Is $9*(1/9) = 0.99999999...$ statement correct?

Question

We can tell that $$0.11111111... \cdot 9 = 0.999999999...$$ And that $$\frac19 = 0.11111111111...$$

Therefore $$\frac19\cdot9 = 0.999999999...$$

However, we know that $$\frac19\cdot9 = \frac99 = 1$$

Note: I'm taking in account that ... are the other rational digits left.

What am I making wrong? What is misunderstood?

Thanks for the help in clearing this problem.

Answer 1

Nothing. What you wrote is correct, and you just proved that $0.\bar 9 = 1$

You can do the same thing with $\frac{1}{3}$ and $3$ for example:

$$1 = \frac{3}{3} = 3\cdot \frac{1}{3} = 3\cdot (0.333333\ldots) = 0.999999\ldots = 0.\bar 9$$

Remark

That holds only for infinite periodic decimals. You cannot, for example, state that $0.999999999999999999999999999999999 = 1$

That is not true!

Answer 2

A more typical way to prove is to subtract $0.1*0.\bar 9$ from $0.\bar 9$:

$$0.\bar 9 - 0.0\bar 9 = 0.9$$

But $X - 0.1*X = 0.9*X = 0.9$ proves $X = 1$