mm-141 > De'ne, for x rational:> f(0) = 0> f(x+1) = (f(x)+1)/2> f(1/x) = 1 - f(x) (for x>0)> Extend to irrationals by limits. The range of f is the half-open [0,1).> f(1/2) = (f(-1/2) + 1)/2; 1/2 - 1 = f(-1/2) = -1/2 ??? The function was only de'ned for NON-NEGATIVE reals, for which the range is> indeed [0,1). > One erratum; you misplace the quanti'er to apply to the third line> instead of at the top where you described the domain.Well, in the text that preceded the recursive de'nition I had mentioned that the function was de'ned for non-negative reals. The x>0 for f(1/x) was just to be pedantic, and (at the time) I simply meant for x non-zero. It turns out however that x>0 is critical -- see below.> In fact, it's not too dif'cult.> One thing we get is: f(x-n) = 2^n f(x) - (2^n - 1) f(x+n) = f(x)/2^n + 1-1/2^n> f(a/b) + f(b/a) = 1, ab /= 0> f(-1/x) = 1 - f(-x)It turns out that the last two (involving inversion) are wrong for negativearguments of f. The rule for f(x+n) can indeed be extended to negative n,but the rule for f(1/n) cannot. Here is why:For n>1 we have: f(1-1/n) = 1 - f(1+1/(n-1)) = f(n-1)/2(The result happens to be ok for n=1 also.)Hence: f(-1/n) = f(n-1)-1 = -1/2^(n-1)On the other hand: f(-n) = 1 - 2^nThese are clearly incompatible with f(x) + f(1/x) = 1 -- that REQUIRES x>0.This function keeps surprising me.Michel. =Because I dont know how to write my idea in the common formal way, Iam going to do it in a non-formal way, but I will do my best to writeit in the clearest way.So here it is:Let us check these lists.P(2) ={{},{0},{1},{0,1}} = 2^2 = 4and also can be represented as:00011011[b:034b32e72f][i:034b32e72f]P[/i:034b32e72f][/b: 034b32e72f](3) ={{},{0},{1},{2},{0,1},{0,2},{1,2},{0,1,2}} = 2^3 = 8and also can be represented as:000001010011100101110111Let us call any full 01 list, combinations list.Now, let us use Cantor's Diagonalization method on some 'nitely longcombinations list, for example, the combinations list of number 3:[b:034b32e72f]0[/b:034b32e72f]000[b:034b32e72f]0[/b: 034b32e72f]101[b:034b32e72f]0[/b:034b32e72f]011100101110111We can change the order of the rows, and then use Cantor'sDiagonalization method, for example:[b:034b32e72f]0[/b:034b32e72f]010[b:034b32e72f]1[/b: 034b32e72f]101[b:034b32e72f]0[/b:034b32e72f]000101100111110The input for Cantor's Diagonalization method in the 'rst example is000 and the output is 111.The input for Cantor's Diagonalization method in the second example is010 and the output is 101.In both examples we 'nd that the result is already in thecombinations list, and this combination, which is already in thelist, is one of the combinations that Cantor's Diagonal does notcover.The number of the combinations, which are out of the range of Cantor'sdiagonal is:[b:034b32e72f][i:034b32e72f]2^n - n[/i:034b32e72f][/b:034b32e72f]Every column, which belongs to some combinations list is a sequence of01 notations, based on some periodic frequency changes, for example:the right column of number 3 combinations list, is based on 2^0(=1).Therefore the periodic frequency changes are 1, and the result in thiscase is:01010101.The result of the middle column is based on 2^1(=2), therefore thesequence is:00110011.The result of the left column is based on 2^2(=4), therefore thesequence is:00001111.and we get the full combinations list of number 3:000001010011100101110111We can get a combinations list of in'nitely many places, by using the[b:034b32e72f][i:034b32e72f]ZF Axiom o'n'nity[/i:034b32e72f][/b:034b32e72f] induction, on the[b:034b32e72f][i:034b32e72f]left[/i:034b32e72f][/b: 034b32e72f] sideof our combinations list, by using the induction on the[b:034b32e72f][i:034b32e72f]power_value[/i:034b32e72f][/b: 034b32e72f]of each column, for example:2^[b:034b32e72f]0[/b:034b32e72f], 2^[b:034b32e72f]1[/b:034b32e72f],2^[b:034b32e72f]2[/b: 034b32e72f], 2^[b:034b32e72f]3[/b:034b32e72f],..In this stage we have proven, by induction, that Cantor's diagonalcannot cover any full 01 combinations list, 'nite or in'nite.Therefore its result is not a new combination (that has to be added tothe list).Because Cantor's diagonal cannot cover the full 01 combinations list(of aleph0 places for each combination) we can conclude that[b:034b32e72f][i:034b32e72f]2^aleph0 >aleph0[/i:034b32e72f][/b:034b32e72f].But, because no diagonal's result is a new combination (and thereforenot added to the list) each in'nitely long sequence of 01 notationscan be mapped with some natural number, for example:..000 <--> 1..001 <--> 2..010 <--> 3..011 <--> 4..100 <--> 5..101 <--> 6..110 <--> 7..111 <--> 8..Therefore we can conclude that [b:034b32e72f][i:034b32e72f]2^aleph0 =aleph0[/i:034b32e72f][/b:034b32e72f], and we come to contradiction.[b:034b32e72f][i:034b32e72f](2^aleph0 >= aleph0) =[/i:034b32e72f][/b:034b32e72f][b:034b32e72f]{}[/b:034b32e72f ], and wehave a proof saying that Boolean Logic cannot deal with in'nitelymany objects in in'nitely many magnitudes.One can say that at least the sequence ...111 is not in the list, forexample:..00[b:034b32e72f]0[/b:034b32e72f] <--> 1..0[b:034b32e72f]0[/b:034b32e72f]1 <--> 2..[b:034b32e72f]0[/b:034b32e72f]10 <--> 3..011 <--> 4 ..100 <--> 5..101 <--> 6..110 <--> 7..111 <--> 8..Let us examine the in'nite from another point of view.When we have ...111 AND ...000 in an[b:034b32e72f]ordered[/b:034b32e72f] combinations list, it means thatthe list is complete.But this is the whole point, in'nitely many objects cannot becompleted, otherwise they are 'nitely many objects.Therefore ...111 AND ...000 are not in the list of in'nity manyobjects.In other words [...000, ...111) XOR (...000, ...111] .There are 2 possible structural types of in'nitely many 01notations:(?...0](?...1]We know how some in'nitely long combination starts, but its left sideisunknown (can be 0 XOR 1) [b:034b32e72f]and this missing information isessential to the existence of the induction.[/b:034b32e72f]Therefore we can 'nd a meaningful missing result by Cantor'sDiagonalization method, only in a 'nite combinations list. For more details please look at:http://www.geocities.com/complementarytheory/ Everything.pdfhttp://www.geocities.com/complementarytheory/ ASPIRATING.pdfhttp://www.geocities.com/complementarytheory/ RiemannsLimits.pdfhttp://www.geocities.com/complementarytheory /LIM.pdfDoron---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =--- = Because Cantor's diagonal cannot cover the full 01 combinations list> (of aleph0 places for each combination) we can conclude that> [b:034b32e72f][i:034b32e72f]2^aleph0 aleph0[/i:034b32e72f][/b:034b32e72f].> Yes, The naturals cannot be put in bijection with its power set. Amongstother things.> But, because no diagonal's result is a new combination (and therefore> not added to the list) each in'nitely long sequence of 01 notations> can be mapped with some natural number, for example: ..000 <--> 1> ..001 <--> 2> ..010 <--> 3> ..011 <--> 4> ..100 <--> 5> ..101 <--> 6> ..110 <--> 7> ..111 <--> 8> ..No it damn well can't. Only those strings with 'nitely many 1's may bemapped in this manner of identifying it with the base 2 expansion. Therefore we can conclude that [b:034b32e72f][i:034b32e72f]2^aleph0 => aleph0[/i:034b32e72f][/b:034b32e72f], and we come to contradiction.No. The whole point is that the strings do not form complete list of theelements of the power set. You've already shown that, so how are yougetting round this.You keep on saying things like ïby in'nite induction'. Whatever this is,you are only getting the 'nite subsets of the natural numbers, not allthe subsets. Induction cannot just be abused like this. [b:034b32e72f][i:034b32e72f](2^aleph0 >= aleph0) => [/i:034b32e72f][/b:034b32e72f][b:034b32e72f]{}[/b:034b32e72f], and we> have a proof saying that Boolean Logic cannot deal with in'nitely> many objects in in'nitely many magnitudes. One can say that at least the sequence ...111 is not in the list, for> example: ..00[b:034b32e72f]0[/b:034b32e72f] <--> 1> ..0[b:034b32e72f]0[/b:034b32e72f]1 <--> 2> ..[b:034b32e72f]0[/b:034b32e72f]10 <--> 3> ..011 <--> 4 > ..100 <--> 5> ..101 <--> 6> ..110 <--> 7> ..111 <--> 8> .. Let us examine the in'nite from another point of view. When we have ...111 AND ...000 in an> [b:034b32e72f]ordered[/b:034b32e72f] combinations list, it means that> the list is complete.> Why must it be? I can just add those elements to the list without alteringits cardinality. The point again is that given any countable set to add tothe list it is still countable... You want it to be true that you can takea countable union of countable objects to be uncountable. Well that isn'ttrue. > But this is the whole point, in'nitely many objects cannot be> completed, otherwise they are 'nitely many objects.> Which in'nitely many objects?completed to what? > Therefore ...111 AND ...000 are not in the list of in'nity many> objects. In other words [...000, ...111) XOR (...000, ...111] . There are 2 possible structural types of in'nitely many 01> notations: (?...0]> (?...1] We know how some in'nitely long combination starts, but its left side> is> unknown (can be 0 XOR 1) [b:034b32e72f]and this missing information is> essential to the existence of the induction.[/b:034b32e72f] Therefore we can 'nd a meaningful missing result by Cantor's> Diagonalization method, only in a 'nite combinations list. > For more details please look at: http://www.geocities.com/complementarytheory/Everything.pdf http://www.geocities.com/complementarytheory/ASPIRATING.pdf http://www.geocities.com/complementarytheory/ RiemannsLimits.pdf http://www.geocities.com/complementarytheory/LIM.pdf Doron ---= 19 East/West-Coast Specialized Servers - Total Privacy viaEncryption =---Doron, cantor's result is simple, elegant and easy to understand, what isit that you are attempting to demonstrate. Forget mathematical equations,just explain in words what is wrong. => Because I dont know how to write my idea in the common formal way, I> am going to do it in a non-formal way, but I will do my best to write> it in the clearest way.> So here it is:> Let us check these lists.> P(2) => {{},{0},{1},{0,1}} = 2^2 = 4> and also can be represented as:> 00> 01> 10> 11> [b:034b32e72f][i:034b32e72f]P[/i:034b32e72f][/b:034b32e72f](3) => {{},{0},{1},{2},{0,1},{0,2},{1,2},{0,1,2}} = 2^3 = 8I have no idea what you are trying to say, because these blocks that looklike [b:034b32e72f] pervade your entire post, making it completelyunreadable. They look like mangled HTML. Can you post a plain textversion?-- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. Dave Seaman>I have no idea what you are trying to say, because these blocks that look>like [b:034b32e72f] pervade your entire post, making it completely>unreadable. They look like mangled HTML. Can you post a plain text>version?Apparently he cut-and-pasted from his PDF proof and the garbageindicates a word that was coloured in blue in the original text.It seems he looks at a set of all 'nite permutations of 'nitely manyobjects, deduces that they are indeed 'nite and then declares twocontradictory results based on this to show that there is an error incore^H^H^H^Hboolean logic.I think he also prooves that naturals are 'nite since any completelist can't be in'nite. This might be groundbreaking work. =Because I dont know how to write my idea in the common formal way, Iam going to do it in a non-formal way, but I will do my best to writeit in the clearest way.So here it is:Let us check these lists.P(2) ={{},{0},{1},{0,1}} = 2^2 = 4and also can be represented as:00011011[b:04da4ee3ca][i:04da4ee3ca]P[/i:04da4ee3ca][/b: 04da4ee3ca](3) ={{},{0},{1},{2},{0,1},{0,2},{1,2},{0,1,2}} = 2^3 = 8and also can be represented as:000001010011100101110111Let us call any full 01 list, combinations list.Now, let us use Cantor's Diagonalization method on some 'nitely longcombinations list, for example, the combinations list of number 3:[b:04da4ee3ca]0[/b:04da4ee3ca]000[b:04da4ee3ca]0[/b: 04da4ee3ca]101[b:04da4ee3ca]0[/b:04da4ee3ca]011100101110111We can change the order of the rows, and then use Cantor'sDiagonalization method, for example:[b:04da4ee3ca]0[/b:04da4ee3ca]010[b:04da4ee3ca]1[/b: 04da4ee3ca]101[b:04da4ee3ca]0[/b:04da4ee3ca]000101100111110The input for Cantor's Diagonalization method in the 'rst example is000 and the output is 111.The input for Cantor's Diagonalization method in the second example is010 and the output is 101.In both examples we 'nd that the result is already in thecombinations list, and this combination, which is already in thelist, is one of the combinations that Cantor's Diagonal does notcover.The number of the combinations, which are out of the range of Cantor'sdiagonal is:[b:04da4ee3ca][i:04da4ee3ca]2^n - n[/i:04da4ee3ca][/b:04da4ee3ca]Every column, which belongs to some combinations list is a sequence of01 notations, based on some periodic frequency changes, for example:the right column of number 3 combinations list, is based on 2^0(=1).Therefore the periodic frequency changes are 1, and the result in thiscase is:01010101.The result of the middle column is based on 2^1(=2), therefore thesequence is:00110011.The result of the left column is based on 2^2(=4), therefore thesequence is:00001111.and we get the full combinations list of number 3:000001010011100101110111We can get a combinations list of in'nitely many places, by using the[b:04da4ee3ca][i:04da4ee3ca]ZF Axiom o'n'nity[/i:04da4ee3ca][/b:04da4ee3ca] induction, on the[b:04da4ee3ca][i:04da4ee3ca]left[/i:04da4ee3ca][/b: 04da4ee3ca] sideof our combinations list, by using the induction on the[b:04da4ee3ca][i:04da4ee3ca]power_value[/i:04da4ee3ca][/b: 04da4ee3ca]of each column, for example:2^[b:04da4ee3ca]0[/b:04da4ee3ca], 2^[b:04da4ee3ca]1[/b:04da4ee3ca],2^[b:04da4ee3ca]2[/b: 04da4ee3ca], 2^[b:04da4ee3ca]3[/b:04da4ee3ca],..In this stage we have proven, by induction, that Cantor's diagonalcannot cover any full 01 combinations list, 'nite or in'nite.Therefore its result is not a new combination (that has to be added tothe list).Because Cantor's diagonal cannot cover the full 01 combinations list(of aleph0 places for each combination) we can conclude that[b:04da4ee3ca][i:04da4ee3ca]2^aleph0 >aleph0[/i:04da4ee3ca][/b:04da4ee3ca].But, because no diagonal's result is a new combination (and thereforenot added to the list) each in'nitely long sequence of 01 notationscan be mapped with some natural number, for example:..000 <--> 1..001 <--> 2..010 <--> 3..011 <--> 4..100 <--> 5..101 <--> 6..110 <--> 7..111 <--> 8..Therefore we can conclude that [b:04da4ee3ca][i:04da4ee3ca]2^aleph0 =aleph0[/i:04da4ee3ca][/b:04da4ee3ca], and we come to contradiction.[b:04da4ee3ca][i:04da4ee3ca](2^aleph0 >= aleph0) =[/i:04da4ee3ca][/b:04da4ee3ca][b:04da4ee3ca]{}[/b:04da4ee3ca ], and wehave a proof saying that Boolean Logic cannot deal with in'nitelymany objects in in'nitely many magnitudes.One can say that at least the sequence ...111 is not in the list, forexample:..00[b:04da4ee3ca]0[/b:04da4ee3ca] <--> 1..0[b:04da4ee3ca]0[/b:04da4ee3ca]1 <--> 2..[b:04da4ee3ca]0[/b:04da4ee3ca]10 <--> 3..011 <--> 4 ..100 <--> 5..101 <--> 6..110 <--> 7..111 <--> 8..Let us examine the in'nite from another point of view.When we have ...111 AND ...000 in an[b:04da4ee3ca]ordered[/b:04da4ee3ca] combinations list, it means thatthe list is complete.But this is the whole point, in'nitely many objects cannot becompleted, otherwise they are 'nitely many objects.Therefore ...111 AND ...000 are not in the list of in'nity manyobjects.In other words [...000, ...111) XOR (...000, ...111] .There are 2 possible structural types of in'nitely many 01notations:(?...0](?...1]We know how some in'nitely long combination starts, but its left sideisunknown (can be 0 XOR 1) [b:04da4ee3ca]and this missing information isessential to the existence of the induction.[/b:04da4ee3ca]Therefore we can 'nd a meaningful missing result by Cantor'sDiagonalization method, only in a 'nite combinations list. For more details please look at:http://www.geocities.com/complementarytheory/ Everything.pdfhttp://www.geocities.com/complementarytheory/ ASPIRATING.pdfhttp://www.geocities.com/complementarytheory/ RiemannsLimits.pdfhttp://www.geocities.com/complementarytheory /LIM.pdfDoron---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =--- =The problem is that you jump from the 'nite to the in'nite case,when (especially in the context of Cantor) that is very dangerous todo without being very careful.You gave the example of the set000001010011100101110111Indeed, Cantor's diagonal produces 111, which is on the list. But theproblem here is that our list has more entries than each entry hasdigits. Try taking precisely 3-- any distinct 3 of your choosing-- ofthose elements as your list, for example,010011110In the above example, the diagonal 101 produces a new element not inthe list, but you may say I just didn't pick those well enough. Well,if you don't take my word, you can try all 8x7x6=336 possible suchlists yourself and I will rest easily knowing that in each of them,the diagonal will lie outside of them.The whole point of the diagonal argument is that there needs to be aone-to-one correspondence between the digits of the diagonal and theentries in the list. Then by de'nition, each element in the originallist has a corresponding digit in the diagonal with which it 'ndsdisagreement- thus, the diagonal cannot possibly be in the list. Thisbreaks down when the diagonal has fewer digits than the list hasentries.Don't be totally downhearted. Just be more careful in the future. You're on a good path here, considering the diagonal argument in*binary* (which is really what you're doing). In the usual decimaleveryone's familiar with, it is not nearly so interesting. But acountably in'nite list of reals produces one *and only one* nextirrational by considering the diagonal when everything's beenconverted to binary (and you make the necessary stipulations to avoid0=.11111... dif'culties). This itself is quite interesting if youstudy it, but not really applicable to anything.Good luck =Please read the attached pdf 'les by their order1) http://www.geocities.com/complementarytheory/ NewDiagonalView.pdf2) http://www.geocities.com/complementarytheory/Everything.pdf3) http://www.geocities.com/complementarytheory/ASPIRATING.pdf4) http://www.geocities.com/complementarytheory/GIF.pdf5) http://www.geocities.com/complementarytheory/ RiemannsLimits.pdfDoron---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =--- => Note that negative zero point energy density means equal and opposite > positive quantum pressure, i.e. w = -1. But the quantum pressure is > three times stronger in its gravity effect in Einstein's GR. So negative > zero point energy density is gravitating dark matter IMHO. Positive > zero point energy density is anti-gravitating dark energy with > dominating negative pressure. Does anyone object to that? If so, why?> Not an objection but an observationThese are inventions of the human mind, speci'cally; zero, negativity and in'nity.R.A.B. =E and F are two normed vector spaces, and U is a convex open subset of E.Consider a sequence (f_n) of functions: U->F which are differentiable on U,such that:(i) (df_n) converges uniformly to a function g:U->F.(ii) there exists a point a in U, (f_n(a)) converges.Then there exists a function f: U->F such that df = g and (f_n) convergesuniformly to f on any bounded subset of U.I had no problem with the proof, but the professor said that result can beextended to the case where U is a connected open subset of E (and U is notnecessarily convex). In this case, I can show (f_n) converges to f(pointwise) and df = g, but I couldn't prove it converges *uniformly* onany bounded subset of U.Another poster from fr.sci.maths suggested me to consider an open set O of Uwhich is maximal for the required property (uniform convergence of (f_n) ->g); from there I could easily conclude using the connexity of U.What I can show is that IF such a maximal set O exists, THEN it is open, butI can't see how to prove the existence of such a O. (the poster was briefabout that, so I guess it is either really easy or false)Any help would be greatly appreciated !--Julien Santini,France. =On normed vector spaces, and U is a convex open subset of E.>Consider a sequence (f_n) of functions: U->F which are differentiable on U,>such that:>(i) (df_n) converges uniformly to a function g:U->F.>(ii) there exists a point a in U, (f_n(a)) converges.>Then there exists a function f: U->F such that df = g and (f_n) converges>uniformly to f on any bounded subset of U.I had no problem with the proof, but the professor said that result can be>extended to the case where U is a connected open subset of E (and U is not>necessarily convex). In this case, I can show (f_n) converges to f>(pointwise) and df = g, but I couldn't prove it converges *uniformly* on>any bounded subset of U.The convergence need not be uniform on bounded subsets of U if Uis not convex. At least it seems clear to me that it should be easy togive a counterexample... right:Take E = R^2. Let U be a thin neighborhood of the spiral s(t) = (1-1/t)*(cos(t), sin(t)) (1 <= t < in'nity);U is a subset of the unit disk consisting of a thin strip thatsort of spirals out towards the boundary of the disk.So U is bounded, but the point is that if we measuredistance by measuring the length of curves _within_ Uthen the distance from the origin to an arbitrary pointof U is unbounded...De'ne f_n on the spiral itself by f_n(s(t)) = t/n,and extend f_n to all of U in a nice conservative way,maybe requiring that f_n be constant on any radialsegment contained entirely in U. Then df_n -> 0uniformly in U, f_n(0) = 0, but f_n does not convergeuniformly on U.>Another poster from fr.sci.maths suggested me to consider an open set O of U>which is maximal for the required property (uniform convergence of (f_n) -g); from there I could easily conclude using the connexity of U.What I can show is that IF such a maximal set O exists, THEN it is open, but>I can't see how to prove the existence of such a O. (the poster was brief>about that, so I guess it is either really easy or false)Any help would be greatly appreciated !************************David C. Ullrich => The convergence need not be uniform on bounded subsets of U if U> is not convex. At least it seems clear to me that it should be easy to> give a counterexample... right:it could be quite dif'cult. After all that was not so easy; your argumentlooks intuitively very convincing (even though it could be very tedious towrite it formally, especially for the thin neighborhood around the spiral,and even more when it comes to de'ne f_n). At least I know now that theuniform convergence doesnt hold !!--J.S =On convergence need not be uniform on bounded subsets of U if U>> is not convex. At least it seems clear to me that it should be easy to>> give a counterexample... right:it could be quite dif'cult. After all that was not so easy; your argument>looks intuitively very convincing (even though it could be very tedious to>write it formally, especially for the thin neighborhood around the spiral,>and even more when it comes to de'ne f_n). At least I know now that the>uniform convergence doesnt hold !!I realized later that one could explain it a little better:If U is a suf'ciently non-convex subset of R^2 then there exists an unbounded function f such that df is bounded. (For exampleif U is that spiral thing then one can get df/dr = 0, df/dtheta = 1in polar coordinates, hence df is bounded.) Let f_n = f/n.Then df_n -> 0 uniformly and f_n -> non-uniformly.************************David C. Ullrich =I am slightly confused. Suppose S_3 = , the symmetricgroup on 3 letters. Suppose h is a simple representation of S_3 ofdimension > 1 in the vector space V.Let e be an eigenvector of s. Then se=pe, i.e. p is the eigenvalueassociated with e.Then let f = te, so f is also an eigenvecotr of s, but with eigenvalue p^2(I omitted all the calculations here).Now consider the subspace U = spanned by e and f. Then U is stableunder s and t, so under S_3. It follows, since h (the simplerepresentation) is simple, that V = U.How does this show that the simple representations of S_3 can have onlydimension 1 and 2?Moshe => I am slightly confused. Suppose S_3 = , the symmetric> group on 3 letters. Suppose h is a simple representation of S_3 of> dimension > 1 in the vector space V.> Let e be an eigenvector of s. Then se=pe, i.e. p is the eigenvalue> associated with e.