mm-241 === Subject: On the gravitation of zero point energyLecture 6 The gravitational influence of virtual zero point exotic vacuum fluctuation stress-energy density.J.S. All zero point vacuum fluctuations (ZPF) from any quantum fieldhasw = -1.J.H. Now you are contradicting yourself again. In between a capacitorwith the same sign charge on each plate there is a positive energydensity in the zpf. This is coming straight from the horse's mouth.Jack: Who? Hal Puthoff? This is irrelevant to what I said. Did you mean hishttp://www.ldolphin.org/zpe.htmlwhich completely misses the important idea of vacuum coherence.Those with a practical bent of mind may be left with yet one more unanswered question. Can this emerging Rosetta Stone of physics be used to translate such lofty insights into mundane application? Could the engineer of the future specialize in vacuum engineering? Could the energy crisis be solved by harnessing the energies of the zero-point sea? After all, since the basic zero-point energy form is highly random in nature, and tending towards of chaos, the, because of the highly energetic nature of the vacuum fluctuations, relatively large effects could in principle be produced. Given our relative ignorance at this point, we must fall back on a quote given by Podolny (12) when contemplating this same issue.Completely random zpf virtual photon fields do have positive zero pointenergy density.Therefore, they have negative pressure because in that case w = -1.pressure = w(energy density)However, the gravity influence of that zpf is(energy density)(1 + 3w)Therefore, in that situation you have repulsive anti-gravity.Charges on plates are completely irrelevant!In contrast the completely random virtual-electron vacuum polarizationhas negative zpf energy densityagain with w = -1 and that has positive pressure hence that will attractively gravitate.See Milonni's text book The Quantum Vacuum.All of the above, including Hal's http://www.ldolphin.org/zpe.html completely neglects the Vacuum Coherence Field which changes thestory!Let too(zpf)* be the total random zero point energy density from ALLphysical fields of all spins.The actual total zero point stress-energy density tensor is thentuv(zpf) = too(zpf)*[(Loop Quantum of Area)^3/2|Vacuum Coherence|^2 -1]guvThe vacuum coherence control parameter X = (Loop Quantum ofArea)^3/2|Vacuum Coherence|^2 has the domain 0 to a large real number1.Therefore X - 1 can be negative, zero or positive.The better way to look at this is the weak field GR Poisson equationlimiting case for the exotic zpf vacuumGrad^2V(exotic vacuum) ~ c^2/zpf/zpf = (Loop Quantum of Area)^-1[(Loop Quantum of Area)^3/2|VacuumCoherence|^2 - 1]Side problem for a sphere of exotic vacuum with uniform /zpf of radius R (e.g. the Galactic Halo when /zpf < 0)V = -GMeff/rmtur > RdV/dr = +GMeff1/r^2d^2V/dr^2 = -2GMeff/r^3GMeff = (4pi/3)c^2/R3There is no electric charge here. There is no Casimir effect here. That is a wrong turn off the true path. All that glitters is not gold. Skim milk masquerades as cream.Therefore /zpf < 0 === and matrices>Let V be a finite dim (say n>2) vector space over a field F.>>Suppose that G is a group of nxn invertible matrices with entries in F>satisfying the following property:>>If S,T are subspaces of V, there is a g in G such that g(S)=T. > Do you want to assume that S,T have the same dimension?>Then G contains all the matrices of determinant one. >>Is this true? I thank any proof or counterexample.> Counterexamples are not hard to find. We could take G to be a subgroup of> order 7 in GL(3,2). Since there are 7 subspaces of dimension 1> and 7 of dimension 2, then G is clearly transitive on both sets.> Derek Holt.I am sorry that I forgot to write that the subspaces S and T have the same dimension, and I thank Derek Holt for providingthe counterexample. Do these groups have a standard name? (Following the analogy with groups of permutations they could be called highly homogeneous linear groups or highly homogeneous classical groups but I couldfind no book speaking of such objects). Are these groups classifyed? A group G of nxn matrices with entries in the field F is said to be proper highly homogeneous linear group if it is an highly homogeneous linear group but does notcontain the special linear group (that is, the group of all nxn matrices of determinant 1). Are the proper highly homogeneous linear groups classifyed? I thank any === following true?Summation (from i = 0 to infinity) of Summation (from n = 0 to infinity) of(u^n i^n)/n!=(Summation (from i = 0 to infinity) of u^i) (Summation (from n = 0 toinfinity) of i^n/n!)I tried to do this term by term but there's no way I can prove it doingthat. Notice that I bring out the u^n and change it to u^i.I would love to be able to convert the first double sum into something ofthe form(Summation (from i = 0 to infinity) of u^i) (something)Is there any way I can do === the following true?>Summation (from i = 0 to infinity) of Summation (from n = 0 to infinity) of>(u^n i^n)/n! >(Summation (from i = 0 to infinity) of u^i) (Summation (from n = 0 to>infinity) of i^n/n!)Why should it be?Hint: u^n i^n = (u i)^n. The series converges only when u is negative (assuming === the following true?>>Summation (from i = 0 to infinity) of Summation (from n = 0 to infinity) of>(u^n i^n)/n!>=>(Summation (from i = 0 to infinity) of u^i) (Summation (from n = 0 to>infinity) of i^n/n!)Hint: u^n i^n = (u i)^n.That hint is OK.Then use: e^x = Summation (from n = 0 to infinity) of x^n /n!to simplify your formulas.> The series converges only when u is > negative (assuming real variables).The first series always diverge,second (geometric === What is JSH's problem?I'm trying to figure out what the fundamental thing is the JSH doesn't seemto grasp, and I *think* it is this, but his argument is so poorly presenthat I'm not sure. Is this really the trivial point he is missing?Essentially, he seems to think that: Given a ring R, with subring S, if ab=c, with a, b, c in S, then if xy=c, x,y in R, x and y must be in S.Is that it, or is there a more === What is JSH's problem?Tim Smith> I'm trying to figure out what the fundamental thing is the JSH doesn'tseem> to grasp, and I *think* it is this, but his argument is so poorlypresen> that I'm not sure. Is this really the trivial point he is missing?> Essentially, he seems to think that:> Given a ring R, with subring S, if ab=c, with a, b, c in S, then if> xy=c, x,y in R, x and y must be in S.> Is that it, or is there a more subtle false statement he thinks is true?See e.g.http://mathquest.com/discuss/sci.math/a/m/363567/ 363567Harris has been doing this sort of thing for eight === Lebesgue measurable subset of R with m(E) < INFTY and let f(x) = m((E+x) INTERSECTION E), where E+x = {x+y : y in E} Show that L.M.T f(x) = 0 x ->INFTY Can anyone give me a hint? === a Lebesgue measurable subset of R with m(E) < INFTY and let > f(x) = m((E+x) INTERSECTION E), where E+x = {x+y : y in E} > Show that L.M.T f(x) = 0> x ->INFTY> Can anyone give me a hint? Choose a bounded interval I such that m(E I) < === Sorry for troubling you guys.> Hey, can anyone help me with a proof of why 2,3,7,8 can't be the lastdigits> of a perfect square? I'm pretty sure that I should start of with the fact> that any number can be written in the form 10a+ b where 0<=b<=9. But that> means that b can't = 2,3,7, or 8. Here, I === anyone help me with a proof of why 2,3,7,8 can't be the last digits> of a perfect square? I'm pretty sure that I should start of with the fact> that any number can be written in the form 10a+ b where 0<=b<=9. But that> means that b can't = 2*2, ... 9*9 mod 10. Or (to explain this without the notion of mod) once you have that any number is written as 10a+b for some digit b, notice that the last digit of the square of 10a+b does not depend *at all* on what value 'a' is. So it should suffice to just look at the possible last digits when b is squared. Since 0<=b<=9, you only need to look at 0*0, 1*1, ..., 9*9, as === difference between Derive, Maple, Mathematica, MathCad,and other mathematical software? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0V4M8b17867;>What is the difference between Derive, Maple, Mathematica, MathCad,>and other mathematical checking out sci.math.symbolic. Just browse thru the title of posts on that forum or do a search. You might find an answer === sin(x)How does sin(x) work? How does it calculate opposite/hypot. by onlyknowing the angle? Is there some special formula for calculating theopposite side of a right triangle using 1 as the length of theadjacent by just knowing === Rocky Dean Pulley<5ae526a4.0401302044.92557c8@ calculate opposite/hypot. by only> knowing the angle? Is there some special formula for calculating the> opposite side of a right triangle using 1 as the length of the> adjacent by just knowing the value of the angle?> I'm not sure how a function definition works; it just is.sin(x) is the ratio of the side opposite over the hypotenuse, givena right triangle whose angle is x, 0 <= x <= pi/2 (or 90 degrees).For all other x, one can use the identities below:sin(pi - x) = sin(2*pi+x) = sin(x)sin(-x) = sin(pi + x) = sin(2*pi - x) = -sin(x)As for the actual computation of sin(x), one can employ theseries expansionsin(x) = x - x^3/3! + x^5/5! - x^7/7! + ...which is an outgrowth of the somewhat famous identitye^(ix) = cos(x) + i * sin(x)and the seriese^z = 1 + z + z^2/2! + z^3/3! + ...which is absolutely convergent for every z in the complex plane.Set z = i*x and collect terms.If you have the side adjacent as 1 you probably shouldn't be usingsin(x); you should be looking at tan(x). tan(x) is the ratio of sideopposite over side adjacent, and can be represen astan(x) = sin(x) / cos(x)Of course cos(x) is the side adjacent over the hypotenuse.It has a series expansion similar to sin's:cos(x) = 1 - x^2/2! + x^4/4! - x^6/6! + ...If you want to do trig another way (this one's harder butarguably more interesting), draw a chord in a unit circle(the hypotenuse or radius is 1) subtending angle 2*x,then bisect it with a line through the circle's center.Center-to-chord is cos(x). Chord-to-circle is 1 - cos(x).Chord length is 2 * sin(x). 2 * sin(x/2) is the hypotenuseof one of the new little triangles; therefore one canwrite the following identity almost by inspection,plus Pythagoras' theorem:4 * sin^2(x/2) = (1 - cos(x))^2 + (sin(x))^2 = (1 - 2*cos(x) + cos^2(x) + sin^2(x)) = 2 - 2*cos(x)thereforesin(x/2) = sqrt(2 - 2*cos(x)) / 2, for 0 <= x <= pi.Since sin^2(x/2) + cos^2(x/2) = 1, we can also derive:4 * (1 - cos^2(x/2)) = 2 - 2*cos(x)4 * (cos^2(x/2) - 1) = 2*cos(x) - 24 * (cos^2(x/2)) = 2*cos(x) + 2and, flipping the sum around for artistic reasons, we getcos(x/2) = sqrt(2 + 2*cos(x)) / 2You can also work out the formulassin(x+y) = sin(x) * cos(y) + cos(x) * sin(y)cos(x+y) = cos(x) * cos(y) - sin(x) * sin(y)using similar diagrams, or by noting e^(ix+iy) = e^(ix) * e^(iy) = (cos(x) + i*sin(x)) * (cos(y) + i*sin(y)) and grinding it out.You now can grind out trig to any desired precision. Say youwan sin(36). You know sin(90) = 1, cos(90) = 0. Expand36/90 = 4/10 in binary, getting4/10 = 0.01100110011...(2)or 36 = 1/4 * 90 + 1/8 * 90 + 1/64 * 90 + 1/128 * 90 + ...,compute the cos and sin of the desired angles, and sum.This is not the way computers do it, but it's interesting anyway.Another way of doing it would be to construct a regularpentagon (subtended angle = 72 degrees) and subtract itsangle from a 75-degree angle: 75 = 45 + 30; 30 = 60/2;60 degrees is the angle subtended by a hexagon. Ultimatelyone gets 3 degrees. Sadly, that's the best one can do witha straighge and compass unless one uses fractions of adegree (e.g. 3/8), or does things with 17- or 65537-gons,which *are* constructible.If one wants to approximately square the circle, onecan use the halving technique to get progressively finer;basically, an inscribed square P_2 in a unit circle has 4slices subtending 90 degrees and with area 2 * sin(pi/4)= 2.828427..., which is getting somewhat close to pialready; for any polygon P_i with area A_i, one can computea polygon P_{i+1} with twice the sides, and an area derivedfrom A_i, or perhaps A_i / 2^i, which is the isoscelestriangle slice. Ultimately one gets a representation ofpi using sums and radicals within suns and radicals.I'll leave the details to the interes reader. There aremore efficient methods of computing pi, but this will get yourfeet wet, and AFAIK closely parallels an argument known tothe ancient Greeks.HTH-- #191, ewill3@earthlink.netIt's === sin(x) work? How does it calculate opposite/hypot. by only> knowing the angle?For similar triangles, the ratio of the sides is the same.Now, suppose you have one right triangle with an angleat a vertex of 30_degrees and another, different sizedtriangle with and angle at one of its vertices of 30_degees.They are similar. The fraction of the length of the side oppositethe 30_degree divided by the length of the hypotenuse, say (or anytwo sides ... there is also the cosine and the tangent and the secantand the cosecant and cotangent, but we are looking at the sine)is the same for the two triangles.If you draw a right triangle with an angle of 30_degrees and verycarefully measure the side opposite the 30_degree angle andthe hypotenuse and divide those two and write it down, you justthat for the ratio of the side opposite the 30_degree angle dividedby the length of the hypotenuse for any right angle with a 30_angle.If you do this for angles of 1_degree, 2_degrees, etc. and writedown those ratios, you have just crea a table of sines.Whenever you have any right triangle, it will be similar to onelarge one for which you measured the sides (well, approximately,since you didn't to it for 30.3_degrees) and the ratio of its sidesSo, your table (which takes awhile to create, since you have to lookat all those triangles and measure their sides) will work for allright triangles.Originally, one could do something like that. Half angle formulasand addition theorems enable one to get approximations withoutactually having to do the measurements. As newer methods weredeveloped (the MacLaurin series for the sine) it becomes eveneasier to create the table, but it is the same table you wouldget by looking at a lot of right triangles so you would have onesimilar (or close to it) to any right triangle you might encounterand could use the ratio of the sides of that similar triangle forthe ratio of the sides of the triangle you encounter. All youhave to do is pick the one similar to the triangle you encounter.To do that, for a right triangle, all you need to know is one ofits other angles.> Is there some special formula for calculating the> opposite side of a right triangle using 1 as the length of the> adjacent by just knowing the value of the angle?The sine of the angle. There are expansions for it, but, onecould simply measure the ratio for a lot of triangles.So, are you interes in how folks figure out the ratio withoutdoing the measurements (I suspect that original tables, Arabian?,were, for a large part based on actual measurements of a lotof triangles, but I am not familiar with the history or methodsthey used)? One can find the value of the sine and cosine forsome angles exactly (e.g. 90_degrees, 30_degrees). One canuse the half angle formula to find the values for sine andcosine for half that (say 15_degrees) and then use the formulaagain to get half that (7.5_degrees) and half that (3.75_degrees)and half that (1.875_degrees) and half that .9375_degrees.Once one has the value for .9375_degrees, one can use additionformulas to get the sine and cosine for twice that, thenthree times it, then four times it, .etc.That will calculate the sines and cosines for a bunch ofangles and any angle you find will be within one degree ofone of those. If that is accurate enough (1 degree), you have it!Of course, it is a lot of calculation, but only involvessquare roots, addition, multiplication and division.Or one can use calculus and various series:(write X = angle_in_degrees*pi/180, a number which is the measure of the angle in RADIANS: sin(the_angle) = X/1-X^3/(3*2*1)+X^5/(5*4*3*2*1)-X^7/(7*6*5*4*3*2*1)+... but that series never ends. One adds up enough terms to get the accuracy one wants (there are more efficient formulas, too)).So ... conceptually, just measure lots of triangles.One could create an accurate table of sines and cosines withjust square roots, addition and dvision without measurements.There are other formulas. If you keep studying math, youwill come === folks figure out the ratio without> doing the measurements (I suspect that original tables, Arabian?,> were, for a large part based on actual measurements of a lot> of triangles, but I am not familiar with the history or methods> they used)? One can find the value of the sine and cosine for> some angles exactly (e.g. 90_degrees, 30_degrees). One can> use the half angle formula to find the values for sine and> cosine for half that (say 15_degrees) and then use the formula> again to get half that (7.5_degrees) and half that (3.75_degrees)> and half that (1.875_degrees) and half that .9375_degrees.