Upper and lower bounds in regards to 0.(9) 
Possible Duplicate:
Does .99999… = 1? 

I'm only doing this at GSCE and I'm really only asking here because of an interesting email conversation between my Grandfather and I regarding the fact that 0.(9) equals 1, so I'd appreciate it if you could make any explanation as simple as possible.
Basically, I have proven to my Grandfather that 0.(9) must equal 1, using the following method:
Let x = 0.(9)
So, 10x will equal 9.(9); 10x - x is  9x which is the same as 9.(9) - 0.(9) = 9, and therefore 9 / 9 is 1!
However, he has questioned the fact that 0.(9) * 9 equals 9, as he rightly stated that it equals 8.(9). I do remember learning in my maths lesson a rule regards to upper and lower bounds that meant that 8.(9) was actually the same as 9, or something along  those lines, but I can not remember the correct statement to inform my Grandfather - so any suggestions would be appreciated.
Thanks in advance
 A: Your "10x = 9.(9) - x = 9x = 9" is a confusing shorthand so let's try: 
$x = 0.(9) \implies 10x=9.(9) \implies 10x-x = 9.(9) - 0.(9) \implies 9x=9 \implies x=1.$
Nowhere do we try to multiply out $9 \times 0.(9)$.  Instead we calculate $9.(9) - 0.(9)$, and I would be surprised if your grandfather thought that this was not $9$.
A: The definite response to the 0.(9) issue is this: By definition, $0.(9)=\sum_{n=1}^\infty\frac{9}{10^n}$. Using the formula for infinite geometric sum, you get $0.(9)=1$. Every other "proof" is simply an appeal to some basic intuition that allows one to avoid going through the formal definition, but when that intuition hits the wall it's time to use the definition.
Now, in general if a real numbers ends in a sequence of digits the form $a999\dots$ where $a\ne 9$ you can replace it by $b000\dots$ where $b=a+1$. This leads to $8.999\dots$ being equal to $9.000\dots$. This is also true in other bases: in base $d$, every number whose representation ends in a sequence of the digit $d-1$ can be changed similarly (adding 1 to the last digit different from $d-1$ and changing the rest of the digits afterwards to 0).
