narc is right

0.99...9 < 1 for any **fixed number** n of 9s beyond the decimal point. Furthermore, 0.99...99 > 0.99...9.

\--|--/ \--|---/ \--|--/

| | |

n 9s (n+1) 9s n 9s

Thus, the sequence {0.99...9} is a monotonicaly increasing and bounded. According to a famous theorem from calculus, it is convergent, meaning it has a unique limit. Let us denote this limit as x. Then we have:

x = 0.99... (notice that the sequence of 9s goes to infinity now)

Multyplying by 10, we get:

10x = 9.99...

Subtracting these two expressions and we have:

10x - x = 9.99... - 0.99...

9x = 9

x - 9/9

x = 1

So, the limit is 1, indicating that, **when the number of decimals goes to infinity** 0.99... = 1.

However, narc is wrong in saying that when a number has infinite number of decimal places, it does not exist. All irrational numbers (like √2 or *π*) have a decimal representation with infinite number of non-repeating decimals. However, they do exist and are used in mathematics all the time.