41

Does $1.0000000000\cdots 1$ (with an infinite number of $0$ in it) exist?

Mikhail Katz
  • 42,112
  • 3
  • 66
  • 131
iMath
  • 2,237
  • 5
    No, no such number exists. – Ittay Weiss Oct 18 '14 at 08:02
  • 25
    That is not the decimal representatlon of a real number. – André Nicolas Oct 18 '14 at 08:05
  • 1
    if you add 1 at the last that means there's space for one more digit implying there's something bigger than infinity – RE60K Oct 18 '14 at 08:12
  • 23
    A decimal representation of a number has digits indexed by natural numbers. Which exactly is the position of that last $1$? Is it the first after the decimal point? The second? The third? Each digit must have its position, which must be a natural number. In other words, when specifying a real number by a decimal representation, you can pick any infinite sequence of digits of your choosing, but it doesn't make any sense to add anything "after the end" of the sequence. – Dan Shved Oct 18 '14 at 08:15
  • 3
    There are contexts in which it can make sense to use expressions like $1+\frac 1{10^{\omega}}$, but they tend to reflect non-standard approaches: interesting to some people, but haven't been persuasive enough or convenient enough to enter general use because the normal way of doing things works well enough. As André Nicolas notes "this is not the decimal representation of a real number". – Mark Bennet Oct 18 '14 at 08:34
  • 1
    Refer to https://math.stackexchange.com/questions/11/is-999999999-1 – Shine Mic Oct 18 '14 at 08:56
  • 4
    @Aditya "Implying there's something bigger than infinity." There is no largest infinity, just as there is no largest finite number. So it's not clear what "bigger than infinity" even means. – David Richerby Oct 18 '14 at 09:28
  • @DavidRicherby: Maybe the cardinality of proper classes? – celtschk Oct 18 '14 at 10:07
  • @Aditya I have the same idea with you . – iMath Oct 18 '14 at 11:15
  • @DavidRicherby I think there is only one infinity, we can think it as the biggest number,so something bigger than infinity seems ridiculous and impossible, thus such number doesn’t exist . – iMath Oct 18 '14 at 11:16
  • @DanShved you mean something bigger than infinity seems ridiculous and impossible? – iMath Oct 18 '14 at 11:17
  • 2
    @celtschk: As you can find on the site, there are models of set theory where different classes can have different cardinalities. In either case the cardinal of a proper class is not an internal notion to set theory, so in some sense it doesn't exist. – Asaf Karagila Oct 18 '14 at 11:17
  • 2
    @user1485853 There are at least two infinities, as Cantor's proof that there are more real numbers than integers shows. In fact, there are infinitely many: see ordinal numbers – David Richerby Oct 18 '14 at 11:51
  • @DavidRicherby if we restrict it in Calculus ,is it right ? – iMath Oct 18 '14 at 12:08
  • 1
    @user1485853 No, I never said that. What I said is that decimal representations are by definition restricted to sequences indexed by natural numbers. It means that in a decimal representation of a number every digit is preceded only by a finite number of digits, which renders the OP's construction "illegal" (i.e. whatever it is, it is not a decimal representation, as already mentioned by André Nicolas). – Dan Shved Oct 18 '14 at 12:18
  • @DanShved I mean in Calculus :I think there is only one infinity , we can think it as the biggest number,so something bigger than infinity seems ridiculous and impossible, thus such number doesn’t exist . Is this right? – iMath Oct 18 '14 at 13:13
  • 3
    @user1485853 It's hard to find a satisfying answer, probably because your phrasing is very vague and unclear. Nevertheless, let me make two notes. First: nowadays, something being "ridiculous" is never used as an argument in a mathematical text. Or, if it is, then only as shorthand for something very precise. Second, consider this: people distinguish between sequences that converge to $\infty$, to $+\infty$, and to $-\infty$. This already looks like three distinct "infinities" to me, and one can encounter all of them in an ordinary Calculus class. – Dan Shved Oct 18 '14 at 14:23
  • @ThePortakal: The real numbers are a field, every number is divisible by any non-zero number. – Asaf Karagila Oct 18 '14 at 14:51
  • 1
    Suppose for the sake of argument there was such a number. Suppose you subtracted one from it. Would the resulting difference be different from zero? If so, you should be able to say by how much is it different -- so, how much? If not, then your number must be equal to one, because two numbers whose difference is zero are equal. – Eric Lippert Oct 18 '14 at 18:18
  • 1
    Actually I think it will be better to close that question as a duplicate of this one, rather than the other way around. – MJD Oct 18 '14 at 19:46
  • RE: "I think there is only one infinity, we can think it as the biggest number..." You might want to get a better grasp on what infinity means; there is no "biggest number." Incidentally, a better way to phrase your title would be to refer to the string of 0's as infinitely many zeroes, not an infinite number of zeroes. – J.R. Oct 19 '14 at 00:34
  • @J.R. While I fully agree with the first part of your comment, the second part ignores the notion of ordinals numbers and the fact that a sequence is often indexed by them. An infinite number of 0's simply means that the initial segment corresponds to an infinite ordinal. In this case, it is implicitly $\omega$, the least infinite ordinal. – Asaf Karagila Oct 19 '14 at 02:23
  • Well. One thing is for certain, no one can say that this question didn't receive enough attention! :-) – Asaf Karagila Oct 20 '14 at 12:00
  • @DanShved I mean there is only one positive infinity in Calculus, we can think it as the one that no number is bigger than it, so something bigger than positive infinity is impossible, thus such number doesn’t exist . Is this right? – iMath Oct 20 '14 at 14:43
  • @ J.R. can I think positive infinity as the one that no number is bigger than it ? – iMath Oct 20 '14 at 14:43
  • @DanShved a decimal representation number is a real number,right ? – iMath Oct 22 '14 at 14:20
  • Perhaps it would be more productive to analyze $1.00000\dots0001 - 1.$ – John Joy Oct 26 '14 at 20:16
  • 1
    I had this exact same question about 4 years ago. The number does not exist –  Nov 08 '15 at 03:52
  • Would that not just be $1$? – asher drummond Jun 01 '16 at 11:46
  • @JohnJoy So... an infitesimal number? – asher drummond Jun 01 '16 at 11:47
  • 1
    @asher drummond $(1.0000...0001) - 1$ would be an infinitesimal, would think. – John Joy Jun 01 '16 at 14:37

