quote:Originally posted by IanO: This thread is so cool.
I have been reading "The Road to Reality" by Roger Penrose, which first gives and mathematical foundation, and then goes on to explain how they explain the physical laws. The problem is, he writes so densly and jumps from a simple subject (at the beginning of the book when he's just trying to bring readers up to speed) to much more complex examples and descriptions. I find much of it is over my head.
But there are two things I would really like to know and, unfortunetly, the online explanations have not really helped.
1) what is a Line integral? I've had up to calculus 2, which included a preliminary on vectors. I don't remember much of that, but I like your explanations.
2) What is complex-analysis?
thanks for anything you can offer.
this thread is awesome.
1. What is a line integral? If you've ever done any integral at all, you've actually been doing a type of line integral.
To think about it properly, lets transition a bit.
You're used to seeing integral from a to b of f(x) dx, and intepreting that as "the area under the curve f(x)".
Let's reinterpret this. Instead of thinking of f(x) as a cuve, I'm going to think of the line from a to be as the "curve". I'm going to think of f(x) as the "density" of the line. In this interpretation, as you integate along the line (at a constant speed), your adding up the total mass you've seen.
To move up to line integrals, we must first move up in dimensions. Thus, f(x) will become a "vector function". Now, pick your favorite curve in this higher dimensional space. Every point on the curve can be described by an equation like r(t) = (x(t), y(t), z(t)). The velocity at any point along this curve is the derivative of r(t) (which is just the derivatives of x,y,and z).
Lets assume, for simplicity, that our curve starts at time 0 and ends at time 1.
Now, for starters, note that we don't really care about what f(x) is off the curve. In fact, to make f lie along the curve, will substitute the curve in for the x values. So instead of f(x), we'll have f(r(t)). All this means is that we'll only be evaluating f(x) along the curve r(t).
Now, lets go back to the integral. I claim that the total mass of our curve is the integral from 0 to 1 of f(r(t)) dot (derivative of r) dt.
Let me explain it. the dot comes from the fact that both f(x) and r(t) are vectors, so we can't just "multiply" them. Also, integration really only works over scalars, so we need to get a scalar. The dot product is the most obvious candidate. The reason we dot with derivative of r (the velocity) is this: If we move for a small amount of time but at a high velocity, we still have a lot of mass being covered, so as our velocity goes up, our mass should go up.
Drawing pictures makes this work a lot better....sorry I can't be of more help.
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I mentioned complex numbers when talking about the Riemann Hypothesis a few posts ago. If you need to brush up on them, I suggest doing so.
Complex analysis is the study of complex functions. That is, functions that accept a complex number as input and spit out a complex number as output. The "analysis" part means that you study things like continuity, differentiability, etc.
It turns out that while requiring a function to be continuous isn't THAT big of a deal, requiring it to be differentiable IS.
In fact, if you've seen calculus before, it certainly seems that almost EVERY continuous function is differentiable (when considered only as real functions). (This is actually VERY false, but it definitely SEEMS that way).
Complex analysis is a whole different ball park.
Let me explain. Since in complex analysis, you have 2 types of numbers "real, imaginary", you can approach any point from an INFINITE number of directions (from the real direction, the imaginary direction, 2/3 real 1/3 imaginary direction, etc). To be differentiable, it has to be differentiable in EVERY direction. Trying to calculate derivatives from different directions yields a pair of equations called the "Cauchy-Riemann" equations. The Cauchy-Riemann equations lead to the harmonic equation. Thus, every differentiable complex function satisfies the harmonic equation. What does this mean? For one, it means that no differentiable function can have a maximum or a minimum value...unless it's at the boundary.
Contrast this with, say, f(x) = x^2, which has a minimum at 0. You just don't have things like this happening in complex analysis.
An easy consequence of this is that ONLY the constant functions are bounded. So, for instance, sin(x), if x is allowedto be complex, can to to infinity (whereas in the real case, it oscillated between -1 and 1).
Another interesting property of differentiable functions: if something has a first derivative everywhere, then it has ALL derivatives everywhere.
Finally, by messing around with complex analysis, you can derive what, according to some, is the most beautiful mathematical equation EVER.
Hold your breath:
e^(pi*i) + 1 = 0. Where e is "Euler's number" = 2.71....., pi = 3.1415...., and i = sqrt(-1).
This one equation relates the 5 most important numbers in all of mathematics!
In conclusion, complex analysis is much more beautiful than real analysis, and in many ways much simpler! Real functions can have some CRAZY behavior that complex functions just can't have.
