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 Not Infinity! By The Writer in ScienceTue Jun 03, 2003 at 04:51:58 PM EST Tags: Focus On... (all tags) Many mathematicians, math geeks, and other strange beings, have jabbered on and on about "infinity", infinite sets, and what-not. They've said a lot about what infinity is; but today, I'd like to tell you what infinity is not. This article hopes to clear up some common confusions and misconceptions surrounding "infinity" which may have led the less mathematically-inclined to think that math is contradictory and confusing. Hopefully, it will also introduce a bit of what infinity is, so that normal people can understand it without their brains turning into a pretzel1. A subsequent article will delve deeper into the subject and clear up even more misconceptions that people may have about infinity. 1Shaped like ∞.

Misconception #1: Infinity is just a "funny-looking number"

Now, I concede that ∞ is funny-looking. It looks like an 8 written by a sideways mathematician (or a sideways 8 written by a normal mathematician). It can also be used for such questionable practices as the following ASCII figure (I mean, Unicode figure):

~o
/∞\
|
/ \

Which I think looks better than K5ARP's sideways "B" in place of the ∞, don't you think? (And that's a swimsuit, not something else, you pervs.)

But aside from the funny-looking part, "infinity" is not a number.

Let me repeat that. "Infinity" is not a number!

That is to say, for any real number x, things like ∞/∞, x/∞, ∞/x, are all nonsensical. When a mathematician says 1/∞ = 0, you have to understand it in the right context (explained in a moment). When a human being says 1/∞ = 0, he's talking nonsense; ignore him.

What context, you ask? First of all, "1/∞ = 0" is just a mathematical shorthand; it is not saying that if you divide 1 by this strange beast (or, lazy beast) called ∞, it magically turns into 0. Remember, ∞ is not a number. What the mathematician means when he (uh, it) says that 1/∞ = 0, is that if you take 1 and divide it by a very large number x, it will be "close to" 0, and that the larger the number, the closer 1/x will be to 0. Note that it is merely "close to 0", but not actually "equal to 0". The "proper" way to write this is:

limx→∞ 1/x = 0
In math-speak, this is read "the limit of 1/x, as x approaches infinity, is 0".

"Aaack, you just fooled me!" I hear you scream. "What do you mean, x approaches infinity? You just said infinity wasn't a number!"

Calm down. When mathematicians say "x approaches infinity", or write "x→∞", they mean "x grows arbitrarily large". And when they say that the limit of something is 0, they mean that it can go as close to 0 as you want it to, but it may or may not be actually equal to 0. They do not mean that x "becomes" ∞ (it cannot, since ∞ is not a number to begin with), or that 1/x becomes equal to 0 (since 1 divided by any number is never equal to 0). When humans (or math students) say or write this, however, they are just talking nonsense; ignore them.

Oh, and did I mention that ∞ is not a number?

Misconception #2: Wash, rinse, repeat (forever)

This is often the high-school teachers' cop-out explanation of infinity.

"Oh, you just repeat this forever and you get infinity."

Bzzzt. No, you do not. You get nothing. Not even zero. Nothing. Nonsense. Nada. "Repeat this forever" is nonsense. Although you could repeat something forever, you won't get anything out of it; you'd just get really, really, tired.

Here's a typical high-school description of the set of natural numbers: "You start with the empty set, { }, and you add 0 to it to get {0}, and then you add 1 to it to get {0,1}, and then you add 2 and get {0,1,2}, and so on, and you repeat this forever, and you get N, the set of all natural numbers."

This description is not only inaccurate, it is just plain wrong on some points. You do not get anything if you repeat this forever; sure, your set will get bigger and bigger, but that's about it. Since you are repeating forever, by definition there is no result yet, and there never will be any result, since you never stop.

So, class, what do you get if you repeat this forever? Repeat after me: "Nothing. Nothing. Nothing. Nothing. Nothing. ..." (repeat this forever)

Mathematically speaking, this description is wrong because:

1. The set of natural numbers is not "built up" by repeating some operation forever. You can never reach infinity by repeating something finite. That's why it's called infinity, duh.
2. Even if it were possible to repeat something forever, you'd still not get the set of natural numbers. If I built a machine that alternately prints 1 and 0 every second, what will it print after infinite time? If you say 1, I can equally argue it should be 0. If you say 0, I can equally argue it should be 1. Or you could say it's both, but "both" is not a number. The real answer: there is no such thing as getting a final result out of "repeating forever". "Forever" means there is no final result. So please get that silly idea out of your head already.

The set of all natural numbers

"Then how do you get the set of natural numbers?!" I hear you ask.

The answer: we get it by definition. For you math buffs, this definition is called the Axiom of Infinity. (And it is called that, precisely because it defines what infinity is.) Don't get fooled by the scary name of this Axiom; it is really rather simple. All it says is this: there exists a set (conventionally called N), which contains the number 0, and if N contains some number x, then N also contains (x+1).

Let's unpack this sentence a bit. First, it says N contains 0. OK, so N looks like {0, ...}. But since it contains 0, it must also contain (0+1), which is what we call '1'. So N must look like: {0,1, ...}. Furthermore, since it contains '1', then it must also contain (1+1), which is what we call '2'; so it looks like {0,1,2, ...}. By the same reasoning, it must also contain what we call '3', so it would look like {0,1,2,3, ...}. And so forth.

"Hold on a second there!" I hear you say. "Isn't this just another case of repeat forever' that you just debunked??"

No. Please notice the subtle, but very important difference: we are not building anything by repeating some operation forever. We are simply defining N to contain every natural number there is. We are not building N piecemeal; we're not saying "add (x+1) to N". We're saying that for every number x, (x+1) is already in N. We're not adding numbers into N to make it N; every number is already in N.

The whole deal with the (x+1) in the Axiom of Infinity is so that we can describe N fully in a finite amount of space and time. Otherwise, we would need an infinitely long paper to list all the members of N, and by definition, we cannot ever finish writing this list, so it is not a proper definition of N, and will never be.

N, the set of natural numbers, contains every number there is by definition. It is not "attained to" by doing some "repeat this forever" nonsense. You cannot attain to infinity; you need to be given an already-completed infinity by a Higher Power. In this case, the Higher Power is the Axiom of Infinity.

Misconception #3: Infinity is nonsense because it doesn't follow the rules of arithmetic

Or, as some people would say:

I was told that 1/∞ = 0. If I multiply both sides by ∞, I get 1 = 0*∞. But since 0*anything = 0, I get 1=0. Argh, my head hurts!

I'm not surprised your head hurts. Besides the fact that infinity is not a number (see misconception #1), this misconception also underlines a faulty assumption people have: that infinite quantities obey the same rules of arithmetic as finite quantities. They do not. If they did, we could prove that every number is 0 using the above argument: let n be any non-zero number. Since 1/∞ = 0, we can multiply both sides by n, and get n/∞ = 0. But that means that n = 0*∞, so we conclude that n=0.

Obviously, this is ridiculous. As I said before, and I'll say again, ∞ is not a number.

But there are such things as infinite quantities, and it is in fact possible to define consistent arithmetic on them. Just don't expect them to behave like finite quantities. (That's why they're called infinite quantities!)

Comparing infinite quantities

Infinite sets cannot be compared by counting the number of elements in them, and then comparing the totals. You cannot ever finish counting elements in an infinite set, because it's infinite! However, it is possible to compare them without ever counting them. Here's how.

Suppose we were to turn back the clock, to the time when we were 2 years old and didn't know how to count up to 5 yet. How would we know if we had the same number of fingers on either hand? We couldn't count the fingers on each hand, and compare the results; because we couldn't count that high. But we could put our hands together, finger to finger, and find that there are no extra or missing fingers that didn't have a matching finger on the other hand. We could then conclude that there are as many fingers on our left hand as our right hand, without knowing how to count how many there are!

Mathematicians call this establishing a 1-to-1 correspondence. In this case, we're establishing a 1-to-1 correspondence between left-hand fingers and right-hand fingers. If there's a 1-to-1 correspondence between two sets A and B, we say that they have the same cardinality. ("Cardinality" is just math-speak for "size".) For infinite sets, we may not be able to count the number of elements in them; but we sure can tell if two of them had the same cardinality (size) by seeing if we can make a 1-to-1 correspondence between their elements. (This, of course, also works perfectly fine with finite sets; so we're not using some "crippled" method of comparison here.)

Here's an example. Take the set of all natural numbers, N, and throw away the odd numbers. Call the result E, the set of even numbers. Question: how many elements does E have? Surely E must only have half the number of elements as N, since we threw away the other half? Not really. Consider this correspondence: for every element x in N, we map it to x*2. Since x*2 is even, x*2 is an element of E. Every number in N is mapped to exactly one element in E, and every element in E has exactly one number in N mapped to it. The fingers match! This means that E and N in fact have the same size.

(Note: we only had to find one possible 1-to-1 correspondence. Just because some correspondences don't work doesn't mean the sets aren't equal in size. Just because the 2-year-old didn't manage to match his index fingers together doesn't mean one of his hands is missing an index finger, as long as there is some way to match them all up correctly.)

"Wait!" I hear you say. "Isn't this a contradiction? We just removed some elements from a set, shouldn't it be smaller now?"

Well, that's why it's called an infinite set. If your computer is connected to a battery that stores enough power to run for 10 hours, then after each hour, the battery has 1 less hour's worth of power left. After 10 hours, it runs out of power. This is your typical finite set: remove something from it, and it contains less than before. But if you plugged your computer into the power outlet, and it continues to consume the same amount of power per hour as before, the power never runs out! How can this be?? Because the power outlet is connected to your local power plant, which is producing more power as your computer consumes it. This is your infinite set: removing something from it doesn't necessarily make it smaller. The power plant is an "infinite battery"; consuming power from it doesn't make it "contain less power".

So when dealing with infinite quantities, don't blindly assume that the rules of finite arithmetic automatically apply to them. They don't. And just because they don't, doesn't make them contradictory.

Misconception #4: There are twice as many integers as there are natural numbers

We've already seen that there are just as many even numbers as there are odd numbers. Now I'm going to show you something even more surprising: the positive and negative integers, taken together, are still not any bigger the set of all natural numbers!

How? Well, consider the set of integers, Z. For every integer z in Z, if z is positive, we map it to (z*2)+1, which is a positive odd number. If z is negative, we map it to -z*2, which is a positive even number. And if z is 0, it stays 0. Note that every odd number is covered, and so is every even number, including 0. Miracle! Did you see what just happened? What looked like two "infinite endpoints" in Z has collapsed into a single endpoint, and we are back to N, the set of natural numbers, yet again!

"OK", I hear you say. "But what about fractions of integers?"

Good question. Fractions of integers, which mathematicians call "rationals" or "rational numbers", are numbers of the form p/q, where p and q are integers. So 1/2, 1/3, 1/4, 2/5, etc., are all rationals. The set of rationals is conventionally called Q (for quotient).

Q is an interesting set that has a property called density. That means that between every two rational numbers, you can always find another rational in between. For example, between 0 and 1, you have 1/2, and between 1/2 and 1 you have 3/4, etc.. It's easy to see that between every two numbers, there is an endless supply of rationals. This is why we say Q is dense; no matter how many times you magnify it, you will still see it jam-packed with rationals.

Now the million-dollar question: is Q larger than N?

Misconception #5: Since the set of rationals is dense, there must be more rationals than integers

On the surface, it sure looks like this is the case. After all, not only do the rationals stretch infinitely to either side of 0, there are also an infinite number of them between every given two. There's no way this isn't bigger than the set of natural numbers!

Well, don't be fooled by the appearance of Q. (That applies in Star Trek, and it applies here, too.) Just like Z, the set of integers, Q is just arranged funny. If we straighten it out the order of its elements, it'll look a lot less scary.

"OK, you've got to be kidding me," I hear you retort. "There is no way you're going to rearrange a dense set into a non-dense one!"

Just wait till you see it. First, to make it less confusing, let's restrict ourselves to Q+, the set of positive rationals. Note that it is still dense, so there's no sleight of hand here. Now, Q+ consists of fractions of the form p/q, where p and q are positive integers. So we make a table, with the columns representing p, and the rows representing q, like this:

1/1 2/1 3/1 4/1 ...
1/2 2/2 3/2 4/2 ...
1/3 2/3 3/3 4/3 ...
...

OK, some of these entries are redundant, since 2/2 is the same as 3/3, etc., but the point is that this table contains every possible rational in Q+. Now, we cut this table up into diagonals. The first diagonal is 1/1; the second is 1/2 and 2/1; the third is 1/3, 2/2, 3/1; etc.. Since each of these diagonals are of finite length, we can glue them together, end to end, to form a list that goes like this: 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, ... . Of course, we can remove the redundant elements from this list, but the point is that we can arrange all the elements of Q+ into a linear, non-dense list.

You can probably guess what's coming next. That's right, we label each element in this list, starting with 0, 1, 2, ..., etc.. It's the set of natural numbers again! So we find out that Q+ is actually the same size as N. But we can do exactly the same thing with Q-, the set of negative rationals. So Q- is also the same size as N.

Now the kicker: when we labelled the elements of Q+, we could've used even numbers instead; and when we labelled the elements of Q-, we could've used odd numbers. Put them together, and BANG! we get N, yet again! So then, there are only as many rationals as there are integers, even though the rationals are dense. Betcha they didn't tell you this in high-school!

You're probably starting to notice a trend here. An awful lot of sets seem to be the same size as N. In fact, N is so special, that mathematicians have come up with a name for the size of N. We say that the cardinality of N is ℵ0. That's pronounced "aleph null" or "aleph zero", by the way; not "squiggly egg". Since Z and Q are also the same size, they all have cardinality ℵ0.

"Why not just call it 'infinity'?" you ask.

That's the topic for the next article. We will see that there isn't just one infinity; there are actually many more! Stay tuned!

 Poll
Infinity is:
 a pretzel 25% a funny-looking number 8% a swimsuit 19% what you get if you repeat something forever 19% nonsense 11% squiggly egg 15%

 Votes: 94 Results | Other Polls

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 Not Infinity! | 419 comments (370 topical, 49 editorial, 0 hidden)
 Can I be the first to say (2.90 / 10) (#3) by starsky on Tue Jun 03, 2003 at 10:06:00 AM EST

 that by assuming the power plant has an infinite supply of energy, you are condoning the continual raping of the worlds natural resources. Also, +1, good stuff, but I'm sure the intello-nazis at k5 will -1 this as it's 'too simple', whereas some gay-ass 'people have sex' article from turmeric is obviously 'very clever'. Fucktards.
 I might have said... (4.00 / 1) (#62) by Erbo on Tue Jun 03, 2003 at 05:19:06 PM EST

 ...that "I can assume the power plant has an infinite supply of energy...but I would have to also make the assumption that I had an infinite supply of money with which to pay my electric bill." --Electric Minds - virtual community since 1996. http://www.electricminds.org[ Parent ]
 sure, why not? (2.00 / 1) (#133) by tang gnat on Tue Jun 03, 2003 at 10:23:49 PM EST

 There aren't any such power plants, so it doesn't matter! [ Parent ]
 +1 FP Fantastic! (4.50 / 6) (#12) by Imperfect on Tue Jun 03, 2003 at 10:24:07 AM EST

 Very well-written, informative, and easy to understand. Withotu trying to sound like I'm insulting you, have you ever considered writing a "...for Dummies" book? You've got the right pacing, patience, and insights to explain complicated theories simply. Just a thought. Not perfect, not quite.
 In perfect agreement with Imperfect (4.00 / 1) (#14) by Control Group on Tue Jun 03, 2003 at 10:49:02 AM EST

 You could probably pull off a "Reasonably Advanced Math For Dummies" book, and it would be fantastic. Especially if you submitted it a piece at a time to k5 for me to read. ;) *** "Oh, nothing. It just looks like a simple Kung-Fu Swedish Rastafarian Helldemon."[ Parent ]
 Writing "for dummies" books (4.00 / 1) (#17) by The Writer on Tue Jun 03, 2003 at 11:03:23 AM EST

 I've never written a book before, for one, so I really have no idea what's involved in writing one. Plus, I'm only "The Writer" on K5, I'm not a writer IRL. How would one go about getting published anyway? [ Parent ]
 Only a vague idea: (4.50 / 2) (#19) by Imperfect on Tue Jun 03, 2003 at 11:10:23 AM EST

 I consider myself something of a writer, too, only I've never really spent much time trying it professionally. The process goes basically like this, however: Come up with an idea. Write a treatment for that idea. Write a sample champter for that treatment. Contact the appropriate publisher(s) for that treatment; see if they accept unsolicted treatments. If not, contact some agents and see if they would be interested in previewing your treatment. Repeat above until successful. Have agent send treatment to publisher(s). Repeat above until successful. Accept advance, quit job or reduce hours at job, write the remainder of the book on the timeline you agree to, and bask in the glory! At least, that's the theory. I have zero actual practice, however. Not perfect, not quite.[ Parent ]
 Fascinating. (3.00 / 1) (#128) by Akshay on Tue Jun 03, 2003 at 09:18:52 PM EST

 Only a sample chapter? And here I was, trying to see how I can schedule the planned 20 chapters for my Great Big Novel! :-| [ Parent ]
 Don't let K5 delude you (1.50 / 2) (#77) by marcos on Tue Jun 03, 2003 at 05:52:48 PM EST

 Get some real critics to read your work, and not a bunch of random computer users. Personally, I find your style a bit too I don't know the english word to be able to read more than a few paragraphs. If you want to write on math, write real math, and not supposedly funny stories. Clarity is much more important than humour when it comes to technical languages like math. And you have to know your audience to rate how it arrives. You don't accuratly know the K5 audience. [ Parent ]
 I know the word (none / 0) (#78) by Roman on Tue Jun 03, 2003 at 05:58:12 PM EST

 "a bit too I don't know the english word to be " - arrogant? :) [ Parent ]
 A bit negative, aren't we? (4.00 / 2) (#147) by poyoyo on Wed Jun 04, 2003 at 12:03:53 AM EST

 I know a well-written piece when I see it. Check out a dummies book: it's written in exactly that kind of style. If this guy has something to write about, he's good enough to get published. Drop your assumption that "writes on K5" = "not good enough to write in a paying publication". [ Parent ]
 Publish on the web (4.00 / 1) (#194) by Viliam Bur on Wed Jun 04, 2003 at 08:11:03 AM EST

 Or just begin with publishing online. Get some feedback, fix errors. When you wrote enough, make a book of it, add something extra and try to publish. After publishing, add "buy the book" URLs on the web pages. [ Parent ]
 No doubt (2.75 / 4) (#18) by marcos on Tue Jun 03, 2003 at 11:09:39 AM EST

 This is clear. However, I still stand by my previous statements that you cannot pick any set of things that exist in the real physical world, and tell me that n+1 exists for every object in that set (1 is not a number, but a member of the set). I say that there must be a limit to the size of the set for everything that exists physically, because you know what it would imply if there wasn't. If anybody can prove otherwise, please do so.
 And furthermore... (5.00 / 6) (#50) by rusty on Tue Jun 03, 2003 at 03:23:41 PM EST

 I assert that you cannot cut off a duck's head with a hatchet and cause that duck to continue living. Your statement may or may not be true but is definitely irrelevant. The article is about math, not ducks. ____Not the real rusty[ Parent ]
 Maybe you missed the last argument (5.00 / 1) (#56) by marcos on Tue Jun 03, 2003 at 04:06:57 PM EST

 On "The Writers" last story, we had a long argument about the paradox, and the crux of my argument was that it was not in any way a paradox, simply becasue the sphere was not real. If you imagine a real sphere, cut in half, and still equal to the original sphere, you get confused, and term it a paradox. However, if you imagine a mathematical sphere, constructed of an infinite number of points, then it is not a paradox, but is clearly visualiseable. I concluded that the paradox was not at all a paradox, because mathematical infinity does not exist in the real world. No sort of infinity exists in the real world actually. His last story tried to link a physically existing object with mathematical definations of extent, and I am addressing both of them in the above comment. What I find quite amazing is how easily most people will reproduce what they have learnt, and will show that they actually understand it, but completely fail to be able to grasp the slightest variation on that very topic, or dismiss it without bothering to understand the variation. [ Parent ]
 I beg to differ (4.75 / 4) (#111) by sigwinch on Tue Jun 03, 2003 at 07:43:48 PM EST

 However, if you imagine a mathematical sphere, constructed of an infinite number of points, then it is not a paradox, but is clearly visualiseable. Maybe not. Imagine your sphere. For each point, imagine multiplying it's distance from the center by sqrt(2). Presto, sphere with twice the volume, no new points needed. Easy to visualize, right? But that's not what the Banach-Tarski paradox is about. Try the same exercise using only the tools of set selection, rotation, and translation. No stretching allowed. If you can trivially visualize the act of ripping something apart into chunks of fractal fog, and slapping the chunks of fog back together into solid objects, then I salute you. --I don't want the world, I just want your half.[ Parent ]
 My turn (none / 0) (#239) by transport on Wed Jun 04, 2003 at 03:08:12 PM EST

 Fixing paradoxes (5.00 / 3) (#243) by The Writer on Wed Jun 04, 2003 at 03:31:37 PM EST

 A physical paradox signifies a problem, a fallacy, a misconception, something "not right", which needs to be fixed. [...] no-one is trying to "fix" the Banach-Tarski, as they would if you could take a physical sphere and do the same trick. While I agree with the rest of your post, I'm not so sure about that "fixing the physical sphere" part. If I knew how to carry out a Banach-Tarski decomposition on a physical sphere, I wouldn't be worrying about "fixing" anything; I'd just invest in little golden ball bearing, and repeatedly reassemble it until it's as large as the moon, sell it for more money than I can name, and subsequently retire to some island somewhere and relax. And perhaps bring a single brick with me, and reassemble it into a large 50-bedroom mansion once I arrive on the island. Heck, I wouldn't even need to buy an island. I could just take some soil from my backyard, and reassemble it into an island. :-) [ Parent ]
 Points of view (5.00 / 1) (#329) by transport on Thu Jun 05, 2003 at 12:58:47 PM EST

 Heh - ok, where I see a paradox in need of "fixing", you see an opportunity :-) But you realise that the reason that you could make money on the scheme is the same that the rest of the world thinks there is a paradox: If it were possible with a gold sphere, it would simply cease to be a paradox. [ Parent ]
 Darn :-) (none / 0) (#333) by The Writer on Thu Jun 05, 2003 at 01:26:51 PM EST

 That's true. If it were possible in the real world, gold wouldn't be very highly-valued at all. At least, its value would not depend on its quantity. [ Parent ]
 Your assertion is false (none / 0) (#118) by scheme on Tue Jun 03, 2003 at 07:59:03 PM EST

 I assert that you cannot cut off a duck's head with a hatchet and cause that duck to continue living. It's been done with a chicken so it's possible with a duck. The actual occurenece stemmed from the amputation of the head missing the chicken's brainstem and the chicken survived for quite a while longer. Here is a link to an account and explanation of what happened. "Put your hand on a hot stove for a minute, and it seems like an hour. Sit with a pretty girl for an hour, and it seems like a minute. THAT'S relativity." --Albert Einstein [ Parent ]
 Ah yes, but... (5.00 / 1) (#138) by rusty on Tue Jun 03, 2003 at 10:47:36 PM EST

 ...I said a duck. And while it may have been done with a chicken, that is no proof it could be done with a duck whatsoever. And regardless, my main point still stands that this is a totally irrelevant conversation to begin with. :-) ____Not the real rusty[ Parent ]
 nobody wants to prove so (5.00 / 6) (#51) by llimllib on Tue Jun 03, 2003 at 03:28:41 PM EST

 Part of the beauty of the system of mathematics is its total removal from the real world. Mathematics is a closed, imaginary system where infinity is extant and other non real-world entities exist. However, it still has the awesome power to map to real-world objects in a meaningful (but still imaginary) way. Thus, the definition of natural numbers exists only in the world of mathematics, which is what I believe the author is talking about. Finally, your assertion (that the set of physical objects is finite, as I see it) is true only if the universe has edges, which (IMHO) is currently unproven. If you can prove that to me, I would be very appreciative. I can't prove to you that it doesn't, so I think we'll have to settle for admitting that we just don't know. Peace.[ Parent ]
 I can prove it easily (none / 0) (#57) by marcos on Tue Jun 03, 2003 at 04:18:33 PM EST

 Here is the logical proof that the universe has got edges. If you can flaw my logic please do so. Accepted presuppositons The universe is defined by the existence of particles that are not equal to nothing. Every particle has got at least one other particle to which it is moving relatively. Because at least two particles are moving, there exists a time required for a particle to move. If a time exists, then the particles are moving at a particular speed. This speed is limited, not only because Einstein said so, but because assuming two particles that are exactly the same in exactly the same circumstances and influenced by the same objects, one cannot move faster than the other. Proof If the speed at which the particles are moving is finite then If there is an edge to the empty space, then the universe has an an edge If there is no edge to the empty space, then the particles are moving in the direction of the empty space with finite speed. This implies that whenever the particles started moving, or even if they have always been moving, the furthermost particle represents the outermost boundery of the universe, and hence the edge of the universe. But I concede a point to you here, while still maintaining my point: the only thing that is actually infinite is "nothing". The "nothing" that the particles are expanding into. There is only one way you can say that the universe does not have edges 1. If you can prove that particles can move instantaneously relative to other particles. [ Parent ]
 Only one way? (5.00 / 4) (#64) by Control Group on Tue Jun 03, 2003 at 05:28:49 PM EST

 D'OH *forehead slap* (5.00 / 1) (#66) by Control Group on Tue Jun 03, 2003 at 05:31:22 PM EST

 That should be: ...at that point you're proving the universe... My most-hated grammatical error, and I'm guilty of it myself. I'm so ashamed... *** "Oh, nothing. It just looks like a simple Kung-Fu Swedish Rastafarian Helldemon."[ Parent ]
 No, I handle that case (1.00 / 1) (#71) by marcos on Tue Jun 03, 2003 at 05:45:29 PM EST

 If you reread my comment. The basis for my theory is that particles are moving! If they are moving, then there is an edge, even if they started moving an infinite time ago. The universe would then be very large, but still have an edge, because it is expanding. On the other hand, the universe simply springing into existence with particles scattered infinitely is not a case that should be considered. 0 != 1, no particle is not equal to particle. I refuse to consider that point as valid, and if you do, you are deluding yourself. [ Parent ]
 You're still assuming it's finite (5.00 / 3) (#79) by Control Group on Tue Jun 03, 2003 at 06:00:37 PM EST

 If the particles have been expanding for an infinite period of time (as distinct from an arbitrarily-long period of time), then there can be no edge. If you presume a finite number of particles, of course, there must be an edge - but that's a presupposition, not a conclusion. If you presume an infinite number of particles, however, then the universe not only doesn't have to have an edge, it mustn't have an edge. Either case is valid, and there's no way to determine which it is (though the finite-particles hypothesis is certainly more intuitively appealing, and neatly avoids the "why isn't the night sky white" problem) from where we are. Again, I'm not saying you're wrong - I'm only pointing out that you're assuming finiteness going in. Unless, of course, I'm misinterpreting your argument, which is entirely possible. After all, I'm not as bright as I think I am. *** "Oh, nothing. It just looks like a simple Kung-Fu Swedish Rastafarian Helldemon."[ Parent ]
 Let me think about that (none / 0) (#81) by marcos on Tue Jun 03, 2003 at 06:12:29 PM EST

 The core issue comes down to this How could something have come out of nothing? If it always existed, how could it have a value, and what was that value? You know, I just had a brainflash. I take back everything I said. I think that the only solution to these problems is to discard the notion of discrete and counteable particles. I don't know what the alternative is yet, but it cannot lie with the particles. I give you the benefit of doubt at the moment, till I have worked out what values fit in with my new thoughts. [ Parent ]
 Answers: (none / 0) (#100) by Kwil on Tue Jun 03, 2003 at 06:57:45 PM EST

 Who said it did? Maybe it always was. Who says it does? A value implies a number. It may be infinite, whatever "it" is. In essence, you're too caught up in the whole cause/effect thing. Think chaos, think quantum. Perhaps there is no cause. That Jesus Christ guy is getting some terrible lag... it took him 3 days to respawn! -NJ CoolBreeze[ Parent ]
 The problem lies here (none / 0) (#169) by marcos on Wed Jun 04, 2003 at 03:12:49 AM EST

 0 never becomes 1. A particle is defined by borders. Why are the borders a particular extent. Why that particular extent? If one can asign a value to the length of a particle, then it could not have always existed. Something must have always existed, but not particles. [ Parent ]
 Why not? (none / 0) (#172) by Kwil on Wed Jun 04, 2003 at 04:07:15 AM EST

 You're still asking why. Still thinking cause and effect. What if there is no why. What if things just are - perhaps because they always have been? Is that so terribly difficult to get your mind around? "Where did the particle come from?" is just another way of saying "What was before the particle?" but the thing is, what if there simply was no 'before' the particle? If the particle has always existed in one form or another (energy/matter/whatever) then the question of "before" is nonsensical. That Jesus Christ guy is getting some terrible lag... it took him 3 days to respawn! -NJ CoolBreeze[ Parent ]
 No, it has to be understandeable (none / 0) (#186) by marcos on Wed Jun 04, 2003 at 06:54:07 AM EST

 Look at it this way - there is always causality. It is the basic law of the universe as we know it. The only way anything can exist without its existence being caused is if it is the basic building block. And if it is the basic building block, then it is free from all constraints. A particle that is clearly defined, and existent in our physical world as we understand it (implying it has mass, or height/length/breadth), MUST have been created by something else. I don't want to run through another logical proof, but it simply holds that it cannot always have existed. The point on which I base my argument is that there is a finite number of these particles, or there is a space between the particles. If either of those two holds, and we claim to know that the second holds, then those particles could not always have existed, because they are defined by borders, and a border cannot always have existed. I understand it, but I don't know if I have made it clear enough to you. Nothing discrete could have existed forever. If you claim so, then what is that particle made up of? Can the particle be split? Over the night, the closest I have come to a solution is that nothing is not actually nothing, but is defined. [ Parent ]
 Ah here's the problem. (none / 0) (#226) by Kwil on Wed Jun 04, 2003 at 01:12:00 PM EST

