Misplaced Pages

This is an old revision of this page, as edited by Oleg Alexandrov (talk | contribs) at 15:57, 8 May 2005 (don't need one thousand proofs). The present address (URL) is a permanent link to this revision, which may differ significantly from the current revision.

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

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

Explanation

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

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

Alternative proof

A less mathematical proof goes as follows. Let x equal 0.999... Then,

10x−x = 9.999... − 0.999...

and so

9x = 9,

which implies that x = 1.

See also

• Alternate proofs that 0.999... equals 1
• Geometric series
• Convergent series
• Infinite series
• Limits

External proofs

• Repeating Nines
• Why does 0.9999... = 1 ?

Template:Mathstub

• Proofs