Misplaced Pages

0.999...: Difference between revisions

Article snapshot taken from Wikipedia with creative commons attribution-sharealike license. Give it a read and then ask your questions in the chat. We can research this topic together.

Browse history interactively ← Previous edit Next edit →Content deleted Content addedVisual WikitextInline

Revision as of 10:56, 7 May 2005 edit68.198.149.91 (talk) substituted k for i to avoid confusion with imaginary unit← Previous edit		Revision as of 15:47, 7 May 2005 edit undoBradBeattie (talk \| contribs)6,888 edits →Explanation: Added see alsoNext edit →
Line 22:		Line 22:
	== Explanation ==		== Explanation ==

	The key step to understand here is that		The key step to understand here is that the infinite geometric series is convergent.

	:<math>\sum_{k=0}^\infty \left( \frac{1}{10} \right)^k = \frac{1}{1 - \frac{1}{10}}.</math>		:<math>\sum_{k=0}^\infty \left( \frac{1}{10} \right)^k = \frac{1}{1 - \frac{1}{10}}.</math>

			== See also ==
	This is the sum of the convergent geometric series. For further information, see ] and ].

			* ]
			* ]
			* ]
			* ]

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

Revision as of 15:47, 7 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

$0.999\ldots$	$={\frac {9}{10}}+{\frac {9}{100}}+{\frac {9}{1000}}+\cdots$
	$=-9+{\frac {9}{1}}+{\frac {9}{10}}+{\frac {9}{100}}+{\frac {9}{1000}}+\cdots$
	$=-9+9\times \sum _{k=0}^{\infty }\left({\frac {1}{10}}\right)^{k}$
	$=-9+9\times {\frac {1}{1-{\frac {1}{10}}}}$
	$=1.\,$

Explanation

The key step to understand here is that the infinite geometric series is convergent.

\sum _{k=0}^{\infty }\left({\frac {1}{10}}\right)^{k}={\frac {1}{1-{\frac {1}{10}}}}.

See also

External proofs

Template:Mathstub

Category:

Proofs