# Is it $0.999999999\ldots = 1$ true?

I'm told by smart people that $0.999999999\ldots = 1$, and I believe them, but is there a proof that explains why this is?

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Seriously people what's with all the duplicates? Check and see if someone has already given your answer first! – Noah Snyder Jul 20 '10 at 20:28
Let me say a word about why 1/3 = .33333... is not obvious at all. What does it mean to say that some number is 1/3? Well division is the operation that undoes multiplication, so it means that it's a number that when you multiply by 3 you get 1. Well what happens when you multiply .3333... times 3? You get .9999... So unless you already know that .99999...=1 then you can't prove that 1/3 = .3333... The fact that .9999...=1 is more basic than the fact that 1/3 = .33333....! – Noah Snyder Jul 20 '10 at 21:32
Between any two different numbers, there is always another number between them. there is no number between 0.999... and 1 – Joao Apr 28 '14 at 3:07
use the identity that if $a+\epsilon > 1$ for all $\epsilon >0$, then $a=1$ – Harish Jan 15 at 11:26
New prospective answerers, please heed Noah Snyder's comment from over five years ago. It's all the more valuable now. – Brian Tung Jan 15 at 23:12

What does it mean when you refer to $.99999\ldots$? Symbols don't mean anything in particular until you've defined what you mean by them.

In this case the definition is that you're taking the limit of $.9$, $.99$, $.999$, $.9999$, etc. What does it mean to say that limit is $1$? Well, it means that no matter how small a number $x$ you pick, I can show you a point in that sequence such that all further numbers in the sequence are within distance $x$ of $1$. But certainly whatever number you chose your number is bigger than $10^{-k}$ for some $k$. So I can just pick my point to be the $k$th spot in the sequence.

A more intuitive way of explaining the above argument is that the reason $.99999\ldots = 1$ is that their difference is zero. So let's subtract $1.0000\ldots -.99999\ldots = .00000\ldots = 0$. That is,

$1.0 -.9 = .1$

$1.00-.99 = .01$

$1.000-.999=.001$,

$\ldots$

$1.000\ldots -.99999\ldots = .000\ldots = 0$

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Finite ones do, but infinite ones don't! The whole reason that this question confuses people is that defining what an infinite decimal means is difficult and confusing! – Noah Snyder Jul 20 '10 at 20:23
I think it's pretty standard to define infinite decimals as the infinite series where the terms are the individual digits divided by the appropriate power of the base. That is, 0.99999... = 9/10 + 9/10^2 + 9/10^3 + ... – Isaac Jul 20 '10 at 20:57
But that's exactly my point, taking the limit of series (and in fact, determining when such a limit exists) is a difficult and confusing concept! – Noah Snyder Jul 20 '10 at 21:30
@Doug: I don't understand what you're talking about. Could you try to clarify? What do you mean by an infinite sum if you don't mean the limit of the partial sums? – Noah Snyder Jul 23 '10 at 3:07
@Doug: Then your reading is incorrect. – user126 Aug 17 '10 at 22:14

Suppose this was not the case, i.e. $0.9999... \neq 1$. Then $0.9999... < 1$ (I hope we agree on that). But between two distinct real numbers, there's always another one (say $x$) in between, hence $0.9999... < x < 1$.

The decimal representation of $x$ must have a digit somewhere that is not $9$ (otherwise $x = 0.9999...$). But that means it's actually smaller – $x < 0.9999...$, contradicting the definition of $x$.

Thus, the assumption that there's a number between $0.9999...$ and $1$ is false, hence they're equal.

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This proof also relies on the assumption that every real number can be represented by a (potentially infinite) decimal, which might or might not be accepted by someone asking the original question. – bryn Jul 22 '10 at 2:21
@bryn, this proof relies on a further assumption that the OP had specifically real numbers in mind when he asked the question. – Mikhail Katz Nov 25 '13 at 17:12
@BogdanAlexandru $\frac{x+y}{2}$ is a number between $x$ and $y$ – Holographer Sep 1 '14 at 15:39
$\neg\neg P\implies P$, yikes. – NikolajK Nov 27 '15 at 3:06
@cat: Not completely sure what you mean, but see Double negative elimination and Constructive logic. – NikolajK May 30 at 21:31

What I really don't like about all the above answers, is the underlying assumption that $1/3=0.3333\ldots$ How do you know that? It seems to me like assuming the something which is already known.

