mm-1199 === Subject: Re: Escultura affair: publication scandal by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iB3GGKw12516; >>Zentralblatt f.9fr Mathematik< 9 publications >>of E. Escultura are listed. For 5 of them the >>Zentralblatt just mentions >not reviewed<, whatever >>that means. >>This is an indexing service like so many others. So it simply notes some contributions unless someone wants to try uncharted course. >I dont agree. The Zentralblatt and the Math Reviews are the >two major indexing services on this planet. Indeed together >they cover all countries. >>>wonder why persons like E. Escultura or J. Harris >>provoke such strong reactions. >>The most likely reason is fear of new ideas. >>E. E. Escultura >>University of the Philippines >This is one possibility. Why is it the most >likely one? Another is feeling of inadequacy. Which one is most likely, I dont know. === Subject: Elementary proof of R(3,7)=23? Is there an elementary proof of Ramsey Number R(3,7)=23? I founded the one with R(3,6)=18 and hope to construct a similar one, but probably need to get the complete classication of all (3,6,17)-good graphs... passed away... === Subject: Re: Elementary proof of R(3,7)=23? > Is there an elementary proof of Ramsey Number R(3,7)=23? I founded the one > with R(3,6)=18 and hope to construct a similar one, but probably need to get > the complete classication of all (3,6,17)-good graphs... I dont know what you call elementary, but perhaps you should look here: S. P. Radziszowski, D. L. Kreher, On R(3,k) Ramsey graphs: theoretical and computational results, Journal of Comb. Math. and Comb. Comp., 4 (1988), 37-52. Also a good reference for the bibliography is: S. Radziszewski, Small Ramsey numbers, Electron. J. Comb., 1 (1994), 1-30. Pawel Gladki === Subject: Re: Elementary proof of R(3,7)=23? > Is there an elementary proof of Ramsey Number R(3,7)=23? I founded the one > with R(3,6)=18 and hope to construct a similar one, but probably need to get > the complete classication of all (3,6,17)-good graphs... > passed away... This news deserves to be brought to the level of a named thread. Dale === Subject: Gamma Function/Mills ratio/Inequalities Let G be the Gamma function, and G(x+0.5) f(x)= -------- , x >= 0 . G(x+ 1) Question: to prove or disprove that for each pair (x,y), 0 =< x < y , there exists u(x,y) in ( 0, 1/2 ), such that f(y) ----- = sqrt{(x+u(x,y))/(y+u(x,y))} . f(x) Note:According to G.N.Watson [1] it is knownn hat for every x>= 0 there exists v(x) in ( 1/4, 1/pi ), such that 1 f(x) = ------------- . sqrt(x+v(x)) Reference: [1] G.N. Watson , ,,A note on gamma functions, Proc.Edinburgh Math.Soc., (2) 11 (1958/59), Edinburgh Math.Notes 42, (1959) 7-9. === Subject: Re: The Size of Grahams Number by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iB3GokK16022; === >>Subject: The Size of Grahams Number >>Message-id: Trying to explain the sheer hugeness of Grahams Number in a popular >>way to someone, I kind of came up short. I have read on the net >>somewhere that even if all matter in the universe were converted to >>pen and paper, it wouldnt be enough to write the number down. But >>thats a pretty difcult concept to grasp as well. Is there an easier >>way to a) estimate the number of digits in Grahams Number and b) >>express this as something slighty less unfathomable (i.e. it >>wouldrequier a hard drive 40 times the size of our galaxy to store it >>or something) >Not likely. Even a tiny number like 10**600 is unfathonable. >See How big is a 2000-bit number? > http://members .aol.com/mensanator666/fun/2000_bit.htmMensanator >Ace of Clubs On the other hand, there are 1 googol (10^100) different ways to arrange 16 balls on a pool table, if the position of each is specied to within 1 mm. To me this is a pretty real world way to grasp a googol. Looked at in this way, a 2000 bit number (=10^602) can be grasped as far less than the numer of ways to arrange the total number of phil === Subject: tting surface I am looking for routines (free and possibly in fortran) to t functions except spline tting. The data is smooth, can provide a sample plot if this would help. Would much appreciate response. Marl === Subject: Re: tting surface >I am looking for routines (free and possibly in fortran) to t functions >except spline tting. The data is smooth, can provide a sample plot >if this would help. Would much appreciate response. >Marl Not really my eld, but have you looked at Shepards quadratic method and variations? Google shepards quadratic fortran gives lots of hits, for example: http://ngwww.ucar.edu/ngdoc/ng/ngmath/shgrid/intro.html might get you started. --Lynn === Subject: Re: need help in understanding Torkels ZFC comment > Summary of the discussion so far: > Frege> http://au.metamath.org/mpegif/2p2e4.html > Charlie-Boo> [6-line informal proof of 0 + 0 = 0] > Charlie-Boo> I beat him by 19,725 steps. > Ullrich> Thats not a formal proof from the axioms of ZFC. > Neither is the MetaMath proof of 2+2=4. > Yes it is. If you believe otherwise then point out an axiom that is used > in the proof but is not part of ZFC. Very few of the terminal nodes of the proof tree are ZFC axioms, because ZFC has nothing to do with the fact that 2+2=4. (Only 1 set is involved, so we need not consider the question of what sets exist in general. 