mm-1389 >Card(IN) = aleph_0 is said to be larger than any natural number. But >the number of natural numbers is counted by natural numbers. No, unless you are using a private definition of natural number. >Define: A_n = {1,2,3,...,n}. Card(A_n) = n. Card(A_n) < aleph_0 for >any n of IN. Card(A_omega) = aleph_0, You haven't defined A_omega. but omega is not a narural number. It does not belong to IN. Hence Card(IN) < aleph_0. That's a non sequitor. The first two statements are correct, but do not imply the third, which is false. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not === Subject: Re: WAR IMMINENT betweeen Mathematicians and Physicists! <41c1dff6$4$fuzhry+tra$mr2ice@news.patriot.net> <41c219d8.1752342@news.ecn.ab.ca> <41c4bd46$11$fuzhry+tra$mr2ice@news.patriot.net> <41c85169.4515008@news.ecn.ab.caWell, yes. I figured that the people reading my post would be >familiar with quaternions, The problem is that the other use of i.j k is more common. >Continuing that trend, I thought to use ! as a unary operator, not >meaning not (~ would do that: bitwise when applied to INTEGER >variables, logical when applied to LOGICAL variables) I like the PL/I convention of treating boolean values as a special case of bit strings. Thus &, | and å apply equally well to BIT(1) expressions, e.g., A & (B=C) and to integers converted to bit strings. Of course, Exclusive Or is useful enough to justify an operator, preferably by overloading the operator for logical Not, e.g., AåB for {A&åB)|(åA&B). >Unary * would be the complex conjugate, Don't forget the use of * for convolution transforms. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not === Subject: Re: domains with pairs of elements having no gcd by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iBN0iVp02364; > [.snip.] >>There has been much study of domains related to GCD domains. >>Below are some of them, in increasing order of generality. >> PID: every ideal is principal >> Bezout: every ideal (a,b) is principal >> GCD: (x,y) := gcd(x,y) exists for all x,y >> SCH: Schreier = pre-Schreier & integrally closed >> SCH0: pre-Schreier: a|bc => a = BC, B|b, C|c >I asked George Bergman about this recently, and he told me that this >property is often refered to by saying that the lattice of ideals of R >has the Riesz Interpolation Property with respect to principal >ideals; at least, when studied from the point of view of properties of >the lattice of ideals. > [.snip.] >>Recall that in an earlier post [1] I mentioned that Schreier domains >>are precisely those domains for which holds true the main result of >>the paper [5] of Magidin and Mckinnon. Reviewing that post reminds >>me that I forgot to supply the reference to Paul Cohn's paper [4] >>of 1968 which includes the main result of M&M. >>Hopefully Arturo will be able to cite Cohn's work >>in his Monthly paper -- it deserves to be better known. >We have submitted the corrected manuscript, but I will try to get my >hand on a copy of Cohn's paper so that an explicit reference can be >It's not denial. I'm just very selective about > what I accept as reality. > --- Calvin (Calvin and Hobbes) >Arturo Magidin >magidin@math.berkeley.edu Guys, I found this post rather late in the day. I gave the name pre-Schreier to the domains whose group of divisibility a Riesz group (i.e. in which a|bc implies that a=xy where x|b and y|c.). I apologize for doing that but there were several good reasons for writing a whole paper on pre-Schreier domains. The paper: On a property of pre-Schreier domains [Comm. Algebra,Comm. Algebra 15(1987), 1895-1920]. If you are in a hurry you can find a copy of that paper on my web page:www.lohar.com (click on Multiplicative Ideal Theory link and go for the lower part of the page.) In this paper I put together several ways of characterizing pre-Schreier domains and show (with an example) that if D is a proper pre-Schreier domain then the polynomial ring D[X] is not. In this paper I also show that while a pre-Schreier domain is characterized by its group of divisibility being a Riesz group, you cannot directly translate results from Riesz groups to Schreier domains. Let us now talk about Arthuro Magidin and David McKinnon's [M&M](reference below) BIG QUESTION: Can every polynomial with integer coefficients be factored into (not necessarily monic) linear terms each with algebraic integer coefficients? There are two answers to it: (1). Yes, because the ring of algebraic integers is a GCD domain with algebraically closed field of fractions. (2). Yes, because the ring of algebraic integers is a Schreier domain with algebraically closed field of fractions. The proofs are very short, if you know your algebra that is. What interests me most is the converse of (2). If every polynomial over D[X] splits (i.e. is linear or is a product of linear polynomials over D) then D is a pre-Schreier domains with algebraically closed field of fractions. The proof is again very simple. I might put the little note that I have written on my web page. Then they use a neat trick used by David Rush [R] without reference to his paper [R] which is to appear in Mathematica. (I have good reason to believe that one of the folks mentioned in the acknowldgements of [M&M] had requested and got a preprint of the Rush paper.) By the way, Paul Cohn's paper [C] has had quite a bit of attention if you care to look. Of my eighty papers nearly half mention it. Dan Anderson and I have written quite a bit using [C] as part of our study of various types of factorizations. Also by the way others such as F.Riesz had studied groups with interpolation property. I asked Paul why he had chosen to name the rings after Schreier rather than after Riesz. He said he chose to do it because Schreier did not have a ring named after him. [M&M]-A.Magidin and D. McKinnon, Gauss' Lemma for Number Fields, to appear in American Math Monthly. [A]. Dan Anderson, GCD domains, Gauss' Lemma, and Contents of polynomials, Non-Noetherian commutative ring theory, 1-32, Math. Appl., 520, Kluwer Acad. Publ., Dordrecht, 2000 [AQ]. Dan Anderson and R. Quintero, Some generalizations of GCD domains,Factorization in Integral Domains (ed. D.D. Anderson) Marcel Dekker, New York, 1997, 189-195. [R] D.E. Rush, Quadratic polynomials, factorization in integral domains ans Schreier domains from Pullbacks, Mathematika (to appear). [C]. Paul Cohn, Bezout rings and their subrings, Proc. Cambridge Philos. Soc. 64(1968) 251-264. Happy holidays and a happy and productive new year to you all, with genuine new results. Muhammad Zafrullah === Subject: product of 2 independent normally distributed random variables Let x and y be two independent, normally distributed random variables. Assume both have zero mean and, respectively, standard deviations sx and sy. What is the distribution of their product z = x y ? Is z also normal? === Subject: Re: product of 2 independent normally distributed random variables > Let x and y be two independent, normally distributed random > variables. Assume both have zero mean and, respectively, > standard deviations sx and sy. > What is the distribution of their product z = x y ? > Is z also normal? z is not normal. See http://mathworld.wolfram.com/NormalProductDistribution.html === Subject: A simple probability clarification Lets say I am picking n random numbers in the range 1-100 (x-y). What is the probabilty that two of the numbers will be the same? is it ((n*(n-1))/2)*(1/100)? using this logic, if i am picking 10 random numbers the probability is .55, which seems awfully high. also if I want three numbers to be the same i would have to do ((n*(n-1)*(n-3))/3)*(1/100), which does not make any sense. I can't really use poisson distribution to model this because there is no mean and the distribution is not gaussian, all the values in the random range are equally likely. Basically, how would i go about modelling this? i would like to see how the probability of two ( or x) random numbers being the same changes as i change the number of random numbers i pick (sample size, i would assume it would be directly proportional) and how it changes with relation to the range of random numbers i choose ( i would assume it would be inversly proportional). I have a feeling there is simple relation that ties it all together that i am missing. Any help would be much appreciated. === Subject: Re: A simple probability clarification >Lets say I am picking n random numbers in the range 1-100 (x-y). What >is the probabilty that two of the numbers will be the same? >is it ((n*(n-1))/2)*(1/100)? using this logic, if i am picking 10 >random numbers the probability is .55, which seems awfully high. That's a rough estimate, and it's .45, which is in the right ballpark. The true number is ~.3718. To get that, you can calculate the total number of outcomes (100^10) and the number of outcomes which have no 2 the same (100!/90!). Divide the latter by the former and you get ~.6282. Subtract that number from 1 for the 3718 value. >also if I want three numbers to be the same i would have to do >((n*(n-1)*(n-3))/3)*(1/100), which does not make any sense. 3 numbers is a little harder. Of the 100^10 outcomes, about choose(10,3) * 100 * 99^7 of them have 3 of the same number. That's roughly .01118. >Basically, how would i go about modelling this? i would like to see >how the probability of two ( or x) random numbers being the same >changes as i change the number of random numbers i pick (sample size, >i would assume it would be directly proportional) and how it changes >with relation to the range of random numbers i choose ( i would assume >it would be inversly proportional). You don't get p=1 until n>y, but it's extremely close for a long time before that. --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. === Subject: Re: A simple probability clarification > Lets say I am picking n random numbers in the range 1-100 > (x-y). What is the probabilty that two of the numbers will be the > same? > is it ((n*(n-1))/2)*(1/100)? using this logic, if i am picking 10 > random numbers the probability is .55, which seems awfully high. I'm just commenting on the feeling that this probability is awfully high. You can sort of test such intuitions by generating samples and seeing how often this happens. Here is a set of ten, sorted, and I've marked the positive cases: (5 6 9 11 32 46 48 55 78 86) (1 17 19 27 39 40 52 69 83 92) (12 14 14 16 20 42 45 84 93 98) * (12 34 45 48 55 66 68 70 88 96) (13 17 24 40 40 57 69 74 81 97) * (16 27 31 39 45 63 64 69 72 88) (7 28 31 34 38 43 46 63 66 99) (1 2 14 30 38 52 59 68 71 71) * (14 20 20 31 33 43 47 58 65 99) * (1 2 18 29 68 85 85 90 94 99) * === Subject: re:A simple probability clarification I was just about to post that your equation should be Prob(n different) = 100!/((100-n)!*(100)^n) when I realized that it actually was. *-----------------------* www.GroupSrv.com *-----------------------* === Subject: re:A simple probability clarification I am not sure where you got your formulas. Also when you say two are the same, does this include more than one pair or other multiples as well? However, for the problem where you are tring to get the probability of at least one pair, the answer is 1 - prob (all different). Prob(all different)=99!/((100-n)!*100^(n-1)) *-----------------------* www.GroupSrv.com *-----------------------* === Subject: Block Upper Triangular Matrix problem by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iBN1G1l04787; I have a block upper triangular matrix problem. How to convert an n*k matrix (n>=k) into a block upper triangular matrix by column operations and row permutations. The block upper triangular matrix has the form: [B_{1,1},B_{1,2},...,B_{1,k} 0, B_{2,2},...,B_{2,k} 0, 0 ,...,B_{3,k} ... ... ... ... 0, 0, ,B_{k,k}] Merry Christmas and Happy New Year to everyone! === Subject: HI-RESOLUTION VIDEOS OF 9/11 HI-RESOLUTION VIDEOS OF THE 9/11 HOAX. Video of the (obvious) demolition of World Trade Center Seven together with video of the (less obvious) demolition of North Tower. Hi-resolution video of the first plane hitting the North tower is also presented (this is footage taken by the Naudet brothers ([color=red:3ea18a2c5c]one of whom quite clearly knew the first aircraft was about to hit the North Tower[/color:3ea18a2c5c])). [color=red:3ea18a2c5c]HI-RESOLUTION VIDEO[/color:3ea18a2c5c] Here [color=blue:3ea18a2c5c]hi-resolution[/color:3ea18a2c5c] video means video recorded with the DivX3.11a codec with the (variable) bitrate set to 6000 and crispness/smoothness set to 100. The hi-resolution videos are much larger files than necessary for good viewing. They are meant to convey as much of the original detail as possible. You can download the hi-resolution videos of the 9/11 hoax from these links. The First Plane Hitting The North Tower (13 MB hi-resolution Codec: DivX3.11a 692x408) The Second Plane Hitting The South Tower (10 MB hi-resolution Codec: DivX3.11a 692x472) [url=http://thunderbay.indymedia.org/uploads/north-tower-demolition.avi]The North Tower Demolition[/url] (2 MB hi-resolution Codec: DivX3.11a 492x408) [url=http://thunderbay.indymedia.org/uploads/north-tower-shake.avi]The North Tower Demolition (longer version)[/url] (13 MB hi-resolution Codec: DivX3.11a 692x472) [url=http://thunderbay.indymedia.org/uploads/wtc-7-demolition.avi]The WTC Building Seven Demolition[/url] (0.9 MB hi-resolution Codec: DivX3.11a 692x408) [url=http://thunderbay.indymedia.org/uploads/wtc7-demolition-3.avi]The WTC Building Seven Demolition (another view)[/url] (0.8 MB hi-resolution Codec: DivX3.11a 692x408) [url=http://thunderbay.indymedia.org/uploads/wtc7-demolition-2.avi]The WTC Building Seven Demolition (yet another view)[/url] (9.5 MB hi-resolution Codec: DivX3.11a 692x408) [url=http://thunderbay.indymedia.org/uploads/north-demolition-pops.avi]Prema ture Detonations in North Tower Demolition[/url] (4.8 MB hi-resolution Codec: DivX3.11a 692x472) [url=http://thunderbay.indymedia.org/uploads/north-tower-pops.avi]More Premature Detonations in North Tower Demolition[/url] (6.1 MB hi-resolution Codec: DivX3.11a 696x472) [url=http://thunderbay.indymedia.org/uploads/woman-waving-close.avi]Woman Codec: DivX3.11a 692x356) The woman is in the lower righthand corner and [url=http://globalresearch.ca.myforums.net/viewtopic.php?t=688]why this is of interest.[/url] [url=http://globalresearch.ca.myforums.net/viewtopic.php?t=685]Premature Detonations.[/url] Here are some hi-resolution videos of the second plane hitting the South Tower. [url=http://thunderbay.indymedia.org/uploads/strike-two-1.avi]View from north-east[/url] (4.9 MB hi-resolution Codec: DivX3.11a 716x480) [url=http://thunderbay.indymedia.org/uploads/strike-two-2.avi]View from north[/url] (13 MB hi-resolution Codec: DivX3.11a 692x472) [url=http://thunderbay.indymedia.org/uploads/strike-two-3.avi]View from east[/url] (1.2 MB hi-resolution Codec: DivX3.11a 692x356) [url=http://thunderbay.indymedia.org/uploads/strike-two-4.avi]Close view from east (includes the 911 In Plane Site flash)[/url] (1.3 MB hi-resolution Codec: DivX3.11a 692x356) [url=http://thunderbay.indymedia.org/uploads/strike-two-5.avi]Another view from north[/url] (0.8 MB hi-resolution Codec: DivX3.11a 692x356) [url=http://thunderbay.indymedia.org/uploads/strike-two-6.avi]Short view from north-east[/url] (1.2 MB hi-resolution Codec: DivX3.11a 696x472) [url=http://thunderbay.indymedia.org/uploads/strike-two-7.avi]Longer view from north-east[/url] (2.8 MB hi-resolution Codec: DivX3.11a 716x480) If you are using windows you need the DivX3.11a codec plug-in for Windows Media Player (many non-Microsoft movie players come standard with the codecs necessary to play DivX movies (eg Mplayer (for Linux))). If you do not already have it, you can find it here: [url=http://public.planetmirror.com/pub/divx/windows/divx_3.11alpha.zip]http ://public.planetmirror.com/pub/divx/windows/divx_3.11alpha.zip[/url] The files from public.planetmirror.com are bit for bit the same (I checked) as those I have had on my system for years (with no harmful effects). You unzip the files and double click on Register_DivX.exe [color=blue:3ea18a2c5c]If you wish to learn about DivX movies, in particular, why they sometimes play upside-down, read this thread:[/color:3ea18a2c5c] [url=http://globalresearch.ca.myforums.net/viewtopic.php?t=599]http://global research.ca.myforums.net/viewtopic.php?t=599[/url] [color=red:3ea18a2c5c]OTHER VIDEO[/color:3ea18a2c5c] Painful Deceptions by Eric Hufschmid (windows media player comes with the necessary codec to play these). [url=http://www.reopen911.org/video/painful_deceptions-an_analysis_of_the_91 1_attack_part1.wmv]Painful Deceptions. Part 1[/url] (39 MB Codec: WMV1 360x240) [url=http://www.reopen911.org/video/painful_deceptions-an_analysis_of_the_91 1_attack_part2.wmv]Painful Deceptions. Part 2[/url] (42 MB Codec: WMV1 360x240) [url=http://www.reopen911.org/video/painful_deceptions-an_analysis_of_the_91 1_attack_part3.wmv]Painful Deceptions. Part 3[/url] (35 MB Codec: WMV1 360x240) [url=http://www.reopen911.org/video/painful_deceptions-an_analysis_of_the_91 1_attack_addendum.wmv]Painful Deceptions. Part 4[/url] (28 MB Codec: WMV1 360x240) [url=http://thunderbay.indymedia.org/uploads/inplanesitedebunked.avi]911 In Plane Site Debunked[/url] and also [url=http://www.flashclub.ch/videos/inplanesitedebunked.avi]here[/url] (25 MB Codec: DivX3.11a 360x240) In Plane Site.[/url] (104 MB Codec: DivX5 352x240) In Plane Site.[/url] (129 MB Codec: Quicktime MOV 352x240) [url=http://thunderbay.indymedia.org/uploads/premature-detonations.avi]Prema ture Detonations in North Tower Demolition.