A proof I really like is:

\begin{align} 0.9999\ldots × 10 &= 9.9999\ldots\\ 0.9999\ldots × (9+1) &= 9.9999\ldots\\ \text{by distribution rule: }\Space{15ex}{0ex}{0ex} \\ 0.9999\ldots × 9 + 0.9999\ldots × 1 &= 9.9999\ldots\\ 0.9999\ldots × 9 &= 9.9999\dots-0.9999\ldots\\ 0.9999\ldots × 9 &= 9\\ 0.9999\ldots &= 1 \end{align}

The only things I need to assume is, that $9.999\ldots - 0.999\ldots = 9$ and that $0.999\ldots × 10 = 9.999\ldots$ These seems to me intuitive enough to take for granted.

The proof is from an old high school level math book of the Open University in Israel.

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You're also assuming that you can multiply 0.9999... by 10 and get 9.9999... (or, rather, that arithmetic with infinite decimals works normally), which is not at all unreasonable to assume. – Isaac Jul 20 '10 at 20:59
0.9999... x 9 = 8.999999... – Sklivvz May 3 '11 at 20:51
@Sklivvz, true, since 8.999....=9. Did you have trouble understanding the algebra I did? – Elazar Leibovich May 4 '11 at 5:55
Sklivvz, but my answer does not rely on the fact that 8.999...=9, did it? Or am I missing something (tried to make it a bit more clear) – Elazar Leibovich May 4 '11 at 5:59
@Michael I never evaluate 9*0.9999 at step 3 or at any step. I leave the expression as it is, and then use the fact that if 9x=9 then x=1 in the last step. – Elazar Leibovich Feb 17 '14 at 9:00

Assuming:

1. infinite decimals are series where the terms are the digits divided by the proper power of the base
2. the infinite geometric series $a + a \cdot r + a \cdot r^2 + a \cdot r^3 + \cdots$ has sum $\dfrac{a}{1 - r}$ as long as $|r|<1$

$$0.99999\ldots = \frac{9}{10} + \frac{9}{10^2} + \frac{9}{10^3} + \cdots$$

This is the infinite geometric series with first term $a = \frac{9}{10}$ and common ratio $r = \frac{1}{10}$, so it has sum $$\frac{\frac{9}{10}}{1 - \frac{1}{10}} = \frac{\frac{9}{10}}{\frac{9}{10}} = 1.$$

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Your method is a simple way of converting the decimal representation of a rational number into a fraction, e.g. $0.150150150...=\sum_{n\geq 1}\frac{150}{10^{3n}}=\frac{0.150}{1-10^{-3}}=\frac{50}{333}$ – Américo Tavares Aug 16 '10 at 22:02
This is exactly how I would answer the question. It is the only correct answer here in a sea of gibberish. +1. – MPW Jul 9 '14 at 16:02

Okay I burned a lot of reputation points (at least for me) on MathOverflow to gain clarity on how to give some intuition into this problem, so hopefully this answer will be at least be somewhat illuminating.

To gain a deeper understanding of what is going on, first we need to answer the question, "What is a number?"

There are a lot of ways to define numbers, but in general numbers are thought of as symbols that represent sets.

This is easy for things like the natural numbers. So 10 would correspond to the set with ten things -- like a bag of ten stones. Pretty straight forward.

The tricky part is that when we consider ten a subset of the real numbers, we actually redefine it. This is not emphasized even in higher mathematics classes, like real analysis; it just happens when we define the real numbers.

So what is 10 when constructed in the real numbers? Well, at least with the Dedekind cut version of the real numbers, all real numbers correspond to a set with an infinite amount of elements. This makes 10 under the hood look drastically different, although in practice it operates exactly the same.

So let's return to the question: Why is 10 the same as 9.99999? Because the real numbers have this completely surprising quality, where there is no next real number. So when you have two real numbers that are as close together as possible, they are the same. I can't think of any physical object that has this quality, but it's how the real numbers work (makes "real" seem ironic).

