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I have read this post which contains many proofs of $0.999\ldots=1$.

Background

The main motivation of the question was philosophical and not mathematical. If you read the next section of the post then you will see that I have asked for a "meaning" of the symbol $0.999\ldots$ other than defining it to be $1$. Now here is a epistemological problem and this is mainly the problem from which the question arose. Suppose you know that $1$ is a real number. Now I give you a symbol, say $0.999\ldots$ which from now on I will denote as $x$. Now I ask you whether $x$ is a real number. To answer this, if you define $x=1$ then you are already attributing the properties of $1$ to $x$ among which one is it being a real number without proving whether to $x$ we can indeed attribute the properties of $1$.

A common response to this question has been to define the symbol $x$ as the limit of the sequence $\left(\displaystyle\sum_{i=1}^n \dfrac{9}{10^i}\right)_{n\in\mathbb{N}}$ and then prove that the limit of this sum is indeed $1$. But again the problem is that you are defining the symbol $x$ to be a real number and hence are assuming a priori that the symbol $x$ denotes a real number.

As per the discussion that has been conducted with Simply Beautiful Art let me state my position again in brief,

Also let me say that I do not disallow $0.999…$ to be a real number. My impression that if you assume $0.999…$ to be a real number then there is no sense in proving that $0.999…$ is indeed equal to $1$ because either you define it to be $1$ or you prove the equality as a theorem. But if you are going to use the limit definition of $0.999…$ then what you are a priori assuming it to be a real number which is an assumption that I don't allow. What can be allowed is that $0.999…$ is a number (but not necessarily a real number).

Question

My question is,

Does it make any sense to prove this equality?

Can one give any "meaning" of the symbol $0.999\ldots$ other than defining it to be $1$?

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    $\begingroup$ Well...you can define $.999...=\sum \frac 9{10^i}$ and then prove that this sum converges to $1$. I'd say that had non-trivial content, no? $\endgroup$ – lulu Jan 5 '16 at 15:41
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    $\begingroup$ The point is to show that $0.999\ldots$ is equal to 1. There is meaning to defining infinitely repeating decimals. $\endgroup$ – Morgan Rodgers Jan 5 '16 at 15:46
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    $\begingroup$ The first answer to the question you link to, with more than 250 upvotes, is likely to be as good an answer as you could possible get here, asking again. $\endgroup$ – Ethan Bolker Jan 5 '16 at 15:47
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    $\begingroup$ $.999\ldots$ is not defined to be equal to $1$; it already has a definition as soon as decimal notation is defined. When we define decimal notation, we agree that $.a_1 a_2 a_2 \ldots \text{ means } a_1/10 + a_2/100 + a_3/1000 + \cdots$. It turns out that $.999... = 1$, but that's only a consequence of the definition of decimal notation. $\endgroup$ – littleO Jan 5 '16 at 15:47
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    $\begingroup$ Otherwise, how can you talk about being equal? $\endgroup$ – Future Jan 5 '16 at 17:33
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The OP asked whether one can assign any meaning to the symbol $0.999\ldots$ other than defining it to be $1$. That question cannot be answered without analyzing what informal pre-mathematical meaning is assigned to $0.999\ldots$, prior to interpreting it in a formal mathematical sense. This of course can only be known to the OP himself but judging from the level of the OP's questions the OP seems to be a student and perhaps a freshman; see, e.g., here.

Now beginning students often informally describe this as "zero, dot, followed by infinitely many $9$s", or something similar. Such a description of course does not refer to any sophisticated number system such as the real numbers, since at this level the students will typically not have been exposed to such mathematical abstractions, involving as they do equivalence classes of Cauchy sequences, Dedekind cuts, and the like.

It is also known that at this level, about $80\%$ of the students feel that such an object necessarily falls a little bit short of $1$. The question is whether such intuitions are necessarily erroneous, or whether they could find a mathematically rigorous interpretation in the context of a suitable number system.

An article by R. Ely in this publication in a leading education journal argues that such intuitions are not necessarily mathematically erroneous because they can find a rigorous implementation in the context of a hyperreal number system, where a number with an infinite tail of $9$s can fall infinitesimally short of $1$ as outlined in a comment by user @GBeau on this page, namely if $H$ is an infinite hypernatural then $\displaystyle\sum_{n=1}^H \frac{ 9}{10} =0.999\ldots9$ where the digit $9$ occurs $H$ times.

This is of course a terminating infinite string of $9$s different from the one usually envisioned in real analysis, but it respects student intuitions and can be helpful in learning the calculus, as argued in Ely's fascinating study.

The existence of such an interpretation suggests that we indeed do assume that such a string represents a real number when we prove that it necessarily equals $1$.

Note I. If one thinks of the infinite string as being represented by the sequence $0.9, 0.99, 0.999, \ldots$ then one can obtain an alternative interpretation as follows. Instead of taking its limit (which is by definition real-valued), one can take what Terry Tao refers to as its ultralimit, to obtain a number than falls infinitesimally short of $1$.

These issues are dealt with in more detail in this recent publication.

The challenging philosophical issue here is the idea that there are distinct ways of formalizing infinity in mathematics, and the possibility of an attendant ambiguity of the symbol in question. These issues were dealt with in more detail in this publication in a leading education journal.

