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Revision as of 16:25, 8 May 2005 editBradBeattie (talk | contribs)6,888 edits See also: Added recurring decimal← Previous edit Revision as of 17:05, 8 May 2005 edit undo213.216.199.18 (talk)No edit summaryNext edit →
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JARGON!
In ], one could easily fall in the trap of thinking that while 0.999... is certainly close to 1, nevertheless the two are not equal. Here's a proof that they actually are.

== Proof ==

{|-
|<math>0.999\ldots</math>
|<math>= \frac{9}{10} + \frac{9}{100} + \frac{9}{1000} + \cdots</math>
|-
|
|<math>= -9 + \frac{9}{1} + \frac{9}{10} + \frac{9}{100} + \frac{9}{1000} + \cdots</math>
|-
|
|<math>= -9 + 9 \times \sum_{k=0}^\infty \left( \frac{1}{10} \right)^k</math>
|-
|
|<math>= -9 + 9 \times \frac{1}{1-\frac{1}{10}}</math>
|-
|
|<math>= 1.\,</math>
|}

== Explanation ==

The key step to understand here is that the following infinite geometric series is convergent:
:<math>\sum_{k=0}^\infty \left( \frac{1}{10} \right)^k = \frac{1}{1 - \frac{1}{10}}.</math>

==Alternative proof==

A less mathematical proof goes as follows. Let ''x'' equal 0.999... Then,

:10''x''&minus;''x'' = 9.999... &minus; 0.999...

and so

:9''x'' = 9,

which implies that ''x'' = 1.

== See also ==

* ]
* ]
* ]
* ]
* ]

== External proofs ==

*
*


{{mathstub}}
]

Revision as of 17:05, 8 May 2005

JARGON!