2+2=4 regardless of what ZFC says.) At these nodes you will nd other axioms, rules that allow arbitrary wffs to be considered proven (unsound) and unveried denitions (also unsound) which must be set up to verify new theorems (a strict no-no in axiomatic systems.) There are also many nodes that have no links or justication, such as ECOPRASS at http://us.metamath.org/mpegif/ecoprass.html . What is the justication for lines 23 and 24 (and the other lines labeled ecoprass)? If you want to argue that certain rules dont count, that is not productive the fact remains that each branch of Mathematics has its own axioms. The assertion that everything comes from ZFC is a lie and I would love to hear all sorts of professors claim that it is so. MetaMath derives almost nothing from any axioms. It uses a number of subterfuges to allow you to declare arbitrary expressions (even meaningless sequences of symbols) to be theorems. An axiomatic system is one that derives everything from a xed set of axioms and rules (and perhaps denitions.) After it is set up, (1) you dont have to add anything to the system to develop a theorem, and (2) only actual theorems can be constructed. (Agree?) Otherwise you have no more than a word processor with a Mathematical font. C-B > Do you have any idea how the theorem was produced? How the proof was > produced? To what extent and how it was veried? > Do you? Its only been explained to you two or three times by now. > For the questions you ask, the web site really does > explain it in detail. Ok, then what are the possible justications for the terminal nodes in a proof tree? === Subject: Re: need help in understanding Torkels ZFC comment >>Summary of the discussion so far: >>Frege> http://au.metamath.org/mpegif/2p2e4.html >>Charlie-Boo> [6-line informal proof of 0 + 0 = 0] >>Charlie-Boo> I beat him by 19,725 steps. >>Ullrich> Thats not a formal proof from the axioms of ZFC. >Neither is the MetaMath proof of 2+2=4. >>Yes it is. If you believe otherwise then point out an axiom that is used >>in the proof but is not part of ZFC. > Very few of the terminal nodes of the proof tree are ZFC axioms, > because ZFC has nothing to do with the fact that 2+2=4. (Only 1 set > is involved, so we need not consider the question of what sets exist > in general. 2+2=4 regardless of what ZFC says.) > At these nodes you will nd other axioms, rules that allow > arbitrary wffs to be considered proven (unsound) and unveried > denitions (also unsound) which must be set up to verify new theorems > (a strict no-no in axiomatic systems.) There are also many nodes that > have no links or justication, such as ECOPRASS at > http://us.metamath.org/mpegif/ecoprass.html . What is the > justication for lines 23 and 24 (and the other lines labeled > ecoprass)? Looks like hypotheses of the theorem to me. > If you want to argue that certain rules dont count, that is not > productive [CapitalEth] the fact remains that each branch of Mathematics has its > own axioms. The assertion that everything comes from ZFC is a lie and > I would love to hear all sorts of professors claim that it is so. > MetaMath derives almost nothing from any axioms. It uses a number of > subterfuges to allow you to declare arbitrary expressions (even > meaningless sequences of symbols) to be theorems. > An axiomatic system is one that derives everything from a xed set of > axioms and rules (and perhaps denitions.) After it is set up, (1) > you dont have to add anything to the system to develop a theorem, and > (2) only actual theorems can be constructed. (Agree?) > Otherwise you have no more than a word processor with a Mathematical > font. > C-B >Do you have any idea how the theorem was produced? How the proof was >produced? To what extent and how it was veried? >>Do you? Its only been explained to you two or three times by now. >>For the questions you ask, the web site really does >>explain it in detail. > Ok, then what are the possible justications for the terminal nodes > in a proof tree? -- Replace Roman numerals with digits to reply by email === Subject: Re: need help in understanding Torkels ZFC comment >>Ullrich> Thats not a formal proof from the axioms of ZFC. >Neither is the MetaMath proof of 2+2=4. >>Yes it is. If you believe otherwise then point out an axiom that is used >>in the proof but is not part of ZFC. Very few of the terminal nodes of the proof tree are ZFC axioms, > because ZFC has nothing to do with the fact that 2+2=4. (Only 1 set > is involved, so we need not consider the question of what sets exist > in general. 