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quote: Like Beethoven's Ninth Symphony, for example, which exists even when nobody is listening to it or performing it and would even if all the copies were destroyed along with everyone who had ever heard it.
Existence must mean something different to you than it does to me.
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posted
Actually, that sort of makes sense. I have a book called "What is Quantum Mechanics" and had gotten through about a 3rd of the books mathematics. But then they started using line integrals and I just didn't get it. Well, later on their using a German E (is that Eiggenvalue?) so I probably need a lot more mathematics to get through it. But it was something I wondered about.
Thanks.
Did you ever read Where Mathematics Comes From? It tries to explain how a physiologically based mentality can come up with higher mathematics. What mental constracts and 'metaphorical blends' does the mind use to conceive of these ideas, like infinities, for instance, when there is no such actual countable (or observable) thing in our experience? The ending, I think, jars with your (and probably mine, as well) view of the platonic existence of mathematics. But the reason I bring it up is that every mathematical explanation you are using begins with a physical metaphor- a transition from a known human action-experience into the abstract. Different concepts, like counting objects, matching, the number line and so on. I thought the book really provided a nice understanding how to TEACH mathematics.
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posted
The complex analysis makes more sense. Thanks.
e^(pi*i) + 1 = 0 is actually discussed at the end of the book I mentioned above. In detail, it explains what, exactly, it means to raise e to (pi*i) and so on. Or at least, it claims to. Made sense when I read it about 3 years. Ago.
Thanks for the answers. This thread rocks.
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quote:Seems harmless, right? Turns out that you can prove a TON of nonintuitive things using this idea. For instance, Banach and Tarski proved, using the Axiom of Choice, that I can start with a sphere, but it into a specific 7 pieces, rotate and shift the pieces, and end up with 2 spheres EXACTLY the same size as the 1 sphere. (Note, there was no stretching, just rotating and moving)
I seem to recall Wikipedia claiming you can do it with just five pieces (when I was looking up paradoxes for the ask-the-next-person-anything thread).
I had to look up Cantor and infinities a while ago when I was reading mathematics-related jokes, because one of them (which now scrolls across one of the computers I use as a screen saver) was: Aleph-null bottles of beer on the wall, Aleph-null bottles of beer, Take one down, Pass it around, Aleph-null bottles of beer on the wall...
quote: Like Beethoven's Ninth Symphony, for example, which exists even when nobody is listening to it or performing it and would even if all the copies were destroyed along with everyone who had ever heard it.
This statement is begging one of the most debated philosphical questions. What does it mean for something to exist? It is the heart of the oft repeated question "If a tree falls in the woods and no one hears it, does it make a sound." I would also note that most of the people I've heard debate this question completely miss the point. Is sound an objective or subjective phenomenon? Is "sound" defined by the physical process we call sound waves, or are sound waves only sound when they are perceived by a sentient being.
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With Beethoven's symphony, you also have to determine if the score is the symphony, a performance is the symphony, or a recording is the symphony. Is a symphony "sound"?
If one decides that a recording is a symphony and the score isn't, then the reason for the distinction needs to be clarified. A recording and a score can both be thought of as instructions for performing the symphony, the former followable by machines, the latter an orchestra.
Same thing with books. Do I have "Ender's Game" on my shelf, or a copy of Ender's Game? My copy is the one with the racial slur still present. Is it still Ender's Game? Did Ender's Game change when OSC changed it for a new reprint?
This is a subject that fascinates me. It comes up on the first day of Contracts in law school. A contract is not a piece of paper. It can't be "ripped up." A writing is evidence of or a memorialization of a contract. A contract comes into existence full-born upon the completion of certain acts by the parties. One instant, no contract. Next instant, contract.
I realize I'm making no point here. You've simply sparked philosophical ramblings that I find interesting. Posts: 26071 | Registered: Oct 2003
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quote:1.a. Vibrations transmitted through an elastic solid or a liquid or gas, with frequencies in the approximate range of 20 to 20,000 hertz, capable of being detected by human organs of hearing.
1.b. Transmitted vibrations of any frequency.
Nice of them to contradict themselves...
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quote:Seems harmless, right? Turns out that you can prove a TON of nonintuitive things using this idea. For instance, Banach and Tarski proved, using the Axiom of Choice, that I can start with a sphere, but it into a specific 7 pieces, rotate and shift the pieces, and end up with 2 spheres EXACTLY the same size as the 1 sphere. (Note, there was no stretching, just rotating and moving)
I seem to recall Wikipedia claiming you can do it with just five pieces (when I was looking up paradoxes for the ask-the-next-person-anything thread).
Lots of fun.