 You're operating with the underlying assumption that everything always has a cause. It may be that way. It certainly doesn't have to be that way We're getting some hints down at the quntum level that, in some cases, it might not be that way. I understand your points, but what I'm trying to get at is that they all rely on that same base assumption that there must be a cause for everything -- ergo, there must be a "before" for everything. Now, none of this means your arguments are wrong. They're logically sound. But they're all based on assumption that may not be true in every case. That Jesus Christ guy is getting some terrible lag... it took him 3 days to respawn! -NJ CoolBreeze[ Parent ]
 Particle borders (none / 0) (#363) by Bernie Fsckinner on Fri Jun 06, 2003 at 03:35:45 AM EST

 But you are assuming that particles have borders. They have position and momentum, more or less measurable, (can someone explain the heisenberg unceratinty principle here?). They have effects on each other (gravity and the various forces of unified field theory - electromagnetism and the strong and weak nuclear forces) which physicists have formed theories to explain. Let's use the example of an electron. How do you define its border? You can't even find out its position to more accuracy than Planck's constant divided by the accuracy of its momentum. An electron does not work like a grain of sand. A grain of sand - you can look at it and say, "This position is inside the grain", or "This position is not inside the grain". [ Parent ]
 Heh, good luck convincing him (none / 0) (#370) by The Writer on Fri Jun 06, 2003 at 10:42:45 AM EST

 I posted this rant about why the word "particle" does not accurately describe the subatomic world, but it was obviously not good enough to convince marcos. I hope you have better luck at this one. My own interpretation of what quantum theory tells us about subatomic particles is that "particles" do not have any boundary; in fact, every particle can each span the entire universe, because the likelihood they exist anywhere in the universe is non-zero, albeit very small. They just have a "focal point" where they are most manifested; and we see it as a "particle" with more-or-less defined boundaries. However, the reality could well be that this "particle" in fact spans the entire universe. It's just that we can't see it anywhere else because its effect on the rest of the universe (outside its focal point) is vanishingly small. Which is why it makes a lot more sense to think of these things as waves rather than particles, although the quantum nature of the waves give it particle-like properties (as opposed to water waves which are not constrained by quantum states---at least not in a way that makes a difference to our observations). [ Parent ]
 Keep both (none / 0) (#117) by levesque on Tue Jun 03, 2003 at 07:57:28 PM EST

 I think that the only solution to these problems is to discard the notion of discrete and counteable particles. The accurate modeling of a non-discrete reality would need a dense set of discrete observations. [ Parent ]
 Holy miss the point batman! (none / 0) (#96) by Kwil on Tue Jun 03, 2003 at 06:49:00 PM EST

 even if they started moving an infinite time ago. Hello? Did you not read the article? Infinity is not a number! Just as things can't end in an infinite amount of time, they can't START from an infinite amount of time ago, because that's puts a number on infinite. Also, your idea of the universe springing into existance is the same thing. You're still trying to put a start time on infinity. But there is no start time because it is, by definition, infinite. That Jesus Christ guy is getting some terrible lag... it took him 3 days to respawn! -NJ CoolBreeze[ Parent ]
 Infinity as a number (none / 0) (#113) by sigwinch on Tue Jun 03, 2003 at 07:49:00 PM EST

 Just as things can't end in an infinite amount of time, they can't START from an infinite amount of time ago, because that's puts a number on infinite. Eternity is just as real as next Tuesday. You just can't reach it by waiting for it (because waiting is a finite act). --I don't want the world, I just want your half.[ Parent ]
 Imagine (none / 0) (#143) by Happy Monkey on Tue Jun 03, 2003 at 11:28:30 PM EST

 Imagine the universe is an infinite sheet of rubber graph paper, with a star at each intersection. Right now, the distance between stars on a line is 1 unit. Moving backward in time, that distance asymptotically approaches zero. Moving forward in time, the distance continues to increase. The universe may or may not work like that, but it is one theoretical way for the universe to expand without expanding into anything, and continuing an infinite amout of time in each direction. ___ Length 17, Width 3[ Parent ]
 Okay.. (none / 0) (#173) by Kwil on Wed Jun 04, 2003 at 04:20:29 AM EST

 But if you're assuming a big-bang for the entire universe (which is essentially what this is) you're also automatically assuming a finite amount of matter. Unless, somehow, there was an infinite amount of matter in that big bang. But since infinity is not a number but instead is a concept, there couldn't be an infinite amount of matter. (Unless you want to argue that matter is simply a concept as well, I guess) So in essence, the point still stands that he's simply shown that if we assume the universe has a finite size, we can then theoretically prove that the size must be finite. That Jesus Christ guy is getting some terrible lag... it took him 3 days to respawn! -NJ CoolBreeze[ Parent ]
 Not finite. (5.00 / 1) (#210) by Happy Monkey on Wed Jun 04, 2003 at 10:30:58 AM EST

 In my example, I was explicitly not assuming finite matter. I was just showing that expansion doesn't imply expansion into something. But if we want to limit ourselves to finite matter, all we have to do is curve space - connect opposing edges of a finite, square piece of graph paper, as if the universe were a big 'Asteroids' game. Assume similar expansion. Again we have expansion without edges, this time with finite matter. So in essence, the point still stands that he's simply shown that if we assume the universe has a finite size, we can then theoretically prove that the size must be finite. Is that a typo? Or were you being sardonic? ___ Length 17, Width 3[ Parent ]
 Whoops! You're right. (none / 0) (#228) by Kwil on Wed Jun 04, 2003 at 01:21:57 PM EST

 My problem was explicitly not assuming finite matter moved into a finite space, but I see what you're getting at now, and I was wrong.  Infinity was confusing me there along the same lines as the size of the natural set being smaller than the size of the set of integers. (It's smaller, so it can't be infinity) And it wasn't a typo, Macros 'proof' relies upon assuming the universe (or particles in it, at least) having a finite beginning point to demonstrate that the universe itself must be finite. That Jesus Christ guy is getting some terrible lag... it took him 3 days to respawn! -NJ CoolBreeze[ Parent ]
 It's OK (none / 0) (#235) by Happy Monkey on Wed Jun 04, 2003 at 02:11:26 PM EST

 I was mostly agreeing with you anyway. Inherent limitations of the medium - my post responded to yours, so you saw it with more expectation of disagreement than was intended. My original post was mostly refuting marcos' post, just as yours was. ___ Length 17, Width 3[ Parent ]
 Big Bang was not an explosion (5.00 / 1) (#167) by ToastyKen on Wed Jun 04, 2003 at 02:24:08 AM EST

 So I asked this friend of mine studying for a Ph.D. in Physics about all this big bang stuff, and he said that (current thinking suggests) the Big Bang was not actually an expanding sphere as people generally think.  There are instead two possibilities generally accepted as likely by physicists, and neither involves an "edge" as I previously also thought: The universe is "closed" and looping.  Think of Asteroids or the original Mario Bros., where if you go in one direction you end up coming back to where you were.  In this case, you'd have a finite amount of stuff in the Universe.  Note the lack of any discernable edge.  (Imagine Asteroids not like looping back at the edge of the screen, but instead as scrolling..) The universe is "open" and infinite in all directions.  That means that there is not only no edge, but that there's an infinite amount of matter in the universe! So doesn't that mean we'd be experiencing an infinite amount of gravitational pull?  Well, no, because gravity, like every other force, has a speed limit, c, the speed of light in a vacuum.  So at any moment in time, we're only affected by the matter that's within a radius of c * age-of-the-Universe, which is finite. Now for the Big Bang..  If the Universe is infinite, is there a center?  Well, no.  Imagine a number line, like mentioned in this article...  And imagine the points in the Universe being really really ... really dense right after the Big Bang.  (I'll ignore the moment of the big bang itself, since "infinitely dense" might be meaningless. :P)  Then all the points spread out... evenly along the entire infinite number line... There is no center.  (Don't think of 0 as 0, but just as yet another meaningless label.) So that's how it might be possible that the Universe contains an infinite amount of matter. In the closed Universe case, you'll have to imagine the Universe as a balloon, with the 3 dimensions we experience collapsed into the 2-dimensional surface (like in those Carl Sagan gravity well things).  Now imagine the balloon starting from a point and then expanding.  Note that the "center of the Universe" is not actually in on the surface of the balloon, and it's thus not actually in any of the space that we perceive. Now, the exact shape of the Universe may not be a sphere... Some recent evidence suggests that it may be a doughnut...  But the general idea holds. I think the latest evidence shows that a closed Universe is more likely than an open one, but it's still a very unresolved question. Oh, and I agree that all this is completely irrelevant to the existence of infinity anyway, because infinity itself is a mathematical concept, and any time we talk about a possible infinity in the real world, like in the case of the Big Bang or black holes, we're really saying, "at this point, all laws of physics go out the window and we know nothing." :) [ Parent ]
 I don't accept your presuppositions. (5.00 / 1) (#68) by hex11a on Tue Jun 03, 2003 at 05:39:59 PM EST

 and you don't accept those of the writer. So you can't play together because you're not playing the same game. Hex [ Parent ]
 What are talking about (1.00 / 1) (#72) by marcos on Tue Jun 03, 2003 at 05:47:28 PM EST

 Of course I accept the presupposuitions of the writer. You have got to accept my presuppositions, because if you did not, you would not exist. I should have termed them "The Facts" instead, and maybe you would have read them differently. If you have a point to make, make it, don't just drive by and yell "I disagree". [ Parent ]
 They're not facts (5.00 / 2) (#112) by hex11a on Tue Jun 03, 2003 at 07:44:44 PM EST

 you just think they are. Case in point; current physical models involve quantum fluctuation which requires the instantaneous change of state of certain particles. This means that they do not move at some finite velocity like you say. I'm not saying that it's a "fact" that this happens - I don't know. Frankly I don't know how you can claim that anything so abstracted from our lives is an a-priori fact - your "facts" require the existence of particles - elements that act as point masses and charges but having no three-space dimensions - height, width, length. Lots of eminent physicists have moved away from this to modelling strings etc, so I don't know why it's a "fact" that particles exist. I don't accept your presuppositions, and I do exist. Bang goes your line of reasoning. What are you trying to say by your first sentence? The whole of maths starts with the word "if" - "If X is an abelian group such that...". We don't state "facts" - I'll leave that for the philosophers to decide, but we do work with axioms which we pick for whatever reasons. To investigate the world, or a system, or game such as mathematics is to accept that we don't know what the "facts" are, we just start with a bunch of axioms and work from there. Greater philosophers than I, for example Wittgenstein, would argue that what you are doing when you call something a "fact" is defining the term "fact" and so not actually saying anything meaningful. After all Descartes got as far as Cogito ergo sum, and that was as much as he was willing to call "fact". Many doubt even that. Hex [ Parent ]
 Hey cool! (5.00 / 2) (#74) by awgsilyari on Tue Jun 03, 2003 at 05:49:18 PM EST

 I just applied your argument to the Earth and proved that the Earth has edges. My geometry teacher gave me an 'A'! The most terrible flaw in your argument is this statement: "This implies that whenever the particles started moving, or even if they have always been moving, the furthermost particle represents the outermost boundery of the universe" Furthermost? Furthest from what? -------- Please direct SPAM to john@neuralnw.com [ Parent ]
 You can't apply it to the earth (none / 0) (#82) by marcos on Tue Jun 03, 2003 at 06:18:12 PM EST

 Because the particles in the earth seem to be moving inwards, which automatically prove that the earth has got edges. Alright, let me take the time to explain my thoughts as the last thing this night: The universe exists because it is the alternative to nothing. If there were no particle, there would be no space. There would simply be nothing. As a result, it is the particle which demarcartes the edge of the universe. Assuming a universe of any shape, every particle has got another particle which is at a position that will take longest to reach assuming a finite speed. The furthest away object. If we find a particle that has got the maximimum distance from any other particle, then both those particles represent the edges of the universe. We don't need to mathematically define any of these objects, why don't you just imagine them? [ Parent ]
 Well (5.00 / 1) (#98) by awgsilyari on Tue Jun 03, 2003 at 06:49:45 PM EST

 The universe exists because it is the alternative to nothing. If there were no particle, there would be no space. That might seem "obvious" to you, but it's a huge assumption that you haven't even attempted to justify. There are two ways of interpretting your assumption: Case 1: Suppose space and time are "created" in the immediate vicinity of a particle. Clearly there must be an "edge of space" near the particle. But if this is true, what separates the galaxies from each other? If space/time are created near particles, at what radius from a particle does space/time end? Why at that particular radius? If space itself depends on the existence of particles, and space only exists near them, then how can it be meaningful to refer to the "distance" between two particles outside of this special radius, since distance is measured through space? Case 2: Suppose space and time are "created" by virtue of the existence of a particle somewhere, regardless of precisely where that particle is. Literally, particles and space/time form a dichotomy, neither existing without the other. But if space/time does not depend on where the particles are, then why should it have an edge? An edge would imply some region of space close to the "furthest" particle, but since it doesn't matter where they are, how can there be an edge? -------- Please direct SPAM to john@neuralnw.com [ Parent ]
 hmm.. letme see if i get it (none / 0) (#156) by truchisoft on Wed Jun 04, 2003 at 01:19:47 AM EST

 what you are saying is that the universe can be analogued to the following in the text: lim(x->inf) 1/x = 0 that means, the universe is finite (0), because it tends to its edges, by expanding itself, only that it will never reach its goal, since it cannot become 0, but it can be as close as it can, everytime closer and closer, never stoping. So in your conception the universe can expand forever to the size it wants, but never surpassing a limit, that can never be reached... Woldnt that become a paradox? the universe is at the same time infinite (it never stops growing, so it has no edges) and finite (it cannot surpase a imaginary line) This sounds like the old tale: If you start decelerating at 1/2 your current speed for an infinite time, where do you stop? The answer is, nowhere, you never stop, because 1/2 of a number always exists... Truchi --- Saludos de Argentina.[ Parent ]
 Aha! (none / 0) (#251) by warrax on Wed Jun 04, 2003 at 04:57:51 PM EST

 You said: Assuming a universe of any shape, every particle has got another particle which is at a position that will take longest to reach assuming a finite speed. Who says so?!? Imagine the universe has wrap-around (the same effect as if it were donut-shaped, but I find it easier to visualize the wrap-around effect). Your argument breaks down, because for any point there is always a point that it takes longer to "get to". (Note, that this says nothing about the total number of particles in the universe being finite or infinite.) -- "Guns don't kill people. I kill people."[ Parent ]
 Truth can never be defined by notation. (none / 0) (#94) by aprosumerKuro5hin on Tue Jun 03, 2003 at 06:45:49 PM EST

 Since you present a logical proof, if we accept only your presuppositons and only that which follows from your presuppositons then we can question the validity of the proof by attacking the assumptions inherent in the proof. 1) If there is an edge to the empty space, then the universe has an an edge Is that not like saying "An apple is what is not an apple" or "The edge is defined by that which is on the other side of the edge". It seems you are assuming that the edge of empty space must have a one to one correlation to the edge of the universe and you don't really define what is "empty space" so that makes part two of your proof questionable also. 2) If there is no edge to the empty space, then the particles are moving in the direction of the empty space with finite speed. This implies that whenever the particles started moving, or even if they have always been moving, the furthermost particle represents the outermost boundery of the universe, and hence the edge of the universe. It seems you are assuming that empty space surrounds the universe and that the two are not a homogeneous mix and also that the quantity of empty space is larger than the quantity of universe. [ Parent ]
 Okay... (5.00 / 1) (#168) by Mudlock on Wed Jun 04, 2003 at 02:43:02 AM EST

 Easy. The universe is the "surface" of a hypersphere (or some other four dimensional construct). BAM, no edges. You can still have your moving particles, you can still measure the distances between them and the time for them to cover it. You can even keep your implied big-bang. -- But everybody wants a rock to wind a piece of string around.[ Parent ]
 Don't be silly (none / 0) (#193) by The Smith on Wed Jun 04, 2003 at 08:03:52 AM EST

 The universe is a real thing, not an abstract metaphysical concept. You cannot discover things about the universe using abstract logical reasoning from arbitrary "first principles", only through scientific measurement and theory. Your reasoning is therefore no more valid than the old "proofs" for the existence of God. [ Parent ]
 A simple example (5.00 / 2) (#91) by gidds on Tue Jun 03, 2003 at 06:34:13 PM EST

 you cannot pick any set of things that exist in the real physical world, and tell me that n+1 exists for every object in that set It depends how you define addition, of course, and what you consider exist' to mean exactly, but there's one fairly simple example which doesn't even involve infinities: Hours on a clock face. There's an initial hour (call it 0, call it 12, it doesn't matter).  And for every hour, the next hour is also there.  (Otherwise we'd need a new clock twice a day!) I'm sure it's not what you had in mind, and it doesn't really help the infinities argument, but it does show that you have to be very careful what you mean.  Mathematics is based on rigorous definitions and steps of logic, which is why it works.  Sometimes correspondence with real-world objects can be very valuable (in both directions); at others, it can simply distract you and lead your intuition astray.  (Look at how much time was spent trying to prove the Parallel Postulate, for example.) (BTW, great article!  Extremely well explained.  I look forward to some Cantor and Hilbert soon :) Andy/[ Parent ]
 possibly.... (none / 0) (#211) by alyosha1 on Wed Jun 04, 2003 at 10:32:53 AM EST

 stars (or particles, or whatever) in the universe. My understanding is that if the universe has a positive or flat curvature, you can travel along a straight line, counting all the stars that you 'pass' (i.e. find within some arbitary distance to the line you're travelling along), then for every one you find (n) there will always be another one (n+1) somewhere further along. If the universe has negative curvature, you may end up back where you started from. [ Parent ]
 Re: possibly... (none / 0) (#391) by Peach on Sat Jun 07, 2003 at 08:41:15 PM EST

 If the curvature is positive, the universe is closed and finite (although with no boundaries - like a sphere). If it's negative it's infinite and Pringle-shaped. Whether you could travel in a straight line and end up back where you started is a much more complicated question, though ;) [ Parent ]
 -1, this town is our town it's so motherfucking (1.00 / 34) (#42) by bayou on Tue Jun 03, 2003 at 02:13:29 PM EST

 glamerous!
 BORING BORING BORING BORING BORING BORING BORING (1.03 / 33) (#43) by bayou on Tue Jun 03, 2003 at 02:47:05 PM EST

 MATH
 Obscure and offtopic comment (4.33 / 3) (#47) by bugmaster on Tue Jun 03, 2003 at 03:06:47 PM EST

 Do you know (or, perchance, are you) a teacher named Mr. Schaak (sp?) ? He was one of my favorite math teacher, and he drilled the "infinity is not a number" concept into us at an early age... by using some extreme intimidation :-) >|<*:=
 Mr Schaak (none / 0) (#48) by The Writer on Tue Jun 03, 2003 at 03:19:51 PM EST

 Sounds like a commendable gentleman. :-) Unfortunately, I do not know who he is, and I'm certainly not him. I picked up the "infinity is not a number" concept the slow way: meaning, I heard about it in highschool (in spite of the fact I poked fun at highschool teachers in the article), but it didn't really sink in. I encountered it again later in college math courses, albeit indirectly (why do we need to use epsilon/delta proofs? 'cos we cannot deal with infinity directly). Later, I picked up set theory, which comprises most of the stuff you see here, but even then it never really struck me quite that hard, until I posted the Banach-Tarski paradox story on K5, and noticed that people were drawing incorrect conclusions by treating infinity as though it were a finite number. Then it dawned on me that it was actually a common misconception that might be worth debunking, for the benefit of those who aren't used to more rigorous modes of mathematical thought. [ Parent ]
 Well done (4.50 / 2) (#61) by StephenThompson on Tue Jun 03, 2003 at 05:14:23 PM EST

 Good job. However, you cut off right when it started to get interesting!  I actually await the followup.  A rare achievement on k5.
 Thanks (none / 0) (#76) by The Writer on Tue Jun 03, 2003 at 05:50:41 PM EST

 Well, I had to cut it at that point, because it was getting too long. Some people were already saying that it's getting too lengthy. Personally, my gut feeling is that 2000 words is the max for a K5 article (this one exceeds that). Longer than that, and people are likely to get bored wading through it. I do want to cover so much stuff which didn't make the final cut, though. Like I said in another post, there's too much good stuff to talk about, but too little space to cram them into. People have already demanded that I do justice to the notion of limits---but I fear that that will turn into its own article, since it's such a big topic. The next article in this series will continue the misconception-debunking line, and go on to talk about the cardinality of the real numbers and Cantor's (in)famous diagonalization proof. And perhaps a bit about why 3-dimensional space (well, in math at least) is only as big as 1-dimensional space. Guaranteed shocker for the uninitiated. ;-) [ Parent ]
 Length (5.00 / 1) (#116) by sigwinch on Tue Jun 03, 2003 at 07:56:16 PM EST

 Long is OK, but I think the discussion is more focused and fun if the topic isn't too big. And perhaps a bit about why 3-dimensional space (well, in math at least) is only as big as 1-dimensional space. Guaranteed shocker for the uninitiated. ;-) No more shocking than the rationals having the same cardinality as the integers. For some reason I expected the Cartesian product to be "bigger". The more I learn, the less I know. P.S. Lovely articles. I look forward to the next. --I don't want the world, I just want your half.[ Parent ]
 Humm.. how about.. (none / 0) (#222) by arcade on Wed Jun 04, 2003 at 12:13:36 PM EST

 Non standard analysis (5.00 / 5) (#65) by kallisti on Tue Jun 03, 2003 at 05:30:39 PM EST

 Even though there isn't such a number as infinity, there is a branch of math called non-standard analysis which treats infiniteismals as numbers. An infinitesimal is defined as: x such that -a < x < a for all positive real values of aThis greatly simplifies some math, but complicates things by creating what is called the hyperreal numbers. John Conway went one better and created a system of Surreal numbers which by recursion can handle things like the square root of omega, which is gibberish in normal mathematics. As a bonus, you also get the ability to define any position in any combinatorial 2 player game, which is quite a lot of them.
 You just defined zero (nt) (none / 0) (#115) by dilap on Tue Jun 03, 2003 at 07:52:40 PM EST

 [ Parent ]
 True (2.33 / 3) (#119) by kallisti on Tue Jun 03, 2003 at 08:18:52 PM EST

 Zero qualifies as an infinitesimal in non-standard analysis. It is, however, not the only one. It is the other ones which make it interesting. [ Parent ]
 huh (none / 0) (#335) by mpalczew on Thu Jun 05, 2003 at 01:54:24 PM EST

 Only zero can fit in this defenition.  If another number x existed where x != 0 then x / 2 would also be a real number and define the absolute value of x/2 as a.  Now  x doesn't fit in the equality.  Therefore such numbers cannot exist by the definition you give. -- Death to all Fanatics![ Parent ]
 New numbers (5.00 / 1) (#386) by cep on Sat Jun 07, 2003 at 02:10:38 PM EST

 The main trick of nonstandard analysis is to define new numbers beside the reals: Your argument only proves after all that there cannot be an infinitesmal number that is real. [ Parent ]
 Very well done! (4.00 / 2) (#69) by Kasifox on Tue Jun 03, 2003 at 05:44:58 PM EST

 Exceptionally well done piece of writing! Had more work like yours existed when I was attending university I might have enjoyed math far more than I did. As it is I'm recommending many of my associates visit K5 to read through your work if only to broaden their horizons. Please continue to post more...I hope that the members here will welcome your work but if they don't I would be very proud to offer you space on my servers from which to host your articles. Thank you for taking a complex concept and making it that more understandable to non-mathematicians.
 Thanks (none / 0) (#108) by The Writer on Tue Jun 03, 2003 at 07:23:57 PM EST

 I seem to get flamed a lot here for continually posting math stuff. Good to know that at least somebody appreciates the work I put into it. :-) [ Parent ]
 Keep posting!!!! (none / 0) (#127) by Julian352 on Tue Jun 03, 2003 at 09:16:43 PM EST

 Your math writings are VERY good. I'm not a math major, rather a CS major so had rudamentary knowledge of infinite sets. But what my professor presented as very short and uninsteresting "proof", you show as very interesting discussion. I showed this (and your previous story) to a couple friends and they both enjoyed them very much. One of them wasn't even in math/science/etc. and he loved the article, as it made him remember what was interesting about math. (Started a debate on .999 = 1 afterword). So please continue writing, so we all might learn the secrets that Math majors hide behind the scary-looking equations ;) [ Parent ]
 Argh! (3.50 / 2) (#70) by tjost on Tue Jun 03, 2003 at 05:45:17 PM EST

 More! More! More!
 Wow! (4.50 / 4) (#73) by Gornauth on Tue Jun 03, 2003 at 05:47:35 PM EST

 I'm not sure if this is a story, a rant, a lecture or a piece of humour, but i loved it! Would you please post more on math like this?
 blah. (4.50 / 2) (#75) by pb on Tue Jun 03, 2003 at 05:49:34 PM EST

 I'm sorry I missed my chance to vote -1 or +1 section on this. Anyhow, I'd just like someone to debunk this for me: Let's say I want to enumerate all of the real numbers between 0 and 1. so I start with a mapping like this: -> 0.1 -> 0.2 -> 0.3 -> 0.4 -> 0.5 -> 0.6 -> 0.7 -> 0.8 -> 0.9 ...then, 10-19 map to 0.01 to 0.09, 20-29 map to 0.11 to 0.19, 30-39 map to 0.21 to 0.29, etc., etc. Now, you say, what about sqrt(2)/2? Well, I say, I could map an integer to 0.7, or 0.707, or .7071067811865475244, or really as far in the expansion of sqrt(2)/2 as you care to go. That's not good enough, you say, sqrt(2)/2 is irrational, therefore, it has an infinite number of digits. So what's the integer that maps to it? Well, I say, I'll find that integer just as soon as you finish writing the decimal expansion of sqrt(2)/2. Yeah, so I live in the real world. So sue me. :) --- "See what the drooling, ravening, flesh-eating hordes^W^W^W^WKuro5hin.org readers have to say." -- pwhysall
 He never said the set of irrationals had ... (5.00 / 1) (#83) by drivers on Tue Jun 03, 2003 at 06:21:07 PM EST

 ... cardinality of aleph zero. How's that, "reality boy"? :) [ Parent ]
 This ties in with the multiple infinites thing (5.00 / 1) (#84) by strech on Tue Jun 03, 2003 at 06:21:51 PM EST

 If I remember correctly, there is an informal proof by contradiction; this may not be entirely mathematically correct, but it should be on the right track: Let's assume I have a list that maps all natural numbers to real numbers. However, I can always create a real number that is not on the list; the nth digit in the new number is the same as the nth digit in the nth number on the list plus one, wrapping 9 around to 0. Since this new number differs from every single real number on the list by at least one digit, we now have a number that is unmapped, and thus the assumption that we have a fully mapped set of real numbers is incorrect. [ Parent ]
 yes, yes it does. (5.00 / 1) (#166) by pb on Wed Jun 04, 2003 at 02:15:05 AM EST

 So what is that number? Have you finished going through that list yet? No? Well, call me when you do! What's that, what about my mapping? I'm still trying to map an integer to it; don't worry, I'm getting closer too! --- "See what the drooling, ravening, flesh-eating hordes^W^W^W^WKuro5hin.org readers have to say." -- pwhysall[ Parent ]
 Constructing a string of infinite decimals (5.00 / 1) (#229) by The Writer on Wed Jun 04, 2003 at 01:40:40 PM EST

 Would you accept it better if, in Cantor's proof, instead of saying you "construct a number x", you say, "consider the number x, which has the property that its nth digit is 3 if the nth digit of the nth number is not 3, and 0 otherwise"? That is to say, you aren't constructing the number per se, but you are simply stating that, given a list of all real numbers, you can think of another real number that isn't on the list, and you are just describing this real number you're thinking of. [ Parent ]
 I have trouble accepting the whole process... (none / 0) (#247) by pb on Wed Jun 04, 2003 at 04:11:59 PM EST

 One of the interesting properties of this construction is that it constructs an n-digit number after looking at n numbers in the list; if you wanted to verify whether or not that number is in the rest of the list, you might have to look through at least the next 10^n numbers. Or, to put it another way, let's say I have a list of all the binary numbers of length 1: Using diagonalization, I can construct another binary number that isn't in this list, namely '11'. So now my list is: And again, I can construct another number that isn't on the list -- '111'. Is it possible to have a list of all the binary numbers? If I did have such a list, couldn't I just use the diagonalization argument to show that my list is incomplete? ...and then, would the natural numbers have the same carinality of the reals? I think there's something fishy about the entire premise of the diagonalization argument.  :) --- "See what the drooling, ravening, flesh-eating hordes^W^W^W^WKuro5hin.org readers have to say." -- pwhysall[ Parent ]
 The essence of Cantor's proof (none / 0) (#260) by The Writer on Wed Jun 04, 2003 at 06:30:34 PM EST