8 Answers8

74

First let me tell you that the idea that an infinite sequence "ends with something" is a solid idea. It's a perfectly natural one. The point is that the sequence is not indexed by $\Bbb N$, anymore, but rather by $\Bbb N\cup\{\infty\}$, where $\infty$ is another point, which lies after all the natural numbers.

The point is that an "infinite sequence" is a very general notion. People just often like to think about sequences which are indexed only using the natural numbers (with their natural ordering, that is). But as you will progress in your studies you might meet other objects which are indexed using other infinite sets.

And the reason people often limit themselves to sequences indexed by the natural numbers is that for the real numbers (and similar concepts), these sequences are enough. In this case, of the real numbers, we have that each real number can be defined as a limit of decimal digits, as others have explained, and therefore $1.\underbrace{000\ldots}_{\text{infinite }0\text{'s}}1$ is not a definition of a real number.

Note that this is not the limit of $1+(\frac1{10})^n$, either, which colloquially might be written as $1+(\frac1{10})^\infty$. That limit would be the limit of $1.1,1.01,1.001,\ldots$ and you can see that at no point in this sequence there is a number with infinitely many $0$'s written after it. And indeed this limit would be equal to $1$.

This is also different from the $0.999\ldots$ situation, since it is a sequence indexed by $\Bbb N$, which can be seen as the limit of its initial segments. Whereas a sequence indexed by $\Bbb N\cup\{\infty\}$ is not the limit of its initial segments, since none of them include information about the last digit.

So does it exist? Yes. It's just not a real number. It's a sequence of digits indexed by something other than $\Bbb N$.


Finally, Let me point out that as far as the concept of infinity goes in calculus, it's not quite unique. There is one infinity which signifies arbitrarily large values, another which signifies arbitrarily large negative values, there are infinities which ignore the sign at all, when you talk about a smooth function that can be differentiated infinitely many times, the infinity here is in fact "infinite sequence" rather than the infinities mentioned before, and it's a completely different type of infinity.

And there are other infinities which you might encounter, even in a calculus class.