> Once one has the value for .9375_degrees, one can use addition> formulas to get the sine and cosine for twice that, then> three times it, then four times it, .etc.I believe that historically, the values for 30 and 45 degreesgave the value for 75 degrees via a sum formula. Then, ananalysis of the pentagon can give the values at 72 degrees.Now using a difference formula, you can get an exact value for3 degrees. For a *long* time, the issue of how to trisect an anglehad the practical aspect of how to compute the trig functionsat 1 degree. This involves solving a cubic if you know the valuesat 3 degrees. Recall that all these values were done by hand andusually base 60. IIRC, al-Kindi calcula sin(1 degree)to several base 60 digits in the early === Is there software for this? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0V4qD320094;>Can the TI-89 do 3D vector field plots? Is there software for this?>What is the difference between Derive, Maple, Mathematica, MathCad,>and other mathematical I've worked with it, I don't see how you could do any vector plots. I think maple probably could though I haven't actually tried that. Why don't check out the web sites for those softwares to get an idea of what they can and can't do. === left to read them.> I think your method works.> It looks like it can be modified to determine if a TMRed flags went up in my mind when you said> you could find every unreachable node.> With a little work you might have a solution to the> halting problem.Russell> - 2 many 2 count> Great! I agree with everything you're saying - except your last> point. After you learned about Turing Machines, wasn't the next thing> that you learned Turing's 1937 proof of the unsolvability of the> Halting Problem? Or do you disagree with Turing?I don't necessarily disagree with him.I don't like proofs by contradiction.Turing's 1937 paper proved that it is impossible to determineif a TM will write an infinite number of ones starting with ablank tape.Your method can be adap to determine ifThis isn't all of the TM's that write an infinite number of 1's,but its a start.Russell- Zeno was right. Motion is === would seem that the Action integral for the world sheets of classical> string theory are presently justified as a higher dimensional version of> the one dimensional case. But I wonder if this formulation would be a> natural description if we were first given the geometry of a world-sheet. > What would be the most natural mathematical description of what is> happening with a world-sheet?of its world-line. The natural analog for a world sheet is simply its four-dimensional area, which is the simplest intrinsic invariant one can constructusing nothing but the geometry of the sheet.> I have a justification for a type of world-sheet geometry, but I am not sure> of the mathematics to descript it. At one instant of time there is a curve> through space with a function evalua at each point along the length of> the curve. This string sweeps out a surface as time passes. If I don't> know the path it takes or the function at each point on the surface, then it> would seem that the best I can do is to describe the situation with a> surface integral of a scalar function of some unspecified path. Such a scalar function would represent an additional variable, extrinsicto the geometry of the string. Since by hypothesis, strings describe_everything_ there is, there is no natural basis for such an extrinsicscalar function.-- Gordon D. === rational and the original sequencing, x_n, contains every> rational in (-1,1), then any translation of (-1,1) by a rational, x ->x> + q, with |q| < 1, produces a new x'_n which still contains every> rational in (-1+q,1+q), including 0.0 is provably not in x'_n.> Then there was some rational p in (-1,0) noy in x_n.Yes. x_n can not contain q.More, there is no linear translation of x_n that includesevery rational in (-1-r, 1-r) where r is irrational.You should really question if q is === Uncountable You should read the abstracts of some of these papers.>> The last paper lis states:>> In fact we prove that if $M$ is an inner model of set theory>> and the set of reals in $M$ is analytic then either>> all reals are in $M$ or else $aleph _1^M$ is countable>> Aha! That's what you thought was funny.>> Russell, you twit, they're talking about models of set theory within set>> theory, not the model of reals. The author is *not* contradicting>> Cantor.>I'm not contradicting Cantor, either.>Obviously, the reals are uncountable if the rationals are.>There are models of ZFC where the reals are countable.Not true. Before you can make such a statement, you first have tounderstand what the statement actually means. There are countable modelsof ZFC. Treating the interpretations of N and R within this model, theyare both countable in the background universe in which the model exists ASA SET. There are bijections, between N within the model and R within themodel, within the background universe in which the model exists as a set. In this manner, we have the countability of the interpretation within themodel of R. But the statement that R is countable is a statement inZFC, and in the interpretation within the model for ZFC, R is countable means that there is a bijection WITHIN THE MODEL between theinterpretation of N in the model and the interpretation of R in the model. So, in order that R is countable be satisfied in a model in ZFC, it isinsufficient that there be a bijection between the interpretation of N andthe interpretation of R - the bijection must also correspond to theinterpretation of a specific mapping within the model, i.e. there must bean element of the model which corresponds to the bijection. The truth ofthe matter is that no bijection between the interpretation of N and andthe interpretation of R has a corresponding element in the model, sowithin the model, R is countable is not satisfied, and R isuncountable is satisfied. To reiterate, when we say that R isuncountable is satisfied in the model, we are not saying that there is nobijection between N and the interpretation of R in the model. We aresaying that there is no bijection IN THE MODEL (i.e. that is an element ofthe model) between N and the interpretation of R in the model.So, although there are countable models of ZFC, R is uncountable in the models since no element of the model represents a bijection between the interpretations of N and R in the model.David McAnally Despite anything you may have heard to the contrary, the rain in Spain === are Uncountable> So, although there are countable models of ZFC, R is uncountable in the> models since no element of the model represents a bijection between the> interpretations of N the explaination.Russell- I don't want to be the hardest === intuition :) 3x+1 does indeed form perfect cubes.3[333] + 1 = 10^33[21]+1 = 4^3etc...Conjecture?: 2*X^X + 1 is an odd number for all positive integers X, X > 1, but 2*X^X + 1 cannot be a === Re: The Lost Proof of Fermat 2*X^X + 1 > is an odd number for all positive integers X, X > 1, but 2*X^X + 1 > cannot be a prime number, for all positive integers X, X > 1 Looking at a couple of examples is usually not good enough to conjecture... X=12, 2*X^X + 1 = 17832200896513 (prime) X=18, 2*X^X + 1 = 78692816150593075150849 (prime) X=251, 2*X^X + 1 = (prime)...and those are the only prime values of 2*X^X+1 with X from 2 to 467When you start looking at a hugely-growing function like X^X, the chance that you hit a prime way up there is really small since primes are fairly sparse in the huge numbers. It might be the case that these are the only prime values... or perhaps it magically hits a prime at some huge X, just by chance.J------------------------------------------------------ -----2*(254^254) + 1 is written below (over 600 digits long) 134522217314440067865082721113431223835597116498348716061446238 375265839912541075884015120618295553923123734598682607107221501 588542232800445122957538781619421535606676907561418735348987909 875230853003242454155623439526558859919442438334798856020852074 609466949570222487832520871038763354402812444586177504493136624 099131686522241644994576393898410721331696879945993957744698060 936312984917929306319261997188386027865023436733496356434270979 546669658860880444563014697473849556086585118658190745867979664 955754279787706976389082709834806488267139123512891635186071718 #1confused. Although our standard model of cosmologyhas been confirmed by recent observations, it stillhas a gaping hole: nobody knows why the expansion of the universeis accelerating.JS: I think I do. http://qedcorp.com/APS/EmergentGravity.pdfGD: If you throw a stone straight up, the pullof Earths gravity will cause it to slow down; it will not accelerateaway from the planet. Similarly, distant galaxies, thrownapart by the big bang expansion, should pull on one anotherand slow down. Yet they are accelerating apart. Researcherscommonly attribute the acceleration to some mysterious entitycalled dark energy, but there is little physics to back up thesefine words.JS: That is false in my opinion. An exotic vacuum phase with net total cumulative random positive zero point vacuum fluctuation energy density, hence negative pressure with w = -1, from all physical quantum fields does the trick. One must realize that the degree of randomness of the zero point vacuum fluctuations is tempered by the local inflation vacuum coherence field whose phase variation is the dominant smooth c-number non-perturbative background-independent geometrodynamic field of Einstein's 1915 general theory of relativity upon which precision cosmology is predica in the equationGuv + /zpfguv = -8pi(G/c^4)TuvGD: The only thing that is becoming clear is that at thelargest observable distances, gravity behaves in a rather strangeway, turning into a repulsive force.JS: Agreed. We can also, I bet, do this on a small scale for exotic warp drive time travel through traversable wormholes.GD: The laws of physics say that gravity is genera by matterand energy, so they attribute a strange sort of gravity to astrange sort of matter or energy. That is the rationale for darkenergy. But maybe the laws themselves need to be changed.JS: That is too cheap as Einstein mistakenly told Bohm. However, I am not mistaken I think in my opinion that a drastic overhaul of the known laws of physics is needed for this problem.GD: Physicists have a precedent for such a change: the law of gravitythat Newton formula in the 17th century, which had variousconceptual and experimental limitations, gave way to Einsteinsgeneral theory of relativity in 1915. Relativity, too, haslimitations; in particular, it runs into trouble when applied toextremely short distances, which are the domain of quantummechanics. Much as relativity subsumed Newtonian physics, aquantum theory of gravity will ultimately subsume relativity.Over the yeaphysicists have come up with a few plausibleapproaches to quantum gravity, the most prominent beingstring theory.JS: Smolin, Ashtekar, Baez, Rovelli et-al will strongly disagree on that against Greene, Witten, et-al. Of course neither string theory nor loop quantum gravity have made hard predictions of any facts nor have they provided compelling explanations of the observational mysteries of respectability and ideological purity reminiscent of Stalinism in the Soviet Union on the LANL Cornell Archive and in mass media like Scientific American and NOVA PBS are quick to embrace these radical speculations which in factare little more than pretty mathematical vaporware.GD: When gravity operates over microscopic distancesfor instance, at the center of a black hole, where a hugemass is packed into a subaic volumethe bizarre quantumproperties of matter come into play, and string theory describeshow the law of gravity changes.Over greater distances, string theorists have generally assumedthat quantum effects are unimportant. Yet the cosmologicaldiscoveries of the past several years have encouraged researchersto reconsider. Four years ago my colleagues and Iasked whether string theory would change the law of gravitynot just on the smallest scales but also on the largest ones. Thefeature of string theory that could bring about this revision iscan roam. The theory adds six or seven dimensions to the usualthree.JS: Theorists today are willing to pay any price to avoid signal nonlocality.GD: In the past, string theorists have argued that the extra dimensionsare too small for us to see or move in. But recent progressreveals that some or all of the new dimensions could actuallybe infinite in size. They are hidden from view not becauseand as a result, the law of gravity changes.Quintessence Even from NothingnessWHEN ASTRONOMERS ENCOUNTERED the cosmic acceleration,their first reaction was to attribute it to the so-called cosmologicalconstant. Notoriously introduced and then retracby Einstein, the constant represents the energy inherent in spaceMaybe cosmic acceleration isnt caused by dark energy after allbut by an inexorable leakage of gravity out of our worldGD: A completely empty volume of space, devoid of all matter,would still contain this energyequivalent to roughly 10[CapitalEth]26kilogram per cubic meter. Although the cosmological constantis consistent with all the existing data so far, many physicists findit unsatisfying. The problem is its inexplicable smallness,JS: This is only a problem because The Pundits have not properly used the idea of vacuum coherence in which the cosmological term in Einstein's equation has a subsidiary equation/zpf = (Quantum of Area)^-1[(Quantum of Area)^3/2|Vacuum Coherence|^2 - 1]Where the tetrad gravity field Cartan 1-form in the sense of Rovelli's book on Quantum Gravity iseuadx^a = Kronecker Deltau^adx^a + (Quantum of Area)(argVacuum Coherence),uA vanishing Vacuum Coherence means maximally random zero point energy fluctuations from all fields in unstable globally flat #1> confused. Although our standard model of cosmology> has been confirmed by recent observations, it still> has a gaping hole: nobody knows why the expansion of the> universe is accelerating.> JS: I think I do.The sin' lunatic is at least worth a few chuckles now and then,> [snipped cloud of primeval gas]The galaxies are accelerating outward because the universe is a bubble inside an infinitely dense mass of packed quantum singularities, each of which is a previous galaxy. That messes up space such that the outward direction is always downhill, so the universe is literally falling apart. This might seem a bit too speculative, or even dumb as hell, but it's a lot shorter === PoserLet a, b, and c = 127[(1+1)(1+1)(1+1)]^2 >= 64 *1*1*1 (1+1+1)^327* [2*2*2]^2 >= 64 * 27okay, let a, b, and c = .527 [1]^2 >= 64*.5*.5*.5 * 1.5^3Dagnabbit.> Show that, for real a,b,c > 0, 27[(a + b)(b + c)(c + a)]^2 >= 64abc(a +> b + === mathbb{Z}_2I am looking for the solution to this for a long time. Usually this isa problem on one of the problem sheets to students. As I am writingsomething about Quantum Field Theory, I don't want to dive in too deepinto this. I just want to understand mathematically precisely whypi_1(SO(3)) = mathbb{Z}_2.On the web, I have just found statements of facts or visualisations,but no rigorous proofs.The motivation is to mathematically, and thus intuitively, understandthe difference between SU(2) and SO(3) (simply connec vs. notsimply connec, etc.). Basically, with the help of this proof, Iwant to understand, what the difference between the structures ofthose two groups is.I would appreciate every explanation you could === being mathbb{Z}_2> I am looking for the solution to this for a long time. Usually this is> a problem on one of the problem sheets to students. As I am writing> something about Quantum Field Theory, I don't want to dive in too deep> into this. I just want to understand mathematically precisely why> pi_1(SO(3)) = mathbb{Z}_2.> On the web, I have just found statements of facts or visualisations,> but no rigorous proofs.Here's a simple, if long-winded, illustration of why SO(3) hasfundamental group Z_2.SO(3) is the group of (proper) rotations of R^3. It isn't difficult toshow that every rotation of R^3 (and more generally, every rotation ofR^(odd)) has an axis of rotation, that is, a 1-dimensional subspace thatis fixed pointwise by the rotation. Further, in SO(3) each rotation hasan angle through which the orthogonal subspace is rota. If you specify an orientation of the axis, that angle is uniquely determined(by the right-hand rule, for instance) as a real number modulo 2 pi. Forconvience's sake, it makes sense to view the representative of theequivalence class (mod 2 pi) as lying between -pi and +pi. Note that onecan take the two characteristics (axis, angle) of a given rotation anddetermine a unique element of the (solid) ball of radius pi, centered atthe origin, with the exception that antipodal points on the boundarysphere represent the same rotation.This model for SO(3) identifies SO(3) with the solid ball of radius pi,with antipodal points on the boundary identified. The quotient space isa standard model for the projective space of dimension 3 over the reals,commonly deno RP^3. It is topologically the same space as thatobtained from the three-dimensional sphere S^3 by identifying antipodalpoints (the equivalence is relatively simple to show).This latter model for RP^3 provides you with a mapping S^3 ---> RP^3,which is locally a homeomorphism (each point of S^3 has a neighborhoodover which the mapping is a homeomorphism), and for which each point ofRP^3 has a preimage comprising two points. In other words, the mappingS^3 ---> RP^3 is a covering map of degree 2. The long exact sequence ofthe mapping: ...