2+2=4 regardless of what ZFC says.) At these nodes you will nd other axioms, rules that allow > arbitrary wffs to be considered proven (unsound) and unveried > denitions (also unsound) which must be set up to verify new theorems > (a strict no-no in axiomatic systems.) There are also many nodes that > have no links or justication, such as ECOPRASS at > http://us.metamath.org/mpegif/ecoprass.html . What is the > justication for lines 23 and 24 (and the other lines labeled > ecoprass)? > Looks like hypotheses of the theorem to me. What are the hypotheses to 2+2=4? > An axiomatic system is one that derives everything from a xed set of > axioms and rules (and perhaps denitions.) After it is set up, (1) > you dont have to add anything to the system to develop a theorem, and > (2) only actual theorems can be constructed. (Agree?) Otherwise you have no more than a word processor with a Mathematical > font. C-B You know, this is kind of like deja vue. All these people yelling and screaming and Im the one who (as is often the case) puts out the formal denition to resolve technical issues - and they all go running for the hills until time for the next skirmish. What a life! === Subject: Re: need help in understanding Torkels ZFC comment >>Ullrich> Thats not a formal proof from the axioms of ZFC. >Neither is the MetaMath proof of 2+2=4. >>Yes it is. If you believe otherwise then point out an axiom that is used >>in the proof but is not part of ZFC. >Very few of the terminal nodes of the proof tree are ZFC axioms, >because ZFC has nothing to do with the fact that 2+2=4. (Only 1 set >is involved, so we need not consider the question of what sets exist >in general. 2+2=4 regardless of what ZFC says.) >At these nodes you will nd other axioms, rules that allow >arbitrary wffs to be considered proven (unsound) and unveried >denitions (also unsound) which must be set up to verify new theorems >(a strict no-no in axiomatic systems.) There are also many nodes that >have no links or justication, such as ECOPRASS at >http://us.metamath.org/mpegif/ecoprass.html . What is the >justication for lines 23 and 24 (and the other lines labeled >ecoprass)? >>Looks like hypotheses of the theorem to me. > What are the hypotheses to 2+2=4? What does that have to do with anything? The theorem in question isnt 2 + 2 = 4. Now that you mention it, why dont you show us a complete listing of a formal proof that 2 + 2 = 4 in the Charlie Boo system, so that we can get an idea of how your formal proof compares to a Metamath formal proof. >An axiomatic system is one that derives everything from a xed set of >axioms and rules (and perhaps denitions.) After it is set up, (1) >you dont have to add anything to the system to develop a theorem, and >(2) only actual theorems can be constructed. (Agree?) >Otherwise you have no more than a word processor with a Mathematical >font. >C-B > You know, this is kind of like deja vue. All these people yelling and > screaming and Im the one who (as is often the case) puts out the > formal denition to resolve technical issues - and they all go > running for the hills until time for the next skirmish. > What a life! -- Replace Roman numerals with digits to reply by email === Subject: How to measure randomness of a deck of cards? I read several papers, there are few approaches such as variation distances, birthday bound and markov chain. I need a method/algorithm to determine the randomness of my deck of cards (permutation) I would be appreciated if anybody can give me some ideas or good solutions. === Subject: Re: How to measure randomness of a deck of cards? Randomness would better be used to describe a specic shufing method. If starting at any particular permutation the probability of getting each permutation is equal I think its safe to say its random. Partial-randomness could probably be measured with variance, but only for methods rather than particular combinations. Dan Rossul ??? > I read several papers, there are few approaches such as variation > distances, birthday bound and markov chain. I need a > method/algorithm to determine the randomness of my deck of cards > (permutation) > I would be appreciated if anybody can give me some ideas or good solutions. === Subject: Re: How to measure randomness of a deck of cards? > I read several papers, there are few approaches such as variation > distances, birthday bound and markov chain. I need a > method/algorithm to determine the randomness of my deck of cards > (permutation) Randomness is a negative property: It is the absence of pattern. Thus you can never prove that there is no pattern in a deck of cards. You can only prove that there _is_ a pattern and that additional shufes have to be done. Thus you will have to rely on non-perfect methods like the ones youve mentioned above, submit the deck to each method and get it approved. And then hope that your deck is sufcently random for your purpose. Sad and simple :/ - Daniel Janzon === Subject: Re: How to measure randomness of a deck of cards? > I need a > method/algorithm to determine the randomness of my deck of cards Cannot be done. Any ordering is as random as any other ordering. Randomness is not a property of