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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 this proof is to recognize 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 proofs

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.

The following proof relies on a property of real numbers. Assume that 0.999... and 1 are in fact distinct real numbers. Then, there must exist infinitely many real numbers in the interval (0.999..., 1). No such numbers exist; therefore, our original assumption is false: 0.999... and 1 are not distinct, and so they are equal.

See also

• Recurring decimal
• Geometric series
• Convergent series
• Infinite series
• Limits

External proofs

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

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