58 ms·

0.999...= 1

• bytedude 6y ago
  Flame wars over this used to be common on the internet. People intuitively have the notion that the left side approaches 1, but never actually equals it. They see it as a process instead of a fixed value. Maybe the notation is to blame.
  • tgv 6y ago
    Repetition can easily be seen as a process, which would indeed approach 1. But I think the idea of infinite repetition is very hard to get.
  • username90 6y ago
    The intuition that there is something in between isn't really wrong, it make sense and they work, otherwise physicists wouldn't be able to work with them. So that intuition is correct, it is mathematicians who just don't understand it fully yet. Maybe fully formalizing this is what unlocks the final piece keeping us from creating a unified theory in physics?
    • steerablesafe 6y ago
      Nonstandard analysis is a rigorous framework for working with infinitesimals (and infinitely large numbers). https://en.wikipedia.org/wiki/Nonstandard_analysis https://en.wikipedia.org/wiki/Nonstandard_analysis
  • jhanschoo 6y ago
    The intuition is right, and the mathematical definition relies on the intuition. It's just that people haven't been exposed to the actual definition when it comes to real numbers. Mathematically, mathematicians prove that there is a unique number that this process goes to, (and not, say, two distinct numbers), and define the notation to represent this unique number.
• orthoxerox 6y ago
  I remember WarCraft 3 official forums being torn apart by this, with probably thousands of comments in the thread. Blizzard even had to post their official stance on the issue, but that didn't calm those who insisted 0.999... was 1 minus epsilon and not exactly 1.
  • jcfields 6y ago
    I'm glad someone else remembers this. To this day, whenever I see 0.999... = 1, I think of the Battle.net forums inexplicably flooded with threads about it for what felt like ages.
• klodolph 6y ago
  An interesting consequence of this in proofs. You’ll see various proofs involving real numbers that must account for the fact that 0.999…=1.0. There are, of course, many different ways to construct real numbers, and often it’s very convenient to construct them as infinite sequences of digits after the decimal. For example, this construction makes the diagonalization argument easier. However, you must take care in your diagonalization argument not to construct a different decimal representation of a number already in your list!
  • rini17 6y ago
    I never understood the fixation on diagonalization. Why can't ever exist another way for mapping any set to countables?
    • voxl 6y ago
      Diagnolization is a pretty deep argument about fixpoints, Godels incompleteness argument is essentially a diagnolization. So why wouldn't there be fascination?
    • deleted 6y ago
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    • rcxdude 6y ago
      The point is that diagonalization works whatever map you have come up with: no matter how you construct your list of the reals, you can come up with another real not on the list.
• cjfd 6y ago
  Ultimately this is more the definition of R than that it is a theorem. One can also work with sets of numbers in which the completeness axiom does not hold. E.g., sets of numbers in which one also has infinitesimals.
• rs23296008n1 6y ago
  I'm actually curious what impact it would have on various proofs if 0.999... wasn't accepted as 1. What gets broken? What consequences do we hit?
  • saagarjha 6y ago
    Arithmetic breaks, as multiplication is no longer the inverse of division. (For example, 1/3 * 3 = 0.999… would no longer work.)
    • tomtomtom1 6y ago
      why? 1/3 * 3 could still be equal to one. but 1/3 != 0.33333... that is, 1/3 is not representable in base 10. Which makes way more sense. I wonder if taking 0.9999.. != 1, that is 0.0000...1 exists would allow us to reslove, the fact that some possible events have probability 0?
      • baddox 6y ago
        Limits would break, which might break or at least cause problems with calculus and other branches of math.
      • atq2119 6y ago
        The crux of the matter is that you have to define what things like "0.3333..." mean in the first place. Any reasonable definition of it as a representation of a real number is going to lead to it being equal to 1/3. If you want to redefine it explicitly as not a real number, you can do that, and maybe even get to some amusing math that way, but you're no longer talking the same language as the rest of the world.
        • tomtomtom1 6y ago
          >"but you're no longer talking the same language as the rest of the world" yes, in the standard real numbers 1 = 0.999.., but people have dealt with numbers like "pi" and "sqrt(2)" before the standard real numbers were defined. Hence the question, if we define such a system such as 0.333... != 1/3. what are the consequences? by 0.3333... I mean a countably infinite sequence of 3s.
• ltbarcly3 6y ago
  This is 'more intuitive' if you think about it this way: If any two real numbers are not equal, then you can take the average and get a third number that is half way between them. Conversely, if the average of two numbers is equal to either of the numbers, then the two numbers are equal. (this isn't a proof, just a way to convince yourself of this) What's the average of .9999... and 1?
  • saagarjha 6y ago
    0.999…5 obviously.
    • fsflover 6y ago
      But at which position that "5" is? Tell the number please.
      • saagarjha 6y ago
        Position 999…, duh.
    • kaetemi 6y ago
      Okay, now multiply that by 2.
      • saagarjha 6y ago
        0.999…
  • edflsafoiewq 6y ago
    Or: if they're different, what is their difference?
    • peanutz454 6y ago
      0.000...1 obviously
• ginko 6y ago
  I remember being doubtful when being presented with this in middle school, but after being shown this as fractions makes it obvious: 1/3 = 0.333.. 3 * 1/3 = 3 * 0.333.. 3/3 = 0.999.. 1 = 0.999..
