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There is a common argument as to whether <math>0.999\ldots = 1</math> or not. It does. There is a common argument as to whether <math>0.999\ldots = 1</math> or not. It does.


Revision as of 18:46, 6 May 2005

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

Proof

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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