[nilkn]

Because that's exactly what the argument above does. It simply subtracts away the decimal part of 9.999....

This is always wrong except in the case of infinitely many repeated digits, and the proof does not explain this.

More rigorously, let 9.999{n} denote an expansion with n 9s after the decimal point, where n can also be infinity. The subtlety with the argument is that it needs X to be the same as everything after the decimal point (so that the result of the subtraction is just 9). This is never true for finite values of n, and the proof does not establish that it's true for an infinite value of n -- indeed, it can't do so without supplying a meaning in the first place.

Another way of phrasing it is that it assumes that if X = 0.999..., then 10X = 9.999..., where there are the "same number" of 9s after the decimal point in 10X as there are in X. This seems intuitive for an infinite repeating sequence of 9s, because "one less than infinity" is still infinity, but it's not very rigorous, and the argument as written certainly doesn't explain this.

[banach2]

Okay, this is a little late but when you say 10X = 9.999... then X = 0.999... not because you have removed the 9 but because that's what X is; a tenth of 10X! So when you say in your argument 10X = 9.9 and therefore 10X - X = 9.9 - 0.9, that's wrong. 10X = 9.9 implies X = 0.99. No wiggle room. No need to consider infinites, just one movement of a decimal point.

[And hence 10X - X = 9.9 - 0.99 = 8.91 = 9X implying that X = 0.99]

Following, as delineated above, from there, you'll see there's no contradiction. It's not so easy to break arithmetic that easily without dividing by zero :)

Sorry for dragging on this meaningless thread.

[JoeAltmaier]

Yeah, yes and no. The premise is, that 0.9999... repeating CAN be represented as a decimal number, so that means it CAN be used in arithmetic. If we deny that 0.999... minus 0.999... is zero, then the premise is broken and the question is kind of moot.