 OK, your analogy with binary numbers doesn't work, because binary numbers are of finite length. I'll elaborate on this, because it will help to get into the crux of Cantor's proof rather than just the surface details. Cantor's proof works only because we are dealing with digit strings of infinite length (countably infinite throughout this post, for the pedants). In fact, it works because the digit strings are of (countably) infinite length, and the entire list is also only of countably infinite length. Cantor's choice of using the diagonal is just for convenience; diagonals are not necessary to make the proof work. I can easily say, mismatch the 2n'th digit on the nth number instead, or the n2'th digit on the nth number, and the proof would still work equally well. The point is, every element of the list contains an infinite number of digits; but because of that, there are an infinite number of digit positions where a mismatch can occur. That is to say, infinite digit strings are inherently able to mismatch every single item on any (countable) list you can come up with. Or to put it another way, there are too many possible places an infinite string of digits can vary, than can be accomodated by a countable list. Digressing to binary numbers momentarily, we see that this argument doesn't work for (non-fractional) binary numbers, simply because every binary number necessarily has a finite number of digits. There are never enough digits for you to mismatch everything on a countably infinite list. Returning to Cantor's proof, maybe it would help to realize that the number that Cantor "constructed" or "visualized" or what-have-you, is only one of many, many, many different possibilities. It's hardly the "one missing number" that the list is short of, so it fails to attain to uncountable infinity. Let's see if we can get an idea of just how many other possibilities there are. Some people may point out that for every n'th digit position, you have 9 possibilities that would mismatch the n'th number; but this is just a consequence of the base 10 notation. If you expand the real numbers in binary notation, there is only one possibility at each digit position. Does that mean there's only one real number missing from the list? Hardly. There is no reason to stick to diagonals. Let's say, for example, that we mismatch the n'th number not on the n'th digit, but on the 2n'th digit. Follow through Cantor's proof, substituting "n'th digit" with "2n'th digit". The proof still works. Notice that a real number that mismatches on every 2n'th digit is different from every real number that mismatches on the diagonal. What about if we mismatched on the n2'th digit? The proof still works. But all the numbers that mismatch on the n2'th digit are different from all the other mismatching numbers before! All of these numbers are not on the list. How such numbers are there? Nothing says I can't define a 1-to-1 function f(n) on the natural numbers, and pick a real number that mismatches the n'th number on the list at the f(n)'th digit. Such a number definitely exists, since it would be an infinite string of digits, which is a proper real number by definition; but it would not be on the list, since it mismatches the n'th number at the f(n)'th digit. Cantor's diagonal, then, would simply correspond to the function f(n)=n, and my alternative mismatching positions above would be f(n)=2n and f(n)=n2. And nothing says f(n) is restricted to being a monotonic function, like we've seen so far. f(n) can be any function, as long as it's 1-to-1, and Cantor's proof still works. From this, we can conclude that there are at least as many mismatching real numbers as there are 1-to-1 functions on the natural numbers, given any list of real numbers. It is a proven fact that the set of all possible functions on a set is strictly larger than the set itself. This is also true of all 1-to-1 functions on a set. In other words, given any list of real numbers, there are more numbers missing from the list than there are numbers on the list! Furthermore, it can be shown that the cardinality of the set of all possible functions on the natural numbers is equal to the cardinality of the set of reals. What this means is that the number of reals missing from any given list of real numbers is effectively the entire set of reals, minus a meager few. There's no way this list actually contains all the real numbers. It doesn't even come close. The crux of the argument is that if your digit strings are (countably) infinitely long, then there is no way a countably infinite list can hope to contain all the possibilities. [ Parent ]
 thanks, (none / 0) (#268) by pb on Wed Jun 04, 2003 at 07:42:55 PM EST

 I'm glad I made someone else actually think about this a bit; I'm a bit sorry it ended up being you, but maybe it'll help you for the next article, eh? :) You'll notice with my binary example that every time I went through the list, I was able to add another item that wasn't on the list yet, which makes what I was doing sort of like counting. That is to say, as long as I can run out of numbers, I can also find a number that wasn't in the list (that I went through so far...); the problem (in both the integers and the reals) is that you never really do run out of numbers. When you look at things this way, of course it's easy to talk about a set of numbers that contains both n=1 and n=n+1, but it's difficult to go through the entire set to see if you missed a number, especially when all you're doing is finding numbers that exist further on in the set... But yes, all I was really doing was questioning the entire premise of the real numbers, and what gives Cantor the right to do his hand-waving in the first place. Once you acknowledge the existence of single numbers that are (countably) infinite all by themselves, then of course you open the door to all sorts of silliness that doesn't play well with our notions of reality. :) Also, I appreciate Cantor's Diagonalization because it's much simpler than some of the earlier proofs that the reals are uncountably infinite. Although I suppose a formal restatement of it would be a bit more complex, but hopefully a bit less nonsensical as well. I looked at the first proof of the uncountably infinite nature of the reals a long time ago, and it was totally incomprehensible with its ratios of irrational numbers and whatever other sorts of similar black magic mathematical rituals. --- "See what the drooling, ravening, flesh-eating hordes^W^W^W^WKuro5hin.org readers have to say." -- pwhysall[ Parent ]
 Don't feel too bad (none / 0) (#271) by The Writer on Wed Jun 04, 2003 at 08:06:06 PM EST

 Actually, I was working on a draft of the next article, and Cantor's proof is a main part of it (at least currently; this is likely to change). I was in the middle of showing how, at every digit position, you have 9 possible choices, which thus leads to an exponentially larger number of "missing reals" as you go down the digits, when the thought struck me: hold on a second here, this doesn't work in binary!! What's wrong? I was grappling with it in my head this morning, when suddenly I realized why in decimal there were 9 choices, whereas in binary there is only one. The reason was simple: you need several digits of binary to encode one digit of decimal (OK, not precisely, but that's the idea), so when you mismatch a diagonal digit in decimal, it is actually mismatching a "spaced out" diagonal in binary. I was in the middle of this realization, when your post came along, and put the icing on the cake. :-) I was in the middle of writing a response, when suddenly it dawned on me that when it comes down to it, there's really nothing special about diagonals. Base 10 is arbitary, after all; if indeed Cantor was right, then it can't just be restricted to base 10. Furthermore, there is no reason why it has to mismatch in a diagonal pattern; it is equally good to mismatch in any arbitrary pattern, as long as the positions are all distinct. So this naturally led to injections over the set of natural numbers (injection = 1-to-1 function whose image doesn't have to be equal to the entire set). And then, it dawned on me that it is precisely because of this that any countable list of reals is always missing almost all the reals: the set of all possible injections over the natural numbers necessarily has a larger cardinality than the set of natural numbers itself, and it has been proven that this cardinality is precisely the cardinality of the real numbers. So I should rather be thanking you for helping me reach this inspiration. :-) [ Parent ]
 here's what I would have said... (5.00 / 1) (#295) by pb on Thu Jun 05, 2003 at 01:29:18 AM EST

 Re: Cantor's proof--essentially he was doing the same thing I was doing, in trying to map the integers to the reals. He numbered the reals by the digits he crossed off, which gives them a 1-to-1 mapping with the integers. However, he simultaneously constructs a real number that doesn't equal any of the reals in the list. Therefore, although he does map all the reals in his list to the integers, he also proves that that list can't contain all the reals, because there is at least one real number that isn't in the list, and simultaneously isn't already mapped to an integer. That's how I'd explain it. And actually, what I was doing was slightly different--I mapped the set of integers (or really, natural numbers) to the set of fixed-point numbers (terminating decimal numbers?) between 0 and 1. That's about the closest thing we have to real numbers on computers (or LBA Turing Machines), so it works for me. :) As for the rest, I think you're right; the bases shouldn't matter, and there shouldn't be anything special about the diagonal either, except that it's neat, and easy to see. And I haven't studied the rest enough to know things like whether the set of integers can map to a single real, or whether the different permutations of the set of integers could map to all the reals, etc., etc.--but I'm sure that's lots of fun too... --- "See what the drooling, ravening, flesh-eating hordes^W^W^W^WKuro5hin.org readers have to say." -- pwhysall[ Parent ]
 What exactly do you want debunked? (5.00 / 2) (#85) by gzt on Tue Jun 03, 2003 at 06:22:36 PM EST

 I can't see what you're asking. [ Parent ]
 I missed the part where he mapped (none / 0) (#87) by porkchop_d_clown on Tue Jun 03, 2003 at 06:30:13 PM EST

 real numbers to integers. He mapped *rational numbers* to integers, which expressly leaves out PI, e, and so on... -- I only read Usenet for the articles.[ Parent ]
 Uh... (none / 0) (#106) by The Writer on Tue Jun 03, 2003 at 07:18:13 PM EST

 π and e are real numbers, but they are not rational numbers. You cannot express them as a fraction of two integers. They are irrational numbers. In fact, not only irrational, but an evil kind of irrational known as "transcendental". [ Parent ]
 (oops) ... rest of reply (none / 0) (#107) by The Writer on Tue Jun 03, 2003 at 07:21:50 PM EST

 I never claimed to map the real numbers to the integers. This is impossible, as I will show in the next article in this series. What I did was just to map the rationals to the integers. The rationals do not include the irrationals like π or e. [ Parent ]
 that was my point; (none / 0) (#190) by porkchop_d_clown on Wed Jun 04, 2003 at 07:46:14 AM EST

 You mapped rationals to integers, not reals. pb apparently doesn't understand the difference. -- I only read Usenet for the articles.[ Parent ]
 Transcendentals are not evil! (none / 0) (#259) by phliar on Wed Jun 04, 2003 at 06:27:16 PM EST

 Take that back! Without the transcendentals we'd be stuck in an ℵ0 world! Faster, faster, until the thrill of...[ Parent ]
 Transcendentals (none / 0) (#275) by The Writer on Wed Jun 04, 2003 at 08:37:32 PM EST

 Well, we're already stuck in a world much smaller than ℵ0, or at least only as large. :-) Anyway, the reason I said that transcendentals are an "evil" kind of irrationals is because while, uh, "well-behaved" irrationals like √2 can be turned into rationals via a finite sequence of algebraic combinations with other rationals, the transcendentals cannot. Being inherently irrational sounds rather evil to me. :-) [ Parent ]
 Evilness (5.00 / 2) (#287) by phliar on Thu Jun 05, 2003 at 12:15:35 AM EST

 I think numbers like √2 are so ordinary. But π and e have that mystique... you know there's some really cool stuff there. Those are the cool kids and √2 is the wallflower. Faster, faster, until the thrill of...[ Parent ]
 e is for Euler... (5.00 / 1) (#298) by pb on Thu Jun 05, 2003 at 02:10:29 AM EST

 The thing about e is that it has such reasonable continued fraction representations, unlike its idiot savant older brother, Pi, who gets all the attention... --- "See what the drooling, ravening, flesh-eating hordes^W^W^W^WKuro5hin.org readers have to say." -- pwhysall[ Parent ]
 e vs. pi (none / 0) (#316) by The Writer on Thu Jun 05, 2003 at 10:57:10 AM EST

 When one first encounters e, one gets the impression it's some faraway, foreign object, (mostly because π occurs more often in "real life", at least in the sense that circles are more readily noticed than exponential attenuation or growth). However, once you look at e up close, it really isn't quite that foreign. I mean, what can be simpler than the limit of the series ∑n:0→∞(1/n!) ? That's just 1/1 + 1/(1*2) + 1/(1*2*3) + ... Sweet and simple; nothing scary at all. But that beast π is just so complicated. I've yet to remember any infinite series for π which doesn't involve multiplying the result by some constant or taking square roots or what-not. Not to mention that the simpler series for π converges so darn slowly they're practically of little use. Why can't I have my π and eat it too? :-P [ Parent ]
 the real world of numbers (5.00 / 1) (#88) by mpalczew on Tue Jun 03, 2003 at 06:30:31 PM EST

 >Yeah, so I live in the real world. So sue me. :) oh the irony -- Death to all Fanatics![ Parent ]
 The real numbers cannot be enumerated in this way (5.00 / 2) (#93) by readams on Tue Jun 03, 2003 at 06:45:28 PM EST

 The very cool proof is called Cantor's Diagonalization or Cantor's Diagonal Argument. I won't steal the thunder of the story submitter by describing it here, but it was a stunning result in mathematics that proved that there are more real numbers than there are integers. Sets with the same cardinality as the natural numbers are called "countable" sets. The real numbers are "uncountable". It turns out that you can apply Cantor's diagonalization repeatedly to show that there must be a infinite progression of ever-larger cardinalities. One question though: is there any set with cardinality between the natural numbers and the real numbers? The surprising answer: yes and no. This result, called the Continuum Hypothesis is an undecidable question. That is, the answer being yes and the answer being no are both consistent with the axioms of number theory. In other words, mathematics cannot answer this question! By the way, the existance of undecidable questions; statements in a formal system which cannot be proven, was shown by Gödel. The Gödel Incompleteness Theorem states that any sufficiently powerful formal system must be either inconsistent or incomplete. [ Parent ]
 yes, but... (none / 0) (#163) by pb on Wed Jun 04, 2003 at 02:06:32 AM EST

 Obviously, my 'proof' works best for numbers with a finite amount of digits (which isn't Z, and really is weaker than Q) but I think it uses one of the same tricks that Cantor's Diagonalization proof does as well, which is something that always bothered me about his proof. To wit: remember that number he constructs from the diagonal in the beginning, to prove that it isn't in the list? Well, isn't that an operation that would require going through every number in the list? Well, my proof requires this too. Therefore, I'll have my integer all ready to map to your irrational number right after Cantor manages to find one real number that isn't in his list of all reals.....  :) And yes, I do know about many of the quirky little theorems in mathematics, which is why I wouldn't have wanted this article to go FP. But if this is new to so many other people, then I guess it's a valuable contribution. I didn't know about that Continuum Hypothesis, though, and it sounds nifty, and perhaps somewhat similar to the different classes of computability (including any extra undiscovered ones that might exist...). --- "See what the drooling, ravening, flesh-eating hordes^W^W^W^WKuro5hin.org readers have to say." -- pwhysall[ Parent ]
 Problem here... (none / 0) (#252) by warrax on Wed Jun 04, 2003 at 05:08:23 PM EST

 To wit: remember that number he constructs from the diagonal in the beginning, to prove that it isn't in the list? Well, isn't that an operation that would require going through every number in the list? No. He simply defines a new number which is not in the list. He does not actually construct the number; he simply shows how to construct it for any instance of a (countably infinite) list such as what you proposed. The mere existence of the number invalidates your list, and, indeed, any such countably infinite list. Whether the number can actually be constructed is irrelevant. -- "Guns don't kill people. I kill people."[ Parent ]
 sounds like hand-waving, doesn't it? (none / 0) (#269) by pb on Wed Jun 04, 2003 at 07:48:11 PM EST

 So he tells you how to construct a number that you might not be able to construct in the first place, and that is supposed to invalidate my list? Please. habeas numerus --- "See what the drooling, ravening, flesh-eating hordes^W^W^W^WKuro5hin.org readers have to say." -- pwhysall[ Parent ]
 Now you're reaching (none / 0) (#274) by p3d0 on Wed Jun 04, 2003 at 08:32:38 PM EST

 Do you expect people to construct every mathematical item they define? Then where do you get off talking about the set of real numbers in the first place? I think the time for Devil's Advocate has ended, and you need to admit you lost this argument. :-) -- Patrick Doyle My comments do not reflect the opinions of my employer.[ Parent ]
 indeed :) (5.00 / 2) (#282) by pb on Wed Jun 04, 2003 at 11:18:39 PM EST

 This argument was lost before it began, which is why I started it by asking people to debunk my claims. I think The Writer made the best attempt at it, really. It's sad how so many people just take things for granted without thinking about them first, especially when the things they are taking for granted aren't necessarily obvious at all, or even very plausible. --- "See what the drooling, ravening, flesh-eating hordes^W^W^W^WKuro5hin.org readers have to say." -- pwhysall[ Parent ]
 A nitpick.... (5.00 / 1) (#175) by Mudlock on Wed Jun 04, 2003 at 04:53:34 AM EST

 But there's a third leg to the Incompleteness theorem: inconsistent, incomplete, or infinite (as in the number of axioms must be infinite). No, it's not a very important or usefull change, but it makes it fit in much better with the "A, B, C: pick any two" form of engineering laws. And if you replace "finite" with "concise", then they even all start with 'c', which makes it even better! Concise, Complete, Consistent: pick any two. -- But everybody wants a rock to wind a piece of string around.[ Parent ]
 Mathematics "cannot" (none / 0) (#254) by phliar on Wed Jun 04, 2003 at 05:33:10 PM EST

 In other words, mathematics cannot answer this question! This is not a good characterisation; it's like saying "Geometry cannot prove the parallel postulate!" The continuum hypothesis or its negation can simply be treated as an axiom, either one will give you a consistent set theory. Just like either the parallel postulate or its (two) negations — choose any of them and you get a consistent and interesting geometry (plane, spherical, hyperbolic). Faster, faster, until the thrill of...[ Parent ]
 Congratulations! (4.50 / 2) (#99) by epepke on Tue Jun 03, 2003 at 06:56:56 PM EST

 You've just rediscovered ordinal counting systems and have described why they're even less powerful than cardinal ones. Your system cannot count more than a finite set of rational numbers and a single irrational number. The truth may be out there, but lies are inside your head.--Terry Pratchett[ Parent ]
 The problem (5.00 / 1) (#103) by Captain Trips on Tue Jun 03, 2003 at 07:09:15 PM EST

 The problem with your mapping is that sqrt(2)/2 has an infinite number of digits, which is OK for real numbers, but not for integers. An integer with infinite number of digits can't be finite (or we should be able to count to it in finite time), and a non-finite integer is an contradiction. So sqrt(2)/2 does not, in fact, map to an integer. -- The fact that cigarette advertising works, makes me feel like maybe, just maybe, Santa Claus is real.—Sloppy[ Parent ]
 Sure it does (5.00 / 2) (#157) by Pseudonym on Wed Jun 04, 2003 at 01:22:05 AM EST

 Here's a mapping which does the job. 1 maps to sqrt(2)/2 2 maps to 1/1 3 maps to 1/2 4 maps to 2/1 etc "But", I hear you reply, "I have a lot more numbers than just that one". OK, then, I'll just tack them on the front. "But I have a countably infinite number to add!", you protest. Fine, I'll use the even integers to encode the rationals and the odd numbers to encode your set. "But I have an uncountably infinite number!", you say. OK, now I'm stuck. See also my previous reply on some numbers which you can't even name. sub f{($f)=@_;print"$f(q{$f});";}f(q{sub f{($f)=@_;print"$f(q{$f});";}f});[ Parent ]
 I think he means ... (5.00 / 5) (#135) by Repton on Tue Jun 03, 2003 at 10:33:32 PM EST

 "There is no way you can specify an irrational number in a finite ammount of time — the best you can do is give a rational approximation. As such, for all practical purposes the irrational numbers are countable." To which I would reply: I can fully specify irrational numbers in a finite amount of space. Eg, I could say "the positive solution to x2 = 2". That's as good a description of the number as any other.. (well, maybe) What do "practical purposes" have to do with pure mathematics anyway? :-) -- Repton. They say that only an experienced wizard can do the tengu shuffle..[ Parent ]
 On Computable Numbers (5.00 / 4) (#155) by Pseudonym on Wed Jun 04, 2003 at 01:15:26 AM EST

 Yes, you can fully specify many irrational numbers in a finite amount of space. You can, for example, write a program which computes the number to any desired precision and encode the text of the program in a large integer. (A text file is just a large integer, let's face it.) That the same goes with English. English text, or mathematical notation encoded into LaTeX is just a big integer. The same goes for any finite message which uses a finite alphabet. It follows that the subset of irrational numbers which you can specify finitely is aleph-0, which is the same as the number of integers. Since there are way more real numbers than there are integers, it follows that the overwhelming majority of real numbers cannot be finitely specified. In particular, the set of numbers which cannot be described using a Turing Machine is called the uncomputable numbers. There are very few uncomputable numbers which we know about because they are very hard to specify finitely, but you can construct some of them from known uncomputable problems. An example is the "busy beaver" number, which is related to the number of Turing Machines of size n which halt on all inputs. Note that this number is uncomputable, but it is finitely specifiable in English prose with some mathematical notation. I had a wonderful specification of an unspecifiable number, but unfortunately the margin is too small to contain it. sub f{($f)=@_;print"$f(q{$f});";}f(q{sub f{($f)=@_;print"$f(q{$f});";}f});[ Parent ]
 Yeah, I always wondered about this (none / 0) (#223) by SpeedBump0619 on Wed Jun 04, 2003 at 12:15:53 PM EST

 I'm not a mathematician, but I have an interest in mathematics...I understand the concept of countably and uncountably infinite. But I have never been able to debunk this mapping of natural numbers to positive reals (which can, of course be trivially extended to all reals). when considering the natural numbers it should be fairly simple to prove that any multiplication involving two members of the set can only result in valid members of the set. So start with the subset {10} and define the set such that for each member of the set the set contains 10*(that member). It should be obvious that this leads to a set containing all the powers of 10. This set being a subset of the natural numbers, can, by definition, never be larger than the set of whole numbers. It should also be intuitively obvious that this set contains a countably infinite number of elements (mapping is 10^n <--> n) The entire point of creating this set is to show that the set of natural numbers contains numbers whose count of digits spans all of the natural numbers, ie the maximum number of digits in a natural number is countably infinite. If this is true then the argument presented by the parent post must be able to present a 1:1 mapping of natural numbers to (positive) real numbers: (numbers represent digits in the natural numbers) ...7531.2468... <--> 87654321 as the number of digits before or after the decimal real number --> inf. so does the length of the whole number --> inf. I don't really know proof theory, so presenting a formal proof is not really possible, but I see no logical errors...someone please show what in this chain of intuition is incorrect. -SpeedBump [ Parent ]
 I know what you mean... (none / 0) (#225) by pb on Wed Jun 04, 2003 at 12:51:20 PM EST

 I think the problem starts because any particular integer necessarily has a finite number of digits, whereas any particular real number doesn't necessarily have this restriction. Therefore, while the decimal representation for 1/9 (0.11111111111111111...) is a perfectly valid real number, an infinite number of 1's stuck together is not an integer. That having been said, integers can still be arbitrarily large. The contents of my hard drive could be expressed as an integer. So, too, the contents of the internet. For all practical purposes, I'd say that any discrete quantity in this Universe can be mapped to an integer, and that might include the entire Universe itself, (ask your local doctorate of physics...) at which point the real numbers just start looking like a bizarre abstract concept. In fact, I'd consider the real numbers to be quite similar to Newtonian physics--they're both concepts that can be used to model the real world, but that doesn't mean that they are entirely accurate, or that they can really exist as anything besides concepts. Or, as Leopold Kronecker once said, "God made the integers; all else is the work of Man". --- "See what the drooling, ravening, flesh-eating hordes^W^W^W^WKuro5hin.org readers have to say." -- pwhysall[ Parent ]
 hmm. countably infinite length of irrationals (none / 0) (#227) by SpeedBump0619 on Wed Jun 04, 2003 at 01:16:15 PM EST

 well, it seems I did leave something out of my statement before. Your explanation is I think on target, but I still am not certain I buy it because: the concept of an infinite series of digits assumes that each digit has an identifiable location in the series. I could therefore describe the count of the digits in an irrational number as countably infinite (mapping digit at 10^-n <--> n). As stated before, the maximum length of a natural number is also countably infinite. (mapping digit at 10^n <--> n) Thus an irrational number with it's countably infinite string of digits should be mapped by one element in the natural numbers, with it's countably infinite length. To disprove this one must prove either: 1) digit count of natural numbers is less than countably infinite...ie finite 2) digit count of irrational numbers is greater than countably infinite...ie uncountably infinite I see no other option. I look forward to further input. thanks. -SpeedBump [ Parent ]
 Be careful there (5.00 / 2) (#231) by The Writer on Wed Jun 04, 2003 at 01:56:07 PM EST

 While it is true that the maximum number of digits an integer can have is countably infinite, it is not true that an actual integer can have a countably infinite number of digits. An integer always has a finite number of digits. Although, when given any particular integer, you can find one that has more digits, you will never be able to find an integer that has countably infinite digits. The upper limit of the number of digits an integer may have is ℵ0, the countable infinity; but no integer actually attains to this limit. But if you look at an irrational number, it has ℵ0 digits already. Although the limit on the number of digits in an irrational number is still countably infinite, an irrational number actually attains to this limit, unlike the integers, which do not attain to the limit. That's why it's not possible to map the digits of an irrational number to an integer. The irrational number already has countably infinite digits, but the integer can only approach countable infinity, it can never actually reach it. [ Parent ]
 Very helpful. Thank you. (4.00 / 2) (#237) by SpeedBump0619 on Wed Jun 04, 2003 at 02:36:31 PM EST

 [ Parent ]
 These are rational. (4.00 / 1) (#244) by Boronx on Wed Jun 04, 2003 at 03:39:07 PM EST

 All the numbers in your list are rational, and your list will never contain an irrational number. Nor will it even contain all of the rational numbers between 1 and 0. Subspace[ Parent ]
 you are correct, sir. (none / 0) (#297) by pb on Thu Jun 05, 2003 at 02:00:10 AM EST

 It only contains non-terminating decimal numbers, which can all be expressed as a quotient with a base of 10n. And actually, I never claimed otherwise in my post. However, for all practical purposes, this set is just about as useful to me as the set of reals is. Man will never be able to fully enumerate even one non-terminating decimal number. Everyone's favorite example seems to be Pi; and although some non-terminating decimal numbers are easier to represent in some ways than Pi, like 1/9, or Phi, or sqrt(2), my statement holds. The reals are a convenient abstraction, but they aren't "real" because of their continuous nature. --- "See what the drooling, ravening, flesh-eating hordes^W^W^W^WKuro5hin.org readers have to say." -- pwhysall[ Parent ]
 Cantor's Diagonalisation (none / 0) (#245) by phliar on Wed Jun 04, 2003 at 03:40:20 PM EST

 Your question is exactly what Cantor was thinking of. You ask:Well, I say, I could map an integer to 0.7, or 0.707, or .7071067811865475244, or really as far in the expansion of sqrt(2)/2 as you care to go. That's not good enough, you say, sqrt(2)/2 is irrational, therefore, it has an infinite number of digits. So what's the integer that maps to it? Here you're speaking of infinity as a number, by saying things like "infinite number of digits." You also bring in rational and irrational numbers, which is an unnecessary complication. Here's Cantor's diagonalisation: You claim you have a list that maps integers to real numbers between 0 and 1. I'm going to construct a number between 0 and 1 by this method: I start by writing 0.. Now look at the first decimal place of the first number in the list; pick a digit other than this. [For definiteness, pick (9 - i).] That is the first decimal place of my number. Its second decimal place is 9 minus the second digit of your second number. For any j, the j-th digit of my number is 9 minus the j-th digit of your j-th number.Here's a picture:   1 → 0.1000...   2 → 0.20...   3 → 0.300... ...  10 → 0.0100000000... ... 238 → 3.14159265358...7... (π) (Note:I don't really know if the 238th digit of π is 7.) My number is 0.89999...2... It cannot be in your list; it is different from every number in your list.I didn't say infinity anywhere; just gave you an algorithm which is guaranteed to produce a counterexample. Faster, faster, until the thrill of...[ Parent ]
 where does his argument differ from mine... (none / 0) (#249) by pb on Wed Jun 04, 2003 at 04:23:43 PM EST

 ah, but 0.89999...2 (238 digits) is in my list; and actually the ordering of my list is much duller, since it takes me more than 10^n numbers to construct an n-digit number; all you're doing is trying to skip ahead. Well, it won't work; it'll take you just as long to go through my entire list as it will take me to construct it. Or, rather, if Cantor can construct a number by going through all the reals, then I can construct my list by going through all the reals. Otherwise (if we can't), then I'd say both proofs are vacuously true. --- "See what the drooling, ravening, flesh-eating hordes^W^W^W^WKuro5hin.org readers have to say." -- pwhysall[ Parent ]
 Sorry, I don't understand! (none / 0) (#255) by phliar on Wed Jun 04, 2003 at 05:53:15 PM EST

 Well, it won't work; it'll take you just as long to go through my entire list as it will take me to construct it. Don't think of it as a construction. The diagonalisation shows that no 1-1 onto map can exist. For every map from integers to reals, there is a real that is not on the right side.Consider the proof that sqrt(2) is irrational. It starts out by saying "assume you can find p and q such that p/q = sqrt(2)." You then show that you reach a contradiction, i.e. since any candidate solution leads to a bogosity, there can be no solution. Here, since every supposed 1-1 onto function from integers to reals leads to a contradiction, no such function can exist. I shouldn't have used the word algorithm earlier, since part of the definition is that an algorithm must terminate. When we're dealing with unbounded lists (as we are here) we shouldn't talk about a process. Obviously no proposition about an infinite set can be verified by a finite enumeration. Faster, faster, until the thrill of...[ Parent ]
 but that's the thing... (none / 0) (#267) by pb on Wed Jun 04, 2003 at 07:25:06 PM EST

 You're right, but what Cantor is doing is indeed a finite enumeration. That is, until he skips ahead and assumes it's possible to create such a number that has those properties. I'd argue that that step isn't possible; or rather, it's just as possible as if I said "ok, I'm done counting the integers now..." However, if it was possible, what's to stop me from doing the same thing to the set of integers? Or, for that matter, just assuming that I can create a number that doesn't fit into it? --- "See what the drooling, ravening, flesh-eating hordes^W^W^W^WKuro5hin.org readers have to say." -- pwhysall[ Parent ]
 I'm fogged (none / 0) (#291) by phliar on Thu Jun 05, 2003 at 12:29:14 AM EST

 However, if it was possible, what's to stop me from doing the same thing to the set of integers? Sorry, I'm lost. If you did the same thing (I assume diagonalise) to the integers, there'd be no contradiction. Any integer has a finite number of digits. You'd "run out"... Please re-phrase your question, I'm interested. Faster, faster, until the thrill of...[ Parent ]
 sort of hard to test this, isn't it? (none / 0) (#296) by pb on Thu Jun 05, 2003 at 01:52:08 AM EST

