# Is $0.999999999… = 1$?

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
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 $.99999\ldots$? 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

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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