# Does .99999… = 1?

I'm told by smart people that `0.999... = 1` and I believe them but is there a proof that explains why?

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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

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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

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Reopened -d this is a valid question – Casebash Jul 21 '10 at 7:43

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@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

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I think it should be tagged "crank magnet". – Tom Stephens Jul 23 '10 at 2:26
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## 19 Answers

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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24
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

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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

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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

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@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

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@Doug: Then your reading is incorrect. – user126 Aug 17 '10 at 22:14
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The 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 demostrandum. 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

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This is going to sound bad, but honestly I think there's a basic intelligence threshold that prevents certain people from grasping that infinity is transcendent, and when you multiply an infinitely repeating sequence by 10 you still have an infinitely repeating sequence (without a zero entering somewhere to the right). Really. I think it's at some level an IQ thing. – ErikE Sep 27 '10 at 5:33
8ε = 0 instead of 10ε = 0 – CutieKrait May 9 at 9:37
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What I really don't like about all the above answers, is the underlying assumption that `1/3=0.3333...`, How do you know that?. It seems to me like assuming the something which is already known.

A proof I really like is:

````0.9999... × 10 = 9.999999...

0.9999... × (9+1) = 9.999999...  [by distribution rule:]

0.9999... × 9 + 0.99999.... × 1 = 9.99999....

0.9999... × 9 = 9.9999....-0.9999.... = 9

0.9999... × 9 = 9

0.9999... = 1
````

The only things I need to assume is, that `9.999.... - 0.9999... = 9` and that `0.999... x 10 = 9.999...`. These seems to me intuitive enough to take for granted.

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

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7
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
@Issac, correct, fixed now. – Elazar Leibovich Jul 21 '10 at 3:58
0.9999... x 9 = 8.999999... – Sklivvz♦ May 3 '11 at 20:51

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@Sklivvz, true, since 8.999....=9. Did you have trouble understanding the algebra I did? – Elazar Leibovich May 4 '11 at 5:55

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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
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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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This should have been a comment; but I post as an answer anyway. I just want to remark that in spite of all the criticisms, all the statements in the most popular proof in the list above are true.

That is, in the real number system, it is indeed true that:

1/3 = 0.333333...

It is again true that:

0.33333.... * 3 = 0.99999....

Etc..

I would still stand by the superiority of Noah Snyder's answer, however. That's the one addressing the subtleties.

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I do not think that 0.999... = 1 unless we're defining the equality operator in some strange way. Assuming you allow N decimals in the 0.999... term, you can take N as large as you want and it will still be less than 1. There is no N for which 0.999... is not less than 1. Sure, if you take the limit as N approaches infinity then you get 1, but "lim 0.999... as N approaches infinity" is not the same as "0.999..." by itself. By itself, there is always some infinitesimal difference.

I think the whole point of writing 0.999... is to indicate that it is NOT equal to 1 and to indicate that there is some infinitesimal difference. Otherwise you would just write 1.

My opinion is that most people say 0.999... = 1 because that's what they were taught in college and they accepted it without really thinking about it. All the arguments I've heard or read so far are just contrived ways of making the equality hold, but there is always some subtle flaw that takes a long time to point out, that the proponents will not accept anyway (for reasons I will not get into here).

From a comment I made:

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.

The current winning response as of 7/22/10 treats 0.999... as being equivalent to lim (0.999...) as the number of decimals goes to infinity. But the sequence and the limit of the sequence are not the same thing.

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The meaning of the symbol "..." is that we can't choose such an N, as you propose. The idea is that 0.999... means that the nines go on forever. The arguments that you've heard or read so far constitute mathematical proof - regardless whether or not you believe in them, or find them to appear contrived. – Tom Stephens Jul 23 '10 at 2:17
I see no proof, mathematical, logical, or otherwise, in the above. – Doug Treadwell Jul 25 '10 at 2:56

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What is "0.999…" according to you if not "lim 0.999...to N digits as N approaches infinity"? Is your "0.999…" a number? If you want to define it as being not a fixed real number but standing for something else (e.g. "0.999… stands for some unspecified number starting with a bunch of 9s, like 0.999999983" or "0.999… stands for 0 followed by some finite number of 9s") then you're right, it won't mean the same thing as 1, but this would be contrary to the usual mathematical convention, and not very useful. – ShreevatsaR Aug 16 '10 at 22:03

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My first downvote. Not only is this response incorrect, it is incorrigible. The problem is well identified by Shreevatsa. If 0.999... does not represent a limit, what DOES it represent? Is there a single real number represented here? This response wants it to represent any rational number in some small neighbourhood of some unspecified rational number that is slightly smaller than one. – yasmar Nov 12 '10 at 21:04

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I downvoted it. But I felt a bit guilty since the answer was sort of useful to me. It is very illuminating to learn how people confuse things. I really like the reasoning that $1 - 0.999\ldots = 0.000\ldots1$. – André Caldas Mar 25 '12 at 14:28
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`.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 at 21:47
– Squirtle Mar 5 at 5:37

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.

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For any digit n = 0,1,2,3,4,5,6,7,8,9 the fraction

````n/9 = 0.nnnnnnnn... (periodically).
````

Therefore 9/9 is equal to 0.999999999.... but also since 9 divided by 9 equals one,

````9/9 = 0.9999999.... = 1
````
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This is a valid proof to me, but pretty similar to the other division-based answers. There are serial down-voters aroudn here, beware! – Noldorin Jul 21 '10 at 8:36
@Noldorin: doesn't matter so much for CW. Although, in case of a down-vote I'd at least like to know what were wrong with this answer beside a "don't like". But thanks for the warning :7 – Tobias Kienzler Jul 29 '10 at 14:26

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

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@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

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@Scott: Might help to stop whining and post what you think is the 'correct' answer then. – Noldorin Jul 20 '10 at 20:17

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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

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@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
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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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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 * r + a * r^2 + a * r^3 + ... has sum a/(1 - r) as long as |r|<1

0.99999... = 9/10 + 9/10^2 + 9/10^3 + ...

This is the infinite geometric series with first term a = 9/10 and common ratio r = 1/10, so it has sum (9/10) / (1 - 1/10) = (9/10) / (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
````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

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@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

Suppose this was not the case, i.e. `0.9999... != 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

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.

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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

Here's a collection of proofs:

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This exhibits why link-only answers are not good. This link has died. – robjohn♦ Mar 20 at 10:52

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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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

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@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

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Whatever proof you could give that 1/3 = .3333... would also prove directly that 1 = .999999... – Noah Snyder Jul 20 '10 at 20:24

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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

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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
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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
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