Asaf Karagila
  • 393,674
  • thank you,it seems you know the final answer , Given the knowledge I have now,I think there is only one infinity, we can think it as the biggest number,so something bigger than infinity seems ridiculous and impossible, thus such number doesn’t exist . – iMath Oct 18 '14 at 11:25
  • 14
    You should be careful when you say that. The notion of infinity in calculus is fickle, and it is incompatible with other notions of infinity in mathematics. Notions you will run into very quickly in your studies, if you won't close your eyes to them. – Asaf Karagila Oct 18 '14 at 11:28
  • The number doesn’t exist in the filed of real numbers and calculus ,right ? – iMath Oct 18 '14 at 12:01
  • Have a look at the Riemann sphere. – mvw Oct 18 '14 at 12:07
  • 4
    @user1485853: Yes, as I said, this doesn't define a real number. It doesn't mean that it doesn't exists. There is more than just real numbers. – Asaf Karagila Oct 18 '14 at 14:26
  • It's stuff like this that I find impossible to wrap my head around.. – Ethan Bierlein Oct 18 '14 at 14:54
  • @Ethan: Why do you find it impossible? – Asaf Karagila Oct 18 '14 at 15:05
  • $1/10^\infty = 0$. So of course it defines a real number, it's just uselessly the same as if it were omitted. It's as wrong to say it doesn't define a real number as it is to say $.999\ldots$ doesn't define a real number just because that slot happens to be taken by $1.000\ldots$. The map from sequences indexed by $\mathbb{N}$ to $\mathbb{R}$ is not injective, so why would you expect the map from $\mathbb{N} \cup {\infty}$ sequences to $\mathbb{R}$ to be? – djechlin Oct 18 '14 at 15:45
  • 4
    @djechlin: $\frac1{10}^\infty$ is a limit of a sequence, and generally it's worth pointing that out; it's not an actual number. Moreover this is not the same as a number whose decimal expansion is infinitely many $0$ and then $1$. It is the limit of $0.1,0.01,0.001$ and so on. At no point there is a number with infinitely many $0$'s, let alone a number with infinitely many $0$ and then $1$ after them. – Asaf Karagila Oct 18 '14 at 16:26
  • 1
    @djechlin: Moreover, I never said anything about injectivity, or about maps (not to mention that $\Bbb{R^N}$ and $\Bbb R$ have the same cardinality, so there is a bijection between them). I just said that having infinitely many $0$ and then $1$ is a sequence of digits which is indexed by $\Bbb N\cup{\infty}$, which is a legitimate object. Despite what some people might say "You can't have an infinite sequence with $1$ at the end". That was my entire point. I repeatedly said that it's not a real number's definition. – Asaf Karagila Oct 18 '14 at 16:28
  • @djechlin: In either case, I've edited to clarify on your comment (and why it is, essentially, mistaken). – Asaf Karagila Oct 18 '14 at 17:07
  • @AsafKaragila For me, at least, the idea of numbers that large, well, I just can't picture it in my head. This may also mean I fail calculus when I get to it... – Ethan Bierlein Oct 18 '14 at 17:18
  • 7
    @Ethan: Then don't picture it in your head. The reason we have definitions in mathematics is exactly so we don't have to picture things in our heads. Work with the definitions, slowly and carefully. You'll see that you'll eventually start having better mental pictures of these objects, and that's true for anything in mathematics, not just very large real numbers. – Asaf Karagila Oct 18 '14 at 17:26
  • Is this a kind of "Surreal" or "Hyperreal" number? – SteveED Oct 18 '14 at 19:58
  • @SteveED: What is? (And probably the answer is no.) – Asaf Karagila Oct 18 '14 at 20:18
  • 5
    This answer helped me put $1$ and $.000...1$ together. – user157227 Oct 19 '14 at 00:22
  • @AsafKaragila http://en.wikipedia.org/wiki/Hyperreal_number – SteveED Oct 19 '14 at 03:00
  • @SteveED: I know what the hyperreal numbers are. You asked "Is this ...", what does "this" refer to? – Asaf Karagila Oct 19 '14 at 03:02
  • does your answer equate the number in the question with something like the hyperreals or surreals? – SteveED Oct 19 '14 at 03:09
  • @SteveED: Ah, now I understand your question! No, my answer doesn't suggest that this string of digits defines a hyperreal number or a surreal number. If it is, what happens when you multiply it by $10$? It seems that the number stays the same. This property is unique to $0$ in any field of characteristics $0$. – Asaf Karagila Oct 19 '14 at 03:12
  • @AsafKaragila
    1)can I think positive infinity as the one that no number is bigger than it ?
    2)Can I say the sequence $\frac1{2},\frac1{2}+(\frac1{2})^2,\frac1{2}+(\frac1{2})^2+(\frac1{2})^3,\frac1{2}+(\frac1{2})^2+(\frac1{2})^3+...,$ ends with $1$ ?
    3)the reason why $1.\underbrace{000\ldots}_{\text{infinite }0\text{'s}}1$ cannot be indexed by the natural numbers is it ends with 1,right ?
    – iMath Oct 20 '14 at 14:44
  • @user1485853: (1) Yes, essentially yes. (2) No, the limit of the sequence is $1$, it doesn't end with $1$. (3) Yes, that is correct. The natural numbers do not have a last element (there is no largest natural number) so a sequence indexed by them cannot have a last element. – Asaf Karagila Oct 20 '14 at 16:55
  • @AsafKaragila you said an infinite sequence "ends with something" is a solid idea. It's a perfectly natural one. Then for question (2) here, does the sequence end with something or it has no end ? – iMath Oct 22 '14 at 13:48
  • 3
    @iMath: No, the sequence you specified in (2) is without a last element. It is sometimes convenient to think about the limit as the last element, and you can find many questions on this site which show that this can be a great source for confusion for people (amateurs, students and even professionals sometimes). But the limit of a sequence doesn't have to be an element of the sequence, that's one of the good things about infinite sequence and convergence, it allows us to talk about objects we don't "have" by talking about things which are more and more similar to them. – Asaf Karagila Oct 22 '14 at 14:06
26

The bad news:

That string of symbols $$ 1.0\cdots01 $$ has no meaning as a real number. Meaning or rather semantics would be a valid mapping from an infinite string $s$, e.g. from $\Sigma^\omega$ (link), to $\mathbb{R}$.