With integers (bag of stones version) this is not the same. When you have two integers as close to each other as possible they are still different, and they are distance one apart.

Put another way, 10 bag of stones are not the same as 9.9999999 but 10 the natural number, where natural numbers are a subset of the real numbers is.

The bottom line is that the real numbers have these tricky edge cases that are hard to understand intuitively. Don't worry, your intuition is not really failing you. :)

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The link doesn't work. – user 170039 Jan 5 at 15:08
x = 0.999...
10x = 9.999...
10x - x = 9.999... - 0.999... = 9
-> 9x = 9
-> x = 1 thus, 0.999... = 1

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This is the most intuitive argument, although some might say "But 10x-x isn't 9, because there's going to be a mismatch all the way to the right" - Noah's more complex deals with that. – Charles Stewart Jul 21 '10 at 10:55
10x - x != 9. 10x - x would be 8.9999...1. However infinite the extent of 9s is in x, if we multiply it by 10, the nines are shifted left by one position and a zero inserted at the "last" place, and then when you subtract the other number there is a nine subtracted from a zero at the far right. Otherwise we'd have to give 0.999.. some unusual properties like automatically increasing the number of nines when it is multiplied. It would not be just an ordinary number. Maybe that's the problem. 0.999... might just not be an ordinary type number as some people are using it. – Doug Treadwell Jul 23 '10 at 2:19
@Doug It's incorrect to talk about the "number of nines" because infinity minus a number = infinity. Infinity is transcendent. It means "uncountable". If you take infinity and slide it left a little bit, it's still infinity long. – ErikE Sep 25 '10 at 7:00

One argument against this is that 0.99999999... is "somewhat" less than 1. How much exactly?

      1 - 0.999999... = ε              (0)


If the above is true, the following also must be true:

9 × (1 - 0.999999...) = ε × 9


Let's calculate:

0.999... ×
9        =
───────────
8.1
81
81
.
.
.

───────────
8.999...


Thus:

     9 - 8.999999... = 9ε              (1)


But:

         8.999999... = 8 + 0.99999...  (2)


Indeed:

8.00000000... +
0.99999999... =
────────────────
8.99999999...


Now let's see what we can deduce from (0), (1) and (2).

9 - 8.999999... = 9ε                      because of (2)
9 - 8.999999... = 9 - (8 + 0.99999...) =  because of (1)
= 9 -  8 - (1 - ε)        because of (0)
=   1    -  1 + ε
=               ε.


Thus:

9ε = ε

8ε = 0

ε = 0

1 - 0.999999... = ε = 0


Quod erat demonstrandum. Pardon my unicode.

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I did my best to avoid 0.00000..., but this made the calculations not as strikingly simple as I'd have liked to. – badp Jul 20 '10 at 21:10
Why was this voted down? It seems reasonable to this amateur math enjoyer. – ErikE Sep 25 '10 at 7:07
@Emtucifor I guess this sounds like "nonsense" to people that disagree on the basic premise of 0.999... = 1 :) – badp Sep 25 '10 at 7:21
8ε = 0 instead of 10ε = 0 – user59671 May 9 '13 at 9:37
@CutieKrait Thanks for that. (You could've suggested that as an edit, btw. You would've got reputation for it, too! :) – badp May 9 '13 at 10:35

.999... = 1 because .999... is a concise symbolic representation of "the limit of some variable as it approaches one." Therefore, .999... = 1 for the same reason the limit of x as x approaches 1 equals 1.

