This might be a silly question, but is it possible at all for n.00000...[infinite zeros]...1 to be the next real number after n? If not, why not?

Firstly, I know (I think) that $$\lim_{x\to \infty} \frac{1}{10^x} = 0$$ but I'm not talking about taking its limit. Surely I'm not required to.

The obvious rebuttal is that n.00000...[infinite zeros]...1 (let's just call this number c) is that c divided by two will be between n and c, which is obviously a closer number to n than c. But why is c necessarily divisible? There are special exceptions for other numbers. For instance, if b=0 in $\frac ab$, we say it's undefined. Why can't $\frac c2$ be similarly undefined? Or something similar to the notion that infinity divided by two is still infinity?

OR, does c equal n.0, similar to how 0.999... = 1?

Apologies if this question has been asked a million times before (I was not able to find it asked quite this way) or if you find it stupid.

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    $\begingroup$ The obvious problem is that something like $1.\overline{0}1$ is not a valid description of a real number in decimal form (at least not in the ordinary context). $\endgroup$ – rschwieb Feb 19 '14 at 20:37
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    $\begingroup$ Oh really? Why not? For the same reasons people mention below...? $\endgroup$ – user124384 Feb 19 '14 at 20:58
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    $\begingroup$ Probably for those reasons and the very detailed reasons I already gave in the solution. If you think that what you described is a "legal" decimal expansion, then I need to encourage you to reexamine how you view and how you define decimal expansions (because your conception appears to be fallacious, at least in the context of ordinary real numbers). $\endgroup$ – rschwieb Feb 19 '14 at 21:03
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    $\begingroup$ Hmm, that's a good question to which I can't immediately offer a suggestion! I'm currently looking to see if this question was previously asked on math.SE. Hopefully someone makes a good suggestion. The wiki page might also get you started. $\endgroup$ – rschwieb Feb 19 '14 at 21:13
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    $\begingroup$ You might enjoy the kaufman decimals $\endgroup$ – Manishearth Feb 20 '14 at 2:09

10 Answers 10


The set of real numbers are (usually) defined in a way that has nothing to do with decimal representations -- they are defined by their arithmetic and geometric properties. e.g. among other things, if $a$ and $b$ are distinct real numbers, then $(a+b)/2$ is a real number that is between them.

The set of decimals are defined as being sequences of digits: there is one place for every integer. e.g. $0$ corresponds to the one's place, $1$ corresponds to the ten's place, $2$ to the hundred's place, $-1$ to the tenth's place, and so forth. Each place gets a single digit (0 through 9) assigned to it. When we write a decimal like


we implicitly mean that all of the remaining positions get filled with zeroes. i.e. in the above numeral, the thousands place contains a zero.

The key point is that each place corresponds to an integer: there aren't any other places. If we write $n.\overline{0}$, meaning that the $0$ to the right of the decimal place should be repeated infinitely, this means that we have written a $0$ in every place corresponding to a negative integer. There aren't any places remaining to the right of the decimal place to insert a $1$! So the notation $n.\overline{0}1$ makes no sense if we try to interpret it as expressing a decimal number.

We could define other sorts of radix notation that extend decimals to have additional places to the right of the decimal place, but then we have to figure out what to do with such things.

The ordinary decimals are useful because we have a way to interpret any decimal number that only has finitely many nonzero digits to the left of the decimal place as a real number. And we also have rules for doing arithmetic with them. There are some ambiguities -- e.g. does $1.\overline{0} + 0.\overline{9}$ add up to $1.\overline{9}$ because there are no carries? Or does it add up to $2.\overline{0}$ because there is a carry in every place to the right of the decimal point? -- but these ambiguities are okay because we are interpreting both possibilities as being the same real number.

But if we extend the decimals, we no longer have the ability to relate them to real numbers. And if we want to do arithmetic with such things, we're going to have to do a lot of work to define the arithmetic operations and figure out if they have any of the familiar algebraic properties we're used to and so forth.

We can construct algebras in this way in which every number has a "next" number, but such things are going to have very little to do with real numbers.

