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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.

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    $\begingroup$ There is nothing wrong: you just proved that $0.99999999... = 1$. $\endgroup$ – Crostul Nov 15 '16 at 13:41
  • $\begingroup$ See also en.wikipedia.org/wiki/0.999... $\endgroup$ – Crostul Nov 15 '16 at 13:43
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    $\begingroup$ $.99999…$ is not irrational. It is rational, just as $.11111…$ is rational. $\endgroup$ – MJD Nov 15 '16 at 13:44
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    $\begingroup$ Similar question here : math.stackexchange.com/questions/11/… $\endgroup$ – Btzzzz Nov 15 '16 at 13:45
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    $\begingroup$ 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). $\endgroup$ – StackTD Nov 15 '16 at 13:54
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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!

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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 $X = 1$

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