your deck of cards. Do you mean a random process of selecting a sequence of cards from your deck? === Subject: Re: How to measure randomness of a deck of cards? > I need a > method/algorithm to determine the randomness of my deck of cards > Cannot be done. Any ordering is as random as any other ordering. Randomness > is not a property of your deck of cards. Do you mean a random process of > selecting a sequence of cards from your deck? referred that cards are likely to be disturbuted randomly by 7 times of Perfect Shufes. I wish to implement an algorithm to test the randomness after different kind of shufing methods, not only Perfect Shufe. Appreciated for help. Looking forward for everybody replies. === Subject: Re: How to measure randomness of a deck of cards? > I wish to implement an algorithm to test the > randomness after different kind of shufing methods... Well, that is a much better question. Rather than asking if a deck is randomly shufed, it makes much more sense to ask if a particular method (i.e. process) of shufing is random, which can be measured statistically. Whether any process in physical reality is truly random is an open question in physics, and it may be true that nothing is truly random (like Einstein said... does god play dice)! If there are any truly random processes, it might require you to resort to quantum physical solutions, which is not practical in a casino, or a typical computer! Most if not all physical processes tend to generate patterns and are therefore to some extent predictable in their output, and thus, are not truly random. In particular, a computer algorithm can absolutely not be truly random! In any case, we can get close enough to a random process on a computer (for most purposes) using a PRNG (pseudo random number generator) algorithm. This is particularly important in cryptographic applications, so most of the PRNG research (including algorithm design and the measurement of randomness) has been done in the crypto community. A truly random process of shufing would mean that every possible card ordering (i.e. all 52! of them) has an equal probability of being the result of shufing the deck. How do you test it? It is actually not possible to measure, in physical reality, since you could test it a trillion times, and get exactly the same sequence every time! So what can you do? Well, you have to apply various statical calculations to see what the probability distribution looks like. There are several metrics that are used in crypto. The full answer to that would take many words, or at least many links to related web sites... and I am too lazy today! But I will make a suggestion. post this question on sci.crypt, and I guarantee that you will get lots of very good quality answers there! === Subject: Re: How to measure randomness of a deck of cards? > Well, you have to apply various statical calculations to see what the probability distribution looks like. May I know what are the various stat calculations for computing the probability distribution please? Secondly, what infomation can I obtain from the probability distribution graph? Is it something like here?: http://www.math.washington.edu/~chartier/Shufe/ simulation.html === Subject: Re: How to measure randomness of a deck of cards? OK... You will get more info on sci.crypt, but to get you going, you can learn about ent and diehard, which are stats tests for PRNG algos. ENT - see: http://www.fourmilab.ch/random/ DIEHARD - http://stat.fsu.edu/~geo/diehard.html === Subject: Re: How to measure randomness of a deck of cards? posting-account=Qiuj5AwAAACmGnmS12qcvqA9IXzD0s4L take the difference between consecutive cards, step 2 take the sum of squares. compare it to a shufed deck. also do it for suits, 0 1 2 3 in mod 4. could also do it for every second card, every 3rd card. or a neural net could be trained to give you a yes or no if its shufed. theres no absolute answer unless youre using a naive interpretation of random, say the deck is 314159, looks stacked to me! step 2: subtract that from the expected difference (I think) Herc === Subject: Re: How to measure randomness of a deck of cards? > take the difference between consecutive cards, > step 2 > take the sum of squares. > compare it to a shufed deck. > also do it for suits, 0 1 2 3 in mod 4. > could also do it for every second card, every 3rd card. > or a neural net could be trained to give you a yes or no if its > shufed. > theres no absolute answer unless youre using a naive interpretation > of random, > say the deck is 314159, looks stacked to me! > step 2: subtract that from the expected difference (I think) > Herc What is this method called? And what is its main purpose? === Subject: Re: How to measure randomness of a deck of cards? posting-account=Qiuj5AwAAACmGnmS12qcvqA9IXzD0s4L The sum of squares (of the differences) is standard for measuring total error of a sample. Im not sure how it will go using the heuristic of an expected difference between consecutive cards, but you should be able to detect a fresh deck from a shufed one. Say you calculate the average difference between consecutive cards is 7. then a fresh deck will have differences of Cards = < 1,2,3,4,5,6,7,8,9,10,11,12,13,1,2,3,4,5,6,7,8,9,10,11,12,13,1, 2..