mm-4679 === Subject: don.mcdonald math WGTON.les miserables EPIPHANY. by Victor 1/ some other messages, mnemonic pppp. Put on your maths safety helmet!! 8/...Pursue parapet, DVD ,ST cubert Sunday, 31-8-2008. Alist (parabola paragraph Don.) 9/ Rob... Guest..,. ;-)) JEAN VALJEAN, lead actor. ñrepays the BishopÍs hospitality by stealing his silver..î Legs leaden limbs strained injury, Olympic XXIX Beijing 8.8.08. Alchemy. Aotea Centre, Auckland, 6.1.1999. January. New Zealand Herald, new year ñ99., don letters, leap second, atmosphere gravity, privileged planet. ñCARRY ï1.î- take up his cross, number plate- haulage, breath heaven, lectionary. Matt 16:24. Proof primality, epiphany Papatoetoe, 619-, performance present, photo purse. Perfect 45-th possibly discovered, Mersenne prime record, pearl price, 16.9, piecewise square [marty F. prayer] Hexadecimal September. NZ Science Monthly, February 1999. ñInteger sequences lookup Hati Honouring a number theorist Alan R BOYD , Hati numbers, N= N.2^a.3^b. Links.. XXXXXXXX hati Mercurial factorisation HG 7817 619 primality. Brit Med Jl. 11/. Pandigital mega puzzle, permutation power pair. ^ Raised to the powder of..;-)) 2^ (7*3 - .9Í recurring) = 1048 576. (9-2)^(2^3) = 576 4801. 12/ Matthew chpt 18 v 22. Forgive your brother 70 times 7== 77 times? Seven up [CapitalEth] again, dominion xx? Evening post? Wellington 1-9-1986. Don. Sally Jessie Raphael ,tv talk show, 24-5-2000. ñMatt 18: Denies he is the father of ErikaÍs son Jacob..î Lectionary, GH 7777, 77 JBM, JB 77. Macdonald Crescent. Prof Ian Stewart, BMJ Brit Medical Journal, Scientific American.., Lamech 777.. ïcircumseptual...???Í Etc. Forgive. Don S. McDonald ... Climate pride# === Subject: don.mcdonald math WGTON.les miserables One a penny, two a penny,**** Hot cross b EPIPHANY. by Victor Hugo, scene theft silver. posting-account=TV2szgkAAACrA1vyuh8IN_0zzgzcwogw 1.1.4322; .NET CLR 2.0.50727; InfoPath.1),gzip(gfe),gzip(gfe) > 8/ æ æles miserables æ æby Victor Hugo, æ scene theft silver. Hot Cross Buns Traditional English a Easter Song at EasterBunny's.Net. *************One a penny, two a penny,**** Hot cross buns! If you have no daughters Give them to your sons One a penny, two a penny, www.easterbunnys.net/hotcrossbuns.htm .87 Cached page maths safety belts! greetings FROM don.mcdonald. NEW ZEALAND. EPN 11... 12 exponential epiphany pepper petrol porsche rego. Cambridge..WGTN. 1 penny wgtn combined. photo. nz herald 5-1-99?? leap second spring //..> 1-9-2008. s hemisphere. 16.09 metres in mile xx. links... Mercurial factorisation (HG) OF 7817 no less (NL.) Group: nz.general welcom.gen.nz (Don Mcdonald) Groups: sci.math Number plate HG 7817 is simply not divisible evenly by 2, 3, 5, 7, 11, 13, 17. I have photographed HG / OF / NL 7817, 3 cars within 5 minutes of where I live. Mercurial factorisation (HG) OF 7817 no less (NL.) Don Mcdonald . (Wellington New Zealand. ... Nov 9 2006 by don.lo...@paradise.net.nz - 1 message - 1 author isprime(431),Isprime(619). Isprime(7817). (fwd) Group: sci.math not divisible by 2 3 5 7 11 13 17 mercurial factorisation - which is based on similar principles and G. 7^510 = 1.000 000E431 PENG DICT CUR + INTG NOS. 1997. and M. smallest integer with 360 factors ... in The Dominion wtd sell K. (new zealand) ... Aug 15 1998 by don - 1 message - 1 author Mersenne numbers, Mp163 Group: nz.general And Mercurial factorisation HG 7817 etc. LN 2718, PI 315, PI 180. All in Newtown. == 5200 -2*52. = 14^2 x(5^2+1.) I have also eliminated some exponents about recent world records squarers (perfect powers 8281..) are rarer than premiers./ sci.math Number plate HG 7817 is simply not divisible evenly by 2, 3, 5, 7, 11, 13, 17. I have photographed HG / OF / NL 7817, 3 cars about 10 mins from home. Mercurial factorisation (HG) OF 7817 no less (NL.) new life NL 8333 = 13*641. one ... Nov 10 2006 by don.lo...@paradise.net.nz - 2 messages - 1 author 44th Mersenne prime discovered..2006.flash. 2^32582657th-1. Group: nz.general S% = -23 factor = 361 FIRST = 19 * 19 contin cs.= 3978 contin cs.= 3984 S% = -10 factor = 21 FIRST = 3 * 7 contin cs.= 3992 S% = 3 factor = 23 FIRST PRIME. (test dup. factors). .... PrimeCabi, end . 17/6/94, 5.01.04. RUN AGAIN, Enter G/go, else end.? donald mcdonald new zealand. 15/9/06. don.lotto. Sep 15 2006 by don.lo...@paradise.net.nz - 5 messages - 2 authors FWD:42nd Mersenne Prime Found Group: alt.math don.lo...@paradise.net.nz alt math nz general nz test don.mcdonald 27.02.05 23:23 23:49 SENT. FWD:42nd Mersenne Prime Found February 26, 2005--Less than a .... 1 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199 211 223 227 ... Feb 28 2005 by don.lo...@paradise.net.nz - 4 messages - 4 authors Primes and factors. [LONG 500 lines.] Group: nz.general Don McDonald don.lo...@paradise.net.nz alt math nz general 19.04.04 00:40. SENT. 21.04.04 00:27 edited an.societ.Science.Mathematic.NZMM.prim +fact an.societ.Science.Mathematic.NZMM. .... 2007835830= 2*3*3*3*5*7*11*13*17*19*23*all 7cs 86) R%(prob-83)..=2111136084= 2^2*3*3*29*31*37*41*43*all 7cs 87) R% (prob-83). ... twin primes about 43 rd known mersenne prime exponent. Group: alt.math don.lo...@paradise.net.nz alt math nz test 26.12.05 23:17 27.12.05 00:10 02:05. sent. Tuesday ....Calc.Factors.FermatMers.Mersenne. ..... 2 * (9) 7600614 LEAST PRIME FACTOR 2 * (10) 3800307 LEAST PRIME FACTOR 3 * (11) 1266769 LEAST PRIME FACTOR 7 * (12) 180967 LEAST PRIME FACTOR 37 * (13) 4891 LEAST PRIME FACTOR 67 ... Dec 27 2005 by don.lo...@paradise.net.nz - 1 message - 1 author factoring Group: nz.general Don McDonald don.lo...@paradise.net.nz alt math sci math nz general factoring what are the factors of 48? 24.01.04 00:04 I'll just briefly comment on the vague .... 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 ENTER POSITIVE INTEGER (EXPRESSION) 0= END, RETN = REPEAT?24 24=24 hit spaces ... XXX apology --Alan BOYD did not have a brother.. very sorry. completely my mistake [confused.] xxxxxxxxxxxxxxxxx Don S. McDonald ... Climate pride# Go Bus! Snap Park, car u shoot.Look after planet,thru roof 2save? Get.. Enrg effic htr 100W incand.[ click.] don.mcdonald math WGTON.les miserables EPIPHANY. æ æby Victor Hugo, > scene theft silver. 1/ some other messages, mnemonic pppp. > Put on your maths safety helmet!! 8/...Pursue parapet, DVD ,ST cubert æ Sunday, 31-8-2008. > Alist (parabola paragraph Don.) 9/ Rob... Guest..,. æ;-)) æ JEAN VALJEAN, lead actor. > ñrepays the BishopÍs hospitality by stealing his silver..î Legs leaden limbs strained injury, Olympic XXIX Beijing 8.8.08. > Alchemy. > Aotea Centre, Auckland, æ 6.1.1999. æJanuary. > New Zealand Herald, new year ñ99., don letters, leap second, > atmosphere gravity, privileged planet. ñCARRY ï1.î- ætake up his cross, number plate- haulage, breath heaven, > lectionary. Matt 16:24. > Proof primality, epiphany Papatoetoe, 619-, performance present, photo > purse. > Perfect 45-th possibly discovered, æMersenne prime record, pearl > price, 16.9, piecewise square [marty F. prayer] > Hexadecimal September. NZ Science Monthly, February 1999. ñInteger sequences lookup Hati > Honouring a number theorist Alan R BOYD æ, Hati numbers, N= N.2^a.3^b. > Links.. XXXXXXXX æ hati > Mercurial factorisation HG 7817 > 619 primality. > Brit Med Jl. 11/. Pandigital mega puzzle, permutation power pair. ^ Raised to the > powder of..;-)) > 2^ (7*3 - .9Í recurring) = 1048 576. > (9-2)^(2^3) = æ576 4801. 12/ æMatthew æchpt 18 v 22. æForgive your brother 70 times 7== æ77 > times? > Seven up [CapitalEth] again, dominion xx? Evening post? Wellington æ1-9-1986. > Don. æSally Jessie Raphael ,tv talk show, 24-5-2000. > ñMatt 18: Denies he is the father of ErikaÍs son Jacob..î > Lectionary, GH 7777, 77 JBM, JB 77. æMacdonald Crescent. > Prof Ian Stewart, BMJ Brit Medical Journal, Scientific American.., > Lamech 777.. ïcircumseptual...???Í > Etc. Forgive. Don S. McDonald ... æClimate pride# === Subject: Another stupid question about the square root of a prime number In a previous post it was shown by those obviously more learned than I that the square root of a prime number is an irrational number. And again I am curious, perhaps again revealing my ignorance, but this irrational number that is the square root of a prime, we could actually start to compute it, and I wonder, what if anything can we say about the actual stream of digits that would be produced by the computation of the square root of a prime number? Would they bear any formal similarity to the stream of digits produced by, say, generating pi? === Subject: Re: idmoochirp1 room sevice 9/7/8 - grandparents day posting-account=GkLPvwoAAABtpmIFdAhqh_Ql7t8dpaFY Gecko/20080702 Firefox/2.0.0.16,gzip(gfe),gzip(gfe) I am reconsidering if i am or am not judgement proof against slander > lawsuits Around Here, Who Cares! > Only totally Insane Foaming Trolls would dare take such actions! Today I am one scared angel. So far past Scared, came back to normal, then became Arthurian! > So bad Nemesis comforted me! > aka Karma Madame Justice! > The Oracle of Delphi who got Raped. Amazing! One guy makes no sense at all and the other makes everything senseless... Saint === Subject: Re: Bases for an Origin Any origin is never an infinitesimal point, but a sphere of small radius > that forms a basis for measurement. When the radius is small, the sphere > approximates an infinitesimal point, and measurements from either the inner > or outer surface of the sphere are the same. Stop right there and explain, please. Please let him on through. He has important appointments elsewhere. > What do you mean by the outer and > inner surface? The surface of a sphere has no thickness. If you're thinking of a sphere made, say, out of physical material, so > that it has some thickness, then the outer and inner surface area are > ALWAYS a little different, regardless of how small the radius gets. You have to explain what you mean. === Subject: Do they still teach complex numbers in high school? Just wondering if the standard curriculum here in the US includes an introduction to complex numbers. Are any of you readers high school teachers? I'm wondering if/when/how complex numbers are introduced to high school students these days. Anyone know? --Lynn === Subject: Re: A lot of Solution Manuals in Electronic (PDF)Format! posting-account=y5oWFAoAAACMSSr5IgcS06ZO0wjPJSdh 1.1.4322; .NET CLR 2.0.50727; .NET CLR 3.0.04506.30; InfoPath.2; .NET CLR 3.0.04506.648),gzip(gfe),gzip(gfe) Introduction to heat transfer, Incropera 5th edition Applied Mathematics and Modeling for Chemical Engineers, R.G. Rice Please let me know if you can get these, as soon as possible would be helpful. JP === Subject: Re: Algebraic integer result with quadratics posting-account=aLpfCwoAAACh4BOs3HOlQBCoxUpEgyxc Gecko/2008070208 Firefox/3.0.1,gzip(gfe),gzip(gfe) > On Aug 30, 5:01æpm, W. Dale Hall A correction: Rick Decker has noted (correctly that my values for > w's here are in error: the sqrt() radicand should be -503, and > not -26. He was kind enough to observe this as the typographical > error that it was. Yeah, I figured it had to be wrong, as the radical ALWAYS must shift > as x shifts or it's trivial to prove that 7 must be a factor of just > one of the w's, which is just kind of one of those funny things. I think math people say it's the class number or something or other > that has to shift based on Galois Theory, so Rotwang didn't realize > it, but he was arguing against GT. It's quite possible. If somebody who actually knows something about Galois theory knows that what I'm trying to do is impossible, by all means speak up. In the mean time here's another stab at the Object ring. I don't know how to compute algebraic integer GCD's myself[1], so let's instead consider another polynomial, which shares many of the features with your original one, but is simple enough that some of the w's can be found by inspection: Q(x) = 25x^2 + 5x + 2 Like your original polynomial this is primitive and irreducible over the rationals. Also like your original polynomial, we can factor 7*Q(x) as 7*Q(x) = (5a 1(x) + 7)(5a 2(x) + 7) where the a's are now the roots of a^2 - (x - 1)a + 7x^2 = 0. Like your original polynomial, at x = 0 we have that one of the a's is equal to zero, so that at x = 0 we can divide the 7 out of one term. So it looks the argument you used with your original polynomial, if correct, should apply to this polynomial