 Well, I've already admitted defeat, (see my most recent replies to The Writer) but basically if you try to diagonalize the integers, you're doing something similar to counting. For any finite set of integers, you can find an integer that isn't in the list. And for the set of all integers... well, you won't run out of integers, of course.  :) As for running out of digits, you can solve that by assuming that every integer has an infinite number of leading zeroes. After all, if you have a list of 0 through 9 and you use diagonalization to show that the number 11 (or 1111111111 for that matter) isn't on the list, that's a perfectly valid result. And at any point in the list, you have a valid integer that hasn't been found yet. However, you never do finish, and you can't just pull a Cantor and claim that your end result (if you could ever get one) is a valid integer--because if you have a countably infinite number of digits, it isn't! I'm sorry if I muddied the waters a bit here; I wanted to see if anyone could give me a clear explanation of my scenario, (which is quite similar to what Cantor was doing insofar as it attempts to map the integers to the reals)--and I wanted to make people have to think about their answers a bit.  :) --- "See what the drooling, ravening, flesh-eating hordes^W^W^W^WKuro5hin.org readers have to say." -- pwhysall[ Parent ]
 Running out (none / 0) (#377) by phliar on Fri Jun 06, 2003 at 07:03:19 PM EST

 Not "run out" of integers, but run out of digits on the left. The list is unbounded; but the candidate "not in the list" integer must be bounded on the left (since any integer has a finite size). Formally, if the mapping is f: Z → Z and the diagonalised integer is j, and k[i] is the i-th digit of k, then we know ∀ i ∈ Z, j[i] = 9 - f(i)[i] — but of course if j is an integer, all j[l] must be 0 for l > N for some positive N. Which is (I think) the same as what you're saying, but your examples leave me befuddled. No matter! Faster, faster, until the thrill of...[ Parent ]
 One more thing (none / 0) (#257) by phliar on Wed Jun 04, 2003 at 05:59:09 PM EST

 ah, but 0.89999...2 (238 digits) is in my list; Note that my number does not have 238 digits after the decimal point. I put an ellipsis after the 238th digit. If you say "hey! You just didn't look far enough, your number is actually in my list at position 1,763,924." I reply "No it isn't; your 1,763,924th real differs from my number in at least the 1,763,924th digit." Faster, faster, until the thrill of...[ Parent ]
 diagonalization (none / 0) (#327) by jejones3141 on Thu Jun 05, 2003 at 12:21:46 PM EST

 Your list will have a lot of approximations to the number that isn't on the list, but it still won't have the number itself, because any number you pick from the list has a position N on the list—that's what it means to have the same cardinality as the natural numbers—and the number generated by diagonalization differs from the Nth number in the Nth digit. [ Parent ]
 Unlimited versus infinite (none / 0) (#272) by p3d0 on Wed Jun 04, 2003 at 08:10:53 PM EST

 It doesn't matter how long you wait for me to "finish" writing the decimal expansion of sqrt(2)/2; the fact is that sqrt(2)/2 has an infinite number of digits, and integers do not. True, there is no limit on the number of digits in an integer. Given an integer, you can always find one with more digits. But no matter how many digits you have, that number is finite, or else it does not represent an integer. The number of digits is finite and unbounded. -- Patrick Doyle My comments do not reflect the opinions of my employer.[ Parent ]
 your "swimsuit" unicode figure (3.33 / 3) (#80) by Hide The Hamster on Tue Jun 03, 2003 at 06:07:03 PM EST

 No, those are lopsided tits. Definitely lopsided jugs. Free spirits are a liability.August 8, 2004: "it certainly is" and I had engaged in a homosexual tryst.
 Kewl! A mathematical rant! (4.85 / 7) (#86) by porkchop_d_clown on Tue Jun 03, 2003 at 06:26:22 PM EST

 How often do you see those anymore?Mathematical rants containing representations of breasts are even more rare! -- I only read Usenet for the articles.
 Not misconceptions (4.66 / 3) (#89) by the on Tue Jun 03, 2003 at 06:33:05 PM EST

 Misconception #4: There are twice as many integers as there are natural numbers There are twice as many integers as naturals. There are aleph0 integers and aleph0 naturals. aleph0=2*aleph0. So the statement is 100% accurate and true! "Oh, you just repeat this forever and you get infinity." Except mathematicians, even set theorists whose field this is, will say things exactly like this. While not strictly correct this does give a good intuition and shouldn't simply be discarded. Anyway...great article. You should post in sci.math where it would make a good antidote to the endless posts about how 1/0=infinity. (But again, I have to add, 1/0=infinity is a good intuition to keep in mind in some contexts.) -- The Definite Article
 Shh! (5.00 / 3) (#104) by The Writer on Tue Jun 03, 2003 at 07:13:29 PM EST

 You aren't supposed to divulge such secrets to the lay people, sheesh! ;-) Jokes aside, misconception #4 is perhaps more accurately understood as a misconception about the arithmetic behaviour of infinite cardinals. Most people, when told something is twice as many, immediately conclude that it can't be equal. That was what I was trying to address: even though it is twice as many in one sense, it is also simultaneously equally many in another sense. And about the whole "repeat this forever" concept that I knocked down, I'm not devaluing the intuitive understanding of infinity; just that there are a lot of caveat emptors one must be aware of when dealing with infinities "intuitively". One example where "repeated infinitely many times" breaks down horribly is, as I pointed out in another post, in sets of order type 2ω or beyond. These sets have multiple "infinitely long stretches"; so when you iterate an "infinite" number of times, it is unclear how many "stretches" you traverse. Some of these order types have an infinite number of such "stretches"; so the ambiguity can get very bad. I submit that it is much, much, safer to stay within the well-defined boundaries of limits (in analysis), and cardinals and ordinals (in set theory), rather than the dangerously vague, blanket term of "infinitely many". [ Parent ]
 Now this is the kind of article worth reading. (4.00 / 2) (#90) by Deus Horribilus on Tue Jun 03, 2003 at 06:33:16 PM EST

 Nice work. I hope that you cover the astrophysical definition(s) of infinity in the next article; namely the infinite universe theorem, quantum singularities and Hubble's law. After all, it is all good for mathematicians to wax lyrical about infinities, but it can all go to hell when they are applied to real world phenomena. Incidentally, I do think infinity's reciprocal is zero, but more for the reason that zero is just as intangible as infinity (you can say you have no money, just in the way you can say you have infinite resources, but you can hold a dollar in your hand.). My thoughts are that they can be reciprocal by the limit proof - in fact, the laws of energy almost beg for it to be true (Anybody familiar with the total E = 0 concept?). _________________________________________"Beliefs are never concrete, they change direction like autumn leaves in a windstorm..."
 unless he has... (none / 0) (#184) by the sixth replicant on Wed Jun 04, 2003 at 06:38:01 AM EST

 ...a degree in cosmology as well as pure mathematics maybe, but the two things are *totally* unrelated. He did spend quite a bit of time trying to show us a mathematical definition of infinity and how unintuitive it can get when we try and use the "physical" definition of infinity as a comparision (which we usually mean to be "very far away" or "the light soming from a star is parallel" etc etc). Summary of article : Infinity is really cool and we play around with it all the time without really understanding it but, actually, mathematicians have a really good hold on it - and this is how they do it..... Ciao [ Parent ]
 1-to-1 mapping (4.75 / 8) (#92) by gzt on Tue Jun 03, 2003 at 06:34:32 PM EST

 Yeah, so, what if some wiseacre asks, "Yeah? There's a bijection from N to Q? What's the 475,475,367,587th rational, then?" Here's a method of enumerating the rational numbers using prime factorization! First, factorize p and q. For p you'll have something like a1ra2s...  where ai is prime and something similar for q, right [let's use b for q]? That's probably not very clear, if I had a chalkboard it'd be easier to write this out. Anyways, p/q will be the ai2r*bi2s-1 rational number. Erm, that's not notationally clear at all, so I'll explain by looking at 15/4, which is 3*5/22. It is the 32*52*23-th rational. What's the 42nd rational? 42 = 2*3*7, so it's 1/42. What's the 45th rational? 45 = 5*32, so the 45th rational is 3/5. doot doot. all you need to do is know how to factor.
 ehh? Explain? (4.00 / 1) (#125) by Julian352 on Tue Jun 03, 2003 at 09:06:52 PM EST

 I'm not sure I undestand your method exactly. Let's look at your example: Erm, that's not notationally clear at all, so I'll explain by looking at 15/4, which is 3*5/22. It is the 32*52*23-th rational. Wouldn't that mean that the 30/2 fraction would also have the same number. Since the factorization of 30/2 leads to: 2*3*5/2 => 22 * 32 * 52 * 2 =>(simplifying) It is the 32*52*23-th. Or am I not understaning it right? [ Parent ]
 I think (5.00 / 1) (#132) by tang gnat on Tue Jun 03, 2003 at 09:45:36 PM EST

 You need to reduce the 30/2 to an equivalent fraction (it's still the same rational number, in a sense). Then you get 15/1, which yields the natural number 215. [ Parent ]
 Definition of the rationals (none / 0) (#140) by gzt on Tue Jun 03, 2003 at 11:10:05 PM EST

 {p/q; p, q in N, p|q=1} There is a unique factorization for every rational. 30/2 isn't a rational number, it is equivalent to one. I'm sorry my notation isn't clear. [ Parent ]
 cool (5.00 / 1) (#131) by tang gnat on Tue Jun 03, 2003 at 09:39:50 PM EST

 Of course, you need to make sure p and q are reduced to "lowest terms" so their factors don't interfere. I presume you would also make it so 0<->0, and the +/- sign carries over? Then it's integers to rationals, of course. [The notation sure was a pain to read! Too bad K5 doesn't support LaTeX.] [ Parent ]
 Sure, whatever (5.00 / 1) (#142) by gzt on Tue Jun 03, 2003 at 11:13:14 PM EST

 This is a mapping of the rationals, which are, by definition, reduced to lowest terms. Yeah, you can extend it like that, I guess I shouldn't've just done N->Q+. [ Parent ]
 Injection, bijection, surjection (4.66 / 3) (#160) by Pseudonym on Wed Jun 04, 2003 at 01:51:15 AM EST

 One thing which should be noted, is that you don't actually need a 1-to-1 correspondence to prove that two sets have the same size. A 1-to-1 correspondence (also known as a bijection) is a total function f : D -> C with two properties: If f(x1) = f(x2), then x1 = x2. In other words, no two elements map onto the same value. We call f an injection if this property holds. For all y in C, there is some x in D such that f(x) = y. In other words, all elements of C are mapped onto by some element of x. We call f a surjection if this property holds. The cardinality of X and Y are the same, by definition, if you can find a bijection from X -> Y (or from Y -> X, which could just be the inverse of f, which is guaranteed to exist). However, there is a theorem (I forget its name) which states that it's also sufficient to find a surjection which maps X to Y and another which maps Y to X. So you actually don't need a unique list to prove that two sets have the same size. All you need to prove is that, for example, there's a mapping f from positive integers to rationals such that every rational is mapped onto from some positive integer, and that there's a mapping g from rationals to positive integers such that every positive integer is mapped onto from some rational. Some suitable functions are: f(i) = r/s, if i = 2^r * 3^s for some positive integers r, s or 0 otherwise g(r) = r if r is a positive integer, 0 otherwise QED sub f{($f)=@_;print"$f(q{$f});";}f(q{sub f{($f)=@_;print"$f(q{$f});";}f});[ Parent ]
 i think this shows (none / 0) (#183) by the sixth replicant on Wed Jun 04, 2003 at 06:31:06 AM EST

 that the cardinality of A is bigger than or equal to the cardinality of B (for f:A -> B, f surjective/onto). And sometimes this is enough (ie make the proof easier!), for instance showing that the cardinality of A is infinite with B=natural numbers. But i could be wrong. Ciao [ Parent ]
 No (none / 0) (#217) by spakka on Wed Jun 04, 2003 at 11:19:10 AM EST

 However, there is a theorem (I forget its name) which states that it's also sufficient to find a surjection which maps X to Y and another which maps Y to X. Intuitive, but false. Your assertion is equivalent to the controversial Axiom of Choice (AC) discussed in the recent Banach-Tarski Paradox article. You can't prove it from the other axioms without assuming AC. [ Parent ]
 Ah (none / 0) (#289) by Pseudonym on Thu Jun 05, 2003 at 12:25:16 AM EST

 Interesting. I wasn't aware of that. I remember the theorem from way back, but not the proof. (I seem to remember the proof being long and complicated, so I'm not surprised that the AC turns up in there somewhere.) sub f{($f)=@_;print"$f(q{$f});";}f(q{sub f{($f)=@_;print"$f(q{$f});";}f});[ Parent ]
 He was thinking of Cantor-Bernstein Theorem (none / 0) (#293) by gzt on Thu Jun 05, 2003 at 12:52:38 AM EST

 Which states: Given any two sets A and B, suppose A contains a subset A1 equivalent to B, and B contains a subset B1 equivalent to A, then A and B are equivalent. Proof a simple exercise, doesn't use AC. But, yeah. Anyways. [ Parent ]
 ah (1.22 / 18) (#95) by TRASG0 on Tue Jun 03, 2003 at 06:46:17 PM EST

 what the FUCK are you people talking about? I love the *idea* of theoretical higher math and all that jooky, but the way you assholes talk about it makes me want to shave my head and declare that math is inherently a demonic lie.  You might as well be screaming gibberish and hurling feces at eachother, for all the enlightenment I got out of all that crazy babbling with the numbers and the letters and shit. And to preempt those of you who will question my intelligence: Fuck you.  Fuck the lot of you. read my diary or I shall turn you into a newt
 ahh come on (5.00 / 1) (#102) by asad on Tue Jun 03, 2003 at 07:00:50 PM EST

 feel the love. [ Parent ]
 I question your intelligence. Fuck me! [nt] (5.00 / 1) (#130) by tang gnat on Tue Jun 03, 2003 at 09:27:24 PM EST

 [ Parent ]
 and what would you prefer then? n/t (none / 0) (#182) by the sixth replicant on Wed Jun 04, 2003 at 06:26:49 AM EST

 [ Parent ]
 Infinity and Hilbert's Hotel (4.00 / 1) (#97) by cinemania on Tue Jun 03, 2003 at 06:49:09 PM EST

 Hilbert's Hotel has an infinite number of rooms. Finite numbers of guests keep coming and going. The manager keeps trying to move the guests around so that no one is sleeping in the lobby (presumably a finite space). By assigning a guest to a room and moving the previous occupant to a vacant room, isn't the manager computing with the room numbers and therefore flirting with computing with infinity?
 Hilbert Hotel (none / 0) (#101) by The Writer on Tue Jun 03, 2003 at 06:58:43 PM EST

 Ah, but you see, the manager already has an infinite number of rooms to begin with. If you have something infinite to start with, then you are surely capable of achieving the same degree of infinity. Just don't ask the Hilton manager to build an infinite number of rooms, though. :-) [ Parent ]
 Infinite number? (nt) (none / 0) (#141) by wurp on Tue Jun 03, 2003 at 11:12:00 PM EST

 --- Buy my stuff[ Parent ]
 Hilbert Hotel explained (5.00 / 2) (#202) by The Writer on Wed Jun 04, 2003 at 09:01:17 AM EST

 OK, I should practise what I preach, so let's be precise here. The Hilbert Hotel is one where there are precisely ℵ0 rooms, numbered, unsurprisingly, Room#0, Room#1, Room#2, etc.. In the original version of the Hilbert Hotel story, the Hotel is always full, because there are ℵ0 guests, each occupying one room in the Hotel. However, even though the Hotel is completely full, every time a new guest arrives, the manager is always able to find space for the guest. Here's how he does it: there is a special bell in the Hotel, which, when rung once, indicates to all the guests that they are to move into the next room. I.e., the guest in Room#n should move into Room#n+1; the guest in Room#n+1 should move into Room#n+2, and so forth. Since there are ℵ rooms in the Hotel, all the current Hotel guests will have a room to move into. Furthermore, once they do so, Room#0 becomes vacant. So, every time a new guest arrives, the manager simply rings the bell once, and assign Room#0 to the guest. When n guests arrive, where n is a finite number, the manager just rings the bell n times, and the current guests would know to move down n rooms, thus vacating Rooms #0 through #n. The new guests can then fill these rooms. Now, suppose one day, a Hilbert Bus arrives at the Hotel, carrying a busload of new guests. The Hilbert Bus carries ℵ0 passengers, you see. So what is the Hotel manager going to do now? He cannot just ring the bell a finite number of times to accomodate the new quests; since there are ℵ0 of them. Well, then he uses another trick: there is a gong in the Hotel, which, when sounded, tells all the guests to move from Room#n to Room#2n. After the gong is rung, all the current guests would move into the even-numbered rooms. Since there are ℵ0 even-numbered rooms, there is enough room to accomodate all of them. But after this move, the odd-numbered rooms are now vacant, and there are exactly ℵ0 of them. So after sounding the gong, the manager simply assigns the guests from the Hilbert Bus to the odd-numbered rooms, and everybody is accomodated again. So you see, the Hilbert Hotel is always full, but at the same time, it always has vacancies. Even if ℵ0 more guests arrive at the same time. [ Parent ]
 Sorry (5.00 / 1) (#219) by wurp on Wed Jun 04, 2003 at 11:44:52 AM EST

 I know about the Hilbert Hotel, I was just harassing you for using "infinite number".  I didn't mean for you to have to write a huge reply to it :) By the way, great story.  I was looking for flaws, and the only thing I could see is that you could have been a little more pointed about saying that there only has to exist one mapping that is 1-1 for two "infinite" sets to be considered of equal cardinality.  There may exist other maps that are into but not onto in either direction, but the existence of one 1-1 map is defined to mean that the cardinality is equal.  You did say it, but it's a point that's easily missed or misunderstood.  Anyway, it's a pretty darned weak complaint that I had, and kudos for a very complete, understandable, and interesting article. --- Buy my stuff[ Parent ]
 Effort not wasted (none / 0) (#340) by Dephex Twin on Thu Jun 05, 2003 at 04:43:08 PM EST

 I didn't follow the intiial description of the Hilbert Hotel and I enjoyed the huge reply. Alcohol: the cause of, and solution to, all of life's problems. -- Homer Simpson[ Parent ]
 I pity the guests in this hotel (none / 0) (#238) by skim123 on Wed Jun 04, 2003 at 02:44:16 PM EST

 Moving one room down isn't a big deal, but... Well, then he uses another trick: there is a gong in the Hotel, which, when sounded, tells all the guests to move from Room#n to Room#2n.Imagine you are in room 1,000 and this damned bell goes off. You have to walk down another 1,000 rooms? Bullocks. At least it's better than the poor sod whose room you're moving into! :-) Money is in some respects like fire; it is a very excellent servant but a terrible master.PT Barnum[ Parent ]
 Moving pains (none / 0) (#242) by The Writer on Wed Jun 04, 2003 at 03:25:13 PM EST

 Ah, but you see, to facilitate such moves, the Hilbert Hotel's rooms are arranged in such a way that every room is at most 1 unit of distance away from every other room. How can this be, you ask? Simple: instead of building the rooms in linear order, stretching away to infinity like the set of natural numbers, the Hilbert Architects who built the Hilbert Hotel came up with a better idea. They arranged the rooms the same way the rational numbers between 0 and 1 are arranged. There are just as many rationals between 0 and 1 as there are natural numbers, you see, so the rooms of the Hilbert Hotel all fit easily into that seemingly small space. I'm in no position to expound further on Hilbert architecture; but the upshot of all this is that every current guest in the Hotel only needs to travel at most 1 unit of distance to reach the destination room. Problem solved. :-) [ Parent ]
 Very well, then, but (5.00 / 1) (#258) by skim123 on Wed Jun 04, 2003 at 06:12:10 PM EST

 I don't think the patrons are going to much enjoy the size of their rooms. Money is in some respects like fire; it is a very excellent servant but a terrible master.PT Barnum[ Parent ]
 Oh, they still do (none / 0) (#273) by The Writer on Wed Jun 04, 2003 at 08:17:16 PM EST

 The patrons are perfectly happy with their rooms, for the simple reason that every room has a volume of precisely 1 cubic unit (let's say 1 unit is equal to 10 metres, so that makes it 1000 cubic metres, which I believe is rather spacious). How can all these rooms be that large, and still be only at most 1 unit away from each other? The answer is very simple, really. The reason the Hilbert Hotel has ℵ0 rooms is because all the rooms are connected to the Hilbert Corridor, which is an ℵ0-dimensional corridor. Since ℵ0 dimensions are really rather lavish, each room gets to have 3 dimensions all to itself, and there are still enough dimensions for the rest of the rooms. The ℵ0 dimensions of the Hilbert Corridor also makes it possible for patrons to simultaneously switch rooms without overcrowding. :-) [ Parent ]
 Alternative metric (none / 0) (#280) by vernondalhart on Wed Jun 04, 2003 at 10:17:33 PM EST

 That's an impressive beast to visualize. Alternatively, couldn't the Hilbert Hotel be designed with the distance between any two rooms, d(x,y) = 1 (for x != y)? Last I checked, that's a perfectly valid (albeit slightly boring) metric... -- "It's like that old saying: A conservative is a liberal who's been attacked by aliens." Simon - mhm27x5[ Parent ]
 Sure (5.00 / 1) (#319) by The Writer on Thu Jun 05, 2003 at 11:14:44 AM EST

 But how would you rationalize such a metric in terms of the analogy with a real hotel? [ Parent ]
 Well... (none / 0) (#321) by vernondalhart on Thu Jun 05, 2003 at 12:03:36 PM EST

 Every room is at the center of the hotel, with all the other rooms surrounding it, one unit away. -- "It's like that old saying: A conservative is a liberal who's been attacked by aliens." Simon - mhm27x5[ Parent ]
 But... (none / 0) (#332) by The Writer on Thu Jun 05, 2003 at 01:24:14 PM EST

 How is this any different from having infinite dimensions where all the rooms are a fixed distance away from each other? [ Parent ]
 Good point. (5.00 / 1) (#336) by vernondalhart on Thu Jun 05, 2003 at 02:21:47 PM EST

 That's a really good question. On thinking on it, while they are (afaik) different ideas, they really would work out to the same thing - although by defining the metric as I had, there is... less a notion of dimension, as I see it. Although I could be wrong. -- "It's like that old saying: A conservative is a liberal who's been attacked by aliens." Simon - mhm27x5[ Parent ]
 Finding your way (5.00 / 2) (#300) by skim123 on Thu Jun 05, 2003 at 02:23:03 AM EST

 I couldn't help but imagine how crowded the wall in that corridor must be, what with all the signs needed: looking for the first dimension exit? Please go this way. Second dimension is over that a way, and so on.In any event, I doubt the hotel excels at customer service. The way you are talking, you make everyone sound like a number. Money is in some respects like fire; it is a very excellent servant but a terrible master.PT Barnum[ Parent ]
 Customer service (5.00 / 1) (#318) by The Writer on Thu Jun 05, 2003 at 11:13:48 AM EST

 Well, I think the fact that patrons have to change rooms at every ring of the bell or sounding of the gong already throws the good customer service bill out the window! :-) If you want good service, you want the Hilton, not the Hilbert. [ Parent ]
 Agreed (5.00 / 1) (#322) by skim123 on Thu Jun 05, 2003 at 12:08:56 PM EST

 I guess the only benefit is not having to worry about no vacancy. Money is in some respects like fire; it is a very excellent servant but a terrible master.PT Barnum[ Parent ]
 haha (4.00 / 1) (#105) by the77x42 on Tue Jun 03, 2003 at 07:16:27 PM EST

 You should have a pretty interesting definition of -‡ ("negative infinity") :) Also, you should explain that 1/0 = ‡ (or relates in some way to infinity). 1/2 means cutting a piece of paper to get 2 parts. 1/4 means cutting a piece of paper to get 4 parts. 1/16, sixteen parts. 1/0... well, how can you get zero parts? A: Cut it an indefinite number of times. "We're not here to educate. We're here to point and laugh." - creature "You have some pretty stupid ideas." - indubitable ‮
 I'll take a stab at this one. (4.00 / 1) (#114) by hex11a on Tue Jun 03, 2003 at 07:52:14 PM EST

 1/0 is not actually defined - division is defined on a field (like the real numbers, for example, or the complex numbers) except for at the zero element. Normal interpretations of 1/0 tend to come from stating that 1/0 is the limit of 1/x as x->0 from above. Which tends to infinity. It also could be the limit as x->0 from below - all these numbers are negative and tend to - infinity. The best thing we can say is that 1/0 is undefined, and actually in complex analysis (don't be fooled by the name - it's not so hard!) the "Great Picard" lemma tells us that an essential singularity of a function (basically a point where you get division by 0 in a special case) takes on every complex value except possibly one of them as you approach the limit. Good enough for an explanation? Hex [ Parent ]
 Projective infinity (5.00 / 1) (#123) by Pseudonym on Tue Jun 03, 2003 at 08:33:46 PM EST

 In some mathematicaly uses of infinity, it has no sign. Consider, for example, two parallel lines. They do not intersect. Some geometers will speak of them meeting "at infinity". Infinity in which direction, though? Talking about positive or negative infinity in this context is like talking about positive or negative zero. It simply makes no sense. Now it could be argued that this is what's happening here; while 1/0 is infinity, it's a projective kind of infinity. This reasoning is incorrect. What this really is is yet another reason why infinity isn't really a number: its definition changes depending on how you want to use it.. sub f{($f)=@_;print"$f(q{$f});";}f(q{sub f{($f)=@_;print"$f(q{$f});";}f});[ Parent ]
 actually (none / 0) (#145) by the77x42 on Tue Jun 03, 2003 at 11:39:05 PM EST

 two parallel lines will meet at some point n such that n < (infinity) because space is curved... or so i've heard. "We're not here to educate. We're here to point and laugh." - creature "You have some pretty stupid ideas." - indubitable ‮ [ Parent ]
 Or... (5.00 / 2) (#152) by Pseudonym on Wed Jun 04, 2003 at 12:51:37 AM EST

 ...depending on how space is curved, they could diverge. sub f{($f)=@_;print"$f(q{$f});";}f(q{sub f{($f)=@_;print"$f(q{$f});";}f});[ Parent ]
 not really (4.00 / 1) (#180) by the sixth replicant on Wed Jun 04, 2003 at 06:22:21 AM EST

 Either you assume this, and you have projective geometry (for free!) or you don't and you get affine geometry. Saying that two parallel lines meet "at infinity" is a "mother's-tale" (the way you use it). It all depends on the topology/geometry of the space you're in.And it's been shown, mathimatically, that your statement is actually an axiom, and not a fact. So it's up you if you want it or not. On the other hand if you want to prove it empirically then go ahead, because that's the only way to "prove" your statement. Ciao P.S. Excellent article. In a drunken-drug filled craze I tried to explain the different types of infinity, starting with the definition of cardinality etc. That person never spoke to me again after that. [ Parent ]
 Curvature of space (none / 0) (#230) by vernondalhart on Wed Jun 04, 2003 at 01:53:19 PM EST

 Also, last I heard (although I can't cite a source off the top of my head), it turns out that space is flat. So two parallel lines in the world around us will never meet. Ever. -- "It's like that old saying: A conservative is a liberal who's been attacked by aliens." Simon - mhm27x5[ Parent ]
 OK... (none / 0) (#178) by hex11a on Wed Jun 04, 2003 at 06:07:04 AM EST

 This is good, I've got a projective geometry exam in a couple of days... In projective geometry people are very cavalier and talk about a "point at infinity" or a line at infinity or some other structure. Now, for the "line at infinity" the way to think about it is that you have two sets of traintracks, and you're standing in the center of them. One pair goes off to the north, one off to the east. As you look down each pair, they appear to converge "At infinity" - the horizon - but the seperate tracks converge at different points. Hence in projective geometry we have a whole structure at infinity. So your 1/0 would need to have a whole structure assigned to it for coping with this in projective space. Where it can be said to work is in short-hand. We ofter work with homogenous coordinates in PG - in the projective line [x,y] becomes [1,y/x] since [Ax,Ay]=[x,y]. This is done to reduce the amount of work done when perfoming operations on this vectore - you only need to work out what happens to its second coordinate, as its first will always go to the same thing. This extends to higher dimensional projective spaces - [x,y,z,t,s...] = [1,y/x,z/x,...]. Our problem lies when x is zero, and in that case we call the points infinity simply because this helps the shorthand to work when carrying out, say, a Mobius transformation (rotation and re-scaling etc). I was being a bit simplistic, I admit in talking about 1/x etc - Mereomorphic functions can cope with this quite easily by projection onto the Riemann sphere, adjoining a point called infinity to the complex plane. The point I was really trying to make is that the definition of division is multiplicative inversion, and multiplication by zero is a many to one map, and so cannot be inverted in a field. Hex [ Parent ]
 question (none / 0) (#129) by tang gnat on Tue Jun 03, 2003 at 09:26:06 PM EST

 Does that lemma work for all complex functions, or just ones that are analytic around the singularity? [ Parent ]
 Ok precise statement: (5.00 / 2) (#179) by hex11a on Wed Jun 04, 2003 at 06:12:15 AM EST

 Picard's Great Theorem (I called it a lemma, sorry Mr P) Every nonconstant entire function attains every complex value with at most one exception. Furthermore, every analytic function assumes every complex value, with possibly one exception, infinitely often in any neighborhood of an essential singularity. An essential singularity is a point at which the Laurent expansion (kind of like the Taylor expansion, but with negative powers of z as well as positive) has an infinite number of coefficients of negative numbers. Normally this is taken to be holomorphic (analytic) around this point, but not at the point itself. I'm not sure if it works with non-isolated singularities - my intuition says no, but I'd have to do some serious checking. Hex [ Parent ]
 Another proof of why x/0 is not defined. (none / 0) (#378) by fifi on Fri Jun 06, 2003 at 07:26:25 PM EST