[/url] (1.1 MB Codec: DivX3.11a 692x472, premature detonation marked, 3fps). Here are frames from a few of the videos (at actual size): http://victoria.indymedia.org/uploads/frame-564-naudet.jpg [img:3ea18a2c5c]http://thunderbay.indymedia.org/uploads/wtc7-frame-40.jpg[/i mg:3ea18a2c5c] [img:3ea18a2c5c]http://thunderbay.indymedia.org/uploads/nt-frame-115.jpg[/im g:3ea18a2c5c] [color=magenta:3ea18a2c5c][size=18:3ea18a2c5c]MORE 9/11 RESOURCES (ANIMATIONS, BOOKS ETC).[/size:3ea18a2c5c][/color:3ea18a2c5c] [color=red:3ea18a2c5c]BOOKS:[/color:3ea18a2c5c] Pearl Harbor[/url] by David Ray Griffin (entire book). [color=red:3ea18a2c5c]ANIMATED GRAPHICS:[/color:3ea18a2c5c] [url=http://thunderbay.indymedia.org/uploads/premature-detonations.gif]Prema ture Detonations in the World Trade Center Demolitions[/url] 0.6 MB. [url=http://thunderbay.indymedia.org/uploads/north-tower.gif]North Tower Demolition[/url] 0.9 MB and [url=http://globalresearch.ca.myforums.net/viewtopic.php?t=313]commentary.[/ url] [url=http://thunderbay.indymedia.org/uploads/south-tower.gif]South Tower Demolition[/url] 2.1 MB and [url=http://globalresearch.ca.myforums.net/viewtopic.php?t=663]commentary.[/ url] [url=http://thunderbay.indymedia.org/uploads/plane-hits-north-tower.gif]Firs t [url=http://globalresearch.ca.myforums.net/viewtopic.php?t=651]why this is of interest.[/url] [url=http://thunderbay.indymedia.org/uploads/jet-blast.gif]What a jets exhaust does to a car[/url] 0.9 MB and [url=http://globalresearch.ca.myforums.net/viewtopic.php?t=682]why this is of interest.[/url] [url=http://thunderbay.indymedia.org/uploads/wtc7-demolition.gif]World Trade Center Seven Demolition[/url] 0.6 MB. [url=http://thunderbay.indymedia.org/uploads/euronews.gif]The dust cloud that gave rise to the WTC6 explosion myth[/url] 0.9 MB and [url=http://globalresearch.ca.myforums.net/viewtopic.php?t=656]why this is of interest.[/url] [color=red:3ea18a2c5c]9/11 WEB-SITES[/color:3ea18a2c5c] These are deliberately suppressed, so I specially mention them. [url=http://911review.org/Wget/members.fortunecity.com/911/]http://911review .org/Wget/members.fortunecity.com/911/[/url] [url=http://911review.org/Wget/www.nerdcities.com/guardian/]http://911review .org/Wget/www.nerdcities.com/guardian/[/url] [url=http://guardian.150m.com]http://guardian.150m.com[/url] (has smaller images for faster downloads) [url=http://guardian.250free.com]http://guardian.250free.com[/url] (has smaller images for faster downloads) [url=http://thewebfairy.com/nerdcities/]http://thewebfairy.com/nerdcities/[/ url] Please add to this list here [url=http://globalresearch.ca.myforums.net/viewtopic.php?t=679]http://global research.ca.myforums.net/viewtopic.php?t=679[/url] [color=red:3ea18a2c5c](No Login Necessary).[/color:3ea18a2c5c] *-----------------------* www.GroupSrv.com *-----------------------* === Subject: Very strange people here, indeed. So explain to me why this is not of interest? Very strange people here, indeed. *-----------------------* www.GroupSrv.com *-----------------------* === Subject: Re: Very strange people here, indeed. > So explain to me why this is not of interest? > Very strange people here, indeed. What does it have to do with math? This is not alt.fan.conspiricy.theory -- Will Twentyman email: wtwentyman at copper dot net === Subject: re:HI-RESOLUTION VIDEOS OF 9/11 Oh here it is. I thought it had been deleted. I meant to post this in the Physics section,.... *-----------------------* www.GroupSrv.com *-----------------------* === Subject: Re: Question about President's Social Security plan Stocks historically outperform ALL other investments. President Bush wants retired Americans to have a more comfortable lifestyle. Why is the liberal media hounding him about that? -- James Slaughter, Professor of Biblical Study Criswell Institute of Creation Science Criswell College Dallas, Texas USA www.criswell.edu === Subject: Re: Question about President's Social Security plan >Stocks historically outperform ALL other investments. President >Bush wants retired Americans to have a more comfortable >lifestyle. Why is the liberal media hounding him about that? >-- >James Slaughter, Professor of Biblical Study >Criswell Institute of Creation Science >Criswell College >Dallas, Texas USA >www.criswell.edu how does one resist commenting? === Subject: Re: Question about President's Social Security plan > Stocks historically outperform ALL other investments. You might want to double check your numbers. > President > Bush wants retired Americans to have a more comfortable > lifestyle. Why is the liberal media hounding him about that? President Bush wants to phase out Social Security. === Subject: Re: Question about President's Social Security plan >> Stocks historically outperform ALL other investments. >You might want to double check your numbers. What do you know that nobody else does? >President Bush wants to phase out Social Security. Which is a good thing. -- A society that robs an individual of the product of his effort... is... a mob held together by institutionalized gang rule. -- Ayn Rand === Subject: Re: Question about President's Social Security plan >> >> Stocks historically outperform ALL other investments. >You might want to double check your numbers. > What do you know that nobody else does? Nobody else? Or just you and Prof Slaughter? >President Bush wants to phase out Social Security. > Which is a good thing. It's a good thing if you don't need it. === Subject: Re: Question about President's Social Security plan >>> >>> Stocks historically outperform ALL other investments. >>You might want to double check your numbers. >> What do you know that nobody else does? >Nobody else? Or just you and Prof Slaughter? Ok, do you have any official numbers that show stocks being outperformed by something else? >>President Bush wants to phase out Social Security. >> Which is a good thing. >It's a good thing if you don't need it. Theft is always a bad thing, whether you need it or not. -- A society that robs an individual of the product of his effort... is... a mob held together by institutionalized gang rule. -- Ayn Rand === Subject: Re: Question about President's Social Security plan > >> Stocks historically outperform ALL other investments. You might want to double check your numbers. > > What do you know that nobody else does? >>Nobody else? Or just you and Prof Slaughter? >Ok, do you have any official numbers that show stocks being >outperformed by something else? First you have to identify a non-prescient buy-hold-sell strategy. >President Bush wants to phase out Social Security. > > Which is a good thing. >>It's a good thing if you don't need it. >Theft is always a bad thing, whether you need it or not. Not when it redresses a theft. -- Roy L === Subject: Re: Question about President's Social Security plan >Stocks historically outperform ALL other investments. That claim is false, and is based on assuming the stock owner makes prescient buy-hold-sell decisions. The fact is, if you bought stocks and held them, you'd go broke. Very few of the big-cap companies of former times even exist today. >President >Bush wants retired Americans to have a more comfortable >lifestyle. Why is the liberal media hounding him about that? Because he _actually_ wants the _current_owners_ of the assets that privatized SS contributions would be buying to have a more comfortable lifestyle -- though I don't know how your average billionaire is going to be any more comfortable with double the asset value. >James Slaughter, Professor of Biblical Study >Criswell Institute of Creation Science Ah. An authority on the science of economics... -- Roy L === Subject: Re: Question about President's Social Security plan >Stocks historically outperform ALL other investments. > That claim is false, and is based on assuming the stock owner makes > prescient buy-hold-sell decisions. > The fact is, if you bought stocks and held them, you'd go broke. Very > few of the big-cap companies of former times even exist today. It's not everything, but the Dow Jones Industrial 30 are 3M, Alcoa, Altria (tobacco), American Express, American International Group (insurance), Boeing, Caterpillar, Citigroup, Coca-Cola, DuPont, Exxon Mobil, General Electric, General Motors, Hewlett-Packard, Home Depot, Honeywell, Intel, IBM, Johnson and Johnson, JP Morgan Chase, McDonald's, Merck, Microsoft, Pfizer, Procter & Gamble, SBC Communications, United Technologies, Verizon, Wal-Mart, and Walt Disney. All are listed by DJ as large cap. (Note that not all are given the same weight. DJ tries to keep a consistent average when replacing one company with another, so they multiply the stock price accordingly.) The typical working career span of an American worker is 45 years. That long ago, some of these companies did not exist (Home Depot, Intel, Microsoft) were not considered to be big (Disney, Mal-Mart) or were known under a different name because of splits and mergers (Exxon Mobil, SBC, Altria). However, tracking the DJIA strategy has been shown to make money over the long run. === Subject: Re: Question about President's Social Security plan >>Stocks historically outperform ALL other investments. >> That claim is false, and is based on assuming the stock owner makes >> prescient buy-hold-sell decisions. >> The fact is, if you bought stocks and held them, you'd go broke. Very >> few of the big-cap companies of former times even exist today. >It's not everything, but the Dow Jones Industrial 30 are 3M, Alcoa, Altria >(tobacco), American Express, American International Group (insurance), >Boeing, Caterpillar, Citigroup, Coca-Cola, DuPont, Exxon Mobil, General >Electric, General Motors, Hewlett-Packard, Home Depot, Honeywell, Intel, >IBM, Johnson and Johnson, JP Morgan Chase, McDonald's, Merck, Microsoft, >Pfizer, Procter & Gamble, SBC Communications, United Technologies, Verizon, >Wal-Mart, and Walt Disney. All are listed by DJ as large cap. (Note that >not all are given the same weight. DJ tries to keep a consistent average >when replacing one company with another, so they multiply the stock price >accordingly.) >The typical working career span of an American worker is 45 years. And he will typically be retired for another 20. 65 years ago, it was 1939, the year IBM was _dropped_ from the DJIA -- and it was not until _1979_, after almost all its growth was finished, that IBM rejoined the Dow. If you followed the Dow and sold IBM in 1939 and bought it in 1979, your investment would not even keep pace with inflation. Very much the same sort of thing has happened with other companies: inclusion in a major index, especially the Dow, is often virtually the kiss of death for a company's growth. What has M$ done since it joined the Dow, compared to what it did before? >That >long ago, some of these companies did not exist (Home Depot, Intel, >Microsoft) were not considered to be big (Disney, Mal-Mart) And McDonald's and H-P. >or were known >under a different name because of splits and mergers (Exxon Mobil, SBC, >Altria). However, tracking the DJIA strategy has been shown to make money >over the long run. Oh, it does make money. Just not more than all other major classes of investment. Land does better than stocks, because aggregate land rent grows faster than GDP, while the major stocks grow slower than GDP (because successful companies rarely grow faster than GDP after they become big). It's no accident that most wealthy people got that way by owning land, not stocks. -- Roy L === Subject: Re: Question about President's Social Security plan > Stocks historically outperform ALL other investments. President > Bush wants retired Americans to have a more comfortable > lifestyle. Why is the liberal media hounding him about that? > -- > James Slaughter, Professor of Biblical Study > Criswell Institute of Creation Science > Criswell College > Dallas, Texas USA > www.criswell.edu Aside from other issues, one thing you did not mention is that if we choose to do so money from the SS trust fund (or the FICA stream) could be invested in equities today much as pension funds do. In other words, private accounts and equities are independent of each other. Bill === Subject: Re: Question about President's Social Security plan > Stocks historically outperform ALL other investments. President Talk to the people in Japan who lost a good chunk of their retirement savings when Japanese stocks lost 2/3s of their value (30,000 to 10,000) about 15 years ago now. Anyone know what the Nikkei index is now? === Subject: Re: Question about President's Social Security plan >> Stocks historically outperform ALL other investments. President >Talk to the people in Japan who lost a good chunk of their retirement >savings when Japanese stocks lost 2/3s of their value (30,000 to 10,000) >about 15 years ago now. Anyone know what the Nikkei index is now? That's why smart people diversify and buy both domestic and international securities. In any case, people still have the option of investing on good old Treasury bonds. -- A society that robs an individual of the product of his effort... is... a mob held together by institutionalized gang rule. -- Ayn Rand === Subject: Re: Question about President's Social Security plan >Stocks historically outperform ALL other investments. President >>Talk to the people in Japan who lost a good chunk of their retirement >>savings when Japanese stocks lost 2/3s of their value (30,000 to 10,000) >>about 15 years ago now. Anyone know what the Nikkei index is now? > That's why smart people diversify and buy both domestic and > international securities. In any case, people still have the option of > investing on good old Treasury bonds. Agreed. But I was referring to the stocks historically outperform ALL other investments comment. Anyways, the decline in Japan stocks would have hit even the smart people hard. === Subject: Re: Question about President's Social Security plan >>Stocks historically outperform ALL other investments. President >Talk to the people in Japan who lost a good chunk of their retirement >savings when Japanese stocks lost 2/3s of their value (30,000 to 10,000) >about 15 years ago now. Anyone know what the Nikkei index is now? >> That's why smart people diversify and buy both domestic and >> international securities. In any case, people still have the option of >> investing on good old Treasury bonds. >Agreed. But I was referring to the stocks historically outperform ALL >other investments comment. Anyways, the decline in Japan stocks would >have hit even the smart people hard. Actually, smart people would have a diversified portfolio, invested in stocks, bonds, precious metals, real estate, specially if you are less than 10 years away from retirement. -- A society that robs an individual of the product of his effort... is... a mob held together by institutionalized gang rule. -- Ayn Rand === Subject: Re: Question about President's Social Security plan > >Stocks historically outperform ALL other investments. President >>Talk to the people in Japan who lost a good chunk of their retirement >>savings when Japanese stocks lost 2/3s of their value (30,000 to 10,000) >>about 15 years ago now. Anyone know what the Nikkei index is now? > > That's why smart people diversify and buy both domestic and > international securities. In any case, people still have the option of > investing on good old Treasury bonds. >>Agreed. But I was referring to the stocks historically outperform ALL >>other investments comment. Anyways, the decline in Japan stocks would >>have hit even the smart people hard. >Actually, smart people would have a diversified portfolio, invested in >stocks, bonds, precious metals, real estate, specially if you are less >than 10 years away from retirement. Japanese investors also got hammered on bonds, real estate and gold. Just how prescient do you have to be, hmmmm? -- Roy L === Subject: Re: Question about President's Social Security plan > >Stocks historically outperform ALL other investments. President >>Talk to the people in Japan who lost a good chunk of their retirement >>savings when Japanese stocks lost 2/3s of their value (30,000 to 10,000) >>about 15 years ago now. Anyone know what the Nikkei index is now? > > That's why smart people diversify and buy both domestic and > international securities. In any case, people still have the option of > investing on good old Treasury bonds. >>Agreed. But I was referring to the stocks historically outperform ALL >>other investments comment. Anyways, the decline in Japan stocks would >>have hit even the smart people hard. >Actually, smart people would have a diversified portfolio, invested in >stocks, bonds, precious metals, real estate, specially if you are less >than 10 years away from retirement. > Japanese investors also got hammered on bonds, real estate and gold. > Just how prescient do you have to be, hmmmm? 