mm-96 === >>Before Genzten proved Con(PA), Wilhelm Ackermann proved it. You have in mind Ackermanns 1940 proof? According to Szabo, >> Ackermann [1940, 1950-1953] adapted Gentzens method of using >> For a paper explaining the technicalities, see www.logic.at/people/moser/publications/ ackermann050901.pdf (Before Gentzen, Ackermann had a supposed proof of the consistency of >> analysis, but this turned out to be incorrect.) Maybe Im confused. Ive noticed. Nice guy one newsgroup / troll in another. Russell -- Are you still here? The message is over. Shoo! Go away! === The slippery gooiness of biology is a consequence of its incredible > complexity, consisting as it does of complex systems based upon chemistry. > And chemistry obeys the rules of physics, which exists because of, and is > consequently best described by, mathematics. Mathematics is the ur-?> R > Reality (gad, how poetic), and our symbollic attempts to represent > mathematics have given us windows through which our mushy grey-matter can > peer, and with which this same mushy grey-matter becomes altered, and we > call this alteration understanding (a frequently generous appellation). > [...to press our noses against some of the windows that look upon biological > systems...] [snip sciolistic brainfarting] Thermodynamics proposes, kinetics disposes. Kinetic control is the wild ride. Nobody can usefully model turbulence through space and over time. Uncle Al likes a universe that actively excludes god. > The book is pleasant and it is written in good Latin American Spanish but to > the readers who do not know the Castilian language (Latin American Spanish):(literate Spanish)::(Ebonics):(literate English) The Real Academia Espa.96ola located in Madrid is entrusted with purifying, clarifying and giving splendor to the language (re similar exercises of jackbooted State lingustic compassion in France, Quebec, and Israel). What a bunch of losers. -- Uncle Al http://www.mazepath.com/uncleal/qz.pdf http://www.mazepath.com/uncleal/eotvos.htm (Do something naughty to physics) === >The slippery gooiness of biology is a consequence of its incredible > complexity, consisting as it does of complex systems based upon chemistry. > And chemistry obeys the rules of physics, which exists because of, and is > consequently best described by, mathematics. Mathematics is the ur-?> R > Reality (gad, how poetic), and our symbollic attempts to represent > mathematics have given us windows through which our mushy grey-matter can > peer, and with which this same mushy grey-matter becomes altered, and we > call this alteration understanding (a frequently generous appellation). > [...to press our noses against some of the windows that look upon biological > systems...] > [snip sciolistic brainfarting] Thermodynamics proposes, kinetics disposes. Kinetic control is the > wild ride. Nobody can usefully model turbulence through space and > over time. Uncle Al likes a universe that actively excludes god. > Then you are of the opinion that the math that has been achieved in the medical sciences, pharmochology, etc... should be revised or eliminated or that this all works into body movements anyway? Not a bad idea, trying to translate robotics into sports medicine. > The book is pleasant and it is written in good Latin American Spanish but to > the readers who do not know the Castilian language (Latin American Spanish):(literate Spanish)::(Ebonics):(literate > English) The Real Academia Espa.96ola located in Madrid is entrusted with > purifying, clarifying and giving splendor to the language (re > similar exercises of jackbooted State lingustic compassion in France, > Quebec, and Israel). What a bunch of losers. -- > Uncle Al > http://www.mazepath.com/uncleal/qz.pdf > http://www.mazepath.com/uncleal/eotvos.htm > (Do something naughty to physics) === > (Latin American Spanish):(literate Spanish)::(Ebonics):(literate hey Uncle Al, arent we getting caustic about something that doesnt even exist? There is no such thing as Latin American Spanish. Youre acting like Thiotimoline ;P === > (Latin American Spanish):(literate Spanish)::(Ebonics):(literate hey Uncle Al, arent we getting caustic about something that doesnt > even exist? There is no such thing as Latin American Spanish. Youre > acting like > Thiotimoline ;P I live in Southern California. Spanish as spoken by literate Spaniards is a lyrical Romance language. Spanish as spoken by local swine is an agrammatical slurred patois that sounds like a gutteral chicken cackling while ?ing a machine gun. New World wetback women are astoundingly ugly as a class. -- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) Quis custodiet ipsos custodes? The Net! === > I need some help with my history essay. > Does anyone know what Archimedes corollary is that was included in his > Measurement of a Circle? Or have any ideas how i could ?d it, as im > having dif?ultity ?ding it in the books i have got. Look for the Great Books of the Western World. This is a 50 or so > volume of translations into English of classic books of the > western world. Alas, the series was out of print for a long time; is it back in? === > I need some help with my history essay. > Does anyone know what Archimedes corollary is that was included in his > Measurement of a Circle? Or have any ideas how i could ?d it, as im > having dif?ultity ?ding it in the books i have got. > Look for the Great Books of the Western World. This is a 50 or so > volume of translations into English of classic books of the > western world. Alas, the series was out of print for a long time; is it back in? Yes. You can buy a copy from my website. James Harris === > Alas, the series was out of print for a long time; is it back in? Yes. You can buy a copy from my website. Im not sure Im willing to trust anything on your website, sorry. === > .... > Look for the Great Books of the Western World.... One volume is partially devoted to the works of Archimedes.... Yes. Pp.447-451 have the standard translation by Heath of Measurement of a Circle without any corollaries. Ken Pledger. === Two points. 1. if f_{n_i}(x_{n_i}) >= s then the closed interval, I_{n_i} containing > x_{n_i} with f_{n_i}(x) >= s for all x in I_{n_i} may only contain > x_{n_i}. This allows len(I_{n_i}) = 0. What you really want is for > f_{n_i}(x_{n_i}) > s so that for some _open_ interval I_{n_i} we have > f_{n_i}(x) > s for all x in I_{n_i} and len(I_{n_i}) > 0. 2. Even assuming that we are considering intervals with positive length, > that the midpoints of those intervals have a limit point does not > imply that any of the intervals intersect. Consider the intervals > { [(1/(2n-1),1/(2n)] : n = 1, 2, 3, ... }. 0 is the limit point of > their midpoints, yet 0 is in none of them, not in?itely many. > ok. I was thinking closed intervals helped, for some reason. rich >Rob Johnson I met the following problem i one book,which can be solved either by >measure theory or by Lebesque Dominated Convergence theorem,but the >author says it is too dif?ult to handle without these means .I think >I need help on it. Let fn:[0,1]--->[0,1] are continuous functions and fn--->0 for each x >in [0,1].Then Int(fn(x)dx,0,1)--->0! The following comes from deconstructing the measure-theoretic proof. Note that any open subset U of [0,1] is the union of at most countably many disjoint open intervals. De?e m(U) as the sum of the lengths of those intervals. If f:[0,1] -> [0,1] is continuous and E(e) = {x: f(x) > e} (which is an open set) for e > 0, note that int_0^1 f(x) dx <= e + m(E(e)). For we can take any partition P of [0,1] and consider the lower Riemann sum L(f;P). The contribution of those intervals not contained in E(e) is bounded above by e, while the contribution of those intervals contained in E(e) is bounded above by m(E(e)). Now for each n let E_n(e) = {x: f_n(x) > e}. It is enough to show that for any e > 0, m(E_n(e)) < e for n suf?iently large. For convenience, Ill ? e and write E_n(e) as E_n. Let G_n = {x: f_n(x) > e but f_m(x) <= e for all m > n}. Since f_n -> 0, E_n = union_{m=n}^in?ity G_n. I claim that there are open sets H_m with G_m contained in H_m and sum_n m(H_n) < in?ity. If so, then taking N so large that sum_{n >= N} m(H_n) < e, E_n is contained in the union of countably many open intervals of total length < e, and therefore m(E_n) < e [ proving this therefore is an exercise left to the reader ]. We can write G_n = F_n F_{n+1} where each F_n = union_{m >= n} E_n is open, and F_{n+1} is a subset of F_n. F_n is the union of at most countably many disjoint open sets V_k of total length m(F_n), and F_{n+1} is the union of at most countably many disjoint open sets U_j of total length m(F_{n+1}); each U_j is contained in exactly one V_k. For each k, if z_k = m(F_(n+1) intersect V_k) is the total length of the U_j contained in V_k, take a ?ite subset of the U_j contained in V_k having total length at least z_k - 2^(-n-k). Take the complement in V_k of the union of this ?ite subset, and fatten up the intervals comprising it to make them open: thus V_k F_{n+1} is contained in an open set W_k of total length at most m(V_k) - z_k + 2^(1-n-k). The union of those W_k is our set H_n, with m(H_n) <= sum_k (m(V_k) - z_k + 2^(1-n-k)) <= m(F_n) - m(F_{n+1}) + 2^(1-n). And in particular, sum_n m(H_n) <= m(F_n) + 2 < in?ity, QED. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === >>Let me take that back. I either have a unplanned proof, or a planned >>unproof. >You all can decide. >>Given f_n:[0,1]-->[0,1], each f_n is continuous, and f_n(x)-->0 for x in >[0,1]. Prove (by elementary means) that int(f_n(x),0,1)-->0. >>Proof: >>Note int(f_n(x),0,1)=f_n(x_n) for some x_n in [0,1]. Also note if >>f_n(x_n)=sup >{ f_n(x) } then f_n(x)=f_n(y) for all x,y in [0,1]. I dont see what that last statement has to do with the proof, but >>its true... >Clearly, {f_n(x_n)} has >some limit point. Suppose L>0 is the largest one. There is some real >>number >s,with 0=s for in?itely many n_i. For each >such n_i there is also some closed interval, I_n_i, containing x_n_i, with >f_n[I_n_i]>=s Slightly informal notation but Im pretty sure I know what you mean. >>Fine so far. >and len(I_n_i)>0. Now let m_n_i be the midpoint of each I_n_i. >The set {m_n_i} has a limit point L and therefore L is in in?itely many >I_n_i. > Whoops. How does it follow that L is in in?itely many I_n_i? > >Because each I_n_i is *closed*? Why post messages on sci.math if youre not going to read the replies? If you look at the example I gave showing that L need not be in in?itely many I_n_i (unless its for some reason you havent explained) youd see that I_n_i being closed doesnt help a bit. >(That they exist follows from the note about >sup{f_n(x_n)} above and the fact L is the largest limit point.) I didnt say the I_n_i doesnt exist... > But then f_n(L) >=s for in?itely many n, contradicting f_n(x)-->0. >So the unique limit point of {f_n(x_n)} is 0, as required. rich ************************ David C. Ullrich === Why post messages on sci.math if youre not going to read the >replies? If you look at the example I gave showing that L need >not be in in?itely many I_n_i (unless its for some reason you >havent explained) youd see that I_n_i being closed doesnt >help a bit. > rich We just started Riemann-Stieltjes Integration and Im completely lost. We did not do Riemann ?st, so I am new to this. I do not understand how to prove this theorem or even what exactly I`m supposed to prove. The d(g1+g2) part is really throwing me off. I guess I expect one g. Im not sure how the existence of 2 of the integrals will be able to ensure the existence of the other 2 in the conclusion. The theorem is (S=integral here) S[b a] f1 dg1 and S [b a] f2 dg2 both exist then S[b a] f1+ f2 d(g1+g2) = S[b a] f1 dg1+ S[b a] f2 dg1 + S [b a] f1 dg2 + S[b a] f2 dg2 === >We just started Riemann-Stieltjes Integration and Im completely lost. We did >not do Riemann ?st, so I am new to this. I do not understand how to prove this theorem or even what exactly I`m >supposed to prove. The d(g1+g2) part is really throwing me off. I guess I >expect one g. Im not sure how the existence of 2 of the integrals will be >able to ensure the existence of the other 2 in the conclusion. The theorem is (S=integral here) S[b a] f1 dg1 and S [b a] f2 dg2 both exist >then S[b a] f1+ f2 d(g1+g2) = S[b a] f1 dg1+ S[b a] f2 dg1 + S [b a] f1 dg2 + >S[b a] f2 dg2 The other integrals do not need to exist. Let f_1 be the characteristic function of [0,1/2] and f_2 be the characteristic function of (1/2,1]. Let g_1(x) = 1/(3-4x) and g_2(x) = 1/(1-4x). Then, |1 |1 2 | f d g = | f d g = - | 0 1 1 | 0 2 2 3 However both of the other integrals diverge. In fact, with a simple change of variables, we get that 2 |1 - + | f d g 3 | 0 2 1 |1 = | (f + f ) d g | 0 1 2 1 |1 4 = | -------- dx | 0 (3-4x)^2 |1 4 = | -------- dx | 0 (1-4x)^2 |1 = | (f + f ) d g | 0 1 2 2 |1 2 = | f d g + - | 0 1 2 3 which obviously diverges. Rob Johnson take out the trash before replying === >We just started Riemann-Stieltjes Integration and Im completely lost. We did >not do Riemann ?st, so I am new to this. >I do not understand how to prove this theorem or even what exactly I`m >supposed to prove. The d(g1+g2) part is really throwing me off. I guess I >expect one g. Im not sure how the existence of 2 of the integrals will be >able to ensure the existence of the other 2 in the conclusion. >The theorem is (S=integral here) S[b a] f1 dg1 and S [b a] f2 dg2 both exist >then S[b a] f1+ f2 d(g1+g2) = S[b a] f1 dg1+ S[b a] f2 dg1 + S [b a] f1 dg2 + >S[b a] f2 dg2 The ?st thing to do is always to get straight exactly what youre asked to prove. I dont know if we can really help you there: youll have to ask your instructor. The existence of S[b a] f1 dg1 and S [b a] f2 dg2 do not imply the existence of any of the other integrals. But maybe the = is to be interpreted as: the left side exists if and only if all integrals on the right side exist, and in that case both sides are equal. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === I am interested in going to a mathematics school in the United States. I live in North Carolina, but where the school is is not important to me, as long as it is affordable and/or has a good scholarship program. I am an A student in math, with an interest primarily in algebra. Where would you recommend I apply? I am interested in pure research, as well as teaching. I am liking physics and may consider furthering my studies there. Katie. === > I am interested in going to a mathematics school in the United States. > I live in North Carolina, but where the school is is not important to > me, as long as it is affordable and/or has a good scholarship program. > I am an A student in math, with an interest primarily in algebra. > Where would you recommend I apply? I am interested in pure research, > as well as teaching. I am liking physics and may consider furthering > my studies there. Katie. You might want to take a look at the North Carolina School of Science and Math for your senior high school year. http://www.ncssm.edu/ === > I am interested in going to a mathematics school in the United States. > I live in North Carolina, but where the school is is not important to > me, as long as it is affordable and/or has a good scholarship program. > I am an A student in math, with an interest primarily in algebra. Where would you recommend I apply? I am interested in pure research, > as well as teaching. I am liking physics and may consider furthering > my studies there. If money is an issue ?st see what you can accomplish with a community college. Check with university before hand to know what theyll accept. === > I am interested in going to a mathematics school in the United States. > I live in North Carolina, but where the school is is not important to > me, as long as it is affordable and/or has a good scholarship program. > I am an A student in math, with an interest primarily in algebra. >Where would you recommend I apply? I am interested in pure research, > as well as teaching. I am liking physics and may consider furthering > my studies there. > If money is an issue ?st see what you can accomplish with a community > college. Check with university before hand to know what theyll accept. Skip the community college. They dont have anything past calculus, except maybe (that is, sometimes but not always) a ?st course in differential equations and/or linear algebra, with emphasis on calculating. You need to be writing proofs as soon as possible. If you go to a large university (say, just to pick one at random, one in calculus. This is a course that emphasizes proofs and understanding rather than just calculation. If youre already ?ished with calculus when you start, then you need to take a course that emphasizes writing proofs. Also, check the catalog. Youre better off at a school that offers a year of analysis and algebra (usually titled abstract or moder algebra, to emphasize the difference between it and college algebra, which is high school algebra) instead of only a semester. You may be better off at a smaller school where you can easily meet the professors and talk to them about things and take independent study courses rather than formal ones. Especially if youre shy. My daughter is at a tiny liberal arts college and she got to analyze DDT levels by gas chromatog raphy of ?h caught in the Pine River (EPA superfund site) her freshman year, just because she talked to the right people and someone was excited about her proposal. On the other hand, I went to a school with about 50,000 students and I didnt have any trouble ?ding professors who were interested in me. Besides, if youre at a school without a graduate program, how will you take graduate courses? When you visit colleges, visit the math departments and talk to people. (Also visit any other department youre interested in.) Look at the math library. Ask about practicing and coaching for the Putnam exam. What the heck, ask about tutoring opportunities or even undergraduate teaching assistantships (thatll help you pay your tuition). And ask about scholarships. For example, Texas Tech has a program that students in the top x% of their graduating class (or with an a.bc gpa) and an SAT above y are automatically granted in-state tuition, with more available for even better scores. Its on their web page. Many other schools have similar things. Your questions to schools should be can I do X? and what can you provide me? Of course, everyone will answer yes, so then you have to ask How? Jon Miller DuBois Double Reversed Dual Earth Model of 4D Space: How to make If someone here knows if these were made before, let me know by whom and I will change the name. I am put a picture of one at http://www.members.aol.com/scandere/ along with some other stuff on 4D. I am not selling them, I am just telling people how to make them and why they represent 4D because I probably wont get to make them myself. I am posting this message just to sci.math, sci.physics.relativity, sci.astro, sci.physics, as I thought it ?s these best. Is there a usenet group devoted to 4D? If so let me know so I know where best to go for this topic. I have read most web sites devoted to 4D space already. Take 1 really big glass sphere and 1 smaller one. Fishbowls are recommended if you can ?d perfectly round ones. Paint a backwards globe (continents only looks best) on the large one. Paint a normal globe on the small one and put it inside the larger one, but ?up side down and line it up backwards or opposite of the continents inside the larger one. You now have the the entire universe in your sights, not to scale, and only if curved for real. I originally intended to make 4 different sculptures of large sizes to demostrate aspects of 4D space. Most were to be about 3 to 4 meters in height. This one, DDRDE was the most obvious and simplest to understand, curved 3D space. Curved 3D space is wherever you go in space you wind up back at the same spot. On a globe it would be like ?ing a bullet so far you shoot yourself in the back of the head. That is curved 2D space, and painful. This represents the same concept with 3D space curved in all directions away in 4D space, and is an accurate representation of how it would line up. Objects curve away and seem completely around you when at equal distances in all directions, the antipole (the south pole on a 3D model, if you shot 4 bullets in directions at once, they would converge twice, once at the southpole and once at the north pole). Objects at either of those two points would, in curved space, not only appear to surround them but those points would be those same distances away, so it is not a true distortion but a mapping of the shape of space. That middle south pole equivalent in this model can be thought of as another sphere in between the two, and also as a single point or spot everywhere along that sphere as well, since it is only a single place like the opposite pole on a globe, but in 3D space opposite of where six bullets would pass each other if shot in curved space from a cube ?g in spaces sides. I did build a small table top model. The full scale (not to the size of the real actual scale model with the really really big lights) models I wanted to make were to be 4 meters built out of a 3 frequency geodesic dome made of glass and steel with a 1 meter center sphere. For aestic purposes the center sphere should always be at least 20% to not more than 25% of the outer sphere. Walking around in one you can get a feel for curved space. The center sphere should also be hollow so you can see through both at the same time with little to no distortion. The invertibility aspect came when I imagined a kid sticking his nose up to the center sphere and seeing how Africa and South America lined up then running around outside it to see the same view again. Thus with glass, the inversion is self- evident. Below is from the 5D notes pertaining to the dual globes. All (of the 4D sculptures) ought to be made of glass to demostrate the invertibility of curved space, or the inverse effect, visually and easily. That becomes so obvious (with glass), that the same view from all outside points (combined) looking inward through both spheres is exactly the same as the outward view from the center sphere looking outward in all directions (from its center point), that few could not be able to glimpse it. The purpose of the sculptures is to enable or help people to spin the Universe on its head, 3 dimensionally speaking. To instead or also see space as folded into objects instead of existing outside or between them. That from a 4th dimensional viewpoint all points away from the center point to the antipole equivalently and equally outside of it and within it, for they are the same spot. This is similar to how the standard idea of an antipole in the sculpture would be another sphere halfway between the other two, yet also (would) be in actuality a single spot or place, not a circular area. This multidimensional viewpoint is a good way to view electrons existing within a set position in folded space instead of orbiting around a circular area. They would not be moving at all, yet seem to be at all points in a (seemingly) circular ball around the nucleus as well, at the same time. Yet another folded dimension around the nucleus could explain how they jump from one orbit to the next without passing the space inbetween. Just because the level of space we percieve between what would seem to be 2 spheres from an outside viewpoint, seems 3 dimensional, that is no reason to believe that on the atomic scale, that space is not de?ed by or exists with more dimensions than we need to deal with out here (nor necessarily curled up so small as to not affect matter). The point of the scupltures is to see that inversible 3D sandwich, or that out there or in here is merely from an external viewpoint, (relative to where you are standing and) potentially equally the same spot. To understand closed 4D space or open curved 3D space, you only need to imagine being 2 places at the same time, which the sculptures do (represent). I could combine 2 of them as 4 different sculptures in which by standing in (and) combining them mentally would simulate being (in) 8 places at the same time, or 4 intersecting 4D worlds at weird angles, representing closed 5D space or 4D open space curved, but few would understand it, though it would look really cool. Jared. === Could someone recommend me rigorous and comprehensive books on mathematical logic and set theory? === > Ive concluded that the problem Im facing is that the human brain > isnt built to handle Mathematics. > > Well, I take it you are now claiming that you dont have a ?human brain? > > Think about it some more. I think James means that he is the only one with a human brain. