I've just happened to read this question on MO (that of course has been closed) and some of the answers to a similar question on MSE.

I know almost nothing of nonstandard analysis and was asking myself if something like the sentence « $1- 0.999 \dots$ is a nonzero positive infinitesimal» could be easily expressed and proved in nonstandard analysis.

First of all, what is 0.999... ? If we take the usual definition as a series or as a limit of a sequence of rationals, then it will still be a real number and equal to $1$ (I guess by "transfer principle", but please correct me if I'm wrong).

Instead, let's define

$$0.9_N:=\sum_{i=1}^N 9\cdot 10^{-i} $$

where $N\in{}^*\mathbb{N}\setminus\mathbb{N}$ is an infinite nonstandard natural number. This $0.9_N$ is a legitimate element of ${}^*\mathbb{R}$, expressed as $0.$ followed by an infinite number of "$9$" digits.

What can be said about $\epsilon_N:=1-0.9_N$ ? Is there an elementary proof that $\epsilon_N$ is a positive infinitesimal of ${}^*\mathbb{R}$ ? (by "elementary" I mean just order and field axioms and the intuitive facts about infinitesimals, like that for $x$ infinite $1/x$ is infinitesimal etc.; no nonprincipal ultrafilters & C).

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    +1 for not making the mistake $0.\overline{9} < 1$, and recognizing that you need a *terminating* decimal to get something less than 1. –  Jan 18 '13 at 16:42

3 Answers3


We can use the geometric series formula:

$$0.9_N = \sum_{i=1}^N 9 \cdot 10^{-i} = 9 \cdot 10^{-1} \cdot \frac{1 - 10^{-N}}{1 - 10^{-1}} = (1 - 10^{-N})$$

Since $N$ is infinite, $\epsilon_N = 10^{-N} = 1 / 10^N$ is infinitesimal.

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    P.S. I hate using the word 'infinite' in this context. But I can't bring myself to write 'transfinite' or 'unlimited'. :( I would write $N \approx \infty$, but I think a lot of people would dislike that notation. –  Jan 18 '13 at 16:47
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    You type faster than I do. :-) *Infinite integer* is the usual term, so it doesn’t bother me. – Brian M. Scott Jan 18 '13 at 16:51
  • But why the formula $\sum_{i=1}^N r^i=(1-r^N)/(1-r)$, for $r$ real, is still valid when $N$ is infinite? (btw, yes I think "infinite" is the technical term used in nostandard analysis) – Qfwfq Jan 18 '13 at 16:51
  • I usually say "a infinite hyperreal number" :) – Ian Mateus Jan 18 '13 at 16:52
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    @Qfwfq Hint: Transfer principle. – Ian Mateus Jan 18 '13 at 16:52
  • We're interested in something like $(1-r)(1+r+r^2+\dots r^N)=1-r^N$. Which property (of ordered fields?) is being transferred? (it would be an occasion for me to understand what this transfer principle is about, in an elementary instance like this) – Qfwfq Jan 18 '13 at 16:57
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    @Qfwfq: One key thing to note is that, internally, this is a *finite* sum; after all, it only has $N$ terms, and $N$ is a (nonstandard) integer! But you can be more bold than that: you can transfer the notion of summation itself (in fact, you have to, to define summations from $1$ to $N$), and thus apply the transfer principle to identities that summation satisfies. –  Jan 18 '13 at 17:02
  • Ok, thank you for the explanation. – Qfwfq Jan 18 '13 at 17:04

$$1-\sum_{k=1}^N9\cdot10^{-k}=\sum_{k\ge N+1}9\cdot 10^{-k}=9\sum_{k\ge N+1}10^{-k}=\frac{9\cdot 10^{-(N+1)}}{1-10^{-1}}=10^{-N}=\frac1{10^N}\;,$$

which is surely intuitively a positive infinitesimal.

Added: There is a nice elementary axiomatization of the hyperreals in Jerry Keisler’s Elementary Calculus: An infinitesimal approach, which is freely available here; it is intended for students in a first calculus course and neatly avoids the transfer principle as such. His Foundations of infinitesimal calculus contains a slightly more sophisticated version, since it’s intended for instructors using the undergraduate text. It’s freely available here, and its version of the axiomatic development can also be found in Section $15$ of this PDF. What he calls the Function Axiom (Axiom $C$ in the PDF) justifies the standard calculation:

For each real function $f$ of $n$ variables there is a corresponding hyperreal function ${}^*f$ of $n$ variables, called the natural extension of $f$.

The function in question here is the function that takes $n$ to $\sum_{k=1}^n9\cdot10^{-k}$.

A slightly different version of this approach is found in Keith Stroyan’s notes here, especially Section $2.3$.