> Then let f = te, so f is also an eigenvecotr of s, but with eigenvalue p^2> (I omitted all the calculations here). Now consider the subspace U = spanned by e and f. Then U is stable> under s and t, so under S_3. It follows, since h (the simple> representation) is simple, that V = U. How does this show that the simple representations of S_3 can have only> dimension 1 and 2? MosheBecause any representation then has a subspace of dimension at most 2 thatis 'xed by the action of the generators. That is those two vectors are'xed by s, and swapped by by t. Although it might be that e is also aneigen vector of t, and you have a one dimensional subspace. In any case ifthe rep is simple, that space spanned by e and f must be all of it. =I have been having some dif'culties with the following problem:It is from Royden (p127 #17). The space throughout is assumed to be[0,1], though I think any 'nite measure space will work. Suppose p > 1. Suppose f_n -> f a.e. , f is in L_p and f_n is inL_p for all n. Suppose there exists M such that || f_n || (p-norm) <= M for all n. Suppose g is in L_q. Show that g*f_n -> g*f in the L_1 norm. The main dif'culty I am having is using the p > 1 hypothesis. My mainstrategy has been to try to establish that |g||f_n - f| is uniformly integrable by using the uniform L_p bound on the f_n. On a small setA we have int(|g||f_n - f|) <= ||g|| * ||f_n - f|| by Holder, but thisbound is not enough for uniform integrability. I suspect I would like[m(A)] ^ (1 - 1/p) somewhere in my inequalities, but I can't seem to insertthat in the right place without changing something important (for instance||f_n - f|| might change from p-norm to q-norm and then I no longer knowthat ||f_n - f|| is bounded for all n). Any hints or ideas on this would be greatly appreciated.Hugh Denoncourt denoncou@euclid.colorado.edu (Hugh>I have been having some dif'culties with the following problem:It is from Royden (p127 #17). The space throughout is assumed to be>[0,1], though I think any 'nite measure space will work. >Suppose p > 1. Suppose f_n -> f a.e. , f is in L_p and f_n is in>L_p for all n. Suppose there exists M such that >|| f_n || (p-norm) <= M for all n. Suppose g is in L_q. >Show that g*f_n -> g*f in the L_1 norm. If epsilon > 0 then there exists delta > 0 such that if m(A) < deltathen the L^q norm of g*chi_A is less than epsilon, because ___.Now there exists a set A of measure less than delta such thatf_n -> f ___ly on the complement of A, by ___'s theorem...>The main dif'culty I am having is using the p > 1 hypothesis. My main>strategy has been to try to establish that |g||f_n - f| is uniformly >integrable by using the uniform L_p bound on the f_n. On a small set>A we have int(|g||f_n - f|) <= ||g|| * ||f_n - f|| by Holder, but this>bound is not enough for uniform integrability. I suspect I would like>[m(A)] ^ (1 - 1/p) somewhere in my inequalities, but I can't seem to insert>that in the right place without changing something important (for instance>||f_n - f|| might change from p-norm to q-norm and then I no longer know>that ||f_n - f|| is bounded for all n). Any hints or ideas on this would be greatly appreciated.Hugh Denoncourt************************David C. Ullrich =>On 1 have been having some dif'culties with the following problem:>>It is from Royden (p127 #17). The space throughout is assumed to be>>[0,1], though I think any 'nite measure space will work. >>Suppose p > 1. Suppose f_n -> f a.e. , f is in L_p and f_n is in>>L_p for all n. Suppose there exists M such that >>|| f_n || (p-norm) <= M for all n. Suppose g is in L_q. >>Show that g*f_n -> g*f in the L_1 norm. If epsilon > 0 then there exists delta > 0 such that if m(A) < delta>then the L^q norm of g*chi_A is less than epsilon, because ___.>Now there exists a set A of measure less than delta such that>f_n -> f ___ly on the complement of A, by ___'s theorem...>The main dif'culty I am having is using the p > 1 hypothesis. My main>>strategy has been to try to establish that |g||f_n - f| is uniformly >>integrable by using the uniform L_p bound on the f_n. On a small set>>A we have int(|g||f_n - f|) <= ||g|| * ||f_n - f|| by Holder, but this>>bound is not enough for uniform integrability. I suspect I would like>>[m(A)] ^ (1 - 1/p) somewhere in my inequalities, but I can't seem to insert>>that in the right place without changing something important (for instance>>||f_n - f|| might change from p-norm to q-norm and then I no longer know>>that ||f_n - f|| is bounded for all n). >>Any hints or ideas on this would be greatly appreciated.It is not mentioned here, but in the problem, is 1/p + 1/q <= 1 assumed?If not, then g*f might not even be in L^1.One has to be careful of which norms are being used in Holder's; forexample, int(|g||f_n - f|) <= ||g|| * ||f_n - f|| should really be | | | | | g (f - f) dx | <= ||g|| ||f - f|| | | n | q n pwhere 1/p + 1/q = 1.One must have p > 1. For example, de'ne f_n for n >= 2 by f_n(x) = n^2 x for x in [0,1/n] = n(2 - nx) for x in [1/n,2/n] = 0 for x in [2/n,1]and let f(x) = 0 and g(x) = 1 for all x. ||f_n||_1 = 1, for all n,f_n(x) -> f(x) for all x in [0,1]. However, ||g*f_n - g*f||_1 = 1for all n, so g*f_n does not tend to g*f in L^1.This fails because we don't have If epsilon > 0 then there existsdelta > 0 such that if m(A) < delta then the L^q norm of g*chi_A isless than epsilon when p = 1 and q = oo (for example, g(x) = 1).Rob Johnson take out the trash before replying => If epsilon > 0 then there exists delta > 0 such that if m(A) < delta> then the L^q norm of g*chi_A is less than epsilon, because ___.> Now there exists a set A of measure less than delta such that> f_n -> f ___ly on the complement of A, by ___'s theorem...ARGH - I had tried variations on the Egoroff theme and never put it all,I have asked about inverse Kronecker product, here is a re'ned problem:That's to say let K=kron(A, B), where K is the Kronecker product of matricesA and B...Now what if I already know K and A, how to 'nd the best approximation of B?Please give me some enlightenment and/or give me some pointers...-Walalla =If K truly is the Kronecker product of A and B, then B can be recoveredexactly.First, partition K into B-sized blocks, then identify a non-zero componentof A, say A(i,j).Go to the (i,j)th block of K and divide each element there by A(i,j).This process recovers B Kronecker product, here is a re'ned problem: That's to say let K=kron(A, B), where K is the Kronecker product ofmatrices> A and B... Now what if I already know K and A, how to 'nd the best approximation ofB? Please give me some enlightenment and/or give me some pointers...> -Walalla => If K truly is the Kronecker product of A and B, then B can be recovered> exactly. First, partition K into B-sized blocks, then identify a non-zero component> of A, say A(i,j).> Go to the (i,j)th block of K and divide each element there by A(i,j).> This thing is that K is not the Kronecker product of A and B... indeed,my current project needs to get the best approximated B when A is known...Could you please give me more hints on it?-Walala => The worst thing is that K is not the Kronecker product of A and B... indeed,> my current project needs to get the best approximated B when A is known... Could you please give me more hints on it?You may consider (see http://www.cs.duke.edu/~nikos/KP/home.html):C. Van Loan, N. Pitsianis, Approximation with Kronecker Products, LinearAlgebra for Large Scale and Real Time Applications (ed. M. S. Moonen andG. H. Golub), Kluwer Publications, 1993, pg 293-314Also, googling kronecker product approximation may reveal some furtherinformation.Daniel =>> If K truly is the Kronecker product of A and B, then B can be recovered>> exactly.>> First, partition K into B-sized blocks, then identify a non-zero component>> of A, say A(i,j).>> Go to the (i,j)th block of K and divide each element there by A(i,j).>> worst thing is that K is not the Kronecker product of A and B... indeed,>my current project needs to get the best approximated B when A is known...Could you please give me more hints on it?You have many hints to the value of each B_i_j, namely one for eachA_i_j, so all you have to do is combine the hints in a sensible fashion.Taking the average is often a sensible 'rst try.-Walala =If G is a 'nite p-group such that Aut(G) is Abelian must G be cyclic ? => If G is a 'nite p-group such that Aut(G) is Abelian must G be cyclic ?> According to Fuchs, it is a result of Chatelet and Baer that Theautomorphism group of a torsion [ abelian ] group G is commutative ifand only if the p-components of G are of rank 1 except the 2-componentwhich is either of rank 1 or is of the form C(2) + C(2^oo).-- Paul SperryColumbia, SC (USA) =For fun, I perfomed the following calculation:Assume a tram, whose track has 10 stops. Each of the 9 'rst stops hasthe same number of passengers. (No one obviously gets in at the laststop - the tram is not going to take them anywhere.)On each stop A, assume that each following stop B has the same amount ofpeople willing to go from stop A to stop B. (No one wants to go from astop to the same stop.)Where during the tram trip is the tram most full of passengers?Turns out it's around stop #7, with approximately 44% of all passengerson board the tram.What I could not do is make a mathematical epxression out of myempirical calculations so that I could easily adjust them to a differentnumber of stops. Can anyone show me how to do it?-- /-- Joona Palaste (palaste@cc.helsinki.') ------------- Finland ---------- http://www.helsinki.'/~palaste --------------------- rules! --------/I am looking for myself. Have you seen me somewhere? - Anon => For fun, I perfomed the following calculation:> Assume a tram, whose track has 10 stops. Each of the 9 'rst stops has> the same number of passengers. (No one obviously gets in at the last> stop - the tram is not going to take them anywhere.)> On each stop A, assume that each following stop B has the same amount of> people willing to go from stop A to stop B. (No one wants to go from a> stop to the same stop.)> Where during the tram trip is the tram most full of passengers?> Turns out it's around stop #7, with approximately 44% of all passengers> on board the tram.> What I could not do is make a mathematical epxression out of my> empirical calculations so that I could easily adjust them to a different> number of stops. Can anyone show me how to do it?> Here's my try at it.Let p = number of passengers at each stopLet s = number of stopsLet a(x) denote the number of passengers in the tram at each stop:At each stop x the new passengers destinations will be evenly distributed among the remaining stops (each remaining stop will get p / (s - x) passengers). Therefore:a(1) = pa(x) for x such that 1 < x < s = a(x-1) + p - sum(i=1 to x-1 : p / (s - i))-- Daniel Sj.9ablom => For fun, I perfomed the following calculation:>> Assume a tram, whose track has 10 stops. Each of the 9 'rst stops has>> the same number of passengers. (No one obviously gets in at the last>> stop - the tram is not going to take them anywhere.)>> On each stop A, assume that each following stop B has the same amount of>> people willing to go from stop A to stop B. (No one wants to go from a>> stop to the same stop.)>> Where during the tram trip is the tram most full of passengers?>> Turns out it's around stop #7, with approximately 44% of all passengers>> on board the tram.>> What I could not do is make a mathematical epxression out of my>> empirical calculations so that I could easily adjust them to a different>> number of stops. Can anyone show me how to do it?>> Here's my try at it. Let p = number of passengers at each stop> Let s = number of stops Let a(x) denote the number of passengers in the tram at each stop: At each stop x the new passengers destinations will be evenly > distributed among the remaining stops (each remaining stop will get p / > (s - x) passengers). Therefore: a(1) = p> a(x) for x such that 1 < x < s> = a(x-1) + p - sum(i=1 to x-1 : p / (s - i))I forgot the remaining case x = sa(x) = a(x-1) - sum(i=1 to x-1 : p / (s - i))-- Daniel Sj.9ablom =...> Assume a tram, whose track has 10 stops. Each of the 9 'rst stops has> the same number of passengers. (No one obviously gets in at the last> stop - the tram is not going to take them anywhere.)> On each stop A, assume that each following stop B has the same amount of> people willing to go from stop A to stop B. (No one wants to go from a> stop to the same stop.)> Where during the tram trip is the tram most full of passengers?> Turns out it's around stop #7, with approximately 44% of all passengers> on board the tram.> What I could not do is make a mathematical epxression out of my> empirical calculations so that I could easily adjust them to a different> number of stops. Can anyone show me how to do it?Perhaps compare sums for the above to sums for e. I think that as the number of stops increases, the high-point percentagesgradually approach 1 - 1/e (~ 0.632121) and 1/e (~ .367879). For example, with 10000 stops, the high point is 36.79%at stop 6321.-jiw => I've ended up in a lot of hostile exchanges where clearly I'm angry,> and clearly people posting are angry, and it seems to me then that the> math gets lost, so I'm going to make a real effort not to post in> anger. With that said, I can move on to considering some math positions which> I say are weird. These positions have been revealed in discussions over some of my> mathematical 'ndings like the following: (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = 49(300125 x^3 - 18375 x^2 - 360 x + 22) where b_3(x) = a_3(x) - 3 and the a's are roots of a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) and when x=0, a_1(0) = a_2(0) = b_3(0) = 0. What I have is a polynomial multiplied by 49 that I factor in a> non-standard way, where you can see some numbers 7, 7 and 22, free and> clear, while you have 5 multiplied by these functions de'ned by the> roots of yet *another* polynomial. It's de'nitely not what you would expect to see in *any* math> textbook, I'd suppose. What I want to talk about now is this ongoing argument about how 49> multiplies through a factorization of 300125 x^3 - 18375 x^2 - 360 x + 22 where I'd think some of you would be dismissive of *anyone* trying to> claim that you can force 49 to multiply through in some particular> manner. After all, how do you choose? Consider 49 with a simpler factorization: 49(x+1)(x+2)(x+3) and, now then, how can you *force* factors of 49 to distribute? You can't. However, let's say I decided to multiply through and> showed you: 49(x+1)(x+2)(x+3) = (7x + 7)(7x + 14)(x+3)? Clearly the situation is different now. I've talked about taking the constant term, where with my example> here, you set x=0 to reveal constant terms 7, 14, and 3, but other> posters have continually challenged doing that claiming that it's a> special case, and when I talk about polynomials, like now, where> it's obvious it's NOT a special case, they claim that my not having a> polynomial factorization with (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = 49(300125 x^3 - 18375 x^2 - 360 x + 22) is the reason that the constants might no longer tell you where the> factors of 49 go. So then, a *reasonable* person might ask them, how then do *they*> claim to determine where the factors of 49 go? Notice, I'm not just asking where, but why. Why would factors of 49> go where these people claim, if they ever bother to ever give a> speci'c example. My position is that mathematicians have a problematic de'nition for> what's called the ring of algebraic integers, where members of that> ring are roots of monic polynomials with integer coef'cients. And my> factorization simply proves that the ring of algebraic integers is too> small to in general include numbers like a_1(x)/7 and a_2(x)/7, as x> varies over the ring of algebraic integers. Now then, not surprisingly, mathematicians would prefer to believe> that they don't have this de'nition that's too small, so I end up in> arguments. However time after time, I 'nd that people I argue with> don't rely on mathematics. For instance, that question I asked above, where would these people> claim that factors of 49 go? As I demonstrated with 49(x+1)(x+2)(x+3) = (7x + 7)(7x + 14)(x+3) you *can* with polynomials see where factors of 7 from 49 go with a> rather basic factorization, so I guess these people are claiming that> if the factorization of a polynomial is a non-standard one then the> constant terms within the factorization, found by setting x=0, can no> longer give you a window into what's happening mathematically. But that de'es mathematical consistency. These people remain vague on details, like refusing to give an example> for a particular x, like x=2, to say *explicitly* where factors of 49> go, while I say, look at the constant terms. So then, to summarize, I found these neat mathematical objects: (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = 49(300125 x^3 - 18375 x^2 - 360 x + 22) where b_3(x) = a_3(x) - 3 and the a's are roots of a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) and when x=0, a_1(0) = a_2(0) = b_3(0) = 0. So I can give all that rather succinctly, and explain the reasoning> how you can 'gure out how 49 multiplies through the factorization of 300125 x^3 - 18375 x^2 - 360 x + 22 with simpler examples like 49(x+1)(x+2)(x+3) = (7x + 7)(7x + 14)(x+3) but in response you have posters who challenge the logical, are vague> on the speci'cs, but who should have to answer basic questions> themselves for their bizarre math positions. For more of my mathematical work in this area you can go to my blog> archives: James Harris The problem here is so simple. You assume a factorization of your polynomial P(x) of the form(*) (5*a1(x) + 7)*(5*a2(x) + 7)*(5*b3(x) + 22). You note that since P(x) is divisible by 49, there must besome way to factor 49 out of the terms in (*). You thinkthat the only possible way such a factorization can occuris of the form 7, 7, 1. But you know that assuming a factorization of the form 7, 7, 1 leads to a contradiction. The contradictioncan be proved in two totally different ways: by using Galoistheory, and by using a basic theorem in algebraic numbertheory. You conclude from this contradiction that there is a problem with the algebraic integers. What you *should* conclude is that your claim that the onlyway the factorization can occur is 7, 7, 1 is wrong. Look hard. You do not really have a proof of that claim.You have a handwaving argument based on your concept ofconstant terms of functions. You think your argument that such a factorization wouldhave to be 7, 7, 1 is ironclad. It is based on the ideathat the ïconstant terms' in (*) are 7, 7, and 22, and in particular22 is coprime to 7. Therefore you conclude that the thirdfactor must be 1. You overlook the fact that when x <> 0, 5*b3(x) + 22 can be non-coprime to 7 even though 22 and 7 have no nonunit factors in common. We say you are wrong about 7, 7, 1. There is another way to factor 49 through the terms in (*). Dik Winter and others havedescribed it explicitly. It is not 7, 7, 1. It isw1(x), w2(x), w3(x), where each of these is a non-unitalgebraic integer factor of 49. None of w1, w2, or w3 areequal to 7. In particular, for example, 5*b3(x) + 22 is divisible by w3(x). No contradiction arises. The arithmetic works out. The constant terms multiple as they should to 49*22. The real root of the problem is that you are doing factorization by inspection, as in high school. You see thoseconstant terms 7, 7, and 22. The temptation to conclude that the factorization of 49 must be as 7, 7, 1 is overwhelming. You cannot imagine it could be any otherway. Your many simplistic examples involving reduciblepolynomials work out as you expect. How could it beotherwise in this case? It is! The key thing here is that for x <> 0 in general,the polynomials involved are irreducible. They simply donot act like the ones in the high-school textbooks. Theyfactor in ways that you do not expect. In some ways theirproperties can defy intuition. A key fact is that 7is not a prime in the algebraic integers. It can be factored in in'nitely many different ways. One suchfactor is w1(x) which Dik has speci'ed exactly as: w1(x) = gcd(a1(x), 49),where a1(x) is a root of (**) a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x).This is an explicit speci'cation in the sense that (1) the roots of the cubic (**) can be computed explicitly, and (2) there is a speci'c algorithm for computing the gcdfunction in the algebraic integers (due to Dedekind). Your factorization leads you to a contradiction. Whatthat should tell you is that you have made a mistake. Neverthelessyou perversely cling to it and refuse to see that anotherfactorization is possible. Our factorization has no such problem. It makes sense and is consistent with existing theory. The factorization itself is dependent onthe value of x. This should not be surprising, sincethe values of a1(x), a2(x) and b3(x) are dependent on x.In no way is this weird or bizarre. math can take you only so far. In this case, not quite farenough. Nora B. => I've ended up in a lot of hostile exchanges where clearly I'm angry,> and clearly people posting are angry, and it seems to me then that the> math gets lost, so I'm going to make a real effort not to post in> anger. With that said, I can move on to considering some math positions which> I say are weird. These positions have been revealed in discussions over some of my> mathematical 'ndings like the following: (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = 49(300125 x^3 - 18375 x^2 - 360 x + 22) where b_3(x) = a_3(x) - 3 and the a's are roots of a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) and when x=0, a_1(0) = a_2(0) = b_3(0) = 0. What I have is a polynomial multiplied by 49 that I factor in a> non-standard way, where you can see some numbers 7, 7 and 22, free and> clear, while you have 5 multiplied by these functions de'ned by the> roots of yet *another* polynomial. It's de'nitely not what you would expect to see in *any* math> textbook, I'd suppose. What I want to talk about now is this ongoing argument about how 49> multiplies through a factorization of 300125 x^3 - 18375 x^2 - 360 x + 22 where I'd think some of you would be dismissive of *anyone* trying to> claim that you can force 49 to multiply through in some particular> manner. After all, how do you choose? Consider 49 with a simpler factorization: 49(x+1)(x+2)(x+3) and, now then, how can you *force* factors of 49 to distribute? You can't. However, let's say I decided to multiply through and> showed you: 49(x+1)(x+2)(x+3) = (7x + 7)(7x + 14)(x+3)? Clearly the situation is different now. I've talked about taking the constant term, where with my example> here, you set x=0 to reveal constant terms 7, 14, and 3, but other> posters have continually challenged doing that claiming that it's a> special case, and when I talk about polynomials, like now, where> it's obvious it's NOT a special case, they claim that my not having a> polynomial factorization with (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = 49(300125 x^3 - 18375 x^2 - 360 x + 22) is the reason that the constants might no longer tell you where the> factors of 49 go. So then, a *reasonable* person might ask them, how then do *they*> claim to determine where the factors of 49 go? Notice, I'm not just asking where, but why. Why would factors of 49> go where these people claim, if they ever bother to ever give a> speci'c example. My position is that mathematicians have a problematic de'nition for> what's called the ring of algebraic integers, where members of that> ring are roots of monic polynomials with integer coef'cients. And my> factorization simply proves that the ring of algebraic integers is too> small to in general include numbers like a_1(x)/7 and a_2(x)/7, as x> varies over the ring of algebraic integers. Now then, not surprisingly, mathematicians would prefer to believe> that they don't have this de'nition that's too small, so I end up in> arguments. However time after time, I 'nd that people I argue with> don't rely on mathematics. For instance, that question I asked above, where would these people> claim that factors of 49 go? As I demonstrated with 49(x+1)(x+2)(x+3) = (7x + 7)(7x + 14)(x+3) you *can* with polynomials see where factors of 7 from 49 go with a> rather basic factorization, so I guess these people are claiming that> if the factorization of a polynomial is a non-standard one then the> constant terms within the factorization, found by setting x=0, can no> longer give you a window into what's happening mathematically. But that de'es mathematical consistency. These people remain vague on details, like refusing to give an example> for a particular x, like x=2, to say *explicitly* where factors of 49> go, while I say, look at the constant terms. So then, to summarize, I found these neat mathematical objects: (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = 49(300125 x^3 - 18375 x^2 - 360 x + 22) where b_3(x) = a_3(x) - 3 and the a's are roots of a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) and when x=0, a_1(0) = a_2(0) = b_3(0) = 0. So I can give all that rather succinctly, and explain the reasoning> how you can 'gure out how 49 multiplies through the factorization of 300125 x^3 - 18375 x^2 - 360 x + 22 with simpler examples like 49(x+1)(x+2)(x+3) = (7x + 7)(7x + 14)(x+3) but in response you have posters who challenge the logical, are vague> on the speci'cs, but who should have to answer basic questions> themselves for their bizarre math positions. For more of my mathematical work in this area you can go to my blog> archives: James Harris The problem here is so simple. You assume a factorization > of your polynomial P(x) of the form (*) (5*a1(x) + 7)*(5*a2(x) + 7)*(5*b3(x) + 22).That's like saying that someone assumes that sqrt(2) is a factor of 2.Factorizations just exist. You note that since P(x) is divisible by 49, there must be> some way to factor 49 out of the terms in (*). You think> that the only possible way such a factorization can occur> is of the form 7, 7, 1. But you know that assuming a factorization And now you claim to know what I *think*?Who cares what I think, it's the mathematical argument that'srelevant.