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bryanmcc
05-02-2003, 08:13 PM
I was just reading "Motion Mountain," an on-line physics textbook by Christoph Schiller (http://motionmountain.dse.nl/welcome.html). In it he says (in Appendix B, page 912) that the graph of the function e^x only goes through one point with two rational coordinates: (0,1). OK, fine -- I'm willing to accept that. But then he says: "If you imagine to paint in blue all points on the plane with two rational coordinates, the plane would look quite bluish..." What? I thought the majority of points on the number line were irrational (and hence the majority of points on a Cartesian plane). No?

The_Peyote_Coyote
05-02-2003, 08:39 PM
Yes, bryanmcc, but consider the sheer number of points with all rational coordinates.

Xema
05-02-2003, 08:44 PM
The "paint the points blue" analogy has the weakness that spots of paint have some area, whereas points don't.

ultrafilter
05-02-2003, 09:29 PM
Qn is dense in Rn. In other words, pick any element r of Rn, and any e > 0. There is a q in Qn such that |q - r| < e.

So up to any resolution you would like, the plane would appear blue, even though the total area of blue points is zero. Freaky, ain't it?

lucwarm
05-03-2003, 05:19 AM
Originally posted by ultrafilter

So up to any resolution you would like, the plane would appear blue, even though the total area of blue points is zero. Freaky, ain't it?

I'm kinda skeptical - what is the probability that a random photon will hit at a point with two rational coordinates?

05-03-2003, 05:24 AM
I want to expand on one point. It is not easy to give a sensible meaning to 'Qn having zero area' or 'picking a real at random has a zero chance of being a rational' but the only sensible way chosen makes these both true.

Obviously 'painting all rational points blue' is an analogy. The point he's making is that 'rational points are dense' (As ultrafilter describes - in any small circle there will be (infinitely many) rational points). You are right that there are 'almost no' rational points compared to irrational. He just picked a metaphor that can be unfortunately misinterpreted.

RM Mentock
05-03-2003, 05:25 AM
Originally posted by lucwarm
I'm kinda skeptical - what is the probability that a random photon will hit at a point with two rational coordinates?
How wide is a photon? :)

Hari Seldon
05-03-2003, 04:03 PM
The trouble is that a photon has size (even ignoring uncertainty) and therefore one cannot ask what "point" it hits.

The sense in which the rationals on the line have 0 measure is this: Take a very small positive number, say 10 to the power
-google. Then I can find a set of intervals whose total length is less than your chosen number, such that every rational number is contained in at least one of those intervals.

Here is the construction: First put the rational numbers in a sequence. Say 0/1, followed by 1/1, followed by 1/2, followed by 2/1, 1/3, 3/1, 1/4, 2/3, 3/2, 4/1,... (first the fractions in lowest terms whose numerator and denominator add up to 1, then 2, then 3 and so on, in order of increasing numerator). Then calling the small number w, I put an interval of size w/2 around the first, w/4 around the second, w/8 around the third and so on. The total lengths of all these intervals in w and every rational is contained in at least one of them. I did only the positive rationals, but it is trivial to get the negative ones too.

ftg
05-03-2003, 04:46 PM
Note that the Real plane is only one such plane you can define. The Rational plane is a plane of less density (Cantor set theory-wise). You can create a plane with numbers of greater density than the Reals. An easy to construct set of greater size than the reals is the set of functions of reals to reals. Those can be mapped into a much denser plane. Go up another level, the cardinality of functions from functions to functions is greater still and results in a denser plane, etc. No stopping it.

It is merely because so much is taught about the Real plane that people think that it is "standard" when it is merely one level in an immense hierarchy.

To me, a plane of blue colored Rational points has just as much claim to be blue as a plane of blue colored Real points. Neither colors "everything".

Xema
05-03-2003, 07:36 PM
So we have a truly huge number of blue points, no one of which is actually visible. Sounds like one hand clapping.

ultrafilter
05-03-2003, 07:39 PM
Originally posted by ftg
Note that the Real plane is only one such plane you can define. The Rational plane is a plane of less density (Cantor set theory-wise). You can create a plane with numbers of greater density than the Reals. An easy to construct set of greater size than the reals is the set of functions of reals to reals. Those can be mapped into a much denser plane. Go up another level, the cardinality of functions from functions to functions is greater still and results in a denser plane, etc. No stopping it.

It is merely because so much is taught about the Real plane that people think that it is "standard" when it is merely one level in an immense hierarchy.

To me, a plane of blue colored Rational points has just as much claim to be blue as a plane of blue colored Real points. Neither colors "everything".

You do end up with sets of higher cardinality, and they are two dimensional, but "denser" there doesn't really mean anything, at least as far as I know. Note that the sense in which the rationals are dense in the reals and both are self-dense is very precise.

Achernar
05-03-2003, 09:23 PM
Originally posted by Hari Seldon
The trouble is that a photon has size (even ignoring uncertainty)Really? What the size of a photon, ignoring uncertainty?

Desmostylus
05-04-2003, 06:34 AM
Originally posted by Achernar
Really? What the size of a photon, ignoring uncertainty? I'm pretty sure you know what he means. Photon size is usually treated as being the wavelength

&lambda; = c / f

Consider a fine metal mesh in the path of a flux of microwave photons.

If the size of the holes in the mesh is ten times the wavelength of the photons, they'll all get through. If the size of the holes is equal to the wavelength, 86% will get through. If the size of the holes if equal to half the wavelength, 63% (1 - 1 / e) will get through.