also. Is this the case? If not, what is the difference which means that your argument no longer applies? Assuming that your argument does apply to Q(x), consider now x = 1. Here we have a = +/- sqrt(-7) so at this value of x, 7 is not a factor of either of the terms (5a i(1) + 7). But we can find w 1(1) and w 2(1) such that w 1(1)*w 2(1) = 7 and w i divides (5a i(x) + 7) such that all coefficients remain in the algebraic integers, namely w 1(1) = w 2(1) = sqrt(7) Is this correct so far? Assuming it is, I believe now that your claim is that the w's must wrappers, meaning that the factorisation 7 = w 1*w 2 is equivalent up to Object ring units to 7 = 7*1; in other words, one of the w's has 7 as a factor in the Object ring. But then the Object ring contains sqrt(7)/7 = 1/sqrt(7), and therefore also contains 1/7. [1] I had a quick look at the Kash documentation but I found myself pretty far out of my depth. If someone can refer me to somewhere I can learn what I need to know to calculate the GCD's I'll be grateful. === Subject: Re: Algebraic integer result with quadratics > On Aug 30, 5:01 pm, W. Dale Hall > A correction: Rick Decker has noted (correctly that my values for > w's here are in error: the sqrt() radicand should be -503, and > not -26. He was kind enough to observe this as the typographical > error that it was. > Yeah, I figured it had to be wrong, as the radical ALWAYS must shift > as x shifts or it's trivial to prove that 7 must be a factor of just > one of the w's, which is just kind of one of those funny things. > I think math people say it's the class number or something or other > that has to shift based on Galois Theory, so Rotwang didn't realize > it, but he was arguing against GT. It's quite possible. If somebody who actually knows something about > Galois theory knows that what I'm trying to do is impossible, by all > means speak up. In the mean time here's another stab at the Object > ring. I don't know how to compute algebraic integer GCD's myself[1], > so let's instead consider another polynomial, which shares many of the > features with your original one, but is simple enough that some of the > w's can be found by inspection: Q(x) = 25x^2 + 5x + 2 Like your original polynomial this is primitive and irreducible over > the rationals. Also like your original polynomial, we can factor > 7*Q(x) as 7*Q(x) = (5a_1(x) + 7)(5a_2(x) + 7) where the a's are now the roots of a^2 - (x - 1)a + 7x^2 = 0. Like your original polynomial, at x = 0 we have that one of the a's is > equal to zero, so that at x = 0 we can divide the 7 out of one term. > So it looks the argument you used with your original polynomial, if > correct, should apply to this polynomial also. Is this the case? If > not, what is the difference which means that your argument no longer > applies? Assuming that your argument does apply to Q(x), consider now x = 1. > Here we have a = +/- sqrt(-7) so at this value of x, 7 is not a factor of either of the terms > (5a_i(1) + 7). But we can find w_1(1) and w_2(1) such that > w_1(1)*w_2(1) = 7 and w_i divides (5a_i(x) + 7) such that all > coefficients remain in the algebraic integers, namely w_1(1) = w_2(1) = sqrt(7) Is this correct so far? Assuming it is, I believe now that your claim > is that the w's must wrappers, meaning that the factorisation 7 = > w_1*w_2 is equivalent up to Object ring units to 7 = 7*1; in other > words, one of the w's has 7 as a factor in the Object ring. But then > the Object ring contains sqrt(7)/7 = 1/sqrt(7), and therefore also > contains 1/7. > [1] I had a quick look at the Kash documentation but I found myself > pretty far out of my depth. If someone can refer me to somewhere I can > learn what I need to know to calculate the GCD's I'll be grateful. Here's how to get started with Kash. I'll compute the w's for x = 4, and give a transcription of my session. Then I'll go through and try to provide some indication of what on Earth I just did. A min[aj]or caveat: I'm no algebraist. Although I've taken adequate coursework in algebra and commutative algebra to be able to decipher the instructions, I've found I'm somewhat behind the curve. Most any practicing algebraist will be able to correct my misconceptions, after recovering from his laughing fit. A couple of key things. X must be capitalized for KASH to recognize the indeterminate in a polynomial. Lists are [thing1, thing2, ..., thingn], where thing.. must have been defined already. Brackets & commas are compulsory, ellipses are not allowed. Help works like this: ?Keyword gives you help on Keyword if that is an exact & completet keyword. ?*eyword gives you help on all completions of eyword It's a good idea to set the window where KASH starts to have a HUGE scrollback range. Be prepared for KASH to complain lots & lots. The .pdf files are more useful if you have a function in mind, and do serve to spell out a few examples explicitly. I'd bet they're a heap more useful to real algebraists. If you have questions, I'll be glad to look into them, but only realize I'm a rank amateur at this program. I'll sign of here and let you continue your reading program... Dale Here we go: *** REMARK: THIS IS THE INTRO TO THE SESSION *** ooooqp oooo oooo .o. .oooooo..o ooooo ooooo .dP' `888 .8P' .888. d8P' `Y8 `888' `888' d88b. 888 d8' .8'888. Y88bo. 888 888 o. )8 88888K .8' `888. `'Y8888o. 888ooooo888 `888P' 888`88b. .88ooo8888. `'Y88b 888 888 888 `88b. .8' `888. oo .d8P 888 888 o888o o888o o88o o8888o 8''88888P' o888o o888o Shell of the KANT V4 Software, Version 3, build: Windows-2005-11-19 Copyright (c) 1994-2005 Prof. Dr. M. E. Pohst, Technische Universitaet Berlin. All rights reserved. For registration and support send an email to kant@math.tu-berlin.de -------------------------------------------------------------------- KANT V4 is based on Magma developed by Prof. J. Cannon, Copyright (c) 2002 Prof. J. Cannon, University of Sydney. Shell is based on GAP developed by Lehrstuhl D Mathematik, RWTH Aachen, Copyright (c) 1992 Lehrstuhl D Mathematik, RWTH Aachen. Enter ? for help and quit; to leave KASH list [0] alist [0] term [0] __DOC [0.148] doc [0.002] docui [0.123] docui [0.0 02] method [0.002] init-methods [0.117] constants [0.007] kash [0.002] matrix [0.005] map [0.008] qaos [0.004] locFact [0.01] *** REMARK: END OF SESSION INTRO *** *** REMARK: First I'll define the coefficients of *** REMARK: a^2 - (7 x - 1)a + (49 x^2 - 14 x): kash% q1 := -(7*X - 1); -7*X + 1 Time: 0 s kash% q0 := 49*X^2 - 14*X; 49*X^2 - 14*X Time: 0 s *** REMARK: set x equal to 4: kash% c1 := Evaluate(q1,4); -27 Time: 0 s kash% c0 := Evaluate(q0,4); 728 Time: 0 s *** REMARK: define the polynomial that sets a's value: kash% p := X^2 + c1*X + c0; X^2 - 27*X + 728 Time: 0.044 s *** REMARK: an order of a number field is a free abelian group ord *** REMARK: that spans the field over Q; in Kash, the equation order *** REMARK: of the polynomial p is the basis for Q[x]/

generated *** REMARK: by the powers of x from x^0 to x^(deg(p)-1). The element *** REMARK: corresponding to x is one root of the polynomial p *** REMARK: *** REMARK: a link for order *** REMARK: http://en.wikipedia.org/wiki/Maximal_order kash% o := EquationOrder(p); Equation Order with defining polynomial X^2 - 27*X + 728 over Z Time: 0.124 s *** REMARK: The maximal order associated to the order ord is an *** REMARK: order Ord, that spans the integers of the field over Z *** REMARK: It may be the same as the order ord, but may be strictly *** REMARK: larger, as it turns out in this case. kash% O := MaximalOrder(o); Maximal Equation Order with defining polynomial X^2 - 27*X + 728 over Z Time: 0.129 s *** REMARK: We need the maximal order to compute the class group *** REMARK: which we do here: kash% ClassGroup(O); Abelian Group isomorphic to Z/42 Defined on 1 generator Relations: 42*_HE.1 = 0, extended by: ext1 := Mapping from: grp^abl: _HE to ids/ord^num: _EJ Time: 0.752 s *** REMARK: Here, I'm trying to find out about the *** REMARK: ideal generated by the root (a) and 7 kash% e1 := Element(O,[0,1]); [0, 1] Time: 0.032 s *** REMARK: First to test whether this is (a) kash% Evaluate(p,e1); [0, 0] Time: 0.002 s *** REMARK: It is. Next to find the ideal <7, a>: kash% Ia := Ideal(O,[7, e1]); Ideal of O Two element generators: [7, 0] [0, 1] Time: 0.091 s *** REMARK: Is there a common divisor here? kash% IsPrincipal(Ia); FALSE, extended by: ext1 := Unassign Time: 0.02 s *** REMARK: nope. Next, I raise Ia to powers that divide 42, *** REMARK: the order of the ideal class group (Lagrange's *** REMARK: theorem says these are the only possible orders *** REMARK: (different meaning of order) of elements of *** REMARK: the class group. kash% IsPrincipal(Ia^2); FALSE, extended by: ext1 := Unassign Time: 0.087 s kash% IsPrincipal(Ia^3); FALSE, extended by: ext1 := Unassign Time: 0.001 s kash% IsPrincipal(Ia^6); FALSE, extended by: ext1 := Unassign Time: 0.001 s kash% IsPrincipal(Ia^7); FALSE, extended by: ext1 := Unassign Time: 0 s kash% IsPrincipal(Ia^14); FALSE, extended by: ext1 := Unassign Time: 0.027 s kash% IsPrincipal(Ia^21); TRUE, extended by: ext1 := [601438915, -31021218] Time: 0.001 s *** REMARK: Oh boy. I only have to take the 21st power *** REMARK: to get a common divisor! *** REMARK: Here's the divisor, in terms of 1 and a: kash% a_gen := IsPrincipal(Ia^21).ext1; [601438915, -31021218] Time: 0.001 s kash% a_gen - (601438915 - 31021218*e1); [0, 0] Time: 0.001 s *** REMARK: Oops, I wanted this in terms of 1 and *** REMARK: sqrt(discriminant(p)): kash% Discriminant(p); -2183 Time: 0 s *** REMARK: Just curious. How does discriminant(p) factor? kash% Factorization(2183); [ <37, 1>, <59, 1> ], extended by: ext1 := 1, ext2 := Unassign Time: 0.042 s *** REMARK: Remind me what p is: kash% p; X^2 - 27*X + 728 Time: 0.001 s *** REMARK: Let's take the extension by just including *** REMARK: the square root of the discriminant (-2183): kash% ox := EquationOrder(X^2 + 2183); Equation Order with defining polynomial X^2 + 2183 over Z Time: 0 s kash% Ox := MaximalOrder(ox); Maximal Order of ox Time: 0.005 s *** REMARK: And here's its class group. We hope I find *** REMARK: the same class group as before: kash% ClassGroup(Ox); Abelian Group isomorphic to Z/42 Defined on 1 generator Relations: 42*_TD.1 = 0, extended by: ext1 := Mapping from: grp^abl: _TD to ids/ord^num: _VY Time: 0.242 s *** REMARK: Goody. Now to write (a) in terms of 1 and sqrt(disc(p)): *** REMARK: First, the generator [0,1] of the order Ox: kash% x1 := Element(Ox, [0,1]); [0, 1] Time: 0 s *** REMARK: Is it really sqrt(-2183)? kash% Evaluate(X^2 + 2183,x1); [1637, 1] Time: 0 s *** REMARK: Sheeeit. Now I gotta dig. *** REMARK: Let's find out what on earth it is: kash% MinimalPolynomial(x1); X^2 - X + 546 Time: 0.041 s *** REMARK: OK, so it's (1 + sqrt(disc(thisthing)))/2 kash% Discriminant(MinimalPolynomial(x1)); -2183 Time: 0 s *** REMARK: or, (1 + sqrt(-2183))/2. *** REMARK: Solving (by hand), I see that *** REMARK: sqrt(-2183) = -1 + 2*root: kash% Evaluate(X^2 + 2183, -1 + 2*x1); [0, 0] Time: 0 s *** REMARK: Yay. Let's give it a name and proceed: kash% r1 := -1 + 2*x1; [-1, 2] Time: 0 s kash% MinimalPolynomial(r1); X^2 + 2183 Time: 0 s *** REMARK: I forgot what p was again, duh: kash% p; X^2 - 27*X + 728 Time: 0.001 s kash% Discriminant(p); -2183 Time: 0 s *** REMARK: and I know r1 = sqrt(-2183), so I *** REMARK: can write (a) out directly: kash% a_1 := (27 + r1)/2; 13/1*_GT.1 + _GT.2 Time: 0.001 s kash% MinimalPolynomial(a_1); _SM.1^2 - 27*_SM.1 + 728 Time: 0 s *** REMARK: OK, it's got the right minimal polynomial *** REMARK: Next check on p itself: kash% Evaluate(p, a_1); 0 Time: 0.001 s *** REMARK: and I proceed to define that ideal again: kash% Ia_1 := Ideal(Ox, [7, a_1]); Ideal of Ox Two element generators: [7, 0] [13, 1] Time: 0.001 s *** REMARK: I have to raise it to the 21st power to *** REMARK: get a principal ideal: *** REMARK: kash% IsPrincipal(Ia_1^21); TRUE, extended by: ext1 := [198163081, -31021218] Time: 0.001 *** REMARK: Are these coefficients the same *** REMARK: I got earlier? kash% a_gen % ; [601438915, -31021218] *** REMARK: Not entirely. the second one is OK, but *** REMARK: not the first. I shouldn't have expected *** REMARK: more, since the two maximal orders have *** REMARK: different generating sets. Time: 0 s kash% a1_gen := IsPrincipal(Ia_1^21).ext1; [198163081, -31021218] Time: 0.001 s *** REMARK: I'll copy