> of the form 7, 7, 1 leads to a contradiction. The contradiction> can be proved in two totally different ways: by using Galois> theory, and by using a basic theorem in algebraic number> theory.Huh?> You conclude from this contradiction that there is a > problem with the algebraic integers.Huh? What you *should* conclude is that your claim that the only> way the factorization can occur is 7, 7, 1 is wrong. Look hard. You do not really have a proof of that claim.> You have a handwaving argument based on your concept of> constant terms of functions.My post goes over important points in the actual argument.You on the other hand went past all the math, went to the end of *my*post and just started making statements.That's taking a bizarre math position, where rather than facing themathematics given in place, you go right past it, and then startasserting.James Harris problem here is so simple. You assume a factorization > of your polynomial P(x) of the form (*) (5*a1(x) + 7)*(5*a2(x) + 7)*(5*b3(x) + 22). That's like saying that someone assumes that sqrt(2) is a factor of 2. Factorizations just exist.> I am just summarizing what you have said. I am notdisagreeing with it, at least not at this point. You note that since P(x) is divisible by 49, there must be> some way to factor 49 out of the terms in (*). You think> that the only possible way such a factorization can occur> is of the form 7, 7, 1. But you know that assuming a factorization And now you claim to know what I *think*?> You have said it many times.> Who cares what I think, You have made a huge error. My interest is in trying toget you to realize it.>it's the mathematical argument that's> relevant.> Your mathematical argument is wrong as shown below.> of the form 7, 7, 1 leads to a contradiction. The contradiction> can be proved in two totally different ways: by using Galois> theory, and by using a basic theorem in algebraic number> theory. Huh?> What a strange thing to say. You yourself have said thatthe 7, 7, 1 factorization leads to the fact that a_1(x)/7cannot be an algebraic integer. If you do not think thatis a contradiction to what you expected, we have no disagreement. If on the other hand your incoherent swinelike grunt meansthat you actually don't know how to PROVE that a_1(x)/7 is not analgebraic integer, let me know. Both of the proofs are short and sweet.> You conclude from this contradiction that there is a > problem with the algebraic integers. Huh?> Have you totally lost your bearings? It is totallyclear from all your recent posts that you think thatthe fact that a_1(x)/7 is not an algebraic integer meansthat the a.i.'s are incomplete or some such nonsense,and that it is necessary for you to complete them byassuming a ring of objects. What you *should* conclude is that your claim that the only> way the factorization can occur is 7, 7, 1 is wrong. Look hard. You do not really have a proof of that claim.> You have a handwaving argument based on your concept of> constant terms of functions. My post goes over important points in the actual argument.> So does mine, virtually all of which you deleted. I appendit below so that others can perhaps understand why you didn'twant them to read it.> You on the other hand went past all the math, went to the end of *my*> post and just started making statements.> *Math* statements. Most of which you deleted. You spend moretime in responding to posts which are simply degrading insults.That must re§ect your preferences: social interaction versusmath.> That's taking a bizarre math position, where rather than facing the> mathematics given in place, you go right past it, and then start> asserting.> Asserting mathematical facts which you are now evading. Appended,the material you deleted.------------------------------------------------------ -------------------------> of the form 7, 7, 1 leads to a contradiction. The contradiction> can be proved in two totally different ways: by using Galois> theory, and by using a basic theorem in algebraic number> theory. You conclude from this contradiction that there is a > problem with the algebraic integers. What you *should* conclude is that your claim that the only> way the factorization can occur is 7, 7, 1 is wrong. Look hard. You do not really have a proof of that claim.> You have a handwaving argument based on your concept of> constant terms of functions. You think your argument that such a factorization would> have to be 7, 7, 1 is ironclad. It is based on the idea> that the ïconstant terms' in (*) are 7, 7, and 22, and in particular> 22 is coprime to 7. Therefore you conclude that the third> factor must be 1. You overlook the fact that when x <> 0, 5*b3(x) + 22 can be > non-coprime to 7 even though 22 and 7 have no nonunit factors > in common. We say you are wrong about 7, 7, 1. There is another way > to factor 49 through the terms in (*). Dik Winter and others have> described it explicitly. It is not 7, 7, 1. It is> w1(x), w2(x), w3(x), where each of these is a non-unit> algebraic integer factor of 49. None of w1, w2, or w3 are> equal to 7. In particular, for example, 5*b3(x) + 22 is > divisible by w3(x). No contradiction arises. The arithmetic > works out. The constant terms multiple as they should > to 49*22. The real root of the problem is that you are doing > factorization by inspection, as in high school. You see those> constant terms 7, 7, and 22. The temptation to conclude > that the factorization of 49 must be as 7, 7, 1 is > overwhelming. You cannot imagine it could be any other> way. Your many simplistic examples involving reducible> polynomials work out as you expect. How could it be> otherwise in this case? It is! The key thing here is that for x <> 0 in general,> the polynomials involved are irreducible. They simply do> not act like the ones in the high-school textbooks. They> factor in ways that you do not expect. In some ways their> properties can defy intuition. A key fact is that 7> is not a prime in the algebraic integers. It can be > factored in in'nitely many different ways. One such> factor is w1(x) which Dik has speci'ed exactly as: w1(x) = gcd(a1(x), 49), where a1(x) is a root of (**) a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). This is an explicit speci'cation in the sense that > (1) the roots of the cubic (**) can be computed explicitly, and > (2) there is a speci'c algorithm for computing the gcd> function in the algebraic integers (due to Dedekind). Your factorization leads you to a contradiction. What> that should tell you is that you have made a mistake. Nevertheless> you perversely cling to it and refuse to see that another> factorization is possible. Our factorization has no > such problem. It makes sense and is consistent with > existing theory. The factorization itself is dependent on> the value of x. This should not be surprising, since> the values of a1(x), a2(x) and b3(x) are dependent on x.> In no way is this weird or bizarre. math can take you only so far. In this case, not quite far> enough. > Nora B.------------------------------------------------------------ ------------------- James Harris is so simple. You assume a factorization > of your polynomial P(x) of the form (*) (5*a1(x) + 7)*(5*a2(x) + 7)*(5*b3(x) + 22). That's like saying that someone assumes that sqrt(2) is a factor of 2. Factorizations just exist. > I am just summarizing what you have said. I am not> disagreeing with it, at least not at this point.Yeah, like what if someone says, let's assume that 6 has 3 as afactor.Technically it's correct.But the use of the word assume looks strange there, as 3 IS a factorof 6.There's no need to assume anything.Now it seems to me that you're playing stupid when the obvious ispointed out to you. > You note that since P(x) is divisible by 49, there must be> some way to factor 49 out of the terms in (*). You think> that the only possible way such a factorization can occur> is of the form 7, 7, 1. But you know that assuming a factorization And now you claim to know what I *think*? You have said it many times.Then give a single quote.Here you're caught, yet again, but rather than acknowledging theobvious--you can't read minds--you instead toss out yet anotherstatement.So I challenge you to give a single quote. > Who cares what I think, > You have made a huge error. My interest is in trying to> get you to realize it.> I don't think so. For instance, a while back in response to yourNow then, why don't you try again and explain what you *really* aretrying to do?James Harris assume a factorization >> of your polynomial P(x) of the form>> (*) (5*a1(x) + 7)*(5*a2(x) + 7)*(5*b3(x) + 22).>> That's like saying that someone assumes that sqrt(2) is a factor of 2.>> Factorizations just exist.>> I am just summarizing what you have said. I am not>> disagreeing with it, at least not at this point.Yeah, like what if someone says, let's assume that 6 has 3 as a>factor.Technically it's correct.But the use of the word assume looks strange there, as 3 IS a factor>of 6.> Typically an expression can be factored lots of ways. You assumed afactorization of the speci'ed form. I see absolutely nothingwrong with my wording and I don't agree that it looks strange.In any case this is an utterly minor issue. >There's no need to assume anything.Now it seems to me that you're playing stupid when the obvious is>pointed out to you.> You're quibbling over a minor wording issue. Meanwhile youdelete and evade the underlying math below. >> You note that since P(x) is divisible by 49, there must be>> some way to factor 49 out of the terms in (*). You think>> that the only possible way such a factorization can occur>> is of the form 7, 7, 1. But you know that assuming a factorization >> And now you claim to know what I *think*?>> You have said it many times.Then give a single quote.> OK, how about one that you posted YESTERDAY in the threadMathematical consistency, courage ? The constant terms are 7, 7, and 22, but when 49 is divided off the resulting constant terms are 1, 1, and 22, which means from BASIC arithmetic, there's only one way to go. There is, in fact, only one way that can happen.>Here you're caught, yet again, but rather than acknowledging the>obvious--you can't read minds--you instead toss out yet another>statement.> See above. That was a direct, unedited quote. I can't read minds but I can read what you post. So can everyone else whosees this. >So I challenge you to give a single quote.> Done. See above. I await your explanation.>> Who cares what I think, >> You have made a huge error. My interest is in trying to>> get you to realize it.I don't think so. For instance, a while back in response to your> I will never, never engage in any nonpublic conversations withyou on any topic.>Now then, why don't you try again and explain what you *really* are>trying to do?> It's simple. I am trying to show you that your factorization of 49 in the 7, 7, 1 pattern is the WRONG FACTORIZATION when x <> 0. Needless to say you are not getting it. Appended again is my post which you keep deleting -clearly you don't want people to read it - and to which youhave not yet given any substantive mathematical response: == > The problem here is so simple. You assume a factorization > of your polynomial P(x) of the form (*) (5*a1(x) + 7)*(5*a2(x) + 7)*(5*b3(x) + 22). You note that since P(x) is divisible by 49, there must be> some way to factor 49 out of the terms in (*). You think> that the only possible way such a factorization can occur> is of the form 7, 7, 1. But you know that assuming a factorization > of the form 7, 7, 1 leads to a contradiction. The contradiction> can be proved in two totally different ways: by using Galois> theory, and by using a basic theorem in algebraic number> theory. You conclude from this contradiction that there is a > problem with the algebraic integers. What you *should* conclude is that your claim that the only> way the factorization can occur is 7, 7, 1 is wrong. Look hard. You do not really have a proof of that claim.> You have a handwaving argument based on your concept of> constant terms of functions. You think your argument that such a factorization would> have to be 7, 7, 1 is ironclad. It is based on the idea> that the ïconstant terms' in (*) are 7, 7, and 22, and in particular> 22 is coprime to 7. Therefore you conclude that the third> factor must be 1. You overlook the fact that when x <> 0, 5*b3(x) + 22 can be > non-coprime to 7 even though 22 and 7 have no nonunit factors > in common. We say you are wrong about 7, 7, 1. There is another way > to factor 49 through the terms in (*). Dik Winter and others have> described it explicitly. It is not 7, 7, 1. It is> w1(x), w2(x), w3(x), where each of these is a non-unit> algebraic integer factor of 49. None of w1, w2, or w3 are> equal to 7. In particular, for example, 5*b3(x) + 22 is > divisible by w3(x). No contradiction arises. The arithmetic > works out. The constant terms multiple as they should > to 49*22. The real root of the problem is that you are doing > factorization by inspection, as in high school. You see those> constant terms 7, 7, and 22. The temptation to conclude > that the factorization of 49 must be as 7, 7, 1 is > overwhelming. You cannot imagine it could be any other> way. Your many simplistic examples involving reducible> polynomials work out as you expect. How could it be> otherwise in this case? It is! The key thing here is that for x <> 0 in general,> the polynomials involved are irreducible. They simply do> not act like the ones in the high-school textbooks. They> factor in ways that you do not expect. In some ways their> properties can defy intuition. A key fact is that 7> is not a prime in the algebraic integers. It can be > factored in in'nitely many different ways. One such> factor is w1(x) which Dik has speci'ed exactly as: w1(x) = gcd(a1(x), 49), where a1(x) is a root of (**) a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). This is an explicit speci'cation in the sense that > (1) the roots of the cubic (**) can be computed explicitly, and > (2) there is a speci'c algorithm for computing the gcd> function in the algebraic integers (due to Dedekind). Your factorization leads you to a contradiction. What> that should tell you is that you have made a mistake. Nevertheless> you perversely cling to it and refuse to see that another> factorization is possible. Our factorization has no > such problem. It makes sense and is consistent with > existing theory. The factorization itself is dependent on> the value of x. This should not be surprising, since> the values of a1(x), a2(x) and b3(x) are dependent on x.> In no way is this weird or bizarre. math can take you only so far. In this case, not quite far> enough. > Nora B. here is so simple. You assume a factorization >> of your polynomial P(x) of the form>> (*) (5*a1(x) + 7)*(5*a2(x) + 7)*(5*b3(x) + 22).>> That's like saying that someone assumes that sqrt(2) is a factor of 2.>> Factorizations just exist.>> I am just summarizing what you have said. I am not>> disagreeing with it, at least not at this point.Yeah, like what if someone says, let's assume that 6 has 3 as a>factor.Technically it's correct.But the use of the word assume looks strange there, as 3 IS a factor>of 6.> Typically an expression can be factored lots of ways. You assumed a> factorization of the speci'ed form. I see absolutely nothing> wrong with my wording and I don't agree that it looks strange.> In any case this is an utterly minor issue. > A person might assume that they're sounding rational, and cogent, whenin fact they're sounding defensive and recalcitrant.You don't just assume a truth. I'll give you another example as I'm somewhat curious about your stateof mind.Let's say that I make it a point to say I assume that the sun willrise tomorrow.Isn't there some implication in that statement to you?There's no need to assume anything.Now it seems to me that you're playing stupid when the obvious is>pointed out to you.> You're quibbling over a minor wording issue. Meanwhile you> delete and evade the underlying math below.> Is that an assumption on your part?>> You note that since P(x) is divisible by 49, there must be>> some way to factor 49 out of the terms in (*). You think>> that the only possible way such a factorization can occur>> is of the form 7, 7, 1. But you know that assuming a factorization >> And now you claim to know what I *think*?>> You have said it many times.Then give a single quote.> OK, how about one that you posted YESTERDAY in the thread> Mathematical consistency, courage ? The constant terms are 7, 7, and 22, but when 49 is divided> off the resulting constant terms are 1, 1, and 22, which means> from BASIC arithmetic, there's only one way to go. There is, in fact, only one way that can happen.> Yup, now then, look back at your claim about what I thought.>Here you're caught, yet again, but rather than acknowledging the>obvious--you can't read minds--you instead toss out yet another>statement.> See above. That was a direct, unedited quote. I can't read > minds but I can read what you post. So can everyone else who> sees this. > Yup. Now then, look carefully back at what *you* said I thought, andI'm curious to hear how you feel that the quote you gave is saying thesame thing.Be quite detailed please.So I challenge you to give a single quote. > Done. See above. I await your explanation.> It's very simple. In fact, I've been rather detailed about myfactorization, and how it works.But you make statements that are attempts at trying to change thefacts.Like even though you gave that quote, you then didn't even bother toreconcile obvious differences!Like you have 7, 7, 1, while I talk of 7, 7 and 22.It's something I've watched before in your postings where you makelittle mistakes, big mistakes, and strange mistakes, but when it'spointed out, you always deny!!!Later, I catch you making the *same* mistakes, and with VERY longposts, as if part of your strategy is burying your behavior with a lotof verbiage.It's kind of interesting behavior, though bizarre.>> Who cares what I think, >> You have made a huge error. My interest is in trying to>> get you to realize it.I don't think so. For instance, a while back in response to your> I will never, never engage in any nonpublic conversations with> you on any topic.So now you *'nally* admit that you need an audience. Why do I haveto drag the truth out of you?You make claims that don't 't with your behavior. Now then, why don't you try again and explain what you *really* are>trying to do?> It's simple. I am trying to show you that your factorization of 49 > in the 7, 7, 1 pattern is the WRONG FACTORIZATION when x <> 0. Needless > to say you are not getting it. Now there's one of the errors YET AGAIN, and I have to admit that it'sa puzzle to me. Do you do that consciously? Or is there somethingdeeper and even more odd going on here? > Appended again is my post which you keep deleting -> clearly you don't want people to read it - and to which you> have not yet given any substantive mathematical response:No, that's not it. I've seen your posting pattern before, and foundthat after spending time and effort to correct your mistakes andexplain to you, you simply deny the bizarre, but now kind of interesting.James Harris => Typically an expression can be factored lots of ways. You assumed a> factorization of the speci'ed form. I see absolutely nothing> wrong with my wording and I don't agree that it looks strange.> In any case this is an utterly minor issue. > A person might assume that they're sounding rational, and cogent, when> in fact they're sounding defensive and recalcitrant. You don't just assume a truth. I'll give you another example as I'm somewhat curious about your state> of mind. Let's say that I make it a point to say I assume that the sun will> rise tomorrow. Isn't there some implication in that statement to you?The rising of the sun tomorrow is of no mathematical consequence.Your refusal to deal with Nora's mathematics on the basis of the mathematics is of consequence mathematically. It demonstrates that you have no interest in the mathematics as mathematics, but only as a tool to aggrandize your pitiful ego. =In sci.math, James Harris<3c65f87.0312310652.3fb872f6@ ended up in a lot of hostile exchanges where clearly I'm angry,>> and clearly people posting are angry, and it seems to me then that the>> math gets lost, so I'm going to make a real effort not to post in>> anger.>> With that said, I can move on to considering some math positions which>> I say are weird.>> These positions have been revealed in discussions over some of my>> mathematical 'ndings like the following:>> (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = >> 49(300125 x^3 - 18375 x^2 - 360 x + 22)>> where b_3(x) = a_3(x) - 3 and the a's are roots of>> a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)>> and when x=0, a_1(0) = a_2(0) = b_3(0) = 0.>> You do realize, of course, that one could just as easily write>> 154 = (x)(7)(7)(22)>> where x is an unknown number, and then wonder where the second 7 went. Huh? I think readers who have doubts about the situation should read> through this entire thread as I just did. My position is that the distributive property holds in general, and> that constants are in fact constant, which is an odd thing to have to> say, but I'm dealing with people with bizarre math positions.Yes it does -- but when are 5/7 * a_1(x) and 5/7 * a_2(x)algebraic integers? > James Harris-- #191, ewill3@earthlink.netIt's still legal to go .sigless. => In sci.math, James Harris> > ended up in a lot of hostile exchanges where clearly I'm angry,>> and clearly people posting are angry, and it seems to me then that the>> math gets lost, so I'm going to make a real effort not to post in>> anger.>> With that said, I can move on to considering some math positions which>> I say are weird.>> These positions have been revealed in discussions over some of my>> mathematical 'ndings like the following:>> (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = >> 49(300125 x^3 - 18375 x^2 - 360 x + 22)>> where b_3(x) = a_3(x) - 3 and the a's are roots of>> a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)>> and when x=0, a_1(0) = a_2(0) = b_3(0) = 0.