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
Explanation
The key step to understanding this proof is to recognize that the following infinite geometric series is convergent:
A note on limits
Bear in mind that the value of any infinite decimal expansion such as
(with the "3" repeating ad infinitum) is the limit of the sequence of its finite truncations
Therefore, to say that 0.999... = 1 is simply to say that the limit of the sequence of finite truncations
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.
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.
Illustration with fractions
When you divide a number by 9, it comes out as a repeating decimal of that number.
etc.
However, any number divided by itself is equal to one. Therefore 0.999... = 1.