#1
Old 09-27-2003, 07:00 PM
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Infinity, squared?

I was reading an article, multiverse, in which the author mentions infinity squared. I've encountered this before, and it would appear that some math geeks believe that infinity and infinity squared can indeed exist seperately.
How so?
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#2
Old 09-27-2003, 07:06 PM
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Infinity squared equals infinity. Just like zero times zero equals zero. I wish math questions were usually as user friendly as that.
#3
Old 09-27-2003, 07:14 PM
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Well I agree, Not In Anger. But I've read where some think differently. I guess I [i]could[i] find a cite.
BTW; One times one equals one. One is closer to infinity than to zero.
#4
Old 09-27-2003, 07:22 PM
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Boring so far.
#5
Old 09-27-2003, 07:30 PM
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I believe infinity squared is a separate entity from infinity because of the fact that it can be either the result of multiplying infinity * infinity or -infinity*-infinity. Since infinity and negative infinity both exist separately, then infinity ^ 2 must be different. Just a guess/hazy memory of what a professor once said.
#6
Old 09-27-2003, 07:38 PM
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As cardinal numbers, they are the same, but as ordinals, they are different.

As you may know, there are different orders of infinity. For simplicity, I'll just take omega (or aleph-naught, as some might call it), the set of natural numbers: {0, 1, 2, 3,...}

omega2 is different from omega--think of it as omega x omega, or omega copies of omega. It looks like this:

{0, 1, 2, 3,...
omega, omega + 1, omega + 2, omega + 3,...
omega+omega, omega+omega+1, omega+omega+2, omega+omega+3,...
omega+omega+omega, omega+omega+omega+1,....
.
.
}

In other words, picture a "copy" of the natural numbers. Next, picture another copy of the natural numbers after that. And another after that, and another after that,... So that, ultimately, you have as many copies of the natural numbers, one after the other, as there were natural numbers themselves in the first place.

That is the ordinal known as omega2.
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#7
Old 09-27-2003, 07:43 PM
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There exists a bijection between the Cantor set, which has outer measure zero, and the Cartesian product of countably infinite copies of the real line, each of which has outer measure infinity. This bijection, sometimes called the Peano space-filling curve, demonstrates that the two sets are in some sense the same size. Both have the cardinality of the continuum, c.

Compared to this craziness, infinity^2 = infinity is considerably easier to wrap your mind around.
#8
Old 09-27-2003, 07:47 PM
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Here's one idea.
#9
Old 09-27-2003, 07:51 PM
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Quote:
Originally posted by Antiochus
Boring so far.
So jazz it up a little.
#10
Old 09-27-2003, 10:00 PM
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amore ac studio, I thought a space filling curve was a continuous map from the reals onto the plane (or some higher dimension).

As for the OP, in addition to what I said above, I suppose it's also possible that the writer confused infinity^2 with 2^infinity.

For any cardinal number (including infinite) x, 2x is always strictly larger than x itself. For example, omega (the natural numbers as I mentioned above). If you take 2omega (which can be described as the collection of all possible subsets of the natural numbers), that's a strictly larger infinity than omega itself. 2omega is also the cardinality of the reals (so the "size" of the set of reals is strictly larger than the "size" of the set of natural numbers).

Unfortunately, we have no idea how much larger 2omega (size of the reals) is than omega (size of the naturals). 2omega could be the next higher cardinal than omega, known as omega-1 (this is known as the Continuum Hypothesis), or it could be much larger. In fact, it's consistent with standard (ZFC) set theory for 2omega to be any cardinal larger than omega, so long as that cardinal has cofinality omega.
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#11
Old 09-27-2003, 10:51 PM
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Quote:
Originally posted by Cabbage
I thought a space filling curve was a continuous map from the reals onto the plane (or some higher dimension).
That's the usual definition, which is why I said the bijection from the Cantor set to Rinfinity was only sometimes called the space-filling curve. You'll note that a bijection from the Cantor set to Rn can be constructed via the composition of the Devil's staircase (which is continuous and monotone increasing) with a bijection from [0,1] to R, followed by the usual space-filling curve for Rn. The construction of a bijection from the Cantor set to Rinfinity is roughly analogous.
#12
Old 09-27-2003, 10:53 PM
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From the Introduction of the page that mangeorge cited:

Quote:
For the purposes of Infinite Math, it must be expressed that infinity can be acted upon by mathematical operations, just as any finite value can be acted upon by mathematical operations. For example, six times infinity is six infinity. It cannot be assumed, as in Finite Math, that since infinity is a concept and not a reachable value, infinity is equivalent to six infinity, nor can it be assumed that infinity is equivalent to infinity plus one. Only infinity plus one is equal to infinity plus one in Infinite Mathematics. It is easily seen by this definition that there are numbers greater than infinity, since infinity plus one is farther along the number line. These are infinite numbers, and they are discussed in Chapter 3.
I'm not about to read Chapter 3 at this hour, and I don't know whether this is just crackpottery or not, but from this paragraph it is clear that his Infinite Mathematics is distinctly not the infinite math that Cantor defined.
#13
Old 09-28-2003, 01:26 AM
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There are several ways to extend the arithmetic operations to apply to infinite numbers. Cabbage has already discussed Cantor's cardinal and ordinal numbers (where infinite numbers are called transfinite). Cantor's cardinal and ordinal numbers are discrete. John Conway's surreal numbers are continuous. You can do just about any operation on his infinite numbers that you can do on finite numbers -- take square roots, subtract them, divide by them. Any surreal number can be squared, and if the number is greater than 1, the square will be greater than the number.
#14
Old 09-28-2003, 01:47 AM
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The author of the page mangeorge cited mentions in the first few paragraphs that it's obvious that 1/infinity = 0. I'm going to look it over tomorrow, but I'm already a bit cautious.
#15
Old 09-28-2003, 02:46 AM
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Frankly, it looks like the author of the page is either:
  • Not using the concept of infinity the way we have used it in all previous threads here, and the way I've always seen it used in the serious texts, or
  • Completely incompetent at the math he professes to teach to others and is making it up as he goes along.
It's a pity that more of the site isn't up yet, but when people begin to say that they've developed a whole new branch of math, I begin to wonder about how many wheels they have on the ground.

Quote of interest:
Quote:
his is an extremely new branch of mathematics that a good friend of mine, JAMES R. WALLACE, and I have recently developed. It has the significance that Calculus has had, since Newton first used it for motion problems. Infinite Math is very complex and is often very difficult to understand. This is partially due to its very recent discovery, but also due to the abstract nature of infinite concepts.
Eh, very few things are as significant to math as the Calculus. Damned few, in fact. Makes one wonder why I've not heard of it, or of these people, before.

Anyway, let's take a little example from the site:

As he says (and has been quoted in this thread), normal mathematical operators work on infinity.

He also states that 1/inf = 0.

Therefore, let's do a little algebra.

inf * 1/inf = 0 * inf
1 = 0

Ah, a little inconsistency! Unless he wants to retract one or more of the above assumptions he's made, he's fallen into a fairly serious trap.
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#16
Old 09-28-2003, 02:52 AM
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I'm not a mathematician but it seems to me that anything other than zero times infinity = infinity.



p.s. I only entered this thread because the board lied and told me it had zero posts. I wanted to be the first with a witty remark after a ham(p)ster eaten OP.
#17
Old 09-28-2003, 04:00 AM
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Quote:
Therefore, let's do a little algebra.

inf * 1/inf = 0 * inf
1 = 0
Derleth, zero * infinity is not necessarily zero. It is usually undefined (or indeterminate).

That is precisely how integral calculus works. Integral calculus shows instances where you can define it. Or more to the point how you can calculate it.

Take for example determining the area under a curve. Calculus takes the area and breaks it up into an infinite number of parts, each with zero thickness (and therefore 0 area).

Totaling up all the areas you get :
0 + 0 + 0 + 0 + 0 + ... (an infinite number of times or [0 * infinity] ) = the area under the curve. Depending on the curve, that area can be zero, +/- infinity, or any number in between.

a visual representation:
http://hyperphysics.phy-astr.gsu.edu/hbase/integ.html
#18
Old 09-28-2003, 06:00 AM
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Those infinite mathemtatic people look like cranks (or jokers, or overenthusiastic high-schoolers) to me. Their method is just defining a symbol oo equal to 1/0, and, if I read the site correctly, assuming that it can then obey all the normal arithmetic rules, which paints them into a corner immediately:

1/oo=0=(1/oo)/2=(1/2oo) rearranging gets 2oo=oo, which they explicitely say it doesn't (alternatively, they say you can cancel oo, so 2=1)

There are number systems that have an infinity obeying normal rules of arithmetic - see hyperreal numbers and surreal numbers - but they have to have infinitessimals (numbers greater than zero, but less than any positive real number) as well. The site gropes at the edges of this by saying they use "a positive and a negative zero."

It's even possible to use infinitessimals in calculus instead of limits, though it was ages before anyone did it rigorously, despite everyone thinking that way
#19
Old 09-28-2003, 07:06 AM
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Quote:
Originally posted by Eleusis
I'm not a mathematician but it seems to me that anything other than zero times infinity = infinity.
I'm not a mathematician either (I'm dabbling in a bit of complex analysis at university, but that's just for fun). But from what I understand, the symbol "infinity" is never actually defined in such a way that makes, for example, "2*inf = inf" any more meaningful than "k) + -& = / )(((()(" (at least not in the more 'traditional' branches of mathematics).

However, "infinity" can be part of another 'symbol'. In this case, an appropriate 'symbol' would be the limit. So we could say something like "lim(x->inf) 2x = inf", which expresses the same idea as "2*inf = inf", except that it's nice, meaningful, and well-defined (so long as we make certain obvious assumptions, like x being a real number). As an added bonus, it's also true!

