mm-123 == Undeniable is silly. Old Man has provided a general closed form> analytic solution, t = t(r), and Undeniable complains because the> inverse, r = r(t) is transcendental, yet, if the situation were reversed,> wherein r = r(t) is analytic, while t = t(r) is transcendental, no> complaint would be heard. Petty personal preferences for a solution> of the form r = r(t) aside, the motion is completely described by> t = t(r) via a straight-forward, non-iterative process.Old Man makes silly statement. Old Man brought another owl to Athensmeaning that his statement of the existence of t=f(r) is well known toeveryone who is not criple and can open a book on classical mechanics.The problem is whether the inverse function r = g(t) exist in closedposed. Old Man should try to understand the difference between solvingthe problem one likes to solve and solving the problem posed.Apparently, Old Man has done the former because it was easy. > Undeniables highly qualied statement is empty of merit and borders> upon being intellectually dishonest as well since Undeniable is now aware> that a closed form solution, t = t(r), exists. [Old Man]> Undeniable is really disturbed to see Old Man resorting to accusationsof dishonesty. Still, Undeniable will refrain from treating Old Man ina similar way. Old Man should look at the two postings made bycame into the game. The second post reads:Undeniable: One can integrate readily dt=dv/v and get t as a functionof thedistance y. This function is analytic, holomorphic and bijective (lookup those terms) Therefore the function t = f(y) has an inverse of theform y = g(t). The inverse function g(t) cannot however be obtained inclose form but it exists for every t and is also analytic (it betterexist). Then, one must use the Langrange Inversion Theorem and getg(t) as a series expansion. For y~R, the secord order term is 1/2gt^2,the rst order v0t and the constant y0.Ok Old Man? Have you gotten any more owl to bring? => This highly informal contest is still open until> contest with some surprising results. (Sorry, no prizes> except for name recognition on my webpage. :-) )Oh, Ghostly one:Although JSH is no longer interested in prime counting, I would beinterested in how my minor change to his integer version stacks up. Iposted this on 9/5. It removes all his calls to sqrt() which in my testsproduced better than 70% improvement in run speed. Sorry, my compilerdoesnt have unsigned long long so I didnt implement it as the projectspecied because I have no means to test that (well I have cygwin and gcc,but only use other cygwin tools).//--- Improved JSH C++ prime counter with removal of all sqrt() callsunsigned S2(unsigned x, unsigned yin);unsigned JSHpi2(unsigned xin, unsigned yin){ return (xin-S2(xin,yin)-1);}//Optionally, use a dene instead of the above function to reduce functioncalls and stack//overhead//#dene JSHpi2 (xin,yin) (xin-S2(xin,yin)-1)//use at the risk of losing type safetyunsigned S2(unsigned x, unsigned yin){ unsigned sum=0, i, sum1, sum2; const unsigned dx=1; for(unsigned i=2; i*i<=yin; i++) { sum2 = ( pi2(i,i) - pi2(i-dx,i-dx)); if (sum2!=0){ int tmp = (i-dx)*(i-dx), tmp2=x/i; sum1 = ( pi2(x/i,min(tmp,tmp2)) - pi2(i-dx,i-dx)) ; sum+=(sum1 * sum2); } } return sum;} =Oops: changed the function name to JSHpi2 but didnt change all the calls inS2... =In sci.math, Stan Gula:> Oops: changed the function name to JSHpi2 but didnt change all the calls in> S2...> Sorry, I cant accept your submission at present as all of the valuesare way off, and I dont see an obvious way of xing it. Perhaps youcan point out the aw?* * */* submitted by sgula@bellatlantic.net.remove.invalid, with xes *///--- Improved JSH C++ prime counter with removal of all sqrt() calls#include #include /* for some reason this wasnt dened -- ed. */#dene min(a,b) ((a) < (b) ? (a) : (b))unsigned S2(unsigned x, unsigned yin);unsigned JSHpi2(unsigned xin, unsigned yin){ return (xin-S2(xin,yin)-1);}//Optionally, use a dene instead of the above function to reduce//function calls and stack overhead//#dene JSHpi2 (xin,yin) (xin-S2(xin,yin)-1)//use at the risk of losing type safetyunsigned S2(unsigned x, unsigned yin){ unsigned sum=0, i, sum1, sum2; const unsigned dx=1; for(unsigned i=2; i*i<=yin; i++) { sum2 = ( JSHpi2(i,i) - JSHpi2(i-dx,i-dx)); if (sum2!=0){ int tmp = (i-dx)*(i-dx), tmp2=x/i; sum1 = ( JSHpi2(x/i,min(tmp,tmp2)) - JSHpi2(i-dx,i-dx)) ; sum+=(sum1 * sum2); } } return sum;}unsigned long long pi(unsigned long long x){ return (unsigned long long) JSHpi2(x, sqrt(x));}// simplied driver routine -- edint main(int argc, char **argv){ if(argc == 2) { unsigned n = atoi(argv[1]); std::cout << n << << pi(n) << std::endl; } else { std::cerr << Usage: << argv[0] << N << std::endl; }}* * *Your routine computes pi(100000) = 18059 but the correct value is 9592.If it helps any to debug this beastie, your routine isalso computing pi(6) = 5, where the correct value is 3.Feel free to resubmit once youve xed this bug. :-)-- #191, ewill3@earthlink.netIts still legal to go .sigless. = Sorry, I cant accept your submission at present as all of the values> are way off, and I dont see an obvious way of xing it. Perhaps you> can point out the aw? unsigned long long pi(unsigned long long x)> {> return (unsigned long long) JSHpi2(x, sqrt(x));> }>Problem is right there... Documentation error!I removed all the usage of sqrt() so the call to JSHpi2 should have the formJSHpi2(x,x) - I should have provided a test harness to document that usage.My harness included stdlib.h which has a template implementation of min() -the macro form is ne as stated. =In sci.math, Stan Gula:>> Sorry, I cant accept your submission at present as all of the values>> are way off, and I dont see an obvious way of xing it. Perhaps you>> can point out the aw?> unsigned long long pi(unsigned long long x)>> {>> return (unsigned long long) JSHpi2(x, sqrt(x));>> } Problem is right there... Documentation error! I removed all the usage of sqrt() so the call to JSHpi2 should have the form> JSHpi2(x,x) - I should have provided a test harness to document that usage. My harness included stdlib.h which has a template implementation of min() -> the macro form is ne as stated.submission. :-) Results will shortly be available athttp://home.earthlink.net/~ewill3/math/primecounters/ index.html .Youre between legendrephi and jamesharris3 in the rankings, so far.-- #191, ewill3@earthlink.netIts still legal to go .sigless. => In sci.math, Stan Gula> :> > Sorry, I cant accept your submission at present as all of the values> are way off, and I dont see an obvious way of xing it. Perhaps you> can point out the aw?> unsigned long long pi(unsigned long long x)> {> return (unsigned long long) JSHpi2(x, sqrt(x));> }> > Problem is right there... Documentation error!> I removed all the usage of sqrt() so the call to JSHpi2 should have theform>JSHpi2(x,x) - I should have provided a test harness to document thatusage.> My harness included stdlib.h which has a template implementation ofmin() ->the macro form is ne as stated. > submission. :-) Results will shortly be available at> http://home.earthlink.net/~ewill3/math/primecounters/ index.html . Youre between legendrephi and jamesharris3 in the rankings, so far.>One more attempt and I have to retire. I am now showing times as good asDaniel A. Jimenez for tests up to 10^8 and Im beating him at 10^9. The keywas running some tests and watching how the recursion works and caching therst 32 values of pi(x). James guess that skipping evaluations at evenvalues didnt help and in fact the time to do the test to see if I couldskip the evaluations took longer than the amount of time I could save...Note that in his S2 function, the evaluation of sum1 is skipped whensum2==0, and it is obvious that this is so for all even values (except 2).//last submission, Stan Gula mods to the JSH prime counter---//removed the sqrt calls, and cached the rst 32 values of pi(x)#include #include #include unsigned S2(unsigned x, unsigned yin);unsigned JSHpi(unsigned xin, unsigned yin){//cache most common answers#dene cache_size 32 static unsigned f[cache_size]= { 0,1,2,2,3,3,4,4,4,4,5,5,6,6,6,6,7,7,8,8,8,8,9,9,9,9,9,9,10,10, 11,11}; if(xin==yin && (xin0 on the commandline<> In sci.math, Stan Gula>> :>> >> Sorry, I cant accept your submission at present as all of the values>> are way off, and I dont see an obvious way of xing it. Perhaps you>> can point out the aw?>> >> unsigned long long pi(unsigned long long x)>> {>> return (unsigned long long) JSHpi2(x, sqrt(x));>> }>> >Problem is right there... Documentation error!>> >I removed all the usage of sqrt() so the call to JSHpi2 should have the>form>>JSHpi2(x,x) - I should have provided a test harness to document that>usage.>> >My harness included stdlib.h which has a template implementation of>min() ->>the macro form is ne as stated.>> > > submission. :-) Results will shortly be available at>> http://home.earthlink.net/~ewill3/math/primecounters/ index.html .>> Youre between legendrephi and jamesharris3 in the rankings, so far.>>One more attempt and I have to retire. I am now showing times as good as>Daniel A. Jimenez for tests up to 10^8 and Im beating him at 10^9. Doh! Ok, heres a tweaked version of my last submission. Its aboutthree times faster for large values.#include #include #include #include unsigned long long pi (unsigned long long v) { const int C[2] = { 1, 5 }; char A[2][65536]; unsigned P[2][8000], Z[2]; unsigned a, b, c, f, i, j, k, l, m, n, o, p, r, w, y; n = 0; memset (Z, 0, sizeof (Z)); w = 1 + v / 6; if (v >= 2) n++; if (v >= 3) n++; if (v >= 5) { n++; P[1][Z[1]++] = 5; } if (v >= 7) { n++; P[0][Z[0]++] = 7; } if (v >= 11) { n++; P[1][Z[1]++] = 11; } r = 65536; if (r > w) r = w; l = 1 + (unsigned) sqrt (r / 6); memset (A, 1, sizeof (A)); for (m=0; m<2; m++) { c = C[m]; for (p=0; p<2; p++) { a = C[p]; for (i=0; i k) r = k; for (i=2; i w) r = w; l = 1 + (unsigned) (sqrt (r / 6)); memset (A, 1, sizeof (A)); for (m=0; m<2; m++) for (p=0; p<2; p++) { a = C[p]; for (i=0; ; i++) { f = P[m][i]; y = f / 6; if (y >= l) break; o = b * 6; o += f - (o % f); for (j=o; ;j+=f) { k = (j - a) / 6; if (k * 6 + a == j) if (k != y) break; } for (j=k; j k) r = k; l = r - b; for (i=0; i >One more attempt and I have to retire. I am now showing times as good as> >Daniel A. Jimenez for tests up to 10^8 and Im beating him at 10^9. Doh! Ok, heres a tweaked version of my last submission. Its about> three times faster for large values.>-> Daniel A. Jim.8enez djimenez@cs.utexas.edu> Assistant Professor djimenez@cs.rutgers.edu> Department of Computer Science> Rutgers UniversityOh, if you want to play that game, I can bump my cache table. Making it8192 values makes it three times faster. Thats the magic of recursion -theres a really easy way to cut branches off by increasing the cachedvalues.By the way, I like your choice of descriptive variable names.Your new version is much faster, however, I have the feeling I can push thecached values as far as necessary...I tested my JSH/SG8192 versus your latest at 10^8 and 10^9 and the resultswere:pmbench 100000000 Prime count (JSH/SG 8192): 5761455 Time (sec) : 0.331 Prime count (DJ): 5761455 Time (sec) : 3.305pmbench 1000000000 Prime count (JSH/SG 8192): 50847534 Time (sec) : 4.707 Prime count (DJ): 50847534 Time (sec) : 38.125I cant currently test over 10^9 because I dont have long long variables inthe compiler I use. => Sorry, I cant accept your submission at present as all of the values>are way off, and I dont see an obvious way of xing it. Perhaps you>can point out the aw?> {>return (unsigned long long) JSHpi2(x, sqrt(x));>}> > Problem is right there... Documentation error! I removed all the usage of sqrt() so the call to JSHpi2 should have the form> JSHpi2(x,x) - I should have provided a test harness to document that usage. My harness included stdlib.h which has a template implementation of min() -> the macro form is ne as stated. Hey Stan, the next speed-up is to skip *evens* in your iteration. Ithink that will at least double the speed. = Hey Stan, the next speed-up is to skip *evens* in your iteration. I> think that will at least double the speed.> How would you suggest doing that? Obviously sum2==0 for all even i notequal to 2, but testing for even i>2 took longer than the amount of cpu timesaved (weird, but thats what the timer said). I found a betteroptimization by caching the low values of the pi(x) function. For largeinput values there are lots of recalculations of low values and thatoverhead cried out to be reduced. I think this is the real aw of usingthis recursive algorithm - mathematically its easy to dene, butcomputation is inefcient.My approach of caching the rst 32 values for pi(x) made a hugeimprovement, but there are obviously more savings by caching more values.The tradeoff is space vs. cpu cycles, but 32 ints isnt a lot of space...Im sure theres some optimal value for the cache size but Im not going todo any more work on this. It was interesting though... = Ive recently demonstrated here that the determination of incomedistribution by the theory of supply and demand makes no sense. Rather than point out any errors in my exposition, certainirrationalists objected, incorrectly, that I was not echoing theliterature. These people claimed that I was being original. For example, poor Chris Auld and JimT asserted, mistakenly,that the literature only applies so-called factor price frontiersreswitching, and capital-reversing to the economy as a whole, notto an individual vertially-integrated industry. Heres some quotes from the literature: ...as Luigi Pasinetti (1969) and Piero Garegnani (1970) point out, the lack of a well-behaved production function, an inverse relation between capital per worker and the rate of prot, and so on, applies at the industry level as well as the economy as a whole... one is not entitled to assume the existence of downward-sloping demand curves for capital in each industry or, as Garegnani argues, downward sloping demand curves for labor. -- Geoffrey M. Hodgson, Sraffa, Value and Distribution: An Expository Essay on the Capital Controversy, British Review of Economic Issues, November 1977. It may also be interesting to notice that the same thing may happen at the level of a single industry. -- Luigi L. Pasinetti, Changes in the Rate of Prot and Switches of Techniques, Quarterly Journal of Economics, V. 80, No. 4 (Nov. 1966), pp. 503-517. To discuss this view of production we must refer to the notions of system of production and integrated industry, applying them to the commodity A(i). [Footnote:] So far, we have used these notions only for the wage-commodity G, but they can be applied to any commodity A(i). A system of production for A(i) is a set of methods of production; one method for A(i), and one for each of the commodities which enter directly or indirectly the production of A(i). Appropriate proportions of these industries will give the integrated industry of A(i). In parallel with the assumptions of continuity inherent in the notions of marginal products, we must assume that the system for the production of A(i) changes continuously with r. What we saw about the relations between r, w, and [net output per head], and their representation by means of the envelope of wage-curves, can then be applied to the relations between r, the wage in terms of A(i), and the net physical product per worker in the production of A(i)... -- P. Garegnani, Heterogeneous Capital, the Production Function and the Theory of Distribution. Review of Economic Studies. 1970.Poor Chris Auld.-- Try http://csf.colorado.edu/pkt/pktauthors/Vienneau.Robert/ Bukharin.html To solve Linear Programs: .../LPSolver.htmlr c A game: .../Keynes.html v s a Whether strength of body or of mind, or wisdom, or i m p virtue, are found in proportion to the power or wealth e a e of a man is a question t perhaps to be discussed by n e . slaves in the hearing of their masters, but highly @ r c m unbecoming to reasonable and free men in search of d o the truth. -- Rousseau => .... > The measure of the 3-dimensional boundary of a 4-dimensional> ball of radius r is 2 pi^2 r^2.... Typo: r^3, not r^2. Ken Pledger. =Though not cited by Mark Ensley, it was secretly John Baez> In fact, I think this is just another way of explaining the > period-24 pattern in the theory of modular forms, but I like> it. I especially like it because SL(2,Z/3) is also known as the > binary tetrahedral group. To get your hands on this group,> take the group of rotational symmetries of the tetrahedron,> also known as A_4. This is a 12-element subgroup of SO(3). > Using the fact that SO(3) has SU(2) as a double cover, take > all the points in SU(2) that map to A_4. You get a 24-element> subgroup of SU(2) which is the binary tetrahedral group. In fact, if you think of SU(2) as the unit sphere in the> quaternions, the binary tetrahedral group becomes the vertices> of a 4-dimensional regular polytope called the 24-cell!Just curious, but do the many occurrences of the 24/8/3 pattern have anything to do with Thurstons conjecture thatall 3-manifolds can be contructed from 8 basic types of geometry?Also, that 2-manifolds can be constructed from the three basic types of 2-geometry (at, elliptic, hyperbolic)?It seems like another occurrence of the 2/3 -- 3/8 pattern, but Im not expert enough in these matters to discern if there really is a deep connection going on.-Mark Ensley =>Just curious, but do the many occurrences of the 24/8/3 >pattern have anything to do with Thurstons conjecture that>all 3-manifolds can be contructed from 8 basic types of geometry?Also, that 2-manifolds can be constructed from the three basic >types of 2-geometry (at, elliptic, hyperbolic)?I know of no way in which these facts are related to anythingIve ever said about the numbers 3, 8, or 24. => For integers k, a, b, and c If k = abc + 1, prove that gcd(k,a) = 1. For integers a, b, and c if a divides c and for any help you can offer.It depends on how much background they gave you before askingyou to solve the problems. If you have at your disposal atheorem about integers a, b: gcd(a,b)=1 if and only if there exist integers x, y satisfying a * x + b * y = 1then you can use it in both directions to solve your problems.If you do not have this theorem yet, wait till it is proved, ornd it in your textbook. If it is not in your textbook,find a better one.Write c = c*a*x + c*b*y and play around with the fact that aand b divide c. =Im just going to give some hints.> For integers k, a, b, and c If k = abc + 1, prove that gcd(k,a) = 1.you have k - abc = 1, or in other notationk + a(-bc) = 1.go back to the denition of gcd (possibly), clearly 1 divides k and a.what if another number d divides a and k...?> For integers a, b, and c if a divides c and be divides c and if> gcd(a,b)=1, prove a*b divides c.>are you aware that gcd(a,b)*lcm(a,b) = a*b?where lcm is the least common multiple of a and b.a divides c and b divides c means that c is a common multiple of a and b.what can you say about the relationship between c and lcm(a,b)?how can you relate this to the equation above?good luckJustin =>9. The source of the claim is people who are well known and in a>position of perceived authority.10. Belief in the assertion gives people a feeling that there is>higher authority that has made great advances for mankind.11. Those who learn of and identify with the assertion are given a>sense of security that this higher authority has things under control.12. Those who know of and identify with the assertion feel that they>are thus part of (they have bonded with) an important authority.Thus, the question of verifying these assertions is removed from the>intellectual realm and becomes a question of psychology. Belief in>the assertion gives people a false sense of security (# 11) and pride>(#12). Facts dont matter - feelings do. The question becomes not>Has it been substantiated? but rather Do you feel better believing>that it is true?Once we realize this important fact, the analogy can be explored>further:13. A small number of people are intellectually (rather than>psychologically) motivated.A much larger number of people are psychologically motivated to believethat they are purely intellectually motivated. Ever tried having aconversation with a Randian Objectivist? Theyre psychologicallyindistinguishable from fundamentalist Christians (or fundamentalistanythings).Also, a large number of people are psychologically motivated to resistauthority in general. Their underlying psychological motivations areusually opaque to them also.If the shoe ts...-- --| BBB b barbara minus knox at iname stop com| B B aa rrr b || BBB a a r bbb | | B B a a r b b | | BBB aa a r bbb | - = ...> now that the Brits have been kicked out. A cavil: the British rented Hong Kong for a xed period of time andleft when that period of time expired. There was no, and no need forany, kicking out.-- G.C.[Moderators note: Ill remove this jocular comment on the websiteversion. Followups set to the more freewheeling newsgroups. - jb] =George Cox, The British rented Hong Kong Ha! a new name for British Colonisationor British occupation.1) A quick web search gives the history:http://208.179.38.218/explore/culture/history/ years.phphttp://www.marimari.com/content/hong_kong/general_ info/history/main.htmlshowing it was taken by force by Queen Victorias gunboats leaving them inposition to force a rental agreement on the Chinese.2) The British were kicked out. If it were not for fear of the militarystrength of China I doubt if the British would have left. Its is interesting that after a 100 years of British rule in Hong Kong viaa British Governor that they suddenly decided they wanted to introducedemocracy for the chinese citizens of Hong Kong before they left.Unbelievable!patrick ...>now that the Brits have been kicked out. A cavil: the British rented Hong Kong for a xed period of time and> left when that period of time expired. There was no, and no need for> any, kicking out.> --> G.C. [Moderators note: Ill remove this jocular comment on the website> version. Followups set to the more freewheeling newsgroups. - jb] =In message , patrickGeorge Cox, The British rented Hong Kong Ha! a new name for British Colonisation>or British occupation.1) A quick web search gives the history:>http://208.179.38.218/explore/culture/history/ years.php>http://www.marimari.com/content/hong_kong/general_ info/history/main.htmlshowing it was taken by force by Queen Victorias gunboats leaving them in>position to force a rental agreement on the Chinese.Is this any worse than an entire nation treating one individual as theirtotal lord and master, for hundreds of years.>2) The British were kicked out. If it were not for fear of the military>strength of China I doubt if the British would have left.Politics my dear boy.> Its is interesting that after a 100 years of British rule in Hong Kong via>a British Governor that they suddenly decided they wanted to introduce>democracy for the chinese citizens of Hong Kong before they left.>Unbelievable!It is interesting that after more than 1000 years of total dominationthe Chinese people still do not have their own voice.>> >...>>now that the Brits have been kicked out.>> A cavil: the British rented Hong Kong for a xed period of time and>> left when that period of time expired. There was no, and no need for>> any, kicking out.>> -->> G.C.>> [Moderators note: Ill remove this jocular comment on the website>> version. Followups set to the more freewheeling newsgroups. - jb]-- http://www.earthpoetry.demon.co.ukRC => True, Coreld describes how in 1996 the > automated theorem prover EQP was the rst to crack the > Robbins conjecture. This states that a Boolean algebra is > the same as a set equipped with an commutative associative > binary operation or together with a unary operation not > for which one mind-numbing axiom holds, namely: not(not(p or q) or not(p or not(q)) = p All the rest of Boolean logic is a consequence!As a side note, a few years ago Stephen Wolfram conducted a computer search for the smallest possible axiom systems for Boolean algebra expressed in terms of nand. The candidates1. (ab)(a(bc))=a, ab=ba2. ((bc)a)(b((ba)b))=a3. (b((ab)b))(a(cb))=a(if I have made no typographical error) turned out to be correct; you can easily nd proofs yourself by plugging them into Waldmeister (an automated theorem prover). See pages 808-9 of A New Kind of Science. =>True, Coreld describes how in 1996 the>automated theorem prover EQP was the rst to crack the>Robbins conjecture. This states that a Boolean algebra is>the same as a set equipped with an commutative associative>binary operation or together with a unary operation not>for which one mind-numbing axiom holds, namely:> not(not(p or q) or not(p or not(q)) = p> All the rest of Boolean logic is a consequence!> As a side note, a few years ago Stephen Wolfram conducted a computer> search for the smallest possible axiom systems for Boolean algebra> expressed in terms of nand. The candidates 1. (ab)(a(bc))=a, ab=ba> 2. ((bc)a)(b((ba)b))=a> 3. (b((ab)b))(a(cb))=a (if I have made no typographical error) turned out to be correct; you> can easilyfind proofs yourself by plugging them into Waldmeister (an> automated theorem prover). See pages 808-9 of A New Kind of Science. You were one of the instigators!http://www.cs.unm.edu/~veroff/BA/http:// www.cs.unm.edu/~veroff/BA/candidates.htmlhttp:// mathworld.wolfram.com/WolframAxiom.html--Uncle Alhttp://www.mazepath.com/uncleal/eotvos.htm (Do something naughty to physics)[Moderators note: followups directed to the most relevantnewsgroup, sci.math. - jb] = >Remember, my past history IS relevant. Ive been labeled gifted since>I was a child. ...>>This business of labeling kids can be very harmful.>>Gib Especially calling them kids. Whatever happened to children?Heh. You sound like my mother.Gib =I can do the math from rst principles and Ive been told that Excel hasfunctions that would do most of the dirty work but since I only need theanswers, not the practice (this is most denitely NOT homework -- Inished with my last assignment over two decades ago), I hope someonesalready done it (and, in the best of all possible worlds, posted theresults on the internet).I need to electronically synthesize signals that are sine waves. There aretwo general ways of doing this: an analog oscillator circuit and a digitalcircuit using a digital-to-analog converter. Analog oscillator circuits canbe designed to yield very high quality signals but usually over a verynarrow range of frequencies. Since the digital approach has a (potentially)much wider frequency range, Id like to use that. The simplest approach isto generate a piecewise constant signal which is ltered to yield anapproximation to a sine wave. There are two basic parameters that need tobe chosen, the number of intervals and the number of bits that determine thevoltage levels. The design of the lter(s) that follow thedigital-to-analog stage in the circuit depends, to some extent, on thequality of the signal thats input to it, SO ...Is there any tabulation (URL, book, paper, ...) of, say, the harmonicdistortion of a piecewise constant approximation to a sine wave as afunction of the # steps and/or the # bits used? Norm => I can do the math from rst principles and Ive been told that Excel has> functions that would do most of the dirty work but since I only need the> answers, not the practice (this is most denitely NOT homework -- I> nished with my last assignment over two decades ago), I hope someones> already done it (and, in the best of all possible worlds, posted the> results on the internet). I need to electronically synthesize signals that are sine waves. There are> two general ways of doing this: an analog oscillator circuit and a digital> circuit using a digital-to-analog converter. Analog oscillator circuits can> be designed to yield very high quality signals but usually over a very> narrow range of frequencies. Since the digital approach has a (potentially)> much wider frequency range, Id like to use that. The simplest approach is> to generate a piecewise constant signal which is ltered to yield an> approximation to a sine wave. There are two basic parameters that need to> be chosen, the number of intervals and the number of bits that determine the> voltage levels. The design of the lter(s) that follow the> digital-to-analog stage in the circuit depends, to some extent, on the> quality of the signal thats input to it, SO ... Is there any tabulation (URL, book, paper, ...) of, say, the harmonic> distortion of a piecewise constant approximation to a sine wave as a> function of the # steps and/or the # bits used? NormCheck out e-panorama.net.Im sure theyll have something for you there.cid ooh => I can do the math from rst principles and Ive been told that Excel has> functions that would do most of the dirty work but since I only need the> answers, not the practice (this is most denitely NOT homework -- I> nished with my last assignment over two decades ago), I hope someones> already done it (and, in the best of all possible worlds, posted the> results on the internet). I need to electronically synthesize signals that are sine waves. Thereare> two general ways of doing this: an analog oscillator circuit and a digital> circuit using a digital-to-analog converter. Analog oscillator circuitscan> be designed to yield very high quality signals but usually over a very> narrow range of frequencies. Since the digital approach has a(potentially)> much wider frequency range, Id like to use that. The simplest approachis> to generate a piecewise constant signal which is ltered to yield an> approximation to a sine wave. There are two basic parameters that need to> be chosen, the number of intervals and the number of bits that determinethe> voltage levels. The design of the lter(s) that follow the> digital-to-analog stage in the circuit depends, to some extent, on the> quality of the signal thats input to it, SO ... Is there any tabulation (URL, book, paper, ...) of, say, the harmonic> distortion of a piecewise constant approximation to a sine wave as a> function of the # steps and/or the # bits used? Norm>You mightfind this discussion by Don Lancaster of interest:http://www.tinaja.com/magsn01.asp#tutorialsIve been a fan of Dons since his Apple II+ days. Hes a veryversatile guy. =Norm:Right off the top of my head:16 bits digital = 96 dbm signalthe sample rate and Nyquist will determine the delity.20 samples/sec means 10 hz is maximum faithful reproduction, so you want 2x asthe sample rate of your highest frequency.20 bit or 24 bit digital should encode the real world quite well => Norm: Right off the top of my head: 16 bits digital = 96 dbm signal the sample rate and Nyquist will determine the delity. 20 samples/sec means 10 hz is maximum faithful reproduction, so you want2x as> the sample rate of your highest frequency. Only if theres sufcient ltering following the A/D stage. Forexample, a 10 KHz square wave is a terrible approximation to a 5 KHz sinewave. Without much ltering, you need many steps in the output. 32-64isnt too much. 20 bit or 24 bit digital should encode the real world quite well> When was the last time you priced 24-bit D/A converters that run at amegahertz or so? Thats why Im asking the question. I need to compromisebetween cost/availability and performance and 12 bits or so is my top end.But Id like to know how much difference 10 bits (or 14 bits) would make. =Nyquist meant the barest minimum. Of course bits are needed, the more thebetter and oversampling at > 2x is better too.> When was the last time you priced 24-bit D/A converters that run at a> megahertz or so? Thats why Im asking the question. I need to compromise> between cost/availability and performance and 12 bits or so is my top end.> But Id like to know how much difference 10 bits (or 14 bits) would make.Just exactly what do you need to do? This will help us gure out what tosuggest.I think a 16 bit convertor pulling 5 mhz is very affordable. Right now, as Itype, I am attempting to land on the Analog Devices web site, which is NOT userfriendly. I do know that AD offers convertors.Tommorrow when I have access to broadband internet and resources, I can morespecically answer your question as I can tap into the engineering catalogs.Randall =Norm:> Right off the top of my head:> 16 bits digital = 96 dbm signal> the sample rate and Nyquist will determine the delity.> 20 samples/sec means 10 hz is maximum faithful reproduction, so you want> 2x as>the sample rate of your highest frequency.> Only if theres sufcient ltering following the A/D stage. For> example, a 10 KHz square wave is a terrible approximation to a 5 KHz sine> wave. Without much ltering, you need many steps in the output. 32-64> isnt too much. 20 bit or 24 bit digital should encode the real world quite well> When was the last time you priced 24-bit D/A converters that run at a> megahertz or so? Thats why Im asking the question. I need to compromise> between cost/availability and performance and 12 bits or so is my top end.> But Id like to know how much difference 10 bits (or 14 bits) would make.What is your highest frequency sine wave?Chuck-- ... The times have been, That, when the brains were out, the man would die. ... Macbeth Chuck Simmons chrlsim@earthlink.net = What is your highest frequency sine wave?