  • iso1631 6y ago
    Another secondary school 'proof' x = 0.9999..... 10x = 9.9999..... (10x -x) = 9x = (9.9999.... - 0.9999....) = 9 x = 9/9 = 1
    • notRobot 6y ago
      or, x = 0.9999... 10x = 9.999... 10x = 9 + 0.999... 10x = 9 + x 9x = 9 x = 1 Presented slightly more clearly https://en.wikipedia.org/wiki/0.999...#Algebraic_arguments https://en.wikipedia.org/wiki/0.999...#Algebraic_arguments
      • hinkley 6y ago
        Yeah that’s way more complicated than it needs to be and I’m tempted to replace that whole section with: x = 0.9999... 2x = 1.9999... 2x - x = 1 x = 1
        • moonchild 6y ago
          It's much more intuitive that 100.9999...=9.9999... than that 20.9999...=1.9999...
          • ColinWright 6y ago
            You've met the formatting limitations. You mean: "It's much more intuitive that 10 * 0.9999... = 9.9999... than that 2 * 0.9999...= 1.9999..."
        • champagneben 6y ago
          How do you jump from x = 0.999... to 2x = 1.999..
• steerablesafe 6y ago
  Maybe the major source of confusion is that our decimal representation for whole numbers is supposed to be unique. Then when we extend it to rationals and reals this property fails at rationals in the form of a/10^n. Arguably the sign symbol ruins it for whole numbers as well, as +0 and -0 could be equally valid representations of the number 0. We just conventionally don't allow -0 as a representation. There are other number representations that don't have this problem.
  • virgilp 6y ago
    Right - I also find it easier to say that really, "1" is just a different/shorthand notation for 0.(9) It's not "two different, but equal numbers" - it's two different notations for the same number. Like how you can write same number in different ways in different bases - this is just writing the same number, in the "infinite number of decimals" vs "natural" way.
    • steerablesafe 6y ago
      Arguably 1 is just as an infinite number of decimals as 0.(9) . 1 is just short for 1.(0) .
  • onion2k 6y ago
    I think the source of confusion is that people can't cope with recurring numbers. When someone says .999...=1 the listener assumes that the 0.999... stops at some point, and if that happens it'll always be below 1 because they can imagine adding another 9. Essentially, people actually ignore the "..." because that's the hard part.
  • baddox 6y ago
    I don’t know, I don’t recall students having much trouble accepting that 1 = 1.0 = 1.000.
• j-pb 6y ago
  I'm not a mathematician, but I would guess that the Surreal numbers developed by John Conway, do contain values that start with an infinite sequence of 0.9_ but "end" with something that makes them != 1. https://en.m.wikipedia.org/wiki/Surreal_number https://en.m.wikipedia.org/wiki/Surreal_number
• deleted 6y ago
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• j7ake 6y ago
  This can be solved by using base 12 rather than base 10 to do the calculation...
  • makotoNagano 6y ago
    Firm believer that adopting base 12 would have had a ripple affect on society, preventing many trials tribulations and wars. Pity we only have base 10 and donald trump
    • rrmm 6y ago
      base 12? Them are fighting words...I call WAR!
      • makotoNagano 6y ago
        We haven't even started the Pi Tau argument yet
• traderjane 6y ago
  Professor N.J. Wildberger is probably among the most well known "ultrafinitist" on YouTube. https://www.youtube.com/watch?v=WabHm1QWVCA https://www.youtube.com/watch?v=WabHm1QWVCA I mention him because I would think he sympathizes with those who have concern over the meaning of this kind of notation.
  • rrmm 6y ago
    Wildberger is great. His lectures that he teaches at UNSW (i think) are interesting, and he usually keeps a clear dividing line between std math and his own predilections. It threads the line between being a kook and legitimate published mathematician very finely. I actually have some sympathies with his contention that real numbers (limit points of infinite series) are somehow a different animal than rational numbers. But it might be easier for me to go there because practically all numbers on computers that we work with are rational, floating point values. On the other hand, it seems like a philosophical distinction in the end because you can fully order them both on a number line.
    • danharaj 6y ago
      If I give you two representatives of real numbers, say turing machines that write out on their tape the binary digits of those real numbers, in general you will not be able to order them.
      • rrmm 6y ago
        Certainly not non-computable ones, but presumably they lie somewhere regardless of my inability to do it on a TM. Which presumably gives rise to all the weirdness uncountable infinities give you. I guess I shouldn't phrase it as "you can fully order it". :D Zermelo's theorem at that point right?
        • danharaj 6y ago
          Even computable reals do not have computable ordering. > but presumably they lie somewhere regardless of my inability to do it on a TM. Why? > Zermelo's theorem at that point right? It is declared by fiat in standard set theory that infinite sets can be well ordered. This is no real mathematical justification. The real justification is social: that it is convenient for mathematicians to not care about the ontology of these nasty infinite objects so long as results are mostly reasonable for objects that mathematicians actually care about. You don't get into too much trouble pretending the reals are nice so long as you don't look too hard.
• russellbeattie 6y ago
  I'll just chime in with my completely ignorant theory that 1 - 0.999... = the infinitely smallest number, but is still, in my mind, regardless of any logic, reason, or educated calculations, greater than 0. I understand and accept this is wrong. However, somewhere in my brain I still believe it. Sort of like +0 and -0, which are also different in my head.
  • thdrdt 6y ago
    Philosophically you are right. Mathematically you are wrong. It is indeed true that if there was an entity that could reason beyond infinity 1-0.999... would be greater than 0.
