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What example demonstrates that a definition

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

necessarily invokes a contradiction?

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

2 Answers 2

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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.

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  • $\begingroup$ This is a really elegant answer (+1) $\endgroup$
    – clathratus
    Commented Nov 14, 2018 at 4:22
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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.

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    $\begingroup$ 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. $\endgroup$ Commented Nov 13, 2018 at 21:56

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