Banach and Tarski proved you could do it with 7 cuts. That has subsequently been improved upon (I don't know if it was improved upon by either Banach or Tarski) to 5 as you mentioned. Curiously, if you ignore the very center of the sphere, you can do it in 4
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To join in on the existence debate, there's 2 kinds of existence going on. For the strictly mathematical one, when I say something exists, what I really mean is this:
When you think of ALL of mathematics being done INSIDE one of the boxes, is there a box with the properties I'm claiming? If so, then such a thing "exists", if not, then such a thing doesn't "exist".
Now, when I talk about "mathematics existing outside of humanity/society", that's a more philosophical notion of exist, one which I really don't feel qualified to talk about - it's more just gut feeling for me.
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The reason Beethoven's symphony would still exist even if all copies were destroyed, and everyone who ever heard it was dead, is that someone might invent a time machine and go back and retrieve it. Or, they could invent a process whereby sounds of the past could be reconstructed by complicated analysis of existing conditions, or something like that. Porter, you are LDS! If none of those things work, then God will remember it, or else we'll remember it ourselves, and we'll have it again after the resurrection! There's just no way to say from the laws of physics that it is definitively gone. So it must still exist.
Mathematician, I probably didn't phrase my question about the AoC very well, because you answered a different question. I guess what I need to know is, what could we do if it were true (state how we can select uniquely exactly one item from each box?) that we couldn't do if it weren't true (?). I need a more concrete example to understand what the AoC means. Like say we had boxes of pencils. If the AoC were true, I could select the longest pencil in each box, and that would uniquely specify a particular pencil in each box. But if it weren't true, there would be some boxes with pencils which were exactly the same length? Is that how it goes? I still feel confused about what the AoC is actually saying.
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quote:1.a. Vibrations transmitted through an elastic solid or a liquid or gas, with frequencies in the approximate range of 20 to 20,000 hertz, capable of being detected by human organs of hearing.
1.b. Transmitted vibrations of any frequency.
Nice of them to contradict themselves...
But in either case, an actual receiver of the sound is not required.
(And there's the same discrepancy with the word "light." Is "light" only the visible (to the average human eye) portion of the EM spectrum, or the entire thing?)
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Ooh, can I ask a question? Speaking of infinities, what is the largest number of non-intersecting topological figure-eights that can be placed in the plane? (And what about topological circles?)
And while we're at it, can you partition 3-space into non-degenerate circles?
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quote:Originally posted by Tatiana: Mathematician, I probably didn't phrase my question about the AoC very well, because you answered a different question. I guess what I need to know is, what could we do if it were true (state how we can select uniquely exactly one item from each box?) that we couldn't do if it weren't true (?). I need a more concrete example to understand what the AoC means. Like say we had boxes of pencils. If the AoC were true, I could select the longest pencil in each box, and that would uniquely specify a particular pencil in each box. But if it weren't true, there would be some boxes with pencils which were exactly the same length? Is that how it goes? I still feel confused about what the AoC is actually saying.
I see, here goes nothing!
First, to clarify. If there are are an infinite number of pencil boxes, and all the pencils are of a different length, we DON'T need the AoC to choose a pencil from each box - we could (for example) just pick the smallest.
The problem comes if they are all the same length. Now, how do you pick one? AoC to the rescue!
Here's the thing. If we have a finite number of pencil boxes, no problem, just go through them one by one and pick a pencil. If you have an infinite number, you'd never get done picking, and that's the problem. The AoC says that I can assume that somehow (I don't care how), something has picked something from every box. The trick is that we don't know what's been picked, we just know that something had been picked from every box.
Let's revisit the pencil boxes with different sized pencils again. I don't have to go through them one by one, picking, I can just say "I've picked the smallest one from each". Thus, every box has one picked from it.
Now, for a more concrete example. Suppose I have an infinite box. Here's the question. Can I section off a chunk of the stuff in the box so that the chunk has the size of the smallest infinity?
Intuition says yes. It turns out, if you accept the axiom of choice, then yes, you can always find such a section (though I'm unsure on HOW exactly you do it). If you assume the AoC is false, then you can find examples of infinite boxes which don't contain sections of the smaller infinite size. (This is SO nonintuitive, I can't even pretend to come close to thinking about it).
This is actually an example where accepting the axiom of choice leads to a more intuitive result than rejecting it. Just goes to show that nonintuive things are, well, nonintuitive!
Posts: 168 | Registered: Jul 2006
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quote:Originally posted by Mike: Ooh, can I ask a question? Speaking of infinities, what is the largest number of non-intersecting topological figure-eights that can be placed in the plane? (And what about topological circles?)
And while we're at it, can you partition 3-space into non-degenerate circles?