    • saagarjha 6y ago
      From the perspective of IEEE-754, they’re pretty close ;)
    • cthalupa 6y ago
      >It is indeed true that if there was an entity that could reason beyond infinity 1-0.999... would be greater than 0. This seems like a common thread in the comments on this article, and I don't quite understand it. Humans can reason about infinity just fine. We have a hard time picturing it, but it's overall a pretty simple concept: It never ends. So there's no such thing as being able to reason "beyond infinity", because beyond infinity doesn't exist. Its very existence is precluded by the definition of infinity.
  • Scarblac 6y ago
    So 0 and that number are two different numbers. What is the difference between them?
    • russellbeattie 6y ago
      One is positive and the other is negative. I thought that was pretty obvious!
  • 6y ago
• berkeleynerd 6y ago
  Perhaps the natural discomfort many face when confronted by this challenging formulation instead indicates that a limitation of the real number system has been perceived? I would encourage those who have this reaction to study hyperreal and other alternative systems as mentioned in the article. If this clicks for them they may help lead us in new direction mathematically and advance the state of the art.
  • jhanschoo 6y ago
    This comes from the fact that most people don't learn limits. Once the definition of the decimal representation is understood to be in terms of a limiting process, the meaning becomes clear. For most purposes hyperreals are too much machinery when learning and manipulating limits would be simpler.
• knzhou 6y ago
  Personally I've always thought "proofs" using "arithmetic" are right, but kind of stated backwards. The point is that in elementary school arithmetic, you define addition, multiplication, subtraction, division, decimals, and equality, but you never define "...". Until you've defined "...", it's just a meaningless sequence of marks on paper. You can't prove anything about it using arithmetic, or otherwise. What the "arithmetic proofs" are really showing that if we want "..." to have certain extremely reasonable properties, then we must choose to define it in such a way that 0.999... = 1. Other definitions would be possible (for example, a stupid definition would be 0.999... = 42), just not useful. What probably causes the flame wars over "..." is that most people never see how "..." is defined (which properly would require constructing the reals). They only see these indirect arguments about how "..." should be defined, which look unsatisfying. Or they grow so accustomed to writing down "..." in school that they think they already know how it's defined, when it never has been!
  • ssivark 6y ago
    > The point is that in elementary school arithmetic, you define addition, multiplication, subtraction, division, decimals, and equality, but you never define "...". Until you've defined "...", it's just a meaningless sequence of marks on paper. You can't prove anything about it using arithmetic, or otherwise. Sure, but the point of "elementary school" arithmetic is not "elementary" arithmetic, as a mathematician would define it :-) The goal is to teach people to reason by matching patterns. Deductive/Inductive reasoning can slowly proceed from that, as they try to frame their intuition for patterns into increasingly more general abstractions.
  • agumonkey 6y ago
    And I thought mathematicians would reject normal arithmetic operation over the domain of `N...` elements. They're always so ultra rigorous to classify what is or is not, what is defined, the domain .. but then they let an infinite sequence be treated as any finite number.
    • saagarjha 6y ago
      > but then they let an infinite sequence be treated as any finite number Technically it’s an infinite series (sum of an infinite sequence), which is a finite number in certain cases like this one.
• vfinn 6y ago
  Sorry for my naivety, but why one couldn't prove by induction that adding 9s never close the gap, or let's say, that by definition the operation is such, that it never closes the gap. If you can always halve the pie, then you can continue eating forever. To me it would be much easier to accept that (1/3)*3 is not 1.
  • saagarjha 6y ago
    Induction can be non-intuitive to people who are troubled by this problem.
    • jhanschoo 6y ago
      Induction has nothing to do with this. It's about a taking the limit of a process.
  • jhanschoo 6y ago
    It's because the rigorous mathematical definition is more subtle. It's defined as the smallest real number such that repeating this process (putting more 9's in the end) can't result in a number that's greater than it. So as long as you have a number that's smaller than 1, there's a gap there, and repeating the process of adding 9's will eventually give you a number that lies in that gap.
    • vfinn 6y ago
      Yeah, it makes sense if the whole foundation of mathematics is laid out in such way that the above has to be true (kinda circular, but still), and this would mean that what I'm talking about is actually not mathematics but something else :).
  • mFixman 6y ago
    Let's try a naïve inductive proof of 0.999… ≠ 1. Base case: Given 𝑎⁰ = 0, 𝑎⁰ ≠ 1. Inductive case: Given 𝑎ⁿ⁺¹ = 1 - (1 - 𝑎ⁿ) / 10, 𝑎ⁿ ≠ 1 ⇒ 𝑎ⁿ⁺¹ ≠ 1 This proves that for every 𝑎ⁿ = 0.999…9, there's an 𝑎ⁿ⁺¹ that's a 9 larger and still different than 1, which is similar to your halving the pie example. However, you can see that it always happens that 𝑎ⁿ⁺¹ > 𝑎ⁿ, so the "last" infinite 0.999… is not part of the sequence of the inductive case. My intuitive way to see this is that infinitely repeating decimals are an abomination that breaks the nice property of decimal notation where each number contains a single representation (without zeroes at the beginning or at the end of the decimal). Fractions are the one true way to represent rational numbers.