The string representation of a floating point real number in base $10$ by convention means e.g. $$ (1.0\cdots)_{10} = (\underbrace{d_0.d_1 d_2 d_3 \cdots }_{\mbox{string}})_{10} = \underbrace{\sum_{k=0}^\infty d_k 10^{-k}}_{\mbox{real number}} $$ Where would you like to add that last $1$ digit?

The good news:

It might be something else.

Of course you can define your own semantics and operations on the strings but you will most likely end up with something that behaves more or less differently from the real numbers (like finite IEEE floats and their operations are not the same as the full set of real numbers and the basic operations on it, but something very close).

And the ugly news: $\tiny \mbox{(c) by Asaf Karagila}$

As fellow user Hyrkel noticed, that finite string I used $$ [1] [.] [0] [.] [.] [.] [0] [1] $$ to suggest an infinite string with two ends (prefix "$1.$" and suffix "$1$") and infinite many $0$ symbols in between is problematic too - interpreted as a string already.

Is this a proper string? is it even a proper infinite string? How would you be able to recognise it?

Next step would be to attach a meaning to it, preferably some number, but I will only argue on the above questions which are not simple already.

Computer scientists stick to mathematical machine models, like the finite automaton or the Büchi automaton to reason about strings. These machines can either accept or reject a string they are presented with. Their recognition process resembles the process of a sensor reading a linear tape or track from left to right. Even the variants for infinite strings act like this.

The infinity here is not so much problematic because of it's sheer size but rather because of it's dullness: what reason should the automaton have to stop the recognition of the infinite part and proceed with the finite suffix? The recognizable infinite strings seem to be of the variant one end finite, one end infinite. (Do not nail me on this)

I am not sure if a non-deterministic (multiple choices possible) Büchi automaton that would accept the string $1.0^\omega$ could be properly extended to recognise $1.0^\omega 1$.

I would attempt it by adding another arc from the final state to itself which is accepting the final $1$ symbol. That would work to accept $1.0^\omega1$ but it would also still accept just $1.0^\omega$. That makes it not much useful, what I can not distinguish is practically the same.

The solution is probably another automaton that starts recognition simultaneously from both ends or some mapping which lists the infinite sequence alternating from both sides at once, something like $$ (1 1) \, (. 0) \, (0 0) \cdots $$ this would resort to established structures but I am not aware of such approaches.