Note II. A certain number of objections have been raised by a colleague who wishes to remain anonymous. Given below are the objections together with my responses.

(0) You have not provided a meaningful syntactic representation of $1/3$ in the hyperreals.

Well $\dfrac13$ is the unending decimal $0.333\ldots$ (indexed by the hypernaturals). If truncated at infinite hypernatural rank $H$ this would produce a hyperrational falling infinitesimally short of a third, similarly to the $0.999\ldots{}$ situation.

(1) Nobody can legitimately disagree that hyperreals can be constructed via the ultraproduct of the reals $\bf{R}$ within $\sf{ZFC}$, which is the mainstream foundations for mathematics.

True, analysis with infinitesimals can be done over the hyperreals, as pointed out by Robinson in 1961. Alternatively, this can be done syntactically in the context of the ordinary real line, following Edward Nelson's approach. Nelson's approach, called Internal Set Theory $(\sf{IST})$, involves enriching the language of set theory by the introduction of a single-place predicate $\textbf{st}$, as well as three additional axiom schemas governing its interaction with the other set-theoretic axioms. Here $\textbf{st}(x)$ reads "$x$ is standard".

(2) Philosophically nobody has provided non-circular ontological arguments justifying $\sf{ZFC}$ (especially with replacement and choice). No logician, whether on Math SE or on Math Overflow or whom I have met, have done anything close to it.

This is a much broader issue. It is possible that $\sf{ZFC}$ has serious flaws. Nonetheless it happens to be currently the standard against which much of modern mathematics is tested. This doesn't mean that we must accept it, but it does mean that such philosophical problems are no smaller for the reals than for the hyperreals (especially in view of Nelson's syntactic approach mentioned above).

I accept various things such as consistency of $\sf{ZF}$ implying consistency of $\sf{ZFC}$, but consistency is quite irrelevant to soundness besides being necessary. Unless you're happy with just $\prod_1$-soundness.

If the sound alternative is predicativism as developed by Sol Feferman and others, then certainly $\sf{ZF}$ is no less problematic than $\sf{ZFC}$. Practically speaking, $\sf{ZF}$ is not enough for some rather standard applications. Consider the following example: it is consistent with $\sf{ZF}$ that there exists a strictly positive real function with vanishing Lebesgue integral; see https://arxiv.org/abs/1705.00493

(3) The construction of the hyperreals is via the ultraproduct of the reals R. If you can construct the hyperreals, then you also can construct $\bf{R}$ and prove the usual second-order real axioms for $\bf{R}$. It would be self-contradictory to say that the properties of $\bf{R}$ (including $0.999... = 1$ suitably interpreted) are not intuitive and then claim that the hyperreals are intuitive. After all, we define an infinitesimal in the hyperreals as a nonzero sequence of reals that converges to zero...

I wouldn't argue that the properties of the reals are not intuitive. Rather, what was explored in several articles in the recent literature is the possibility that there may be multiple approaches to interpreting the business with "a tail with an infinite number of $9$s", some of which may be helpful in harnessing student intuitions in a productive direction rather than merely declaring them to be erroneous.

Incidentally, your definition of a hyperreal infinitesimal is not quite correct.

An important distinction here is between procedures taught in a calculus class and set-theoretical justification (ontology of the entities involved) usually treated in an analysis course. This applies both to the reals and the hyperreals.

(4) Let $\bf{R}^\ast$ be the hyperreals and $\varepsilon = 1 - 0.999\ldots$. You claim that $\varepsilon$ is nonzero in a suitable interpretation of $0.999\ldots$ Ignoring the fact that you cannot represent $1/3$ meaningfully in similar decimal form, I now present you another fact that you can't represent $\varepsilon/2$, not to say $\sqrt{\varepsilon}$. Wait, what does the latter even mean in the hyperreals. Can your students figure that out? Are you sure hyperreals are so intuitive now?

I am not sure what you mean. Both $1/3$ and $\sqrt{\varepsilon}$ are well-defined in the hyperreals, simply by the transfer principle. As far as teaching the set-theoretic justification of the hyperreals in terms of the ultrapower, as I mentioned this belongs in a more advanced course, just like set-theoretic justification of the reals.

In contrast, asymptotic expansion can happily deal with $\sqrt{x}$ for any asymptotic expression $x$ that is non-negative. No trouble at all. $x^{1+x}$ for positive $x$? No problem.

All of these are well-defined over the hyperreals by the transfer principle.