I'll answer the second 2 first, because I actually have seen the proofs with regards to those. The maximum number of circles that can be fit in the plane? The same number as the number of real numbers!
In fact, it's possible (this UBER uses the AoC) to partition 3-space into non-degenerate circles (and therefore all higher n-spaces). Interestingly, it is NOT possible to partition 2-space into non-degenerate non-intersecting circles.
As far as the topological figure 8's, the answer is the same as with the circles. I can fit the same number as the number of reals. Here's quick outline of the proof.
Let "N" be the number of topological figure 8's I can fit. Let c be the number of real numbers.
To see that c is less than or equal to N, note that there are c distinct planes, and I can easily just put a single figure 8 in each one. They certainly are nondegenerate and obviously don't intersect (since the planes don't intersect). Thus, I've stuck AT LEAST c figure 8's, and thus, N is at least c, that is, c is less than or equal to N.
To see the reverse, notice that each figure 8 is really just a curved line, and so has c points. If there where more than c figure 8's possible, and all of them were non-interesection (say, d of them), then I'd have number of points in each figure 8 * number of figure 8's points. That is, I'd have c*d > c points, an obvious contradiction (since I know there are only c points).
From this I can conclue that N is less than or equal to c.
By the Schroder-Cantor-Bernstein theroem, N = c.
Thus, I can place c nonintersection figure 8's. That is, I can place as many non-intersecting figure 8's as there are real numbers.
(I still haven't worked out whether or not 3-space can be partitioned into figure 8's. My gut instinct is yes, but that probably means the answer is no ;-). I KNOW that I can't partition 2-space into figure 8's, for more or less the same reason I can't do it with circles).
Edited to add a line or 2
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Oh no, Mike, you said we're sticking the topological figure 8's in the PLANE, not in 3-space. Give me some time to work on that....
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How do you figure that you can partition 3-space into non-degenerate circles? (And what does non-degenerate mean?) And why can't you partition 2-space into non-degenerate non-intersecting circles? And why can't you do it with figure 8s either?
-Gwen, who still doesn't see how this will help her count sheep or pile pebbles.
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posted
Gwen, the proof (at least, the one I've seen) that you can partition 3-space into non-degenerate circles is quite difficult. Non-degenerate simply means the circles aren't just points; they have some finite but non-zero radius.
The proof that you CAN'T do it for 2-space isn't too bad. Draw a circle (call it C1). If we're gonna be partitioning all of the plane, the center of C1 must be contained in some circle which is completely contained in C1 (otherwise, they'd interesect). But C2 has a center which must be contained, so suppose C3 contains it. Rinse repeat. Now consider the sequence of the centers of the circles. It turns out, the centers converge to a specific point (I'll call it p). What convergence means is this: Pick your favorite positive number n. I can find a point in my sequence so that all the centers after than are a distance of less than n from the point we're calling p. In otherwords, our centers get closer and closer to p (in fact, infinitely close, in some sense).
But we're covering the entire plane in circles. So p must be contained in a circle C of a postive radius. It will turn out that since the centers of the other circles get arbitrarly close to this one, our circle C MUST have intersected one of our other circles, so we don't actually have a partition. This contradiction means that we could never partition it in the first place.
Figure 8's work the same way. Look at the sequence of points which are the intersection points of the the figure 8s.
Back to work on the number of figure 8's in a plane. It's kicking my butt!
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posted
I think I got the number of figure 8's. I had actually heard before that it was the same as the number of natural numbers, but I just worked out a proof (I think).
It's surprisingly simple....
Pick your 2 favorite ordered pairs, then there can be only 1 figure 8 that goes around them both (that is, 1 point in one of the circles, another point in the other). I don't really know how to PROVE this, though it seems quite obvious.
Lets look at all the ordered pairs of rational numbers (i.e., fractions). Since the rationals are countable (meaning same size as the naturals), so are the the ordered pairs of rationals. The rationals are important because they are dense, meaning they're practically everywhere (in between any 2 numbers, there is a rational number)
Lets assume for a contradiction, that we have MORE than a natural number of fig 8's. Pair each of these up with their 2 "defining" pair of ordered pairs of real numbers.
Now, fix a figure 8 and it's 2 defining ordered pairs. Since the rationals are dense, I can find a 2 rational ordered pairs so that each ordered pair is contained in one of the 2 circles of the figure 8. But then I can use these 2 rational ordered pairs as my defining ordered pairs!
