# We can tell that 0.11111111...*9=0.999999999... And that 1/9=0.11111111111... Therefore 1/9*9=0.999999999...However, we know that 1/9⋅9=9/9=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.

Is $9\ast \left(1/9\right)=0.99999999...$ statement correct?
We can tell that
$0.11111111...\cdot 9=0.999999999...$
And that
$\frac{1}{9}=0.11111111111...$
Therefore
$\frac{1}{9}\cdot 9=0.999999999...$
However, we know that
$\frac{1}{9}\cdot 9=\frac{9}{9}=1$
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.
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

nuramaaji2000fh
Nothing. What you wrote is correct, and you just proved that $0.\overline{9}=1$
You can do the same thing with $\frac{1}{3}$ and $3$ for example:
$1=\frac{3}{3}=3\cdot \frac{1}{3}=3\cdot \left(0.333333\dots \right)=0.999999\dots =0.\overline{9}$
Remark
That holds only for infinite periodic decimals. You cannot, for example, state that $0.999999999999999999999999999999999=1$
That is not true!

Cristofer Graves
A more typical way to prove is to subtract $0.1\ast 0.\overline{9}$ from $0.\overline{9}$
$0.\overline{9}-0.0\overline{9}=0.9$
But $X-0.1\ast X=0.9\ast X=0.9$ proves $X=1$