If x is the hole size, the proportion that passes through is given by:

1 - e-.5x/&lambda;

Achernar
05-04-2003, 06:44 AM
Okay, right. I guess that makes a whole lot of sense, but I'm not used to treating photon size like that. At least, it seems counter-intuitive to me that we have meter-sized photons flying around for radio broadcasts. But I get it.

Desmostylus
05-04-2003, 06:46 AM
Originally posted by Desmostylus
1 - e-.5x/&lambda; [/B]Oops, make that

1 - e-2x/&lambda;

bryanmcc
05-04-2003, 10:59 AM
ultrafilter, Rn is, I assume, the set of real numbers. So Qn is what, the set of rational numbers? And what the heck is epsilon? It appears that you are saying that for any real, there is a rational arbitrarily close to it. Fine, but isn't there also an irrational arbitrarily close to it? Why do the rationals "win" to color the plane? Could you explain this in language understandable to someone who is bright enough, but only has two semesters of calculus?

I want to expand on one point. It is not easy to give a sensible meaning to 'Qn having zero area' or 'picking a real at random has a zero chance of being a rational' but the only sensible way chosen makes these both true.

Obviously 'painting all rational points blue' is an analogy. The point he's making is that 'rational points are dense' (As ultrafilter describes - in any small circle there will be (infinitely many) rational points). You are right that there are 'almost no' rational points compared to irrational. He just picked a metaphor that can be unfortunately misinterpreted.

What is the mathematical definition of "dense"? Does it just mean "an infinite number within any arbitrarily small area"? (If so, it sounds like your typical mathematical understatement. ;)) If so, aren't the irrationals also dense? I.e., aren't there also an infinite number of irrational numbers in any arbitrarily small interval? If there are almost no rational points compared to irrational, won't the plane appear red?

Achernar
05-04-2003, 11:11 AM
Originally posted by bryanmcc
ultrafilter, Rn is, I assume, the set of real numbers. So Qn is what, the set of rational numbers?Close. They're the n-dimensional reals and rationals, respectively (not the technical terms). The OP was talking about the plane, so we want R2 and Q2, but the idea is the same for any n.It appears that you are saying that for any real, there is a rational arbitrarily close to it. Fine, but isn't there also an irrational arbitrarily close to it? Why do the rationals "win" to color the plane?You're absolutely right the irrationals are dense too. And in at least one sense, they're "denser" than the rationals. The reason they don't win is because they don't get any paint. If we painted all rational spots blue and all irrationals red, it would be a different story. I don't know what story it would be, exactly, but it would not just look like a blue plane.

bryanmcc
05-04-2003, 11:26 AM
Ah, and herein lies the problem: apparently, I can't even correctly read the material that I am quoting. Somehow I had it in my head that the passage in the OP said paint all rational-rational pairs blue _and_ all irrational-irrational pairs red, in which case I couldn't figure out why the plane wouldn't be red. I now have no problem with the quote in the OP (in fact, it's bloody obvious), but I do think I might need to invest in some reading glasses.

In an attempt to redeem myself, I'll ask what I thought I was asking originally: if you paint the rational-rational points blue and the irrational-irrational points red, the reds will vastly outnumber the blues, no?

DrMatrix
05-04-2003, 12:21 PM
There are more irrational numbers than rational. The rational numbers are countable. That is, you can pair up the rational numbers with the integers. The irrational numbers are uncountable. If you try to pair up the integers (or the rational numbers) with the irrational numbers, you will have irrational numbers left over.

ultrafilter
05-04-2003, 10:22 PM
If you paint every rational pair blue, and every irrational red, you'd get a red plane. I don't know what you'd get if you painted (Q X (R - Q)) U ((R - Q) X Q) yellow.

Orbifold
05-04-2003, 11:07 PM
Well, (Q X (R-Q)) U ((R-Q) X Q) is measure 0 in the plane, unlike (R-Q) X (R-Q), even though both sets are dense and have the same cardinality. So perhaps the plane would still be red.

Now, if you let C be a non-measurable dense subset of R, and coloured every point in (R-C)X(C U Q) with fuschia paint mixed with day-glo glitter...

Hari Seldon
05-05-2003, 07:20 AM
If (and that is a big if) you could paint all the rational points blue and all the irrational points red, then the plane would be red since the probability of a randomly chosen point being rational is 0. So a random light ray would almost certainly hit an irrational point. But this whole exercise, unless it is there to illustrate measure theory is, no pun intended, pointless.

To answer another point raised, yes the irrationals are also dense. A set of points is dense if for every element x and every positive number r, some element of the set is within distance r of x. In other words, every element can be approximated arbitrarily well by elements of the set. The rationals are dense, the irrationals are dense, the integers are not dense.

ultrafilter
05-05-2003, 09:20 AM
Originally posted by Orbifold
Well, (Q X (R-Q)) U ((R-Q) X Q) is measure 0 in the plane, unlike (R-Q) X (R-Q), even though both sets are dense and have the same cardinality. So perhaps the plane would still be red.

Now, if you let C be a non-measurable dense subset of R, and coloured every point in (R-C)X(C U Q) with fuschia paint mixed with day-glo glitter...

I think the plane might be orange in the first case. I knew the set had measure 0, but I wasn't sure if it was dense (though I strongly suspected as much).