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    $\begingroup$ @Mikhail It might prove helpful to fully cite the mentioned articles so that readers have an immediate clue that they refer to scholarly publications (as opposed to web pages authored by cranks). Also it might prove helpful to establish this scholarly context at the beginning of the answer. This may help to discourage any further downvotes from users who have not invested proper effort to distinguish this fine answer from the many cranky answers that are often given for such FAQs. $\endgroup$ – Bill Dubuque Jun 9 '17 at 14:22
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    $\begingroup$ I want to emphasize my +1 on this; it's great that this post emphasizes that it's discussing a terminating decimal with infinitely many $9$'s. So many expositions neglect to give any indication that they're talking about something different from a nonterminating decimal, which I think sabotages the education of people who haven't yet conceived just what limits (and similar) are. $\endgroup$ – Hurkyl Jun 9 '17 at 18:52
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    $\begingroup$ For me, this is the answer that hits the nail on the head, namely that whether or not such a number equals one depends on the axiomatic system in which we are working. Therefore the answer is YES. In a given axiomatic system in which both symbols define elements of a set, one can legitimately ask for a proof that the elements they define are equal, when this is the case. $\endgroup$ – Daniel Moskovich Jun 16 '17 at 10:50
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    $\begingroup$ What do you mean by "meaningful" here @user21820? (I thought that for a concept to be meaningful, having an instantiation of it was both necessary and sufficient but since you write, "...but it does not imply that the notion is meaningful and has an instantiation." - I presume that you use the terms in different sense. Feel free to correct me if I am wrong.) $\endgroup$ – user 170039 Sep 12 '17 at 8:34
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    $\begingroup$ @user21820: If we have some prior notion of "the reals", then we don't know that Z is enough to construct them. Z constructs an object it calls $\mathbb{R}$ that we can identify in any model of Z, but there is no reason whatsoever that that object should be "the reals". It may even be possible that there isn't any model of Z whose $\mathbb{R}$ is "the reals". $\endgroup$ – Hurkyl Sep 16 '17 at 1:15
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We have to agree about what the symbols

$$ 0.99999\dots $$

are supposed to mean. The symbols capture an intuitive idea, but it doesn't have meaning unless we agree on what that meaning is. When you write these symbols down, everyone will agree that what you mean is the following:

$$ .9 + .09 + .009 + \dots = \frac{9}{10} + \frac{9}{100} + \frac{9}{100} + \dots = \sum_{i = 1}^\infty \frac{9}{10^i} $$

There is no proof of that -- it is an agreement.

If you wrote this, perhaps similar-looking, string of symbols down

$$ \begin{aligned} 0.00000\dots1\\ \end{aligned} $$

then there is no agreement on what you mean. You would seem to be talking about a real number which is smaller than all other real numbers -- an object that doesn't exist.

So what string of symbols means what, rigorously, is a matter of agreement. Usually we elide that fact when it seems intuitive what we mean, but we do not all share the same intuition (or the same knowledge of how to express that intuition), so we must drag it out on occasions like now.

Anyway, the content of a proof of $0.9999\dots = 1$ is not that we must agree to define $0.9999\dots$ as $1$. The content is to define $0.9999\dots$ as the sum above, then by deduction show this sum is equal to $1$.

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    $\begingroup$ So, we are assuming beforehand that $0.999\ldots$ is a real number, right? $\endgroup$ – user 170039 Jan 5 '16 at 16:57
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    $\begingroup$ @user170039: Yep. We're assuming it to mean $\sum_{i=0}^\infty \frac{9}{10^i}$, an infinite sum, which is a real number because it converges. $\endgroup$ – Eli Rose Jan 5 '16 at 17:05
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    $\begingroup$ We aren't assuming. Before we even think about this we define what real numbers are and prove the real number system exists (very analytical and obscure). By the proof and construction of what real numbers are, it comes out that all bounded sets of rational numbers have limits and these limits are always real numbers. {.9, .99,.999, etc.} is one of these bounded sets or rational numbers and $.999\overline9$ is the notation of its limit. And therefore we know $.999\overline9$ is a real number because all limits are real numbers and all bounded sets have limits. $\endgroup$ – fleablood Jan 14 '16 at 10:53
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    $\begingroup$ @EliRose I believe you meant to start your sum at $i=1$. $\endgroup$ – GPhys Jan 14 '16 at 11:14
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    $\begingroup$ @user72694: I agree -- this question was marked as duplicate, but the question it's putatively a duplicate of is about proving that $0.999\dots = 1$, not about whether it makes sense to do so. Voted to reopen. $\endgroup$ – Eli Rose Jan 14 '16 at 18:52
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Many of the OP's questions in the comments (both to his own question and to Eli Rose's answer) keep circling back to the question "Are you assuming that $0.999\dots$ is a real number"?

The answer is no, we are not assuming it -- it can be proven. More generally, the following theorem can be proven:

Let $(a_1,a_2,a_3,\dots)$ be any sequence of numbers where each $a_i$ is chosen from the set $\{0,1,2,\dots,9\}$. Then the sequence $$0.a_1, \space 0.a_1a_2, \space 0.a_1a_2a_3,\dots$$ converges to a unique real number.

Again I want to stress that the theorem above is not assumed; it can be proven.

The notation $0.999\dots$ denotes the unique real number that is the limit of the sequence $$0.9, \space 0.99, \space 0.999, \space 0.9999\dots$$ This is just an individual instance of the general case considered in the theorem. We know that such a limit exists by the theorem , so there is no need to assume that $0.999\dots$ is a real number.

Once we know that $0.999\dots$ is a real number, and that in particular it is the limit of the sequence above, we can observe that this particular sequence converges to $1$. Since the theorem says that the limit of the sequence is unique, that proves that $0.999\dots \space = 1$.