>I will have available a digital clock at (a minimum of) 4 MHz. The goal ofthe instrument is twofold: 1. Output arbitrary waveforms using this clock. Theres an optionaloutput stage which can be made to contain any desired lter, but the rawoutput will be from a (probably) 12-bit D/A converter clocked (at a maximumof) 4 MHz. 2. Output high quality sine waves of as high a frequency as possiblegiven the constraints of the output D/A stage. This stage can be expandedto be from 12-bits to 16-bits, depending on how much money I want to spend.If I use a 4 MHz clock and assume a 32-point sine wave, the maximumfrequency is 125 KHz. An 8-point sine wave (which obviously takes a lotmore ltering) goes to 500 KHz. Norm = I can do the math from rst principles and Ive been told that Excel has> functions that would do most of the dirty work but since I only need the> answers, not the practice (this is most denitely NOT homework -- I> nished with my last assignment over two decades ago), I hope someones> already done it (and, in the best of all possible worlds, posted the> results on the internet). I need to electronically synthesize signals that are sine waves. There are> two general ways of doing this: an analog oscillator circuit and a digital> circuit using a digital-to-analog converter. Analog oscillator circuits can> be designed to yield very high quality signals but usually over a very> narrow range of frequencies. Since the digital approach has a (potentially)> much wider frequency range, Id like to use that. The simplest approach is> to generate a piecewise constant signal which is ltered to yield an> approximation to a sine wave. There are two basic parameters that need to> be chosen, the number of intervals and the number of bits that determine the> voltage levels. The design of the lter(s) that follow the> digital-to-analog stage in the circuit depends, to some extent, on the> quality of the signal thats input to it, SO ... Is there any tabulation (URL, book, paper, ...) of, say, the harmonic> distortion of a piecewise constant approximation to a sine wave as a> function of the # steps and/or the # bits used? NormTry posting in sci.electronics.design. This topic comes up thereoccasionally and rhere are a few posters who know the answers you need.Chuck-- ... The times have been, That, when the brains were out, the man would die. ... Macbeth Chuck Simmons chrlsim@earthlink.net => First I will call the rotation Q*U*Q=M instead of R, in order not to mix> it up with R coming from the QR-factorization [Q,R]=qr([a b]). (a and b> column vectors).To start, I wasnt familiar with QR decomposition, so I looked up a paper onit (and the paper had numerous errors as I started working the explanations)but the basic idea is that we start with 1 vector from a set and createorthogonal vectors to the set, by gradually decomposing our original set ofvectors.The key idea: B2 = a2 - b1*b1T*a2where b1 is the rst vector of magnitude 1 derived from a1. We then subtractoff the projection onto b1 of a2 to get an orthogonal vector to b1. This isrepeated as we slowly add to our pile of orthogonal vectors.I worked out the [Q,R] decomposition asa = [-2sqrt(6/25),0,0,0,-1/5]~b = [-sqrt(6/25),-sqrt(6/25),-sqrt(6/25),-sqrt(6/25),-1/5]~q1 = a = [-sqrt(24/25),0,0,0,-1/5]~q2 = [sqrt(1/1900),-sqrt(25/76),-sqrt(25/76),-sqrt(25/76),-sqrt(6/ 475)]~q1 and q2 are magnitude 1 and orthogonal to each other.So I get Q = [-sqrt(24/25) sqrt(1/1900) 0 -sqrt(25/76) 0 -sqrt(25/76) 0 -sqrt(25/76) -1/5 -sqrt(6/475)] and checking the multiplication Q~*a = [1,0]~ which shows rotation correctlyinto the x plane as desired. Also Q~*b = [13/25, sqrt(456/625)]~ as we want.but what happened to the missing columns??? Later on they seem to becomemandatory.I also get R = [1 13/25 0 sqrt(456/625)]and [a b] = Q*R> Im not familiar with the GP-Pari program but I gure it will be pretty> messy to gure out how the rotation M=Q*U*Q depends on rot-vw,wx,xy.agreed!> In case you dont know the following I can explain how M rotates a (and> b). Maybe you can gure out a dependence. M can be seen as three> rotations, rst Q then U and nally Q. Observe that Q is the inverse> rotation of Q and vice versa.> Q is (always) the rotation such that Q*a rotates a into> the x-axis [1,0,0,0,0] and Q*b rotates b into the x-y plane> [1,1,0,0,0] relative to Q*a.> So call a2=Q*a=[1,0,0,0,0] and b2=Q*b=[13/25,sqrt(456/625),0,0,0]. The> angle between a2 and b2 is still the same as between a and b. a | / b ______ a2 | / Q |/ --> b2> Now we only need to rotate a2 to b2, then use the inverse rotation Q of Q> in order to get back to the original form. The rotation U, which depends> on just one angle, is really the only rotation angle that is needed to> rotate a2 into b2 (or a into b).> U(phi)=[cos(phi) -sin(phi) 0 0 0> sin(phi) cos(phi) 0 0 0> 0 0 1 0 0> 0 0 0 1 0> 0 0 0 0 1]I get then for U: U(phi)=[ 13/25 -sqrt(456/625) 0 0 0 sqrt(456/625) 13/25 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1]> We get the that, since a2=[1,0,0,0,0], cos(phi)=13/25 and> sin(phi)=sqrt(456/625).> So U*a2=b2 but we wish to return to the original coordinate system, we> must perform the inverse rotation of Q, that is Q. Q*Q*b2=b So Q*U*a2=b> which gives M=Q*U(phi)*Q.huh?? Doesnt Q*U*a2 = a? didnt you mean Q*U*b2 = b ?> M is only dependent on one angle phi. I regard Q as just an orthogonal> base for R^5 telling us how/in what direction we should rotate phi> radians in order to rotate a-->b.Well, Ive stopped at Q. I have 2 orthogonal basis, but apparently I need 5?Must Q be extended?I tried to multiply QU(phi)Q~, but since they are NOT square matrices, thingshave hung.Just what are the other 3 missing columns of Q?I played with Q = [-sqrt(24/25) sqrt(1/1900) 0 0 0 0 -sqrt(25/76) 0 0 0 0 -sqrt(25/76) 0 0 0 0 -sqrt(25/76) 0 0 0 -1/5 -sqrt(6/475)] 0 0 0]and got M = [949/1900, -37/76, -37/76, -37/76, sqrt(3/451250) 899/1900, 13/76, 13/76, 13/76, sqrt(7803/451250) 899/1900, 13/76, 13/76, 13/76, sqrt(7803/451250) 899/1900, 13/76, 13/76, 13/76, sqrt(7803/451250) sqrt(17787/451250), -sqrt(3/722), -sqrt(3/722), -sqrt(3/722), 13/475]which does work and seems to correctly rotate a-->bIs this correct??? The missing 3 columns for Q bother me.Randall =>> First I will call the rotation Q*U*Q=M instead of R, in order not to mix>> it up with R coming from the QR-factorization [Q,R]=qr([a b]). (a and b>> column vectors).>....I worked out the [Q,R] decomposition as>a = [-2sqrt(6/25),0,0,0,-1/5]~>b = [-sqrt(6/25),-sqrt(6/25),-sqrt(6/25),-sqrt(6/25),-1/5]~>....I played with Q = [-sqrt(24/25) sqrt(1/1900) 0 0 0> 0 -sqrt(25/76) 0 0 0> 0 -sqrt(25/76) 0 0 0> 0 -sqrt(25/76) 0 0 0> -1/5 -sqrt(6/475)] 0 0 0]and got M => [949/1900, -37/76, -37/76, -37/76, sqrt(3/451250)> 899/1900, 13/76, 13/76, 13/76, sqrt(7803/451250)> 899/1900, 13/76, 13/76, 13/76, sqrt(7803/451250)> 899/1900, 13/76, 13/76, 13/76, sqrt(7803/451250)> sqrt(17787/451250), -sqrt(3/722), -sqrt(3/722), -sqrt(3/722), 13/475]which does work and seems to correctly rotate a-->bIs this correct??? The missing 3 columns for Q bother me.RandallIf what you want is only to calculate a rotation matrix that will takevector a to vector b, here is a way that always works in anydimension. Calc the following:w = (a+b)/norm(a+b)M1 = Id - 2*column(a)*row(a)M2 = Id - 2*column(w)*row(w)where Id is the 5x5 identity matrix, norm(a+b) is the length of vector a+b,column(a) means to consider a as a column vector,row(a) means to consider a as a row vector,so column(a)*row(a)results in a 5x5 matrix (the outer product).The rotation matrix is then M=M2*M1 and therefore the rotation of a tob is written as b=(M2*M1)*aThe way this works is by using 2 reections in the plane thatcontains a and b: M1 reects a to -aM2 reects -a to bw is the vector that bisects the angle between a and b. You can do adrawing to see how this method works.In another message you talked about decomposing into 3 rotations inplanes like vw, wx, xy, etc. I dont know nothing about that, but ifyou know the standard rotation matrices for these planes, it might bepossible, given the matrix M calculated above, to extract the anglesnecessary for each plane rotation. I know how to do it in 3d (forexample, extracting Euler angles for rotating around the axes x,y,z inthat order, from the matrix), but I dont know if its possible in 5d.PS: with this method I get the matrix M=(1/1900) * [1024, -900, -900, -900, -148*sqrt(6),924, 1600, -300, -300, 52*sqrt(6),924, -300, 1600, -300, 52*sqrt(6),924, -300, -300, 1600, 52*sqrt(6),4*sqrt(6), -100*sqrt(6), -100*sqrt(6), -100*sqrt(6), 1852]I used vectors a={-2*sqrt(6/25.0),0,0,0,-1/5.0}b={-sqrt(6/25.0),-sqrt(6/25.0 ),-sqrt(6/25.0), -sqrt(6/25.0),-1/5.0};and your matrix doesnt seem to work correctly. =>and got M =>> [949/1900, -37/76, -37/76, -37/76, sqrt(3/451250)>> 899/1900, 13/76, 13/76, 13/76, sqrt(7803/451250)>> 899/1900, 13/76, 13/76, 13/76, sqrt(7803/451250)>> 899/1900, 13/76, 13/76, 13/76, sqrt(7803/451250)>> sqrt(17787/451250), -sqrt(3/722), -sqrt(3/722), -sqrt(3/722), 13/475]>>which does work and seems to correctly rotate a-->b>>Is this correct??? The missing 3 columns for Q bother me.>>Randall>..and your matrix doesnt seem to work correctly.Correction, your matrix does work. =>Here is an excellent analysis:>http://www.u.arizona.edu/~chalmers/papers/ envelope.htmlYup, thats good. (And it shows how even experienced philosophers can getthis wrong.)> To resolve this paradox, separate the superous information.> There are two envelopes, call them A and B.> Able chose A, then switched to B.> Baker chose B and stood pat.> Charley chose B, then switched to A.> David chose A and stood pat.> Was Able better, or worse off than Baker?> Was Charley better, or worse off than David?> I answer no to both those questions, therefore there seems to be no> advantage for switching, or not switching.I think youre missing the point. Everybody knows, via simple symmetry,that there is no advantage for switching. So we dont need yet another proofthat switching has no advantage. Nobody is confused about what to do werethe actual situation to arise.The puzzle is, what is wrong with the simple probability calculation thatswitching has an expected value of 50%x1/2 + 50%x2 = 1.25, i.e. more thanyou have in the rst envelope. That seem a very compelling proof. Yet,through symmetry, we know the proof is wrong.The puzzle is, why is that probability proof wrong?The Chalmers paper addresses the issue. Your new proof doesnt help. -- Don__________________________________________________________ _____________________Don Geddis http://don.geddis.org/ don@geddis.orgSometimes I lie awake at night and ask Why me? Then a voice answersNothing personal, your name just happened to come up. -- Charlie Brown, _Peanuts_ [Charles Schulz] =Please assist in sketching out the proofs of the following assertions:1. Any closed subset of R x R with nonzero area has Cardinality = c.2. The Cardinality of the the number of closed sets in R x R is also c.Seems much easier in R.THXHarry = Please assist in sketching out the proofs of the following assertions: 1. Any closed subset of R x R with nonzero area has Cardinality = c.> 2. The Cardinality of the the number of closed sets in R x R is also c.> For 1, as long as the area is nite, its sufcient to consider Sq =[0,1] x [0,1] and its relatively easy to construct a bijection from Sq onto[0,1]. => Please assist in sketching out the proofs of the following assertions:> 1. Any closed subset of R x R with nonzero area has Cardinality = c.>2. The Cardinality of the the number of closed sets in R x R is also c. For 1, as long as the area is nite, its sufcient to> consider Sq = [0,1] x [0,1] and its relatively easy to construct a> bijection from Sq onto [0,1].Why is that sufcient? =>Please assist in sketching out the proofs of the following assertions:1. Any closed subset of R x R with nonzero area has Cardinality = c.>2. The Cardinality of the the number of closed sets in R x R is also c.>For 2, the cardinality of the collection of closed sets is the same as that of the topology, which is second countable. (A big hint, not a complete answer.)Cite sources (if only the newgroup) of assistance in submitted homework.-- Stephen J. Herschkorn herschko@rutcor.rutgers.edu =>> Please assist in sketching out the proofs of the following assertions:>> 1. Any closed subset of R x R with nonzero area has Cardinality = c.>> 2. The Cardinality of the the number of closed sets in R x R is also c. For 2, the cardinality of the collection of closed sets is the same as > that of the topology, which is second countable. (A big hint, not a > complete answer.) Im confused. Surely the assertion depends on the topology. If all sets are open (and therefore closed), the number of closed sets in R x R is 2^c, which is greater than c. Similarly, if only the empty set and the universal set are open (and therefore closed), then the cardinality of closed sets in R x R is 2, which is less than c.Is the standard topology assumed for this problem?> Cite sources (if only the newgroup) of assistance in submitted homework. Why? It seems to me that if all you got was a hint, that doesnt need to be footnoted for homework. Although discussing homework is dangerous, because you very well might get lots of practice watching math get done, which is much less valuable than getting practice doing math.Do I need to footnote that? The idea comes to me from Harold Diamond. He didnt footnote it, so I dont know if it was original with him or plagiarized. Uh, borrowed. Used.Jon Miller =>> Please assist in sketching out the proofs of the following assertions:>> 1. Any closed subset of R x R with nonzero area has Cardinality = c.> 2. The Cardinality of the the number of closed sets in R x R is also c. >> For 2, the cardinality of the collection of closed sets is the same >> as that of the topology, which is second countable. (A big hint, not >> a complete answer.) > Im confused. [...] Is the standard topology assumed for this problem? Of course. That is the convention when one talks of closed sets in R^n without mention of a specic topology.>> Cite sources (if only the newgroup) of assistance in submitted homework. > Why? It seems to me that if all you got was a hint, that doesnt need > to be footnoted for homework.In this case, the hint does most of the work. Not citing a source comes close to plagiarism, similar tofinding the answer in another book, even if you write it in your own words. One should *never* present anothers work as ones own.The same ethical rules apply to homework as well as to work submitted for publication. You should always cite the contributions of others to note that they are not original to you. Henry Helson told the story once of how he (unintentionally) got his name attached to a theorem because he just commented in passing, Hmm, have you tried Baire category?> Although discussing homework is dangerous, because you very well might > get lots of practice watching math get done, which is much less > valuable than getting practice doing math. Do I need to footnote that? The idea comes to me from Harold Diamond. > He didnt footnote it, so I dont know if it was original with him or > plagiarized. Uh, borrowed. Used. Jon Miller You did exactly the right thing, though attributions for this kind of conventional wisdom are not really so necessary, particularly in newsgroups. You quoted the source you know. If Diamond mentioned that in conversation, there was no need for him to cite a source. If it was in publication and he knew the idea came from someone else, he might have cited the source. (As so and so has pointed out,...) Perhaps he did and you do not recall.-- Stephen J. Herschkorn herschko@rutcor.rutgers.eduX-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Punge: Micro$oft = at 01:26 PM, ArtDodgr said:>Reason: the parallelogram law>|x+y|^2 + |x-y|^2 = 2|x|^2 + 2|y|^2>fails for x = (1,0,0,0,...) and y = (0,1,0,0,...).x+y =(1,1,0,0, . . .)x-y =(1,-1,0,0, . . .)|x+y|^2=|x-y|^2=2|x+y|^2 + |x-y|^2 = 4|x|^2=|y|^2=12|x|^2 + 2|y|^2 =4-- Shmuel (Seymour J.) Metz, SysProg and JOATto = Shmuel (Seymour J.) Metz said: >Reason: the parallelogram law >|x+y|^2 + |x-y|^2 = 2|x|^2 + 2|y|^2 >fails for x = (1,0,0,0,...) and y = (0,1,0,0,...). x+y =(1,1,0,0, . . .) |x+y| = |x-y| = 2, so |x+y|^2 = |x-y|^2 = 4. > |x+y|^2 + |x-y|^2 = 4 |x|^2=|y|^2=1 2|x|^2 + 2|y|^2 =4-- A. =Consider Limit[Sum[i=1,n,1/(a*n^k+b*i)],n->Innity].Prove that when k=1, the limit in question equals (Ln[(a+b)/b]-Ln[a/b])/b,when k>1, the limit=0, and when k<1, the limit=Innity. => Consider Limit[Sum[i=1,n,1/(a*n^k+b*i)],n->Innity]. Prove that when k=1, the limit in question equals (Ln[(a+b)/b]-Ln[a/b])/b,> when k>1, the limit=0, and when k<1, the limit=Innity.I assume a and b are positive. The above sum is Sum_(i=1,n) {1/[a*n^(k-1) + b*(i/n)]}*(1/n).When k = 1, were looking at a Rimeann sum for int_[0,1] [1/(a + bx)] dx. The limit is the value of this integral.k < 1: a*n^(k-1) -> 0. So let eps > 0. For large n, the above sum is greater than Sum_(i=1,n) {1/[eps + b*(i/n)]}*(1/n),which is a Riemann sum for int_[0,1] [1/(eps + bx)] dx. Evaluate this integral to see the sum can be made arbitrarily large.k > 1: Same approach. This time a*n^(k-1) -> oo. =4d squares (cubes, whatever) are called hybercubes, and ive also>heard them called tesseracts, so... what are 4d circles (spheres)>called, if theyre called anything at all?also, what would the formulae for the area of a 4d sphere be?If I remember correctly:In 2-D the boundry of a circle has length (2 Pi Radius)In 2-D the circle has area (2 Pi) ((Radius^2)/2) In 3-D the boundry of a circle (sphere) has area (4 Pi Radius^2)I 3-D the sphere has volume (4 Pi) ((Radius^3)/3)In 4-D the boundry of a circle (hypersphere)has volume (4 Pi^2 Radius^3)In 4-D the hypersphere has hypervolume (4 Pi^2 (Radius^4)/4)adam =As Ive considered various explanation for the surprising lack of> >agreement with my explanation for the problem with algebraic integers,> >Ive concluded that symbol confusion may indeed be the problem, as> >though mathematicians *say* that they can just follow what theyre> >told a symbol means, it seems to me that the best explanation for the> >scope of disagreement with me, is symbol confusion.> >> >So Im going to change some symbols, and see what happens.> >> >Editorial note: > >> >Several posters had a problem with the following statement:> >> >In the ring of algebraic integers, given a factor G(X) of a polynomial> >P(X), R(X) and C exists such that G(X)=R(X) + C, where C=G(0), and> >R(X)=G(X)-C.> >> >As they noted that a polynomial is not a number, and rather than argue> >about such details Im reposting with changes. Your lemma 2 is still a mess, and your lemma 1 still has no actual> content; you never use your hypothesis in the proof of the lemma. Why> include them?> >Huh? Your statement of Lemma included G is a factor of P. Your proof of> Lemma 1 never used the fact that G is a factor of P; in fact, the> conclusion from your statement of Lemma 1 was just that G can be> written as R(x)+G(0), where R(x)=G(x)-G(0); so you never use that G is> a factor of P for that. You only use that ->later<-, when you conclude> that if G(x) is a factor of P(x) (in A^A), then G(0) must be a factor> of P(0) in A. Since you never use the hypothesis in the body of Lemma 1, there is no> reason to include them.The point of lemma 1 is that a factor of a *polynomial* can be brokenup into a varying and non-varying portion, which sort of connects itto a polynomial.So its material that G is a factor of P.Otherwise, what is it?There is no otherwise, as it is a factor of a polynomial, as stated. > >In what follows variables, unless otherwise noted, are in the ring ofalgebraic integers. So you are wroking in the ring of algebraic integer-valued functions> of algebraic integer variables. How many variables? Lemma 1 and> apparently Lemma 2 are for functions of 1 variable; but you try to> apply them to functions of several variables. You cannot do that.> >Why not? Because you say so? Are you kidding? A lemma/theorem/corollary is a statement of the form If A then B. In> order to apply a lemma/theorem/corollary, the hypothesis must be> met. If your lemma is about functions of one variable, and you try to> apply them to functions of more than one variable, then your> hypothesis are not met, and therefore the lemma cannot be applied. Basic logic, James.Im not interested in a debate on the subject, nor with distractions.Well here its mostly about appearances as in fact the key variable isX, while you have continually tried to add confusion by acting as ifthe other variables, used to factor the expression into non-polynomialfactors, are more than just placemarkers for the non-polynomialfactorization.Lemma 1:> >> >Given a factor G(X) of a polynomial P(X), R(X) and C exists such that> >G(X)=R(X) + C, where C=G(0), and R(X)=G(X)-C. You REALLY should label the beginning of a proof. This has no content, and you never use, in this proof, the fact that> G(X) is a factor of P(X).> >G(X) is a factor of the polynomial P(X), so G(0) is a factor of P(0). If that is your conclusion, then it should be in the body of the> below.Are you serious? Given that G(X) is a factor of the polynomial P(X),it must be the case that G(0) is a factor of P(0), as one is moregeneral, while the other is specic.> Though not necessary, you might want to note that you are considering> P(X) as a function from A to A, not as a polynomial with coefcients> in A.> >Im not interested in not necessary. Are you interested in correctness and clarity? Well, if something is not necessary then its not necessary.Consider that C is a factor of the constant term P(0), which follows> >as G(X) is a factor of P(X), so G(0) is a factor of P(0), but C=G(0);> >therefore, C is a factor of the constant term P(0). Proof Complete. You never stated in the lemma that c is a factor of the constant term> P(0). So if you wanted to do that, you should have stated it. I think> what you were trying to say is: LEMMA 1. Let A be the ring of all algebraic integer, and let A^A be> the ring of functions from A to A. Let P(X) be a polynomial with> coefcients in A, considered as an element of A^A, and let G(X) be an> element of A^A which is a factor (in A^A) of P(X). Then we can write> G(X)=R(X)+C, where C is the constant G(0); moreover, C is a factor in> A of P(0). Proof. We can certainly write G(X) as G(X) = (G(X)-G(0))+G(0), so> setting R(X)=G(X)-G(0) accomplishes the rst part of the lemma. For> the second, let Q(X) be a function from A to A such that> G(X)*Q(X)=P(X) in A^A. Then in particular G(0)*Q(0)=P(0) in A, so G(0)> is a divisor of P(0) in A. This proves the Lemma. Proof complete.Hmmm...is this cooperation nally? Sigh. Ive been suggesting changes for clarity and xed to your> arguments for YEARS. This particular change I suggested before, but> you ignored it and decided instead to insult me. So get off your moral> high horse; it doesnt actually exist.Facts not in evidence.For those who dont know, and in case I messed up with my usage, thatmeans that the poster has said something that requires outsideverication. That is, tofind out if hes telling the truth, youhave to go do research.Its a nasty little trick that Arturo Magidin likes to play.> >Lemma 2:> >Now consider P(X) such that it has b^2 as a factor, and for a factor> >G(X), C has b as a factor.> >> >Further consider that P(0)/b^2 is coprime to b; therefore, as C is afactor of P(0) a factor of b must divide off of C, when b^2 divides> >off of P(m). Proof Complete. Once again, you have no hypothesis, and no conclusion. How can you> have a proof? How can you have a lemma? What does divide off mean?> What is m? What is b^2? A function or a constant? Is b^2 a factor of> P(X) in the sense that every coefcient of P(X) is a multiple of b^2> (that is, b^2 is a factor of P(X) in A[x]), or in the sense that b^2> is a factor of P(X) in A^A? Different things. Presumably: LEMMA 2. Let P(X) be a polynomial with coefcients in A[x],> and let b be an algebraic integer. Let G(X) in A^A be a> factor of P(X) in A^A, and write G(X)=R(X)+G(0). Suppose that b^2> divides every coefcient of P(X) in A. Then G(0) is a multiple of b> in A. Moreover, if P(0)/b^2 is coprime to b in A, then ____________________ > (I was unable to gure out the blank; please ll it).> >Well if you divide b^2 off from the polynomial then a factor must> >divide off of G(0) such that what remains is coprime to b. Still doesnt make sense. And you do not answer any of the questions I> asked. What is m? Is b^2 a factor of each coefcient of P(X), or do> you mean that the VALUES of P(X) are divisible by b^2 for every valid> value of the variable? And in the latter case, what are the valid> values of the variable?You ask what is m? Well, as m is not in the argument, I must say Idont know what you mean by the question. Possibly you refer to X, asin past arguments Ive used m where I now use X.Its already given that the variables are in the ring of algebraicintegers.Given b^2, and that b^2 is a factor of P(X), it follows that it is afactor of P(X), which is whats mathematically necessary.> Divide b^2 off from the polynomial. Is that for the value? Or as> polynomials? Or as functions? What does dividing off of G(0) mean?The polynomial has a factor b^2, so Im talking of dividing it off.For instance, P(x) = 2x^2 + 4x + 2, has a factor of 2, and dividing itoff gives P(x)/2 = x^2 + 2x + 1.Now then, as P(X) has b^2 as a factor, P(0) has b^2 as a factor, andG(0) is a factor of P(0) as well. So dividing off b^2 will dividesome factor off of G(0), if G(0) isnt already coprime to b, sinceP(0)/b^2 IS coprime to b. > If this were what you had in mind, thben it is certainly wrong. For> instance, P(x)=(2x+2)(x+1), G(x)=x+1; then 2 divides every coefcient> of P(x), but does not divide G(0).> >If you divide off 2 from P(x) here then the constant terms with> >respect to x of both of its factors are coprime to 2. But it is still false that b divides off G(x). So you need to state> your Lemma CLEARLY AND COMPLETELY. Stop using your own personal> language such as divides off.Here you have P(x) with 2 as a factor, but dividing off 2 leaves theconstant terms of both factors of P(x)/2 coprime to 2.Its trivial.> >Thats obvious enough as then you have (x+1)(x+1), so the constant> >term is 1. Nonetheless, the conclusion you claim in the body of your lemma> seems false in this case. So either your lemma is badly written, is> not what you meant, or is just plain wrong. Which is it?Well, it could very well be badly written.But then all that matters is xing it.You dont have to get so hostile, as if its badly written, thenagain, its just a matter of xing it.> Of course, I probably interpreted the lemma wrong, so please write it> correctly, clearly specifying the hypothesis and the conclusion. Also> try to label where the proof begins and ends.> >Lemma 2:> >Now consider P(X) such that it has b^2 as a factor, and for a factor> >G(X) of P(X), from lemma 1, G(X) = R(X)+ C. You still failed to answer an important question here. When you say> P(x) such that it has b^2 as a factor does that mean that (a) b^2 divides each coefcient of P(X); or (b) The values of P(X) are always divisible by b^2? Different things: x^2+x is a multiple of 2 in the latter sense, but> not in the former. Which is it?Your statement that x^2 + x is a multiple of 2 is false, as the ringis algebraic integers.Now it is the case that b^2 being a factor of P(X) means that it is afactor of each coefcient of P(X).However, Ifind it troubling that you *continue* to try and throw indistracting information. Now if P(0)/b^2 is coprime to b, then if C is not already coprime to> >b, there must exist a factor of C that is divided off when b^2 divides> >off such that the result is coprime to b, since C is a factor of P(0). Ah, a new clause (if C is not already coprime to b). So, basically,> what you are claiming here is (I am assuming the weakest of the two> possibilities above; feel free to change them): LEMMA 2. Let P(X) be a polynomial in A[X], b an element of A, such> that b^2 divides P(X) in A^A. Let G(X) be a factor of P(X) in A^A, and> write G(X)=R(X)+G(0). If P(0)/b^2 is coprime to b in A, then either> (a) G(0) is coprime to b; or> (b) There exists a divisor r of b^2 in A such that r divides G(0)> and G(0)/r is coprime to b. Is that it?Yup, that looks ok to me. Still youre rather tricky, so I say thatcautiously.> >Proof:> >C is a factor of P(0). So there exists D, such that P(0) = CD. Then> >P(0)/b^2 = CD/b^2, and given that P(0)/b^2 is coprime to b, CD/b^2> >must be coprime to b as well, so if C shares some factor with b^2 that> >factor divides off, leaving a result coprime to b.> >Proof Complete. Much better. Of course, this is just an application of the fact that> the algebraic integers are a Bezout domain: if you know that r*c=a*b> and gcd(r,c)=1, then r=gcd(r,a)*gcd(r,b), and gcd(a/gcd(r,a),r)=1. >Now consider> >> P(X) = b^2((b^4 X^3 - 3b^2 X^2 + 3X) t^3 - > > 3(-1 + b^2 X )t u^2 + b u^3)> >where the odd grouping is so that I can factor P(X) intonon-polynomial factors. And here we run into trouble. This is an expression in at least 2> variables; your lemmas are for expressions on 1 variable. Hence, your> lemmas do not apply.> >How about this to help you out?> P(X) = b^2((b^4 X^3 - 3b^2 X^2 + 3X) 11^3 - 3(-1 + b^2 X )11 u^2 + b u^3)> >so now I took away a variable, feel better? Its not a question of feeling better or not. It is a question of> go away. Sleight of hand doesnt x problems with your arguments.Its not hiding the variable if its NOT a variable, as 11 is NOT avariable, now is it?Heres where it might help to use my little non-polynomialfactorization example:Consider P(x) = 11x + 123,and now consider 11x + 123 = 11^2 + 11x + 2where I can use the 11s as placemarkers, so I can *act* like its thepolynomial y^2 + xy + 2, to get the factorization P(x) = (11 + (x+sqrt(x^2-8))/2)(11 + (x-sqrt(x^2-8))/2).> You still havent said what b and u are, though I suspect they are> meant to be specic, xed, algebraic integers satisfying some> mysterious properties you have not bothered to spell out just yet.I *did* say what b and u are as at the beginning I have the statement:In what follows variables, unless otherwise noted, are in the ring ofalgebraic integers. > >If youre still lost I can take away u as a variable as well, as> >follows. What, u was a variable? Oh my.In algebra, letter symbols are variables.> P(X) = b^2((b^4 X^3 - 3b^2 X^2 + 3X) 11^3 - 3(-1 + b^2 X )11 13^2 + b 13^3)> >So there you go Arturo Magidin, as your objection is now handled, in a> >way that shows it wasnt a valid objection in the rst place. You have not handled any objections. You have merely gone off the> deep end.Actually Im demonstrating that certain symbols are placemarkers,where because thats what they are, I can put in numbers for them.And again, my simple non-polynomial factorization example highlightshow it works: P(x) = 11x + 123,and now consider 11x + 123 = 11^2 + 11x + 2where I can use the 11s as placemarkers, so I can *act* like its thepolynomial y^2 + xy + 2, to get the factorization P(x) = (11 + (x+sqrt(x^2-8))/2)(11 + (x-sqrt(x^2-8))/2).Its a neat way to get a non-polynomial factorization.> >You tried to make up a bogus math rule. The bogus math rule being: In order to be able to apply a> lemma/theorem/corollary to a particular situation, the hypothesis of> that lemma/theorem/corollary must be satised in that situation?The key variable is X. That doesnt change.Your bogus objection is trying to act as if there are other importantvariables, when those variables are just there to get thenon-polynomial factorization.So the polynomial is P(X), the key variable is X, and just because Imfactoring P(X) into non-polynomial factors it doesnt mean that P(X)is changed, any more than factoring 6 in different ways makes itanything other than 6. > >Doing so I have the factorization> >> P(X) = (r_1(X,t) t + bu)(r_2(X,t) t + bu)(r_3(X,t) t + bu)> >> >where again the purpose above with the special grouping was to get> >that factorization. What are the properties of r_1, r_2, and r_3? Are they elements of> ->what<- ring? Is this an equality that has to hold in A^{AxA}?> >They are roots of a monic cubic which has coefcients that give> >algebraic integer values. So they are CONSTANTS? They have the same value for each value of X> and t? For each value of X? No, you are certainly just mouthing off> words whose meaning you do not understand. Presumably, you meant that their VALUES for any given X and/or t are> algebraic integers. So they are FUNCTIONS of algebraic integer> values. Which means that they are elements of A^{AxA}. So again I ask: is this an equality that holds WHERE? Is this an> equality of specic values, or an equality of FUNCTIONS? If the right> hand side is clearly a function of two variables, why is the right> hand said a function of just one variable? You are talking nonsense.Given that the variables are in algebraic integers, as stated at thebeginning, and further as the rs are dened by monic cubics usingthose variables, it follows that they are themselves, algebraicintegers.Its not nonsense; its mathematics.> >If youre still worried about t and u, then stick in 11 and 13. That is pure nonsense.No, it emphasizes that those are placemarkers for the non-polynomialfactorization.And again, its not nonsense; its mathematics.> >Now letting G(X) = r_1(X,t) t + bu, consider that X=0, gives> >> P(0) = b^2 (3t u^2 + b u^3) = b^2 u^2 (3t + bu), so at X=0,> >>(r_1(0,t) t + bu)(r_2(0,t) t + bu)(r_3(0,t) t + bu) = > >> b^2 u^2 (3t + bu)> >> >which shows that at least two of the rs go to 0, when X=0, andchoosing r_1 to be one of them, as the indices are arbitrary, I have> >from lemma 1 that> >> G(0) = bu, so C=bu. You cannot apply Lemma 1 to expressions of several variables. Please> specify what plays the role of X, the role of P, and the role of G in> this expression.> >Well to help you out, Ive given t the value of 11, and u the value of> >13. You are not helping out. You are just being condecending. Which is> mighty ironic, given that you are just showing that you do not> understand what it is you are talking about.Well I understand in such detail that I can give you a simpler exampleto help you out. P(x) = 11x + 123,and now consider 11x + 123 = 11^2 + 11x + 2where I can use the 11s as placemarkers, so I can *act* like its thepolynomial y^2 + xy + 2, to get the factorization P(x) = (11 + (x+sqrt(x^2-8))/2)(11 + (x-sqrt(x^2-8))/2).> >X is X. The polynomial is P(X), and> P(X) = b^2((b^4 X^3 - 3b^2 X^2 + 3X) t^3 - 3(-1 + b^2 X )t u^2 + b u^3).> G(X) = r_1(X,t) t + bu) So then, C is r_1(0,t)t + bu?Yup.> So that means that the equality must be an equality in A^{A}. So t> is what? Not a variable, not a xed value. What is it, then?Its a placemarker, and otherwise irrelevant as the key variable is X.I realized that using r_1(X,t) was confusing as it indicated that t isotherwise important, so Ive switched to r_1(X).> >Or, if you prefer to help you with all those variables that seemed to> >be bugging you out: They are not bugging me out; they are screwing up your argument. P(X) = b^2((b^4 X^3 - 3b^2 X^2 + 3X) 11^3 - 3(-1 + b^2 X )11 13^2 + b 13^3).> G(X) = r_1(X) 11 + 13b) Should that not be G(X) = r_1(X,11)11 + 13b?Ive dropped the r_1(X,t) usage as confusing.> And then the C in your lemma is G(0)=r_1(0,11)11 + 13b; NOT 13b. Right?Well, it turns out that r_1(0) = 0, so you are wrong.> I cant really follow anything else. I have the u, and should go> home. I will wait for you to reply to these questions for now. [.rest deleted for now.]> >Thats ok as youre not actually needed, and I know from your history> >over a period of years that youre probably just looking for yet> >another route to fooling the sci.math audience. In other words, you are just going to go ahead an insult me regardless> of what I say?Well, as long as youre not telling the truth, you cant expect to betreated like someone who tells the truth. =>> >As Ive considered various explanation for the surprising lack of>> >agreement with my explanation for the problem with algebraic integers,>> >Ive concluded that symbol confusion may indeed be the problem, as>> >though mathematicians *say* that they can just follow what theyre>> >told a symbol means, it seems to me that the best explanation for the>> >scope of disagreement with me, is symbol confusion.>>So Im going to change some symbols, and see what happens.>>Editorial note: >>Several posters had a problem with the following statement:>>In the ring of algebraic integers, given a factor G(X) of a polynomial>> >P(X), R(X) and C exists such that G(X)=R(X) + C, where C=G(0), and>> >R(X)=G(X)-C.