>> You do realize, of course, that one could just as easily write>> 154 = (x)(7)(7)(22)>> where x is an unknown number, and then wonder where the second 7 went. Huh? I think readers who have doubts about the situation should read> through this entire thread as I just did. My position is that the distributive property holds in general, and> that constants are in fact constant, which is an odd thing to have to> say, but I'm dealing with people with bizarre math positions. Yes it does -- but when are 5/7 * a_1(x) and 5/7 * a_2(x)> algebraic integers?The major research 'nd is that in general they are not.I'll give an example I used recently in another thread.Consider 2. It's an integer, right? Now then, let's say I write:x^2 = 2and *prove* that x is NOT an integer!!!What have I just done?I've shown an element outside the ring of integers!!!That's what my result is about, while what I've done is morecomplicated as I've used a factorization of a polynomial intonon-polynomial factors, but it's still basically a decomposition,which shows that there's more out there beyond algebraic integers.Understand?James Harris =... > Yes it does -- but when are 5/7 * a_1(x) and 5/7 * a_2(x) > algebraic integers? > The major research 'nd is that in general they are not.Indeed. The suspicion is that it is only true (for integer x) whenx = 0. > I'll give an example I used recently in another thread. > Consider 2. It's an integer, right? Now then, let's say I write: > x^2 = 2 > and *prove* that x is NOT an integer!!! > What have I just done? > I've shown an element outside the ring of integers!!! > That's what my result is about, while what I've done is more > complicated as I've used a factorization of a polynomial into > non-polynomial factors, but it's still basically a decomposition, > which shows that there's more out there beyond algebraic integers.Well, that is already known a long time. What is so shocking aboutit? > Understand?No.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ =In sci.math, James sci.math, James Harris>> sci.math, James Harris> > ended up in a lot of hostile exchanges where clearly I'm angry,> and clearly people posting are angry, and it seems to me then that the> math gets lost, so I'm going to make a real effort not to post in> anger.>> With that said, I can move on to considering some math positions which> I say are weird.>> These positions have been revealed in discussions over some of my> mathematical 'ndings like the following:>> (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = >> 49(300125 x^3 - 18375 x^2 - 360 x + 22)>> where b_3(x) = a_3(x) - 3 and the a's are roots of>> a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)>> and when x=0, a_1(0) = a_2(0) = b_3(0) = 0.>> You do realize, of course, that one could just as easily write>> 154 = (x)(7)(7)(22)>> where x is an unknown number, and then wonder where the second 7 went.>> Huh?>> I think readers who have doubts about the situation should read>> through this entire thread as I just did.>> My position is that the distributive property holds in general, and>> that constants are in fact constant, which is an odd thing to have to>> say, but I'm dealing with people with bizarre math positions.>> Yes it does -- but when are 5/7 * a_1(x) and 5/7 * a_2(x)>> algebraic integers? The major research 'nd is that in general they are not. I'll give an example I used recently in another thread. Consider 2. It's an integer, right? Now then, let's say I write: x^2 = 2 and *prove* that x is NOT an integer!!! What have I just done? I've shown an element outside the ring of integers!!! That's what my result is about, while what I've done is more> complicated as I've used a factorization of a polynomial into> non-polynomial factors, but it's still basically a decomposition,> which shows that there's more out there beyond algebraic integers. Understand?So you've found a mildly interesting pair of sets of algebraic numbers,given a function argument (namely, x).Whoopee.Now how does this relate to your ring with units -1 and +1?Show me the webblog; presumably you have one. James Harris-- #191, ewill3@earthlink.netIt's still legal to go .sigless. =>> In sci.math, James Harris>> sci.math, James Harris> > ended up in a lot of hostile exchanges where clearly I'm angry,> and clearly people posting are angry, and it seems to me then that the> math gets lost, so I'm going to make a real effort not to post in> anger.>> With that said, I can move on to considering some math positions which> I say are weird.>> These positions have been revealed in discussions over some of my> mathematical 'ndings like the following:>> (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = >> 49(300125 x^3 - 18375 x^2 - 360 x + 22)>> where b_3(x) = a_3(x) - 3 and the a's are roots of>> a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)>> and when x=0, a_1(0) = a_2(0) = b_3(0) = 0.>> You do realize, of course, that one could just as easily write>> 154 = (x)(7)(7)(22)>> where x is an unknown number, and then wonder where the second 7 went.>> Huh?>> I think readers who have doubts about the situation should read>> through this entire thread as I just did.>> My position is that the distributive property holds in general, and>> that constants are in fact constant, which is an odd thing to have to>> say, but I'm dealing with people with bizarre math positions.>> Yes it does -- but when are 5/7 * a_1(x) and 5/7 * a_2(x)>> algebraic integers?The major research 'nd is that in general they are not.I'll give an example I used recently in another thread.Consider 2. It's an integer, right? Now then, let's say I write:x^2 = 2and *prove* that x is NOT an integer!!!What have I just done?I've shown an element outside the ring of integers!!!Wow! There's an element outside the ring of integers.>That's what my result is about, while what I've done is more>complicated as I've used a factorization of a polynomial into>non-polynomial factors, but it's still basically a decomposition,>which shows that there's more out there beyond algebraic integers.Understand?We all understand that not everything is an algebraic integer.What we don't understand is why you think that that's bignews (or why you think you're proving more than that.)>James Harris************************David C. Ullrich => In sci.math, James Harris>> sci.math, James Harris> > ended up in a lot of hostile exchanges where clearly I'm angry,> and clearly people posting are angry, and it seems to me then that the> math gets lost, so I'm going to make a real effort not to post in> anger.>> With that said, I can move on to considering some math positions which> I say are weird.>> These positions have been revealed in discussions over some of my> mathematical 'ndings like the following:>> (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = >> 49(300125 x^3 - 18375 x^2 - 360 x + 22)>> where b_3(x) = a_3(x) - 3 and the a's are roots of>> a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)>> and when x=0, a_1(0) = a_2(0) = b_3(0) = 0.>> You do realize, of course, that one could just as easily write>> 154 = (x)(7)(7)(22)>> where x is an unknown number, and then wonder where the second 7 went.>> Huh?>> I think readers who have doubts about the situation should read>> through this entire thread as I just did.>> My position is that the distributive property holds in general, and>> that constants are in fact constant, which is an odd thing to have to>> say, but I'm dealing with people with bizarre math positions.>> Yes it does -- but when are 5/7 * a_1(x) and 5/7 * a_2(x)>> algebraic integers?The major research 'nd is that in general they are not.I'll give an example I used recently in another thread.Consider 2. It's an integer, right? Now then, let's say I write:x^2 = 2and *prove* that x is NOT an integer!!!What have I just done?I've shown an element outside the ring of integers!!! Wow! There's an element outside the ring of integers.Yes sqrt(2) is outside the ring of integers. >That's what my result is about, while what I've done is more>complicated as I've used a factorization of a polynomial into>non-polynomial factors, but it's still basically a decomposition,>which shows that there's more out there beyond algebraic integers.Understand? We all understand that not everything is an algebraic integer.> What we don't understand is why you think that that's big> news (or why you think you're proving more than that.)> I think by We you mean you, and I don't think you really care aboutunderstanding any more, but simply can't stop obsessively replying tomy posts.James Harris =In sci.math, James Harris<3c65f87.0312310652.3fb872f6@ ended up in a lot of hostile exchanges where clearly I'm angry,>> and clearly people posting are angry, and it seems to me then that the>> math gets lost, so I'm going to make a real effort not to post in>> anger.>> With that said, I can move on to considering some math positions which>> I say are weird.>> These positions have been revealed in discussions over some of my>> mathematical 'ndings like the following:>> (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = >> 49(300125 x^3 - 18375 x^2 - 360 x + 22)>> where b_3(x) = a_3(x) - 3 and the a's are roots of>> a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)>> and when x=0, a_1(0) = a_2(0) = b_3(0) = 0.>> You do realize, of course, that one could just as easily write>> 154 = (x)(7)(7)(22)>> where x is an unknown number, and then wonder where the second 7 went. Huh? I think readers who have doubts about the situation should read> through this entire thread as I just did. My position is that the distributive property holds in general, and> that constants are in fact constant, which is an odd thing to have to> say, but I'm dealing with people with bizarre math positions.Constants are constant, yes. The question is: what does it mean todivide a non-integer by 7? > James Harris-- #191, ewill3@earthlink.netIt's still legal to go .sigless. => In sci.math, James Harris> > ended up in a lot of hostile exchanges where clearly I'm angry,>> and clearly people posting are angry, and it seems to me then that the>> math gets lost, so I'm going to make a real effort not to post in>> anger.>> With that said, I can move on to considering some math positions which>> I say are weird.>> These positions have been revealed in discussions over some of my>> mathematical 'ndings like the following:>> (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = >> 49(300125 x^3 - 18375 x^2 - 360 x + 22)>> where b_3(x) = a_3(x) - 3 and the a's are roots of>> a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)>> and when x=0, a_1(0) = a_2(0) = b_3(0) = 0.>> You do realize, of course, that one could just as easily write>> 154 = (x)(7)(7)(22)>> where x is an unknown number, and then wonder where the second 7 went. Huh? I think readers who have doubts about the situation should read> through this entire thread as I just did. My position is that the distributive property holds in general, and> that constants are in fact constant, which is an odd thing to have to> say, but I'm dealing with people with bizarre math positions. Constants are constant, yes. The question is: what does it mean to> divide a non-integer by 7?>Huh?James Harris = >I have given speci'c exmaples with other polynomials (that are easier >to handle) to show that for different x the factors of 49 go in >different places. The basic idea of the why comes from Galois theory, >which you are apparently unwilling (or unable) to understand. Arturo >has shown that given an irreducible, primitive cubic polynomial: > x^3 + a.x^2 + b.x + c >*all* the roots have a factor in common with c. > Well, I don't know if has shown is the right expression. It's > just baby Galois Theory applied to baby Algebraic Number > Theory...newsgroup. I think baby Galois Theory is still a bit too high for James.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ =>I have given speci'c exmaples with other polynomials (that are easier>to handle) to show that for different x the factors of 49 go in>different places. The basic idea of the why comes from Galois theory,>which you are apparently unwilling (or unable) to understand. Arturo>has shown that given an irreducible, primitive cubic polynomial:> x^3 + a.x^2 + b.x + c>*all* the roots have a factor in common with c. Well, I don't know if has shown is the right expression. It's> just baby Galois Theory applied to baby Algebraic Number> Theory...newsgroup. Fair enough; it just seemed like you were giving me too muchcredit. Perhaps has explained why, or has noted that...But the clari'cation that all roots have nontrivial common factorswith any proper divisor of c (excluding 1 and -1) was important.-- =It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidinmagidin@math.berkeley.edu... > Yup, you say so. You never really eacplain why. > I rely on the distributive property: a(b+c) = ab + ac, and on > constants being constant.Nope, you rely on the inverse of the distributive property, which is falsein general. The distributive property says: If a, b and c are elements of a ring then a(b+c) = ab + ac,you are assuming something like: When a and ab + ac are in a ring, a(b + c) = ab + ac should be true in that ring.The latter is false.However, before that assumption you make already a basic error. Youassume that some factor is divisible by 7: (a_1(x) + 7), which is *not*justi'ed.> These positions have been revealed in discussions over some of my> mathematical 'ndings like the following:> (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = > 49(300125 x^3 - 18375 x^2 - 360 x + 22)> where b_3(x) = a_3(x) - 3 and the a's are roots of> a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) > I have shown that when you replace 49x by y you get: > (5 a_1(y) + 7)(5 a_2(y) + 7)(5 b_3(y) + 22) = > (125 y^3 - 375 y^2 - 360 y + 1078) > where b_3(y) = a_3(y) - 3 and the a's are roots of > a^3 + 3(-1 + y)a^2 - (y^3 - 3 y^2 + 3y)> and when x=0, a_1(0) = a_2(0) = b_3(0) = 0. > Yup. Check. (And I not in passing that b_3(x) + 22 = a_3(x) + 7.) So you can get a similar factorisation when the original > polynomial is *not* divisible by 49. Now, when y is a multiple of > 7, the original polynomial is a multiple of 7. So, how does that > 7 divide among the three factors? Do you know? > 125 y^3 - 375 y^2 - 360 y + 1078 > shares values with > 49(300125 x^3 - 18375 x^2 - 360 x + 22), if y=49x. > So what is the signi'cance?It not only shares values in that case, it is *identical* in that case. > What I have is a polynomial multiplied by 49 that I factor in a> non-standard way, where you can see some numbers 7, 7 and 22, free and> clear, while you have 5 multiplied by these functions de'ned by the> roots of yet *another* polynomial. > And I factor a polynomial not multiplied by 49 in an exactly similar > way. > Why is that signi'cant to you?Your polynomial is identical to my polynomial when you change variable.My polynmial is only divisible for a few values of y (e.g. when it isa multiple of 49).> What I want to talk about now is this ongoing argument about how 49> multiplies through a factorization of> 300125 x^3 - 18375 x^2 - 360 x + 22> where I'd think some of you would be dismissive of *anyone* trying to> claim that you can force 49 to multiply through in some particular> manner. > I have given (using gcd's) explicit formula's to calculate the three > factors. You dismiss them simply out-of-hand. You never have shown > they are *wrong*. > Then give the factorization for x=2.I *have*. Please do the calculations involved to satisfy your needs.The factorisation I gave is: [(5 a_1(x) + 7)/w_1(x)] [(5 a_2(x) + 7)/w_2(x)] [(5 b_3(x) + 22)/w_3(x)]where I gave speci'c de'nitions for w_1, w_2 and w_3, and where I haveshown that the product is 49 for *all* algebraic integer x, and thatthe three factors are algebraic integers for *all* algebraic integer x.No need for a rede'nition of the concept algebraic integer. For somereason you appear to think that result is false, but you have failed tonotice any part in the de'nition of them that is wrong.Also, I have said before that actually doing the calculations is aformidable amount of work. > Now then, I say that at x=2, from the constant terms you have that two > factors are multiplied by 7 to give you > (5 a_1(2) + 7)(5 a_2(2) + 7), where you can *see* the constant 7, > while the third is not leaving 5 b_3(2) + 22. Why should you *see* the constant term? Take the function: (5 z(x) + 7)/w(x)where z(0) = 0 and w(x) = 1. The constant term is 7 (according toyour de'nition). Do we see it in (5 z(2) + 7)/w(2)?Moreover, what is the constant term of: cbrt(x + 1) + cbrt(x + 8)(where cbrt is taken to be the real cube root of its argument)? Dowe see that constant term?> After all, how do you choose?> Consider 49 with a simpler factorization:> 49(x+1)(x+2)(x+3)> and, now then, how can you *force* factors of 49 to distribute? > Non-sequitor. This one is factorisable over the integers for all > integer x. Your polynomial is not. Worse, your polynomial is > factorisable over the integers if and only if x = 0. > I'm relying on the distributive property: a(b+c) = ab + ac.*No* you are not. And if so please state in what way you are relyingon it here. > Are you claiming that it only applies to polynomials?You either do not read what I write or only skim over it. When you stayin the integers there are exactly 6 ways you can distribute 49 overthe factors, and in none of those you get a distribution among threefactors. When you are going to the algebraic integers there arein'nitely many ways to distribute 49 over the factors, and there areexactly also in'nitely many way to distribute over the three factors.So far so good. > You can't. However, let's say I decided to multiply through and> showed you:> 49(x+1)(x+2)(x+3) = (7x + 7)(7x + 14)(x+3)?> Clearly the situation is different now. > Yup, it is factorisable over the integers. > Which indicates that you're taking the position that the distributive > property applies here because they are polynomial factors.Where do you use the distributive property? O, in showing that 7(x + 1) = 7x + 7 and 7(x + 2) = 7x + 14.But this does *not* apply in your case. You have: 49(300125 x^3 - 18375 x^2 - 360 x + 22)and there you can not factor the second term in such a way. So it doesnot apply. You can probably factor it as: 300125 x^3 - 18375 x^2 - 360 x + 22 = (g_1(x) + 1)(g_2(x) + 1)(g_3(x) + 22)but you will 'nd that *none* of g_1 and g_2 are algebraic integer functionsfor all algebraic integer x.> I've talked about taking the constant term, where with my example> here, you set x=0 to reveal constant terms 7, 14, and 3, but other> posters have continually challenged doing that claiming that it's a> special case, and when I talk about polynomials, like now, where> it's obvious it's NOT a special case, they claim that my not having a> polynomial factorization with> (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = > 49(300125 x^3 - 18375 x^2 - 360 x + 22)> is the reason that the constants might no longer tell you where the> factors of 49 go. > Note my polynomial with y and its factorisation above? The constant > terms do not even tell you where a single factor 7 goes. > I never said they did. Why do you make that statement?You state that constant terms tell you where factors go. I have apolynomial in y (which is identical to yours with a change of variable)which is divisible by 7, for some values of y, but the constant termsdo not tell me anything. > What is the signi'cance of that polynomial to you?It is your polynomial with a change of variable.> So then, a *reasonable* person might ask them, how then do *they*> claim to determine where the factors of 49 go? > Yes, and a reasonable person would show how that is calculated. And > actually I provided the calculations needed. > Then produce for x=2.I *have* produced the way they should be calculated, and have shown thatthis gives exactly what is needed. If you doubt that either1. Show which step is wrong, and whyor2. Do the calculations and show they give a wrong result.*Moreover*, with a similar polynomial, and similar steps I have shownhow exactly factors of 49 did distribute. You have not yet shown whyyour polynomial is different. > I have given speci'c exmaples with other polynomials (that are easier > to handle) to show that for different x the factors of 49 go in > different places. The basic idea of the why comes from Galois theory, > which you are apparently unwilling (or unable) to understand. Arturo > has shown that given an irreducible, primitive cubic polynomial: > x^3 + a.x^2 + b.x + c > *all* the roots have a factor in common with c. > Huh? It sounds like you're claiming that while you've been unable to > produce for the polynomial *I* use, you claim that different > polynomials support your claim, and then you claim that it comes from > Galois Theory, make assertions about what *I* know, and then say that > Arturo has shown some result.Yup. Although I cannot say yet unable. Unwilling is more like it. Ihave said multiple times already that it involves long and tediouscalculations. > Can you relate all of those claims to the situation at hand?You claim something is true for your polynomial which is shown falsefor other polynomials. Nevertheless, you fail to state why yourpolynomial is special. But, eh, do you claim now that Arturo has*not* shown that result? Bizarre. > How do you get back to *my* polynomial? What speci'cally does > Arturo's work have to do with *my* polynomial? Can you actually give > a calculation to support your case at x=2, with *my* polynomial?I do not understand anymore what you are now wanting. You have saidthat indeed a_1(x)/7 and a_2(x)/7 are not algebraic integers. Moreoveryou have said they should be. It still remains a question why. Youclaim that exactly two of the factors should be divisible by 7, youfail to show in what way your polynomial is different from thepolynomial I gave where similar reasoning was clearly false. I haveshown a way how you could calculate the values involved, but you refuseto say which part of that is false, you merely request from me to dolong and tedious calculations (which you would not believe anyhow). > Yup. a_2(x)/7 and a_2(x)/7 are not algebraic integers in general. I > fail to see the problem. Division properties are erratic, even when > you stay within the integers. > Now it seems that division properties are erratic is supposed to > explain something, but what?That the constant term does not tell you *anything* about divisibilityof a term.> For instance, that question I asked above, where would these people> claim that factors of 49 go? > I have *shown* giving formula's including gcd's. > Then support your claim with a demonstration with x=7.Go through the formula's and sea where it lands you. > It is the difference between reducible polynomials and irreducible > polynomials. For x = 0, your polynomial is reducible, and you draw > from that conclusions for the irreducible case. > You're just making a claim, where's the math to back up your position?Look up Galois theory. It is about that amongst other things.> But that de'es mathematical consistency. > Nope, it has been shown with other polynomials that you indeed can not > draw conclusions for the irreducible case from the reducible case. > So your claim is that the distributive property no longer applies if > the polynomial is irreducible over rationals?Darn. I have said this already about two months before, you are usingthat property in a way it can not be used.> These people remain vague on details, like refusing to give an example> for a particular x, like x=2, to say *explicitly* where factors of 49> go, while I say, look at the constant terms. > I have given the formula's. It is up to you to calculate the gcd's. > My formula's work, provided Dedekind was correct. > Yet you can't give a demonstration at x=2.Neither can you falsify it. > Ok, I'll give a different example. Set 7x = z: > So then, to summarise, I found these neat mathematical objects: > (5 a_1(z) + 7)(5 a_2(z) + 7)(5 b_3(7) + 22) = > 7(6125 z^3 - 2625 x^2 - 360 z + 154) > where b_3(z) = a_3(z) - 3 and the a's are roots of > a^3 + 3(-1 + 7z)a^2 - 7(49 z^3 - 21 x^2 + 3z) > and when x=0, a_1(0) = a_2(0) = b_3(0) = 0.> So I can give all that rather succinctly, and explain the reasoning> how you can 'gure out how 49 multiplies through the factorization of > Pray explain how 7 multiplies through in my factorisation above. Well, I can assume that you have just a typo in what you presented, > but I'm not interested in giving you the bene't of the doubt. > Why don't you explain the signi'cance you see in your other > polynomials?They are your polynomials, with a change of variable.