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === >Ive concluded that the problem Im facing is that the human brain >isnt built to handle Mathematics. > > Well, I take it you are now claiming that you dont have a ?human brain? > > Think about it some more. I think James means that he is the only one with a human brain. Think about it some more. James Harris === The quick brown fox jumps over the lazy dog. c bfgn zwgmh hel zaclfv. === >The quick brown fox jumps over the lazy dog. Cwm fjord bank glyphs vext quiz. -- Richard -- FreeBSD rules! === Richard Nixon > The quick brown fox jumps over the lazy dog. > c bfgn zwgmh hel zaclfv. The ?st sentence is a cipher key for the second. The plaintext begins I know. LH === > Richard Nixon > The quick brown fox jumps over the lazy dog. > c bfgn zwgmh hel zaclfv. > The ?st sentence is a cipher key for the second. > The plaintext begins I know. And continues aoout the alieno ? -- Clive Tooth http://www.clivetooth.dk === The Last Danish Pastry > Larry Hammick > Richard Nixon > The quick brown fox jumps over the lazy dog. > c bfgn zwgmh hel zaclfv. > The ?st sentence is a cipher key for the second. > The plaintext begins I know. And continues aoout the alieno ? Apparently so :) === On 14 Jan 2004 19:26:08 -0800, K00L-Aid@excite.com (Richard Nixon) >The quick brown fox jumps over the lazy dog. > Hey there tricky Dick! I thought you were dead. It means you used all the letters of the alphabet in your sentence. I remember practicing that sentence when learning to type. --Lynn === > On 14 Jan 2004 19:26:08 -0800, K00L-Aid@excite.com (Richard Nixon) >The quick brown fox jumps over the lazy dog. > Hey there tricky Dick! I thought you were dead. It means you used all > the letters of the alphabet in your sentence. I remember practicing > that sentence when learning to type. --Lynn But the redundancy can be lowered a bit by making one of the words the into the word a. And I always preferred: Pack my box with ?e dozen liquor jugs which achieves the same goal with fewer letters and fewer words. There are, I believe, even better, though odder, ways to do it. === There are, I believe, even better, though odder, ways to do it. An odder and better example: Mr. Jock, TV quiz Ph.D., bags few lynx. -------- Tad === >I would like to know if this is a mathematical proof of the Cantors >goof, or conversely it is just another mathematical goof about the >Cantors proof. Its a goof. count the reals. Thus demonstrating that Mr. Mathman is not a genuine mathematician. >HC: I dont know it, but if we put a simple analogy perhaps we could >clarify our thoughts. Lets try it. >We may take the following equivalence table > N <==> M = A set of linear units of measure > R <==> A = The air in a room >Being our goal to ?d out whether A is measurable or not, we have >the following analogy: >POINT 1: A cannot be measured with linear unities, so we cannot >assume M -> A. >CRITERION: A set D will be lineally measurable if and only if it is >possible M -> D. >POINT 2: We cant use CRITERION to ?d out whether A is measurable >or not, because POINT 1 tells us that A cant be measured in this way. >PROOFS: We have found out that A is not measurable, because CRITERION >case. >PROPOSAL: We can admit CRITERION to measure A, and consequently A is >not measurable. Well, no one will accept this, because POINT 1 tells >us that we cannot apply CRITERION to measure A, and also because A >would be able to be measured by other means. >Good, Mr. Mathman, this is the end of our analogy. >MR. MATHMAN: And? >HC: What? >MR. MATHMAN: Are there other criteria to ?d out whether R is >countable or not? >HC: Well, in our analogy A cannot be measured with linear units, but >if we arrange orthogonally three of them, then we can. Perhaps the >reals have an undiscovered property which may be used as criterion to >count them. Think about it Mr. Mathman, think about it. >MR. MATHMAN: Ill do it. Are you suggesting the possibility that there is a bijection between N^3 and R? We know that there is a bijection between N^3 and N, and we also know that there is no bijection between N and R. It follows that there is no bijection between N^3 and R. Furthermore, for ANY nonzero natural number k, there is a bijection between N^k and N, and so for all natural numbers k, there is no bijection between N^k and R. The upshot is that your analogy fails miserably. And Mr. Mathman proves once again that he is not a genuine mathematician. The de?ition for countability of an in?ite set is known. Perhaps if you spent more time studying mathematics, you might see that we were right all along. Or is there nothing that can shake your conviction that you are correct? >*************************** >Proof of Point 2 As I noted earlier, your proof of Point 2 was based on the unstated assumption that there is no bijection between N and a proper subset of N, an assumption which is known to be false. So, the proof that there is no bijection between N and R is not as trivial as you thought. David McAnally Despite anything you may have heard to the contrary, the rain in Spain stays almost invariably in the hills. === > Prove that > ((a % m) + (b % m))%m = (a + b)%m > where a,b,m are integers. > Assuming your ns should be ms Whats an imteger? :-) Phil A member of a set which satis?s Peamos Postulates? -- Paul Sperry Columbia, SC (USA) === I have a question about provability and unprovability results that must have been studied by now. I was hoping someone could tell me a summary of what is known about it. For example, let us work in ZF, and let us choose a dif?ult question, say the Continuum Hypothesis, as the statement S. We know that S is not provable or disprovable (in ZF), that is, it is undecidable. If we now let S1 be the statement S is undecidable in ZF, then we know that S1 is decidable, indeed, it is true. Question: Are there statements S for which S1 (S is undecidable) is *also* undecidable? We can go further, and for any statement S construct S1 = S is undecidable, S2 = S1 is undecidable, S3, etc. Question: How far can we progress down this series? Dale === > We know that S is not provable or disprovable (in ZF), that is, it is > undecidable. If we now let S1 be the statement S is undecidable in ZF, then we > know that S1 is decidable, indeed, it is true. True, yes, but decidable in what theory? Not in ZF. No statement of the form A is undecidable in ZF is provable in ZF. So you get your series: CH, CH is undecidable in ZF, ?CH is undecidable in ZF is undecidable in ZF,... of statements undecidable in ZF. === On 15 Jan 2004 08:30:19 +0100, Torkel Franzen > We know that S is not provable or disprovable (in ZF), that is, it is >> undecidable. If we now let S1 be the statement S is undecidable in ZF, then we >> know that S1 is decidable, indeed, it is true. True, yes, but decidable in what theory? Not in ZF. No statement of >the form A is undecidable in ZF is provable in ZF. ??? Now you have me confused again. What _do_ we need in addition to ZF in order to show that this and that is undecidable in ZF? Oh, duh: For _any_ given P, showing that ZF does not imply P requires Con(ZF); if ~Con(ZF) then ZF certainly does imply P. (So when anyone says that this or that is undecidable in ZF what they really mean is that if ZF is consistent then this or that is undecidable in ZF; in particular your True, yes is assuming Con(ZF) (the word assuming is not meant to say anything about whether youre justi?d in assuming this.)) >So you get your >series: CH, CH is undecidable in ZF, ?CH is undecidable in ZF is >undecidable in ZF,... of statements undecidable in ZF. ************************ David C. Ullrich === I have a question about provability and unprovability results that > must have been studied by now. I was hoping someone could tell me a > summary of what is known about it. Not really, but I can give a perspective on it and see if the common-sense approach has any holes in it. :) > For example, let us work in ZF, and let us choose a dif?ult > question, say the Continuum Hypothesis, as the statement S. We know that S is not provable or disprovable (in ZF), that is, it is > undecidable. If we now let S1 be the statement S is undecidable in ZF, then we > know that S1 is decidable, indeed, it is true. So far so good. > Question: Are there statements S for which S1 (S is undecidable) is > *also* undecidable? [My intuition:] No. Suppose S is decidable. Then S1 is decidably false, since we can exhibit an actual proof or disproof of S, at least in theory. So (S decidable) -> (S1 decidably false) Heres where my ability to express my intuition fails, and Im forced into tricks with notation. ;-) Taking the contrapositive of this statement yields ~(S1 decidably false) -> ~(S decidable) Now, suppose for the sake of argument that S1 were actually undecidable. (S1 undecidable) -> ~(S1 decidable) -> ~(S1 decidably false) where each line represents a weaker proposition than the line before it. But by the last proposition, -> ~(S decidable) Which is exactly equivalent to the statement (S undecidable) So, if we assume (S1 undecidable), we are led to the inescapable conclusion that (S undecidable), which is precisely the statement (S1 true). So if S1 is undecidable, then it is the case that S1 is true; which looks an awful lot like a contradiction! Caveat: The contradiction depends on our being able to express the above argument entirely in the system under discussion. Its quite possible to have true, undecidable, statements. Im pretty sure that Godel proved that we *must* have such statements in *any* system. But that doesnt automatically mean that this proof is wrong. :) > We can go further, and for any statement S construct S1 = S is > undecidable, S2 = S1 is undecidable, S3, etc. Question: How far can we progress down this series? I think not even the ?st step holds water, unfortunately; but Im eager to see what subtleties Ive missed. -Arthur === On Fri, 9 Jan 2004, Mark Grif?h asked: > Anyone know of any recent work on the odd perfect number > question? Oddly enough, a paper claiming to prove the conjecture has appeared, dated Thu, 8 Jan 2004, at http://www.arxiv.org/abs/hep-th/0401052/ I havent got the background to tell whether its good or bad. === > On Fri, 9 Jan 2004, Mark Grif?h asked: > Anyone know of any recent work on the odd perfect number > question? Oddly enough, a paper claiming to prove the conjecture has > appeared, dated Thu, 8 Jan 2004, at > http://www.arxiv.org/abs/hep-th/0401052/ > I havent got the background to tell whether its good or bad. It doesnt look very promising to me, although admittedly Ive only skimmed through it. isnt relevant, or at least I didnt see in the paper any explanation of why it might be, and that doesnt inspire much con?ence at the outset. (It could be he feels obliged to add it because his institution or department specializes in physics. Seems odd, but I guess that would account for it.) Also, in the paper itself, despite being reasonably formatted and having some sensible looking references, the proof itself is at the very least abominably badly presented. He launches straight into in intricate series of manipulations with no prior introduction or summary, and no explanation of what ?k is in the ?st equation, for example. That in itself wouldnt be enough to condemn it out of hand; but just at a ?st glance there are even more worrying features. For example, throughout the proof, fractions pop up which it appears he may be assuming without justi?ation are integers (unless that is taken care of in one of the references, but if so why not mention it?) It may all be impeccably correct; but for what its worth Id bet a fair amount that the whole thing is nonsense. John Ramsden === >> On Fri, 9 Jan 2004, Mark Grif?h asked: >> Anyone know of any recent work on the odd perfect number >> question? Oddly enough, a paper claiming to prove the conjecture has >> appeared, dated Thu, 8 Jan 2004, at >> http://www.arxiv.org/abs/hep-th/0401052/ >> I havent got the background to tell whether its good or bad. It doesnt look very promising to me, although admittedly Ive only > skimmed through it. isnt relevant, or at least I didnt see in the paper any explanation > of why it might be, and that doesnt inspire much con?ence at the > outset. (It could be he feels obliged to add it because his institution or > department specializes in physics. Seems odd, but I guess that would > account for it.) It would also account for the loathsome practice (common amongst ?zisists) of omitting titles of papers in references [9, 12]. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === > On Fri, 9 Jan 2004, Mark Grif?h asked: > Anyone know of any recent work on the odd perfect number > question? > > Oddly enough, a paper claiming to prove the conjecture has > appeared, dated Thu, 8 Jan 2004, at > http://www.arxiv.org/abs/hep-th/0401052/ > I havent got the background to tell whether its good or bad. If I correctly read what is written in that paper, the ?st error is around the top of the ?st page. I do not know whether 1.1 can be satis?d. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === >On Fri, 9 Jan 2004, Mark Grif?h asked: >Anyone know of any recent work on the odd perfect number >question? > > Oddly enough, a paper claiming to prove the conjecture has >appeared, dated Thu, 8 Jan 2004, at > http://www.arxiv.org/abs/hep-th/0401052/ >I havent got the background to tell whether its good or bad. If I correctly read what is written in that paper, the ?st error is > around the top of the ?st page. I do not know whether 1.1 can be > satis?d. Its certainly one of the worst-written maths papers Ive seen for a while. He really is most allergic to actually explaining his notations. :-( I nearly got to the end of section 1. Abstract. He might mention that its easy to prove that an odd perfect number (indeed any odd number n with sigma(n) = 2 (mod 4)) has the form n = q^{4m+1} p_1^{2 r_1} ... p_s^{2 r_s} where q, p_1, ..., p_s are distinct primes, and q = 1 (mod 4). (He really should say that his 4k+1 is a *prime* distinct from any of his q_i s). Page 1. (1.1) is just a de?ition --- hes saying that if []/[] is put into lowest terms, its a_i/b_i. (Here [] and [] are those gruesome fractions on either side which I wont copy). (1.2) is really an if and only if. As it stands its just a complicated way of saying that N is *not* perfect. (This is a bit loopy: his section heading talks of one for a number being not perfect :-) ) Im not sure why all those square roots are lying around... why didnt he write this as 2(4k+1)(b_1 ...b_l)/() []^{l+1}.... =/= ... ? (1.4) Now here his notation starts to get a bit gothic. He has things like a_{13}. I reckon that doesnt mean the thirteenth a_i but rather a_1 a_3. Also he has .Square . It took me a while to realize that .Square means times a square (presumably of a rational number). But realizing that, the formula looks even more batty. Why include brackets like (a_1 a_3 b_2/b_1 b_3 a_2) when he could have had (a_1 a_2 a_3/b_1 b_2 b_3) without any difference in meaning? After that he talks about these things, but in his notation he shoves overlines onto his subscripts ... Why? He then claims that some of these fractions arent square multiples of 2(4k+1) referring to his previous paper ..... In (1.5) he introduces some notation and talks about fractions being square-free.... why doesnt he keep things simple, and represent these quantities as a square-free integer times a square of a rational? Next page... (1.7) lacks a parenthesis, but (1.6) through to (1.9) are unreadable as they stand. E.g., (1.9) contains rho-hat_{2i,2} which has never been de?ed. ((1.5) de?es rho-hat_{3j+2}). At this stage, I wonder whether there is any worth trying to keep second-guessing this geezer as to what he means. Anyway my guess as to the import of section 1 is that he reckons with N of the form given in the abstract, sigma(N)/N cant even be twice a square of a rational and he reckons hes proved that in the sequel. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === >On Fri, 9 Jan 2004, Mark Grif?h asked: >Anyone know of any recent work on the odd perfect number >question? > > Oddly enough, a paper claiming to prove the conjecture has >appeared, dated Thu, 8 Jan 2004, at > http://www.arxiv.org/abs/hep-th/0401052/ >I havent got the background to tell whether its good or bad. If I correctly read what is written in that paper, the ?st error is > around the top of the ?st page. I do not know whether 1.1 can be > satis?d. What about A_i = (q_i-1)((4k+1)^(4m+2)-1) B_i = (4k) (q_i^(2^alpha_i+1)-1) g=gcd(A_i,B_i) a_i=A_i/g b_i=B_i/g ? However, Im not putting any money on 1.2 being meaingful or not. Hmmm... I thought General Math was the kooks hangout? Im glad I normally surround myself with mathematical matters, as this physics stuff, particularly the stuff Simon Davis is into, looks pretty heavy: http://www.iop.org/EJ/S/UNREG/G6Lj4JvumVUVaKYOUtyVpw/abstract /-search=522350 5.2/0264-9381/18/17/305 <<< Abstract. String propagation in ten-dimensional Minkowski space or the direct product of Minkowski space and a six-dimensional K.8ahler manifold or orbifold might be regarded as an approximation to a theory which allows for the local curvature of spacetime by the energy-momenta of the component ?lds. String scattering in the interaction region might then be based on quantum ?ld theory in a local region with a curved geometry. Special emphasis is given to ?ld theory in anti-de Sitter space, as it represents a maximally symmetric spacetime of constant curvature which could arise in the description of matter interactions in local regions of spacetime. Curvature shifts in the momentum and squared mass are evaluated for scalar ?lds in anti-de Sitter space, and it is shown that the shift in p2 + m2 compensates the ground-state contribution to the bosonic string Hamiltonian, implying the consistency of computing the scattering entirely in ?ace. Dual space rules for evaluating Feynman diagrams in Euclidean anti-de Sitter space are initially de?ed using eigenfunctions based on generalized plane waves. Loop integrals can be evaluated even more easily using momentum space rules in conformally ?ordinates for anti-de Sitter space, which admits ?ree-dimensional sections that are analytic continuations of horospheres in hyperbolic space H4. An additional argument in favour of the model of string propagation described in this paper is based on the removal of re?e boundary conditions on quantum ?lds interacting in a locally anti-de Sitter region without spatial in?ity, implying the existence of a one-parameter family of O(3,2)-invariant vacua in this region consistent with the degree of freedom in de?ing the string theory vacuum. > http://www.iop.org/EJ/abstract/0264-9381/11/5/007 <<< Abstract. The divergences that arise in the regularized partition function for closed bosonic string theory in ?ace lead to three types of perturbation series expansions, distinguished by their genus dependence. This classi?ation of in?ities can be traced to geometrical characteristics of the string worldsheet. Some categories of divergences may be eliminated in string theories formulated on compact curved manifolds. > http://www.iop.org/EJ/S/UNREG/G6Lj4JvumVUVaKYOUtyVpw/abstract /-search=522350 5.1/0264-9381/20/13/331 <<< Abstract. The quantum cosmological wavefunction for a quadratic gravity theory derived from the heterotic string effective action is obtained near the in?ary epoch and during the initial Planck era. Neglecting derivatives with respect to the scalar ?ld, the wavefunction would satisfy a third- order differential equation near the in?ary epoch which has a solution that is singular in the scale factor limit a(t) -> 0. When scalar ?ld derivatives are included, a sixth-order differential equation is obtained for the wavefunction and the solution by Mellin transform is regular in the a 0 limit. It follows that inclusion of the scalar ?ld in the quadratic gravity action is necessary for consistency of the quantum cosmology of the theory at very early times. > He seems in a very specialist ?ld - quantum cosmological wavefunction yields only 19 hits in google, of which 18 pertain to Simon Davis own papers. And what is the research foundation of southern california? Googles not really heard of it except for discussions of whether it exists or not in the context of tracing the author of another paper (by a B. Davis, rather than S. Davis). Weird. Whatever. Phil -- Unpatched IE vulnerability: history.back method caching Description: cross-domain scripting, cookie/data/identity theft, command execution Exploit: http://www.safecenter.net/liudieyu/RefBack/RefBack-MyPage.HTM === Hmmm... I thought General Math was the kooks hangout? Im glad I normally surround myself with mathematical matters, as this > physics stuff, particularly the stuff Simon Davis is into, looks > pretty heavy: > He seems in a very specialist ?ld - quantum cosmological wavefunction > yields only 19 hits in google, of which 18 pertain to Simon Davis own > papers. And what is the research foundation of southern california? > Googles not really heard of it except for discussions of whether > it exists or not in the context of tracing the author of another > paper (by a B. Davis, rather than S. Davis). Davis has 21 items on MathSciNet, all on stringy stuff