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    $\begingroup$ By first proving a general theorem and then introducing a notational convention. $\endgroup$ – mweiss Jan 6 '16 at 15:30
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    $\begingroup$ (1) Prove the theorem I highlighted in my box. We now know that there is a real number that represents the limit of a certain sequence. $\endgroup$ – mweiss Jan 7 '16 at 16:07
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    $\begingroup$ (2) Introduce the notational convention that an "infinite decimal" represents the limit of that sequence. $\endgroup$ – mweiss Jan 7 '16 at 16:08
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    $\begingroup$ (3) Now prove that the limit of the particular sequence under consideration is 1. $\endgroup$ – mweiss Jan 7 '16 at 16:08
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    $\begingroup$ At no point in this process is anything being assumed to exist. $\endgroup$ – mweiss Jan 7 '16 at 16:08
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I understand that this question is more than 1.5 years old, and I presume you now know how the structure of the real numbers is constructed in mathematics and proven to satisfy the second-order completeness axiom, which can then be used to define $0.99\overline9$ and prove that it is equal to $1$. But in the interest of future readers, here is a similar looking question that may give insight into why this question stems from a conceptual misunderstanding:

I have seen many proofs of $1+2+\cdots+n = \frac12 n(n+1)$ for natural $n$. Now suppose you know that $\frac12 n(n+1)$ is an integer. Now I give you an expression, say $1+2+\cdots+n$ which from now on I will denote as $x$. Now I ask you whether $x$ is an integer. To answer this, if you define $x = \frac12 n(n+1)$ then you are already attributing the properties of $\frac12 n(n+1)$ to $x$ among which one is it being an integer without proving whether to $x$ we can indeed attribute the properties of $\frac12 n(n+1)$.

A natural response to this question is to define the expression $x$ as the sum of all the integers from $1$ to $n$, and then prove that this sum is indeed $\frac12 n(n+1)$. But again the problem is that you are defining the expression $x$ to be an integer and hence are assuming a priori that the expression $x$ denotes an integer.

My impression that if you assume $1+2+\cdots+n$ to be an integer then there is no sense in proving that $1+2+\cdots+n$ is indeed equal to $\frac12 n(n+1)$ because either you define it to be $\frac12 n(n+1)$ or you prove the equality as a theorem. But if you use the summation definition of $1+2+\cdots+n$ then you are a priori assuming it to be an integer, an assumption that I don't allow. What can be allowed is that $1+2+\cdots+n$ is a number (but not necessarily an integer).

My question is, does it make any sense to prove this equality? Can one give any "meaning" of the expression $1+2+\cdots+n$ other than defining it to be $\frac12 n(n+1)$?

I hope it is clear where this goes wrong:

  • We indeed define the expression "$1+2+\cdots+n$" to have a certain value, in a certain way (using recursion). For this we need to work in a foundational system that can construct the needed recursion and then prove the existence of a function that satisfies the recursion, and then define the value of "$1+2+\cdots+n$" according to that function. It then is non-trivial to prove that this value is $\frac12 n(n+1)$.

  • We never assume a priori that the expression "$1+2+\cdots+n$" denotes an integer. Defining "$1+2+\cdots+n$" by the summation does not assume that the sum is an integer. As in the first point, the "summation" here is the recursive construction, which we can easily prove yields a function from naturals to integers, and so in particular "$1+2+\cdots+n$" has integer value.

  • It does not make sense to say "What can be allowed is that $1+2+\cdots+n$ is a number (but not necessarily an integer).". Why? Because even if you have defined what "number" means, how can you 'allow' some arbitrary expression to be a number? As in the first point, we define the value of the expression, and whether or not it is an integer is not up to us.

Similarly:

  • We can define "$0.99\overline9$" to denote the unique real $x$ that is a lowest upper bound for the set $\{0,0.9,0.99,\cdots\}$. You can ask how we define "$\{0,0.9,0.99,\cdots\}$". Exactly the same kind of recursion as in the previous analogy. You can further ask how we define things like "0.99". Again, by some suitable recursion. After all, we cannot define decimal notation without recursion. This definition of "$0.99\overline9$" is valid, because after the standard construction of the structure of the reals, we can prove the second-order completeness axiom, which gives us the theorem that there is indeed a unique such $x$...

  • One might ask whether it is valid to assign values to expressions based on objects that we have proven to exist. We can do so with no qualms if we can uniquely identify each object that we wish to assign to each expression. In logic this is equivalent to asking whether we can extend a first-order theory by a function-symbol (for the value assignment function) if we can prove that there is a $2$-parameter property $P$ such that for every input $i$ from the desired domain there is a unique object $j$ such that $P(i,j)$ is true. This is not only possible (see here for the technical details), but also yields a conservative extension, so we are not making any more philosophical commitment than we already did in using the original system.

  • If you think that the above definition of "$0.99\overline9$" is not intuitive, here is another one. Define "$0.99\overline9$" as the unique real $x$ that lies in all the intervals $[0,1],[0.9,1.0],[0.99,1.00],\cdots$. After all, any layman that has read about $π$ 'knows' that "$3.14...$" denotes something that lies between $3.14$ and $3.15$ inclusive, and that having more digits will narrow down that interval. It turns out that this 'more intuitive' definition is 'more demanding' than the earlier one, since ignoring the upper endpoints of the sequence of intervals gives the earlier one.

If anyone objects that my analogy is inaccurate because "everyone knows what $1+2+\cdots+n$ means", it just shows that they themselves do not know how to precisely define it. Every rigorous definition of "$1+2+\cdots+n$" must invoke recursion. Every rigorous definition of "$0.99\overline9$" that captures the notion of the infinite decimal expansion must likewise invoke recursion.

