# Additive inverse of infinity [closed]

What example demonstrates that a definition

$$\infty-\infty=\infty+\big( -\infty \big)=0$$

• $\infty+\infty=\infty$ Commented Nov 13, 2018 at 21:42
• Do you also want to have $2 \cdot \infty = \infty$? Commented Nov 13, 2018 at 21:43

Consider the expression below:

$$1 + \infty - \infty$$

Let us assume $$\infty - \infty = 0$$; in the above expression, we can get, through the associativity of addition, this:

$$1 + \infty - \infty = 1 + (\infty - \infty)= 1 + (0)= 1$$

However, note that $$1 + \infty = \infty$$. (Though this depends on the number system and kind of numbers we're working with, so this might not be valid. Context would help.) Then, we also have

$$1 + \infty - \infty = (1 + \infty) - \infty = \infty - \infty = 0$$

Note that the associativity of addition guarantees equality:

$$A + B + C = (A+B)+C = A+(B+C)$$

Thus, we have shown that, depending on how you associate the sum $$1+\infty-\infty$$, you get either $$1$$ or $$0$$. Since $$1\neq0$$, one of our assumptions must be wrong.

The sticking point here is ... well, what's the assumption?

That $$\infty - \infty = 0$$?

That $$1 + \infty = \infty$$?

Or, indeed, that once you introduce infinities, that associativity even holds? Who's to say that there doesn't exist a system in which you can add up infinities, but it just doesn't work as it does in the real numbers.

It's not a trivial question, either.

There are systems in which each of these are interpreted differently.

For example, consider the ordinal numbers. In these numbers, there is a sense in which you can say $$\omega +1 \neq \omega$$ (and indeed is the next ordinal after $$\omega$$) but $$1+\omega=\omega$$. (In this discussion, $$\omega$$ is sort of like $$\infty$$ - it is the first/smallest ordinal number such that all finite ordinals come prior to it.)

That's the only one that really comes to mind, to be honest, but it's to hammer in a point: that context is always, always, always important. There's actually a system of numbers - the surreals, I believe they're called - which extends the real numbers to allow infinities and infinitesimals, for example. But then, $$\infty \notin \mathbb{R}$$, for example. So to discuss the notion of an additive inverse of $$\infty$$ in the reals (since you tagged this with real analysis) doesn't even make sense because it's not a real number, and need not inherently obey the same axioms that $$\mathbb{R}$$ does.

• This is a really elegant answer (+1) Commented Nov 14, 2018 at 4:22

What you wrote doesn't make any sense. Infinity isn't a number and thus you cannot apply arithmetic operations to it. However in the spirit of the question, consider:

$$\lim_{x\rightarrow\infty} (x^2 - x).$$

In isolation, both of those terms go to infinity, and the difference goes to infinity.

• Wow that's crazy how what I wrote makes no sense but you understand exactly what I meant and even gave me the exact example that I had seen before but forgotten about. Thanks for making the effort to try to decipher my ambiguous garbage. Commented Nov 13, 2018 at 21:56