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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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21 Answers 21

up vote 213 down vote accepted




$1.0 -.9 = .1$

$1.00-.99 = .01$




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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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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

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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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












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x = 0.999...
10x = 9.999...
-> 9x = 9
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      1 - 0.999999... = ε              (0)



Let's calculate:

0.999... ×
9        =



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




8.00000000... +
0.99999999... =


                =   1    -  1 + ε         
                =               ε.


9ε = ε

8ε = 0

ε = 0

1 - 0.999999... = ε = 0

Quod erat demonstrandum. Pardon my unicode.

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Here's a very simple proof:




Therefore 0.999... = 1
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Given (by long division):

Multiply by 3:



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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

Another approach is the following:


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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.

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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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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

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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0.9 0.99 0.999 0.9999


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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

protected by Alex Becker Mar 27 '14 at 2:58


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