Quote Originally Posted by Tavrobel
Have you people completely forgotten that .999... is an approximation?
Intergers, radicals, and fractions are exact values.
.999... vs. 1 is comparing an infinte concept (an approximation, at that) to an exact value. That is like asking someone to divide by 0.

Ok, so 1/3 == .333..., but .333... is still a approximation. You cannot have an exact value for 1/3 in decimal notation because it does not end. So stop saying that 1/9 * 9 == .999... == 1 or 1/3 * 3 == .999... == 1 is true. 1/7 * 7 is also an infinite set of numbers, yes, even though it does not equal .999..., but rather, a smaller number, it still would equal one.
Any number divided by itself is 1 (except 0/0). This the FIRST grand assumption that people miss/omit.

However, the "pick out a number between the two and you define them as different numbers", that one, I couldn't attack even if I were smart enough to explain it.
It's not an approximation. It's a method of describing a number as an infinite geometric series. So let's rewrite this without decimals.

Been a while since I've had calc, so forgive me if I mess this up. A decimal number

0.x1x2x3x4x5x6...xn

(those numbers are meant to be subscripts) is actually a way of writing

0 + x1/10 + x2/10^2 + x3/10^3 + x4/10^4 + x5/10^5 + ... + xn/(10^n)

It so happens in the case of .99999999 (however many 9's you'd like) that all of x1...xn all are 9.

9/10 + 9/100 + 9/1000 + ... + 9/10^n

Easier to write it like this:

SUM(k = 1 to n) of [9/(10^k)]

Or you could rewrite it equivalently as this:

SUM(k = 0 to n) of [(9/10) / (10^k)]

or

SUM(k = 0 to n) of [(9/10) * (1/10)^k]

This so happens to be a geometric series.

Now, imagine that we let n approach infinity. In the case of .999.... n does approach infinity. In this case we have an infinite geometric series.

Some infinite series add up to (converge to) a finite sum. It seems non-intuitive, but that's calculus for you. If it wasn't possible to add up an infinite number of numbers and (sometimes) get a finite result, we'd be screwed. See <a href="http://en.wikipedia.org/wiki/Zeno%27s_paradoxes">Zeno's paradoxes</a>.

It so happens that this infinite geometric series is one of the types of infinite series that converges to a finite sum. But you have to use limits.

lim(n -> infinity) SUM(k = 0 to n) of [a * r^k] = a / (1 - r)

The proof of this is simple and can be seen <a href="http://mathworld.wolfram.com/GeometricSeries.html">here</a> for example. This is true when |r| < 1, and clearly|1/10| < 1.

Substituting from above, a = 9/10 and r = 1/10, we have

.999...
= lim(n -> infinity) SUM(k = 0 to n) of [(9/10) * (1/10)^k]
= (9/10) / (1 - (1/10))
= (9/10) / (9/10)
= 1

So no approximations or decimals needed.