I don't know whether the proof of $.999... = 1$ means it's true as a limit or whether the number $.999... = 1$, if that makes any sense.

In the basic algebra proof, no limits are ever taken; yet, there are plenty of proofs that use the infinite series expansion and by definition, the limit of partial sums would necessarily be used.


  • $\begingroup$ Please provide the proof to which you refer... $\endgroup$ Commented Sep 22, 2017 at 1:04
  • $\begingroup$ If you define it as a real number, then it is necessarily define-able as a limit (as per construction of the reals). $\endgroup$ Commented Sep 22, 2017 at 1:04
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    $\begingroup$ Depends on what you "define" to be a number. What is a number?! $\endgroup$ Commented Sep 22, 2017 at 1:08
  • $\begingroup$ @SimplyBeautifulArt, right -- but then the limit of 0.999... = 1 seems unsurprising, right? It seems that only when I consider 0.999... a number and not a limit of a sequence, and then show it is exactly equal to the number 1 with basic algebra does it become a surprising fact. Do you think it's also surprising when viewing 0.999... as a limit, and that it is equal to 1? I don't see it ... $\endgroup$ Commented Sep 22, 2017 at 3:15
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    $\begingroup$ $0.999\ldots$ represents the number that is the limit of the sequence $0, 0.9, 0.99, 0.999, 0.9999, \ldots$. Thus you can think of $0.999\ldots$ as both a number and a limit at the same time. Since the sequence converges to $1$, the number represented by $0.999\ldots$ is $1$, which can also be represented as $1.000\ldots$. $\endgroup$
    – md2perpe
    Commented Sep 22, 2017 at 14:46

2 Answers 2


This really depends on your definition of number.

That said, the naive definition of real number usually goes along the lines of the following.

  1. The natural numbers $\mathbb{N}$ are intuitive enough that we assume them to be understood.
  2. We introduce the concept of additive inverses and get the integers $\mathbb{Z}$.
  3. We introduce the concept of multiplicative inverses and get the rationals $\mathbb{Q}$ -- at this point, thought of as numbers of the form $\frac{p}q$ with $p,q\in\mathbb{Z}$.
  4. Now, we introduce the concept of decimal expansions. We observe that rational numbers either have terminating decimal expansions or infinite but periodic expansions. We will say a number is irrational if it is given by a decimal expansion that does not fit into those cases -- that is, if its decimal expansion is infinite and non-periodic.
  5. By combining the rational and irrational numbers, we get the reals $\mathbb{R}$.

Of course, the notion of limit is implict in steps $(4)$ and $(5)$ at least in the sense of 'How does $\mathbb{R}$ inherit an order/metric from $\mathbb{Q}$?'. That said, when the reals are introduced to children, this is inuitive enough to be usually omitted without any problem.

At this point, it should be clear that under our definition $.999\dots$ gets to be a number in its own right -- we don't even know what limits are! It just so happens that a number's decimal expansion need not be unique.

At step $(3)$, one typically learns that the representation of a rational number as a fraction of the form $\frac{p}q$ with $p,q\in\mathbb{Z}$ is also not unique. Although obviously not the same case, this precedent should make it so that the idea

$$``\text{The representation of a real number via decimal expansion might not be unique"}$$

should not be unthinkable or completely alien.

In the very least, it should come as no surprise from our 'construction' that $.999\dots$ is rational: after all, it has an infinite periodic decimal expansion.

  • $\begingroup$ By the way, if anyony knows how to MathJax opening quotation marks into LaTeX, I'd be grateful to know. $\endgroup$ Commented Sep 22, 2017 at 5:08
  • $\begingroup$ I believe you can use `` for opening marks. Like $``xyz"$ $\endgroup$
    – Jam
    Commented Sep 22, 2017 at 12:46

If we don't define $0.999\ldots$ as a limit, how else do we define it? Of course, the decimal expansion of a number, $x$, is the digits we need such that $x=\frac{a_1}{1}+\frac{a_2}{10}+\frac{a_1}{100}+\ldots$ is as close as possible to $x$. So then, what else can we define $0.999\ldots$ as? Except for $x=\frac{9}{10}+\frac{9}{100}+\frac{9}{1000}+\ldots$ but this definition is a limit. Sure, we could define the number as $3\cdot\frac13$, but this doesn't tell us anything about the decimal expansion. We could try to define it as $3\cdot(0.333\ldots)$, but we'd need to define that decimal expansion as a limit too.

There may be algebraic proofs of $0.999\ldots=1$, but there's an implicit use of limits in making the number on the left hand side. If we didn't use limits, we'd just call it $1$ and be done with it, since there'd be no confusion. It's only when we use limits (albeit behind the scenes) that we can actually show numbers as non-terminating decimal expansions.

Whether you use sequences or series, the result is the same: $0.999\ldots$ is the limit of $0.9\rightarrow 0.99\rightarrow 0.999\rightarrow\ldots$

Regarding your question about "just a number". What's stopping the value of a sequence or series from being a number? All numbers are defined by successive operations. You can only get $2$ if you can make $1+1$. You can only get $0.5$ if you can make $1\div2$. Likewise, $0.999\ldots$ is just as much the limit of a sequence as is a number.

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    $\begingroup$ It's not a surprising fact if you understand limits. But it is a surprising fact if you don't. People who don't understan limits, think that $.9< 1$ and $.99 < 1$ and $.999999.......9999999 < 1$ so therefore $.99999999...... < 1$. What they are missing is that if $0.999999.......$ means anything at all (how can you have an infinity digit number??? That's not an sarcastic question; seriously how can you?) that it means a limit as an infinite series. And if you understand limits, you are unsurprised that the limit is the number 1. $\endgroup$
    – fleablood
    Commented Sep 22, 2017 at 5:38

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