In this answer to the question Is it true that $0.999999999\dots=1$?, Noah Snyder points out that

Symbols don't mean anything in particular until you've defined what you mean by them.

This question is motivated by the quote above. In real analysis, the symbol $0.99\cdots$ is defined as a real number, and $0.99\cdots=1$. One such definition can be seen in Principles of Mathematical Analysis (3rd edition) by Walter Rudin.1

While $0.99\dots<1$ is a "false" intuition in the real number system, it is suggested by this Wikipedia article that the notion of "a number that falls short of $1$ by an infinitesimal amount" can be rigorously defined. See also this blog post by Terry Tao on ultralimit analysis.

Could anyone come up with an "introductory" reference (textbooks/papers) in nonstandard analysis and explain briefly how, in that reference, $0.99\cdots$ is defined in a way that is fundamentally different from the standard real analysis?



  1. In his Principles of Mathematical Analysis (3rd edition), Rudin introduced the set of real numbers by the following theorem (page 8 of Chapter 1):

    There exists an ordered field $\mathbf{R}$ which has the least-upper-bound property. Moreover, $\mathbf{R}$ contains $\mathbf{Q}$ as a subfield.

    In section 1.22, he what "decimal expansion" means for a given positive real number $x>0$ as follows.

    Let $n_0$ be the largest integer such that $n_0\le x$. (The existence depends on the archimedean property of $\Rbf$.) Having chosen $n_0,n_1,\cdots,n_{k-1}$, let $n_k$ be the largest integer such that $$ n_0+\frac{n_1}{10}+\cdots+\frac{n_k}{10^k}\le x. $$ Let $E$ be the set of these number $$ n_0+\frac{n_1}{10}+\cdots+\frac{n_k}{10^k} \quad (k=0,1,2,\cdots).\tag{1} $$ Then $x=\sup E$. The decimal expansion of $x$ is $$ n_0\cdot n_1n_2n_3\cdots\tag{2} $$ Conversely, for any infinite decimal (2) the set $E$ of number (1) is bounded above, and (2) is the decimal expansion of $\sup E$.

    This gives one definition of $0.99\cdots$ by (1) and (2).

  • Mainly it says that there is a number, say x, that is greater than zero but less than any positive number. So in this sense 0.9999... is 1-x. – user12986714 May 01 '20 at 14:18
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    Simply speaking, NSA does not lead to conclusions about $\mathbb{R}$ that differ from those of standard analysis. Giving alternate definitions of $0.99\dots$ is not really the concern of most researchers working in NSA: we roll with the usual definition (as in Rudin above), and so $0.999\dots = 1$. – Z. A. K. May 01 '20 at 21:35
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    That said, "simply speaking" is necessarily inaccurate: questions of mathematical culture tend to be subtle rather than from simple. There are many possible alternative semantics for decimal notation that nonstandard analysis can formalize, and some good (social) reasons for adopting them. The reference you are looking for is the article [When is .999... less than 1?](https://arxiv.org/abs/1007.3018) by K. U. Katz and M. Katz. One of the authors, [Mikhail Katz](https://math.stackexchange.com/users/72694/mikhail-katz) is both an expert on NSA and a frequent contributor to this website. – Z. A. K. May 01 '20 at 21:40
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    @Z.A.K. Except for perhaps commentary from Katz this is probably as good of an answer I can imagine for this question – GPhys May 02 '20 at 06:08
  • Related: https://math.stackexchange.com/questions/281492/about-0-999-1 – Hans Lundmark Sep 24 '21 at 20:08
  • Does this answer your question? [About 0.999... = 1](https://math.stackexchange.com/questions/281492/about-0-999-1) – amWhy Oct 28 '21 at 17:11
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    **NO**. This question is about nonstandard analysis. –  Oct 28 '21 at 17:16
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    @Z.A.K.: May I suggest you turn your comments into an answer? –  Oct 28 '21 at 17:18
  • @user: Thanks. Alas, there's a reason I never expanded these comments into an answer: I know from experience that sometimes, potential postdoc supervisors/employers check out my Math.SE user page, and the last thing I want is something related to $0.999\dots$ showing up there. You or anyone else should feel free to reuse my comments without attribution as part of their answer. – Z. A. K. Oct 29 '21 at 00:14

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