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.

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
The result you get to is correct, whatever the form you write it in. Making the multiplication like that for numbers with infinitely many digits is not exactly the technique I would recommend. — , Nov 15 '16 at 13:45
Similar question here : http://math.stackexchange.com/questions/11/is-it-true-that-0-999999999-ldots-1 — Btzzzz, Nov 15 '16 at 13:45
Googling "0.99999 site:math.stackexchange.com" comes up with a lot of other people asking this exact question on this exact site. — Arthur, 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
@Pichi Wuana The decimal system is a remarkably clever way of denoting real numbers, but it does have a small confusing flaw:although each irrational number has only one representation as a decimal, there are certain fractions that can be written as decimals in two different ways. I was not crazy enough to want to be a psychiatrist, but I assure you it's comforting to figure out which fractions these are. — Airymouse, Nov 15 '16 at 14:08

score 6 · Accepted Answer · answered Nov 15 '16 at 13:44

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!

score 0 · Answer 2 · answered Nov 15 '16 at 14:02

0

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 $X = 1$

answered Nov 15 '16 at 14:02

Pieter21

2 Answers2