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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. –  Hurkyl Jan 18 '13 at 16:42

3 Answers 3

up vote 6 down vote accepted

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. –  Hurkyl Jan 18 '13 at 16:47
    
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
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@Qfwfq Hint: Transfer principle. –  Ian Mateus Jan 18 '13 at 16:52
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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. –  Hurkyl Jan 18 '13 at 17:02

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

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 $2.3$.

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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. –  user72694 May 20 '13 at 8:44
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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

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 here: http://www.jstor.org/stable/2316619?origin=crossref (see especially page 246).

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