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Revision as of 19:19, 6 May 2005 editBradBeattie (talk | contribs)6,888 editsm Whoops. Proof fix.← Previous edit Revision as of 19:20, 6 May 2005 edit undoMichael Hardy (talk | contribs)Administrators210,264 editsNo edit summaryNext edit →
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{|- {|-
|<math>0.999\ldots</math> |<math>0.999\ldots</math>
|<math>= \frac{9}{10} + \frac{9}{100} + \frac{9}{1000} + \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} + \ldots</math> |<math>= -9 + \frac{9}{1} + \frac{9}{10} + \frac{9}{100} + \frac{9}{1000} + \cdots</math>
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|- |-
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|<math>= 1\,</math> |<math>= 1.\,</math>
|} |}


== Explanation == == Explanation ==


The key step to understand here is that 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> :<math>\sum_{i=0}^\infty \left( \frac{1}{10} \right)^i = \frac{1}{1 - \frac{1}{10}}.</math>
⚫ For further information, read up on ] and ].

⚫ For further information, see ] and ].


== External proofs == == External proofs ==

Revision as of 19:20, 6 May 2005

In mathematics, 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.

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

Explanation

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}}}}.}

For further information, see geometric series and convergence.

External proofs


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