# Difference between revisions of "0.999..."

 This is an AoPSWiki Word of the Week for April 25-May 2

$0.999\ldots$ (or $0.\overline{9}$) is an equivalent representation of the real number $1$.

It is intuitively clear what the dots after the nines mean. However, to resolve the problem, one needs mathematics beyond the elementary school level math that is needed to understand the question.

What do the dots after the $9$s actually mean?

One potential definition is to say that $.999...$ is the limit of the sequence $.9,.99,.999,...$. That is to say, $a_n$ is $.999...9$ where there are $n$ nines. Then we can say that $.999....$ does in fact equal $1$.

But in some senses, this is not the most satisfying definition. When we say that the limit of $a_n$ as $n$ goes to infinity is equal to $L$, that means that for all $\epsilon>0$, there exists an integer $N>0$ such that $|a_n-L|<\epsilon$ for all integers greater than $N$. So in fact, we had to resort to limit definitions to resolve this problem. If you think about it, there is no $n$ such that $a_n=1$.

Another mathematical tidbit: how do we know that $.999...$ is a real number? Intuitively, a real number line seems to be continuous, so such a limit seems to be a real number. But how do we know that there are no 'gaps' so it could possibly not be a real number. But lets prove it.

Let $S=\{a_n|n\in\mathbb{Z}\}$. $a_1\in S$, so $S$ is nonempty. It is easy to see that $a_n=1-(.1)^n$ since it is a finite geometric series. But then $1-(.1)^n<1$. Hence $S$ has an upper bound of $1$. But by the least upper bound axiom of real numbers, if a set of real numbers has a upper bound, then it has a least upper bound (that is a real number).

Suppose that $S$ had a least upper bound, $L'$ such that $L'<1$. But then we have $1-(.1)^n\le L'<1\implies 0< 1-L\le (.1)^n$. But that is nonsense since $\log_{.1} (1-L)\ge n$, which would be a least upper bound for the natural numbers. Hence $1$, the limit of the sequence, is a least upper bound. Hence that expression is in fact a real number.

The proofs below show how to evaluate the limit, but they implicitly use the definition given above.

## Proofs

### Fractions

Since $\frac 13 = 0.\overline{3} = 0.333\ldots$, multiplying both sides by $3$ yields $1 = 0.999\ldots$

Alternatively, $\frac 19 = 0.\overline{1} = 0.111\ldots$, and then multiply both sides by $9$.

### Algebraic Manipulation

Let $x = 0.999\ldots$ Then

\begin{align*} 10x &= 9.999\ldots\\ x &= 0.999\ldots \end{align*} (Error compiling LaTeX. ! Package amsmath Error: \begin{align*} allowed only in paragraph mode.)

Subtracting,

\begin{align*} 9x &= 9\\ x &= 1 \end{align*} (Error compiling LaTeX. ! Package amsmath Error: \begin{align*} allowed only in paragraph mode.)

### Infinite series

$0.999\ldots = 0.9 + 0.09 + 0.009 + \ldots = \frac{9}{10} + \frac{9}{100} + \frac{9}{1000} + \ldots$

This is an infinite geometric series, so

$0.999\ldots = \frac{\frac{9}{10}}{1 - \frac{1}{10}} = 1$

### Limits

$0.999\ldots = \lim_{n\to\infty}0.\underbrace{ 99\ldots9 }_{n} = \lim_{n\to\infty}\sum_{k = 1}^n\frac{9}{10^k} = \lim_{n\to\infty}\left(1-\frac{1}{10^n}\right) = 1-\lim_{n\to\infty}\frac{1}{10^n} = 1$