--> pi_i (S^3) --> pi_i (RP^3) --> pi_(i-1)(Z_2) --> pi_(i-1) (S^3) --> pi_(i-1) (RP^3) --> ...is trivial for dimensions > 1, since pi_i (Z_2) = 0 for i>0, and itends in the sequence: pi_1(Z_2) --> pi_1(S^3) --> pi_1(RP^3) --> pi_0(Z_2) --> pi_0(S^3) --> pi_0(RP^3)The fact that both S^3 and RP^3 are path-connec makes the two finalsets (pi_0 doesn't usually have a group structure) one-point sets, andthe map is a bijection; this forces the homomorphism pi_1(RP^3) --> pi_0(Z_2)to be a surjection (pi_0(Z_2) = Z_2, with the group structure, and themap above is a homomorphism). Further, S^3 is simply-connec, soyou get a monomorphism; together these facts force the homomorphismto be an isomorphism.Thus pi_1(SO(3)) = Z_2.> The motivation is to mathematically, and thus intuitively, understand> the difference between SU(2) and SO(3) (simply connec vs. not> simply connec, etc.). Basically, with the help of this proof, I> want to understand, what the difference between the structures of> those two groups is.Simple-connectivity for SU(2) follows from the following argument: SU(1) = {1}, since U(1), the unitary group in dimension 1 is the circle group S^1, and SU(1) is the set of elements of U(1) with determinant 1. SU(2) acts by multiplication on complex space of dimension 2, and preserves the norm of vectors; it thus acts on the unit sphere of that space, which happens to be S^3. By evaluating this action at your favorite point x in S^3, you get a map SU(2) ---> S^3. It is also worth noting that for any other point y of S^3, there is an element of SU(2) that maps x to y. That is, the map is a surjection. The isotropy subgroup I_x of the point x (that is, the set of all A in SU(2) that map x to itself) fixes the complex line Cx, passing through x in C^2, pointwise, and also maps its hermitian complement to itself. The restriction of that map is again a unitary mapping. Since every element A of I_x has determinant 1, and since A is the identity along Cx, the determinant of the restriction to the complementary subspace is also 1. In other words, the group I_x fixing the element x in S^3 is a copy of SU(1) (since it's unitary and has determinant 1). However, SU(1) is the trivial group. The implication is that the quotient mapping I_x ---> SU(2) ---> S^3 has each point-preimage equal to a single point. Thus, the evaluation mapping described above is a homeomorphism. So, SU(2) is homeomorphic to S^3, and so is simply-connec.> I would appreciate every explanation you could give meI hope this helps. Sorry if the arguments were too topological, or === fundamental group of SO(3) being mathbb{Z}_2|I am looking for the solution to this for a long time. Usually this is|a problem on one of the problem sheets to students. As I am writing|something about Quantum Field Theory, I don't want to dive in too deep|into this. I just want to understand mathematically precisely why|pi_1(SO(3)) = mathbb{Z}_2.||On the web, I have just found statements of facts or visualisations,|but no rigorous proofs.|||The motivation is to mathematically, and thus intuitively, understand|the difference between SU(2) and SO(3) (simply connec vs. not|simply connec, etc.). Basically, with the help of this proof, I|want to understand, what the difference between the structures of|those two groups is.||I would appreciate every explanation you could give me1. su(2) is homeomorphic to the 3-sphere.2. the fundamental group of the 3-sphere is trivial.3. su(2) is a connec 2-fold covering space of so(3).4. whenever a space has a connec 2-fold covering space withtrivial fundamental group, the fundamental group of the original spaceis z/2.is there one of these that you have trouble with? or do you havetrouble with how 1-4 together imply that the fundamental group ofso(3) is === sqrt(-1)=0/0In sci.logic, ZZBunker> 0/0 is every number, 0 is their placeholder.> 0/0=undefined, so 0 cannot be a place holder for all x in N, I or R. > 0> denotes the absence of quantity which is instrinsically a quantity> (ie. nothing is something).> Nothing has no constituent parts, and therefore cannot be divided by> or multiplied against...*> Do not equate 'nothing' with 'zero'.earle> *Zero denotes the absence of quantity which is intrisically a quantity,> since osophically, nothing is something. *>> 'Nothing' is the opposite of 'something'. 'Zero' is a point on the number line, like 6, -23, pi, and 1,931.>> Pedant point: the number line doesn't exist either, at least not>> as a physical object. Then again, the only reason zero needs>> to be specially trea anyway is because it *is* the arithmetic>> identity and therefore X * 0 = 0 for all X in the field.>> No other number in the field has that property.> Double pendant point. That's not true, > since zero is not handled in any proof.> Only it's current day dork symbol is > handled. It's handled like gravity.> Gently, gently, move it a little bit to the left, > move it a little bit to right, Then move it faster. > Then all of sudden, vailo, it's magic. The result is > a Quantum Chemist Dork gibbering about the holistic > CONSCIOUSNESS of parallel universes, and cold fusion.You're not making a whole lot of sense here.How are we supposed to handle numbers?How are we supposed to handle zero?[rest snipped]-- #191, ewill3@earthlink.netIt's still legal to go === Abhi:>> In sci.math, Abhi>> built this Action Device. It is on floor of my room. I>> pulled the springs to generate enough restoring force.>> I marked the position of upper end of bolt. Closing one eye, I poin>> on this mark and I rota the nut.>> The upper end of bolt i.e. point D is moving in upward direction as I>> predic here..>> http://www.geocities.com/inertial_propulsion>> First Result: It Does Work!>> That particular device looks like a bunch of vectors pointing>> every whichaway and maybe a V and a very distor W.>> Do you have a picture of this device?>> I could see crawling devices working without too much>> trouble; ABCG is on the ground; pull F and H apart>> while holding point G off the ground. (A and C are>> castor rollers.) Then put G on the ground while FGH is>> a straight line; lift B off. Push F and H together (or>> let them spring back); they'll move backwards. Lift G>> off the ground and repeat.>> It won't be the fastest of crawls, admitly; you'd get>> better results from a treadmill.>> [rest hoke bula lo mujhe chaaho jis waqt> maiN gaya waqt naheeN hooN ke fir aa bhee na read gibberish. Or is that Sanskrit?Please transmit in English.> -Abhi.-- #191, ewill3@earthlink.netIt's still legal === sci.math, hinoon:> 1+1/8+1/4+1/32+1/16+1/128+1/64+...> I try to check this by detecting n-th term and applying> root test.> But, I don't know what is n-th term.It would appear that this series is:t_0 = 1t_n = 1/2^(n+2) if n is odd = 1/2^n if n is evenSince this series is absolutely convergent (0 <= t_n <= 1/2^n,terms of a known absolutely convergent series),we can play all sorts of games. The simplest is arguably tosplit it into three parts:11/8 + 1/32 + ... = 1/8 * (1 + 1/4 + 1/16 + ...) = 1/241/4 + 1/16 + ... = 1/4 * (1 + 1/4 + 1/16 + ...) = 1/12and therefore the sum is 1 + 1/24 + 1/12 = 1 + 1/8 = 1.125.> If someone know how can I know convergence(or divergence) of> ewill3@earthlink.netIt's still legal to go === The Machine escribi.97:> In sci.math, hinoon> :>> 1+1/8+1/4+1/32+1/16+1/128+1/64+...>> I try to check this by detecting n-th term and applying>> root test.>> But, I don't know what is n-th term.> It would appear that this series is:> t_0 = 1> t_n = 1/2^(n+2) if n is odd> = 1/2^n if n is even> Since this series is absolutely convergent (0 <= t_n <= 1/2^n,> terms of a known absolutely convergent series),> we can play all sorts of games. The simplest is arguably to> split it into three parts:> 1> 1/8 + 1/32 + ... = 1/8 * (1 + 1/4 + 1/16 + ...) = 1/24> 1/4 + 1/16 + ... = 1/4 * (1 + 1/4 + 1/16 + ...) = 1/121/8 + 1/32 + ... = 1/8 * (1 + 1/4 + 1/16 + ...) = 1/8(1/(1 - 1/4) = 1/61/4 + 1/16 + ... = 1/4 * (1 + 1/4 + 1/16 + ...) = 1/4(1/(1 - 1/4) = 1/3> and therefore the sum is 1 + 1/24 + 1/12 = 1 + 1/8 = 1.125.and therefore the sum is 1 + 1/6 + 1/3 = 1 + 1/2 = 3/2.Alternativally, as the series is absolutely convergent, its sum isSum(1/2^n, n, 0, inf) - 1/2 = 1/(1 - 1/2) - 1/2 = 2 - 1/2 = 3/2-- Ignacio Larrosa Ca.96estroA Coru.96a === Is this series converge ?> 1+1/8+1/4+1/32+1/16+1/128+1/64+...> I try to check this by detecting n-th term and applying> root test.> But, I don't know what is n-th term.> If someone know how can I know convergence(or divergence) of> this series, please post reply. Well, if you compare it to 1 + 1/2 + 1/4 + 1/8 + 1/16 + ... you will see that a(n) of your series is less than b(n) of this second series. And once you have established convergence (and thus absolute convergence) the sum is fairly easy to find (look at all the odd terms of your series, and === quotientan integer?> for k, p, m, n in N*, let> B(k,p,m,n)=((k+p+1)m)!((k+p+1)n)! / [(pm)!(pn)!(km+n)!(kn+m)!]> Can somebody proove that it is an help.> Amically,sB(1,3,4,5) = 1372552324000/3 (Note this is not an integer.)Well, I tried B(k,p,m,n) for the 10^4 values of (k,p,m,n) with each variabletaking values from 1 to 10 and exactly 9334 times obtained an integer. Inthe 4-tuples not === n in N*, let> B(k,p,m,n)=((k+p+1)m)!((k+p+1)n)! / [(pm)!(pn)!(km+n)!(kn+m)!]> Can somebody proove that it is an integer (if not, is it for some values> of k)?> B(1,3,4,5) = 1372552324000/3 (Note this is not an integer.)> Well, I tried B(k,p,m,n) for the 10^4 values of (k,p,m,n) with each variable> taking values from 1 to 10 and exactly 9334 times obtained an integer. In> the 4-tuples not yielding an integer I don't see a pattern.OK I ask now :for k=1,2,3 and m,n in N*, m escribi.97:>> for k, p, m, n in N*, let>> B(k,p,m,n)=((k+p+1)m)!((k+p+1)n)! / [(pm)!(pn)!(km+n)!(kn+m)!]>> Can somebody proove that it is an any help.>> Amically,>s> B(1,3,4,5) = 1372552324000/3 (Note this is not an integer.)> Well, I tried B(k,p,m,n) for the 10^4 values of (k,p,m,n) with each> variable taking values from 1 to 10 and exactly 9334 times obtained> an integer. In the 4-tuples not yielding an integer I don't see a> pattern.Yes I get the same result. But Georges made the problem too general. Theinitial question, from other newsgroup, was with k = 3 and p = 1. I.e., showthatB(m, n) = (5m)!(5n)!/(m!n!(3m + n)!(3n + m)!)is integer for all m, n >= 1 and integers.It is true for all m, n <= 25-- Ignacio Larrosa Ca.96estroA Coru.96a === headed S-curvesFor those who won't do their own homework. Take the inverse function off(x)=x^3 and transform it linearly ieI'm going to use f over again:f(x)=x^(1/3) the cuberoot of xg(x)=((x-1)^(1/3))+1 makes an S between 0 and 1g(0)=0,g(1)=1let h(x)=g((x-a)/(b-a)) then h(a)=0 and h(b)=1the inflection point is steep though.also cos(x) makes a nice S between pi and 2*piso [1+cos(pi*(x-2a+b)/(b-a))]/2 makes === Codes and parity-check matrix?I want to extend these principles to codes over GF(2^m). I tried an> example using codes over GF(2^4). I facorized x^13 + 1 in GF(2^m),> and obtained the following factors (using MAGMA) (these will each> generate a (13,10) cyclic code with d_{min} = 4):> Why should these codes necessarily have d_min=4? May be they do, I'm> just curious, because I didn't see how this would follow from the BCH-bound> or any bound that I know of?> f1 = x^3 + z^7*x^2 + z^13*x + 1;> f2 = x^3 + z^11*x^2 + z^14*x + 1;> f3 = x^3 + z^13*x^2 + z^7*x + 1;> f4 = x^3 + z^14*x^2 + z^11*x + 1;(z = alpha)> What exactly is alpha here? My guess is that it is a root of > x^4+x+1=0, but unless you know that you are fumbling in the dark,> and it is hard to guess/reproduce what might have gone wrong.> When I factored f1 over GF(2^12), I got the following:> x + z^568> x + z^3625> x + z^3997> What is z this time? Is it again something called 'alpha'. Surely> you are aware of the fact that this 'alpha' is different from the> 'alpha' in the construction of GF(16)? Your notation suggests that> you always want alpha to be a primitive element of the field. So,> if beta=z is a primitive element of GF(2^12), then alpha=z^273 is> a primitive element of GF(2^4) (caveat: this alpha may have a different> minimal polynomial, i.e. it may now be that z^273 is a zero of> x^4+x^3+1=0 instead of x^4+x+1=0. (273 here = 4095/15).> Something very fishy about these zeros anyway. If f1 is a factor> of x^13+1, then all these roots should have order 13, i.e. the exponents> should be multiples of 4095/13=315, but they aren't.However, when I multiply a polynomial p(x) with f1 in GF(2^4) to> obtain c(x) and then substitute the roots (z^568, z^3625, z^3997) into> c(x) in GF(2^12), the results are not zero. Why doesn't this work?> No wonder! I guess that what happened is that you have named two different> things 'alpha' (or z), and this has lead to a confusion.> How can I change this? (I know that the fields are genera using> Conway polynomials -> Should I try to use another polynomial to> generate the GF(2^12) field?) Also, multiplying f1*p in GF(2^4)> results in a different polynomial than when I multiply f1*p in> GF(2^12). Why does this work for the binary case, but doesn't work> for the non-binary case?> In binary case no powers of alpha appear as coefficients, so you didn't> have a chance to make this mistake.> What you should do, is to > 1) generate the field GF(2^12), call a primitive element alpha=z> 2) identify the subfield GF(2^4) within this field: let beta=z^273,> then GF(2^4)=GF(2)(beta)> 3) you may want to find out the minimal polynomial of beta, so that> you can properly map elements of this subfield into a copy of GF(2^4)> genera by some other means.> If any of this is not crystal clear, then follow the earlier advice> and study a book on abstract algebra.> Good luck!> Jyrki Lahtonen, Turku, FinlandLast year I did a third year Abstract Algebra course. It was anintroduction to Abstract Algebra, so we only touched the concepts. Wedid learn about subgroups and subrings, and maybe this ideas can beextended to fields as well... (We used Contempory Abstract Algebrafrom J.A. Gallian. We covered only the first 15 chapters in thesemester. The book was excellent, for me anyways.) Can you recommendany book that will help me with the above problem.I know that GF(2^12) is genera by a binary primitive polynomial,(don't know which one was used by MAGMA, but I can find out). Also,for an m there are many fields GF(2^m), each genera by a differentprimitive polynomial of degree m -> thus, there exists an isomorphismbetween the fields, to use the correct mathematical terminology.If I understand you correctly, I need to find the primitive polynomialused to generate GF(2^4) such that GF(2^4) will be a subfield of theGF(2^12) that was used with MAGMA. The reason why the above examplefailed was because the GF(2^4) was genera by a different primitivepolynomial than the GF(2^4) appreciate it greatlyJacops.I can be reached at: JV at ING dot === manifolds> I have a question about the cross product of two orientable smooth manifolds.> Proof that M and N are two orientable smooth manifolds if and only if MxN is an answeJorge.Surely you don't mean to suggest that if MxN is a manifold, that M and Nmust be manifolds?If I recall correctly, there are spaces K such that KxR is R^4, but suchthat K is not a manifold. Bing's dogbone space is an === smooth manifolds>>> I have a question about the cross product of two orientable smooth>> manifolds.>> Proof that M and N are two orientable smooth manifolds if and only if MxN>> answeJorge.> If you're dealing with the case of _closed_, connec manifolds> M and N, you can give a relatively simple proof of the fact that> M, N orientable implies MxN orientable -- the result follows from> the fact that H_n(M^n;Z) = Z if M is orientable and it's trivial> otherwise. Combine that with the fact that the Kuenneth Formula> implies that> H_m(M^m) tensor H_n(N^n) = H_{m+n}(MxN)> if M, N are orientable (this also needs the fact that H_{m-1}(M^m)> is free abelian if M is orientable).But as we are only concerned with smooth manifolds, then we don'tneed this apparatus of homology at all. A smooth n-manifold is orientableiff the n-th exterior power of the tangent bundle has a nonzero(smooth) section. I'll call this line bundle the orientation bundle O(M).If M and N are orien, then O(M) and O(N) define subbundles of O(M x N).Just exterior multiplying nonzero sections of these gives a nonzerosection of O(MxN).Now suppose that M is nonorientable. Then there will be a closed path P in Mand a section of O(M) along P such that the sign of the section at the ends of the path is negative. One can consider this as a path in M x {n}and multiply the orientation field by a nonzero element of O(N) at ngives such a path for M x N showing that that's nonorientable.