apart from his cited paper whose review I append. MR1979400 (2004b:11007) Davis, Simon(5-SYD-SM) A rationality condition for the existence of odd perfect numbers. (English. English summary) 11A25 (11B39 11D41 11D61) Review in linked PDF Add citation to clipboard Document Delivery Service Journal Original Article More links More links If an odd perfect number exists, it must exceed $10^{300}$ ref[R. P. Brent, G. L. Cohen and H. J. J. te Riele, Math. Comp. 57 (1991), no. 196, 857--868; MR 92c:11004]. The author observes that an odd integer $N$ can be perfect only if a certain product depending on its prime divisors has a rational square root. This product contains factors which are repunits, that is, of the form $(q^n-1)/(q-1)$, for the primes $q$ appearing to an even power in the canonical decomposition of $N$. By using this rationality condition and properties of repunits the author studies the existence of odd perfect numbers $N$. He obtains an upper bound for the density of such $N$ in any ?ed interval (above $10^{300}$) and proves the nonexistence of $N$ in some special classes of integers. An important role in his discussions is played by various Diophantine equations. Reviewed by T. Mets?nkyl? The maths department at Sydney University is certainly reputable. The journal Int. J. Math. Math. Sci. is certainly legit (they did once publish a paper by me, albeit a potboiler even by my standards :-) ) A couple of the reviews of his stringy stuff include some barbed comments: corresponding style. ... The mathematical de?itions of the above terms previous work ... ... The author shows the modular covariance of the volume form on Siegels upper half-space. (This is of course a well-known mathematical fact.) ... -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === >> Ha. You dont know what you are talking about. Poincare conjecture? >> And even Goldbach is not recursive, but recursively enumerable. To >> stay in simple number theory : there exist an in?ity of twin >> primes is not even recursively enumerable (nor corecursively >> enumerable, of course). You didnt answer that So if I make a TM that for input x=0, tries to asnwer the yes/no question, if there is an in?ity of twin primes, you say it doesnt halt and you cant prove it, but the point is, that maybe it does halt, you just dont know, and currently cant prove neither. >> How do you know? Undecidable problems exists. What about the >> Continuum hypothesis? You didnt answer that either. sorry, dont know about them. > No. This time, you dont understand the rules (and are trying to disprove > G.9adel in some sense) What is outside the *whole* system of proofs? ok you are right. let me formalize your argument in my words, so I will understand better. Bascially, you say, that mathematics has more True claims, that it can prove(by Godel). Since mathemaics captures the notion of algorithm(?) as we know it, and if we take the Church thesis, TM is equivalent to algorithms, there are problems (sets of numbers) or there are yes/no quesions, that TM will never be able to answer. That is the TM may halt or not halt, but we dont know it. === >> Ha. You dont know what you are talking about. Poincare conjecture? >> And even Goldbach is not recursive, but recursively enumerable. To >> stay in simple number theory : there exist an in?ity of twin >> primes is not even recursively enumerable (nor corecursively >> enumerable, of course). You didnt answer that So if I make a TM that for input x=0, tries to asnwer the yes/no > question, How on earth do you do that? The meaning of my sentence is that the brden is on you to make such a TM, and it is not possible by enumerating anything... > if there is an in?ity of twin primes, you say it doesnt halt and > you cant prove it, but the point is, that maybe it does halt, you > just dont know, and currently cant prove neither. >> How do you know? Undecidable problems exists. What about the >> Continuum hypothesis? You didnt answer that either. sorry, dont know about them. Time to learn some maths? > No. This time, you dont understand the rules (and are trying to >> disprove G.9adel in some sense) What is outside the *whole* system >> of proofs? ok you are right. let me formalize your argument in my words, so I > will understand better. Bascially, you say, that mathematics has more > True claims, > that it can prove(by Godel). Since mathemaics captures the notion of > algorithm(?) as we know it, and if we take the Church thesis, TM is > equivalent to algorithms, there are problems (sets of numbers) or > there are yes/no quesions, > that TM will never be able to answer. That is the TM may halt or not > halt, > but we dont know it. Mmmm... Something like that. But actually mathematics capture much more than algorithms. How do you propose to capture some truth like the pythagore theorem by some TM? Before Godel codes, no one had any idea about that... === Grab this perl script: http://www.farviolet.com/~entropy/decompose.txt Here is what it does, dont know if this is any use, but it has an interestingly odd result. Recursive decomposition of prime numbers, for the purpose of this discussion 0 is counted as a prime #. Any number 0 - N can be represented in the form: 0^a * 2^b * 3^c * 5^d * 7^e ... Or more simply the array: (a, b, c, d, e, ...) For example: 1 -> (0 0 0 0 0 ...) 0 -> (1 0 0 0 0 ...) 12 -> (0 2 1 0 0 0 ...) 16 -> (0 4 0 0 0 ...) Or to simplify, leave off trailing zeros: 1 -> () 0 -> (1) 12 -> (0 2 1) 16 -> (0 4) Now the recursive part, represent the powers in the array the same way: 1 -> () 0 -> (1) -> (()) 12 -> (0 2 1) -> ((1) (0 1) ()) -> ((())((())())()) 16 -> (0 4) -> (0 (0 2)) -> ((())((())((())()))) Your left with the ability to represent any number as a set of grouping symbols after repeated recursive prime decomposition. Dont know if this is of any use at all, but its neat :) Notice that for the numbers 0 and 1, the expression ?0^n is equivalent to ?NOT n. -- Laughter-Confusion, Pleasure-Pain, Happyness-Sadness, Excitement-Fear, Love-Hate, etc. The true primary emotions, a modi?r makes each into two. This modi?r is acceptance/unacceptance. Let go, surrender, accept... Be a counter terrorist perpetrate random senseless acts of kindness === > Grab this perl script: http://www.farviolet.com/~entropy/decompose.txt Here is what it does, dont know if this is any use, but it has an > interestingly odd result. Recursive decomposition of prime numbers, for the purpose of this > discussion > 0 is counted as a prime #. Any number 0 - N can be represented in the form: 0^a * 2^b * 3^c * 5^d * 7^e ... Or more simply the array: (a, b, c, d, e, ...) For example: 1 -> (0 0 0 0 0 ...) > 0 -> (1 0 0 0 0 ...) > 12 -> (0 2 1 0 0 0 ...) > 16 -> (0 4 0 0 0 ...) Or to simplify, leave off trailing zeros: 1 -> () > 0 -> (1) > 12 -> (0 2 1) > 16 -> (0 4) Now the recursive part, represent the powers in the array the same way: 1 -> () > 0 -> (1) -> (()) > 12 -> (0 2 1) -> ((1) (0 1) ()) -> ((())((())())()) > 16 -> (0 4) -> (0 (0 2)) -> ((())((())((())()))) Your left with the ability to represent any number as a set of grouping > symbols after repeated recursive prime decomposition. Dont know if this > is of any use at all, but its neat :) Notice that for the numbers 0 and 1, the expression ?0^n is equivalent to > ?NOT n. I myself posted something along these lines myself recently (amongst the cr**) at: http://groups.google.com/groups?hl=en&lr=&ie=UTF-8&safe=off& threadm=b4be2fdf .0312061711.15da0b01%40posting.google.com&rnum=6&prev= Your contribution was (I gather by introducing the zero) to do away with the primes (and 1), leaving just parentheses. At the very least, this is an easy proof that the set of lists of parentheses grouped this way has a cardinality of aleph-null (because of the bijection with the integers). thanks, Leroy Quet === > Grab this perl script: http://www.farviolet.com/~entropy/decompose.txt GP script into tree form: e(n)=local(s,p,c,q);Str((,if(n==0,(),if(n==1,,s=(());forprim e(p=2,99999,c=0;while(n%p==0,c++;n/=p);s=Str(s,e(c));if(n==1, break));s)),) ) > Here is what it does, dont know if this is any use, but it has an > interestingly odd result. ... > Now the recursive part, represent the powers in the array the same way: 1 -> () > 0 -> (1) -> (()) > 12 -> (0 2 1) -> ((1) (0 1) ()) -> ((())((())())()) > 16 -> (0 4) -> (0 (0 2)) -> ((())((())((())()))) Your left with the ability to represent any number as a set of grouping > symbols after repeated recursive prime decomposition. Dont know if this > is of any use at all, but its neat :) It is indeed neat. It appeals to the Lisp programmer in me! Whod have thought that the numbers 5 ((())(())(())()) 12 ((())((())())()) 18 ((())()((())())) 42 ((())()()(())()) 64 ((())((())()())) 70 ((())()(())()()) 105 ((())(())()()()) 2310 ((())()()()()()) would have had something in common. Phil -- Unpatched IE vulnerability: dragDrop invocation Description: Arbitrary local ?e reading through native Windows dragDrop invocation. Reference: http://msgs.securepoint.com/cgi-bin/get/bugtraq0302/12.html Exploit: http://kuperus.xs4all.nl/security/ie/x?es.htm === I have this homework question that Ive been struggling for quite a while. I hope someone will give me some hint on how to do this proof. For a, b natural numbers, consider the set of numbers ar + bs for all integers r, s so that ar + bs >= 1. Since this set is nonempty, by well-ordering it has a least element Show that the least element of this set is the greatest common disvisor of a and b. First of all, i can tell that (a, b) exists in the set because (a, b) = ar + bs for some r, s. Now, how can I prove it has to be the least element? Gavin === >I have this homework question that Ive been struggling for quite a while. I >hope someone will give me some hint on how to do this proof. For a, b natural numbers, consider the set of numbers ar + bs for all >integers r, s so that ar + bs >= 1. Since this set is nonempty, by >well-ordering it has a least element Show that the least element of this set >is the greatest common disvisor of a and b. Let c be the least positive element of {ar+bs : r,s in Z}. 1. Show (a,b) | c (remember that (a,b) | a and (a,b) | b). 2. Show c | a and c | b (that is, c is a common divisor of a and b). If all else fails, see http://www.whim.org/nebula/math/bezout.html >First of all, i can tell that (a, b) exists in the set because (a, b) = ar + >bs for some r, s. Now, how can I prove it has to be the least element? I could be wrong here, but isnt one of the things you are trying to prove that (a,b) = ar+bs for some integers r and s? If so, you are not allowed to use this fact in your proof. Rob Johnson take out the trash before replying === > Are the following proofs correct? 2) Let R be the ring of all real-valued functions of > a single variable under pointwise addition and multiplication. > The subset S of R of functions whose graphs pass through > the origin forms a subring of R. Prove S is a subring of R. > f in R passes thru the origin when f(0) = 0. Thus for f,g in S, f(0) = g(0) = 0 and (f+g)(0) = f(0) + g(0) = 0 (-f)(0) = -f(0) = 0 (f*g)(0) = f(0) * g(0) = 0 showing f+g, -f and f*g are in S. Does your text require rings to have identities? Mind doesnt. > Proof: > It suf?es to show that S is closed under substraction and > multiplication. > Clearly S is non-empty since f(x)=x is in S. Ok. So is the additive identity f(x) = 0. > So assume f(x)=x*h(x) and g(x)=x*z(x) for some h(x) > and z(x) in R[x]. Just a nanosec. R[x] isnt the ring of real functions, its the ring of polynomials with real coef?ients. > But then f(x)-h(x)=x[h(x)-z(x)]=x*q(x) hence x=0 is a root > thus S is closed under substraction. > Similarly f(x)*h(x)=x^2*h(x)*z(x)=x*p(x) thus S is closed under Whycramequationstogetherwithoutspaces? > multiplication and by the subring test this implies that S > is a subring of R. > Makes no sense in the context of ring of real functions. === I was wondering if anyone knows how to set restrictions on integers or simply how to generate the 8 integers that satisfy the following two conditions: 1) a8^2 + a1^2 = a5^2 + a4^2 = a7^2 + a2^2 = a6^2 + a3^2 2) (a8^2 - a1^2)^2 + (a5^2 - a4^2)^2 = (a7^2 - a2^2)^2 + (a6^2 - a3^2)^2 or to put it another way: 1) a8^2 + a1^2 = x a5^2 + a4^2 = x a7^2 + a2^2 = x a6^2 + a3^2 = x 2) (a8^2 - a1^2)^2 + (a5^2 - a4^2)^2 = y (a7^2 - a2^2)^2 + (a6^2 - a3^2)^2 = y [No relation between x and y] [0 < a1 < a2 < a3 < a4 < a5 < a6 < a7 < a8] Now, I know how to solve for the ?st condition, but Im at a loss on how to mathematically generate a set of 8 integers that satisfy both conditions simultaneously. Ive used brute force methods and have found two sets of 8 integers that satisfy the above two conditions. But I would like to know if I can _generate_ more (instead of searching for more). Or, maybe these are the only two solutions? The two sets of solutions are: {a1, a2, a3, a4, a5, a6, a7, a8} {11, 77, 101, 131, 343, 353, 359, 367} {139, 317, 541, 719, 827, 953, 1049, 1087} I would greatly appreciate any help that anyone can offer in this -David C. === I dont have any speci? solutions, but I can put the problem into a general context for you. You really have 4 equations in 8 unknowns: a8^2 + a1^2 = a5^2 + a4^2 a5^2 + a4^2 = a7^2 + a2^2 a7^2 + a2^2 = a6^2 + a3^2 (a8^2 - a1^2)^2 + (a5^2 - a4^2)^2 = (a7^2 - a2^2)^2 + (a6^2 - a3^2)^2 There are three equations that are homogeneous of degree 2, and the other one is homogeneous of degree 4. They de?e a variety V in projective 7-space, and since the four equations are independent, your variety V has dimension 3. Your integer solutions correspond to rational points on the projective variety V. Your variety V is probably nonsingular(?), though I havent checked. Lets assume for the moment that it is nonsingular. Then its canonical sheaf is O(-7-1+2+2+2+4) = O(2). (In general, if you have a smooth variety in projective n-space given by the intersection of homogeneous polynomials of degree d_1,d_2,...,d_r, then its canonical sheaf is O(-n-1+d_1+...+d_r). Of course, Im assuming that it has dimension n-r.) Anyway, the fact that V has canonical sheaf O(2) means that it is a variety of general type. There is a famous conjecture of Bombieri and Lang saying that the rational points on a variety of general type are not Zariski dense. In less fancy language, what that means for your system of equations is that there is a homogeneous polynomial F(a1,a2,...,a8) which is independent of your equations so that every integer solution of your equations is also a solution to F=0. Unfortunately, the Bombieri-Lang conjecture is still just a conjecture, and in any case, even conjecturally it gives no practical procedure for ?ding the polynomial F. OTOH, it does suggest that the non-obvious solutions to your equations are likely to be few and far between. N.B. All of this is contingent on your equations de?ing a nonsingular variety. If they dont, then all of this may not apply. Finally, on a positive note, the way that people typically produce lots of integer solutions to equations like this (when they do manage) is by ?ding an elliptic curve of positive rank that sits on the variety. Thats how Elkies solved the famous problem of sums of fourth powers; but Elkies variety had trivial canonical sheaf, so it wasnt of general type. Joe Silverman I was wondering if anyone knows how to set restrictions on integers or > simply how to generate the 8 integers that satisfy the following two > conditions: > 1) > a8^2 + a1^2 = a5^2 + a4^2 = a7^2 + a2^2 = a6^2 + a3^2 > 2) > (a8^2 - a1^2)^2 + (a5^2 - a4^2)^2 = (a7^2 - a2^2)^2 + (a6^2 - a3^2)^2 > > two sets of 8 integers that satisfy the above two conditions. But I > would like to know if I can _generate_ more (instead of searching for > more). Or, maybe these are the only two solutions? The two sets of solutions are: > {a1, a2, a3, a4, a5, a6, a7, a8} > {11, 77, 101, 131, 343, 353, 359, 367} > {139, 317, 541, 719, 827, 953, 1049, 1087} I would greatly appreciate any help that anyone can offer in this -David C. === > I am going crazy trying to ?d a formula/equation for the >equivalent resistance asross diagonally opposite ends of a >resistive mesh. Can someone help. X > o--+--Rh--+--Rh--+--Rh--(n columns)--Rh--+--Rh--+ > | | | | | | | > Rv Rv Rv | | Rv Rv > | | | | | | | > +--Rh--+--Rh--+--Rh--(n columns)--Rh--+--Rh--+ > | | | | | | | > Rv Rv Rv | | Rv Rv > | | | | | | | > . . . .(m rows) .--Rh--+--Rh--+ > | | | | | | | > Rv Rv Rv | | Rv Rv > | | | | | | | > +--Rh--+--Rh--+--Rh--(n columns)--Rh--+--Rh--+ > | | | | | | | > Rv Rv Rv | | Rv Rv > | | | | | | | > +--Rh--+--Rh--+--Rh--(n columns)--Rh--+--Rh--+--o Y > > Imagine a rectangular mesh with resistors at every small >segment. The horizontal segment resistances are Rh and >vertical ones are Rv. Need to get a formula to calculate >the effective resistance between X and Y. Any help welcome! > I do have a method for solving this. Very nice problem. The bad news is, that my method is rather involved and I do not have much time available at the moment. So, I will write it up and post it, but it will take another week or so. I hope you can manage from going totally crazy for that long. Michael. -- &&&&&&&&&&&&&&&&#@#&&&&&&&&&&&&&&&& Dr. Michael Ulm FB Mathematik, Universitaet Rostock michael.ulm@mathematik.uni-rostock.de === f:[0,1] -> R f(x) = (x^2)sin(1/x^2) if x != 0 0 if x = 0 int f(x)dx 0~1 decide that this intergral of f(x) is exist or not exist?? -------------------- i think..... f(x) = (2x)sin(1/x^2) - (2/x)cos(1/x^2) thus f(x) is conti except x=0 i cant progress any more about it i wait your response === > f:[0,1] -> R f(x) = (x^2)sin(1/x^2) if x != 0 0 if x = 0 int f(x)dx > 0~1 decide that this intergral of f(x) is exist or not exist?? -------------------- i think..... f(x) = (2x)sin(1/x^2) - (2/x)cos(1/x^2) Its easy to see f is not bounded on (0,1) (look along the sequence 1/sqrt(2nPi)), therefore it is not Riemann integrable on [0,1]. Does int_[0,1] f(x) dx exist as an improper Riemann integral? Sure. Let a > 0. Then int_[a,1] f(x)dx = f(1) - f(a), by the Fundamental Theorem of Calculus. But f is continuous at 0, so the last term has a limit as a -> 0+. === > f:[0,1] -> R f(x) = (x^2)sin(1/x^2) if x != 0 0 if x = 0 int f(x)dx > 0~1 decide that this intergral of f(x) is exist or not exist?? -------------------- i think..... f(x) = (2x)sin(1/x^2) - (2/x)cos(1/x^2) thus f(x) is conti except x=0 i cant progress any more about it i wait your response f is unbounded in every interval (0,eps), thus as a Riemann integral, int_0^1 f(x) dx is at best improper (at worst, doesnt exist). Since f is continuous on [eps,1], we have f(1) - f(eps) = int_eps^1 f(x) dx. But f is continuous, hence as eps decreases to 0, f(1) - f(0) = int_0^1 f(x) dx (the right-hand side being BY DEFINITION the limit as eps decreases to 0 of the earlier RHS). As Rob Johnson points out, f is not Lebesgue integrable. --Ron Bruck === > f:[0,1] -> R f(x) = (x^2)sin(1/x^2) if x != 0 0 if x = 0 int f(x)dx > 0~1 decide that this intergral of f(x) is exist or not exist?? -------------------- i think..... f(x) = (2x)sin(1/x^2) - (2/x)cos(1/x^2) calculation error: ^ should be x^2*(-2/x)cos(1/x^2) thus f(x) is conti except x=0 ^ possibly But now you can easily check for continuity at x=0. Jon Miller === En el mensaje:m7idnTGF7MvKVJvd4p2dnA@comcast.com, Jonathan Miller escribi.97: >> f:[0,1] -> R f(x) = (x^2)sin(1/x^2) if x != 0 0 if x = 0 int f(x)dx >> 0~1 decide that this intergral of f(x) is exist or not exist?? -------------------- i think..... f(x) = (2x)sin(1/x^2) - (2/x)cos(1/x^2) calculation error: ^ should be x^2*(-2/x)cos(1/x^2) >> Not should be x^2*(-2/x^3)cos(1/x^2) = - (2/x)cos(1/x^2), as the OP write ... -- Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com === > En el mensaje:m7idnTGF7MvKVJvd4p2dnA@comcast.com, > Jonathan Miller escribi.97: >> f:[0,1] -> R >> f(x) = (x^2)sin(1/x^2) if x != 0 >> 0 if x = 0 >> int f(x)dx >> 0~1 >> decide that this intergral of f(x) is exist or not exist?? >> -------------------- >> i think..... >> f(x) = (2x)sin(1/x^2) - (2/x)cos(1/x^2) >calculation error: ^ should be x^2*(-2/x)cos(1/x^2) > > Not should be x^2*(-2/x^3)cos(1/x^2) = - (2/x)cos(1/x^2), as the OP write Duh. Thats the second time in a week Ive made a stupid mistake because I wouldnt take the time to write down what I was thinking. Must be getting old. Move on folks. Nothing to see here, just a guy whose brain is wandering away from him. Jon Miller === >f:[0,1] -> R f(x) = (x^2)sin(1/x^2) if x != 0 0 if x = 0 int f(x)dx >0~1 decide that this intergral of f(x) is exist or not exist?? -------------------- i think..... f(x) = (2x)sin(1/x^2) - (2/x)cos(1/x^2) thus f(x) is conti except x=0 2x sin(1/x^2) is continuous in [0,1] so it is integrable. Let u = 1/x^2 so that du/u = -2 dx/x. Then |1 2 1 | - cos( --- ) dx | 0 x x^2 |oo 1 = | - cos(u) du [1] | 1 u For each integer k >= 0, |5pi/4 + k pi 1 | | - cos(u) | du | 3pi/4 + k pi u pi/2 1 >= ------------ ------- 5pi/4 + k pi sqrt(2) sqrt(2) = ------- [2] 5+4k Therefore, |oo 1 | | - cos(u) | du | 1 u oo --- sqrt(2) >= > ------- [3] --- 5+4k k=0 and [3] diverges by comparison to the harmonic series. Thus, [1] does not exist as a Lebesgue integral. As a Riemann integral, [1] |oo 1 = | - d sin(u) | 1 u |oo 1 = -sin(1) + | --- sin(u) du [4] | 1 u^2 The integral in [4] converges absolutely. Therefore, [1] exists as a Riemann integral but not a Lebesgue integral. Rob Johnson take out the trash before replying === Suppose you have a -A and a A V B operator de?ed, and A ^ B = -(-A V -B) as usual, and commutative and associative law hold for ^ and V, and --A=A, and 0,1 exists (0=-1, 0 V A = A etc.)...i.e. standard Boolean and DeMorgan, only the distributive law fails. Does this structure already have a name? -- Hauke Reddmann <:-EX8 fc3a501@uni-hamburg.de als man ankam wollte man werden, die geschichte schreiben, die doofen sollen sterben, der plan als man damals nach hamburg kam (Kettcar) === > Suppose you have a -A and a A V B operator de?ed, > and A ^ B = -(-A V -B) as usual, and commutative and > associative law hold for ^ and V, and --A=A, and 0,1 > exists (0=-1, 0 V A = A etc.)...i.e. standard Boolean > and DeMorgan, only the distributive law fails. Does this structure already have a name? > A non distributive complemented lattice. For example the complete atomic (not uniquely) complemented non-modular lattice of topologies of a set. cf A.K.Steiner (1966) ?The Lattice of Topologies === > Suppose you have a -A and a A V B operator de?ed, > and A ^ B = -(-A V -B) as usual, and commutative and > associative law hold for ^ and V, and --A=A, and 0,1 > exists (0=-1, 0 V A = A etc.)...i.e. standard Boolean > and DeMorgan, only the distributive law fails. I guess complemented Lattice, satisfying the DeMorgan identities is not exactly, what you want. Have you already looked through Birkhoffs Lattice theory? Marc === * Hauke Reddmann > Suppose you have a -A and a A V B operator de?ed, > and A ^ B = -(-A V -B) as usual, and commutative and > associative law hold for ^ and V, and --A=A, and 0,1 > exists (0=-1, 0 V A = A etc.)