Is $9*(1/9) = 0.99999999…$ statement correct? [duplicate]

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.

• There is nothing wrong: you just proved that $0.99999999... = 1$. – Crostul Nov 15 '16 at 13:41
• $.99999…$ is not irrational. It is rational, just as $.11111…$ is rational. – MJD Nov 15 '16 at 13:44
• Similar question here : math.stackexchange.com/questions/11/… – Btzzzz Nov 15 '16 at 13:45
• As others have pointed out, there is no problem. Careful though: the statement in the title is wrong, the 'dots' are important: $9 \cdot \tfrac{1}{9} = 1$ does not equal $0.99999999$ (with a finite number of decimals); but it is the same as $0.999\ldots$ (an infinite number of 9's). – StackTD Nov 15 '16 at 13:54

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!

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 $X = 1$