the coefficients, since I *** REMARK: need to do some arithmetic: kash% coeff1 := 198163081; 198163081 Time: 0.001 s kash% coeff2 := -31021218; -31021218 Time: 0 s *** REMARK: I'll rewrite a1_gen: kash% coeff1 + coeff2*x1; [198163081, -31021218] Time: 0 s kash% coeff1 + coeff2*x1 - a1_gen; [0, 0] Time: 0.001 s *** REMARK: I forgot how r1 ( = sqrt(-2183)) *** REMARK: came out in its basis kash% r1 := -1 + 2*x1; [-1, 2] Time: 0 s *** REMARK: so I adjust coefficients to write *** REMARK: the generator in terms of 1 & sqrt(-2183): kash% k2 := coeff2/2; -15510609 Time: 0 s kash% k1 := coeff1 + k2; 182652472 Time: 0.001 s *** REMARK: here it should be: kash% k1 + k2*r1; 198163081/1*_GT.1 - 31021218/1*_GT.2 Time: 0.001 s *** REMARK: Let's check: kash% k1 + k2*r1 - a1_gen; 0 Time: 0 s *** REMARK: I'll give it a new name: kash% gen1 = k1 + k2*r1; Error, the variable 'gen1' must have a value *** REMARK: Darn it. definitions are :=, not = kash% gen1 := k1 + k2*r1; 198163081/1*_GT.1 - 31021218/1*_GT.2 Time: 0 s *** REMARK: And I'll find the complex conjugate: kash% gen2 := k1 - k2*r1; 167141863/1*_GT.1 + 31021218/1*_GT.2 Time: 0 s *** REMARK: Do these numbers give the correct product? kash% gen1*gen2; 558545864083284007/1*_GT.1 Time: 0 s kash% 7^21; 558545864083284007 Time: 0 s *** REMARK: Looks OK, but I never trust my eyes any more: kash% gen1*gen2 - 7^21; 0 Time: 0 s kash% *** REMARK: Looks like my work is finished for now. *** REMARK: END OF KASH SESSION And there we have it: for x = 4, the factorization of 7 comes out like this: w(4) = (182652472 +/- 15510609 sqrt(-2183))^(1/21) That's it. I'm sure it would have been simpler if I knew what I was doing. Dale === Subject: Re: Algebraic integer result with quadratics [cut] > I don't know how to compute algebraic integer GCD's myself[1], [cut] > [1] I had a quick look at the Kash documentation but I found myself > pretty far out of my depth. If someone can refer me to somewhere I can > learn what I need to know to calculate the GCD's I'll be grateful. I'm not sure myself, but I think it goes something like this. Suppose we want to find the GCD of the two algebraic integers a and b. Consider the ideal (a, b) generated by a and b. By the finiteness of the class number, there is a finite natural number h and an algebraic integer c such that the ideal (c) generated by c is equal to (a, b)^h. I am being sloppy here, since I am not specifying the ring I am working in. For simplicity, I will assume that the class number h of the ideal generated by a and b is equal to 2. Thus, (a, b)^2 = (c). Find an algebraic integer e such that e^2 = c. I will again be sloppy here, since I am now working in an even larger ring containing a, b, c and e. Thus, I have (a, b)^2 = (e^2). I claim that e is a GCD of a and b. That is, I claim that GCD(a, b) = e. Note that (a, b)^2 = (a^2, a*b, b^2). Thus, a^2 and b^2 are multiples of e^2. Thus, a and b are multiples of e. I don't know if this is obvious or not, but at least it follows from the unique factorization of ideals (I think). Hence, e is a common divisor of a and b. [In more detail: the ideal (a^2) is a subset of the ideal (e^2); so, the ideal (a) times the ideal (a) is a subset of the ideal (e) times the ideal (e); so, the ideal (a)*(a) = (e)*(e)*J for some ideal J; by unique factorization of ideals, the number of times a prime ideal factor occurs in the ideal (e) must be at least the number of times that prime ideal occurs as a prime ideal factor of the ideal (a); so it follows that the ideal (a) is a subset of the ideal (e).] Conversely, let d be a common divisor of a and b. Then, the ideals (a^2), (a*b), and (b^2) are subsets of the ideal (d^2). Since (e^2) equals the ideal (a^2, a*b, b^2), the ideal (e^2) is a subset of the ideal (d^2). Like above, it follows that the ideal (e) is a subset of the ideal (d). Thus, e is a multiple of d. Hence, e is a greatest common divisor of a and b. I guess all the above can be shortened to just noting that (a, b) = (e) follows from (a, b)^2 = (e)^2 by using the unique factorization of ideals. The hard part is figuring out what rings you are working in. Also, I think there was some discussion of what the definition of GCD is in this situation. Also note that I am using the unique factorization of ideals into prime ideals (which is relatively easy to prove, although there is one very tricky step in doing so) and the finiteness of the class number (which is a hard thing to prove). -- Bill Hale === Subject: Re: Algebraic integer result with quadratics posting-account=aLpfCwoAAACh4BOs3HOlQBCoxUpEgyxc Gecko/2008070208 Firefox/3.0.1,gzip(gfe),gzip(gfe) [cut] I don't know how to compute algebraic integer GCD's myself[1], > [cut] > [1] I had a quick look at the Kash documentation but I found myself > pretty far out of my depth. If someone can refer me to somewhere I can > learn what I need to know to calculate the GCD's I'll be grateful. I'm not sure myself, but I think it goes something like this. up Ideal class on planetmath) so I have a few questions and I hope you will forgive me if I say anything thick. Suppose we want to find the GCD of the two algebraic integers a and b. > Consider the ideal (a, b) generated by a and b. By the finiteness of the > class number, there is a finite natural number h and an algebraic > integer c such that the ideal (c) generated by c is equal to (a, b)^h. > I am being sloppy here, since I am not specifying the ring I am working > in. At this point I suppose we can take the ring R to be the ring of algebraic integers in Q(a,b), which is a finite extension so that the class number is finite. According to what I've read, the class number is the number of equivalence classes of ideals, where two ideals I and J are equivalent (written I~J) if there exist non-zero ring elements alpha and beta such that (alpha)I = (beta)J. I think it is easy to show that an ideal I is principal iff I is equivalent to the whole ring (1). The statement you give above then follows from the finiteness of the class number together with the fact that the ideal class group is a group, since we must have (a, b)^n = (a, b)^m for some n > m from finiteness, and therefore that (a, b)^{n - m) = (1) from general properties of groups. Is it easy to show that the ideal class group is indeed a group? > For simplicity, I will assume that the class number h of the ideal > generated by a and b is equal to 2. Thus, (a, b)^2 = (c). Find an > algebraic integer e such that e^2 = c. I will again be sloppy here, > since I am now working in an even larger ring containing a, b, c and e. I think this is OK. Let's now consider the whole ring A of algebraic integers, and define a map f from the set of ideals of R to those of A, such that f(I) is the smallest ideal of A containing I. Then we can show that f((x,y,z,...) R) = (x,y,z,...) A (using a notation whose meaning I hope is obvious), and f(I)f(J) = f(IJ) so that the relation (a, b)^2 = (c) still holds in A. > Thus, I have (a, b)^2 = (e^2). I claim that e is a GCD of a and b. That > is, I claim that GCD(a, b) = e. Note that (a, b)^2 = (a^2, a*b, b^2). Thus, a^2 and b^2 are multiples of > e^2. Thus, a and b are multiples of e. I don't know if this is obvious > or not, but at least it follows from the unique factorization of ideals > (I think). Much simpler than that, I think - the algebraic integers are closed under extraction of roots, so if e.g. a^2/e^2 is in A then so is a/e. > Hence, e is a common divisor of a and b. [In more detail: the > ideal (a^2) is a subset of the ideal (e^2); so, the ideal (a) times the > ideal (a) is a subset of the ideal (e) times the ideal (e); so, the > ideal (a)*(a) = (e)*(e)*J for some ideal J; by unique factorization of > ideals, the number of times a prime ideal factor occurs in the ideal (e) > must be at least the number of times that prime ideal occurs as a prime > ideal factor of the ideal (a); so it follows that the ideal (a) is a > subset of the ideal (e).] Conversely, let d be a common divisor of a and b. Then, the ideals > (a^2), (a*b), and (b^2) are subsets of the ideal (d^2). Since (e^2) > equals the ideal (a^2, a*b, b^2), the ideal (e^2) is a subset of the > ideal (d^2). Like above, it follows that the ideal (e) is a subset of > the ideal (d). Thus, e is a multiple of d. Hence, e is a greatest common > divisor of a and b. Looks good. > I guess all the above can be shortened to just noting that (a, b) = (e) > follows from (a, b)^2 = (e)^2 by using the unique factorization of > ideals. The hard part is figuring out what rings you are working in. Also, I > think there was some discussion of what the definition of GCD is in this > situation. My understanding is that a GCD of a and b is defined to be an e such that e|a, e|b and ((d|a & d|b) -> d|e). I thought this was standard. > Also note that I am using the unique factorization of ideals into prime > ideals (which is relatively easy to prove, although there is one very > tricky step in doing so) and the finiteness of the class number (which > is a hard thing to prove). -- Bill Hale === Subject: Re: Algebraic integer result with quadratics > [cut] > I don't know how to compute algebraic integer GCD's myself[1], > [cut] > [1] I had a quick look at the Kash documentation but I found myself > pretty far out of my depth. If someone can refer me to somewhere I can > learn what I need to know to calculate the GCD's I'll be grateful. > I'm not sure myself, but I think it goes something like this. up Ideal class on planetmath) so I have a few questions and I hope > you will forgive me if I say anything thick. > Suppose we want to find the GCD of the two algebraic integers a and b. > Consider the ideal (a, b) generated by a and b. By the finiteness of the > class number, there is a finite natural number h and an algebraic > integer c such that the ideal (c) generated by c is equal to (a, b)^h. > I am being sloppy here, since I am not specifying the ring I am working > in. At this point I suppose we can take the ring R to be the ring of > algebraic integers in Q(a,b), which is a finite extension so that the > class number is finite. According to what I've read, the class number is the number of > equivalence classes of ideals, where two ideals I and J are equivalent > (written I~J) if there exist non-zero ring elements alpha and beta > such that (alpha)I = (beta)J. I think it is easy to show that an ideal I is principal iff I is > equivalent to the whole ring (1). The statement you give above then > follows from the finiteness of the class number together with the fact > that the ideal class group is a group, since we must have (a, b)^n = > (a, b)^m for some n > m from finiteness, and therefore that (a, b)^{n > - m) = (1) from general properties of groups. Is it easy to show that > the ideal class group is indeed a group? > For simplicity, I will assume that the class number h of the ideal > generated by a and b is equal to 2. Thus, (a, b)^2 = (c). Find an > algebraic integer e such that e^2 = c. I will again be sloppy here, > since I am now working in an even larger ring containing a, b, c and e. I think this is OK. Let's now consider the whole ring A of algebraic > integers, and define a map f from the set of ideals of R to those of > A, such that f(I) is the smallest ideal of A containing I. Then we can > show that f((x,y,z,...)_R) = (x,y,z,...)