mvw
  • 34,562
  • this is the right answer! :) – Ant Oct 18 '14 at 08:34
  • 1
    It might be more to the point to note that $1.0\dots 01$ doesn't even have an interpretation as a string comprised of digits and a decimal point; the problem comes even before you can start wondering about numbers! –  Oct 18 '14 at 08:50
  • 2
    Actually, as a string from $\Sigma^{\omega+1}$. – Asaf Karagila Oct 18 '14 at 08:52
  • Funny, but Hurkyl is right I actually cheated by writing the string in that way. It is the same problem. I need to revisit automata on infinite words but I guess I can not model this. – mvw Oct 18 '14 at 08:58
  • Of course, one could define pairs of "strings" $(A,B)$ where $A$ is of the type $"a_1,a_2,\dots"$ and $B$ is of the type $"\dots, b_2,b_1"$. But... – Lehs Oct 18 '14 at 09:33
  • 7
    Do you also have some ugly news? :-) – Asaf Karagila Oct 18 '14 at 09:50
  • 2
    Tudu dudu du, duuuduuuuu.. – mvw Oct 18 '14 at 10:56
  • 1
    In what sense are IEEE floats "not real numbers, only something very close"? They're real numbers, just not the entire infinite range of e.g., 0.0 - 1.0, and the result of a calculation is always predictable using only real numbers. – goldilocks Oct 18 '14 at 11:28
  • @AsafKaragila: No, there string mentioned in this answer, and the comment, is (at least) in $\Sigma^{\omega+2}$: there is a digit $0$ just before the $1$. But that digit is absent in the question, so this answer is not really talking about the same string as the question. – Marc van Leeuwen Oct 18 '14 at 11:59
  • Their representation has to work with limited memory space so there are reals with no exact representation and operations with deviations from the exact result. – mvw Oct 18 '14 at 12:04
  • The number doesn’t exist in the filed of real numbers and calculus , the number exist in the field of surreal numbers , right ? – iMath Oct 18 '14 at 12:04
  • 1
    @mvw That there are real numbers without representation does not mean that IEEE floats represent anything other than real numbers. You are looking at it backward. All IEEE 754 floating point values are real numbers, but not all real numbers have an exact representation. What you've said is the equivalent of saying standard C ints "are not whole numbers" because the entire (infinite) range of such are not representable that way, either. – goldilocks Oct 18 '14 at 14:04
  • I look not backwards but at the whole field with it's properties of closedness and exact arithmetics. I used the word "behave" on purpose, with that operational aspect in mind. – mvw Oct 18 '14 at 14:11
  • @Marc: You are right. Maybe other ordinals, we don't know how many digits are transfinite! Just that at least two of them are! – Asaf Karagila Oct 18 '14 at 14:19
  • 1
    @goldilocks: Actually no. In addition to a useful inventory of honest real numbers, IEEE 754 floating-point values can also represent "positive infinity", "negative infinity" and a variety of "not-a-number"s, neither of which are real numbers. – hmakholm left over Monica Oct 18 '14 at 14:37
  • 1
    I fail to see the connection of automaton with the problem at hand.. – nicolas Oct 18 '14 at 14:49
  • I can put some funny symbols on paper or screen which is kind of pointless if I can not define what it means and how to deal with it. An automaton is a model to define a set of strings. That ability to distinguish in set or not is a first useful thing one could do and is already not that easy. Continuing with arithmetics is probably harder. Here it was an example why the handling of that notation is already troubling at interpreting it as string (job for automata) not only for interpreting it as some kind of number. – mvw Oct 18 '14 at 14:56
  • @HenningMakholm Interesting point. I still think mvw's analogy is a frivolous one though -- even including "the whole field", part of the automation here is in the operators (+, *, etc.) which have special rules attached to them dictating what happens when. This is all determinable using normal arithmetic and real numbers (excluding NaN, etc.), just as ints wrapping around is determinable with normal arithmetic and whole numbers. – goldilocks Oct 18 '14 at 15:30
  • You don't see my point: it is designed to be as good as real numbers in all their glory would be, but some aspects are compromises. I am perfectly aware that $1$ is represented exactly and $10^{42}$ as well, but already $1/3$ will loose precision. It is not to make the floats look bad, just to state you get something different, while still familiar. – mvw Oct 18 '14 at 15:49
  • +1 For getting my point (the loss of precision is predictable with real number arithmetic) and editing the previous reading whereby such floats involved things that "are no real numbers only something very close" ;) I enjoyed your answer. – goldilocks Oct 18 '14 at 18:38
  • Thanks. I try to learn here and am happy for any opinion. – mvw Oct 18 '14 at 18:47
13

In real analysis, $$\lim_{n\rightarrow\infty}\left( 1+\frac{1}{10^n}\right )=1$$ Even if it does not mean anything to say "infinite number of zeros" in real analysis, we can suppose this number is equal to $1$.

But in the field of surreal numbers, it's not the same. This number exists and will be equals to $1+\frac{1}{\omega}$, if you consider the infinity number of zeros to be $\omega$.