...i.e. standard Boolean > and DeMorgan, only the distributive law fails. Does this structure already have a name? Beats me, but do you have an example? A smallest set that satis?s your requirements? Should be interesting enough. It is not a ring, so it avoids any elementary algebra books... -- Jon Haugsand === > * Hauke Reddmann > >>Suppose you have a -A and a A V B operator de?ed, >>and A ^ B = -(-A V -B) as usual, and commutative and >>associative law hold for ^ and V, and --A=A, and 0,1 >>exists (0=-1, 0 V A = A etc.)...i.e. standard Boolean >>and DeMorgan, only the distributive law fails. Does this structure already have a name? Beats me, same here. nondistributive boolean-like algebra? Nearboolean? > but do you have an example? A smallest set that satis?s > your requirements? Should be interesting enough. facts courtesy of mace (http://www-unix.mcs.anl.gov/AR/mace4) A model doesnt exist when restricted to 2 elements (that is using the boolean axioms, there is exactly one model, and using the axioms above (over just - and V), there is exactly one model and it is equivalent to the boolean algebra on 2 elements. For 3 elements, there is a single algebra that satis?s your axioms but is not a boolean algebra (there are 3 models in total, only 2 also satisfying the boolean algebra axioms (all 3 are equivalent when restricted to 0 and 1). Here is the nondistributive model: not : 0 1 2 --------- 1 0 2 or : | 0 1 2 --+------ 0 | 0 1 2 1 | 1 1 1 2 | 2 1 0 and : | 0 1 2 --+------ 0 | 0 0 0 1 | 0 1 2 2 | 0 2 1 One distributivity offending term is or(2,and(2,2)) = 1 != 0 = and(or(2,2),or(2,2)) But theres a bit of a catch (I realize after doing the brute calculation). I left out idempotence: or(a,a) = a, and absorption: or(x,and(x,y))=x Then, the two axiom sets have the same (unique) model. Same with 4 elts (both have 4 models), but with 5 elts there are 6 boolean algebras and 16 models of the -,V algebras. -- Mitch Harris (remove q to reply) === >[snip Dinkys trash] >Quite a CV {:-)) >Franz > Apart from your subjective opinions of any people involved, Franz, do you > have any > objective opinion on the mathematics at > http://www.androc1es.pwp.blueyonder.co.uk/Fundamental_rv_2.0. htm > or are you completely enraptured by Dinkys silly game of one-up-manship? > Androcles > http://groups.google.com/groups?&as_umsgid=TCVIb.109057$ Jl3.4986923@phobos.te lenet-ops.be >Dirk Vdm >I can testify that among relativists you said the truest thing ever > about the 1905 paper,you didnt recognise the excerpt from the 1905 > paper and this makes it all the more funny. >Perhaps Androcles recognises that it is from the kinematics section of > the 1905 paper and I am delighted that you agree with him that it is > ?shit that is not worth reading. > ------------------------------------------------------------- ---------------- > >Let us take a system of co-ordinates in which the equations of > Newtonian mechanics hold good. In order to render our presentation > more precise and to distinguish this system of co-ordinates verbally > from others which will be introduced hereafter, we call it the > ``stationary system. >What is this? > Some kind of quote of some post? > An introduction to the shit you produce later on? > Shit that you expect someone will bother reading? >Dirk > ------------------------------------------------------------- ---------------- I already said this on > http://groups.google.com/groups?&as_umsgid=3f71af17@usenet01. boi.hp.com > but I will repeat it here: You still dont seem to understand why I asked you > these 4 questions. > So let me try to explain in simple words. > Lets have a close look at the message you are referring > to here: > http://groups.google.com/groups?&threadm= 273f8e06.0207281123.57d80819@posting .google.com | >Let us take a system of co-ordinates in which the equations of > | Newtonian >mechanics hold good.2 In order to render our presentation > | more precise and to >distinguish this system of co-ordinates verbally > | from others which will be >introduced hereafter, we call it the > | ``stationary system. > | > | Dear Al, > | ... This is what you gave us. > As you see, > - There is something severely wrong with the format. > - If this was a quotation from Einstein, you left out the quotes. > - You write a paragraph followed by Dear Al... > - In what you have included here, you have reformatted > and left out the failed introduction Dear Al. So I will kindly ask again, and I will clarify what I mean: 1) What is this? > 2) Some kind of quote of some post? > Clari?ation: > Something you want us to believe you invented? > Something you found somewhere? > Something you want to tell us? > Something you want to tell us something about? > Something you forgot to delete when you started > with the beginning of your message Dear Al,? 3) An introduction to the shit you produce later on? > 4) Shit that you expect someone will bother reading? > Clari?ation: > The ?shit in question 4 is a reprise of the ?shit in > question 3. This is what we call a ?style ?ure. Didnt they teach you to write English in Germany? > How old are you? Dirk Vdm It still remains a classic in my book and there is not a blessed thing you can do about it. Any Englishman will recognise the title of that thread - ?A memorable Fancy comes from William Blake who uses historical characters to carry his points unfortunately you missed the point,didnt recognise Alberts 1905 work and all in all made a jackass out of yourself. That you try to defend yourself makes you twice as dumb as you already are however it is always useful to point out that relativists havent the foggiest idea what the original relativity concept means,at least thanks for that. === > >escribi.97 en el mensaje > message > [snip Dinkys trash] >Quite a CV {:-)) >Franz > Apart from your subjective opinions of any people involved, > Franz, > do you > have any > objective opinion on the mathematics at > http://www.androc1es.pwp.blueyonder.co.uk/Fundamental_rv_2.0. htm > or are you completely enraptured by Dinkys silly game of > one-up-manship? > Androcles > > http://groups.google.com/groups?&as_umsgid=TCVIb.109057$ Jl3.4986923@phobos.t > elenet-ops.be >Dirk Vdm >Androcles will never understand that c-v and c+v in Einsteins paper > are algebraic expressions which are not physical speeds measured by > any observer, but rates of travelled ** relative ** distances by the > light rays. >Yep, but Id call them rates of change of distances between some > object and the front of a lightray, distances as calculated by a third > party observer who is assumed to measure c to be the speed of the > front of the lightray w.r.t. himself. >Dirk Vdm >As this: >But the ray moves relatively to the initial point of k, when measured > in > the stationary system, with the velocity c-v, so that ... >in which we have the observer (k), and the initial point of k. > But that is somehow confusing and for a similar calculation I got a zero > in > an exam, as the teacher believed that I meant: >But the ray moves relatively to [observer] k with the velocity c-v >which ignores when measured in the stationary system. >It would have been easier just to make c*t = v*t + L with L the initial > separation among lightray front and origin of k, than write directly L = > (c-v)*t and talk of c-v as a speed (it is a speed dimensionally, but not > physicall, i.e., cannot be adscribed to any observer). >Hm, to me the initial separation between lightray front and origin > of k seems to be 0. It grows to x, and upon re?n, shrinks > back to 0. > It is clear on the spacetime diagram: > http://users.pandora.be/vdmoortel/dirk/Stuff/tau-equation.gif > There you clearly see from the geometry that the ?horizontal distances > (the dashed blue lines) between the red k-worldline (tau-axis) and > the grey light-worldline ?st grow from 0 at time t0 (event Flash) > to x at time t1 (event Re?n), and then shrink back to 0 at > time t2 (event Flashecho). >Dirk Vdm >Yes, I was thinking of the mirror as the origin of k (I didnt have the > paper with me). > In any case, Einstein made a quite dif?ult and lenghty derivation. >It would have been much easier to derive the Lorentz transforms by using the > time dilation and length contraction effects. However, I guess that > psychologically it WAS more convenient the other way: get the Lorentz > transform. and from them, derive these effects. >You provided a reference with such a kind of derivation > http://www.courses.fas.harvard.edu/~phys16/Textbook/ch10.pdf > but one thing I am at odds with, namely that the relative speeds of the two > systems of reference are not necessarily equal and opposite (a strange > wording, btw). Without a proof of such a thing, the demo is not completely > correct. As far as I can see the phrase equal and opposite does not appear > in the derivation, but merely in an example. > Notice how Einstein was very careful about it, and instead of assuming it (I > know PhDs who make that), he showed that the relative velocities were v > and -v respectively [ phi(v)*phi(-v)=1] And as far as I can see here, he did not *show* that the relative > velocities are v and -v, but rather *used* that trivial fact to show > that phi(v)*phi(-v)=1. Dirk Vdm Besides my parallel post, I have found this interesting reference: http://arxiv.org/PS_cache/physics/pdf/9703/9703006.pdf in which the concept of different relative speeds among two inertial observers is considered (see the text near the equation 4). However, if we accept the principle of relativity, the velocities are equal and opposite in the formula. By using a 3rd observer, two observers moving with v and -v will observe the same relative speeds. As this is true for any v [ Does any one know where these following sequences of numbers came from(the 4 > below) ? Who was the ?st person to discover this? And how did they come > about? There must be some reason besides the fact they are neat. Do any > others exist? 11 x 11 = 121 > 111 x 111 = 12321 > 1111 x 1111 = 1234321 > 11111 x 11111 = 123454321 > 111111 x 111111 = 12345654321 > 1111111 x 1111111 = 1234567654321 > 11111111 x 11111111 = 123456787654321 > 111111111 x 111111111 = 12345678987654321 > ----------------------------------- > 1 x 8 + 1 = 9 > 12 x 8 + 2 = 98 > 123 x 8 + 3 = 987 > 1234 x 8 + 4 = 9876 > 12345 x 8 + 5 = 98765 > 123456 x 8 + 6 = 987654 > 1234567 x 8 + 7 = 9876543 > 12345678 x 8 + 8 = 98765432 > 123456789 x 8 + 9 = 987654321 > ----------------------------------- > 0 x 9 + 1 = 1 > 1 x 9 + 2 = 11 > 12 x 9 + 3 = 111 > 123 x 9 + 4 = 1111 > 1234 x 9 + 5 = 11111 > 12345 x 9 + 6 = 111111 > 123456 x 9 + 7 = 1111111 > 1234567 x 9 + 8 = 11111111 > 12345678 x 9 + 9 = 111111111 > 123456789 x 9 + 10 = 1111111111 > ----------------------------------- > 9 x 9 + 7 = 88 > 9 x 98 + 6 = 888 > 9 x 987 + 5 = 8888 > 9 x 9876 + 4 = 88888 > 9 x 98765 + 3 = 888888 > 9 x 987654 + 2 = 8888888 > 9 x 9876543 + 1 = 88888888 > 9 x 98765432 + 0 = 888888888 > 9 x 987654321 - 1 = 8888888888 > ----------------------------------- Stephen You asked Do any others exist? Heres something which Newton used to help explain the binomial theorem. 11^1 = 11 11^2 = 121 11^3 = 1331 11^4 = 14641 After that the pattern of binomial coef?ients is obscured by the carrying of decimal digits, but even these few powers of (10 + 1) can be enlightening to learners. Ken Pledger. Ken Pledger. === > Does any one know where these following sequences of numbers came from(the 4 > below) ? Who was the ?st person to discover this? And how did they come > about? There must be some reason besides the fact they are neat. Do any > others exist? >11 x 11 = 121 > 111 x 111 = 12321 > 1111 x 1111 = 1234321 > 11111 x 11111 = 123454321 > 111111 x 111111 = 12345654321 > 1111111 x 1111111 = 1234567654321 > 11111111 x 11111111 = 123456787654321 > 111111111 x 111111111 = 12345678987654321 > ----------------------------------- > 1 x 8 + 1 = 9 > 12 x 8 + 2 = 98 > 123 x 8 + 3 = 987 > 1234 x 8 + 4 = 9876 > 12345 x 8 + 5 = 98765 > 123456 x 8 + 6 = 987654 > 1234567 x 8 + 7 = 9876543 > 12345678 x 8 + 8 = 98765432 > 123456789 x 8 + 9 = 987654321 > ----------------------------------- > 0 x 9 + 1 = 1 > 1 x 9 + 2 = 11 > 12 x 9 + 3 = 111 > 123 x 9 + 4 = 1111 > 1234 x 9 + 5 = 11111 > 12345 x 9 + 6 = 111111 > 123456 x 9 + 7 = 1111111 > 1234567 x 9 + 8 = 11111111 > 12345678 x 9 + 9 = 111111111 > 123456789 x 9 + 10 = 1111111111 > ----------------------------------- > 9 x 9 + 7 = 88 > 9 x 98 + 6 = 888 > 9 x 987 + 5 = 8888 > 9 x 9876 + 4 = 88888 > 9 x 98765 + 3 = 888888 > 9 x 987654 + 2 = 8888888 > 9 x 9876543 + 1 = 88888888 > 9 x 98765432 + 0 = 888888888 > 9 x 987654321 - 1 = 8888888888 > ----------------------------------- >Stephen You asked Do any others exist? Heres something which Newton > used to help explain the binomial theorem. 11^1 = 11 > 11^2 = 121 > 11^3 = 1331 > 11^4 = 14641 After that the pattern of binomial coef?ients is obscured by the > carrying of decimal digits, but even these few powers of (10 + 1) can > be enlightening to learners. Ken Pledger. Ken Pledger. I also liked the fact that you can work with powers of (100 + 1) to go even further than what the (10 + 1) can get you. As in: 101^1 = 101 101^2 = 10201 101^3 = 1030301 101^4 = 104060401 101^5 = 10510100501 101^6 = 1061520150601 101^7 = 107213535210701 101^8 = 10828567056280801 Then the carry kills the pattern, but then you could just use (1000 + 1) to go even further. -David C. === On a related topic, a non-mathematical friend of mine once asked me, Why is it that 1/7 = .142857142857...? He elaborated: ?st you have 14 = 2 x 7, then 28 = 4 x 7, then 56 = 8 x 7 (plus a carried 1 from the sequel), then 112 = 16 x 7, etc. He wasnt asking about a proof; he was asking what *is* it about the number 7 that makes this so? The best answer I could come up with is because 7^2 + 1 = (10^2)/2. And it was hard to come up with similar examples in other bases, though I did point out that 1/9 = .09 + .018 +.0027 + ... What would you have told him? -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === >Does any one know where these following sequences of numbers came from(the 4 >below) ? Who was the ?st person to discover this? And how did they come >about? There must be some reason besides the fact they are neat. Do any >others exist? >11 x 11 = 121 >111 x 111 = 12321 ... (sum_{j=0}^n r^j)^2 = sum_{k=0}^{2n} sum_{j=max(0,k-n)}^{min(n,k)} r^k = sum_{k=0}^{2n} min(k+1,2n+1-k) r^k >1 x 8 + 1 = 9 >12 x 8 + 2 = 98 ... (r-2) sum_{j=0}^{n-1} (n-j) r^j + n = sum_{j=0}^{n-1} (r-n+j) r^j which is equivalent to (r-1) sum_{j=0}^{n-1} (n-j) r^j = -n + r sum_{j=0}^{n-1} r^j >0 x 9 + 1 = 1 >1 x 9 + 2 = 11 >12 x 9 + 3 = 111 ... (r-1) sum_{j=0}^{n-1} (n-j) r^j + n + 1= sum_{j=0}^n r^j which is the same as the previous one. >9 x 9 + 7 = 88 >9 x 98 + 6 = 888 >9 x 987 + 5 = 8888 ... (r-1) sum_{j=0}^{n-1} (j+r-n) r^j + (r-n-2) = (r-2) sum_{j=0}^n r^j which is again equivalent. How do these grab you? ((9*1-1)^2+10)* 1 - 9*(9*1-1)* 1 = 1*2 ((9*2-1)^2+10)* 21 - 9*(9*2-1)* 41 = 2*3 ((9*3-1)^2+10)* 321 - 9*(9*3-1)* 941 = 3*4 ((9*4-1)^2+10)*4321 - 9*(9*4-1)*16941 = 4*5 ((99*1-1)^2+100)* 1 - 99*(99*1-1)* 1 = 1*2 ((99*2-1)^2+100)* 201 - 99*(99*2-1)* 401 = 2*3 ((99*3-1)^2+100)* 30201 - 99*(99*3-1)* 90401 = 3*4 ((99*4-1)^2+100)* 4030201 - 99*(99*4-1)* 16090401 = 4*5 ((99*5-1)^2+100)* 504030201 - 99*(99*5-1)* 2516090401 = 5*6 ((99*6-1)^2+100)* 60504030201 - 99*(99*6-1)* 362516090401 = 6*7 ((99*7-1)^2+100)* 7060504030201 - 99*(99*7-1)* 49362516090401 = 7*8 ((99*8-1)^2+100)* 807060504030201 - 99*(99*8-1)* 6449362516090401 = 8*9 ((99*9-1)^2+100)*90807060504030201 - 99*(99*9-1)*816449362516090401 = 9*10 Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === thusly: > Does any one know where these following sequences of numbers came from(the > 4 below) ? Who was the ?st person to discover this? And how did they > come about? There must be some reason besides the fact they are neat. Do > any others exist? 11 x 11 = 121 > 111 x 111 = 12321 > 1111 x 1111 = 1234321 > 11111 x 11111 = 123454321 > 111111 x 111111 = 12345654321 > 1111111 x 1111111 = 1234567654321 > 11111111 x 11111111 = 123456787654321 > 111111111 x 111111111 = 12345678987654321 > ----------------------------------- > 1 x 8 + 1 = 9 > 12 x 8 + 2 = 98 > 123 x 8 + 3 = 987 > 1234 x 8 + 4 = 9876 > 12345 x 8 + 5 = 98765 > 123456 x 8 + 6 = 987654 > 1234567 x 8 + 7 = 9876543 > 12345678 x 8 + 8 = 98765432 > 123456789 x 8 + 9 = 987654321 > ----------------------------------- > 0 x 9 + 1 = 1 > 1 x 9 + 2 = 11 > 12 x 9 + 3 = 111 > 123 x 9 + 4 = 1111 > 1234 x 9 + 5 = 11111 > 12345 x 9 + 6 = 111111 > 123456 x 9 + 7 = 1111111 > 1234567 x 9 + 8 = 11111111 > 12345678 x 9 + 9 = 111111111 > 123456789 x 9 + 10 = 1111111111 > ----------------------------------- > 9 x 9 + 7 = 88 > 9 x 98 + 6 = 888 > 9 x 987 + 5 = 8888 > 9 x 9876 + 4 = 88888 > 9 x 98765 + 3 = 888888 > 9 x 987654 + 2 = 8888888 > 9 x 9876543 + 1 = 88888888 > 9 x 98765432 + 0 = 888888888 > 9 x 987654321 - 1 = 8888888888 > ----------------------------------- Stephen Thats how your money is supposed to grow when you subscribe to a pyramid scheme :) -- Paul Townsend I put it down there, and when I went back to it, there it was GONE! Interchange the alphabetic elements to reply === > Thats how your money is supposed to grow when you subscribe to a pyramid > scheme :) > -- I dont mind that in this scheme the people down the bottom make the money. === >Does any one know where these following sequences of numbers came from(the 4 >below) ? Who was the ?st person to discover this? And how did they come >about? There must be some reason besides the fact they are neat. Do any >others exist? 11 x 11 = 121 >111 x 111 = 12321 >1111 x 1111 = 1234321 >11111 x 11111 = 123454321 >111111 x 111111 = 12345654321 >1111111 x 1111111 = 1234567654321 >11111111 x 11111111 = 123456787654321 >111111111 x 111111111 = 12345678987654321 >----------------------------------- >1 x 8 + 1 = 9 >12 x 8 + 2 = 98 >123 x 8 + 3 = 987 >1234 x 8 + 4 = 9876 >12345 x 8 + 5 = 98765 >123456 x 8 + 6 = 987654 >1234567 x 8 + 7 = 9876543 >12345678 x 8 + 8 = 98765432 >123456789 x 8 + 9 = 987654321 >----------------------------------- >0 x 9 + 1 = 1 >1 x 9 + 2 = 11 >12 x 9 + 3 = 111 >123 x 9 + 4 = 1111 >1234 x 9 + 5 = 11111 >12345 x 9 + 6 = 111111 >123456 x 9 + 7 = 1111111 >1234567 x 9 + 8 = 11111111 >12345678 x 9 + 9 = 111111111 >123456789 x 9 + 10 = 1111111111 >----------------------------------- >9 x 9 + 7 = 88 >9 x 98 + 6 = 888 >9 x 987 + 5 = 8888 >9 x 9876 + 4 = 88888 >9 x 98765 + 3 = 888888 >9 x 987654 + 2 = 8888888 >9 x 9876543 + 1 = 88888888 >9 x 98765432 + 0 = 888888888 >9 x 987654321 - 1 = 8888888888 I believe the Babylonians used these sequences in drawing up plans for ziggurats... >----------------------------------- Stephen > ************************ David C. Ullrich === If f:B^2 ---> B^2 is the identity mapping, then any map g within epsilon > of f must contain the closed ball of radius 1-epsilon. Yes, but this needs a proof, in fact this is the hard part of the problemm. observe that the restrictions f|S^1 and g|S^1 are homotopic as maps S^1-->B^2{x}, so g|S^1 is homotopic to identity. Then the index of point x with respect to g|S^1 equals 1 and therefore x must be covered by g(B^2). Simeon === (note 1) If you are unaware of my posts or are aware of my posts and opened this for the purpose of ridicule, you would best consider searching the post for quoted material. There are long quotes describing the formal implications and axiomatizations for quantum logic. You should ?d those to be of interest independent of any purposes I might have. (note 2) If you are aware of my posts and the ? that had been raging on sci.logic about the foundations of mathematics, then you will recognize that there is a context in the later material. It was originally an off-group correspondence to George Greene. I decided to make an independent post because of the quoted materials. References concerning Georges statements are at a minimum--one sentence per section. The extent to which you will be able to follow it will depend on what posts you did or did not read over the last year. Nevertheless, I am just posting for informative purposes. I spent the time transcribing the excerpts and this is material of which most people are unaware. :-) mitch ------------- ----- You apparently thought everything I was doing could be reduced to Boolean algebras (or Heyting algebras). The class OML of orthomodular lattices forms a variety which includes the class BA of Boolean algebras. Of course, I became aware of orthomodularity from my investigations of set theory. I had no way to state it speci?ally and I had no reference with a simple declarative statement like that above. --- You apparently thought we were in agreement about the problematic nature of models. I could not get the document http://www.uni.torun.pl/~jacekm/latoml.pdf to open at the time of this response. The paper is entitled The Lattice of Orthomodular Logics. I had hoped to avoid the convoluted explanation of a diagram that follows. If you can get it to open, you will ?d a lattice (c) with two isolated paths between TOP and BOTTOM on the right side and several interlaced paths on the left side. The two paths on the right have only one point each between TOP and BOTTOM. All paths on the left side are interlaced among six points. Let me call the two points on the isolated paths R2 and the six points with the interlaced paths L6. The interlaced paths may be explained by the form, TOP / / / / / * * * / / / / / * * * | | | | | BOTTOM There is a path exactly opposite in sense running from TOP to BOTTOM and passing through the three points not connected in this diagram. Moreover, there are uniform diagrams for all six points. So, if the edge running from BOTTOM was shifted to the L6 vertex on the bottom left, the two edges above that vertex would shift to the bottom left vertex as well. Continuing, the two edges incident to the L6 vertex on the top left would be shifted to the top center. One of my earliest questions concerned the nature of the complete connectives. Given that ?st-order logic has a soundness theorem and a completeness theorem that are converse, it seemed natural to refer to the structure I had discerned in terms of completeness connectives and soundness connectives. What I mean by the structure I had discerned is the result of deliberations about classical truth-functional connectivity that would have associated the complete connectives with R2. Those deliberations led to a recognition of L6 as a pattern without any clear concept of how the pattern should be described. That is how idiolectic usage begins. You dont have words for things, so you make the best associations you can in order to work. I had no way to explain why I thought the Steiner 1-factorization was important when I did the deduction with the Steiner quasigroup. But, if you look at the factorization, {{#,0},{1,3},{2,6},{4,5}} {{#,1},{0,3},{2,4},{6,5}} {{#,2},{0,6},{1,4},{3,5}} {{#,3},{0,1},{2,5},{4,6}} {{#,4},{0,5},{1,2},{3,6}} {{#,5},{0,4},{2,3},{1,6}} {{#,6},{0,2},{3,4},{1,5}} you can see that it segregates eight symbols into pairs. Moreover, where ?0 and ?