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    $\begingroup$ Aha, very interesting. Is it just me or would it be fun to determine what the properties of numbers with "additional places to the right of the decimal place" might be? At any rate, I think this sufficiently answers my question. Thank you! $\endgroup$ – user124384 Feb 19 '14 at 21:37
  • $\begingroup$ It's not just you! One possible way to do this (quite naturally) is by constructing the "surreal numbers": en.wikipedia.org/wiki/Surreal_number. The clean cut definitions are a bit technical though, so you might want to look at some of the references at the end of the article. $\endgroup$ – cody Feb 20 '14 at 1:17
  • $\begingroup$ Yay, thank goodness. Another commenter mentioned those too, and yes, I'm going to have to check out the references. They're really mind-boggling. Loving it. $\endgroup$ – user124384 Feb 20 '14 at 1:35
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    $\begingroup$ (The surreal numbers don't have next numbers either) $\endgroup$ – user14972 Feb 20 '14 at 9:29

The "first one after"

To say that there is a "first real number after an integer" implicitly says that they are distinct numbers and that there aren't any other real numbers between the number and the integer, but of course, the midpoint of these two (distinct) numbers lies between the two numbers.

On decimals

I'd like to address your conception of a decimal expansion of a real number.

Think of the ordinary decimal representation of a real number this way.

Aside from the whole part above the decimal, there are infinitely many slots below the decimal where digits can go. There are not too many slots open: they are each labeled by the natural number in $\Bbb N$ so that we can talk about them being "the second decimal place, the third, the fourth... " etc.

So you are free to do two things:

  1. look at slot $n$ and see what the digit occupies it

  2. look at a digit and see which slot $n$ it occupies.

There is no slot for "after infinitely many $0$'s" The $1$ you insert falls at a specific decimal place, for which you can see the natural number labeling it.

So by definition, every digit in the decimal expression has only finite many digits between it and the decimal point. There is simply no place to put a digit such that infinitely many digits will fall between it and the decimal point.

  • $\begingroup$ I like this way of looking at it. Another way to phrase this is that a decimal representation of $x\in[0,1]$ is given by a map $\varphi:\mathbb N\to\{0,1,\dots,9\}, n\mapsto a_n$, where $a_n$ is the $n$-th digit of $x=0.a_1 a_2 a_3 \dots$. For which $n\in\mathbb N$ should we have $\varphi(n)=1$ in the case of "infinitely many zeros" between the decimal point and $1$? $\endgroup$ – Christoph Feb 19 '14 at 21:04
  • $\begingroup$ I understand, and this is very clearly explained, but I just can't agree. Oh well, guess I just have to live with it. :\ $\endgroup$ – user124384 Feb 19 '14 at 21:05
  • $\begingroup$ Dear @user124384 : What do you disagree with, I wonder? If you're speaking of the feeling of disorientation between ideas that appear to contradict each other, then you're probably experiencing cognitive dissonance. That's a very normal thing, and nothing to be worried about. If you're patient and think carefully, then things will make sense again. The real numbers are by no means a trivial mathematical idea. $\endgroup$ – rschwieb Feb 19 '14 at 21:08
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    $\begingroup$ What does this mean? Again: If you use non-convential terms (like "a number is unbounded"), define them. It's your concept to begin with, we have pointed out the flaws in it, now you want us to define your terms for you? $\endgroup$ – Christoph Feb 19 '14 at 21:25
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    $\begingroup$ @rschwieb Yes, that's exactly what I'm asking. I think the idea that this basically violates the "rules" of real numbers (and their derivation) is what's enabling me to accept this. Thanks for your help! $\endgroup$ – user124384 Feb 19 '14 at 21:40

The sequence $(0.1, 0.01, 0.001, 0.0001, \dots)$ converges to $0$ as you said. There is no other useful definition of what $0.\overline 0 1$ could mean except from this limit. That's why we won't even write any number behind a period in a decimal expansion, because it amounts to nothing after taking the limit that the period implies. You also noticed correctly that this is related to the limit of $(0.9, 0.99, 0.999, 0.9999, \dots)$ being $1$, so $0.\overline 9 =1 $.

