Difference between revisions of "0.999..."

Revision as of 00:48, 27 April 2008

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 $a_n=1$ .

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. Unknown error_msg)

Subtracting,

$\begin{align*}

9x &= 9\\ x &= 1

\end{align*}$ (Error compiling LaTeX. Unknown error_msg)

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

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

Related Threads

<url>viewtopic.php?t=201302 ".9 rep = 1?"</url>
<url>viewtopic.php?t=188041 "0.999999... =1?"</url>

@@ Line 8: / Line 8: @@
 One potential definition is to say that <math>.999...</math> is the limit of the sequence <math>.9,.99,.999,...</math>. That is to say, <math>a_n</math> is <math>.999...9</math> where there are <math>n</math> nines. Then we can say that <math>.999....</math> does in fact equal <math>1</math>.
-But in some senses, this is not the most satisfying definition. When we say that the limit of <math>a_n</math> as <math>n</math> goes to infinity is equal to <math>L</math>, that means that for all <math>\epilon>0</math>, there exists an integer <math>N>0</math> such that <math>|a_n-L|<\epsilon</math> for all integers greater than <math>N</math>. So in fact, we had to resort to limit definitions to resolve this problem. If you think about it, there is no <math>n</math> such that <math>a_n=1</math>.
+But in some senses, this is not the most satisfying definition. When we say that the limit of <math>a_n</math> as <math>n</math> goes to infinity is equal to <math>L</math>, that means that for all <math>\epsilon>0</math>, there exists an integer <math>N>0</math> such that <math>|a_n-L|<\epsilon</math> for all integers greater than <math>N</math>. So in fact, we had to resort to limit definitions to resolve this problem. If you think about it, there is no <math>n</math> such that <math>a_n=1</math>.
 The proofs below show how to evaluate the limit, but they implicitly use the definition given above.

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