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There are genuine conceptual difficulties implicit in this question. The transition from the rational numbers to the real numbers is a difficult one, and it took a long time and a lot of thought to make it truly rigorous. It has been pointed out in other answers that the notation $0.999999\ldots$ is just a shorthand notation for the infinite geometric series $\sum_{n=1}^{\infty} \left( \frac{9}{10} \right)^{n},$ which has sum $1.$ This is factually correct, but still sweeps some of the conceptual questions under the carpet. There are questions to be addressed about what we mean we we write down ( or pretend to) an infinite decimal, or an infinite series. Either of those devices is just a shorthand notation which mathematicians agree will represent some numbers, given a set of ground rules. Let me try to present an argument to suggest that if the notation $0.99999\ldots$ is to meaningfully represent any real number, then that number could be nothing other than the real number $1$, if we can agree that some truths are "self-evident". Surely we can agree that the real number it represents can't be strictly greater than $1$, if it does indeed represent a real number. Let's now convince ourselves that it can't be a real number strictly less than $1,$ if it makes any sense at all. Well, if it was a real number $r < 1,$ that real number would be greater than or equal to $\sum_{n=1}^{k} \left( \frac{9}{10} \right)^{n}$ for any finite integer $k.$ (This last number is the decimal $0.99 \ldots 9$ which terminates after $k$ occurrences of $9,$ and differs from $1$ by $\frac{1}{10^{k}}.$ Since $0 < r <1,$ there is a value of $k$ such that $\frac{1}{10^{k}} < 1-r,$ so $1 - \frac{1}{10^{k}} >r.$ Hence $\sum_{n=1}^{k} \left( \frac{9}{10} \right)^{n} > r.$ But this can't be, because we agreed that $r$ should be greater than or equal to each of those truncated sums. Have I proved that the recurring decimal is equal to $1$? Not really- what I have proved is that if we allow that recurring decimal to meaningfully represent any real number, that real number has to be $1,$ since it can't be strictly less than $1$ and can't be strictly greater than $1$. At this point, it becomes a matter of convention to agree that the real number $1$ can be represented in that form, and that convention will be consistent with our usual operations with real numbers and ordering of the real numbers, and equating the expression with any other real number would not maintain that consistency.

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This is the best answer here; it's a shame it was posted four years later so it's gotten so little attention. – Eric Wofsey Oct 4 '15 at 0:44
If I am not mistaken then this answer also answers the question of this post. – user 170039 Jan 6 at 14:32
By the way, I was thinking that maybe you could write an answer for the linked post as well (or else, can you just tell me how can I give a link to this answer in a comment below my post?). – user 170039 Jan 6 at 14:40
I'm not sure how to set up links on here ( maybe the "cite") button. I am basically agreeing with what you say in your post. – Geoff Robinson Jan 6 at 15:03

If you take two real numbers x and y then there per definition of the real number z for which x < z < y or x > z > y is true.

For x = 0.99999... and y = 1 you can't find a z and therefore 0.99999... = 1.

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You can visualise it by thinking about it in infinitesimals. The more $9's$ you have on the end of $0.999$, the closer you get to $1$. When you add an infinite number of $9's$ to the decimal expansion, you are infinitely close to $1$ (or an infinitesimal distance away).

And this isn't a rigorous proof, just an aid to visualisation of the result.

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Indeed this is true. The underlying reason is that decimal numbers are not unique representations of the reals. (Technically, there does not exist a bijection between the set of all decimal numbers and the reals.)

Here's a very simple proof:

1 / 3 = 0.333... (by long division)

=> 0.333 * 3 = 0.999... (multiplying each digit by 3)

But then we already know 0.333... * 3 = 1

Therefore 0.999... = 1

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-1. This is not a proof at all! Why is 1/3=0.333...? Seriously folks, for the private beta, let's try to maintain a little correctiness. – Scott Morrison Jul 20 '10 at 20:03
@Scott: Sure it is. You can prove it easily by long division. This is about algorithms for mathematical methods really. – Noldorin Jul 20 '10 at 20:12
@Scott: Might help to stop whining and post what you think is the 'correct' answer then. – Noldorin Jul 20 '10 at 20:17
Just to nitpick, there is a bijection between the set of decimal expansions and reals because they are sets with the same cardinality. It's just that the natural map taking expansions to real numbers isn't injective. – Simon Nickerson Jul 20 '10 at 21:15
@Scott, I see it would not be obvious that 1/3=0.333..., but as by Noldorins comment regarding long division, what would be wrong with this as a proof, if the first line is annotated with 'by long division' ? – Sami Jul 21 '10 at 4:58

Given (by long division):
$\frac{1}{3} = 0.\bar{3}$

Multiply by 3:
$3\times \left( \frac{1}{3} \right) = \left( 0.\bar{3} \right) \times 3$

Therefore:
$\frac{3}{3} = 0.\bar{9}$

QED.

