Although some people find it counterintuitive, the decimal expansions and $1$ represent the same real number.

Informal proofs

These "proofs" can help give insight, but be careful; a similar technique can "prove" that $1 + 2 + 4 + 8 + \dots = - 1$ . They work in this case because the series corresponding to $0.999 \dots$ is absolutely_convergent.

$\begin{matrix} x & = 0.999 \dots 10 x & = 9.999 \dots 10 x - x & = 9.999 \dots - 0.999 \dots 9 x & = 9 x & = 1 \end{matrix}$

$\begin{matrix} \frac{1}{9} & = 0.111 \dots 1 & = \frac{9}{9} = 9 \times \frac{1}{9} = 9 \times 0.111 \dots = 0.999 \dots \end{matrix}$

The real numbers are dense, which means that if $0.999 \dots \neq 1$ , there must be some number in between. But there's no decimal expansion that could represent a number in between $0.999 \dots$ and $1$ .

Formal proof

This is a more formal version of the first informal proof, using the definition of decimal notation.

Show proof

$0.999 \dots$ is the decimal expansion where every digit after the decimal point is a $9$ . By definition, it is the value of the series $\sum_{k = 1}^{\infty} 9 \cdot 10^{- k}$ . This value is in turn defined as the limit of the sequence $(\sum_{k = 1}^{n} 9 \cdot 10^{- k})_{n \in N}$ . Let $a_{n}$ denote the $n$ th term of this sequence. I claim the limit is $1$ . To prove this, we have to show that for any $ε > 0$ , there is some $N \in N$ such that for every $n > N$ , $| 1 - a_{n} | < ε$ .

Let's prove by induction that $1 - a_{n} = 10^{- n}$ . Since $a_{0}$ is the sum of {$0$ terms, $a_{0} = 0$ , so $1 - a_{0} = 1 = 10^{0}$ . If $1 - a_{i} = 10^{- i}$ , then

$\begin{matrix} 1 - a_{i + 1} & = 1 - (a_{i} + 9 \cdot 10^{- (i + 1)}) = 1 - a_{i} - 9 \cdot 10^{- (i + 1)} = 10^{- i} - 9 \cdot 10^{- (i + 1)} = 10 \cdot 10^{- (i + 1)} - 9 \cdot 10^{- (i + 1)} = 10^{- (i + 1)} \end{matrix}$

So $1 - a_{n} = 10^{- n}$ for all $n$ . What remains to be shown is that $10^{- n}$ eventually gets (and stays) arbitrarily small; this is true by the Archimedean_property and because $10^{- n}$ is monotonically decreasing.

Arguments against $0.999 \dots = 1$

These arguments are used to try to refute the claim that $0.999 \dots = 1$ . They're flawed, since they claim to prove a false conclusion.

$0.999 \dots$ and $1$ have different digits, so they can't be the same. In particular, $0.999 \dots$ starts " $0.$ ," so it must be less than 1.

Why is this wrong?

Decimal expansions and real numbers are different objects. Decimal expansions are a nice way to represent real numbers, but there's no reason different decimal expansions have to represent different real numbers.

If two numbers are the same, their difference must be $0$ . But $1 - 0.999 \dots = 0.000 \dots 001 \neq 0$ .

Why is this wrong?

Decimal expansions go on infinitely, but no farther. $0.000 \dots 001$ doesn't represent a real number because the $1$ is supposed to be after infinitely many $0$ s, but each digit has to be a finite distance from the decimal point. If you have to pick a real number to for $0.000 \dots 001$ to represent, it would be $0$ .

$0.999 \dots$ is the limit of the sequence $0.9, 0.99, 0.999, \dots$ . Since each term in this sequence is less than $1$ , the limit must also be less than $1$ . (Or "the sequence can never reach $1$ .")

Why is this wrong?

The sequence gets arbitrarily close to $1$ , so its limit is $1$ . It doesn't matter that all of the terms are less than $1$ .

In the first proof, when you subtract $0.999 \dots$ from $9.999 \dots$ , you don't get $9$ . There's an extra digit left over; just as $9.99 - 0.999 = 8.991$ , $9.999 \dots - 0.999 \dots = 8.999 \dots 991$ .

Why is this wrong?

There are infinitely many $9$ s in $0.999 \dots$ , so when you shift it over a digit there are still the same amount. And the "decimal expansion" $8.999 \dots 991$ doesn't make sense, because it has infinitely many digits and then a $1$ .