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ => ...> Yup, you say so. You never really eacplain why.> I rely on the distributive property: a(b+c) = ab + ac, and on> constants being constant. Nope, you rely on the inverse of the distributive property, which is false> in general. The distributive property says:> If a, b and c are elements of a ring then a(b+c) = ab + ac,> you are assuming something like:> When a and ab + ac are in a ring, a(b + c) = ab + ac should be true in> that ring.> The latter is false.Your statements are misleading and untrue.> However, before that assumption you make already a basic error. You> assume that some factor is divisible by 7: (a_1(x) + 7), which is *not*> justi'ed.Those are more false statements.James Harris = > ...> Yup, you say so. You never really eacplain why.> I rely on the distributive property: a(b+c) = ab + ac, and on> constants being constant. > Nope, you rely on the inverse of the distributive property, which is false > in general. The distributive property says: > If a, b and c are elements of a ring then a(b+c) = ab + ac, > you are assuming something like: > When a and ab + ac are in a ring, a(b + c) = ab + ac should be true in > that ring. > The latter is false. > Your statements are misleading and untrue.What part of these statements is misleading and/or untrue? > However, before that assumption you make already a basic error. You > assume that some factor is divisible by 7: (a_1(x) + 7), which is *not* > justi'ed. > Those are more false statements.What part is false?Oh, indeed, you state that indeed (a_1(x) + 7) should be divisible by7. However it has been shown again and again that the non-divisibilityis *no* problem.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ => ...> Yup, you say so. You never really eacplain why.> I rely on the distributive property: a(b+c) = ab + ac, and on> constants being constant. Nope, you rely on the inverse of the distributive property, which is false> in general. The distributive property says:> If a, b and c are elements of a ring then a(b+c) = ab + ac,> you are assuming something like:> When a and ab + ac are in a ring, a(b + c) = ab + ac should be true in> that ring.> The latter is false. Your statements are misleading and untrue. However, before that assumption you make already a basic error. You> assume that some factor is divisible by 7: (a_1(x) + 7), which is *not*> justi'ed. Those are more false statements. James HarrisSuppose we have ïa' = 49, ïb' = 3/7 and ïc' = 1. Then, for a(b + c) = ab + ac we have:49(3/7 + 1) = 21 + 49.Please explain how it is that ïa', ïab' and ïac' are *in the ring of algebraic integers* while neither'b' nor ïb + c' is in the ring. All quantities on the right hand side are algebraic integers, but notall quantities on the left hand side are algebraic integers.Otherwise please clarify how your use of the distributive property can be used to establish theproperties of the elements involved in the distribution. Your explanation would be much moreinformative than merely declaring some statement to be false.--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 = ...> Yup, you say so. You never really eacplain why.> I rely on the distributive property: a(b+c) = ab + ac, and on> constants being constant. Nope, you rely on the inverse of the distributive property, which is false> in general. The distributive property says:> If a, b and c are elements of a ring then a(b+c) = ab + ac,> you are assuming something like:> When a and ab + ac are in a ring, a(b + c) = ab + ac should be true in> that ring.> The latter is false. Your statements are misleading and untrue. However, before that assumption you make already a basic error. You> assume that some factor is divisible by 7: (a_1(x) + 7), which is *not*> justi'ed. Those are more false statements. James Harris Suppose we have ïa' = 49, ïb' = 3/7 and ïc' = 1. Then, for a(b + c) = ab + ac we have: 49(3/7 + 1) = 21 + 49. Please explain how it is that ïa', ïab' and ïac' are *in the ring of algebraic integers* while neither> ïb' nor ïb + c' is in the ring. All quantities on the right hand side are algebraic integers, but not> all quantities on the left hand side are algebraic integers.No. Otherwise please clarify how your use of the distributive property can be used to establish the> properties of the elements involved in the distribution. Your explanation would be much more> informative than merely declaring some statement to be false.> No.James Harris 49, ïb' = 3/7 and ïc' = 1. Then, for a(b + c) = ab + ac we have: 49(3/7 + 1) = 21 + 49. Please explain how it is that ïa', ïab' and ïac' are *in the ring of algebraic integers* while neither> ïb' nor ïb + c' is in the ring. All quantities on the right hand side are algebraic integers, but not> all quantities on the left hand side are algebraic integers. No.> Otherwise please clarify how your use of the distributive property can be used to establish the> properties of the elements involved in the distribution. Your explanation would be much more> informative than merely declaring some statement to be false.> No. James HarrisVERY REVEALING! (The only logical explanation is that you CANNOT do so.)--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 = [.snip.]> I have given speci'c exmaples with other polynomials (that are easier> to handle) to show that for different x the factors of 49 go in> different places. The basic idea of the why comes from Galois theory,> which you are apparently unwilling (or unable) to understand. Arturo> has shown that given an irreducible, primitive cubic polynomial:> x^3 + a.x^2 + b.x + c> *all* the roots have a factor in common with c. [.snip.]>Yup. Although I cannot say yet unable. Unwilling is more like it. I>have said multiple times already that it involves long and tedious>calculations. [.snip.]>You claim something is true for your polynomial which is shown false>for other polynomials. Nevertheless, you fail to state why your>polynomial is special. But, eh, do you claim now that Arturo has>*not* shown that result? Bizarre. [.snip.]>I do not understand anymore what you are now wanting. You have said>that indeed a_1(x)/7 and a_2(x)/7 are not algebraic integers. Moreover>you have said they should be. It still remains a question why. You>claim that exactly two of the factors should be divisible by 7, you>fail to show in what way your polynomial is different from the>polynomial I gave where similar reasoning was clearly false. I have>shown a way how you could calculate the values involved, but you refuse>to say which part of that is false, you merely request from me to do>long and tedious calculations (which you would not believe anyhow).The obvious way to relate this to the situation at hand is to notethat since the a_i(x) are de'ned as being roots of a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)then for any integer value of x for which this polynomial isirreducible Q, since the constant term is divisible by 7, all roots willbe divisible by 7. When x=0, we havea^3 -3a^2which is reducible over Q, so the result that all roots will havenontrivial algebraic integer factors in common with any rational primedividing the constant term does not hold.When x=2, we havea^3 + 3(-1+98)a^2 - 40 (2401*8 - 147*4 + 6) = a^3 +291*a^2 - 745040If this is irreducible over Q, then all roots (hence a_1(2), a_2(2),and a_3(2)) will all have nontrivial common factors in the algebraicintegers with 7.-- =It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidinmagidin@math.berkeley.edu give an example> for a particular x, like x=2, to say *explicitly* where factors of 49> go, while I say, look at the constant terms. On the contrary, it is *you* who refuses to give an example for a> particular x, like x=2, preferring instead to say look at the constant> terms. Many readers *have* looked at the so-called ïconstant terms'> (coef'cients of the x^n term where n=0), and found that they have no> general application to the value of the expression which results when x is> not 0. Constant terms aren't dependent on the value of x, which is my point. No one ever disputed that point. You completely misunderstood the criticisms. You yourself put in your post when x is not 0, which indicates that> you're making a claim dependent on x's value. But constant terms> aren't dependent on x's value. The quoted passage refers to the value of the expression when x is not 0. I did *not* say> anything about the value of the constants in the expression. You have, again, misrepresented the> criticism. The expression contains constants *and* terms dependent on x.> Well, that's an accusation that can be gone over rather thoroughly.You statement:Many readers *have* looked at the so-called ïconstant terms'(coef'cients of the x^n term where n=0), and found that they have nogeneral application to the value of the expression which results whenx is not 0.It sounds to me like you're mentioning the value of x, speci'callyits value when not 0, as being relevant. However, if *constants* arebeing discussed how can you possibly believe that the value of x isrelevant?That is, why care if x is not 0, or if it is, if *constants* are beingdiscussed?Now in fact the factorization is(5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = 49(300125 x^3 - 18375 x^2 - 360 x + 22)where b_3(x) = a_3(x) - 3 and the a's are roots ofa^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)and when x=0, a_1(0) = a_2(0) = b_3(0) = 0.The constant terms here are 7, 7 and 22, which are in the factors(5 a_1(x) + 7), (5 a_2(x) + 7), and (5 b_3(x) + 22)and in fact, since you worry about x not zero, I can give x=2, to have(5 a_1(2) + 7)(5 a_2(2) + 7)(5 b_3(2) + 22) = 49(300125 2^3 - 18375 2^2 - 360 (2) + 22).Now then, it should be quite clear what *I* am talking about when Italk about constant terms.James Harris vague on details, like refusing to give an example> for a particular x, like x=2, to say *explicitly* where factors of 49> go, while I say, look at the constant terms. On the contrary, it is *you* who refuses to give an example for a> particular x, like x=2, preferring instead to say look at the constant> terms. Many readers *have* looked at the so-called ïconstant terms'> (coef'cients of the x^n term where n=0), and found that they have no> general application to the value of the expression which results when x is> not 0. Constant terms aren't dependent on the value of x, which is my point. No one ever disputed that point. You completely misunderstood the criticisms. You yourself put in your post when x is not 0, which indicates that> you're making a claim dependent on x's value. But constant terms> aren't dependent on x's value. The quoted passage refers to the value of the expression when x is not 0. I did *not* say> anything about the value of the constants in the expression. You have, again, misrepresented the> criticism. The expression contains constants *and* terms dependent on x.> Well, that's an accusation that can be gone over rather thoroughly. You statement: Many readers *have* looked at the so-called ïconstant terms'> (coef'cients of the x^n term where n=0), and found that they have no> general application to the value of the expression which results when> x is not 0. It sounds to me like you're mentioning the value of x, speci'cally> its value when not 0, as being relevant. However, if *constants* are> being discussed how can you possibly believe that the value of x is> relevant? That is, why care if x is not 0, or if it is, if *constants* are being> discussed?Simple. You made claims about the properties of the expressions between parentheses which apply to thegeneral case in which x is not 0. It is *your* claims about those properties which you use to supportyour arguments. If the value of x has no signi'cance in your further arguments, why fabricate anarbitrary polynomial in x to begin your ïproof'? After all, if you are simply going to set x to zeroand derive important properties from there, set it to zero to begin with.> Now in fact the factorization is (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = 49(300125 x^3 - 18375 x^2 - 360 x + 22) where b_3(x) = a_3(x) - 3 and the a's are roots of a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) and when x=0, a_1(0) = a_2(0) = b_3(0) = 0. The constant terms here are 7, 7 and 22, which are in the factors (5 a_1(x) + 7), (5 a_2(x) + 7), and (5 b_3(x) + 22) and in fact, since you worry about x not zero, I can give x=2, to have (5 a_1(2) + 7)(5 a_2(2) + 7)(5 b_3(2) + 22) = 49(300125 2^3 - 18375 2^2 - 360 (2) + 22). Now then, it should be quite clear what *I* am talking about when I> talk about constant terms.Everyone knows what a constant is. What's your point?--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 remain vague on details, like refusing to give an example> for a particular x, like x=2, to say *explicitly* where factors of 49> go, while I say, look at the constant terms. On the contrary, it is *you* who refuses to give an example for a> particular x, like x=2, preferring instead to say look at the constant> terms. Many readers *have* looked at the so-called ïconstant terms'> (coef'cients of the x^n term where n=0), and found that they have no> general application to the value of the expression which results when x is> not 0. Constant terms aren't dependent on the value of x, which is my point. No one ever disputed that point. You completely misunderstood the criticisms. You yourself put in your post when x is not 0, which indicates that> you're making a claim dependent on x's value. But constant terms> aren't dependent on x's value. The quoted passage refers to the value of the expression when x is not 0. I did *not* say> anything about the value of the constants in the expression. You have, again, misrepresented the> criticism. The expression contains constants *and* terms dependent on x.> Well, that's an accusation that can be gone over rather thoroughly. You statement: Many readers *have* looked at the so-called ïconstant terms'> (coef'cients of the x^n term where n=0), and found that they have no> general application to the value of the expression which results when> x is not 0. It sounds to me like you're mentioning the value of x, speci'cally> its value when not 0, as being relevant. However, if *constants* are> being discussed how can you possibly believe that the value of x is> relevant? That is, why care if x is not 0, or if it is, if *constants* are being> discussed? Simple. You made claims about the properties of the expressions between parentheses which apply to the> general case in which x is not 0. It applies to ALL values of x within the ring of algebraic integers!!!Now then can you see in any small way that you continually make upthings to try and make your points?*You* made a claim that mentioned the value of x, while I've pointedout that x's value is irrelevant to constants.When I pointed out what you were doing, you accused me of malice!So I focused on your own statements yet again, and again pointed outthat *you* are making a claim relying on the value of x, and in replyyou make a claim insinuating that *I* am the one!!!The reality is that the value of x doesn't matter to the constantterms as they are just that, constant.>It is *your* claims about those properties which you use to support> your arguments. If the value of x has no signi'cance in your further arguments, why fabricate an> arbitrary polynomial in x to begin your ïproof'? After all, if you are simply going to set x to zero> and derive important properties from there, set it to zero to begin with.Consider a polynomial and its factorization like,x^2 + 5x + 6 = (x+2)(x+3)where on the left you have the constant term 6, and on the right youhave the constant terms 2 and 3.Now then, let's say someone starts getting upset if you say that you'nd the constant terms by setting x=0, as that reveals the constantterms, and then when you repeatedly remind them that the value of x isirrelevant, they then start hollering about having x at all!My point is that the constant terms are just that, constant. Thevalue of x is irrelevant to the value of the constant terms, however,given an expression where you're not sure what the constant terms are,you can 'nd them by setting x=0.It's not complicated, and using that technique does not changeconstants into variables!!! > Now in fact the factorization is (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = 49(300125 x^3 - 18375 x^2 - 360 x + 22) where b_3(x) = a_3(x) - 3 and the a's are roots of a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) and when x=0, a_1(0) = a_2(0) = b_3(0) = 0. The constant terms here are 7, 7 and 22, which are in the factors (5 a_1(x) + 7), (5 a_2(x) + 7), and (5 b_3(x) + 22) and in fact, since you worry about x not zero, I can give x=2, to have (5 a_1(2) + 7)(5 a_2(2) + 7)(5 b_3(2) + 22) = 49(300125 2^3 - 18375 2^2 - 360 (2) + 22). Now then, it should be quite clear what *I* am talking about when I> talk about constant terms. Everyone knows what a constant is. What's your point?Constant terms are constant. The number 7 is a constant and it's nota variable dependent on x, so it's not relevant, when constant termsare being discussed to claim that the value of x is important.Also notice that to help you out even more, I used x=2, so that youcan *still* see the constant terms.The mathematics here is rigid C. Bond, as constants are constant, sothe value of x is irrelevant to them.James Harris Consider a polynomial and its factorization like, x^2 + 5x + 6 = (x+2)(x+3) where on the left you have the constant term 6, and on the right you> have the constant terms 2 and 3. Now then, let's say someone starts getting upset if you say that you> 'nd the constant terms by setting x=0, as that reveals the constant> terms, and then when you repeatedly remind them that the value of x is> irrelevant, they then start hollering about having x at all!Well, why have an x at all? What's the point of constructing a polynomial simply to show that setting thepolynomial variable to zero reduces its value to the constant term? Everyone knows you can evaluate apolynomial at x = 0 and the surviving term will be the constant. But you keep making the point that'constants are constant' (see below) when no one has challenged that. Is that all you have to offer?> My point is that the constant terms are just that, constant. The> value of x is irrelevant to the value of the constant terms, however,> given an expression where you're not sure what the constant terms are,> you can 'nd them by setting x=0. It's not complicated, and using that technique does not change> constants into variables!!!Of course not. No one said Constant terms are constant. The number 7 is a constant and it's not> a variable dependent on x, so it's not relevant, when constant terms> are being discussed to claim that the value of x is important.No one, including me, did that. You cannot 'nd any quotable passage from any of my posts which suggest thatthe values of the constants depend on the value of ïx'. My claim was that the values of the expressionsbetween parentheses, which include functions of x plus a constant, will vary with ïx' and those values maynot have the same divisibility properties as those of the constant alone. Please stop insisting that I havesomehow claimed that constants vary.> Also notice that to help you out even more, I used x=2, so that you> can *still* see the constant terms. The mathematics here is rigid C. Bond, as constants are constant, so> the value of x is irrelevant to them.What's your problem, James Harris? If the purpose of all your arguments and derivations is simply toestablish that constants are constant, you needn't have bothered. That has never been an issue. The bone ofcontention here is whether you have proven that there are numbers which *should* be algebraic integers, butwhich are not. Proclaiming (correctly) that constants are constant is not enough to establish that.Somehow I thought that your so-called ïadvanced polynomial factorization' had some higher goal than simplyshowing that constants are constant. If not, you've wasted even more bandwidth than I though.--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 =>Message-id: signi'cance?...>Why is that signi'cant to you?...>Why do you make that statement?...>Can you be more speci'c ...>That's my personal concern, not yours....James HarrisHas James been replaced by an Eliza program?--MensanatorAce of Clubs = I've ended up in a lot of hostile exchanges where clearly I'm angry,> and clearly people posting are angry, and it seems to me then that the> math gets lost, so I'm going to make a real effort not to post in> anger. I welcome the chance to get back to a discussion of mathematics. I think Mr. Harris is observing something, namely, non-uniqueness of> factorization. This is not new, but it certainly is interesting, being> in fact a major concern of algebraic number theory since the nineteenth> century.Well, I *do* depend on non-uniqueness of *polynomial* factorization,meaning that polynomials can be factored in an in'nite number ways.I'd be surprised if anyone else has presented work that so depends.But if you know of anyone--any mathematician in recorded history--whohas presented work dependent on non-uniqueness of polynomialfactorization, it might be helpful if you'd give a citation. > With that said, I can move on to considering some math positions which> I say are weird. These positions have been revealed in discussions over some of my> mathematical 'ndings like the following: (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = 49(300125 x^3 - 18375 x^2 - 360 x + 22) where b_3(x) = a_3(x) - 3 and the a's are roots of a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) and when x=0, a_1(0) = a_2(0) = b_3(0) = 0. Let us stipulate that the equations are valid. (And let us agree that> the distributive law is not controversial.) The question remains: so> what? Mr. Harris says that the polynomial is being factored in a> non-standard way. I think I know what this means. Here: a_1(x), a_2(x), and b_3(x), along with x itself, may all be understood> as elements of an algebraic function 'eld. Speci'cally, it is the> splitting 'eld of a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) = 0 over the 'eld Q(x) , x being an indeterminate . The elements a_1,> a_2, a_3, and x generate a ring, within which unique factorization does> not seem to hold. speci'cally, in this ring, the element 14706125 x^3 - 900375 x^2 - 17640 x + 1078 can be factored in two ways. One is (A) (7)(7)(300125x^3-18375x^2-360x+22) and the other is> (B) (5a_1+7)(5a_2+7)(5a_3+7). Parenthetical note: 5a_3+7 = 5b_3+22. I don't know, but I think it probable, that there is no way to break> down these factors further within this ring. However, look at the full expression:5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = 49(300125 x^3 - 18375 x^2 - 360 x + 22)It's not a great leap to consider dividing both sides by 49, since therather basic polynomial on the right side has 49 as a factor.Now I emphasize that point because it's possible to prove rathereasily that the result pushes you out of the ring of algebraicintegers. > What I have is a polynomial multiplied by 49 that I factor in a> non-standard way, where you can see some numbers 7, 7 and 22, free and> clear, while you have 5 multiplied by these functions de'ned by the> roots of yet *another* polynomial. It's de'nitely not what you would expect to see in *any* math> textbook, I'd suppose. Look for textbooks on algebraic function 'elds. Come back with> titles and authors; someone here will have an informed opinion as to> which ones are best.I've done better by putting my work on Usenet, communicating withMcKenzie at Vanderbilt University.The results of extensive research on my part over a period now ofalmost two years is that there are no mathematicians to cite.And again, it might be useful if *anyone* can cite resarch thatexplores the non-uniqueness of polynomial factorization. > A good place to start is the paper on Gauss's Lemma accessible from> http://www.math.umt.edu/magidin/preprints/preprints.html .