The problem
Wow, my classmate told me that $0{,}\overline{9}$ is equal to $1$. What do you think?
$0{,}\overline{9} = 0{,}9999\ldots$, with infinitely many nines.
See it
Think of $1$ as a whole rectangle (its proportions are chosen so that each blank piece left over is the same shape as the rectangle, only smaller). For $0{,}9$ we split it into $10$ equal parts and colour $9$: one is left blank. For $0{,}99$, the blank part is split into $10$ again and $9$ are coloured. And so on with each nine. If $0{,}\overline{9}$ and $1$ were different, some piece of the rectangle would have to stay blank. Where is it?
With any number of nines the blank part is there, but each nine makes it $10$ times smaller: soon you cannot see it without the magnifier, and the magnifier has to zoom in more and more. $0{,}\overline{9}$ has infinitely many nines: no piece of the rectangle is left unpainted, however small you try to make it. There is no difference between $0{,}\overline{9}$ and $1$.
What do you think?
Is it true that $0{,}\overline{9} = 1$? Justify your answer.
See the solution
In the numerator we write the number without the decimal point up to the end of the first repeating block ($09 = 9$) and subtract the part that does not repeat ($0$). In the denominator, as many nines as there are digits in the repeating block: one.
$0{,}\overline{9} = 1$
This is where the generating-fraction rule comes from. Let $N = 0{,}999\ldots$ and multiply it by $10$: there are still infinitely many nines after the decimal point.
$N = 1$
Why is it hard to believe? Because with a finite number of nines something is always missing (you saw it in the figure). With infinitely many nines, what is missing is smaller than any positive number you can think of, and the only number like that is $0$. There is no number between $0{,}\overline{9}$ and $1$: they are the same number written in two ways.