The second definition of "$0.99\overline9$" using narrowing intervals has the pedagogical advantage that it is far easier to understand each decimal as an approximation scheme for reals. A decimal could be understood as a oracle that spits out one digit at a time, each of which puts a more precise bound on the 'real' value. Furthermore, it is natural to consider computable decimals, namely those digit oracles that are programs. One can then see that there is a crucial distinction between decimals and the reals they represent; clearly "$0.99\overline9$" and "$1.00\overline0$" are represented by different digit oracles, and whether they approximate the same 'real' value is a totally separate question.

By the way, to explicitly address the notion that the idea that $0.99\overline9 < 1$ can be 'correct' with a different definition of "$0.99\overline9$", note that if it were so, then one would naturally expect to have $0.33\overline3 = 0.99\overline9 \div 3 < 1 \div 3 = \frac13$ since there is exact division of each digit. But then $\frac13$ would not have a decimal representation. How weird... It may be possible to 'fix' this somehow, but any 'fix' is going to be weirder and less intuitive than $0.99\overline9 = 1$. Try it!

Finally, mathematicians do not define "$0.99\overline9$" to be $1$, because it is as meaningless as defining it to be $2$. However, if you choose to define "$0.99\overline9$" as $1$ then it becomes non-trivial that there is no upper bound for $\{0,0.9,0.99,\cdots\}$ that is smaller than $0.99\overline9$. So whatever way you pick, the fact remains that there is a non-trivial theorem that corresponds to $\sum_{k=1}^n 9·10^{-k} \to 1$ as $n \to \infty$.

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  • $\begingroup$ Concerning rigorous recursion, see here. $\endgroup$ – user21820 Sep 8 '17 at 13:31
  • $\begingroup$ You write "By the way, to explicitly address the notion that the idea that $0.99\overline9 < 1$ can be 'correct' with a different definition of "$0.99\overline9$", etc." This is quite correct: one cannot define the unending decimal $0.99\overline9$ in any other way, whether in the reals or the hyperreals. However, it seems to me that you are changing the subject--or moving the goalpost--here. The OP is clearly not referring to the unending decimal $0.99\overline9$ which is certainly uniquely defined and equals $1$, but rather to a broader discussion of what kind of meaning to be assigned... $\endgroup$ – Mikhail Katz Sep 10 '17 at 14:47
  • $\begingroup$ ... to "a tail of an infinite number of 9s". Given that there is a whole literature devoted to this subject and its usefulness in harnessing students' intuitions in a constructive direction, your stomping the ground a little harder does little to advance the debate. @user170039 $\endgroup$ – Mikhail Katz Sep 10 '17 at 14:48
  • $\begingroup$ @MikhailKatz: You said in your own answer "Instead of taking its limit (which is by definition real-valued), one can take what Terry Tao refers to as its ultralimit, to obtain a number than falls infinitesimally short of 1.", which explicitly states that there is an interpretation of "0.999..." that makes "0.999... < 1" true. I did not move any goalpost; I explicitly mention that to show that any such alternative interpretation is not intuitive. Nowhere did I say in my answer that the asker believed that, so what you thought (seems to be moving goalpost) is simply false. $\endgroup$ – user21820 Sep 11 '17 at 11:54
  • $\begingroup$ What's involved here is the distiction between procedure on the one hand and (set-theoretic) ontology, on the other, as I mentioned in the extended version of my answer. I can elaborate if you like. $\endgroup$ – Mikhail Katz Sep 12 '17 at 7:39
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Note that $0.99999\dots$ means $\dfrac 9 {10} + \dfrac 9 {100} + \dfrac 9 {1000} + \dots = \sum \limits _{n=1} ^\infty \dfrac 9 {10^n}$.

Now, in your opinion, does it make sense to "define" this series to be $1$? Of course not, because otherwise we may "define" anything to be anything. For instance, we may "define" $\sum \limits _{n=1} ^\infty \dfrac 1 n$ to be $3$; does this look correct to you?

Does it make sense to show that the above series converge and that its limit is $1$? Yes, of course, it"s a simple exercise in real analysis. Therefore, it does make sense to prove that $0.99999\dots = 1$.

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  • $\begingroup$ Surely "we may "define" $\sum \limits _{n=1} ^\infty \dfrac 1 n$ to be $3$". But I don't understand what you wanted to mean when you said "does this look correct to you?" What do you mean by looking "correct" here? $\endgroup$ – user 170039 Jan 15 '16 at 13:49
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    $\begingroup$ @user170039: No, of course you can't "define" it, because $\sum \limits _{n=1} ^\infty x_n$ is already defined as $\lim \limits _{N \to \infty} \sum \limits _{n=1} ^N x_n$. Given that $\lim \limits _{N \to \infty} \sum \limits _{n=1} \frac 1 n = \infty$, defining $\lim \limits _{N \to \infty} \sum \limits _{n=1} ^\infty \frac 1 n$ to be $3$ will lead to the contradiction $3 = \infty$. To conclude: given that $\sum \limits _{n=1} ^\infty \dfrac 9 {10^n}$ already has a meaning as a convergent series, it is equal to $1$, not defineable to be $1$! $\endgroup$ – Alex M. Jan 15 '16 at 15:31
  • $\begingroup$ I can "define" it. What you are saying is that whether my definition is "consistent" or not. But that has no (at least it seems so to me) relation to the "definition" itself. $\endgroup$ – user 170039 Jan 15 '16 at 15:39
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    $\begingroup$ Also, why "$0.999\ldots$ means $\dfrac{9}{10}+\dfrac{9}{10}+\dfrac{9}{10}+\ldots=\displaystyle\sum_{n=1}^\infty \dfrac{9}{10^n}$"? $\endgroup$ – user 170039 Jan 15 '16 at 15:41
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As far as I understand the original question, the question has not been answered. Not that the other posts are wrong but they always touch another topic. Actually two questions were asked:

Does it make sense to prove $0.999\ldots=1$?