> Why?I'm actually rather curious about your reference, and mathematical work in this area you can go to my blog> archives:> Neat link. Check it out.James Harris = I've ended up in a lot of hostile exchanges where clearly I'm angry,> and clearly people posting are angry, and it seems to me then that the> math gets lost, so I'm going to make a real effort not to post in> anger. I welcome the chance to get back to a discussion of mathematics. I think Mr. Harris is observing something, namely, non-uniqueness of> factorization. This is not new, but it certainly is interesting, being> in fact a major concern of algebraic number theory since the nineteenth> century. Well, I *do* depend on non-uniqueness of *polynomial* factorization,> meaning that polynomials can be factored in an in'nite number ways. I'd be surprised if anyone else has presented work that so depends. But if you know of anyone--any mathematician in recorded history--who> has presented work dependent on non-uniqueness of polynomial> factorization, it might be helpful if you'd give a citation.such as Z[a_1, a_2, a_3, x] which I suggested are not Dedekind domains. Therefore factorization questions in these rings are likely to be hard,with unsatisfying answers. I think this is why people don't work much with factorization inalgebraic function 'elds.Non-unique factorization in alebraic *number* 'elds (more precisely,in the rings of integers within such 'elds)... expression: 5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = > 49(300125 x^3 - 18375 x^2 - 360 x + 22) It's not a great leap to consider dividing both sides by 49, since the> rather basic polynomial on the right side has 49 as a factor. Now I emphasize that point because it's possible to prove rather> easily that the result pushes you out of the ring of algebraic> integers.> I suppose it does, but I don't see that this is important. > What I have is a polynomial multiplied by 49 that I factor in a> non-standard way, where you can see some numbers 7, 7 and 22, free and> clear, while you have 5 multiplied by these functions de'ned by the> roots of yet *another* polynomial. It's de'nitely not what you would expect to see in *any* math> textbook, I'd suppose. Look for textbooks on algebraic function 'elds. Come back with> titles and authors; someone here will have an informed opinion as to> which ones are best. I've done better by putting my work on Usenet, communicating with> McKenzie at Vanderbilt University. The results of extensive research on my part over a period now of> almost two years is that there are no mathematicians to cite. And again, it might be useful if *anyone* can cite resarch that> explores the non-uniqueness of polynomial factorization.> A good place to start is the paper on Gauss's Lemma accessible from> http://www.math.umt.edu/magidin/preprints/preprints.html . > Why?It's informative, and well written. It brie§y describes an importantpart of the history of mathematics, some of which is motivated byefforts to prove FLT. And it's not the ïdead past by any means I'm actually rather curious about your reference, and request that you> give more details.You can see for in this area you can go to my blog> archives: > Neat link. Check it out.I just did. I see the phenomenon in question, which is that (5 a_1(x)/7 + 1) etc. are probably not algebraic integers for allnon-zero integer x. But this does not seem to me to be a problem. 7is not a unit in the ring of algebraic integers, so it is possible fordivision by 7 to lead from inside that ring to outside. We'd beembarrassed as hell if we found ourselves outside te *'eld* ofalgebraic *numbers*, but that doesn't seem to be happening here.The case x=0, where you were able to take a_1(x) = a_2(x) = 0, makesdivision easy. But is is a very special case. > James Harris-- Chris HenrichBallerinas are always on their toes. Why don't they just get tallerballerinas? -- (?)Joshua Burton = > I've ended up in a lot of hostile exchanges where clearly I'm angry,> and clearly people posting are angry, and it seems to me then that the> math gets lost, so I'm going to make a real effort not to post in> anger. I welcome the chance to get back to a discussion of mathematics. I think Mr. Harris is observing something, namely, non-uniqueness of> factorization. This is not new, but it certainly is interesting, being> in fact a major concern of algebraic number theory since the nineteenth> century. Well, I *do* depend on non-uniqueness of *polynomial* factorization,> meaning that polynomials can be factored in an in'nite number ways. I'd be surprised if anyone else has presented work that so depends. But if you know of anyone--any mathematician in recorded history--who> has presented work dependent on non-uniqueness of polynomial> factorization, it might be helpful if you'd give a citation.> such as Z[a_1, a_2, a_3, x] which I suggested are not Dedekind domains. > Therefore factorization questions in these rings are likely to be hard,> with unsatisfying answers. So? > I think this is why people don't work much with factorization in> algebraic function 'elds. Non-unique factorization in alebraic *number* 'elds (more precisely,> in the rings of integers within such 'elds)... now that *is*> history.Why? So far what you're saying lacks relevance. I'm at a loss tryingto understand what your point is.My work *depends* on there being non-uniqueness of polynomialfactorization, which I call factorizations of polynomials intonon-polynomial factors.I really doubt that *any* mathematicians out there have published workin that area, but think it'd be VERY INTERESTING if any do!!!So if you know of any, I think it would be helpful if you provided expression: 5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = > 49(300125 x^3 - 18375 x^2 - 360 x + 22) It's not a great leap to consider dividing both sides by 49, since the> rather basic polynomial on the right side has 49 as a factor. Now I emphasize that point because it's possible to prove rather> easily that the result pushes you out of the ring of algebraic> integers.> I suppose it does, but I don't see that this is important. Well I have an example for you!!!You know that 2 is an integer, right? Well let's say just for fun Iwrite:x^2 = 2and then people argue and 'ght and go to a lot of trouble, but manageto prove that x is NOT an integer!!!Now then, if that were new information, are you saying that's NOT abig deal?My point of course is that mathematical progress comes about whenpeople make 'nds like mine, as before, when people discoveredsqrt(2), it was kind of a big deal!Or would you disagree? > What I have is a polynomial multiplied by 49 that I factor in a> non-standard way, where you can see some numbers 7, 7 and 22, free and> clear, while you have 5 multiplied by these functions de'ned by the> roots of yet *another* polynomial. It's de'nitely not what you would expect to see in *any* math> textbook, I'd suppose. Look for textbooks on algebraic function 'elds. Come back with> titles and authors; someone here will have an informed opinion as to> which ones are best. I've done better by putting my work on Usenet, communicating with> McKenzie at Vanderbilt University. The results of extensive research on my part over a period now of> almost two years is that there are no mathematicians to cite. And again, it might be useful if *anyone* can cite resarch that> explores the non-uniqueness of polynomial factorization.> A good place to start is the paper on Gauss's Lemma accessible from> http://www.math.umt.edu/magidin/preprints/preprints.html . > Why?> It's informative, and well written. It brie§y describes an important> part of the history of mathematics, some of which is motivated by> efforts to prove FLT. And it's not the ïdead past by any meansSo? In case you missed it this thread is about my *current*mathematical research and bizarre math positions that some people havetaken in trying to deny it.What's with you and mentioning this Magidin paper? What's yourmotivation? I'm actually rather curious about your reference, and request that you> give more details. You can see for yourself from the URL. Why should *I* go through that effort!!!You're being remarkably dif'cult. *You* gave a link. *You* seem tothink it relevant! But when pressed as to why, you want *me* to go tomore effort?What's your reasoning here? go to my blog> archives: > Neat link. Check it out.> I just did. I see the phenomenon in question, which is that > (5 a_1(x)/7 + 1) etc. are probably not algebraic integers for all> non-zero integer x. But this does not seem to me to be a problem. 7> is not a unit in the ring of algebraic integers, so it is possible for> division by 7 to lead from inside that ring to outside. We'd be> embarrassed as hell if we found ourselves outside te *'eld* of> algebraic *numbers*, but that doesn't seem to be happening here.Don't go running off into 'elds!!! The case x=0, where you were able to take a_1(x) = a_2(x) = 0, makes> division easy. But is is a very special case. Why? I'm curious about that statement, as in fact, mentioning thevalue of x, when talking about *constants* like 7 and 22 is a bizarremath position to take!James Harris *You* gave a link. *You* seem to> think it relevant! But when pressed as to why, you want *me* to go to> more effort?> What's your reasoning here?>Samuel Johnson -- Sir, I have found you an argument, I cannot give you an understanding.--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 remarkably dif'cult. *You* gave a link. *You* seem to> think it relevant! But when pressed as to why, you want *me* to go to> more effort?> What's your reasoning here?> Samuel Johnson -- Sir, I have found you an argument, I cannot give you an understanding.-->There's nothing of mathematical interest in your post.James Harris remarkably dif'cult. *You* gave a link. *You* seem to>> think it relevant! But when pressed as to why, you want *me* to go to>> more effort? What's your reasoning here?>> Samuel Johnson -- Sir, I have found you an argument, I cannot give you an understanding.-->> There's nothing of mathematical interest in your post.There was, however, something of general intellectual interest, whichis why it went right over your head. Perhaps you'd understand if youhad been educated, rather than merely trained.-- Wayne Brown (HPCC #1104) | When your tail's in a crack, you improvisefwbrown@bellsouth.net | if you're good enough. Otherwise you give | your pelt to the trapper.e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock an argument, I cannot give you an understanding. There's nothing of mathematical interest in your post. James HarrisNor in any of yours.--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 = [.snip.]>such as Z[a_1, a_2, a_3, x] which I suggested are not Dedekind domains. >Therefore factorization questions in these rings are likely to be hard,>with unsatisfying answers. I think this is why people don't work much with factorization in>algebraic function 'elds.Nitpicking: people ->do<- work with factorizations in (algebraic)function 'elds; the point is that rings like Z[a_1,a_2,a_3,x] are->not<- function 'elds, so they are unlikely to be Dedekind domains. [.snip.]-- =It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidinmagidin@math.berkeley.edu =>[...]>>My position is that mathematicians have a problematic de'nition for>>what's called the ring of algebraic integers, where members of that>>ring are roots of monic polynomials with integer coef'cients. And my>>factorization simply proves that the ring of algebraic integers is too>>small to in general include numbers like a_1(x)/7 and a_2(x)/7, as x>>varies over the ring of algebraic integers.>>Now then, not surprisingly, mathematicians would prefer to believe>>that they don't have this de'nition that's too small, so I end up in>>arguments. >> Uh, no. People don't deny that the de'nition is too small, meaning>> that a_1(x)/7 is not an algebraic integer. The argument is over your>> assertion that it should be an algebraic integer, for some reason>> that only you know. In particular when you claim that this shows>> the algebraic integers are incomplete you're not even wrong,>> _until_ you give a de'nition of complete, which you've never>> done.Actually no, as can be seen by the *mathematics* in my post which>readers should notice this poster deleted out. Uh, you're gibbering here. I said something about what people> said about things you've said - whether what I said was true or> false, there's no way that that can be demonstrated by the> mathematics in your post - whether it's true or false depends> on the content of other's posts in the past.Mathematics isn't about what people *say* but about mathematical truthand consistency.Now then, clearly if someone feels the need to delete out the *math*in a reply to a mathematical post on sci.math, only to then claim todepend on what other people said, it's rather clear that persondoesn't have the capability or mathematical con'dence necessary tohandle that mathematics.After all, people can *say* anything. Mathematics is about theability to do more than just say, but also to demonstratemathematically. > Duh.> Mathematics is distinguished from many other disciplines by theability to determine truth based on logic and mathematicalconsistency.>A point I made is that if you have a constant times a polynomial, like49(x+1)(x+2)(x+3) = (7x + 7)(7x + 14)(x+3)you can set x=0, to see how factors of 49 distributed, which is>trivial here as you can *see* how it distributed easily enough.So the short answer is that my position is that the distributive>property *always* holds, while these posters are trying to argue that>there are ways to break the distributive property: a(b+c) = ab + ac. You say this a lot, although it's never happened even once> that anyone has tried to break the distributive property.> I've used the *constant* terms of a factorization, and changes tothose constant terms when dividing by 49 to prove a mathematicalresult, which follows rather *easily* from the distributive property.> Which is in any case totally irrelevant to what I said about what> you said about what people say about the fact that a_1/7 is> not an algebraic integer.> Huh?>When I've given examples like49(x+1)(x+2)(x+3) = (7x + 7)(7x + 14)(x+3)in the past, they have claimed that it matters that those are>polynomials versus other expressions I use that are not polynomials,>but the reality is that they're STILL attacking the distributive>property, which is a bizarre math position to take. No, what's bizarre is that you think that people are attacking the> distributive property.> I've explained clearly how and in what way posters are attacking thedistributive property.However, I can explain again.James Harris =>>[...]>>My position is that mathematicians have a problematic de'nition for>what's called the ring of algebraic integers, where members of that>ring are roots of monic polynomials with integer coef'cients. And my>factorization simply proves that the ring of algebraic integers is too>small to in general include numbers like a_1(x)/7 and a_2(x)/7, as x>varies over the ring of algebraic integers.>>Now then, not surprisingly, mathematicians would prefer to believe>that they don't have this de'nition that's too small, so I end up in>arguments. >> Uh, no. People don't deny that the de'nition is too small, meaning> that a_1(x)/7 is not an algebraic integer. The argument is over your> assertion that it should be an algebraic integer, for some reason> that only you know. In particular when you claim that this shows> the algebraic integers are incomplete you're not even wrong,> _until_ you give a de'nition of complete, which you've never> done.>Actually no, as can be seen by the *mathematics* in my post which>>readers should notice this poster deleted out.>> Uh, you're gibbering here. I said something about what people>> said about things you've said - whether what I said was true or>> false, there's no way that that can be demonstrated by the>> mathematics in your post - whether it's true or false depends>> on the content of other's posts in the past.Mathematics isn't about what people *say* but about mathematical truth>and consistency.But my comments _were_ about what you said about what peoplesaid about what you say...>Now then, clearly if someone feels the need to delete out the *math*>in a reply to a mathematical post on sci.math, only to then claim to>depend on what other people said, it's rather clear that person>doesn't have the capability or mathematical con'dence necessary to>handle that mathematics.And clearly, if a person can read English, he sees that I includedall the math that was relevant to my point. You saidAnd my>factorization simply proves that the ring of algebraic integers is too>small to in general include numbers like a_1(x)/7 and a_2(x)/7, as x>varies over the ring of algebraic integers.>>Now then, not surprisingly, mathematicians would prefer to believe>that they don't have this de'nition that's too small, so I end up in>arguments. and I pointed out that you were misrepresenting what people have beensaying about that - nobody disputes that a_1/7 is not an algebraicinteger as you imply, people instead say so what, why is that aproblem?It's not a problem unless you _prove_ that a_1/7 _is_ analgebraic integer as well as showing that it's not one. Butyou haven't done that - the fact that it seems to you likeit should be an algebraic integer proves nothing.See, that's _mathematics_ I'm discussing. Wow.>After all, people can *say* anything. Mathematics is about the>ability to do more than just say, but also to demonstrate>mathematically.> Duh.Mathematics is distinguished from many other disciplines by the>ability to determine truth based on logic and mathematical>consistency.>>A point I made is that if you have a constant times a polynomial, like>>49(x+1)(x+2)(x+3) = (7x + 7)(7x + 14)(x+3)>>you can set x=0, to see how factors of 49 distributed, which is>>trivial here as you can *see* how it distributed easily enough.>>So the short answer is that my position is that the distributive>>property *always* holds, while these posters are trying to argue that>>there are ways to break the distributive property: a(b+c) = ab + ac. >> You say this a lot, although it's never happened even once>> that anyone has tried to break the distributive property.I've used the *constant* terms of a factorization, and changes to>those constant terms when dividing by 49 to prove a mathematical>result, which follows rather *easily* from the distributive property.> Which is in any case totally irrelevant to what I said about what>> you said about what people say about the fact that a_1/7 is>> not an algebraic integer.Huh?>When I've given examples like>>49(x+1)(x+2)(x+3) = (7x + 7)(7x + 14)(x+3)>>in the past, they have claimed that it matters that those are>>polynomials versus other expressions I use that are not polynomials,>>but the reality is that they're STILL attacking the distributive>>property, which is a bizarre math position to take.>> No, what's bizarre is that you think that people are attacking the>> distributive property.I've explained clearly how and in what way posters are attacking the>distributive property.However, I can explain again.>James Harris************************David C. Ullrich =>[...]>>My position is that mathematicians have a problematic de'nition for>what's called the ring of algebraic integers, where members of that>ring are roots of monic polynomials with integer coef'cients. And my>factorization simply proves that the ring of algebraic integers is too>small to in general include numbers like a_1(x)/7 and a_2(x)/7, as x>varies over the ring of algebraic integers.>>Now then, not surprisingly, mathematicians would prefer to believe>that they don't have this de'nition that's too small, so I end up in>arguments. >> Uh, no. People don't deny that the de'nition is too small, meaning> that a_1(x)/7 is not an algebraic integer. The argument is over your> assertion that it should be an algebraic integer, for some reason> that only you know. In particular when you claim that this shows> the algebraic integers are incomplete you're not even wrong,> _until_ you give a de'nition of complete, which you've never> done.>Actually no, as can be seen by the *mathematics* in my post which>>readers should notice this poster deleted out.>> Uh, you're gibbering here. I said something about what people>> said about things you've said - whether what I said was true or>> false, there's no way that that can be demonstrated by the>> mathematics in your post - whether it's true or false depends>> on the content of other's posts in the past.Mathematics isn't about what people *say* but about mathematical truth>and consistency. But my comments _were_ about what you said about what people> said about what you say...> That is, your comments were not mathematical, and you not only don'tdeny that fact, you emphasize it.>Now then, clearly if someone feels the need to delete out the *math*>in a reply to a mathematical post on sci.math, only to then claim to>depend on what other people said, it's rather clear that person>doesn't have the capability or mathematical con'dence necessary to>handle that mathematics. And clearly, if a person can read English, he sees that I included> all the math that was relevant to my point. You said And my>factorization simply proves that the ring of algebraic integers is too>small to in general include numbers like a_1(x)/7 and a_2(x)/7, as x>varies over the ring of algebraic integers.>>Now then, not surprisingly, mathematicians would prefer to believe>that they don't have this de'nition that's too small, so I end up in>arguments. and I pointed out that you were misrepresenting what people have been> saying about that - nobody disputes that a_1/7 is not an algebraic> integer as you imply, people instead say so what, why is that a> problem?The mathematics tells why I'm talking about a_1(x)/7 and a_2(x)/7, butdeleting out the mathematics, and simply talking about what people*say* is actually a misrepresentation.Your posts and replies indicate that you're taking a *social*position, trying to parse words in some particular way to claimmeaning *after* deleting out the mathematical statements which givecontext and meaning.That is a rather bizarre math position to take, as I think DavidUllrich is fairly unique in pushing what people *say* over mathematicsitself!!!James Harris =>[...]>>My position is that mathematicians have a problematic de'nition for>>what's called the ring of algebraic integers, where members of that>>ring are roots of monic polynomials with integer coef'cients. And my>>factorization simply proves that the ring of algebraic integers is too>>small to in general include numbers like a_1(x)/7 and a_2(x)/7, as x>>varies over the ring of algebraic integers.>>Now then, not surprisingly, mathematicians would prefer to believe>>that they don't have this de'nition that's too small, so I end up in>>arguments. >> Uh, no. People don't deny that the de'nition is too small, meaning>> that a_1(x)/7 is not an algebraic integer. The argument is over your>> assertion that it should be an algebraic integer, for some reason>> that only you know. In particular when you claim that this shows>> the algebraic integers are incomplete you're not even wrong,>> _until_ you give a de'nition of complete, which you've never>> done.>>Actually no, as can be seen by the *mathematics* in my post which>readers should notice this poster deleted out.>> Uh, you're gibbering here. I said something about what people> said about things you've said - whether what I said was true or> false, there's no way that that can be demonstrated by the> mathematics in your post - whether it's true or false depends> on the content of other's posts in the past.>>Mathematics isn't about what people *say* but about mathematical truth>>and consistency.>> But my comments _were_ about what you said about what people>> said about what you say...That is, your comments were not mathematical, and you not only don't>deny that fact, you emphasize it.Nope. My comments were 100% mathematical.>>Now then, clearly if someone feels the need to delete out the *math*>>in a reply to a mathematical post on sci.math, only to then claim to>>depend on what other people said, it's rather clear that person>>doesn't have the capability or mathematical con'dence necessary to>>handle that mathematics.>> And clearly, if a person can read English, he sees that I included>> all the math that was relevant to my point. You said>> And my>>factorization simply proves that the ring of algebraic integers is too>>small to in general include numbers like a_1(x)/7 and a_2(x)/7, as x>>varies over the ring of algebraic integers.