But the set of all pairs of ordered pairs of rational numbers is countable. So, I have more than countable fig 8's matched up with a countable set. That means that at least 2 different fig 8's have the SAME defining 2 ordered pairs of rational numbers. But we already argued that any 2 fig 8's encompassing the same 2 ordered pairs MUST intersect. This contradicts our assumption that they are non interesecting.
The fact that we were able to derive a contradiction from assuming we had more than a natural number of fig 8's means that that assumption was wrong. Thus, there can be at most a countable number of fig 8's.
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I guess I could've been clearer — the two questions aren't really related to each other. It's just that one made me think of the other.
Yeah, the figure eights in a plane is a tricky one. (Now if they were snakes...) I'll let you stew on it for a bit.
Partitioning 3-space with circles is actually not too difficult. I know a couple of different constructive proofs, one of which is really slick. I'll describe it if anyone's interested in hearing it. Unless you guys want to figure it out yourselves.
Hmm. How about this: can you partition 4-space into 2-spheres? (A 2-sphere is the object that most of us know as the surface of a three-dimensional ball.) I haven't given any thought to this one, but I suspect the answer is no.
Posts: 1810 | Registered: Jan 1999
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posted
I'd definitely be interested in a constructive proof. The one I know relies on transfinite induction.
4-space into 2-spheres, huh...
If a 2-sphere not lying entirely in a particular 3-space can only intersect that 3-space in a finite number of places (or countable), then I can do it. The problem is, I can't convince myself one way or another.
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posted
Glad you like it! Keep the questions coming! (But I'm going to bed now, I'll answer whatever I can tomorrow morning).
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OK, for this construction we'll build things up step by step. We'll start with a spherical shell (a 2-sphere) which we'll call A: can you partition it into circles? Well, no, but you can come pretty close. Take, eg., the set of planes parallel to the xy-plane. The intersection of these planes with A is a set of circles, plus two points at the north and south poles.
It turns out you can do the same thing with any two distinct points p and q on A. (To do this explicitly, note that the tangent planes of A at p and q intersect in a line L that lies outside of A. The set of all planes containing L, intersected with A form a set of circles that partition A - {p,q}.)
OK, now that we have a way of dealing with these shells, let's try to construct a solid sphere, or ball. A ball can be thought of as a union of shells, plus the point at the center. Each shell can be partitioned into circles except for two points, which we can choose arbitrarily. Consider the ball B centered at the origin with radius 1. Let C be a circle of radius 1/2 that is tangent to B and contains the center of B (eg. the circle in the xy-plane centered at (1/2, 0, 0)). Let's also choose a point p on the surface of B, say (-1, 0, 0). Can we partition B - C - p into circles? Sure we can: every "shell" of B - C - p is missing exactly two points, and so can be partitioned into circles; these shells themselves form a partition of B - C - p. Of course, it's a piece of cake to add C back in (it's a circle after all), so B - p is partitionable into circles. Pretty nice.
Now we're ready to extend to all of 3-space. Consider the set of circles in the xy-plane of radius 1/2 centered at x = ..., -11/2, -7/2, -3/2, 1/2, 5/2, 9/2, ..., y = 0. Let U be the union of these circles.
U:
code:
... O O O O O O O ...
For every positive r, the sphere S_r of radius r centered at the origin intersects U at exactly two points, and so S_r - U can be partitioned into circles. U itself can of course be partitioned into circles. And the union of U and the S_r is R^3. (Ta-da.) There are other sets U that work this way, like the one containing all circles centered at (1/2, 0, 0) with radius 1/2, 3/2, 5/2, ....
There is also a construction involving nested torus-like objects, but it's a bit harder to describe rigorously.
Posts: 1810 | Registered: Jan 1999
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quote:1.a. Vibrations transmitted through an elastic solid or a liquid or gas, with frequencies in the approximate range of 20 to 20,000 hertz, capable of being detected by human organs of hearing.
1.b. Transmitted vibrations of any frequency.
Nice of them to contradict themselves...
It is a different thing for vibrations to be "capable of being detected by human hearing" and to actually be detected by human hearing. These definitions both adhere to the objective definition of sound. Neither recognizes the possibility for a subjective definition.
[ August 18, 2006, 05:19 PM: Message edited by: The Rabbit ]
Posts: 12591 | Registered: Jan 2000
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posted
Just thought I'd attempt resurrect this (I confess, I'm enjoying this, perhaps a bit too much), and try to turn it more basic ideas. (That last bit about filling 3-space with circles and all....It's extraordinarily hard stuff).
I wanna start with an interesting tidbit. We've already talked about how the "infinity" of the natural numbers and the "infinity" of the reals are different sizes (with the reals bigger than the natural numbers).
