There front cover of New Scientist today read “Infinity isn’t the biggest thing any more.”
OK, the article was a decent introduction to larger infinities, and it’s useful to tell people this, but it just looked like this was some sort of breakthrough, not a standard mathematical result being jazzed up because there was a play called ‘infinities’. Cantor wasn’t that recent.
People looked at me strangley when I laughed out loud in the newsagent.
All you have to do is mention “infinity” and sensible people go all soft and squishy and you get the posts above. First off, it cannot be said too often, There is no mathematical entity as infinity. Say it. Good. Now say it again.
On the other hand, there are infinite sets, many, many of them. So many that one cannot assign a size to them. You cannot even say (legitimately, in most versions of set theory), the “set of all sets”. (You cannot say in any version, the “set of all sets that are not elements of themselves”.)
Now what about infinity squared. Well that would usually be understood to mean the cartesian product of an infinite set with itself. In the most usual axiomatization of set theory, called ZFC, for Zermelo-Frankel with choice, the cartesian product of any infinite set with itelf has the same cardinality (or size) as the set itself, whatever that is. Without choice that can break down. However, for the set of natural numbers (0,1,2,3,4,…) that is easy to prove without choice. You can enumerate the set of pairs (which is the cartesian product of the set with itself) in a perfectly definite way. Begin with (0,0), the pair whose coordinates add up to 0, then follow with (0,1), (1,0), the two whose coordinates add up to 1, then (0,2),(1,1),(2,0) and so on. It is quite easy to write a simple formula that implements this.
Leaving sets aside, what about 1/0? Well 1/0 makes just as much sense as the set of all sets that are not elements of themselves. In other words, it is just not there. But there are kinds of arithmetic that allow infinite entities. One way of doing calculus (which used to be called infinitesimal calculus and still is in French) is to introduce positive numbers called infinitesimals that are non-zero but smaller than 1/n for any natural number n. If i is such an infintesimal, then 1/i is called an infinite number. This use of the word “infinite” is basically unrelated to the use in set theory, even though some fairly sophisticated set theory (called ultrafilters) is used to find a good model of infinitesimals.
From this point of view, a derivative is essentially a difference quotient: f’(x) is the “ordinary part” (every finite number is the sum of an ordinary number and an infinitesimal number) of (f(x+i)-f(x))/i for an infinitesimal i, provided that ordinary part is independent of i. And integrals are defined as sums over an infinite index of the usual kinds of terms.
Then there are the Conway numbers that have their own variety of infinite numbers and their infinitesimal reciprocals. And there may be other uses of “infinity” that I don’t know about. But my main point is that the word itself has no single meaning and neither does the question. Oh yes, if e is an infinte number, e^2 is another infinite number, infinitely larger than e since e^2/e is still infinitely large. And not only is 1/e^2 infinitesimal, it is infinitesimally smaller than 1/e.
Great first post, Mbossa, and I for one don’t blame you for leaving out the “way too many epsilons and deltas”. But then again, I’m a physicist, and our idea of a proof is to point at the board and say “Well, isn’t it obvious?” :).
On the first question, moriah, the most sensible way to discuss “The infinite root of infinity” would be lim[sub]x --> oo[/sub](xsup[/sup] (that is, the limit for large x of the xth root of x), and that’s equal to 1. On your second question, there are, indeed, different infinite numbers, and infinity^infinity would, in fact, be interpreted as a larger infinity. More particularly, if c is any infinite number, and x is any number (infinite or finite) with 1 < x <= c, then x[sup]c[/sup] is an infinite number which is strictly larger than c. Furthermore, if 1 < x[sub]1[/sub] <= c and 1 < x[sub]2[/sub] <= c, then x[sub]1[/sub][sup]c[/sup] = x[sub]2[/sub][sup]c[/sup], and if the generalized continuum hypothesis is true (it can’t be proven one way or another), then that number is the next larger infinite number. As a specific example (and this one is true regardless of the Continuum Hypothesis), if we let N be the cardinality of the integers (which happens to be the smallest infinite number), then N[sup]N[/sup] = 2[sup]N[/sup] = googol[sup]N[/sup] = the cardinality of the real numbers.