> Delta = < 1,1,1,1,1,1,1,1,1,1,1,1,1,1,13,1,1,1,1,1,1,1,1,1,1,1,1,1,13,1, 1....> Errors = <6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,...> A random deck will have Delta = <2,4,1,10,12,4,2,6...> Errors = <5,1,3,5,1,5,1....> The standard deviation will pick it up (on Delta), thats a function of the sum of squares, so you dont need an expected difference or Errors. heres another technique : use GPS (General Problem Solver AKA Search algorithm) to sort the cards *into* order count the number of card swaps and you have a measure of the *distance* from an ordered list you could weight the cards swaps, *distance, *sqrt(distance), or just use adjacent swaps. Deck one took 5 card swaps to be ordered, deck two took 20, so deck two was more shufed. Some problems... a reverse ordered deck will appear shufed! Herc === Subject: Re: How to measure randomness of a deck of cards? > say the deck is 314159, looks stacked to me! Actually, the digits of pi look not too bad when you do PRNG stats tests... Not as good as SHA-1, but not too bad... === Subject: Re: How to measure randomness of a deck of cards? >> I need a >> method/algorithm to determine the randomness of my deck of cards > Cannot be done. Any ordering is as random as any other ordering. Randomness > is not a property of your deck of cards. Do you mean a random process of > selecting a sequence of cards from your deck? This is about as helpful as looking at the sequence: 1, 2, 3, 4, 5, ... and being unable to gure out the next number. Sure, if every arrangement is equiprobable, every arrangement is just as random as any other. But its not always the case that all arrangements are equiprobable. If I get a deck of cards, break the shrink wrap and look at it, the sequence will be pretty predictable. What is true is that there is no one, true, absolute, objectively correct way of looking at the order of the deck and saying that deck is 50% random. Randomness is a property of the distribution of possible shufes, not of the shufe resulting from one particular trial. And you cant measure a complete distribution with a single sample. What is also true is that there are useful metrics that can produce gures of merit for how thoroughly the deck appears to have been shufed. A single sample does provide some information about the underlying distribution. And it can provide a lot of information if you already know something about the set of distributions that are likely. If I see a card sharp rife-shufing a deck and see that the resulting deck order is consistent with a sequence of four perfect rife shufes starting from the store bought starting order, I wouldnt call that resulting order random. I might be wrong. Maybe the sharp was playing fair and it was just a one in fty-two factorial freak occurrence. John Briggs === Subject: Re: How to measure randomness of a deck of cards? >If I see a card sharp rife-shufing a deck and see that the resulting >deck order is consistent with a sequence of four perfect rife shufes >starting from the store bought starting order, I wouldnt call that >resulting order random. >I might be wrong. Maybe the sharp was playing fair and it was just >a one in fty-two factorial freak occurrence. > John Briggs If I recall correctly a programming exercise I worked out many years ago, in fact seven perfect shufes leaves the deck as it started. --Lynn === Subject: Re: How to measure randomness of a deck of cards? >>I need a >>method/algorithm to determine the randomness of my deck of cards > Cannot be done. Any ordering is as random as any other ordering. Randomness > is not a property of your deck of cards. Do you mean a random process of > selecting a sequence of cards from your deck? I suspect the OP meant how far (in some sense) a particular arrangement of cards is from the canonical arrangement. There are lots of ways to do this. A good starting point is _The Art of Computer Programming_, vol. 2. Rick === Subject: Re: .99999... still=/= 1 You can do it simpler, x = 0.9_ > 10x = 9.9_ > = 9 + x > 9x = 9 > x = 9/9 > = 1 What youre showing with this argument is that, IF the limit exists, > then it MUST be 1. What you are not showing is that the limit exists > (which is not very difcult). > It only requires the distributive axiom a(b+c)= ab + ac, which is a given. No, you are wrong. What I was referring to is the fact that the original limit x = lim_{n->infty} SUM_{k=1}^{n} 9/(10)^k exists. The distributive property has nothing to do with this --- rather, we use the statement that every bounded monotone sequence of real numbers has a limit. However, this fact uses deep properties of the real numbers. It is actually EASIER to show directly that the limit is 1 --- this can be done strictly in the realm of rational numbers. > The downside is that this denition may