20-20 hindsight prescient. === Subject: Re: Question about President's Social Security plan > Stocks historically outperform ALL other investments. President >>Talk to the people in Japan who lost a good chunk of their retirement >>savings when Japanese stocks lost 2/3s of their value (30,000 to 10,000) >>about 15 years ago now. Anyone know what the Nikkei index is now? >That's why smart people diversify and buy both domestic and >international securities. In any case, people still have the option of >investing on good old Treasury bonds. Most people, frankly, ain't that bright. That's one of the nice things about social security right now - you don't have to be that bright or know how to manage an investment to get money out of it. I'm disinclined to punish those people to provide a benefit to people like you and me (since I know at least one isn't relying on social security for his retirement). Alan -- Defendit numerus === Subject: Re: Question about President's Social Security plan >> Stocks historically outperform ALL other investments. President >Talk to the people in Japan who lost a good chunk of their retirement >savings when Japanese stocks lost 2/3s of their value (30,000 to 10,000) >about 15 years ago now. Anyone know what the Nikkei index is now? >>That's why smart people diversify and buy both domestic and >>international securities. In any case, people still have the option of >>investing on good old Treasury bonds. >Most people, frankly, ain't that bright. That's one of the nice things about >social security right now - you don't have to be that bright or know how to >manage an investment to get money out of it. I'm disinclined to punish those >people to provide a benefit to people like you and me (since I know at least one >isn't relying on social security for his retirement). And I'm disinclined in dumbing down everything and, in essence, being the one punished for other people's inability to manage their own life. -- A society that robs an individual of the product of his effort... is... a mob held together by institutionalized gang rule. -- Ayn Rand === Subject: Re: Question about President's Social Security plan Quoth Professor James Slaughter in > Stocks historically outperform ALL other investments. Except when they don't. > President Bush wants retired Americans to have a more comfortable > lifestyle. B.S. Bush wants to default on the treasuries that the SS Trust Fund is tied up in. He wants to pretend his record-breaking deficits don't exist. He's using fuzzy math to pretend that there's a problem with Social Security. There' isn't a problem with Social Security -- but there IS a crisis in the General Fund, a crisis of George Bush's making. And that's a crisis he doesn't want to fix, because it would mean rolling back the tax cuts he gave his corporate billionare buddies. And there is no liberal media to speak of. -- George Bush's War of Choice on Iraq is a totally unnecessary war. Every life lost, every limb lost, every disfigurement, every disability caused there is more blood on George W. Bush's hands, and on the hands of everyone who voted for George W. Bush. The more you know, the less likely you were to vote for Bush. Feeling a draft? For the facts on Iraq, see . === Subject: Re: Question about President's Social Security plan >I've seldom seen anyone mention what I think is the >real reason for the Bushist Social Security plan. I've explained it a few times here on sci.econ. >As social security money moves from the Trust Fund >(U.S. Treasury bonds) to private accounts (stocks), >there will be a strong upward pressure on U.S. stock >prices. This will be a windfall for those who already >own stocks -- i.e. Bush's constituency -- though those >the program supposedly helps will be buying stocks at >inflated prices. Right. It's purely a plan to shovel vast unearned wealth into the pockets of those who currently own what the privatized SS contributions would be buying. >Every dollar diverted to the private accounts, whether >subtracted from the Trust Fund or from current retirement >payouts, will represent a dollar of increased (non Trust >Fund) Treasury borrowing, so interest rates will almost >certainly rise and the supposed reason for the plan >(stocks outperform bonds) will prove tragically false. Well, one good thing about privatizing SS: it would finally eliminate the superstition that stocks are always the best long-term investment. >(Even after the 2000 crash, stocks are always a good >long-term investment is prattled as a law of nature, >with NYSE history cited as proof. Try the same long-term >historic study with German stocks!) Or Japanese or British or French or etc. ones. While a diverse portfolio of stocks in a politically stable capitalist country has usually been a reasonably good investment -- and occasionally a spectacularly good one -- I have never seen an explanation of how stocks yield better long-term returns than all other investments that did not assume some sort of prescient buy-hold-sell strategy. -- Roy L === Subject: Re: Question about President's Social Security plan > Ideally you take care of this by increasing the contributions slightly > well in advance and restraining the growth of the program to > inflation. If we had done this back in, oh, 1983 the system would be > solvent. Unfortunately we only did *half* of this: We raised FICA > taxes and declared we had a surplus. But Congress never reined in > entitlements and then spent the FICA surplus on other things. > Bottom line is the surplus disappears from the books around 2011 and > from that point on the system runs in the red. And it only gets > *redder* the further out you go, with too few workers paying into the > system to cover benefits going out. I am puzzled by the politics of this debate. Bush wants to (in effect) reduce the FICA now with the restriction that the tax reduction be invested/saved for future retirement. Why do those who recognize that FICA is simply a regressive tax on wages not support Bush on this? now collects much more than current SS benefits paid out. What about future SS benefits? They would be reduced for those who took the FICA reduction now. Why are stocks and bonds a better source of income in your retirement than a FICA tax on the wages of American workers? Because the US population is aging faster than the world population. A demographic crisis will hit the US at least 30 years before it hits the majority of the human race (depending of course on many assumptions). Retirement plans based on corporate stocks and bonds are based on the increasingly multi-national/ international global business community and the world population. Sure the same demographics will get to that population eventually. While I may worry about a 2020 shortfall, I don't lose sleep over a 2050 shortfall. Not yet, anyway. Consider an analogy: The population in Florida or Arizona is aging because of an influx of retirees from all over the US. If they were supported ONLY by taxes on residents of their state, they would be a drag on the state economy. Instead they are a boom to those states. Why? I say because they are being supported by money from a larger and younger population through pensions, SS, IRA's, etc. Replace Arizona with USA, and larger and younger population with global economy. >... The most basic issue is >whether the nation's productive assets will continue to grow at a rate >that will render the issue less threatening. >sjfromm I agree, if you change that to world's productive assets. I say invest now to provide capital for the future. ,,,,,,, _______________ooo___(_O O_)___ooo_______________ (_) jim blair (jeblair@facstaff.wisc.edu) Madison Wisconsin USA. This message was brought to you using biodegradable binary bits, and 100% recycled bandwidth. For a good time call: http://www.geocities.com/capitolhill/4834 === Subject: Re: Question about President's Social Security plan >> Ideally you take care of this by increasing the contributions slightly >> well in advance and restraining the growth of the program to >> inflation. If we had done this back in, oh, 1983 the system would be >> solvent. Unfortunately we only did *half* of this: We raised FICA >> taxes and declared we had a surplus. But Congress never reined in >> entitlements and then spent the FICA surplus on other things. >> Bottom line is the surplus disappears from the books around 2011 and >> from that point on the system runs in the red. And it only gets >> *redder* the further out you go, with too few workers paying into the >> system to cover benefits going out. > I am puzzled by the politics of this debate. Bush wants to (in effect) > reduce the FICA now with the restriction that the tax reduction be > invested/saved for future retirement. Why do those who recognize that > FICA is simply a regressive tax on wages not support Bush on this? Because the thieving Republican is simply directing funds into the pockets of the Wall Street types that financed his campaign. There is no reduction in the regressive tax called FICA, it is merely a way to rob the middle class -- again. > now collects much more than current SS benefits paid out. No. Privatization of SS is very easy: You do away with the FICA tax and raise taxes across the board to make up for whatever is needed to pay benefits. And you never look back. > What about future SS benefits? They would be reduced for those who took > the FICA reduction now. There is no FICA reduction. Just a shoveling of funds over to Wall Street. > Why are stocks and bonds a better source of income in your retirement than > a FICA tax on the wages of American workers? Why do you think that ONLY workers should have to pay to support the retired people? In fact, we the people, end up paying one way or another: We pay inflated prices so that the retired owners of stocks and bonds can get their income, or we pay taxes so that the retired non owners of stocks and bonds can get their income. <<<<<<< -- I know no safe depository of the ultimate powers of society but the people themselves; and if we think them not enlightened enough to exercise their control with a wholesome discretion, the remedy is not to take it from them, but to inform their discretion by education. - Thomas Jefferson. http://GreaterVoice.org === Subject: Re: Question about President's Social Security plan > I am puzzled by the politics of this debate. Bush wants to (in effect) > reduce the FICA now with the restriction that the tax reduction be > invested/saved for future retirement. Why do those who recognize that FICA > is simply a regressive tax on wages not support Bush on this? Because a simple way to fix the regressivity problem is to lift the cap and cut the rate. Dubya is not proposing that. So his goal is something different and, to some of us, less desirable than even the status quo. > ... > Replace Arizona with USA, and larger and younger population with > global economy. Japan and Europe are aging at least as much as the US. The places growing larger and younger populations are the Arab world and the Muslim world more generally. Your analogy requires a certain level of goodwill between the graying Arizonas and the productive Winsconsins of the world. So replace global economy with a gray Japan, a gray Europe, and however productive an Iraq as you dare to imagine. See if you think you can retire on _those_ returns. -- TP === Subject: Re: Question about President's Social Security plan > I am puzzled by the politics of this debate. Bush wants to (in effect) > reduce the FICA now with the restriction that the tax reduction be > invested/saved for future retirement. Why do those who recognize that > FICA > is simply a regressive tax on wages not support Bush on this? > Because a simple way to fix the regressivity problem is to lift the cap and > cut the rate. But regressivity is not the major problem with the US Social Security system. Sure the FICA is regressive, but the benefit payments are progressive in the sense of being a better deal for the lower incomes. But that point is debatable. But the big problem with US SS is demographic. Your suggestion does nothing to address that. >...Dubya is not proposing that. So his goal is something > different .... Yes, he wants to deal with the large and clear problem, not the minor and debatable one. >...and, to some of us, less desirable than even the status quo. The status quo is an ever higher tax on the wages an ever smaller fraction of the population to fund ever smaller benefits for and ever growing fraction of the population. And likely an ever later retirement age. Is that really the less desirable option? > ... > Replace Arizona with USA, and larger and younger population with > global economy. > Japan and Europe are aging at least as much as the US. Yes. And collectively they are what fraction of the world population? >...The places growing > larger and younger populations are the Arab world and the Muslim world > more generally. And China and India and Mexico and South America. >...Your analogy requires a certain level of goodwill between > the graying Arizonas and the productive Winsconsins of the world. Well there are still some people In Wisconsin who speak English. >...So > replace global economy with a gray Japan, a gray Europe, and however > productive an Iraq as you dare to imagine. See if you think you can retire > on _those_ returns. > -- TP I think the question is how likely are people in China, India, Mexico, Asia, Central & South America and yes even the Middle East to be included in a global economy? While that may not be a sure thing, the destruction of Social Security by demographics is about as certain as any future prediction can be. The only alternative to massive numbers of young foreign workers moving to the USA, is expanding the global economy to include them in their home countries. ,,,,,,, _______________ooo___(_O O_)___ooo_______________ (_) jim blair (jeblair@facstaff.wisc.edu) Madison Wisconsin USA. This message was brought to you using biodegradable binary bits, and 100% recycled bandwidth. For a good time call: http://www.geocities.com/capitolhill/4834 === Subject: Re: Question about President's Social Security plan >> I am puzzled by the politics of this debate. Bush wants to (in effect) >> reduce the FICA now with the restriction that the tax reduction be >> invested/saved for future retirement. Why do those who recognize that >> FICA >> is simply a regressive tax on wages not support Bush on this? >> Because a simple way to fix the regressivity problem is to lift the cap > and >> cut the rate. > But regressivity is not the major problem with the US Social Security > system. That is, of course, quite wrong indeed. > Sure the FICA is regressive, but the benefit payments are > progressive in the sense of being a better deal for the lower incomes. > But that point is debatable. The obfuscation will be ignored. > But the big problem with US SS is demographic. Your suggestion does > nothing to address that. The big problem with SS is that the Republicans (conservatives all) refuse to invest in anything that would improve infrastructure. They go in search of EEEEEEEEEEEEEEEEEEEEEVIL around the world so we can burn the production of the USA in a war furnace. If we invested in our future as opposed to making war so as to keep Pinocchio Bush in office as commander in thief, we would have plenty of tax revenue with which to pay SS benefits. >>...Dubya is not proposing that. So his goal is something >> different .... > Yes, he wants to deal with the large and clear problem, not the minor and > debatable one. Nope. Bush wants to shovel more money to the rich people of Wall Street who will waste it on their own opulence. Same as the war furnace. >>...and, to some of us, less desirable than even the status quo. > The status quo is an ever higher tax on the wages an ever smaller > fraction > of the population to fund ever smaller benefits for and ever growing > fraction of the population. And likely an ever later retirement age. > Is that really the less desirable option? The guy GAVE you the proper options Blair. He said remove the cap and cut the rate. You see, B-Liar, that actually addresses the problem as opposed to compromising the system of making matters worse. >> ... >> Replace Arizona with USA, and larger and younger population with >> global economy. >> Japan and Europe are aging at least as much as the US. > Yes. And collectively they are what fraction of the world population? Irrelevant. They do not seem to be terribly worried about their retirement systems. Why is Bush so very interested in this subject and not the least concerned about the unfunded obligations he is creating in the form of deficits? >>...The places growing >> larger and younger populations are the Arab world and the Muslim world >> more generally. > And China and India and Mexico and South America. The B-Liar plan is to collect rent form the third world and use that to pay the bills. Just a little more Repugnican slavery. >>...Your analogy requires a certain level of goodwill between >> the graying Arizonas and the productive Winsconsins of the world. > Well there are still some people In Wisconsin who speak English. Republican to the core. >>...So >> replace global economy with a gray Japan, a gray Europe, and however >> productive an Iraq as you dare to imagine. See if you think you can > retire >> on _those_ returns. >> -- TP <<< deletia >> Just can't stomach any more....... -- I know no safe depository of the ultimate powers of society but the people themselves; and if we think them not enlightened enough to exercise their control with a wholesome discretion, the remedy is not to take it from them, but to inform their discretion by education. - Thomas Jefferson. http://GreaterVoice.org === Subject: Re: Question about President's Social Security plan >> Ideally you take care of this by increasing the contributions slightly >> well in advance and restraining the growth of the program to >> inflation. If we had done this back in, oh, 1983 the system would be >> solvent. Unfortunately we only did *half* of this: We raised FICA >> taxes and declared we had a surplus. But Congress never reined in >> entitlements and then spent the FICA surplus on other things. >> Bottom line is the surplus disappears from the books around 2011 and >> from that point on the system runs in the red. And it only gets >> *redder* the further out you go, with too few workers paying into the >> system to cover benefits going out. >I am puzzled by the politics of this debate. Bush wants to (in effect) >reduce the FICA now with the restriction that the tax reduction be >invested/saved for future retirement. WHOA ! saved for future retirement?? How does we do that? Put $100 bills in a safe? Piles of gold? Personally I'd prefer they freeze lots of corn and beans -- that I could eat in the future. (I'll read no further. Sorry, Jim.) Mason C === Subject: Re: Diophantine equation > This is an exercise in a textbook: > Use Gaussian integers to solve the Diophantine equation > x^2+y^2=z^3. > Hmmmm....I haven't seen this one before, and I'm a bit > puzzled. It seems like there are lots of solutions, > since if z is representable as the sum of two squares > then so is z^3 and we should be able to find a > parameterization along those lines. I haven't been able > to get anywhere with it, so I'm wondering what y'all > might say. If x +iy is a cube, say (a +ib)^3, this will give you the parameterization you want, subject to appropriate conditions > Another, unrelated, question is that I need to express > a prime p = 1 (mod 3) in the form u^2+3v^2. Certainly > -3 is a quadratic residue mod p, which gives me > something like a^2+3=kp. But I'm not getting anywhere > after that. If k is expressible in the form u^2+3v^2 and k