(OK one can do all this for general manifolds but the construction ofan appropriate orientation bundle is harder, requiring local === Lebesgue measurable subset of R with m(E) < INFTY; and let> f(x) = m((E+x) INTERSECTTION E), where E+x = {x+y : y IN E}> Show that L.M.T f(x) = 0> x -> INFTY> Could you please give me any hint...? Thanx!Recall the definition of the outer measure: $m^*(E)=infsum_n|I_n|$ wherethe $inf$ runs over all countable collections of intervals ${I_n}$ suchthat $Esubsetcup_n I_n$. By the construction of the Lebesgue measure andbecause $E$ is measureable, $m(E)=m^*(E)$. So by the definition of theinfimum, for any $delta>0$ there exists a collection of intervals${J_n}$ such that $Esubset cup_n J_n$ and $m(E)+delta> sum_n |J_n|$.In particular, $sum_n|J_n|$ converges. So there exists $N$ such that$sum_{ngeq N}|J_n|M-m$: First note$$(cup_{n Let E be a Lebesgue measurable subset of R with m(E) < INFTY; and let> f(x) = m((E+x) INTERSECTTION E), where E+x = {x+y : y IN E}> Show that L.M.T f(x) = 0> x -> INFTY> Could you please give me any hint...? Thanx!Given e>0, there exists a compact set K such that the measure of the === Derived group> I can't sleep because of one problem:> We have nonabelian group G of order 24,> and we know that it has> a normal subgroup H of order 8.> Problem: Find derived subgroup G'> (aka [G:G] - commutator subgroup)> Any ideas?> Is it possible to definitely answer> this question at all?> (Of course without checking by hand> all 12 nonabelian groups of order 24 :) )Well, actually there are only 4 non-abelian groups G of order 24with a normal subgroup H of order 8, so checking them all byhand isn't too hard.> Surely it must be subgroup of H,> (|G/H|=3, so G/H is abelian)> but I think it's really smaller,> probably cyclic group of order 2.No. For example, G = SL(2,3), which is the semi-direct productQ_8 x| C_3 with presentation , has derived group [G,G] = which is its normal subgroup H =~ Q_8. (This is clearsince none of the 3 subgroups of order 4 is normal in G, and theonly subgroup of order 2 is , but G/ =~ PSL(2,3) =~A_4 is non-abelian.)The other counter-example, with |[G,G]| = 4, is the directproduct G = A_4 x C_2.The remaining 2 groups, namely the direct products G = Q_8 x C_3and G = D_8 x C_3, do both have [G,G] cyclic of order 2.> Are we able to say something === inven?According to Georges Ifrah in his monumental Histoire universelle deschiffres, zero was discovered 4 times independently through mankindshistory.1.- By the Babylonians at the beginning of the second millenium B.C.2.- By the Chinese shortly before the beginning of the Christian era.3.- Sometime between the III and V centuries c.E.4.- By the Indians sometime about the V century c.E.However, only the Babylonians, Mayas and Indians did actuallyimplement positional number systems. The chinese did only so === zero inven?> One is a number for all practical purposes.> Because with one, comes minus one.> Zero is not even a number.> It's a set of NOTHING, idiots.>> As has been mentioned in this thread already, zero IS a number and>> is very different from the empty set (that is, it is NOT a set of>> NOTHING, as you say.)> In set theory, zero is usually defined as the empty set.Mmmm - zero, about what your trolling is worth.Russell--Are you still here? The message is over. === number for all practical purposes.> Because with one, comes minus one.> Zero is not even a number. > It's a set of NOTHING, idiots. As has been mentioned in this thread already, zero IS a number and is > very different from the empty set (that is, it is NOT a set of NOTHING, as > you say.)An example you might be able to understand: When someone says it is zero> degrees outside today it does NOT mean that there is no temperature> today.> That follows trivially from your use of the > words, degrees, outside, today idiot, not from your idiot> suggestion that there is an absolute zero temperature.It seems that there is an absolute zero === number for all practical purposes.> Because with one, comes minus one.> Zero is not even a number. > It's a set of NOTHING, idiots. As has been mentioned in this thread already, zero IS a number and is > very different from the empty set (that is, it is NOT a set of NOTHING, as > you say.)An example you might be able to understand: When someone says it is zero> degrees outside today it does NOT mean that there is no temperature> today. That follows trivially from your use of the > words, degrees, outside, today idiot, not from your idiot> suggestion that there is an absolute zero temperature.> It seems that there is an absolute zero temperature. But, since it only seems that way in classical physics, it also seems that Einstein is a wanker. Which is well-known, but also worth repeating to math idiots. If you can't find an eraser for that === Non-standard analysis >> Well, yes. Hence one of your assumptions must be erroneous. How do>> you express finite set in first-order logic? It's well known that>> in second-order logic (finite set is your clue) there's only on>> model of the reals, and hence of the naturals.>> If you change your sentences to Ex (x>n), (clearly expressible in>> first-order logic, since it's expressed in first-order logic), then>> the contradiction goes away.> [S finite implies by definition that there exists some natural> number n, such that S can be mapped 1-1 onto In. But S larger than> n implies, again by definition, that S can be mapped 1-1 onto Im> for some natural number m > n]>> But how do you state this in the language you're considering? > 1. Is it fair to say that the nub of the problem is that the> definitions of finite and larger that I have sugges cannot be> formula in first-order logic?> 2. Robinson also shows compactness for higher order structures and> correspondiing languages. So my definitions of finite and larger> cannot be formula in a higher order language?> 3. Presumably similar problems occur with the following set of> sentences S?:> S1. There exists a set, T, of cardinality greater than 1 and strictly> less than aleph0.> S2. There exists a set, T, of cardinality greater than 2 and> strictly less than aleph0.> ...> Sn. There exists a set, T, of cardinality greater than n and strictly> less than aleph0.> ...> Any finite subset of S is consistent, therefore there exists a model> of S with set T of cardinality greater than n for any n but strictly> less than aleph0.I didn't follow the rest of it, but yes. In Nelson's IST, it is an easyconsequence of the idealization axiom that for all E, there exist a finiteset F containing all the standard elements of E ... If you use Von Neumanndefiniton of ordinals, you can take as T === <>sSHfTy;{Dhe&:+?b`9fUj5A~$gIYlYT0/$-asR-K~3S3[]q.R3YSmpR|$- GiZp>UN2a}!Fmw+%h}YL`!h_XXr5Q>_nGsY2_> 1. Is it fair to say that the nub of the problem is that the> definitions of finite and larger that I have sugges cannot be> formula in first-order logic?I think you would really benefit from reading Chapter 1 ofRobinson's book NON-STANDARD === Re: Choosing a Math Grad school> Hi guys,> I'd like some info about math PhD programs, if anyone can please help. A> concern I have are rankings, in particular: how much do they matter when you> make *final* decisions?> I've my hopes fixed on one among: Berkeley, UCLA, Penn State, Chicago,> Texas-Austin, Princeton, Illinois-Urbana. My area of interest is operator> algebras (thus the first three) or Harmonic analysis and PDE (thus the> latter four). Depending on place-offers (and assuming that one of them can> be excluded), do any of you guys know reasons for preferring one to the> other? (modulo that these are all great places for analysis research).> For instance, suppose you get an offer from UCLA and Illinois-Urbana or Penn> State, do you obviously go for UCLA because it is higher ranked? My> intention is to follow an academic career (professorship), so future> placement particularly from mathematicians that have been or> have visi the above mentioned universities and may be able to give me> some hands-on impressions! Well if you object to gravity research pick Princeton over Penn State. If you to object to Computer Research pick Illinois-Urbana over Princeton. If you object to Energy Research, pick UCLA over all of them. > Please email comments to:> hcanab@yahoo.com> === clarify: I would eventually like to become a professor at aresearch university. So I presume that rankings matter to a certain degree.What I am unsure of is if, say, it would be much easier to get aprofessorship having got a PhD degree from Berkeley or having got it from,say, Illinois-Urbana or Penn State (assuming all else is equal, i.e.publications, good advisoetc.). In other words, how much is the systembiased?When looking at faculty lists at top universities, profs usually obtainedtheir PhD at top universities, but maybe that's just because in the pastthere was less math research and it was more concentra at the oldschools, while today it's more spread === support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0V31Kq11613;>> I'd like some info about math PhD programs, if anyone can>> please help. A concern I have are rankings, in particular:>> how much do they matter when you make *final* decisions?> The rankings matter a whole lot less than the fit with your> interests, particularly the faculty.that is, unless the lad expects an academic job === publication of research should not peer review becalled rivals' review? A referee may read a paper and benefit fromit and then decide that others should not have the same privilege. Consider also the story of Darwin and Wallace who arrived at thetheory of natural selection at the same time. Wallace published firstand prior publication is supposed to be what counts. But he publishedto Darwin and although Darwin gave him some credit the theory todaybears Darwin's name and not Wallace's. We must be grateful that after many years we have this forum wherewe may be contradic but cannot be === also the story of Darwin and Wallace who arrived at the> theory of natural selection at the same time. Wallace published first> and prior publication is supposed to be what counts. But he published> to Darwin and although Darwin gave him some credit the theory today> bears Darwin's name and not Wallace's.Do you have a reference for this story.-- Jesse HughesSuch behaviour is exclusively confined to functions inven bymathematicians for the sake of causing trouble. -Albert Eagle's _A Practical === ReviewX-DMCA-Notifications: http://www.giganews.com/info/dmca.html>In paper-based publication of research should not peer review be>called rivals' review? A referee may read a paper and benefit from>it and then decide that others should not have the same privilege.You know of any specific examples where this has happened?> Consider also the story of Darwin and Wallace who arrived at the>theory of natural selection at the same time. Wallace published first>and prior publication is supposed to be what counts. But he published>to Darwin and although Darwin gave him some credit the theory today>bears Darwin's name and not Wallace's.> We must be grateful that after many years we have this forum where>we may be contradic but cannot be === In paper-based publication of research should not peer review be> called rivals' review? A referee may read a paper and benefit from> it and then decide that others should not have the same privilege.(preprints) to his colleagues in the field, and nowadays postsit on his web page. So if the paper is rejec for reasons such asthose you fear, everyone in the field will still know he was first.The other side is this: When a bad paper is submit, then rejec,it is natural and common for the author to blame the rejection onanything he can, such as incompetent referees. While there is surelysome incompetent refereeing, that is minuscule compared to the number ofactual bad papers submit to mathematics journals.> Consider also the story of Darwin and Wallace who arrived at the> theory of natural selection at the same time. Wallace published first> and prior publication is supposed to be what counts. But he published> to Darwin and although Darwin gave him some credit the theory today> bears Darwin's name and not Wallace's.> We must be grateful that after many years we have this forum === Re: Please advise! what kind of training I am lackfor math?> So I am at a very akward stage now: if give me undergraduate math to read, I> feel too easy and boring, if give me difficult math to read as those in> Information Theory Transactions, I feel dizzy...I don't know about information theory, but I did once take an intro classon real analysis. I find that to be a great aid in reading contemporary === problem... please help me! let me try> to explain my problem to you:> For years I have been headache about reading math notations. A concept, if> it is put in straightforward way, I can understand. But if it is put in a> math notation, or even advanced math notation, I will feel dizzy when I read> it. I cannot avoid feeling dizzy if I meet math formulars having more than 3> lines. But I am in graduate school and must accquaint myself with maths. For> some reason, in my research field, information theory and signal processing,> if there is no math, the research may be regarded as low quality; hence high> quality journals are full of maths. I can never fully understand a paper> full of maths. Well Information Theory is not even math anymore. A bunch of [high|low]-on-Recursive-LSD2pH-9 Psychologists took over the entire field about 30 years ago, so don't worry about too much. Signal Processing is the same way too these days. It's been 50 years since anybody in signal processing outside the US Air Force, so it's not even signal processing. And since the people doing the really advanced math don't even publish === advise! what kind of training I am lackfor math?Originator: rhn@mauve.rahul.net (Ronald H. Nicholson Jr.)>The way to read math is slowly, one line at a time.Well, everybody works differently, but I would disagree with this.One should read the math stuff quickly, but in multiple passes.On the early passes, try to figure out what notation is being used andwhy, where the math is going, what methods were used and the basic ideaof how they got to the results. Hit your textbooks after each stepif you find a particular notation, assumption or method too opaque.Perhaps look into some of the citations to see if they explain some ofthe concepts better. Then, after thinking about the stuff for awhile,and maybe writing out some of equations in your own formulations, thelater passes won't be nearly as slow or frustrating.I remember checking a book out of the library to read about some DSPtopic. The chapters on that topic were completely unreadable, so I'dturn the book back in to the library and check out a different book.Finally after trying 4 or 5 different unreadable books in succession,there were no other books on the subject to be found. So I went backto the first book I had tried, and found that the words and equationsmust have rearranged themselves on the pages while the book was sittingon the library shelf, because the chapter of interest was now completelyreadable.IMHO. YMMV.-- Ron Nicholson rhn AT nicholson DOT com http://www.nicholson.com/rhn/ #include === long been headache about this problem... please help me! let me try> to explain my problem to you:> For years I have been headache about reading math notations. A concept, if> it is put in straightforward way, I can understand. But if it is put in a> math notation, or even advanced math notation, I will feel dizzy when I read> it. I cannot avoid feeling dizzy if I meet math formulars having more than 3> lines. But I am in graduate school and must accquaint myself with maths. For> some reason, in my research field, information theory and signal processing,> if there is no math, the research may be regarded as low quality; hence high> quality journals are full of maths. I can never fully understand a paper> full of maths.> I guess my problem is my lack of exposure to math and lack of systematical> education in math. I thought I will be a programmer and all my past time was> devo to programming. But later I found programming is kind boring so I> end up need to use math.> In fact I am quite good in terms of grades in math classes; but I have only> studied very few math courses: calculas, linear algebra, complexity> analysis, probability, all in undergraduate and introductory graduate level.