# are separated from one another (recall that they were special symbols added to the symbols labeling my columns), the other pairs of symbols are segregated from one another. This is the same kind of uniform interlaced relationship for L6 in the lattice diagram. But, the speci? labeling introduces additional complications. I cannot yet make all the connections that I need to appear coherent. But, what lies behind this is a very simple concept, ... The class H of all characteristic functions of theories of C determines the consequence of C. This somewhat trivial fact was independently observed by several logicians - see Scott[1971], Routley[1976], and Suszko[1977a], cf. also van Fraassen[1973]. It gave rise to discussions on two-valuedness and the scope of the principle of bivalence. Suszko seems to be the one who has drawn the most far-reaching conclusions from this slogan: Every logic is two-valued. What seems to be a source of dif?ulty is that Suszkos thesis does not provide any manageable description of truth-valuations that determine a given logic C. Whatever else may be true, this thesis is immensely complicated. The text which follows goes on to say, There is no general and satisfactory de?ition of an admissible truth-valuation for a given logic C. Nevertheless, having a sense through which classical truth-functional connectivity relates to the structure of an orthomodular logic does not seem trivial with regard to the questions. The only way I could get away from the model question was to resort to threshold logic and geometric intuitions leading to linear fractional transformations. I had no way to explain what I was doing since what I was trying to demonstrate was even unclear to myself. Nevertheless, it seems to me that it what is expressed here would lead one to threshold logic and linear separability as an alternative to the usual alternative to model-theoretic consequence. --- You apparently thought my linear order was trivial. I will transcribe the necessary de?itions, but here is a theorem that seem to explain why I thought otherwise, Theorem 5.5.1. Up to identity over OML, p_1,..., p_5 are the only formulas p in two variables having the following property: for any algebra (A in OML) and all (a, b in A), p(a,b)=1 iff a<=b. We recall that the orthomodular quantum logic OML[|=] is de?ed semantically by the class OML of orthomodular lattices with the unit element 1 designated. Thus: (a in OML[|=](X)) iff for every (A in OML) and every homomorphism h:S->A, h(X) subset {1} implies h(a)=1 OML[|=] is thus the assertional logic of the class OML. Consider the following sentences of S in two variables x and y: p_0(x,y) := ~x / y p_1(x,y) := (~x / y) / (~x / ~y) / (x / (~x / y)) p_2(x,y) := (~x / y) / (x / y) / ((~x / y) / ~y) p_3(x,y) := ~x / (x / y) p_4(x,y) := y / (~x / ~y) p_5(x,y) := (~x / y) / (x / y) / (~x / ~y) Theorem 5.5.1. [stated above]... The proof of Theorem 5.5.1 is omitted. The proof makes use of the free algebra F_OML(2) on two generators. It has 96 elements and it is known to be isomorphic with the direct product MO2 x 2^4 of the Chinese lantern MO2 with the 16-element Boolean algebra, denoted 2^4. 1 // / / / / / / MO2 a ~a b ~b / / / / / / // 0 F_OML(2) is ?ite but the free algebra F_OML(3) is in?ite. Theorem 5.5.1 iimplies: Corollary 5.5.2 The logic OML[|=] is implicative and each of the above (de?able) connectives p_i (i=1,...,5) is its implication. Furthermore, the class Mod*(OML[|=]) coincides with OML. It follows from the above corollary that each of the sets {p_i(x,y),p_i(x,y)}, i=0,...,5 is an equivalence for OML[|=]. The consistent strengthenings of the logic OML[|=] are called quantum logics. Every quantum logic C is thus regularly algebraizable and, if C is ?itary, its equivalent algebraic semantics coincides with the quasivariety Mod*(C) which is clearly a quasivariety of orthomodular lattices. The classical consequence K is a limit case - it is the strongest quantum logic. The class BA of Boolean algebras is the smallest non-trivial variety of orthomodular lattices. The consequence OML[|=] has been axiomatized by Kalmbach [1974], [1981]. Let MP_i be the detachment rule determined by the implication p_i, i.e., MP_i is the rule x, p_i(x,y) ----------- y for i=0,...,5. De?e the binary connective R by: (a R b) := (a / b) / (~a / ~b) for any a, b. Then de?e the following axioms: A1 ~(p R q) / (~p / q) A2 p R p A3 ~(p R q) / (~(q R r) / (p R r)) A4 ~(p R q) / (~p R ~q) A5 ~(p R q) / ((p R r) R (q R r)) A6 (p / q) R (q / p) A7 (p / (q / r)) R ((p / q) / r) A8 (p / (p / q)) R p A9 (~p / p) R ((~p / p) / q) A10 ~(p / q) R (~p / ~q) A11 p R ~~p A12 (p / (~p / (p / q))) R (p / q) A13 (p R q) R (q R p) A14 (~p / q) ->_1 (p ->_1 (p ->_1 q)) A15 ~(p ->_1 q) / (~p / q) (In A14 and A15 we write (a ->_1 b) instead of p_1(a,b).) Theorem 5.5.3. Each of the sysetems {A1,...,A13,MP_0} and {A1,...,A15,MP_1} forms an inferential base for OML[|=]. The proof is omitted. The logics determined by the bases OML[|=](emptyset) cup MP_i for i in {2,...,5} are known to be weaker than OML[|=]. Quantum logics give rise to many counterexamples to some metalogical properties which hold for classical logic and for a large class of weaker logics. We mention here one result: Theorem 5.5.4. If C is a logic such that OML[|=]<=C<=K, and ~(C=K), then C does not admit the parameter-free Local Deduction Theorem; in particular, C does not admit the Deduction Theorem in the sense of Section 2.6 The above theorem has a simple algebraic interpretation: under the hypotheses of the theorem, the class Mod*(C) fails to have the C-?ter extension property. This result implies that BA is the only variety of orthomodular lattices with the congruence extension property. There is a lot here. I hope you noticed the special cases involved with Boolean algebras and classical logic. I have no way to explain how or why I saw things so differently from everyone else. And, I have no way to say that any material I have ever presented constitutes evidence of a rational thought process. If you look at the relation used in the axioms, (a R b) := (a / b) / (~a / ~b) you can see why I had been so adamant about the ?edness of the projection connectives and their complements under DeMorgan conjugation. Moreover, it seemed particularly important given the unate recognition problem and the symmetries I was seeing that led to the discussion of Steiner quasigroups. --- You apparently interpreted my discussion about identity as intuitionism in spite of my assertions that it was more subtle than that. There seems to be a non-linear hierarchy of logics. Fregean Protoalgebraic Logics | | | | Regularly Regularly Weakly Algebraizable Algebraizable ------- Logics Logics | | | | | | | Weakly Algebraizable Algebraizable ------- Logics Logics | | | | | | | | | Equivalential | Logics / / / / / / / / Protoalgebraic Logics I believe the concept of an equality-free logic refers to logics that are not equivalential. Here are some details: Two connectives of special interest in metalogical investigations of classical logic: the connective of implication, strictly tied to the Deduction Theorem, and the connective of equivalence. The latter expresses, in the material sense, the fact that two sentences have the same logical value while in the strict sense it expresses the property that two sentences have the same logical value while in the strict sense it expresses the property that two sentences are mutually interderivable on the basis of a given logic. The process of identi?ation of equivalent sentences relative to the theories of a logic C de?es a class of abstract algebras. The members of the class are called Lindenbaum-Tarski algebras of the logic C. One may abstract from the origin of these algebras and examine them by means of purely algebraic methods. This approach, based on Lindenbaum-Tarski algebras, turns out to be particularly important because it bridges the gap between logic and algebra, and therefore makes it possible to apply the powerful methods of contemporary algebra in metalogic. If C is the classical consequence (logic), then the relation (1) a ==_T b iff a<->b in C(T) de?es for any theory T, a congruence on the algebra of sentences (formulas). The resulting class of Lindenbaum-Tarski algebras coincides with the class of Boolean algebras. In turn, the class of Lindenbaum-Tarski algebras corresponding to intuitionistic logic coincides with the class of Heyting algebras. We may also use the implication connective ->. The logic C with the property that ==_T, de?ed by a ==_T b iff a->b, b->a in C(T) is the largest congruence on the formula algebra compatible with C(T) are called implicative. They are extensively studied in Rasiowas monograph [1974]. The question arises about the scope of the above method. There are numerous examples of logics which are intractible by this method because there may not exist connectives in the language which, according to the above formulas, de?e a congruence on the formula algebra. This is a typical situation for intensional logics. Prucnal and Wronski [1974] have proposed a generalization of the Lindenbaum-Tarski method by replacing the equivalence connective by a possibly in?ite set of sentential formulas which collectively possess many properties of the equivalence. Any logic which has such a set is called equivalential (or congruential). In a more formal rendering, a logic C is equivalential (?itely equivalential) if there exists a set (a ?ite set) E(p,q) of sentential formulas in two variables p and q such that the relation ==_T, where (2) a ==_T b iff E(a,b) subset C(T) is a congruence on the language compatible with C(T), for all theories T. The notion of an equivalential logic turns out to be very useful in the analysis of intensional logics such as modal, tense, or dynamic logics. It is actually the modal logic that seems to be relevant to the structures I had been considering. To see, why, I need to invoke a structure very similar to L6 discussed above, TOP / / / / / * * * / / / / / * * * | | | | | BOTTOM The actual result involved has to do with two modal logics, E[->] and RE[->] that are not equivalential. The proof of this is based on an algebra whose reduct is an eight-element Boolean algebra, 1 a c b ~a ~c ~b 0 I cannot draw the diagram. Here is the adjacency matrix: | 0 1 a b c ~a ~b ~c ------------------------------- 0 | 0 0 0 0 0 1 1 1 1 | 0 0 1 1 1 0 0 0 a | 0 1 0 0 0 0 1 1 b | 0 1 0 0 0 1 0 1 c | 0 1 0 0 0 1 1 0 ~a | 1 0 0 1 1 0 0 0 ~b | 1 0 1 0 1 0 0 0 ~c | 1 0 1 1 0 0 0 0 The actual proof refers to RE[->] since that result would suf?e for E[->]. It involves certain necessitation relations, []1=[]a=[]c=1 []b=~b []~a=[]~b=[]~c=[]0=0 The proof depends on two de?ed sets from the vertices in the graph, {a,1} and {0,a,~a,1}. I realize that neither of us can make sense of the proof at this point. But, the paragraph that follows the proof gives some insight as to what is pertinent here, The answer to the question whether a logic is equivalential or not depends on the non-axiomatic inference rules of the logic. For instance, the system E(=E[->](emptyset)) of the logic E[->] is closed with respect to the rule of extensionality; but other theories of this logic need not be closed with respect to RE. If E[->] is strengthened by adjoining the rule RE to the rules of E[->] (and so RE can be applied in arbitrary proofs), the resulting logic is equivalential and {p<->q} is its equivalential system. What is going on here is that the modal systems are de?ed with respect to invariant subsets of formulas containing all of the classical tautologies, CL. E is the smallest modal system and its de?ing condition is E := Sb(CL,MP,RE) where MP:= p,p->q ------ q RE:= p<->q --------- []p<->[]q The logic E[->] derived from E admits only MP as a rule of inference on arbitrary collections of sentences from the language. So closure with respect to RE only holds for the emptyset by default for E[->]. The discussion above says that if one examines the logic E[->,RE], the logic is equivalential and the formula, p<->q serves as its equivalence. This is, of course, exactly what is involved in forming a Tarski-Lindenbaum algebra for a classical system of logic. I could probably now make reference to Kant and the relation between possibility and necessity in characterizing the transcendental logic. Then I would point out his remarks on negation with a redirect to Frege. But, that would not be a rational argument. Or, at least, it would not be considered coherent by modern standards of peer review. In any case, there is a clear notion of equality-free logic, and, it does relate to intensional interpretation of quanti?rs. --- You apparently never realized what I meant in referring to almost universality or why this might be important in a discussion of set theory. The origin of non-Fregean logics is strictly connected with the abolition of the so-called Fregean Axiom by Suszko [1975a]. Let us quote the following passage by Wojcicki [1984], p.326: According to Frege, denotations (Bedeutung) of sentences are logical values. Thus, each sentence denotes either Truth or Falsehood. Suszko, who sought support for his ideas in Wittgenstein, rejects this point of view. For him, denotation of a sentence is what the sentence says about: a certain situation. This term was chosen by Suszko to interpret Wittgensteins Sachlage - the state of affairs. Situations which exist create positive facts, those which do not exist create negative facts. not denote the same. It is a certain fact that Wittgenstein knew Frege just like it is a fact that Wittgenstein exchanged letters with Russell, but these two facts are quite different, and thus two sentences stating these two facts have different denotations although their truth value is the same. Obviously, Frege was not of the opinion that all true (or false) sentences ?say the same either. In Suszkos apprehension the differences lay in the sense (Sinn) of sentences and not in their denotations. For comparison of Suszkos ideas with those of Frege it is essential that neither Sinn itself nor any of its components is an element of the objective world. Sinn is a way in which sentences are assigned their logical values (one is tempted to repeat after Ajdukiewicz ?the way of how the sentence is understood), or - which also can be found in Freges works - ?the thought conveyed by the sentence. The thought (...) understood as a certain abstract object and not an individual mental experience. The differences between Suszkos and Freges approaches are by no means of verbal character: among the concepts used by Frege there is no counterpart for the notion of a situation. In the Suszkos times the situational theory of meaning did not exist. Thus the principle that the meaning of a sentence coincides with the situation described by this sentence had a purely postulative character at that time - building a situational semantics was a task for the future. This task was performed by Wojciciki who developed foundations of situational semantics for Suszkos non-Fregean logic with identity. [...] ... We do not discuss these issues here because we would have to begin with a general account of what a situation is. Instead, we shall focus our attention on some formal aspects of the sentential logic with the identity connective. Let (S_^, /, /, ->, <->, ~, ^) be the extension of the language (S, /, /, ->, <->, ~) of classical sentential logic obtained by adjoining to it a new binary connective ^ called the identity connective: (^) a^b in C(T) iff (A(phi) in S_^)(A(p) in Var(phi)) C(T,phi(p/a))=C(T,phi(p/b)) for all (a,b in S_^) and (T subset S_^) Note that (a^b) -> (a<->b) is a thesis of K_^; the sentence that if a and b are identical, they have the same truth value. The Frege Principle in the logical form (alias Frege axiom) is the converse of the above implication. More precisely, the Fregean Axiom states that the identity of two sentences is identical to their material equivalence: (F) (p^q)^(p<->q) (F) is not a theorem of K_^ The categoreal error I made here, of course would have involved confusing zero-order logic with ?st-order logic. But, to be honest, I see no real difference since I can turn to the Blackwell Guide to ?d a discussion of motivation that led to generalized quanti?ation. If I make a uniform substitution in the sentences above, I obtain (p^q)^(p<->q) Au(u in p <-> u in q) <-> p=q (a^b) -> (a<->b) Au(u in a <-> u in b) -> a=b I will not try to justify this mistake either. Since I had been unaware of anything I am sending you here, it is hard for me to call this a mistake, though. I spent a long time thinking about extensionality in relation to identity (both eliminable and not eliminable) and ended up concluding that there was something more than was explained by what I had been taught. No one could answer my questions. --- I look forward to reading about logics discussed in the book from which these many quotes have been taken: Protoalgebraic Logics Janusz Czelakowski ISBN: 0-7923-6940-8 <3c65f87.0401131920.67d8e399@posting.google.com> <1g7ivag.1praqrrqra0psN%see.sig@for.addy> <3c65f87.0401140753.3bcde45f@posting.google.com> === > And we may speculate on JSHs deeper motivations. So far, greed and > egocentricity seem primary. And they have corrupted his judgement > woefully. speculate that greed is a motivation. Hes been explicit on that point. -- I dont know why I live in a world with so many supposed mathematicians who are all so dumb AND rude. Why oh why couldnt someone like Gauss or Dedekind still be around? Shoot, Id even take someone like Hardy at this point. -- James S Harris compromises === > And we may speculate on JSHs deeper motivations. So far, greed and > egocentricity seem primary. And they have corrupted his judgement > woefully. speculate that greed is a motivation. Hes been explicit on that > point. But he has also said that he regularly lies, and that he is using this NG as a psychological laboratory, so what do you _know_ about him (other than that he is a world champion kook)? === > On second thought... > Tomorrow morning (07.01.2004), I will go UP on terrace and toss one > Rupee coin in blue sky.. > Head: I will prepare and ?e detailed patent application. > Tail: I will NOT ?e any patent application. I will start execution > sequence of this Action Device at the same instantaneous moment I see > Tail. > There will be only one toss.. > It is UP to Gravity NOW! > ............ > My guess: Head and you cant afford to ?e the patent, therefore you > can hold on to your phantasy a bit longer. Greetings! > Volker Last night, at about 00.55 AM, I went up on terrace and with full moon right over my head, I tossed one rupee coin. It was Head. Today morning, I tossed again one rupee coin in blue sky. Again, it was Head. So God is telling me to ?e one more patent application. I have already ?ed 7 patent applications. He has taught me that Love can trigger Logic. But Logic can not trigger Love. So my beloved God, Gravity, if you are telling me to patent You so that I sell You to these people on earth, then it is simply not acceptable. I will not patent God. I will not patent Gravity, Supreme Force. I will not patent Future of Mankind, Planets, Stars, Galaxies, this Universe and I will not patent Love. I am handing over Supreme Force, this Action Device to United Nations provided that India get Veto Power. -Abhi. === > >Oh, yes. I got my instructions, script to enter in this new character >on universal stage. I have done lot of rehearsal in real life and I >am aware of my melodramatic tragedy. That is why I am shouting that, >those rehearsals were setup, a plot. I was controlled to do those >rehearsals in real life. > >In those rehearsals, I never tried to commit suicide. This is why I >cant die. So I will have to play this character. You have probably seriously considered suicide. Good thing you didnt try, though. It might have worked, and then youd have found out that you *can* die. Of course you can! I know youre the chosen one and all, but everybody dies. > >Whatever I do, it is already scripted anyway. No it isnt. Youre making it up as you go along. You may not be consciously aware of the process, but you are making it up. I realize the convenience of thinking that you are without personal will and accountability - an instrument of a higher power - but it is a delusion. Your genius has been mocked, so now you feign a sudden transformation and promise doom for all. Sorry, but I dont think anyone is buying it. More than likely, you are psychotic. Seek the kind of help that involves medication. > >-Abhi. > > Why is it always about you? Because I am Zero, feeling of zeroness, cause of creation of this universe. Clever. So, because you feel like a zero, you are essentially an omnipotent being? And now you are tasked to undo the universe, right? Do you realize how common it is for megalomania to manifest in people with catastrophically low self esteem? It is equally common for persons under such delusions of grandeur to make threats of annihilation towards everyone who ever doubted their greatness. So far, you ? the pro?e very well. Because I am truth. No. You are deluded. Its probably pointless to try to help you, but you really should seek help. You may be a danger to yourself and others. Unless, of course, you are just acting all weird on usenet without anything being wrong with you.... If you really believe the things you have been writing, you are mentally ill. === > >Oh, yes. I got my instructions, script to enter in this new character >on universal stage. I have done lot of rehearsal in real life and I >am aware of my melodramatic tragedy. That is why I am shouting that, >those rehearsals were setup, a plot. I was controlled to do those >rehearsals in real life. > >In those rehearsals, I never tried to commit suicide. This is why I >cant die. So I will have to play this character. You have probably seriously considered suicide. > Good thing you didnt try, though. It might have worked, and then youd have > found out that you *can* die. Of course you can! > I know youre the chosen one and all, but everybody dies. > >Whatever I do, it is already scripted anyway. No it isnt. Youre making it up as you go along. You may not be consciously > aware of the process, but you are making it up. > I realize the convenience of thinking that you are without personal will and > accountability - an instrument of a higher power - but it is a delusion. > Your genius has been mocked, so now you feign a sudden transformation and > promise doom for all. Sorry, but I dont think anyone is buying it. > More than likely, you are psychotic. Seek the kind of help that involves > medication. > >-Abhi. > >Why is it always about you? >Because I am Zero, feeling of zeroness, cause of creation of this > universe. Clever. So, because you feel like a zero, you are essentially an omnipotent > being? And now you are tasked to undo the universe, right? > Do you realize how common it is for megalomania