One divided by Infinity?

Okay, I'm not much of a mathematician (I'm an 8th grader in Algebra I), but I have a question about something that's been bugging me.

I know that $0.999 \cdots$ (repeating) = $1$. So wouldn't $1 - \frac{1}{\infty} = 1$ as well? Because $\frac{1}{\infty} $ would be infinitely close to $0$, perhaps as $1^{-\infty}$?

So $1 - 1^{-\infty}$, or $\frac{1}{\infty}$ would be equivalent to $0.999 \cdots$? Or am I missing something? Is infinity something that can even be used in this sort of mathematics?

You could say that $\frac{1}{\infty} = 0$, so $1-\frac{1}{\infty} = 1$. But then, you're stretching the definition of division past breaking point - division as you know it isn't defined for infinity, so the answer is undefined. Otherwise, you can quickly get yourself into a pickle and end up saying 1=2.

Arithmetic operators - add, subtract, divide, multiply, raise to the power of - are defined on a particular set of numbers: such as real numbers, or complex numbers.

The set you use for definition, will determine what you can and can't say meaningfully. Typically (but not always), infinity is excluded from that set.

If we take the set of real numbers, and look at "raise to the power of", then $1^x$ is equal to 1 for any x, as x -> infinity. So in that case, you could have a convention of saying that $1^\infty = 1$. But $\frac{1}{1} = 1$, so $1^{-\infty}$ would also equal 1. However, when you go about defining these new conventions, you have to be extremely careful - sometimes, a convention will seem obvious, but if you run with it, you end up seeming to prove 1=2, which means that your convention wasn't that helpful.

Let's compare with raising to the power 0.5, i.e. taking the square root. $-1^{0.5}$ is undefined when we are working on the reals - so, just as dividing by infinity, you can't include it in your arithmetic. Only when you expand to the complex numbers and extend your definition of the arithmetic operators to cope, can you say something meaningful about $(-1)^{0.5}$

Similarly, the reals and the complex numbers each exclude infinity, so arithmetic isn't defined for it.

You can extend those sets to include infinity - but then you have to extend the definition of the arithmetic operators, to cope with that extended set. And then, you need to start thinking about arithmetic differently. If you want to learn more about that, then there are lots of friendly places on the web to get into the work of Cantor on the different types of infinity (of which there are an infinite number of different infinities).

Whether one defines $1/\infty$ may be a matter of convention, but one can say that $1/x$ approaches $0$ as $x$ approaches $\infty$, and what that means is that $1/x$ can be made as close as desired to $0$ by making $x$ big enough. How big is big enough depends on how close you want to make $x$ to $0$. If that's what you mean by $1/\infty = 0$, then that statement seems unobjectionable. More specifically, if you want the distance between $1/x$ and $0$ to be less than a small positive number $\varepsilon$, then that is so whenever $x > 1/\varepsilon$.