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Revision as of 18:46, 6 May 2005 editOleg Alexandrov (talk | contribs)Administrators47,244 edits VfD← Previous edit Revision as of 18:53, 6 May 2005 edit undoOleg Alexandrov (talk | contribs)Administrators47,244 edits I have big doubts as to ever this article is useful or encyclopedic, but I cleaned it up a bit anyway.Next edit →
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In ], one could easily fall in the trap that while 0.999... is certainly close to 1, nevertheless the two are not equal. Here's a proof that they actually are.
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There is a common argument as to whether <math>0.999\ldots = 1</math> or not. It does.

== Proof ==


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== Explaination == == Explaination ==


The key step to understand here is that <math>\sum_{i=0}^\infty \left( \frac{1}{10} \right)^i = \frac{1}{1 - \frac{1}{10}}</math>. This rests on a ] ] if the common ratio is between -1 and 1 exclusive. The key step to understand here is that
:<math>\sum_{i=0}^\infty \left( \frac{1}{10} \right)^i = \frac{1}{1 - \frac{1}{10}}.</math>
This rests on a ] ] if the common ratio is between -1 and 1 exclusive.


{{mathstub}} {{mathstub}}
]

Revision as of 18:53, 6 May 2005

In mathematics, one could easily fall in the trap that while 0.999... is certainly close to 1, nevertheless the two are not equal. Here's a proof that they actually are.

0.999 … {\displaystyle 0.999\ldots } = 9 × 0.111 … {\displaystyle =9\times 0.111\ldots }
= 9 × ( 1 10 + 1 100 + 1 1000 + … ) {\displaystyle =9\times \left({\frac {1}{10}}+{\frac {1}{100}}+{\frac {1}{1000}}+\ldots \right)}
= 9 × ( − 1 + 1 1 + 1 10 + 1 100 + 1 1000 + … ) {\displaystyle =9\times \left(-1+{\frac {1}{1}}+{\frac {1}{10}}+{\frac {1}{100}}+{\frac {1}{1000}}+\ldots \right)}
= 9 × ( − 1 + ∑ i = 0 ∞ ( 1 10 ) i ) {\displaystyle =9\times \left(-1+\sum _{i=0}^{\infty }\left({\frac {1}{10}}\right)^{i}\right)}
= 9 × ( − 1 + 1 1 − 1 10 ) {\displaystyle =9\times \left(-1+{\frac {1}{1-{\frac {1}{10}}}}\right)}
= 9 × ( − 1 + 10 9 ) {\displaystyle =9\times \left(-1+{\frac {10}{9}}\right)}
= 1 {\displaystyle =1\,}

Explaination

The key step to understand here is that

∑ i = 0 ∞ ( 1 10 ) i = 1 1 − 1 10 . {\displaystyle \sum _{i=0}^{\infty }\left({\frac {1}{10}}\right)^{i}={\frac {1}{1-{\frac {1}{10}}}}.}

This rests on a geometric series converging if the common ratio is between -1 and 1 exclusive.

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