Now, here's where it gets hard to wrap your head around (in case it's not there already!), how many sentences (in english) are there (where we require that sentences be of finite length)?
Lets work it out. First, let's order the sentences alphabetically (we'll define a space as being less than a. That is, "aba" comes after "a ba"). Let's just start counting them and see what happens. Can you convince yourself that given any sentence, after some finite number of steps, we'll reach that one (since there are only a finite number of letters/spaces)? It turns out, that this means that the box having all possible sentences in will have the same size as the natural numbers.
Note that we haven't even talked about the fact that sentences like "a dog" or "akbla kasld" aren't even sentences! Let's assume, just for ease, that these ARE "valid" sentences.
Anyone see where we're going with this?
We have at most a natural numbers number of sentences, but we have MORE real numbers than that. What this means is that even if EVERY sentence describes a unique real number (which, of course, they don't), there will be some real numbers which are inherently indescribable. They transcend all possible description. This rocks my world!
Posts: 168 | Registered: Jul 2006
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posted
Except that in English, sentences can, oh I don't remember the word Language Log used in that one post, loop or nest or iterate or something...so you could potentially describe every number with something like "this is the number you get when you add one and one and one and one and one..." And if there's a real number which is inherently indescribable, you could just call it "this is the first real number which is inherently indescribable," which is a unique description. Except it would then be false, so it would be something like "this is the first real number which would be inherently indescribable if it weren't for the sentence 'this is the first real number which is inherently indescribable'" and then the next number would do it, and so on. Come on, Mathematician: 27 digits, if we count only letters and spaces, vs. 11 digits, if we count decimals and numbers--which do you think will win?
Posts: 283 | Registered: Jul 2006
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Gwen: no, there are lots of real numbers that go on forever with no pattern, something that definitely can't be described by listing numbers one at a time, since any given sentence is, by nature, finite.
Now, some of that variety of number can be described in other ways (pi, anyone?), but the theorem does conclusively prove there are others that can't.
Once we know that any given sentence is finite, we know that the total (infinite) number of sentences (even arbitrarily long ones) is the same as the number of natural numbers, by Mathematician's proof.
Since there are many more real numbers than natural numbers, we know there are oodles of real numbers out there indescribable with sentences.
edit: and there is no beginning to the real numbers (excluding the natural numbers), nor would there exactly be a 'first' one even if there were, so no, your sentence doesn't work.
Posts: 15770 | Registered: Dec 2001
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The term is escaping me as well. I believe in GEB they're referred to as strange loops. The thing is, irrelevent of intended meaning, there are always less sentences than real numbers, period. It doesn't matter if the sentence has self references (even if every sentence is allowed to refer to as many (but finite) sentences as you'd like)
With regards to "this is the number you get when you add one and one and one and....", I agree, we can generate all the natural numbers (using exactly those types of sentences), but there will still be numbers you can't get to using that method (for instance, fractions, negative numbers, real numbers which are neither).
As far as calling a real number "the first real number which is inherently indescribable", the real numbers aren't well ordered. What this means is that I can pick collections of real numbers that DON'T have a smallest number. Consider this example:
Think of all x between 0 and 1, but not including 0 or 1. Does this have a smallest number in it? No! How do I know? If you hand me a number and claim "this is the one!", I can halve it. It's still between 0 and 1, but it's definitely smaller. Thus, no smallest number.
Because the real numbers aren't ordered in a nice way, there may be no "first number which is inherently dindescribable", meaning that sentence DOESN'T describe a real number.
I'll admit that if we're only trying to describe countable things, sentences such as "the first number not describable in 47 words or less" would create havoc (because then I have described it in less than 47 words). When doing hardcore mathematics, it turns out that statements like this simply can't be formulated. (And so if we restrict to PURELY mathematical sentences, written in the language of logic, my arguement goes through, no rebuttals are really possible).
As far as the 27 digits vs 11, you're forgetting one important detail. Looking at numbers (decimals), we allow for an infinite number of decimal places. I stressed at the begining that our sentences must be of finite length. It is exactly restriction that allows the proof to go through. If I allow sentences of infinite length, I can easily describe any real number - just write out it's digits one by one.
Why restrict sentences to be finite length, while nubmers can be any length? Because we understand how to interpret decimals of infinite length - the length just gives us more and more precision. How do you interpret a sentence of infinite length? Is it possible that given an arbitary (but finite) length sentence, I can construct a larger one which starts with that sentence, but which has a totally different meaning? I'd suggest that it's yes. And this means letting our length go to infinity doesn't help us intepret the sentence- in fact, it may make it worse (depending on the sentence).
posted
BTW, here's what I mean by there not really being a 'first' number even if the real numbers had a beginning.