_A (using a notation whose meaning I hope is obvious), and f(I)f(J) = f(IJ) so that the relation (a, b)^2 = (c) still holds in A. > Thus, I have (a, b)^2 = (e^2). I claim that e is a GCD of a and b. That > is, I claim that GCD(a, b) = e. > Note that (a, b)^2 = (a^2, a*b, b^2). Thus, a^2 and b^2 are multiples of > e^2. Thus, a and b are multiples of e. I don't know if this is obvious > or not, but at least it follows from the unique factorization of ideals > (I think). Much simpler than that, I think - the algebraic integers are closed > under extraction of roots, so if e.g. a^2/e^2 is in A then so is a/e. > Hence, e is a common divisor of a and b. [In more detail: the > ideal (a^2) is a subset of the ideal (e^2); so, the ideal (a) times the > ideal (a) is a subset of the ideal (e) times the ideal (e); so, the > ideal (a)*(a) = (e)*(e)*J for some ideal J; by unique factorization of > ideals, the number of times a prime ideal factor occurs in the ideal (e) > must be at least the number of times that prime ideal occurs as a prime > ideal factor of the ideal (a); so it follows that the ideal (a) is a > subset of the ideal (e).] > Conversely, let d be a common divisor of a and b. Then, the ideals > (a^2), (a*b), and (b^2) are subsets of the ideal (d^2). Since (e^2) > equals the ideal (a^2, a*b, b^2), the ideal (e^2) is a subset of the > ideal (d^2). Like above, it follows that the ideal (e) is a subset of > the ideal (d). Thus, e is a multiple of d. Hence, e is a greatest common > divisor of a and b. Looks good. > I guess all the above can be shortened to just noting that (a, b) = (e) > follows from (a, b)^2 = (e)^2 by using the unique factorization of > ideals. > The hard part is figuring out what rings you are working in. Also, I > think there was some discussion of what the definition of GCD is in this > situation. My understanding is that a GCD of a and b is defined to be an e such > that e|a, e|b and ((d|a & d|b) -> d|e). I thought this was standard. > Yes, but if the class number is greater than 1, then unique factorization can fail, and there may be no GCD. It's when is not principal, that you go to powers of the ideal , trying to reach a power that is principal. Some folks may use GCD(a,b) to denote the ideal which, when principal, does yield the older notion [as any of its generators]. > Also note that I am using the unique factorization of ideals into prime > ideals (which is relatively easy to prove, although there is one very > tricky step in doing so) and the finiteness of the class number (which > is a hard thing to prove). > -- Bill Hale Dale === Subject: Re: JSH: The Futility of Engagement > I looked into that with Mizar. Google it. trying to teach Mizar about algebraic integers. Did you ever get past this > point? Even if someone had already defined algebraic integers in Mizar, according > to you, the definition would have been suspect, wouldn't it? You would have > had to put in the correct definition of the algebraic integers anyway, > right? I'm not sure what else you were expecting. Were you hoping to be > able to generate a computer proof without having to do any work? > I've now lost interest in working with the mathematical community and > have decided it's simpler to re-form it completely and along the way > reform modern academia as well. Working with the mathematical community is one thing, but working with a > computer system is another. Do you object to Mizar per se? And if so, > are you working on an alternative system for computer checking of proofs? > It is hypocritical for you to argue that computer checking is necessary, > yet do nothing to advance the state of the art. What, you think JSH is hypocritical!? Wellll, in a teensy way if by that you mean that every criticism he levels at others is a detailed account of his own workings, then yes. -- Michael Press === Subject: Re: JSH: The Futility of Engagement > At this rate, computers will be having real conversations with people > before they can read even basic mathematical proofs, which I think is > about a will in math society to not have computers checking as the > current system is so, friendly, and buddy-buddy. It's called job security. How's yours? -- Michael Press === Subject: Re: JSH: The Futility of Engagement On Aug 30, 9:26 am, Jens Stueckelberger Why do you people waste time with guys like JSH, Basti and a few > others? These are individuals with pathological mental problems, who > obviously thrive on attention - no matter what kind of attention. You seem to be implying that denying them the attention > they crave will help them. Or are you just being selfish, > that it will only help you by lessening the chaff in this > newsgroup? What if you're wrong, that denying them the attention, > no matter what kind of attention, makes their condition > worse? What if the next time JSH visits a math professor, > heated words are exchanged in person? That insinuation is beyond the pale. Not an insinuation, those are your words. > For one thing I don't bother any more. ? You say that now. University of California at > Berkeley is nearby but I have no interest in going to the math > department there You say that now. though I have visited the campus Oh-what-a-giveaway! (ended up oddly > enough hanging out with some tennis people who were getting lessons > from the Berkeley tennis coach). Female tennis players? With short skirts? Can't fault you for that. But then he was warned off. -- Michael Press === Subject: Re: JSH: The Futility of Engagement > Why do you people waste time with guys like JSH, Basti and a few > others? These are individuals with pathological mental problems, who > obviously thrive on attention - no matter what kind of attention. Why should sociopaths have all the fun? Riddle me that. -- Michael Press === Subject: Re: JSH: The Futility of Engagement > And now you're starting threads about him. You, sir, are a JSH addict. Of course you're right, but I'm trying to recover. After all, I took the > pledge! Please let us know when they start the JSH twelve step program. That'll be just as soon as they discover a Higher Power than JSH. The Hammer. -- Michael Press === Subject: This Week's Finds in Mathematical Physics (Week 269) Originator: bergv@math.uiuc.edu (Maarten Bergvelt) Also available at http://math.ucr.edu/home/baez/week269.html August 30, 2008 This Week's Finds in Mathematical Physics (Week 269) John Baez No fancy math today. I've been working hard with Aristide Baratin, Laurent Freidel and Derek Wise on infinite-dimensional representations of 2-groups. It's a gnarly mix of higher category theory and analysis. But I won't bother you with that until the paper is done. I need a break! So, today I want to talk about the sulfur geysers on Io, honeycombs, the work of Kelvin, the Weaire-Phelan structure, and gas clathrates. I already teased you with pictures of Jupiter's moon Io in week266 and week268. There's a reason. Tortured by powerful tidal forces, Io is the most geologically active object in the Solar System! It has mountains taller than Mount Everest, and over 400 active volcanos. These put out the hottest lava ever seen - and a lot of it, too. A big eruption in 1997 produced more than 3500 square kilometers of the stuff! Most moons in the outer Solar System are pale and icy. Io looks like an evil pizza. The ghastly red ring in one of the views here is sulfur spewed out by Pele, the biggest volcano on Io: 1) NASA Photojournal, Three views of Io, http://photojournal.jpl.nasa.gov/catalog/PIA00292 Io is big on sulfur. It has lakes of molten sulfur... pale sulfur dioxide snow... and geysers that spew plumes of sulfur dioxide up to 500 kilometers high! Here's a picture of two such geysers, taken by the Galileo spacecraft in 1979: 2) Astronomy Picture of the Day, Io: the Prometheus Plume, http://apod.nasa.gov/apod/ap070211.html At top you see a bluish plume rising 140 kilometers above a massive volcano called Pillan Patera. If you look carefully you can also see the Prometheus plume, dead center. This thing has been active at least since 1979 - but it's *moved 85 kilometers west* during that time! Scary. Ironically, all this chaotic activity on Io may be caused by the music of the spheres. Io is locked in a 2:1 orbital resonance with the moon Europa, and a 4:1 resonance with Ganymede. These keep Io's orbit a bit eccentric, which causes tidal heating - to the tune of about 100 trillion watts. Another interesting thing is that the red spots on Io are made of sulfur, but so are the yellow plains. Of course there are lots of compounds involving sulfur, but even the pure element has a lot of different forms, or allotropes. I've always been fascinated by those. Back here on Earth, the Weaire-Phelan structure was in the news recently! If you watched the Olympics in Beijing, you may have seen the National Aquatics Center - a building also called the Water Cube: 3) Spectacular mathematical bubble design at the Olympics, Math in the News, Mathematical Association of America, August 8, 2008, http://mathdl.maa.org/mathDL?pa=mathNews&sa=view&newsId=392 The story behind the design of this building goes back to 36 BC, when Marcus Terentius Varro described two competing theories for why bees have hexagonal honeycombs. The first said: because bees have six legs! The second said: efficiency! You see, a hexagonal lattice lets you divide the plane into cells of equal area with the least possible perimeter per cell. So if the bees want to save wax, that's the pattern they'll pick. The second theory seems more plausible. But is it true? I'm not even sure how to resolve that question. But it's worth noting that honeycomb cells are actually 3-dimensional - and the *end* of each cell consists of three rhombi that meet at the same angle as bubbles in soap suds! Now, soap films minimize surface area subject to whatever constraints they encounter. So, a single bubble that holds a given amount of air will form a sphere. But soap suds with lots of bubbles do more complicated things. Take a bubble bath and pay careful attention! You'll see that three bubble faces meet along each edge, at precisely 120 degree angles. And when four bubbles meet at a vertex, they form a pattern with tetrahedral symmetry, with edges meeting at an angle of arccos(-1/3), or about 109.5 degrees. So if honeycombs display these patterns, we can guess area is being minimized. But as usual, things get more complicated when you look deeper. First, while everyone *believed* for a long time that a hexagonal honeycomb is the way to divide the plane into equal-area cells with minimal perimeters, this was only *proved* much later: in 1999, by Thomas Hales. It's an interesting story. Hales had just finished his epic proof of Kepler's conjecture about the densest way to pack equal-sized spheres - a proof so complicated that the referees ran out of energy trying to check it. That's quite a tale in itself... but to avoid an infinite sequence of nested digressions, I'll refer you to these: Reviewed by Frank Morgan in Notices Amer. Math. Soc. 52 (2005), 44-47. Also available at http://www.ams.org/notices/200501/rev-morgan.pdf 5) Thomas Hales, The Kepler Conjecture, http://www.math.pitt.edu/~thales/kepler98/ Anyway, after Hales proved the Kepler Conjecture, Denis Weaire suggested that he tackle the Hexagonal Honeycomb Conjecture - and Hales promptly solved that too! He said, In contrast with the years of forced labor that gave the Kepler Conjecture, I felt as if I had won the lottery. 6) Thomas C. Hales, The Honeycomb Conjecture, http://www.math.pitt.edu/~thales/kepler98/honey/ Hales wasn't the first to make progress on the Hexagonal Honeycomb Conjecture. A guy named Fejes T.97th had already proved it's true if we assume the cells are polygons: 7) L. Fejes T.97th, Regular Figures, Macmillan Co, New York, 1964. So what Hales had to do is rule out cells with curved edges. This is harder than you might think. In fact, for clusters of finitely many cells, the optimal shapes can be curved, even near the middle! You can see pictures in the review by Frank Morgan above, or here: 8) S. J. Cox, M. Fatima Vas, C. Monnereua-Pittet and N. Pittet, Minimal perimeter for N identical bubbles in two dimensions: Another thing T.97th did is carefully define the 3d optimization problem that bees might be trying to solve, and find a slightly better solution: 9) L. Fejes T.97th, What the bees know and what the bees do not know, Bull. Amer. Math. Soc. 70 (1964), 468-481. Also available at http://projecteuclid.org/euclid.bams/1183526078 I won't try to describe his results in detail, since the paper is freely available and well-written. But here's the basic idea. The end of a bee's honeycomb cell looks just like the corner of a rhombic dodecahedron. This is a 12-sided solid that you can pack to completely fill space. This makes sense, because the ends of one layer of cells in a honeycomb should neatly fit against those of the next layer. However, it's been known since the work of Kelvin that there's another solid you can use to pack space more efficiently: that is, with less surface area per cell. This is the truncated octahedron. Using this, T.97th found a design for the end of a honeycomb cell that would be more efficient than what bees use! How much more efficient? How much area did T.97th manage to shave off? Almost 0.35% of the area of cell's opening! In the eternal battle of man against bee, we triumph yet again! It makes