This is an exercise in a textbook: >Use Gaussian integers to solve the Diophantine equation >x^2+y^2=z^3. >Hmmmm....I haven't seen this one before, and I'm a bit >puzzled. It seems like there are lots of solutions, >since if z is representable as the sum of two squares >then so is z^3 and we should be able to find a >parameterization along those lines. I haven't been able >to get anywhere with it, so I'm wondering what y'all >might say. Since z < 0 is impossible and z = 0 easy, I'll assume z > 0. Suppose z^3 = x^2 + y^2 = (x+yi)(x-yi), and we factor x+yi into Gaussian primes. Of course each prime factor must occur in the factorization of z; if we have p+qi with index j in x+yi and p-qi with index k, then p+qi and p-qi occur with indices k and j respectively in x-yi, and j+k must be divisible by 3. So what you can do is look at a factorization of z into Gaussian primes: z = product_j p_j^(d_j) where u is a Gaussian unit. Since z > 0, this can be done so that whenever a non-rational Gaussian prime p_j occurs here, so does conjugate(p_j) with the same index. For each rational p_j, if d_j is odd the representation is impossible. If d_j is even, let q_j = p_j^(3 d_j/2). For each j such that p_j = a + bi with b > 0, let q_j = (a+bi)^{m_j} (a-bi)^(3 d_j - m_j) where 0 <= m_j <= 3 d_j. For each j such that p_j = a + bi with b < 0, let q_j = 1. Then if x+iy = (+/-) prod_j q_j, we have z^3 = (x+iy)(x-iy) = x^2 + y^2. And this should obtain every possible representation of z^3 as the sum of two squares. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Diophantine equation > This is an exercise in a textbook: > Use Gaussian integers to solve the Diophantine equation > x^2+y^2=z^3. > Hmmmm....I haven't seen this one before, and I'm a bit > puzzled. It seems like there are lots of solutions, > since if z is representable as the sum of two squares > then so is z^3 and we should be able to find a > parameterization along those lines. I haven't been able > to get anywhere with it, so I'm wondering what y'all > might say. I agree with you - there are lots of solutions, and no obvious parametrisation of them. But there is unique factorisation in the ring of gaussian integers, so it does actually follow that you get all the solutions in the way you suggest, from an expression z = a^2 + b^2. Maybe that is what was wanted. > Another, unrelated, question is that I need to express > a prime p = 1 (mod 3) in the form u^2+3v^2. Certainly > -3 is a quadratic residue mod p, which gives me > something like a^2+3=kp. But I'm not getting anywhere > after that. Actually, it is quite closely related, as the ring of integers Z[w], where w = (1 + sqrt{-3})/2, is also a unique factorisation domain. So if you could find a,b such that a^2 + 3b^2 = kp, with k < p, you would get near to the solution you want. I would suggest Hardy and Wright, An Introduction to the Theory of Numbers, for a nice introduction to quadratic fields like these, if you can get hold of it. -- Timothy Murphy e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie tel: +353-86-2336090, +353-1-2842366 s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland === Subject: Re: mathematicians utter contempt for common sense Whoops, of course I meant richer, not righter. :-) 'cid 'ooh === Subject: Re: mathematicians utter contempt for common sense | Many years ago, at school, my senior (head of department) mathematics |master used to quote (and, no, I haven't look up the source) the idea that, |in mathematics, the key is not truth but _consistency_. I disagree with that. We want to know whether there are infinitely many twin primes, which is a matter of truth. It can be thought of as a matter of consistency, if we take consistency in a relatively broad sense. In particular, it's not a matter of *formal* consistency. One can give a list of properties that characterize the natural numbers, including in particular the inductive property. Whether there are infinitely many twin primes can then be thought of as being a matter of these properties being consistent with the existence of infinitely many twin primes in that structure. But this is a somewhat eccentric way of describing the situation, since those axioms characterize the natural numbers up to isomorphism. Saying that some structural property is consistent with them is equivalent to saying that that property holds *true* in the natural numbers. Saying that some structural property is inconsistent with them is equivalent to saying that that property is *false* in the natural numbers. |Consistency is the |key to so many proofs (or proofs, or proofs, and so on ...), provided |that one does not utterly reject /tertium non datur/, and even if one does, |can still be adjusted in subtle ways to complete proofs. I don't know what this remark is supposed to mean. In classical logic, where tertium non datur is assumed, a proof, of B from premises A1,...,An say, can always be thought of as a proof of a contradiction from A1,...,An and not-B. So searching for a proof can be thought of as searching for a contradiction, there. One might say for this reason that looking for contradictions holds a prominent position in the doing of classical mathematics. I still would be reluctant to call it key. In intuitionist logic, where tertium non datur is not assumed, proving A->B and proving that (A & not-B) leads to a contradiction are not the same thing. I don't know what kind of adjustment to the notion of consistency would make it more relevant. Keith Ramsay === Subject: Re: A periodic matrix representation of 17^(1/3)? Curious fact: if f(x,y,z)=x^3+17*y^3+17^2*z^3-3*17*x*y*z then f(324,126,49)=1, > f(833,324,126)=17 and f(2142,833,324)=17^2! If r = real cubrt(k) k<> a^3, w = primitive cube root unity f(x,y,z)= (X + rY + rrZ)(X + wrY +wwrrZ)(X +wwrY +wrZ) = x^3 + kY^3 +k^2Z^3 -3kXYZ So if you multiply factors of f(P,Q,R) = 1 by r, rr ,wr, wwrr, etc resp. you get a cycling effect. === Subject: Re: A periodic matrix representation of 17^(1/3)? >If r = real cubrt(k) k<> a^3, w = primitive cube root unity >f(x,y,z)= (X + rY + rrZ)(X + wrY +wwrrZ)(X +wwrY +wrZ) = >x^3 + kY^3 +k^2Z^3 -3kXYZ >So if you multiply factors of f(P,Q,R) = 1 by r, rr ,wr, wwrr, etc >resp. you get a cycling effect. While we are on this topic, I am curious: Say I can find P,Q,R with f(k;P,Q,R)=1. How can I test if this is the smallest solution? Rich === Subject: Re: A periodic matrix representation of 17^(1/3)? > While we are on this topic, I am curious: Say I can find P,Q,R with > f(k;P,Q,R)=1. How can I test if this is the smallest solution? There is a method that will tell you if a cubic unit is an nth powerof a fundamental unit, but it's rather tedious. I'm sure there must be better methods. Say you find that f(2:19,15,12) =1 and you want to determine if it is the power of a fundamental unit. Find the cubic that 19 + 15r + 12rr satisfies: -w + 19 + 15r + 12rr =0 multiply by r and rr to get two other equations and eliminate 1,r,rr by taking determinant | (19 -w) 15 12 | =0 | 24 (19 -w) 15 | | 30 24 (19-w)| => w^3 -57w^2 +3w -1=0 (1) Say w = v^2 and v satisfies the cubic v^3 +sv^2 +qv -1 =0 v = sqrt(w) => w^3 +(2q +s^2)W^2 +(q^2 +2s)w -1 =0 Equate coefficients, elimiante q and get a quartic in s. Check if there are integer roots using rational root theorem. If not, w is not an even power. Repeat to check if fourth power and so on. You can now check if w is a odd prime power. w+1 = v^P +1 =(v+1)... w-1 = v^p -1 =(v-1)... (v-1) divides (w-1) (v+1) divides (w+1) norm(v-1) divides norm(w-1) norm(v+1) divides norm(w+1) Make substitutions v -> v+/-1, w -> w +/-1 in equations for v and w => 1 + s + q -1 = s+q divides 1 -57 + 3 -1 = -54 -1 +s -q -1 = s-q-2 divides -62 Possible values are s+q = -1,-2,-3,-6,-9,-27,-54 s-q -2 = -1,-2,-31,-62 Possible odd prime values compatible with s+q are -1 : 53 -2 : 13 -3 : 3,11 -6 : 3 -9 : 3,5 -27 : 3 For s-q -2 -1 : 61 -2 : 3,5 -31 :31 s+q = -6 s -q -2 = 2 => s = -3, q= -3 => v^3 -3v^2 -3v -1 =0 w^3 -57w^2 +3w -1= (w -v^3)(w-v'^3)(w-v''^3) Check that v^3 -3v^2 -3v -1 divides v^9 -57v^6 +3v^3 -1 Repeat process on v^3 -3v^2 -3v -1 This tells you that x = 1. y,z can be obtained solving equation or by trial and error as 1:y:z a.e rr:r:1 === Subject: Re: A periodic matrix representation of 17^(1/3)? > While we are on this topic, I am curious: Say I can find P,Q,R with > f(k;P,Q,R)=1. How can I test if this is the smallest solution? There is a method that will tell you if a cubic unit is an nth powerof a fundamental unit, but it's rather tedious. I'm sure there must be better methods. Say you find that f(2:19,15,12) =1 and you want to determine if it is the power of a fundamental unit. Find the cubic that 19 + 15r + 12rr satisfies: -w + 19 + 15r + 12rr =0 multiply by r and rr to get two other equations and eliminate 1,r,rr by taking determinant | (19 -w) 15 12 | =0 | 24 (19 -w) 15 | | 30 24 (19-w)| => w^3 -57w^2 +3w -1=0 (1) Say w = v^2 and v satisfies the cubic v^3 +sv^2 +qv -1 =0 v = sqrt(w) => w^3 +(2q +s^2)W^2 +(q^2 +2s)w -1 =0 Equate coefficients, elimiante q and get a quartic in s. Check if there are integer roots using rational root theorem. If not, w is not an even power. Repeat to check if fourth power and so on. You can now check if w is a odd prime power. w+1 = v^P +1 =(v+1)... w-1 = v^p -1 =(v-1)... (v-1) divides (w-1) (v+1) divides (w+1) norm(v-1) divides norm(w-1) norm(v+1) divides norm(w+1) Make substitutions v -> v+/-1, w -> w +/-1 in equations for v and w => 1 + s + q -1 = s+q divides 1 -57 + 3 -1 = -54 -1 +s -q -1 = s-q-2 divides -62 Possible values are s+q = -1,-2,-3,-6,-9,-27,-54 s-q -2 = -1,-2,-31,-62 Possible odd prime values compatible with s+q are -1 : 53 -2 : 13 -3 : 3,11 -6 : 3 -9 : 3,5 -27 : 3 For s-q -2 -1 : 61 -2 : 3,5 -31 :31 s+q = -6 s -q -2 = 2 => s = -3, q= -3 => v^3 -3v^2 -3v -1 =0 w^3 -57w^2 +3w -1= (w -v^3)(w-v'^3)(w-v''^3) Check that v^3 -3v^2 -3v -1 divides v^9 -57v^6 +3v^3 -1 Repeat process on v^3 -3v^2 -3v -1 This tells you that x = 1. y,z can be obtained solving equation or by trial and error as 1:y:z a.e rr:r:1 === Subject: Re: A periodic matrix representation of 17^(1/3)? >> While we are on this topic, I am curious: Say I can find P,Q,R with >> f(k;P,Q,R)=1. How can I test if this is the smallest solution? >There is a method that will tell you >if a cubic unit is an nth powerof a >fundamental unit, but >it's rather tedious. I'm sure there >must be better methods. >Say you find that f(2:19,15,12) =1 >and you want to determine if it is >the power of a fundamental unit. I was thinking yesterday that the following might work: Given, say f(2;19,15,12)=1, where f(k;x,y,z)=x^3+ky^3+k^2z^3-3kxyz, solve f(2;x,19,15)=2 and then solve f(2;y,x,19)=4. Let A=[19,15,12;x,19,12;y,x,19], and following the suggestion of Jon Slaughter given earlier, express A as A=S*D*S^(-1) where D is a diagonal matrix. Assuming(!) the smallest solution to f(2;x,y,z)=1 is in fact the first row of A^(1/n) for some n, finding the desired n should not be too hard given A^(1/n)=S*D^(1/n)*S^(-1). Unfortunately, I have not yet figured out how to find the required S and D, hence can't give a worked example. This method may not even work, although my guess is it does. Rich === Subject: Re: A periodic matrix representation of 17^(1/3)? > I was thinking yesterday that the following might work: > Given, say f(2;19,15,12)=1, where f(k;x,y,z)=x^3+ky^3+k^2z^3-3kxyz, solve > f(2;x,19,15)=2 and then solve f(2;y,x,19)=4. > Let A=[19,15,12;x,19,12;y,x,19], and following the suggestion of Jon Slaughter > given earlier, express A as A=S*D*S^(-1) where D is a diagonal matrix. > Assuming(!) the smallest solution to f(2;x,y,z)=1 is in fact the first row of > A^(1/n) for some n, finding the desired n should not be too hard given > A^(1/n)=S*D^(1/n)*S^(-1). > Unfortunately, I have not yet figured out how to find the required S and D, > hence can't give a worked example. This method may not even work, although my > guess is it does. Coincidentally, I have also been thinking about using matrix factorizations to determine if a cubic unit is fundamental. If r = cubrt(2), say, you can associate a unit, say, 1+r + rr with a matrix whose determinant is 1 by multiplying the column vector [rr, r, 1]^T by the unit. (1 + r + rr)[rr, r, 1]^T = [2 + 2r + rr, 2 + r + rr, 1 + r + rr]^T So in matrix form | 1 2 2|)[rr, r, 1]^T = | 1 1 2| | 1 1 1| (1 + r + rr)[rr, r, 1]^T So this is some kind of inverse eigenvector problem where you are given an eigenvalue , (1 + r + rr),and an eigenvector, [rr, r, 1]^T, and you find the associated positive unimodular matrix. I think you can express any positive unimodular matrix in powers of the following matrices | 1 1 1| = S | 0 1 1| | 0 0 1| and | 0 1 0| = P | 1 0 0| | 0 0 1| The fundamental unit would be represented by a long string of Ps and Ss so if you expand any unit that is a power of the fundamental unit you would expect a repeating pattern of Ps and Ss that would show that it was a power of the fundamental unit. But this does not seem to work as the factorisation does not appear to be unique. Although there may be a way of reducing certain products to simpler ones that gives uniqueness. Of course, if every associated matrix could be factorised in a predictable way, this would mean that the product applied to the vector [rr, r, 1]^T would provide a method of finding a unit, s^(-1) reducing the vector and P permuating it. === Subject: Calculating probability distribution of weighted die Hi all, I have racked my brains out trying to solve this problem. It is not a homework problem. It is just a puzzle that occured to me in the course of an online discussion. The question is, how do I calculate the probabilities of each number coming up on a weighted six-sided die, not necessarily in the shape of a cube, but being a polyhedron with six faces, and not necessarily the same topology as a cube or rectangular prism or parallelepiped, but convex and genus 0, and knowing the location of its centre of mass ? My physical intuition tells me that configurations which have the centre of mass at the same height should come up with the same probability. I may be wrong and I am open to discussion of this conjecture. My second intuition is that the lower the centre of mass of a configuration, the greater the probability of that configuration appearing. Thirdly, and I can confidently say this with certainty, all six configurations of the die have to be a stable equilibrium. Here are some ideas I have come up across , some generated with my own head, some by googling, and some inspired by discussion with others : (i) the ratio of the probabilities of each of the six configurations appearing is directly linearly proportional to the solid angle formed at the centre of mass to the six faces (assumes convex and genus 0 polyhedron) [my own idea, inspired by discussions with others] I would like to think this is the correct answer, but I will scientifically test it by constructing different wooden dies and rolling them and compiling results then trying to adduce a mathematical relationship. (ii) try to use the kinetic theory gases, either replacing spherically symmetric molecules with loaded die and re-deriving the equations of kinetic theory, or somehow equate the different potential energies of each configuration [=mgh, h = height of centre of mass] with proportion of atoms moving at different velocites and find probabilit using integration of Maxwell-Boltzmann distribution of velocities of molecules at a certain news group sci.math] (iii) try to quantify how much energy is required to tip die from one configuration to another [I see this idea in the following thesis also]. I noticed thesis suggests Markov chain, I did not come up with that, but it seems logical to me. http://www.geocities.com/dicephysics/3sgeom.htm I wanted to design a die with probabilities 1,2,3,4,5 = 1/7, 6 = 2/7. I like the idea I came up with of a regular pentagonal pyramid with apex on normal to base through centre of pyramid, uniform mass distribution, and experiment with height. Number defined as face touching table (6 is on pentagonal face). Mt intuition tells me chance of 6 is between 0.5 (flat coin limiting case) and 0 (chopstick limiting case). Anyhow, I eventually found the Pegg irregular dice page and his thesis. I would like to ask if anyone knows any other references ? Roy Banicevic === Subject: Re: Calculating probability distribution of weighted die > (i) the ratio of the probabilities of each of the six configurations > appearing is directly linearly proportional to the solid angle > formed at the centre of mass to the six faces I suspect not. In the limit of a large number of faces, you could approach a half cylinder very closely. I suspect that the probability of rolling onto the big flat face