  • $\begingroup$ Yeah, I guess it's not useful other than taking its limit, but does it not still exist (and serve as the first real number after 0)? $\endgroup$ – user124384 Feb 19 '14 at 20:50
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    $\begingroup$ What does $0.\overline 0 1$ mean, other than the limit I suggested? If you use this notation, you need to define what it is. Otherwise you could as well ask why don't we invent a number $\widetilde 0$ which servers as the "next number after $0$", but as others have pointed out, this contradicts the axioms of the real line, since there will always be $\widetilde 0/2$ between the two. $\endgroup$ – Christoph Feb 19 '14 at 20:53

If you are interested in such questions you might explore the surreal numbers of JH Conway, which fill gaps in the number line you never knew existed.

My eldest daughter once wanted to make a distinction between $0.99999999 \dots$ and $1$ on the same basis. I put a question up on the subject, which got closed, because although there is an intuition there, it doesn't really go very far mathematically, and there are better ways of capturing that intuition to mathematical effect (for example by thinking about open and closed intervals).

The Conway thing is serious, by the way.

  • $\begingroup$ Funny enough, reading about surreal numbers (and Cantor stuff related to them) is what got me thinking about this. :) I didn't finish my reading on them though, and intend to do so. $\endgroup$ – user124384 Feb 19 '14 at 20:51
  • $\begingroup$ @user124384 There are just more productive ways through, which have proven fruitful. $\endgroup$ – Mark Bennet Feb 19 '14 at 21:04

Even if your construct $0.\overline 01$ made sense, couldn't I still pick a number between those two numbers and say $0.\overline 005$

It's a number that (in your world where your construct made sense) is probably halfway between the number you picked and zero.


It isn't necessary that $0. \overline 01$ describes a process for constructing a real number in countable steps. The real numbers that can be so defined are a subset with measure zero. If we allow $0. \overline 01$ to be notation for a nonzero real number, there is indeed a nonzero real number exactly half that, which might be denoted $0. \overline 005$ (as above). There is another which is half that, and so on, because multiplication of a real number by a real number in $(0, 1)$ is defined to be another real number of smaller magnitude ($\mathbb R$ is closed under multiplication, a vitally important property).

In fact the open interval $(0, 0. \overline 01)$ contains more real numbers than $\overline 0$ contains zeros — an uncountable infinitude. Given any symbol for a nonzero real number (say $x$), a dense interval on $[0, 1]$ can always be mapped to an equally dense interval on $[0, x]$ (e.g., multiply by $x$). Likewise, $[1, \infty)$ can always be mapped to $(0, x]$. There is no smallest difference between distinct real numbers, and no definable next consecutive real number.


It looks as though you're trying to extend an algebraic structure. Would you like Clippy to help you with that? ;-)

why is c necessarily divisible? There are special exceptions for other numbers.

There is one special exception in the real numbers, which is that you can't divide by 0. Whether you define the reals axiomatically or via a construction from the rationals, this is a property of the reals. It's a property of any "field", that the only special case like this for division is the additive identity. The reason an exception is made is because of the way that addition interacts with multiplication in a field: to have an additive identity at all you must accept that multiplying by 0 has no inverse operation.

Therefore if your candidate value c is not 0 then is not part of the real numbers. That's OK, it's perfectly legitimate to invent new values and explore the consequences, but you are not exploring the real numbers. You're exploring a new thing.

For example if you extend the real numbers by inventing a square root for -1 then you eventually arrive at the complex numbers. But there are a lot more complex numbers than just the reals and the square root of -1. How do the rest come about? The answer is that people want to do arithmetic involving complex numbers. In fact they wanted the complex numbers to be a field, although the first conception of imaginary numbers preceded the formal definition of a field by some centuries. If we told them that they can't divide i by 2 because it's a special case, then they simply wouldn't use our construction, because (among other problems) it would contain a solution to x^2 + 1 = 0 but not contain a solution to 4x^2 + 1 = 0. So it wouldn't be very general.

OK, so back to your extension of the real numbers. You've added one special value, but you've said we can't necessarily do arithmetic on the resulting structure. It is not a field. So this is all very well, you've introduced a single "infinitesimal", but by inventing a structure with the property that the set of all values greater than 0 has a least element, you have lost the property of the reals, of being an algebraic field. Which property is more interesting and useful? So which structure are people going to use? There may be special purposes for which your structure is just the thing, but it's not generally useful enough to take over from the real numbers.