>>As they noted that a polynomial is not a number, and rather than argue>> >about such details Im reposting with changes.>> Your lemma 2 is still a mess, and your lemma 1 still has no actual>> content; you never use your hypothesis in the proof of the lemma. Why>> include them?>>Huh?>> Your statement of Lemma included G is a factor of P. Your proof of>> Lemma 1 never used the fact that G is a factor of P; in fact, the>> conclusion from your statement of Lemma 1 was just that G can be>> written as R(x)+G(0), where R(x)=G(x)-G(0); so you never use that G is>> a factor of P for that. You only use that ->later<-, when you conclude>> that if G(x) is a factor of P(x) (in A^A), then G(0) must be a factor>> of P(0) in A. >> Since you never use the hypothesis in the body of Lemma 1, there is no>> reason to include them.The point of lemma 1 is that a factor of a *polynomial* can be broken>up into a varying and non-varying portion, which sort of connects it>to a polynomial.I cannot speak for what the point of your lemma is. I can only speakbe the point of the lemma. So, the obvious conclusion is that you cannot write what youmean. Which is no big surprise.>So its material that G is a factor of P.Good for you. One has to wonder, then, why your proof of the lemmanever used the material fact that G is a factor of P. Maybe you canexplain that?>Otherwise, what is it?There is no otherwise, as it is a factor of a polynomial, as stated.But you NEVER used that fact in your proof. It only occurred when Ithat conclusion later (and claiming it followed from Lemma 1, despitethe fact that Lemma 1 did not include any such conclusion).>> >In what follows variables, unless otherwise noted, are in the ring of>> >algebraic integers.>> So you are wroking in the ring of algebraic integer-valued functions>> of algebraic integer variables. How many variables? Lemma 1 and>> apparently Lemma 2 are for functions of 1 variable; but you try to>> apply them to functions of several variables. You cannot do that.>>Why not? Because you say so?>> Are you kidding?>> A lemma/theorem/corollary is a statement of the form If A then B. In>> order to apply a lemma/theorem/corollary, the hypothesis must be>> met. If your lemma is about functions of one variable, and you try to>> apply them to functions of more than one variable, then your>> hypothesis are not met, and therefore the lemma cannot be applied.>> Basic logic, James.Im not interested in a debate on the subject, nor with distractions.In other words Oops, I guess hes right. Rather than admit it, I willjust claim it is a distraction.>Well here its mostly about appearances as in fact the key variable is>X, while you have continually tried to add confusion by acting as if>the other variables, used to factor the expression into non-polynomial>factors, are more than just placemarkers for the non-polynomial>factorization.Its not about appearances, its about proper mathematics. If yourargument is powerful enough to include other situations, AND you wantto ->use<- the lemma in those other situations, then the lemma MUST bestated in such a way as to include those situations. If your abilitiesdo not extend so far as to enable you to do so, then ask for help.>> >Lemma 1:>>Given a factor G(X) of a polynomial P(X), R(X) and C exists such that>> >G(X)=R(X) + C, where C=G(0), and R(X)=G(X)-C.>> You REALLY should label the beginning of a proof.>> This has no content, and you never use, in this proof, the fact that>> G(X) is a factor of P(X).>>G(X) is a factor of the polynomial P(X), so G(0) is a factor of P(0).>> If that is your conclusion, then it should be in the body of the>> below.Are you serious? Yes, I am. If the point is that G(0) is a factor of P(0), then WHYdid you only rst mention it in the middle of a proof of aproposition which did not mention it in its body? If that is thepoint, then it should be written in the body of the statement, likeI did later:>> LEMMA 1. Let A be the ring of all algebraic integer, and let A^A be>> the ring of functions from A to A. Let P(X) be a polynomial with>> coefcients in A, considered as an element of A^A, and let G(X) be an>> element of A^A which is a factor (in A^A) of P(X). Then we can write>> G(X)=R(X)+C, where C is the constant G(0); moreover, C is a factor in>> A of P(0).>> Proof. We can certainly write G(X) as G(X) = (G(X)-G(0))+G(0), so>> setting R(X)=G(X)-G(0) accomplishes the rst part of the lemma. For>> the second, let Q(X) be a function from A to A such that>> G(X)*Q(X)=P(X) in A^A. Then in particular G(0)*Q(0)=P(0) in A, so G(0)>> is a divisor of P(0) in A. This proves the Lemma. Proof complete.You see the difference? The conclusion, C is a factor in A of P(0)is in the body of the statement, not just lost somewhere in the middleof the proof.I am not challenging your conclusion here: I am trying to explain toyou HOW to write a coherent mathematical statement that you can use.> Given that G(X) is a factor of the polynomial P(X),>it must be the case that G(0) is a factor of P(0), as one is more>general, while the other is specic.>> Though not necessary, you might want to note that you are considering>> P(X) as a function from A to A, not as a polynomial with coefcients>> in A.>>Im not interested in not necessary.>> Are you interested in correctness and clarity? Well, if something is not necessary then its not necessary.Bollocks. You have plenty of unnecessary statements all throughoutyour arguments. It is only those that would help in clarity that areommitted, perhaps because with clarity your arguments would becomeclearly bogus even to you.Are clarity and correctness necessary or not?>> >Consider that C is a factor of the constant term P(0), which follows>> >as G(X) is a factor of P(X), so G(0) is a factor of P(0), but C=G(0);>> >therefore, C is a factor of the constant term P(0). Proof Complete.>> You never stated in the lemma that c is a factor of the constant term>> P(0). So if you wanted to do that, you should have stated it. I think>> what you were trying to say is:>> LEMMA 1. Let A be the ring of all algebraic integer, and let A^A be>> the ring of functions from A to A. Let P(X) be a polynomial with>> coefcients in A, considered as an element of A^A, and let G(X) be an>> element of A^A which is a factor (in A^A) of P(X). Then we can write>> G(X)=R(X)+C, where C is the constant G(0); moreover, C is a factor in>> A of P(0).>> Proof. We can certainly write G(X) as G(X) = (G(X)-G(0))+G(0), so>> setting R(X)=G(X)-G(0) accomplishes the rst part of the lemma. For>> the second, let Q(X) be a function from A to A such that>> G(X)*Q(X)=P(X) in A^A. Then in particular G(0)*Q(0)=P(0) in A, so G(0)>> is a divisor of P(0) in A. This proves the Lemma. Proof complete.>>Hmmm...is this cooperation nally?>> Sigh. Ive been suggesting changes for clarity and xed to your>> arguments for YEARS. This particular change I suggested before, but>> you ignored it and decided instead to insult me. So get off your moral>> high horse; it doesnt actually exist.Facts not in evidence.Because you insist on deleting what I write, misinterpreting it, orjust closing your eyes, sticking your ngers in your ear, and hummingreal loud.>For those who dont know, and in case I messed up with my usage, that>means that the poster has said something that requires outside>verication. That is, tofind out if hes telling the truth, you>have to go do research.And that research will bear me out, you little insulting twerp.>Its a nasty little trick that Arturo Magidin likes to play.Go to hell, James. You libel me, you accuse me of things I did notsay, you put words in my mouth, and you have the chutzpah of claimingthat I pull nasty little tricks on you when I state what is on therecord?>> >Lemma 2:>>Now consider P(X) such that it has b^2 as a factor, and for a factor>> >G(X), C has b as a factor.>>Further consider that P(0)/b^2 is coprime to b; therefore, as C is a>> >factor of P(0) a factor of b must divide off of C, when b^2 divides>> >off of P(m). Proof Complete.>> Once again, you have no hypothesis, and no conclusion. How can you>> have a proof? How can you have a lemma? What does divide off mean?>> What is m? What is b^2? A function or a constant? Is b^2 a factor of>> P(X) in the sense that every coefcient of P(X) is a multiple of b^2>> (that is, b^2 is a factor of P(X) in A[x]), or in the sense that b^2>> is a factor of P(X) in A^A? Different things.>> Presumably:>> LEMMA 2. Let P(X) be a polynomial with coefcients in A[x],>> and let b be an algebraic integer. Let G(X) in A^A be a>> factor of P(X) in A^A, and write G(X)=R(X)+G(0). Suppose that b^2>> divides every coefcient of P(X) in A. Then G(0) is a multiple of b>> in A. >> Moreover, if P(0)/b^2 is coprime to b in A, then ____________________>> (I was unable to gure out the blank; please ll it).>>Well if you divide b^2 off from the polynomial then a factor must>> >divide off of G(0) such that what remains is coprime to b.>> Still doesnt make sense. And you do not answer any of the questions I>> asked. What is m? Is b^2 a factor of each coefcient of P(X), or do>> you mean that the VALUES of P(X) are divisible by b^2 for every valid>> value of the variable? And in the latter case, what are the valid>> values of the variable?You ask what is m?I ask a lot of things. Why do you restrict yourself to that?> Well, as m is not in the argument, I must say I>dont know what you mean by the question. >> >factor of P(0) a factor of b must divide off of C, when b^2 divides>off of P(m). Proof Complete.Do you see the m now?>Possibly you refer to X, as>in past arguments Ive used m where I now use X.>Its already given that the variables are in the ring of algebraic>integers.Which is NONSENSE. What you mean, presumably, is that it is a variablethat takes values in the ring of algebraic integers. That is VERYDIFFERENT from saying that it is a variable which IS in the ring ofalgebraic integers. Why dont you get the terminology straight for once?>Given b^2, and that b^2 is a factor of P(X), it follows that it is a>factor of P(X), which is whats mathematically necessary.Sigh. There are at least two ways in which b^2 can be a factor ofP(X). I asked you to specify which one it is:(1) P(x) is a polynomial, and each coefcient is a multiple of b^2. Example: 2x^3 + 4x + 6 is a multiple of 2 in Z[x].(2) P(x) is a polynomial, and it is a multiple of b^2 as a FUNCTION, but not necessarily as a polynomial. Example: x^2+x is a multiple of 2 as a function, but not as a polynomial.Which kind of factor do you have in mind? Type (1) implies type (2),but not vice versa.> Divide b^2 off from the polynomial. Is that for the value? Or as>> polynomials? Or as functions? What does dividing off of G(0) mean?The polynomial has a factor b^2, so Im talking of dividing it off.Sigh. I ask you what the term means, and you reply by repeating theterm. Does dividing it off mean I divide by it?>For instance, P(x) = 2x^2 + 4x + 2, has a factor of 2, and dividing it>off gives P(x)/2 = x^2 + 2x + 1.Yes, it means dividing by 2. Sigh. Yet more Harrispeak in an attemptto confuse and confound.>Now then, as P(X) has b^2 as a factor, P(0) has b^2 as a factor, and>G(0) is a factor of P(0) as well. So dividing off b^2 will divide>some factor off of G(0), if G(0) isnt already coprime to b, since>P(0)/b^2 IS coprime to b.Your Lemma is empty, as Nora noted. You are saying that either G(0) isnot coprime to b, or else it is coprime to b. Big surprise.>> If this were what you had in mind, thben it is certainly wrong. For>> instance, P(x)=(2x+2)(x+1), G(x)=x+1; then 2 divides every coefcient>> of P(x), but does not divide G(0).>>If you divide off 2 from P(x) here then the constant terms with>> >respect to x of both of its factors are coprime to 2.>> But it is still false that b divides off G(x). So you need to state>> your Lemma CLEARLY AND COMPLETELY. Stop using your own personal>> language such as divides off.>Here you have P(x) with 2 as a factor, but dividing off 2 leaves the>constant terms of both factors of P(x)/2 coprime to 2.Its trivial.Once you use proper mathematical terminology, maybe.>> >Thats obvious enough as then you have (x+1)(x+1), so the constant>> >term is 1.>> Nonetheless, the conclusion you claim in the body of your lemma>> seems false in this case. So either your lemma is badly written, is>> not what you meant, or is just plain wrong.>> Which is it?Well, it could very well be badly written.But then all that matters is xing it.You dont have to get so hostile, as if its badly written, then>again, its just a matter of xing it.->I<- dont have to get hostile? Why dont you look up where you claimI pull nasty little tricks, where you call me one of the mostevil people in the world?If I looked up hypocrite in the dictionary there would be picturesof you and your address, wouldnt there?I cannot x your results for you because I do not know what it is youwant to do with them. It is up to you.>> Of course, I probably interpreted the lemma wrong, so please write it>> correctly, clearly specifying the hypothesis and the conclusion. Also>> try to label where the proof begins and ends.>>Lemma 2:>>Now consider P(X) such that it has b^2 as a factor, and for a factor>> >G(X) of P(X), from lemma 1, G(X) = R(X)+ C.>> You still failed to answer an important question here. When you say>> P(x) such that it has b^2 as a factor does that mean that>> (a) b^2 divides each coefcient of P(X); >> or>> (b) The values of P(X) are always divisible by b^2?>> Different things: x^2+x is a multiple of 2 in the latter sense, but>> not in the former. Which is it?Your statement that x^2 + x is a multiple of 2 is false, as the ring>is algebraic integers.Sigh.IN THE RING OF INTEGERS, x^2 + x is a multiple of 2 in the lattersense, but not in the former.>Now it is the case that b^2 being a factor of P(X) means that it is a>factor of each coefcient of P(X).Ah. So NOW you are considering P(x) and b^2 as polynomials, not asfunctions. So in the same lemma, you have two different meanings ofthe word factor, with no indication that they are different meanings.>However, Ifind it troubling that you *continue* to try and throw in>distracting information.Ifind it troubling that you do not see that having two meanings ofthe same word in the same statement could possibly lead to confusion.>> >Now if P(0)/b^2 is coprime to b, then if C is not already coprime to>> >b, there must exist a factor of C that is divided off when b^2 divides>> >off such that the result is coprime to b, since C is a factor of P(0).>> Ah, a new clause (if C is not already coprime to b). So, basically,>> what you are claiming here is (I am assuming the weakest of the two>> possibilities above; feel free to change them):>> LEMMA 2. Let P(X) be a polynomial in A[X], b an element of A, such>> that b^2 divides P(X) in A^A. Let G(X) be a factor of P(X) in A^A, and>> write G(X)=R(X)+G(0). If P(0)/b^2 is coprime to b in A, then either>> (a) G(0) is coprime to b; or>> (b) There exists a divisor r of b^2 in A such that r divides G(0)>> and G(0)/r is coprime to b.>> Is that it?Yup, that looks ok to me. Still youre rather tricky, so I say that>cautiously.So who was it that was exhibiting hostility, again?Since the rest of your post is just the same old reheated crap, withthe same old errors, I will remove it. [.snip.]>> You still havent said what b and u are, though I suspect they are>> meant to be specic, xed, algebraic integers satisfying some>> mysterious properties you have not bothered to spell out just yet.I *did* say what b and u are as at the beginning I have the statement:In what follows variables, unless otherwise noted, are in the ring of>algebraic integers.Which is nonsense as written. Either write coherent sense, or bewilling to answer questions.Are they variables which take values in the ring of algebraicintegers? Are they parameters? Are they being considered xed values,or are they being considered variables? >> >If youre still lost I can take away u as a variable as well, as>> >follows.>> What, u was a variable? Oh my.In algebra, letter symbols are variables.You dont know much algebra, then. [.snip.]>> >You tried to make up a bogus math rule.>> The bogus math rule being: In order to be able to apply a>> lemma/theorem/corollary to a particular situation, the hypothesis of>> that lemma/theorem/corollary must be satised in that situation?The key variable is X. That doesnt change.Your arguments look like the variable on one side is X and thevaraible on the other is t. In either case, if there are more than onevariable, your lemmas, as written, do not apply. Either x yourlemmas, or write new ones.>Your bogus objection is trying to act as if there are other important>variables, when those variables are just there to get the>non-polynomial factorization.My bogus objection is to state that the hypothesis of a Lemma mustbe satised for the lemma to be applied. You seem to think this is not the case. Good for you. Thats certainlynew mathematics. [.snip.]>> G(X) = r_1(X,t) t + bu)>> So then, C is r_1(0,t)t + bu?Yup.> So that means that the equality must be an equality in A^{A}. So t>> is what? Not a variable, not a xed value. What is it, then?Its a placemarker, and otherwise irrelevant as the key variable is X.Placemarker is not a mathematical term I am familiar with. You meanit is a variable or a paramenter, presumably. Choose and stick to it. [.snip.]>> >Thats ok as youre not actually needed, and I know from your history>over a period of years that youre probably just looking for yet>> >another route to fooling the sci.math audience.>> In other words, you are just going to go ahead an insult me regardless>> of what I say?Well, as long as youre not telling the truth, you cant expect to be>treated like someone who tells the truth.Please provide ONE veriable instance of me not telling the truth.You, on the other hand, have lied about me countless times. There isonly one liar in this conversation, but it isnt me.You are truly a despicable individual. [Gabriele Rossetti] has left a vast body of writings... in which he has attempted to prove the truth of his unorthodox interpre- tation of medieval literature. They present a formidable record of unsystematic research in which we see an enthusiast plunging farther and farther and farther from the logic of facts and good sense until truth is lost in the dreadful nightmare of an idee xe. There is no real evolution of the Theory although it grows and expands until it embraces ever wider horizons. The numerous inaccuracies of deduction, mis-statements of historical fact, and self-contradictions...have caused critics to turn away from them in disgust... [...] It is impossible to read far... without realizing that we have to deal with a work of faith and imagination rather than of reasoning. There is an appearance of reason, for the author is set on proving by logic the truth of what he already believes by intuition. The truth is plain to him and he cannot comprehend why others do not immediately accept it, but as they desire demonstration he has multiplied his proofs. It is the redundancy and confusion of a prophet expounding by a familiar method the truth revealed to his own simple soul in a ash of inspiration... In such work as this... it is idle to look for the calm reasoning of a scholar; we do notfind it, and there is little or no advantage in attacking the obvious inconsistencies and absurdities that abound. -- E.R. Vincent, _Gabriele Rossetti in England_, quoted in _The Shakespearan Ciphers Examined_, by William F. Friedman and Elizebeth S. Friedman == ==Arturo Magidinmagidin@math.berkeley.edu> >As Ive considered various explanation for the surprising lack of> >> >agreement with my explanation for the problem with algebraic integers,> >> >Ive concluded that symbol confusion may indeed be the problem, as> >> >though mathematicians *say* that they can just follow what theyre> >told a symbol means, it seems to me that the best explanation for the> >> >scope of disagreement with me, is symbol confusion.> >> >> >So Im going to change some symbols, and see what happens.> >> >> >Editorial note: > >> >Several posters had a problem with the following statement:> >> >> >In the ring of algebraic integers, given a factor G(X) of a polynomial> >> >P(X), R(X) and C exists such that G(X)=R(X) + C, where C=G(0), and> >> >R(X)=G(X)-C.> >> >As they noted that a polynomial is not a number, and rather than argue> >> >about such details Im reposting with changes.>> Your lemma 2 is still a mess, and your lemma 1 still has no actual> >> content; you never use your hypothesis in the proof of the lemma. Why> >> include them?> >Huh? Your statement of Lemma included G is a factor of P. Your proof of> Lemma 1 never used the fact that G is a factor of P; in fact, the> conclusion from your statement of Lemma 1 was just that G can be> written as R(x)+G(0), where R(x)=G(x)-G(0); so you never use that G is> a factor of P for that. You only use that ->later<-, when you conclude> that if G(x) is a factor of P(x) (in A^A), then G(0) must be a factor> of P(0) in A. Since you never use the hypothesis in the body of Lemma 1, there is no> reason to include them.The point of lemma 1 is that a factor of a *polynomial* can be broken> >up into a varying and non-varying portion, which sort of connects it> >to a polynomial. I cannot speak for what the point of your lemma is. I can only speak> be the point of the lemma. So, the obvious conclusion is that you cannot write what you> mean. Which is no big surprise.Why is that an obvious conclusion?The point is that G(X) = R(X) + C, where C=G(0). > >So its material that G is a factor of P. Good for you. One has to wonder, then, why your proof of the lemma> never used the material fact that G is a factor of P. Maybe you can> explain that?Why?The point is that G(X) = R(X) + C, where C=G(0).> >Otherwise, what is it?> >There is no otherwise, as it is a factor of a polynomial, as stated. But you NEVER used that fact in your proof. It only occurred when I> that conclusion later (and claiming it followed from Lemma 1, despite> the fact that Lemma 1 did not include any such conclusion).The point is that G(X) = R(X) + C, where C=G(0).> >In what follows variables, unless otherwise noted, are in the ring of> >> >algebraic integers.>> So you are wroking in the ring of algebraic integer-valued functions> >> of algebraic integer variables. How many variables? Lemma 1 and> apparently Lemma 2 are for functions of 1 variable; but you try to> >> apply them to functions of several variables. You cannot do that.> >> >Why not? Because you say so? Are you kidding? A lemma/theorem/corollary is a statement of the form If A then B. In> order to apply a lemma/theorem/corollary, the hypothesis must be> met. If your lemma is about functions of one variable, and you try to> apply them to functions of more than one variable, then your> hypothesis are not met, and therefore the lemma cannot be applied. Basic logic, James.Im not interested in a debate on the subject, nor with distractions. In other words Oops, I guess hes right. Rather than admit it, I will> just claim it is a distraction.Actually youre wrong as I mistakenly used expressions like r_1(X,t)when t is not even in them, as the rs are roots of r^3 + 3(-1 + b^2 X)r - b^2(b^4 X^3 - 3b^2 X^2 + 3X)=0.So it should have been r_1(X), which is what is current.> >Well here its mostly about appearances as in fact the key variable is> >X, while you have continually tried to add confusion by acting as if> >the other variables, used to factor the expression into non-polynomial> >factors, are more than just placemarkers for the non-polynomial> >factorization. Its not about appearances, its about proper mathematics. If your> argument is powerful enough to include other situations, AND you want> to ->use<- the lemma in those other situations, then the lemma MUST be> stated in such a way as to include those situations. If your abilities> do not extend so far as to enable you to do so, then ask for help.The key variable is X. Im using r_1(X), as r_1(X,t) was a mistake,which is obvious enough from the dening cubic.> >Lemma 1:> >> >> >Given a factor G(X) of a polynomial P(X), R(X) and C exists such that> >> >G(X)=R(X) + C, where C=G(0), and R(X)=G(X)-C.>> You REALLY should label the beginning of a proof.>> This has no content, and you never use, in this proof, the fact that> >> G(X) is a factor of P(X).> >> >G(X) is a factor of the polynomial P(X), so G(0) is a factor of P(0). If that is your conclusion, then it should be in the body of the> below.> >Are you serious? Yes, I am. If the point is that G(0) is a factor of P(0), then WHY> did you only rst mention it in the middle of a proof of a> proposition which did not mention it in its body? If that is the> point, then it should be written in the body of the statement, like> I did later:Given that G(X) is a factor of P(X), then it follows that G(0) is afactor of P(0), and does not require proof.That is trivially true.I was not proving that G(0) is a factor of P(0), as it is triviallytrue, given that G(X) is a factor of P(X).> >> LEMMA 1. Let A be the ring of all algebraic integer, and let A^A be> >> the ring of functions from A to A. Let P(X) be a polynomial with> >> coefcients in A, considered as an element of A^A, and let G(X) be an> >> element of A^A which is a factor (in A^A) of P(X). Then we can write> >> G(X)=R(X)+C, where C is the constant G(0); moreover, C is a factor in> >> A of P(0).>> Proof. We can certainly write G(X) as G(X) = (G(X)-G(0))+G(0), so> >> setting R(X)=G(X)-G(0) accomplishes the rst part of the lemma. For> >> the second, let Q(X) be a function from A to A such that> >> G(X)*Q(X)=P(X) in A^A. Then in particular G(0)*Q(0)=P(0) in A, so G(0)> >> is a divisor of P(0) in A. This proves the Lemma. Proof complete. You see the difference? The conclusion, C is a factor in A of P(0)> is in the body of the statement, not just lost somewhere in the middle> of the proof. I am not challenging your conclusion here: I am trying to explain to> you HOW to write a coherent mathematical statement that you can use.Well then, are you no longer trying to claim that the conclusion ofthe entire argument is incorrect?Thats whats relevant.>Given that G(X) is a factor of the polynomial P(X),> >it must be the case that G(0) is a factor of P(0), as one is more> >general, while the other is specic.I knew I had put that in my last reply.> >> Though not necessary, you might want to note that you are considering> >> P(X) as a function from A to A, not as a polynomial with coefcients> >> in A.> >> >Im not interested in not necessary. Are you interested in correctness and clarity? > >Well, if something is not necessary then its not necessary. Bollocks. You have plenty of unnecessary statements all throughout> your arguments. It is only those that would help in clarity that are> ommitted, perhaps because with clarity your arguments would become> clearly bogus even to you. Are clarity and correctness necessary or not?If you say something is not necessary then why would you then arguethat it IS necessary?> >> >Consider that C is a factor of the constant term P(0), which follows> >> >as G(X) is a factor of P(X), so G(0) is a factor of P(0), but C=G(0);> >> >therefore, C is a factor of the constant term P(0). Proof Complete.>> You never stated in the lemma that c is a factor of the constant term> >> P(0). So if you wanted to do that, you should have stated it. I think> >> what you were trying to say is:>> LEMMA 1. Let A be the ring of all algebraic integer, and let A^A be> >> the ring of functions from A to A. Let P(X) be a polynomial with> >> coefcients in A, considered as an element of A^A, and let G(X) be an> >> element of A^A which is a factor (in A^A) of P(X). Then we can write> >> G(X)=R(X)+C, where C is the constant G(0); moreover, C is a factor in> >> A of P(0).>> Proof. We can certainly write G(X) as G(X) = (G(X)-G(0))+G(0), so> >> setting R(X)=G(X)-G(0) accomplishes the rst part of the lemma. For> >> the second, let Q(X) be a function from A to A such that> >> G(X)*Q(X)=P(X) in A^A. Then in particular G(0)*Q(0)=P(0) in A, so G(0)> >> is a divisor of P(0) in A. This proves the Lemma. Proof complete.> >> >Hmmm...is this cooperation nally? Sigh. Ive been suggesting changes for clarity and xed to your> arguments for YEARS. This particular change I suggested before, but> you ignored it and decided instead to insult me. So get off your moral> high horse; it doesnt actually exist.> >Facts not in evidence. Because you insist on deleting what I write, misinterpreting it, or> just closing your eyes, sticking your ngers in your ear, and humming> real loud.Your claim is that youve suggested changes for clarity and xed myarguments for years, with the additional claim that youve suggested aparticular change but were ignored and instead insulted by me, areclaims which require that readers go do an intensive search if theywish to verify. > >For those who dont know, and in case I messed up with my usage, that> >means that the poster has said something that requires outside> >verication. That is, to nd out if hes telling the truth, you> >have to go do research. And that research will bear me out, you little insulting twerp.Its a nasty little trick because who is going to go do all thatresearch to check?So Arturo Magidin tosses something out that requires a lot of effortpossibly hoping that most readers will not make the effort and justtrust him.> >Its a nasty little trick that Arturo Magidin likes to play. Go to hell, James. You libel me, you accuse me of things I did not> say, you put words in my mouth, and you have the chutzpah of claiming> that I pull nasty little tricks on you when I state what is on the> record?And now that hes caught he starts whining.> >> >Lemma 2:> >> >> >Now consider P(X) such that it has b^2 as a factor, and for a factor> >> >G(X), C has b as a factor.> >> >Further consider that P(0)/b^2 is coprime to b; therefore, as C is a> >> >factor of P(0) a factor of b must divide off of C, when b^2 divides> >> >off of P(m). Proof Complete.Oh, my mistake as I put in P(m) instead of P(X).>> Once again, you have no hypothesis, and no conclusion. How can you> >> have a proof? How can you have a lemma? What does divide off mean?> >> What is m? What is b^2? A function or a constant? Is b^2 a factor of> >> P(X) in the sense that every coefcient of P(X) is a multiple of b^2> >> (that is, b^2 is a factor of P(X) in A[x]), or in the sense that b^2> >> is a factor of P(X) in A^A? Different things.>> Presumably:>> LEMMA 2. Let P(X) be a polynomial with coefcients in A[x],> and let b be an algebraic integer. Let G(X) in A^A be a> factor of P(X) in A^A, and write G(X)=R(X)+G(0). Suppose that b^2> >> divides every coefcient of P(X) in A. Then G(0) is a multiple of b> >> in A. >> Moreover, if P(0)/b^2 is coprime to b in A, then ____________________> >> (I was unable to gure out the blank; please ll it).> >> >Well if you divide b^2 off from the polynomial then a factor mustdivide off of G(0) such that what remains is coprime to b. Still doesnt make sense. And you do not answer any of the questions I> asked. What is m? Is b^2 a factor of each coefcient of P(X), or do> you mean that the VALUES of P(X) are divisible by b^2 for every valid> value of the variable? And in the latter case, what are the valid> values of the variable?> >You ask what is m? I ask a lot of things. Why do you restrict yourself to that?I apologize as I did have P(m) instead of P(X).> Well, as m is not in the argument, I must say I> >dont know what you mean by the question. > Well I did just now and see where I have P(m) instead of P(X), so itwas my mistake.> >> >factor of P(0) a factor of b must divide off of C, when b^2 divides> >> >off of P(m). Proof Complete. Do you see the m now?Yup. It should be P(X).Possibly you refer to X, as> >in past arguments Ive used m where I now use X.> Now readers can see that Ive still managed to explain my own mistake,as in previous arguments Ive used m where I now use X, so it wasa simple mistake; however, notice how Arturo Magidin reacted.> >Its already given that the variables are in the ring of algebraicintegers. Which is NONSENSE. What you mean, presumably, is that it is a variable> that takes values in the ring of algebraic integers. That is VERY> DIFFERENT from saying that it is a variable which IS in the ring of> algebraic integers. Why dont you get the terminology straight for once?The variables are in the ring of algebraic integers, so X is analgebraic integer. As X is an algebraic integer, it is in the ring ofalgebraic integers.> >Given b^2, and that b^2 is a factor of P(X), it follows that it is a> >factor of P(X), which is whats mathematically necessary. Sigh. There are at least two ways in which b^2 can be a factor of> P(X). I asked you to specify which one it is: (1) P(x) is a polynomial, and each coefcient is a multiple of> b^2. Example: 2x^3 + 4x + 6 is a multiple of 2 in Z[x]. (2) P(x) is a polynomial, and it is a multiple of b^2 as a FUNCTION,> but not necessarily as a polynomial. Example: x^2+x is a multiple> of 2 as a function, but not as a polynomial.That is not generally true in the ring of algebraic integers, as, forinstance, it could be that x^2 + x = 1, and 1 is not a multiple of 2.> Which kind of factor do you have in mind? Type (1) implies type (2),> but not vice versa.Your question is specious. There arent two possible interpretationsin the ring of algebraic integers, as there is only one possibility.> >> Divide b^2 off from the polynomial. Is that for the value? Or as> polynomials? Or as functions? What does dividing off of G(0) mean?> >The polynomial has a factor b^2, so Im talking of dividing it off. Sigh. I ask you what the term means, and you reply by repeating the> term. Does dividing it off mean I divide by it?I gave an example which follows.> >For instance, P(x) = 2x^2 + 4x + 2, has a factor of 2, and dividing it> >off gives P(x)/2 = x^2 + 2x + 1. Yes, it means dividing by 2. Sigh. Yet more Harrispeak in an attempt> to confuse and confound.Who would get confused about something that simple?> >Now then, as P(X) has b^2 as a factor, P(0) has b^2 as a factor, andG(0) is a factor of P(0) as well. So dividing off b^2 will divide> >some factor off of G(0), if G(0) isnt already coprime to b, since> >P(0)/b^2 IS coprime to b. Your Lemma is empty, as Nora noted. You are saying that either G(0) is> not coprime to b, or else it is coprime to b. Big surprise.Well its not empty and its not complicated. Its like how with P(x) = 2x^2 + 4x + 2, for G(x) = 2x+2,you know that dividing the 2 off of P(x) divides it off of G(x).Or if you have G(x) = x+1, nothing divides off, but you know that G(0)is coprime to 2.Ive broken things up into *very* simple pieces to take away room forspecious objections.> >> If this were what you had in mind, thben it is certainly wrong. For> >> instance, P(x)=(2x+2)(x+1), G(x)=x+1; then 2 divides every coefcient> of P(x), but does not divide G(0).