Is it possible?

$x=0.9999.....$ $10x=9.99999...$ $10x-x=9.999999...-0.999999...$ $9x=9$ $x=1$

0.333333333333... = 1/3

Multiply both sides by three

0.999999999999... = 1

$0.999\ldots = 0.\dot{9}$ is a recurring decimal. It can be written in the form of a sum of a geometric progression as below:

$0.999\ldots = 0.9 + 0.09 + 0.009 + \ldots$

where the first term, $a=0.9$ and the common ratio, $r = \dfrac{0.09}{0.9} = 0.1$

The sum goes to infinity, therefore

$0.999 \ldots = \dfrac{0.9}{1-0.1} = \dfrac{0.9}{0.9} = 1$

$0.999 \ldots = 1$ , therefore the statement is $\boxed{\text{True}}$