lemmy.dbzer0.com ⁨10⁩ ⁨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

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      • ColeSloth@discuss.tchncs.de ⁨10⁩ ⁨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.

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    • yetAnotherUser@discuss.tchncs.de ⁨10⁩ ⁨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.

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      • Valthorn@feddit.nu ⁨10⁩ ⁨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.

        a0=b0, thus

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

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    • ColeSloth@discuss.tchncs.de ⁨10⁩ ⁨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.

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      • blue@ttrpg.network ⁨10⁩ ⁨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.

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      • Redex68@lemmy.world ⁨10⁩ ⁨months⁩ ago

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

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      • lazyViking@lemmy.world ⁨10⁩ ⁨months⁩ ago

        Quick maffs

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      • force@lemmy.world ⁨10⁩ ⁨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.

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    • sp3tr4l@lemmy.zip ⁨10⁩ ⁨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.

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      • Tomorrow_Farewell@hexbear.net ⁨10⁩ ⁨months⁩ ago

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

        The ellipsis notation generally refers to repetition of a pattern. Either as infinitum, or up to some terminus. In this case we have a non-terminating decimal.

        In the case of 0.999…, it can be shown to converge toward 1

        0.999… is a real number, and not any object that can be said to converge. It is exactly 1.

        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

        In what way is it distinct?
        And what is a ‘repeating number’? Did you mean ‘repeating decimal’?

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  • jerkface@lemmy.ca ⁨10⁩ ⁨months⁩ ago

    Okay, but it equals one.

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    • lauha@lemmy.one ⁨10⁩ ⁨months⁩ ago

      No, it equals 0.999…

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      • Karcinogen@discuss.tchncs.de ⁨10⁩ ⁨months⁩ ago

        2/9 = 0.222… 7/9 = 0.777…

        0.222… + 0.777… = 0.999… 2/9 + 7/9 = 1

        0.999… = 1

        No, it equals 1.

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      • jerkface@lemmy.ca ⁨10⁩ ⁨months⁩ ago

        THAT’S EXACTLY WHAT I SAID.

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  • pruwybn@discuss.tchncs.de ⁨10⁩ ⁨months⁩ ago

    I thought the muscular guys were supposed to be right in these memes.

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    • myslsl@lemmy.world ⁨10⁩ ⁨months⁩ ago

      He is right. 1 approximates 1 to any accuracy you like.

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      • pruwybn@discuss.tchncs.de ⁨10⁩ ⁨months⁩ ago

        Is it true to say that two numbers that are equal are also approximately equal?

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    • RustyNova@lemmy.world ⁨10⁩ ⁨months⁩ ago

      Nah. They are supposed to not care about stuff and just roll with it without any regrets.

      It’s just like the wojak crying with the mask on, but not crying behind it.

      There’s plenty of cases of memes where the giga chad is just plainly wrong, but they just don’t care. But it’s not supposed to be in a troll way. The giga chad applies what it believes in. If you want a troll, there’s troll face, who speak with the confidence of a giga chad, but know he is bullshiting

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  • LodeMike@lemmy.today ⁨10⁩ ⁨months⁩ ago

    0.9<overbar.> is literally equal to 1

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    • UnderpantsWeevil@lemmy.world ⁨10⁩ ⁨months⁩ ago

      There’s a Real Analysis proof for it and everything.

      Basically boils down to

      • If 0.(9) != 1 then there must be some value between 0.(9) and 1.
      • We know such a number cannot exist, because for any given discrete value (say 0.999…9) there is a number (0.999…99) that is between that discrete value and 0.(9)
      • Therefore, no value exists between 0.(9) and 1.
      • So 0.(9) = 1
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      • LodeMike@lemmy.today ⁨10⁩ ⁨months⁩ ago

        Even simpler: 1 = 3 * 1/3

        1/3 =0.333333…

        1/3 + 1/3 + 1/3 = 0.99999999… = 1

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      • Swedneck@discuss.tchncs.de ⁨10⁩ ⁨months⁩ ago

        the explanation (not proof tbf) that actually satisfies my brain is that we’re dealing with infinite repeating digits here, which is what allows something that on the surface doesn’t make sense to actually be true.

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      • beejboytyson@lemmy.world ⁨10⁩ ⁨months⁩ ago

        That actually makes sense, thank you.

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    • jonsnothere@beehaw.org ⁨10⁩ ⁨months⁩ ago

      0.9 is most definitely not equal to 1

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      • mathemachristian@lemm.ee ⁨10⁩ ⁨months⁩ ago

        Hence the overbar. Lemmy should support LaTeX for real though

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  • magic_lobster_party@kbin.run ⁨10⁩ ⁨months⁩ ago

    If 0.999… < 1, then that must mean there’s an infinite amount of real numbers between 0.999… and 1. Can you name a single one of these?