daOnlyBG
  • 2,711
Xoff
  • 10,310
  • 1
    Are decimal expansions of general surreal numbers really well-defined? There's the sign-expansion as an ordinal-length sequence of plus and minus signs, which is similar to, but not quite the same thing as, a binary expansion. But decimal? – Hans Lundmark Oct 18 '14 at 08:38
  • @HansLundmark That's a good point. You can see usual expansion as a binary one, and use the usual modulo algorithm to get the decimal one, using the fact that $2^{-\omega}=10^{-\omega}=\frac{1}{\omega}$... But I admit this must be defined more accurately ! – Xoff Oct 18 '14 at 08:42
  • How are $2^{-\omega}$ and $10^{-\omega}$ defined? I'm not an expert on surreal numbers, but isn't there an exponential function $x \mapsto \exp(x)$ which is strictly increasing (and nontrivial to define)? If we take $2^x=\exp(x \ln 2)$ and $10^x=\exp(x \ln 10)$, then it would seem that $10^x<2^x$ for $x<0$. – Hans Lundmark Oct 18 '14 at 08:54
  • Actually, you don't need $10^{-\omega}$ to be well defined, since that doesn't occur in the answer. What you need is a well defined limit. – celtschk Oct 18 '14 at 09:52
  • I use $2^{-\omega}$ as a shorthand for the $\omega^{th}$ digit. This is not defined from the exponential function, even if it coincides for the finite positions... – Xoff Oct 18 '14 at 10:04
  • The number doesn’t exist in the filed of real numbers and calculus , the number exist in the field of surreal numbers , right ? – iMath Oct 18 '14 at 12:13
  • thanks for mentioning non standard analysis. that is what physicists are using everyday... – nicolas Oct 18 '14 at 14:46
  • @user1485853. Yes, surreals are an extension (with ordinals) of reals. – Xoff Oct 18 '14 at 16:01
  • Maybe it's worth noting that the sequence of infinitely many $0$'s followed by a $1$ is not the limit of $\frac1{10^n}$. – Asaf Karagila Oct 18 '14 at 18:57
  • @AsafKaragila When you talk about limit, you talk about topology. So it depends of your topology. Of course, on reals, the limit is 1, and there is no limit on surreals (except for constant sequences). So yes, this is not the limit !! :) – Xoff Oct 18 '14 at 19:20
  • Xoff, then you and I work with very different limits when concerning the real numbers (I'm not even talking about the surreal numbers here). How did the $1$ jump all the way to the $\omega$-th coordinate? The limit of a sequence, especially one where longer and longer initial segments are fixed, is the eventual stabilization at each digit. None of the $\frac1{10^n}$ have a digit (let alone $1$) on the $\omega$-th place. None whatsoever! So where did it get to the limit? No, the limit is the sequence $0.000\ldots$ – Asaf Karagila Oct 18 '14 at 19:27
  • @AsafKaragila Sorry I was talking about the initial sequence of the OP, that is $1+\frac{1}{10^n}$. – Xoff Oct 18 '14 at 19:33
  • And I simply noted that $1.000\ldots 1$ is not the limit of $1+\frac1{10^n}. – Asaf Karagila Oct 18 '14 at 19:38
  • @AsafKaragila It's not horrendous to say that $\lim_{n\in\mathbb N\rightarrow\infty}n=\omega$... – Xoff Oct 18 '14 at 19:50
  • Xoff, it is, depending on the context. In the context of analysis? VERY horrible. In the context of set theory, this is even correct. But here's the thing, the order type of a sequence which is the limit of finite sequences is $\omega$. Not $\omega+1$. – Asaf Karagila Oct 18 '14 at 20:06
  • @AsafKaragila I do agree with you ! – Xoff Oct 18 '14 at 20:40
  • Xoff, your answer is neither here nor there because the OP was not talking about the real numbers. – Mikhail Katz May 31 '16 at 17:30
  • Decimal expansions of surreal numbers are defined using the Conway normal form of base 10, generalizing Cantor's normal form. – Mike Battaglia Mar 20 '22 at 00:53
9

If a sequence $1.00000\dots $ is infinite it can't have an end $\cdots 0001$. Infinite means endless.


A finite sequence is a line of (mathematical) objects $a_0,a_1,a_2,\dots, a_n$. But it seems to be some disagreement what an infinite sequence is. At least I disagree.

Obviously, the object $a_k$ represent a function $k\mapsto a_k$ with ordered indices $k$, but could it be any function? Due to Wikipedia:

Most precisely, a sequence can be defined as a function whose domain is a countable totally ordered set, such as the natural numbers.

In that case a function $\mathbb N\cup\{\infty\}\rightarrow A$ (for some set $A$) is a sequence $a_0,a_1,a_2,\dots$ with a last element $a_\infty$, without an immediately preceding element in the sequence.

In my intuition and in my opinion any element in a sequence, except the first, has an immediately preceding element.

Latin: sequentia (“a following”).

However, it's possible to generalize to "bi-sequences" $(S_1,S_2)$ when $S_1=(1,0,0,\dots)$ is the initial sequence and $S_2=(\dots,0,0,1)$ is the termimal sequence, and define an arithmetic for "numbers" defined by bi-sequences as $(S_1,S_2)$.

