Recently, an AT&T commercial has been running on TV where a moderator asks some children about the largest number they could think of.

At the end, one kid replies “∞ times ∞”, which of course is simply ∞^{2}.

Natually, one can instantly think of a larger number: ∞^{∞}. But then, that got me to thinking about ∞.

Now, by non-standard, but logical, mathematical rules of conventions regarding reciprocal and powers notation, it can be argued that:

∞ = 1/0 = 0^{-1}

and

-∞ = -1/0 = -0^{-1}

This naturally leads to the formula:

∞^{∞} = 0^{(0-1)}

However, due to those rules governing mixtures of powers, I quickly realized that:

∞^{∞} ≠ 0^{(0-1)}

Because

0^{(0-1)} = 0^{∞} = 0 ! (& no, I’m not intending that to represent a factorial) :tongue:

Therefore, to properly notate the equation, it must go like this:

∞^{∞} = 0^{-(0-1)} = 0^{-∞} !! (& no, I’m not intending that to represent a double-factorial) :tongue2:

Which results in the surprising conclusion:

∞ = -∞

Furthermore:

1/0 = -1/0

And

1 = -1:surprised

Spoiler

(Of course, there is an error in the above—the negative in the exponent of 0 is the act of making the reciprocal of 0, NOT ∞.

It’s essentially the same error that folks often make when proving 1=0. Besides, even if there was no math error in my original calculation, there’s still an error in logic, as the entire exercise is merely a matter of mathematical notation, which is not the same as actual numbers.

But, ironically, it is quite interesting that the graphing of nearly any function that involves the value of ∞, there is a corresponding value to those functions that implies that ∞ does indeed equal -∞.)

First, we would like to define what ##\infty## actually is. Contrary to what most laymen believe, there is no unique form of infinity.

You use the relation ##\infty = \frac{1}{0}##. This relation is actually false for most definitions of ##\infty## that you use. Except for the projective real line.

In the projective real line, you add one symbol ##\infty## to ##\mathbb{R}##. And this symbol is both larger than each real number and smaller (in fact, the usual ordering relation breaks down). In this case, we can indeed define division by ##0## by ##\infty##. And interestingly enough, ##\infty = -\infty## holds in the projective real line.

And by the way, in the projective real line, it doesn't hold that ##\frac{\infty}{\infty} = 1##. So you can't go from ##\infty = - \infty## to ##1=-1##. So there is no contradiction.

Oh, okay. I'm not as up on some of the current heiroglyphics used nowadays.

a is a subset of the Regular Reals, but ∞ is beyond in the Regular. (Sorry, but it also seems I'm not having much fun with the Latex Reference symbols--I have to use IE at my office)