would tend toward one, instead of less than half. > (ii) try to use the kinetic theory gases, [...] > (iii) try to quantify how much energy is required to tip die [...] These look like better bets, though there are still going to be all sorts of problems matching theory with practice. You could perhaps devise a physical model for the die and rolling surface, and simulate it through brute-force. My suspicion is that messy details of initial conditions, elasticity, and friction will strongly affect the final distribution. For example, you may well get noticeably different probabilities when rolling highly non-symmetric dice on felt vs polished hardwood. - Tim === Subject: Re: Calculating probability distribution of weighted die >> (i) the ratio of the probabilities of each of the six configurations >> appearing is directly linearly proportional to the solid angle >> formed at the centre of mass to the six faces >I suspect not. In the limit of a large number of faces, you could >approach a half cylinder very closely. I suspect that the probability >of rolling onto the big flat face would tend toward one, instead of >less than half. There has been useful and considerable (relatively) earlier discussion of this problem, largely in the context of coins, not dice. /group/alt.math.recreational/msg/d56f13b1a90662f2?dmode=source In this earlier thread on alt.math.recreational the quoted post (mine) points out that to be fair a coin or die must be symmetrical in both the solid angle AND its potential energy at rest in all landing configurations. Else, it becomes a hidden variable where the unstated condition becomes the skill and methods of the thrower/tosser. The idea that your die design would be a cylinder did not occur until I read this post. If your design requirements do not allow a fully symmetrical solid shape, I think a small solid angle and high center of gravity could be found such that the die never remains on the flat ends. See also: /sci.math/msg/81d9ee606a12f0dc?dmode=source rec.puzzles/msg/e650c70b9a7e4168?dmode=source alt.math.recreational/msg/6598dc2ff9b5a5a5?dmode=source >> (iii) try to quantify how much energy is required to tip die [...] >These look like better bets, though there are still going to be all >sorts of problems matching theory with practice. You could perhaps >devise a physical model for the die and rolling surface, and simulate >it through brute-force. My suspicion is that messy details of initial >conditions, elasticity, and friction will strongly affect the final >distribution. For example, you may well get noticeably different >probabilities when rolling highly non-symmetric dice on felt vs >polished hardwood. Please, please--if you get results, post a note as to what you find. John Bailey http://home.rochester.rr.com/jbxroads/mailto.html === Subject: Re: Calculating probability distribution of weighted die John Bailey > http://home.rochester.rr.com/jbxroads/mailto.html I certainly will post any results I find. This is so exciting for me. Doing scientific experiments for the cost of a couple of blocks of wood and some wood glue. Sure cheaper than those atom smashers. I will do a scientific study of this over the next year, on low priority. Roy === Subject: How to obtain the matrix with optimal eigenvectors? Given two 2-by-2 matrices A and B, where A is known, but B is partially known (e.g. we only know the components b_11 and b_22 of B). Now we need to determine the matrix B (i.e. to find the missing components b_12 and b_21), to ensure that the eigenvectors of B are as close to those of A as possible, i.e. we need to find B to minimize the problem min || v_b1 - v_a1 ||, or maybe the problem min | v_b1 . v_a1 | (The dot product of v_b1 and v_a1 ) where v_b1 and v_a1 are eigenvectors corresponding to the larger eigenvalues of B and A, respectively. How to solve the above problem to obtain the matrix B? Zhaozhong Wang === Subject: Re: How to obtain the matrix with optimal eigenvectors? > Given two 2-by-2 matrices A and B, where A is known, but B is > partially known (e.g. we only know the components b_11 and b_22 of B). > Now we need to determine the matrix B (i.e. to find the missing > components b_12 and b_21), to ensure that the eigenvectors of B are as > close to those of A as possible, i.e. we need to find B to minimize > the problem > min || v_b1 - v_a1 ||, > or maybe the problem > min | v_b1 . v_a1 | (The dot product of v_b1 and v_a1 ) > where v_b1 and v_a1 are eigenvectors corresponding to the larger > eigenvalues of B and A, respectively. > How to solve the above problem to obtain the matrix B? > Zhaozhong Wang If u and v are eigenvectors of A, solve for the offdiagonal elements of B and for scalars p and q such that B.u = u.p and B.v = v.q === Subject: Re: How to obtain the matrix with optimal eigenvectors? > Given two 2-by-2 matrices A and B, where A is known, but B is > partially known (e.g. we only know the components b_11 and b_22 of B). > Now we need to determine the matrix B (i.e. to find the missing > components b_12 and b_21), to ensure that the eigenvectors of B are as > close to those of A as possible, i.e. we need to find B to minimize > the problem > min || v_b1 - v_a1 ||, > or maybe the problem > min | v_b1 . v_a1 | (The dot product of v_b1 and v_a1 ) > where v_b1 and v_a1 are eigenvectors corresponding to the larger > eigenvalues of B and A, respectively. > How to solve the above problem to obtain the matrix B? > Zhaozhong Wang First of all, since eigenvectors are of arbitrary length or modulus, you need to restrict your search to unit eigenvectors, those of length 1. === Subject: Dividing by Zero and Axiomatic Set Theory One of the major questions in the philosophy of mathematics is whether or not mathematical objects are real, that is, having an independent pre-existence, or whether they are human creations. A recent thread, begun by someone asking if zero was even or odd, led to a discussion of division by zero. I noted that one could indeed say that 0/0 equals just about anything, and a finite number divided by zero equals both positive and negative infinity - but doing so is generally viewed by mathematicians as pointless and unproductive. Thus, mathematicians concern themselves with the real numbers. This seems to relate to the philosophical question above in this way: one can perhaps think of a wild and wooly reality of quantity or number which is pre-existing, and which is manifest in many physical processes in nature - and then one can also think of a subset of that reality, abstracted from it, which is chosen by mathematicians to serve their purposes in efficiently finding information which is applicable both to the formal system of arithmetic over the real numbers, and to the original wild and wooly reality from which it was abstracted. I don't think it settles the philosophical question here, though, because the real numbers are a very natural choice of abstraction from the fundamental natural notion of quantity. As we know, there are many small villages in the real world. In some of these villages, there is only one barber. In some of those villages, the barber is an adult male. In some of those villages, every adult male in the village is clean-shaven. And in some of those villages, the adult males are creatures of habit in how they arrange to be shaved; some will always shave themselves with a safety razor at home, looking in their bathroom mirror, and others will go to the barber shop, and sit down in the barber's chair, and be shaved by the barber who stands behind him. In such a village, of course, the barber, since he cannot both be sitting in the barber's chair and standing behind it at the same time, must instead shave himself at home, before his bathroom mirror. Thus, Bertrand Russel's barber paradox is due to a slight error in phrasing: what one should say is that every man in the village either shaves himself, or is shaved by the barber _in his capacity as barber_. But that doesn't change the fact that there definitely is a paradox when one talks about the set of all sets that are not members of themselves. Here, again, we begin with a wild-and-wooly reality; the world of sets that naive set theory attempts to describe. There are objects; these can be numbers, physical objects, or other abstractions such as names, words, or dates. There are sets; a set has the property that for every object, the question is this object a member of that set has a yes or no answer. Sets are objects. Clearly, here, the set of all objects, and the smaller set of all sets, are perfectly meaningful. But while the statement Set X is a member of itself is in itself meaningful and non-paradoxical, it still can't be used to define a set. While it is nice to know that each set leads to the statement Object A is a member of set S, it would be very useful to have a rule for going from statements to sets. Obviously, the statement x is a real number, and it is less than 3 leads to the set of real numbers less than 3; but if the statement X is a set that does not contain itself does not lead to a set, we need to know why. One possible rule that might be tried is a statement which is either true or false of all objects leads directly to a corresponding set of objects for which that statement is true if the truth of that statement for an object is not affected by that object's membership in the set corresponding to that statement to eliminate self-reference. Note that the conditional is if, not if and only if. But that implies we _can_ have the set of all _other_ sets that do not contain themselves. This set doesn't contain itself. But then, can we call that set Joe, and have another set Fred that is the set of all _other_ sets that do not contain themselves, which doesn't contain Fred, but which does contain Joe? Or, what about the set consisting of 'the set of all _other_ sets that do not contain themselves' and all other sets that do not contain themselves, except this set? To avoid these pitfalls, a well-behaved portion of naive set theory was abstracted out, known as axiomatic set theory, based on the Zermelo-Frankel axioms, with or without the axiom of choice. So, if one takes the position that there is a pre-existing wild and wooly reality, and the attempt to describe it by naive set theory failed, from which a part called axiomatic set theory was extracted... then one can meaningfully ask whether or not the axiom of choice, or the Continuum Hypothesis, or the Generalized Continuum Hypothesis, is true. What that would mean is that, if one tries to bite off a *larger* chunk of the wild-and-wooly reality people have in mind when they talk about objects and the sets that contain them, and one succeeds in biting of such a chunk in which these questions are no longer undecidable - how will they be decided? One can extend axiomatic set theory _either_ way for any of the questions undecidable within it, but the right answer to those questions is the one consistent with that vague thing which naive set theory attempted to describe. Unlike the real numbers, which are a highly satisfactory abstraction from quantity, leaving us with no real motivation for extending the real numbers to allow division by zero, axiomatic set theory is not a large enough chunk of naive set theory to be fully satisfying. Because a wild and wooly reality is not tamed with axioms and postulates, nothing can be directly deduced about it - but the wild and wooly realities are still real. They motivate our choice of which formal systems to talk about. Physical reality led us to talk about the properties of numbers, and about the properties of sets. Thus, is the continuum hypothesis true means will the physical reality of objects and collections motivate us to extend axiomatic set theory to include postulates leading to the continuum hypothesis being true. John Savard http://home.ecn.ab.ca/~jsavard/index.html === Subject: Re: Dividing by Zero and Axiomatic Set Theory >One of the major questions in the philosophy of mathematics is >whether or not mathematical objects are real, that is, having an >independent pre-existence, or whether they are human creations. That has nothing to do with the question of Division by zero. That question relates to two other questions: 1. What do you mean by number? 2. What do you mean by quotient. If the answer to 1. is real number, whether defined axiomatically or constructed, and the answer to 2. is a/b is the unique number c such that b*c=a, then 0/0 is not defined. If the answers are different then it is incumbent on the poster to explain his private nomenclature. >Thus, Bertrand Russel's barber paradox is due to a slight error in >phrasing: There is no error in the phrasing; Russel posed the question for the purpose of illustrating the paradox, and the phrasing that he used was correct for that purpose. >Clearly, here, the set of all objects, and the smaller set of all >sets, are perfectly meaningful. That's not at all clear, and in, e.g., GBN, ZFC, it's false. Whether it is true is na.95ve set theory depends on your conceptual model. >Because a wild and wooly reality is not tamed with axioms and >postulates, nothing can be directly deduced about it But the resulting logical systems often turn out to be surprisingly useful; the Wigner problem. >They motivate our choice of which formal systems to talk about. Only to a limited extent; there is a huge body of Mathematics that was not inspired by external reality. >Physical reality led us to talk about the properties of numbers, and >about the properties of sets. But external reality did not lead us to talk about, e.g., holomorphic functions. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not === Subject: Re: Dividing by Zero and Axiomatic Set Theory Message-id: <41ca6f2d.60372@news.ecn.ab.caOne of the major questions in the philosophy of mathematics is whether >or not mathematical objects are real, that is, having an independent >pre-existence, or whether they are human creations. > [...] Thus, mathematicians concern themselves with >the real numbers > [...] the real numbers are a very natural choice of abstraction from >the fundamental natural notion of quantity. >As we know, there are many small villages in the real world. Not real numbers as in the famous set 'R' in Arithmetic, I presume. > [...] So, if one takes the position that there is a pre-existing wild and >wooly reality, [...] one can meaningfully ask whether or not the axiom of choice, or the >Continuum Hypothesis, or the Generalized Continuum Hypothesis, is true. > [...] Because a wild and wooly reality is not tamed with axioms and >postulates, nothing can be directly deduced about it - but the wild and >wooly realities are still real. > [...] Thus, is the continuum hypothesis true means will the physical >reality of objects and collections motivate us to extend axiomatic set >theory to include postulates leading to the continuum hypothesis being >true. true as in the famous constant 'T' in Mathematical Logic, I presume - or are you suggesting that mathematicians agreeing to use a mathematical axiom might cause the axiom (an otherwise undecidable statement) to be true in the reality of small villages in the real world, when previously it might even have been false? If the continuum hypothesis were suddenly to become permanently true in reality, would it be a safe event to happen in the universe? - the rest of abstract mathematics is unaffected one way or the other by the truth or falsity of the continuum hypothesis (that is why it is an axiom), so everyday real objects such as small villages would presumably not be in any danger (of the universe collapsing around them?) - perhaps mathematical axioms in Chaos Theory could similarly be decided upon by mathematicans, to prevent the risk of butterfly wing movements causing real storms (or wars). On the other hand, if Goldbach's conjecture is unprovable (as might be the case), the undecidability of Goldbach's conjecture would not fall victim even to the godlike powers that mathematicians may have in their definitions of acceptable mathematical axioms, because Goldbach's conjecture cannot be an axiom - if a computer existed that could perform simple arithmetic infinitely quickly, Goldbach's conjecture could be verified merely by checking every even number (to infinity and beyond!), so within mathematics the conjecture must be either true or false, and cannot be neither (or both, or even undecidable other than because of the limitations of technology). According to the philosophical position of theoretical physics, that unprovability would leave Goldbach's conjecture in an uncollapsed wave function state (like Schroedinger's cat) of being both true and false, in reality, forever and everywhere in the cosmos, furthermore, mathematicians are unable to alter its fundamental unknowability by inventing a few convenient axioms, and they also know that the unprovability cannot be proven! As such, the unprovability of the conjecture has similar properties to many of the important because the verification and proof of falsifiablity has been performed by the very mathematicians whose inventions physicists rely on as tools of their own trade! Physicists have recently discovered that, contrary to their previous beliefs, a mysterious 'Dark Energy' forms 99% of the universe. Surely that is just a mirage of physics notation, or possibly of physics communication - perhaps theoretical physicists have been staring too long at the weird equations of string theories (of everything), and have suddenly come to the conclusion that the universe consists mainly of a genuinely (to them) unusual form of energy; they should of course call the mysterious energy 'unprovability'. The answer to the 'major question in the philosophy of