> >> >If you divide off 2 from P(x) here then the constant terms with> >respect to x of both of its factors are coprime to 2. But it is still false that b divides off G(x). So you need to state> your Lemma CLEARLY AND COMPLETELY. Stop using your own personal> language such as divides off.> >Here you have P(x) with 2 as a factor, but dividing off 2 leaves the> >constant terms of both factors of P(x)/2 coprime to 2.> >Its trivial. Once you use proper mathematical terminology, maybe.Whats so wrong with saying that youre dividing off 2, when you havesomething like 2x^2 + 4x + 2, and divide it by 2?Thats obvious enough as then you have (x+1)(x+1), so the constant> >term is 1. Nonetheless, the conclusion you claim in the body of your lemma> seems false in this case. So either your lemma is badly written, is> not what you meant, or is just plain wrong. Which is it?> >Well, it could very well be badly written.> >But then all that matters is xing it.> >You dont have to get so hostile, as if its badly written, then> >again, its just a matter of xing it. ->I<- dont have to get hostile? Why dont you look up where you claim> I pull nasty little tricks, where you call me one of the most> evil people in the world?Well, you are, which can be seen with what youre doing now, where themathematical truth is clearly irrelevant to you, while you keep tryingto nd anything you can use to cast doubt on the truth.If you werent balanced towards evil, you would simply look for thetruth, rather than try tofind ways to hide the truth.> If I looked up hypocrite in the dictionary there would be pictures> of you and your address, wouldnt there? I cannot x your results for you because I do not know what it is you> want to do with them. It is up to you.And there you go again. Who needs you to x anything?Thats what shows youre balanced towards evil Arturo Magidin, yourcontinuing efforts to hide the truth rather than follow the math.> >> Of course, I probably interpreted the lemma wrong, so please write it> >> correctly, clearly specifying the hypothesis and the conclusion. Also> >> try to label where the proof begins and ends.> >> >Lemma 2:> >> >Now consider P(X) such that it has b^2 as a factor, and for a factor> >G(X) of P(X), from lemma 1, G(X) = R(X)+ C. You still failed to answer an important question here. When you say> P(x) such that it has b^2 as a factor does that mean that (a) b^2 divides each coefcient of P(X); or (b) The values of P(X) are always divisible by b^2? Different things: x^2+x is a multiple of 2 in the latter sense, but> not in the former. Which is it?> >Your statement that x^2 + x is a multiple of 2 is false, as the ring> >is algebraic integers. Sigh. IN THE RING OF INTEGERS, x^2 + x is a multiple of 2 in the latter> sense, but not in the former.You didnt say in the ring of integers, and even if you did, itsirrelevant to an argument in the ring of algebraic integers.> >Now it is the case that b^2 being a factor of P(X) means that it is a> >factor of each coefcient of P(X). Ah. So NOW you are considering P(x) and b^2 as polynomials, not as> functions. So in the same lemma, you have two different meanings of> the word factor, with no indication that they are different meanings.If you wish to consider b^2 to be a polynomial, that is to be of theform 0*X + b^2, you can.Im not using different meanings.> >However, I nd it troubling that you *continue* to try and throw indistracting information. Ifind it troubling that you do not see that having two meanings of> the same word in the same statement could possibly lead to confusion.Youre the one trying to confuse, not me.> >Now if P(0)/b^2 is coprime to b, then if C is not already coprime to> >b, there must exist a factor of C that is divided off when b^2 divides> >off such that the result is coprime to b, since C is a factor of P(0). Ah, a new clause (if C is not already coprime to b). So, basically,> what you are claiming here is (I am assuming the weakest of the two> possibilities above; feel free to change them): LEMMA 2. Let P(X) be a polynomial in A[X], b an element of A, such> that b^2 divides P(X) in A^A. Let G(X) be a factor of P(X) in A^A, and> write G(X)=R(X)+G(0). If P(0)/b^2 is coprime to b in A, then either> (a) G(0) is coprime to b; or> (b) There exists a divisor r of b^2 in A such that r divides G(0)> and G(0)/r is coprime to b. Is that it?> >Yup, that looks ok to me. Still youre rather tricky, so I say that> >cautiously. So who was it that was exhibiting hostility, again? Since the rest of your post is just the same old reheated crap, with> the same old errors, I will remove it. [.snip.]So readers are supposed to just trust you Arturo Magidin?> You still havent said what b and u are, though I suspect they are> meant to be specic, xed, algebraic integers satisfying some> mysterious properties you have not bothered to spell out just yet.> >I *did* say what b and u are as at the beginning I have the statement:> >In what follows variables, unless otherwise noted, are in the ring ofalgebraic integers. Which is nonsense as written. Either write coherent sense, or be> willing to answer questions.The variables are algebraic integers.> Are they variables which take values in the ring of algebraic> integers? Are they parameters? Are they being considered xed values,> or are they being considered variables? In algebra, letter symbols are variables. In the argument, thevariables are in the ring of algebraic integers unless notedotherwise.> >If youre still lost I can take away u as a variable as well, asfollows. What, u was a variable? Oh my.> >In algebra, letter symbols are variables. You dont know much algebra, then. [.snip.]Youre claiming that letter symbols in algebra are NOT variables?Ill give an example. Consider P(x) = ax^2 + bx + cwhere you have several letter symbols, like a and x, which are ALLvariables.However, x has been designated the key variable.One can *say* that the others are held constant, but in algebra, theyare called variables.> >You tried to make up a bogus math rule. The bogus math rule being: In order to be able to apply a> lemma/theorem/corollary to a particular situation, the hypothesis of> that lemma/theorem/corollary must be satised in that situation?> >The key variable is X. That doesnt change. Your arguments look like the variable on one side is X and the> varaible on the other is t. In either case, if there are more than one> variable, your lemmas, as written, do not apply. Either x your> lemmas, or write new ones.Thats back to the specious claim that the factorization changes thepolynomial.The polynomial is P(X), while I use a special grouping to factor itinto non-polynomial factors.And as for other variables, the same thing happens with P(x) = ax^2 +bx + c.> >Your bogus objection is trying to act as if there are other important> >variables, when those variables are just there to get the> >non-polynomial factorization. My bogus objection is to state that the hypothesis of a Lemma must> be satised for the lemma to be applied.> > You seem to think this is not the case. Good for you. Thats certainly> new mathematics. [.snip.]The key variable is X. That is the key variable in the lemmas, soyour objection is specious.> G(X) = r_1(X,t) t + bu) So then, C is r_1(0,t)t + bu?> >Yup.> So that means that the equality must be an equality in A^{A}. So t> is what? Not a variable, not a xed value. What is it, then?> >Its a placemarker, and otherwise irrelevant as the key variable is X. Placemarker is not a mathematical term I am familiar with. You mean> it is a variable or a paramenter, presumably. Choose and stick to it. [.snip.]Heres an example.Consider P(x) = 11^2 + 11x + 2where you have a special grouping for the non-polynomialfactorization, as you can see that P(x) is also P(x) = 11x + 123and I just use the 11s as placemarkers, so I can *act* like its thepolynomial y^2 + xy + 2, to get the factorization P(x) = (11 + (x+sqrt(x^2-8))/2)(11 + (x-sqrt(x^2-8))/2)which is an example of a non-polynomial factorization.> >Thats ok as youre not actually needed, and I know from your history> >over a period of years that youre probably just looking for yet> >another route to fooling the sci.math audience. In other words, you are just going to go ahead an insult me regardless> of what I say?> >Well, as long as youre not telling the truth, you cant expect to betreated like someone who tells the truth. Please provide ONE veriable instance of me not telling the truth. You, on the other hand, have lied about me countless times. There is> only one liar in this conversation, but it isnt me. You are truly a despicable individual.Youre the one refusing to follow the math, possibly at least partlyout of pride.Ive refuted you point-by-point more than once but you keep comingback to try and convince others of false things.Mathematics isnt about debating or trying to convince others, butabout what actually follows from mathematical logic. =>> >> >As Ive considered various explanation for the surprising lack of>> >> >agreement with my explanation for the problem with algebraic integers,>> >> >Ive concluded that symbol confusion may indeed be the problem, as>> >though mathematicians *say* that they can just follow what theyre>> >> >told a symbol means, it seems to me that the best explanation for the>> >> >scope of disagreement with me, is symbol confusion.>> > >So Im going to change some symbols, and see what happens.>> > >Editorial note: >> Several posters had a problem with the following statement:>> > >In the ring of algebraic integers, given a factor G(X) of a polynomial>> >> >P(X), R(X) and C exists such that G(X)=R(X) + C, where C=G(0), and>> >R(X)=G(X)-C.>> > >As they noted that a polynomial is not a number, and rather than argue>> >> >about such details Im reposting with changes.>> >> >> Your lemma 2 is still a mess, and your lemma 1 still has no actual>> >> content; you never use your hypothesis in the proof of the lemma. Why>> >> include them?>>Huh?>> Your statement of Lemma included G is a factor of P. Your proof of>> Lemma 1 never used the fact that G is a factor of P; in fact, the>> conclusion from your statement of Lemma 1 was just that G can be>> written as R(x)+G(0), where R(x)=G(x)-G(0); so you never use that G is>> a factor of P for that. You only use that ->later<-, when you conclude>> that if G(x) is a factor of P(x) (in A^A), then G(0) must be a factor>> of P(0) in A. >> Since you never use the hypothesis in the body of Lemma 1, there is no>> reason to include them.>>The point of lemma 1 is that a factor of a *polynomial* can be broken>> >up into a varying and non-varying portion, which sort of connects it>> >to a polynomial.>> I cannot speak for what the point of your lemma is. I can only speak>> be the point of the lemma. >> So, the obvious conclusion is that you cannot write what you>> mean. Which is no big surprise.Why is that an obvious conclusion?Because:(a) You claim that statement S is the point of your lemma; and yet(b) Your statement of the lemma does not include statement S.Therefore:(c) You cannot write what you mean. You meant for Lemma 1 to make point S, yet it somehow failed to do so. That is a failure in your writing ability.Clear enough for you now?>The point is that G(X) = R(X) + C, where C=G(0).This does not depend on P(X). This is true for ANY function withcomplex values which has 0 in its domain.So, when you say:>> >So its material that G is a factor of P.Then, as I said, one has to wonder why something which is completelyirrelevant to the point of your Lemma can be said to also bematerial to your lemma.>> Good for you. One has to wonder, then, why your proof of the lemma>> never used the material fact that G is a factor of P. Maybe you can>> explain that?Why?Because Ifind it difcult to understand why something which isirrelevant can at the same time be material.>The point is that G(X) = R(X) + C, where C=G(0).Then it is irrelevant and immaterial that G is a factor of P, yes?Then when you said>> >So its material that G is a factor of P.you were either confused or lying, yes?>Otherwise, what is it?>>There is no otherwise, as it is a factor of a polynomial, as stated.>> But you NEVER used that fact in your proof. It only occurred when I>> that conclusion later (and claiming it followed from Lemma 1, despite>> the fact that Lemma 1 did not include any such conclusion).The point is that G(X) = R(X) + C, where C=G(0).Then the fact taht G is a factor of P is irrelevant, and when youclaimd that it was material, you were either confused or lying. Whichwas it? [.snip.]>> Are you kidding?>> A lemma/theorem/corollary is a statement of the form If A then B. In>> order to apply a lemma/theorem/corollary, the hypothesis must be>> met. If your lemma is about functions of one variable, and you try to>> apply them to functions of more than one variable, then your>> hypothesis are not met, and therefore the lemma cannot be applied.>> Basic logic, James.>>Im not interested in a debate on the subject, nor with distractions.>> In other words Oops, I guess hes right. Rather than admit it, I will>> just claim it is a distraction.Actually youre wrongI am? Which of the following statements that you are replying to isfalse? Be specic:(1) A lemma/theorem/corollary is a statement of the form IF A then B.(2) In order to apply a lemma/theorem/corollary, the hypothesis must be met.(3) If your lemma is about functions of one variable, and you try to apply it to functions of more than one variable, then your hypothesis are not met.(4) If you hypothesis are not met, then the lemma cannot be applied.> as I mistakenly used expressions like r_1(X,t)>when t is not even in them, as the rs are roots of r^3 + 3(-1 + b^2 X)r - b^2(b^4 X^3 - 3b^2 X^2 + 3X)=0.So it should have been r_1(X), which is what is current.So, what you are saying is that I was ->RIGHT<-, and that you weremistaken when you claimed that you were applying the lemma tofunctions of more than one variable?That is, that my mistake was in thinking that what you said was true?That I should have realized that you said something wrong, and insteadto write? [.rest deleted.]>> If that is your conclusion, then it should be in the body of the>> below.>>Are you serious? >> Yes, I am. If the point is that G(0) is a factor of P(0), then WHY>> did you only rst mention it in the middle of a proof of a>> proposition which did not mention it in its body? If that is the>> point, then it should be written in the body of the statement, like>> I did later:Given that G(X) is a factor of P(X), then it follows that G(0) is a>factor of P(0), and does not require proof.If it does not require proof, then why include it in the proof? Ifthat is the POINT, then why not mention it explicitly?>That is trivially true.Yes. Yet you are making a big deal out of it. If it is the point,then it should be made explicitly. If it is trivially true and doesnot require a proof, then it cannot be the point.Easy, yes?>I was not proving that G(0) is a factor of P(0), as it is trivially>true, given that G(X) is a factor of P(X).And yet you felt the need to mention it explicitly even though it wasnot necessary?What happened to that if it is not necessary Im not interested in it? [.snip.]>> >> LEMMA 1. Let A be the ring of all algebraic integer, and let A^A be>> >> the ring of functions from A to A. Let P(X) be a polynomial with>> >> coefcients in A, considered as an element of A^A, and let G(X) be an>> >> element of A^A which is a factor (in A^A) of P(X). Then we can write>> >> G(X)=R(X)+C, where C is the constant G(0); moreover, C is a factor in>> >> A of P(0).>> >> Proof. We can certainly write G(X) as G(X) = (G(X)-G(0))+G(0), so>> >> setting R(X)=G(X)-G(0) accomplishes the rst part of the lemma. For>> >> the second, let Q(X) be a function from A to A such that>> >> G(X)*Q(X)=P(X) in A^A. Then in particular G(0)*Q(0)=P(0) in A, so G(0)>> >> is a divisor of P(0) in A. This proves the Lemma. Proof complete.>> You see the difference? The conclusion, C is a factor in A of P(0)>> is in the body of the statement, not just lost somewhere in the middle>> of the proof.>> I am not challenging your conclusion here: I am trying to explain to>> you HOW to write a coherent mathematical statement that you can use.Well then, are you no longer trying to claim that the conclusion of>the entire argument is incorrect?Thats whats relevant.Depends on what you mean by the entire argument, James, so dont tryto pull a fast one. The trivial conclusions from Lemma 1 and fromLemma 2 are not in dispute if correctly written (despite your claimsto the contrary). Your applications of these lemmas are nonsense, andyour claims about what should and should not be an algebraic integerare provably false and have been explicitly proven false manytimes. So of COURSE I still CONCLUDE, based on solid mathematical workthat you do not want to try to understand, that your conclusions tothe entire argument is incorrect.>>Given that G(X) is a factor of the polynomial P(X),>> >it must be the case that G(0) is a factor of P(0), as one is more>> >general, while the other is specic.I knew I had put that in my last reply.Talking to yourself again?>> Though not necessary, you might want to note that you are considering>> >> P(X) as a function from A to A, not as a polynomial with coefcients>> >> in A.>>Im not interested in not necessary.>> Are you interested in correctness and clarity? >>Well, if something is not necessary then its not necessary.>> Bollocks. You have plenty of unnecessary statements all throughout>> your arguments. It is only those that would help in clarity that are>> ommitted, perhaps because with clarity your arguments would become>> clearly bogus even to you.>> Are clarity and correctness necessary or not?If you say something is not necessary then why would you then argue>that it IS necessary?If your writing were uniformly clear, it would not be necessary. Ifyou were to adopt standard mathematical terminology, then it would notbe necessary. SINCE YOU REFUSE TO CLEAN UP YOUR PROSE, SINCE YOUREFUSE TO USE STANDARD MATHEMATICAL TERMINOLOGY AND RELY INSTEAD ONYOUR OWN PRIVATE LANGUAGE, SINCE YOU INSIST ON USING CONTEXT-DEPENDENTTERMS AS IF THEY WERE CONTEXT-INDEPENDENT, then it becomes necessaryfor clarity. [.snip.]>> Sigh. Ive been suggesting changes for clarity and xed to your>> arguments for YEARS. This particular change I suggested before, but>> you ignored it and decided instead to insult me. So get off your moral>> high horse; it doesnt actually exist.>>Facts not in evidence.>> Because you insist on deleting what I write, misinterpreting it, or>> just closing your eyes, sticking your ngers in your ear, and humming>> real loud.Your claim is that youve suggested changes for clarity and xed my>arguments for years, with the additional claim that youve suggested a>particular change but were ignored and instead insulted by me, are>claims which require that readers go do an intensive search if they>wish to verify.Or to have a bit of memory. Just because you forget everything exceptfor your false and wild accusations does not mean everyone else does.You really are a terrible hypocrite, James.But, just for YOU, going back ONLY to 2000:CORRECTION OF 241%40agate.berkeley.edu (September 40agate.berkeley.edu (December 2002)CORRECTIONS FOR 241%40agate.berkeley.edu (July 40agate.berkeley.edu (July UTF-8&threadm=9moa7f%24r3k%241%40agate.berkeley.edu (August 2001)So do we need to go back to your errors with the denitions of ring,of object? Of rational and irrational part a couple of years back?Are you going to drop the crap?>> >For those who dont know, and in case I messed up with my usage, that>> >means that the poster has said something that requires outside>> >verication. That is, tofind out if hes telling the truth, you>> >have to go do research.>> And that research will bear me out, you little insulting twerp.Its a nasty little trick because who is going to go do all that>research to check?I suspect you meant not to include the question mark.It is not a nasty little trick because people have enough memory toknow. YOU seem to think that people have no memory, which is whatallows you to continue to make false claims, like that I always lie,or that you have caught me in lies before, or that I have been wrongbefore. I guess those are not nasty little tricks, since you dothem. This one is, because it is both true and to yourdetriment. Yes, I can see why you wouldfind it nasty.>So Arturo Magidin tosses something out that requires a lot of effort>possibly hoping that most readers will not make the effort and just>trust him.Not at all. If you will stop pretending to be a mind reader, you mightstop assigning to others the base motives that you so enjoy yourself.>> >Its a nasty little trick that Arturo Magidin likes to play.>> Go to hell, James. You libel me, you accuse me of things I did not>> say, you put words in my mouth, and you have the chutzpah of claiming>> that I pull nasty little tricks on you when I state what is on the>> record?And now that hes caught he starts whining.I wasnt caught, and YOU were whining, so drop the crap.>> >> >Lemma 2:>> > >Now consider P(X) such that it has b^2 as a factor, and for a factor>> >> >G(X), C has b as a factor.>> Further consider that P(0)/b^2 is coprime to b; therefore, as C is a>> >> >factor of P(0) a factor of b must divide off of C, when b^2 divides>> >> >off of P(m). Proof Complete.Oh, my mistake as I put in P(m) instead of P(X).Oh, yes, your mistake. And yet you berated me when I asked what m was.Couldnt even be bothered to look up in the same post. No wonder youthink that referring to well-known history is a nasty little trick.You cant even remember what you said 24 hours ago. [.snip.]>> >Possibly you refer to X, as>> >in past arguments Ive used m where I now use X.>> Now readers can see that Ive still managed to explain my own mistake,>as in previous arguments Ive used m where I now use X, so it was>a simple mistake; however, notice how Arturo Magidin reacted.How did I react?I asked you what m was. You replied by accusing me of dishonesty andof bringing up irrelevancies. Turns out, I did neither. Turns out, YOUhad made a mistake. When you started accusing me of all sorts ofthings BECAUSE YOU HAD MADE A MISTAKE, I replied by pointing outEXACTLY what had happened.This is something to be noted?>> >Its already given that the variables are in the ring of algebraic>> >integers.>> Which is NONSENSE. What you mean, presumably, is that it is a variable>> that takes values in the ring of algebraic integers. That is VERY>> DIFFERENT from saying that it is a variable which IS in the ring of>> algebraic integers. Why dont you get the terminology straight for once?The variables are in the ring of algebraic integers, so X is an>algebraic integer. Again, no. The ring of algebraic integers ONLY CONTAINS ALGEBRAICINTEGERS. It does not contain (nonconstant) polynomials, it does notcontain functions, it does not contain variables.The statement the variables are in the ring of algebraic integers isincorrect nonsense. What you mean is that the variables TAKE VALUES fromor in the ring of algebraic integers.> As X is an algebraic integer, it is in the ring of>algebraic integers.No, X is a variable, according to you, so it is not an algebraicinteger. It cannot be both. What you mean is that X is a variablewhich takes value in the ring of algebraic integers. I told you how tox the mistake, and why it was a mistake, and you simply repeated thesame mistake again.>> >Given b^2, and that b^2 is a factor of P(X), it follows that it is a>> >factor of P(X), which is whats mathematically necessary.>> Sigh. There are at least two ways in which b^2 can be a factor of>> P(X). I asked you to specify which one it is:>> (1) P(x) is a polynomial, and each coefcient is a multiple of>> b^2. Example: 2x^3 + 4x + 6 is a multiple of 2 in Z[x].>> (2) P(x) is a polynomial, and it is a multiple of b^2 as a FUNCTION,>> but not necessarily as a polynomial. Example: x^2+x is a multiple>> of 2 as a function, but not as a polynomial.That is not generally true in the ring of algebraic integers, as, for>instance, it could be that x^2 + x = 1, and 1 is not a multiple of 2.context, it is clear that the statement in the example in (2) is ALSOa statement about Z[x], not about A[x].You have still failed to answer the question.>> Which kind of factor do you have in mind? Type (1) implies type (2),>> but not vice versa.Your question is specious. Then it should be rather easy to answer.>There arent two possible interpretations>in the ring of algebraic integers, as there is only one possibility.Then it would be amazingly simple to answer: you could just say (1) oryou could say (2), or you could say both. Why did you fail to do so.There arent two possible interpretations? Really? Are you claimingthat: Let P(x) be a polynomial with algebraic integer coefcients, and let b be an algebraic integer. If for every algebraic integer a, P(a) is a multiple of b in the ring of all algebraic integers, then each coefcient of P(x) is a multiple of b in the ring of algebraic integers.?If so, then please prove it. I honestly do not know whether it is trueor not.>> Divide b^2 off from the polynomial. Is that for the value? Or as>> polynomials? Or as functions? What does dividing off of G(0) mean?>>The polynomial has a factor b^2, so Im talking of dividing it off.>> Sigh. I ask you what the term means, and you reply by repeating the>> term. >> Does dividing it off mean I divide by it?I gave an example which follows.Examples are no substitute for denitions and explanations.>> >For instance, P(x) = 2x^2 + 4x + 2, has a factor of 2, and dividing it>off gives P(x)/2 = x^2 + 2x + 1.>> Yes, it means dividing by 2. Sigh. Yet more Harrispeak in an attempt>> to confuse and confound.Who would get confused about something that simple?Because you insist on using your own private language, instead ofuniversally accepted language. I can only assume that you do itbecause YOU WANT to be confusing, to give you something to complainabout when others dont understand you.>> >Now then, as P(X) has b^2 as a factor, P(0) has b^2 as a factor, and>> >G(0) is a factor of P(0) as well. So dividing off b^2 will divide>> >some factor off of G(0), if G(0) isnt already coprime to b, since>> >P(0)/b^2 IS coprime to b.>> Your Lemma is empty, as Nora noted. You are saying that either G(0) is>> not coprime to b, or else it is coprime to b. Big surprise.Well its not empty and its not complicated. The conclusion of your Lemma is: either G(0) is coprime to b, or elseit is not coprime to b. Thats the excluded middle, a logicaltautology. Therefore, it is empty.> Its like how with P(x) = 2x^2 + 4x + 2, for G(x) = 2x+2,you know that dividing the 2 off of P(x) divides it off of G(x).Or if you have G(x) = x+1, nothing divides off, but you know that G(0)>is coprime to 2.Again, doesnt matter what your example are. You are saying thateither G(0) is coprime to b, or else that it is NOT coprime to b. Youare simply applying the law of the excluded middle.>Ive broken things up into *very* simple pieces to take away room for>specious objections.Youve tied yourself into knots and stated vacuous theorems.> >> If this were what you had in mind, thben it is certainly wrong. For>> >> instance, P(x)=(2x+2)(x+1), G(x)=x+1; then 2 divides every coefcient>> >> of P(x), but does not divide G(0).>>If you divide off 2 from P(x) here then the constant terms with>> >respect to x of both of its factors are coprime to 2.>> But it is still false that b divides off G(x). So you need to state>> your Lemma CLEARLY AND COMPLETELY. Stop using your own personal>> language such as divides off.>> >Here you have P(x) with 2 as a factor, but dividing off 2 leaves the>> >constant terms of both factors of P(x)/2 coprime to 2.>>Its trivial.>> Once you use proper mathematical terminology, maybe.Whats so wrong with saying that youre dividing off 2, when you have>something like 2x^2 + 4x + 2, and divide it by 2?Dividing off is a neologism you invented. Divide by 2 is a propermathematical terminology universally understood. Whats wrong withusing neologism invented by yourself is that in general people dontknow what you mean unless you dene these neologisms, and you have arather poor and decient track record in explaining your neologisms.>> >Thats obvious enough as then you have (x+1)(x+1), so the constant>> >term is 1.>> Nonetheless, the conclusion you claim in the body of your lemma>> seems false in this case. So either your lemma is badly written, is>> not what you meant, or is just plain wrong.>> Which is it?>>Well, it could very well be badly written.>>But then all that matters is xing it.>>You dont have to get so hostile, as if its badly written, then>> >again, its just a matter of xing it.>> ->I<- dont have to get hostile? Why dont you look up where you claim>> I pull nasty little tricks, where you call me one of the most>> evil people in the world?Well, you are, Bollocks. Thats just you being a lying hypocrite. [.rest of crap deleted.]>> Ah. So NOW you are considering P(x) and b^2 as polynomials, not as>> functions. So in the same lemma, you have two different meanings of>> the word factor, with no indication that they are different meanings.If you wish to consider b^2 to be a polynomial, that is to be of the>form 0*X + b^2, you can.Im not using different meanings.You use factor for b^2 and P(x) in terms of polynomials. Theconstant polynomial b^2 divides the polynomial P(x) in the ring ofpolynomials A[x].You use factor for P(x) and G(x) in terms of functions. Thepossibly non-polynomial factor G(x) is a factor of the polynomialP(x). This occurs in the ring A^A.I realize that it hasnt sunk in that factor of is a term which iscontext dependent, which depends on the ring in which you are working(and, no, you are not working in the ring of algebraic integers,because the latter does not include nonconstant polynomials nornonconstant functions). You are using factor to mean factor inA[x] and to mean factor in A^A.Therefore, you are using the same word to mean TWO DIFFERENT MEANINGS. [.snip.]>> >Yup, that looks ok to me. Still youre rather tricky, so I say that>> >cautiously.>> So who was it that was exhibiting hostility, again?>> Since the rest of your post is just the same old reheated crap, with>> the same old errors, I will remove it.>> [.snip.]So readers are supposed to just trust you Arturo Magidin?No, they can go back and read whathever they want. I did not removeyour post from their server. I did not remove your post from Google.Only someone with a memory disorder like you are trying to exhibitwould think that I have somehow pulled the wool over someones eyes. [.snip.]>> You still havent said what b and u are, though I suspect they are>> meant to be specic, xed, algebraic integers satisfying some>> mysterious properties you have not bothered to spell out just yet.>>I *did* say what b and u are as at the beginning I have the statement:>>In what follows variables, unless otherwise noted, are in the ring of>> >algebraic integers.>> Which is nonsense as written. Either write coherent sense, or be>> willing to answer questions.The variables are algebraic integers.Which is nonsense as written. The CORRECT way of saying what you meanis that the variables TAKE VALUES IN THE RING OF ALGEBRAIC INTEGERS.>> Are they variables which take values in the ring of algebraic>> integers? Are they parameters? Are they being considered xed values,>> or are they being considered variables? In algebra, letter symbols are variables.You dont know much algebra, then.> In the argument, the>variables are in the ring of algebraic integers unless noted>otherwise.No, the variables are NOT IN the ring of algebraic integers. Whatyou mean to say is that the variables TAKE VALUES in the ring ofalgebraic integers.>> >If youre still lost I can take away u as a variable as well, as>> >follows.>> What, u was a variable? Oh my.>>In algebra, letter symbols are variables.>> You dont know much algebra, then.>> [.snip.]Youre claiming that letter symbols in algebra are NOT variables?Not always. Sometimes they are parameters. Sometimes they are elementsof a set; sometimes they are elements of a group. Just because you gotstuck at baby algebra does not mean that the only things that existare the things you know.Ill give an example. You are trying to prove the universal statement IN algebra, everyletter is a variable, by giving an example?You dont know much logic. [.snip.]>Consider P(x) = ax^2 + bx + cwhere you have several letter symbols, like a and x, which are ALL>variables.However, x has been designated the key variable.One can *say* that the others are held constant, but in algebra, they>are called variables.Not always. Calling them variables here causes complications that yousimply do not know how to deal with. It is much better to call themwhat they are, which is parameters.Weve gone through this before, James. Stop projecting your ignoranceonto reality.>> >You tried to make up a bogus math rule.>> The bogus math rule being: In order to be able to apply a>> lemma/theorem/corollary to a particular situation, the hypothesis of>> that lemma/theorem/corollary must be satised in that situation?>>The key variable is X. That doesnt change.>> Your arguments look like the variable on one side is X and the>> varaible on the other is t. In either case, if there are more than one>> variable, your lemmas, as written, do not apply. Either x your>> lemmas, or write new ones.Thats back to the specious claim that the factorization changes the>polynomial.No, its back to the simple fact that your lemmas are written forfunctions of one variable, and you are trying to apply them tofunctions of several variables. If you want to apply the lemmas,either change the lemmas so they are about functions of severalvariables, or change the setting so they are about functions of onevariable.This factorization changes the polynomial is your misunderstandingand misrepresentation. Every time you repeat it you are quite simply,lying your ass off.>The polynomial is P(X), while I use a special grouping to factor it>into non-polynomial factors.And as for other variables, the same thing happens with P(x) = ax^2 +>bx + c.And then you try to apply a lemma written for functions in onevariable to a function in 4 variables. That is not valid mathematics.>> >Your bogus objection is trying to act as if there are other important>variables, when those variables are just there to get the>non-polynomial factorization.>> My bogus objection is to state that the hypothesis of a Lemma must>> be satised for the lemma to be applied.>> >> You seem to think this is not the case. Good for you. Thats certainly>> new mathematics.>> [.snip.]The key variable is X. That is the key variable in the lemmas, so>your objection is specious.Your lemmas are for functions of one variable. You are trying to applythem in situations where the hypothesis are not met. This is not aspecious objection, this is simple logic. If you can modify yourtheorems so they apply to the situation you want, YOU MUST DO SOBEFORE APPLYING THEM. [.snip.]> G(X) = r_1(X,t) t + bu)>> So then, C is r_1(0,t)t + bu?>>Yup.