Misplaced Pages

This is an old revision of this page, as edited by 4.250.177.162 (talk) at 15:00, 8 May 2005 (→See also: just trying to help.). 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:

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

Three alternative ways of explaining this truth

x = 0.999... 10x-x = 9.999... - 0.999... = 9x = 9 x = 1

What is 1-0.99999... ? You get 0.000000... which is the same as zero.

Divide one by three (one third) and you get .333333(an unending series of threes). Three one-thirds is one so three times .3333(an unending series of threes) is .99999999999(an unending series of nines).

If you don't have a problem with 1.00000(an unending series of zeros), why should there be a problem with 0.9999(an unending series of nines) ?

See also

• Geometric series
• Convergent series
• Infinite series
• Limits

External proofs

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

Template:Mathstub

• Proofs