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I think the long division precisely involves the proof of a limit like the sum mentioned above... – Alexander Frei Aug 27 '14 at 21:07

Another approach is the following:

$$0.\overline9=\lim_{n \to \infty} 0.99.....9 = \lim_{n \to \infty} \sum\limits_{k=1}^n \frac{9}{10^k}=\lim_{n \to \infty} 1-\frac{1}{10^n}=1-\lim_{n \to \infty} \frac{1}{10^n}=1$$

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This approach appears in Isaac's answer from 4 years earlier. – Jonas Meyer Aug 26 '14 at 2:15

Often times people who ask this question are not very convinced by a proof. Since they may not be particularly math inclined, they may feel that a proof is a sort of sleight-of-hand trick, and I find the following intuitive argument (read "don't down-vote me for lack of rigor, lack of rigor is the point") a bit more convincing:

STEP 1) If $.99...\neq1$, everyone agrees that it must be less than $1$. Let $\alpha$ denote $.99...$, this mysterious number less than $1$.

STEP 2) Using a number line, you can convince them that since $\alpha<1$, there must be another number $\beta$ such that $\alpha<\beta<1$.

STEP 3) Since $\alpha<\beta$, one of the digits of $\beta$ must be bigger than the corresponding digit of $\alpha$.

STEP 4) However it is usually intitively clear that you cannot make any digit of $.99...$ bigger without making the resulting number (ie $\beta$) bigger than $1$.

STEP 5) Thus no such $\beta$ can exist, and thus $.99...$ cannot be less than $1$.

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The real number system is defined as an extension of the rationals with the property that any sequence with an upper bound has a LEAST upper bound. The expression " 0.9-repeated" is defined to be the least real-number upper bound of the sequence 0.9. 0.99, 0.999,..... , which is 1. The rationals (and the reals) can also be extended to an arithmetic system (an ordered field) in which there are positive values which are less than every positive rational. In such systems the expression ".9-repeated" has no meaning.

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There are some situations in which something like $0.99999\ldots < 1$ indeed holds. Here is one coming from social choice theory.

Let $w_1>w_2>\ldots$ be an infinite sequence of positive numbers, and let $T$ be a number in the range $(0,\sum_i w_i)$. Pick an index $i$. Choose a random permutation $\pi$ of the positive integers, and consider the running totals $$w_{\pi(1)}, w_{\pi(1)} + w_{\pi(2)}, w_{\pi(1)} + w_{\pi(2)} + w_{\pi(3)}, \cdots$$ The Shapley value $\varphi_i(T)$ is the probability that the first time that the running total exceeds $T$ is when $w_i$ is added.

We will particularly be interested in the case in which the sequence $w_i$ is super-increasing: for each $i$, $w_i \geq \sum_{j=i+1}^\infty w_j$. The simplest case is $w_i = 2^{-i}$. Every number $T \in (0,1)$ can be written in the form $$T = 2^{-a_0} + 2^{-a_1} + \cdots, \qquad a_0 < a_1 < \cdots.$$ In this case we can give an explicit formula for $\varphi_i(T)$: $$\varphi_i(T) = \begin{cases} \sum_{t\colon a_t>i} \frac{1}{a_t \binom{a_t-1}{t}} & \text{if } i \notin \{a_0,a_1,\ldots\}, \\ \frac{1}{a_s \binom{a_s-1}{s}} - \sum_{t\colon a_t>i} \frac{1}{a_t \binom{a_t-1}{t-1}} & \text{if } i = a_s. \end{cases}$$

The first two functions are plotted here:

What happens for different sets of weights? The same formula applies, for $$T = w_{a_0} + w_{a_1} + \cdots, \qquad a_0 < a_1 < \cdots.$$ In general not all $T$ will be of this form; for $T$ not of this form, we take the lowest upper bound which is of this form. What we get for $w_i = 3^{-i}$ is:

Notice all the horizontal parts, for example the blue line at $y=1$ at $x \in (1/6,1/3)$. Where does this stem from? Note that $1/3 = 3^{-1} = w_1$, whereas $1/6 = \sum_{i=2}^\infty 3^{-i} = \sum_{i=2}^\infty w_i$. If we substitute $w_i = 2^{-i}$, then $1/3$ corresponds to $0.1$ (in binary), whereas $1/6$ corresponds to $0.011111\ldots$. So in this case there is a (visible) gap between $0.011111\ldots$ and $0.1$!