> So I am at a very akward stage now: if give me undergraduate math to read, I> feel too easy and boring, if give me difficult math to read as those in> Information Theory Transactions, I feel dizzy...> Can anybody recommend some procedures/reference books to give me a treatment> for my current sickness? Can anybody give a booklist that any big guy in the> field would recommend as a must-read as a systematic treatment to math you very much,> -WalalaIn the UK and NZ it's Maths === am lackfor math?> In the UK and NZ it's Maths (plural)Dunno about the Kiwis but in the UK it's maths (singular).-- === problem... please help me! let me try> to explain my problem to you:> For years I have been headache about reading math notations. A concept, if> it is put in straightforward way, I can understand. But if it is put in a> math notation, or even advanced math notation, I will feel dizzy when Iread> it. I cannot avoid feeling dizzy if I meet math formulars having more than3> lines. But I am in graduate school and must accquaint myself with maths.For> some reason, in my research field, information theory and signalprocessing,> if there is no math, the research may be regarded as low quality; hencehigh> quality journals are full of maths. I can never fully understand a paper> full of maths.> I guess my problem is my lack of exposure to math and lack of systematical> education in math. I thought I will be a programmer and all my past timewas> devo to programming. But later I found programming is kind boring so I> end up need to use math.> In fact I am quite good in terms of grades in math classes; but I haveonly> studied very few math courses: calculas, linear algebra, complexity> analysis, probability, all in undergraduate and introductory graduatelevel.> So I am at a very akward stage now: if give me undergraduate math to read,I> feel too easy and boring, if give me difficult math to read as those in> Information Theory Transactions, I feel dizzy...> Can anybody recommend some procedures/reference books to give me atreatment> for my current sickness? Can anybody give a booklist that any big guy inthe> field would recommend as a must-read as a systematic treatment to math for> researchers in math-rela engineering/science area?I agree with the other posts.I must say that your problem is very familiar to me too! So don't bedisheartened.The idea that this is simply language is a particularly powerful idea. Theproblem is that the language part you're having trouble with is that it'sshorthand! So you try to read at your own normal rate and the shorthandjust doesn't allow that - it's too compact. Yet, it's still language with afew new terms thrown in.The suggestions of others are good ones. Slow down. Practice.Finally, understand this: many papers are not well written for clarity orto be understood. So, in addition to there being issues of language andshorthand, some stuff is just crap from a presentation point of view andmaybe from a real content point of view as well. Skip those papers unlesssomeone tells you they're really important. You will never read*everything* anyway - so you may as well be selective. And, by this I meanyou will never read everything in either: information theory, in circuittheory, in DSP, in control systems because each topic is just too broad andmostly much easier to read than others. Put your time into the classics andinto the more recent ones that are being often referenced. Then pick asingle topic and research it very well - using the widest variety of searchmethods possible - including Scientific Abstracts if that's still around. Acareful review of the papers on that one narrow subject will reveal papersin categories something like this:- the classics- a much better notion of what the often-used references are - which shouldbecome part of your research base.- some very interesting ideas that you haven't seen before- quite a few papers that aren't very clear and don't seem to mean much incontext of your research - they are off-topic- some interesting facts and proofs that may come in handy in yourresearch of this topic.So, you can see from this that it's likely OK to skip the ones that are justtoo obscure. And, you will still have to work at understanding the goodones! (but probably not nearly as hard as trying to understand the obscurepapers).It's also likely that doing this research will cause you to expand yourknowledge of some things: maybe it will be some parts of set theory,approximation theory or approaches, linear spaces, etc. As you add a levelof understanding in these areas, it will be easier to read some of thosepapers.It may seem not good enough to be so narrow but it usually helps becauseyou become more and more familiar with the ideas and will start noticingvariations in notation - oh! that's just the same old idea but look howstrangely it's written here ... that sort of thing. Then, the surprisesof notation will teach you something and will allow you to read other thingsmore easily, etc.Now, if you want references, then in what area? You need to help focus thehelpers === am lackfor math?> I must say that your problem is very familiar to me too! So don't be> disheartened.Me too. Early on, math came with great difficulty to mealso and being in an electrical engineering program I wasconstantly challenged. It takes perserverence but thefacility comes with usage. I am amazed today at how easilyI can move into a new mathematical area and how quickly itmakes sense. I'da never thought. Sometimes difficulty is a sign that you need totalunderstanding and perspective to carry on and that can makeit seem harder than it might to someone who can settle forits utility. You might think you are in the latter schoolwhen you are in fact in the former. That's a good thing inthe long run.> The idea that this is simply language is a particularly powerful idea. The> problem is that the language part you're having trouble with is that it's> shorthand! So you try to read at your own normal rate and the shorthand> just doesn't allow that - it's too compact. Yet, it's still language with a> few new terms thrown in.But with a big difference. To me, visualization was anessential component in learning to swim in that ocean. I'dnever make a theoretical or mathematical physicist. > Finally, understand this: many papers are not well written for clarity or> to be understood. So, in addition to there being issues of language and> shorthand, some stuff is just crap from a presentation point of view and> maybe from a real content point of view as well. Skip those papers unless> someone tells you they're really important. On the other hand, much of my learning was in puzzlingthrough exactly such papers and filling in the gaps in themand in my own knowledge to come to an understanding of thetopic that was sufficient for implementation or variation. I'd say don't skip a paper that you feel you should be ableto understand but don't. Rather, buckle down and work itthrough. I can think of some papers that I spent monthscoming back to, setting aside, looking at things that mightclarify them until the final aha moment that made it allworth it. Bob-- Things should be described as simply as possible, but nosimpler. A. === this problem... please help me! let me try> to explain my problem to you:> For years I have been headache about reading math notations. A concept, if> it is put in straightforward way, I can understand. But if it is put in a> math notation, or even advanced math notation, I will feel dizzy when Iread> it. I cannot avoid feeling dizzy if I meet math formulars having more than3> lines. But I am in graduate school and must accquaint myself with maths.For> some reason, in my research field, information theory and signalprocessing,> if there is no math, the research may be regarded as low quality; hencehigh> quality journals are full of maths. I can never fully understand a paper> full of maths.> I guess my problem is my lack of exposure to math and lack of systematical> education in math. I thought I will be a programmer and all my past timewas> devo to programming. But later I found programming is kind boring so I> end up need to use math.> In fact I am quite good in terms of grades in math classes; but I haveonly> studied very few math courses: calculas, linear algebra, complexity> analysis, probability, all in undergraduate and introductory graduatelevel.> So I am at a very akward stage now: if give me undergraduate math to read,I> feel too easy and boring, if give me difficult math to read as those in> Information Theory Transactions, I feel dizzy...> Can anybody recommend some procedures/reference books to give me atreatment> for my current sickness? Can anybody give a booklist that any big guy inthe> field would recommend as a must-read as a systematic treatment to math you very much,> -WalalaI guess I am lack of math thinking... or the buzzword: thinking in math. Arethere any books/courses that === all,I have long been headache about this problem... please help me! let metry> to explain my problem to you:For years I have been headache about reading math notations. A concept,if> it is put in straightforward way, I can understand. But if it is put ina> math notation, or even advanced math notation, I will feel dizzy when Iread> it. I cannot avoid feeling dizzy if I meet math formulars having morethan 3> lines.> ...> Can anybody recommend some procedures/reference books to give me atreatment> for my current sickness? Can anybody give a booklist that any big guy inthe> field would recommend as a must-read as a systematic treatment to mathfor> researchers in OK, Buddy: we're much alike. My problem with math is not doing it -- I> find that straightforward -- but reading it. I tend to skim, and that's> fatal. When I need details and not just gist (which is most of the time,> unfortunately), I decipher it instead of reading it. I use the symbols> as a guide to the pathway that the writer tries to lead me down, and I> walk it, every step. Most of the time, that seems to work.No, Jerry buddy, we aren't alike. You feel doing math is straightforward foryou... But I am not. Maybe I will need more math courses to catch up withyou. But currently I feel eitherr reading and doing math are difficult forme... Have you tried some paper in === the arc length of a curveI have to calc the arc length of the graph of the functionf(x) = (2/3) x^(3/2)within the interval [0,1].As a tip I shall describe the function as a parametrized curve.Okay, I tried to get a parametrized description with the following approach:y(t) is already given with:(1) y(t) = (2/3) * t^(3/2)---------------------x(t) determined as follows:y = (2/3) x^(3/2)<=> (3/2) y = x^(3/2)<=> ((3/2)y)^(2/3) = y<=> x = ((3/2)y)^(2/3)(2) x(t) = ((3/2)t)^(2/3)Does (1),(2) describe the function as a parametrized curve or have Imisunderstood what I shall do ?Is this the general way to find a parametrized description of a function ?Calculating the arc length should then be easy... there is some === arc length of the graph of the function> f(x) = (2/3) x^(3/2)> within the interval [0,1].> As a tip I shall describe the function as a parametrized curve.> Okay, I tried to get a parametrized description with the followingapproach:> y(t) is already given with:> (1) y(t) = (2/3) * t^(3/2)> ---------------------> x(t) determined as follows:> y = (2/3) x^(3/2)> <=> (3/2) y = x^(3/2)> <=> ((3/2)y)^(2/3) = y> <=> x = ((3/2)y)^(2/3)> (2) x(t) = ((3/2)t)^(2/3)> Does (1),(2) describe the function as a parametrized curve or have I> misunderstood what I shall do ?> Is this the general way to find a parametrized description of a function ?> Calculating the arc length should then be easy... there is some formula> given...One way to test this is to plot the parametric functions y(t) against y(x)on a graph. Then compare this with the y=f(x). Stick in a few values toconvince === http://www.giganews.com/info/dmca.html>I'll admit that I'm still hoping on that math journal or some contacts>that I have out there who are supposed to get back to me soon, Guffaw. A little while back you said that you'd discovered somemathematicians out there who were much nicer than the meanieson sci.math and the big names you'd contac. Evidently thesenice guys you've discovered have realized that (i) the thingsyou say make no sense, and much more important (ii) youhave no interest in trying to understand explanations of _why_they make no sense, all you care about is finding someonesomewhere who will acknowledge the fact that you're amisunderstood genius.Whoever those nice guys are, they've realized that, and hencethey're not going to be calling back (because when they tryto talk to you about the math as though you were a rationalhuman being it doesn't work, and being nice guys they'renot interes in repeating their explanations of why thethings you say make no sense.)>but I'm>bored, so I'll see what happens here.Guffaw. I know, let's post exactly the same nonsense as we'vebeen posting for the last few months. So far nobody's beenbuying it, but maybe this time will be === of xIn sci.math, > I'll admit that I'm still hoping on that math journal or some contacts> that I have out there who are supposed to get back to me soon, but I'm> bored, so I'll see what happens here.> Recently Rick Decker, a professor at Hamilton College, apparently> trying to refute my research came up with a quadratic example, which I> like because it's a quadratic, and easier to manipulate than the> cubics I've used before.> If you wish to see his original post here are some headers which also> show that he posts from Hamilton College:> === Subject: Re: Mathematical consistency, courage> Decker put forward the quadratic> (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) > where his a's are roots of > a^2 - (x - 1)a + 7(x^2 + x).> The factors (5a_1(x) + 7) and (5a_2(x) + 7) are examples of> non-polynomial factors.> Notice that despite not being polynomials they are algebraic integers> if x is an algebraic integer because a_1(x) and a_2(x) are the two> roots of> a^2 - (x - 1)a + 7(x^2 + x).> However, there's something odd here as if you let a=0, you have one of> the roots equals 0, but the other equals -1, so it makes sense to use> b_2(x), where> a_2(x) = b_2(x) - 1> which gives> (5a_1(x) + 7)(5b_2(x) + 2) = 7(25x^2 + 30x + 2)> where a_1(0) = b_2(0) = 0.> But that implies that a_1(x) and b_2(x) *both* have some factor in> common with x with algebraic integer x.You're doing it the odd way, as usual; perhaps my mind istoo straightforward but what's wrong with explicitly computingthe a's?a_1(x) = ( (x - 1) + sqrt( (x-1)^2 - 4 * 1 * 7 *(x^2 + x)) ) / 2a_2(x) = ( (x - 1) - sqrt( (x-1)^2 - 4 * 1 * 7 *(x^2 + x)) ) / 2It's probably a bit easier if one expands out the sqrt() contents,getting:a_1(x) = ( (x - 1) + sqrt( -27 * x^2 - 30 * x + 1) ) / 2a_2(x) = ( (x - 1) - sqrt( -27 * x^2 - 30 * x + 1) ) / 2> Now given that > a_1(x) a_2(x) = 7(x^2 + x) = 7x(x+1)> that's especially odd, as what about the one that does versus the one> that doesn't? Or doesn't it? Do *both* somehow have factors in> common with x?> It might seem too esoteric with x there, so let's put in a value for> x, and let x=13. ThenOK.a_1(13) = ( 12 + sqrt(-4952) ) / 2 = 6 + sqrt(-1238)a_2(13) = 6 - sqrt(-1238)b_2(13) = 7 - sqrt(-1238)These are algebraic integers. a_1(13) * a_2(13) = 1274 = 7 * 13 * 14> a^2 - 12a + 7(13)(14)> and we already know that *one* of the a's, is coprime to 13, or wait,> do we?Neither a is divisible by 13. Let c = a/13, or a = 13*c,and then the c's satisfy the equation.(13*c)^2 - (12)*(13)*c + 7(13)(14) = 013*13*c^2 - 12*13*c + 7*13*14 = 0Since 13 is a common factor we can divide:13*c^2 - 12*c + 7*14 = 0This is an irreducible quadratic which is not of the requisite form.QED.If we want to test whether b = a+1 is divisible by 13, we can setb = 13*d = a+1, or a = 13*d-1, and we grind some more:(13*d - 1)^2 - 12*(13*d - 1) + 7*14 = 0169*d^2 - 182*d + 111 = 0It turns out this is irreducible, too.> Can any of you explain how it all works?> Note that the a's are (12 + sqrt(-4952))/2, and I didn't put indices> on them as how do you know?That's one of them, yes.-- #191, ewill3@earthlink.netIt's still legal to === forward the quadratic> (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) > where his a's are roots of > a^2 - (x - 1)a + 7(x^2 + x).> The factors (5a_1(x) + 7) and (5a_2(x) + 7) are examples of> non-polynomial factors.> Notice that despite not being polynomials they are algebraic integers> if x is an algebraic integer because a_1(x) and a_2(x) are the two> roots of> a^2 - (x - 1)a + 7(x^2 + x).> However, there's something odd here as if you let a=0, you have one of> the roots equals 0, but the other equals -1, so it makes sense to use> b_2(x), where> a_2(x) = b_2(x) - 1> which gives> (5a_1(x) + 7)(5b_2(x) + 2) = 7(25x^2 + 30x + 2)> where a_1(0) = b_2(0) = 0.> But that implies that a_1(x) and b_2(x) *both* have some factor in> common with x with algebraic integer x.Certainly, but you presumably want a non-unit factor of x, in whichcase when x = 1, a_1(1) = sqrt(-14) and b_2(1) = 1 + sqrt(-14) andI don't see either of those numbers having a non-unit factorin common with x (= 1).