Can one give any "meaning" of the symbol $0.999\ldots$ other that defining it to be 1?

Regarding the first question: I'll understand it in the way "Why are we interested in this?"

Mathematicians are often interested if something is unique in this case the decimal representation. $0.999\ldots=1$ proves by example that the decimal representation is not unique.

Now let's consider the second question. Yes we could give it another meaning, bacause we can define anything we want. But if we define it in another way without changing other definitions we obtain a contradiction.

Could we change the definitions in a way that $0.999\ldots \neq 1$ and they still are reasonable in a way? I would say probably yes but I'm no expert in this matter. The Wikipedia article about $0.999\ldots$ states following:

The equality of 0.999… and 1 is closely related to the absence of nonzero infinitesimals in the real number system, the most commonly used system in mathematical analysis. Some alternative number systems, such as the hyperreals, do contain nonzero infinitesimals. In most such number systems, the standard interpretation of the expression 0.999… makes it equal to 1, but in some of these number systems, the symbol "0.999…" admits other interpretations that contain infinitely many 9s while falling infinitesimally short of 1.

You might want to read the wikipedia article in more detail.

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Can we assign a meaning to $0.999...$?

$1=0.999...$ may be written as

$$\begin{align} 1&=\frac{9}{10}\sum_{k=0}^\infty \frac{1}{10^k}\\ &=\left(1-\frac{1}{10}\right) \sum_{k=0}^\infty \frac{1}{10^k}\\ \end{align} $$

This is case $r=\frac{1}{10}$ of the more general

$$1=(1-r)\sum_{k=0}^\infty {r}^k, |r|<1$$

The product $(1-p)p^k$ has probabilistic interpretations. For instance, in the context of Nondeterministic finite automata, it is the probability of exiting a state that has self-transition probability $p$ after exactly $k$ self-transitions. Summing for all nonnegative integers represents the probability of exiting after any possible number of self-transitions. Therefore, the question about $.999...=1$ may be rewritten as

What is the probability that a state with self-transition probability $p=\frac{1}{10}$ is exited?

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  • $\begingroup$ Indeed, it might be an infinitesimal probability, as explored in a number of recent articles. Your approach answers the question in the negative only if you assume that there are no infinitesimals in the number system being used. $\endgroup$ – Mikhail Katz Jun 7 '17 at 7:53
  • $\begingroup$ An infinitesimal probability that it is never exited, right? $\endgroup$ – Jaume Oliver Lafont Jun 7 '17 at 8:08
  • $\begingroup$ Right. If your model of the process is that there is an infinite number (in the sense of Robinson's framework, for instance) of simultaneous tries then indeed there will be a positive infinitesimal probability that it does not exit. $\endgroup$ – Mikhail Katz Jun 7 '17 at 8:13
  • $\begingroup$ I admit I should read those articles before, but just for intuition, let me ask: is that infinitesimal probability any similar to the probability of picking exactly $\frac{1}{2}$ when choosing randomly a real number in $(0,1)$ under uniform distribution? It is possible, not impossible as picking, say, $\frac{3}{2}$. $\endgroup$ – Jaume Oliver Lafont Jun 7 '17 at 8:22
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    $\begingroup$ Jaume, very good question. The answer is negative: it is not similar. Rather, it is similar to picking a number in an infinitesimal interval centered on $\frac{1}{2}$. If you are intrigued by these ideas I would recommend Keisler's excellent textbook Elementary Calculus as a first try. $\endgroup$ – Mikhail Katz Jun 7 '17 at 8:25
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We don't simply define $0.9999.... $ to be equal to 1.

We do a lot of background analysis first which involves constructing the real numbers out of the rational numbers.

In very short summary:

Consider this basic fact about rational numbers: no matter how close together two rational numbers are we can find a rational number between them (which also means for any value greater than zero, no matter how small, you can find a smaller one between it and zero).

This is nice. It means we can get as close as we like to any value using rational numbers.

Now consider this monkey wrench: There are many values we can't express (for example: pi and the square root of 2). [This is weird because we can get as close as we like to to these values by the statement above.]

And consider this horrible result: Between any two rationals no matter how close we get, there is always one of the inexpressible holes between them!

So how can we think of these ... irrational values...

Well, a lot of subtle analysis and picayune debating we notice that in the rational numbers we can have an infinite set of rationals, where all the rationals are within a range, but the set needn't have a biggest value. For example: {all the rational numbers less than pi}; all these values are less than 3 1/3 so it's bounded, but there is no precise rational upper bound to this set.

(Another such set is all the numbers $.9, .99, .999, .9999$ etc. It's infinite but each term is rational. It has no biggest value. And all values are less than 1.)