me proud to be human. T.97th is more modest: We must admit that all this has no practical consequence.... Besides, the building style of the bees is definitely simpler than that described above. So we would fail in shaking someone's conviction that the bees have a deep geometrical intuition. I doubt intuition is the right word for it, but they're definitely good at what they do. Now, back to Kelvin! When he bumped into the truncated octahedron, he was actually studying the 3d version of the 2d Hexagonal Honeycomb Conjecture. In other words, he was trying to chop 3d space into cells of equal volume with the least surface area per cell. And he conjectured that the answer was very similar to filling space with truncated octahedra, as shown here: 10) Bitruncated cubic honeycomb, Wikipedia, http://en.wikipedia.org/wiki/Bitruncated_cubic_honeycomb I say very similar because it's actually more efficient if you let the hexagonal faces in this structure be slightly *curved*. So, the possibility Hales ruled out in the 2d case actually matters here! No shading could show satisfactorily the delicate curvature of the hexagonal faces, though it may be fairly well seen on the solid model made as described in Section 12. But it is shown beautifully, and illustrated in great perfection, by making a skeleton model of 36 wire arcs for the 36 edges of the complete figure, and dipping it in soap solution to fill the faces with film, which is easily done for all the faces but one. The curvature of the hexagonal film on the two sides of the plane of its six long diagonals is beautifully shown by reflected light. I think this is a nice passage. We may remember Kelvin for his profound work on electromagnetism and thermodynamics - or his 1900 lecture on two dark clouds hanging over physics: the Michelson- Morley experiment (which foreshadowed special relativity) and black body radiation (which foreshadowed quantum mechanics). We may not imagine him playing around with soap bubbles! But it shows that good science stems from curiosity, and curiosity knows no bounds. You can read Kelvin's paper here: 11) Lord Kelvin, On the division of space with minimum partitional area, Phil. Mag. 24 (1887), 503. Also available at http://zapatopi.net/kelvin/papers/on_the_division_of_space.html His lively mind is evident from the selection of papers on this site. For example: On Vortex Atoms, where he unsuccessfully tried to build atoms out of knotted electromagnetic field lines, and wound up giving birth to knot theory. Some others I hadn't heard of: On the origin of life, The sorting demon of Maxwell, and Windmills must be the future source of power. Anyway: for over a century the so-called Kelvin structure was believed to be the best solution to the problem of chopping space into equal volume cells with minimal surface area. But in 1993 two physicists at Trinity College in Dublin - Denis Weaire and Robert Phelan - found a solution that has 0.3% less surface area! It looks like this: 12) Weaire-Phelan structure, Wikipedia, http://en.wikipedia.org/wiki/Weaire-Phelan_structure It's built from *two* kinds of cells - a 12-sided one and a 14-sided one. Was that allowed in Kelvin's original puzzle? I can't tell! In fact, this so-called Weaire-Phelan structure was no bolt out of the blue. The basic pattern had already been seen in certain cage-like crystals called clathrates. For example, down at the bottom of cold oceans, there are about 6 trillion tons of methane hydrate, a funny substance in which methane molecules are trapped in polyhedral cages formed by water molecules. It looks like ice, but you can ignite it with a cigarette lighter! If you think global warming is bad now, just wait until people figure out how to mine this stuff.... Anyway, methane hydrate is just one of a collection of gas hydrates with different geometries. And in a so-called type I gas hydrate, the water molecules form cages in the pattern of the Weaire-Phelan structure! 13) Clathrate hydrate, Wikipedia, http://en.wikipedia.org/wiki/Clathrate_hydrate So, it's nicely appropriate to use the Weaire-Phelan structure for a building called the Water Cube! Since this struture is perfectly periodic, the engineer for the Water Cube cut it at an odd angle to Made of a plastic known as ethylene tetrafluoroethylene and filled with air, the bubbles are attached to a steel framework outlining the bubble edges. Surface tension holds the bubbles together and tends to pull them into a structure with least surface area. The building really looks like nothing else in the world, Tristam Carfrae told the New York Times. It's a box made of bubbles. Carfrae is the structural engineer who designed the center. On the math side of things, there's plenty left to be done. Nobody has proved that the Weaire-Phelan structure is the best solution to Kelvin's problem. According to Frank Morgan, the expert on minimal surface who reviewed Spziro's book on Kepler's problem, Proving the Weaire-Phelan structure optimal looks perhaps a century beyond current mathematics to me, but I understand that Hales is already thinking about it. More generally, minimal surface theory is a lively subject that uses a lot of deep tools. Morgan is really big on explaining math, so his book is probably the place to start if you want to dig deeper: 14) Frank Morgan, Geometric Measure Theory: a Beginner's Guide, Academic Press, New York, 2000. Personally, I'm more in love with symmetry than minimization. So, I want to learn more about the 28 convex uniform honeycombs - ways of uniformly packing 3d space with uniform solids: 15) Convex uniform honeycomb, Wikipedia, http://en.wikipedia.org/wiki/Convex_uniform_honeycomb They're related to Coxeter groups and crystallographic groups, which are the 3d analogues of the wallpaper groups I discussed back in week267. In fact, we can study honeycombs and their symmetry groups in any dimension, both in flat space and in positively curved (spherical) and negatively curved (hyperbolic) space. A lot is known about them.... ----------------------------------------------------------------------- Quote of the Week: My suggestion is that Aepinus' fluid consists of exceedingly minute equal and similar atoms, which I call electrions, much smaller than the atoms of ponderable matter. Lord Kelvin, from Aepinus Atomized ----------------------------------------------------------------------- mathematics and physics, as well as some of my research papers, can be obtained at http://math.ucr.edu/home/baez/ For a table of contents of all the issues of This Week's Finds, try http://math.ucr.edu/home/baez/twfcontents.html A simple jumping-off point to the old issues is available at http://math.ucr.edu/home/baez/twfshort.html If you just want the latest issue, go to http://math.ucr.edu/home/baez/this.week.html === Subject: Re: Algebraic integer result with quadratics posting-account=n1ZfDgkAAABbCs44qOtz8dP-RkWuEBif Gecko/2008070208 Firefox/3.0.1,gzip(gfe),gzip(gfe) > A rather remarkable bit of mathematics casts a darker view on the ring > of algebraic integers where quadratics thankfully are useful for a > somewhat subtle result, and the distributive property is almost > bizarrely the linchpin of the proof. Consider in an integral domain P(x) = 175x^2 - 15x + 2 and 7*P(x) = (5a 1(x) + 7)*(5a 2(x)+ 7) where the a's are roots of a^2 - (7x-1)a + (49x^2 - 14x) = 0, So I have a non-monic quadratic with integer coefficients which is > multiplied times 7 and then factored using functions given by the > roots of a monic quadratic expression, so I've entangled one quadratic > with a quadratic generator in such a way as to allow myself to probe > at the ring of algebraic integers in a way never before done without > such remarkable tools. Of course with a regular polynomial factorization like 7*(x^2 + 3x + 2) = (7x + 7)*(x+2) it's trivial to just divide the 7 off to get (x^2 + 3x + 2) = (x + 1)*(x+2) but by entangling the two expressions and making the functional values > roots of monic quadratics with integer coefficients when x is an > integer, I've managed to FORCE a division of the 7 across non-rational > solutions, and in that way remove the ability to in general divide it > off from 7*P(x) = (5a 1(x) + 7)*(5a 2(x)+ 7) in the ring of algebraic integers. Really the key points are covered at that point. So it's worth clearly showing how arguments have raged for years when you ignore the possibility of unit factors. After all, it's trivial to note that if you have a polynomial P(x) with linear functions f 1(x) and f 2(x) such that 7*P(x) = (5f 1(x) + 7)*(5f 2(x)+ 7) that the 7 must divide through only one of the f's, which of course, posters have routinely acknowledged as then you have a polynomial factorization, so there's no denial possible. So you have P(x) = (5f 1(x)/7 + 1)*(5f 2(x)+ 7) or P(x) = (5f 1(x) + 7)*(5f 2(x)/7 + 1). But change the functions to non-linear ones by making them roots of a quadratic generator as I did with the a's and the arguments start up again, but what changed? The TYPE of the functions is what changed!!! But if a*(f(x) + b) = a*f(x) + a*b and the value of the function doesn't matter, then how can the type of function change the distributive property? Of course it cannot. For years posters dance around this issue, but it is a mathematical fact. A function is a function. If one function works one way based on the distributive property then how can mathematics be consistent and changing the type of FUNCTION shift the distributive property? It cannot, if mathematics is consistent. So what actually happens? Well, I don't have a case where you have something like a*(f(x) + b) = a*f(x) + a*b but instead have cases like 7*P(x) = (5f 1(x) + 7)*(5f 2(x)+ 7) where you can have unit factors u 1*u 2 = 1, and get P(x) = (5u 1*f 1(x)/7 + u 1)*(5*u 2*f 2(x)+ 7*u 2) or P(x) = (5u 1*f 1(x) + 7*u 1)*(5*u 2*f 2(x)/7 + u 2) where the u's are not units in the ring of algebraic integers because they are not roots of a monic polynomial with integer coefficients with a unit last coefficient. And mathematical consistency remains. But those factorization cannot exist in the ring of algebraic integers in that form as 7*u 1 is not available in the ring of algebraic integers which does not allow that 7 be a factor in that ring, and u 1 and u 2 are NOT units in the ring of algebraic integers. Notice here that the problem is, how can the distributive property vary depending on the TYPE of function? The answer is it cannot. But then you have to resolve the fact that 7 cannot be a factor in the ring of algebraic integers of the a's when 7*P(x) = (5a 1(x) + 7)*(5a 2(x)+ 7) where the a's are roots of a^2 - (7x-1)a + (49x^2 - 14x) = 0. So for those looking for an absolute crux point where posters fail in their reasoning then it is in their position that the type of function can shift the distributive property. Once you dismiss that illogical belief then you have a mystery to resolve. To preserve mathematical consistency the resolution of the mystery requires unit factors, but you can quickly find that units are not available in the ring of algebraic integers. Which proves that the ring is incomplete, or mathematical consistency fails. Disputes here no matter what posters say, or what they delete out in reply reduce to asserting that the TYPE of function shifts a*(f(x) + b) = a*f(x) + a*b or the result that is clearly correct with linear functions would not be an issue with non-linear ones cleverly forced to be the roots of a quadratic generator as the a's are with a^2 - (7x-1)a + (49x^2 - 14x) = 0. It is frustrating to have a logical argument, which is being refuted by people who are arguing against the distributive property who simply say they are not, and get accepted by a society that claims it is logical. But actions speak louder than words. James Harris === Subject: Re: Algebraic integer result with quadratics posting-account=wVv_VwoAAAAVTfUuyxLzug5SzYWCgHj1 Gecko/20080702 Firefox/2.0.0.16,gzip(gfe),gzip(gfe) A rather remarkable bit of mathematics casts a darker view on the ring > of algebraic integers where quadratics thankfully are useful for a > somewhat subtle result, and the distributive property is almost > bizarrely the linchpin of the proof. Consider in an integral domain P(x) = 175x^2 - 15x + 2 and 7*P(x) = (5a_1(x) + 7)*(5a_2(x)+ 7) where the a's are roots of a^2 - (7x-1)a + (49x^2 - 14x) = 0, So I have a