Imagine we're only looking at positive numbers, so there's sort of beginning to the numbers; at least, there's a point where there's nothing before and all the numbers are immediately after (0).
Lets try to pick the first number. Clearly it must be a specific number.
Now divide that number by two -- the new number is clearly before the previous number, and also positive, therefore the previous number cannot be the first number. Perhaps this new one is? But no, the trick works again and again.
We never reach a first number. There is no first number.
edit: and then I'm beat to my explanation
BTW, there's a way out of your paradox. The sentence doesn't describe a number, even though it says it does.
edit again: that is, this paradox -- "the first number not describable in 47 words or less", with the same resolution for the other paradox tackled.
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posted
If we are already Anne Kate's acolytes, and she becomes and acolyte of Methematician, would this make us sub-acolytes.
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posted
If I understand correctly, the number of rational numbers is the same as the number of natural numbers. The rational numbers between 0 and 1 can be arranged in order using the following pattern, 0 , 1, 1/2, 1/3, 2/3, 1/4, 2/4, 3/4, 1/5, 2/5, 3/5, 4/5. We could then create a similar pattern to order all the rational numbers between 1 and 2, if we repeat this to infinity we will have all the rational numbers. The number of irrational numbers, however, can not be counted and is therefore a larger infinite set than the number of rational numbers.
Also, there are an infinite and uncountable number of irrational numbers between any two rational numbers. So the number of irrational numbers between for example (1/100) and (1/101) is greater than the total number of rational numbers.
So now here is my problem. Is the number of irrational numbers between (1/100) and (1/101) the same size infinite set as the set of all irrational numbers? Since neither set can be arranged in a counting order, is there anyway to tell whether the two sets are the same size? Are there any infinite sets which are known to be larger than the set of irrational numbers? Is so, how can we know that they are larger when they can not be counted?
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posted
My favorite Mathematical "proof", is the proof that there are no known uninteresting numbers.
First we shall prove that there are no uninteresting rational numbers. We begin by arranging the rational numbers into some logical order, such as the one I described in my previous post.
The first number, 0, has all kinds of interesting properties. The second number, 1, also has numerous interesting properties. Now let us assume that there are rational numbers that have no interesting properties. As we progress along our ordered list of rational numbers, we will eventually come to the first rational number that has no interesting properties. This would, however, be in itself an interesting property. Therefore, there can be no uninteresting rational numbers.
This can be extended to all known numbers, both rational and irrational. Consider the set of known numbers. We can order this set based on the time the number was discovered. We would start with the oldest known number and move up to the most recently discovered number. Now let us assume that numbers have been discovered which have no interesting properties. These would of course have to be irrational numbers since we have already demonstrated that there can be no uninteresting rational numbers. As we searched through the list of known numbers, we would eventually come to the first irrational number discovered which had no interesting properties, which would of course be a very interesting property.
Ergo, we will never discover an uninteresting number.
[ August 21, 2006, 08:44 PM: Message edited by: The Rabbit ]
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posted
But--but--doesn't fugu13's and Mathematician's reason for mine being wrong count for yours too?
Or do I just have to define what I mean by "first" to be the way you meant it (the number whose existence we learned of chronologically first)?
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posted
Gwen, My definition is not a definition of the set of all irrational numbers. It can only work as a definition of the set of all known irrational numbers. The set of known irrational numbers, is a countable infinte set. It took me a minute to persuade myself that it is indeed an infinite set but consider the following example. Take the irrational number pi. If I multiply pi by any rational number, I will get an irrational number. This would create an infinite set of irrational numbers all of which are known. Since the set of irrational numbers is countable, this sub set of the irrationals (pi*x) where x is any rational number, is also countable.
The point is, that the set of known irrational numbers is not and can never be the same as the set of irrational numbers. There will always be irrational numbers that we don't know about.
What this means is that we can never determine whether or not there are uninteresting irrational numbers. All we can prove is that we will never know of any uninteresting numbers.
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quote:Originally posted by The Rabbit: So now here is my problem. Is the number of irrational numbers between (1/100) and (1/101) the same size infinite set as the set of all irrational numbers? Since neither set can be arranged in a counting order, is there anyway to tell whether the two sets are the same size? Are there any infinite sets which are known to be larger than the set of irrational numbers? Is so, how can we know that they are larger when they can not be counted?
Good to see this thread alive and kicking!
To answer your question, the irrational numbers between 1/100 and 1/101 have the same size as ALL irrational numbers.