• upofadown 6y ago
  > ...infinitely many 9s... How about we prove that an infinite number of 9s is impossible? Assume that we have a finite number of 9s. Add a 9. The result is not infinite. Add another 9. The result is still not infinite. We can repeat this process for an infinite amount of time and still not have an infinite number of nines. Any process that can not be completed in a finite amount of time can not complete and can not have a valid result based on that completion. Any process that can not be completed in an infinite amount of time is also bogus, but is in a sense even more bogus. Added: Note that this is different than the case where we are asked to contemplate infinity with respect to continuous functions. By defining the number of 9s as a discrete (integer) value it opens things up to a discrete argument. These pointless navel gazing exercises always end up as a war of what everyone things things are defined as.
  • eof 6y ago
    I tried this route as a counter to Cantor's diagonal argument, and got chastised by my then professor. I hope you have better luck, as I was never able to convince myself otherwise.
    • upofadown 6y ago
      By that point you were already working in a universe where infinity was in some sense real. The universe of mathematics is different than the universe we live in.
  • Faaak 6y ago
    > We can repeat this process for an infinite amount of time Well, no, you cant. How about I prove it to you: Assume that we have a finite number of processes. Repeat the process. The number of processes is still not infinite. /s
  • gspr 6y ago
    > How about we prove that an infinite number of 9s is impossible? First you need to define what you even mean by this statement. The rest of what you wrote makes no sense either.
  • 6y ago
• ristos 6y ago
  I don't think that 0.999... = 1 is actually provable. I think this and all of calculus is actually axiomatic, which has the following axiom: Given ε = 1/∞ then: ε = 0 Am I wrong in thinking this way? It seems as though there's no way to actually truly prove that an infinite series converging towards zero actually hits zero (from a constructivist pov)
  • qppo 6y ago
    I think the mistake in your thinking is that infinity is a value or reachable destination. It's a concept and it behaves differently than a number. Also go read Generatingfunctionology and Concrete Mathematics
    • ristos 6y ago
      That's the point I'm trying to make actually. 1/infinity is an infinite series, which would be a computation that takes infinite time to compute. Saying that 0.9999...=1 is saying that the infinite series is the same as that concrete value, which don't actually know and can't prove.
      • qppo 6y ago
        I think the hiccup here is the notion that an open form expression doesn't have a closed form representation, which is not always the case. Closed forms exist for infinite series and vice versa. 1 - sum{k=1}^{\ifnty} 10^-k is one example.
  • SpicyLemonZest 6y ago
    A mathematician would quibble with your notation, but you're basically right. The fact that 0.999... = 1 depends on the fact that 0 is the only number smaller than 1/n for every integer n. This isn't exactly an arbitrary axiom, it encodes some of our natural intuitions about what a number is, but you can construct a system known as the "hyperreal numbers" where it isn't true and 0.999... doesn't converge.
  • jmull 6y ago
• smlckz 6y ago
  Reminds me of the paradoxes of Zeno [1], especially the paradox of Achilles and the tortoise. At least one can simply prove that 0.999... = 1 without much hard work. Maybe less controversial than the following: 1 + 2 + 3 + ... [somehow] = -1/12 {{Riemann's zeta(-1)?}} 1 + 2 + 4 + 8 + 16 + ... [somehow] = -1 As well as the weird prime product (Product of 1/(1-(p^-2)) for p prime) and the sum of x^-2 from x=1 to [sigh] being equal to (pi^2)/6 are some example of infinite beauty of mathematics that I remember. [1]: https://en.wikipedia.org/wiki/Zeno's_paradoxes https://en.wikipedia.org/wiki/Zeno's_paradoxes
  • cochne 6y ago
    It's less controversial than those because it's actually true. The latter statements are not true with the traditional definition of a sum.
  • gilbetron 6y ago
    1 + 2 + 3 + ... does not equal -1/12, except if you redefine what you mean by +. https://www.youtube.com/watch?v=YuIIjLr6vUA https://www.youtube.com/watch?v=YuIIjLr6vUA
  • frogpelt 6y ago
    Your two examples can be debunked though. See here: https://www.youtube.com/watch?v=YuIIjLr6vUA https://www.youtube.com/watch?v=YuIIjLr6vUA
    • smlckz 6y ago
      ah, Mathologer video. Have seen that. Another one by 3b1b on that topic: https://youtube.com/watch?v=sD0NjbwqlYw https://youtube.com/watch?v=sD0NjbwqlYw
• undecisive 6y ago
  There is no proof that will ever satisfy a person dead-set against this. Ever since I brought this home from school as a child, my whole family ribbed me mercilessly for it. If you tell a person that 3/6 = 1/2, they'll believe you - because they have been taught from an early age that fractions can have multiple "representations" for the same underlying amount. People mistakenly believe that decimal numbers don't have multiple representations - which, in a way is correct. The bar or dot or ... are there to plug a gap, allowing more values to be represented accurately than plain-old decimal numbers allow for. It has the side effect of introducing multiple representations - and even with this limitation, it doesn't cover everything - Pi can't be represented with an accurate number, for example. But it also exposes a limitation in humans: We cannot imagine infinity. Some of us can abstract it away in useful ways, but for the rest of the world everything has an end. I wonder if there's anything I can do with my children to prevent them from being bound by this mental limitation?
  • traderjane 6y ago
    You must go up to something like limits to make ... meaningful.