 Consider the following assertion: [a] For any b != 0, a * b = b implies a = 1 It is demonstrably true. This assertion expresses that division by zero is not defined. If we hypothesize ab absurdum that division by zero has a meaning, then we can reformulate the assertion as: [b] For any b, a * b = b implies a = 1 Now it is also true (and demonstrable) that: [c] For any x, x * 0 = 0 If you assume [b] is true, and combine it with [c], you've just proven that [d] For any x, x = 1 ...which is a contradiction. Hence [b] is false and you can't divide by zero. [ Parent ]
 That's good (none / 0) (#384) by hex11a on Sat Jun 07, 2003 at 11:42:21 AM EST

 what you're using there - don't know if you know it - are the rules defining a field (which most things used in every day life are), or even special types of integral domain . However what is fun about abstract algebra is that we can create different strutures that disobey these rules (rings without unity, for example) and get a whole boat load of different systems, each with its own set of rules. Hex [ Parent ]
 Powersets (4.75 / 4) (#109) by Arevos on Tue Jun 03, 2003 at 07:35:34 PM EST

 IIRC, there is a theorum that states that the powerset of a set is always of a greater cardinality than the original set, which implies that there are an infinite variety of infinities, to put it bluntly. Have I remembered correctly? (Math isn't my best subject, I'm afraid).
 I still remember prooving this (5.00 / 3) (#121) by bgalehouse on Tue Jun 03, 2003 at 08:24:41 PM EST

 Once took a college course which either required us to show this or just mentioned this fact and pointed out that it is "easily prooved". Anyway, the proof is to take an arbitrary function f, say, between a set and its power set. Then consider the set of points which are not contained in their image under the function -the set of points x such that x is not in f(x). Then note that no x can possibly map to this set. (think about it) Therefore f is not a bijection - it doesn't line the sets up. The argument holds for all f without restriction, since we assumed nothing about it. So no such bijection exists, and the sets must have different cardinalities. [ Parent ]
 Yes (none / 0) (#122) by celeriac on Tue Jun 03, 2003 at 08:30:14 PM EST

 I hope the next installment goes into some detail over the continuum hypothesis: Does the power set of the integers have the same cardinality as the set of real numbers? It turns out to that you can't prove the continuum hypothesis from the axioms of set theory, but you can't disprove it either. You can add another axiom, postulating the answer to be "yes" or "no," with some interesting effects in either case. [ Parent ]
 the continuum hypothesis (4.33 / 3) (#126) by tang gnat on Tue Jun 03, 2003 at 09:16:08 PM EST

 IIRC, CH states that taking the powerset of a set with cardinality Aleph[0] gives you one with a cardinality of Aleph[1] (the next highest level of infinity). In fact, Cantor (the one who proposed CH) showed that the power set of the integerset has the same cardinality as the realset. link That combined with CH would then tell us that there are no sets which are "in-between" the integerset and the realset, in terms of "size". nb: Apparently, using the Axiom of Choice, one can show that infinite countable sets (eg integers) are the "smallest" kind of infinite set. [ Parent ]
 Small infinite sets (5.00 / 3) (#146) by bgalehouse on Tue Jun 03, 2003 at 11:55:50 PM EST

 It is trivialy provable that the infinite countable set is the "smallest" kind of infinite set. I don't think that the Axiom of choice is required, but maybe implicitly somewhere. Given an arbitrary set, assign to each positive integer in turn an arbitrary element from the set, different than all previously assigned elements. If no such assigment is possible for the integer i, then the set must be finite of size i-1. If we can continue then we have a way to assign a unique element to each positive integer. Even though it would take forever to finish the calculation this is still a valid definition of a bijection between the positive integers and a subset of our infinite set. Therefore any infinite set has cardinality at least as large as the integers. [ Parent ]
 AC (3.00 / 1) (#207) by Lev Black on Wed Jun 04, 2003 at 09:48:11 AM EST

 Given an arbitrary set, assign to each positive integer in turn an arbitrary element from the set, different than all previously assigned elements Thats the axion of choice [ Parent ]
 Not Obviously (none / 0) (#220) by bgalehouse on Wed Jun 04, 2003 at 11:49:35 AM EST

 Or at least, not obviously to me. To quote www.math.vanderbilt.edu Let C be a collection of nonempty sets. Then we can choose a member from each set in that collection. In other words, there exists a function f defined on C with the property that, for each set S in the collection, f(S) is a member of S. Now, where exactly is the collection of sets in my example? The sets that I need to choose from do not exist before the function is partially defined, so one cannot apply the axiom of choice directly. Also note that sometimes one can explicitly construct such a function without the axiom of choice. In such cases the axiom is not required - the point of the axiom is that it can be applied to any collection. Now consider that recursive definitions of say, factorial or fibonacci series do define complete functions on the integers. Given any integer n one can find the nth term of the fibonacci series, and so you will agree that it is defined on all the integers even if you don't happen to know that there is a closed form expression for it. Does this require the axiom of choice, and is it fundamentaly different than my argument about countability? [ Parent ]
 No need for AC. n/t (none / 0) (#292) by gzt on Thu Jun 05, 2003 at 12:47:18 AM EST

 [ Parent ]
 Great article! (4.66 / 3) (#110) by Pholostan on Tue Jun 03, 2003 at 07:42:13 PM EST

 I just wish that my "high school" teacher had been able to explain this that good. It would have saved me humongus amount of time spent at trial and error.   - And blood tears I cry Endless grief remained inside
 no need (1.00 / 1) (#136) by dh003i on Tue Jun 03, 2003 at 10:34:39 PM EST

 The average high school students has no need to understand mathematical theory. The most important thing to be learned from math by most people is statistical analysis, so that we can do CBA on insurance plans, and risk/reward analysis on investments, not to mention figure out the best way to manipulate one's money, given taxes.
 "I agree" (5.00 / 9) (#150) by cgibbard on Wed Jun 04, 2003 at 12:49:10 AM EST

 Yeah, and while we're at it, why bother with any of that music or art or fine literature stuff. Why, that's just a waste of time and money that could be spent learning practical things. Learning about things that don't help you be a better worker is a waste of time, and a pointless drain on the resources of the economic machine. Of course, I'm being sarcastic here, as I'm a student of Mathematics at the University of Waterloo. In all seriousness though, people like you needn't worry about the math front - little to no actual mathematics is presently taught in high-schools anyway. At least in North America. Mostly it's just arithmetic and how to do simple repetitive calculations. In my opinion, highschool (and elementary school) math needs a complete overhaul. Mathematics to me is about looking for patterns and abstracting and generalizing them - taking just the essence of something, pulling it out, and seeing what we can find out about it. Taking a set of assumptions that creates some structure, and then exploring that structure. It's also about the beauty naturally occuring in these structures - the proofs, the myriad kinds of constructions of thought. There's something to be gained from that appreciation alone. Life is not all entirely about the GNP. [ Parent ]
 ok (1.00 / 1) (#241) by dh003i on Wed Jun 04, 2003 at 03:24:37 PM EST

 So, just because statistics is an extremely useful branch of mathematics, and number theory and extremely useless one, number theory is the parallel to beauty, statistics the parallel to utilitarianism? I'm not saying that I find number theory uninteresting. I find it interesting. I also find singularities extremely interesting. Neither of them have any need of being mandatory learning for high school students. Even though you and I may be interested by such things, the vast vast majority of people are not. So, they spend time being forced to learn about stuff that is not only completely uninteresting to them, but that is also useless; while they could be learning about principles which would help them determine which stocks are solid and which are good value-buys. Things that have both theoretical basis' and practical implications should be required and come first. Statistics should be taught in high school, not number theory and other abstract mathematics that has no (or limited) practical applications. Rather than wasting time teaching about things which have no impact on our lives (like history), the teachers should teach students about important laws, new laws that affect their lives, government processes in the judicial, legislative, and executive branches, and how to find and read laws.
 I disagree (5.00 / 2) (#248) by The Writer on Wed Jun 04, 2003 at 04:19:14 PM EST

 The point of school is not, as many believe, to give information to the student so that the student can survive out in the cold, harsh world. The point of school is to train the students to think, so that in the event the information they did pick up from school is found useless in the real world, they are adequately prepared to pick up new, useful information on their own. A lot of stuff I learned in school is "useless" --heck, even what I learned in grad school was 95% useless (and that's being generous). Everything I need in my job right now, I've learned on my own during my school years, outside of any curriculum. But would I be where I am today if I never went to school? Of course not. The information I got from school was mostly inapplicable to the real world, but they trained me to think, they sharpened my ability to comprehend, to analyse, and to derive results. Without this training, I'd still be writing stone-age computer programs in a basement somewhere, instead of actually being paid for real software development. I do not agree that school should merely focus on teaching students how to count, how governmental procedures work, etc.. This is information which will one day become outdated, and then what will these students do? No, you want to train them how to think, how to analyse, and how to evaluate things. Sure, a lot of the information you use to train them may be "useless", but the point is, once trained to analyse and think critically, you can throw them in anywhere in the world, and they will be able to apply their skills, pick up the information relevant to that locale, and survive. They will also have the potential to develop much further than those who only learned specific tasks. [ Parent ]
 I really hate to do this (5.00 / 1) (#256) by Control Group on Wed Jun 04, 2003 at 05:57:34 PM EST

 Because, in a sense, I agree with you completely. Education should be a matter of teaching students how to think, so that they can properly assimilate and analyze varied information, rather than simply know responses to canned data. HOWEVER As a practical matter, teaching someone who doesn't want to think how to think is futile, and the current state of (public, at least) education in the US is a testament to that problem. Given where we're at right now, I am forced to disagree with you on a practical basis. If given the choice, I would much rather every high school student understood the legislative and judicial systems than number theory. And you've no idea how it pains me to say that. If I had my druthers, of course, the one would not preclude the other—but as it now stands, neither is achieved. I'd rather focus on civics first. I'm sorry. *** "Oh, nothing. It just looks like a simple Kung-Fu Swedish Rastafarian Helldemon."[ Parent ]
 I still disagree (5.00 / 2) (#262) by The Writer on Wed Jun 04, 2003 at 06:46:23 PM EST

 As a practical matter, teaching someone who doesn't want to think how to think is futile, and the current state of (public, at least) education in the US is a testament to that problem. While I understand that given the current state of affairs it is prudent to teach civics, I have a hard time believing that that many students "doesn't want to think". From my own observations (which I admit may not be reflective of the true state of things, but nevertheless), it is often the teachers who turn the students off. Now I'm not putting all the blame on the teachers; many of them are overworked and underpaid, and there are too many students and not enough teachers, etc.. But the fact of the matter is that most students will be interested to learn when coached the right way. I'm afraid that what happens is the teacher, being overworked and underpaid, gives a less-than-desirable lesson, and the students are turned off. They look around, and see that this appears to be the way things are supposed to be, so they say, "to heck with this, I hate school, why must my parents make me come here, I'm just going to do the least I can get away with". Furthermore, without accusing anyone in particular, I notice that some teachers just plain aren't interested to teach. Their excuse may be perfectly legitimate---it's necessary supplementary income, it's a last-ditch job, or whatever the case may be. I'm sorry, but if the teacher is not interested in the subject, is it any surprise that the students aren't, either? And this is just talking about the teaching materials. Don't even talk about teaching them how to think, analyse, etc.. I personally would rather settle for less teachers, but better teachers, than for more mediocre, uninterested teachers. The standards for hiring educators are way too low. I'm not an American, but I tell you, you are shooting yourselves in the foot by taking care of the immediate need and forgetting the long-term effects of such bad decisions. If this continues for a few more generations, I will not be surprised the US becomes a backwater country. [ Parent ]
 You're spot on, as usual (5.00 / 3) (#313) by Control Group on Thu Jun 05, 2003 at 10:20:46 AM EST

 As far as the underlying causes of general disaffection with secondary education go. Though I'd also add that the US' decision that a degree in teaching is more important than an education in what you're teaching is also a contributing factor towards teachers who make subjects boring for their students. The problem is that it's been decided education through high school is mandatory, which means the bar has to be lowered for teachers (both in terms of salary to, and ability of) simply to meet the numbers necessary. The other consequence is that classes often face a lowest-common-denominator problem: if someone in the class isn't keeping up, it's incumbent upon the teacher to make sure they do. This slows the class as a whole, and is guaranteed to make the subject boring for anyone in the class who is actually good with the material. At any rate, I find that I don't disagree with anything you say. Rather, I think the difference between your POV and mine is that I've set my sights lower in terms of what I hope for. I'd rather see your ideas come to fruition, but I'm stuck believing that even my more limited goals are all but unattainable. One more thing: my fear isn't so much of becoming a backwater, my fear is of a growing morlock/eloi style divide. If you pursue an education in the US, you can certainly get one. Money helps this, of course, but even without much, a dedicated student can still do a lot of learning. Unfortunately, most students need to be convinced that education is interesting/important/useful/fun, and the system is bad at that. What I see is a growing discrepancy in both functional skills and cognitive skill (note that I use skill—learned—as distinct from talent—innate) between the minority who have gone out of their way to develop them, and the majority who did just enough to get through the required schooling. Bearing in mind, of course, that for far too many students, college is perceived as yet more "required schooling." *** "Oh, nothing. It just looks like a simple Kung-Fu Swedish Rastafarian Helldemon."[ Parent ]
 one thing before another (5.00 / 1) (#278) by martingale on Wed Jun 04, 2003 at 09:38:19 PM EST

 Allow me to but in for a second. Civics is useless if the students don't have a grasp of logic (and I'm not talking esoteric logics here). I can't think of a better, smoother first training program in logic than simple euclidean geometry. So I'd have to say reading, writing, arithmetic and euclidean geometry are the fundamentals. Now these days, students spend an awful lot of years in high school, partly to keep the unemployment numbers down. Be that as it may, euclidean geometry doesn't have enough material for so many years of study. Keeping the kids focused on the basics of logic is necessary, lest they revert to thinking in "concrete" terms, ie case by case. I think abstract mathematics in high school serves this purpose, by focusing on the thought patterns (ie proofs). Teaching applications such as statistics in preference to number theory is bad, in the same way teaching law or civics in preference to number theory is bad: it encourages the student to think of the specifics, and to miss the general patterns because of the specific cases. By all means, teach civics and statistics too, but if there has to be a choice, drop them before you drop number theory. [ Parent ]
 please (1.50 / 2) (#283) by dh003i on Wed Jun 04, 2003 at 11:24:23 PM EST

 Ok, let me get this straight. You talk about the need to think. Then you mention something about logic, which is the basis of geometry. Yet, one can easily understand geometry while having little sense of logic. Statistics is also based -- very strongly, I might add -- on logic, and logic is usually the first thing taught in any statistics course. Statistics also requires intense thought by individuals, and one can delve into the theory of statistics as well, by indulging in proofs and abstracts. Many branches of mathematics have both practical applications, as well as being worthy of pursuit in-and-of themselves. Statitistics is one of these, as is logic, geometry, calculus, differential equations, and so-on and so-forth. Number theory, however, is most certainly not one of those branches of mathematics. I can think of no practical applications of number theory. Worse yet, unlike paintings and music, number theory is most often only appreciated as something beautiful by mathematicians. The average joe can listen to the "Moonlight Sonata" and hear beauty; he cannot, however, see beauty in number theory. P.S.: I'm sure that all the people who lost thousands of dollars -- or their life savings -- after the stock-market crash will be rejoiced in that they at least know something about number theory. Maybe if just a little bit was taught about investing in high school, most people would know that the dot-com companies were bullshit, and that the stock-market was outrageously over-valued. However, again, I'm sure they're happy that they at least had an education in "art".
 that's upside down (5.00 / 3) (#294) by martingale on Thu Jun 05, 2003 at 01:09:08 AM EST

 your arguments are backwards (1.50 / 2) (#346) by dh003i on Thu Jun 05, 2003 at 06:29:00 PM EST

 Your arguments that statistics is useless in practical applications is fallicious. It is known that many people mis-use statistics. That does not negate the value of proper statistics. Statistics can help one make many useful decisions. Should you buy partial disability coverage, or full disability coverage? That depends on the probability of becoming "differently abled" in your field of work. Statistics. If you want to invest in stocks or bonds over a 10-year period, should you invest in a stock-index portfolio or a bond-index portfolio? There has never been a 10-year period in which stocks have not significantly outperformed all competing investments. Statistics. What if you want to invest money for growth, but might need it quickly? The stock-market has great volatility, and is not a reliable investment over the short-term. Also statistics. This is something that everyone can apply, at least rudimentarily, in regular every-day life. It's something that can be useful to us all in every-day life, something the benefits of which we can reap only if we understand it. Likewise with knowledge of investing. However, number theory is of no use to us in every-day life. Even if you argue that number theory is necessary for cryptology -- and I argue that it is not -- so what? I don't need to understand how a car works to use it, nor need I understand number theory to use cryptology.
 !lla ta ton (5.00 / 2) (#356) by martingale on Thu Jun 05, 2003 at 10:28:02 PM EST

 Flaws in your statements (none / 0) (#357) by scheme on Thu Jun 05, 2003 at 10:50:04 PM EST

 Should you buy partial disability coverage, or full disability coverage? That depends on the probability of becoming "differently abled" in your field of work. Statistics. This is assuming that your historical data is accurate and that there isn't any anomalies with your workplace and that previous conditions still hold. If any of those things don't hold then all bets are off and statistics don't help you out. If you want to invest in stocks or bonds over a 10-year period, should you invest in a stock-index portfolio or a bond-index portfolio? There has never been a 10-year period in which stocks have not significantly outperformed all competing investments. Statistics. Actually, that's historical knowlege. Knowing that in the last 80 years, stocks outperformed competing investments says nothing about how they will perform in the future. Sure you can say that stocks will probably continue to out perform other investments but since you don't know the the underlying probabilities of a given investment returning a certain amount, you can't do much. Sure you can calculate this for historical data but as the SEC says, past performance does not guarantee future results. "Put your hand on a hot stove for a minute, and it seems like an hour. Sit with a pretty girl for an hour, and it seems like a minute. THAT'S relativity." --Albert Einstein [ Parent ]
 Music analogy also doesn't work (5.00 / 2) (#380) by phliar on Fri Jun 06, 2003 at 07:45:39 PM EST

 The average joe can listen to the "Moonlight Sonata" and hear beauty I'd argue that Joe Average hears not its beauty, but its prettiness. Will Joe Average care about the difference between Beethoven and an ad for high-end chocolate?I'm not saying Joe Average should care about beauty; but if he does, he will find that he needs to learn a lot of "useless crap" no matter the field. Joe Average certainly will not like most twentieth century composers. I'd also like to add that I was able to give at least one layperson an idea of how beautiful number theory is, with Fermat's Little Theorem and its elegant proof using group theory. Faster, faster, until the thrill of...[ Parent ]
 ok, so let's see (1.00 / 1) (#281) by dh003i on Wed Jun 04, 2003 at 11:16:33 PM EST

 So, your argument is that learning about number theory gives one more training in thinking than learning about statistics? I highly doubt that. And learning about history gives one more training in thinking than learning about how to read and understand laws? Again, highly doubt that. I agree that current factual information on many laws will some-day become outdated. However, individual's need to be educated about the law none-the-less, because while it may become outdated one day, not knowing it now can land them in jail (it is very easy break a law that would land one in jail, all-the-while doing what one considered to be the right thing, and what could be defended as being the right thing by any reasonable person). Also, because it may become outdated one day, they should learn about resources on the law, how to find laws, and how to read laws, as well as court-case decisions (obviously, in a limited scope). Furthermore, are you syaing that statistics will one day become outdated, and have no bearing on investment decisions and insurance decisions?
 Ah useless number theory... (5.00 / 3) (#311) by hex11a on Thu Jun 05, 2003 at 07:59:20 AM EST

 that's why most American's are financially fucked (3.00 / 2) (#345) by dh003i on Thu Jun 05, 2003 at 06:19:20 PM EST

 Firstly, number theory is hardly necessary for creating encryption. Encryption is simply a mathematical algorithm for tangling information in a seemingly random fashion, such that the knot can only be untied using a key. Even if number theory did have uses in computer-science and encrypion -- which I argue it does not, since most computer science relies on algorithms, which have very little to do with number theory -- so what? It's something that's so specialized that ordinary people need not know about it. I don't need to know how my car works to use it. Secondly, this type of reasoning is why most American's are financially fucked. Everyone will need to be managing their financial situation. Everyone will need to fill out tax-forms, manage a check-book, manage their debt, and preferrably invest their money at least to compensate for inflation, and perhaps further if they're aggressive. Everyone will also need to obey the laws, which means knowing about laws that are here to stay, knowing how to find laws, and knowing how to read laws. Despite not encouraging much thought, we still have Health classes in high school. Thirdly, if you think managing your financial life is something that can be done by an automotons, that is obviously a measure of your complete and total ignorance of the financial world. Financial decisions require extensive thought, self-analysis, risk-analysis, and general analysis of the character and quality of investments, as well as the potential gain vs. risk.
 OK, clue (5.00 / 1) (#349) by hex11a on Thu Jun 05, 2003 at 07:28:59 PM EST

 I quote "Firstly, number theory is hardly necessary for creating encryption." As someone who's studied it, trust me it is. How do you know your encryption is safe? Prime numbers hold the answer. You obviously don't know much about the subject (I don't mean to be demeaning here, but I don't think you know much about it), so I suggest we drop this one. My ignorance of the financial world is terrible. I only live with a few actuaries and some investment bankers, and a lot of people who've managed their own accounts very nicely, despire having studied NO stats at all. In my country it's not necessary. The average person here knows enough to know what a good interest rate is, why they shouldn't go into debt etc. And it works very nicely. A lot of americans are screwed up because the system is designed that way. You don't need to be a financial wizard to understand that dropping taxes for the rich and benefits for the poor fucks people. You talk of investments! Hah! Really, do the poor people of america invest a lot? Or are you just talking about your middle class friends not making their 5% on their money and crying about numbers on a machine somewhere that mean very little? Get an appreciation for life - there's a lot more to life than money! Hex [ Parent ]
 clue yourself (2.00 / 1) (#353) by dh003i on Thu Jun 05, 2003 at 09:25:20 PM EST

 As I clearly stated, even if number theory is necessary for encryption, so the fuck what? Those who need to learn it will; to the rest of us, knowing about it is useless knowledge. I don't need to know how my car works to use it. Likewise with encryption, and basically any other tool. I'm curious as to what country you live in, where no knowledge of statistics is necessary for making good financial decisions, since statitistics are the only way you can say that one investment option is significantly better than another? As for taxes, you can drop your liberal dribble. Poor people get plenty of free rides by parasiting off of the middle class and the rich; it's really the middle-class people who get nothing, and are punished for working hard by having 15% of their salaray stolen at the federal level. Why don't you get an appreciation for life. Learning about number theory isn't going to benefit the average person. The average person will sleep through a lecture on number theory. Learning about financial management and the law is going to benefit the average person -- in fact, it will benefit anyone. When you talk about there being more to life than money, most of the "more" is only possible if you have enough money to start out with. It's difficult to enjoy the higher pleasures of life when you're in thousands of dollars of debt, can't get loans, have no retirement savings, and are saddled with various other financial problems. Before you go off on a tirade against the rich the next time, remember that the richest man in the world (Bill Gates) is only that rich because the average individual -- who obviously hasn't been educated enough on CBA -- insists on giving him lots of money for inferior software. The second richest man in the world (Warren Buffet) is that rich due idiotic irrational exuberance in times of bull markets, and idiotic irrational paranoia in times of bear markets, of the average investor.
 Ok (5.00 / 2) (#366) by hex11a on Fri Jun 06, 2003 at 06:29:20 AM EST

 hmpf (5.00 / 1) (#372) by adiffer on Fri Jun 06, 2003 at 01:24:12 PM EST

 You both are right and clueless at the same time. In each of your subject areas, you are both right to point out the value of knowing about it.  By treading on each others areas, though, you are both being silly at best. Anyone who has learned about the complexities of financial planning and investing knows just how clueless most of the rest of the population really is.  Many of us take advantage of it daily.  Anyone who has learned some of the complexities of Number Theory knows just how useful it is in our lives every day.  That knowledge gives deeper meaning and connectivity to a lot of things around us. You don't have to know how a car really works to use it.  Knowing how a car works, though, can save your life in a tight spot. You don't have to know how a refriderator works to use it.  Knowing, though, would allow you to make more efficient use of your resources on the environmental and financial levels. I could go on, but I won't.  Bashing each other's knowledge is silly and unproductive.  If you don't care to learn a subject, that's fine.  No one has the time to learn everything, so pick and choose what you want to make you happy. --BE The Alien![ Parent ]
 Thanks... (5.00 / 1) (#375) by hex11a on Fri Jun 06, 2003 at 05:28:24 PM EST

 I guess I have drifted into bashing somewhat - sorry, I blame finals! My main point isn't that one things is useful and the other useless, but that we shouldn't have education being completely utilitarian - we should teach people how to appreciate life and all its wonders rather than just tell them "This is how you write a cheque, this is how you...". Something of this is necessary, obviously, but I don't think that we should drop things that aren't entirely necessary just because someone doesn't think they'll ever use it. Hex [ Parent ]
 Well (5.00 / 2) (#212) by Pholostan on Wed Jun 04, 2003 at 10:34:09 AM EST

 For me, it would have been very nice to have really understood the concept of infinity in high school. I did sorta understand it, but I did have many of the common misconceptions. This turned out to be a big problem in my higher studies and I were forced to relearn, obviously. As I see it, a lot of my time have been wasted for nothing. But then again, I might not be a prime example of "the average high school student". I don't think many are, actually. I want to aim higher.   - And blood tears I cry Endless grief remained inside [ Parent ]
 Dense vs non-dense sets (4.66 / 6) (#120) by Pseudonym on Tue Jun 03, 2003 at 08:23:47 PM EST

 Every person who works in text knows that you can very easily turn a dense set into a non-dense set. Consider, for example, the set of ASCII strings. The usual ordering that we use is lexicographic so that, for example: a < aardvark < able < ace < ada This set is dense. If you don't restrict yourself to English words, you can always find a string which lies between two existing strings. To get a non-dense set, you just pick a non-dense ordering. If, for example, you order by length first and then lexicographic ordering: a < aca < ada < able < aardvark The ordering is now dense. The point of this exercise is that density is not a property of a set as such, but rather a property of a combination of a set and an ordering. This is obvious when you look at the definition: To choose an element "between" two existing elements, you first need a definition of "between". Of course there are sets for which there is no dense ordering (e.g. the real numbers), but I'm sure that will come up next article. sub f{($f)=@_;print"$f(q{$f});";}f(q{sub f{($f)=@_;print"$f(q{$f});";}f});
 *non*-dense ordering? (5.00 / 3) (#137) by The Writer on Tue Jun 03, 2003 at 10:41:56 PM EST

 You wrote: To get a non-dense set, you just pick a non-dense ordering. If, for example, you order by length first and then lexicographic ordering: a < aca < ada < able < aardvark The ordering is now dense. I believe you meant, "the ordering is now non-dense"? The point of this exercise is that density is not a property of a set as such, but rather a property of a combination of a set and an ordering. This is obvious when you look at the definition: To choose an element "between" two existing elements, you first need a definition of "between". Correct. The fact that by reordering Q, I can essentially get N, shows that the density is just a result of how elements in the set are ordered, not necessarily an inherent property of the set. I alluded to this with the "it's just ordered funny" comment, although I admit that wasn't very clear at all. Thanks for pointing it out. Of course there are sets for which there is no dense ordering (e.g. the real numbers), but I'm sure that will come up next article. I believe you meant "non-dense ordering" again? And yes, this is a candidate for being in the next article. Only thing is, there is so much to talk about w.r.t. to the real numbers that I'm at a loss as to how to put it all into a single article. I've started on a draft of it already, but 6 pages later, I haven't even finished talking about Cantor's Diagonalization proof! :-( There's still the Continuum Hypothesis (although I probably will pass up on that one, since it will involve ordinals, which will at least fill another full-length article on their own), the powerset theorem, cardinals beyond c, etc.. And beyond that, I need to at least touch on other types of infinity, outside set theory, such as the infinity at the extreme ends of the real line (which is not at all the same as cardinal infinity, although it appears that many people confuse the two). *sigh* So many theorems, so little time. :-P [ Parent ]
 Agh!!! dense ordering??? (5.00 / 4) (#144) by bgalehouse on Tue Jun 03, 2003 at 11:38:12 PM EST

 Of course, density is a property of the topology of some extrinsic set. Not of ordering or of cardinality. You probably realize all this, but it seems to me that those who don't might become confused by your terminology. Avoiding math terminology adds so much to the fun of discussing math in general discussion boards. To that end, I think that it would make math people's eyes bleed less if you used "embedding" or even "labeling" to describe what is different between the set of rationals and the set of integers. Seriously, ordered is such a dangerous term. The rationals are ordered. However, they are not well ordered, which makes it impossible to even start writing them down in "order". For the unintiated, in math speak, according to, say, Kolmogorov and Fomin, "ordered" means always comparable - any two elements are either equal or one is less than the other. "Well ordered" means that every subset contains a lowest element. The positive integers are well ordered, the positive rationals are not. Both are, however, ordered. I vote for the diagonalization argument, against getting into ordinals. You got me thinking about doing one on Godel and the axiom of choice, though I need to check the archives - I vaguely remember him coming up in some context or another on this site. The proof in the abovementioned book that cardinalities are ordered is way cool, but probably too much for this board. (Actually, I think it is better to not use ordinals but to instead use the axiom of choice directly to do the last bit, but maybe I just have a thing against ordinals). [ Parent ]
 Ordering vs topology (5.00 / 1) (#303) by flo on Thu Jun 05, 2003 at 03:55:44 AM EST