> Now given that > a_1(x) a_2(x) = 7(x^2 + x) = 7x(x+1)> that's especially odd, as what about the one that does versus the one> that doesn't? Or doesn't it? Do *both* somehow have factors in> common with x?Yes. In case x = 1, for example, a_1(1) = sqrt(-14), a_2(1) = -sqrt(-14)and a_1(1) * a_2(1) = 14 = 7(1)(2). Both of a_1(1) and a_2(1) havea factor in common with x (1) (and with 7 (sqrt(7)) and with x + 1(sqrt(2)). This behavior (sharing factors) holds in general,though the specific factors in other cases are messy. > It might seem too esoteric with x there, so let's put in a value for> x, and let x=13. Then> a^2 - 12a + 7(13)(14)> and we already know that *one* of the a's, is coprime to 13, or wait,> do we?No. Both a's have factors in common with 13.> Can any of you explain how it all works?Dik got in ahead of me; his === === Notice that despite not being polynomials they are algebraic integers > if x is an algebraic integer because a_1(x) and a_2(x) are the two > roots of > a^2 - (x - 1)a + 7(x^2 + x). > However, there's something odd here as if you let a=0, you have one of > the roots equals 0, but the other equals -1, so it makes sense to use > b_2(x), where > a_2(x) = b_2(x) - 1 > which gives > (5a_1(x) + 7)(5b_2(x) + 2) = 7(25x^2 + 30x + 2) > where a_1(0) = b_2(0) = 0. > But that implies that a_1(x) and b_2(x) *both* have some factor in > common with x with algebraic integer x.Yes. b_2(x) is a root of the polynomial: (b-1)^2 - (x - 1)(b - 1) + 7(x^2 + x) b^2 - 2b + 1 - (x - 1)b + x - 1 + 7(x^2 + x) b^2 - (x + 1)b + (7x^2 + 2x)So what? Both a's and both b's have a factor in common with x.Nothing strange here. Oh, you are wondering how it is possiblethat both a_2(x) and b_2(x) = a_2(x) + 1 can have a factor incommon with x. A much more simple example: Let a_1 and a_2 be the roots of a^2 + a + x we have: a_1(0) = 0 and a_2(0) = -1, quite similar, so define b_2 as above, which is the root of: b^2 - b + x Let's calculate the the roots: a_1(2) = [-1 + sqrt(-7)]/2 a_2(2) = [-1 - sqrt(-7)]/2 b_1(2) = [+1 + sqrt(-7)]/2 b_2(2) = [+1 - sqrt(-7)]/2 Because all of a_1(2), a_2(2), b_1(2) and b_2(2) are divisors of 2, it all fits. The main point is that the factor a_2(2) has in common with 2 is quite different from the factor b_2(x) = a_2(x) + 1 has in common with 2.And something similar is true in your example. The factor a_1(x) has incommon with x is the same as the factor b_2(x) has in common with x, andis in general coprime to the factor a_2(x) has in common with x.BTW, also something similar occurs in the integers. Both 15 and 16 havea factor in common with 6. In the algebraic integers the situation isa bit more complica, because there are so many divisors and no primes. > Now given that > a_1(x) a_2(x) = 7(x^2 + x) = 7x(x+1) > that's especially odd, as what about the one that does versus the one > that doesn't? Or doesn't it? Do *both* somehow have factors in > common with x?Yup, has already been shown. Given an irreducible, primitive, monic polynomial with integer coefficients: x^n + c_(n-1).x^(n-1) + ... + c1.x + c0each of the roots has a factor in common with each of the prime divisorsof c0. > It might seem too esoteric with x there, so let's put in a value for > x, and let x=13. Then > a^2 - 12a + 7(13)(14) > and we already know that *one* of the a's, is coprime to 13, or wait, > do we?No, we don't, *both* are not coprime to 13. > Note that the a's are (12 + sqrt(-4952))/2, and I didn't put indices > on them as how do you know?the mathematical definition of sqrt as a function. (Yes, I know thereis a dichoy between i and -i, but when you consistently change i to-i and the reverse, you end up with exactly the same.)But disregarding the wrong notation, yes, both of the a's are not coprimeto 13. And allowing further disregard for notation, also both of (14 + sqrt(-4952))/2are not coprime to 13. But I wonder what you === (indexed sets)X-DMCA-Notifications: http://www.giganews.com/info/dmca.html>This came up in a discussion with a friend, he found it in a topology>textbook:>cap_k { A_k: k in I }>should have the meaning>{ x | forall k in I, x in A_k }>which seems ok, but for I = O {the empty set), reads>{ x | forall k in O, x in A_k }>obviously there is something absurd about it. We think (and I think>the book says so), that every x satisfies it. That's correct. This is why the intersection of the empty set(ie the intersection of the elements of the empty set) is_undefined_ - one only talks about the intersection of anon-empty collection of sets.>I think the reason is,>the opposite of it is:>{ x | exists k in O, x not in A_k }>which is manifestly false. My friend easily concludes then, the>desired set is U, the universal set, but having read (the first third>of) Foundations of Set Theory, I am confused what U would mean in>this context.In actual official set theory there _is_ no universal set (which iswhy that empty intersection has to be undefined - the only thingit could be is the universal set, and there's no such thing.)>But the *basic* question which I know no answer of is: is what we did>indeed legitimate? Sta alternatively, does this kind of treatment>invites any of the === question (indexed sets)Without a specific universal set specified, the empty intersection is undefined: As you no, every set would be in such an empty interescetion, since x in {} implies y in x for all x and y, since the hypothesis is always false. In ZFC, the set of all sets does not exist.However, sometime one works in the context of a specifc universal set. (Think of the rectangle one sometimes sees around a Venn diagram.) In such cases, some authors define the empty intersection to be this universal set. For example, with such a convention, one can define a topology on a set (the universal set) as any collection of subsets thereof which is closed under arbitrary union and finite intersection. No need to specify that the empty set ant the whole set is in the topology, since you get these from the empty union and (here's === Basic set theory question (indexed sets)> Without a specific universal set specified, the empty intersection is > undefined: As you no, every set would be in such an empty > interescetion, since x in {} implies y in x for all x and y, > since the hypothesis is always false. In ZFC, the set of all sets does > not exist.> However, sometime one works in the context of a specifc universal set. > (Think of the rectangle one sometimes sees around a Venn diagram.) In > such cases, some authors define the empty intersection to be this > universal set. It's not a definition, it can be proven. Let {X_a}, a in the empty set, be a collection of sets indexed by the empty set. Let X be the intersection of all the X_a. Then if x is any element in the universe, I claim that x is in X. Proof: In order for x to fail to be in X, there must be some a in the empty set such that x fails to be in X_a. If you can't find such an a, then you must agree that for any a, x is in X_a, therefore by the definition of intersection, x is in X. But you *CAN'T* find such an a, because there is nothing at all in the empty set! If you can't find find a in the empty set with x not in X_a, then you must agree that x is in X.Once you get used to reasoning like this, you will be enlightened in === =?ISO-8859-1?Q?Goldbach_and_G=F6del?=> No. I don't mean that I can't recognise a counter-example.> Suposse someone prove that the primes are random distribu, then the> Goldbach Conjecture is not an anlytical event but a lucky> occurrence.Simply, until now the actual density of primes has been> sufficient for that its pairing two by two produce all the even> numbers < 10^20.> In this case the conjecture is undecidable that is: not true neither> false and its necessary to wait an eternity for finding a> counterexample. L.Rodriguez.> If a counterexample exists, that means the proposition is provably false.> If there are no counterexamples then the proposition is true, though we> may not be able to prove it. Summary = The === reference>> where a_1(0) = b_2(0) = 0.> Then there must exist a_1(x) = 7b_1(x), and that substitution gives>> Proof? Why must 7 divide a_1(x) for all x?> Are you assuming a ring? If so, what ring?> Why ask what I'm assuming? It's your claim 7 must divide a_1 for all x inthe algebraic integers. Prove === referenceX-DMCA-Notifications: http://www.giganews.com/info/dmca.html>[...]>Oh yeah, and why don't you s off as I strongly sugges before?Right. That's a pretty convincing argument.>I'm sort of tired of you chasing after my posts like a lost puppy.That's tough. The rest of us are tired of you posting the samenonsense over and === example, referenceX-DMCA-Notifications: http://www.giganews.com/info/dmca.html>> where a_1(0) = b_2(0) = 0.> Then there must exist a_1(x) = 7b_1(x), and that substitution gives>> Proof? Why must 7 divide a_1(x) for all x?>Are you assuming a ring? You'd be much better off if you learned to make sense.For example, there's no such thing as assuming a ring;one assumes _statements_, and a ring is not a statement.One assumes that various things are elements of variousrings. (One _specifies_ this when one assumes this -then one has the ability to assume that one thing isin one ring and another thing is in another ring. Whenone says the ring is the algebraic integers one isnot making much sense.)> === Modified Decker example, reference> Recently Rick Decker, a professor at Hamilton College, apparently> trying to refute my research > *Successfully* refuting your research - and why don't you at> least quote Decker's conclusion? Is this a propaganda tactic -> not stating the opponent's argument because other people might> actually see that it makes sense ? > No. So why have you never quo Decker's conclusion? Clearlyyou are very concerned about the example. You have star something like 12 threads on this topic. One would think you were concerned about what Decker said. But you never mentionhis core result. Why not?came up with a quadratic example, which I> like because it's a quadratic, and easier to manipulate than the> cubics I've used before.If you wish to see his original post here are some headers which also> show that === he posts from Hamilton College:Subject: Re: Mathematical consistency, courage> Why is it so important to keep noting that he posts from> Hamilton College? Might it not be more important to quote> the *substance* of what he said, rather than where he posts> from ???> Why ask why? Why not explain? > Decker put forward the quadratic(5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) where his a's are roots of a^2 - (x - 1)a + 7(x^2 + x).The factors (5a_1(x) + 7) and (5a_2(x) + 7) are examples of> non-polynomial factors. Just to clarify: these *are* polynomials in the number 5. > Irrelevant. No. Using 5 as a number rather than using a variable in itsplace renders the equation[**] a^2 - (x - 1)*a + 7*(x^2 + x) = 0meaningless. The origin of this was a polynomial in a variableformerly called x, in the place of 5. 5 is a fixed constant;when you use that, you take the entire factorization problem outof the realm of polynomial factorization and into the realm ofnumerical factorization. The restriction that -7/a_1(x) is a solution to [**] no longer applies. You can, for example, havefactorizations like the following when x = 1: (5*10 + 7)*(5*1 + 2) = 7*57 = 7*[25*1^1 + 30*1 + 2],and you note that a_1(1) = 10, a_2(1) = 1, and BOTH are coprime to 7. Or when x = 2: (5*4 + 7)*(5*8 + 2) = 7*162 = 7*(25*2^2 + 30*2 + 2)where a_1(2) = 4 and a_2(2) = 8 are BOTH coprime to 7. Again, I know you are not going to understand this, but replacingthe original x with 5 threw you out of the situation ofneeding that a_1(x) and a_2(x) were roots of [**]. > What's important for the discussion is that the everything works.> Here if you take the roots of> a^2 - (x - 1)a + 7(x^2 + x)> and substitute one in for a_1(x), and the other for a_2(x), then you have>[***] (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) > as asser. And you can get the same form [***] of the factorization with integers as I no above.> Oh yeah, and why don't you s off as I strongly sugges before? Right. Since I know that your intention is that we are to thinkof 5 as a polynomial variable even though for some reason it is generallyand widely regarded as a fixed integer even by nonmathematicians, let's assume [**] holds. For x = 2, we obtain a_1(2) = (1 + sqrt(-167))/2 and a_2(x) = (1 - sqrt(-167))/2.I note that a_1(2)*a_2(2) = 7*(2^2 + 2) = 7*6,so the product of a_1(2) and a_2(2) is a multiple of 7. Thismeans that at least one of a_1(2) or a_2(2) must have a factorin common with 7. You claim that that factor must be 7 itself.Therefore let's assume a_1(2) is divisible by 7 and see what happens: a_1(2) = 7*A = (1 - sqrt(-167))/2,where A must be an algebraic integer. Thus 14*A - 1 = -sqrt(-167), or 14^2*A^2 - 28*A + 1 = -167, or 14^2*A^2 - 28*A + 168 = 0. Factoring out 28 from each term, we obtain 7*A^2 - A + 6 = 0.This is an irreducible nonmonic quadratic. Therefore A is not analgebraic integer. Therefore a_1(2) is not divisible by 7.Similarly, a_2(2) is not divisible by 7. Yet their product is divisible by 7. Therefore BOTH of them must have nonunitalgebraic integer factors in common with 7. How do you explain AdvancedPolynomial Factorization, where both Dale Hall and I providedindependent proofs that your claim was wrong. You said thatif P(x) = 65*x^3 - 12*x + 1is factored in the form P(x) = (a_1*x + 1)*(a_2*x + 1)*(a_3*x + 1),then one of a_1, a_2, or a_3 is coprime to 5. I don't think you ever retrac this. You just abandoned theargument. Just curious: Do you still maintain you were right aboutthis? Is it still part of the conclusion to the paper you submit to the Southwest Journal? Have you ever heard back from them ? Don't you think it's about time they gave you an answer? Would youlike for us to send them an inquiry, along with a copy of someof your pithier recent posts on this and other topics? > I'm sort of tired of you chasing after my posts like a lost puppy. Really? Too bad. You mean this endless string of posts demonstrating how worried you are about the Decker example, so worried that you don't dare even say what Decker's conclusion was? All these recentnew threads rehashing the same idea, in which you refuse to answer anysubstantive mathematics? Why does that make you sort of tired? Certainly no mental effort has been involved on === presen with the basic> algebra claim that it's the sum (5b_2(x) + 2) that mattebut that> is refu by noting that the 2 in that expression is a factor of 7(2)> in 7(25x^2 + 30x + 2)as the constants are NOT a product of the full sums, but are in fact> products of the constants 7 and 2 within those sums, so by trying to> include variable expressions they are relying on claims insupportable> mathematically.> Consider (4/3 + 2/3) (9/2 + 3/2) = (4/3)(9/2) + (4/3)(3/2) + (2/3)(9/2) + (2/3)(3/2)> = 6 + 2 + 3 + 1 = 12> That's ok in the field of algebraic numbebut doesn't exist in the> ring of algebraic integers.> Here we have (a+b)(c+d) = ab + ac + bc + bd = g> The algebra doesn't care what the ring is.> and (a+b), (c+d), ab, ac, bc, bd, and g are all integers.> However none of a, b, c or d is an integer.> You sum first for your example, or you're out of the ring.Actually, the whole point of the example is that whether yousum first or multiply first you get integer results.> But what about (x+2)/3?In general it is not an algebraic integer. However, the important fact isthat for at least one value of x it is an algebraic integer. And the factthat 2/3 is not an algebraic integer does not mean that (x+2)/3 isnever an algebraic integer. - William === the basic> algebra claim that it's the sum (5b_2(x) + 2) that mattebut that> is refu by noting that the 2 in that expression is a factor of 7(2)> in 7(25x^2 + 30x + 2)as the constants are NOT a product of the full sums, but are in fact> products of the constants 7 and 2 within those sums, so by trying to> include variable expressions they are relying on claims insupportable> mathematically.> Consider (4/3 + 2/3) (9/2 + 3/2) = (4/3)(9/2) + (4/3)(3/2) + (2/3)(9/2) + > (2/3)(3/2)> = 6 + 2 + 3 + 1 = 12> That's ok in the field of algebraic numbebut doesn't exist in the> ring of algebraic integers.> Here we have (a+b)(c+d) = ab + ac + bc + bd = g> The algebra doesn't care what the ring is.> and (a+b), (c+d), ab, ac, bc, bd, and g are all integers.> However none of a, b, c or d is an integer.> You sum first for your example, or you're out of the ring.> But what about (x+2)/3?> Now here's a pop quiz, for algebraic integer x, is (x+2)/3 in the ring> of algebraic integers?> Can you *add* first to say that it always is?