to manifest in people with > catastrophically low self esteem? > It is equally common for persons under such delusions of grandeur to make > threats of annihilation towards everyone who ever doubted their greatness. > So far, you ? the pro?e very well. > Because I am truth. No. You are deluded. > Its probably pointless to try to help you, but you really should seek help. > You may be a danger to yourself and others. > Unless, of course, you are just acting all weird on usenet without anything > being wrong with you.... > If you really believe the things you have been writing, you are mentally > ill. To all your queries like, mentally ill, psychotic, schizophrenia, megalomania etc, I replied on 22 November. But I am ?ding that you never gave any response to that post. That post is given below. --------------------------------------------------- > You know, you really should pay heed to that post on schizophrenia. Paranoid > schizophrenics often think that really big events are set in motion just for > them. It seems to me that you exhibit that kind of megalomania. He also trained me to what to do in such circumstances. He trained me to execute Logic. If switch is ON, it just cant be OFF at exactly same moment. Yes # No. I am talking about generation of unidirectional force which will change course of history, physics and will open gateway to entire universe in future. Now we make a statement that this device does work. (1)Above statement is TRUE statement. Then looking at the simplicity of this device, any person who know physics can see that mechanism of this device does not need my theory. I have not talked a single word about my theory in this mechanism. It can be explained on the basis of Newtonian mechanics and Pythagoras Theorem. But millions of scientists, physicists could not think of it in at least last 350 years. Only conclusion, I can draw is that thinking ability of millions of people in last 350 years was controlled by someone. If we look further closely, we ?d that this device is so simple that it could have been discovered accidently by someone who does not know even basics of physics. And this accident could have happened at any time in last 5000 years. But it never happened. Only conclusion, I can draw is that actions of billions of people in last 5000 years was controlled by someone for speci? purpose. And if that someone can control the actions of billions of people in last 5000 years, He can also control my actions, my mind, absolutely every event happened around me from the moment I was born. Is there any ? this logic, Laura? (2) Above statement is FALSE statement. Then Laura, you are right. I am suffering from Paranoid schizophrenia. TRUTH, what is that and where is it..... -Abhi. --------------------------------------------------- Laura, you never replied to my above post. this Time Theory and Action Device), only way to show me that I am show me and all of you some miracles. BTW, I have ?ed 7 patent application and I have sent no.8 by ground make it. Because if I make it and it works, then it means that the statement I made in above post is TRUE and all the logic which follows it, is also TRUE. Those lights were not my delusion. It was God. In such situation, I will not have any moral right to hide this device in my room and prepare patent application, ?e it etc. No, I must disclose it to mankind immediately because I have seen God in action through this device. [And as soon as I disclose it to media, it destroys absolute novelty part of patent rules!] Till this date, I have not prepared or even tried to prepare this device. I have sent my patent application no.8 and now what is holding me back, apart from my personal problems, is that all these 8 applications might not be enough to secure this device. But I am going to prepare this device now and disclose it to media even if it is not secured. I have tried my best in last 6 and half months. Thats All. -Abhi. === > He also trained me to what to do in such circumstances. He trained me > to execute Logic. If switch is ON, it just cant be OFF at > exactly same moment. Yes # No. I am talking about generation of unidirectional force which will > change course of history, physics and will open gateway to entire > universe in future. Now we make a statement that this device does work. (1)Above statement is TRUE statement. Then looking at the simplicity of this device, any person who know > physics can see that mechanism of this device does not need my theory. Heres a logical mistake: The simplicity of the device does not depend on it working. In fact, from your drawings any person who know physics can see that it does not work. And anyone who has posted here has seen it. So, the two possibilities I see are: - The device does not work. - The device does work and looks different from your drawings. I hesitate to link device work not to deluded because its obvious anyway that you are deluded by you seeing yourself as vastly important which is a classical symptom. There are lots of people who discovered that people did something wrong all along but they never bragged on about them being specially chosen. So, build it and post a picture (please not more drawings!). > I have not talked a single word about my theory in this mechanism. It > can be explained on the basis of Newtonian mechanics and Pythagoras > Theorem. But millions of scientists, physicists could not think of it > in at least last 350 years. Only conclusion, I can draw is that > thinking ability of millions of people in last 350 years was > controlled by someone. Wrong logic, see above, wrong conclusion. Very simple. > Laura, you never replied to my above post. this Time Theory and Action Device), Well, have you? Whats it look like? Jesus there or Echnaton? > BTW, I have ?ed 7 patent application and I have sent no.8 by ground > make it. Because if I make it and it works, then it means that the > statement I made in above post is TRUE and all the logic which follows > it, is also TRUE. Those lights were not my delusion. It was God. > In such situation, I will not have any moral right to hide this > device in my room and prepare patent application, ?e it etc. No, I > must disclose it to mankind immediately because I have seen God in > action through this device. [And as soon as I disclose it to media, it > destroys absolute novelty part of patent rules!] Logical mistake number two: Why are you always posting long explanations of how and why it has to work? Right now I could take any of those, build it and sell it. If you see usenet as medium it *has* been disclosed by now. eetimes or so about some funny guy in sci.astro and youd have your disclosure. Till this date, I have not prepared or even tried to prepare this > device. I have sent my patent application no.8 and now what is holding > me back, apart from my personal problems, is that all these 8 > applications might not be enough to secure this device. So, you have ?ed seven patents? You should have gotten reference numbers for this, right? What are they? But I am going to prepare this device now and disclose it to media > even if it is not secured. I have tried my best in last 6 and half > months. Right. If it shows up in the news, post a link, please? Lots of Greetings! Volker === > On Mon, 12 Jan 2004 08:56:19 -0800, Uncle Al >Note that the US Enviro-whiner 3-gallon ?and smaller) is a >horror. We had a plumber replace the ?echnism on one of our >American Standard 5-gallon jobs. He offered us $300 cash for our >pudgy porcelain goddess plus a brand new eco-john free, with free >installation. No way! 5-gallon jobs are being smuggled in from >Canada as you read this. They go for $1000+ each. > > Im still not sure why we cant get some kind of standardized > 2-stage terlets in this country. Push the handle down, you get the > 2-gallon piss-rinser mode, pull up on the handle, you get the full > 5-gallon power ?ode. Is this beyond our engineering > capabilities? Is the irony in this entire message thread deliberate? The original post is total sh*t, and the responses discuss toilets. Too funny!!!!! ROFL! === > Vanilla Gorilla (Monkey Boy) On Mon, 12 Jan 2004 08:56:19 -0800, Uncle Al >> >>Note that the US Enviro-whiner 3-gallon ?and >>smaller) is a horror. We had a plumber replace the ?mec hnism on one of our American Standard 5-gallon jobs. >>He offered us $300 cash for our pudgy porcelain goddess >>plus a brand new eco-john free, with free installation. >>No way! 5-gallon jobs are being smuggled in from Canada >>as you read this. They go for $1000+ each. Im still not sure why we cant get some kind of >> standardized 2-stage terlets in this country. Push the >> handle down, you get the 2-gallon piss-rinser mode, pull >> up on the handle, you get the full 5-gallon power ? mo de. Is this beyond our engineering capabilities? Is the irony in this entire message thread deliberate? The > original post is total sh*t, and the responses discuss > toilets. Too funny!!!!! ROFL! Might as well add this...The downtrodden slaves of the primitive cult of Islam have to ask the imam for permission for every moment of their lives, including how to take a dump.... http://www.islam.tc/ask-imam/view.php?q=2325 ========= Do we have to use a squatpan? I have dif?ulty using a squatpan? Can one use a high toilet when taking the necessary precautions not to let any water or najaasat splash on oneself? Jazakallah Khair Answer: One should relieve oneself in a squatting posture. However, if one is unable or experiences dif?ulty in squatting, he may use the western toilet and avoid the impurities splashing on the body or clothes. === <... Might as well add this...The downtrodden slaves of the primitive > cult of Islam have to ask the imam for permission for every > moment of their lives, including how to take a dump.... Most dont, you know. See below. > http://www.islam.tc/ask-imam/view.php?q=2325 ========= > Do we have to use a squatpan? I have dif?ulty using a > squatpan? Can one use a high toilet when taking the necessary > precautions not to let any water or najaasat splash on oneself? > Jazakallah Khair Answer: > One should relieve oneself in a squatting posture. However, if > one is unable or experiences dif?ulty in squatting, he may > use the western toilet and avoid the impurities splashing on > the body or clothes. Gimme a fucking break, John. This is no worse than the Judeo-Christian losers who call in to Laura Schlessinger every goddamn day asking for moral guidance on issues even more trivial. Jim === <... >> Might as well add this...The downtrodden slaves of the >> primitive cult of Islam have to ask the imam for >> permission for every moment of their lives, including how >> to take a dump.... Most dont, you know. See below. >> http://www.islam.tc/ask-imam/view.php?q=2325 ========= >> Do we have to use a squatpan? I have dif?ulty using a >> squatpan? Can one use a high toilet when taking the >> necessary precautions not to let any water or najaasat >> splash on oneself? Jazakallah Khair Answer: >> One should relieve oneself in a squatting posture. >> However, if one is unable or experiences dif?ulty in >> squatting, he may use the western toilet and avoid the >> impurities splashing on the body or clothes. Gimme a fucking break, John. This is no worse than the > Judeo-Christian losers who call in to Laura Schlessinger > every goddamn day asking for moral guidance on issues even > more trivial. Give yourself a break. No Judeo-Christian losers ask is it permissible, or do we have to... Around 80% of the questions to the brainwashed imam on that web site are of that type. Do we have to redo the woodhoo if wind escapes the hind passage...? (Just a hilarious example, slightly paraphrased.) Mohammedans are slaves. By now, everyone should realize that. They moslems themselves beginning to realize it, and since their penalty for even mildly questioning their 7th century cults moment-to-moment mind control is a death sentence, the sorry bastards are trying to release their anger, humiliation and frustration by murdering innocent citizens of the civilized world whose freedom they resent so intensely. Furthermore, moslems are REQUIRED to try to overthrow a non- islamic government in any country they infest. Islam should fold up its tents and get the fuck out of the way. Modern humans dont need shit like that, and its slaves could easily become modern humans if given the chance. === > <... >> Might as well add this...The downtrodden slaves of the >> primitive cult of Islam have to ask the imam for >> permission for every moment of their lives, including how >> to take a dump.... > Most dont, you know. See below. >> http://www.islam.tc/ask-imam/view.php?q=2325 >> ========= >> Do we have to use a squatpan? I have dif?ulty using a >> squatpan? Can one use a high toilet when taking the >> necessary precautions not to let any water or najaasat >> splash on oneself? Jazakallah Khair >> Answer: >> One should relieve oneself in a squatting posture. >> However, if one is unable or experiences dif?ulty in >> squatting, he may use the western toilet and avoid the >> impurities splashing on the body or clothes. > Gimme a fucking break, John. This is no worse than the > Judeo-Christian losers who call in to Laura Schlessinger > every goddamn day asking for moral guidance on issues even > more trivial. Give yourself a break. No Judeo-Christian losers ask is it > permissible, or do we have to... Around 80% of the questions > to the brainwashed imam on that web site are of that type. Ok, islam is a more strict religion. Buddhism is maybe more relaxed than christianism and Im sure any religion calling for a regular supply of virgins would be regarded as more strict than islam. So what? They are all religions with the same basic property of relying on some more or less unfathomable outside in?. The rest is detail and will be decided as soon as someone discoveres it (which is IMHO never). I read that guys website regularly. He and his team really answer *all* questions (they limit the number of questions per day), and yes, they get loads of stupid and sometimes even abusive ones, like how allahs ass looks like or so. But then you get questions like http://www.islam.tc/ask-imam/view.php?q=10295 where the answer is radically different from whan you expect and you get the feel that hes working toward changes too. I respect him. For me its a very valuable window into the muslim world and their mindset and not only from some know-your-enemy point of view. Lots of Greetings! Volker === > <... >> Might as well add this...The downtrodden slaves of the >> primitive cult of Islam have to ask the imam for >> permission for every moment of their lives, including how >> to take a dump.... >Most dont, you know. See below. > >> http://www.islam.tc/ask-imam/view.php?q=2325 >> ========= >> Do we have to use a squatpan? I have dif?ulty using a >> squatpan? Can one use a high toilet when taking the >> necessary precautions not to let any water or najaasat >> splash on oneself? Jazakallah Khair >> Answer: >> One should relieve oneself in a squatting posture. >> However, if one is unable or experiences dif?ulty in >> squatting, he may use the western toilet and avoid the >> impurities splashing on the body or clothes. >Gimme a fucking break, John. This is no worse than the > Judeo-Christian losers who call in to Laura Schlessinger > every goddamn day asking for moral guidance on issues even > more trivial. Give yourself a break. No Judeo-Christian losers ask is it > permissible, or do we have to... Around 80% of the questions > to the brainwashed imam on that web site are of that type. Listen to her show before you say nonsense like that. Jim === > message >> <...>> Might as well add this...The downtrodden slaves of the >>> primitive cult of Islam have to ask the imam for >>> permission for every moment of their lives, including >>> how to take a dump.... >> Most dont, you know. See below. >>> http://www.islam.tc/ask-imam/view.php?q=2325 >> ========= >>> Do we have to use a squatpan? I have dif?ulty using a >>> squatpan? Can one use a high toilet when taking the >>> necessary precautions not to let any water or najaasat >>> splash on oneself? Jazakallah Khair >> Answer: >>> One should relieve oneself in a squatting posture. >>> However, if one is unable or experiences dif?ulty in >>> squatting, he may use the western toilet and avoid the >>> impurities splashing on the body or clothes. >> Gimme a fucking break, John. This is no worse than the >> Judeo-Christian losers who call in to Laura Schlessinger >> every goddamn day asking for moral guidance on issues >> even more trivial. Give yourself a break. No Judeo-Christian losers ask >> is it permissible, or do we have to... Around 80% of >> the questions to the brainwashed imam on that web site are >> of that type. Listen to her show before you say nonsense like that. Ive listened to people who listen to her show. Most of them need to loosen up and suck a dick, eat some pussy, or just get laid. Theyre mis?s. Only the dozen most fucked up of all Christians regularly ask their church for permission for anything. The ordinary Christian might ask for advice, but since they wont be stoned, beaten with clubs, have a stone wall toppled on them or thrown off the top of a tall building for the crime of thinking for themselves, as those downtrodden slaves of the primitive islam cult would, they dont think of it as permission. === >> Gimme a fucking break, John. This is no worse than the >> Judeo-Christian losers who call in to Laura Schlessinger >> every goddamn day asking for moral guidance on issues even >> more trivial. Give yourself a break. No Judeo-Christian losers ask is it >> permissible, or do we have to... Around 80% of the questions >> to the brainwashed imam on that web site are of that type. Listen to her show before you say nonsense like that. Jim > Most of the people who call the Dr. Laura show are an embarrassment to the entire species. -- V.G. People are more violently opposed to fur than leather, because it is easier to harrass rich women than it is motorcycle gangs. - Bumper Sticker (This sig ?e contains not less than 80% recycled SPAM) Sarcasm is my sword, Apathy is my shield. === >> Gimme a fucking break, John. This is no worse than the > Judeo-Christian losers who call in to Laura Schlessinger > every goddamn day asking for moral guidance on issues even > more trivial. >> Give yourself a break. No Judeo-Christian losers ask is it > permissible, or do we have to... Around 80% of the questions > to the brainwashed imam on that web site are of that type. Listen to her show before you say nonsense like that. Jim > >Most of the people who call the Dr. Laura show are an embarrassment >to the entire species. Which species would that be, VG? -- Jimmy Snibbler === in alt.fan.art-bell: > Gimme a fucking break, John. This is no worse than the >> Judeo-Christian losers who call in to Laura Schlessinger >> every goddamn day asking for moral guidance on issues even >> more trivial. >> Give yourself a break. No Judeo-Christian losers ask is it >> permissible, or do we have to... Around 80% of the questions >> to the brainwashed imam on that web site are of that type. >>Listen to her show before you say nonsense like that. >>Jim >Most of the people who call the Dr. Laura show are an embarrassment >>to the entire species. Which species would that be, VG? Whichever one they belong to. -- V.G. People are more violently opposed to fur than leather, because it is easier to harrass rich women than it is motorcycle gangs. - Bumper Sticker (This sig ?e contains not less than 80% recycled SPAM) Sarcasm is my sword, Apathy is my shield. === My analysis continues to indicate that my research on counting prime numbers *should* be more accessible than my other math research which is more abstract, and clearly more dif?ult. Still I also recognize that signi?ant parts of my prime counting research are beyond a lot of people simply because that research extends into partial differential equations. To me the starting point is simple enough: dS(x,y) = [p(x/y, y-1) - p(y-1, sqrt(y-1))][ p(y, sqrt(y)) - p(y-1, sqrt(y-1))], S(x,1) = 0, p(x, y) = ?) - S(x, y) - 1, and S(x,y) is the sum of dS from dS(x,2) to dS(x,y). But Im increasingly aware that what looks simple to me, leaves mathematicians all over the world befuddled. Not surprisingly, faced with a dif?ult mental challenge, people cling to whats known, and with the less sophisticated audience of sci.math that has usually meant looking towards Legendres Method, while when Ive contacted mathematicians more expert, it has usually meant looking towards Meissels formula. And in fact in contacts with mathematicians at universities, for instance my alma mater Vanderbilt University, I heard that what Id found was some variant on Meissels formula. However, anyone who actually looks over known methods will ?d that what I have above is astonishing in its conciseness. Its just amazingly short for a way to count prime numbers. And then you might notice that it has a partial difference equation at its heart, but thats something Ive emphasized only to see it apparently sail over the heads of readers, so Im less interested in emphasizing it now, as hey, its just a tad bit beyond most of you. I can get some sense why even experts, like mathematicians by de?ition, would ?d it intimidating and dif?ult to comprehend, but then again, I ?d myself surprised at how clear it is that the expression is completely overwhelming. It gets more interesting, and rather than move into the calculus by talking about the partial differential equation that follows, Ill talk more about practical matters. For instance, facing a daunting expression, Ive seen a tendency by posters to try and simplify it, as if their minds are desperate to ?d something more familiar. Beyond what I already mentioned, for instance, many posters would keep deleting off the second factor and emphasizing using primes! So theyd keep pushing something like dS(x,p_j) = [p(x/p_j, p_j - 1) - j-1], where using primes makes things look simpler, though youre shifted from a mathematical formula--the partial difference equation--to an algorithm. That behavior is in line with what I mentioned before where posters grasping for the familiar looked to Legendres method if novice in the ?ld, or Meissels formula if more sophisticated in their knowledge. Now then, on to the more expert commentary on my work looking like Meissels formula, which has helped me to understand that yes, my work somehow is beyond most mathematicians ability to handle, as indeed you can relate from it back to Meissels formula, but you cant get to it from Meissels formula. Its actually easy to show what I mean, as if you know anything about Meissels formula, you know that theres one aspect of it where you sum something like pi(x/p_j) - (j-1) where you iterate through primes p. That follows from the root prime counting function which I discovered easily enough with dS(x,y) = [p(x/y, y-1) - p(y-1, sqrt(y-1))][ p(y, sqrt(y)) - p(y-1, sqrt(y-1))], by considering