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.

%%hidden(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. %%

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.

%%hidden(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 $~$1-0.999\dotsc=0.000\dotsc001\neq0$~$.

%%hidden(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 $~$1$~$.")

%%hidden(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 $~$9.99-0.999=8.991$~$, $~$9.999\dotsc-0.999\dotsc=8.999\dotsc991$~$.

%%hidden(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 $~$1$~$. %%