0.999...

In mathematics, one could easily fall in the trap of thinking that the recurring decimal 0.999... is not equal to 1. The following proof shows that the two numbers are actually equal.

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 understanding this proof is to recognize that the following infinite geometric series is convergent:

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

A note on limits

Bear in mind that the value of any infinite decimal expansion such as

0.333\dots \,

(with the "3" repeating ad infinitum) is the limit of the sequence of its finite truncations

0.\overbrace {333\dots 3} .\,

Therefore, to say that 0.999... = 1 is simply to say that the limit of the sequence of finite truncations

0.\overbrace {999\dots 9} \,

is 1. Since this is an increasing sequence, that is the same as saying that 1 is the smallest number that none of the finite truncations can exceed.

Alternative proofs

Algebraic manipulation

While this proof fails to address the converging series, it may be simpler to understand.

$x$	$=0.999\ldots$
$10x-x$	$=9.999\ldots -0.999\ldots$
$9x$	$=9$
$x$	$=1$

Property of real numbers

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