It's an inevitable property of an ordered field that it doesn't have "next values", precisely because between a and b in the field you can always place (a + b) / 2. So if you introduce "next values" then you take away division. You can't have both, and for the real numbers we want division, except in the single special case we can't avoid, where we have to accept that there's no division by 0.

  • $\begingroup$ This is great; thanks for your thorough explanation. Now I feel motivated to tinker with the potential purposes of this structure... I think it'd be fun. And yeah, I'm totally not even trying to compete with the reals! That would clearly be silly. $\endgroup$ – user124384 Feb 22 '14 at 0:51

You can define a number like this one if you consider hyperreal numbers: and it would be written $n + \epsilon$, which is greater than n but less than any other number greater than n.

Indeed, if you could define n.00000...[infinite zeros]...1 , then you should be able to define n.00000...[infinite zeros]...2 and many other numbers. This 0.000[infinite zeros]1 is called $\epsilon$ and is the first infinitesimal part. It can be multiplied by any real and would still be infinitesimal. $\epsilon^2$ is an even smaller infinitesimal part than the first one.

But because of the hierarchy, you would also have $n < ... < n + \epsilon^3 < n + \epsilon^2 < n + \epsilon$ and even with ordinals you could not get around this problem.

So even in this theory you cannot find "the" one after n, only one in a subhierarchy.

  • $\begingroup$ Thanks for that; I actually just looked at the hyperreals last night to see how they fit into all this. In "my" system, however, $\epsilon^2 = \epsilon$, since you can't have more zeros than infinite zeros. Yes, I know this violates Cantor and the surreal numbers and stuff. But why not? $\endgroup$ – user124384 Feb 20 '14 at 18:48
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    $\begingroup$ indeed, if you don't think there are different classes of infinity, then it's ok with $\epsilon^2 = \epsilon$. But then you lose a lot of properties. For example, $\epsilon/2 = \epsilon$ since it is supposed to be the smalled number. You cannot however divide by epsilon, else it would be $1/2 = 1$. But you cannot multiply by two, else you would have $\epsilon = 2*\epsilon$. Unless you specify that $\epsilon*n = \epsilon$ if $n \ne 0$. Why not indeed. $\endgroup$ – Mikaël Mayer Feb 21 '14 at 23:12
  • $\begingroup$ Yes, that's exactly how I see it. Really interesting to know this isn't just completely crazy. Thanks for humoring me. $\endgroup$ – user124384 Feb 22 '14 at 0:46

A simple proof that may or may not work:

.999... = 1

.999... + c = 1

c = 0

n + c = n + 0

n + c = n

Just an idea that is, at the very least, easy to understand.

  • $\begingroup$ That's interesting. I was wondering about some sort of connection to .999... like that, but couldn't formulate one. Looks like a pretty compelling case to me, although I'm sure some would object to it on grounds laid out in other comments. On the other hand, if it's not valid, it sort of proves the non-existence or lack of meaning of c. $\endgroup$ – user124384 Feb 20 '14 at 2:26
  • $\begingroup$ @user124384 Wow. I never really thought of it in that much depth. Thanks, you made me feel a lot better about myself... $\endgroup$ – KnightOfNi Feb 20 '14 at 3:12

Just look at this in the first place: the reason why probably, .9999... = 1, no rounding! {Perhaps one will argue about '0.9999...' being just a teensy amount less than 1. Well, but isn’t '.9999…' just 3 times '.3333…'? Everyone will agree with this one. And isn’t '.3333…' the same as 1/3? Yes, of course. And isn’t 3 times 1/3 equal to 1?}

Back to our problem: can n.00000...[infinite zeros]...1 be the next real number after n? Impossible! When you see a number like '0.9999...' (i.e. a decimal number with an infinite series of 9s), there is no end to the number of 9s. Therefore, you cannot say "but what happens if it ends in an 8?", because it simply does not end. (This is also why 0.9999... equals 1). Similarly, 'AAAA...', an infinite series of "A"s followed by a "B" would NEVER have a "B", not in actual Mathematics, not in physical Mathematics. That is why you can never reach the end of all numbers by listing them. You can try out this! But this is possible only in potential Mathematics, Mathematics full of imaginations, like the world has an end; time has a beginning and an end, e.t.c, potential Mathematics, that is it.


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