seem somewhat articial. Of > course it is easy to show that, when dened like this, decimal > expansions have all the usual properties --- but that brings us back > to having to discuss convergence, which we were trying to avoid in the > rst place. > No. No to having to discuss convergence, or no to trying to avoid it? In any case, you are wrong: the usual denition involves convergence, which thus has to be discussed, and the original argument was clearly intended to avoid doing this. If you disagree, please elaborate. Lasse --- === Subject: Re: .99999... still=/= 1 >> You cant justify the >> step 10x = 9.999... without appealing to a theorem that says you can >> multiply each term of a convergent series by a constant, and the >> resulting series will converge to the constant times the sum of the >> original series. > Such a theorem would be a tautology of little value, since the > statement follows from the directly from distributive axiom a(b+c)= ab > + ac. > It does not follow that > a(b_1 + b_2 + b_3 + b_4 + ...) = a b_1 + a b_2 + a b_3 + a b_4+ ... > The distributive axiom generalizes to arbitrary nite sums, but not > to arbitrary countable sums. The above theorem can be proved for > convergent sums, but it does *require* proof. Not really. a(b_1 + b_2 + b3 + ...) = a(b_1 + (b_2 + b3 + ...)) {associative} = a b_1 + a(b_2 + b_3 + ...) {distributive} > Hey, why dont you prove it and call it Schoenfelds theorem? Since it is simple tautology. === Subject: Re: .99999... still=/= 1 <87acsxynol.fsf@phiwumbda.org> posting-account=HPGXOQ0AAAB8Vqh61Q9SABrM_YhXJe__ Theorem 3.47 in Rudin Principles of mathematical analysis: If Sum a_n = A and Sum b_n = B then Sum (a_n+b_n) = A+B and Sum( c*a_n ) = c*A for any xed c. If the series is convergent, the distributive property holds. To apply the distributive property to a series, we therefore rst need to know if it converges. === Subject: Re: .99999... still=/= 1 In sci.math, John Schoenfeld step 10x = 9.999... without appealing to a theorem that says you can > multiply each term of a convergent series by a constant, and the > resulting series will converge to the constant times the sum of the > original series. >> Such a theorem would be a tautology of little value, since the >> statement follows from the directly from distributive axiom a(b+c)= ab >> + ac. >> It does not follow that >> a(b_1 + b_2 + b_3 + b_4 + ...) = a b_1 + a b_2 + a b_3 + a b_4+ ... >> The distributive axiom generalizes to arbitrary nite sums, but not >> to arbitrary countable sums. The above theorem can be proved for >> convergent sums, but it does *require* proof. > Not really. > a(b_1 + b_2 + b3 + ...) = a(b_1 + (b_2 + b3 + ...)) {associative} > = a b_1 + a(b_2 + b_3 + ...) {distributive} Then 1 + 2 + 4 + 8 + ... = -1. Proof. Let a = 1 + 2 + 4 + 8 + ... Then 2*a = 2 + 4 + 8 + .... = a - 1. Since 2*a = a-1, a = -1. QED. The aw in this proof is admittedly rather subtle, though the biggest clue is that 1 + 2 + 4 + 8 + ... is in fact divergent, tending towards no real number, though one might make a case that it tends towards positive innity. Even with conditionally convergent series one can play some games, though Id have to look up the details now. >> Hey, why dont you prove it and call it Schoenfelds theorem? > Since it is simple tautology. Not quite that simple, methinks. -- #191, ewill3@earthlink.net Its still legal to go .sigless. === Subject: Re: .99999... still=/= 1 > 1 + 2 + 4 + 8 + ... is in fact divergent, tending towards no real > number, That statement is mostly true in the sense that its divergent in the real metric, so tends toward no real number, but its convergent in the 2-adic metric, so it *does* tend toward a 2-adic number. > Even with conditionally convergent series one can play > some games, though Id have to look up the details now. Yeah. For example 1 - 1/2 + 1/3 - 1/4 + 1/5 -+... If you rearrange the terms, still hitting every term once each, you can get a different sequence of partial sums that diverges to plus or minus innity, or which oscillates wildly, or which converges to some (nite real) number other than what the original series converges to. Its actually rather easy to construct an algorithm that will converge to any desired value: Separate the positive and negative terms into independent input streams, then whenever the partial sum is below your goal you take in the next positive term, and whenever the partial sum is above your goal you take in the next negative term. Every term in each of the two input streams is guaranteed to be used exactly once, and the partial sums are guaranteed to oscillate back and forth across your chosen goal, and satisfy the condition for convergence to the goal you chose. Or if you want it to oscillate wildly, or diverge to positive innity, etc., you merely set up a sequence of goals that force it in a particular direction or through a particular oscillating pattern, and make sure you not only pass (cross) each goal but also use at least one term from each input stream