> So the fact that the partial sums as well as full sums are integers,> does not mean that a,b,c, and/or d must be integers.In particular, the fact that the product of the constant terms,> b and d, is an integer does not mean that b and d must be integers. - William Hughes> Think about (x+2)/3.Yes indeed, lets. Whether (x+2) is divisible by 3 depends on x. For you to keeps claiming that it doesn't depend === example, reference[non-math dele]Decker put forward the quadratic(5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) where his a's are roots of a^2 - (x - 1)a + 7(x^2 + x).The factors (5a_1(x) + 7) and (5a_2(x) + 7) are examples of> non-polynomial factors.Notice that despite not being polynomials they are algebraic integers> if x is an algebraic integer because a_1(x) and a_2(x) are the two> roots ofa^2 - (x - 1)a + 7(x^2 + x).Using the substitution a_2(x) = b_2(x) - 1 with the factorization> gives(5a_1(x) + 7)(5b_2(x) + 2) = 7(25x^2 + 30x + 2)where a_1(0) = b_2(0) = 0.Then there must exist a_1(x) = 7b_1(x), and that substitution gives(5(7)b_1(x) + 7)(5b_2(x) + 2) = 7(25x^2 + 30x + 2)and the 7 can now easily be removed from both sides giving(5b_1(x) + 1)(5b_2(x) + 2) = 25x^2 + 30x + 2.But, it is rather easy to show that in general with algebraic integer> x, b_1(x) is not an algebraic integer.At this point it would seem obvious to conclude that the desire to > replace a_1(x) with 7b_1(x) is not a valid replacement.> No it's not obvious. After all constants like 7 normally multiply> through in a rather basic way. So it makes sense that it does here as> well.But JSH argues that if (a + b) is in a ring and c is in that ring and (a+b)/c is also in that ring, then a/c and b/c must also be in that ring.Which arguement is demonstrably not true in the ring of integeconsidering (6 + 8)/7, versus 6/8 and 8/7, for instance.> It actually takes an illogical leap to conclude that somehow, someway> the mathematics decides to behave in an inconsistent way.And the inconsistent way is JSH's way, which === Decker example, reference> Why ask why? Answer your own question!> Decker put forward the quadratic(5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) where his a's are roots of a^2 - (x - 1)a + 7(x^2 + x).The factors (5a_1(x) + 7) and (5a_2(x) + 7) are examples of> non-polynomial factors. Just to clarify: these *are* polynomials in the number 5. > Irrelevant.> What's important for the discussion is that the everything works.Not when trying to do things your way. > Here if you take the roots of> a^2 - (x - 1)a + 7(x^2 + x)> and substitute one in for a_1(x), and the other for a_2(x), then you have> (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) > as asser.> Oh yeah, and why don't you s off as I strongly sugges before?> I'm sort of tired of you chasing after my posts like a lost puppy.The brilliance of mathematical insight included in these last two sentences outdoes all of JSH's previous === example> In an attempt at questioning an important conclusion of mine a Rick> Decker, a professor at Hamilton College, made a post with his own> example of a non-polynomial factorization:> (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) > where his a's are roots of > a^2 - (x - 1)a + 7(x^2 + x).> Making the substitution a_2(x) = b_2(x) - 1 to get> (5a_1(x) + 7)(5b_2(x) + 2) = 7(25x^2 + 30x + 2) > allows both a_1(0) = 0 and b_2(0) = 0, so let that be a requirement.> Therefore, both a_1(x) and b_2(x) have some factor of x itself, which> implies a_1(x) has a factor of 7, so using a_1(x) = 7b_1(x) you haveTry x=1. a^2=-14, a= +/-sqrt(-14). I don't see that seven divideseither of those. > (5(7) b_1(x) + 7)(5b_2(x) + 2) = 7(25x^2 + 30x + 2) > and dividing both sides by 7 gives> (5b_1(x) + 1)(5b_2(x) + 2) = 25x^2 + 30x + 2.> At this point it's all straightforward except that you can actually> check at x=1, you find that doesn't work in the ring === modified Decker exampleIn sci.math, <3c65f87.0401301610.7ae35186@ important conclusion of mine a Rick> Decker, a professor at Hamilton College, made a post with his own> example of a non-polynomial factorization:> (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) > where his a's are roots of > a^2 - (x - 1)a + 7(x^2 + x).> Making the substitution a_2(x) = b_2(x) - 1 to get> (5a_1(x) + 7)(5b_2(x) + 2) = 7(25x^2 + 30x + 2) > allows both a_1(0) = 0 and b_2(0) = 0, so let that be a requirement.> Therefore, both a_1(x) and b_2(x) have some factor of x itself,You cannot draw that conclusion.a_1(x) = ( (x - 1) + sqrt(-27*x^2 - 30*x + 1) ) / 2a_2(x) = ( (x - 1) - sqrt(-27*x^2 - 30*x + 1) ) / 2(or one can reverse them; it doesn't matter).> which> implies a_1(x) has a factor of 7, so using a_1(x) = 7b_1(x) you have> (5(7) b_1(x) + 7)(5b_2(x) + 2) = 7(25x^2 + 30x + 2) > and dividing both sides by 7 gives> (5b_1(x) + 1)(5b_2(x) + 2) = 25x^2 + 30x + 2.> At this point it's all straightforward except that you can actually> check at x=1, you find that doesn't work in the ring of algebraic> integers.It doesn't work for most x.> -- #191, ewill3@earthlink.netIt's still legal to go === === understanding Torkel's ZFC comment> So are my questions not well-defined? Unclear? Too hard? Too> embarrassing? Why can't anyone answer a few simple questions about> ZFC? After all, people talk about it enough!> Alas. You just peddle your alternative.?? I'm afraid you've lost me on that one. Peddle sounds like asynonym for sell, which is what you advoca. In any case, the 6questions that I ask don't actually refer to my alternative. Why notaddress them? You for one talk about ZFC a lot, so presumably youknow a lot about the subject. I'm certainly willing to discuss thepros and cons of ZFC, my system, and any alternative that anyone wantsto - day or night. (Proof: See the times on my posts!)Once more:1. Would ZFC be better if it included its rules of inference?2. Does ZFC violates Occam's Razor to have axiom schemes when goodold rules of inference will do?3. Is it better to replace a set of axioms with a smaller set ofaxioms and rules of inference that can be used to derive them astheorems?4. And what if these same rules of inference could be used to generatetheorems from the Theory of Computation and to auatically constructcomputer programs that implement functions from number theory? Wouldthat enhance its usefulness?5. Is ZFC an attempt to formalize Set Theory (which also avoids theparadoxes)? Where have authors been able to construct formal proofsin ZFC? Would it be better if they did?6. Where do people talk about using rules of inference instead ofaxiom schemes in ZFC, or in general?Any takers? (I do realize that it's one thing to be able to quoteother people, and another thing altogether to be === understanding Torkel's ZFC commentX-DMCA-Notifications: http://www.giganews.com/info/dmca.html>> Could you address these points?>> Better...violates...better...enhance...better... What points? If>> you don't like ZFC, make up something else and try to sell it.>Of course I did and I am.>So are my questions not well-defined? Unclear? Too hard? Too>embarrassing? Why can't anyone answer a few simple questions about>ZFC? People have - the fact that you don't believe the answers doesn'tmean that the questions have not been answered. (For example,you continue to ask what the inference rules of ZFC are, aftera detailed explanation of why the question makes no sense;why ZFC does not contain inference rules, any more thanthere are inference rules of group theory.)>After all, people talk about === EqQHHwZvYeCZkqgnsB79oKdN3hETlUcKkOKq2r-MOYXLN2gZKSHzYl> 5. Is ZFC an attempt to formalize Set Theory (which also avoids the> paradoxes)? Where have authors been able to construct formal proofs> in ZFC?Get a grip, idiot:Inspired by Whitehead and Russell's monumental Principia Mathematica,the Metamath Proof Explorer has over 3,000 completely worked out proofsin logic and [ZFC] set theory, interconnec with more than 200,000hyperlinked cross-references. Each proof is pieced together withrazor-sharp precision using a simple substitution rule. Withpoint-and-click links, every step can be drilled down deeper and deeperinto the labyrinth until axioms will ultimately be found at the bot.You could spend literally days exploring the complex tangle of logicleading, say, from 2 + 2 = 4 back to the axioms of set theory. The proofcollection includes many famous theorems of elementary set theory.Essentially everything that is possible to know in mathematics can bederived from a handful of axioms known as Zermelo-Fraenkel set theory,which is the culmination of many years of effort to isolate theessential nature of mathematics and is one of the most profoundachievements of mankind. The Metamath Proof Explorer starts with these axioms to build up itsproofs. There may be symbols that are unfamiliar to you, but we show indetail how they are manipula in the proofs, and in principle youdon't have to know what they mean. In fact, there is a osophy calledformalism which says that mathematics is a game of symbols with nointrinsic meaning. With that in mind, Metamath lets you watch the gamebeing played and the pieces manipula according to the rules. As humans, we observe interesting patterns in these meaningless symbolstrings as they evolve from the axioms, and we attach meaning to them.One result is the set of natural or counting numbewhose propertiesmatch those we observe when we count everyday objects. (Of course,numbers were discovered centuries before set theory, and historicallythey were reversed engineered back to the axioms of set theory.) Atthe other extreme of abstraction is the theory of infinite sets(transfinite cardinal numbers), sometimes called Cantor's paradise.Some of the world's most brilliant mathematicians have given us deepinsight into this mysterious and wondrous universe that exists only inthe mind. Metamath's formal proofs are much more detailed than the proofs you seein textbooks. They are broken down into the most explicit detailpossible. Each proof step represents a microscopic increment towards thefinal goal. But each step is derived from previous ones with a verysimple rule, and you can verify any proof for yourself with very littleskill. All you need is patience. You can jump into the middle of anyproof, from the most elementary to the most advanced, and understandimmediately the symbol manipulations with no prior knowledge of advancedmathematics or even any mathematics at all. In the next section we showyou how.Taken from theMetamath Proof Explorer Home Page === Re: need help in understanding Torkel's ZFC comment> The user types in any expression they want.> And if the user makes a mistake in her proof, she gets an error message.> Nothing is derived formally or auatically. Notice that there's> nothing about how the system constructs the proofs?> Norm Megill's 175-page _Metamath Book_ describes -- in great detail --> the Metamath system and how to construct formal proofs in it.> http://au.metamath.org/downloads/metamath.pdfHow the user constructs proofs, not how the system generates them. But if the system does verify them, then fine - that's half of thejob. But I don't see offhand how to get a copy or access it over theinternet to try it out. Does anybody know how? I can cite a numberof web sites that purport to have auatic theorem-proving but can beshown to not. (Will look them up and post here.)And when I looked for a proof using ZFC axioms, I saw tons of otherrules being used. This verifies my point that ZFC per se reallydoesn't do it. You need a lot more - or, in the case of my system, asmaller system that does more. It looks like they use the old trickof hard-wiring the rules to the theorem, rather than having a fixedset of axioms and rules that works in every case. Remember howZermelo said that all of mathematics could be derived from his 7axioms?How about if you point me to the best example or two of proofs usingZFC that is given - preferably not just proving part of ZFC withanother part? Then we can talk about specifics.> I'll have to get ahold of the paper ci. [...] (Feel free to email> me a copy, anyone.)> There's a link to download a copy in the View or === Relativity>> It basically says all observers will MEASURE the one-way speed of light to be>> c, irrespective of the source's speed.>>Sorry Henri, that's wrong. Only _inertial_ instruments>measure the speed as c according to SR.> Yes. OK. You have to include the great SRian escape route. inertial.But Henry knows that there is no such thing as an inertial frame,its only a myth crea by the stupid SRists to use as an escape route.That's why gyros don't work.They get confused because they don't know whatto compare the rotation to.Isn't === Synchronization Clocks in Relativity> SR says that the speed of light is c relative to its source. Therefore if a> ring laser is analysed in the co-rotating frame, no fringe shifts should be> expec according to SR.> Sagnac clearly refutes SR.>>We both know that you know nothing about and don't understand SR.>>It wasn't really necessary to remind us.>> Do you only pull out the above statement as a last resort, ?>>The above demonstration of your understanding of SR>make it obvious that it is a simple statement of fact.>> Well come on . Tell me what SR says about ring lasers.Do you pretend to have === of Synchronization Clocks in Relativity >>SR says the speed will be c only as measured in the inertial>>(lab) frame. The source and detector on the table move in the>>lab frame during the time the light is in flight, therefore>>the time taken for the beams to travel from source to detector>>is not the same in the two directions (same sense as the table>>and the opposite), and varies with the speed of rotation.>> That is a perfect example of why SR is just a disguised aether theory.>> An aether with an 'arbitrary inertial frame'.>> I'm not impressed!> to point out how how silly you sound, consider a completely remote ring>> laser with a built in fringe detector. It has no lab or any other reference>> frame.>> Are you trying to tell me that it wont detect rotation?>>So what does 'rotation' mean?>Remember that there is no ether with an 'arbitrary inertial frame',>no lab or any other reference frame.>>Henry, I look forward to see your acrobatics to divert>the attention from the fact that you shot yourself in the foot.>>Will you invent a new manoeuvre, or will you repeat an old one?>> Answer the question .> Consider yourself flying through remote space with only a ring laser, an oxygen> bottle and a tin of sardines.> If you rotate the gyro, does it work?> If so, why, - since there is no lab frame to be used as a reference.I have got it now, Henry.You claim that gyros doesn't work because there is no ether withan 'arbitrary inertial frame', no lab or any other reference === <1075557524.25099.0@ersa.uk.clara.net>SR says that the speed of light is c relative to its source.>>Ritzian theory says it is c relative to the source, SR says>>the speed is independent of the speed of the source but will>>be c relative to the instrument measuring the speed if that>>instrument is moving inertially.>> It basically says all observers will MEASURE the one-way speed of lightto>be>> c, irrespective of the source's speed.>>Sorry Henri, that's wrong. Only _inertial_ instruments>measure the speed as c according to SR.> Yes. OK. You have to include the great SRian escape route. inertial.Try this experiment: stand up, turn round then sit down.Now consider a flash of light moving away from a recentsupernova in the Andromeda galaxy. It is 2 million lightyears away so in the frame that co-rota with you as youturned, that light moved 2*pi*R or a distance of 13 millionlight years in just a few seconds. Obviously the speed oflight in a rotating frame cannot possibly be c.>> In this case, the speed of the instrument is the same as the speed ofthe>> source/receiver.>> Analysis using the co-rotating frame shows that fringe shift should not>occur>> according to SR.>>A co-rotating instrument is not inertial. Do the analysis>properly and you will find it predicts the observed shift.> I said a co-rotating frame. You were the one who first introduced such tothis> argument remember.No, you replied to Anderson:>I am stating the fact that in a ring laser, the speed of light isisotropic in>the inertial frame, EVEN when the source (lasing gas) is moving in thisframe.>This is proven by the mere fact that the ring laser gyro works.>This falsifies the ballistic light theory.> SR says that the speed of light is c relative to its source. Therefore ifa> ring laser is analysed in the co-rotating frame, no fringe shifts shouldbe> expec according to SR.> Sagnac clearly refutes SR.SR does NOT imply that the speed of light is c relative toits source in the co-rotating frame, only that it is isotropicin the inertial frame as sta.> I have done the analysis and it shows that there will be a fringe shiftwith> source dependency.That analysis is incorrect, see my other reply.> That means that there will also be one according to SR which> clearly states that light moves at c relative to its source.You are right, there will be a shift according to SR butyou have been arguing the opposite.> Source and observer are the same in a ring gyro.Both are rotating and accelerating, hardly inertial.