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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 ?

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• Proofs