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  • tom-thistime 6y ago
    (STATEMENT OF PERSONAL IGNORANCE [SOPI]: Anyone who actually understands this stuff please correct my mistakes below. Thanks.) In the real numbers, which are not always simple or intuitive, 0.99... = 1. That's true and I seem to understand the proof. But the real numbers aren't the only system that might be sitting behind "0.99..." and "1" when I write those symbols down and talk intuitively to people in my family. The reals are just the system we're taught first. I believe there are other systems (I think the surreals are an example) that work just as well for everyday purposes, but where ( my understanding is that) there are numbers that differ from 1 by a value that approaches zero, yet those numbers are not equal to 1. (I've played with the surreals but only as a hobbyist.) If you do calculus in these other numbers, I think physically meaningful problems will still yield the same answers. (For example Zeno's Paradoxes are still not an excuse for failure to attend school.) But it isn't a law of nature, I think, that all number systems that can hold 0.999... and 1 must make them equal.
• amelius 6y ago
  How about the expression: 0.9999... < 1 And consider that if a < b then a != b.
  • Scarblac 6y ago
    It's false.
  • OskarS 6y ago
    Both are false. 0.99999... is not less than 1. It is the same as 1.
    • amelius 6y ago
      Yes, because someone defined it that way. It is because the "limit" in 0.999... = lim[eps->0] 1-eps is implicit and defined as being applied before anything else. But you might as well define that implicit limit as applying over the entire expression. UPDATE: So instead of interpreting the expression as: (lim[eps->0] 1-eps) < 1 which is indeed false, you can also interpret the expression as: lim[eps->0] ((1-eps) < 1) which is true (assuming that -> denotes a limit from above). Note that here the "lim" has been taken out and acts over the entire expression.
      • gspr 6y ago
        > But you might as well define that implicit limit as applying over the entire expression. What is this supposed to even mean?
      • OskarS 6y ago
        It's still not true. No no sense is it true. It is true that 0.999... = lim[eps->0] (1 - eps), but it is ALSO true that lim[eps->0] (1 - eps) = 1. That's how limits work. If you accept both of those two (which you should, because they're correct), then since equality is transitive, 0.999... = 1. Therefore it is not less than 1, it is equal to 1.
        • 6y ago
• jefftk 6y ago
  What if you have 0.9̅4? Can we say 0.9̅5 > 0.9̅4 > 0.9̅3? More on what happens if you allow this: https://mathwithbaddrawings.com/2013/08/13/the-kaufman-decimals/ https://mathwithbaddrawings.com/2013/08/13/the-kaufman-decim...
  • ubercow13 6y ago
    That's fun but it's not clear how interesting those numbers are compared to real numbers, which have turned out to be pretty interesting over the years
  • cthalupa 6y ago
    >What if you have 0.9̅4? Well, you fundamentally can't. If the 9s go on for forever then you never reach a point where you can add the 4. The definition of infinity precludes anything after infinity, because it never ends so you can never get there.
    • jefftk 6y ago
      In this alternate number system you can think of it as a tuple: (.9̅, 4). Which is larger than (.9̅, 3) and smaller than (.9̅, 5).
    • ubercow13 6y ago
      That is, until Cantor showed otherwise
      • cthalupa 6y ago
        Are you saying that Cantor showed you can have 0.9..4? I'm not even remotely an expert here, so I might certainly be wrong, but I don't understand how Cantor's theorems show an ability to stick a finite number and stop on the end of an infinite number.
        • ubercow13 6y ago
          Yes, one thing Cantor showed is that it is somehow meaningful to define numbers like infinity+1, where there are an ordered infinity of elements followed by one more element. Sets like this are called ordinals https://en.wikipedia.org/wiki/Ordinal_number https://en.wikipedia.org/wiki/Ordinal_number So you could, if you like, define 0.9...4 to be a bunch of digits indexed by the ordinal ω+1. However the thing you have now defined isn't really a representation of a real number any more, unless you just ignore all the bits after the ... I guess.
• seyz 6y ago
  "In other words, "0.999..." and "1" represent the same number." - Ok, I'm done with this world.
• dwheeler 6y ago
  A formally rigorous proof of this (in Metamath) is here: http://us.metamath.org/mpeuni/0.999....html http://us.metamath.org/mpeuni/0.999....html Unlike typical math proofs, which hint at the underlying steps, every step in this proof only uses precisely an axiom or previously-proven theorem, and you can click on the step to see it. The same is true for all the other theorems. In the end it only depends on predicate logic and ZFC set theory. All the proofs have been verified by 5 different verifiers, written by 5 different people in 5 different programming languages. You can't make people believe, but you can provide very strong evidence.
  • 317070 6y ago
    It depends on more than just ZFC, also on the definitions of the real/complex numbers. The crux of the proof is that 0.99999... is being constructed within the real/complex numbers, and in that system it is equal to 1. And at the point where students see this, the whole concept of real numbers and infinity is usually ill-defined. I actually understand the scepsis for this theorem and where it comes from. The proof relies on the existence of a supremum, which is non-trivial.
    • _Donny 6y ago
      I think this is spot on, at least for me personally. I am not very good at mathematics, so I never questioned my professors when they said that "You cannot treat infinites as regular numbers". Perhaps due to that statement, I did not really pursue these kinds of equations. For instance, I do not really see how the algebraic argument on the Wiki is any different from: 2 * inf = inf inf + inf = inf (subtract inf from both sides) inf = 0
      • shawnz 6y ago
        That causes a contradiction which is why infinity can't be used that way. But what is the contradiction with 0.999... = 1?
