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EDIT: This question is NOT about 0.9999... !!! It is clearly stated in the heading. It is about definition of numbers. The confusion in all the comments and the answer - assuming it is about 0.9999... - is related to the confusion about that number. I admit the question is fairly broad. However, philosophical questions tend to be broad.

Let us consider again - perhaps legal in this rule-laden environment, perhaps not - the question: ” Is 0.9999… a number, and if it is, is it perhaps =1?

This is actually not my question but I am surprised I don’t se the answer that comes to my mind: 0.9999… is not a number but a series with the limit 1. But if you insist: Can we accept the notion of numbers such as those depending on an infinite series of decimals? Obviously we can get around some repetitive numbers by representing them with the limit of a series, but how about infinite numbers in general?

I am intrigued with the notion of the definition of real numbers, and I note that quite a few questions is related to this issue (even without my own follow up question "Constructivism implied or not").

1 All real numbers can be expressed as a limit if rational numbers? 2 0.9999…=1 in several versions – related in my mind but perhaps not explicitly. 3 Are real numbers “a joke”?

I have met two conflicting points of view

i) is my own after having consulted (as I believe) Gödel-specialist Prof Podnieks who informs me that

“Any Goedel-style enumeration can cover only those real numbers that are definable by formulas (in some fixed language). Thus, if in one's set theory, there are uncountably many real numbers, then some of these numbers must be undefinable (by formulas)” (private communication).

ii) Hamkins in http://mathoverflow.net/questions/44102/ referred to in question 3 above: “The concept of definable real number, although seemingly easy to reason with at first, is actually laden with subtle metamathematical dangers to which both your question and the Wikipedia article to which you link fall prey. In particular, the Wikipedia article contains a number of fundamental errors and false claims about this concept./ …. / The naive account continues by saying that since there are only countably many such descriptions φ, but uncountably many reals, there must be reals that we cannot describe or define. But this line of reasoning is flawed in a number of ways and ultimately incorrect. The basic problem is that the naive definition of definable number does not actually succeed as a definition.”

My problem is that Gödel’s enumeration looks very similar to me, and that has succeeded as a definition.

Many articles deals with theproblem, such as “How real are real numbers?” from http://www.cs.auckland.ac.nz/~chaitin/olympia.pdf but I am not sure it provides much of an answer.

In any case, I speculate that these - seemingly – conflicting views are the root of problem. It would be valuable to have an answer or something in that direction. Better understanding would perhaps not settle all these questions, but it would probably take some of the confusion out of the debate.

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closed as not a real question by Thomas Andrews, Pedro Tamaroff, Alexander Gruber Jun 19 '13 at 23:22

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

What is your actual question? –  Qiaochu Yuan Jun 19 '13 at 22:18
$0.999...$ is not a number, it is one of several representations of a number (also known as $1$) –  Hagen von Eitzen Jun 19 '13 at 22:24
0.999999... represents the real number that is the limit. It does not represent the series. Thus it is a real number. –  Thomas Andrews Jun 19 '13 at 22:56
This closure seems a little quick to me, but I'm not up on the recent cultural standards of the site. –  Pete L. Clark Jun 19 '13 at 23:34
@Mikael: The business about $0.9999\ldots$ is so far away from set theoretic issues of definability so as to be distracting when you talk about both at once. Also you called attention to a (minor) misconception you have about decimal expansions of real numbers. It is a common and useful pedagogical technique to address the most basic issues first, hence my answer. With all respect, if you think there's something fishy about decimal expansions then you're very far away from being able to engage with what's in Joel Hamkins's answer. –  Pete L. Clark Jun 21 '13 at 3:02

1 Answer 1

By "real numbers" I will assume you mean the standard mathematical object by that name, as described e.g. here and in thousands of texts. That is, the real numbers are a set $\mathbb{R}$ endowed with two binary operations $+$ and $\cdot$ and a total order relation $\leq$ (which can be expressed in terms of $+$ and $\cdot$: $x \leq y \iff \exists z \ | \ y-x=z^2$) which form an ordered field in which every nonempty subset which is bounded above has a least upper bound. This structure is essentially unique in the sense that if we had a different such structure $(\mathbb{S},+,\cdot,<)$ then there is a unique bijection $\Phi: \mathbb{R} \rightarrow \mathbb{S}$ preserving $+$ and $\cdot$ and (thus) $<$.

"Is $0.9999\ldots$ a real number?"