>> > So that means that the equality must be an equality in A^{A}. So t>> is what? Not a variable, not a xed value. What is it, then?>>Its a placemarker, and otherwise irrelevant as the key variable is X.>> Placemarker is not a mathematical term I am familiar with. You mean>> it is a variable or a paramenter, presumably. Choose and stick to it.>> [.snip.]Heres an example.Why are you incapable of giving DEFINITIONS instead at hadn?>Consider P(x) = 11^2 + 11x + 2where you have a special grouping for the non-polynomial>factorization, as you can see that P(x) is also P(x) = 11x + 123and I just use the 11s as placemarkers, so I can *act* like its the>polynomial y^2 + xy + 2, to get the factorization P(x) = (11 + (x+sqrt(x^2-8))/2)(11 + (x-sqrt(x^2-8))/2)which is an example of a non-polynomial factorization.Lets remove all your obfuscation, shall we?You are taking a polynomial in TWO variables, P(y,x) = y^2 + yx + 2.Then you are considering x as a parameter, and factoring the resultingfamily of polynomials P_x(y) = y^2 + xy + 2 using the quadraticformula:y = (-x + sqrt(x^2 - 8))/2y = (-x - sqrt(x^2 - 8))/2so thatP_x(y) = (y + (x-sqrt(x^2-8))/2)(y + (x+sqrt(x^2-8))/2)Then you are coming back to think of P_x(y) as a FUNCTION of twovariables,P(x,y), and taking the factorsG_1(x,y) = y+ (x-sqrt(x^2-8))/2G_2(x,y) = y+ (x+sqrt(x^2-8))/2as functions of two variables.Then you are mapping from A^{AxA} to A^A by evaluation at y; in thiscase, y=11.You are not using 11 as a placeholder, you are using TWO VARIABLESand then evaluating at one of them to map from A^{AxA} to A^A. Allwell known stuff, if you drop all of your obfuscation.Stop using your own private language. [.snip.]>> In other words, you are just going to go ahead an insult me regardless>> of what I say?>>Well, as long as youre not telling the truth, you cant expect to be>> >treated like someone who tells the truth.>> Please provide ONE veriable instance of me not telling the truth.>> You, on the other hand, have lied about me countless times. There is>> only one liar in this conversation, but it isnt me.>> You are truly a despicable individual.Youre the one refusing to follow the math, possibly at least partly>out of pride.In other words, you cannot produce any single instance of me lying,except perhaps for me currently saying your argument is wrong when youclaim it is right.Same as it always has been: I have never told anything but the truth,except that I disagree with you on your current argument, and so thatmakes me a liar; even though your current argument will eventually bedropped.You never remember that all those other times you called me a liarusing werewrong; it turned out I was not lying. All you remember is that youhave been calling me a liar for years. And if you have been saying itfor years, it must be true.THAT is the factual content of your libelous accusation. As valid asyour argument.>Ive refuted you point-by-point more than once but you keep coming>back to try and convince others of false things.No, you have refuted nothing except the strawmen you have erected forlack of comprehension.I dont need ot try and convince others of anything. The people whofollow these threads have known that you are wrong for over 2years. The only person who still needs convincing is YOU. == == [Gabriele Rossetti] has left a vast body of writings... in which he has attempted to prove the truth of his unorthodox interpre- tation of medieval literature. They present a formidable record of unsystematic research in which we see an enthusiast plunging farther and farther and farther from the logic of facts and good sense until truth is lost in the dreadful nightmare of an idee xe. There is no real evolution of the Theory although it grows and expands until it embraces ever wider horizons. The numerous inaccuracies of deduction, mis-statements of historical fact, and self-contradictions...have caused critics to turn away from them in disgust... [...] It is impossible to read far... without realizing that we have to deal with a work of faith and imagination rather than of reasoning. There is an appearance of reason, for the author is set on proving by logic the truth of what he already believes by intuition. The truth is plain to him and he cannot comprehend why others do not immediately accept it, but as they desire demonstration he has multiplied his proofs. It is the redundancy and confusion of a prophet expounding by a familiar method the truth revealed to his own simple soul in a ash of inspiration... In such work as this... it is idle to look for the calm reasoning of a scholar; we do notfind it, and there is little or no advantage in attacking the obvious inconsistencies and absurdities that abound. -- E.R. Vincent, _Gabriele Rossetti in England_, quoted in _The Shakespearan Ciphers Examined_, by William F. Friedman and Elizebeth S. Friedman == ==Arturo Magidinmagidin@math.berkeley.edu =>>[...]>>Hmmm...is this cooperation nally?>> Sigh. Ive been suggesting changes for clarity and xed to your>> arguments for YEARS. This particular change I suggested before, but>> you ignored it and decided instead to insult me. So get off your moral>> high horse; it doesnt actually exist.Facts not in evidence.For those who dont know, and in case I messed up with my usage, that>means that the poster has said something that requires outside>verication. That is, tofind out if hes telling the truth, you>have to go do research.Its a nasty little trick that Arturo Magidin likes to play.Actually the facts you claim are not in evidence are quite apparentto anyone whos been following the discussion. A nasty trick _you_like to play is to pretend that tens or hundreds of posts dont existunless theyre explicitly quoted in the current reply.every post.>> [...]>> Nonetheless, the conclusion you claim in the body of your lemma>> seems false in this case. So either your lemma is badly written, is>> not what you meant, or is just plain wrong.>> Which is it?Well, it could very well be badly written.But then all that matters is xing it.You dont have to get so hostile, as if its badly written, then>again, its just a matter of xing it.This is incredible. All he says is that something is badly written orwrong and you complain about hostility? _You_ complain abouthostility? Wow.[...]**************************As far as Im concerend youre trying to wait until I die, so I guremaybe you should die instead. How about that, eh? Wouldnt that be abetter twist?You refuse to follow the math, so the great Powers that controlreality and *speak* in mathematics decide to kill you instead of me.So what do you think about that, eh? Oh, cant hear Them talking?Well, I guess thats because you dont really understand Mathematics,the true language, which is THE language.Theyre talking about you now, and They agree with my assessment, andwill not penalize me as They allowed the others like Galois and Abelto be penalized.They will kill you instead. speaking on Weird factorization, genius =...This is still not the way of a mathematical exposition. In detail: > Lemma 1: > > Given a factor G(X) of a polynomial P(X), R(X) and C exists such that > G(X)=R(X) + C, where C=G(0), and R(X)=G(X)-C.This is completely unclear. You do not state in what ring G(X) is a factorof P(X). However, the lemma as stated is completely trivial, and it is notnecessary for the conclusion that G(X) is a factor of P(X). Also P is notused in the conclusion of the lemma. Moreover, it is not even needed thatC = G(0) for the conclusion. So I wonder what you wish to accomplish withthis lemma. I think something along the lines of: Given a factor G(X) of a polynomial P(X) (over ring B) in ring A, R(X) and C exist such that G(X) = R(X) + C with C = G(0) and so that C is a factor of P(0) in ring B.However, even with this version there are problems as it is not clearwhat factor for G(X) means here. If it means that for each X inring B, the value of G(X) is a factor (in ring B) of P(X), the result isagain trivial.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ =>4d squares (cubes, whatever) are called hybercubes, and ive also>>heard them called tesseracts, so... what are 4d circles (spheres)>>called, if theyre called anything at all?>>also, what would the formulae for the area of a 4d sphere be?If I remember correctly:>In 2-D the boundry of a circle>has length (2 Pi Radius)>In 2-D the circle has area (2 Pi) ((Radius^2)/2)In 3-D the boundry of a circle (sphere)>has area (4 Pi Radius^2)>I 3-D the sphere has volume (4 Pi) ((Radius^3)/3)In 4-D the boundry of a circle (hypersphere)>has volume (4 Pi^2 Radius^3)>In 4-D the hypersphere has hypervolume (4 Pi^2 (Radius^4)/4)adam>Oops, that last part should be:In 4-D the boundry of a circle (hypersphere)has volume (2 Pi^2 Radius^3)In 4-D the hypersphere has hypervolume (2 Pi^2 (Radius^4)/4And in general:In N+1-D the boundry of a circle has n-area:((2 Pi^(N+1/2)) / Gamma[(N+1/2]) Radius^(N)In N+1-D the circle has n-volume:((2 Pi^(N+1/2)) / Gamma[(N+1/2]) Radius^(N+1)adam =This is addicting! Every time I nish a leveland want to stop that little voice whispersthe next level will only take 26% more timeand 28% more disk space, if you did this oneyou can do the next one.I cant go any farther without resorting to le splitting. Level 84 is a 3,014,822,741 text leand took 2.5 hours to run.So here is the nal table to level 84Level 6 2Level 7 2Level 8 4Level 9 4Level 10 6Level 11 6Level 12 8Level 13 10Level 14 14Level 15 18Level 16 24Level 17 29Level 18 36Level 19 44Level 20 58Level 21 72Level 22 91Level 23 113Level 24 143Level 25 179Level 26 227Level 27 287Level 28 366Level 29 460Level 30 578Level 31 732Level 32 926Level 33 1174Level 34 1489Level 35 1879Level 36 2365Level 37 2988Level 38 3780Level 39 4788Level 40 6049Level 41 7628Level 42 9635Level 43 12190Level 44 15409Level 45 19452Level 46 24561Level 47 31025Level 48 39229Level 49 49580Level 50 62680Level 51 79255Level 52 100144Level 53 126542Level 54 159930Level 55 202085Level 56 255455Level 57 322869Level 58 408002Level 59 515542Level 60 651407Level 61 823238Level 62 1040490Level 63 1315036Level 64 1661989Level 65 2100226Level 66 2654294Level 67 3354459Level 68 4239037Level 69 5357022Level 70 6769885Level 71 8555671Level 72 10812460Level 73 13664435Level 74 17268507Level 75 21823498Level 76 27579913Level 77 34854067Level 78 44048399Level 79 55667055Level 80 70349192Level 81 88903628Level 82 112352268Level 83 141987524Level 84 179441377--Mensanator2 of Clubs http://members.aol.com/mensanator666/2ofclubs/2ofclubs.htm => This is addicting! Every time I nish a level> and want to stop that little voice whispers> the next level will only take 26% more time> and 28% more disk space, if you did this one> you can do the next one. I cant go any farther without resorting to le > splitting. Level 84 is a 3,014,822,741 text le> and took 2.5 hours to run. So here is the nal table to level 84 Level 6 2> Level 7 2> Level 8 4> Level 9 4> Level 10 6> Level 11 6> Level 12 8> Level 13 10> Level 14 14> Level 15 18> Level 16 24> Level 17 29> Level 18 36> Level 19 44> Level 20 58> Level 21 72> Level 22 91> Level 23 113> Level 24 143> Level 25 179> Level 26 227> Level 27 287> Level 28 366> Level 29 460> Level 30 578> Level 31 732> Level 32 926> Level 33 1174> Level 34 1489> Level 35 1879> Level 36 2365> Level 37 2988> Level 38 3780> Level 39 4788> Level 40 6049> Level 41 7628> Level 42 9635> Level 43 12190> Level 44 15409> Level 45 19452> Level 46 24561> Level 47 31025> Level 48 39229> Level 49 49580> Level 50 62680> Level 51 79255> Level 52 100144> Level 53 126542> Level 54 159930> Level 55 202085> Level 56 255455> Level 57 322869> Level 58 408002> Level 59 515542> Level 60 651407> Level 61 823238> Level 62 1040490> Level 63 1315036> Level 64 1661989> Level 65 2100226> Level 66 2654294> Level 67 3354459> Level 68 4239037> Level 69 5357022> Level 70 6769885> Level 71 8555671> Level 72 10812460> Level 73 13664435> Level 74 17268507> Level 75 21823498> Level 76 27579913> Level 77 34854067> Level 78 44048399> Level 79 55667055> Level 80 70349192> Level 81 88903628> Level 82 112352268> Level 83 141987524> Level 84 179441377No, I didnt expect you to go this far.It is also interesting how close you can determine the next levelcount of seeds by using the constant (3+ sqrt(21))/6. The differencefrom actual counts have a very small percentage error +/-. As theselevels grow larger in general the error percentages will be less thenthe preceeding error percentages.The next projected level 85 = 179441377*((3 + sqrt(21))/6) = 226771304Dan => > For All = for every 1>> Exists = there is at least 1>> The = there is 1 and only 1>???> Actually, we sometimes DO use a _special_ quantier for there is 1 and>only 1 x such that...: E!x.> sort of term forming operator. (Ever heard of Russells theory of>denite descriptions?)> F.>Isnt saying that there exists a unique x such that P(x) equivalent tosaying that there exists an x such that for all y, P(y) iff y=x?Dan => The only denition of quantier that I have ever seen is the> denition of the two, there exists and for all. In other words,> there is no denition of what a quantier is in general. Hm, youve never seen such a thing, so, in other words, there is no such> thing.I thought that I have ever seen reduces the scope of my commentsconcerning what has been published to what I personally am aware of,rather than it implying that I am omniscient.There is none that Ive ever seen, yes. Ive always felt that Googlegroups is a forum for individuals to contribute their analysis orquote what they have read, for a collective effort at addressingissues. If you have something to contribute, then please do!> This sort of inference seems pretty common in your posts. Often> followed by your laying claim to already well-known theorems.I would only ask that you substantiate your claims. Could you quoteinstead of making generalized assertions?> Id recommend googling generalized quantier theory before matters> get out of hand here, should you actually care about the history and> content of this rich and extensively researched eld of study.At the risk of ridicule, I will continue to qualify my assertions withthat Ive ever seen. I was referring to expositions of RicesTheorem which dene an Index Set as a set of programs that areincluded according to the function computed by the program. Moreformally, if two programs X and Y compute the same function, and theIndex Set is P, then P(X) <=> P(Y). (e.g. The Language of Machines,Floyd & Beigel, pg. 558)What I offered is a formalism that doesnt dene a set of programwith a certain additional property, but rather a function over the setaccepted by any given program. Then instead of referring to thisproperty of the set, we merely say any f(yes(I)) meaning any functionf of the set of inputs on which program I halts yes. That is, insteadof a predicate over programs, we have a predicate (function ingeneral, actually) over sets, and the extra condition is not needed. We can see that if X and Y compute the same function/set, thenf(yes(X))=f(yes(Y)) because X and Y computing the same function/setmeans that yes(X)=yes(Y).Every exposition that Ive ever seen states Rices Theorem in termsof a set of programs, not a function over the set accepted by anygiven program. My approach allows us to directly manipulate thisfunction, rather than drawing inferences from this property of theset. Taking a function of a set also allows us to dene thequantiers using functions, so that we can manipulate PredicateCalculus wffs as pure functions as well. I personally have found thisapproach very useful, so I posted it.If you have seen this published before, then please quote it!Charlie VolkstorfCambridge, MA> Chris Menzel =>> The only denition of quantier that I have ever seen is the>> denition of the two, there exists and for all. In other words,>> there is no denition of what a quantier is in general. Hm, youve never seen such a thing, so, in other words, there is no>such thing.> > I thought that I have ever seen reduces the scope of my comments> concerning what has been published to what I personally am aware of,> rather than it implying that I am omniscient.> > There is none that Ive ever seen, yes. Sorry, no. That may have been what you meant, but thats not what yousaid. In English, There is no X simply does not mean I am notpersonally aware of any X. And saying the latter does not reduce thescope of the former in the manner you suggest.> Ive always felt that Google groups is a forum for individuals to> contribute their analysis or quote what they have read, for a> collective effort at addressing issues.That is certainly one of its functions.> If you have something to contribute, then please do!collective effort at addressing the issue at hand, is to point youtoward literature of which you are evidently unaware.>This sort of inference seems pretty common in your posts. Often>followed by your laying claim to already well-known theorems.> > I would only ask that you substantiate your claims. Could you quote> instead of making generalized assertions?I have especially in mind your claims to have discovered the onlyexample of a Quine atom (i.e., a set that contains only itself), whenin fact such constructions have been well-known for years innon-wellfounded set theory. However, I really dont have any interestat all in combing through your posts for further examples, so Illwithdraw the general remark and simply say that you appear to be on thesame course with respect to quantication as you embarked upon withrespect to non-wellfoundedness. I provided several standard referencesin that case which you did not acknowledge, nonetheless I was hoping --in the seless spirit of collective effort, of course -- to steer youtoward similarly useful literature in this case.Chris Menzel =Im using a statistics textbook at Berkeley (A Course in Probability andStatistics, by Charles J. Stone). I was very confused by some of theformulas in the book, until I found the ne print that said this book:- treats the letter pi like any other variable (e.g. x, y)- log means the same thing as lnIsnt this a bad idea?? Or is this normal in the world of statistics? => Im using a statistics textbook at Berkeley (A Course in Probability and> Statistics, by Charles J. Stone). I was very confused by some of the> formulas in the book, until I found the ne print that said this book: - treats the letter pi like any other variable (e.g. x, y) - log means the same thing as ln Isnt this a bad idea?? Or is this normal in the world of statistics?No idea about statistics, but Real Mathematicians use log, not ln.-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html The League of Gentlemen =>Im using a statistics textbook at Berkeley (A Course in Probability and>Statistics, by Charles J. Stone). I was very confused by some of the>formulas in the book, until I found the ne print that said this book:- treats the letter pi like any other variable (e.g. x, y)- log means the same thing as lnIsnt this a bad idea?? Or is this normal in the world of statistics?>Robert Israel has already given you most of the answer. I add that the Greek letter pi often used as a variable in higher mathematics. For example, as a map or permutation in abstract algebra, as well as elsewhere. I remember encountering this for the rst time in my undergraduate abstract algebra course (using Herstein) and thinking, Is nothing sacred?-- Stephen J. Herschkorn herschko@rutcor.rutgers.edu =>Im using a statistics textbook at Berkeley (A Course in Probability and>Statistics, by Charles J. Stone). I was very confused by some of the>formulas in the book, until I found the ne print that said this book:>- treats the letter pi like any other variable (e.g. x, y)>- log means the same thing as ln>Isnt this a bad idea?? Or is this normal in the world of statistics?Log without a subscript generally means ln (or sometimes a complex versionof ln) in almost all post-secondary mathematics. If you want the base-10log (and very few people do nowadays), you say log_10.Using pi as a variable is standard in some areas of probability (in particular for probability vectors in a Markov chain). This is somewhatdangerous if the formulas might happen to involve the constant pi, butusually whats meant is fairly obvious from the context. It comes fromwanting to use p for probability; since there are already too manyps oating around we settle for pi which is the Greek version of p.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 => Log without a subscript generally means ln (or sometimes a complex version> of ln) in almost all post-secondary mathematics. If you want the base-10> log (and very few people do nowadays), you say log_10.Wow ok, thats a change from the time I learned logarithms in high school inthe late 1980s. Even math does change its conventions.> Using pi as a variable is standard in some areas of probability (in> particular for probability vectors in a Markov chain). This is somewhat> dangerous if the formulas might happen to involve the constant pi, but> usually whats meant is fairly obvious from the context. It comes from> wanting to use p for probability; since there are already too many> ps oating around we settle for pi which is =>> Log without a subscript generally means ln (or sometimes a complex version>> of ln) in almost all post-secondary mathematics. If you want the base-10>> log (and very few people do nowadays), you say log_10.> Wow ok, thats a change from the time I learned logarithms in high school in> the late 1980s. Even math does change its conventions.mathematics beyond high school.-- Dave SeamanJudge Yohns mistakes revealed in Mumia Abu-Jamal ruling. => Log without a subscript generally means ln (or sometimes a complex version>> of ln) in almost all post-secondary mathematics. If you want the base-10>> log (and very few people do nowadays), you say log_10.Wow ok, thats a change from the time I learned logarithms in high school in>the late 1980s. But I learned math and physics in the mid 70s, and that was theconvention.> Even math does change its conventions.No, the convention has to do with which courses youre taking. In mostapplications in math and physics beyond calculus, the logarithm thatcomes up most often is the natural log. Therefore it is common to uselog to stand for ln. As I recall in calculus, the notation ln wasstill used for the natural log, but thats the last time I saw it as aconvention.This is reected in every computer math library I know, where thereis a log(x) and a log10(x) function. The rst one is the naturallog. It has probably been so since at least the early days of FORTRAN,which would take you back to the early 60s or so.The only place I commonly see the base-10 log is in engineeringcalculations, where decibels (20 times the base-10 log of amplitude)are common. And there, because of the connection to computercalculation, people are used to writing log10 as well. - Randy => >Log without a subscript generally means ln (or sometimes a complex version>of ln) in almost all post-secondary mathematics. If you want the base-10>log (and very few people do nowadays), you say log_10.> >Wow ok, thats a change from the time I learned logarithms in high school in>>the late 1980s. >> >But I learned math and physics in the mid 70s, and that was the>convention.>At least in the non-math-research world, the convention once upon a time was that log is log_10 and ln is log_e.The reason was that common logs (base 10) were used in calculations (when slide rules were not accurate enough) and natural logs are what comes up in applications. (Hey! Theyre natural!)The advent of the computer and then the calculator changed that. Common logs are now practically irrelevant, so there is no pressing need to indicate strongly the difference between common and natural logs. So log meaning log_10 died out very quickly. (Of course, some people refuse to change, so Ill bet you can stillfind references written within the last year or so.)Of course, some people had always (well, not before 1612) used log to mean logarithm in the context they were using it. Napiers logarithms were just plain weird (from the modern point of view) and didnt even satisfy the logarithm laws, as we know them today. (There is an easy translation.) Fortunately, Briggs invented common logs for calculation. Natural logarithms werent even invented until (choose one: a) 1647, Gregoire de St. Vincent; b) 1649, Alfonso Antonio de Sarasa; c) sometime before 1676, which is when Newton mentioned them in a letter to Henry Oldenburg). Newton called them hyperbolic logarithms (the area under the curve 1/x), so the ln abbreviation had to occur after that. According to Cajori, ln was introduced by Irving Stringham in 1893. I guess thats approximately when both logarithms for calculation (common logs) and logarithms for mathematical modelling (natural logs) were widespread enough to make the distinction immediately apparent. (Lowest common denominator communication.) So the effective life of ln was less than 100 years.Ive also seen log used to mean log_2. Usually theres a strong underlying (binary) reason for this.>>Even math does change its conventions.>> >No, the convention has to do with which courses youre taking. In most>applications in math and physics beyond calculus, the logarithm that>comes up most often is the natural log. Therefore it is common to use>log to stand for ln. As I recall in calculus, the notation ln was>still used for the natural log, but thats the last time I saw it as a>convention.This is reected in every computer math library I know, where there>is a log(x) and a log10(x) function. The rst one is the natural>log. It has probably been so since at least the early days of FORTRAN,>which would take you back to the early 60s or so.The only place I commonly see the base-10 log is in engineering>calculations, where decibels (20 times the base-10 log of amplitude)>are common. And there, because of the connection to computer>calculation, people are used to writing log10 as well.> >Seismology, for the same reason. A man-made scale.Jon Miller = In a 2-dimensional lattice random walk with balanced probabilities,> i.e. four of 1/4 each, starting at <0,0>; what is the probabil;ity> that it gets to <0,1> before returning to <0,0> ? The answer was given as exactly 1/2, and my crude numerical estimations> seem to bear this out. At the time, I think a full proof was given,> but it seemed a rather long one, and it struck me that for such a simple> result, there ought to be a simple proof. But I never found one. So, can anyonefind a quick proof of the result?Oh, thats easy! One moment...[ several hours and a few dozen wasted sheets of paper later ]Well, okay, maybe its not so easy, but sometimes the key to doinggood mathematics is knowing which good mathematician to ask for help.I was fortunate enough to mention your problem to Vlada Limic, a UBCprobabilist, just as she was heading home after a particularlyexhausting day. Well, I guess she solved it halfway to the bus stop,because the next day the following (effectively three-line) proof wason my desk.Let p be the probability, starting at <0,0> that we return to <0,0>before hitting <1,0>. Of course, the number of times we hit <1,0>before returning to <0,0> is one less than a Geom(p) random variable.Therefore, it has expectation: E (# of hits of <1,0> before return to <0,0>) = 1/p - 1On the other hand, consider the number of visits Y we make to *any* of<0,0>s neighbours before nally returning to <0,0>. On our rststep from the origin, we make one visit (to whichever neighbour weland on). If we successfully go right back (with probability 1/4),then Y=1. Otherwise, we fail (probability 3/4) to go right back, makean excursion of some duration, and eventually return to one of <0,0>sneighbours. Once again, we either succeed in returning to the origin(with probability 1/4), so Y=2, or we fail (with probability 3/4), andso on.Thus, Y is really the number of trials up to and including the rstsuccessful return to the origin where trials are independent withprobability of success 1/4. So, Y is a Geom(1/4) random variable, andits expectation is 4. That is: E (# of hits of <1,0>, <0,1>, <-1,0>, or <0,-1> before return to <0,0>) = 4But this expectation is equal to the sum of the expected number ofhits to each of these four neighbours before return to <0,0>. Bysymmetry, they each have the expectation 1/p - 1 calculated above, so: 4 = 4 * (1/p - 1)and p=1/2, as required.P.S., If anyones seen this neat little proof elsewhere, please letus know.-- Kevin =>These ancient denitions are not the way we use the word>denition today. A more apt term for Euclids denitions>is descriptions; the idea being that everybody knew what>the elements of geometry were, like lines and planes. Euclid>just described them. There was only one geometry, the one we>can represent accurately on a backboard (representation of>plane) using ruler (representation of line) and compass>(representation of circle).Point well taken, but I think that it addresses only my vicious circleremark below and not quite the hidden presence (or not) of tangentiality inEuclids descriptions. Anyhow, a major motivation for my interpretation(s)that I did not explicitly mention is the reference to a surfaces straightlines in #7: although this may still refer to the straight line joiningany two points on a surface (as mainstream interpretations advocate), thereis an imbalance between a curves points (which certainly do exist) in #4and a surfaces straight lines (which generally do not) in #7; in anotherdirection, while the traditional convex interpretation requires *two*points (joined by a straight line lying evenly with the curve or surface),my tangential interpretation requires only *one* point on the curve orsurface (through which one (#4) or innitely many (#7) tangent lines pass). ^^^ ^^^^^^^^^^^^^^^[Observe that the difference underlined above may well be responsible forEuclids mention of *points* in #4 but *straight lines* in #7!]>> #4: A straight line is a line which lies evenly with the points on itself.>> #7: A plane surface is a surface which lies evenly with the straight lines>> on itself.>> [Denitions listed as in Heath, p. 153; of course line = curve in>#4.]>> My interpretation(s): lies evenly with the points on itself = lies>evenly>> with the [straight] tangent line drawn at each of the points on itself>(#4);>> lies evenly with the straight lines on itself = lies evenly with any>> straight tangent line (drawn at each of the points) on itself (#7).>> Justication: denitions #8 (A plane angle is the inclination to one>> another of two lines in a plane which meet one another and do not lie in a>> straight line) and #9 (And when the lines containing the angle are>straight,>> the angle is called rectilineal) make it clear that Euclid had no qualms>> about discussing the angle between two curves (lines), a strong>indication>> (if not proof) that tangentiality did cast its shadow on his denitions>> *and* that vicious circularity -- certainly a problem in the>interpretation>> of #4 above -- was not absent from them.>> I do not see these possibilities mentioned in Heaths 1925 edition,>perhaps>> they have been raised elsewhere since then? Any such information or other>> comments would be appreciated!>> baloglouAToswego.edu>> P.S. Speaking of angles, I suspect that the term right angle instead of>> upright angle in denition #10 is due to the double meaning of Greek>> orthos/orthi as both vertical and correct :-) Distribution: Keywords: Cc: >These ancient denitions are not the way we use the word>denition today. A more apt term for Euclids denitions>is descriptions; the idea being that everybody knew what>the elements of geometry were, like lines and planes. Euclid>just described them. There was only one geometry, the one we>can represent accurately on a backboard (representation of>plane) using ruler (representation of line) and compass>(representation of circle).Point well taken, but I think that it addresses only my vicious circleremark below and not quite the hidden presence (or not) of tangentiality inEuclids descriptions. Anyhow, a major motivation for my interpretation(s)that I did not explicitly mention is the reference to a surfaces straightlines in #7: although this may still refer to the straight line joiningany two points on a surface (as mainstream interpretations advocate), thereis an imbalance between a curves points (which certainly do exist) in #4and a surfaces straight lines (which generally do not) in #7; in anotherdirection, while the traditional convex interpretation requires *two*points (joined by a straight line lying evenly with the curve or surface),my tangential interpretation requires only *one* point on the curve orsurface (through which one (#4) or innitely many (#7) tangent lines pass). ^^^ ^^^^^^^^^^^^^^^[Observe that the difference underlined above may well be responsible forEuclids mention of *points* in #4 but *straight lines* in #7!]>> #4: A straight line is a line which lies evenly with the points on itself.>> #7: A plane surface is a surface which lies evenly with the straight lines>> on itself.>> [Denitions listed as in Heath, p. 153; of course line = curve in>#4.]>> My interpretation(s): lies evenly with the points on itself = lies>evenly>> with the [straight] tangent line drawn at each of the points on itself>(#4);>> lies evenly with the straight lines on itself = lies evenly with any>> straight tangent line (drawn at each of the points) on itself (#7).>> Justication: denitions #8 (A plane angle is the inclination to one>> another of two lines in a plane which meet one another and do not lie in a>> straight line) and #9 (And when the lines containing the angle are>straight,>> the angle is called rectilineal) make it clear that Euclid had no qualms>> about discussing the angle between two curves (lines), a strong>indication>> (if not proof) that tangentiality did cast its shadow on his denitions>> *and* that vicious circularity -- certainly a problem in the>interpretation>> of #4 above -- was not absent from them.>> I do not see these possibilities mentioned in Heaths 1925 edition,>perhaps>> they have been raised elsewhere since then? Any such information or other>> comments would be appreciated!>> baloglouAToswego.edu>> P.S. Speaking of angles, I suspect that the term right angle instead of>> upright angle in denition #10 is due to the double meaning of Greek>> orthos/orthi as both vertical and correct :-) =>These ancient denitions are not the way we use the word>denition today. A more apt term for Euclids denitions>is descriptions; the idea being that everybody knew what>the elements of geometry were, like lines and planes. Euclid>just described them. There was only one geometry, the one we>can represent accurately on a backboard (representation of>plane) using ruler (representation of line) and compass>(representation of circle).Point well taken, but I think that it addresses only my vicious circleremark below and not quite the hidden presence (or not) of tangentiality in Euclids descriptions. Anyhow, a major motivation for my interpretation(s)that I did not explicitly mention is the reference to a surfaces straightlines in #7: although this may still refer to the straight line joiningany two points on a surface (as mainstream interpretations advocate), thereis an imbalance between a curves points (which certainly do exist) in #4and a curves straight lines (which generally do not) in #7; in another direction, while the traditional convex interpretation requires *two* points (joined by a straight line lying evenly with the curve or surface), my tangential interpretation requires only *one* point on the curve orsurface (through which one (#4) or innitely many (#7) tangent lines pass). ^^^ ^^^^^^^^^^^^^^^[Observe that the difference underlined above may well be responsible forEuclids mention of *points* in #4 but *straight lines* in #7!]