mathematics' which you posed could therefore be: mathematical objects are not real, but physical objects are real, however, unprovability is a physical object and not a mathematical object. BSP === Subject: The Smart Multi-Universal Body Model And More... The Smart Multi-Universal Body Model As you look out into outer space you may notice trillions of stars. ANd you probably have even noticed that the Solar System we exist in in the Milky Way Galaxy looks like a speck of dust, compared to what seems to be countless stars in this universe. It looks mighty big don't you think? Well, you haven't seen anything yet. According to the Smart Multi-Universal Body Model, there are about, (ok, sit down, I don't want to shock you too badly... - just humor) 7,000,000,000,000,000,000,000,000,000 Universes According to the Smart Model they exist in such a way that they have practically no collisions ( except for galactic levels which is also kept to a minimum) , and exist in spiral type parallel universal fields in a unified intellegent logical order. You may ask how did the Smart Model determine this estimate. According to the Smart Model the universe is continuously growing, with a modified NEW Conservation Law of Matter and Energy. Which basically states that matter and energy can be created in such a way that maintains mass and gravity relatively constant. In other words, as the universe grows, we grow with it. It's just we don't notice it. The Smart Multi-Universal Body Model IS a human form body. We exist like children inside a universal womb, just waiting to be born into the universe. Here is the current estimate of the number of atoms in the human body. See Reference LINK http://education.jlab.org/qa/mathatom_04.html As was mentioned before, the universe is like a large chemical interacting body, with intelligence from the micro level to the macra level. So you thought this universe was large. What does 7 X 10^27 universes sound like to you, with countless stars and planets in each universe? Various topics from the unposted part of the online book, The Smart Model . You are welcome to join in and learn what really exists here in the universe. The Smart Model or The Creation Model, is a model where all people are taken care of and can proper together with their families, with LOVE. You can even be a creator too. The Smart Model shows the foundations of creation principles, that have perpetual motion in it, too. Why was it developed with practically no references? To try to keep it as free as possible, so that no just a few people would proper, but all can proper and have life to the fullest. It's a shame this world has turned out this way. Dog eat dog. So much fighting and wars. Only a few extremely rich, with billions of poor. But through enlightment of the Smart Model, hopefully one day ALL will be happy and live happily ever after. Does The Creator exist? I think so. And I think things have been revealed to me a humble servant human, to help people to learn new things, and show them that there is more to life than just this planet. We are creators like, The Creator.. We can have our own world too. But it takes time to get there. You may say what's the delay? Well many people want to keep the Smart Model or Creation Model out of it, which is causing a delay in the fine tuning of the universe. Some people want hell not heaven because they are misinformed or misunderstand. The goal of the Smart Model is to find a way that ALL people can be saved from self destruction and be lead to their Paradise. Where does it begin can we visit these places now? Yes, in the Astral Projection state within the dream world, you can visit various new worlds RIGHT NOW for a few minutes, and then you have to come back here, to try to enlighten others of the countless other better places ( worlds) that exist. I personally have visited some of my friends and relatives that have passed away, in other worlds, and they seem to be doing alright. They have things they always wanted in their world, that they could never find here. I even spoke to some friends that are alive today, that said they too have been to other places that look different from this planet for a few minutes, even to areas called Heaven. Well why not check it out. It's hard to predict when we are allowed to visit these worlds, they appear when you least expect in your dreams. Alot of people say if they died they would like to die in their sleep. Well, things look so good in some worlds, sometimes you just don't want to come back here. But we have work to do here. To help and lead others to the truth and enlightenment and give hope and help to those that suffer now, ( including myself). I struggle too. But I believe there are better places we can go to in time in a new body. Many have observed what appears to be an Astral Projected body. Well the Smart Model does describe this level too. It is what could be a different level except this one is free to roam. This body can then be placed into another more perfect solidified body, that will never die or suffer. Smart's Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: A Quantum Poem for Xmas <41ca3dca.51159512@netnews.att.net> A Quantum Poem for Xmas >> --------------- >> I wonder if science shall ever see, >> a quantum discrete as a tree, >> Alas, the answer no must be, >> For what science views is energy, >> Where, as discrete, Quantum Mystics see >> in their symmetric reverie >> the integral over r of nm0c. > Actually pretty excellent especially where it came from. America is a > country of unsophisticated and certainly unpretentious tastes as yet. this is a polite description. I will learn it by heart. it will come in handy in many ways. the last time I was gasping for words was when I saw a collection of US Xmas cards for sale on the net. > This is a takeoff on a Bennett Cerf parody of a Joyce (?) poem. >let's say it should be admired, not for its rhymes, not for any reasons >of form, but for its *content*, the depth of its meaning. > The purpose of the poem was not its artistic merit but its content and > depth of meaning. remember this >is probably from the same country as jingle bells and dreaming of a >white krismes which are infinitely, immensely, unspeakably more >stupid. don't you think so? > Remembering that America is the country that pulled Europe's cujones > out of the fire in the first and second world wars at enormous cost to > itself and its people, you had to. your *markets* were being re-destributed. it took Churchill to figure out what had to be done in case the US came in too late: give them the British fleet for them to make it alone. > I consider that its tastes in popular music can > be forgiven in the worst cases and considerably enjoyed in best cases. !!! I really like what you call country and the banjo and Dylan and lots of other things like blue jeans and the Herald Tribune. I thought the US would conquer the world culturally, without a single shot to be fired. well, I was wrong. I wish I could slap myself I was so wrong. however, that dreaming of a white Christmas sounds more like what they mean by religion in the US. it is awful. === Subject: Lemma Bersayt's Hi! Anybody know something about lemma Bersayt's??? Where I can read more about it at WWW? === Subject: Re: Lemma Bersayt's >Anybody know something about lemma Bersayt's??? I suspect this is a transliteration to English of a transliteration to Russian of a name in some other language, and it's rather hard to say what the original would be. Maybe Burnside's lemma? It might help if you said what area of mathematics this is from. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Topology question Call a topological space Alexandroff if arbitrary intersections of open sets are open. Let S be the Sierpinski space, that is S={0,1} with open sets empty, {1} and {0,1}. If X is a topological space, write SX for the final topology induced by all continuous functions p:S->X, that is the finest topology on X such that all these p's are continuous. I want to show that X is Alexandroff <=> X = SX. I already have the <= part, but I can't find the => part. Any suggestions? The paper I'm reading says it's an 'easy exercise', so I seem to be missing something obvious. Pieter === Subject: Re: Topology question === Subject: Topology question > Call a topological space Alexandroff if arbitrary intersections of > open sets are open. Let S be the Sierpinski space, that is S={0,1} > with open sets empty, {1} and {0,1}. If X is a topological space, > write SX for the final topology induced by all continuous functions > p:S->X, that is the finest topology on X such that all these p's are > continuous. I want to show that > X is Alexandroff <=> X = SX. > I already have the <= part, but I can't find the => part. Any > suggestions? The paper I'm reading says it's an 'easy exercise', > so I seem to be missing something obvious. Indeed, something is missing. I get SX is indiscrete topology while noting discrete X is Alexandroff. U open iff for all f:S -> X, f^-1(U) open iff for all f:S -> X, f^-1(U) = { 1 } or { 0,1 } fxy(1) = x, fxy(0) = y fxy^-1(U) = {1} if x in U, y not in U fxy^-1(U) = { 0,1 } if x,y in U fxy^-1(U) = {0} if x not in U, y in U (not open) fxy^-1(U) = nulset if x,y not in U fxx^-1(U) = { 0,1 } if x in U fxx^-1(U) = nulset if x not in U Now assume U open and x not in U. Thus by fxy, y not in U Conversely by fyx, if y not in U, then x not in U. Hence open U ==> for all x,y (x in U iff y in U) ---- === Subject: Re: Topology question > Indeed, something is missing. I get SX is indiscrete > topology while noting discrete X is Alexandroff. If X is discrete, then the only continuous maps from S to X are the constant maps. Now, to find out what SX is, we have to figure out what is the finest topology on X for which all those constant maps are continuous. I'll leave that as an exercise for you: given a topological space Z, and a set Y, what is the finest topology on Y with the property that every constant map from Z to Y is continuous with respect to that topology? (Hint: it's not the indiscrete topology.) === Subject: Re: Topology question > Call a topological space Alexandroff if arbitrary intersections of open > sets are open. Let S be the Sierpinski space, that is S={0,1} with open > sets empty, {1} and {0,1}. If X is a topological space, write SX for the > final topology induced by all continuous functions p:S->X, that is the > finest topology on X such that all these p's are continuous. I want to > show that > X is Alexandroff <=> X = SX. > I already have the <= part, but I can't find the => part. If X is Alexandroff then each point x in X has a least neighborhood, call it N(x). Assume for a contradiction that some subset A of X is SX-open but not X-open. Since A is not X-open, there is a point a in A such that N(a) is not a subset of A. Choose a point b in N(a)A. Let p(0) = a, p(1) = b. The function p:S->X is continuous, since p(1) is in N(p(0)); but p^{-1}(A) = {0} is not open in S, contradicting the assumption that A is SX-open. === Subject: Re: Topology question >> Call a topological space Alexandroff if arbitrary intersections of open >> sets are open. Let S be the Sierpinski space, that is S={0,1} with open >> sets empty, {1} and {0,1}. If X is a topological space, write SX for >> the final topology induced by all continuous functions p:S->X, that is >> the finest topology on X such that all these p's are continuous. I want >> to show that >> X is Alexandroff <=> X = SX. >> I already have the <= part, but I can't find the => part. > If X is Alexandroff then each point x in X has a least neighborhood, > call it N(x). Assume for a contradiction that some subset A of X is > SX-open but not X-open. Since A is not X-open, there is a point a in A > such that N(a) is not a subset of A. Choose a point b in N(a)A. Let > p(0) = a, p(1) = b. The function p:S->X is continuous, since p(1) is in > N(p(0)); but p^{-1}(A) = {0} is not open in S, contradicting the > assumption that A is SX-open. Ok, now show the converse, to assure X = SX, that if A is X-open, then A is SX-open. === Subject: Re: Topology question > Ok, now show the converse, to assure X = SX, > that if A is X-open, then A is SX-open. I didn't think I needed to discuss the trivial direction. OK, for the benefit of the blue states: The topology of SX is defined as the finest topology in a certain set of topologies; the topology of X belongs to that set; hence the topology of SX is a refinement of the topology of X. -- The Kerry electors in NY cast their 31 electoral votes for John L. Kerry. === Subject: Re: Topology question > Call a topological space Alexandroff if arbitrary intersections of open > sets are open. Let S be the Sierpinski space, that is S={0,1} with open > sets empty, {1} and {0,1}. If X is a topological space, write SX for > the final topology induced by all continuous functions p:S->X, that is > the finest topology on X such that all these p's are continuous. I want > to show that > X is Alexandroff <=> X = SX. > I already have the <= part, but I can't find the => part. >> If X is Alexandroff then each point x in X has a least neighborhood, >> call it N(x). Assume for a contradiction that some subset A of X is >> SX-open but not X-open. Since A is not X-open, there is a point a in A >> such that N(a) is not a subset of A. Choose a point b in N(a)A. Let >> p(0) = a, p(1) = b. The function p:S->X is continuous, since p(1) is in >> N(p(0)); but p^{-1}(A) = {0} is not open in S, contradicting the >> assumption that A is SX-open. > Ok, now show the converse, to assure X = SX, > that if A is X-open, then A is SX-open. Doesn't this follow from the definition of SX? Perhaps I haven't been clear with the definition, but see my reply to Ron Sperber's post to see what I mean. I think SX is finer than X. === Subject: Re: Topology question >> Call a topological space Alexandroff if arbitrary intersections of open >> sets are open. Let S be the Sierpinski space, that is S={0,1} with open >> sets empty, {1} and {0,1}. If X is a topological space, write SX for >> the final topology induced by all continuous functions p:S->X, that is >> the finest topology on X such that all these p's are continuous. I want >> to show that >> X is Alexandroff <=> X = SX. >> I already have the <= part, but I can't find the => part. > If X is Alexandroff then each point x in X has a least neighborhood, > call it N(x). Assume for a contradiction that some subset A of X is > SX-open but not X-open. Since A is not X-open, there is a point a in A > such that N(a) is not a subset of A. Choose a point b in N(a)A. Let > p(0) = a, p(1) = b. The function p:S->X is continuous, since p(1) is in > N(p(0)); but p^{-1}(A) = {0} is not open in S, contradicting the > assumption that A is SX-open. >> Ok, now show the converse, to assure X = SX, >> that if A is X-open, then A is SX-open. > Doesn't this follow from the definition of SX? Perhaps I haven't been > clear with the definition, but see my reply to Ron Sperber's post to see > what I mean. I think SX is finer than X. Yes, I overlooked the limitation of the the functions f:S -> X to C(S,X). === Subject: Re: Topology question > Yes, I overlooked the limitation of the the functions f:S -> X to C(S,X). If that's how you read the question, you must have wondered what the topology on X has to do with anything. === Subject: Re: Topology question > If X is Alexandroff then each point x in X has a least neighborhood, > call it N(x). Assume for a contradiction that some subset A of X is > SX-open but not X-open. Since A is not X-open, there is a point a in A > such that N(a) is not a subset of A. Choose a point b in N(a)A. Let > p(0) = a, p(1) = b. The function p:S->X is continuous, since p(1) is in > N(p(0)); but p^{-1}(A) = {0} is not open in S, contradicting the > assumption that A is SX-open. Pieter === Subject: Re: Topology question > Call a topological space Alexandroff if arbitrary intersections of open > sets are open. Let S be the Sierpinski space, that is S={0,1} with open > sets empty, {1} and {0,1}. If X is a topological space, write SX for the > final topology induced by all continuous functions p:S->X, that is the > finest topology on X such that all these p's are continuous. I want to > show that > X is Alexandroff <=> X = SX. > I already have the <= part, but I can't find the => part. Any > suggestions? The paper I'm reading says it's an 'easy exercise', so I > seem to be missing something obvious. > Pieter Unless I'm misunderstanding the statements, I don't think its true. Let X be the space {0,1} with the discrete topology. This is certainly Alexandroff. Let p:S->X be the identity function. p is not continuous since {0} is open in X and p^{-1}({0})={0} is not open in S. So it would seem to me that the given topology on X is finer than SX unless I misunderstand the definition of SX. === Subject: Re: Topology question > Unless I'm misunderstanding the statements, I don't think its true. Let > X be the space {0,1} with the discrete topology. This is certainly > Alexandroff. Let p:S->X be the identity function. p is not continuous > since {0} is open in X and p^{-1}({0})={0} is not open in S. So it would > seem to me that the given topology on X is finer than SX unless I > misunderstand the definition of SX. I think you've misunderstood the definition of SX. The topology is constructed as follows: start with the given topology on X. Then consider C(S,X), the set of all continuous functions S->X (in your example, the identity clearly isn't). Then add sets to the topology on X, such that for each f in C(S,X), f is still continuous with respect to the new topology on SX) is finer than the original topology. Pieter === Subject: induzione N={0, 1, 2, 3, 4, ...