        • jl2718 6y ago
• ping_pong 6y ago
  My 5 year old stumped me with this, and I had to look it up. He asked me why 1/3 + 1/3 + 1/3 = 1, since it's equal to 0.333... + 0.333... + 0.333... which is 0.999... How can that possibly equal 1.000...? And is 0.66... equal to 0.67000...? I didn't have a good enough answer for him, so I had to look it up and found this page. I tried to explain it to him but since I'm a terrible teacher and he's only 5, it was hard for me to convince him. Luckily he has many years before it matters!
  • bognition 6y ago
    No .666666 is not equal to .6700000 0.666... is equal to 0.666...7
    • hilbertseries 6y ago
      Not sure if you’re joking, but 0.666...7 is not a real number. Can you define it?
      • dacohenii 6y ago
        I'm pretty sure it's a real number in that it falls within ℝ (the set of all real numbers).
        • RHSeeger 6y ago
          I doesn't, because 0.666....7 is not a valid representation for a number. The ... means goes on forever, and you cannot put something "after forever". It's not different than saying "0.0j" is a real number in base 10; it's not, that string does not represent a number in our number base 10 number system.
        • hilbertseries 6y ago
          There’s not a way to define it, typically you’d define .666... as the sum of 6/10^n from 1 to infinity. This decimal representation does not terminate, so you can’t put a 7 at the “end of it” because there is no end.
• ttonkytonk 6y ago
  Basically an infinite series of 9's just means they're all maxed so the .999... = 1 makes sense to me (kind of anyway).
• 2OEH8eoCRo0 6y ago
  Yes it does but it seems like a less correct way of writing it. Like you could represent the number 10 as 10/1 (ten over one) but why would you? Why would you represent 1 as .9 repeated?
• sebringj 6y ago
  The thing that helps me "understand" it is that the universe has finite sizes of things like the Planck length for example being a theoretical thing at the smallest distance I would imagine. Now imagine it going smaller than the Planck length (finite) in terms of the difference of .9 repeating and 1 since infinitely small differences can do that. Essentially there is no way to tell the difference between .9 repeating and 1 then from a practical or theoretical perspective of measurement. So not imagining infinity lets us at least imagine smaller than the smallest measurable thing.
  • cthalupa 6y ago
    >The universe has finite sizes of things like the Planck length for example being a theoretical thing at the smallest distance I would imagine The Planck length might or might not be a physical limit of the universe. We don't have any specific proof that it's the smallest, just that we will not be able to observe any of that size or smaller. To look at something, we need to use light, and we must use a wavelength smaller than the details we wish to resolve. For something of the Planck length or smaller, this ultimately results in a photon that would have more energy in that area than can exist without a black hole forming... so one does, which then prevents us from measuring it, much less anything smaller. Space and time might very well be discrete and not continuous - certainly the Loop Quantum Gravity folks would agree there. But there are widely supported theories that take both sides. (I tend to lean towards them being discrete, but I would hesitate to call myself even an amateur hobbyist when it comes to theoretical physics...)
• deleted 6y ago
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• zests 6y ago
  Is 1 a prime number? No, because we define it not to be. Why do we define it not to be a prime number? That's the real question. Is 0.999... = 1? Yes, because we define decimal numbers to behave that way. Why do we define them to behave that way? That's the real question.
• calibas 6y ago
  And this is why I prefer hyperreals. 0.999... = 1 - 1/∞ We talk about infinity all the time in mathematics, teachers use the concept to introduce calculus in a way that people can more easily understand, but using infinity directly is almost universally banned within classrooms. Nonstandard analysis is a much more intuitive way of understanding calculus, it's the whole "infinite number of infinitely small pieces" concept, but you're allowed to write it down too.
  • __s 6y ago
    I think what's important here is that if you're making that claim, 0.333... = 1/3 - 1/∞ Which implies 3/∞ = 1/∞
    • calibas 6y ago
      Apologies, I should have explained it differently. 0.999... implies a number infinitesimally smaller than 1. You wouldn't use 0.999... in a hyperreal system because you can represent it directly. I shouldn't have mixed different systems and claimed they're mathematically equivalent, you've proven that doesn't work.
      • __s 6y ago
        Agreed, I think the main lesson from this topic is that decimal numbers are a pain. Stick to integers & symbolic operations. You can spit out a decimal approximation at the end Computers agree: never trust precision to floats
• heavenlyblue 6y ago
  So is then ‘0.(0...)1 = 0’?
• heinrichhartman 6y ago
  0.9999 = 1 is a consequence of the way we define rational and real numbers and limits. There are alternative definitions of numbers where this equality does not hold: Non Standard Analysis https://en.wikipedia.org/wiki/Nonstandard_analysis https://en.wikipedia.org/wiki/Nonstandard_analysis being the most famous one. But for the sake of argument, let's just define numbers as sequences of digits with a mixed in period somewhere: MyNumber := { a = (a_1, a_2, ...) -- list of digits a_i = 0 .. 9; a_1 != 0. e -- exponent (integer) s -- sign (+/- 1) } Each such sequence corresponds to the (classical) real number: s * \sum_i a_i * 10^{i + e}. We can go on and define addition, subtraction, multiplication and division in the familiar way. Problems arise only when we try to establish desireable properties, e.g. (1/3) * 3 = 1 Does NOT hold here, since 0.9999... is a difference sequence than 1.000.... So yes, you can define these number systems, and you will have 0.999... != 1. But working with them will be pretty awkward, since a lot of familiar arithmetic breaks down.