special cases. First if you speci?ally use primes for y, you get dS(x,p_j) = [p(x/p_j, p_j - 1) - j-1], where you can see a lot of information from the root mathematics can be dropped. It also is true that if p_j-1 >= sqrt(x/p_j), you can use [p(x/p_j, sqrt(x/p_j)) - j-1], and since with my function p(x,sqrt(x) = pi(x), you can ?ally get to [pi(x/p_j) - j-1]. Now though, notice that you cant go back the other way!!! So my work in less space encodes more information that relates back to what mathematicians already discovered! However, in looking at it, even experts seem to get lost from what Ive gathered in communicating with mathematicians worldwide since May 2002. Now looking at dS(x,y) = [p(x/y, y-1) - p(y-1, sqrt(y-1))][ p(y, sqrt(y)) - p(y-1, sqrt(y-1))], Now heres something for fun. For a while I pursued fast prime counting programs to see if I couldnt get progress by that route, and eventually I really just got bored with ?uring out fast algorithms, as Im more interested in the math, but along the way I found some of the fastest expressions possible for certain counts: With even N, N/2 - ?N-4)/6) - ?N-16)/10) + ?N-16)/30) - ?N-36)/14) + ?N-22)/42) is basically the explicit result of summing an algorithmic form of the dS(x,y) function and subtracting the resulting S(x,y) from N, where evens are automatically handled, from 2 up to and including 10, which represents the primes 2, 3, 5 and 7. That formula counts primes from N=36 up to N=174. For instance, N=100, gives 50 - 16 - 8 + 2 - 4 + 1 = 25 as expected. But beyond counting primes in a small range a slightly longer formula works to give N minus the count of composites up to and including N that have 2, 3, 5, or 7 as a factor out to positive in?ity for even N: N/2 - ?N-4)/6) - ?N-16)/10) + ?N-16)/30) - ?N-36)/14) + ?N-22)/42) + ?N-106)/70) - ?N-106)/210) + 2 Modern mathematicians cant ?d anything of their own without my research of similar length that works out to in?ity. Yup, despite the research done on prime numbers through the entire math history of the world, mathematicians as a group can only produce *longer* expressions if they use their own research on counting prime numbers! Want more on prime counting? Then see my blog archives: James Harris === > My analysis continues to indicate that my research on counting prime > numbers *should* be more accessible than my other math research which > is more abstract, and clearly more dif?ult. Still I also recognize > that signi?ant parts of my prime counting research are beyond a lot > of people simply because that research extends into partial > differential equations. To me the starting point is simple enough: dS(x,y) = [p(x/y, y-1) - p(y-1, sqrt(y-1))][ p(y, sqrt(y)) - p(y-1, > sqrt(y-1))], S(x,1) = 0, p(x, y) = ?) - S(x, y) - 1, and S(x,y) is the sum of dS from dS(x,2) to dS(x,y). It doesnt get more interesting when you repost it. === [snip the usual self-aggrandizing rubbish] The Mad Russian: Ive invented a new type of television, but you dont see anything - you just hear it. I call it Hear-o-vision. Eddie Cantor: But Russian - thats radio! The Mad Russian: Well what do you know? I invented radio too! === > My analysis continues to indicate that my research on counting prime > numbers *should* be more accessible than my other math research which > is more abstract, and clearly more dif?ult. Still I also recognize > that signi?ant parts of my prime counting research are beyond a lot > of people simply because that research extends into partial > differential equations. To me the starting point is simple enough: dS(x,y) = [p(x/y, y-1) - p(y-1, sqrt(y-1))][ p(y, sqrt(y)) - p(y-1, > sqrt(y-1))], You have previously repudiated that method on the grounds that it relies on the *inherently* ambiguous ?sqrt. It fails for the negative values returned which are, according to you, impossible to avoid. > S(x,1) = 0, p(x, y) = ?) - S(x, y) - 1, and S(x,y) is the sum of dS from dS(x,2) to dS(x,y). But Im increasingly aware that what looks simple to me, leaves > mathematicians all over the world befuddled. On the contrary. The method has left *you* befuddled, as you repeatedly confessed in previous posts. Your inability to ?d any means of con?ing the sqrt to positive numbers renders the method useless. [snip redundant material previous posted] -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === > My analysis continues to indicate that my research on counting prime > numbers *should* be the most accessible than my other math research That should have been that it should be more accessible than my other > math research. > which is more abstract, and clearly more dif?ult. Still I also > recognize that signi?ant parts of my prime counting research are > beyond a lot of people simply because that research extends into > partial differential equations. > > With even N, > N/2 - ?N-4)/6) - ?N-16)/10) + ?N-16)/30) - > ?N-36)/14) + ?N-22)/42) Thats ?e within a certain range. > is basically the explicit result of summing an algorithmic form of the > dS(x,y) function, where evens are automatically handled, from 2 up to > and including 10, which represents the primes 2, 3, 5 and 7, so it > gives a count of primes up to and including 120. > For instance, N=100, gives > 50 - 16 - 8 + 2 - 4 + 1 = 25 > as expected. >But beyond counting primes in a small range it works to give N minus > the count of composites up to and including N that have 2, 3, 5, or 7 > as a factor out to positive in?ity for even N. Actually I forgot that beyond 106 you have more terms, and the > expression then is N/2 - ?N-4)/6) - ?N-16)/10) + ?N-16)/30) - > ?N-36)/14) + ?N-22)/42) + ?N-106)/70) - > ?N-106)/210) + 2. > > It is the smallest expression possible for that job. You can roll that 2 back into it--carefully--but its just as easy to > leave it hanging out there. > > Want more on prime counting? Then see my blog archives: > > > > James Harris You have got to be kidding. You have posted this over and over again and you still cannot get it right? What is wrong with you? It is *your* amazing discovery and you cant type it in correctly the ?st time. Amazing indeed. You cannot even cut and paste from your previous garbage posts. Pathetic. Have a nice day, Jay === > Why do you think _anyone_ wants more of _any_ of this stuff? I mean really: has _anyone_ expressed any interest? I mean > even one person? There would be things that I would ?d interesting: For example, since there is a very fast rough estimate for pi (x), namely li (x) = gamma + log log x + sum ((log x)^k) / k! / k, 1 <= k <= inf) and a rather slow algorithm taking about O (x^(2/3)) steps to calculate pi (x) exactly, is there any approximation for pi (x) that is substantially more precise than li (x) but can be calculated substantially faster than O (x^(2/3))? Another question: pi (M) - pi (N) can easily be calculated in O (min (M - N, N^(2/3))) steps by using a sieve if M is close to N, and using the Extended Meissel-Lehmer algorithm to calculate pi (M) and pi (N) of the difference between M and N is large. Is there any substantially faster algorithm? Another problem which I think I have solved: Given k >= 2 and values x(i) for 1 <= i <= k. Can the Extended Meissel-Lehmer algorithm be modi?d so that it calculates pi (x (i)) for all given values x (i) substantially faster than O (sum (x(i)^(2/3), 1 <= i <= k) ? === > There would be things that I would ?d interesting: For example, since > there is a very fast rough estimate for pi (x), namely li (x) = gamma + log log x + sum ((log x)^k) / k! / k, 1 <= k <= inf) and a rather slow algorithm taking about O (x^(2/3)) steps to calculate > pi (x) exactly, is there any approximation for pi (x) that is > substantially more precise than li (x) but can be calculated > substantially faster than O (x^(2/3))? Given my own trend for being grossly inaccurate, this thread may be of no use whatsoever... http://tinyurl.com/3cjlh ...but some may ?d it of interest. Carl === >> automatic decision support system for my strategy game, Im looking for >> an equation for estimating the monopolization factor. > I think it depends on whether youre the top hat or the little racing > car. Also, if you have Boardwalk and Park Place with hotels, those > factors just dont mean the same thing as when you are sitting there > with Baltic and Mediterranean Avenues. ROTFL. I prefer sunny beaches of Puerto Rico, however I also like CA governed by Mr Arnie... (-; >> Company1 produces 3 units of goods, Company2 - 2, Company3 - 2. What is >> the market monopolization factor for Company1, and how will it change >> when it increases its production from 3 to 4 units? > BTW, you might be more likely to get good answers if you were to de?e > your terms. Either that, or ask an economics newsgroup. percentage of the market has any potential traps. First situation: C1 produces 3 units, total production = 7 units; market monopolization = 0.43; second situation: C1 produces 4 units, total production = 8 units; market monopolization = 0.5. I personally dont see any drawbacks here (yet), but I thought someone with more mathematical experience would. If not, I will be trying it in practice and let you know. :)) -- BB === Dear all, Ive tried various approaches, but none seem convincing. Heres the problem: suppose t is a real number, de?e the following polynomial on the complex numbers: f(z)=z^3-(4+t^2)z^2-2tz+1. My goal is to show that this polynomial has one and only one zero in (0,1) (so we can actually restrict f to be de?ed only on the real numbers). The fact that there is a zero is not dif?ult, since f|R(0)=1 and f|R(1)=-1-(1+t)^2<0 (by f|R I mean f restricted to the real numbers). By the continuity theorem(?) there has to be a zero between 0 and 1. Now, how would you go about showing that there is only one zero. And how would you show that this is not a multiple zero of f? === > Dear all, Ive tried various approaches, but none seem convincing. Heres the > problem: suppose t is a real number, de?e the following polynomial on > the complex numbers: f(z)=z^3-(4+t^2)z^2-2tz+1. My goal is to show that this polynomial has one and only one zero in > (0,1) (so we can actually restrict f to be de?ed only on the real > numbers). The fact that there is a zero is not dif?ult, since f|R(0)=1 > and f|R(1)=-1-(1+t)^2<0 (by f|R I mean f restricted to the real > numbers). By the continuity theorem(?) there has to be a zero between 0 > and 1. Now, how would you go about showing that there is only one zero. > And how would you show that this is not a multiple zero of f? Did you check its value at -1 ? Jim Buddenhagen -- To reply copy jbuddenh@REMOVEtexas.net to address bar and edit out REMOVE === > Dear all, > Ive tried various approaches, but none seem convincing. Heres the > problem: suppose t is a real number, de?e the following polynomial on > the complex numbers: > f(z)=z^3-(4+t^2)z^2-2tz+1. > My goal is to show that this polynomial has one and only one zero in > (0,1) (so we can actually restrict f to be de?ed only on the real > numbers). The fact that there is a zero is not dif?ult, since f|R(0)=1 > and f|R(1)=-1-(1+t)^2<0 (by f|R I mean f restricted to the real > numbers). By the continuity theorem(?) there has to be a zero between 0 > and 1. Now, how would you go about showing that there is only one zero. > And how would you show that this is not a multiple zero of f? > > Did you check its value at -1 ? > and its sign as z -> in?ity ? === >Dear all, >>Ive tried various approaches, but none seem convincing. Heres the >problem: suppose t is a real number, de?e the following polynomial on >the complex numbers: >>f(z)=z^3-(4+t^2)z^2-2tz+1. >>My goal is to show that this polynomial has one and only one zero in >(0,1) (so we can actually restrict f to be de?ed only on the real >numbers). The fact that there is a zero is not dif?ult, since f|R(0)=1 >and f|R(1)=-1-(1+t)^2<0 (by f|R I mean f restricted to the real >numbers). By the continuity theorem(?) there has to be a zero between 0 >and 1. Now, how would you go about showing that there is only one zero. >And how would you show that this is not a multiple zero of f? >> Did you check its value at -1 ? > and its sign as z -> in?ity ? Ooops... This turned out to be a very silly question. :-( My apologies for taking up everyones time for reading the original question. For the two responders, thanks for the subtle hints. === -T Approved: ebunn@richmond.edu (sci.physics.research moderator) === This section of TWF was of particular interest to me: > The classic example, familiar to all physicists, is the Galois group > of the complex numbers, C, over the real numbers, R. This group has > two elements: the identity transformation, which leaves everything > alone, and complex conjugation, which switches i and -i. Since the > only group with 2 elements is Z/2, we have Gal(C/R) = Z/2 Where does complex conjugation come from? It comes from the fact that > C from R by throwing in a solution of the quadratic equation x^2 = -1. We say C is a quadratic extension of R. But as soon as we throw in > one solution of this equation, we inevitably throw in another, namely > its negative - and theres no way to tell which is which. And complex > conjugation is the symmetry that switches them! One thing I have yet to read much about is symmetry groups and quaternions. I am sure one could de?e Gal(H/R), the question is what would it equal? One would need four automorphisms: the identity transformation, one that switches i, j, k and -i, -j, -k, and two others. In a hotel lobby as I was leaving Rome from a conference on quaternions, I had the idea for what I called the ?st and second conjugates, symbolized by q*1 and q*2, which ?e signs of all except the ?st and second members of the quaternion 3-vector, so (t, x, y, z)*1 = (-t, x, -y, -z) and (t, x, y, z)*2 = (-t, -x, y, -z). These are related to the standard conjugation using a rotation around i and j: q*1 = (i q i)* and q*2 = (j q j)*. Since quaternions do not commute, Z cannot be used. This might be a logical place to begin non-Abelian ?ld theory. Non-Abelian physics appears to be every, from bicycle tires to spin. Any references would be appreciated. doug quaternions.com === One thing I have yet to read much about is symmetry groups and > quaternions. I am sure one could de?e Gal(H/R), the question is what > would it equal? One would need four automorphisms: the identity > transformation, one that switches i, j, k and -i, -j, -k, and two > others. Thats not an automorphism. Its an anti-atomorphism: denoting it by phi, phi(z w) = phi(w) phi(z) not phi(z) phi(w). H has uncountably many automorphisms, all ?ing R pointwise. A typical one takes i to a_{11} i + a_{12} j + a_{13} k, j to a_{21} i + a_{22} j + a_{23} k, k to a_{31} i + a_{32} j + a_{33} k where (a_{pq}) is a matrix in SO(3). -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === > One thing I have yet to read much about is symmetry groups and > quaternions. I am sure one could de?e Gal(H/R), the question is what > would it equal? One would need four automorphisms: the identity > transformation, one that switches i, j, k and -i, -j, -k, and two > others. Thats not an automorphism. Its an anti-atomorphism: > denoting it by phi, phi(z w) = phi(w) phi(z) not phi(z) phi(w). H has uncountably many automorphisms, all ?ing R pointwise. > A typical one takes > i to a_{11} i + a_{12} j + a_{13} k, > j to a_{21} i + a_{22} j + a_{23} k, > k to a_{31} i + a_{32} j + a_{33} k > where (a_{pq}) is a matrix in SO(3). Not unexpected. But are these the only automorphisms ?ing reals? I think I asked the same question somewhere before...but my real concern is of course the sensible (those I can construct) automorphisms. === >> One thing I have yet to read much about is symmetry groups and >> quaternions. I am sure one could de?e Gal(H/R), the question is what >> would it equal? One would need four automorphisms: the identity >> transformation, one that switches i, j, k and -i, -j, -k, and two >> others. Thats not an automorphism. Its an anti-atomorphism: >> denoting it by phi, phi(z w) = phi(w) phi(z) not phi(z) phi(w). H has uncountably many automorphisms, all ?ing R pointwise. >> A typical one takes >> i to a_{11} i + a_{12} j + a_{13} k, >> j to a_{21} i + a_{22} j + a_{23} k, >> k to a_{31} i + a_{32} j + a_{33} k >> where (a_{pq}) is a matrix in SO(3). Not unexpected. But are these the only automorphisms ?ing reals? As I said automorphisms, all ?ing R pointwise. To expand: if f: H -> H is a ring automorphism, then f(a) = a for all a in R. Proof: exercise for reader. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Heres another example of a Galois group that physicists should like. > Let C(z) be the ?ld of rational functions in one complex variable z - > in other words, functions like f(z) = P(z)/Q(z) where P and Q are polynomials in z with complex coef?ients. You can > add, subtract, multiply and divide rational functions and get other > rational functions, so they form a ?ld. And they contain C as a > sub?ld, because we can think of any complex number as a *constant* > function. So, we can ask about the Galois group of C(z) over C. > Whats it like? Its the Lorentz group! To see this, its best to think of rational functions as functions not > on the complex plane but on the Riemann sphere - the complex plane > together with one extra point, the point at in?ity. The only > conformal transformations of the Riemann sphere are fractional linear > transformations: Could you please explain why an element of the Galois group must be a conformal transformation? At ?st sight one would think it can be any rational substitution that is bijective. So why does bijective imply conformal? Arnold Neumaier === There are a few dollars to be had here: http://faculty.evansville.edu/ck6/integer/unsolved.html === When I cared about ?ding fast prime counting algorithms versus concentrating on the core mathematics, I found formulas like the following, which counts primes up to a given even N over a certain range: N/2 - ?N-4)/6) - ?N-16)/10) + ?N-16)/30) - ?N-36)/14) + ?N-22)/42) + ?N-106)/70) - ?N-106)/210) + 2. And it counts primes for an even N from N=38 to N=120. Using pi(N) = ?) - S(N) + 1, where S(N) is the count of composites, and solving for S(N), gives N/2 + ?N-4)/6) + ?N-16)/10) - ?N-16)/30) + ?N-36)/14) - ?N-22)/42) - ?N-106)/70) + ?N-106)/210) - 1 which is the count of composites that have 2, 3, 5 or 7 as a factor with an even N in the range N=38 to positive in?ity. So thats 9 terms. Up to N=120, in contrast, Legendres Method has 15 terms. Which continues a trend started much earlier as theyre closest up to 8 where for both you have N/2 - 1 while up to 24, Legendres Method gives N/2 + ?/3) - ?/6) - 2 while my work gives N/2 + ?-4)/6 + 1 for the count of composites with 2 or 3 as a factor. James Harris === In sci.math.num-analysis, James Harris on 15 Jan 2004 07:31:20 -0800 <3c65f87.0401150731.73acc418@posting.google.com>: > When I cared about ?ding fast prime counting algorithms versus > concentrating on the core mathematics, I found formulas like the > following, which counts primes up to a given even N over a certain > range: N/2 - ?N-4)/6) - ?N-16)/10) + ?N-16)/30) - > ?N-36)/14) + ?N-22)/42) + ?N-106)/70) - > ?N-106)/210) + 2. And it counts primes for an even N from N=38 to N=120. Using pi(N) = ?) - S(N) + 1, where S(N) is the count of > composites, and solving for S(N), gives N/2 + ?N-4)/6) + ?N-16)/10) - ?N-16)/30) + > ?N-36)/14) - ?N-22)/42) - ?N-106)/70) + > ?N-106)/210) - 1 which is the count of composites that have 2, 3, 5 or 7 as a factor > with an even N in the range N=38 to positive in?ity. So thats 9 terms. Up to N=120, in contrast, Legendres Method has 15 terms. Which continues a trend started much earlier as theyre closest up to > 8 where for both you have N/2 - 1 while up to 24, Legendres Method gives N/2 + ?/3) - ?/6) - 2 while my work gives N/2 + ?-4)/6 + 1 for the count of composites with 2 or 3 as a factor. > James Harris Whoop-te-do. Youre going to have to work hard to beat Christian Baus implementation of Meissel-Lehmer. http://www.cbau.freeserve.co.uk/ (Ill admit Im tempted to code it in Java, myself. Just for the perversity... :-) ) -- #191, ewill3@earthlink.net Its still legal to go .sigless. === > When I cared about ?ding fast prime counting algorithms versus > concentrating on the core mathematics, I found formulas like the > following, which counts primes up to a given even N over a certain > range: N/2 - ?N-4)/6) - ?N-16)/10) + ?N-16)/30) - > ?N-36)/14) + ?N-22)/42) + ?N-106)/70) - > ?N-106)/210) + 2. If you no longer care about ?ding fast prime counting algorithms, why are you posting this now? [snip exposition of method cryptically posted in spite of James above suggested lack of current interest] -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === > When I cared about ?ding fast prime counting algorithms versus > concentrating on the core mathematics, I found formulas like the > following, which counts primes up to a given even N over a certain > range: >N/2 - ?N-4)/6) - ?N-16)/10) + ?N-16)/30) - > ?N-36)/14) + ?N-22)/42) + ?N-106)/70) - > ?N-106)/210) + 2. If you no longer care about ?ding fast prime counting algorithms, why > are you posting this now? > In this case Im leaving a note to myself covering a slightly different approach. But hey, as usual, theres always the possibility that someone else might give a useful comment. I have lots of areas where I can go over my own research, so I get to pick and choose depending on my mood and purpose. I ?d it useful to sharpen my focus at times by posting in a certain direction. James Harris === > >> When I cared about ?ding fast prime counting algorithms versus >> concentrating on the core mathematics, I found formulas like the >> following, which counts primes up to a given even N over a certain range: N/2 - ?N-4)/6) - ?N-16)/10) + ?N-16)/30) - >> ?N-36)/14) + ?N-22)/42) + ?N-106)/70) - >> ?N-106)/210) + 2. If you no longer care about ?ding fast prime counting algorithms, why > are you posting this now? If it is mathematically correct and interesting, why does it matter? Unlike pretty much all of his earlier postings, this one appears to be mathematically correct, uses standard terminology, and is either interesting or suggests interesting things. Why are you picking on him? -- --Tim Smith === thats what I was going to say, Hear, here!, but I was going to try to edit-out all of the super?arke ting, or what ever it is that looks just like whining (for going-on Decade Two .