non-monic quadratic with integer coefficients which is > multiplied times 7 and then factored using functions given by the > roots of a monic quadratic expression, so I've entangled one quadratic > with a quadratic generator in such a way as to allow myself to probe > at the ring of algebraic integers in a way never before done without > such remarkable tools. Of course with a regular polynomial factorization like 7*(x^2 + 3x + 2) = (7x + 7)*(x+2) it's trivial to just divide the 7 off to get (x^2 + 3x + 2) = (x + 1)*(x+2) but by entangling the two expressions and making the functional values > roots of monic quadratics with integer coefficients when x is an > integer, I've managed to FORCE a division of the 7 across non-rational > solutions, and in that way remove the ability to in general divide it > off from 7*P(x) = (5a_1(x) + 7)*(5a_2(x)+ 7) in the ring of algebraic integers. Really the key points are covered at that point. So it's worth > clearly showing how arguments have raged for years when you ignore the > possibility of unit factors. After all, it's trivial to note that if you have a polynomial P(x) > with linear functions f_1(x) and f_2(x) such that 7*P(x) = (5f_1(x) + 7)*(5f_2(x)+ 7) that the 7 must divide through only one of the f's, which of course, > posters have routinely acknowledged as then you have a polynomial > factorization, so there's no denial possible. So you have P(x) = (5f_1(x)/7 + 1)*(5f_2(x)+ 7) or P(x) = (5f_1(x) + 7)*(5f_2(x)/7 + 1). But change the functions to non-linear ones by making them roots of a > quadratic generator as I did with the a's and the arguments start up > again, but what changed? The TYPE of the functions is what changed!!! But if a*(f(x) + b) = a*f(x) + a*b and the value of the function > doesn't matter, then how can the type of function change the > distributive property? Of course it cannot. For years posters dance around this issue, but > it is a mathematical fact. A function is a function. If one function works one way based on the > distributive property then how can mathematics be consistent and > changing the type of FUNCTION shift the distributive property? It > cannot, if mathematics is consistent. So what actually happens? Well, I don't have a case where you have something like a*(f(x) + b) = a*f(x) + a*b but instead have cases like 7*P(x) = (5f_1(x) + 7)*(5f_2(x)+ 7) where you can have unit factors u_1*u_2 = 1, and get P(x) = (5u_1*f_1(x)/7 + u_1)*(5*u_2*f_2(x)+ 7*u_2) or P(x) = (5u_1*f_1(x) + 7*u_1)*(5*u_2*f_2(x)/7 + u_2) > Right there, in both equations, you are not using the distributive property. The distributive property is about multiplication. It says: a * (b + c) = a*b + a*c. But you are DIVIDING by 7. Division is not a ring operation. Division works in fields. In general, you cannot talk about a/b in a ring. The ring may not contain a/b. The version of the distributive property that you want is (b + c) / a = b/a + c/a. This is NOT a statement of the distributive property in rings. Look it up. In your case, b/a is 5a_1(x)/7. But you don't even know that 5a_1(x)/7 is in your ring. If you assume it is, then you are assuming what you want to prove. > where the u's are not units in the ring of algebraic integers because > they are not roots of a monic polynomial with integer coefficients > with a unit last coefficient. And mathematical consistency remains. But those factorization cannot exist in the ring of algebraic integers > in that form as 7*u_1 is not available in the ring of algebraic > integers Of course it is. You can let u_1 = 1. > which does not allow that 7 be a factor in that ring, and u_1 > and u_2 are NOT units in the ring of algebraic integers. Notice here that the problem is, how can the distributive property > vary depending on the TYPE of function? The answer is it cannot. > Of course it can. Look: a_1(x) is a variable function. Right? What you need is the greatest common divisor of a_1(x) and 7. Since a_1(x) is a variable function of x, there is EVERY REASON to believe that the greatest common divisor of a_1(x) and 7 must ALSO be a variable function of x. You SHOULD assume it is a function of x until you have proven otherwise. And for sure here, wiener, you have not proven otherwise. And in fact, as shown dramatically by Dale Hall's computations, the correct factors DO depend on x. > But then you have to resolve the fact that 7 cannot be a factor in the > ring of algebraic integers of the a's when 7*P(x) = (5a_1(x) + 7)*(5a_2(x)+ 7) where the a's are roots of a^2 - (7x-1)a + (49x^2 - 14x) = 0. > That side of the resolution is YOUR problem. It you who is claiming this, based on your incorrect logic where you regard the distributive property as a statement about division rather than about multiplication. > So for those looking for an absolute crux point where posters fail in > their reasoning then it is in their position that the type of function > can shift the distributive property. > Assuming that 7 = w_1(x) * w_2(x), etc., does not shift or otherwise violate the distributive property. It is safely preserved. > Once you dismiss that illogical belief then you have a mystery to > resolve. > No, the mystery is all on your side. You erroneously think your flawed logic implies something which is not true. > To preserve mathematical consistency the resolution of the mystery > requires unit factors, but you can quickly find that units are not > available in the ring of algebraic integers. > Units have nothing to do with the central problem. The central problem is that you think there is only one, CONSTANT way to divide 7 out of (5a_1(x) + 7)*(5a_2(x) + 7), even though both a_1(x) and a_2(x) are ***variable*** functions of x. > Which proves that the ring is incomplete, or mathematical consistency > fails. Disputes here no matter what posters say, or what they delete out in > reply reduce to asserting that the TYPE of function shifts a*(f(x) + b) = a*f(x) + a*b or the result that is clearly correct with linear functions would not > be an issue with non-linear ones cleverly forced to be the roots of a > quadratic generator as the a's are with a^2 - (7x-1)a + (49x^2 - 14x) = 0. It is frustrating to have a logical argument, You are in no position to know that. You don't have one. Marcus. > which is being refuted > by people who are arguing against the distributive property who simply > say they are not, and get accepted by a society that claims it is > logical. But actions speak louder than words. James Harris === Subject: Re: Algebraic integer result with quadratics posting-account=aLpfCwoAAACh4BOs3HOlQBCoxUpEgyxc Gecko/2008070208 Firefox/3.0.1,gzip(gfe),gzip(gfe) On 1 Sep, 07:46, W. Dale Hall [1] I had a quick look at the Kash documentation but I found myself > pretty far out of my depth. If someone can refer me to somewhere I can > learn what I need to know to calculate the GCD's I'll be grateful. Here's how to get started with Kash. I'll compute the w's for x = 4, > and give a transcription of my session. Then I'll go through and try > to provide some indication of what on Earth I just did. [...] your post, so I may have some questions later. === Subject: Re: idmoochirp1 room sevice 9/7/8 - grandparents day posting-account=8rj5DwkAAACk-Rg_RjMoORLoau6kdpX- Gecko/2008070208 Firefox/3.0.1,gzip(gfe),gzip(gfe) On Sep 1, 12:06æam, David St. Albans lawsuits Around Here, Who Cares! > Only totally Insane Foaming Trolls would dare take such actions! Today I am one scared angel. So far past Scared, came back to normal, then became Arthurian! > So bad Nemesis comforted me! > aka Karma Madame Justice! > The Oracle of Delphi who got Raped. Amazing! One guy makes no sense at all and the other makes everything > senseless... Saint While you like yer sycophants simply throw it all away! NO matter who posts, ya'll have nothing but negative to say! Yer a loser, get lost already.. ;-) Bruce === Subject: Re: Another stupid question about the square root of a prime number >In a previous post it was shown by those obviously more learned than I that >the square root of a prime number is an irrational number. And again I am >curious, perhaps again revealing my ignorance, but this irrational number >that is the square root of a prime, we could actually start to compute it, >and I wonder, what if anything can we say about the actual stream of digits >that would be produced by the computation of the square root of a prime >number? Would they bear any formal similarity to the stream of digits >produced by, say, generating pi? > What do you mean by similarity? What is the difference between ordinary similarity and formal similarity? Why are you limiting yourself to the square root of prime numbers? All irrational numbers, whether transcendental (such as pi), the square root of a prime, the square root of any other non-perfect square integer, etc, share certain properties. When expressed in decimal (or other integer base): they never terminate and they never form an infinitely repeating pattern. What more are you looking for? -- Remove del for email === Subject: Re: recipe ingredients SQL ... entity to entity relationships ... let the computer do it YUM YUM... HAH HAH > Listen to this... > I recently got rid of my old blender... and got a new one and > rediscovered smoothie drinks made with multiple ingredients. > I know enough about combination and matrix to know that between all > the ingredient elements there is a matrix that describes every combo. > If I have juice, fruit, and other as three elements each with 3 > juices and 3 fruit and 5 other elements... > How many smoothies can I make in combination? > I feel like an idiot... after opening Excel and just typing the combos > I knew the manual matrix combos would be overwhelmed... > WHAT LINE of applied math is this? Matrix or Factorial? I don't > think this is an algebraic system thing... The Cartesian Product. Number of juice * number of fruit * number of other. 3*3*5 = 45 Like that returned by a simple query from a SQL database >such as MS-Access: SELECT juices.juice, fruits.fruit, others.other >FROM fruits, juices, others; juice fruit other >--------- ----------- ---------- >grape cherry walnuts >grape plum walnuts >grape strawberry walnuts >apple cherry walnuts >apple plum walnuts >apple strawberry walnuts >cranberry cherry walnuts >cranberry plum walnuts >cranberry strawberry walnuts >grape cherry chocolate >grape plum chocolate >grape strawberry chocolate >apple cherry chocolate >apple plum chocolate >apple strawberry chocolate >cranberry cherry chocolate >cranberry plum chocolate >cranberry strawberry chocolate >grape cherry yogurt >grape plum yogurt >grape strawberry yogurt >apple cherry yogurt >apple plum yogurt >apple strawberry yogurt >cranberry cherry yogurt >cranberry plum yogurt >cranberry strawberry yogurt >grape cherry honey >grape plum honey >grape strawberry honey >apple cherry honey >apple plum honey >apple strawberry honey >cranberry cherry honey >cranberry plum honey >cranberry strawberry honey >grape cherry pistachios >grape plum pistachios >grape strawberry pistachios >apple cherry pistachios >apple plum pistachios >apple strawberry pistachios >cranberry cherry pistachios >cranberry plum pistachios >cranberry strawberry pistachios If None is a possibility for any category, you need >to add it as a potential ingredient which will increase >the number of possibilities. Or you may want multiple occurences of a category, such >as 8oz of grape juice with 8oz of apple juice. To do >that, you have two juice categories, each representing >8oz (grape grape being 16oz of grape). For example, two juices (None not allowed), three fruits >(None allowed as a possibility) and three others (None >allowed as a possibility) gives you: (3 * 3) * (4 * 4 * 4) * (6 * 6 * 6) = 124416 possibilites, one of which is cranberry >grape >None >None >None >honey >yogurt >None And many wouldn't make a lot of sense, for instance >honey None honey is simply honey as the other, so the >124416 combinations don't necessarily result in 124416 >distinct flavors. > I can't remember what I learned in School... Maybe you never learned it. === Subject: Re: JSH: Simply brutal, algebraic integers undone No, that's just plain wrong. It's difficult to compute, > but there is an algorithm which will work for any integer > value of x. It was proven probably 150 years ago. > If you happen to know it off the top of your