} A = for all E = is an element of c = is a subset of / = and I have a little problem I have this a0) BcN / 0EB / [Ab bEN/( Aa aEN/a<=b => aEB ) => (b+1)EB] => B=N and this a1) BcN / 0EB / [Ab bEB => (b+1)EB] => B=N How to prove a0 => a1 ? How to prove a1 => a0 ? === Subject: Re: induzione > N={0, 1, 2, 3, 4, ...} > A = for all > E = is an element of > c = is a subset of > / = and > I have a little problem > I have this > a0) BcN / 0EB / [Ab bEN/( Aa aEN/a<=b => aEB ) => (b+1)EB] => B=N > and this > a1) BcN / 0EB / [Ab bEB => (b+1)EB] => B=N > How to prove a0 => a1 ? > How to prove a1 => a0 ? Notice that the only difference is: [Ab bEN/( Aa aEN/a<=b => aEB ) => (b+1)EB] vs [Ab bEB => (b+1)EB] Can you show that they are equivalent? Hint: b<=b -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: Focus of a Conic >> There surely already exist a formula for the foci >> of a conic, given in general form Ax^+...+F=0? >> (Or is it kept secret for homework reasons? :-) > You know what Eigenvalues and Eigenvectors are? SVD decomposition: yes. I could transform to standard form...and surely introduce an error somewhere. (The fact that I'm very idle might be the other reason - but I DID try Google :-) -- Hauke Reddmann <:-EX8 fc3a501@uni-hamburg.de His-Ala-Sec-Lys-Glu Arg-Glu-Asp-Asp-Met-Ala-Asn-Asn === Subject: Re: Focus of a Conic >There surely already exist a formula for the foci >of a conic, given in general form Ax^+...+F=0? >(Or is it kept secret for homework reasons? :-) >>You know what Eigenvalues and Eigenvectors are? > SVD decomposition: yes. I could transform to standard > form...and surely introduce an error somewhere. > (The fact that I'm very idle might be the other reason - > but I DID try Google :-) Alex Bogomolny, The Parabola, which is quite entertaining and has references: http://www.maa.org/editorial/knot/Parabola.html David Bernier === Subject: Re: Focus of a Conic the focus of a conic... may be You can work it out from this formula for the polar coordinates of a conic, where one focus is in the center: S=R(1+p) /(1+p*cos@), the distance and @ the angle R is the minimal distance, the peri p is the form or excentricity Have fun Hero === Subject: Re: Focus of a Conic As I said, *doing* it wasn't the real problem (finally I did - the result doesn't look like anything sensible to put on a website :-) Anyway, THX to all. In any case, here is what I needed it for: The parabola through the corners ABC of a triangle and the Euler infinity point of ABC has no focus listed by Kimberling, but at least its intersection with the circumcircle is the Tixier point of ABC. -- Hauke Reddmann <:-EX8 fc3a501@uni-hamburg.de His-Ala-Sec-Lys-Glu Arg-Glu-Asp-Asp-Met-Ala-Asn-Asn === Subject: Advent calendars Re: Was the Phaistos disc an Advent calendar? Merry Christmas Hagen http://home.gvdnet.dk/~hagen/phaistos.htm === Subject: Re: Advent calendars === >Subject: Advent calendars >Message-id: <41caccb3.7051579@news.tiscali.dkRe: Was the Phaistos disc an Advent calendar? No, it's a pizza menu. > Merry Christmas > Hagen >http://home.gvdnet.dk/~hagen/phaistos.htm -- Mensanator Ace of Clubs === Subject: Re: Advent calendars === >>Subject: Advent calendars >>Message-id: <41caccb3.7051579@news.tiscali.dk>Re: Was the Phaistos disc an Advent calendar? >No, it's a pizza menu. >> Merry Christmas >> Hagen >>http://home.gvdnet.dk/~hagen/phaistos.htm >-- >Mensanator >Ace of Clubs It is the old-fashioned kind made by pixels. Don't eat ! Hagen === Subject: Ten months - - - . by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iBNEL4Y32581; Since I came to Mathforum... What seemed important to me was 'exchanging'. As an amateur in mathematics I wanted to discuss the matter I was working in: solving functional equations using iterated functions. (maybe you'd noticed!). not always very rigourous and cannonical ,my purpose being to settle and adapt mecchanisms to resolution of them. Firstly I travelled on the site and picked up (from year2000 upwards) Time and again I tried to give some elements to help posters and chiefly to keep myself into math bath... All that time I did enjoy this open website, Friendly,Alain. === Subject: Re: Ten months - - - . > Since I came to Mathforum... > What seemed important to me was 'exchanging'. > As an amateur in mathematics I wanted to discuss the matter > I was working in: > solving functional equations using iterated functions. > (maybe you'd noticed!). > not always very rigourous and cannonical ,my purpose being to settle > and adapt mecchanisms to resolution of them. > Firstly I travelled on the site and picked up (from year2000 upwards) > Time and again I tried to give some elements to help posters and chiefly to keep myself into math bath... > All that time I did enjoy this open website, > Friendly,Alain. I'm not trying to be rude or anything, but did you have a point you wanted to make? Breckin === Subject: Kernel of the composed operator by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iBNEL4H32597; It is easy to see that if A,B are diagonalizable linear operators defined on the Euclidean space X and AB=BA then ker(AB)=ker(A)+ker(B). What is the situation when X in infinite dimensional (Hilbert or Banach)? Valeriu === Subject: Re: Kernel of the composed operator >It is easy to see that if A,B are diagonalizable linear operators defined on the Euclidean space X and AB=BA then ker(AB)=ker(A)+ker(B). That's true if A and B are _simultaneously_ diagonalizable. >What is the situation when X in infinite dimensional (Hilbert or Banach)? With an appropriate definition of simultaneously diagonalizable it seems like it should be more or less the same. >Valeriu ************************ David C. Ullrich === Subject: Galois groups and A_4.... The following question has been bugging me : I am trying to come up with a quartic polynomial whose Galois group is A_4. I know I can go about the business of finding a discriminant that is the square of a rational number, and the resolvent cubic is irreducible. But I was wondering if there was a more illuminating way to geometrically come up with a quartic polynomial. In particular, we know A_4 has no transpositions and no 4-cycles. So, what can I say about the quartic? Wouldn't it be true that since there are no transpositions, the quartic cannot have exactly 2 real roots? What can I deduce about the fact that the Galois group has no 4-cycles? Tony === Subject: Re: Galois groups and A_4.... > The following question has been bugging me : I am trying to come up with a > quartic polynomial whose Galois group is A_4. I know I can go about the > business of finding a discriminant that is the square of a rational number, > and the resolvent cubic is irreducible. But I was wondering if there was a > more illuminating way to geometrically come up with a quartic polynomial. > In particular, we know A_4 has no transpositions and no 4-cycles. So, what > can I say about the quartic? > Wouldn't it be true that since there are no transpositions, the quartic > cannot have exactly 2 real roots? > What can I deduce about the fact that the Galois group has no 4-cycles? > Tony I'll let someone else comment on theoretical aspect, and will just demonstrate how that construction can be done using PARI/GP software: (23:01:35) gp> p3=polsubcyclo(163,3,y) %35 = y^3+y^2-54*y-169 (23:01:41) gp> p6=rnfequation(bnf=bnfinit(p3,1),rnfkummer(bnrinit(bnf,1,1),[2,0;0,1])) %36 = x^6-42*x^4+425*x^2-625 (23:01:44) gp> p4=nfsubfields(polcompositum(p6,p6)[3],4)[1][1] %37 = x^4-84*x^2-200*x+64 p4 is the polynomial whose Galois group is A_4 Igor === Subject: Re: Galois groups and A_4.... > The following question has been bugging me : I am trying to come up with a > quartic polynomial whose Galois group is A_4. I know I can go about the > business of finding a discriminant that is the square of a rational > number, > and the resolvent cubic is irreducible. But I was wondering if there was > a more illuminating way to geometrically come up with a quartic > polynomial. > In particular, we know A_4 has no transpositions and no 4-cycles. So, > what can I say about the quartic? > Wouldn't it be true that since there are no transpositions, the quartic > cannot have exactly 2 real roots? Yes, that's true. > What can I deduce about the fact that the Galois group has no 4-cycles? Beats me. -- Jim Heckman === Subject: substitute matrix in a function with MATLAB Post before in matlab groups, but get no good answer. For the function ft = t/(t^2+1) and matrix A. when using matlab function subs(ft, t, A), then the result is based on the element operation. But this is not I want. I am hoping to get the result: A * (A^2 + I)^(-1). That is, t is totally replaced by matrix A and constant becomes indentity matrix. How to program in MATLAB? Hoping that I can get some suggestions. -- Yan ZHANG http://www.nict.com.sg/zhang/ === Subject: Random triangle problem Consider the area K of a triangle whose vertices' x- and y-coordinates are given by six independent draws from a normal distribution. Show that: (a) The expected value of K is m = v sqrt(3)/2, where v is the variance of the normal distribution. (b) K is exponentially distributed: i.e., the probability density of K is e^(-K/m)/m. [This is essentially a repost of something I posed inside a sci.math thread.] -- -Jim Ferry Metron, Inc. f rr @m tsc .c m e y e i o === Subject: Re: Random triangle problem [spoiler] >Consider the area K of a triangle whose vertices' x- and y-coordinates >are given by six independent draws from a normal distribution. >Show that: >(a) The expected value of K is m = v sqrt(3)/2, where v is the variance >of the normal distribution. >(b) K is exponentially distributed: i.e., the probability density of K >is e^(-K/m)/m. Let Z0, Z1, and Z2 be the three random vertices. Let (R_i, T_i) be the polar cooridnates of Z_i - Z0 for i = 1, 2. Let S = T1 - T2. Then the area K = R1 R2 |sin S| / 2. After much laborious integration using repeated changes of variable, I derive the joint distribution of R1^2, R2^2, and S and verify the expectation. It looks like determining P{K>k} will take more of the same. Did you have a slicker method in mind? -- Stephen J. Herschkorn sjherschko@netscape.net === Subject: Re: Kernel of the composed operator by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iBNHxTY18798; >>It is easy to see that if A,B are diagonalizable linear operators defined on the Euclidean space X and AB=BA then ker(AB)=ker(A)+ker(B). >That's true if A and B are _simultaneously_ diagonalizable. >>What is the situation when X in infinite dimensional (Hilbert or Banach)? >With an appropriate definition of simultaneously diagonalizable it >seems like it should be more or less the same. >>Valeriu >************************ >David C. Ullrich If A,B are diagonalizable and AB=BA then they are automalically simultaneously diagonalizable! Valeriu === Subject: Re: Kernel of the composed operator >It is easy to see that if A,B are diagonalizable linear operators defined on the Euclidean space X and AB=BA then ker(AB)=ker(A)+ker(B). >>That's true if A and B are _simultaneously_ diagonalizable. >What is the situation when X in infinite dimensional (Hilbert or Banach)? >>With an appropriate definition of simultaneously diagonalizable it >>seems like it should be more or less the same. [...] >If A,B are diagonalizable and AB=BA then they are >automalically simultaneously diagonalizable! I missed the hypothesis AB = BA, sorry. >Valeriu ************************ David C. Ullrich === Subject: Re: How to prove that Fibonacci sequence is primitive recursive? > Look up Tarski on satisfaction. > What do you take to be the relevance of Tarski on satisfaction? I take it that fitting a definitions means satisfying a definition. === Subject: Re: How to prove that Fibonacci sequence is primitive recursive? >I take it that fitting a definitions means satisfying a >definition. And what does that have to do with Tarski on satisfaction? === Subject: The Meaning of Einstein's General Relativity Paul Wake up and smell the coffee. General relativity is a mature well understood theory. There are important issues today with interesting questions and you are not asking them. I started out on this with you very open-minded. Basically you are garbling Newton's force idea with Einstein's geometrical ideas. Your idea is not even wrong. You cannot compute anything. You cannot explain anything that needs explaining. You have not asked the right question. The Question is: What is The Question? (Wheeler) Your entire thesis is ill-posed. There are 3 analogies of some interest (electromagnetic, fluid & elastic) none of which you have even thought of. The GR (LC) connection is analogous to the EM vector potential connection. The curvature is the NONLINEAR NON-ABELIAN (LC) curl of itself analogous to the ABELIAN magnetic tield or the circulation of a fluid. Any 3D-vector field A has a gradient (divergence) part and a curl (circulation part). Roughly and non-rigorously to start (Kiehn can say it better more rigorously) A = GradF + eCurlB CurlGrad = 0 DivCurl = 0 In the language of Cartan forms 1-form = exact 1-form + non-exact 1-form Here we are in 3-space. The Cartan exterior derivative on a 0-form is a one-form. Neglecting anholonomic singularities d^2 = 0 Exact 1-form = dF F = 0 form scalar B is a 1 form curlB = dB is a 2-form e converts the 2-form to a dual 1 form in 3-space or a dual 2-form in 4-space of GR depending on the context. The point is that CURVATURE is a 2-form like EM field tensor, and like VORTICITY in a fluid, and like disclination topological defect densities in a crystal lattice (H. Kleinert). Your COORDINATE PART is the analog to the DIVERGENCE or GRAD exact 1-form part. The INTRINSIC PART is the analog to the Curl or non-EXACT 1-form part. However, your ERROR is to think that the non-EXACT 1-form part is a GCT tensor T ALL BY ITSELF. What we have in fact is (LC) 1-form = exact 1-form + non-exact 1-form = N NOT A GCT TENSOR in the metric space. Curvature 2-form = d(LC) = d(non-exact 1-form) Now in Minkowski space (LC) = exact 1-form Note that even in this case, you can still get the Vilenken-Taub curvature without curvature effect from global non-trivial topology like the quantized vortices in the superfluid irrotational flow and in Type II superconductors. The dark energy is the normal fluid of the macro-quantum coherent holographic vacuum. Bottom line, the only GCT tensor you can get in 1916 GR (no torsion) from the metric tensor is its (LC) covariant derivative which is the ZERO nonmetricity tensor. (LC) is always a non-GCT tensor that under the GCT X transforms as (LC) = N -> (LC)' = XXX(LC) + XY = N' The only possible tensor T here is guv;w = 0 where ;w is the (LC) covariant derivative. Y is a derivative of X so (LC) is a quasi-tensor under the limited group of linear transformations, i.e. a subgroup of GCT. The pure gravity energy pseudo-tensor has this same kind of property as (LC). We want gravity energy to be nonlocal. That is a good thing not a bad thing to be eliminated with Rube Goldberg devices. Curvature is like the breakdown of irrotational flow in a superfluid. There is an analogy to Bohm-Aharonov effect because of the local quantum vacuum coherence that is a giant single-valued wave function. Indeed, this explains the Pioneer 10-11 anomaly as a hedgehog exotic vacuum zero point dark energy topological defect centered at the Sun and maybe ALL stars, the Galactic Halo and other interesting observations that appear mysterious. PS Note in the Newtonian mechanics of rotating non-inertial frames ,t' (non inertial) = ,t(inertial) + Wx = Galilean relativity analog to the (LC) covariant derivative of GR W = instantaneous rotation axial vector of the non-inertial frame (common origin with inertial frame) r' = r v' = v + Wxr a' = a + Wxv + W,txr + Wxv + WxWxr = a + 2Wxv + WxWxr + W,txr 2Wxv = INERTIAL CORIOLIS acceleration of the non-inertial rotating frame WxWxr = INERTIAL CENTRIFUGAL acceleration ... W,txr = INERTIAL TORQUE acceleration In Gennady's Shipov's torsion theory extension of Einstein's 1916 GR this is part of a TORSION FIELD . Note that there are no-translational inertial accelerations in this particular problem. If P is fixed to the rotating frame S' then in this REST rotating non-inertial frame, obviously v' = 0 therefore v = -Wxr a' = 0 0 = a + WxWxr + W,txr When W,t = 0, conservation of angular momentum L a + WxWxr = 0 But F = ma F/m + WxWxr = 0 is compensation of the centrifugal inertial force in the rest rotating non-inertial force by the applied force F measured in the inertial frame. In Newton's theory, unlike Einstein's, gravity is an external force in the inertial frame F = GM/r^2 Giving, for a circular orbit Kepler's law GM = r^3/T^2 T = period of orbit r = radius of orbit in flat Euclidean space where the geodesics are traced out by point test In contrast, the Newtonian non-inertial non-geodesic motion of this test in which there is no gravity force. That is, in Einstein's theory F = 0 and W = 0 in the above problem. What we have instead is the geodesic equation D^2x^u/ds^2 = 0 In a non-inertial LNIF this becomes d^2x^u/ds^2 + (LC)^uvw(dx^v/ds)(dx^w/ds) = 0 Where (LC) is computed from guv e.g. gtt = (1 - 2GM/c^2r) = - grr^-1, gtheta,theta = gphi,phi = -1 in local orthogonal basis (cdt)et, drer, rdthetaetheta, rsinthetadphiephi r >> GM/c^2 gives same answers as Newton's theory. This particular REPRESENTATION guv of a CURVED SPACE-TIME is only for REST LNIF observers at fixed r because of some non-gravity force! In a coincident LIF the EEP gives (LC) = 0 and d^2x^u/dt^2 = 0. In general for COINCIDENT LNIF & LNIF' gu'v' = Xu'^uXv'^vguv Where X is the Jacobian matrix for an element of the GCT symmetry group. EEP mean the existence of tetrads Eu^a and their inverses where guv(LNIF) = Eu^anabEv^b nab = Minkowski metric Where LNIF and LIF are COINCIDENT at same physical event E. Physics is simple, when it is local. Wheeler * Curved space-time physics is local because Einstein's gravity is More is different emergence of the LOCAL macro-quantum cohering of the nonlocal micro-quantum zero point pre-inflationary false vacuum fluctuations. The cohering is only partial so there is some dark energy/matter remnants in our post-inflationary universe. Cartan Tetrad 1-form ~ (1 + Lp^2(Goldstone Phase),u) dx^u Vacuum Coherence = (Higgs Field)e^i(Goldstone Phase) G/c^4 comes from Goldstone Phase Rigidity (P.W. Anderson) X Jacobian matrix of GCT comes from canonical transformation generating function THETA(x^u,x^u') Where Goldstone Phase(x^u) -> Goldstone Phase(x^u) + THETA(x^u,x^u') That is Einstein's GR is really a local gauge theory from a post-inflationary macro-quantum LOCAL vacuum coherence ODLRO parameter of zero entropy setting the direction of the Arrow of Time of The Second Law of Thermodynamics in the same direction as the accelerating expansion of the universe. Note that Curvature (tidal stretch-squeeze) = (LC) Covariant Curl of (LC) = Ricci Part + Conformal Part Guv = Ruv - (1/2)Rguv = (8piG/c^4)Tuv(non-gravity source) When Tuv = 0 Ricci Part of curvature = 0 In classical vacuum there is only conformal curvature with 10 independent parameters at each point. Obviously if Cuvwl =/= 0 in any LNIF, then Cabcd =/= 0 in any COINCIDENT LIF. === Subject: Re: Nested roots > Then, f(x)^2 = x + f(x+1) (1) > The equation f(x)^2=f(x+1) has got a simple solution f(x)=A^(2^x) , > So if p(x) is a polynomial we may perhaps try approximate (1) > with f1(x)=A^(2^x)+p(x)*A^(-2^x) . > and compute p(x)... This should be taken as the beginning of a series. With A^(-2^x) = w I get f(x) = 1/w + 1/2*x*w + (-1/8*x^2+1/4*x+1/4)*w^3 + (-1/8*x^2-1/8*x+1/16*x^3)*w^5 + (5/32+3/32*x^3-1/16*x-5/128*x^4)*w^7 + ... It looks like there will be a formal series solution f(x) = sum_{j=0}^infinity p_j(x)*w^(2*j-1) where each p_j is a polynomial of degree j. Of course this is only formal, and I don't know about convergence (though my guess is that for A > 1 the series will converge for every x). Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Nested roots >> Then, f(x)^2 = x + f(x+1) (1) >> The equation f(x)^2=f(x+1) has got a simple solution f(x)=A^(2^x) , >> So if p(x) is a polynomial we may perhaps try approximate (1) >> with f1(x)=A^(2^x)+p(x)*A^(-2^x) . >> and compute p(x)... >This should be taken as the beginning of a series. With A^(-2^x) = w >I get >f(x) = 1/w + 1/2*x*w + (-1/8*x^2+1/4*x+1/4)*w^3 >+ (-1/8*x^2-1/8*x+1/16*x^3)*w^5 >+ (5/32+3/32*x^3-1/16*x-5/128*x^4)*w^7 + ... >It looks like there will be a formal series solution >f(x) = sum_{j=0}^infinity p_j(x)*w^(2*j-1) >where each p_j is a polynomial of degree j. >Of course this is only formal, and I don't know about convergence >(though my guess is that for A > 1 the series will converge for >every x). Slightly more generally: if w is a solution of w(x)^2 = w(x+1) and f(x) = sum_{j=0}^infinity p_j(x)*w^(2*j-1) with p_0 = 1 and p_1 = x/2, the equation f(x)^2 = x + f(x+1) becomes the system of equations sum_{j=0}^{n+1} p_j(x) p_{n+1-j}(x) = p_n(x+1) for n from 1 to infinity. This is satisfied if p_{n+1}(x) = 1/2 * (p_n(x+1) - sum_{j=1}^n p_j(x) p_{n+1-j}(x)) By induction, each p_j is a polynomial of degree <= j. The leading coefficient is the coefficient of x^j in the Taylor series of sqrt(1+x). The coefficient of x^(j-1) (for j > 1) is the coefficient of x^(j-1) in the Taylor series of -1/(2*sqrt(1+x)). I don't know about the other coefficients. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Cooperative Human-Robot Learning System using a Virtual Reality Telerobotic Interface System using a Virtual Reality Telerobotic Interface. Conference on Advances in Internet Technologies and Applications, Purdue University, West Lafayette, Indiana, U.S.A. http://www.compactech.com/kartoun/ === Subject: hysterical gift idea Need a unique gift idea for that picky guy or gal who's impossible to shop for? Why not a Dead Leprechaun?!? Underground comedian and all-around malcontent T.G. Fleming has finally gathered together his greatest hits to form the hilarious collection Dead Leprechaun in a Tire-Swing. Filled with the fierce social commentary, biting sexual satire, strange stories, and bizarre ruminations that Fleming is known (and often reviled) for, Dead Leprechaun in a Tire-Swing will spice up any given bowel movement or trip to the beach! Click the link below to read excerpts and learn how to order your very own Dead Leprechaun! http://www.fallingskyhazard.com/catalog.htm === Subject: Re: Probability m out of n people meeting??? >Let's say that you have the above situation. and let's say that mWhat is the probability that m of those n people will meet, and the >remaining n-m people don't meet anyone? ... the probability that >two people meet and the 3rd person just gets lost is >1 - (5/32) - (4/32) = 23/32. Not so. You've overlooked one of the four distinct possible outcomes, namely, the case when A overlaps with B, and B overlaps with C, but C does not overlap with A. There's a web page on the site you referenced that discusses the various possible meeting combinations for different values of n. >I thought that at one time, I'd read a treatment of this problem which >compared it to the birthday problem in which you compute the probability >of at least two people in a room having the same birthday, but I can't >seem to find it... The web site you referenced does indeed contain a page on the generalized birthday problem, giving the explicit formula for the probability of any set of coincidences, but of course that's a much simpler problem because, in that case, overlap is a transitive relation. === Subject: Re: Probability m out of n people meeting??? >>Let's say that you have the above situation. and let's say that m>What is the probability that m of those n people will meet, and the >>remaining n-m people don't meet anyone? ... the probability that >>two people meet and the 3rd person just gets lost is >>1 - (5/32) - (4/32) = 23/32. > Not so. You've overlooked one of the four distinct possible outcomes, > namely, the case when A overlaps with B, and B overlaps with C, but > C does not overlap with A. There's a web page on the site you > referenced that discusses the various possible meeting combinations > for different values of n. You're right! I should've thought of that. This reminds me of the time that I tried to generalize the Gamblers Ruin problem for more than 2 people to calculate the probability that a particular player wins. When I posed that problem, the answer I got was that the problem wasn't fully defined. When a player exits the game after going broke, the rules and scoring must change in order to acommodate n-1 people. This gives me more to think about. Instead of just the probability of m out of n people, meeting, there is also the question of what is the probability that A meets B, B meets C, C meets D, etc. >>I thought that at one time, I'd read a treatment of this problem which >>compared it to the birthday problem in which you compute the probability >>of at least two people in a room having the same birthday, but I can't >>seem to find it... > The web site you referenced does indeed contain a page on the > generalized birthday problem, giving the explicit formula for the > probability of any set of coincidences, but of course that's a much > simpler problem because, in that case, overlap is a transitive > relation. Yeah, I found it. I'll have to read it through. A long time ago, Id gotten some e-mails from the author of this website, and I thought I'd seen him mention the birthday problem in the context of the meeting problem. The e-mails are in a place where I can't lay my hands on them right now. I couldn't find the exact phrase that I was looking for on the generalized birthday problem page, but I guess I'll have to read it through and see if I can get an answer. I know that there is at least an integral for doing this. The only reason that I'm hesitant to use that is because I'd like to write a program which can do any kind of meeting problem, and if I'm going to do integration, that can be time consuming, unless I have a very fast integration method. Anyone else? Anybody knowledgable in qeueing theory? -- ------------------------------- Patrick D. Rockwell === Subject: Re: Probability m out of n people meeting??? probability theory, but I'm no expert.:-) In the two dimensional case, you can picture a 1 X 1 plane with the lines y=x+d and y=x-d cutting through it. The arrival times where the two people meet are between x+d and x-d. Does anyone else have any ideas? Can you apply Ed's idea? :-) -- ------------------------------- Patrick D. Rockwell > Perhaps you can approach this problem with some queue theory. A M|D|N|N > queue (markov arrivals, deterministic service time, N servers, N sized > population) should work well with a deterministic service time of d*60 and > a relevant arrival rate. This would allow you to use all the queue theory > formulas without concern for integrands. You could then calculate the > probability that the queue has exactly m (0<= m <= N-1) arrivals in time t > = d*60. You could also calculate the average length of the queue = > average number of people expected to meet. > I used (0<= m <= N-1) since I am taking the Probability(m arrivals in time > t = d*60 given 1 has just arrived) and since Poisson is memory-less that > works out to be just Probability(m arrivals in time t = d*60). This > works out to the Probablity(m+1 people meet a once). > Hope this helps > --Robby >> Most of you know that if N people agree to meet between say, 3:00 P.M. >> and 4:00 P.M., and each agree to wait >> d amount of time with (0<=d<=1) and d*60 minutes is how long they are >> willing to wait, then the probability >> that ALL will meet is >> n*d^(n-1)-(n-1)*d^ n >> and the probability that NONE of them meet is >> (1-(n-1)d)^n >> Ostensibly, this is assuming that their arrival times are randomly >> distributed. >> Here is a different take one the problem. Let's say that you have the >> above situation. and let's say that >> m> What is the probability that m of those n people will meet, and the >> remaining n-m people don't meet >> anyone? >> for the case of n=3, it's relatively easy. Let's say that d=0.25 which >> means that the 3 people are willing >> to wait 15 minutes each. P(ALL 3 meet) is >> 3(.25)^2-2(.25)^3=(3/16)-(2/64)=(6-1)/32=5/32. >> P(NONE meet) = (1-2(0.25))^3=1/8=(4/32). The case of 1 person meeting >> himself is an anomalous >> case that we won't count so, then the probability that two people meet >> and the 3rd person just gets >> lost is >> 1-(5/32)-(4/32)=23/32. >> What I'd like to find is a formula or algorithm which will help me >> compute the probability of >> m out of n people meeting in the above described problem, for any value >> of n and m where >> m> whatever I get. >> There is some interesting probability stuff on http://www.mathpages.com >> and I thought that >> at one time, I'd read a treatment of this problem which compared it to >> the birthday problem >> in which you compute the probability of at least two people in a room >> having the same >> birthday, but I can't seem to find it, if it was even there, and the >> author of the website >> doesn't seem to show his e-mail address. Does anyone know his name? I'm a >> bit hesitant >> to write him if he prefers to be anonymous right now. >> -- >> ------------------------------- >> Patrick D. Rockwell === Subject: Re: New Famous Proof > Nonsense. I talk to myself all the time, and on occasion I > respond. >> For example: >> Me: Now where did I put my car keys? [1] >> Me: Did you look on the night table? [2] >> Me: No I didn't, I think I will. [3] >> Me: Hey, twarn't nothin'. [5] >> I ask you. Was that a conversation? > Any utterance could be classified as a conversation, but we take a > more methodical approach, > the definition is you learnt something from what was said. > A 'null conversation' is just a series of utterances where nothing > is learnt, we don't > include this type since the general obversation of conversations is > extra knowledge. > stated by you > so you didn't learn anything from this statement. > Similarly for [2] and [3]. > [4] You looked on the table. This new information came from your > senses, > eyes, hearing, tactile, not from any of the statements made in [1] > [2] or [3]. [5] This is correct, since a remedial plan of action > could find the keys the utterences were superfluous, nothin'. > Herc >> Have you folks spotted the connexion between what you are saying >> and the notorious private language argument of Wittgenstein? (For >> those interseted, it is all in the /Philosophical Investigations/.) >> I might summarize it thus: if it's a language, it _cannot_ be >> private; and if it's private, it _cannot_ be a language. > this is the fact I'm relying on for the proof to hold, to show how > 'redundant' proofs still stand up mathematically. in practice more > than one person converses, but that has no theoretical merit. Pardon me: yer what? Sorry, M'r Fawlty: I no understan', M'r Fawlty! >> (To avoid the anticipated nit-picking: >> /private/ here means logically private.) You'll also find it >> discussed *phenomenally* deeply in the works of Saul Kripke ... >> whose analysis is the post-Wittgenstein /locus classicus/. >> And remember too that Wittgenstein explicitly made it plain that >> what he was putting forward was not in any sense a theory: just a >> set of remarks to bear in mind when one's thinking seems to be going >> astray. Here is a possible example of the kind of thing that I have >> in mind: In Broken Images, by Robert Graves: >> He is quick, thinking in clear images; >> I am slow, thinking in broken images. >> He becomes dull, trusting to his clear images; >> I become sharp, mistrusting my broken images. >> Trusting his images, he assumes their relevance; >> Mistrusting my images, I question their relevance. >> Assuming their relevance, he assumes the fact, >> Questioning their relevance, I question the fact. >> When the fact fails him, he questions his senses; >> When the fact fails me, I approve my senses. >> He continues quick and dull in his clear images; >> I continue slow and sharp in my broken images. >> He in a new confusion of his understanding; >> I in a new understanding of my confusion. >> John >> johnDOTmorrisonATtescoDOTnet >> -- >> Anything which can be understood can also be misunderstood. > An observation in its atomic form can't be negated can it? > Herc It can: by contradicting the observer. However, I don't think that is quite what you mean ... John johnDOTmorrisonATtescoDOTnet -- There are two sorts of fool: the fool who thinks that what is old is good, and the fool who thinks that what is new is better. - [Ancient Chinese proverb] === Subject: Re: New Famous Proof >> Anything which can be understood can also be misunderstood. > An observation in its atomic form can't be negated can it? > Herc > It can: by contradicting the observer. However, I don't think that is > quite what you mean ... of course, that just leaves... a non-observation in its atomic form. This is my solution to absolute truth, what can a brain in a vat actually deduce? Ex there is something. Herc === Subject: Re: New Famous Proof >> Anything which can be understood can also be misunderstood. > An observation in its atomic form can't be negated can it? > Herc >> It can: by contradicting the observer. However, I don't think >> that is quite what you mean ... > of course, that just leaves... a non-observation in its atomic form. > This is my solution to absolute truth, what can a brain in a vat > actually deduce? > Ex > there is something. > Herc Sorry, mate: it's a silly time of the year to go into such matters at such depth. How about in January - when both of us have had /more/ time to think about all the implications? I'm not Welsh - but, if I were, I would add: nadolig llawen y blwyddyn newydd da! When *both* our brains are not in vats - then, let us continue! I await a sensible response in about two weeks - or, as we Brits would have it, a fortnight. I wish you and _all_ your friends all the best now, and into 2005 ... John johnDOTmorrisonATtescoDOTnet -- Now that you know that 'George Bush' is an anagram of 'shoe bugger', you can at last stop wondering what the 'W' stands for. - Mark F. Hunter === Subject: Re: New Famous Proof >> Anything which can be understood can also be misunderstood. An observation in its atomic form can't be negated can it? Herc >> It can: by contradicting the observer. However, I don't think >> that is quite what you mean ... > of course, that just leaves... a non-observation in its atomic form. > This is my solution to absolute truth, what can a brain in a vat > actually deduce? > Ex > there is something. > Herc > Sorry, mate: it's a silly time of the year to go into such matters at > such depth. How about in January - when both of us have had /more/ time to > think about all the implications? > I'm not Welsh - but, if I were, I would add: nadolig llawen y blwyddyn > newydd da! When *both* our brains are not in vats - then, let us continue! > I await a sensible response in about two weeks - or, as we Brits would have > it, a fortnight. I wish you and _all_ your friends all the best now, and > into 2005 ... no problem! actually that's where my philosophy begins and ends. Exists X is true! i should look at some of the formal provers, introducing Ex is one of the primitive steps so I could see what the others are. no comments on theory please during the 'fall off a seey prac! Herc