-) much better, if he did it himself. maybe he doesnt have a wordprocessor? one mans inability to grok mathematics, is another mans masturbation ... I *said* Id get outta here ... unless theres some actual math involved. > If it is mathematically correct and interesting, why does it matter? Unlike > pretty much all of his earlier postings, this one appears to be --Give the Gift of Trickier Dick Cheeny -- out of of?e, at last! http://www.wlym.com/pages/music.html http://members.tripod.com/~american_almanac http://www.benfranklinbooks.com/ http://www.wlym.com/PDF-68-76/CAM7606.pdf http://www.rand.org/publications/randreview/issues/rr.12.00/ === >> When I cared about ?ding fast prime counting algorithms versus >> concentrating on the core mathematics, I found formulas like the >> following, which counts primes up to a given even N over a certain range: >> N/2 - ?N-4)/6) - ?N-16)/10) + ?N-16)/30) - >> ?N-36)/14) + ?N-22)/42) + ?N-106)/70) - >> ?N-106)/210) + 2. > If you no longer care about ?ding fast prime counting algorithms, why > are you posting this now? If it is mathematically correct and interesting, why does it matter? It doesnt, and everyone has said so. > Unlike > pretty much all of his earlier postings, this one appears to be > mathematically correct, That has never been an issue. > uses standard terminology, Nor has that. > and is either interesting Youre getting warmer... > or suggests interesting things. Now you got it! The interesting things that are being suggested are in dispute as being uninteresting, ?t wrong and outright lies. > Why are you picking on him? Because hes a troll. === > When I cared about ?ding fast prime counting algorithms versus >> concentrating on the core mathematics, I found formulas like the >> following, which counts primes up to a given even N over a certain range: >> N/2 - ?N-4)/6) - ?N-16)/10) + ?N-16)/30) - >> ?N-36)/14) + ?N-22)/42) + ?N-106)/70) - >> ?N-106)/210) + 2. >If you no longer care about ?ding fast prime counting algorithms, why > are you posting this now? If it is mathematically correct and interesting, why does it matter? Unlike > pretty much all of his earlier postings, this one appears to be > mathematically correct, uses standard terminology, and is either interesting > or suggests interesting things. Why are you picking on him? -- > --Tim Smith The material has been posted before, has been repudiated by him, and was placed here for no other reason than to use this newsgroup as a vehicle for leaving a note to himself. See James response for a con?mation. Havent you been paying attention? -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === I need to know a method to ?d the rate of convergence of any iterative solution. Say suppose, I need to ?d the convergence rate of Newton-Raphson method or secent method of ?ding roots of an equation, how should I go for it. I know (from other posting on this site) that the rate of convergence of Newton Method is 2, but I dont know the mathematics behind this. Can anyone explain me how the rates are calculated, and can anybody point me to some site/book/eBook which will teach me these things in detail. Satya === Can anyone tell me what a Schur Complement is and what are its properties? I think it has something to do with the correlation between two signals, for example if x and y have auto- and cross-correlation matrices Rxx, Ryy, Rxy and Ryx then the Schur complement is Rx|y = Rxx - Rxy*inv(Ryy)*Ryx or Ry|x = Ryy - Ryx*inv(Rxx)*Rxy a million... ************************** === have a look at Mersenne Twister website http://www.math.keio.ac.jp/~matumoto/emt.html > Copyright(c)2004 by Hermann Samso. > All Rights Reserved. > Good evening. First test at a RNG with a Motorola > MC68000 processor. > First try was to code the Lewis, Goodman > and Miller algorithm from 1969, but then > realized that a long division opcode was > necessary, and MC68000 can only divide > by words. So I gave up, and tried desperately with > a 32bit long dividend described in above > mentioned algorithm and a 16 bit number > as divisor, I chose 65533 that is nearly > the whole of 16bits. The algorithm here described is probably > not much random, and hasnt yet been > studied or benchmarked respectly. Anyways I found the results good enough > to be included in my last retro scene > demo intro for Atari ST. So, I have > decided to include it here, and maybe > get some advisings and/or proofs. The code I present here is in assembler > and can be easily compiled. It is only > 4 instructions long! It bases on the machine code of the > MC68000 which delivers remainder in the > high word, quotient in the low word of > a long word as result of an unsigned > division. > The only step inbetween is to rotate > this results around in the long word > by a ?ed amount. I chose 8 bits, > but probably 7 or 23 are also good > values. > saludos, > Hermann Samso # Copyright(c)2004 by Hermann Samso. > # All Rights Reserved. > # RND1.s > # An easy way ? RND1 move.l D,d0 > divu d,d0 > ror.l #8,d0 > move.l d0,D dc.l D 2147483647 ;dividend > dc.w d 65533 ;divisor > http://members.tripod.com/so_o2 === >>|question: are there important results in Analysis or Topology that >>|depend on axioms for large cardinals? Not so far as I know, but I wouldnt take my word for it if I were >>you. My understanding is that certain aspects of point-set >>topology have become relatively set-theoretical, involving >>propositions independent of ZF. (Isnt there a volume titled >>Set-Theoretical Topology?) But I dont know that large >>cardinal axioms play a role. Heres an example of a statement of analysis independent >>of ZFC. Say that a set S of reals has _absolute measure >>zero_ if for any sequence a0,a1,... of positive reals, there >>exist intervals of length a0, a1,... respectively whose union >>contains S. That every set of absolute measure zero is >>countable is independent of ZFC. But as far as I can recall >>theres no large cardinal axiom which decides it. > Huh - this is news to me. Not that it has anything to do with large cardinals, but > the result in analysis that I know thats independent > of ZFC is this: Every homomorphism from one Banach > algebra to another is continuous. (For context, recall > that every homomorphism from a Banach algebra to > C is automatically continuous...) > While were on this subject, do you have a reference for that? Ive heard its true as well, and was looking for a discussion of the subject a while ago but was unable to ?d one. David === On Thu, 15 Jan 2004 17:38:15 +0000, David R MacIver >|question: are there important results in Analysis or Topology that >|depend on axioms for large cardinals? >>Not so far as I know, but I wouldnt take my word for it if I were >you. My understanding is that certain aspects of point-set >topology have become relatively set-theoretical, involving >propositions independent of ZF. (Isnt there a volume titled >Set-Theoretical Topology?) But I dont know that large >cardinal axioms play a role. >>Heres an example of a statement of analysis independent >of ZFC. Say that a set S of reals has _absolute measure >zero_ if for any sequence a0,a1,... of positive reals, there >exist intervals of length a0, a1,... respectively whose union >contains S. That every set of absolute measure zero is >countable is independent of ZFC. But as far as I can recall >theres no large cardinal axiom which decides it. >> Huh - this is news to me. Not that it has anything to do with large cardinals, but >> the result in analysis that I know thats independent >> of ZFC is this: Every homomorphism from one Banach >> algebra to another is continuous. (For context, recall >> that every homomorphism from a Banach algebra to >> C is automatically continuous...) >> While were on this subject, do you have a reference for that? Ive >heard its true as well, and was looking for a discussion of the subject >a while ago but was unable to ?d one. Sorry. The only reason I know about it is someone who knew even less about logic than I did gave me a preprint - that was about a thousand years ago, I have no idea what happened to my copy. Um, I think that some time in the last century there was an about the corresponding question for Frechet algebras - maybe you can ?d that? >David ************************ David C. Ullrich <87smiudtrj.fsf@phiwumbda.org> <3ffadc1f$19$fuzhry+tra$mr2ice@news.patriot.net> <87smisrg0b.fsf@phiwumbda.org> <3ffd41a6$22$fuzhry+tra$mr2ice@news.patriot.net> <871xqa1n9s.fsf@phiwumbda.org> <40009ae6$11$fuzhry+tra$mr2ice@news.patriot.net> <87smim7rqf.fsf@phiwumbda.org> <4001e592$19$fuzhry+tra$mr2ice@news.patriot.net> <87brp98ys1.fsf@phiwumbda.org> <40044d8d$31$fuzhry+tra$mr2ice@news.patriot.net> <87lloby0ht.fsf@phiwumbda.org> <4005a73b$31$fuzhry+tra$mr2ice@news.patriot.net> === > In <87lloby0ht.fsf@phiwumbda.org>, on 01/13/2004 > at 10:58 PM, jesse@phiwumbda.org (Jesse F. Hughes) said: >And yet, as it turns out, it was an apt question. No. What does that paper have to do with it? would have been an apt > question. Youre an idiot, plain and simple. -- Jesse Hughes And a journal can beg me for the right to publish it [...] because Id rather see it in People magazine [...] --James Harris on his simple proof of Fermats last theorem <87smiudtrj.fsf@phiwumbda.org> <3ffadc1f$19$fuzhry+tra$mr2ice@news.patriot.net> <87smisrg0b.fsf@phiwumbda.org> <3ffd41a6$22$fuzhry+tra$mr2ice@news.patriot.net> <871xqa1n9s.fsf@phiwumbda.org> <40009ae6$11$fuzhry+tra$mr2ice@news.patriot.net> <87smim7rqf.fsf@phiwumbda.org> <4001e592$19$fuzhry+tra$mr2ice@news.patriot.net> <87brp98ys1.fsf@phiwumbda.org> <40044d8d$31$fuzhry+tra$mr2ice@news.patriot.net> <87lloby0ht.fsf@phiwumbda.org> <4005a73b$31$fuzhry+tra$mr2ice@news.patriot.net> <87u12x1uju.fsf@phiwumbda.org> X-Cise: tanbanso@iinet.net.au X-CompuServe-Customer: Yes X-Coriate: admin@interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Pose: george_cox@btinternet.com X-Punge: Micro$oft === In <87u12x1uju.fsf@phiwumbda.org>, on 01/15/2004 at 09:32 AM, jesse@phiwumbda.org (Jesse F. Hughes) said: >Youre an idiot, plain and simple. *PLONK* -- Shmuel (Seymour J.) Metz, SysProg and JOAT not reply to spamtrap@library.lspace.org === > In <87u12x1uju.fsf@phiwumbda.org>, on 01/15/2004 > at 09:32 AM, jesse@phiwumbda.org (Jesse F. Hughes) said: >Youre an idiot, plain and simple. *PLONK* Ah, a response that only idiots give, in my experience. The Usenet equivalent of sticking your ?gers in your ears and saying nyah nyah nyah I cant hear you! A mature person ?ds a polite way to stop responding, and doesnt think its necessary to announce the fact. Thomas === > In , on 01/13/2004 > at 03:43 PM, Russell Easterly said: >If this is your de?ition of a TM then I can write a TM that >performs the operations I have described. Please do. OK This TM will ?d a natural number larger than any natural number on the input tape. Assume the input tape is: 01011011101111011111.... Each natural number is represented in unary and followed by a 0. Starting at the beginning of the tape and reading right: 1) Find a zero 2) Find a second zero 3) Backup and write a 1 on the previous zero Repeat steps 1-3. I can give you a state transition table if you want. There will always be one zero on the tape when this TM ?ishes reading it. The tape will contain a unary natural number larger than any on the input tape. It doesnt matter how long the input tape is. You can say that the TM will never ?ish reading the tape. This doesnt matter, since at any point in time the TM will still have produced a natural number larger than any number it has read from the input tape. Russell - 2 many 2 count === > In , on 01/13/2004 > at 03:43 PM, Russell Easterly said: > >If this is your de?ition of a TM then I can write a TM that >performs the operations I have described. >Please do. > OK > This TM will ?d a natural number larger than any natural number on the > input tape. Since, below, you assume an input tape contains every natural number, you are saying that a TM can do what humans can prove cannot be done, namely, ?d a natural number larger than any natural number. Assume the input tape is: 01011011101111011111.... > Each natural number is represented in unary and followed by a 0. Starting at the beginning of the tape and reading right: 1) Find a zero > 2) Find a second zero > 3) Backup and write a 1 on the previous zero > Repeat steps 1-3. I can give you a state transition table if you want. > There will always be one zero on the tape when this TM ?ishes reading it. But the TM does not ?ish. Ever. Since there is always another natural on teh tape that has not yet been touched. > The tape will contain a unary natural number larger than any on the input > tape. > It doesnt matter how long the input tape is. Unless the tape ends after a ?ite number of positions, NO. You can say that the TM will never ?ish reading the tape. > This doesnt matter, since at any point in time the TM > will still have produced a natural number larger than > any number it has read from the input tape. What happens at completion of a ?ite number of steps does not determine the effect of completing in?itely many steps. Is your TM tape ?ite or in?ite? If ?ite, then it is of no use in analyzing the in?ite case any more than a ?ite natural can be larger that all other ?ite naturals. === > in > In , on 01/13/2004 > at 03:43 PM, Russell Easterly said: > >If this is your de?ition of a TM then I can write a TM that >performs the operations I have described. >Please do. > > OK > This TM will ?d a natural number larger than any natural number on the > input tape. Since, below, you assume an input tape contains every natural number, > you are saying that a TM can do what humans can prove cannot be done, > namely, ?d a natural number larger than any natural number. This TM does not ?d the largest natural number. My TM will ?d a natural number larger than any natural number on the input tape. This only proves the input tape does not contain every natural number. >Assume the input tape is: 01011011101111011111.... > Each natural number is represented in unary and followed by a 0. >Starting at the beginning of the tape and reading right: >1) Find a zero > 2) Find a second zero > 3) Backup and write a 1 on the previous zero > Repeat steps 1-3. >I can give you a state transition table if you want. > There will always be one zero on the tape when this TM ?ishes reading it. But the TM does not ?ish. Ever. Since there is always another natural > on teh tape that has not yet been touched. Using this argument the TM that produces the input tape never ?ishes either. So the input tape cant contain every natural number because it was never completed. Lets assume that both TMs can perform an in?ite number of operations. > The tape will contain a unary natural number larger than any on the input > tape. > It doesnt matter how long the input tape is. Unless the tape ends after a ?ite number of positions, NO. >You can say that the TM will never ?ish reading the tape. > This doesnt matter, since at any point in time the TM > will still have produced a natural number larger than > any number it has read from the input tape. What happens at completion of a ?ite number of steps does not > determine the effect of completing in?itely many steps. How can the result be different after an in?ite number of steps? > Is your TM tape ?ite or in?ite? > If ?ite, then it is of no use in analyzing the in?ite case any more > than a ?ite natural can be larger that all other ?ite naturals. The output of my TM must be ?ite. There must be a ?ite number of 1s followed by a 0. The proof shows that the input tape was ?ite even though we assumed it was in?ite. Russell - 2 many 2 count <3ffadc52$20$fuzhry+tra$mr2ice@news.patriot.net> <1g75muz.1d66n4i14gjvdlN%panoptes@iquest.net> <1g75yr4.1si6z861lg1554N%panoptes@iquest.net> <97adneZXNdeRbWCi4p2dnA@comcast.com> <40009fc8$16$fuzhry+tra$mr2ice@news.patriot.net> <40044b66$29$fuzhry+tra$mr2ice@news.patriot.net> <4005a65f$29$fuzhry+tra$mr2ice@news.patriot.net> === > Since, below, you assume an input tape contains every natural number, >> you are saying that a TM can do what humans can prove cannot be done, >> namely, ?d a natural number larger than any natural number. This TM does not ?d the largest natural number. > My TM will ?d a natural number larger than any natural number > on the input tape. This only proves the input tape does not > contain every natural number. No, no, no. You said it correctly the ?st time. ,---- | You can say that the TM will never ?ish reading the tape. | This doesnt matter, since at any point in time the TM | will still have produced a natural number larger than | any number it has read from the input tape. `---- Clearly, the two statements below are not equivalent. (1) at each point[1], the machine will have produced a number larger than any number its read. (2) after an in?ite number of steps, the machine will have written a number larger than any number on the input tape. Why do you persist in such obviously bad arguments? Footnotes: [1] Not to be taken literally. More explicitly, there is a state that the machine hits in?itely often and each time it enters that state the claim is true --- or something like that. -- Jesse Hughes Depression hits more people than thought. --headline in Lexington, KY newspaper, as reported on NPRs Morning Edition <3ffadc52$20$fuzhry+tra$mr2ice@news.patriot.net> <1g75muz.1d66n4i14gjvdlN%panoptes@iquest.net> <1g75yr4.1si6z861lg1554N%panoptes@iquest.net> <97adneZXNdeRbWCi4p2dnA@comcast.com> <40009fc8$16$fuzhry+tra$mr2ice@news.patriot.net> <40044b66$29$fuzhry+tra$mr2ice@news.patriot.net> <4005a65f$29$fuzhry+tra$mr2ice@news.patriot.net> === >This TM will ?d a natural number larger than any natural number on the > input tape. >Since, below, you assume an input tape contains every natural number, > you are saying that a TM can do what humans can prove cannot be done, > namely, ?d a natural number larger than any natural number. This TM does not ?d the largest natural number. > My TM will ?d a natural number larger than any natural number > on the input tape. This only proves the input tape does not > contain every natural number. Thats ?e. There exist many (*in?itely* many, but lets not get into that) tapes that do not contain every natural number. So its perfectly possible that your particular input tape does not contain every natural number. No contradiction. > Using this argument the TM that produces the input tape never ?ishes > either. True; no TM (no machine, *period*) can ever ?ish printing *all* of the natural numbers. Thats true. No contradiction. > So the input tape cant contain every natural number because it was > never completed. Lets assume that both TMs can perform an in?ite > number of operations. Lets not. Its a stupid thing to assume, since its not only physically impossible but also poorly de?ed (*how big* an in?ite number of steps are you assuming we can perform in ?ite time?) and in fact Id go so far as to say its meaningless. Moreover, Turing Machines cannot perform an in?ite number of operations in a ?ite time. (I de?e time to mean number of computational steps, or what a modern scientist might call CPU ticks.) Thats a direct result of Turings de?ition, and is not debatable -- remember back when you were still quoting Turings paper? I presume were still discussing that paper. > What happens at completion of a ?ite number of steps does not > determine the effect of completing in?itely many steps. How can the result be different after an in?ite number of steps? Consider the set {0}. It has a ?ite number of elements: 1. Consider the set {0,1}. It has a ?ite number of elements: 2. Consider the set {0,1,2}. It has a ?ite number of elements: 3. Consider the set {0,1,2,3}. It has a ?ite number of elements: 4. ... Consider the set {0, ..., n}. It has a ?ite number of elements: n+1. Now take that to in?ity: Consider the set {0, 1, 2,...}, a.k.a. the natural numbers. It has an in?ite number of elements: aleph-null. Thus the properties of the set at the limit are qualitatively different from the properties of each set on the way to the limit. The former has in?itely many elements; each of the latter has only ?itely many elements. Q.E.D.; the result is different after an in?ite number of steps. > Is your TM tape ?ite or in?ite? > If ?ite, then it is of no use in analyzing the in?ite case any more > than a ?ite natural can be larger that all other ?ite naturals. The output of my TM must be ?ite. > There must be a ?ite number of 1s followed by a 0. > The proof shows that the input tape was ?ite even > though we assumed it was in?ite. All right, so the input tape was ?ite. Whoop-de-doo. Stop the presses; someones found a tape with only ?itely many symbols on it. Thats NOT NEW, and its NOT A CONTRADICTION. I suppose I dont see what youre trying to prove; obviously its no great shakes that some Turing machine operates on, or produces, a ?ite tape. Thats the usual state of things. -Arthur === > So the input tape cant contain every natural number because it was > never completed. Lets assume that both TMs can perform an in?ite > number of operations. Lets not. Its a stupid thing to assume, since its not only > physically impossible but also poorly de?ed (*how big* an in?ite > number of steps are you assuming we can perform in ?ite time?) and > in fact Id go so far as to say its meaningless. > Moreover, Turing Machines cannot perform an in?ite number of > operations in a ?ite time. (I de?e time to mean number > of computational steps, or what a modern scientist might call CPU > ticks.) Thats a direct result of Turings de?ition, and is not > debatable -- remember back when you were still quoting Turings paper? > I presume were still discussing that paper. > You should read the paper. Turing clearly states that a computable sequence is produced by a circle free machine. Circle free is de?ed as a TM that Circle free machines do not halt. There is a whole ?ld of mathematics called hypercomputation that assumes that a machine can perform an in?ite number of operations in ?ite time. For example, there are accelerated Turing Machines. Such a machine performs the ?st operation in one unit of time. The second operation takes 1/2 unit of time, the third operation takes 1/4 unit of time, etc. After 2 units of time this machine will have performed an in?ite number of operations. Turing invented Oracle TMs. A TM with an oracle is also assumed to be able to perform an in?ite number of operations. > What happens at completion of a ?ite number of steps does not > determine the effect of completing in?itely many steps. >How can the result be different after an in?ite number of steps? Consider the set {0}. It has a ?ite number of elements: 1. > Consider the set {0,1}. It has a ?ite number of elements: 2. > Consider the set {0,1,2}. It has a ?ite number of elements: 3. > Consider the set {0,1,2,3}. It has a ?ite number of elements: 4. > ... > Consider the set {0, ..., n}. It has a ?ite number of elements: n+1. Now take that to in?ity: Consider the set {0, 1, 2,...}, a.k.a. the natural numbers. It has > an in?ite number of elements: aleph-null. Prove it. > Thus the properties of the set at the limit are qualitatively different > from the properties of each set on the way to the limit. The former > has in?itely many elements; each of the latter has only ?itely many > elements. > Q.E.D.; the result is different after an in?ite number of steps. My proof shows that no set can contain every computable natural number. There is no such thing as the set of all natural numbers. There will always be a computable natural number larger than every member in the set. Russell ===