>>Now then, not surprisingly, mathematicians would prefer to believe>>that they don't have this de'nition that's too small, so I end up in>>arguments. >> and I pointed out that you were misrepresenting what people have been>> saying about that - nobody disputes that a_1/7 is not an algebraic>> integer as you imply, people instead say so what, why is that a>> problem?The mathematics tells why I'm talking about a_1(x)/7 and a_2(x)/7, but>deleting out the mathematics, and simply talking about what people>*say* is actually a misrepresentation.Nope. _You_ started by saying something about what people say:mathematicians would prefer to believe that they don't have this de'nition that's too small - I was explaining how it is thatthat's not the issue; it's _not_ what mathematicians believe.>Your posts and replies indicate that you're taking a *social*>position, trying to parse words in some particular way to claim>meaning *after* deleting out the mathematical statements which give>context and meaning.That is a rather bizarre math position to take, as I think David>Ullrich is fairly unique in pushing what people *say* over mathematics>itself!!!>James Harris************************David C. Ullrich => [.snip.]>>a_1(x), a_2(x), and b_3(x), along with x itself, may all be understood>>as elements of an algebraic function 'eld. Speci'cally, it is the>>splitting 'eld of>> a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) = 0>>over the 'eld Q(x) , x being an indeterminate . >> This is not a function 'eld, is it? As I understand it, function>> 'elds are 'elds of transcendence degree 1 over an ->algebraically>> closed<- 'eld. So Q(x) does not qualify; you would have to consider>> C(x), C the complex numbers.I think you are correct.I have a book on Number Theory over Function Fields sitting in myof'ce, on the books I would love to read, but don't really have timeto category... But I'm away this week and next, so I can't verify,but this is what I understand to be the case.>> But let's set that aside for a second...>>The elements a_1,>>a_2, a_3, and x generate a ring, within which unique factorization does>>not seem to hold.>>speci'cally, in this ring, the element>> 14706125 x^3 - 900375 x^2 - 17640 x + 1078>>can be factored in two ways. One is>>(A) (7)(7)(300125x^3-18375x^2-360x+22)>>and the other is>>(B) (5a_1+7)(5a_2+7)(5a_3+7).>>Parenthetical note: 5a_3+7 = 5b_3+22.>>I don't know, but I think it probable, that there is no way to break>>down these factors further within this ring. >> I assume that by this ring you mean the ring of elements of the>> splitting 'eld which are integral over Q(x). In that ring, 7 is a>> unit (since the ring contains Q(x), and therefore contains Q, the>> rationals). So factorization (A) is not a factorization into>> irreducibles: the element>> 300125x^3-18375x^2-360x+22 >> is divisible by each of 5a_1+7, 5a_2+7, and 5a_3+7 (since 7 is a unit).Does it make sense (and lead to interesting results) to look for>factorizations over Z[a_1,a_2,a_3,x]?Well, it certainly makes sense, but you run into several problems inthat sort of rings, which may lead to dif'culties. First and foremost is that, in general, they are not Dedekind Domains,since not every nonzero prime is maximal (I would guess that (x) is,in general, prime and not maximal). So you will not have uniquefactorization into prime ideals. Right there, you lose a lotalready. You can certainly do ring theory over them, but I would besurprised if you can do much meaningful number theory.-- =It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidinmagidin@math.berkeley.edu =|>|>> |>> [.snip.]|>> |>>a_1(x), a_2(x), and b_3(x), along with x itself, may all be understood|>>as elements of an algebraic function 'eld. Speci'cally, it is the|>>splitting 'eld of|>>|>> a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) = 0|>>|>>over the 'eld Q(x) , x being an indeterminate . |>> |>> This is not a function 'eld, is it? As I understand it, function|>> 'elds are 'elds of transcendence degree 1 over an ->algebraically|>> closed<- 'eld. So Q(x) does not qualify; you would have to consider|>> C(x), C the complex numbers.|>|>I think you are correct.||I have a book on Number Theory over Function Fields sitting in my|of'ce, on the books I would love to read, but don't really have time|to category... But I'm away this week and next, so I can't verify,|but this is what I understand to be the case.A book titled Number Theory over Function Fields is liable to beusing function 'eld to mean a 'nite separable extension of k(t)where k is a 'nite 'eld. Such 'elds, together with number 'elds(which are 'nite extensions of Q) are known as global 'elds.There's an extensive analogy between number 'elds and function'elds in this sense. One can de'ne a Riemann hypothesis forfunction 'elds analogous to the one for number 'elds, but whichwe already know how to prove.I think function 'eld also is used in other ways, however.Keith Ramsay => Actually if to be speaking along the lines of exotic power/energy> resources for anything interplanetary worthy and further, this one is> way more than convincing (his video tape is worth the price and then> some): David Sereda: EVIDENCE the case for NASA UFOs> http://www.ufonasa.com/> Besides the before mentioned evidence that was extracted from NASA, asoffered by David Sereda, I've got a little something other that'scooking on the moon, as for obtaining a lunar gateway or pitstop forgoing to places like Mars and Venus.http://guthvenus.tripod.com/lunar-space-elevator.htmhttp ://guthvenus.tripod.com/gv-lse-he3.htmhttp:// guthvenus.tripod.com/moon-04.htm = > Besides the before mentioned evidence that was extracted from NASA, as> offered by David Sereda, I've got a little something other that's> cooking on the moon, as for obtaining a lunar gateway or pitstop for> going to places like Mars and Venus.> Crank Information = not a real maths question but i hope im right here!My question ist about the number of legal chessboard positions.The German book schach und Zahl and other sources (even this board)say that the number of positions after the 'rst move from both sides is at 400. Thats logical.After the second move from white it is at 5362!After blacks second move the number rises to 71824.(with somecorrection to 72084)now my question: The number was calculated by counting the positionswhen only white is moving twice and black does nothing. So it gets268 different positions. this number multiplicated with itselfgets to the 71824.But it goes on: to get the number of positions after the third move fromwhite the authors count the positions of 3white moves(black does nothing)which gets to 3022 positions and thy multiplicate this number again with 286so that they become 809896.(positions after whites third move)3022 mltiplicated wiht 3022 gets 9132484 which is the number after black 3. move.So. does anyone have an idea how they come to this 268??Greeting and a happy new yearGuido = not a real maths question but i hope im right here! My question ist about the number of legal chessboard positions.> The German book schach und Zahl and other sources (even this board)> say that the number of positions after the 'rst move from both > sides is at 400. Thats logical.> After the second move from white it is at 5362!> After blacks second move the number rises to 71824.(with some> correction to 72084)> now my question: The number was calculated by counting the positions> when only white is moving twice and black does nothing. So it gets> 268 different positions. this number multiplicated with itself> gets to the 71824.> But it goes on: to get the number of positions after the third move from> white the authors count the positions of 3white moves(black does nothing)> which gets to 3022 positions and thy multiplicate this number again with 286> so that they become 809896.(positions after whites third move)> 3022 mltiplicated wiht 3022 gets 9132484 which is the number after > black 3. move. So. does anyone have an idea how they come to this 268?? Greeting and a happy new year> Guidonobody got an idea?or just a stupid question...... = not a real maths question but i hope im right here! My question ist about the number of legal chessboard positions.> The German book schach und Zahl and other sources (even this board)> say that the number of positions after the 'rst move from both> sides is at 400. Thats logical.> After the second move from white it is at 5362!> After blacks second move the number rises to 71824.(with some> correction to 72084)> now my question: The number was calculated by counting the positions> when only white is moving twice and black does nothing. So it gets> 268 different positions. this number multiplicated with itself> gets to the 71824.> But it goes on: to get the number of positions after the third move from> white the authors count the positions of 3white moves(black doesnothing)> which gets to 3022 positions and thy multiplicate this number again with286> so that they become 809896.(positions after whites third move)> 3022 mltiplicated wiht 3022 gets 9132484 which is the number after> black 3. move. So. does anyone have an idea how they come to this 268?? Greeting and a happy new year> Guido nobody got an idea? or just a stupid question......Number of positions...After exactly two pawn moves:... after two moves by the same pawn: 16... after two pawns each moving 1 step: 28... after two pawns each moving 2 steps: 28... after two pawns, one moving 1 and one moving 2: 56TOTAL: 128After a pawn move and a non-pawn move:... after a3: 4... after a4: 6... after b3: 6... after b4: 6... after c3: 6... after c4: 7... after d3: 12... after d4: 13... after e3: 15... after e4: 15... after f3: 4... after f4: 5... after g3: 6... after g4: 6... after h3: 4... after h4: 6TOTAL: 121After a knight move and a rook move: 4After two moves by same knight: 11After two moves by different knights: 4TOTAL: 19Grand total: 128+121+19=268-- Clive Toothhttp://www.clivetooth.dk =>> not a real maths question but i hope im right here!>> My question ist about the number of legal chessboard positions.>> The German book schach und Zahl and other sources (even this board)>> say that the number of positions after the 'rst move from both >> sides is at 400. Thats logical.>> After the second move from white it is at 5362!>> After blacks second move the number rises to 71824.(with some>> correction to 72084)>> now my question: The number was calculated by counting the positions>> when only white is moving twice and black does nothing. So it gets>> 268 different positions. this number multiplicated with itself>> gets to the 71824.>> But it goes on: to get the number of positions after the third move from>> white the authors count the positions of 3white moves(black does nothing)>> which gets to 3022 positions and thy multiplicate this number again with 286>> so that they become 809896.(positions after whites third move)>> 3022 mltiplicated wiht 3022 gets 9132484 which is the number after >> black 3. move.>> So. does anyone have an idea how they come to this 268??>> Greeting and a happy new year>> Guidonobody got an idea?or just a stupid question......Use a computer and count everything, make sure not to overcount transpositions. After the 'rst two half-moves you have interaction between black and white, and no simple multiplicationof possibilities can give the number of moves.-- Wim Benthem =http://www.quantum'elds.com/469Maclay.pdfYes, this goes back to the Chickering letter to Richard De Lauerreproduced in my book Destiny Matrix and also to Harold Chipman'sefforts in the 80's also shown in the book.Curious it is Sandia isn't it? ;-)Nothing much new in what they say. Interesting that they are interested,Chickering's letter is alluded to inThe Quark and The Jaguar by Murry Gell-MannThe Quantum Engineers in New Yorker and a book by Jeremy Bernstein on JS Bell,In a Physics Today Op Ed by N.D. MerminAntony Valentini's work gives new life to this issue in that orthodox quantum theorywith signal locality is to a more general quantum theory with signal nonlocality as special relativity is togeneral relativity IMHO. The latter may go against Lenny Susskind's idea, but I am not sure yet.I have basically 'nished Physics Meets Philosophy at The Planck Scale - a good book.I am now reading Three Roads to Quantum Gravity by Lee Smolin - also a good book.The key is Lenny Susskind's Holographic Principle which I think is essentially correct.It is clear that no one yet really understands why.Curious con§uence in Universe as BIT in that the dual stringy uncertainty principle with a minimum length fuses the two meanings of string both as a vibrating string of pure energy (Brian Greene on NOVA) and as a BIT-string of pure information.Note that the discrete BIT linked 1-dim string is dual to 2-dim area in the 3D uncompacti'ed spacelike surfaces. So this seems to suggest the Hologram Principle as well?The affair with Ginsparg on the Cornell archive re Carlos Castro's work, which is good IMHO, plus all thespeculative hype in NOVA's The Elegant Universe, not clearly labeled as speculation, plus Brian Greene inNY Times today (not that he says anything objectionable there) all suggest that physics today has sold out toBig Money - not that that is not understandable of course. John Brockman, the literary agent in New York,has gotten very rich in making this a veritable cottage industry launched by The Dancing Wu Li Masters.I do not begrudge him. At least it gets the public interested.On the other hand Carlos Castro and Pavsic may have a powerful formalism that explains M theory in a relatively simply way and for Ginsparg tosuppress open debate on that is a very bad sign. Castro is not a crackpot in any sense as spelled out in John Baez's set of criteria. I am not saying Castro and Pavsic have it right. Maybe not, but their ideas should not be censored out of the Cornell e-print archive. =Statement:Equation (1) has no solutions under the given conditios.(a^2)x -(b^2)y = (x - y)^5 (1)Conditions: (a, b) = 1; (x, y) = 1; x and y are non-square integers > 1.Any comment upon the correctness of the statement will be appreciated. =En el K. Deb escribi.97:> Statement: Equation (1) has no solutions under the given conditios. (a^2)x -(b^2)y = (x - y)^5 (1) Conditions: (a, b) = 1; (x, y) = 1; x and y are non-square integers 1. Any comment upon the correctness of the statement will be appreciated.What with a = b = 1, x = y + 1?-- Ignacio Larrosa Ca.96estroA Coru.96a (Espa.96a)ilarrosaQUITARMAYUSCULAS@mundo-r.com => En el K. Deb escribi.97:> Statement: Equation (1) has no solutions under the given conditios. (a^2)x -(b^2)y = (x - y)^5 (1) Conditions: (a, b) = 1; (x, y) = 1; x and y are non-square integers 1. Any comment upon the correctness of the statement will be appreciated. What with a = b = 1, x = y + 1?If a nontrivial solution is desired: a = 31, b = 39, x = 8, and y = 3David =I read some messages on this group for the 'rst time yesterday,prompted by my musings while installing a combination lock on a door.I wondered, if I was the manufacturer of this lock, how many differentcombinations would I say it had?I'm wans't a math major, butI think I know a fairly laborious way todoit. (shown below) But I'm wondering if there is an easier way. Anytakers?The lock has 've buttons, which are pressed in some sequence to openit. No button may be pressed more than once. The combination canconsist of any number of buttons. So, there are obviously:5 one-button combinations20 two-button combinations60 three-button combinations120 four-button combinations and, 120 've-button combinations (the same 120 combos as the 4 buttonones, but with the remaining button pushed last)BUTThe lock also allows you to press 2 buttons simultaneously. So, forcombinations in which the 'rst press consists of 2 buttonssimultaneously, I 'gure the following:20 two-buttons pressed simultaneously without any other pressescombosplus:A: 20 two-button presses x 3= 60 two-press combosB: 20 two-button presses x 3 x 2= 120 three-press combosC: and another 120 four-press combosNow it seems to me that the number of combinations shouldn't beaffected by where in the sequence you use your double button, sothere should be 2 times as many As, 3 times as many Bs and 4 timesas many Cs. But, of course, you can use TWO double button entries in yourcombination, so that should be 20 ('rst double button permutations) x6 (second double button permutations)= 120, all of which are repeatedif you choose to press the remaining button. I don't know if the lock allows for triple-presses, so let's ignorethat. I could now go back and add up all of the above, but I'm notthat interested in the actual answer. I'm just curious to know asimpler way to get it, if one exists. =* gguarino@pipeline.com> 20 two-buttons pressed simultaneously without any other presses> combosSorry, I get only 10 here. Please explain.> plus:> A: 20 two-button presses x 3= 60 two-press combos?> B: 20 two-button presses x 3 x 2= 120 three-press combos?> C: and another 120 four-press combos?-- Jon Haugsand Haugsand > 20 two-buttons pressed simultaneously without any other presses>> combosSorry, I get only 10 here. Please explain.Sorry. You're right. Result of thinking it out WHILE writing it. I'mafraid. > plus:>> A: 20 two-button presses x 3= 60 two-press combos?> B: 20 two-button presses x 3 x 2= 120 three-press combos?> C: and another 120 four-press combos? = I'm having trouble trying to solve a problem in Rudin's Principlesof Mathematical Analysis book. The problem is:Let L_n = 1/(2*PI)*integrate(|D_n(t)|, t, -PI, PI) (n=1, 2, 3, ...)where D_n(t) is the Dirichlet kernel. D_N(t) = Sum(E^(i*n*t), n, -N, N)=sin((N+1/2)*t)/sin(1/2*t) for t!=0Prove that there exists a constant C>0 such thatL_n > C*Log(n) (n = 1, 2, 3, ...), or more precisely that the sequence {L_n - 4/PI^2*log(n)} is bounded.______________________________________________________ _____I can show that L_n > 4/PI^2*log(n), but I don't know how to show thatthesequence {L_n - 4/PI^2*log(n)} is bounded above.Here's my proof for the 'rst part:1/(2*PI)*integrate(|D_n(t)|, t, -PI, PI) = 1/PI*integrate(|D_n(t)|, t,0, PI)= 1/PI*integrate(|sin((n+1/2)*t)|/sin(1/2*t), t, 0, PI)put A = n+1/2. Then L_n =1/PI*integrate(|sin(t)|/sin(t/(2A)) * 1/A, 0, PI*A) >1/PI*integrate(|sin(t)|/(t/(2A)) * 1/A, 0, PI*A) =2/PI*integrate(|sin(t)|/t, 0, PI*A) >Sum(2/PI*integrate(|sin(t)|/t, t, (k-1)*PI, k*PI), k, 1, [A]) >Sum(2/PI*2/(k*PI), k, 1, [A]) =4/PI^2*Sum(1/k, k, 1, [A]) > 4/PI^2*log(n)Now how do you prove that the sequence is bounded above? => I'm having trouble trying to solve a problem in Rudin's Principles> of Mathematical Analysis book. The problem is: Let L_n = 1/(2*PI)*integrate(|D_n(t)|, t, -PI, PI) (n=1, 2, 3, ...) where D_n(t) is the Dirichlet kernel. D_N(t) = Sum(E^(i*n*t), n, -N, N)=sin((N+1/2)*t)/sin(1/2*t) for t!=0 Prove that there exists a constant C>0 such that L_n > C*Log(n) (n = 1, 2, 3, ...), > or more precisely that the sequence {L_n - 4/PI^2*log(n)} is bounded.> ___________________________________________________________> I can show that L_n > 4/PI^2*log(n), but I don't know how to show that> the> sequence {L_n - 4/PI^2*log(n)} is bounded above. Here's my proof for the 'rst part: 1/(2*PI)*integrate(|D_n(t)|, t, -PI, PI) = 1/PI*integrate(|D_n(t)|, t,> 0, PI)> = 1/PI*integrate(|sin((n+1/2)*t)|/sin(1/2*t), t, 0, PI) put A = n+1/2. Then L_n => 1/PI*integrate(|sin(t)|/sin(t/(2A)) * 1/A, 0, PI*A) 1/PI*integrate(|sin(t)|/(t/(2A)) * 1/A, 0, PI*A) => 2/PI*integrate(|sin(t)|/t, 0, PI*A) > Sum(2/PI*integrate(|sin(t)|/t, t, (k-1)*PI, k*PI), k, 1, [A]) Sum(2/PI*2/(k*PI), k, 1, [A]) => 4/PI^2*Sum(1/k, k, 1, [A]) > 4/PI^2*log(n) Now how do you prove that the sequence is bounded above? 2/(Pi*(k+1)) < int_[Pi*k, Pi*(k+1)] |sin(t)/t| dt < 2/(Pi*k),k = 1, ... =Paul you should come to London March 8 - 13 to iron this out.You cannot reduce the gravitational 'eld to g-forces, as Einsteinonce thought.JS; Why do you think that? Quote Einstein on that.PZ: I mean as to the essential nature of the phenomenon.JS: Justify your conclusion with evidence from Einstein's writing - date important of course.PZ: I am not suggesting that Einstein thought that Riemann curvature could be ignored.Of course the metric description of permanent physical 'elds entails non-vanishingcurvature.I think you're missing the point. I don't agree with Einstein's view, but I think Iat least understand his idea.JS: Yes, I do miss your point. I cannot 'nd it as I try to pin it down. Your point then continues to elude me as I try to get you to pin it down in a clear enough way for me to understand. Look, take a globally §at space-time. An arbitrary curvilinear coordinate transformation corresponds physically, in principle, to a non-rigid lattice of generically accelerating LNIF tiny observer-detectors. You can never in that situation get Hawking's nice example of Alice and Bob both LNIF on timelike non-geodesics accelerating away from each other BUT at a FIXED unchanging spatial separation.Also you need overlapping coordinate patches in really curved spacetime, you cannot use a single global curvilinear coordinate transformation. Look at the Penrose diagram for the SSS Ruv = 0 non-exotic vacuum Einstein-Bridge non-traversable Mass without mass Wheeler geon for example. See Matt Visser's Lorentzian Wormholes. Of course Puthoff ignores all that in his PV model. Puthoff says no event horizons which completely throws away all the Hawking-Penrose theorems on singularities, the Bekenstein bound from black hole thermodynamics, the Hawking radiation, Unruh's work, the t'Hooft-Susskind holographic principle, the computation of black hole entropy from string theory. No wonder the Pundits dismiss Puthoff's suggestion. He does not justify his engineering approach adequately.PZ: This is an example of where a naive Machian empiricism leaves youbarking upthe wrong tree -- the wrong tree in this case being strict equivalence.Of course Einstein knew about Riemann curvature and tidal effects, butthen tackedon a doctrine that all physics is local, insisting that *only* thetranslational g-forcesobserved at each point are fundamentally de'nitive of thegravitational 'eld.JS; Why do you think that? Quote Einstein on that.PZ: Jack, that doesn't mean you can *ignore* Riemann curvature.JS: Good, but your essential point still eludes me.BTW see Hawking's nice example I mentioned earlier today of 2 observersaccelerating AWAY from each other at opposite podes of Earth's surfaceyet staying FIXED distance from each other. This demands non-EuclideanSPACE-TIME CURVATURE!PZ: I agree this is a nice example, but *of course* you need Riemann curvaturein GR to describe such a 'eld.JS: Then what is it you object to? Some early remark Einstein may have made before he understood completely what he was doing?PZ:: In my POV this was simply an *ad hoc* stratagem for protectinghis equivalencehobbyhorse from legitimate criticism. There is nothing deep here.JS: Hawking gives a counter-example to what you just said IMHO.PZ: A counterexample to what?JS: To what I thought was your thesis, but apparently is not. :-)PZ: Jack, I don't think you have understood what I've been saying. You certainlyhaven't understood Einstein. Einstein's thinking was very subtle.I am NOT suggesting that Einstein thought that Riemann curvature did not exist,or that non-vanishing Riemann curvature was not *technically necessary* for describingparticular permanent gravitational 'elds in terms of the metric g_uv. Obviously it is --and of course Einstein was fully aware of that.JS: Good. So what is your point?PZ: I am saying that he regarded only the *pointwise local equivalence* of the inertialand gravitational 'eld to be of fundamental importance: he did not regard non-vanishingRiemann curvature as *de'nitive* of the essential nature of the gravitational 'eld. Thequestion of vanishing vs. non-vanishing Riemann curvature was viewed as merelyaccidental to this fundamental nature.JS: Justify what you just said with relevant quotes from Einstein and the dates he made them.the approximation thatthe idea of the point represents. This is obviously an approximate macro idea. It breaks downfor quantum gravity and also even above that if you include very sensitive tidal curvature detectors.This