# Is $.\overline{9} = 1$?

I'm told by smart people that $.\overline{9} = 1$ 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
Reopened -d this is a valid question –  Casebash Jul 21 '10 at 7:43
@Harry tagging this question [cauchy-sequences] is about as useful as tagging the question "What's the best color for a Porsche?" as [combustion-engine]. –  balpha Jul 21 '10 at 19:44
I think it should be tagged "crank magnet". –  Tom Stephens Jul 23 '10 at 2:26

What does it mean when you refer to $.99999...$? 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 $.99999... = 1$ is that their difference is zero. So let's subtract $1.0000... -.99999... = .00000... = 0$. That is,

$1.0 -.9 = .1$

$1.00-.99 = .01$

$1.000-.999=.001$,

$...$

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

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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

well the simplest is 3x(1/3) = 1 where 1/3 = 0.33·

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-1. Uhh, everyone knows this answer is nonsense, right? And is just upvoting it for a lark? –  Scott Morrison Jul 20 '10 at 19:55
Why this answer is nonsense ? to me it looks good, that's why I would appreciate more explication. thanks, –  idm Aug 9 '14 at 22:50
Isn't this just the $3(.333...)=.999...$ argument? Seems a bit harsh to call that nonsense. –  Grumpy Parsnip Nov 13 '14 at 22:19

The proof I've always seen was that if 1/3 == 0.333... then 0.333... x 3 must be equal to 1, but at the same time, calculating it on a digit level gives 0.999...

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-1. Not a proof, see my comments on all the other duplicate answers. –  Scott Morrison Jul 20 '10 at 20:03
@Scott Also, that proof happens to appear on the Wikipedia page for 0.999... too, so I'd suggest you "fix it" there, after you've provided your own answer. –  Rowland Shaw Jul 20 '10 at 20:11
Whatever proof you could give that 1/3 = .3333... would also prove directly that 1 = .999999... –  Noah Snyder Jul 20 '10 at 20:24
But how do you know that long division works? How do you know the pattern goes on forever? You can do long division for 1/1 and get .999999... Try it out! –  Noah Snyder Jul 20 '10 at 20:33
This isn't just philosophical, why isn't 1/3 = ...33333367. (That's going infinitely to the left and with the decimal point at the right. Decimals going infinitely to the left work just as well formally as ones going to the right.) –  Noah Snyder Jul 20 '10 at 22:18

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... –  Freeze_S Aug 27 '14 at 21:07

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

.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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ok, so based on this method, you could say that any number is equal to another number, because I could argue, that 8.999999 = 9, and then argue that 9.0000= 9.1 and so on so if 8.9999 = 9 and 9 = 9.1 therefore 8.999 must be also equal to 9.1, you can also state that 3=6 as 3=7, so that thinking must be wrong in so many levels :) –  Val Jan 24 '13 at 21:47
–  Squirtle Mar 5 '13 at 5:37
@Val $8.999999 \neq 9$ but $8.\bar9 = 9$.Notice the difference ? –  ARi Aug 2 '13 at 15:25

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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I like this answer better because it says they are not Absolutely the same but same enough for our day to day measurements... If you split $1 in 3 (absolutely) it means you will have to count down to nearest atom, then if all 3 where to say we want absolute equal do we also start counting to the last sub-atomics, I feel this way of thinking means there isn't such thing as infinite it's just a very large (comprehensiveness) of a number. – Val Aug 5 '13 at 20:59 What I really don't like about all the above answers, is the underlying assumption that$1/3=0.3333\ldotsHow do you know that? It seems to me like assuming the something which is already known. A proof I really like is: \begin{align} 0.9999\ldots × 10 &= 9.9999\ldots\\ 0.9999\ldots × (9+1) &= 9.9999\ldots\\ \text{by distribution rule: }\Space{15ex}{0ex}{0ex} \\ 0.9999\ldots × 9 + 0.9999\ldots × 1 &= 9.9999\ldots\\ 0.9999\ldots × 9 &= 9.9999\dots-0.9999\ldots\\ 0.9999\ldots × 9 &= 9\\ 0.9999\ldots &= 1 \end{align} The only things I need to assume is, that9.999\ldots - 0.999\ldots = 9$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. - 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 If you want to go the calculus route, then basically you're gonna follow this pattern and use mathy magic and limits to follow it up to its logical conclusion: 0.9 0.99 0.999 0.9999 And so on. This pattern is a sum of 9*10^-i from i=1 to i=infinity. Every iteration, the difference between the pattern and 1 is a bit less. So at i=infinity, the difference is so small that it is literally zero. - This is closest to the proof I had in mind. First, $$.999... = 9 \sum_{k=1}^{\infty} 10^{-k}$$ This seems rather noncontroversial. Let's call this series$S$. This is a geometric series with common ratio$1/10$. Therefore, $$S = 9 \cdot \frac{1/10}{1 - 1/10} = 9 \cdot \frac{1}{9} = 1$$ What's wrong with this proof? – user41583 Sep 18 '12 at 22:40 It is the best one. – ARi Aug 2 '13 at 15:29 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. - 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. – user72694 Nov 25 '13 at 17:12 Why is it that if the two numbers are distinct, then there must be another number in between them. Why can we not say that they are "neighbors"? Why is the notion of "neighbors" incorrect for real numbers? – Bogdan Alexandru Sep 1 '14 at 14:56 @BogdanAlexandru$\frac{x+y}{2}$is a number between$x$and$y$– Holographer Sep 1 '14 at 15:39 x = 0.999... 10x = 9.999... 10x - x = 9.999... - 0.999... = 9 -> 9x = 9 -> x = 1 thus, 0.999... = 1  - 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. - 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 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 $$\frac{\frac{9}{10}}{1 - \frac{1}{10}} = \frac{\frac{9}{10}}{\frac{9}{10}} = 1.$$ - 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 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. - 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. :) I didn't feel confident answering until I got this Terence Tao link: http://www.google.com/buzz/114134834346472219368/RarPutThCJv/In-the-foundations-of-mathematics-the-standard. - We know that all repeating decimals are rational numbers. So is 0.9999. What we need to do is to convert it to a fraction: Let x=0.(9), than 10x=9.(9) and 10x-x=9.(9)-0.(9) or 9x=9 from where x=1. - This is the same as this earlier answer, no? – user1729 May 7 '14 at 8:02 With the difference that I've stressed out the fact that all repeating decimals (so is 0.999...) are rational numbers. – Emin May 8 '14 at 17:04 Otherwise, most of the given answers are equivalent with that one that you've highlighted. – Emin May 8 '14 at 17:04 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 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$1,$sine 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. - Another approach is the following: $$0.\overline9=\lim_{n \to \infty} 0.99.....9 = \lim_{n \to \infty} \sum\limits_{k=1}^n \frac{9}{10^k}=\lim_{n \to \infty} 1-\frac{1}{10^n}=1-\lim_{n \to \infty} \frac{1}{10^n}=1$$ - This approach appears in Isaac's answer from 4 years earlier. – Jonas Meyer Aug 26 '14 at 2:15 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. - 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$1$. - let \begin{eqnarray} 0.\dot 9 = x \end{eqnarray} over 10 on each side \begin{eqnarray} 0.0\dot 9 = 0.1x \end{eqnarray} \begin{eqnarray} 0.9 = 0.9x \end{eqnarray} we solve the problem$ x = 1 \$

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