 [diclaimer: I use the term "topology" without definition here, sorry. Basically, its a very weak way of defining when two elements of a set are "near" to each other. Except that even this is not entirely accurate... the full defintion would be long, and possibly incomprehensible to the layman.] Whether or not a subset A of a set B is dense in B depends on the topology of the set B. In our case, A is the set of rationals, and B is the set of reals, and here A is indeed dense in B. However, as pointed out in the article, you can find another embedding of A into B, which is not the inclusion function, whose image is not dense in B. So density really is not an intrinsic property of A, but of the chosen embedding into B, and the topology of B. Now, the usual topology on the real numbers is defined in terms of the ordering of the reals (the distance between numbers a and b being |a-b|), which is why people have written in other comments that density depends on the ordering. But this is inconvenient, as there are other conceivable topologies of the reals (e.g. in the discrete topology [where ALL distances are the same], the only dense subset is the whole set iteslf). But what is worse is the fact that one can apply the notion of density and all that to other sets, like the complex numbers, which do NOT admit any (useful) ordering. This is also why I don't like Dedekind cuts (despite their elegance) - they depend on the ordering. I prefer the definition of the reals from the rationals via Cauchy sequences - which do not require an ordering, but a choice of absolute value (in this case, the usual absolute value "|x|"). In particular, the same construction can be applied to other absolute values of the rationals (or any other valued fields), such as the p-adic absolute values: Pick a prime number p. Let x be a rational number. Then we may write x=pn a/b, where n is some integer (positive, negative or zero), and p does not divide a or b. Then we set |0|p=0 and |x|p=p-n if x is not zero. You'll see that this behaves similar to an absolute value: |x|p = 0 if and only if x=0. |x|p > 0 if x is not zero |xy|p = |x|p|y|p for any x and y |x+y|p is less than or equal to |x|p + |y|p for any x and y In fact, we have a stronger property, called the ultrametric inequality: |x+y|p = min(|x|p,|y|p). Now we can also define Cauchy sequences in terms of this new absolute value, and define the completion of Q with respect to it, and obtain the so-called p-adic numbers, Qp, which are really cool. --------- "Look upon my works, ye mighty, and despair!"[ Parent ]
 Construction of the reals (none / 0) (#324) by vernondalhart on Thu Jun 05, 2003 at 12:11:58 PM EST

 I've only read two different ways to construct the Reals - via the Dedekind cuts, or by defining the Reals as a complete ordered field. Can you reccomend a good source to read on the construction via Cauchy Sequences? -- "It's like that old saying: A conservative is a liberal who's been attacked by aliens." Simon - mhm27x5[ Parent ]
 Try any general metric space book (5.00 / 1) (#339) by bgalehouse on Thu Jun 05, 2003 at 03:15:54 PM EST

 The argument is actually quite short. The completion of any metric space can be seen as the space of equivalence classes of Cauchy sequences where two sequences are equivalent iff the distance between the two nth elements of the sequences goes to zero with n. The paragraph above is perhaps hampered by the lack of subscripts in ASCII, and the fact that I'm feeling lazy today. There is a perfectly decent presentation at www.wikipedia.org. Scroll to the section called Completion. [ Parent ]
 Stupid mistake (none / 0) (#362) by flo on Fri Jun 06, 2003 at 12:23:07 AM EST

 I got the ultrametric inequality wrong, in several ways in fact. my brain must have been in screensaver mode. Sorry. It should read: |x+y| <= max(|x|,|y|), with equality if |x| != |y|. (Here | | denotes the p-adic absolute value). --------- "Look upon my works, ye mighty, and despair!"[ Parent ]
 The problem with completing reals... (5.00 / 1) (#390) by gzt on Sat Jun 07, 2003 at 06:12:00 PM EST

 ...from the rationals via Cauchy sequences is that the usual method of completing a metric space depends on the completeness of the reals! It's subtle, you know. Q/{0} is the way to go. [where {0} is the set of null sequences] Though, you probably meant that, but so often students try to complete the reals the same way they complete all other metric spaces, which is a grave sin. I love you for bringing up the p-adics, though. They're so cute! Have you ever had to prove they aren't ordered? And the integers are dense in the unit ball! And Bolzano-Weierstrass! So many sweet memories... [ Parent ]
 I'm not sure I see the problem... (5.00 / 1) (#406) by bgalehouse on Wed Jun 11, 2003 at 03:04:08 PM EST

 It seems to me that the only place where you used the completness of the reals in the standard completion strategy is in showing that the resulting metric is well defined. The argument usualy goes along the lines of: The sequence of distances between two Cauchy sequences is itself Cauchy. So by completness of the reals, this converges to a real value. However, if we are in the middle of applying this construction to the reals then we already have defined them as equivalence classes of Cauchy sequences and can therefore map our sequence to it's equivalence class. This strikes me as a sufficiantly trivial "fix" that I'm left wondering if I missed some aspect of the issue. [ Parent ]
 Shouldn't be too hard (none / 0) (#355) by scheme on Thu Jun 05, 2003 at 10:17:56 PM EST

 The proof in the abovementioned book that cardinalities are ordered is way cool, but probably too much for this board. (Actually, I think it is better to not use ordinals but to instead use the axiom of choice directly to do the last bit, but maybe I just have a thing against ordinals). As I recall ( I lost my K&F so I can't check) this proof is within the first 30 pages or so right? So it shouldn't be to hard to explain it :). Assuming of course you start getting fairly terse in your explanation and throw in a exercises for the reader. "Put your hand on a hot stove for a minute, and it seems like an hour. Sit with a pretty girl for an hour, and it seems like a minute. THAT'S relativity." --Albert Einstein [ Parent ]
 Blerg (5.00 / 1) (#151) by Pseudonym on Wed Jun 04, 2003 at 12:51:14 AM EST

 You're right, I got "dense" and "non-dense" round the wrong way. But you got what I meant. :-) sub f{($f)=@_;print"$f(q{$f});";}f(q{sub f{($f)=@_;print"$f(q{$f});";}f});[ Parent ]
 Cantor's diagonalization proof (5.00 / 3) (#153) by cgibbard on Wed Jun 04, 2003 at 12:53:03 AM EST

 A quick version of Cantor's diagonalization proof based on decimal representations of numbers is as follows. We can put the real numbers in bijection with the real numbers in (0,1). (This is an easy exercise.) Suppose we had a list putting all real numbers in (0,1) into correspondence with N. It might start out something like this: 0.532891.... 0.772189.... 0.233219.... 0.321743.... ... We can create a new number in (0,1) by defining it to be the following: The nth digit of x is 3 if the nth digit of the nth number in the list is not 3, and it's 7 otherwise. Now x is not equal to the first number in the list, or the second number, or any number we choose, since it differs from the nth in at least one digit (that being the nth). So x is not in the list. Contradiction. Note that we don't have to worry about the problem of multiple representations of real numbers (i.e.  0.9999... = 1.0000...) here, for the obvious reason that zeros and nines do not occur in the decimal expansion of x. [ Parent ]
 top notch (4.00 / 1) (#124) by scoby on Tue Jun 03, 2003 at 08:58:19 PM EST

 Excellent article. I have a fear of maths and I really enjoy seeing interesting maths subjects explained in english. Hope to see more tús maith leath na hoibre
 getting into tricky areas (2.00 / 7) (#134) by dh003i on Tue Jun 03, 2003 at 10:31:46 PM EST

 Your article is very good, but you're getting into some tricky areas when you try to compare different sets of numbers that are infinite. Mathematicians consider some sets of infinite numbers to be "larger" than others. There's a reason why I quote "larger": that's because I'm using it loosely. When I say that -- for example -- the set of infinite numbers represented by all integers is "larger" than the set of integers represented by all prime numbers, what I mean is that the set of all integers approaches infinity faster than does the set of prime numbers. That is, if if you plot a count of the number of integers numbers vs. prime numbers, the set of integers will grow at a faster rate than the set of prime numbers. (note, because both of these sets of numbers are discrete, I can't talk of slopes and derivatives, unless I say that you connect the dots with a line or curve). Whenever matheticians say one infinite set of numbers is "larger" than another infinite set of numbers, this underlying definition is assumed. The problem only occurs when people who don't understand that try to figure out wtf's going on.
 Your definition doesn't match mine... (5.00 / 2) (#139) by zml on Tue Jun 03, 2003 at 10:50:35 PM EST

 Whenever we talked about "larger" in terms of infinite sets (in all the math classes I had through college), we were relating cardinality and not this fuzzy growth rate you're talking about. I would not say that the set of integers is larger than the srt of primes, they're both countably infinite. I'm pretty sure you have to take things in context, so I wouldn't nitpick his use of the term "larger" here. His use is fairly canonical and also helps people understand (since the word "cardinality" begins to scare people away). [ Parent ]
 I'd dispute that (5.00 / 2) (#177) by zakalwe on Wed Jun 04, 2003 at 05:31:24 AM EST

 Whenever matheticians say one infinite set of numbers is "larger" than another infinite set of numbers, this underlying definition is assumed. If you said the primes were "larger", I would normally interpret that as "has larger cardinality than", which is clearly wrong. I don't think that your definition is very common, and it has a few serious flaws if its to be generally applicabile. That is, if if you plot a count of the number of integers numbers vs. prime numbers, the set of integers will grow at a faster rate than the set of prime numbers. But this isn't terribly useful, since we're talking about sets - things which are not ordered. You've introduced an extra bit of information with the "count of integers." (Presumably indicating the "natural" ordering of the integers) Present this count in a different order ( say interleaving a single non-prime after every 100 primes, and the "growth rate" difference is only 1% of what it was. If the sets are at least partially disjoint (eg even integers and prime numbers), the "count" can be arranged in different orders to show that either set is "larger" by this definition. Even if we take take definition as "using the ordering 1,2,3,4,...", that means its completely removed any generality, and can only apply specificly to Natural numbers. Whats the "natural ordering" for the set of rationals or complex numbers? Even an enumeration of the Integers probably woundn't seem that natural to non-mathematicians. [ Parent ]
 So close... (2.00 / 1) (#361) by dipierro on Fri Jun 06, 2003 at 12:02:56 AM EST

 Even integers and prime numbers are not disjoint ;). [ Parent ]
 I know (none / 0) (#365) by zakalwe on Fri Jun 06, 2003 at 05:11:05 AM EST

 Which is why I said "partially disjoint." [ Parent ]
 Damn (none / 0) (#368) by dipierro on Fri Jun 06, 2003 at 08:55:08 AM EST

 I shouldn't have been posting last night... [ Parent ]
 You are wrong. (none / 0) (#199) by joto on Wed Jun 04, 2003 at 08:39:03 AM EST

 No, you are wrong. The set of primes and the set of natural numbers have exactly the same size in mathematics. While there do exist different sizes of infinite sets, it is not their "growth rate" that determines their size. For more on sizes of infinite sets, look at this comment that I originally intended to put here, but moved to a top-level comment since I thought maybe more people should notice it. [ Parent ]
 is there a reason this is here? (1.40 / 5) (#148) by modmans2ndcoming on Wed Jun 04, 2003 at 12:12:19 AM EST

 the people that know what you are talking about will be to board to finish it, and the people who have no idea of what you are talking will be to board to finish it OR to intimidated to start it.
 Yes. (none / 0) (#189) by porkchop_d_clown on Wed Jun 04, 2003 at 07:41:24 AM EST

 Because people voted for it. -- I only read Usenet for the articles.[ Parent ]
 And (none / 0) (#232) by vernondalhart on Wed Jun 04, 2003 at 01:56:14 PM EST

 Because The Writer does a damn fine job of writing these articles. -- "It's like that old saying: A conservative is a liberal who's been attacked by aliens." Simon - mhm27x5[ Parent ]
 also (1.50 / 2) (#312) by Liet on Thu Jun 05, 2003 at 08:28:40 AM EST

 This story is a subtle hint that the author believes that he is smarter than the average K5er (ie you) and wants to show off. BTW whats the best way to factorise a quadratic when there is a number infront of the x-squared. I use the times the independent term by the co-efficient of the x-squared and devide the whole thing by that same coefficent the factorise method. But I don't really like it. Any suggestions. [ Parent ]
 Smarter? (5.00 / 2) (#325) by vernondalhart on Thu Jun 05, 2003 at 12:13:41 PM EST

 So you're saying that anytime anyone provides an informative article, that its really because they're just showing off how smart they are, and how dumb the rest of the world is? Well, its good you told me that. I'll start posting articles right away. -- "It's like that old saying: A conservative is a liberal who's been attacked by aliens." Simon - mhm27x5[ Parent ]
 Im glad you like to make stupid assumetions (none / 0) (#392) by modmans2ndcoming on Sun Jun 08, 2003 at 01:22:29 AM EST

 by your Math lingo I see you don't know what you are talking about...either than or you do not speak english well. [ Parent ]
 Which is worse: (3.50 / 8) (#149) by Kasreyn on Wed Jun 04, 2003 at 12:46:52 AM EST

 Fools who say that 1 divided by infinity equals zero, or fools who say that 1, followed by a decimal point and an infinite number of 9's, is equal to 2? -Kasreyn "Extenuating circumstance to be mentioned on Judgement Day:We never asked to be born in the first place."R.I.P. Kurt. You will be missed.
 Recurring nines (5.00 / 2) (#154) by Pseudonym on Wed Jun 04, 2003 at 12:59:01 AM EST

 I personally never found the standard way of handling "recurring nines" very convincing. Here's the way I visualise it now. Consider 0.1999..., which we might believe might be a different number than 0.2. Now convert both these numbers to binary. They have the same representation. Since there's nothing special about the decimal system, I can conclude that these numbers are the same, and just happen to have two different representations in the decimal number system. It's only a small step from there to convince you that 0.999... = 1.0. sub f{($f)=@_;print"$f(q{$f});";}f(q{sub f{($f)=@_;print"$f(q{$f});";}f});[ Parent ]
 Er (3.00 / 2) (#159) by DJBongHit on Wed Jun 04, 2003 at 01:41:24 AM EST

 Now convert both these numbers to binary. They have the same representation. They do in IEEE floating point binary format, sure, but only because it's an imprecise way of storing numbers. That's why programmers generally don't check floating point numbers for equality, and instead check for "close enough." It's not possible to represent an infinite set of numbers using 32 (or 64, or 80) bits. Unfortunately, the substandard way that CPUs represent floating point numbers in binary does not in any way prove that 0.99999 == 1.0, homeboy. ~DJBongHit -- GNU GPL: Free as in herpes.[ Parent ]
 They do regardless (5.00 / 1) (#161) by Pseudonym on Wed Jun 04, 2003 at 01:53:49 AM EST

 Precision doesn't matter. Go ahead and try it. Hint: 0.2 is a recurring number in binary.. sub f{($f)=@_;print"$f(q{$f});";}f(q{sub f{($f)=@_;print"$f(q{$f});";}f});[ Parent ]
 I'll try to explain the standard way for everyone. (5.00 / 3) (#164) by cgibbard on Wed Jun 04, 2003 at 02:09:06 AM EST

 The usual way of dealing with this problem is to use one of the actual definitions of the reals, either as limits of Cauchy sequences or as Dedekind cuts. There happens to be an actual definition of the reals in terms of decimal expansions, but it simply excludes values ending in infinite strings of 9s, or declares them to be equal to their matching strings that happen to end in all 0s. If you think of the reals as limits of Cauchy sequences, we just need to prove that the limit of the sequence {9/10, 9/10 + 9/100, 9/10 + 9/100 + 9/1000,...} is 1. A limit of a sequence of real numbers is defined as follows: If {a_1,a_2,...,a_n,...} is a sequence of real numbers (indexed by the naturals), we say L is the limit of the sequence if for every epsilon > 0 there exists a natural number N such that for all n > N, the distance from a_n to L is less than epsilon. That is to say that no matter how close you want the sequence to get to L, say you pick a positive number, call it epsilon, and you want it to be at least within (L-epsilon, L+epsilon), if you go far enough out (to N) it will be within epsilon of L at any point after that point in the sequence. Okay, so we get an epsilon and we have to prove that we can choose a number of 9s that makes 1 - 0.9999...9 < epsilon. To figure out how many 9s we're going to need, look at epsilon. Since it's positive, it must have at least one digit that's not zero. (0.0000... = 0 < epsilon) Say it's the kth digit. 1 - 0.99...99 (n nines) = 0.00...01 where the 1 occurs at the nth digit after the decimal point. Then take N = k, and we see that 1 - 0.99...99 (n nines) < epsilon for any n > N. So the sequence of rational numbers that 0.99999... represents has a limit of 1, and so does 1.00000... for the same (or simpler) reasons. [ Parent ]
 It's too easy! ;-) (none / 0) (#195) by Viliam Bur on Wed Jun 04, 2003 at 08:24:00 AM EST

 1.99999... = 2.0  0.99999... = 1.0 -1.99999... = 0.0 etc [ Parent ]
 Nines (none / 0) (#265) by Boronx on Wed Jun 04, 2003 at 07:18:43 PM EST

 1/9 = .111... 2/9 = .222... 3/9 = .333... Guess how many ninths 0.999... makes? Subspace[ Parent ]
 A simple way to think about it (5.00 / 1) (#176) by zakalwe on Wed Jun 04, 2003 at 05:01:32 AM EST

 In the spirit of the article, here's a nicely intuitive way to think of it. 10x - x = 9x  right? So let x = 1.999... 10x = 19.999... (with the same infinite sequence of decimals). So:   19.999...  - 1.999... = 18.0 So 9x = 18 therefore x=2, so 1.999... = 2 [ Parent ]
 NO! (5.00 / 3) (#185) by hex11a on Wed Jun 04, 2003 at 06:47:50 AM EST

 Hmm (none / 0) (#187) by zakalwe on Wed Jun 04, 2003 at 07:32:47 AM EST

 You talk about "infinite sequence of decimals" - the whole point of this article is that you can't define that! I' using 1.999... as shorthand for:       ∞ x=1 + ∑   9/10i       i=1 Which does seem reasonably well defined - working with infinite sequences and their sums isn't that unusual.  I'll admit that my original comment isn't very precice, and is quite possibly flawed but I'm not too clear on what the flaw is. I'll write it out a bit more rigorously in the hope that you'll clarify:          ∞ 10x=10 + ∑   10*9/10i          i=1          ∞    =10 + ∑   9/10i-1          i=1              ∞    =10 + 9 + ∑   9/10i-1              i=2          ∞    =19 + ∑   9/10i          i=1 What is the flaw?  I'm guessing the last step, but why? [ Parent ]
 Your very first line (none / 0) (#203) by hex11a on Wed Jun 04, 2003 at 09:01:34 AM EST

 1 + sum(9/10)^i = 2, as it is 1 + the sum to infinity of a geometric progression (=a/1-r, here a=0.9, r=0.1) So what you've said is that you're using 1.999... as shorhand for 2 and then proved from that that 1.999... = 2. Think about it like this: What is 2- 1.9? What is 2-1.99 etc. You find that the limit point of your sum is actually 2. So you've proved very nicely that your sum is 2, and then said that you're using 1.999... as shorthand for your sum. You can do that, but all you're doing therefore is defining 1.999... to be a series that sums to 2. Hex [ Parent ]
 OK (none / 0) (#215) by zakalwe on Wed Jun 04, 2003 at 11:08:41 AM EST

 I think I see. I thought you were saying there was something wrong with declaring the progression 1/10i =2, but you're actually saying that 1.999... should not be considered equal to the progression. I would have though that 1.9999... is just decimal notations way of writing that progression, but presumably there are areas where this is wrong. Thanks for clearing this up. [ Parent ]
 No, not wrong (5.00 / 2) (#290) by awgsilyari on Thu Jun 05, 2003 at 12:27:13 AM EST

 I would have though that 1.9999... is just decimal notations way of writing that progression, but presumably there are areas where this is wrong. You aren't "wrong," you've just discovered that there is more than one way to represent a number in decimal, if you are allowed to use an infinite number of digits. The thing is, you can "get hold of" the number 2, but you can't really "get hold of" 1.999... since you can't write down an infinite number of nines. In other words, when you allow the "..." notation (which is dubious), out pops these "weird" facts, like 1.999... = 2. It's just an oddity of the way we represent numbers. It's easier to see it in binary. 0.111111... = 1 in binary. 0.111... is just 1/2 + 1/4 + 1/8 + 1/16 + ... which is the well known series summing to 1. -------- Please direct SPAM to john@neuralnw.com [ Parent ]
 THANK YOU. (5.00 / 1) (#246) by Kasreyn on Wed Jun 04, 2003 at 04:00:35 PM EST

 I have been struggling with people who say "1.99... == 2" for years, and I've never had the tools with which to properly refute it. I'm not a maths major, the farthest I got was trig and that stretched my brain like taffy. =P Yet I just knew, somehow, that it was very very wrong to say that 1.99... was equal to 2. To me, there's a clear difference between a number which merely *approaches* another number, even if it is impossible to define how close it gets, and the actual other number. Thank you for giving me the tools to back up my gut feeling. I always hated conceding such arguments due to my math ignorance, even when I was sure I was right. -Kasreyn "Extenuating circumstance to be mentioned on Judgement Day:We never asked to be born in the first place."R.I.P. Kurt. You will be missed.[ Parent ]
 Well (none / 0) (#350) by hex11a on Thu Jun 05, 2003 at 07:34:22 PM EST

 In my ideal world you'd get taught about the beauty of certain areas of maths in high school to encourage you to go on and gain an interest in it - this stuff isn't conceptually too hard to grasp, but of course some ignoramuses would prefer it if school were simply a training factory. Ah well, c'est la guerre. Hex [ Parent ]
 It is! (5.00 / 1) (#188) by porkchop_d_clown on Wed Jun 04, 2003 at 07:40:53 AM EST

 for small values of 2. -- I only read Usenet for the articles.[ Parent ]
 Fools (5.00 / 1) (#224) by debillitatus on Wed Jun 04, 2003 at 12:30:03 PM EST

 Fools who say that 1 divided by infinity equals zero, or fools who say that 1, followed by a decimal point and an infinite number of 9's, is equal to 2? Well, I'd say that the first set of fools were worse, because they'd be wrong, whereas the second set wouldn't be fools at all, because they'd be correct. Damn you and your daily doubles, you brigand![ Parent ]
 How so? (none / 0) (#360) by dipierro on Thu Jun 05, 2003 at 11:52:33 PM EST

 They're both merely shorthand for something which is correct. [ Parent ]
 Wrong (5.00 / 1) (#315) by raygirvan on Thu Jun 05, 2003 at 10:54:29 AM EST

 "fools who say that 1, followed by a decimal point and an infinite number of 9's, is equal to 2?" Do you consider it possible that your intuition, rather than the result, could be wrong? Look in a school mathematics textbook and find the formula for the sum to infinity of a converging geometric series: a/(1-r). 1.999... (or 1.99 recurring if you like) = 1 + (0.9 + 0.09 + 0.009 + ...) The bracketed term is a converging geometric series. Apply the formula (hint: a=0.9, r=1/10). You're disputing a result that is so established in mathematics as to be a matter of fact, not opinion. No mathematician will agree with you. [ Parent ]
 Also... (5.00 / 1) (#326) by vernondalhart on Thu Jun 05, 2003 at 12:16:10 PM EST

 Another way to consider this is that in the Reals (or the rationals for that matter), given any two x != y, you can always find a number between the two of them. However, there is not a single number between 1.999... and 2. -- "It's like that old saying: A conservative is a liberal who's been attacked by aliens." Simon - mhm27x5[ Parent ]
 No, because (5.00 / 1) (#343) by Kasreyn on Thu Jun 05, 2003 at 05:39:48 PM EST

 1.999... is not a real number, as it cannot be defined. 1.999 IS a real number, being between 1.9989 and 1.9991. But it's of course obvious that 1.999 != 2. 1.999... !=2 because 1.999... is not and cannot be defined. The infinite repetition makes it impossible to define the number. It's not a real number. -Kasreyn "Extenuating circumstance to be mentioned on Judgement Day:We never asked to be born in the first place."R.I.P. Kurt. You will be missed.[ Parent ]
 Infinite series (5.00 / 1) (#354) by raygirvan on Thu Jun 05, 2003 at 09:38:34 PM EST

 > The infinite repetition makes it > impossible to define the number Sorry (actually, I'm not terribly sorry) but that is total pig-thick garbage. Infinite series are very common - fundamental, even - in mathematics. For instance, in the summation of a converging geometric series, and the definition of many trigonometric and exponential functions. If you have a problem with this, it's in your understanding, not in the mathematics. [ Parent ]
 The decimal expansion of 1/3 (5.00 / 2) (#376) by i on Fri Jun 06, 2003 at 06:50:59 PM EST

 is fascinated by your ideas and wishes to subscribe to your newsletter. —and we have a contradicton according to our assumptions and the factor theorem[ Parent ]
 How to believe infinity isn't a number (4.66 / 6) (#158) by sigwinch on Wed Jun 04, 2003 at 01:40:01 AM EST

 Confusion is too easy when you use monotonic equations. When x gets larger, 1/x gets closer to zero. So somebody can easily write 1/∞=0 by doing an unconscious limit process. "If x gets bigger forever, 1/x must get infinitely close to zero. Q.E.D." A solution might be to show students ∞ in a non-monotonic context like cos(∞). That expression has no obvious value. Even the "goes on forever" approach shows the correct path. --I don't want the world, I just want your half.
 gah (3.60 / 10) (#162) by lester on Wed Jun 04, 2003 at 01:55:52 AM EST

 re: gah (5.00 / 1) (#192) by The Smith on Wed Jun 04, 2003 at 07:57:43 AM EST

 Don't be too hard on him; he's given a very good introduction to a difficult and much-misunderstood branch of maths. Any slight misuse of words in the article was perfectly justifiable, given that it was written to explain transfinite concepts to people with only high-school maths background. [ Parent ]
 re: gah (none / 0) (#419) by fmazoit on Mon Dec 15, 2003 at 09:54:56 AM EST

 That does not allow him to say things that are not true. [ Parent ]
 What are you talking about? (5.00 / 1) (#240) by the on Wed Jun 04, 2003 at 03:16:49 PM EST

 nothing more than a convention among mathematicians I would have read your response but you chose to write in in English. This is merely a convention used by people in the US and other one time British colonies. Pity I didn't read it because if I had I would have pointed out how ignorant you are to deride someone for using the Axiom of Infinity when it is the axiom of choice (no pun intended) by mathematicians the world over. -- The Definite Article [ Parent ]
 plz read thx (5.00 / 1) (#279) by lester on Wed Jun 04, 2003 at 09:47:09 PM EST

 I would have pointed out how ignorant you are to deride someone for using the Axiom of Infinity when it is the axiom of choice (no pun intended) by mathematicians the world over. i did not deride him for using this axiom, i derided him for putting together a nonsense justification for it. you can use the axiom if you like, but it doesn't save you from the "problem" that constructive methods only give you potential infinites, it only substitutes one kind of potential infinity for another [ Parent ]
 so what instead of cardinality? (5.00 / 1) (#263) by city light on Wed Jun 04, 2003 at 07:13:50 PM EST

 I'm wondering what other sensible definition there could possibly be for comparing the 'size' of two arbitrary infinite sets? arbitrary being the operative word. We can't rely on things like one being a proper subset of the other in the general case. Or on things like ordinals, as the sets may have no order relations. It all has to end up relying on mappings. I mean, cardinality is really at the heart of all maths - in a way it's the most basic kind of isomorphism, like an isomorphism of sets before any structure (orders, relations etc) is added to them. So dismissing it as merely a convention is a bit misleading. Yeah it is just a convention, but in a way it's the only sensible choice for the general case. Mathematicians don't just pluck conventions out of the air... [ Parent ]
 maths and rhetoric (5.00 / 2) (#284) by lester on Wed Jun 04, 2003 at 11:49:51 PM EST

 I'm wondering what other sensible definition there could possibly be for comparing the 'size' of two arbitrary infinite sets? what about none? what hangs on R being "larger" than N anyway? (and do not give an answer which explicitly or tacitly replaces "being larger" with "having bigger cardinality") of course cardinality has its uses in math. i'm not pointing out any problem with that. what i'm pointing out is this haughty attitude that makes mathematicians continuously present themselves as having discovered some profound, unexpected truth here: "So you, vile ordinary person, thought the set of rational numbers was larger than the naturals? You fool, it is not! Here's the proof. Ain't we math people smart?" this is of course nonsense. the mathematician has not discovered some surprsing fact about the "sizes" of sets, she has defined a technical concept and applied it. the *rhetoric* she uses however is designed to disguise the banality of the "accomplishment" [ Parent ]
 intuition (5.00 / 1) (#351) by city light on Thu Jun 05, 2003 at 08:43:47 PM EST

 Mm. Well I'd go this far in defending cardinality - it's the definition which comes closest to expressing our intuitive notion of size without sacraficing generality. Obviously we can't unambiguously extend our intuitions about the meaning of 'size' to infinite sets because we only have a definite grasp on its common usage (& intuition about its 'meaning') for finite quantities. But we can look at various possible ways of extending our notion of size for finite sets, up and beyond, and it turns out cardinality is the most general and natural such extension. And so I do think there is some justification in going back to people and saying, yes, R is 'bigger than' N, according to the most general and abstract sense of the word that mathematicians can sensibly define. With |Q|=|N|, I'd say you need more of a disclaimer before saying they're 'the same size' - that this comparison ignores their structure and just compares them as sets. That one might draw a different conclusion from taking into account their order structures and the fact Q is dense, but that one can't in general rely on such structure to compare sets. It depends really if you think mathematics is just a game with axioms or if you think it's something higher than that, something that exists independantly of you or I and is discovered not invented. There's a name for both these positions (platonism/intuitionism vs constructivism? anyone? I'm not too sure on the names) You can argue well for either but I tend towards the latter - you get more of a sense of beauty from studying maths if you see it that way I think. Otherwise it just becomes another geek toy. [ Parent ]
 Game with axioms? (5.00 / 1) (#367) by raygirvan on Fri Jun 06, 2003 at 07:59:33 AM EST

 In a sense, it doesn't matter. Platonism says mathematics underlies the real world and is there to be discovered. Alternatively, a constructivist view of the history of mathematics might be that axioms have been chosen, from many possibilities, that steer mathematics toward usefulness on real-world problems. I get very tired of defending mathematics against the 'intuition' view: i.e. if it doesn't match someone's uninformed gut feeling, then the whole edifice must be fake or a conspiracy. When you routinely use Taylor series expansions, sums-to-infinity, integrals with infinite limits, etc, what do you say to someone who asserts that mathematics has no way to define an infinite series? [ Parent ]
 cardinality is cardinality, no more (5.00 / 1) (#389) by lester on Sat Jun 07, 2003 at 06:04:56 PM EST