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    • ns1@feddit.uk ⁨10⁩ ⁨months⁩ ago

      Sure 0.999…95

      Just kidding, the guy on the left is correct.

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      • magic_lobster_party@kbin.run ⁨10⁩ ⁨months⁩ ago

        You got me

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    • Drewfro66@lemmygrad.ml ⁨10⁩ ⁨months⁩ ago

      (0.999… + 1) / 2

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      • WldFyre@lemm.ee ⁨10⁩ ⁨months⁩ ago

        That number happens to be exactly 1

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  • rustyfish@lemmy.world ⁨10⁩ ⁨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.

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    • Wxnzxn@lemmy.ml ⁨10⁩ ⁨months⁩ ago

      To everyone who might not have heard about that before: It was an attempt to introduce it as a bill in Indiana:

      en.m.wikipedia.org/w/index.php?title=Indiana_pi_b…

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      • ValiantDust@feddit.de ⁨10⁩ ⁨months⁩ ago

        the bill’s language and topic caused confusion; a member proposed that it be referred to the Finance Committee, but the Speaker accepted another member’s recommendation to refer the bill to the Committee on Swamplands, where the bill could “find a deserved grave”.

        An assemblyman handed him the bill, offering to introduce him to the genius who wrote it. He declined, saying that he already met as many crazy people as he cared to.

        I hope medicine in 1897 was up to the treatment of these burns.

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    • roguetrick@lemmy.world ⁨10⁩ ⁨months⁩ ago

      I prefer my pi to be in duodecimal anyway. 3.184809493B should get you to where you need to go.

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    • fin@sh.itjust.works ⁨10⁩ ⁨months⁩ ago
      [deleted]
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      • rbesfe@lemmy.ca ⁨10⁩ ⁨months⁩ ago

        You didn’t even read the first paragraph of that article LMAO

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    • myslsl@lemmy.world ⁨10⁩ ⁨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.

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  • humdrumgentleman@lemmy.world ⁨10⁩ ⁨months⁩ ago

    This is why we can’t have nice things like dependable protection from fall damage while riding a boat in Minecraft.

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  • HexesofVexes@lemmy.world ⁨10⁩ ⁨months⁩ ago

    Reals are just point cores of dressed Cauchy sequences of naturals (think of it as a continually constructed set of narrowing intervals “homing in” on the real being constructed). The intervals shrink at the same rate generally.

    1!=0.999 iff we can find an n, such that the intervals no longer overlap at that n. This would imply a layer of absolute infinite thinness has to exist, and so we have reached a contradiction as it would have to have a width smaller than every positive real (there is no smaller real >0).

    Therefore 0.999…=1.

    However, we can argue that 1 is not identity to 0.999… quite easily as they are not the same thing.

    This does argue that this only works in an extensional setting (which is the norm for most mathematics).

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    • pumpkinseedoil@sh.itjust.works ⁨10⁩ ⁨months⁩ ago

      Easiest way to prove it:

      1/3 = 0.333…

      1/3 * 3 = 3/3 = 1

      0.333… * 3 = 0.999…

      1 = 3/3 = 1/3 * 3 = 0.333… * 3 = 0.999…

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      • HexesofVexes@lemmy.world ⁨10⁩ ⁨months⁩ ago

        Ehh, completed infinities give me wind…

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    • lemmy_99c4zb3e3@reddthat.com ⁨10⁩ ⁨months⁩ ago
      [deleted]
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      • HexesofVexes@lemmy.world ⁨10⁩ ⁨months⁩ ago

        Thanks for the bedtime reading!

        I mostly deal with foundations of analysis, so this could be handy.

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  • someacnt_@lemmy.world ⁨10⁩ ⁨months⁩ ago

    Are we still doing this 0.999… thing? Why, is it that attractive?

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    • Tlaloc_Temporal@lemmy.ca ⁨10⁩ ⁨months⁩ ago

      The rules of decimal notation don’t sipport infinite decimals properly. In order for a 9 to roll over into a 10, the next smallest decimal needs to roll over first, therefore an infinite string of anything will never resolve the needed discrete increment.

      Thus, all arguments that 0.999… = 1 must use algebra, limits, or some other logic beyond decimal notation. I consider this a bug with decimals, and 0.999… = 1 to be a workaround.

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      • Leate_Wonceslace@lemmy.dbzer0.com ⁨10⁩ ⁨months⁩ ago

        don’t sipport infinite decimals properly

        Please explain this in a way that makes sense to me (I’m an algebraist). I don’t know what it would mean for infinite decimals to be supported “properly” or “improperly”. Furthermore, I’m not aware of any arguments worth taking seriously that don’t use logic, so I’m wondering why that’s a criticism of the notation.

