Interesting Infinite Powers Paradox

[Ryuu]

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)

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)

Which results in the surprising conclusion:

∞ = -∞

Furthermore:

1/0 = -1/0

And

1 = -1

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 -∞.)

 

[micromass]

Interesting.

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.

So, what you described, is no more and no less that the projective real line.
http://en.wikipedia.org/wiki/Real_projective_line

 

[micromass]

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.

 

[Ryuu]

Thanks, Micromass.

I've known about the concept for a long time, but I've never known its formal name.

Although I see that the first 3 lines of the Arithmatic Operations defined conflict with the first 2 lines of those undefined, if one sets a = ∞

 

[micromass]

Thanks, Micromass.

I've known about the concept for a long time, but I've never known its formal name.

Although I see that the first 3 lines of the Arithmatic Operations defined conflict with the first 2 lines of those undefined, if one sets a = ∞

It doesn't conflict, since it says that $a\in \mathbb{R}$. And $\infty$ is not an element of $\mathbb{R}$.

 

[Ryuu]

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)

 

[micromass]

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)

Not "subset", but "element". That is: $a$ is an element of the real numbers, but $\infty$ is not a real number.

 

[FOIWATER]

a^2-a^2 = a^2-a^2

a(a-a) = (a+a)(a-a)

a(a-a) divided by a-a = (a+a)(a-a) divided by a-a

a=2a

1=2

... but I divided by zero

 

[mfb]

∞ = -∞ is a very natural result if you extend real numbers, as micromass mentioned. It gets even better with complex numbers.