 I get very tired of defending mathematics against the 'intuition' view: i.e. if it doesn't match someone's uninformed gut feeling, then the whole edifice must be fake or a conspiracy. if you present the edifice as being correct where intuitions aren't you should expect this. a.k.a. "cardinality isn't size, dammit, cardinality is cardinality" and you ought to take "uninformed gut feelings" more seriously. when somebody questions statements such as "the set of rational numbers is just as large as the set of natural numbers", the "gut feeling" in question is quite sophisticated: it involves a conventional identification of members of the two sets, and one which the notion of cardinality simply ignores. sure, you can give a procedure that establishes a one-to-one correspondence between the sets, but in general that correspondence won't assign a member of N the "same" number in Q (it *can't*) [ Parent ]
 Breakdown (5.00 / 1) (#398) by DrH0ffm4n on Mon Jun 09, 2003 at 09:10:09 AM EST

 if you present the edifice as being correct where intuitions aren't you should expect this It's more suprising to me that people don't expect their common intuition to fail them when dealing with abstract notions such as infinite sets, sub atomic particles and other supernatural entities. But I do enjoy watching the arguments that ensue all the same. --- The face of a child can say it all, especially the mouth part of the face. [ Parent ]
 The poor rationals (4.75 / 8) (#165) by arvindn on Wed Jun 04, 2003 at 02:11:02 AM EST

 As if being put into 1-1 correspondence with the integers weren't humiliation enough for the rationals, students of analysis get to see even more: the set of rational numbers has measure zero: which means that the set can be covered by a sequence of intervals the sum of whose lengths can be made arbitrarily small. If that surprised you, once you see the construction it will be obvious: let ε be any positive quantity. Label the rationals q1, q2, q3, ... Cover qn by an interval of length ε/2^n. Therefore the sum of the lengths is ε(1/2 + 1/4 + 1/8 + ...) = ε. Now make epsilon go to zero. Voila! Worse yet, there are sets of real numbers which are denser than the rationals (and hence can be put in 1-1 correspondence with the reals) which still have measure zero. The "Cantor Set" is an example. I guess The Writer will be taking up things like that in the next part. There is nothing more bruising to the ego than the realization that "obvious" things about the world around us are hopelessly wrong :-) So you think your vocabulary's good?
 #4-#5 (3.50 / 2) (#171) by ultimai on Wed Jun 04, 2003 at 03:42:02 AM EST

 Reading your article, it makes sense. But I had a hard time from Misconception #4 to the end because you write as if we have fully compriheneded everything before and now we can put all the backround into understanding #4 and #5. Repeating the same concept again somewhat more fully in relation to what you are proving, would help in comprihenshion. At least for me.
 In regards to your story about infinity (1.40 / 5) (#174) by Ta bu shi da yu on Wed Jun 04, 2003 at 04:38:58 AM EST

 I can see why you can write such an excellent article on infinity. It's because you go on, and on, and on, and on... Yours humbly (and forever), Ta bù shì dà yú ---AdTIה"the think tank that didn't".ה
 Ta bu shi da yu? (5.00 / 1) (#234) by Ruidh on Wed Jun 04, 2003 at 02:07:15 PM EST

 They who are not gentlemen/cultured are greatly amusing?   "Laissez-faire is a French term commonly interpreted by Conservatives to mean 'lazy fairy,' which is the belief that if governments are lazy enough, the Good Fairy will come down from heaven and do all their work for them."[ Parent ]
 Uh, no (5.00 / 1) (#236) by The Writer on Wed Jun 04, 2003 at 02:26:03 PM EST

 Not sure how you arrived at that, but if you look at his sig (which has tone marks on it---very important to avoid ambiguity), it means "He/she is not a big fish." [ Parent ]
 I guessed (5.00 / 1) (#250) by Ruidh on Wed Jun 04, 2003 at 04:23:59 PM EST

 Zhongwen offers a basic dictionary and I think I did respect the tonal marks (I trued to), but I'm obviously completely unable to chose from all of the alternatives presented. FOr example, I see 25 characters for shi. At least I've learned that be (verb) is a common choice. I really need to take lessons.   "Laissez-faire is a French term commonly interpreted by Conservatives to mean 'lazy fairy,' which is the belief that if governments are lazy enough, the Good Fairy will come down from heaven and do all their work for them."[ Parent ]
 Ambiguities (5.00 / 1) (#261) by The Writer on Wed Jun 04, 2003 at 06:31:22 PM EST

 One difficulty with Zhongwen is that the same syllable/tone combination can potentially mean a large number of things when taken in isolation. You have to interpret them in context; if you learn the grammatical rules, this will become much easier. [ Parent ]
 Why there are no tones in my username (5.00 / 1) (#304) by Ta bu shi da yu on Thu Jun 05, 2003 at 04:07:44 AM EST

 Kuro5hin doesn't allow me to use tone marks in the username. The other problem is that even my signoff is incorrect as I can't find an appropriate tone marker for the "a". The Writer is correct... I had a bit of a laugh at Ruidh's translation though. Quite apt - have a 5. Yours humbly, Ta bù shì dà yú BTW, try Zhõngwén. Again a bit of a hack... that õ should actually be a "o" with a line above it. ---AdTIה"the think tank that didn't".ה[ Parent ]
 Tone for "a"? (5.00 / 1) (#317) by The Writer on Thu Jun 05, 2003 at 11:07:43 AM EST

 You're referring to the "a" in "Ta"? That's just the ping tone, but the "a" is nasalized. Conventionally, a superscript tilde ~ is used for marking nasality, so you'd need both a ~ and a bar over the "a". But I've yet to learn how to combine multiple diacritics in HTML (or Unicode for that matter). Another thing I find annoying about Pinyin is that the "u" in "yu" is a different vowel from the "u" in "hu" (lake). And the "i" in "shi" (is) and the "i" in "si" (fine) are different, too. It makes the result really hard to read. OTOH, I don't blame the people who came up with Pinyin; there are just not enough vowels in the Roman alphabet to represent all those different vowels. Why can't those Europeans shape up their writing system... :-P [ Parent ]
 Interesting! (5.00 / 1) (#373) by Ta bu shi da yu on Fri Jun 06, 2003 at 01:33:07 PM EST

 I can't comment on the "nasality" of it. Well, I can... but in the interests of politeness towards the Chinese language I won't. :) What I do know is that ta in the context of ta bù shì dà yú (which btw, I've been told isn't actually gramatically correct) uses the 1st tone. "ta" by itself would make this a neutral tone, and to tell you the truth I don't actually think that it's a word in Mandarin! Alas, I am but a beginning xuésheng who has not progressed very far in the language for quite some time. This is despite the fact that I go to a Chinese church. (oops, I'm letting facts slip!) My only defense is that Mandarin is hard!!!! This being said, learning can sometimes be embarassing. I once said to a Ms Dong, "Dong laoshi" and instead of calling her "Teacher Dong" I ended up calling her "Dong rat". Yours humbly, Ta bù shì dà yú ---AdTIה"the think tank that didn't".ה[ Parent ]
 You lost me (3.14 / 7) (#181) by SanSeveroPrince on Wed Jun 04, 2003 at 06:26:08 AM EST

 I was following you, until you did not mention Bush, Iraq or individual rights even a little bit. Jokes aside, I am a bit miffed that the best article to grace K5 in the past 5-6 weeks passes me by and I don't get a chance to vote it up. For once the K5 readership showed good taste :) Postumous +1 ---- Life is a tragedy to those who feel, and a comedy to those who think
 What about omega numbers? (4.00 / 2) (#191) by galen on Wed Jun 04, 2003 at 07:47:32 AM EST

 There's another kind of infinite number that does follow the laws of arithmentic. Define omega such that omega is an element of N, but omega!=0, omega!=1, omega!=2, ... "But wait," you say, "that's inconsistent!" Even though the definition looks inconsistent, any proof that shows that the definition is inconsistent must be infinitely long, and thus not a valid proof. Since omega is a natural number, you also get other interesting numbers like omega+1, 2*omega, omega^2, ... Time flies like an arrow. Time arrows with a stopwatch.
 That's not very fair (3.00 / 1) (#198) by gazbo on Wed Jun 04, 2003 at 08:31:19 AM EST

 Define omega such that omega is an element of N, but omega!=0, omega!=1, omega!=2, ... any proof that shows that the definition is inconsistent must be infinitely long, and thus not a valid proof.So you're allowing infinitely long definitions, but not infinitely long proofs? Seems a bit unreasonable...
 Definition of omega (none / 0) (#213) by The Writer on Wed Jun 04, 2003 at 10:36:17 AM EST

 Well, his definition of ω isn't exactly correct. ω is, if he's referring to what I think he's referring to, identical to the set of natural numbers. That's the definition of ω. Therefore, no natural number is equal to it. ω is the first infinite member of the "ordinals". See my other comment for more details. [ Parent ]
 Yes (none / 0) (#214) by gazbo on Wed Jun 04, 2003 at 10:46:37 AM EST

 Your comment made sense - although I'm extremely rusty, I do have some set theory under my belt.
 Omega vs. transfinite ordinals (none / 0) (#286) by galen on Thu Jun 05, 2003 at 12:15:23 AM EST

 The definition I gave for ω is usually set up as an axiom schema as an exercise in logic. This ω isn't exactly the same as that of the transfinite ordinals, but it ends up having more or less the same properties. The big difference here is that my definition has ω being a finite natural number with indefinite value. Time flies like an arrow. Time arrows with a stopwatch.[ Parent ]
 Oh I get it (none / 0) (#320) by The Writer on Thu Jun 05, 2003 at 11:21:27 AM EST

 You're saying ω forms a different "line" of numbers which is in some sense "parallel" to the finite numbers? (I.e., since ω∈N, ω+1 must be also, etc., so the ω's form a series of numbers independent of the finite numbers.) I can see why this would be the same as the transfinite ordinals, since ω would essentially be a "limit ordinal" that begins a new series closed under the successor function. However, wouldn't ω be incomparable with all the finite integers? The definition of <, at least under Cantor's construction, is identical to ∈; so unless you also define ω to contain all the integers, it would not be "greater than" any integer. [ Parent ]
 Uncountable number of models? (none / 0) (#399) by DrH0ffm4n on Mon Jun 09, 2003 at 09:30:17 AM EST

 There are potentially an infinite number of alternative models for arithmetic. Some with just one parallel series, others with a larger finite number, yet others have countably infinite. This does not even depend on which axiom system that you use, but follows from Löwenheim/Skolem compactness & Gödel I. Complex naturals are one example. I've yet to find a proof from PA or ZFC why you couldn't have a model with uncountably infinite  parallel series and hence indeterminate cardinality of N. I'm leaning towards the fact that you could. I'm pretty sure it follows from the independence of GCH in ZFC (or is at least related). --- The face of a child can say it all, especially the mouth part of the face. [ Parent ]
 Ordinals! (5.00 / 2) (#200) by The Writer on Wed Jun 04, 2003 at 08:43:40 AM EST

 I believe you're talking about ordinals. I didn't want to get into ordinals in the article, 'cos it would take up too much space, and the article is a bit too lengthy already as it is. But if you want to deal with ordinals, the simplest way to do it is to define the natural numbers as follows: let '0' denote the empty set, and define the successor operation s(x) to be x∪{x}. (That is, x union the set that contains x.) It is easy to see that s(x) = {0}. We call this '1'. s(1) then is {0,1}, or {{},{{}}}, if you expand the sets within it. We call {0,1} the number '2'. We can apply s(x) successively, and we will get any natural number we want. Now notice that given any number n, if m
 Nah. (none / 0) (#323) by i on Thu Jun 05, 2003 at 12:11:28 PM EST

 You can't define omega such, because it would be inconsistent with Peano's axioms. In a very finite way. —and we have a contradicton according to our assumptions and the factor theorem[ Parent ]
 Need proof for existence of Omega. (none / 0) (#418) by Nebu Pookins on Wed Nov 12, 2003 at 05:00:33 PM EST

 any proof that shows that the definition is inconsistent must be infinitely long, and thus not a valid proof. Let's say for now that we accept your definition of proof (that it must be finite), and so we cannot prove that this number does not exist. I claim that you cannot prove that this number DOES exist either. Your definition is not a proof that it exists (it isn't a proof by construction, because you're simply stating properties of omega, not stating how to construct it). It is trivial to define something that doesn't exist, e.g.: Let P be a prime number from the set of natural numbers such that P is even and P != 2. I've define P without any ambiguity, so it's a perfectly valid definition, but that doesn't mean P actually exists. [ Parent ]
 Thank you. (3.66 / 3) (#196) by StephenGilbert on Wed Jun 04, 2003 at 08:24:42 AM EST

 I found this article very helpful. I've been trying to unlearn high school math for a while, which for me consisted of, "Use this formula, apply these rules, and voila: the Right Answer." It hasn't been hard, since I've forgotten most of it. As I suspected, math is far more interesting than most people have been led to believe. -------------------------------- Wikipedia: The Free Encyclopedia
 Math is easily made uninteresting (5.00 / 1) (#359) by obvious on Thu Jun 05, 2003 at 11:36:57 PM EST

 Judging by impression (none / 0) (#369) by The Writer on Fri Jun 06, 2003 at 10:32:53 AM EST

 [...] you really can't judge math by your impression of it in school (pre-college, that is). Just pre-college? I've had bad math teachers in undergrad, too. It didn't help that it was discrete math, which was an area I was weak in at the time. I couldn't understand why one has to be bothered with all the nitpicky details of logic and proof and what-not. Esp. not after I saw how the lecturer just walks in, doesn't wait for everyone to quiet down, and just rambles on and on in what sounded like a deliberately soft voice. He just brushes any questions aside, and often doesn't manage to finish the material he seemed to have planned to cover, but never bothering to go back and make up for it afterwards, and at the same time expecting us to learn the stuff ourselves. Needless to say, I didn't do very well in the course. Now, in 2nd year, though, I had a great discrete math teacher. He made me interested in the whole thing again. [ Parent ]
 Advice in such situations (none / 0) (#387) by cep on Sat Jun 07, 2003 at 03:05:15 PM EST

 Be a bad student -- as lazy as you can afford. Use the saved time for something else. Otherwise you will ruin your brain. [ Parent ]
 more about countable sets and the real numbers... (4.40 / 5) (#197) by joto on Wed Jun 04, 2003 at 08:28:18 AM EST

 The article only considered countable sets. But there do exist infinite sets that do have different sizes. I originally intended to post this as an answer to a different comment, but as it grew longer, I thought I'd put it in as a top-level comment. A countable set is one that is either finite, or enumerable by the set of natural numbers (0, 1, 2, 3, ...). The natural numbers is thus, by definition countable (it's what we use to count, silly!) Let's look at a few examples: The set of whole numbers (...-2, -1, 0, 1, 2...) is countable because you can count them. For example, as this 0: 0, 1: -1, 2: 1, 3: -2, 4: 2... The set of prime number is countable, because you can count them in the same way: 0: 2, 1: 3, 2: 5, 3: 7, 4: 11... (well, as long as you can find a proof that there are infinitely many primes, which I'm sure exists) The set or rational numbers is countable, because you can enumerate them in a sequence by weaving through them (skipping duplicates) such as this (I'm dropping the naturals now to save space): 0, 1/1, -1/1, 2/1, -2/1, (skipping +/- 2/2) 1/2, -1/2, 3/1, -3/1, 3/2, -3/2, (skipping +/- 3/3) 4/1, -4/1, (skipping +/-4/2) 4/3, -4/3, (skipping +/-4/4) 1/4, -1/4, (skipping +/-2/4) 3/4, -3/4, ... The set of letters in the alphabet is countable, because we can enumerate them in the same way: 0: a, 1: b, 2: c, ..., 25: z. A set most certainly doesn't have to be infinite to be countable (on the other hand, being finite and not countable sounds weird, and is of course not possible). And so on... I'm sure you can come up with more examples. All sets that are countable and infinite are considered to have the same size. There are (of course) people who disagree with this. Normally, we tend to label them as either not knowledgeable about mathematics, or in a few cases as someone that might be clever enough to reinvent it in a new and better way. But this is the standard view. The question then is: Does there exist sets thare are infinite but not countable? This is an interesting question, because so far all infinite sets seems to have the same size (even the rationals). As it turns out the answer is 'yes'. One example is the real numbers, or to be more precise: any countinuos interval of real numbers (without getting technical, the word continuos here means, any interval of neighbouring real numbers, the interval [1] consisting of only one number is not good enough, but [1, 1.01] is.) To show you why, consider the interval [0, 1> (the right bracket means that we don't need the final 1 now). We can try make a (hypothetical) table putting those numbers (in numerical order) under each other, with their decimal expansion (which is infinite) to their right. For simplicity, let's restrict ourselves to the binary number system, using only the digits 0 and 1, such as this: 0.000000000000000000000... some missing numbers here 0.000000000000000000001 some missing numbers here 0.000000000000000000010... some missing numbers here 0.000000000000000000011... lots of missing numbers here 0.111111111111111111111... If this was possible, then we had shown that this interval of real numbers was countable. As it turns out, there is a clever counter-argument to why this is not possible. We can find a missing number in our table by a trick known as diagonalization. Assume we have the full table. We shall construct a number starting with 0, that is not in the table. Before the decimal point, we start with a 0, as usual. Then we pick each consecutive digit by taking the opposite of what the digit at that place (diagonally from left top to right bottom) in the table has. Our first digit after the decimal point will then be 1 (since the first number starts with 0.00...(because the first digit after the decimal point in the first number. The second number will also be 1 of course (since the second number in the table starts with 0.0000... Eventually we'll end up with a an infinite string, starting with 0. and having an infinite sequence of 0's and 1's behind it. Now, this is a number that can't be in our table! Because of the way it is constructed, it can't be first in the table (it differs in the first digit after the decimal point), and in general, it can't be in the nth place in the table (since it differs in the nth digit after the decimal point. Our hypothetical table enumerating all the real numbers in [0,1> was not possible to construct after all! And since we couldn't do that, the real numbers in that interval is (by definition) not countable. Ok, if you are still with me, we have now shown that there is at least two different sizes of infinite sets. The countable, and the uncountable (the real numbers have the same size as the set just discussed, but I'm not going to show you why). Does there exist more? Maybe? I'm not sure whether mathematicians have been able to construct sets with more elements than the real numbers, but it doesn't seem too unlikely. But you'd better ask a real mathematician about that. But what I do know, is that we are still looking for a proof of whether there exists or don't exist something in between the sizes of countable sets, and sets of the same size as the real numbers. This is still an unsolved problem.
 Primes and the continuum hiptothesis (none / 0) (#204) by Lev Black on Wed Jun 04, 2003 at 09:29:58 AM EST

 firstly,congartulations on a great article. now. there is an infinite number of primes proof: suppose you could enumerate all the primes P={P1,P2,....Pn} look at the number Qn=P1*P2*...*Pn + 1. either Qn is a new prime, or Qn is a composite, in which case there exists a prime Pn+1 which is not in the set P, which divides Qn. in either case we have found a new prime which is not in the original set. second. for each set A we can define the power set P(A) as the set of all the subsets of A. it is easy tho show that card(P(A))>card(A) therefore, there is an "infinity" of different types of infinities.... sometimes in the early 1900's (dun remember the year exactly, think it was 1908) Cantor stated the hypothesis that Alef one (the next cardinality after alef zero) is equal to the cardinality of 2^A (a shorthand for P(A)) later Godel & Paul Cohen (separately, and well actually they showed different things) showed that it is impossible to prove OR disprove this hypothesis from the axioms we have. therefore, you can't answer the question whether there is something "between" the rationals and the real numbers :(|Q|<|A|<|R|) [ Parent ]
 Thanks.. I had forgotten about that. (none / 0) (#208) by joto on Wed Jun 04, 2003 at 10:11:03 AM EST

 The existence proof of an infinite number of primes is quite clever. I thought I should be able to whip up something, but thought about it the wrong way (you don't have to find them all). When you mention it, I now remember the power sets. Do you remember the "simple" proof for their cardinality being higher too? This infinity stuff is really tricky... I've never really learned about this stuff of alephs and so on, but I guess it's just a funny notation for the insight given by using power sets to construct sets of higher cardinality. The proof that the continuos-hypothesis must remain a hypothesis is also beyond me. One more reason to keep watching kuro5hin, hoping that "The Writer" will pump out some more of his excellent stuff. [ Parent ]
 Cantor's Theorem (5.00 / 3) (#221) by QED Duh on Wed Jun 04, 2003 at 12:01:12 PM EST

 Terminology used: Power Set - For any set A, the power set of A, denoted P(A), contains every possible subset of A.  For finite sets with n elements, the power set contains 2^n elements.  For example, consider A = {0, 1}.  P(A)={ {}, {0}, {1}, {1,2} } Note: Every element in P(A) is a subset of A, but not necessarily an element of A.  So {0} is an element of P(A), and {0} is a subset of A, but not an element of A. Surjective Map - A function f from A to B such that every element in B has at least one element of A that maps to it.  Example:  Let A = { 1, 2 } and B = { 1, 2, 3 }.  There isn't a surjective map from A to B (exercise left to the reader), though there is at least one from B to A by the pigeonhole principle. Image of an element - For some map from a set A to a set B, the image of an element x in A is what it maps to in B.  That is the image of x is f(x) which is in B. Cantor's Theorem: There does not exist a surjective map from a set A to P(A). Assume that there was such a map called f from A to P(A).  Consider the set D which contains of all elements who are not contained in their image.  That is, for a in A, f(a) is an element of P(A) and thus f(a) is a subset of A.  Since f(a) is a subset of A we can ask ourselves the question, "Is a in its own image?", that is, is a in f(a). Now D is a subset of A by construction, which means that D is an element of P(A).  Since our map f is surjective, then there must be an element x such that f(x) = D.  Now comes the fun part.  There are two possibilities for x; either x is in D, or x is not in D.  This is where we derive a contradiction. If x is in D, then by definition x is not in f(x).  But f(x) is D.  x cannot be in D and not in D at the same time, so this case yields a contradiction. If x is not in D, then x is not contained in the image of x under f.  That is x is not in f(x).  However this is precisely the condition to be a member of D, so x is in D.  Again, this yields a contradiction. Since both cases give us a contradiction, we're forced to conclude that our original premise is faulty and that there cannot be a surjective map from A to P(A). Now that we know there cannot be a surjective map, we know that there is at least one more element in P(A) than there is in A, so P(A) is strictly bigger than A.  This lets us construct arbitrarily large sets once you assume the Axiom of Infinity.  If you have N, the natural numbers, then you know that P(N) has a larger cardinality than N.  By the same argument, P(P(N)) has a larger cardinality than P(N), and so on. [ Parent ]
 I'm not a mathematician by any means (none / 0) (#205) by Control Group on Wed Jun 04, 2003 at 09:33:57 AM EST

 (IANAM?) I'm pretty sure, though, that there are either an infinite number of cardinalities of infinite sets, or at least an arbitrarily large number. As I understand it, the power set of an infinite set is of higher cardinality than the set (this makes intuitive sense). So, if the set of the reals is cardinality c, the reals' power set is of greater cardinality than c. Then, of course, the power set of the power set of the reals is a yet higher cardinality, etc. Like I said, I'm by no means good with math, so if someone would like to chime in with a more formal/more accurate explanation (or a debunking of my pseudo-explanation), please do. *** "Oh, nothing. It just looks like a simple Kung-Fu Swedish Rastafarian Helldemon."[ Parent ]
 Cardinalities of Infinite Sets (none / 0) (#331) by bwcbwc on Thu Jun 05, 2003 at 01:13:54 PM EST

 One problem with uncountably infinite (like the reals) vs. countably infinite sets ( sets that are 1-1 with N), is that uncountably infinite sets don't really have a cardinality, since cardinality implies some way of "counting" the elements or the existence of a 1-1 mapping from the set to N. Uncountable sets like the reals, or an interval of the reals (for example [0,1) ) have a generalization of cardinality called measure. Generically, the measure of a set is just a function that somehow defines the size of a set, so cardinality is a valid measure for countable sets. There are certain conditions that the function has to fill (for example if set A contains set B, then meas(A) >= meas(B) ). The most commonly used measure other than cardinality is probably the "Lebesque" measure used to measure uncountable sets and perform calculus operations on sets that can't be handled by normal college calculus (Riemann). The Lebesque measure of some uncountable sets is actually closer to everyday experience than some of the other concepts discussed here. For example the measure of any interval [0, n) is simply n, just like on a ruler. However, there are complications (there always are, aren't there). One reason for the complications is because the Lebesque measure of any countable set (finite or infinite) is zero. This means, for another example, the measure of the IRRATIONAL numbers (Reals that aren't rational) over an interval [0,n) is also n, because the rational numbers are countable, so the measure of rational numbers over any interval is zero. [ Parent ]
 Uh... (none / 0) (#334) by The Writer on Thu Jun 05, 2003 at 01:35:19 PM EST

 I'm sorry, but "cardinality" applies to arbitrary sets. Your definition of cardinality would only ever allow one infinite cardinal, which is ℵ0 itself. As for measure, Lebesgue measure is only intuitive over the real numbers. You can, of course, have other kinds of measures. I don't know enough measure theory to come up with one off the top of my head, but conceivably you could come up with a measure that is non-zero over the rational numbers, yet always "larger" over any uncountable subset of the real numbers. Also, I suspect that measures may not quite behave the way you expect them to when applied to sets larger than the real numbers (for example, ℘(ℜ)). I might be wrong on this, though. [ Parent ]
 Lebesque is wrong (5.00 / 1) (#379) by fifi on Fri Jun 06, 2003 at 07:39:43 PM EST

 That would be Lebesgue... [ Parent ]
 proof that card(R) = card(P(N)) = aleph_1 (5.00 / 1) (#266) by Sacrifice on Wed Jun 04, 2003 at 07:18:47 PM EST

 Represent any real number in binary positional notation. Construct a subset of Z (integers) containing the positions of the one digits (for the digit with value 2^n, n is in the set) For example: ... 00101.100 ... = 2^2 + 2^0 + 2^-1 <-> {-1,0,2} This one-to-one mapping shows that the reals have the same cardinality as the powersets of integers, which have the same cardinality as powersets of natural numbers, or aleph_1. [ Parent ]
 minor correction (none / 0) (#270) by Sacrifice on Wed Jun 04, 2003 at 07:54:19 PM EST

 Infinite binary strings with a point in the middle aren't exactly one-to-one with reals, since the sum defined with infinite preceding ones doesn't converge to a finite number, and infite trailing ones are redundant, that is, 0.111111.... = 1.0000.... However, you can easily show "legal infinite binary strings" and "infinite binary strings" have the same cardinality.  For every legal string there are three other illegal infinite strings (with infinite leading zeros, trailing zeros, or both).  To map an arbitrary infinite binary string (with a binary point in the middle), legalize it by chopping off all but one of the infinite preceding and trailing ones, and then insert two marker bits somewhere recording the presence or absence of infinite leading and trailing ones (say, to the left and right of the binary point, respectively). For example: ...11111111111101.0 -> ...01011.00... ...0.0111111110... -> ...00.1010... ...01.0.. -> ...010.00... Of course, this is just a funny way of showing that the cartesian product of Powerset(Z) with {00,01,10,11} is the same as the cardinality of Powerset(Z). In general, you can diagonalize to show that card(A x B) = max(card(A),card(B)).  (Like how positive rationals are NxN, and have the same cardinality as N).  Or, three dimensional real space (RxRxR) has the same cardinality as reals (R). [ Parent ]
 aleph_1 (none / 0) (#277) by The Writer on Wed Jun 04, 2003 at 08:47:42 PM EST

 Be careful about claiming that card(℘(N))=ℵ1. This is the Continuum Hypothesis, which has been proven to be orthogonal to the axioms of set theory (i.e., it cannot be proven true or false from the axioms of set theory). Most mathematicians in fact incline against CH, so I wouldn't go around claiming that the cardinality of ℜ is ℵ1 without qualifications. [ Parent ]
 i like this! (5.00 / 1) (#309) by the sixth replicant on Thu Jun 05, 2003 at 05:59:29 AM EST

 "...to be orthogonal to the axioms..." it made my head spin in a very abstract way :) Ciao [ Parent ]
 Orthogonality (none / 0) (#314) by The Writer on Thu Jun 05, 2003 at 10:45:27 AM EST

 I guess I'm just making an abstraction of the axioms as "basis vectors" with which one constructs the mathematical universe ("vector space"). In that sense, CH is orthogonal to the axioms; assuming CH or its negation results in a slightly different "vector space". :-) [ Parent ]
 It's an abstracted concept: (5.00 / 1) (#330) by hex11a on Thu Jun 05, 2003 at 01:00:10 PM EST

 Orthogonal to really can be taken to mean "independent of". We can talk about mass being orthogonal to lengths, or whatever without any incoherence this way - it just sounds neat :) Hex [ Parent ]
 Well done (3.60 / 5) (#201) by rweba on Wed Jun 04, 2003 at 08:59:28 AM EST

 I think this is your best work yet. You have a real knack for simplifying math. I was already somewhat familiar with the ideas in this article, but I enjoyed the way you presented them. It has inspired me to make a plan to learn more about this stuff(Set theory? Real Analysis?) at some  point in the future. I am looking forward to your future articles on similar topics. Ignore the nay sayers. You submit them, and us math freaks will vote them up!
 I agree! (5.00 / 1) (#206) by halo64 on Wed Jun 04, 2003 at 09:38:57 AM EST

 Although I'm not a "math freak," I have always been fairly proficient at it and wish my teachers and professors had taken the time to explain things in this manner. Maybe if instructors put more thought and effort into their lessons such as this essay, kids in school would be more likely to enjoy learning rather than looking at it as a dull requirement. /* begin sig hereI don't have one because I'm lamefinish sig here */[ Parent ]
 Math teaching (5.00 / 1) (#209) by The Writer on Wed Jun 04, 2003 at 10:29:20 AM EST