Incidentally, before any of the mathematicians nitpick, this is in terms of “numbers” as cardinals, which I believe is the most common way of discussing infinite numbers. What I said above may or may not be true for ordinals, surreals, or any other model of infinite numbers.
I know that, you know that, but the yahoo who put the page up apparently didn’t. I just showed a trivial way in which he falls flat on his face, given the assumptions he stated.
Read my first post to see what those two assumptions were.
Yeah, but it looks like on the right hand side of your equation, you simplified 0 × inf to just 0. If not, what did you do?
I did that based on his assumptions. It was an example of why he was wrong.
Did you read any other part of my post, perchance? I think you should at least read his two assumptions I listed in my post. Without that much, my equations lack context.
The two assumptions are: mathematical operations can be performed on infinity, and 1/infinity = 0. Am I reading you right?
How do you get 0 × infinity = 0 based on those assumptions?
Scritchscritchscritch…
Maybe you’ve noticed that mangeorge hasn’t jumped back in here. 
I’m starting to think that calculus could be made more intuitive if we removed the infinity symbol from it. It’s not necessary. We never actually have to deal with infinity in calculus. “The limit as x goes to infinity of 2x is infinity” I think causes more problems to understanding than “The limit as x goes up of 2x is positive unrestricted” or something like that would. They both mean the same thing, that for any N > 0, there is an M > 0 such that f(M) > N, where f(x) = 2x.
Infinity squared = minus one
Cite? 
Perhaps Derleth considers ‘normal mathematical operations’ implies any number multiplied by zero is zero?
I did skim their site, expecting them to have something of the sort, but they don’t seem to have blundered into that one.
I think you can do the derivation with those assumptions though: 0/oo=0*1/oo=0 so 0=0.oo
But this isn’t completely self-evident, I don’t think it would be fair to expect people to read it into his proof.
Of course, I think we’ve shown their system is self-contradictory (for reasonable values of ‘normal math ops’), so obviously you can proove 0=0.o somehow 
I read the quotes others posted to this thread and the pages themselves (all there are of them) and I noted those two statements.
I assumed that by `normal mathematical operators’ they mean what they say: 0 * n = 0 for any n, 1 * n = n for any n, and x * n = xn for any x and n neither equal to zero. In fact, he mentions that 6 * inf = 6inf, so presumably 0 * inf = 0. I believe this presumption to be justifiable.
He mentions specifically that 1/inf = 0. This is really non-negotiable: It’s on the site and the relevant pages have been linked to.
Using nothing more than the two assumptions above, I stated that since multiplication `works’ (is defined for all values and does not produce unexpected results), you can multiply both sides of the above equality by inf to derive the obvious fallacy 1 = 0.
Now, reductio ad absurdum comes into play: If a logical foundation leads to a contradiction, the foundation must be flawed. Since 1 = 0 is the most obvious possible contradiction, the foundation must be deeply flawed.
From what I can see, the only way to save the system is for him to specify 0 * inf != 0. Maybe his bizarre notions of +0 and -0 will come into play, but I don’t see precisely how. But such a specification would negate (or seriously modify) his statement that normal mathematical operators work on inf in the normal way.
So you assumed that “normal mathematical operators” implies the field axioms? Or something like that? I’m not saying it’s unreasonable, I just wanted to know what you were saying.
As a side note, there is a manifold C* which is the union of the complex numbers with an infinity. I’m not sure how much algebra you can do on C*, but I don’t think it’s a field.
I assumed the author of the site probably hadn’t thought about it in that way and that he was aiming his exposition at the non-mathematician (i.e., someone with a knowledge of trivial algebra but not the more advanced forms of algebra).
In other words, I just applied the `reasonable person’ test to a pseudo-mathematical website. Does that mean I’m going to hell?
It’s not, because the cancellation law doesn’t hold, and infinity doesn’t have a multiplicative inverse.
I had always thought that “infinity” is not actually a quantity and is thus not amenable to trivial calculation.
It’s more of a “way-way-way big whatsitoosie” or “way-way-way small whatsitoosie”, but said in a more hoity-toity fashion.
That’s pretty much the correct interpretation for anyone who isn’t doing graduate level mathematics.