Difference between revisions of "0.999..."

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<math>0.999\ldots</math> (or <math>0.\overline{9}</math>) is an equivalent representation of the [[real number]] <math>1</math>.  
 
<math>0.999\ldots</math> (or <math>0.\overline{9}</math>) is an equivalent representation of the [[real number]] <math>1</math>.  
  
It is often mistaken that <math>0.999\ldots \neq 1</math> for various reasons (that there can only be a finite number of <math>9</math>s, that there is a <math>\frac 1{\infty}</math> term left over at the end).   
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It is often mistaken that <math>0.999\ldots \neq 1</math> for various reasons (that there can only be a finite number of <math>9</math>s, that there is a <math>\frac 1{\infty}</math> term left over at the end, etc.).   
  
 
== Proofs ==
 
== Proofs ==
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=== Manipulation ===
 
=== Manipulation ===
 
Let <math>x = 0.999\ldots</math> Then
 
Let <math>x = 0.999\ldots</math> Then
 
 
<center><math>
 
<center><math>
 
\begin{align*}
 
\begin{align*}
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\end{align*}
 
\end{align*}
 
</math></center>
 
</math></center>
 
 
Subtracting,  
 
Subtracting,  
 
 
<center><math>
 
<center><math>
 
\begin{align*}
 
\begin{align*}
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== Related threads ==
 
== Related threads ==
*[http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1651108580&t=201302]
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*[http://www.artofproblemsolving.com/Forum/viewtopic.php?t=201302]

Revision as of 18:59, 24 April 2008

$0.999\ldots$ (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 $9$.

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

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

Related threads

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