Brian M. Scott
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  • (Thank you for your valid answer. I accepted the other one because of the comment thread following it) – Qfwfq Jan 18 '13 at 17:08
  • @Qfwfq: You’re welcome. I added a couple of references that you may find interesting or useful. – Brian M. Scott Jan 18 '13 at 17:36
  • Sir, apologize for interrupting you with an old question.Your answer seems to be a little bit weird to me, since it follows that the decimal expansion ranged over $^\ast \Bbb N$ but restricted to $\Bbb N$ wouldn't be identical to the decimal expansion ranged over $\Bbb N$. Thus I think it might be more reasonable to add a standard predicate to "prevent" such restriction. – Metta World Peace May 19 '13 at 21:19
  • It's not that it "wouldn't be identical" to the decimal expansion ranged over $\mathbb{N}$. Rather, a "decimal expansion ranged over $\mathbb{N}$" does not exist since $\mathbb{N}$ is not an internal set. Hyperfinite sums can only be taken over hyperfinite sets. – Mikhail Katz May 20 '13 at 08:44
  • ...(i.e. hyperfinite, and in particular internal, sets). – Mikhail Katz May 20 '13 at 09:23
  • Here I am talking of course about HYPERREAL decimal expansions, which apriori range over $^\star\mathbb{N}$. – Mikhail Katz May 20 '13 at 09:30
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    @MettaWorldPeace: As user72694 says, the problem doesn’t arise: working within ${}^*\Bbb R$, you can’t talk about a sum over $\Bbb N$, because you can’t even talk about $\Bbb N$: it’s an external set, not an element of the model, so you can only ‘see’ it and talk about it from outside the model. – Brian M. Scott May 20 '13 at 16:54
  • Thank you for your reply, sir. As you said, working within ${}^{\ast}{\Bbb N}$, we can't form a set $\Bbb N$ or ${}^{\ast}{\Bbb N} \setminus \Bbb N $. But can we talk about a proper subset of ${}^{\ast}{\Bbb N} \setminus \Bbb N $ without using the predicate "standard"(in your example, it's $\{N\}$)? It seems to me it's also an illegal set formation. – Metta World Peace May 20 '13 at 17:11
  • @MettaWorldPeace: There’s no problem talking about $N$: it’s just some element of ${}^*\Bbb N$, say $\langle n_k:k\in\omega\rangle_{\mathscr{U}}$, where each $n_k\in\Bbb N$. And $\{N\}$ is then $\langle\{n_k\}:k\in\omega\rangle_{\mathscr{U}}$, so it’s not problematic either, though I’m not in fact using it. – Brian M. Scott May 20 '13 at 17:20
  • Sorry for being like a devil's advocate. But if you permit using ultrapower construction to find an nonstandard element $N$ , you must have the set, $\Bbb N$ in the first place. After all it's a sequence ranged over $\Bbb N$. – Metta World Peace May 20 '13 at 17:28
  • @MettaWorldPeace: The construction of the model is external to the model. You use $\Bbb N$ to construct the model, but the set $\Bbb N$ is not an element of the model. – Brian M. Scott May 20 '13 at 17:32
  • That's really weird to me. It goes like, when you finished constructing a model, you immediatly throw the tools away. – Metta World Peace May 20 '13 at 17:39
  • @MettaWorldPeace: You don’t throw them away; you use them in reasoning about the model. But the tools themselves are not part of the model. – Brian M. Scott May 20 '13 at 17:44
  • Thank you very much for enlightening me. Hope you don't mind. I still don't get used to these internal, external views, though heard of them at times. Is there some reference that clarify these matters? – Metta World Peace May 20 '13 at 17:48
  • @MettaWorldPeace: You’re welcome. You might find the first three or four chapters of [this master’s thesis](http://infohost.nmt.edu/~lenth/thesis.pdf‎) helpful as a start: they go into considerable detail that would be taken for granted at a higher level. [This incomplete draft](http://www.math.tamu.edu/~saichu/HyperrealsBook.pdf‎) is a little more sophisticated, and [this](http://www.math.tamu.edu/~saichu/ModelTheoreticNonstandard.pdf‎) is a brief introduction to the relevant model theory. – Brian M. Scott May 20 '13 at 18:01
  • Just want to let you know. The stuff in your link are really helpful, I do become less muddled. Thank you, sir. You're really a good teacher. – Metta World Peace May 20 '13 at 20:16
  • @MettaWorldPeace: Glad it helped, and thank you. – Brian M. Scott May 21 '13 at 17:04

In addition to the fine answers given earlier, I would like to mention also the somewhat unsung hero, A. H. Lightstone, and his extended decimal notation in which your infinitesimal $\epsilon_N$ can be written as $0.000\ldots;\ldots 0001$, where the first nonzero digit occurs precisely at infinite rank $N$. The notation is explained in his article in the American Mathematical Monthly (see especially page 246).

Mikhail Katz
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