>> #4: A straight line is a line which lies evenly with the points on itself.>> #7: A plane surface is a surface which lies evenly with the straight lines>> on itself.>> [Denitions listed as in Heath, p. 153; of course line = curve in >#4.]>> My interpretation(s): lies evenly with the points on itself = lies>> evenly>> with the [straight] tangent line drawn at each of the points on itself >(#4);>> lies evenly with the straight lines on itself = lies evenly with any>> straight tangent line (drawn at each of the points) on itself (#7).>> Justication: denitions #8 (A plane angle is the inclination to one>> another of two lines in a plane which meet one another and do not lie in a>> straight line) and #9 (And when the lines containing the angle are>straight,>> the angle is called rectilineal) make it clear that Euclid had no qualms>> about discussing the angle between two curves (lines), a strong>indication>> (if not proof) that tangentiality did cast its shadow on his denitions>> *and* that vicious circularity -- certainly a problem in the>interpretation>> of #4 above -- was not absent from them.>> I do not see these possibilities mentioned in Heaths 1925 edition,>perhaps>> they have been raised elsewhere since then? Any such information or other>> comments would be appreciated!>> baloglouAToswego.edu>> P.S. Speaking of angles, I suspect that the term right angle instead of>> upright angle in denition #10 is due to the double meaning of Greek>> orthos/orthi as both vertical and correct :-) =How would I convert standard cartesian coordinates (x,y) intocoordinates based on hexagons? Im sure teh formula has already beenworked out, but I cant do it from rst principles.The hex coordiantes are wanted such that the x and y coordinates followthe exact same directions as the cartesoian coordinates. However,fractions of a hex will be ignored, and every alternate hex on the xaxis will obviously be shifted vertically halfd a hex.Solutions?-- --FabianOnce you get over the initial panic stage, oxygen starvation is actuallya rather pleasant sensation, almost like falling asleep. => How would I convert standard cartesian coordinates (x,y) into> coordinates based on hexagons? Im sure teh formula has already been> worked out, but I cant do it from rst principles. The hex coordiantes are wanted such that the x and y coordinates follow> the exact same directions as the cartesoian coordinates. However,> fractions of a hex will be ignored, and every alternate hex on the x> axis will obviously be shifted vertically halfd a hex. Solutions?Use offset squares instead of hexagons+---+---+---+---+---+---+| | | | | | |+-+-+-+-+-+-+-+-+-+-+-+-+ | | | | | |--+-+-+-+-+-+-+-+-+-+-+-+| | | | | | |+-+-+-+-+-+-+-+-+-+-+-+-+ | | | | | |--+-+-+-+-+-+-+-+-+-+-+-+| | | | | | |+-+-+-+-+-+-+-+-+-+-+-+-+ | | | | | |--+-+-+-+-+-+-+-+-+-+-+-+| | | | | | |+-+-+-+-+-+-+-+-+-+-+-+-+ > -- => a, b, c, d are positive numbers so that ab + cd =1.> (x1, y1), (x2, y2), (x3, y3), (x4, y4) are points on the circle x^2 + y^2>=1.> Prove that:> (a*y1 + b*y2 + c*y3 + d*y4)^2 + (a*x4 + b*x3 + c*x2 + d*x1)^2 <= 2*(a/b +> b/a + c/d +d/c)> Here is an outline of the proof for this inequality.First, note that we may assume that x_i >= 0 and y_i >= 0.Then do a variable transformation:a b = u^2, c d = v^2, a/b = t^2, d/c = s^2and we introduce the notation w = (u,v) (so, w is the vectorwith components u and v). This transforms the condition a b + c d = 1 to the condition |w| = 1and the lhs of the inequality to |A w|^2, where A is thepositive matrix ( y1 t + y2/t, y3/s + y4 s )A = ( ) ( x4 t + x3/t, x2/s + x1 s )(view with xed width font). Thus the problem now transforms toestimating the 2-norm of the matrix A (which is the square rootof the spectral radius of the matrix A^T A (where A^T denotesthe matrix A transposed)). We do this by noting, thatthe square of the spectral radius of A^T A is less or equalthe trace of A^T A (because the determinant of A^T A is nonnegative), and the trace of A^T A is just the sum of the entries of A. Putting all this together we obtain|A w|^2 <= (y1 t + y2/t)^2 + (y3/s + y4 s)^2 + (x4 t + x3/t)^2 + (x2/s + x1 s)^2 Finally, it is easy to prove that (y1 t + y2/t)^2 + (x2/s + x1 s)^2 <= t^2 + 1/t^2 + s^2 + 1/s^2and (x4 t + x3/t)^2 + (y3/s + y4 s)^2 <= t^2 + 1/t^2 + s^2 + 1/s^2which concludes the proof. It is of course easy to remove the linear algebra part of theproof, and use just standard inequalities. However, IMO as itstands the proof gives some insight of an underlying ideaof how to proove the inequality. It should also be easy nowto produce a generalisation of the original fact.HTH,Michael.-- &&&&&&&&&&&&&&&&#@#&&&&&&&&&&&&&&&&Dr. Michael UlmFB Mathematik, Universitaet Rostockmichael.ulm@mathematik.uni-rostock.de => I hope this kind of question ts this forum, if not maybe you can> tell me where to ask.> In blackjack, the dealer always end up with a hand thats 17 to 26,> because he must> stand on 17 or higer and draw on any hand lower than 17, lets say if> 16, he then might get a 10 which gives him 26. Anyway, im looking for a way to exactlyfind out the probability for> the dealers 10 different hands (17,18,19...25,26), which of course> depends on which cards are left in the deck at that time, which> changes from game to game.> The most efcient method would be to use dynamic programming, i.e., > recursion. With a computer, this is doable assuming an innite deck > - i.e., probability of drawing each rank is unaffected by previous > cards. Then, letting p(i) = the probability of the dealers ever > hitting i during the deal, p(26) = p(16)* 4/13> p(25) = p(16)/13 + p(15) * 4/13> etc. I am *so* embarrassed - those equations are completely wrong. The correct ones are derived as follows.Let q(x,k) the probability of reaching state k starting in state x. (A state includes the information of hard vs. soft hand when necessary). We then want q(0,k) for k = 17, 18, 19,..., 26 (assuming the dealer stands on soft 17 so we need not distinguish) and 0 indicates the state, no card dealt yet. The q(x,k) are determined by conditioning on the next card dealt:q(x,k) = sum(y, p(x,y)q(y,k))where the sum is over all ten possible successive states. (For x < 17, p(x,y) will be either 1/13 or 4/13 accordingly.) These equations can be solved by backwards recursion.Or, letting P be the 30 x 30 one-step transition matrix of the Markov chain, you can get the answer asr P^14,where r is the row vector representing the distribution of the state after the dealing of the rst card. Actually, a power smaller than 14 will do, but we would need to think about how large we need.[I have left out many details here.]-- Stephen J. Herschkorn herschko@rutcor.rutgers.edu =Another reference is Grifn PA, The Theory of Blackjack.-- Stephen J. Herschkorn herschko@rutcor.rutgers.edu =Tis needed to prove by induction the following statement:n^(n+1) > (n+1)^n for any n=> 3I couldnt. => Tis needed to prove by induction the following statement:> n^(n+1) > (n+1)^n for any n=> 3> I couldnt.Start with your base case. Is 3^4 > 4^3?Now, suppose n>=3 and n^(n+1) > (n+1)^nTry to demonstrate that (n+1)^(n+2) > (n+2)^(n+1)This much is basic. If you know how induction works you shouldhave gotten this far.Look at the left hand side. When you go from n to n+1you have two things going on. Youve increased the base term. That contributes a factor of ( (n+1)/n ) ^ (n+1)And youve increased the exponent. That contributes a factor of n+1Now look at the right hand side. When you go from n to n+1you have the same two things going on. Youve increased thebase term. That contributes a factor of ( (n+2)/(n+1) ) ^ nAnd youve increased the exponent. That contributes a factor of n+2Now compare those two increases. See if you can show that the increaseon the left hand side is larger than the increase on the right hand side.Key is to note that (n+1)/n > (n+2)/(n+1). since 1 + 1/n > 1 + 1/(n+1)Try dividing by ( (n+1)/n ) ^ n on the left andby ( (n+2)/(n+1) ) ^ n on the right. Were dividing the lefthand side by something bigger and the right hand side by somethingsmaller. If the inequality still holds, were all set.Now all you have to do is work it through, formalize and turn it in.One general approach to problems of this sort is to split aninequality you want to demonstrate into two inequalities thatcan each be demonstrated and which can be added (or multiplied)together to get the desired result.Youfind those two inequalities by looking for an inequality that isreasonably tight and which, when subtracted (or divided as in this case)out of the target inequality, leaves something simple enough to dealwith. John Briggs => Tis needed to prove by induction the following statement:> n^(n+1) > (n+1)^n for any n=> 3> I couldnt.>Show n^r >= nCrn^(n+1) > (n-1)n^n + n + 1 >= sum(r=0..n-2) nCr n^(n-r) + n + 1 = (n+1)^n =Whats mean nCr?William Elliot [NonBreakingSpace].83/[IGrave][NonBreakingSpace].83 [Times] [Times].84 .81[NonBreakingSpace]201.81:Tis needed to prove by induction the following statement:>n^(n+1) > (n+1)^n for any n=> 3>I couldnt.> Show n^r >= nCr n^(n+1) > (n-1)n^n + n + 1> >= sum(r=0..n-2) nCr n^(n-r) + n + 1> = (n+1)^n =Please dont top-post. Sposibo.> William Elliot ???????/???????? ? ???????? ?????????:> > Tis needed to prove by induction the following statement:>> n^(n+1) > (n+1)^n for any n=> 3>> I couldnt.>Show n^r >= nCr> n^(n+1) > (n-1)n^n + n + 1>>= sum(r=0..n-2) nCr n^(n-r) + n + 1>= (n+1)^n Whats mean nCr?n choose r, also sometimes written C(n,r) or(n)(r).Its hard to write the latter in Usenet.nCr = n!/[r!(n-r)!] - Randy =n!/r!(n-r)!> Whats mean nCr?> William Elliot [CapitalYAcute]/[CapitalIDoubleDot][CapitalYAcute][ CapitalIGrave] [Times] [Times][CapitalOAcute] [CapitalOTilde][CapitalYAcute]:> > Tis needed to prove by induction the following statement:>> n^(n+1) > (n+1)^n for any n=> 3>> I couldnt.>Show n^r >= nCr> n^(n+1) > (n-1)n^n + n + 1>>= sum(r=0..n-2) nCr n^(n-r) + n + 1>= (n+1)^n => * A prophet is without honor in his own country.> The Mentifex AI Mind project is not faring well here in the> United States -- a country full of ignorance and war-mongering.Arthur, I want you to read this document very carefully,consider your posting habits over the past several years,and give yourself a fair score on each and every item, asit would be scored by someone completely impartial; justsay: have I ever done _anything_ where someone might thinkthis describes me:http://math.ucr.edu/home/baez/crackpot.htmlYou dont need to tell anyone but you the result, but you_do_ need to understand that you just boosted that score,and that if you cannot gain some perspective on yourself,and on how your work is perceived by those with some realknowledge in the eld, you are going to join of sci.math as one of the really obnoxious and over the edgemental cases of Usenet, and not just the fairly harmless,obsessed, and blind to his own ideas trivialities net.kookyou are right now.Recommending your self-published book as a serious work fora classroom was a much more severe symptom of progress upthe scale; that starts to impinge on the realworld and cause real harm to others.Claiming AI had been solved, and that YOU, YOU, YOU had doneit, was a _much_ worse symptom again, and having instantly torecant the claim to the point of meaninglessness should havewarned you that you were shooting youself a little closer yetto .James doesnt have a life left, or any judgement at all; orthe respect of anyone at all; try not to go there. All hedoes any more is add blunder after blunder to his list, whileclaiming to be misunderstood, persecuted, and all the restyour above statement starts to have you saying, and rantingat and attacking anyone who tries to point out the weaknessesexample he is.I dont have a life either, for other reasons all of my owndevising, and it is no fun at all to be only a horrible exampleto others. Dont go there.If you want to contribute to the AI mind problem, and notjust brag falsely that you already have but are universallymisunderstood, start by bringing yourself up to speed onwhat _real_ researchers, who submit their work for peer reviewAND LISTEN TO THE REVIEWS and FIX THEIR ERRORS IN RESPONSEand SUBMIT THEIR WORK FOR REVIEW AGAIN, AND AGAIN, AND AGAIN,UNTIL THE REVIEWERS SAY IT IS CORRECT, in purist humility thatanything less isnt science, have already discovered and setas the next level from which to advance.This will take you literal years of formal education and hardstudy, the same years the other real researchers all cheerfullyinvested just to be able to add one new advance to the effort,if very lucky. You may wellfind that you arent smart enough,after all, and need to make other plans. It happens to many,but you owe it to yourself at least to try.[My daughter just started the next ve years of such a slog,after working a while and deciding that with her brains shecould contribute more if shed learn more to have the tools tomake the contributions, so this has special meaning to metoday. Hers is whats known as the sane approach, and whatIve been doing, hoping the years since since she left schoolfor a prestigeous job that shed someday choose it, is as closeas an atheist is allowed to get to praying.]Dont pretend that you are going to be the architect, becontent to add a few bricks to the edice and call that alifes work. Thats the only way the real work of scienceever gets done.Almost everything Einstein rst published was derived frompieces rst found by others, he was as much the jigsawpuzzle assembler as anything, bringing just his one morepiece that was key to the work of seeing how the other partsbest t, not springing the whole thing forth from sheertestosterone poisoning derived genius, as your attemptsreect. The media should better explain this, not pretendhe worked in isolation.Try to be at least that humble.xanthian, who has a little GA project that sometimes has alittle piece that can help the next guy take the next babystep, and is otherwise mostly just a way to ll time andtry to be useful and stay out of jail.When it doesnt go well for a while, I use the output tonger paint, and Im no artist.This is all I _can_ do, because I failed to nish myeducation, wasting a pretty good mind on a pretty pointlesscareer.Im hoping my offsprings better choices let her do more.I hope that for other people as well. The motto of theUnited Negro College Fund, my unbelievably racistgrandfathers nevertheless only regular charity check,is exactly right: a mind _is_ a terrible thing to waste.[It frankly doesnt matter the color or style of wrapperin which it comes, its _still_ a terrible thing to wastea good mind.]Thus this quite possibly futile and long past too late note.-- =I would be grateful if, in vector space theory, anyone could say, what, ifany, difference there is between a scalar product and an inner product.Ron Jones => I would be grateful if, in vector space theory, anyone could say, what, if> any, difference there is between a scalar product and an inner product.As far as I know, the terms are interchangeable, though whenused to dene a norm Ive only seen the term inner productspace, not scalar product space. - Randy =>I would be grateful if, in vector space theory, anyone could say, what, if>any, difference there is between a scalar product and an inner product.They mean the same thing. (Note that students sometimes confuse thephrases scalar product and scalar multiple, which alas mean totally different things.)>Ron Jones>************************ =Suppose you are given a set of vectors v_1,...,v_k in Q^n (exactrational arithmetic!!!) and another vector x. The task is tofind thevector on the hyperplane spanned by the v_i which has the shortesteuclidean distance to x.I implemented this with Gram-Schmidt-orthogonalization, but Iencounter arithmetic problems because nominator and denominator of thenumbers become quite large, although in the result they are not toobig. So during calculation there is an explosion of size thatdisappears again towards the end.So are there any alternatives to Gram-Schmidt? The whole procedureworks inductively, that is we have v_1,...,v_k (condition is:calculate (it does not matter how) the next vector x which will thenbe orthogonalized to yield v_(k+1).Are there versions of singular value or QR decomposition applicable toexact rational arithmetic? What I saw so far always used orthogonalmatrices, so square roots are involved. => Suppose you are given a set of vectors v_1,...,v_k in Q^n (exact> rational arithmetic!!!) and another vector x. The task is tofind the> vector on the hyperplane spanned by the v_i which has the shortest> euclidean distance to x.> I implemented this with Gram-Schmidt-orthogonalization, but I> encounter arithmetic problems because nominator and denominator of the> numbers become quite large, although in the result they are not too> big. So during calculation there is an explosion of size that> disappears again towards the end.> So are there any alternatives to Gram-Schmidt? The whole procedure> works inductively, that is we have v_1,...,v_k (condition is:> calculate (it does not matter how) the next vector x which will then> be orthogonalized to yield v_(k+1).> Are there versions of singular value or QR decomposition applicable to> exact rational arithmetic? What I saw so far always used orthogonal> matrices, so square roots are involved.> A better forum would be sci.math.num-analysis.Here is my speculation: what you need is the action of theprojector onto span of {v_j : j=1 to k} in exact rationalarithmetic.The issue is excessive memory requirements to keep up with theincrease in size of numerators/denominators. Conditioning andaccuracy is not an issue.So: if v_1,...,v_k are linearly independent (displayed as columnvectors) then the Gram matrix G: G(i,j) = is non-singular, rational, and k-by-k. (The smaller k iscompared with n, the better.)G can be inverted exactly, and the resultingnumerators/denominators are only as big as the nature of theproblem demands (much smaller than in G-S which is a form ofCholesky decomposition of G in disguise).Denote H = inv(G), V = [v_1, ... , v_k]then the projector P is P = V * H * Vwhich may as well be kept factored, rather than evaluated as ann-by-n matrix. But you be the judge of that.So, if the vector to be projected is w, then the projection isP*w, and the error vector is w-P*w.It may be faster to calculate V * (H * (V * w)) rather than (V*H*V) * w.It is possible to make it into a recursive procedure, just digup methods for inverting a bordered matrix (arising from G bygluing on a row and a column), taking advantage of the known H. => Suppose you are given a set of vectors v_1,...,v_k in Q^n (exact> rational arithmetic!!!) and another vector x. The task is tofind the> vector on the hyperplane spanned by the v_i which has the shortest> euclidean distance to x.The sought vector y satises y.v_j = x.v_j for all k.Write y = sum a_i v_i. Then sum_i a_i v_i.v_j = x.v_j.This is a system of k equations which can be solved for thea_i.Dunno if this is practical, but there are tricks to doinglinear algebra over Q, e.g., using modular arithmetic instead.> I implemented this with Gram-Schmidt-orthogonalization, but I> encounter arithmetic problems because nominator and denominator of the> numbers become quite large, although in the result they are not too> big.nominator?-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html The League of Gentlemen =Oops, forgot something.What did you mean with using modular arithmetic? Could you pleaseexplain in a little more detail because I do not see how this could beuseful.Tobias = The sought vector y satises y.v_j = x.v_j for all k.> Write y = sum a_i v_i. Then sum_i a_i v_i.v_j = x.v_j.> This is a system of k equations which can be solved for the> a_i. Dunno if this is practical, but there are tricks to doing> linear algebra over Q, e.g., using modular arithmetic instead.> Already thought about this, could be worth a try. But solving a linearequation system in general needs O(k^2) operations, (k=number of givenvectors v_i), whereas Gram-Schmidt would only need O(k) operationsbecause we are given an orthogonal basis. I did not try this so farbecause the orthogonalization is the bottleneck of my program so I didnot want to use an O(k^2) algorithm.The dimension of the vector space is typically 20 to 50, the number ofvectors may vary from just a few to (dimension - 1). =>Suppose you are given a set of vectors v_1,...,v_k in Q^n (exact>>rational arithmetic!!!) and another vector x. The task is tofind the>>vector on the hyperplane spanned by the v_i which has the shortest>>euclidean distance to x.>> >The sought vector y satises y.v_j = x.v_j for all k.>Write y = sum a_i v_i. Then sum_i a_i v_i.v_j = x.v_j.>This is a system of k equations which can be solved for the>a_i.Dunno if this is practical, but there are tricks to doing>linear algebra over Q, e.g., using modular arithmetic instead.>I implemented this with Gram-Schmidt-orthogonalization, but I>>encounter arithmetic problems because nominator and denominator of the>>numbers become quite large, although in the result they are not too>>big.>> >nominator?> >Dont be picky. This is clearly the codenominator.Jon Miller = P(X) = b^2((b^4 X^3 - 3b^2 X^2 + 3X) t^3 - > > 3(-1 + b^2 X )t u^2 + b u^3) where the odd grouping is so that I can factor P(X) into> non-polynomial factors. Doing so I have the factorization P(X) = (r_1(X,t) t + bu)(r_2(X,t) t + bu)(r_3(X,t) t + bu)> James,you claim that there are functions r_1(X,t) , r_2(X,t) and r_3(X,t)such that the two terms are equal for all X, t, b and u. Right?It is very easy to see that this is wrong in general. So observe thatthe second term does only depend on the *product* bu, but not on b oru alone. This is not the case for the rst term, for example you mayconsider X=1 and t=1 so you get for the rst termb^2 * ( b^4 - 3b^2 + 3 - 3u^2 + 3(bu)^2 + b u^3 )So for b=1 and u=2 you get 9.And for b=2 and u=1 you get 72.But if you substitute X=1, t=1, b=1 and u=2 into the second term, youwill get the same as for X=1, t=1, b=2 and u=1 no matter what thefunctions r_j(X,t) are. =P(X) = b^2((b^4 X^3 - 3b^2 X^2 + 3X) t^3 - 3(-1 + b^2 X )t u^2 + b u^3)where the odd grouping is so that I can factor P(X) into>non-polynomial factors.Doing so I have the factorization P(X) = (r_1(X,t) t + bu)(r_2(X,t) t + bu)(r_3(X,t) t + bu) > James,> you claim that there are functions r_1(X,t) , r_2(X,t) and r_3(X,t)> such that the two terms are equal for all X, t, b and u. Right?OOPS! I just realized that r_1(X,t) is wrong, as the rs are denedby the cubic r^3 + 3(-1 + b^2 X)r - b^2(b^4 X^3 - 3b^2 X^2 + 3X)=0.Ive switched to just emphasizing the key polynomial variable by usingr_1(X).> It is very easy to see that this is wrong in general. So observe that> the second term does only depend on the *product* bu, but not on b or> u alone. This is not the case for the rst term, for example you may> consider X=1 and t=1 so you get for the rst term b^2 * ( b^4 - 3b^2 + 3 - 3u^2 + 3(bu)^2 + b u^3 ) So for b=1 and u=2 you get 9.> And for b=2 and u=1 you get 72. But if you substitute X=1, t=1, b=1 and u=2 into the second term, you> will get the same as for X=1, t=1, b=2 and u=1 no matter what the> functions r_j(X,t) are.Well that sounds like youve managed to confuse yourself by trying tojust *look* at the expressions.However, its easier to consider that the rs are determined by r^3 + 3(-1 + b^2 X)r - b^2(b^4 X^3 - 3b^2 X^2 + 3X)=0to see that you cant be correct. =>> It has occurred to me that possibly the paper Advanced Polynomial>> Factorization is *too* advanced for many of you, as some people have>> taken it upon themselves to confuse you on various aspects of advanced>> polynomial factorization, so Ill back up a bit to clear things up.>[...]It has occurred to me that yo, the gods themselves mightfind your>paper Advanced Polynomial Factorization beyond Their ken ...David Bernier So Im simplifying for those of you who got lost. Consider P(x) = 11^2 + 11x + 2 where you have a special group simply to allow the non-polynomial> factorization, as you can see that P(x) is also P(x) = 11x + 123 and for many of you, as for most of mathematics that has been it. However, now consider that I just use the 11s as placemarkers, so I> can *act* like its the polynomial y^2 + xy + 2, to get the> factorization P(x) = (11 + (x+sqrt(x^2-8))/2)(11 + (x-sqrt(x^2-8))/2) which is an example of a non-polynomial factorization. Here, notice that its *still* P(x) = 11x + 123, and that doesnt> change just because of the factorization, any more than 6 changes> depending on how you factor it. Several posters unfortunately got> away with convincing many of you that the factorization can change the> polynomial, which is the kind of bogus thinking that allowed me to say> they were either incompetent or liars. I think they were lying to many of you, and only now have been forced> to back off because I can use P(x) = 11x + 123, but some of them like> Nora Baron have still been ghting to push the bogus thinking,> possibly believing that many of you will just toss out math simply> because you dislike me so much that mathematical truth itself is> irrelevant to you if it means admitting Im right.> What is really going on here is a charade. Mr. Harriswants people to believe that I have said he cannot factora polynomial using non-polynomial factors. I have neversaid any such thing. This incredible repetition of thesubstitution of 11 for t example is his pretended refutation of my bogus rules. The problem is, I havestated no such bogus rules. This is his excuse for an extended, self-righteous rant. There are very denite andbasic errors in his arguments, but they have nothing todo with what he continues to rave about here. There isnothing wrong with his factorization above. It is irrelevant to his main argument.> Such a poster as Nora Baron shows by their continued posting a> certain kind of arrogance, where the posters shows a belief that math> is politics, so since youve voted for the poster before, in support> of bogus math, the poster possibly believes youll just go along now. Now consider P(X) = b^2((b^4 X^3 - 3b^2 X^2 + 3X) t^3 - > > 3(-1 + b^2 X )t u^2 + b u^3) where the odd grouping is so that I can factor P(X) into> non-polynomial factors. Doing so I have the factorization P(X) = (r_1(X,t) t + bu)(r_2(X,t) t + bu)(r_3(X,t) t + bu) and nally I hope that you will not be susceptible to posters> claiming that suddenly you have a new polynomial, or that its really> P(t), or P(X,t) just because of the factorization.> No one has EVER said that you get a new polynomial. As for writing P(X, t), that is perfectly reasonable: clearly what you have is a function of both X and t, not a function of X alone. The factors in your factor-ization are of the form A * t + B,so it LOOKS like you are factoring it as a polynomialin t. That is the usual form of such factorizations.Here however the difference is that A is taken tobe not a constant, but a function of both X and t, e.g., A = r_1(X, t). You have been doing factorizations in this form for some time now (though in general r_1(X, t) was not a function of t) and we have not objected to it.> What some of you may not appreciate is the reality that these posters> have worked to teach you bogus math. Theyve been dedicated, and> since Ive seen lots of posts in their favor, some of you are running> around out there with bogus math that you probably think is correct,> which was taught to you by these posters. Given that there are probably people not reading my posts who trusted> these people, they are out there believing false mathematics taught to> them by people like Arturo Magidin and Nora Baron, and I think> thats wrong. Now will Arturo Magidin or Nora Baron at least step forward to> retract their objections now?> I cannot speak for Arturo, though I can guess he will notproduce any such retraction. THERE IS NO REASON FOR HIM TODO SO. AS IN VIRTUALLY ALL OTHER ARGUMENTS YOU HAVE HAD WITH ARTURO OVER THE YEARS, HE IS RIGHT AND YOU ARE, FOR THE And as for myself: NO! Absolutely no retraction. I would be untrue to my own thoughts to retract anything I havesaid in this discussion. No, I do not acquiesce. No,I do notfind merit in your proof. This current proos just as wrong as your previous faulty argument that thefactorization when m <> 0 must be similar to that which holds when m = 0, and which, apparently, you havenow abandoned. Here is some more detail on where your present argument goes wrong: You start with two lemmas. In both, you assume that a polynomial P(X) has a factor G(X). You let C be the constant term in G(X), i.e., C = G(0). You note that C must divide the constant term P(0) of P(X). You dene R(X) = G(X) - C = G(X) - G(0). It is important to note here that[0] R(0) = G(0) - G(0) = 0. That is the kind of function R(X) that you need in yourlemmas, right ? And the kind of function G(X) that you intend to consider in your factorization of your polynomial, right? Then you consider your favorite polynomial,[1] P(X, t) = b^2((b^4 X^3 - 3b^2 X^2 + 3X) t^3 - 3(-1 + b^2 X )t u^2 + b u^3) You let X = 0 to get the constant term. It is[2] P(0) = b^2*u^2*(3*t + b*u). Save that for reference below. Now you assume a factorization of the form P(X) = (r1(X,t)*t + b*u)*(r2(X,t)*t + b*u)*(r3(X,t)*t + b*u). Choose one of the factors: say G1(X) = r1(X,t)*t + b*u. This is, as you intend, of the form G1(X) = R1(X) + C1. As in the lemmas that you will apply here, C1 = G1(0) [because, by [0] above, R1(0) = 0], and C1 must divide P(0, t). Also, obviously, C1 = b*u. Similarly for C2 and C3: C2 = b*u and C3 = b*u. Notice here that I am not saying anything about G1(X), G2(X), or G3(X) being a *polynomial* factor. That is not important atthe moment. I dont care whether they are or not. Now of course the product of the constant terms of the three factors, G1(X), G2(X), and G3(X) must equal the constantterm of the original polynomial, P(X). That is, C1*C2*C3 = (b*u)*(b*u)*(b*u) = P(0), or, from [2] above,[3] (b*u)^3 = b^2*u^2*(3*t + b*u). So there is a problem. The right side of [3] is not constant with respect to t. See that 3*t in there? But the left side IS constant with respect to t.Do you see any factors of t on the left side? Any tsin there anywhere? I certainly do not see any. This is the heart of the problem with yourpresent argument. It has nothing to do with thet vs. 11 factorization stuff. That is just irrelevant uff, your own little straw man, a distractionfrom your real troubles. The problem above is what you really need to deal with, and, so far, have not. Nora B.PS: There are, in addition, proofs by Arturo,W. Dale Hall and me which show that your main claim in thisrecent argument [the same claim as in previous arguments]is wrong. You have not produced any refutation whatsoever of those proofs. They are independent and dissimilar. They are simple enough to be understood by an undergrad. Instead of dealing with them in a straightforward way, you continue to evade them and deny. You have said essentially, I dont believe your proofs because I cannot see an error in my own proof. You have to tell me where my error is before I will listen to you. We accordingly pointed out *exactly* where the error was in your own proof: in your utter lack of justication for generalizing from m = 0 to m <> 0. [Incredibly, for several weeks you claimed that you couldfind 3 nonunit algebraic integers a, b, and c such that a*b*c = p and none of a, b, or c had nonunit factors in common with p ! An astonishingly stupid statement, but you made it the basis for saying that mathematicians were teaching awed math to unsuspecting students. The irony is so thick here you have to cut it with a chainsaw! Your nal admission of error on this was very, very quiet, not repeated *quite* as often as the irrelevant t vs. 11 uff. I wonder why.]So nally, after months of bluster and rant and sheer denial, you suddenly abandon that proof and, in the last 3 days, invent another one. And again we have pinpointed the error. And AGAIN you start the cycle of distraction, denial, rage, accusations, paranoia, even veiled threats of death anddestruction. You have already contacted the FBI aboutour lying and refusal to recognize your accomplishments. How far have you lost touch with reality? Do you really not see the madness in your actions? What will come next?And still, in the background, three solid, simple proofsthat you have not been able to refute; three proofs that say, no matter how you twist and turn and kickand scream, no matter what B.S. you spew, no matterhow you try to ll the gap in your argument, YOUWILL NOT BE ABLE TO DO IT. The issue is not, really, your present argument, or any of your preceding arguments. Our proofs show that as longas you keep claiming the same thing, NO ARGUMENTIS GOING TO WORK. The best bet for you is to either(1) try to point out where our arguments are wrong, or(2) give up and start over. Your choice. One of theproofs is given below. Its short, its simple - even shorter and simpler than your own proofs -its almost entirely self-contained. See if youcan refute it. Or Dales. Or Arturos. = ==Given the polynomial P(x) = (v^3 + 1)*x^3 - 3*v*x*(u*f)^2 + (u*f)^3,where v = -1 + m*f^2, and m, u, and f are integers,with f prime and m coprime to f, says that P(x)/f^2 can be factored in the form P(x)/f^2 = (b1*x + u)*(b2*x + u)*(b3*x + u*f) [1]where b1, b2, and b3 are algebraic integers.I say not. Let m = 1, f = 5, and u = 1. Thenv = 24, and v^3 + 1 = 13825 = 25*553. It is easily veried that P(x)/f^2 = 553*x^3 - 72*x + 5.If this is factored in the form [1] as claimed byHarris, then -u/b1 = -1/b1 is a root of Q(x) = P(x)/f^2. That is, Q(-1/b1) = 553*(-1/b1)^3 - 72*(-1/b1) + 5 = 0.Multiply through by b1^3: 5*b1^3 + 72*b1^2 - 553 = 0.The expression on the left is a *non-monic*polynomial in b1 with integer coefcients,and it is *irreducible* over the rationals.Therefore b1 cannot be an algebraic integer.Therefore Harriss claim is false. wad some powr the giftie gie us to see oursels as ithers see us. --- Robert Burns [Ode to a Mouse???]