For more, take a look at this question and this manuscript.

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Here's my favorite reason why $.999\ldots$ should equal $1$: \begin{align*} .999\ldots + .999\ldots &= (.9 + .09 + .009 + \cdots) + (.9 + .09 + .009 + \cdots) \\ &= (.9 + .9) + (.09 + .09) + (.009 + .009) + \cdots \\ &= 1.8 + .18 + .018 + .0018 + \cdots \\ &= (1 + .8) + (.1 + .08) + (.01 + .008) + (.001 + .0008) + \cdots \\ &= 1 + (.8 + .1) + (.08 + .01) + (.008 + .001) + \cdots \\ &= 1 + .9 + .09 + .009 + \cdots \\ &= 1 + .999\ldots \end{align*} It follows subtracting $.999\ldots$ from both sides that $.999\ldots = 1$.$^{*}$

The reason I like this explanation best is that addition of (positive) infinite decimal expansions (defined in a particular way) is both commutative and associative even if you insist that $.999\ldots$ and $1$ are different numbers. That is, it forms a commutative monoid. But the cancellation property fails: if $a + b = a + c$, then we can't necessarily conclude $b = c$. The example of this is above, and the most fundamental reason why $.999\ldots = 1$ is arguably so that the cancellation property can hold.

$^{*}$The calculation given here (using rearrangmenet and regrouping of terms) is informal, and not intended to be a proof, but rather to give some idea of how you can add infinite decimal expansions in the monoid where $.999\ldots \ne 1$. It does end up being true that $.999\ldots + .999\ldots = 1 + .999\ldots$ in this monoid.

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It doesn't feel convincing to me to assume that blithely that you can rearrange the infinite sums the way you do here, without changing their value. It is true that it is safe in this particular case, but I don't see how one can show that except by developing enough of a theory of limits that we might as well just sum $\sum_{n=1}^\infty 9/10^n$ directly. – Henning Makholm Jul 17 at 2:46
@HenningMakholm Right--it's not a convincing proof that $.999\ldots = 1$ unless you accept that infinite sums of real numbers can be rearranged and regrouped in any way (which is true of course but much more nontrivial than the basic result that the limit of $.999\ldots$ is $1$). – 6005 Jul 17 at 4:19
My actual point is that you can define the addition of infinite expansions even while you insist that $.999\ldots \ne 1$, and if you do so (although I don't give the definition) you get $.999\ldots + .999\ldots = 1 + .999\ldots$. The aligned equation is just informal rearrangements but is intended to sort of show how the definition of addition of infinite decimal expansions would work, in this case at least. – 6005 Jul 17 at 4:20
It is not true in general that "infinite sums of real numbers can be rearranged and regrouped in any way". It works in this particular case because all of the terms being rearranged are all positive -- but in order to argue that this is safer than the general case (rearranging a conditionally convergent series such as $\sum \frac{(-1)^n}{n}$ fails spectacularly) it seems that you need to develop so much material that you cannot really keep open an option for $0.999...$ to differ from $1$. – Henning Makholm Jul 17 at 9:12
@HenningMakholm Oh I'm sorry, that was a typo. Of course I'm aware. I meant positive real numbers. – 6005 Jul 17 at 14:43

Adding zero to a number doesn't change it's value, right? So let's add them, properly accounting for the carry that occurs at every place to the right of the decimal, because $1+9+0 = 10$:

$$\begin{matrix} & \mathop{0}^1 &.& \mathop{9}^1 & \mathop{9}^1 & \mathop{9}^1 & \mathop{9}^1 & \mathop{9}^1 & \mathop{9}^1 & \mathop{9}^1 & \ldots \\+ & 0 &.& 0 & 0 & 0 & 0 & 0 & 0 & 0 & \ldots \\\hline \\ & 1 &.& 0 & 0 & 0 & 0 & 0 & 0 & 0 & \ldots \end{matrix}$$

It's actually ambiguous in this and other similar examples whether or not carry happens all or none of the trailing places, which is another good reason to insist that the two different decimals you could get as the answer should refer to the same number.