So the issue was can we come up with a bigger system of numbers than the rationals, where every set will have a precise upper bound? Mathematicians decided they could and it was called the Real numbers[1].

Now here is the definition of the real numbers and how the were created (it's very subtle): Every real number is the least upper bound limit of a bounded set of rational numbers. And every bounded set of rational numbers has a real number as an upper bound.

That's the definition of the real numbers.

So {.9, .99, .999, .9999,.....} is an bounded infinite set of rational numbers. By definition it has a real number least upper bound limit. We call that real number .999999999999......

Okay, we have defined .99999...... and we DIDN'T define it as 1. So now we can prove that it does equal 1.

The general idea is that if .99999......= c < 1 then can find a finite number in {.9, .99, .999, .9999,.....} that is bigger than c. So we were wrong about c being an upper bound of {.9, .99, .999, .9999,.....} So .9999....... $\ge$ 1. But 1 is bigger than any of the {.9, .99, .999, .9999,.....}. So .99999....... $\not >$ 1. The only consistent option left is .99999.... = 1.

[1] Well, the didn't just wave their hands and declare it. They had to prove constructing such a number system was possible. The proof is ... abstract. And tedious. And abstract.

===== 2nd answer: Different approach and philosophy====

The OP has a point that we don't really "need" to prove .999.... = 1. If .999... equals anything at all, showing that it must by 1 is trivially easy. The idea of proving .999.... = 1 misses the point. The point really is how do we know that .999.... equals anything at all.

The OP isn't entirely accurate in stating we defined .999... to be 1. We defined .999.... to be something that can be shown to be 1. (Slight difference but a difference with ramifications.) Other comments claim we make an assumption via definitions that .999... is a limit of bounded sequence of rational numbers and another assumption via definition that limits of bounded sequence of rational numbers are real numbers.

These aren't entirely accurate either. The definitions weren't assumptions. They were analysis as to what the real numbers are and the discoveries came along the way.

Suppose we know nothing about the real numbers. ... well, not nothing but suppose we have no real sense of them.

With or without real numbers we do have to define $ 0.9999....$ as $ \sum_{n = 1}^{\infty} \frac{9}{10^n}$. But we don't really know anything about it. It's possible that it's unbounded and "blows up". (Actually, we can show that can't happen but I don't want to get into that yet.) It's possible that it adds up to an actual number. It could be rational or it could be one of those numbers that can't be written as a rational like square root of two or pi.

But the natural concern is that it might simply never resolve into anything.

To me, the surprising and subtle thing about analyzing the real numbers is the realization that the very nature of the real numbers means that this "simply never resolving into anything" is impossible. Everything in the reals that is bounded resolves to something, and every real is a resolution of something. This really astounded me when I finally wrapped my head ahead. I mean it really really surprised me! (Yes, I'm that much a geek.)

Okay, so this leads to the fundamental question: Real Numbers. What the Heck are they?

We know that we can measure discrete units via integers, like $n$. And we know we can chop these discrete units into $m$ pieces as $1/m$ and as $m$ can be arbitrarily large, these $1/m$ can be arbitrarily precise and this collection of ratios, so all the possible $n/m; m \ne 0$ form a system of arbitrarily precise measurements that span every possible range. So these Rationals can measure anything to arbitrary precision.

That ought to be enough.

But it isn't. Because we know there are always numbers like $\sqrt{2}$ and $\pi$ that can't be written as any Rational = $m/n$ where $m$ and $n$ are integers.

So w have gone for a thousand years with a number system that we don't really understand and cant describe. What are these "holes" and what do we know about them?

Enter the Dedekind cut.

An example: Let's "cut" the rationals into two sets. Set $A = ${all the rational numbers, $r$, such that $r^2 < 2$} and $B = ${all the rational numbers, $s$, such that $s^2 < 2$}. Several things we note about these sets:

1) they are completely disjoint and neither are empty; 2) all the elements or 1 are smaller than all the elements of the other; 3) A does not have a largest element and B does not have a smallest element; and 4) They are arbitrarily close together. (That is, for any number $e > 0$ no matter how small we can always find an $a \in A$ and a $b \in B$ so that $b - a < e$. Always.)

Any "procedure" that can cut the Rational numbers into two sets with those 4 properties is called a "cut" and we can, for sake of vocabulary, refer to such cut as $\overline x$ and the two sets as $A\overline x$ and $B\overline x$.

And we can consider the collection of all such cuttings. One thing is about these cuts is that we don't necessarily need to know how to describe them for them to exist.

One such cut could be $\overline \omega$ such that $A\overline \omega = $ {all rationals less than some .99...9} and $B\overline \omega = $ {all rationals larger than all .99...9}. [That's not actually true and I pulled a fast one on you. A brownie point to any one who can figure out my deception.]

Now the astute reader might have noticed that although these cuts cut the rational in two, the two sets don't always contain all the rationals. Ex: A = {all rationals < 27 1/2} and B = {all rationals > 27 1/2} is a cut. But the rational number 27 1/2 is not in either A nor B. This is okay. The sets don't have to contain all the rationals. But notice, 27 1/2 is the only rational that is not in either set.

We can say a cut like that one cuts the rationals "on" a rational number and others cut the rationals "between" rational numbers. There is a correspondence between the rational numbers and the cuts that cut on the rational numbers and we can refer to such a cut as $\overline{1/2}$ as the cut that cuts on 1/2.