>In the 'lab' frame SR predicts the speed of the light>will be c while the ballistic theory predicts it will be c + v' = v * cos(a) + c * cos(b)> Not so. If the source is moving in the direction v, then light speed inthe> horizontal direction will be c+[v*cos(a)]>Obviously they are not equivalent.Either way, SR says it is c in the inertial frame so theyare not equivalent. See my other post for the details.>> Are you trying to tell me that it wont detect rotation?>>No, I said that SR predicts the time taken for the beams>to travel from source to detector is not the same in the>two directions .. and varies with the speed of rotation.>so I said SR predicts that it _will_ detect the rotation.> You also explained the R 'expolanation' by refering to a 'lab frame'.> I have cp,letely removed the lab frame. Where is your explanation now?It remains the same, I said:> SR says the speed will be c only as measured in the inertial> (lab) frame.SR says the speed is c in any inertial frame, whether thereis a 'lab' handy to define that or not. You have removed thelab but not the inertial frame.>Perhaps you should read my posts more carefully, it is the>_Ritzian_ theory that predicts that it won't be detec.> You people are really funny.>>Well you're the one pushing a theory that predicts there>will be no shift.> There will be a shift under source dependency.See my other post where I have shown that there is no speedchange on the reflections and the path lengths vary as (v/c)^2for v< You are becoming quite confused here George.Well you have been saying SR doesn't predict a shift but aboveyou said ... there will be a fringe shift with sourcedependency. That means that there will also be one accordingto SR .. and in the previous post you asked Are you trying totell me that it wont detect rotation? when I had just said itwould.> Remember the source and observer are the same in a ring gyro. Therefore> according to both you (and me), the light travels at the same speed 'c'> relative to both.No, remember SR says the light travels at c ONLY in an inertialframe. The co-rotating frame is not inertial. Ballistic theoryon the other hand says the light is emit a c relative to thesource and each mirror on subsequent reflections.> I really don't see where SR comes into this at all.It didn't until you introduced it in your reply to . Yousaid SR says that the speed of light is c relative to itssource. which it appears you know is untrue, SR says it is cin any inertial frame but the rotating table is not inertial.When you bear that in mind you will find === in Relativity Sorry, I forgot to put it on the web page.> It downloads for me now.It's fine I was just too early.>> For instance, can it be assumed that a ray, which is aligned with thecentre of>> the first mirror, strikes that mirror at exactly the same point whenthe>> apparatus is rotating. If not, I don't think your interpretation ofwhat>> happens in the co-rotating frame can be correct.>>You have to remember that a 'ray' is just the path formed by>the normal to the wavefront. In reality any light beam has a>finite width.> And it isn't perfectly parallel.No, but it is the fact that the wavefront is normal to theray that answers your question.>> I understand your point but it is hardly convincing. You have notconsidered>> where the rays will strike the moving mirrors.>>The rays must hit the mirrors at exactly the same points as>in the non-rotating case because, in the co-rotating frame,>the mirrors are static and the light eventually has to hit>the same receiver (telescope, screen or whatever.) The point>of reflection could move if the angles turned along the curved>paths differed from leg to leg, but since they must be>symmetrical, there can be no change.> None of that is obvious. I don't know how you can make those claims.Sometimes things may not be obvious but they may still becorrect. The symmetry about the midpoint of a path ensuresthe angles of approach at each end of a light path frommirror to mirror are the same. That the wavefront is normalto the tangent to the path ensures the angle of incidence isequal the angle of reflection. Once you know those, it meansthe point where the ray hits the mirror has to be the same.>>The travel time is barely affec but more importantly it is>>increased by exactly the same amount for both rays since it>>depends on v^2, not v, so there should be _no_ path difference.>>However, the observed fringe shift is exactly as if the speed>>was c-v and c+v in the co-rotating frame.>> Again this is hardy convincing. When you say the time depends on v^2(actually>> 1/v^2) you are forced to include a constant (k) that is notdimensionless,>> (T/L^2). I'm not sure what that implies.>>Actually it depends on (v/c)^2 which is dimensionless but that>is beside the point. the key is that any curve based on even>powers is symmetrical about zero hence the path length change>for +v is the same as for -v and the path difference is>therefore zero.> Well, I'll have to think a lot more about this. The curve can besymmetrical> but need not strike the same point on the mirror for all rotationalspeeds.It takes some thought but the symmetry argument doesgive you an answer.>The rays must follow identical paths when the apparatus is>stationary from basic optics, I'm sure you know that. Since>the angles change symmetrically, they do not change the point>of reflection.> This is where the second order factor comes in. The travel times betweenthe> mirrors is different for the two paths because, with source dependency therays> get a velocity 'kick' each time they are reflec. The mirrors act assources.> You are ignoring that.No, we have looked at the equations several times. Ihave expanded the sketches to try to answer your nextpoint as well:>> You must also consider velocity variations that occur during reflectionat each>> moving mirror. Each acts as a source.> Under source dependency, the velocity of light leaving the sourceappears to be>> (c+vsin45).>> But is it?>> Is it c+vsin(45+x) for the red ray and c+vsin(45-x) for the green one?>>The angle 'a' in this diagramhttp://www.briar.demon.co.uk/Henri/speed.gif>>is 45+x for the red ray at the top right mirror in this diagram:http://www.briar.demon.co.uk/Henri/paths.gif>>The trick though is that you than have to find the speed at>which the light approaches the mirror at the top left. The>angle there is also 45+x so when you subtract the motion of>the mirror which is moving away from the light, the nett>result is a relative incident speed of exactly c.> That is true only for the first mirror.>The same applies to the green ray but you need to take a>mirror image of the speed.gif sketch and the angle then, as>you say, is 45-x at both ends but again since this:> www.users.bigpond.com/hewn/sagnac.jpgI have copied that and added two small blue lines at themid-point of two paths. The paths are S->A->B->C->S' andS->C'->B->A'->S'http://www.briar.demon.co.uk/Henri/ lab.gifThe diagram on the right concentrates on the A/B section andhas the two aligned on the mid-point. This confirms what youwere asking about whether the angles were exactly 45 degrees.For the green path, the angle turned by the ray is slightlymore than 90 degrees at each end so the angle between themirror and the horizontal ray is slightly less than 45degrees. For the red path, the angle is slightly more at bothends.You said in another post:[GD]> c + v' = v * cos(a) + c * cos(b)[HW]> Not so. If the source is moving in the direction v, then light speed inthe> horizontal direction will be c+[v*cos(a)]I have added a bit more to this diagram:http://www.briar.demon.co.uk/Henri/speed.gifObviously for the normal situation of v<Fringes are cused by the relative angle between the beams>which stays the same as both rays rotate the same way. The>beam has to be wide enough to fall on the eyepiece so>sideways movement doesn't have an effect, the actual shift>is very small and much less than the size of the beam.but the sideways movement is also indicative of an angulardifference>>between> the two beams.>>No, see the green lines on the diagram. The rays turn the same>>way by the same amount so the angle between them is unaffec.>> But your diagram is not indicative of what actually happens..>>Why not? you seemed happy with it a few posts ago.> I was happy with the diagram, not its significance.http://www.briar.demon.co.uk/Henri/paths.gifThe intention was to show that the two rays turned in thesame direction hence the angle between changes by thedifference, not the sum. What you cannot get from thediagram is that the derivative of the angle for each rayat v=0 must be the same for both rays so the angle betweenthem must be unchanging at === in Relativity......... ...First of all, it would be as it should be an essentialnecessity to determine, a specification of any Clock. However, out of anyfalling along any amalgam, which it can takes to an attempt along the finalconfusion.The define as the definite reason, for what, for instance, to know, thatthe sun along its highest point in the sky, is absolutely a matter ofmetamorphosis along its variations over an orbit as over its rotation.Something, which it does allows the time of a being just an average alongany appearance among any solar days as along an orbital year.Therefore, there is a definitely also, the ultimate Clock, which is along anutilisation of an aic Clock, among a several other clocks. However, theaic Clock, which is usually along any measure of any vibration along anyenergy over any cesium along any a. That, usually it is used along anaddition of a fraction of a seconds every more or less decades.However, a finally, there is an universal time, used as knowninternationally as a Greenwich time. Therefore, basically used along therotation of the earth, whether, it does a definitely have as a major pillar,which is the aic Clock. Therefore, from that point to arrive to determineany Synchronization along any Clock. It would be as it should be, something,which it could takes to an infinite calculations, which it would as shouldalways and a definitely takes to the departure case, and this is what is allabout, definitely as a matter a fact!!!!!!!!!!!!!!!!!!..................-- Ahmed Ouahi, kirjoitti viestiss.8a> It downloads for me now.> It's fine I was just too early.>> For instance, can it be assumed that a ray, which is aligned with the> centre of>> the first mirror, strikes that mirror at exactly the same point when> the>> apparatus is rotating. If not, I don't think your interpretation of> what>> happens in the co-rotating frame can be correct.>>You have to remember that a 'ray' is just the path formed by>the normal to the wavefront. In reality any light beam has a>finite width.And it isn't perfectly parallel.> No, but it is the fact that the wavefront is normal to the> ray that answers your question.>> I understand your point but it is hardly convincing. You have not> considered>> where the rays will strike the moving mirrors.>>The rays must hit the mirrors at exactly the same points as>in the non-rotating case because, in the co-rotating frame,>the mirrors are static and the light eventually has to hit>the same receiver (telescope, screen or whatever.) The point>of reflection could move if the angles turned along the curved>paths differed from leg to leg, but since they must be>symmetrical, there can be no change.None of that is obvious. I don't know how you can make those claims.> Sometimes things may not be obvious but they may still be> correct. The symmetry about the midpoint of a path ensures> the angles of approach at each end of a light path from> mirror to mirror are the same. That the wavefront is normal> to the tangent to the path ensures the angle of incidence is> equal the angle of reflection. Once you know those, it means> the point where the ray hits the mirror has to be the same.>>The travel time is barely affec but more importantly it is>>increased by exactly the same amount for both rays since it>>depends on v^2, not v, so there should be _no_ path difference.>>However, the observed fringe shift is exactly as if the speed>>was c-v and c+v in the co-rotating frame.>> Again this is hardy convincing. When you say the time depends on v^2> (actually>> 1/v^2) you are forced to include a constant (k) that is not> dimensionless,>> (T/L^2). I'm not sure what that implies.>>Actually it depends on (v/c)^2 which is dimensionless but that>is beside the point. the key is that any curve based on even>powers is symmetrical about zero hence the path length change>for +v is the same as for -v and the path difference is>therefore zero.Well, I'll have to think a lot more about this. The curve can be> symmetrical> but need not strike the same point on the mirror for all rotational> speeds.> It takes some thought but the symmetry argument does> give you an answer.>The rays must follow identical paths when the apparatus is>stationary from basic optics, I'm sure you know that. Since>the angles change symmetrically, they do not change the point>of reflection.This is where the second order factor comes in. The travel times between> the> mirrors is different for the two paths because, with source dependencythe> rays> get a velocity 'kick' each time they are reflec. The mirrors act as> sources.> You are ignoring that.> No, we have looked at the equations several times. I> have expanded the sketches to try to answer your next> point as well:>> You must also consider velocity variations that occur duringreflection> at each>> moving mirror. Each acts as a source.> Under source dependency, the velocity of light leaving the source> appears to be>> (c+vsin45).>> But is it?>> Is it c+vsin(45+x) for the red ray and c+vsin(45-x) for the greenone?>>The angle 'a' in this diagramhttp://www.briar.demon.co.uk/Henri/speed.gif>>is 45+x for the red ray at the top right mirror in this diagram:http://www.briar.demon.co.uk/Henri/paths.gif>>The trick though is that you than have to find the speed at>which the light approaches the mirror at the top left. The>angle there is also 45+x so when you subtract the motion of>the mirror which is moving away from the light, the nett>result is a relative incident speed of exactly c.That is true only for the first mirror.>>The same applies to the green ray but you need to take a>mirror image of the speed.gif sketch and the angle then, as>you say, is 45-x at both ends but again since this:> www.users.bigpond.com/hewn/sagnac.jpg> I have copied that and added two small blue lines at the> mid-point of two paths. The paths are S->A->B->C->S' and> S->C'->B->A'->S'> http://www.briar.demon.co.uk/Henri/lab.gif> The diagram on the right concentrates on the A/B section and> has the two aligned on the mid-point. This confirms what you> were asking about whether the angles were exactly 45 degrees.> For the green path, the angle turned by the ray is slightly> more than 90 degrees at each end so the angle between the> mirror and the horizontal ray is slightly less than 45> degrees. For the red path, the angle is slightly more at both> ends.> You said in another post:> [GD]> c + v' = v * cos(a) + c * cos(b)> [HW]> Not so. If the source is moving in the direction v, then light speed in> the> horizontal direction will be c+[v*cos(a)]> I have added a bit more to this diagram:> http://www.briar.demon.co.uk/Henri/speed.gif> Obviously for the normal situation of v< is negligible but this is just simple vector addition.> although less than 45 degrees for the green path and more for> the red path.> Regardless of that, it should be clear that v*cos(a) adds at> one end and subtracts at the other and the factor of cos(b)> multiplies at one and divides at the other so the apparent> speed of the light at the left end of the diagram must be> exactly c regardless of angle a. The result is that since,> in your source-dependent theory, the light is re-emit> at c, there is no a velocity kick in this case. I am not> ignoring this aspect, I have calcula it and found it to> be zero.> The same diagram can be applied when considering the leg> from the source or to the detector.>Fringes are cused by the relative angle between the beams>which stays the same as both rays rotate the same way. The>beam has to be wide enough to fall on the eyepiece so>sideways movement doesn't have an effect, the actual shift>is very small and much less than the size of the beam.but the sideways movement is also indicative of an angular> difference>>between> the two beams.>>No, see the green lines on the diagram. The rays turn the same>>way by the same amount so the angle between them is unaffec.>> But your diagram is not indicative of what actually happens..>>Why not? you seemed happy with it a few posts ago.I was happy with the diagram, not its significance.> http://www.briar.demon.co.uk/Henri/paths.gif> The intention was to show that the two rays turned in the> same direction hence the angle between changes by the> difference, not the sum. What you cannot get from the> diagram is that the derivative of the angle for each ray> at v=0 must be the same for both rays so the angle between> them must be unchanging at v=0.