head, > what's the name of the theorem? A statement of the theorem is: the number of ideal classes of > a number field (of finite degree) is finite. I don't know > that the theorem has a name. This was all discussed in-depth in 2002/4 and 2005/3 when JSH made identical claims. E.g. see my post [1] and its links. Note esp. that [1] also discusses the important related property of root-closure. --Bill Dubuque === Subject: Re: Algebraic integer result with quadratics > A rather remarkable bit of mathematics casts a darker view on the ring > of algebraic integers where quadratics thankfully are useful for a > somewhat subtle result, and the distributive property is almost > bizarrely the linchpin of the proof. Consider in an integral domain P(x) = 175x^2 - 15x + 2 and 7*P(x) = (5a_1(x) + 7)*(5a_2(x)+ 7) where the a's are roots of a^2 - (7x-1)a + (49x^2 - 14x) = 0, So, a1_(x) = (7x-1+sqrt(1+42x-147x^2))/2 [1] and a2_(x) = (7x-1-sqrt(1+42x-147x^2))/2 [2] .... > So as they are mathematically the same in all key ways, not > surprisingly, in each case you simply have 7 multiplied rather simply, > which can be seen with the more complex example by letting x=0 as then a^2 - (7x-1)a + (49x^2 - 14x) = 0 gives a^2 + a = 0 so one of the a's is 0 and the other is -1, so I have from 7*P(x) = (5a_1(x) + 7)*(5a_2(x)+ 7) that 7*P(0) = (7)*(2) or (2)*(7). Or, using [1] and [2], plugging in 0 gives 0 and -1, which gives 7*2 for 7*P(0). So one of (5a_1(x) + 7) or (5a_2(x)+ 7) is coprime to 7 at that value > just like with 7*(x^2 + 3x + 2) = (7x + 7)*(x+2) one of them is coprime to 7 at x=0, and since I've mentioned > coprimeness I'll go ahead and say that we'll try to be in the ring of > algebraic integers. I say try because you may already know that now there is a HUGE > problem as if integer x generates a quadratic that is irreducible over > Q, then you already know that NEITHER of the a's can have 7 as a > factor in that ring!!! For example with x=1, the result would need that for Plugging 1 into [1] and [2] gives 3+sqrt(-26) and 3-sqrt(-26). That gives us 7*P(1) = 7*162 = (22 + 5*sqrt(-26)) * (22 - 5*sqrt(-26)). Why do you think there is something broken about being able to write 7*162 as the product of two irrational complex numbers? -- --Tim Smith === Subject: Re: Algebraic integer result with quadratics posting-account=aLpfCwoAAACh4BOs3HOlQBCoxUpEgyxc Gecko/2008070208 Firefox/3.0.1,gzip(gfe),gzip(gfe) Really the key points are covered at that point. æSo it's worth > clearly showing how arguments have raged for years when you ignore the > possibility of unit factors. After all, it's trivial to note that if you have a polynomial P(x) > with linear functions f 1(x) and f 2(x) such that 7*P(x) = (5f 1(x) + 7)*(5f 2(x)+ 7) that the 7 must divide through only one of the f's, which of course, > posters have routinely acknowledged as then you have a polynomial > factorization, so there's no denial possible. So you have P(x) = (5f 1(x)/7 + 1)*(5f 2(x)+ 7) or P(x) = (5f 1(x) + 7)*(5f 2(x)/7 + 1). [...] So what actually happens? Well, I don't have a case where you have something like a*(f(x) + b) = a*f(x) + a*b but instead have cases like 7*P(x) = (5f 1(x) + 7)*(5f 2(x)+ 7) (1) where you can have unit factors u 1*u 2 = 1, and get P(x) = (5u 1*f 1(x)/7 + u 1)*(5*u 2*f 2(x)+ 7*u 2) or P(x) = (5u 1*f 1(x) + 7*u 1)*(5*u 2*f 2(x)/7 + u 2) (2) where the u's are not units in the ring of algebraic integers because > they are not roots of a monic polynomial with integer coefficients > with a unit last coefficient. James, yesterday I posted a new polynomial and a factorisation satisfying the equation I have labelled (1), and showed that at x = 1 the assumption that the u's in (2) are units in the object ring means that the object ring contains 1/7. Did you see that post? If not, it can be read here: It looks to me like I followed your method exactly. Can you (or anyone else) explain where, if anywhere, I went wrong? === Subject: Re: Algebraic integer result with quadratics posting-account=n1ZfDgkAAABbCs44qOtz8dP-RkWuEBif Gecko/2008070208 Firefox/3.0.1,gzip(gfe),gzip(gfe) > A rather remarkable bit of mathematics casts a darker view on the ring > of algebraic integers where quadratics thankfully are useful for a > somewhat subtle result, and the distributive property is almost > bizarrely the linchpin of the proof. Consider in an integral domain P(x) = 175x^2 - 15x + 2 and 7*P(x) = (5a 1(x) + 7)*(5a 2(x)+ 7) where the a's are roots of a^2 - (7x-1)a + (49x^2 - 14x) = 0, So I have a non-monic quadratic with integer coefficients which is > multiplied times 7 and then factored using functions given by the > roots of a monic quadratic expression, so I've entangled one quadratic > with a quadratic generator in such a way as to allow myself to probe > at the ring of algebraic integers in a way never before done without > such remarkable tools. Of course with a regular polynomial factorization like 7*(x^2 + 3x + 2) = (7x + 7)*(x+2) it's trivial to just divide the 7 off to get (x^2 + 3x + 2) = (x + 1)*(x+2) but by entangling the two expressions and making the functional values > roots of monic quadratics with integer coefficients when x is an > integer, I've managed to FORCE a division of the 7 across non-rational > solutions, and in that way remove the ability to in general divide it > off from 7*P(x) = (5a 1(x) + 7)*(5a 2(x)+ 7) in the ring of algebraic integers. Really the key points are covered at that point. æSo it's worth > clearly showing how arguments have raged for years when you ignore the > possibility of unit factors. After all, it's trivial to note that if you have a polynomial P(x) > with linear functions f 1(x) and f 2(x) such that 7*P(x) = (5f 1(x) + 7)*(5f 2(x)+ 7) that the 7 must divide through only one of the f's, which of course, > posters have routinely acknowledged as then you have a polynomial > factorization, so there's no denial possible. So you have P(x) = (5f 1(x)/7 + 1)*(5f 2(x)+ 7) or P(x) = (5f 1(x) + 7)*(5f 2(x)/7 + 1). But change the functions to non-linear ones by making them roots of a > quadratic generator as I did with the a's and the arguments start up > again, but what changed? The TYPE of the functions is what changed!!! But if a*(f(x) + b) = a*f(x) + a*b and the value of the function > doesn't matter, then how can the type of function change the > distributive property? Of course it cannot. æFor years posters dance around this issue, but > it is a mathematical fact. A function is a function. æIf one function works one way based on the > distributive property then how can mathematics be consistent and > changing the type of FUNCTION shift the distributive property? æIt > cannot, if mathematics is consistent. So what actually happens? Well, I don't have a case where you have something like a*(f(x) + b) = a*f(x) + a*b but instead have cases like 7*P(x) = (5f 1(x) + 7)*(5f 2(x)+ 7) where you can have unit factors u 1*u 2 = 1, and get P(x) = (5u 1*f 1(x)/7 + u 1)*(5*u 2*f 2(x)+ 7*u 2) or P(x) = (5u 1*f 1(x) + 7*u 1)*(5*u 2*f 2(x)/7 + u 2) æ Right there, in both equations, you are not using the > distributive property. Yes, I am. If you have 7*(x+3x+2) = (7x + 7)*(x+2) and divide off the 7, are you not using the distributive property? Given that a*(b+c) = a*b + a*c it must be true that 'a' is a factor. So the point is that 7 is a factor based on the distributive property. If it is a factor then it can be removed, which is normally described as dividing it off. > æ The distributive property is about multiplication. æIt > says: æ æ æ æa * (b + c) = a*b + a*c. Yes. > æ But you are DIVIDING by 7. æDivision is not a ring > operation. æDivision works in fields. æIn general, you > cannot talk about a/b in a ring. æThe ring may not contain > a/b. æThe version of the distributive property that you want > is æ æ æ æ (b + c) / a = b/a + c/a. No. The reality can be seen with linear functions: 7*(x^2 + 3x + 2) = (7x + 7)*(x+2) where dividing off the 7 is trivial. I pointed out that there is no critical difference mathematically between that example and P(x) = 175x^2 - 15x + 2 7*P(x) = (5a 1(x) + 7)*(5a 2(x)+ 7) where the a's are roots of a^2 - (7x-1)a + (49x^2 - 14x) = 0, while there is the difference of linear versus non-linear function being used, as 7x is a linear function, or do you disagree? If a*(f(x) + b) = a*f(x) + a*b without regard to whether you have a linear or a non-linear function how can the behavior change just because the a's are non-linear versus 7*(x^2 + 3x + 2) = (7x + 7)*(x+2) where they are? > æ This is NOT a statement of the distributive property > in rings. æLook it up. æ In your case, b/a is 5a 1(x)/7. æBut you don't even > know that 5a 1(x)/7 is in your ring. æIf you assume it is, > then you are assuming what you want to prove. The full statement is 7*P(x) = (5a 1(x) + 7)*(5a 2(x)+ 7) where the a's are roots of a^2 - (7x-1)a + (49x^2 - 14x) = 0. So 7 is multiplied times P(x), so it's not about assuming. > where the u's are not units in the ring of algebraic integers because > they are not roots of a monic polynomial with integer coefficients > with a unit last coefficient. And mathematical consistency remains. But those factorization cannot exist in the ring of algebraic integers > in that form as 7*u 1 is not available in the ring of algebraic > integers æ Of course it is. æYou can let u 1 = 1. Nope, as then 7 would be a factor in that ring, and it is not. The crucial logical point here is that a*(f(x) + b) = a*f(x) + a*b WITHOUT REGARD to the TYPE of function. While your argument and that of Arturo Magidin, and W. Dale Hall, and Rick Decker among others is that if f(x) is a non-linear function then with 7*P(x) = (5a 1(x) + 7)*(5a 2(x)+ 7) where the a's are roots of a^2 - (7x-1)a + (49x^2 - 14x) = 0 somehow, someway the value of FUNCTIONS shift where factors of 7 go, which is in direct contradiction to the reality that a*(f(x) + b) = a*f(x) + a*b WITHOUT REGARD to the TYPE of function. So your position can be shown to directly contradict basic algebra. You are arguing that functional type defines the output of the distributive property when that is trivially wrong. It is a bad math mistake. James Harris === Subject: Re: Algebraic integer result with quadratics posting-account=1lE9SQkAAADFrJsDv61dh1YXcJ_ahy5I > After all, it's trivial to note that if you have a polynomial P(x) > with linear functions f_1(x) and f_2(x) such that 7*P(x) = (5f_1(x) + 7)*(5f_2(x)+ 7) that the 7 must divide through only one of the f's, which of course, is not the distributive property. The distributive property does not tell you if the 7 divides through one f or two. > posters have routinely acknowledged as then you have a polynomial > factorization, so there's no denial possible. So you have P(x) = (5f_1(x)/7 + 1)*(5f_2(x)+ 7) or P(x) = (5f_1(x) + 7)*(5f_2(x)/7 + 1). But change the functions to non-linear ones by making them roots of a > quadratic generator as I did with the a's and the arguments start up > again, but what changed? The 7 divides through two f's rather than one. Note that this change has nothing to do with the distibutive property. - William Hughes === Subject: Re: Algebraic integer result with quadratics posting-account=n1ZfDgkAAABbCs44qOtz8dP-RkWuEBif Gecko/2008070208 Firefox/3.0.1,gzip(gfe),gzip(gfe) After all, it's trivial to note that if you have a polynomial P(x) > with linear functions f 1(x) and f 2(x) such that 7*P(x) = (5f 1(x) + 7)*(5f 2(x)+ 7) that the 7 must divide through only one of the f's, which of course, is not the distributive property. æThe distributive property does > not tell you if the 7 divides through one f or two. Correct. Because it's possible that each has sqrt(7) as a factor or an infinity of other variations, even with linear functions. But how could you determine whether or not that is true, rigorously? > posters have routinely acknowledged as then you have a polynomial > factorization, so there's no denial possible. So you have P(x) = (5f 1(x)/7 + 1)*(5f 2(x)+ 7) or P(x) = (5f 1(x) + 7)*(5f 2(x)/7 + 1). But change the functions to non-linear ones by making them roots of a > quadratic generator as I did with the a's and the arguments start up > again, but what changed? The 7 divides through two f's rather than one. > Note that this change has nothing to do with the > distibutive property. æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ- William Hughes Incorrect. It's