Lehs
  • 13,791
  • 4
  • 25
  • 77
  • 10
    Because $\omega+1$ is not a thing. – Asaf Karagila Oct 18 '14 at 08:32
  • 1
    @Asaf I just focused on the language: an infinite sequence is an endless sequence and an endless sequence can't end at 1. – Lehs Oct 18 '14 at 08:38
  • 1
    I guess that $\omega+1$ is just not one of them "infinite ordinals", then, because it has an end. – Asaf Karagila Oct 18 '14 at 09:00
  • I just don't understand your objections(?) @Asaf. I haven't written anything about ordinals, I have just noted that endless "sequences" can't have an end. In general language! – Lehs Oct 18 '14 at 09:07
  • Yes, but the point is that we are doing mathematics, so there's a good deal of non-natural language involved. – Asaf Karagila Oct 18 '14 at 09:11
  • 2
    @Asaf: But OP wasn't really doing mathematics and I ventured to respond in the same style. – Lehs Oct 18 '14 at 09:13
  • @AsafKaragila what does ω+1 mean ? – iMath Oct 18 '14 at 09:33
  • @user1485853: It denotes a linear order which looks like the natural numbers, with an additional point at the end. It's an infinite order, which has a last element. – Asaf Karagila Oct 18 '14 at 09:49
  • 3
    @Lehs: I'm confused. Is the banner on the top of the site saying something other than "MATHEMATICS" for you? – Asaf Karagila Oct 18 '14 at 09:49
  • @AsafKaragila: The banner at the top of the site contains, besides a set of images, "StackExchange" and some numbers, as well as the words "review", "help" and "search", but not the word "Mathematics". ;-) – celtschk Oct 18 '14 at 09:57
  • Well, @Asaf Karagila, mathematics has it's roots in human language and human thinking (as logic also has) and I made a mathematically related reasoning. What's your problem? I normally really appreciate your knowledge, but... – Lehs Oct 18 '14 at 09:58
  • 1
    @celtschk: The topbar is not the banner. :-) – Asaf Karagila Oct 18 '14 at 10:06
  • 1
    @Lehs: I don't really have a problem, and I'm sorry if I come off as being aggressive. It's not my intention, and I apologize if that is the case. My point is, however, that despite the natural language roots, we wish to remove ambiguity. And while "infinite sequence" is not without ambiguity, it still doesn't mean "endless sequence". And since this is asked on a mathematics website, it deserves to be treated as a mathematical question. – Asaf Karagila Oct 18 '14 at 10:08
  • I have the same idea with you . I think there is only one infinity in Calculus , we can think it as the biggest number,so something bigger than infinity seems ridiculous and impossible, thus such number doesn’t exist . – iMath Oct 18 '14 at 12:09
  • 2
    @user1485853: actually, in Calculus, infinity is never used on its own (and is not s number!), it is always part of an expression like "$x$ approaches $\infty$"... – Taladris Oct 18 '14 at 13:37
  • 2
    @Taladris Unless his elementary calculus class uses the extended reals (but who would do that to their students?!) :) –  Oct 18 '14 at 14:11
  • @Taladris can I think positive infinity as the one that no number is bigger than it ? – iMath Oct 20 '14 at 14:47
  • @Bye_World: +1. I was thinking of this when writing my comment. There are excellent reasons to use the extended real line: existence of limsup/liminf, Monotonic Sequence Theorem,... but this is more an analysis concept (I admit that the boundary between calculus and analysis is quite fuzzy). In particular, the objection I would have against using the extended real line in a Calculus class is that it does not unify the usual $\epsilon-\delta$ definitions of limits. – Taladris Oct 20 '14 at 16:40
  • @user1485853: you have to define $\infty$ first. In particular, in which set does it belong? It is not a real number (otherwise, what would be the result of $0\cdot\infty$?), so you have to consider the extended real line $\overline{\mathbb R}=\mathbb R\cup{\infty,-\infty}$ and extend the usual order relation $<$ of $\mathbb R$ to this new set: $-\infty<x<\infty$ for any $x$ in $\mathbb R$. But you have to keep in mind this is an arbitrary construction. Being arbatrary is not a problem per se, and the extended real line is actually a useful concept. But if you start to define $\infty$,(..) – Taladris Oct 20 '14 at 16:45
  • (cont) then nothing prevent you to defined a "super infinite" $\alpha$ that is greater than $\infty$. The problem with $1.0000\dots1$ is that it is not defined as a real number: if ${d_n}{n\in\mathbb N}$, we know the meaning of $0.d_1d_2\dots d_n\dots$; it is $\sum{j=1}^\infty\frac{d_j}{10^j}$. But the decimals of $1.000\dots 1$ would be indexed by $\mathbb N\cup{\infty}$. What is the meaning of $0.d_1d_2\dots d_n\dots d_{\infty}$ in this context? This has no widely agreed meaning as a real number. (As someone pointed out, you can give it a meaning in the context of surreal numbers). – Taladris Oct 20 '14 at 16:52
  • @Taladris thanks very much ! – iMath Oct 22 '14 at 13:48
  • Lehs, your error is assuming that the number is real. – Mikhail Katz May 31 '16 at 17:29
  • @MikhailKatz. Where did I wrote that the "number" was real? You can define "numbers" like 1.0000...00001 but that takes two sequences: a left part running to the right and a right part running to the left. – Lehs May 31 '16 at 17:52
  • What I contest is the literal intepretation of your claim that "infinite" means "endless". On the contrary, Leibniz spoke about terminaing infinities and in modern mathematics it is possible to formalize that idea, as I outlined in my answer. – Mikhail Katz Jun 01 '16 at 09:09
  • @MikhailKatz: I'm confident with the idea of formalizing mathematical concepts, but an infinite sequence will always be an endless sequence. As I wrote before one could define bi-sequences with the desired properties. – Lehs Jun 01 '16 at 10:12
  • A bi-sequence is not enough actually. To make this work you will have more complicated behavior in the "middle". – Mikhail Katz Jun 01 '16 at 12:00
  • @MikhailKatz: can you elaborate on that or leave a reference? – Lehs Jun 01 '16 at 12:03
  • The best reference for the extended decimal notation is the article by Lightstone in the Monthly mentioned in my answer. – Mikhail Katz Jun 01 '16 at 13:29
6