> Well, think about the kind of people who would go on such a campaign> in the rst place, what makes you think they are also the kind of> people who would suddenly start telling the truth? > =>>It has occurred to me that possibly the paper Advanced Polynomial>>Factorization is *too* advanced for many of you, as some people have>>taken it upon themselves to confuse you on various aspects of advanced>>polynomial factorization, so Ill back up a bit to clear things up.>> [...]>> It has occurred to me that yo, the gods themselves mightfind your>> paper Advanced Polynomial Factorization beyond Their ken ...>> David BernierSo Im simplifying for those of you who got lost.Consider P(x) = 11^2 + 11x + 2where you have a special group simply to allow the non-polynomial>factorization, as you can see that P(x) is also P(x) = 11x + 123and for many of you, as for most of mathematics that has been it.However, now consider that I just use the 11s as placemarkers, so I>can *act* like its the polynomial y^2 + xy + 2, to get the>factorization P(x) = (11 + (x+sqrt(x^2-8))/2)(11 + (x-sqrt(x^2-8))/2)which is an example of a non-polynomial factorization.Here, notice that its *still* P(x) = 11x + 123, and that doesnt>change just because of the factorization, any more than 6 changes>depending on how you factor it. Several posters unfortunately got>away with convincing many of you that the factorization can change the>polynomial, which is the kind of bogus thinking that allowed me to say>they were either incompetent or liars.I think they were lying to many of you, and only now have been forced>to back off because I can use P(x) = 11x + 123, but some of them like>Nora Baron have still been ghting to push the bogus thinking,>possibly believing that many of you will just toss out math simply>because you dislike me so much that mathematical truth itself is>irrelevant to you if it means admitting Im right. > What is really going on here is a charade. Mr. Harris> wants people to believe that I have said he cannot factor> a polynomial using non-polynomial factors. I have never> said any such thing. This incredible repetition of the> substitution of 11 for t example is his pretended > refutation of my bogus rules. The problem is, I have> stated no such bogus rules. This is his excuse for > an extended, self-righteous rant. There are very denite and> basic errors in his arguments, but they have nothing to> do with what he continues to rave about here. There is> nothing wrong with his factorization above. It is > irrelevant to his main argument.Well readers, if Nora Baron is telling the truth, then that ripsaway any protests claiming that my polynomial P(X) is actually P(t),or P(X,t), and leaves it as it is, P(X). >Such a poster as Nora Baron shows by their continued posting a>certain kind of arrogance, where the posters shows a belief that math>is politics, so since youve voted for the poster before, in support>of bogus math, the poster possibly believes youll just go along now.Now considerP(X) = b^2((b^4 X^3 - 3b^2 X^2 + 3X) t^3 - 3(-1 + b^2 X )t u^2 + b u^3)where the odd grouping is so that I can factor P(X) into>non-polynomial factors.Doing so I have the factorization P(X) = (r_1(X,t) t + bu)(r_2(X,t) t + bu)(r_3(X,t) t + bu)and nally I hope that you will not be susceptible to posters>claiming that suddenly you have a new polynomial, or that its really>P(t), or P(X,t) just because of the factorization. No one has EVER said that you get a new polynomial. > As for writing P(X, t), that is perfectly reasonable: > clearly what you have is a function of both X and t, > not a function of X alone. The factors in your factor-And here we go yet again readers as Nora Baron is still at it.The polynomials is P(X).Its like if you have P(x) = ax^2 + bx + c, and some person keepshollering at you that its really P(x,a) or P(x,b) you can rightlytell that person to just stuff it.> ization are of the form A * t + B, so it LOOKS like you are factoring it as a polynomial> in t. That is the usual form of such factorizations.> Here however the difference is that A is taken to> be not a constant, but a function of both X and t, e.g., > A = r_1(X, t). You have been doing factorizations in > this form for some time now (though in general r_1(X, t) > was not a function of t) and we have not objected to it.Its time to address this poster.Yes, you have Nora Baron, and by we I take it you mean yourfellows, like Arturo Magidin and Will Twentyman.Well let me explain it yet again.If you have P(X) then it is P(X), and its no more P(X,t) just becauseof the factorization than P(x) = 11x + 123 suddenly becomes P(11) or P(x,11) because of P(x) = (11 + (x+sqrt(x^2-8))/2)(11 + (x-sqrt(x^2-8))/2)where the point Nora Baron is that the factorization does not changethe polynomial.Now if youre going to be reasonable and follow the mathematics, youneed to admit that if a polynomial is P(X), then it remains that,independent of how its factored.Your argument is like claiming that P(x) = ax^2 + bx + c, that youreally have P(x,a) or P(x,b).Algebra *does* allow additional variables along with the polynomialvariable, which is why people say things like P(X), which *tells* youwhat the key variable is.Do you understand?>What some of you may not appreciate is the reality that these posters>have worked to teach you bogus math. Theyve been dedicated, and>since Ive seen lots of posts in their favor, some of you are running>around out there with bogus math that you probably think is correct,>which was taught to you by these posters.Given that there are probably people not reading my posts who trusted>these people, they are out there believing false mathematics taught to>them by people like Arturo Magidin and Nora Baron, and I think>thats wrong.Now will Arturo Magidin or Nora Baron at least step forward to>retract their objections now? > I cannot speak for Arturo, though I can guess he will not> produce any such retraction. THERE IS NO REASON FOR HIM TO> DO SO. AS IN VIRTUALLY ALL OTHER ARGUMENTS YOU HAVE HAD > WITH ARTURO OVER THE YEARS, HE IS RIGHT AND YOU ARE, FOR THEWell there you have it readers.The poster Nora Baron clearly IS too proud to just follow the math.Now Ive been wrong lots of times and learned the important lessonthat bucking against mathematics is just stupid, as the math alwayswins.Now Nora Baron may be going through gut-wrench time facing havingsaid bogus stuff for a while, but hey, thats just the way it is.Refusing to simply tell the truth out of ego or to protect ones ownfeelings is just weak.You cannot ght mathematical truth.> And as for myself: NO! Absolutely no retraction. I > would be untrue to my own thoughts to retract anything I have> said in this discussion. No, I do not acquiesce. No,> I do notfind merit in your proof. This current proof> is just as wrong as your previous faulty argument that the> factorization when m <> 0 must be similar to that > which holds when m = 0, and which, apparently, you have> now abandoned.And now you see the refuge of the intellectually weak--denial.If Nora Baron could just deal with admitting being wrong, the postermight feel bad for a while, but then an *enormous* weight would belifted.I know, as Ive been there.> Here is some more detail on where your present > argument goes wrong: > You start with two lemmas. In both, you assume that > a polynomial P(X) has a factor G(X). You let C be the > constant term in G(X), i.e., C = G(0). You note that C > must divide the constant term P(0) of P(X). You dene Yup. > R(X) = G(X) - C = G(X) - G(0). It is important to note here that [0] R(0) = G(0) - G(0) = 0.Yup. > That is the kind of function R(X) that you need in your> lemmas, right ? And the kind of function G(X) that you intend to > consider in your factorization of your polynomial, right?Well, if you want to call it a function, thats ok with me.It is a function and is actually expressible as the root of a moniccubic.> Then you consider your favorite polynomial, [1] P(X, t) = b^2((b^4 X^3 - 3b^2 X^2 + 3X) t^3 > > - 3(-1 + b^2 X )t u^2 + b u^3)There you go again.How many times do I have to tell you that it is P(X), for you not tojust try and sneak in a change?Do you NOT understand that other readers can see that as well?The polynomial P(X), and your *wishing* it to be something else DOESNOT CHANGE IT.Just like you cant go around claiming that P(x) = ax^2 + bx + c isreally P(x,a) or P(x,b), as the key variable is x. > You let X = 0 to get the constant term. It isGiven P(X), its math that the constant term is P(0). No quotes arenecessary. > [2] P(0) = b^2*u^2*(3*t + b*u). Save that for reference below. Now you assume a factorization of the formThe factorization exists, no assumption necessary.Its math you know. No need to play word games. > P(X) = (r1(X,t)*t + b*u)*(r2(X,t)*t + b*u)*(r3(X,t)*t + b*u). Choose one of the factors: say G1(X) = r1(X,t)*t + b*u. This is, as you intend, of the form G1(X) = R1(X) + C1. > As in the lemmas that you will apply here, C1 = G1(0) [because, > by [0] above, R1(0) = 0], and C1 must divide P(0, t). There you go again.The polynomial is P(X), so its P(0).> Also, obviously, C1 = b*u. Similarly for C2 and C3: C2 = b*u and C3 = b*u.Thats not true. > Notice here that I am not saying anything about G1(X), G2(X), > or G3(X) being a *polynomial* factor. That is not important at> the moment. I dont care whether they are or not.Well they cant be polynomial factors as its a non-polynomialfactorization. > Now of course the product of the constant terms of the > three factors, G1(X), G2(X), and G3(X) must equal the constant> term of the original polynomial, P(X). That is,HEY, you nally said it right, except theres only *one* polynomialso no need to call it the original polynomial.Remember, a *factorization* doesnt change the expression any morethan different factorizations of 6 change it from being 6. > C1*C2*C3 = (b*u)*(b*u)*(b*u) = P(0), or, from [2] above, [3] (b*u)^3 = b^2*u^2*(3*t + b*u). So there is a problem. The right side of [3] is not > constant with respect to t. See that 3*t in there?Well, you just screwed up, and probably did so deliberately, as I havegone over this part.Actually two of the rs are 0 when X=0, while one is 3.> But the left side IS constant with respect to t.> Do you see any factors of t on the left side? Any ts> in there anywhere? I certainly do not see any.Yeah, you deleted it out with your bogus claims about the Cs.Correctly it is that C_1 = bu, C_2=bu, and C_3 = 3t+bu.> This is the heart of the problem with your> present argument. It has nothing to do with the> t vs. 11 factorization stuff. That is just > irrelevant uff, your own little straw man, a distraction> from your real troubles. The problem above is what > you really need to deal with, and, so far, have not. Nora B.Well Ive answered every point youve brought up, so the ball is inyour court.Will you *nally* swallow your pride and tell the truth?> PS: There are, in addition, proofs by Arturo,> W. Dale Hall and me which show that your main claim in this> recent argument [the same claim as in previous arguments]> is wrong. You have not produced any refutation whatsoever > of those proofs. They are independent and dissimilar. > They are simple enough to be understood by an undergrad. > Instead of dealing with them in a straightforward way, you > continue to evade them and deny. Well thats false as Ive answered all their claims, and repudiatedthem.What happens is, like youve done in your post, your crew just deniesthe mathematical logic, keeps making the same claims, and from whatIve seen people give show a simpler non-polynomialfactorization, Ive not only managed to show how you operate, but alsoto show your disdain for the truth.As well as for those poor slobs who trusted and believed in NoraBaron. > You have said essentially, I dont believe your proofs > because I cannot see an error in my own proof. You have > to tell me where my error is before I will listen to you. > We accordingly pointed out *exactly* where the error was > in your own proof: in your utter lack of justication > for generalizing from m = 0 to m <> 0. There you go again.Here in this post is the reality. You make bogus claims, and I shootthem down.The math says one thing, but you say another.I show you the math, and you refuse to follow the math.> [Incredibly, for several weeks you claimed that you > couldfind 3 nonunit algebraic integers a, b, and c such > that a*b*c = p and none of a, b, or c had nonunit factors in common> with p ! An astonishingly stupid statement, but you> made it the basis for saying that mathematicians were> teaching awed math to unsuspecting students. The> irony is so thick here you have to cut it with a > chainsaw! Your nal admission of error on this was > very, very quiet, not repeated *quite* as often as the> irrelevant t vs. 11 uff. I wonder why.]Oh yeah, that was just off. It should be that given algebraicintegers a, b, and c, with different prime numbers p_1 and p_2,you can have abc = p_1 p_2where neither a, b, nor c share *any* non-unit algebraic integerfactors with p_1.It was silly of me to get that wrong before. My apologies for anyconfusion.> So nally, after months of bluster and rant and sheer > denial, you suddenly abandon that proof and, in the last > 3 days, invent another one. And again we have pinpointed the > error. And AGAIN you start the cycle of distraction, denial, > rage, accusations, paranoia, even veiled threats of death and> destruction. You have already contacted the FBI about> our lying and refusal to recognize your accomplishments. > How far have you lost touch with reality? Do you really > not see the madness in your actions? What will come > next?Well at least you admitted lying.Is that a Perry Mason moment?> And still, in the background, three solid, simple proofs> that you have not been able to refute; three proofs > that say, no matter how you twist and turn and kick> and scream, no matter what B.S. you spew, no matter> how you try to ll the gap in your argument, YOU> WILL NOT BE ABLE TO DO IT. The issue is not, > really, your present argument, or any of your > preceding arguments. Our proofs show that as long> as you keep claiming the same thing, NO ARGUMENT> IS GOING TO WORK. The best bet for you is to either> (1) try to point out where our arguments are wrong, or> (2) give up and start over. Your choice. One of the> proofs is given below. Its short, its simple - > even shorter and simpler than your own proofs -> its almost entirely self-contained. See if you> can refute it. Or Dales. Or Arturos.Ok. Sounds like fun.> the polynomial P(x) = (v^3 + 1)*x^3 - 3*v*x*(u*f)^2 + (u*f)^3, > where v = -1 + m*f^2, and m, u, and f are integers,> with f prime and m coprime to f, says > that P(x)/f^2 can be factored in the form > P(x)/f^2 = (b1*x + u)*(b2*x + u)*(b3*x + u*f) [1] > where b1, b2, and b3 are algebraic integers.Well thats not what I say, as the very *problem* is that you canfindcases where they arent algebraic integers, though they should be oryou have that wacky thing with abc=p_1 p_2.> I say not. Let m = 1, f = 5, and u = 1. Then> v = 24, and v^3 + 1 = 13825 = 25*553. It is > easily veried that > P(x)/f^2 = 553*x^3 - 72*x + 5. > If this is factored in the form [1] as claimed by> Harris, then -u/b1 = -1/b1 is a root of Q(x) = P(x)/f^2. > That is, > Q(-1/b1) = 553*(-1/b1)^3 - 72*(-1/b1) + 5 = 0. > Multiply through by b1^3: > 5*b1^3 + 72*b1^2 - 553 = 0. > The expression on the left is a *non-monic*> polynomial in b1 with integer coefcients,> and it is *irreducible* over the rationals. Therefore b1 cannot be an algebraic integer.That is correct.> Therefore Harriss claim is false.That is incorrect, as it was your claim, not mine.> some powr the giftie gie us> to see oursels as ithers see us. --- Robert Burns [Ode to a Mouse???] >Well, think about the kind of people who would go on such a campaign>in the rst place, what makes you think they are also the kind of>people who would suddenly start telling the truth?Well readers, Nora Baron still seems to be playing to math societyas if denial is a mathematical argument like its all political.But, mathematics is NOT a democracy. => It has occurred to me that possibly the paper Advanced Polynomial> Factorization is *too* advanced for many of you, as some people have> taken it upon themselves to confuse you on various aspects of advanced> polynomial factorization, so Ill back up a bit to clear things up.>>[...]>>It has occurred to me that yo, the gods themselves mightfind your>>paper Advanced Polynomial Factorization beyond Their ken ...>>David Bernier>> So Im simplifying for those of you who got lost.>> Consider>> P(x) = 11^2 + 11x + 2>> where you have a special group simply to allow the non-polynomial>> factorization, as you can see that P(x) is also>> P(x) = 11x + 123>> and for many of you, as for most of mathematics that has been it.>> However, now consider that I just use the 11s as placemarkers, so I>> can *act* like its the polynomial y^2 + xy + 2, to get the>> factorization>> P(x) = (11 + (x+sqrt(x^2-8))/2)(11 + (x-sqrt(x^2-8))/2)>> which is an example of a non-polynomial factorization.>> Here, notice that its *still* P(x) = 11x + 123, and that doesnt>> change just because of the factorization, any more than 6 changes>> depending on how you factor it. Several posters unfortunately got>> away with convincing many of you that the factorization can change the>> polynomial, which is the kind of bogus thinking that allowed me to say>> they were either incompetent or liars.>> I think they were lying to many of you, and only now have been forced>> to back off because I can use P(x) = 11x + 123, but some of them like>> Nora Baron have still been ghting to push the bogus thinking,>> possibly believing that many of you will just toss out math simply>> because you dislike me so much that mathematical truth itself is>> irrelevant to you if it means admitting Im right.>> What is really going on here is a charade. Mr. Harris>wants people to believe that I have said he cannot factor>a polynomial using non-polynomial factors. I have never>said any such thing. This incredible repetition of the>substitution of 11 for t example is his pretended >refutation of my bogus rules. The problem is, I have>stated no such bogus rules. This is his excuse for >an extended, self-righteous rant. There are very denite and>basic errors in his arguments, but they have nothing to>do with what he continues to rave about here. There is>nothing wrong with his factorization above. It is >irrelevant to his main argument. > Well readers, if Nora Baron is telling the truth, then that rips> away any protests claiming that my polynomial P(X) is actually P(t),> or P(X,t), and leaves it as it is, P(X). Oh, nonsense. Your function P is clearly a function ofat least two variables. It is perfectly reasonable to call it P(X, t), and anyway, it doesnt do any harm toyour argument. Here again you are protesting something just so you can sound self-righteous. > >> Such a poster as Nora Baron shows by their continued posting a>> certain kind of arrogance, where the posters shows a belief that math>> is politics, so since youve voted for the poster before, in support>> of bogus math, the poster possibly believes youll just go along now.>> Now consider>> P(X) = b^2((b^4 X^3 - 3b^2 X^2 + 3X) t^3 - >> >> 3(-1 + b^2 X )t u^2 + b u^3)>> where the odd grouping is so that I can factor P(X) into>> non-polynomial factors.>> Doing so I have the factorization>> P(X) = (r_1(X,t) t + bu)(r_2(X,t) t + bu)(r_3(X,t) t + bu)>> and nally I hope that you will not be susceptible to posters>> claiming that suddenly you have a new polynomial, or that its really>> P(t), or P(X,t) just because of the factorization.>> > No one has EVER said that you get a new polynomial. >As for writing P(X, t), that is perfectly reasonable: >clearly what you have is a function of both X and t, >not a function of X alone. The factors in your factor- And here we go yet again readers as Nora Baron is still at it. The polynomials is P(X). Its like if you have P(x) = ax^2 + bx + c, and some person keeps> hollering at you that its really P(x,a) or P(x,b) you can rightly> tell that person to just stuff it.> You can, but clearly there are several symbols on theright side. It is a function of multiple variables. There is nothing *wrong* with writing P(x, a, b, c), especially if you are considering all of them asvariables. In some situations (as with P(X, t)) itclaries things to do this.>ization are of the form A * t + B,so it LOOKS like you are factoring it as a polynomial>in t. That is the usual form of such factorizations.>Here however the difference is that A is taken to>be not a constant, but a function of both X and t, e.g., >A = r_1(X, t). You have been doing factorizations in >this form for some time now (though in general r_1(X, t) >was not a function of t) and we have not objected to it. Its time to address this poster. Yes, you have Nora Baron, and by we I take it you mean your> fellows, like Arturo Magidin and Will Twentyman. Well let me explain it yet again. If you have P(X) then it is P(X), and its no more P(X,t) just because> of the factorization than P(x) = 11x + 123 suddenly becomes P(11) or P(x,11) because of P(x) = (11 + (x+sqrt(x^2-8))/2)(11 + (x-sqrt(x^2-8))/2) where the point Nora Baron is that the factorization does not change> the polynomial. Now if youre going to be reasonable and follow the mathematics, you> need to admit that if a polynomial is P(X), then it remains that,> independent of how its factored. Your argument is like claiming that P(x) = ax^2 + bx + c, that you> really have P(x,a) or P(x,b). Algebra *does* allow additional variables along with the polynomial> variable, which is why people say things like P(X), which *tells* you> what the key variable is. Do you understand?> Of course. If t were a constant, it might make senseto omit it. But it isnt. In your FLT argument,t corresponds to x. You assume x is part of a solutionto the Fermat equation. You dont know what x is, soyou consider your function as a polynomial in the indeterminate x [now converted to t for some reason].Thus you NEED to consider t as a variable. Thereforeit makes sense to write P(X, t). Do you understand? >> What some of you may not appreciate is the reality that these posters>> have worked to teach you bogus math. Theyve been dedicated, and>> since Ive seen lots of posts in their favor, some of you are running>> around out there with bogus math that you probably think is correct,>> which was taught to you by these posters.>> Given that there are probably people not reading my posts who trusted>> these people, they are out there believing false mathematics taught to>> them by people like Arturo Magidin and Nora Baron, and I think>> thats wrong.>> Now will Arturo Magidin or Nora Baron at least step forward to>> retract their objections now?>> I cannot speak for Arturo, though I can guess he will not>produce any such retraction. THERE IS NO REASON FOR HIM TO>DO SO. AS IN VIRTUALLY ALL OTHER ARGUMENTS YOU HAVE HAD >WITH ARTURO OVER THE YEARS, HE IS RIGHT AND YOU ARE, FOR THE Well there you have it readers. The poster Nora Baron clearly IS too proud to just follow the math. Now Ive been wrong lots of times and learned the important lesson> that bucking against mathematics is just stupid, as the math always> wins. Now Nora Baron may be going through gut-wrench time facing having> said bogus stuff for a while, but hey, thats just the way it is.> So you think writing P(X, t) instead of P(X) is bogus ???> Refusing to simply tell the truth out of ego or to protect ones own> feelings is just weak. You cannot ght mathematical truth. And as for myself: NO! Absolutely no retraction. I >would be untrue to my own thoughts to retract anything I have>said in this discussion. No, I do not acquiesce. No,>I do notfind merit in your proof. This current proof>is just as wrong as your previous faulty argument that the>factorization when m <> 0 must be similar to that >which holds when m = 0, and which, apparently, you have>now abandoned. And now you see the refuge of the intellectually weak--denial.> You have the annoying habit of accusing people of the same things they have just described of you - e.g., inthis case, denial. It shows a lack of imagination, averbal impoverishment. Not a good omen for the oldwriting career.> If Nora Baron could just deal with admitting being wrong, the poster> might feel bad for a while, but then an *enormous* weight would be> lifted.> ???> I know, as Ive been there. *Are* there.> Here is some more detail on where your present >argument goes wrong:> You start with two lemmas. In both, you assume that >a polynomial P(X) has a factor G(X). You let C be the >constant term in G(X), i.e., C = G(0). You note that C >must divide the constant term P(0) of P(X). You dene Yup. > R(X) = G(X) - C = G(X) - G(0). It is important to note here that[0] R(0) = G(0) - G(0) = 0. Yup.> > That is the kind of function R(X) that you need in your>lemmas, right ? And the kind of function G(X) that you intend to >consider in your factorization of your polynomial, right? Well, if you want to call it a function, thats ok with me.> When you write G(X), it is clear you are thinking of itas a function.> It is a function and is actually expressible as the root of a monic> cubic. Then you consider your favorite polynomial,[1] P(X, t) = b^2((b^4 X^3 - 3b^2 X^2 + 3X) t^3 - 3(-1 + b^2 X )t u^2 + b u^3) There you go again. How many times do I have to tell you that it is P(X), for you not to> just try and sneak in a change?> I see two variables on the right side: X and t. Dont you?Or are you now denying that?> Do you NOT understand that other readers can see that as well?> Certainly not. Other readers probably see the same thingthat I see: X and t, two variables in your expression.> The polynomial P(X), and your *wishing* it to be something else DOES> NOT CHANGE IT.> Its not a matter of wishing. Its cold fact. There aretwo variables in your expression: X and t.> Just like you cant go around claiming that P(x) = ax^2 + bx + c is> really P(x,a) or P(x,b), as the key variable is x.> Who gets to say what the key variable is?> You let X = 0 to get the constant term. It is Given P(X), its math that the constant term is P(0). No quotes are> necessary.> >[2] P(0) = b^2*u^2*(3*t + b*u). Save that for reference below. Now you assume a factorization of the form The factorization exists, no assumption necessary.> Fair enough. > Its math you know. No need to play word games.> > P(X) = (r1(X,t)*t + b*u)*(r2(X,t)*t + b*u)*(r3(X,t)*t + b*u). Choose one of the factors: say G1(X) = r1(X,t)*t + b*u. This is, as you intend, of the form G1(X) = R1(X) + C1. >As in the lemmas that you will apply here, C1 = G1(0) [because, >by [0] above, R1(0) = 0], and C1 must divide P(0, t). There you go again. The polynomial is P(X), so its P(0).> The function is P(X, t). When X = 0, its P(0, t).Note that t is still in there, even when you evaluateat X = 0. It is still a variable in the expression.There is no reason to *suppress* it.>Also, obviously, C1 = b*u.> Similarly for C2 and C3: C2 = b*u and C3 = b*u. Thats not true.> You are right about this. I had overlooked your previousclaim that C3 = 3*t + b*u.> Notice here that I am not saying anything about G1(X), G2(X), >or G3(X) being a *polynomial* factor. That is not important at>the moment. I dont care whether they are or not. Well they cant be polynomial factors as its a nonpolynomial> factorization.> With respect to X, it is not a polynomial factorization.With respect to t, it clearly *is*.> Now of course the product of the constant terms of the >three factors, G1(X), G2(X), and G3(X) must equal the constant>term of the original polynomial, P(X). That is, HEY, you nally said it right, I should have said P(X, t).> except theres only *one* polynomial> so no need to call it the original polynomial.> No. G1(X), etc., are all polynomials in the variable t.> Remember, a *factorization* doesnt change the expression any more> than different factorizations of 6 change it from being 6.> Who ever said it did? Another Harris straw man.> C1*C2*C3 = (b*u)*(b*u)*(b*u) = P(0), or, from [2] above,[3] (b*u)^3 = b^2*u^2*(3*t + b*u). So there is a problem. The right side of [3] is not >constant with respect to t. See that 3*t in there? Well, you just screwed up, and probably did so deliberately, as I have> gone over this part. Actually two of the rs are 0 when X=0, while one is 3.> I will go along with that.> But the left side IS constant with respect to t.>Do you see any factors of t on the left side? Any ts>in there anywhere? I certainly do not see any. Yeah, you deleted it out with your bogus claims about the Cs. Correctly it is that C_1 = bu, C_2=bu, and C_3 = 3t+bu.> I can go along with this part. I had overlooked yourprevious statement regarding C_3.> This is the heart of the problem with your>present argument. It has nothing to do with the>t vs. 11 factorization stuff. That is just >irrelevant uff, your own little straw man, a distraction>from your real troubles. The problem above is what >you really need to deal with, and, so far, have not. Nora B. Well Ive answered every point youve brought up, so the ball is in> your court.> Yes. I concede you are right that you said earlieressentially that C3 = 3*t + bu. I had overlooked it. You have answered satisfactorily to this part. The problem now is your Lemma 2 and its application. In theend, the problem is going to come down to the same error youhave made before, assuming that the factorization has thesame properties for X = 0 [formerly m = 0] as it does forX <> 0. What you need to do now is state Lemma 2 withmuch greater precision and try to give a valid proof. As you know and as others have pointed out, it iscompletely decient as it stands. We can discuss thisagain after you have tried to x it. There are alsoquestions about how you apply it to your factorization, but try xing it rst.> Will you *nally* swallow your pride and tell the truth?> See above. >PS: There are, in addition, proofs by Arturo,>W. Dale Hall and me which show that your main claim in this>recent argument [the same claim as in previous arguments]>is wrong. You have not produced any refutation whatsoever >of those proofs. They are independent and dissimilar. >They are simple enough to be understood by an undergrad. >Instead of dealing with them in a straightforward way, you >continue to evade them and deny. Well thats false as Ive answered all their claims, and repudiated> them.> That is absolutely, demonstrably false.> What happens is, like youve done in your post, your crew just denies> the mathematical logic, keeps making the same claims, and from what> Ive seen people give you all the benet of the doubt.> simpler Simpler than what? It assuredly is not simple. The functionsr_i(X, t) [your notation, by the way - why did you put that tin there ???] are related to inverses of roots of a cubic. NOT simple. Try writing them down explicitly.> non-polynomial> factorization, Ive not only managed to show how you operate, but also> to show your disdain for the truth. As well as for those poor slobs who trusted and believed in Nora> Baron.You have said essentially, I dont believe your proofs >because I cannot see an error in my own proof. You have >to tell me where my error is before I will listen to you. >We accordingly pointed out *exactly* where the error was >in your own proof: in your utter lack of justication >for generalizing from m = 0 to m <> 0. There you go again.> Are you really that much a follower of our former President ???> Here in this post is the reality. You make bogus claims, and I shoot> them down. The math says one thing, but you say another. I show you the math, and you refuse to follow the math. [Incredibly, for several weeks you claimed that you > couldfind 3 nonunit algebraic integers a, b, and c such > that a*b*c = p and none of a, b, or c had nonunit factors in common> with p ! An astonishingly stupid statement, but you> made it the basis for saying that mathematicians were> teaching awed math to unsuspecting students. The> irony is so thick here you have to cut it with a > chainsaw! Your nal admission of error on this was > very, very quiet, not repeated *quite* as often as the> irrelevant t vs. 11 uff. I wonder why.] Oh yeah, that was just off. It should be that given algebraic> integers a, b, and c, with different prime numbers p_1 and p_2,> you can have abc = p_1 p_2 where neither a, b, nor c share *any* non-unit algebraic integer> factors with p_1.> This contradicts basic theorems in algebraic number theory.It does not show that the algebraic integers are incomplete.It shows that you have arrived at a *contradiction*. Thesource of the contradiction is obviously your claim [see below]that b1, b2, or b3 are all algebraic integers. Therefore thatclaim is false. > It was silly of me to get that wrong before. My apologies for any> confusion.> How about retracting what you said about teaching awedmath, etc.? Corrupting youth, etc.>So nally, after months of bluster and rant and sheer >denial, you suddenly abandon that proof and, in the last >3 days, invent another one. And again we have pinpointed the >error. And AGAIN you start the cycle of distraction, denial, >rage, accusations, paranoia, even veiled threats of death and>destruction. You have already contacted the FBI about>our lying and refusal to recognize your accomplishments. >How far have you lost touch with reality? Do you really >not see the madness in your actions? What will come >next? Well at least you admitted lying.> Saying that you contacted the FBI about our lyingis by no stretch an admission of our lying. It is simplya description of what you did and said. The main point here: loss of normal healthy perspective.> Is that a Perry Mason moment? If it doesnt t, you must acquit!And still, in the background, three solid, simple proofs>that you have not been able to refute; three proofs >that say, no matter how you twist and turn and kick>and scream, no matter what B.S. you spew, no matter>how you try to ll the gap in your argument, YOU>WILL NOT BE ABLE TO DO IT. The issue is not, >really, your present argument, or any of your >preceding arguments. Our proofs show that as long>as you keep claiming the same thing, NO ARGUMENT>IS GOING TO WORK. The best bet for you is to either>(1) try to point out where our arguments are wrong, or>(2) give up and start over. Your choice. One of the>proofs is given below. Its short, its simple - >even shorter and simpler than your own proofs ->its almost entirely self-contained. See if you>can refute it. Or Dales. Or Arturos. Ok. Sounds like fun.> This is the FIRST TIME you have tried to address this.Did it not sound like fun before, on the many occasionsit was presented?> =Given the polynomial P(x) = (v^3 + 1)*x^3 - 3*v*x*(u*f)^2 + (u*f)^3,>where v = -1 + m*f^2, and m, u, and f are integers,>with f prime and m coprime to f, says >that P(x)/f^2 can be factored in the form> P(x)/f^2 = (b1*x + u)*(b2*x + u)*(b3*x + u*f) [1]>where b1, b2, and b3 are algebraic integers. Well thats not what I say, In Advanced Polynomial Factorization (previous version)that is exactly what you said. You never publicly retractedthat. In the PRESENT version, you say that b1, b2, andb3 SHOULD BE algebraic integers. That can only mean thatyou still think you have a proof. That is what has causedyou to conclude, bizarrely, that the ring of algebraicintegers is incomplete [a so-far undened term]. Main point: there is no basis for saying SHOULD BE unless you still believe you have a proof.