not determined. The functional type does NOT impact the statement a*(f(x) + b) = a*f(x) + a*b which is true regardless of whether or not the function is linear or non-linear. Possibly understanding this point will help you understand the full result? Clearly you and other posters believe that linear versus non-linear functions make a difference. James Harris === Subject: Re: Algebraic integer result with quadratics posting-account=1lE9SQkAAADFrJsDv61dh1YXcJ_ahy5I > After all, it's trivial to note that if you have a polynomial P(x) > with linear functions f_1(x) and f_2(x) such that 7*P(x) = (5f_1(x) + 7)*(5f_2(x)+ 7) that the 7 must divide through only one of the f's, which of course, is not the distributive property. The distributive property does > not tell you if the 7 divides through one f or two. Correct. So, the important point, does 7 divide through one f or two? has nothing to do with the distributive property. In particular, something can change whether 7 divides through one f or two without changing the distributive property. - William Hughes === Subject: Re: Algebraic integer result with quadratics posting-account=n1ZfDgkAAABbCs44qOtz8dP-RkWuEBif Gecko/2008070208 Firefox/3.0.1,gzip(gfe),gzip(gfe) After all, it's trivial to note that if you have a polynomial P(x) > with linear functions f 1(x) and f 2(x) such that 7*P(x) = (5f 1(x) + 7)*(5f 2(x)+ 7) that the 7 must divide through only one of the f's, which of course, is not the distributive property. æThe distributive property does > not tell you if the 7 divides through one f or two. Correct. So, the important point, does 7 divide through one f or two? has > nothing to do with the distributive property. æ æIn particular, > something can change whether 7 divides through one f or > two without changing the distributive property. æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ - William Hughes Incorrect. The proper answer is that you can simply check at some value of the functions. For instance, with *linear* functions, if at x=6 in the ring of algebraic integers, you have that with 7*P(x) = (5f 1(x) + 7)*(5f 2(x)+ 7) (5f 2(6)+ 7) has 7 as a factor then you have that 7 is the factor for all x. Understand? James Harris === Subject: Re: Algebraic integer result with quadratics posting-account=1lE9SQkAAADFrJsDv61dh1YXcJ_ahy5I > For instance, with *linear* functions, if at x=6 in the ring of > algebraic integers, you have that with 7*P(x) = (5f_1(x) + 7)*(5f_2(x)+ 7) (5f_2(6)+ 7) has 7 as a factor then you have that 7 is the factor for all x. Understand? Yes. For *linear* functions 7 must divide through one f for all x. If the functions are not linear, 7 may divide through one f for some values of x and divide through two f's for other values of x. However, for non-linear functions the distributive property must still hold. The question of whether 7 divides through one f or two f's has nothing to do with the distibutive property. - William Hughes === Subject: Re: Algebraic integer result with quadratics posting-account=n1ZfDgkAAABbCs44qOtz8dP-RkWuEBif Gecko/2008070208 Firefox/3.0.1,gzip(gfe),gzip(gfe) For instance, with *linear* functions, if at x=6 in the ring of > algebraic integers, you have that with 7*P(x) = (5f 1(x) + 7)*(5f 2(x)+ 7) (5f 2(6)+ 7) has 7 as a factor then you have that 7 is the factor for all x. Understand? Yes. æFor *linear* functions 7 must divide through one f for all x. Wrong. What I said was incorrect as maybe at x=6, P(6) has 7 as a factor. Correctly stated it should be that given 7*P(x) = (5f 1(x) + 7)*(5f 2(x)+ 7) if at x=6, (5f 2(6)+ 7) has 7 as a factor and 5f 1(6) + 7 is coprime to 7, then 5f 1(x) + 7 has 7 as a factor for all x. > If the functions are not linear, 7 may divide through one f for some > values of x and divide > through two f's for other values of x. æHowever, for non-linear Incorrect. Consider again 7*P(x) = (5f 1(x) + 7)*(5f 2(x)+ 7) where P(x) is a polynomial function, does the distributive property care about the type of the function? a*(f(x) + b) = a*f(x) + a*b Does the TYPE of the function have any impact? Think carefully. You've already answered more than one question wrong. James Harris === Subject: Re: Algebraic integer result with quadratics posting-account=1lE9SQkAAADFrJsDv61dh1YXcJ_ahy5I Consider again 7*P(x) = (5f_1(x) + 7)*(5f_2(x)+ 7) where P(x) is a polynomial function, does the distributive property > care about the type of the function? The important question of whether 7 divides through one f or two f's does care about the type of the function. As you have noted, this question has nothing to do with the distributive property. - William Hughes === Subject: Re: Algebraic integer result with quadratics posting-account=n1ZfDgkAAABbCs44qOtz8dP-RkWuEBif Gecko/2008070208 Firefox/3.0.1,gzip(gfe),gzip(gfe) Consider again 7*P(x) = (5f 1(x) + 7)*(5f 2(x)+ 7) where P(x) is a polynomial function, does the distributive property > care about the type of the function? The important question of whether 7 divides through one f > or two f's does care about the type of the function. Nope. Let's say that P(x) = x^2 + 3x + 2 = (x+1)(x+2) so you multiply 7*P(x), how many ways can you do that in the ring of algebraic integers? Ans: An infinity of ways. Is there any way given 7*P(x) = (5f 1(x) + 7)*(5f 2(x)+ 7) to determine *which* way was used. > As you have noted, this question has nothing to do with > the distributive property. æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ - William Hughes Of course it does because if you have 7*(x+1)*(x+2) = (7x + 7)(x + 2) that's all about the distributive property as you *distribute* the 7 through the factor. Now it's trivial algebra but the problem here is that posters like you deny trivial algebra and then just claim that's not what you're doing because the stakes are high, and you actually clearly suck at basic math. Now if P(x) = x^2 + 3x + 2 = (x+1)(x+2) and 7*P(x) = (5f 1(x) + 7)*(5f 2(x)+ 7) there are an infinity of ways to distribute factors of 7 across the factorization of P(x), but there's one clear way to determine which one was used as let's say, where still the f's are linear functions, that f 1(0) = 0. In that case HOW did the 7 distribute? Not a trick question but answering honestly will reveal that yes, it is about the distributive property. Given that answer with linear functions can you deny that your claims that non-linear functions are somehow different are non-mathematical and completely bogus efforts to deny very basic mathematics? James Harris === Subject: Re: Algebraic integer result with quadratics > Consider again > 7*P(x) = (5f_1(x) + 7)*(5f_2(x)+ 7) > where P(x) is a polynomial function, does the distributive property > care about the type of the function? > The important question of whether 7 divides through one f > or two f's does care about the type of the function. Nope. Let's say that P(x) = x^2 + 3x + 2 = (x+1)(x+2) so you multiply 7*P(x), how many ways can you do that in the ring of > algebraic integers? Ans: An infinity of ways. Is there any way given 7*P(x) = (5f_1(x) + 7)*(5f_2(x)+ 7) to determine *which* way was used. > As you have noted, this question has nothing to do with > the distributive property. > - William Hughes Of course it does because if you have 7*(x+1)*(x+2) = (7x + 7)(x + 2) that's all about the distributive property as you *distribute* the 7 > through the factor. Now it's trivial algebra but the problem here is that posters like you > deny trivial algebra and then just claim that's not what you're doing > because the stakes are high, and you actually clearly suck at basic > math. Now if P(x) = x^2 + 3x + 2 = (x+1)(x+2) and 7*P(x) = (5f_1(x) + 7)*(5f_2(x)+ 7) there are an infinity of ways to distribute factors of 7 across the > factorization of P(x), but there's one clear way to determine which > one was used as let's say, where still the f's are linear functions, > that f_1(0) = 0. In that case HOW did the 7 distribute? Not a trick question but answering honestly will reveal that yes, it > is about the distributive property. > Not really. It has to do with the ring where the factorization occurs. The distributive property has nothing to say about the distribution of factors over a product. Instead, it has everything to say about the distribution of a factor over a sum. In the case you've described P(x) = (x + 1)(x + 2), if you write 7P = (ax + a)(bx + 2b) the factors a and b multiply to give 7, of course, and if your factorization is over the integers, or the rational numbers, there are only two options (disregarding sign). If you are factoring over Z[sqrt(7)], you have three options (again disregarding sign), the additional option being a = b = sqrt(7). Factoring over the full ring of algebraic integers, you get an infinite array of options, amounting to the full set of factorizations of 7 that are available in that ring. Now, having specified a particular form of factorization: (*) 7 P = (5 f_1 + 7)(5 f_2 + 7) I run into a bit of a problem as to how you maintain that a factorization of this sort exists. In the preceding case, you held correctly that your a_1 & a_2 satisfied an algebraic equation, namely the quadratic equation a^2 - (7 x - 1)a + (49 x^2 - 14 x) = 0. I note that for integer x, this has integer coefficients, and so (being monic) resulted in algebraic integer solutions. One might well say that your a's are integral over the ring Z[x] of polynomials with integer coefficients. At least I'd say that. In this case, I don't see a quadratic polynomial that the f's satisfy, in the same fashion. Expanding the product on the right side of the above equation (*), we find 25 f_1 f_2 + 35 (f_1 + f_2) + 49 and if the f's satisfy a monic quadratic polynomial f^2 + p_1 f + p_0 with the p's being polynomials in x, for the above product to hold, you'd need f_1 f_2 = p_0 f_1 + f_2 = -p_1 Since I've assumed the p's to be polynomials, the equation 7 x^2 + 21 x + 14 = 25 p_0 + 35 p_1 + 49 suggests that they be of no greater than 2nd degree. Let p_0 = r x^2 + s x + t p_1 = u x^2 + v x + w and we'll get 7 x^2 + 21 x + 14 = 25 (r x^2 + s x + t) + 35 (u x^2 + v x + w) + 49 This is supposed to be an identity, so we equate powers of x: x^2: 25 r + 35 u = 7 x : 25 s + 35 v = 14 1: 25 t + 35 w + 49 = 14 or 25 t + 35 w = -35 So, we have these constraints on the coefficients: 25 r + 35 u = 7, 25 s + 35 v = 14, 25 t + 35 w = -35. Integrality of these coefficients leads to the following: 25 r = 7(1 - 5u) Note that for integer r,u, the right side must be congruent to 2 mod 5, while the left side is divisible by 5. I obtain similar incongruities for s and v: 25 s = 7(2 - 5 v) in which the right side is congruent to 4 mod 5, the left divisible by 5. For the final coefficients, t and w: 25 t = - 35( 1 + w) I note that setting w congruent to 4 mod 5 (e.g., w = -1, 4, 9, ...) is sufficient to obtain a solution for t. Did you intend your functions f_1, f_2 to satisfy a quadratic equation with polynomial coefficients? Was that polynomial intended to be monic? What else have I assumed that is not the case here? > Given that answer with linear functions can you deny that your claims > that non-linear functions are somehow different are non-mathematical > and completely bogus efforts to deny very basic mathematics? > James Harris Again, it's not the linearity vs. non-linearity that is at issue. I could (and did) factor the above polynomial over Z[sqrt(7)] with a different distribution of factors than having 7 residing in one factor or the other. In exactly the same fashion, in the non-polynomial factorization example, the w's that are obtained are really only the minimal examples, because one could also extend the ring in which the factors reside, to further vary the range of factors of 7 available for the factorization. In closing, it's not necessary to castigate the efforts of others in understanding the situation. Calling these efforts non-mathematical and completely bogus does nothing to enhance your argument, and merely suggests your own case to be based on emotion. Given the fact that I have exhibited factors that you maintained to be impossible (and disregarding your so-called wrapper theorem, which I have examined and found to be somewhat incoherent, followed by an argument that is, if read literally, patently false), it is erroneous to suggest that my correct computations are either non-mathematical or bogus. Dale