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Revision as of 19:20, 6 May 2005 editMichael Hardy (talk | contribs)Administrators210,264 editsNo edit summary← Previous edit Revision as of 19:22, 6 May 2005 edit undoBradBeattie (talk | contribs)6,888 edits Proof headerNext edit →
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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. 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 ==


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Revision as of 19:22, 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.

Proof

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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