 I have to agree. I was fortunate enough to have high-school math teachers who knew what they were doing, and who knew how to make the subject interesting. But I also had bad teachers in subjects like history. That scarred me for life, at least as far as those subjects were concerned. Even today, I still can't shed the connotations of "boring", "tedious", "painful rote memorization" every time I hear the word "history". I'm starting to believe that in fact every subject is interesting; it's just a matter of how it's taught. In spite of my acquired aversion to historical subjects, I do find it very interesting to learn about historical developments once in a while. One of my memorable experiences was when I took a math elective in undergrad, entitled "The History of Mathematics". The lecturer was awesome; he knew the stuff like the back of his hand, and the course felt like a joy ride over the fascinating landscape of the historical development of natural numbers, the acceptance of zero as a number, negative numbers, rationals, real numbers, imaginary numbers, and even quaternions and octanions, which are usually rather dreary subjects. [ Parent ]
 History and math (5.00 / 2) (#307) by arvindn on Thu Jun 05, 2003 at 05:12:08 AM EST

 I couldn't agree with you more. I'm from India, and high school history here are as interesting as a telephone directory. I remember kicking my history books across the room when I was done with my exams. In university, I took a history course because the prof was reputed to be good, and man it was amazing. Before the end of that sem Indian history became my favorite subject, and I've read a lot of history since then. OTOH, math was my favorite subject in school (I was in IMO99). I'm majoring in CS, but enrolled for a summer math program. That pretty much killed my interest. They were supposed to feed you a year's worth of higher mathematics in two months in the summer, and so it was basically an assload of abstractions and theorems in topology, algebra, functional analysis and other things and you were supposed to motivate yourself. It was so bad I almost hate math now :-( Indeed, I'm now working in cryptography, and when my prof reviews my research papers he says "This looks like a novel! Where are the _theorems_?". It's all f'ing incredibly ironical. So you think your vocabulary's good?[ Parent ]
 Fukc u all (1.03 / 29) (#299) by diab0lus on Thu Jun 05, 2003 at 02:22:38 AM EST

 fuck u all [ Parent ]
 there was something we said when i was a child (4.50 / 2) (#216) by ironfroggy on Wed Jun 04, 2003 at 11:09:41 AM EST

 and lately im finding it is still very relavent... "Only fools are positive." -- Question
 You know it's good stuff when . . (4.00 / 4) (#218) by Dphitz on Wed Jun 04, 2003 at 11:35:14 AM EST

 a math moron like myself can enjoy the article (and even understand much of it).  Perhaps if my high school teachers had the skill to explain math in this fashion I would have passed it the first time around (skipping class didn't help much either).   The standard method of teaching is to say, "just plug in this formula and solve it".  Well that simply teaches you how to remember what formula to use, not how it works.  Which is why I forgot most of it 10 minutes after I graduated. I'm looking forward to the next part of this.  I'll set aside a +1FP for it. God, please save me . . . from your followers
 well... (none / 0) (#383) by valar on Sat Jun 07, 2003 at 03:15:19 AM EST

 Having gone through the traditional high school mathematics (up to a college level true-multivarible class) as well as modern algebra, linear algebra and topology, I can tell you why they don't teach you this in normal high school mathematics classes (algebras, geometry, trig, etc). They are trying to teach you math you might need as an adult in a non-mathematics related job. They get to the mathematicians in college. Most people don't need to know this kind of thing (though it is very neat). What most people (business people, surveyors, and even less math related professionals) is the ability to solve formulas that they can look up in a book (and let's face it, in the "real world," you can always use the book). So the traditional general education style focuses on this skill. [ Parent ]
 actually (none / 0) (#394) by cronian on Sun Jun 08, 2003 at 11:49:38 AM EST

 I recall they taught the same class about four times in my high school. First it was Algebra then Algebra II then Trignometry and then the first quarter of BC Calculus. Actual Trig was mixed in there, but it only took about a week. We perfect it; Congress kills it; They make it; We Import it; It must be anti-Americanism[ Parent ]
 High school "mathematics" (5.00 / 1) (#408) by sheafification on Thu Jun 12, 2003 at 05:45:15 AM EST

 Mathematics involves making clear and unambiguous arguments, dealing with multiple levels of abstraction, and problem solving skills. This is the mathematics you will need as an adult in any job where you are actually required to think. Abraham Lincoln studied Euclid's _Elements_(which was essentially a geometry text book) to become a better lawyer. He certainly didn't do this because he thought that he could use geometry as a lawyer. He did it to pick up the mathematical skills which I listed above. Think about the distinction between traditional high school mathematics and your undergraduate mathematics major courses. Typical problems from these courses would be "find all of the roots of -2x^2 + 5x + 7" and "prove that every subgroup of a free group is free". The distinction between the problems is not just that most people will never have to consider free groups in their daily lives and that they might need to find the roots of a polynomial. The solution to the first problem is essentially the application of an algorithm. The solution to the second problem is much more involved. Just to be clear, I am not saying that learning how to use formulas that you can look up in a book is not important for some professions. What I am saying is that studying mathematics offers an opportunity to pick up far more important skills. In my opinion, these skills are not emphasized in high school courses because they are very hard to teach, and even harder to evaluate, especially in a national or statewide system. I think that if you transition the high school courses to a format which emphasizes these skills then you provide much more opportunity to investigate more interesting areas of mathematics. [ Parent ]
 Encore, Encore! (3.33 / 3) (#233) by melia on Wed Jun 04, 2003 at 01:57:49 PM EST

 I have to study some maths next year, I only learn stuff i'm interested in, and this makes maths seem interesting. I love it when you get to enjoy doing something constructive. Disclaimer: All of the above is probably wrong
 hmmm (4.00 / 5) (#253) by myvitriol on Wed Jun 04, 2003 at 05:32:00 PM EST

 sorry to be arsey, but you've got to expect it from a math major (actually maths is the only subject i do since I'm in the UK). Though it's useful to remember that infinity is not a real 'number', it is used all over the place in mathematics, in perfectly legitimate ways. I.e. you can perform well defined calculations on them. Someone has already mentioned, surreal numbers, where a whole new class of numbers are constructed from the real numbers, and intuitively represent infinitessimals. Though this is really more of a curiosity these days, and there are many more useful examples. Another class is the ones referred to in the article, the cardinal numbers which provided you accept the axiom of choice have a well-defined structure, which I'm guessing he's gonna talk about next time. However, another important one is C_\infty which is the complex plane adjoined with infinity. It's actually homeomorphic (can be wrapped onto smoothly) to he surface of a sphere, and is an elegant way of looking at complex analytic functions. Basically wherever a complex-valued function has a pole, we say it takes the value infinity. This simplifies the discussion of, for example, mobius maps which are smooth bijections of the complex plane. These are often used in conformal mapping problems, which are useful for solving laplaces equation, which inturn is used for solving electrostatic problems! (If you're looking for an application) Another nice example might be [0,\infty) u \infty which is the positive real line adjoined with the point at infinity. We can endow it with a topology that gives it certain nice properties, like a sequence tends to \infty if and only if it tends to \infty in the traditional sense (i.e. it _stays_ arbitrary large after a certain time). Anyway, don't really know what the point of that was. Just to say, infinity is in a very well-defined sense a number. Ultimately the problem with the teaching of maths in school is it's too ad-hoc. If they taught limits even vaguely properly, there wouldn't be such misunderstanding.. Also, here's a cool proof about infintiy and countability. Suppose X is a set and P(X) is the set of all subsets of X (the powerset). We wish to prove that in the sense of the article, P(X) is bigger than X. Namely, there is no surjection from X to P(X) (a function which maps to every point in P(X)). Suppose there were - call it f - then consider a subset of X - call it Y - of elements that are not contained in their image (remember the image of an element will be some subset of X). Then since we claimed that this function was surjective, there is an x such that f(x) = Y. Then we ask whether or not x is in Y. if x is in Y then it's not in f(x) = Y! Further if x is not in Y then it is in f(x) = Y! Clearly a contradiction and there is no such surjection. cheers, vitriol
 Only a curiosity? (5.00 / 1) (#306) by arvindn on Thu Jun 05, 2003 at 04:49:07 AM EST

 Someone has already mentioned, surreal numbers, where a whole new class of numbers are constructed from the real numbers, and intuitively represent infinitessimals. Though this is really more of a curiosity these days, and there are many more useful examples. I assume you are referring to nonstandard analysis? I'm sorry to hear its only a curiosity, because I once read a layman's intro to nonstd anal., and from what I understood it looked like a very cool thing, a very nice way of avoiding loads of formalism by going outside the set of reals. So can you please tell me why it isn't used more often? So you think your vocabulary's good?[ Parent ]
 Nothing wrong with formalism (5.00 / 3) (#310) by jonathan_ingram on Thu Jun 05, 2003 at 07:19:56 AM EST

 When the formalism is already in place, and you're confident that it works, there's no real reason to not use it. Particularly because in normal life you *can* mostly ignore the formalism, and go back to the intuitive ways of calculating limits, etc. Basically, the first year or so of undergraduate mathematics consists of destroying all your A-level preconceptions, then rebuilding the framework to demonstrate that the A-level methods work correctly, but for deeper reasons. After that, you can, if you want, forget about the formalism 95% of the time, but it's there to support you if you meet an edge case. In any case, 'surreal numbers' are not nonstandard analysis (very little progress has been made on integrating surreal numbers, for example). -- Jon[ Parent ]
 curiosity (4.50 / 2) (#342) by myvitriol on Thu Jun 05, 2003 at 05:35:37 PM EST

 i was indeed referring to non-standard analysis (actually i think they're called hyperreals, not surreals, oops). Of course this is just my opinon, but i think there are a few reasons why it's not part of mainstream analysis. One is the construction relies on some very advanced (and very non-constructive!) set theory, whereas the construction of the reals is comparatively simple and can be understood by a first year maths undergraduate. Then once you've proved theorems in non-standard analysis, you need a fairly advanced theorem in mathematical logic (probably graduate level here!) to pull it back to a meaningful result in standard analysis. You could take the view that you just take the formalism you get from non-standard analysis for granted, and use it to prove results. But that's quite unsatisfying for a math major! and it's arguably easier to use the epsilon-delta formalism, or whatever you want to use. Not that I want to detract from the beauty of non-standard analysis, I don't think it will really ever be an integral part of mathematics, in the way real analysis is. [ Parent ]
 Agreed (none / 0) (#371) by raygirvan on Fri Jun 06, 2003 at 01:07:16 PM EST

 Yep: there are plenty of circumstances where infinity can be plugged into functions that take numerical arguments (for instance, limits of summations or integrals). Granted, it's a shorthand for a limit, but for practical purposes it behaves as a number with special properties. I'm not terribly keen on misconception #2 either: I don't see any problem with the "goes on forever" view, except that it seems to faze some people who don't grasp that a series can be infinite conceptually even if you can't write it out in full. Infinite series expansions are massively commonplace in mathematics. [ Parent ]
 hmmmmmm (none / 0) (#407) by sheafification on Thu Jun 12, 2003 at 04:48:48 AM EST

 You should have said that infinity is not a 'real number'. It is perfectly reasonable to consider \infty and -\infty as numbers. Just to consider another example, it is possible to define the extended reals R\union{-\infty,\infty}. This is a perfectly valid numbering system. If you want to actually perform arithmetic, then you are going to have to define what happens when you involve those \inftys in operations, but this is rather straight forward. Of course, you don't get a field, but in many situations you really don't need a field. This construction is used throughout big Rudin(I think he actually formulates it in chapter 0). There are also quite a few other errors in the article unfortunately. I would suggest looking in the wikipedia for a more accurate description of infinity, transfinites, construction of the naturals, etc. I imagine that one of the mainstream mathematics writers(Paulos, Devlin, Peterson, etc.) has covered these topics in an approachable manner, but I am not prepared to give a specific referenece. [ Parent ]
 Infinity has no value... (5.00 / 1) (#413) by malfunct on Fri Jun 20, 2003 at 09:18:49 PM EST

 I think thats what the author should have said. 6 has a value, well defined, we can add 3 to it and get another value (9). Infinity has no value, what do you get if you add 3 to infinity? Hopefully a slap from your math teacher because its nonsense. Anyways thats the point the author was trying to get across (or thats what I got from the authors explanation). [ Parent ]
 Definiton of "value" (none / 0) (#417) by Nebu Pookins on Wed Nov 12, 2003 at 04:45:49 PM EST

 I think the problem with this definition is that we then have to define what we mean by "having value". That's the way it is in formal maths, you always have to define all the terms you use. =/ Note that I could define a set which includes all the integers and which also includes infinity, and define operations appropriately so that the set is closed over addition and multiplication. (I believe that such sets are called rings, though perhaps I'm confusing them with fields or some other term.) Those operations work in the usual way when they involved two integers. When these operations involves my element "infinity", I use these definitions: n + inf = inf inf + inf = inf n * inf = inf inf * inf = inf I believe that my above proposed definitions are very compatible with our "intuitive" sense of what infinity is. [ Parent ]
 cardinals versus "infinity" squiggle (4.33 / 6) (#264) by jejones3141 on Wed Jun 04, 2003 at 07:15:14 PM EST

 Good article, but it confuses infinite cardinal numbers with the little "figure eight on a side" squiggle ∞ that gets used in analysis. The author nails it—∞ is not a number; it's just that mathematicians define some uses of that squiggle in terms of limits. If f(x) is some function of x, then when we say f(x0) = ∞, we really mean the limit of f(x) as x → x0 = ∞, which in turn just means "No matter how big a number M you pick, I can hand you an epsilon such that for any x within epsilon of x0, but not equal to x0, f(x0) > M." If we say f(∞) = a, we really mean the limit of f(x) as x → ∞ = a, which in turn just means "Tell me how close to a you want to get, and I can hand you an M so that for any x > M, f(x) is that close to a." So, ∞ is just a convenient shorthand; it's not a number at all. Transfinite cardinal numbers, OTOH, are numbers.
 Good article (4.66 / 3) (#276) by elvstone on Wed Jun 04, 2003 at 08:46:45 PM EST

 I don't know alot of math and I can't say I enjoyed my math classes in school very much, but this is kind of neat stuff. And to all those "Why don't they teach like this in highschool / Just learning how to use a formula is no fun"-people: What they teach us in highscool is "hands-on" math, things that are useful to most people. I know this is much more interesting, but I don't think there's room for it in highschool classes. --- I need a job, email me if you have one in Linköping, Sweden
 University course you might be interested in (none / 0) (#416) by Nebu Pookins on Wed Nov 12, 2003 at 04:31:26 PM EST

 This kind of stuff is typically covered in university courses with names like "Abstract Algebra". If you're really interested in learning more, you need not actually register in a university to learn any of this. It turns out that most university professors don't take attendance, so you can just go to your local university's website, find the schedule for the courses you're interested in, and just sit in. [ Parent ]
 All infinities the same size? (3.33 / 3) (#285) by Vader82 on Thu Jun 05, 2003 at 12:08:12 AM EST

 It would seem to me that this is incorrect.  I see where you are going with that, but the reals has a starting point.  The integers do not. One set has a starting point, the other does not.  How can they have the same size? Also, from physics when integrating from 0 to infinity we get an answer half as large as when we integrate from negative infinity to infinity.  How does that work? Need food? Like sharing? http://reciphp.vader82.net/
 Starting point? (5.00 / 2) (#288) by vernondalhart on Thu Jun 05, 2003 at 12:22:42 AM EST

 but the reals has a starting point. The integers do not. What on earth do you mean by that? Do you maybe mean that the Natural Numbers {0, 1, 2, ...} have a starting point? Also, from physics when integrating from 0 to infinity we get an answer half as large as when we integrate from negative infinity to infinity. How does that work? But that completely depends on what you're integrating. If you, say integrated f(x) = 0 from 0 to infinity, or from -infinity to +infinity, you would end up with the same result. So that question doesn't make any sense with respect to the size of the real line. -- "It's like that old saying: A conservative is a liberal who's been attacked by aliens." Simon - mhm27x5[ Parent ]
 if you have an even function [nt] (none / 0) (#308) by the sixth replicant on Thu Jun 05, 2003 at 05:52:16 AM EST

 [ Parent ]
 Integrals and sums in finity (5.00 / 1) (#344) by Vann on Thu Jun 05, 2003 at 05:51:58 PM EST

 If you're integrating f(x) from 0 to infinity, e.g., int(0,inf,f(x)), then this doesn't make any sense per-se. However, we define it to mean int(0,N,f(x)) as N->inf. The same goes for "infinite" sums -- they're all just limits in disguise. ____________ Sex is tedious all year except on Arbor Day. -- Rusty[ Parent ]
 Also: (none / 0) (#348) by LilDebbie on Thu Jun 05, 2003 at 07:14:55 PM EST

 Not all infinities are equal, which I believe the author is going to go into on his next article. An example is the the set of natural numbers and the set of irrational numbers (numbers that cannot be represented by a fraction of two natural numbers). The set of irrational numbers is bigger, making for a bigger infinity since the set of natural numbers is infinitely large. Yay! My name is LilDebbie and I have a garden. - hugin -[ Parent ]
 Re: All infinities the same size? (none / 0) (#415) by Nebu Pookins on Wed Nov 12, 2003 at 04:28:59 PM EST

 It would seem to me that this is incorrect. I see where you are going with that, but the reals has a starting point. The integers do not. One set has a starting point, the other does not. How can they have the same size? It turns out that the size of the Real numbers and the size of the Integers is not equal, though they are both infinite. Typically, the size of the Integers is called Aleph zero (as mentioned in the article.) The size of the Irratitional numbers (which is bigger than the size of Integers) is called Aleph one, and the size of Real numbers is called Aleph two. It is not yet known if Aleph one = Aleph two. Note, however, that you could argue that the Real numbers have a "starting point" in the same sense that the integers have a starting point: That starting point is zero. For the integers, your list might be written: 0, 1, -1, 2, -2, 3, -3, 4, -4, etc. Similarly, you could have 0 be the starting point for Real numbers, and alternate between positive and negative numbers. However, it turns out that you can't ever make a list of Real Numbers, because they are "uncountable". Sets of size Aleph zero are said to be countable sets, and sets of size Aleph one or two are said to be uncountable sets. If you're interested in reading the proof that the real numbers are uncountable, look up "Cantor's Diagonal Slash Argument" on Google. Also, from physics when integrating from 0 to infinity we get an answer half as large as when we integrate from negative infinity to infinity. How does that work? That's only true if the function is symmetric around zero. Consider the function f(x) = x. If you integrate this from negative infinity to positive infinity, the answer is zero. If you integrate this from zero to infinity, the answer is infinity (or the answer is the limit of y as y approaches infinity, to be precise). Obviously zero is not half of infinity. - Nebu Pookins [ Parent ]
 ehem (3.66 / 3) (#301) by relief on Thu Jun 05, 2003 at 03:25:32 AM EST

 just because people-who-call-themselves-mathematitians say the above, doesn't mean its the only math. one could easily define math to be a computational/time-relational science. ---------------------------- If you're afraid of eating chicken wings with my dick cheese as a condiment, you're a wuss.
 And of course... (5.00 / 1) (#328) by the on Thu Jun 05, 2003 at 12:45:56 PM EST

 ...just because English speakers call a cat a cat it doesn't stop you calling one a hat. -- The Definite Article [ Parent ]
 The Sun did not shine. (none / 0) (#400) by DrH0ffm4n on Mon Jun 09, 2003 at 09:43:41 AM EST

 It was too wet to play. So we sat in the house All that cold, cold, wet day. --- The face of a child can say it all, especially the mouth part of the face. [ Parent ]
 "N word" Jim speaks (none / 0) (#409) by kfg on Sat Jun 14, 2003 at 05:16:25 AM EST

 Ah, but if you knew it was a cat, how could you call it anything else? With my apologies to Mark Twain. KFG [ Parent ]
 it's not the only math. (none / 0) (#382) by valar on Sat Jun 07, 2003 at 03:08:05 AM EST

 it just happens to make calculus, and therefore, physics easy. [ Parent ]
 rational part gets a bit messy (4.00 / 1) (#302) by m a r c on Thu Jun 05, 2003 at 03:34:24 AM EST

 From reading the article it seemed like you were saying (correct me if wrong), that you get some infinite set and apply some transform and if the size of the set is the same then they are both 'equally infinite'. What is the tranform for the rational part.. your logic is 'write down all the numbers in a certain way and count them up'. The counting process is defined by the N set so as long as you can write any set down couldn't you just say it's the same size as N using this idea? I got a dog and named him "Stay". Now, I go "Come here, Stay!". After a while, the dog went insane and wouldn't move at all.
 Exactly (5.00 / 2) (#305) by arvindn on Thu Jun 05, 2003 at 04:40:55 AM EST

 That's why sets equinumerous with N are called countable sets. Lots of sets are countable, such as the set of all rational points in n-dimensional space, or even the set of all finite sequences of integers. (Can you see how to count that one? Hint: if you start with the subset consisting of sequences of length 1, you'll never get to sequences of length 2! But there's a way.) So you think your vocabulary's good?[ Parent ]
 Bingo (none / 0) (#352) by Vann on Thu Jun 05, 2003 at 08:46:39 PM EST

 That's exactly right. Any set which you can put in one-to-one correspondence with the natural numbers is countable. To expand on the other response, take the real numbers, which are not countable. This means that for any list of real numbers you create I can find another number that isn't in your list. The canonical argument is Cantor's diagonalization argument, if you're interested, and shows that the irrational numbers are uncountable (hence the union of the rationals and irrationals -- the real numbers -- is also uncountable). You can make some pretty large, countable sets, though. For example, the set of all numbers which are roots of polynomials with rational coefficients is countable. That is, take any polynomial function of arbitrary degree f(x) = a_n *x^n + a_n-1 * x^(n-1) + ... + a_1 * x + a_0 where all the a_i are rational numbers. Any number u such that f(u)=0 is called a root. The set of all these u is countable. Any number which can be described like this is called an algebraic number, and so the set of all algebraic numbers is countable. This set includes a lot of irrational numbers, like sqrt(2) and so on, but doesn't include things like pi and e (they're called transcendental numbers). ____________ Sex is tedious all year except on Arbor Day. -- Rusty[ Parent ]
 N<->Q (5.00 / 1) (#397) by avery on Mon Jun 09, 2003 at 08:24:39 AM EST

 By the way: The proof of the one-to-one mapping between N and Q demonstrates another interesting property of countably infinite sets: Each rational q in Q corresponds to at least one pair (n,m) in Z x Z, the two-dimensional integer-space. "The Writer" proved that there are as many elements in Z as in Z2! By induction, this applies to any dimension - there are exactly as many elements in Z as in Zn for any finite dimension n! In fact, something similar applies to the continuum. There exists a continuous surjection from the interval [a;b] to any hyper-volume [c;d] x [e;f] x ... x [y;z]. This means, that one can map any line-segment continously onto an entire volume of any dimension, such that to any point in the volume corresponds at least one point on the line-segment [a;b]! How's that for counter intuitive? [ Parent ]
 Another way of saying this (none / 0) (#414) by fizbin on Tue Aug 12, 2003 at 12:54:42 PM EST

 Another way of saying all of this is: "any infinte set such that one can uniquely name every element in the set is only as large as N."  (where a "name" is a finite string of characters drawn from some finite set) A consequence of this (which, for some reason, saddens me a bit) is that almost all the numbers from 0 to 1 cannot be named.  (well, by us mere mortals) [ Parent ]
 infinities and limits (4.00 / 1) (#337) by jolt rush soon on Thu Jun 05, 2003 at 02:38:53 PM EST

 i just wondering if i'm going wrong somewhere here so any comments would be appreciated. if i have a set of all rational numbers, greater or equal to 1. let's call it x. and then i pair those numbers with 1/x, does the original set have one end point (1) and the mapping of the set have two end points (1 to 1/infinity) or is 1/infinity not a valid end point. apologies if i've driven off the track here. -- Subosc — free electronic music.
 End p;oints (5.00 / 1) (#347) by Vann on Thu Jun 05, 2003 at 06:37:23 PM EST

 What do you mean by end points? I think you mean something like infimum and supremum, but I'm not sure so I don't want to type a big message about bounded, ordered sets. Care to elaborate? ____________ Sex is tedious all year except on Arbor Day. -- Rusty[ Parent ]
 range (none / 0) (#385) by jolt rush soon on Sat Jun 07, 2003 at 12:53:29 PM EST

 i just meant that the new set would be limited in range, yes also contained within the original set and possibly the same site as the original set as it is also infinite (comes from a 1:1 mapping of an infinite set). -- Subosc — free electronic music.[ Parent ]
 As the author points out (none / 0) (#374) by GoStone on Fri Jun 06, 2003 at 03:07:59 PM EST

 infinity is not a number, so there can be no end point 1/infinity. Cut first, ask questions later[ Parent ]
 limits and end points (5.00 / 1) (#403) by btherl on Tue Jun 10, 2003 at 12:57:22 AM EST

 1/infinity would only be in the mapping if infinity was in the original set.  But infinity is not in the set of rational numbers, so 1/infinity is not in the mapping. However you can say this: The limit as x -> infinity of x is infinity (note that this isn't one of the elements of the original set) The limit as x -> infinity of 1/x is 0 (and this isn't an element of the target set, it's the limit of the elements) These could still be described as "end points" since they are where the sets end, but they are not included in the sets themselves.  On the other hand, the end point 1 is included in both sets. [ Parent ]
 interesting (3.40 / 5) (#338) by anonymous pancake on Thu Jun 05, 2003 at 03:07:27 PM EST

 I learned all of this in high school calculus, though. --- . <---- This is not a period, it is actually a very small drawing of the prophet mohhamed.
 Are you sure? (5.00 / 1) (#358) by dipierro on Thu Jun 05, 2003 at 11:28:53 PM EST

 They taught you about the cardinality of the set of rational numbers in High School? That's a pretty advanced school. [ Parent ]
 yeah (5.00 / 2) (#364) by anonymous pancake on Fri Jun 06, 2003 at 03:54:44 AM EST

 I live in canada --- . <---- This is not a period, it is actually a very small drawing of the prophet mohhamed.[ Parent ]
 It's not just canada. (none / 0) (#381) by valar on Sat Jun 07, 2003 at 03:04:36 AM EST

 I learned it in high school too (in Lousiana, of all places) at LSMSA. [ Parent ]
 Ditto... (none / 0) (#401) by silent on Mon Jun 09, 2003 at 11:52:51 AM EST

 High school calc in NH...and I never heard infinity described as "do this forever and you get infinity", while I did hear "to do something infinitely is to do it over and over again forever." --- silence is poetry[ Parent ]
 its a little different... (1.33 / 3) (#341) by relief on Thu Jun 05, 2003 at 05:00:53 PM EST

 when you have to consider consistency. ---------------------------- If you're afraid of eating chicken wings with my dick cheese as a condiment, you're a wuss.
 best description of infinity (4.00 / 1) (#388) by mincus on Sat Jun 07, 2003 at 05:28:44 PM EST

 "Did you know there are twice as many numbers as numbers?" <3 feynman
 Definition of Infinity (4.00 / 1) (#393) by Noboldogonorce on Sun Jun 08, 2003 at 09:35:15 AM EST

 Your expose is very good and easy to follow. I would like to add another definition that's used in arithmetics (the part of mathematics dealing with numbers). Here it is: "Let A be as big an integer as we want; infinity is greater than A." Looks stupid but it's not. This definition is used, amongst other things, to prove that the set of prime numbers is infinite. The more I look around, the better my eyesight grows.
 exposé versus exposition (4.50 / 2) (#395) by kubalaa on Sun Jun 08, 2003 at 03:20:50 PM EST

 I'm not 100% sure, but I think that one would only call this an exposé if there were a conspiracy to hide the existence of infinity. [ Parent ]
 In Mathematica (none / 0) (#396) by gmol on Sun Jun 08, 2003 at 10:37:19 PM EST

 1/Infinity=0 Convient in many cases, but should it not the expression unevaluated?
 Math Texts (none / 0) (#402) by welkin on Mon Jun 09, 2003 at 04:30:50 PM EST

 Why can't you write my math texts? You're much more fun to follow.
 top article, but one bit i didn't (none / 0) (#404) by werner on Tue Jun 10, 2003 at 08:51:58 AM EST

 follow: when you put Q+ in the table, why are the diagonals finite if Q+ is infinite? if Q+ is not infinite, what did i miss?
 Each diagonal is finite (5.00 / 2) (#405) by LodeRunner on Tue Jun 10, 2003 at 04:23:05 PM EST

 The first diagonal contains 1 element, the second contains 2, and so on. There is an infinite number of diagonals, but each of them is finite. --- "dude, you can't even spell your own name" -- Lode Runner[ Parent ]
 Misconception #4 - a possible error? (5.00 / 2) (#410) by ascension on Thu Jun 19, 2003 at 04:25:06 PM EST

 Enjoyable article. However, I am not sure that you showed equivalent cardinality between Z and N. Using your formula of (z*2) + 1 for positive integers and -z*2 for negative integers, how do you map to 1. The only way to map to 0 and to map to 1 is to have z=0 be both postive and negative. sincerely, ascension
 Darn, you're right! (none / 0) (#411) by The Writer on Thu Jun 19, 2003 at 05:03:09 PM EST

 I must've missed it. It should be (x*2)-1 for all x>0. I hate those off-by-1 errors. Thanks for pointing it out. I'm surprised nobody else caught it. :-) [ Parent ]
 Its always the small ones that go by unnoticed (5.00 / 1) (#412) by ascension on Thu Jun 19, 2003 at 05:09:29 PM EST

 It is always the small ones that get unnoticed. Only after I had posted my comment did I realize that you had actually shown that N was larger than Z (by 1) :) Look forward to your next posting. [ Parent ]
 Not Infinity! | 419 comments (370 topical, 49 editorial, 0 hidden)
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