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Revision as of 18:26, 6 May 2005 editBradBeattie (talk | contribs)6,888 edits Sorry, edit conflict. Readded mathstub.← Previous edit Revision as of 18:30, 6 May 2005 edit undoBradBeattie (talk | contribs)6,888 editsNo edit summaryNext edit →
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There is a common argument as to whether 0.9999~ equals 1 or not. It does. ] is the primary reason as to why. There is a common argument as to whether <math>0.999\ldots = 1</math> or not. It does. ] is the primary reason as to why.


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|<math>= 1\,</math> |<math>= 1\,</math>
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The key step to understand here is that <math>\frac{1}{1} + \frac{1}{10} + \frac{1}{100} + \frac{1}{1000} + \ldots = \sum_{i=0}^\infty \left( \frac{1}{10} \right)^i\,\!</math>


{{mathstub}} {{mathstub}}

Revision as of 18:30, 6 May 2005

There is a common argument as to whether 0.999 … = 1 {\displaystyle 0.999\ldots =1} or not. It does. Convergence is the primary reason as to why.

The key step to understand here is that 1 1 + 1 10 + 1 100 + 1 1000 + … = ∑ i = 0 ∞ ( 1 10 ) i {\displaystyle {\frac {1}{1}}+{\frac {1}{10}}+{\frac {1}{100}}+{\frac {1}{1000}}+\ldots =\sum _{i=0}^{\infty }\left({\frac {1}{10}}\right)^{i}\,\!} Template:Mathstub
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\,}