Again I have to assume that you mean the standard definition. If $a_0$ is a non-negative integer and $a_1,a_2,\ldots,a_n,\ldots \in \{0,1,2,3,4,5,6,7,8,9\}$, then the standard definition of $a_0.a_1a_2\ldots a_n \ldots$ is the sum of $a_0$ and the limit of the convergent infinite series $\sum_{n=1}^{\infty} \frac{a_n}{10^n}$. (That the series is convergent is a piece of calculus addressed in many places, including on wikipedia. You need to establish the convergence of the geometric series $\sum_{n=1}^{\infty} \frac{1}{10^n}$ and then use the fact that the partial sums of the given series are weakly increasing and have $10 \sum_{n=1}^{\infty} \frac{1}{10^n}$ as an upper bound, so that the least upper bound of the sequence of partial sums is the limit.)

Thus the answer to your question is yes. You say that you thought that it was an infinite series rather than a number. That's a fair question, but it's just a question of notation, and the standard answer is the one I've given above: the notation $a_0.a_1 \ldots a_n \ldots$ does allow you to associate an infinite series, but by definition we mean the sum of that series rather than the series itself. Note that this is related to the slightly unfortunate but standard ambiguity of the notation $\sum_{n=1}^{\infty} a_n$ which is used both for an infinite series -- i.e., a sequence of partial sums $S_1 = a_1, S_2 = a_1 + a_2,\ldots, S_n = a_1 + \ldots + a_n$ -- and for the limit of that series, if it exists.

Why is this unfortunate? Because then by definition $\sum_{n=1}^{\infty} a_n = S$ is a real number, so referring to e.g. "the $n$th term of $\sum_{n=1}^{\infty} a_n$" is strictly speaking meaningless, as there are many, many infinite series which converge to any given real number $S$.

Indeed this happens sometimes even with the form of restricted infinite series associated with the decimal notation. The answer to your question

"Is $0.9999\ldots" = 1$?"

is yes. The series $\sum_{n=1}^{\infty} \frac{9}{10^n}$ is $9$ times the geometric series $\sum_{n=1}^{\infty} \frac{1}{10^n}$, so its sum is $9 \cdot (\frac{1}{10} \cdot \frac{1}{1-\frac{1}{10}}) = 1$.

Since $1.0000\ldots$ denotes $1 + \sum_{n=1}^{\infty} \frac{0}{10^n} = 1$, we have

$0.9999\ldots = 1.0000\ldots$.

The only place where even a tiny amount of philosophy comes into this is in the distinction between notation and the thing which the notation denotes. When I write $1.0000\ldots = 1$, I mean that the real numbers that these symbols denote are the same, just as when I write $2+2 = 4$. The symbol (or string of symbols) "$2+2$" is not the same as the symbol "$4$", obviously: this is a general property of symbols. So when one wants to refer to the symbol rather than the object that the symbol signifies, you should put quotation marks around it (or use some other such convention). Thus "$0.999\ldots$" and "$1.0000\ldots$" are not the same -- obviously -- but the real numbers they denote are the same. (For that matter, notice that "$0.999\ldots$" is not the same as "$0.9999\ldots$".)

Finally, it is not hard to show that every non-negative real number is of the form $a_0.a_1 a_2 \ldots a_n \ldots$. This uses the fact that the least upper bound axiom implies the Axiom of Archimedes: for every real number, there is an integer $n > x$. (For if not, the set of integers would be nonempty and bounded above but have no least upper bound.) In particular every non-negative real number is the limit of a sequence of rational numbers, and multiplying by $-1$ we get this for negative real numbers as well.

I have the suspicion that you meant to ask a more profound question than this, but this was the only unambiguous question I could wring out from your post. If this answer is off the mark, please clarify by stating a single precise question. If you want to ask about "definability of real numbers", please include a statement or reference to a statement of what you mean by that.

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Just a question that I'm sure I'm wrong about. Isn't it true that rational numbers have the same properties you described in the first parragraph, but you can't make a bijection between them and the reals because the first are countable? –  MyUserIsThis Jun 19 '13 at 23:13
@MyUserIsThis: The rational numbers have almost all of the same properties. They do not have the least upper bound property. –  Cameron Buie Jun 19 '13 at 23:21
My question is related but not equal to the 0.9999…vs 1 question. It is elaborated a bit more here ocf.berkeley.edu/~wwu/cgi-bin/yabb/… –  Mikael Jensen Jun 25 '13 at 3:49

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