In the hyperreal number system which is an extension of the real number system you have infinite integers and the corresponding extended decimal expansions where it is meaningful to talk about digits at infinite rank (more precisely, rank defined by an infinite integer). In this system your decimal makes sense.

Extended decimals were discussed in detail in an article by Lightstone:

Lightstone, A. H. Infinitesimals. Amer. Math. Monthly 79 (1972), 242–251.

Mikhail Katz
  • 42,112
  • 3
  • 66
  • 131
6

In the real number system $\mathbb R$, there is no such number.

P.S. There is no such number in the complex numbers $\mathbb C$, either.

P.P.S. Sorry.

3

As @Lehs said, infinite means endless - every integer value $n$ is finite, so if you assume infinite sequence of 'zero' digits, then whatever number $n$ you think, the $n$-th position holds digit 0. Consequently there's no space where you can append 'one'.

CiaPan
  • 13,049
  • 1
    Yes, we are right! :) Even in the OP case, which is about a decimal string "1.000000..." If it is endless, it cant end with "..0001". – Lehs Oct 18 '14 at 09:28
  • 4
    You need to be very careful with this kind of argument. You could just as well argue (a la Zeno) that the number 1 doesn't exist because there is an infinite sequence of numbers $\frac12,\frac34,\frac78,\frac{15}{16}, \dots$ that come before it and nothing can "come after" an infinite sequence. – David Richerby Oct 18 '14 at 09:33
  • I have the same idea with you . I think there is only one infinity in Calculus , we can think it as the biggest number,so something bigger than infinity seems ridiculous and impossible, thus such number doesn’t exist . – iMath Oct 18 '14 at 12:10
  • 2
    @DavidRicherby Nonsense, Zeno confused the infinity in the sequence length with the set of values. We don't. – CiaPan Oct 18 '14 at 21:51
  • There is indeed such a space unless you assume the number is real, which the OP is not assuming. – Mikhail Katz May 31 '16 at 17:28
  • 1
    @MikhailKatz They also do not assume they might use anything unreal or hyperreal... – CiaPan Jun 01 '16 at 06:38
  • @CiaPan, I think you are confusing two meanings of the word "real": the adjective with the generic meaning of "something that truly exists", on the one hand, and the technical meaning of the entity obtained as an equivalence class of Cauchy sequences of rational numbers, on the other. – Mikhail Katz Jun 01 '16 at 07:14
  • 1
    @MikhailKatz Could be interesting to learn which of the two meanings you had in mind in your comment above.... – CiaPan Oct 29 '20 at 21:47
-5

Looking at the answers seems to exhibit a rather chaotic mixup of concepts.

Of course, the question is not uninvolved since it talks about whether a particular number representation "exists". Obviously, in its informal manner, it exists or could not have been written down. Since we are talking about calculus here, the next question would be whether some real number value can be associated with that loose description. It would appear that it would mean something akin to

$$\lim_{n\to\infty} \left(1 + 10^{-n}\right)$$

It turns out that this number perfectly well exists and is the same as the real number $1$. So far, so straightforward.

The really interesting thing is that the majority of answers diverge from this. If you asked whether a number like $2-1$ exists, few people would deny its existence on the basis that its specification looks different from $+1$. Or would refuse to acknowledge the existence of octal $10_8$ because it would be the same as $8_{10}$.

So given that the majority of answers look different, it would appear that the question is not as much about calculus as it is about number psychology. Is this the same as numerology?

Next question.

user26486
  • 11,331
  • 1
    Note that you interpreted the OP's 1.0000000000……1 as – mvw Oct 18 '14 at 18:20
  • 1
    @mvw has hit the nail on the head. You've assumed a definition in your answer that may not necessarily be valid. (And numerology is totally different, btw.) – apnorton Oct 18 '14 at 18:27
  • You made your decision to interpret the given string. it is necessarily non-standard. In your case you added zero in a complicated way. A suitable interpretation should give reasons what it is and why to introduce it next to $1$. – mvw Oct 18 '14 at 18:32