  • astrobe_ 6y ago
    1 = 0.9... is the consequence of purposely ambiguous and questionable notation. That's an old teachers' trick to make students talk and listen about mathematics.
    • heinrichhartman 6y ago
      This has nothing to do with notation. It's perfectly possible to define infinite sequences without using dots. In particular if they are constant. In the above case: a_i = 9 for i \in \IZ and i < 0 a_i = 0 for i \in \IZ and i >= 0 Where \IZ are the integers.
• alpple 6y ago
  Why do we accept .999... as a valid notation. Why not only allow 1 to denote this concept?
  • gspr 6y ago
    > Why do we accept .999... as a valid notation. Why not only allow 1 to denote this concept? That would be adding a special rule for purely cosmetic reasons. It's typically not done in mathematics, where concise rules are usually more cherished than special-casing things.
    • deleted 6y ago
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• shrimpx 6y ago
  A dumb consequence of the axiom of choice? The reals are like a membrane with no atomic pieces. You can move in either direction infinitely and you can zoom in infinitely without reaching any “Planck unit” so to speak. So what does it even mean to pick out a “real number”? To me anything built on this concept is nonsense.
• deleted 6y ago
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• flerchin 6y ago
  I'd have figured "approaches 1 from the left" would be more accurate.
• JJMcJ 6y ago
  Usually the concept of a limit, which assigns a meaning to 0.999..., isn't studied until calculus. There are approaches to mathematics that avoid infinite constructions, and a "strict finitist" would not assign 0.999... a meaning. The stunning success of limit based mathematics makes finitism a fringe philosophy. Remember, class, for every epsilon there is a delta.
• jl2718 6y ago
  The proof relies on the assertion that the supremum of an increasing sequence is equal to the limit. This is mathematical dogma, and should be introduced as such. Once that is accepted, it becomes obvious. This is illustrative of what I see as a fundamental problem in mathematics education: nobody ever teaches the rules. In this case, the rules of simple arithmetic hit a dead end for mathematicians, so they invented a new rule that allowed them to go further without breaking any old rules. This is generally acceptable in proofs, although it can have significant implications, such as two mutually exclusive but otherwise acceptable rules causing a divergence in fields of study. When I was taught this, it was like, “Look how smart I am for applying this obtusely-stated limit rule that you were never told about.” This is how you keep people out of math. The point of teaching it is to make it easy, not hard.
  • andrewprock 6y ago
    This is in large part due to the difficulty with reasoning about infinite representations. You do have to add axioms to your system to be able to reason about 0.9999... Stating that 0.9999... = 1 without exposing these new tools meant to grapple with concepts that physically cannot be grappled with is a huge mistake.
  • supercasio 6y ago
    What? 0.999... = 1 is not dogma. Please don't spread misinformation. And at least read the link before commenting on something.
    • dvt 6y ago
      Did you read what @jl2718 posted? Namely: > the supremum of an increasing sequence is equal to the limit -- this is not misinformation (and to anyone familiar with some introductory analysis, correct[1]). Of course, calling it "dogma" is a bit inflammatory, but not technically wrong. It's kind of of a made-up rule to help us work with infinities (particularly in ℝ -- but it happens all the time in set theory, as well). But to agree with GP, touting it as "intuitive" or "mind-blowing" is indeed silly. [1] http://www.math.toronto.edu/ilia/Teaching/MAT378.2010/Limits%20and%20real%20numbers.pdf http://www.math.toronto.edu/ilia/Teaching/MAT378.2010/Limits...
• fourseventy 6y ago
  The best way to think about it is that 1 - 0.999... = 0.000... The result of 1 minus 0.999... is 0.000 with zeroes that go to infinity. And I think its easier to reason that 0.000 with repeating zeroes forever is in fact equal to zero.
• novacole 6y ago
  So 99.999..% of the speed of light is just the speed of light?
  • cthalupa 6y ago
    Yes. How much faster than 99.999..% the speed of light would you need to go to get as fast as the speed of light? If your answer is 0.00..1% the speed of light, this answer is nonsensical because infinity never ends, and you never ever have the ability to add that 1 at the end. So, the only answer can be 0.00.., and 0.00.. = 0, so 99.99.. has to = 1.
    • novacole 6y ago
      Hmm, This reminds me a lot like Zenos paradox. Also, I’d like to understand how if we can say 99.999... is = 100, wouldn’t we also be able to say 99.9999888999... = 99.9999..., and therefore also just 100? And so on?
      • gspr 6y ago
        > Also, I’d like to understand how if we can say 99.999... is = 100, wouldn’t we also be able to say 99.9999888999... = 99.9999..., and therefore also just 100? And so on? You've just introduced a novel symbol into the discussion. What is the definition of 99.9999888999...? Before anyone can answer what it equals, you must define it. Recall that if the digits repeat, we already have a definition, so that case is fine. Your case is not à priori well-defined.
• grensley 6y ago
  What is the largest number smaller than 1?
  • iiv 6y ago
    0, if we're talking natural numbers or integers. There isn't one if we're talking real numbers. Simple proof: Assume x is the largest number smaller than 1. (x + 1)/2 is a number larger than x but smaller than 1. Our assumption that x is the largest number smaller than 1 must be wrong. QED.