before you change to the next goal. === Subject: Re: .99999... still=/= 1 > In sci.math, John Schoenfeld > step 10x = 9.999... without appealing to a theorem that says you can > multiply each term of a convergent series by a constant, and the > resulting series will converge to the constant times the sum of the > original series. >> Such a theorem would be a tautology of little value, since the >> statement follows from the directly from distributive axiom a(b+c)= ab >> + ac. >> It does not follow that >> a(b_1 + b_2 + b_3 + b_4 + ...) = a b_1 + a b_2 + a b_3 + a b_4+ ... >> The distributive axiom generalizes to arbitrary nite sums, but not >> to arbitrary countable sums. The above theorem can be proved for >> convergent sums, but it does *require* proof. > Not really. > a(b_1 + b_2 + b3 + ...) = a(b_1 + (b_2 + b3 + ...)) {associative} > = a b_1 + a(b_2 + b_3 + ...) {distributive} > Then 1 + 2 + 4 + 8 + ... = -1. > Proof. Let a = 1 + 2 + 4 + 8 + ... > Then 2*a = 2 + 4 + 8 + .... = a - 1. > Since 2*a = a-1, a = -1. QED. 2a = a. > The aw in this proof is admittedly rather subtle, > though the biggest clue is that 1 + 2 + 4 + 8 + ... > is in fact divergent, tending towards no real number, > though one might make a case that it tends towards > positive innity. > Even with conditionally convergent series one can play > some games, though Id have to look up the details now. >> Hey, why dont you prove it and call it Schoenfelds theorem? > Since it is simple tautology. > Not quite that simple, methinks. The variable a is transnite and not an element of an ordered ring. http://mathworld.wolfram.com/Ring.html http://mathworld.wolfram.com/TransniteNumber.html === Subject: Re: .99999... still=/= 1 <87acsxynol.fsf@phiwumbda.org> Discussion, linux) > Not really. > a(b_1 + b_2 + b3 + ...) = a(b_1 + (b_2 + b3 + ...)) {associative} > = a b_1 + a(b_2 + b_3 + ...) {distributive} And after a nite number of iterations, you still have a term of the form a (b_n + b_{n+1} + ...) Theres no such thing as a derivation that involves innitely many steps, so you still havent proved a damn thing. >> Hey, why dont you prove it and call it Schoenfelds theorem? > Since it is simple tautology. Wrong. -- When you go to class today, if your professor talks about algebraic number theory, or misuses Galois Theory[,] I want you to carefully notice how you feel. Hold on to that feeling so that you never forget it. --James S. Harris, on channeling rage via Galois theory. === Subject: Re: .99999... still=/= 1 > Not really. > a(b_1 + b_2 + b3 + ...) = a(b_1 + (b_2 + b3 + ...)) {associative} > = a b_1 + a(b_2 + b_3 + ...) {distributive} > And after a nite number of iterations, you still have a term of the > form a (b_n + b_{n+1} + ...) > Theres no such thing as a derivation that involves innitely many > steps, so you still havent proved a damn thing. Trivially false. >> Hey, why dont you prove it and call it Schoenfelds theorem? > Since it is simple tautology. > Wrong. === Subject: Re: .99999... still=/= 1 <87acsxynol.fsf@phiwumbda.org> <87k6rz9rb5.fsf@phiwumbda.org> Discussion, linux) >> Not really. >> a(b_1 + b_2 + b3 + ...) = a(b_1 + (b_2 + b3 + ...)) {associative} >> = a b_1 + a(b_2 + b_3 + ...) {distributive} >> And after a nite number of iterations, you still have a term of the >> form >> a (b_n + b_{n+1} + ...) >> Theres no such thing as a derivation that involves innitely many >> steps, so you still havent proved a damn thing. > Trivially false. You must be talking about Schoenfeld derivations. Theyre not in the books yet. Maybe tomorrow. -- Jesse F. Hughes History will hate you and love me. Im the misunderstood and persecuted genius. Youre the assholes. -- James Harris === Subject: Re: .99999... still=/= 1 >> Not really. >> a(b_1 + b_2 + b3 + ...) = a(b_1 + (b_2 + b3 + ...)) {associative} >> = a b_1 + a(b_2 + b_3 + ...) {distributive} >> And after a nite number of iterations, you still have a term of the >> form >> a (b_n + b_{n+1} + ...) >> Theres no such thing as a derivation that involves innitely many >> steps, so you still havent proved a damn thing. > Trivially false. > You must be talking about Schoenfeld derivations. Theyre not in > the books yet. Maybe tomorrow. Recursion. === Subject: Re: .99999... still=/= 1 <87acsxynol.fsf@phiwumbda.org> <87k6rz9rb5.fsf@phiwumbda.org> <87fz2mnzj0.fsf@phiwumbda.org> Discussion, linux) Not really. a(b_1 + b_2 + b3 + ...) = a(b_1 + (b_2 + b3 + ...)) {associative} > = a b_1 + a(b_2 + b_3 + ...) {distributive} And after a nite number of iterations, you still have a term of the > form a (b_n + b_{n+1} + ...) Theres no such thing as a derivation that involves innitely many > steps, so you still havent proved a damn thing. >> Trivially false. >> You must be talking about Schoenfeld derivations. Theyre not in >> the books yet. Maybe tomorrow. > Recursion[...] aint got a damn thing to do with anything here. Hint: your tautology is plainly false without some assumptions (like the sum converges). It doesnt follow directly from distributivity. You