Try not to think of "lim(x->inf) 2x = inf" as an equation, despite the equals sign. In particular, don't try anything silly like rewriting it as "(lim(x->inf) 2x)/inf = 0", or we'll be back where we started. Another way of writing "lim(x->inf) 2x = inf" is "As x->inf, 2x->inf", which doesn't have an equals sign. This way is probably nicer, but it won't impress the girls/guys/whatever. The formal definition (and proof) involves more epsilons and deltas than is appropriate for my first post to this message board . But basically it says that "as a variable x increases without bound, 2x also increases without bound." Note the absence of any mention of equality.

A great thing about using this limit notation is that it doesn't foul up the rules of arithmetic any more than necessary. If infinity was a symbol we could plug into equations willy-nilly, we couln't even be sure that obvious statements like "x = x" are true anymore!! Granted, mathematics is already quite dirty with all the problems that zero causes, but why make things worse than they already are?

So yeah, basically you're right. Another way of saying what you want to say is that as a variable x increases without bound, x multiplied by any non-zero number also increases without bound. Well, I guess that's not quite right, because if that non-zero number is negative, it decreases without bound. So, in a nutshell,

lim(x->inf) nx = inf, where n>0
lim(x->inf) nx = -inf, where n<0

These can be proven, but just not by me right now.

(now I'd better brace myself for my post to be ripped apart by people who actually know what they're talking about)
#20
Old 09-28-2003, 11:04 AM
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What's the infinite root of infinity? Or infinity to the infinite power?
#21
Old 09-28-2003, 11:40 AM
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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.
#22
Old 09-28-2003, 11:54 AM
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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.
#23
Old 09-28-2003, 01:40 PM
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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?" .
Quote:
What's the infinite root of infinity? Or infinity to the infinite power?
On the first question, moriah, the most sensible way to discuss "The infinite root of infinity" would be limx --> oo(x(1/x) (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 xc is an infinite number which is strictly larger than c. Furthermore, if 1 < x1 <= c and 1 < x2 <= c, then x1c = x2c, 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 NN = 2N = googolN = 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.
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#24
Old 09-28-2003, 01:48 PM
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Quote:
Originally posted by Sqwert
Derleth, zero * infinity is not necessarily zero. It is usually undefined (or indeterminate).
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.
#25
Old 09-28-2003, 02:08 PM
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Yeah, but it looks like on the right hand side of your equation, you simplified 0 &times; inf to just 0. If not, what did you do?
#26
Old 09-28-2003, 02:28 PM
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Quote:
Originally posted by Achernar
Yeah, but it looks like on the right hand side of your equation, you simplified 0 &times; 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.
#27
Old 09-28-2003, 02:49 PM
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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 &times; infinity = 0 based on those assumptions?
#28
Old 09-28-2003, 03:29 PM
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Welcome, Mbossa.

Scritchscritchscritch.....
Maybe you've noticed that mangeorge hasn't jumped back in here.
#29
Old 09-28-2003, 03:42 PM
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Quote:
Originally posted by Mbossa
However, "infinity" can be part of another 'symbol'. In this case, an appropriate 'symbol' would be the limit. So we could say something like "lim(x->inf) 2x = inf", which expresses the same idea as "2*inf = inf", except that it's nice, meaningful, and well-defined (so long as we make certain obvious assumptions, like x being a real number). As an added bonus, it's also true!...

A great thing about using this limit notation is that it doesn't foul up the rules of arithmetic any more than necessary. If infinity was a symbol we could plug into equations willy-nilly, we couln't even be sure that obvious statements like "x = x" are true anymore!! Granted, mathematics is already quite dirty with all the problems that zero causes, but why make things worse than they already are?
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.
#30
Old 09-28-2003, 03:46 PM
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Infinity squared = minus one
#31
Old 09-28-2003, 04:31 PM
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Quote:
Originally posted by AcidKid
Infinity squared = minus one
Cite?
#32
Old 09-28-2003, 04:40 PM
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Quote:
Originally posted by Achernar
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 &times; infinity = 0 based on those assumptions?
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
#33
Old 09-28-2003, 11:17 PM
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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.
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#34
Old 09-29-2003, 01:17 AM
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Quote:
Originally posted by Derleth
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.
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.
#35
Old 09-29-2003, 01:57 AM
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Quote:
Originally posted by Achernar
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.
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?
#36
Old 09-29-2003, 10:32 AM
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Quote:
Originally posted by Achernar
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.
It's not, because the cancellation law doesn't hold, and infinity doesn't have a multiplicative inverse.
#37
Old 09-29-2003, 01:58 PM
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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.
#38
Old 09-29-2003, 02:54 PM
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That's pretty much the correct interpretation for anyone who isn't doing graduate level mathematics.
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