It's really hard to tell things about sizes of irrational numbers, in general. In fact, I've never seen a proof that says "here's how I can match these up with all real numbers". All proofs about the size of irrationals I've seen are more indirect.
Here's an example. Suppose I have a box with 5 things in it, and another one with 2 things in it. I make a new box by dumping the contents of the first 2 in it. Then that thing clearly has 7 things in it. We've just demonstrated how addition works.
Similarly, if I have 2 boxes with 5 things in them each, and I dump the 2 boxes worth into a new box, this one has 10. This demonstrates multiplication
With infinite things, things are actually FAR easier (suprise surprise). Suppose I have 2 boxes, and at least one of them has an infinite (doesn't matter which infinity we're talking about). How do I put their contents together? It turns out (this is a midly difficult proof), that if I add them, or multiply them.
1) I get the same answer and 2) The answer is whichever the biggest one was.
That is, if I add to infinite boxes (say of sizes a and b), or multiply them, I get the same answer - either a or b, depending on which is bigger.
Using this, lets get back to the irrationals.
We know that the Rationals + Irrationals = Reals.
Using what we know about addition, we know that the Irrationals must be the same size as the reals.
Now, since all the numbers between 0 and 1 (or between any 2 specific points, including 1/100 and 1/101) have the same size as ALL the real numbers, we can use the same line of reasoning as above:
rationals between 0 and 1 + irrationals between 0 and 1 = reals between 0 and 1. We know the size of the reals between 0 and 1 (same as the size as all the reals). Thus, the irrationals between 0 and 1 have the same size of the reals.
This reasoning applies to all the irrationals between any 2 numbers.
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posted
I did a bit of research and I found an explicit matching up between the reals and the irrationals.
First, we must establish that sqrt(2) is irrational.
This isn't too difficult. Assuming for a contradiction that sqrt(2) is rational. That is, sqrt(2) = p/q for some integers p and q. We may as well assume that p and q have no common divisors (in other words, our fraction is in it's lowest terms).
Let's square both sides. We get 2 = p^2/q^2 and so p^2 = 2*q^2. Here's comes the somewhat tricky part. EVERY natural number can be broken down into primes. For instance, 42 = 7*3*2. But we're SQUARING it, so that means we actually have 42^2 = 7^2*3^2*2^2 = 7*7*3*3*2*2. Notice that no matter what we square, we'll have 2 of every prime it breaks down into. This means it has an even number of each prime contained in it (since 2*anything is even).
Still with me?
Let's go back to our equation. p^2 = 2*q^2.
Now, we know that if we break q^2 into it's primes, it has an even number of 2's. But this tells us p^2 has one more than that! In otherwords, p^2 has an ODD number of 2s in it. This is our desired contradiction. And that means that our assumption that sqrt(2) was rational was foolish. In otherwords, sqrt(2) is irrational.
Allright, now that we have the irrationalisty of sqrt(2) established, we just need a few more facts.
Rational + irrational is irrational. nonzero rational * irrational is irrational.
With that in hand, we're ready for our matching.
Let q be ANY rational number (i.e. fraction), and n be ANY integer (i.e. ...-2,-1,0,1,2...)
Consider numbers of the form q + n*sqrt(2). It may not be easy to determine if a number is of this form, but we can at least agree that every number is either of this form or it isn't.
Notice first, that if we let n = 0, we get EVERY possible rational number (since q can be any irrational).
Allright, we're ready to match things up.
We're gonna start with a real number and end up getting just irrationals.
First, if a real number ISN'T of the form above, we'll just match it up with itself. Notice again that ALL rationals ARE of that form, so we're just matching a large chunk of irrationals back to themselves.
What about things that ARE of that form?
Well, if something looks like q + n*sqrt(2), we match it up with q + (n+1)*sqrt(2).
Notice that this takes rationals (n=0) to irrationals (since products and sums with irrationals yields irrationals.
Now, lets first convince ourself that we match up with EVERY irrational. First, if a irrational is of the form q+n*sqrt(2), then the number q+(n-1)*sqrt(2) matches with it. So numbers like that are taken care of. If it's not of that form, we just matched it with itself, so those are taken care of.
Thus, we do actually get every irrational.
The other thing we need to check is that we don't have more than one number matching up to each irrational number. This is actually the harder part (and is a bit beyond the scope of this post). So, I guess you'll just have to take my word for it.
There are a few technical details that need filling in. (We must be certain that if a number can be written as q + n*sqrt(2), there's only 1 choice of q and n that give it, and a few others odds and ends). But they can be filled in (I won't do it).
Assuming these get filled in, we do have a matching up with ALL the real numbers and just the irrationals. Thus, the reals and the irrationals have the same size.
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