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One cool way I learned to prove this is that, assuming by $0.99999...$ you mean $0.\bar{9}$. Well we can say that $$0.\bar{9}=\sum_{n=1}^{\infty}9\cdot 10^{-n}=9\sum_{n=1}^{\infty}\frac{1}{10^n}$$ Which we know converges by fact that this is a geometric series with the ratio between terms being less than $1$. So we know that $$9\sum_{n=1}^{\infty}\frac{1}{10^n}=9\left(\frac{1}{1-\frac{1}{10}}-1\right)=10-9=1$$ Note that we subtract off the $1$ in the parentheses because we started indexing at $1$ rather than at $0$, so we have to subtract of the value of the sequence at $n=0$ which is $1$.

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"Which we know converges by fact that this is a geometric series with the ratio between terms being less than 1." I don't think anyone who does know this would be questioning whether .9999.... = 1. The only people who are confused on the concept most assuredly do not know this. – fleablood Jan 15 at 20:03
Just because you have studied series doesn't mean you have seen this proof. Hence why I have added it as a wiki answer on such a highly up voted answer, which people visiting it do not necessary have the same mathematical knowledge as the OP. – Will Fisher Jan 15 at 21:55
It is a pretty cool proof. – fleablood Jan 15 at 22:52

I remember the teacher explained this case for us :

$$0.9999... = 0.9 + 0.09 + 0.009 + 0.0009 + ...$$

$$0.99999\ldots = \frac{9}{10} + \frac{9}{100} + \frac{9}{1000} + \cdots$$

$$0.99999\ldots = \frac{9}{10} + \frac{9}{10}(\frac{1}{10}) + \frac{9}{10} (\frac{1}{100})+ \cdots$$

$a = \frac{9}{10}$ and common ratio $r = \frac{1}{10}$, so it has sum

$$\frac{\frac{9}{10}}{1 - \frac{1}{10}} = \frac{\frac{9}{10}}{\frac{9}{10}} = 1.$$

I hope it is the right answer cause I don't remember what the teacher did exactly

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This is almost the same as what is done in Will Fisher's answer. The only difference being how you set up the geometric series. – Michael Albanese Jan 14 at 21:49
well sorry because I wasn't helpful I should've read the comments first. My apologies – Manal Bouabdallaoui Jan 16 at 20:36

The problem isn't proving that $0.9999... = 1$. There are many proofs and they all are easy.

The problem is being convinced that every argument you are making actually is valid and makes sense, and not having a sinking feeling you aren't just falling for some parlor trick.

$0.99...9;$ (with $n$ 9s) is $\sum_{i= 1}^n \frac 9 {10^i}$ so "obviously" $0.999....$ (with an infinite number of 9s) is $\sum_{i = 1}^{\infty} \frac 9{10^i}$.

The obvious objection is: does it even make sense to talk about adding an infinite number of terms? How can we talk about taking and adding an infinite number of terms?

And it's a legitimate objection.

So when we learn math in elementary school we are told: Every real number can be written as a decimal expansion (maybe infinite) and every possible decimal expansion is a real number. And this is true. But we are not told why and we are expected to take it on faith, and we usually do.

IF we take this on faith then a proof is very easy:

$0.9999.... = \sum_{i = 1}^{\infty} \frac 9{10^i}$

$10*(0.9999....) = 10*\sum_{i = 1}^{\infty} \frac 9{10^i}= \sum_{i = 1}^{\infty} \frac {90}{10^i}=$

$\sum_{i = 1}^{\infty} \frac 9{10^{i-1}} = 9/10^0 + \sum_{i = 2}^{\infty} \frac 9{10^{i-1}}= 9 + \sum_{i = 1}^{\infty} \frac 9{10^i}$ (Look at the indexes!)