> as the very *problem* is that you canfind> cases where they arent algebraic integers, though they should be There it is! The old should be. Yes, you still thinkyou *can* prove that the b_i are algebraic integers. This removes all doubt.> or you have that wacky thing with abc=p_1 p_2.> If you have that wacky thing, then you havea contradiction. It is well known from establishedtheory that no such wacky thing can happen. Ifyou have a contradiction, there are only twopossibilities: (1) your proof is wrong, or (2) mathematics is inconsistent.I believe you have rejected (2). So do I. Thatleaves only (1). Your proof is wrong. Yourassertion in it that b1, b2, and b3 are algebraicintegers is wrong.>I say not. Let m = 1, f = 5, and u = 1. Then>v = 24, and v^3 + 1 = 13825 = 25*553. It is >easily veried that > P(x)/f^2 = 553*x^3 - 72*x + 5.>If this is factored in the form [1] as claimed by>Harris, then -u/b1 = -1/b1 is a root of Q(x) = P(x)/f^2. >That is,> Q(-1/b1) = 553*(-1/b1)^3 - 72*(-1/b1) + 5 = 0.>Multiply through by b1^3:> 5*b1^3 + 72*b1^2 - 553 = 0.>The expression on the left is a *non-monic*>polynomial in b1 with integer coefcients,>and it is *irreducible* over the rationals.Therefore b1 cannot be an algebraic integer. That is correct.> Good. We agree. The SHOULD BE phrase is thereforewrong. If you think you have a proof of SHOULD BE,clearly you are incorrect. Your proof is wrong.>Therefore Harriss claim is false. That is incorrect, as it was your claim, not mine.> No. When you say SHOULD BE you are clearly either(1) claiming you can prove it, or (2) saying itSHOULD BE because you cannot stand to contemplatethe alternative; because basically, you cannot admityou are wrong; because you want it so bad that facingup to the truth that it ISNT is just unbearable. I really think you are saying SHOULD BE onlybecause you desperately WANT it to be. I note that you still have found absolutely no errorin my argument. In fact, you now agree with it! We SHOULD BE in agreement that claiming that b1, b2and b3 are algebraic integers produces a contradiction.We SHOULD BE done at this point. Your turn to try to weasel out of it again. Nora B.> some powr the giftie gie us> to see oursels as ithers see us. --- Robert Burns [Ode to a Mouse???]> Ode to a Louse, actually.> Well, think about the kind of people who would go on such a campaign>> in the rst place, what makes you think they are also the kind of>> people who would suddenly start telling the truth? Well readers, Nora Baron still seems to be playing to math society> as if denial is a mathematical argument like its all political. But, mathematics is NOT a democracy.> And youre not the President, either. =>It has occurred to me that possibly the paper Advanced Polynomial>Factorization is *too* advanced for many of you, as some people have>taken it upon themselves to confuse you on various aspects of advanced>polynomial factorization, so Ill back up a bit to clear things up.> [...]>> It has occurred to me that yo, the gods themselves mightfind your> paper Advanced Polynomial Factorization beyond Their ken ...>> David Bernier>>So Im simplifying for those of you who got lost.>>Consider>> P(x) = 11^2 + 11x + 2>>where you have a special group simply to allow the non-polynomial>>factorization, as you can see that P(x) is also>> P(x) = 11x + 123>>and for many of you, as for most of mathematics that has been it.>>However, now consider that I just use the 11s as placemarkers, so I>>can *act* like its the polynomial y^2 + xy + 2, to get the>>factorization>> P(x) = (11 + (x+sqrt(x^2-8))/2)(11 + (x-sqrt(x^2-8))/2)>>which is an example of a non-polynomial factorization.>>Here, notice that its *still* P(x) = 11x + 123, and that doesnt>>change just because of the factorization, any more than 6 changes>>depending on how you factor it. Several posters unfortunately got>>away with convincing many of you that the factorization can change the>>polynomial, which is the kind of bogus thinking that allowed me to say>>they were either incompetent or liars.>>I think they were lying to many of you, and only now have been forced>>to back off because I can use P(x) = 11x + 123, but some of them like>>Nora Baron have still been ghting to push the bogus thinking,>>possibly believing that many of you will just toss out math simply>>because you dislike me so much that mathematical truth itself is>>irrelevant to you if it means admitting Im right.>> What is really going on here is a charade. Mr. Harris>> wants people to believe that I have said he cannot factor>> a polynomial using non-polynomial factors. I have never>> said any such thing. This incredible repetition of the>> substitution of 11 for t example is his pretended >> refutation of my bogus rules. The problem is, I have>> stated no such bogus rules. This is his excuse for >> an extended, self-righteous rant. There are very denite and>> basic errors in his arguments, but they have nothing to>> do with what he continues to rave about here. There is>> nothing wrong with his factorization above. It is >> irrelevant to his main argument.>Well readers, if Nora Baron is telling the truth, then that rips>away any protests claiming that my polynomial P(X) is actually P(t),>or P(X,t), and leaves it as it is, P(X). Oh, nonsense. Your function P is clearly a function of> at least two variables. It is perfectly reasonable to > call it P(X, t), and anyway, it doesnt do any harm to> your argument. Here again you are protesting something > just so you can sound self-righteous. Posters like yourself have claimed my argument is wrong for MONTHS,and Ive changed variables to keep you from continuing to get awaywith your false claims.So any little trick that you try, like trying to force it to be P(X,t)when X is the polynomial variable, is important to stop.>Such a poster as Nora Baron shows by their continued posting a>>certain kind of arrogance, where the posters shows a belief that math>>is politics, so since youve voted for the poster before, in support>>of bogus math, the poster possibly believes youll just go along now.>>Now consider>>P(X) = b^2((b^4 X^3 - 3b^2 X^2 + 3X) t^3 - >> 3(-1 + b^2 X )t u^2 + b u^3)>>where the odd grouping is so that I can factor P(X) into>>non-polynomial factors.>>Doing so I have the factorization>> P(X) = (r_1(X,t) t + bu)(r_2(X,t) t + bu)(r_3(X,t) t + bu)>>and nally I hope that you will not be susceptible to posters>>claiming that suddenly you have a new polynomial, or that its really>>P(t), or P(X,t) just because of the factorization.>> No one has EVER said that you get a new polynomial. >> As for writing P(X, t), that is perfectly reasonable: >> clearly what you have is a function of both X and t, >> not a function of X alone. The factors in your factor-And here we go yet again readers as Nora Baron is still at it.The polynomials is P(X).Its like if you have P(x) = ax^2 + bx + c, and some person keeps>hollering at you that its really P(x,a) or P(x,b) you can rightly>tell that person to just stuff it. > You can, but clearly there are several symbols on the> right side. It is a function of multiple variables. > There is nothing *wrong* with writing P(x, a, b, c), > especially if you are considering all of them as> variables. In some situations (as with P(X, t)) it> claries things to do this.If there werent posters like yourself who were dedicated to producingconfusion, then maybe that would be true.But for months posters like yourself have gotten away with falseclaims, and Im not interested in giving you more fodder to keepgetting away with them.It really does hurt the discussion when there is a dedicated group ofpeople who are willing to say just about anything to push their ownagenda against the truth. >> ization are of the form>> A * t + B,>> so it LOOKS like you are factoring it as a polynomial>> in t. That is the usual form of such factorizations.>> Here however the difference is that A is taken to>> be not a constant, but a function of both X and t, e.g., >> A = r_1(X, t). You have been doing factorizations in >> this form for some time now (though in general r_1(X, t) >> was not a function of t) and we have not objected to it.Its time to address this poster.Yes, you have Nora Baron, and by we I take it you mean your>fellows, like Arturo Magidin and Will Twentyman.Well let me explain it yet again.If you have P(X) then it is P(X), and its no more P(X,t) just because>of the factorization than P(x) = 11x + 123 suddenly becomes P(11) or P(x,11) because of P(x) = (11 + (x+sqrt(x^2-8))/2)(11 + (x-sqrt(x^2-8))/2)where the point Nora Baron is that the factorization does not change>the polynomial.Now if youre going to be reasonable and follow the mathematics, you>need to admit that if a polynomial is P(X), then it remains that,>independent of how its factored.Your argument is like claiming that P(x) = ax^2 + bx + c, that you>really have P(x,a) or P(x,b).Algebra *does* allow additional variables along with the polynomial>variable, which is why people say things like P(X), which *tells* you>what the key variable is.Do you understand? > Of course. If t were a constant, it might make sense> to omit it. But it isnt. In your FLT argument,> t corresponds to x. You assume x is part of a solution> to the Fermat equation. You dont know what x is, so> you consider your function as a polynomial in the > indeterminate x [now converted to t for some reason].> Thus you NEED to consider t as a variable. Therefore> it makes sense to write P(X, t). Do you understand? Ive changed variable names because of posters like Nora Baron whohave dedicated themselves to confusing other readers, and have beenrather successful.The t is just a placemarker for the non-polynomial factorization.>>What some of you may not appreciate is the reality that these posters>>have worked to teach you bogus math. Theyve been dedicated, and>>since Ive seen lots of posts in their favor, some of you are running>>around out there with bogus math that you probably think is correct,>>which was taught to you by these posters.>>Given that there are probably people not reading my posts who trusted>>these people, they are out there believing false mathematics taught to>>them by people like Arturo Magidin and Nora Baron, and I think>>thats wrong.>>Now will Arturo Magidin or Nora Baron at least step forward to>>retract their objections now?>> I cannot speak for Arturo, though I can guess he will not>> produce any such retraction. THERE IS NO REASON FOR HIM TO>> DO SO. AS IN VIRTUALLY ALL OTHER ARGUMENTS YOU HAVE HAD >> WITH ARTURO OVER THE YEARS, HE IS RIGHT AND YOU ARE, FOR THEWell there you have it readers.The poster Nora Baron clearly IS too proud to just follow the math.Now Ive been wrong lots of times and learned the important lesson>that bucking against mathematics is just stupid, as the math always>wins.Now Nora Baron may be going through gut-wrench time facing having>said bogus stuff for a while, but hey, thats just the way it is. > So you think writing P(X, t) instead of P(X) is bogus ???Your agenda is to confuse the issue by getting readers to think thatthe factorization is a linear one where t is the primary variable, butthat is not the case, as in fact, its a non-polynomial factorizationwhere t is just there for the factorization.>Refusing to simply tell the truth out of ego or to protect ones own>feelings is just weak.You cannot ght mathematical truth.> And as for myself: NO! Absolutely no retraction. I >> would be untrue to my own thoughts to retract anything I have>> said in this discussion. No, I do not acquiesce. No,>> I do notfind merit in your proof. This current proof>> is just as wrong as your previous faulty argument that the>> factorization when m <> 0 must be similar to that >> which holds when m = 0, and which, apparently, you have>> now abandoned.And now you see the refuge of the intellectually weak--denial. > You have the annoying habit of accusing people of the > same things they have just described of you - e.g., in> this case, denial. It shows a lack of imagination, a> verbal impoverishment. Not a good omen for the old> writing career.You and posters like you have worked for months to convince others ofthings that are false.Now that Ive found yet another way to highlight the falsity of yourarguments, you still refuse to just follow the math, and STILL areworking to convince readers of things that are false.Now as I shoot down your claims, you just deny. >If Nora Baron could just deal with admitting being wrong, the poster>might feel bad for a while, but then an *enormous* weight would be>lifted. > ??? >I know, as Ive been there. *Are* there.And there you go again.>> Here is some more detail on where your present >> argument goes wrong:>> You start with two lemmas. In both, you assume that >> a polynomial P(X) has a factor G(X). You let C be the >> constant term in G(X), i.e., C = G(0). You note that C >> must divide the constant term P(0) of P(X). You dene Yup. >> R(X) = G(X) - C = G(X) - G(0). >> It is important to note here that>> [0] R(0) = G(0) - G(0) = 0.Yup.> That is the kind of function R(X) that you need in your>> lemmas, right ? And the kind of function G(X) that you intend to >> consider in your factorization of your polynomial, right?Well, if you want to call it a function, thats ok with me. > When you write G(X), it is clear you are thinking of it> as a function.And thats why previously I did NOT write similar expressions in thatmanner, but then posters whined on that account!!!>It is a function and is actually expressible as the root of a monic>cubic.> Then you consider your favorite polynomial,>> [1] P(X, t) = b^2((b^4 X^3 - 3b^2 X^2 + 3X) t^3 >> >> - 3(-1 + b^2 X )t u^2 + b u^3)There you go again.How many times do I have to tell you that it is P(X), for you not to>just try and sneak in a change? I see two variables on the right side: X and t. Dont you?> Or are you now denying that?There are MORE than two variables, as u is in there as well, so whyhavent you talked about u then? >Do you NOT understand that other readers can see that as well? > Certainly not. Other readers probably see the same thing> that I see: X and t, two variables in your expression.Well then other readers can see MORE than two variables, and therebysee you picking and choosing while trying to push your choices overothers, when Im saying that the key variable is X.That arbitrariness in your posting is clearly part of an agenda and isnot mathematics.Youve picked on t as the point of your attack, so you dont mentionb and u as variables in the same way, though they are, whichclearly reveals how inconsistent you are. >The polynomial P(X), and your *wishing* it to be something else DOES>NOT CHANGE IT. > Its not a matter of wishing. Its cold fact. There are> two variables in your expression: X and t.There are more variables than that Nora Baron, as u and b are inthere as well.You clearly have a contempt for mathematical truth. >Just like you cant go around claiming that P(x) = ax^2 + bx + c is>really P(x,a) or P(x,b), as the key variable is x. > Who gets to say what the key variable is?If you see P(x) then x is the key variable. Thats it. No room fordiscussion.>> You let X = 0 to get the constant term. It isGiven P(X), its math that the constant term is P(0). No quotes are>necessary.> [2] P(0) = b^2*u^2*(3*t + b*u).>> Save that for reference below.>> Now you assume a factorization of the formThe factorization exists, no assumption necessary.> > Fair enough. >Its math you know. No need to play word games.> P(X) = (r1(X,t)*t + b*u)*(r2(X,t)*t + b*u)*(r3(X,t)*t + b*u).>> Choose one of the factors: say>> G1(X) = r1(X,t)*t + b*u.>> This is, as you intend, of the form G1(X) = R1(X) + C1. >> As in the lemmas that you will apply here, C1 = G1(0) [because, >> by [0] above, R1(0) = 0], and C1 must divide P(0, t). There you go again.The polynomial is P(X), so its P(0). > The function is P(X, t). When X = 0, its P(0, t).> Note that t is still in there, even when you evaluate> at X = 0. It is still a variable in the expression.> There is no reason to *suppress* it.Then why are you suppressing u and b?Inconsistency is a hallmark of someone who doesnt care about themathematical logic.You are clearly inconsistent, pushing your agenda of confusion.>> Also, obviously, C1 = b*u.>> Similarly for C2 and C3: C2 = b*u and C3 = b*u.Thats not true. > You are right about this. I had overlooked your previous> claim that C3 = 3*t + b*u.Well that takes away your objection!So why arent you posting that youre no longer disagreeing with theproof?>> Notice here that I am not saying anything about G1(X), G2(X), >> or G3(X) being a *polynomial* factor. That is not important at>> the moment. I dont care whether they are or not.Well they cant be polynomial factors as its a nonpolynomial>factorization. > With respect to X, it is not a polynomial factorization.> With respect to t, it clearly *is*.And you reveal your agenda yet again, which is to push the notion thatt is the polynomial variable, so that you can try and claim that theargument is false, though it doesnt make sense to say so, as fromwhat Ive seen, you sow confusion by pushing that notion and then just*claim* that the math isnt true.>> Now of course the product of the constant terms of the >> three factors, G1(X), G2(X), and G3(X) must equal the constant>> term of the original polynomial, P(X). That is,HEY, you nally said it right, > I should have said P(X, t).And again you reveal an inconsistency by trying to push *your* choicewhile arbitrarily ignoring b and u.Your aim is to confuse readers.>except theres only *one* polynomial>so no need to call it the original polynomial. > No. G1(X), etc., are all polynomials in the > variable t.Thats irrelevant but true. So what?>Remember, a *factorization* doesnt change the expression any more>than different factorizations of 6 change it from being 6. > Who ever said it did? Another Harris straw man.Well then, whats your point of disagreement with my proof?>> C1*C2*C3 = (b*u)*(b*u)*(b*u) = P(0), >> or, from [2] above,>> [3] (b*u)^3 = b^2*u^2*(3*t + b*u).>> >> So there is a problem. The right side of [3] is not >> constant with respect to t. See that 3*t in there?Well, you just screwed up, and probably did so deliberately, as I have>gone over this part.Actually two of the rs are 0 when X=0, while one is 3. I will go along with that. > But the left side IS constant with respect to t.>> Do you see any factors of t on the left side? Any ts>> in there anywhere? I certainly do not see any.Yeah, you deleted it out with your bogus claims about the Cs.Correctly it is that C_1 = bu, C_2=bu, and C_3 = 3t+bu. > I can go along with this part. I had overlooked your> previous statement regarding C_3. >> This is the heart of the problem with your>> present argument. It has nothing to do with the>> t vs. 11 factorization stuff. That is just >> irrelevant uff, your own little straw man, a distraction>> from your real troubles. The problem above is what >> you really need to deal with, and, so far, have not.>> Nora B.Well Ive answered every point youve brought up, so the ball is in>your court. > Yes. I concede you are right that you said earlier> essentially that C3 = 3*t + bu. I had overlooked it. > You have answered satisfactorily to this part. > The problem now is your Lemma 2 and its application. In the> end, the problem is going to come down to the same error you> have made before, assuming that the factorization has the> same properties for X = 0 [formerly m = 0] as it does for> X <> 0. So youre attacking lemma 2?For those who wonder thats why Im glad its separated out!!!> What you need to do now is state Lemma 2 with> much greater precision and try to give a valid proof. As > you know and as others have pointed out, it is> completely decient as it stands. We can discuss this> again after you have tried to x it. There are also> questions about how you apply it to your factorization, > but try xing it rst.What lemma 2 basically states is that if the constant term P(0) iscoprime to an algebraic integer, then a factor C of the constant termis coprime to that number as well.>Will you *nally* swallow your pride and tell the truth? > See above. > PS: There are, in addition, proofs by Arturo,>> W. Dale Hall and me which show that your main claim in this>> recent argument [the same claim as in previous arguments]>> is wrong. You have not produced any refutation whatsoever >> of those proofs. They are independent and dissimilar. >> They are simple enough to be understood by an undergrad. >> Instead of dealing with them in a straightforward way, you >> continue to evade them and deny. Well thats false as Ive answered all their claims, and repudiated>them. > That is absolutely, demonstrably false.Then demonstrate. >What happens is, like youve done in your post, your crew just denies>the mathematical logic, keeps making the same claims, and from what>Ive seen people give you all the benet of the doubt. > simpler > Simpler than what? It assuredly is not simple. The functions> r_i(X, t) [your notation, by the way - why did you put that t> in there ???] are related to inverses of roots of a cubic. > NOT simple. Try writing them down explicitly.Thats not necessary.>non-polynomial>factorization, Ive not only managed to show how you operate, but also>to show your disdain for the truth.As well as for those poor slobs who trusted and believed in Nora>Baron.> > You have said essentially, I dont believe your proofs >> because I cannot see an error in my own proof. You have >> to tell me where my error is before I will listen to you. >> We accordingly pointed out *exactly* where the error was >> in your own proof: in your utter lack of justication >> for generalizing from m = 0 to m <> 0. There you go again. > Are you really that much a follower of our former President ???Trust, but verify.>Here in this post is the reality. You make bogus claims, and I shoot>them down.The math says one thing, but you say another.I show you the math, and you refuse to follow the math.> [Incredibly, for several weeks you claimed that you >> couldfind 3 nonunit algebraic integers a, b, and c such >> that >> a*b*c = p>> and none of a, b, or c had nonunit factors in common>> with p ! An astonishingly stupid statement, but you>> made it the basis for saying that mathematicians were>> teaching awed math to unsuspecting students. The>> irony is so thick here you have to cut it with a >> chainsaw! Your nal admission of error on this was >> very, very quiet, not repeated *quite* as often as the>> irrelevant t vs. 11 uff. I wonder why.]Oh yeah, that was just off. It should be that given algebraic>integers a, b, and c, with different prime numbers p_1 and p_2,>you can have abc = p_1 p_2where neither a, b, nor c share *any* non-unit algebraic integer>factors with p_1. > This contradicts basic theorems in algebraic number theory.> It does not show that the algebraic integers are incomplete.> It shows that you have arrived at a *contradiction*. The> source of the contradiction is obviously your claim [see below]> that b1, b2, or b3 are all algebraic integers. Therefore that> claim is false. Its a, b, and c.>It was silly of me to get that wrong before. My apologies for any>confusion. > How about retracting what you said about teaching awed> math, etc.? Corrupting youth, etc.The ring of algebraic integers is screwed up, as Ive proven. Mathematicians should not continue to teach the awed mathematics.> So nally, after months of bluster and rant and sheer >> denial, you suddenly abandon that proof and, in the last >> 3 days, invent another one. And again we have pinpointed the >> error. And AGAIN you start the cycle of distraction, denial, >> rage, accusations, paranoia, even veiled threats of death and>> destruction. You have already contacted the FBI about>> our lying and refusal to recognize your accomplishments. >> How far have you lost touch with reality? Do you really >> not see the madness in your actions? What will come >> next?Well at least you admitted lying. > Saying that you contacted the FBI about our lying> is by no stretch an admission of our lying. It is simply> a description of what you did and said. The main point here: loss of normal healthy perspective.Heres your statement:You have already contacted the FBI about our lying and refusal torecognize your accomplishments. How far have you lost touch withreality?>Is that a Perry Mason moment? > If it doesnt t, you must acquit! And still, in the background, three solid, simple proofs>> that you have not been able to refute; three proofs >> that say, no matter how you twist and turn and kick>> and scream, no matter what B.S. you spew, no matter>> how you try to ll the gap in your argument, YOU>> WILL NOT BE ABLE TO DO IT. The issue is not, >> really, your present argument, or any of your >> preceding arguments. Our proofs show that as long>> as you keep claiming the same thing, NO ARGUMENT>> IS GOING TO WORK. The best bet for you is to either>> (1) try to point out where our arguments are wrong, or>> (2) give up and start over. Your choice. One of the>> proofs is given below. Its short, its simple - >> even shorter and simpler than your own proofs ->> its almost entirely self-contained. See if you>> can refute it. Or Dales. Or Arturos.Ok. Sounds like fun. > This is the FIRST TIME you have tried to address this.> Did it not sound like fun before, on the many occasions> it was presented?Its not a debate.> Given the polynomial>> P(x) = (v^3 + 1)*x^3 - 3*v*x*(u*f)^2 + (u*f)^3,>> where v = -1 + m*f^2, and m, u, and f are integers,>> with f prime and m coprime to f, says >> that P(x)/f^2 can be factored in the form>> P(x)/f^2 = (b1*x + u)*(b2*x + u)*(b3*x + u*f) [1]>> where b1, b2, and b3 are algebraic integers.Well thats not what I say, > In Advanced Polynomial Factorization (previous version)> that is exactly what you said. You never publicly retracted> that. In the PRESENT version, you say that b1, b2, and> b3 SHOULD BE algebraic integers. That can only mean that> you still think you have a proof. That is what has caused> you to conclude, bizarrely, that the ring of algebraic> integers is incomplete [a so-far undened term]. Main point: there is no basis for saying SHOULD BE > unless you still believe you have a proof.The primary conclusion of the paper Advanced Polynomial Factorizationhas *always* been that one of the as is coprime to 5.Heres the last sentence from the paper, which readers canfind at mywebsite.(Formatted for text.)Therefore, with the factorization 65x^3− 12x + 1 = (a_1x + 1)(a_2x + 1)(a_3x + 1)one of the as is coprime to 5.>as the very *problem* is that you canfind>cases where they arent algebraic integers, though they should be > There it is! The old should be. Yes, you still think> you *can* prove that the b_i are algebraic integers. This > removes all doubt.The paper says that one of the as is coprime to 5, which is proven.Youre making things up.>or you have that wacky thing with abc=p_1 p_2. > If you have that wacky thing, then you have> a contradiction. It is well known from established> theory that no such wacky thing can happen. If> you have a contradiction, there are only two> possibilities: (1) your proof is wrong, or (2) mathematics is inconsistent. > I believe you have rejected (2). So do I. That> leaves only (1). Your proof is wrong. Your> assertion in it that b1, b2, and b3 are algebraic> integers is wrong.There is another possibility, which is that youre wrong. >> I say not. Let m = 1, f = 5, and u = 1. Then>> v = 24, and v^3 + 1 = 13825 = 25*553. It is >> easily veried that >> P(x)/f^2 = 553*x^3 - 72*x + 5.>> If this is factored in the form [1] as claimed by>> Harris, then -u/b1 = -1/b1 is a root of Q(x) = P(x)/f^2. >> That is,>> Q(-1/b1) = 553*(-1/b1)^3 - 72*(-1/b1) + 5 = 0.>> Multiply through by b1^3:>> 5*b1^3 + 72*b1^2 - 553 = 0.>> The expression on the left is a *non-monic*>> polynomial in b1 with integer coefcients,>> and it is *irreducible* over the rationals.>> Therefore b1 cannot be an algebraic integer.That is correct. > Good. We agree. The SHOULD BE phrase is therefore> wrong. If you think you have a proof of SHOULD BE,> clearly you are incorrect. Your proof is wrong.The bs arent even in the paper!!! >> Therefore Harriss claim is false.That is incorrect, as it was your claim, not mine. > No. When you say SHOULD BE you are clearly either> (1) claiming you can prove it, or (2) saying it> SHOULD BE because you cannot stand to contemplate> the alternative; because basically, you cannot admit> you are wrong; because you want it so bad that facing> up to the truth that it ISNT is just unbearable. I really think you are saying SHOULD BE only> because you desperately WANT it to be.Ive gone step-by-step through the proof repeatedly. > I note that you still have found absolutely no error> in my argument. In fact, you now agree with it! > We SHOULD BE in agreement that claiming that b1, b2> and b3 are algebraic integers produces a contradiction.> We SHOULD BE done at this point. Your turn to try to weasel out of it again. Nora B.Well I must say Nora Baron is persistent. =[cut]> Its like if you have P(x) = ax^2 + bx + c, and some person keeps> hollering at you that its really P(x,a) or P(x,b) you can rightly> tell that person to just stuff it.You cant rightly tell that person to just stuff it.Or, better phrased, if that person asks you What area, b, and c, then you cant tell him to stuff it.Something has to be said about what the a, b, c are.Also, something has to be said about what the x is.This is what the others are talking about with you.Of course, the answer depends upon what you want to do withthe x, a, b, and c. This also has been mentioned several times.Offhand, the suggestion of treating the x, a, b, c on equalfooting as indeterminate variables is ok, so that you wouldwrite P(x,a,b,c), and have P(x,a,b,c) in the polynomialring A[x,a,b,c].However, this will not work for the normal use of derivingthe quadratic formula for P(x) = ax^2 + bx + c. This isbecause the roots of the equation ax^2 + bx + c = 0 aregiven by -b/2a +- sqrt(b^2 - 4ac) / 2a, where divisionby 2a doesnt make sense in the ring A[x,a,b,c].If you want to derive the quadratic formula forthe equation ax^2 + bx + c = 0, at least twodifferent approaches can be used to specifywhat the x, a, b, and c are.Here is the rst approach, which I think is themore common. Let C be the eld of complex numbers.Let a, b, c be complex numbers, so that a, b, care elements of C. Let C[x] be the ring of polynomialover the complex numbers C.Then, P(x) = ax^2 + bx + c is a polynomial in C[x].Now, I can derive the quadratic formuala for theroots because division by 2a now makes sense sincea is a complex number (assumed to be non-zero).Here is the second approach, which can be foundin an algebra book by Richard A. Dean (and probablymany other algebra books). The idea is to treatthe a, b, c as indeterminates. Thus, let a, b, cbe indeterminate variables. Let C[a,b,c] be thering of polynomials in a,b,c over the complexnumbers C. Note C[a,b,c] is an integral domain.Let K = C(a,b,c) be its quotient eld.Note a, b, c are elements of the eld K.Let K[x] be the ring of polynomials over the eld K.Then, P(x) = ax^2 + bx + c is a polynomial in K[x].Now, I can derive the quadratic formuala for theroots because division by 2a now makes sense sincea is a non-zero element of the eld K.To handle the square root, you need to takea nite extension E of K of degree 2.-- Bill Hale =>Let G be a nite group with a Sylow p-subgroup P that lies in>Z(G), the center of G. Is there an elementary way to show that P>has a normal complement in G?I know that since P lies in Z(N_G(P))), the center of the>normalizer in G of P, this can be shown by invoking the Burnside>Normal p-Complement Theorem, but Im wondering if the much>stronger condition that P is in Z(G) allows for a proof that>doesnt involve transfer.>I do not know of a proof that does not involve transfer or the Schur-Zassenhaus Theorem. They are of similar difculty in that they bothinvolve a summation argument over the elements of a group. All youneed to show is that the elements of order coprime to p in your groupform a subgroup, but I cannot see how to do that without using asummation argument!Anyway, here is a reasonably short proof of your special case, which isreally just proving the S.-Z. Theorem in the case of trivial action.(As Anvita pointed out, you could reduce to the case when P has orderp, but I still dont see how that helps!)Choose a transversal T of P in G, and for g in G, let r(g) denote theelement in T with gP = r(g)P. Now dene a map phi:G -> P by:phi(g) = Pi_{t in T} gt r(gt)^-1(Here Pi_{t in T} means the product over all t in T.)Claim phi is a homomorphism:phi(gh) = Pi_{t in T} ght r(ght)^-1 = Pi_{t in T} g ht r(ht)^-1 r(ht) r(ght)^-1 = (by centrality of P) Pi_{t in T} g r(ht) r(ght)^-1 ht r(ht)^-1 = Pi_{t in T} g r(ht) r(ght)^-1 Pi_{t in T} ht r(ht)^-1 = phi(g) phi(h)because, for xed h, a product over t in T is the same (but in a differentorder) as a product over r(ht) in T.Now, for g in P, we have r(gt) = t, so phi(g) = Pi_{t in T} g = g^|G:P|.Since |P| is coprime to |G:P|, phi restricted to P is an isomorphism ofThe proof of the general S.Z. Theorem reduces to the case when the normalsubgroup P is abelian. The argument above does not require muchmodication for the case when G does not act trivially on P.phi is dened as before. phi is no longer a homomorphism, but acrossed homomorphism (or derivation), which means that it satises phi(gh) = phi(g) (g.phi(h).g^{-1})The kernel of a crossed homomorphism is still a subgroup (although notnecessarily normal) and is still the required complement.Derek Holt. => Let G be a nite group with a Sylow p-subgroup P that lies in> Z(G), the center of G. Is there an elementary way to show that P> has a normal complement in G? I know that since P lies in Z(N_G(P))), the center of the> normalizer in G of P, this can be shown by invoking the Burnside> Normal p-Complement Theorem, but Im wondering if the much> stronger condition that P is in Z(G) allows for a proof that> doesnt involve transfer.I thought a little about your question. Since P is normal in G, all you need to show is that P has _any_ complement, say A, which will automatically make it a normal p-complement.Of course, we could use the Schur-Zassenhaus theorem for that, but you are looking for an elementary proof... I think some reduction is possible here to the case when P is a cyclic group of order p.First, we could factor out the Frattini subgroup of P to assume thatP is elementary abelian. Then assume that G acts irreducibly on P.Since P is in the center this ensures that P is cyclic.>> many years ago, Asplund & Bungart A First Course in Integration,