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      • pyre@lemmy.world ⁨10⁩ ⁨months⁩ ago

        i don’t think any number system can be safe from infinite digits. there’s bound to be some number for each one that has to be represented with them. it’s not intuitive, but that’s because infinity isn’t intuitive. that doesn’t mean there’s a problem there though. also the arguments are so simple i don’t understand why anyone would insist that there has to be a difference.

        for me the simplest is:

        1/3 = 0.333…

        3×0.333… = 3×1/3

        0.999… = 3/3

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    • Leate_Wonceslace@lemmy.dbzer0.com ⁨10⁩ ⁨months⁩ ago

      People generally find it odd and unintuitive that it’s possible to use decimal notation to represent 1 as .9~ and so this particular thing will never go away. When I was in HS I wowed some of my teachers by doing proofs on the subject, and every so often I see it online. This will continue to be an interesting fact for as long as decimal is used as a canonical notation.

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      • someacnt_@lemmy.world ⁨10⁩ ⁨months⁩ ago

        Welp, I see. Still, this is way too much recurting of a pattern.

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  • uis@lemm.ee ⁨10⁩ ⁨months⁩ ago

    0.(9)=0.(3)3=1/33=1

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    • fin@sh.itjust.works ⁨10⁩ ⁨months⁩ ago

      Beautiful. Is that true or am I tricked?

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      • cordlesslamp@lemmy.today ⁨10⁩ ⁨months⁩ ago

        0.3 is definitely NOT equal 1/3.

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  • bluewing@lemm.ee ⁨10⁩ ⁨months⁩ ago

    Meh, close enough.

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  • dylanTheDeveloper@lemmy.world ⁨10⁩ ⁨months⁩ ago

    I wish computers could calculate infinity

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    • jerkface@lemmy.ca ⁨10⁩ ⁨months⁩ ago

      Computers can calculate [infinite series](en.wikipedia.org/wiki/Series_(mathematics\)) as well as anyone else

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    • ILikeBoobies@lemmy.ca ⁨10⁩ ⁨months⁩ ago

      As long as you have it forget the previous digit, you can bring up a new digit

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  • BachenBenno@feddit.de ⁨10⁩ ⁨months⁩ ago

    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.

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    • Sop@lemmy.blahaj.zone ⁨10⁩ ⁨months⁩ ago

      That’s not an axiom or definition, it’s a consequence of the axioms that define arithmetic and can therefore be proven.

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    • aoidenpa@lemmy.world ⁨10⁩ ⁨months⁩ ago

      That’s not how it’s defined. 0.99… is the limit of a sequence and it is precisely 1. 0.99… is the summation of infinite number of numbers and we don’t know how to do that if it isn’t defined. (0.9 + 0.09 + 0.009…) It is defined by the limit of the partial sums, 0.9, 0.99, 0.999… The limit of this sequence is 1.

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  • Th4tGuyII@fedia.io ⁨10⁩ ⁨months⁩ ago

    0.999... / 3 = 0.333...
    1 / 3 = 0.333...
    Ergo 1 = 0.999...

    (Or see algebraic proof by @Valthorn)

    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.

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    • themeatbridge@lemmy.world ⁨10⁩ ⁨months⁩ ago

      Excelt it isn’t infinitesimally smaller at all. 0.999… is exactly 1, not at all less than 1. That’s the power of infinity. If you wanted to make a wooden board exactly 0.999… m long, you would need to make a board exactly 1 m long (which presents its own challenges).

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    • Feathercrown@lemmy.world ⁨10⁩ ⁨months⁩ ago

      Math doesn’t care about physical limitations like the planck length.

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  • WolfLink@sh.itjust.works ⁨10⁩ ⁨months⁩ ago

    en.wikipedia.org/wiki/0.999…

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  • ettyblatant@lemmy.world ⁨10⁩ ⁨months⁩ ago
    [deleted]
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    • EatATaco@lemm.ee ⁨10⁩ ⁨months⁩ ago

      I think the … here is meant to represent repeating digits.

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  • Tomorrow_Farewell@hexbear.net ⁨10⁩ ⁨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.

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  • cumskin_genocide@lemm.ee ⁨10⁩ ⁨months⁩ ago

    The only sources I trust are the ones that come from my dreams

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  • CodexArcanum@lemmy.world ⁨10⁩ ⁨months⁩ ago

    Even the hyperreal numbers *R, which include infinitesimals, define 1 == .999…

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  • Unbeelievable@beehaw.org ⁨10⁩ ⁨months⁩ ago

    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

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  • LordCrom@lemmy.world ⁨10⁩ ⁨months⁩ ago

    I can honestly say I learned something from the comment section. I was always taught the .9 repeating was not equal to 1 but separated by imaginary i … Or infinitely close to 1 without becoming 1.

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  • Etterra@lemmy.world ⁨10⁩ ⁨months⁩ ago

    Sort of how 0.0000001 = 0

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    • null@slrpnk.net ⁨10⁩ ⁨months⁩ ago

      No, not like that.

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