Although some people find it counterintuitive, the decimal expansions $0.999\dotsc$ and $1$ represent the same real number.

Informal proofs

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

\begin{align} x &= 0.999\dotsc \newline 10x &= 9.999\dotsc \newline 10x-x &= 9.999\dotsc-0.999\dotsc \newline 9x &= 9 \newline x &= 1 \newline \end{align}
\begin{align} \frac 1 9 &= 0.111\dotsc \newline 1 &= \frac 9 9 \newline &= 9 \times \frac 1 9 \newline &= 9 \times 0.111\dotsc \newline &= 0.999\dotsc \end{align}
The real numbers are dense, which means that if $0.999\dots\neq1$, 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 $(\sum_{k=1}^n 9 \cdot 10^{-k})_{n\in\mathbb N}$. 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 $\varepsilon>0$, there is some $N\in\mathbb N$ such that for every $n>N$, $|1-a_n|<varepsilon$.

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{align} 1 - a{i+1} &= 1 - (ai + 9 \cdot 10^{-(i+1)}) \newline &= 1-a_i − 9 \cdot 10^{-(i+1)} \newline &= 10^{-i} − 9 \cdot 10^{-(i+1)} \newline &= 10 \cdot 10^{-(i+1)} − 9 \cdot 10^{-(i+1)} \newline &= 10^{-(i+1)} \end{align}

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. <div><div>

Arguments against $0.999\dotsc=1$

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

$0.999\dotsc$ and $1$ have different digits, so they can’t be the same. In particular, $0.999\dotsc$ 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 $1-0.999\dotsc=0.000\dotsc001\neq0$.

Why is this wrong?

Decimal expansions go on infinitely, but no farther.$0.000\dotsc001$ 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 $0.000\dotsc001$ to represent, it would be $0$.

$0.999\dotsc$ is the limit of the sequence $0.9, 0.99, 0.999, \dotsc$. 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 $0.999\dotsc$ from $9.999\dotsc$, you don’t get $9$. There’s an extra digit left over; just as $9.99-0.999=8.991$, $9.999\dotsc-0.999\dotsc=8.999\dotsc991$.

Why is this wrong?

There are infinitely many $9$s in $0.999\dotsc$, so when you shift it over a digit there are still the same amount. And the “decimal expansion” $8.999\dotsc991$ doesn’t make sense, because it has infinitely many digits and then a $1$.

0.999...=1

Informal proofs

Formal proof

Arguments against \(0.999\dotsc=1\)