The other cuts that cut between rational numbers are extra cuts.

So... back to the collection of all possible cuts...

They form a number system.

For two cuts $\overline x$ and $\overline y$ we say $\overline x < \overline y$ if $A\overline x \subset A\overline y$. We note that if we define $\alpha$ = $A\overline x + A\overline y = $ {all rationals that are sums of two rationals one in $A\overline x$ and the other in $A\overline y$} and $\beta$ = {all the rationals that are bigger than all the rationals in $\alpha$}, then $\alpha$ and $\beta$ perform a cut.

[Again, I'm pulling a fast one. Another brownie point to whoever finds it. It's the same as the fast one I pulled above.]

We define $\overline x + \overline y$ as finding that type of cut.

Similarly we define subtraction, multiplication, and division. (Divisions kind of a pain to define but we do it.)

So the collection of all cuts form a number system. What's more, the cuts that cut "on" rational numbers behave equivalently as the rational numbers do.

These "cuts" are the real numbers. Every real number is a point at which we "cut" the rational numbers into two sets. If we cut "on" a rational number that point is the rational number, if we cut "between" rational numbers that point is some irrational number.

That's it. That's what the heck Real Numbers are.

... deep sigh and coffee break....

Okay, WHAT THE #@&! DOES THAT HAVE TO DO WITH .9999....?

Notice that we now have discovered (NOT "defined"; NOT "assumed") that every real number cuts the rationals in two. So every bounded sequence of rationals can be fit into the $A\overline x$ set of some cut $\overline x$ and we can find precisely the lowest cut that will do that. And that cut point is a precise and unique real number.

Thus the FUNDAMENTAL PROPERTY OF REAL NUMBERS!!!! Every real number is a limit of a sequence of rational numbers and every bounded sequence of rational numbers has a real number limit. And this real number is the point that cuts the rational numbers at precisely the "edge" of the sequence.

So let's look at $.9999...... = \sum_{n = 1}^{\infty} \frac{9}{10^n}$ and let's look at this bounded sequence of rational numbers: {.9; .99; .999; etc.} and let's look at the number $1$ which is larger than all the elements of the sequence.

So we know:

1)There is a cut $\overline {\omega}$ where $A\overline{\omega}$ = {all the rationals that are less than or equal to sum .999...9}

2)This cut occurs at a real number, $\omega$.

3)$\omega = \lim $ of the sequence {.9;.99;.999; etc.}

4).999.... "lingers" between the elements of $A\overline {\omega}$ and the elements of $B\overline{\omega}$.

Therefore we can conclude $.999.... = \omega$.

And that's it:

$.9999..... = \sum 9/10^n = \lim \{.9'.99;.999;...\}$ is a real number.

Now it's trivial to prove $.99999..... = 1$.

And there was not a single assumption or circular definition.

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  • $\begingroup$ Well, to be subtle... analysis has proven that all bounded sequences of rationals get closer and closer to one and precisely one real number. And analysis has shown an infinite sum of decimals is the limit of the sequence. So we know that the sequence gets closer to exactly one number and we know that what the number is is .9999... What remains is to show that this number happens to be 1. $\endgroup$ – fleablood Jan 14 '16 at 18:26
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    $\begingroup$ To say ".9, .99, .9999 gets closer and closer to 1 so if we do it an infinite number of times we say that is 1" is a bit like saying "Well, we put a cat in a box and we don't know if the cat is alive or dead so we say the cat is both". They miss the basic point and mislead into thinking everything is an estimate. $\endgroup$ – fleablood Jan 14 '16 at 18:29
  • $\begingroup$ Put much simpler: x = 0.999..., 10x = 9.999..., 9x = 9, x = 1. $\endgroup$ – user117644 Jan 14 '16 at 22:06
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    $\begingroup$ "With or without real numbers we do have to define $0.9999$ as $\displaystyle\sum \dfrac{9}{10^n}$"-why? $\endgroup$ – user 170039 Jan 15 '16 at 14:11
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    $\begingroup$ One important thing. We never started talking about infinite decimals until AFTER analysts such as Dedekind had defined and proven the reals were such a system with the least upper bound so that all sequences of rationals that infinitely approach have real limits. It's only AFTER that, that is was determined that infinite decimals were meaningful and well-defined and !*useful*! They were shown to be a method to thoroughly describe the reals. Unfortunately, they became a magic mantra to elementary school teachers and befuddled students. $\endgroup$ – fleablood Jan 15 '16 at 19:01
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My answer is not essentially different from some of the ones presented above, except for some notation. My apologies beforehand. Let $X = \{0, 1,2,3, \cdots , 9\}^{\mathbb{N}}$ ($\mathbb{N} = \{1, 2, \dots\}$) where $X$ is equipped with the product topology. $X$ is a compact (metrizable) space, the space of "decimals". Define a ``canonical'' continuous surjective map $\varphi: X \rightarrow [0, 1]$ by $\{r_i\}_{i \in \mathbb{N}}\ \mapsto \sum_{k=1}^\infty \ \ r_k/10^k$. The map $\varphi$ is not injective since for instance $\varphi$ ( 5,0,0,0, $\ldots$)=$\varphi$ ( 4,9,9,9, $\ldots$)

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protected by J. M. is a poor mathematician May 26 '17 at 19:09

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