• sleepyams 6y ago
  There is a nice characterization of decimal expansions in terms of paths on a graph: Let C be the countable product of the set with ten elements, i.e. {0, 1, 2, ..., 9}. The space C naturally has the topology of a Cantor set (compact, totally disconnected, etc). Furthermore, for example, in this space the tuples (1, 9, 9, 9, ...) and (2, 0, 0, 0, ...) are distinct elements. The space C can also be described in terms of a directed graph, where there is a single root with ten outward directed edges, and each child node then has ten outward directed edges, etc. C can be thought of as the space of infinite paths on this graph. A continuous and surjective map from C to the unit interval [0, 1] can be constructed from a measure on these paths. For any suitable measure, this map is finite-to-one, meaning at most finitely many elements of C are mapped to a single element in the interval. For example there is a map which sends (1, 9, 9, ...) and (2, 0, 0,....) to the element "0.2". The point is that all decimal expansions of elements of [0, 1] can be described like this, and we can instead think of the unit interval not as being composed of numbers _instrinsically_, but more like some kind of mathematical object that _admits_ decimal expansions. The unit interval itself can be described in other ways mathematically, and is not necessarily tied to being represented as real numbers. Hope this helps someone!
• juanmacuevas 6y ago
  # on Python (3.7.4) 1 == 0.99999999999999994448884876874217297882 # but 1 != 0.99999999999999994448884876874217297881
  • maxnoe 6y ago
    Or any other language using IEEE 754 64 bit floats
    • juanmacuevas 6y ago
      I see... Nice!
• gigatexal 6y ago
  I hate to say it but I still don't believe this, it just goes against all intuition that I have, but people much smarter than I have proven it so I take it on faith for doing things like calculus etc just my lizard brain won't let me accept something that looks like less than 1 being 1 the same way that the limit of 1/x as x goes to infinity is zero but it doesn't seem like ti should be. The number gets infinitesimally small but it's still some non-zero number -- I dunno this is probably proving my ignorance it's just what it is.
  • ben509 6y ago
    Maybe try this: do longhand division of 7 by 7 or 9 by 9 or whatever, but instead of putting 1 in the first place, start with 0. 0.99 7 | 7.00000 0 7 0 6 3 70
    • gigatexal 6y ago
      I follow that but then I could see it just continues to .9999 forever never converting to 1 it’s always that tiny fraction less than 1...
  • MrManatee 6y ago
    Don't feel too bad. The notation "0.999..." looks non-threatening, which tricks people into believing that they understand what it means. We could make "0.999... = 1" look scarier by writing it as [n ↦ 1 - 10^(-n)] = [n ↦ 1], where [n ↦ a_n] denotes the equivalence class of a Cauchy sequence of rational numbers. These statements mean the same thing, but with the scarier notation much fewer people would mistakenly believe that they understand what it says. I would expect mathematics majors to learn what 0.999... means during their undergraduate university courses. But then there's still the question of why mathematicians chose to define it that way. To really understand that, you need to be able to come up with alternative definitions and to investigate the consequences of those definitions. And for most undergraduates, it might still take a few years to build that level of mathematical maturity. For anyone who is not a math major, I certainly don't want to discourage any curiosity about this subject. Just don't be discouraged if you feel you can't fully understand what's going on. Understanding what 0.999... means and why mathematicians chose to define it that way is quite subtle.
• edanm 6y ago
  I think if you're trying to "prove" this using axioms, you've already lost. The problem isn't that you can't come up with axioms to convince people you have a proof - the problem is with people not understanding that 0.99999.... is not a number - it's one representation of an abstract entity called a number. The problem is, the maths required to actually define the concept of a number is fairly complicated, so it's hard to explain to someone why all of these axioms make sense in the first place.
• fluganator 6y ago
  Can someone help me out here with least upper bounds? Generally the proofs of .9...=1 rely on the fact there is no number that exists that can be between .9.. and 1 and therefore .9... is equal to 1. .9... is the least upper bounds of the set. My question is if .9... was removed from the set what would be the new least upper bounds. Another way of asking the question is if we define it in this context doesn't any set bounded by a real number have a least upper bounds and aren't all real numbers equal to each other? Thanks!
• deleted 6y ago
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• jdashg 6y ago
  I think this is a notation and definition problem. To me, it behaves differently in `Y = 1 / X`, which distinguishes quite strongly between `X = 1 - 0.9999` and `X = 0.9999 - 1`! If 0.9999 ought be exactly equivalent to 1.0, there ought to be no difference between `1 / (1 - 0.9999)` and `1 / (0.9999 - 1)`. To me, 0.9999 indicates a directional limit, which can't necessarily be evaluated and substituted separately from its context.
  • gspr 6y ago
    Luckily your taste doesn't factor into whether it's true or not :-)
• sv_h1b 6y ago
  0.999...=1 is true in the mathematical sense, period. However as a representation of physical world, there is a caveat. What we understand is physical world appears and behaves discretely, because at planck scale (approx. 10^-35) the distances seem to behave discretely. Although common people don't know/ understand planck scale, they do grasp this concept intuitively. What they are really saying is that in physical world there's some small interval (more precisely, about[1 - 10^-35, 1]) which can't be subdivided further, based on our current knowledge. Same thing applies to planck time (approx. 5 * 10^-43) too. So people are arguing two different things - the pure maths concept, or the real world interpretation.
• clevbrown 6y ago
  Given infinity that doesn’t make sense.