is not news and no one thinks that that refutes Einstein's fundamental idea of general relativity.I think ALL key ideas in theoretical physics are metaphors and correspondence approximationsthat have mathematical formulation with operational meanings at least at the gedanken level, whichexplain according to our subjective standards and which also must be in accord with, and better yetpredict, real measurement numbers. Does the physics in NOVA's Elegant Universe meet these standards?Does it have a chance of doing so? Actually, I am optimistic on that.of equivalencewas about doing justice (Einstein 1921, 1949) to the weak equivalence principle byidentifying the essential natures of the two classes of phenomena, and the fundamentalphysical nature of the two kinds of 'eld. Riemann curvature may or may not vanish forparticular real gravitational 'elds -- but this is *accidental*, not essential, in the Einsteinianview.JS: What exactly is wrong with that? It is defensible operationally. Physics is essentially operational in its basic nature.If you take the connection 'eld as being the gravity 'eld then with that de'nition everything is 'ne. The gravity 'eldis eliminated in the LIF. Fine. One can also see if there is local gravity 'eld vanishes for BOTH of them, nevertheless, there is a tidal effect, there is local curvature at the next deeper gradient level.So I see no conceptual con§ict here?PZ: (Here essential does not mean or imply technically necessary; it means essential tothe de'nition of the fundamental nature of the gravitational 'eld -- which is consistentwith philosophical usage )I am not saying I *agree* with this Einsteinian view. In fact, it's quite the opposite: I don't.The point I am trying to drive home with the covariant metric description of inertial forces in§at spacetime is not that Riemann curvature is irrelevant to Einstein's theory of gravitation;but Kretschman's point that general covariance in itself does not amount to general relativity.JS: Is this news? Who rejects that? Again look at Hawking's key example.PZ: This should be now quite obvious, since you can apply a general covariant kinematical metricdescription to *'ctitious* forces, in terms of connection 'elds, *entirely within Newtoniantheory*. Yet this is hardly general relativistic.JS: The dynamical equations are thesame in the sense of being covariant equations. It is true that thetidal curvature tensor is not zero in an LIF if it is not zero in aCOINCIDENT LNIF at same point event P.PZ: Yes, of course. Everyone knows this technical fact, but few seemwilling to draw theobvious conclusions. There is no deep thinking behind the*pseudo-operationalist*principle all physics is local. The gravitational 'eld is notlocal; and no actualmeasurements at all can be performed inside an in'nitesimalneighborhood.JS: All that EEP, yes only acorrespondence principle, asserts, is that under most conditions, onecan manage to make the relative tidal acceleration between twobreaks down on the approach to a space-time singularity or if one isprobing the Planck scale Lp. For practical problems here on Earth thetidal effects are very tiny because, as I recall, radius of curvatureof the Earth at its surface is ~ 1 AU ~ 10^13 cm. I need to computeit for Sun at Earth as well.PZ: But that is a sleight of hand, which works by tacitly assumingthat the only wayThere areother measurement devices that are not scale-sensitive in the mannerof EEP.So this is yet another MTW red herring.JS: Huh? Show me.PZ: Read Ohanian & Ruf'ni Chapter 1, Section 1.9.Here's their conclusion:we see that there exist several methods for measuring the tidal 'eld *locally*,in an arbitrarily small neighborhood of a given point.JS: Good.PZ: The limitations on the minimum size of the neighborhood that is neededto perform measurements of a given precision do not arise from any intrinsicproperties of the gravitational 'eld; rather, these limitations arise from thequantum nature of matter, which prevents us from constructing an apparatusof arbitrarily small size.JS: Fine, I agree, always did.PZ: The tidal 'eld is no less a local quantity than, say, the electric 'eld.-- Gravitation and Spacetime, p 53.JS: Of course. That is what I have been saying. So?JS: All EEP says is that in the presence of gravitational 'elds, SRworks as anapproximate model if we ignore all but a restricted class ofmeasurements --PZ: Of course here I meant it works in free §oat.JS: This is too vague.PZ: But this principle is in truth inherently vague, since it doesn't, and cannot, specifywhich measurements are to be excluded (ignored).JS: It does not. SR has the local curvature tensor as well!For exampleRuv(LNIF) = eu^a(P)e^vb(P)Rab(LIF)where the e's are the tetrad components. So what is the problem?Einstein's 'eld equation in the SR LIF is simply, for /zpf = 0 non-exotic vacuum limitGab(Einstein) = -[(QED dimensionless coupling)/(Witten's String Tension)]Tab(Matter)How do you like my cool Post-Modern deconstructing of Einstein's GR showing its deeper meaning? ;-)Similarly for ALL the local differential equations for standard physics.This even works for local quantum 'eld theory using second-quantization.Carlos Castro & Pavsic have a Clifford Algebra generalization of all this, which naturally includes extended strings and branes with the points as centers of the extended geometrodynamical objects.PZ: Or perhaps you like random precision?The whole scale-dependence aspect of EEP is a red herring IMO. Of course ifyou ignore all local signs of roundness, then things locally *look* §at.JS: Yes, but I think I showed you the essential point above i.e.Gab(Einstein) = -[(QED dimensionless coupling)/(Witten's String Tension)]Tab(Matter)in the LIFalso its Penrose spinor form deeper than tensor and perhaps generalized to Castro-Pavsic poly-vectors in Clifford Space (like Grassmann's algebra and Cartan's exterior forms?). Spinors as ideals of Clifford Algebras etc.PZ: All EEP really says is that if you look only at translational effects, then (usingSR) in free fall you can model spacetime as §at, since in that case zero resultant g-forces.PZ: In order to do this locally you have to ignore the locally detectable effects ofRiemann curvature.JS: Not really, i.e.Gab(Einstein) = -[(QED dimensionless coupling)/(Witten's String Tension)]Tab(Matter)in the LIFPZ: Yes, with *some* tidal measuring devices this becomes easier (you get a betterapproximation) as the volume of spacetime in which measurements areperformed shrinks. But with alternative devices this is not necessarily the case(Ohanian & Ruf'ni 1.9)JS: I would like the details of these alternative devices. Is this ET technology from Colonel Corso? ;-)PZ: So this epsilon-delta aspect of EEP is a red herring IMO. You might as well say:If you ignore evidence of curvature, then you will notice no evidence of curvature.Vacuous, really. The *appearance* of epsilon-delta precision is, well -- just that.Cardboard-cutout....like a tale told by an idiot, full of sound and fury, signifying nothing.JS: The translational g-force vanishes in an LIF -astronauts free §oat in space with rockets off.PZ: As they do in Newtonian physics. So what? Where is the general relativity here?There *is* none -- unless you add it. Einstein added it by pretending inertial 'eldswere, fundamentally, real gravitational 'elds (Einstein 1907, 1911, 1916, 1921, 1949).JS: No, I think the basic idea is adding non-Euclidean geometry as in Hawking's nice example.PZ: I believe that is why he erroneously implicitly concluded that the inertial 'eld makes acontribution to the gravitational vacuum energy -- yielding the Einstein conservation law,the Einstein 'eld equations, and the Einstein stress-energy pseudotensor. This is allbuilt into his gravitational conservation law, based on the uni'ed connection 'eld takenseriously.JS: I do not understand what you are saying. It's too compressed. Flesh it out with equations.Are you saying thatGuv + 8pi(G/c^4)Tuv = 0is wrong?Why?What do you replace it with?I see no need for the pseudo-tensor locally. You only need it to get global Pu integrals inasymptotically §at space-time as in MTW(ch 19-20?) asking tricky questions on gravity waves.The problem there is to separate near and far 'eld dynamical degrees of freedom of the guv 'eld.BTW Wheeler-Feynman/Hoyle-Narklikar type elimination of 'eld dynamical degrees of freedomONLY applies to FAR FIELD, i.e. real quanta not to NEAR FIELD, i.e. virtual quanta.The NEAR FIELD is always the relative BACKGROUND reference for the waves.For example in SSS solution of Ruv = 0ds^2 = (1 - 2GM/c^2r)(cdt)^2 - (1 - 2GM/c^2r)^-1dr^2 - r^2(dpolar)^2r > 2GM/c^2this is all NEAR guv 'eld not guv(gravity waves) for which Pu with pseudo-tensor is relevant -- if I recall correctly from fuzzy memory. ;-)Do you object toGuv^;u = 0 ?That is an approximation also IMHO that is violated in real practical metric engineering by advanced super-technological life forms of the Q-class in sense of Star Trek. That's what the real UFO stuff is about IMHO and that why Hal Puthoff does what he does, mistakenly IMHO. :-)May the best idea win. That ain't always necessarily so as we see recently with the Castro-Ginsparg Case.PZ: And this conservation principle 'gured large in Einstein's heuristic derivation of thestandard GR 'eld equations. So if this is questionable, then so are the 'eld equationsin their classic form, for deep physical reasons.JS: You need to be more speci'c for this to be more than couch potato idle speculation.So howGuv^;v = 0can fail? ENGINEERING PROBLEMthat Hal Puthoff has never properly posed in his publications with his group ever IMHO.PZ: It just super'cially seems to preserve the physical content ofEinstein equivalence-- but in fact it doesn't. That's why I say EEP is a grosslymisleading acronym.It should be called WCP -- for Wheeler Correspondence Principle.PZ: This is Einstein's basic extension of the restricted physicalprinciple of relativity to mutuallyaccelerating frames.My point here is that the whole general covariant formal apparatusof metric tensors,connection 'elds, etc. based onds^2 = g_uv dx^u dx^vcan be applied just as well under the Newtonian or the Einsteinianinertial models: so far, this generalcovariant formulation is completely neutral as to Newton v.Einstein.The difference here is ratherin the *interpretation* of the inertial metric tensor g_uv --whetherit is to be understood as a purelykinematical quantity (Newtonian), or as an integral aspect of adynamic uni'ed gravitational-inertial'eld (Einsteinian).JS: Since Newton's gravity is the slow speed weak curvature limit ofEinstein's GRFor example in SSS where r > 2GM/c^2ds^2 = (1 - 2GM/rc^2)(cdt)^2 - (1 - 2GM/rc^2)^-1dr^2 + r^2(dtheta^2 +sin^2theta dphi^2)Therefore, one CAN seamlessly interpret Newton's gravity in exactlythe same geometrodynamic way that Einstein does.PZ: Yes, you can get to this by correspondence -- but you can also doit *ab initio*within a purely Newtonian theory, and without any reference togravity. Thisfollows directly from the de'nition of the invariant interval on asmooth manifold.JS: But you get no post-Newtonian effects like gravi-magnetism.PZ: Sure, but this is not relevant to my argument.JS: This part of your argument was done years ago I think by Cartan and is not really new.PZ: I am not suggesting that we go back to the Newtonian theory -- just to the Newtonianmodel of inertial compensation (by analogy with inertial compensation of electrostaticforces in GR, as in my example) applied within the GR framework -- subject tocertain modi'cations as to the de'nition of gravitational vacuum stress-energy andthe formulation of the matter-'eld conservation principle.JS: You lost me again. Inertial compensation the way I understand the words mean setting the torsion-free connection 'eld = 0 de'ning the LIFs on timelike geodesics. It says nothing about local curvature.JS: Therefore, there isno necessary difference of interpretation since Einstein's theoryeliminating gravity force is the covering theory of Newton's usingforce acting at a distance. Newton's intepretation is clearly lessfundamental and is superceded by Einstein's even in Newton's originaldomain of relevance.PZ: This is really beside the point, which is that there is nothingspeci'cally Einsteinianabout covariantly associating inertial forces with a space-timeconnection 'eld, sinceyou can also do this in a purely Newtonian theory (of course, you canalso do it in SR).JS: OK explain Hawking's above example then. I am waiting.PZ: See above. Of course you need 4D Riemann curvature to describe a sphericallysymmetric 'eld in GR. I never suggested otherwise. I don't know of anyonewho has. Einstein certainly didn't, and I am not saying he did.Einstein wanted to identify the *fundamental natures* of inertial and gravitationalphenomena, and conceive of the inertial 'eld as a real gravitational 'eld (Einstein 1949).He did not consider non-vanishing Riemann curvature to be *essential* to the nature ofthe real gravitational 'eld.JS: I think I showed above that Einstein is correct about this. Indeed the geodesic deviation equation for thecompletely IMHO justi'es what Einstein said in 1949according to your above description.PZ: That most physical g-'elds exhibit such curvature was notfundamentally signi'cant in Einstein's de'nition of the gravitational 'eld.JS. Correct, with GOOD REASON! I think I now see your point and the error in your thinking. You have simply not really understood the meaning of the geodesic deviation equation! IMHO.PZ: That does NOT imply that you don't technically need Riemann curvature to describe aparticular permanent 'eld, as in Hawking's example. Obviously you do.JS: Well then you have answered your own question about Kretschmann may He Rest In Peace and not roll over in his grave - at least not until Halloween when all the Dead Physicists return.Hey that's a name for a Rock Band Dead Physicists.Looks like Usama Bin Laden was thwarted at the pass - no New Years attacks on scale he allegedly boasted of. Rumor is that he is cornered. => Paul you should come to London March 8 - 13 to iron this out.> You cannot reduce the gravitational 'eld to g-forces, as Einstein> once thought. JS; Why do you think that? Quote Einstein on that.Seems to me that you can -http://www.geocities.com/physics_world/gr/grav_force.htmPmb== ==There's one type of attempt at disputing my work that I've seen pop upregularly, and it popped up today from Rick Decker, a professor atHamilton University, so I thought I'd talk about it in detail. Hereare some headers so you can 'nd the post:Now then, consider what happens if you divide both sides of(5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) by 7, as then you end up with something like(5b_1(x) + 1)(5b_2(x) + 2) = 25x^2 + 30x + 2where the b's are roots of some unknown quadratic, though the 'rstand last coef'cients ARE known:b^2 + ? b + (x^2 + x).Now then, it's just a quadratic people. SOME mathematician in all theworld should be able to give what the middle coef'cient is, right?Let's see.Moving on, there's the question of why did Decker pick his example. My guess is that it's because at x=1 *both* a's have sqrt(7) as afactor.However, does the *second* factorization exist or not? Given that itexists, where it follows from dividing both sides of the initialexample by 7, what is the missing coef'cient?Now my point is that there *are* factorizations where you're forcedinto rings where integers other than -1 and 1 are units, but even withsuch a factorization, like Decker's example, I can show you howlimited your own mathematics is, even when it comes to a quadratic.Or at least that's the idea.James Harris => There's one type of attempt at disputing my work that I've seen pop up> regularly, and it popped up today from Rick Decker, a professor at> Hamilton University, To be precise, the legal name is Hamilton *College*. We do havea university--Colgate--just down the road, but we're a college.(I'll let maky make of that what he will.) In fairness, it'sa common error, especially in most of the rest of the world,where college means something entirely different than itdoes in the US.> so I thought I'd talk about it in detail. Here> are some headers so you can 'nd the post: > In his post Decker claimed to mirror my argument using a quadratic> instead of a cubic, where he has (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) where his a's are roots of a^2 - (x - 1)a + 7(x^2 + x). Checking at x=0 reveals that the actual constant terms of the> factorization are 7 and 2, where Decker picked a_1(0) = 0 at x=0. Now then, consider what happens if you divide both sides of (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) by 7, as then you end up with something like (5b_1(x) + 1)(5b_2(x) + 2) = 25x^2 + 30x + 2 where the b's are roots of some unknown quadratic, though the 'rst> and last coef'cients ARE known: b^2 + ? b + (x^2 + x). Now then, it's just a quadratic people. SOME mathematician in all the> world should be able to give what the middle coef'cient is, right?> Right. The middle coef'cient in this case is (-3x + sqrt(-7x^2 - 8x))/4> Let's see. Moving on, there's the question of why did Decker pick his example. > My guess is that it's because at x=1 *both* a's have sqrt(7) as a> factor.>Precisely. My point being that the constants don't serve asindicators of how things are to be divided in any but the x = 0 case.It's worth noting, by the way, that if we take x = 2we have a situation where you can't even split the factorof 7 into anything as nice as sqrt(7) * sqrt(7). Rick = There's one type of attempt at disputing my work that I've seen pop up> regularly, and it popped up today from Rick Decker, a professor at> Hamilton University, > To be precise, the legal name is Hamilton *College*. We do have> a university--Colgate--just down the road, but we're a college.> (I'll let maky make of that what he will.) In fairness, it's> a common error, especially in most of the rest of the world,> where college means something entirely different than it> does in the US.Ok, my mistake, so it's Hamilton College then. > so I thought I'd talk about it in detail. Here> are some headers so you can 'nd the post: > In his post Decker claimed to mirror my argument using a quadratic> instead of a cubic, where he has (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) where his a's are roots of a^2 - (x - 1)a + 7(x^2 + x). Checking at x=0 reveals that the actual constant terms of the> factorization are 7 and 2, where Decker picked a_1(0) = 0 at x=0. Now then, consider what happens if you divide both sides of (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) by 7, as then you end up with something like (5b_1(x) + 1)(5b_2(x) + 2) = 25x^2 + 30x + 2 where the b's are roots of some unknown quadratic, though the 'rst> and last coef'cients ARE known: b^2 + ? b + (x^2 + x). Now then, it's just a quadratic people. SOME mathematician in all the> world should be able to give what the middle coef'cient is, right? > Right. The middle coef'cient in this case is (-3x + sqrt(-7x^2 - 8x))/4Well I put that up kind of curious that someone might 'll it in, butyou failed badly here Decker, as my 'rst check was at x=1.Even if his quadratic 'ts as a factorization, it's not THEfactorization that was being looked for as I've just shown. Remember,there is no uniqueness of polynomial factorization here, so there arean in'nity of factorizations.Still Decker might seemingly get *some* credit, if he found one ofthem.He fails at the task at hand though for not 'nding what follows fromthe a's in his *own* example.Remember the question I'm raising is what happens when 7 is dividedfrom both sides of(5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2)and I'm using Decker's own example, primarily because it's dramatic todo so, and because it's a *quadratic* so that I can show how completehis failure is, with something so seemingly simple. > Moving on, there's the question of why did Decker pick his example. > My guess is that it's because at x=1 *both* a's have sqrt(7) as a> factor.> Precisely. My point being that the constants don't serve as> indicators of how things are to be divided in any but the x = 0 case.But Decker, how do you suppose *constants* like 7 can change dependenton x?Now then, why don't you explain where you got your factorization from,like the techniques you used to generate it? > It's worth noting, by the way, that if we take x = 2> we have a situation where you can't even split the factor> of 7 into anything as nice as sqrt(7) * sqrt(7).Why? Can you elaborate more on why you think that is the case, andwhy you think it's worth noting?James Harris = a^2 - 7(1^2 + 1) = a^2 - 14, so a=+/-sqrt(-14).If Harris means for a^2 - 14 to be zero, he should takea = +/-sqrt(14), not sqrt(-14) Now then, dividing off 7, should then give b=+/-sqrt(-2).In true mathematics, sqrt(-14)/7 is not equal to sqrt(-2),nor is sqrt(14)/7 equalo to sqrt(2).If Harris means to divide sqrt(-14) by sqrt(7), he must say so unambiguously.And it would be helpful if he were to clean up his own Augean stables before sneering at others stables. =My research can be dif'cult to understand, so I thought I'd try outyet another way of explaining it. Some of you may have 'gured outthat I test out explanations on Usenet for use elsewhere, to re'ne myown understanding, or just in case someone out there might 'nally getit.Now then, again here's my discovery:(5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = 49(300125 x^3 - 18375 x^2 - 360 x + 22)where b_3(x) = a_3(x) - 3 and the a's are roots ofa^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)and when x=0, a_1(0) = a_2(0) = b_3(0) = 0.In that form it's hard to understand what follows next unless you payattention to what you have, speci'cally that cubic de'ning the a's.I can get it because of the symmetry of(5 a_1(x) + 7)(5 a_2(x) + 7)(5 a_3(x) + 7) = 49(300125 x^3 - 18375 x^2 - 360 x + 22)where I've gone ahead and substituted a_3(x) back in to replaceb_3(x), and it's important that you focus on that symmetry.It's that symmetry which allows the cubica^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)to de'ne ALL the a's, but something happens when I divide by 49.Then the symmetry is broken. Without that symmetry it's impossible to'nd a SINGLE cubic to handle what results when you divide both sidesby 49.That's important because it's why the functions are NOT algebraicinteger functions!!!Now then, I'll recap. Symmetry allows the a's to be de'ned by acubic, which shows them to be algebraic integer functions, butdividing by 49 *breaks* that symmetry, taking away the ability to 'ndsome cubic to de'ne the results, which proves that the resultingfunctions are not algebraic integer functions.(5 b_1(x) + 1)(5 b_2(x) + 1)(5 b_3(x) + 22) = 300125 x^3 - 18375 x^2 - 360 x + 22where the b's are roots ofb^3 + ? b^2 + ? b - (2401 x^3 - 147 x^2 + 3x)and when x=0, b_1(0) = b_2(0) = b_3(0) = 0.My point is that the second and third coef'cients are impossible tode'ne in general.You may 'nd them for some particular x, but in general, they areforever hidden from you.Notice that doing that substitution with a_3(x) for b_3(x) gives me(5 b_1(x) + 1)(5 b_2(x) + 1)(5 a_3(x) + 7) = 300125 x^3 - 18375 x^2 - 360 x + 22but you have broken symmetry since the other constant terms are 1 and1, so you're still stuck.Now by emphasizing what happens *after* 49 is divided from both sidesI'm trying to get at least some of you to face the mathematicalrealities here, and I've made other posts pointing it out as well.Courage in mathematics might sound new to you as an idea, but thosewho came before you had courage. That's how they learned of sqrt(2),and sqrt(-1), and much other mathematics that some refused, fought,and bitterly attacked as they lacked that courage.Of course, the mathematics should win, assuming that humanity surviveslong enough, but regardless each of you now faces an individual testof your own courage. It's not yet in the history books this time,where you can read and just imagine that you'd never have fought oversqrt(2) or refused to accept sqrt(-1) as an imaginary number.Here it's the present, and your test of courage is now.Oh, and be sure to check my blog archives:And hey, if you're moping and miserable because mathematics tests you,then maybe, if you think you're a mathematician, you might want to trya different 'eld. =