So...

$10*(0.999...) - (0.9999...) = (10 - 1)*0.9999.... = 9*0.99999.... =$

$9 + \sum_{i = 1}^{\infty} \frac 9{10^i} - \sum_{i = 1}^{\infty} \frac 9{10^i} = 9$.

So...

$0.9999.... = 9/9 = 1$.

Easy! !!!!!!!IF!!!!!!! we take it on faith that: Every real number can be written as a decimal expansion (maybe infinite) and every possible decimal expansion is a real number.

So why can we take that on faith? That's the issue: why is that true and what does it mean?

So....

We've got the Integers. We use them to count discrete measurements. We can use an integer to divide a unit 1 into $m$ sub-units to measure measurements of $1/m$. As the $m$ can be as large as we want the $1/m$ can be as precise as we want and the system of all possible $n/m; m \ne 0$ can measure any possible quantity with arbitrary and infinite precision.

We hope. We call these $n/m$ numbers the Rationals and everything is fine until we discover that we can't actually measure measurements such as the square root of two or pi.

But the Rationals still have infinite precision. We can get within 1/10 away from pi. We can get within 1/100 away from pi. Within $1/10^n$ for any possible power of 10.

At this point, we hope we can say "we can't measure it with any finite power of 10 but we can always go one more significant measure, so if we go through infinite powers of 10 we will measure it to precision" and we hope that explanation will be convincing.

But it isn't really. We have these "missing numbers" and we can get infinitely close the them, but what are they really?

Well, we decide to become math majors and in our senior year of college we take a Real Analysis course and we find out.

We can view numbers as sets of rational numbers. We can split the rational numbers at any point into two sets. We can split the rational numbers so that all the rational numbers less than 1/2 are in set A and all the rational numbers greater than or equal to 1/2 are in set B (which we ignore; we are only interested in set A.)

These "cuts" can occur at any point but they must follow the following rules:

--the set A of all the smaller rational numbers is not empty. Nor does it contain every rational number. Some rational number is not in it.

--if any rational number (call it q) is in A, then every rational number smaller than q is also in A. (This means that if r is a rational not in A, then every rational bigger than r is also not in A.)

-- A does not have a single largest element. (So it can be all the elements less than 1/2 but it can't be all the elements less than or equal to 1/2).

And we let $\overline R$ be the collection of all possible ways to "cut" the rational numbers in half that way.

Notice sometimes the cut will occur at a rational number (all the rationals less than 1/2), but sometimes it will occur at points "between" the rational numbers. (All the rationals whose squares are less than 2). So the collection $\overline R$ is a larger set than the set of Rational numbers.

It turns out we can define the Real numbers as the points of $\overline R$ where we can cut the rationals in two.

We need to do a bit or work to show that this is actually a number system. We say $x, y \in \overline R; x < y$ if the "Set A made by cutting at x" $\subset$ "Set A made by cutting at y". And we say $x + y =$ the point where we need to cut so that the set A created contains all the sums of the two other sets created by cutting at x and y. And we have to prove math works on $\overline R$. But we can do it. And we do.

But as a consequence we see that every real number is the least upper bound limit of a sequence of rational numbers. That's pretty much the definition of what a "cut point" is; the point that separate all the rationals less than it from all the other rationals.

I like to say (somewhat trivially) that: the real number $x$ is the least upper bound of all the rational numbers that are smaller than $x$. And it's true!

In the real numbers, every real number is the limit of some sequence of rational numbers. And every bounded sequence of rational numbers will have a real number least upper bound limit.

...

Let that sink in for a minute.

=====

Okay, so given a sequence {3, 3.1, 3.14, 3.141,....} = {finite decimals that are less than pi} is a bounded sequence of rational numbers so $\pi =$ the limit of the sequence which is also the limit of the infinite sequence 3.1415926....

It now makes sense to talk of $0.9999.... = \sum_{i=1}^{\infty}9/10^i = \lim\{\sum_{i=1}^n9/10^i\}$ = a precise and real number.

And from there we can say with confidence that that number is $1$. (By any of these proofs.)

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## protected by Alex BeckerMar 27 '14 at 2:58

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