Difference between revisions of "0.999..."

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Alternatively, <math>\frac 19 = 0.\overline{1} = 0.111\ldots</math>, and then multiply both sides by <math>9</math>.  
 
Alternatively, <math>\frac 19 = 0.\overline{1} = 0.111\ldots</math>, and then multiply both sides by <math>9</math>.  
  
=== Manipulation ===
+
=== Algebraic Manipulation ===
 
Let <math>x = 0.999\ldots</math> Then
 
Let <math>x = 0.999\ldots</math> Then
 
<center><math>
 
<center><math>
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0.999\ldots = \frac{\frac{9}{10}}{1 - \frac{1}{10}} = 1
 
0.999\ldots = \frac{\frac{9}{10}}{1 - \frac{1}{10}} = 1
 
</math></center>
 
</math></center>
 +
 +
===Limits===
 +
<math>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</math>
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==See Also==
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*[[Geometric sequence]]
  
 
== Related threads ==
 
== Related threads ==
 
*[http://www.artofproblemsolving.com/Forum/viewtopic.php?t=201302]
 
*[http://www.artofproblemsolving.com/Forum/viewtopic.php?t=201302]

Revision as of 20:28, 24 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 often mistaken that $0.999\ldots \neq 1$ for various reasons (that there can only be a finite number of $9$s, that there is a $\frac 1{\infty}$ term left over at the end, etc.).

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

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

See Also

Related threads

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