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

  • 5
    $\begingroup$ No, no such number exists. $\endgroup$ – Ittay Weiss Oct 18 '14 at 8:02
  • 24
    $\begingroup$ That is not the decimal representatlon of a real number. $\endgroup$ – André Nicolas Oct 18 '14 at 8:05
  • 22
    $\begingroup$ 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. $\endgroup$ – Dan Shved Oct 18 '14 at 8:15
  • 3
    $\begingroup$ 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". $\endgroup$ – Mark Bennet Oct 18 '14 at 8:34
  • 4
    $\begingroup$ @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. $\endgroup$ – David Richerby Oct 18 '14 at 9:28

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.

  • 14
    $\begingroup$ 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. $\endgroup$ – Asaf Karagila Oct 18 '14 at 11:28
  • 4
    $\begingroup$ @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. $\endgroup$ – Asaf Karagila Oct 18 '14 at 14:26
  • 3
    $\begingroup$ @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. $\endgroup$ – Asaf Karagila Oct 18 '14 at 16:26
  • 6
    $\begingroup$ @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. $\endgroup$ – Asaf Karagila Oct 18 '14 at 17:26
  • 4
    $\begingroup$ This answer helped me put $1$ and $.000...1$ together. $\endgroup$ – user157227 Oct 19 '14 at 0:22

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.

  • 1
    $\begingroup$ 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! $\endgroup$ – user14972 Oct 18 '14 at 8:50
  • 2
    $\begingroup$ Actually, as a string from $\Sigma^{\omega+1}$. $\endgroup$ – Asaf Karagila Oct 18 '14 at 8:52
  • 6
    $\begingroup$ Do you also have some ugly news? :-) $\endgroup$ – Asaf Karagila Oct 18 '14 at 9:50
  • 2
    $\begingroup$ Tudu dudu du, duuuduuuuu.. $\endgroup$ – mvw Oct 18 '14 at 10:56
  • 1
    $\begingroup$ 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. $\endgroup$ – goldilocks Oct 18 '14 at 11:28

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 $\omega$.

  • 1
    $\begingroup$ 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? $\endgroup$ – Hans Lundmark Oct 18 '14 at 8:38
  • $\begingroup$ @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 ! $\endgroup$ – Xoff Oct 18 '14 at 8:42
  • $\begingroup$ 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 $x<0$. $\endgroup$ – Hans Lundmark Oct 18 '14 at 8:54
  • $\begingroup$ 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. $\endgroup$ – celtschk Oct 18 '14 at 9:52
  • $\begingroup$ 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... $\endgroup$ – Xoff Oct 18 '14 at 10:04

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 $(S_1,S_2)$.

  • 9
    $\begingroup$ Because $\omega+1$ is not a thing. $\endgroup$ – Asaf Karagila Oct 18 '14 at 8:32
  • 1
    $\begingroup$ @Asaf I just focused on the language: an infinite sequence is an endless sequence and an endless sequence can't end at 1. $\endgroup$ – Lehs Oct 18 '14 at 8:38
  • 1
    $\begingroup$ I guess that $\omega+1$ is just not one of them "infinite ordinals", then, because it has an end. $\endgroup$ – Asaf Karagila Oct 18 '14 at 9:00
  • 2
    $\begingroup$ @Asaf: But OP wasn't really doing mathematics and I ventured to respond in the same style. $\endgroup$ – Lehs Oct 18 '14 at 9:13
  • 3
    $\begingroup$ @Lehs: I'm confused. Is the banner on the top of the site saying something other than "MATHEMATICS" for you? $\endgroup$ – Asaf Karagila Oct 18 '14 at 9:49

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.

  • $\begingroup$ Did the OP say anything about "real"? $\endgroup$ – Mikhail Katz May 31 '16 at 17:28

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.


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'.

  • $\begingroup$ 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". $\endgroup$ – Lehs Oct 18 '14 at 9:28
  • 4
    $\begingroup$ 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. $\endgroup$ – David Richerby Oct 18 '14 at 9:33
  • $\begingroup$ 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 . $\endgroup$ – iMath Oct 18 '14 at 12:10
  • 1
    $\begingroup$ @DavidRicherby Nonsense, Zeno confused the infinity in the sequence length with the set of values. We don't. $\endgroup$ – CiaPan Oct 18 '14 at 21:51
  • $\begingroup$ There is indeed such a space unless you assume the number is real, which the OP is not assuming. $\endgroup$ – Mikhail Katz May 31 '16 at 17:28

Yes, if the infinite quantity of zeros are placed both before the decimal point and before the leading $1$. Those six adjacent dots appear to me to just be small zeros filled in. I'd like to use the notation that ... placed before a sequence means that it repeats indefinitely in the left direction. So then we would have:

...0$1.00000000000000001$ with the bold stuff being fixed in position.


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.

  • 1
    $\begingroup$ Note that you interpreted the OP's 1.0000000000……1 as $\endgroup$ – mvw Oct 18 '14 at 18:20
  • 1
    $\begingroup$ @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.) $\endgroup$ – apnorton Oct 18 '14 at 18:27
  • $\begingroup$ 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 $1$. $\endgroup$ – mvw Oct 18 '14 at 18:32

Not the answer you're looking for? Browse other questions tagged or ask your own question.