are confused on (at least) two points. (1) The innite summation notation is just a shorthand for a particular limit. What you have is: a (lim_{n -> oo} Sum_{i=1}^n b_n) Distributivity doesnt tell you a damn thing about how multiplication (2) Even if you werent confused about the meaning of the summation notation, you cannot prove what you want because derivations are nite. Your proof would require an innite number of steps, which is just hogwash. -- Jesse F. Hughes Yesterday was Judgment Day. Howd you do? -- The Flatlanders === Subject: Re: .99999... still=/= 1 <87acsxynol.fsf@phiwumbda.org> <87k6rz9rb5.fsf@phiwumbda.org> <87fz2mnzj0.fsf@phiwumbda.org> <87hdn1dkv2.fsf@phiwumbda.org> posting-account=I8cafwwAAACoOYfL9BiKocZY4Rsgl4L7 It has to do with your objection to the sentence 10*0.9_ = 9.9_ despite it following directly from the axioms of the implied ordered ring. a(b+c) = ab + ac {Distributive} a(b+c+d) = a(b+(c+d)) {Associative} It follows that a(b_1+ ... + b_n) = a b_1 + a (b_2 + ... + b_n). === Subject: Re: .99999... still=/= 1 Originator: richard@cogsci.ed.ac.uk (Richard Tobin) >a(b+c) = ab + ac {Distributive} >a(b+c+d) = a(b+(c+d)) {Associative} >It follows that a(b_1+ ... + b_n) = a b_1 + a (b_2 + ... + b_n). Very good. Now prove the limit case. -- Richard === Subject: Re: .99999... still=/= 1 >a(b+c) = ab + ac {Distributive} >a(b+c+d) = a(b+(c+d)) {Associative} >It follows that a(b_1+ ... + b_n) = a b_1 + a (b_2 + ... + b_n). > Very good. Now prove the limit case. > -- Richard Tell us which denition of limit you prefer, and well prove that a (lim_{n -> oo} k_n) = lim_{n -> oo} (a k_n) The desired result then follows from k_n = Sum_{i=1}^n b_n -- Daniel W. Johnson panoptes@iquest.net http://members.iquest.net/~panoptes/ 039 53 36 N / 086 11 55 W === Subject: Re: .99999... still=/= 1 <90ji72-pun.ln1@sirius.athghost7038suus.net> <87acsxynol.fsf@phiwumbda.org> <87k6rz9rb5.fsf@phiwumbda.org> <87fz2mnzj0.fsf@phiwumbda.org> <87hdn1dkv2.fsf@phiwumbda.org> <1gohk2g.e7iaoy1ucb2xcN%panoptes@iquest.net> Discussion, linux) >>a(b+c) = ab + ac {Distributive} >>a(b+c+d) = a(b+(c+d)) {Associative} >>It follows that a(b_1+ ... + b_n) = a b_1 + a (b_2 + ... + b_n). >> Very good. Now prove the limit case. >> -- Richard > Tell us which denition of limit you prefer, and well prove that > a (lim_{n -> oo} k_n) = lim_{n -> oo} (a k_n) *Given* that the limit on the left hand side exists. No one doubts the truth of this theorem. The doubt is whether the theorem is obvious from the distributive law alone, with no extra argument needed. -- Jesse F. Hughes Thats whats brutal about mathematics! When youre wrong, you can have spent years, and lots of effort, and come out at the end with nothing. -- James S. Harris on the path of self-discovery (?) === Subject: Re: .99999... still=/= 1 > No one doubts the truth of this theorem. The doubt is whether the > theorem is obvious from the distributive law alone, with no extra > argument needed. Obviousness is in the eye of the beholder. Someone who considers the identity a (lim_{n -> oo} k_n) = lim_{n -> oo} (a k_n) to be obvious (assuming the existence of the limit on the left hand side, of course) will not need anything beyond the nite distributive law to reach the desired conclusion. -- Daniel W. Johnson panoptes@iquest.net http://members.iquest.net/~panoptes/ 039 53 36 N / 086 11 55 W === Subject: Re: .99999... still=/= 1 <90ji72-pun.ln1@sirius.athghost7038suus.net> <87acsxynol.fsf@phiwumbda.org> <87k6rz9rb5.fsf@phiwumbda.org> <87fz2mnzj0.fsf@phiwumbda.org> <87hdn1dkv2.fsf@phiwumbda.org> <1gohk2g.e7iaoy1ucb2xcN%panoptes@iquest.net> <87mzwoxud0.fsf@phiwumbda.org> <1gohtfb.13q81fsq6e2eN%panoptes@iquest.net> Discussion, linux) >> No one doubts the truth of this theorem. The doubt is whether the >> theorem is obvious from the distributive law alone, with no extra >> argument needed. > Obviousness is in the eye of the beholder. Someone who considers the > identity a (lim_{n -> oo} k_n) = lim_{n -> oo} (a k_n) to be obvious > (assuming the existence of the limit on the left hand side, of course) > will not need anything beyond the nite distributive law to reach the > desired conclusion. Look, maybe you dont realize the extent of the stupidity in this thread. Heres what the great Schoenfeld said: ,---- | Such a theorem would be a tautology of little value, since the | statement follows from the directly from distributive axiom a(b+c)= ab | + ac. `---- This is utter nonsense. I dont care whether some mathematicians would claim the proof is obvious. *Of course* its an easy proof. Duh. I didnt say it was hard. I said it needs more argument than just by the distributive law. I am dubious that any mathematician will disagree with my doubt that the theorem is obvious from the distributive law alone, *with no extra argument needed*. [emphasis added] Now, to refute my position, I think youll have to assume that John Schoenfeld knows the proof of the theorem and that he is mathematically procient enough to view it as obvious by the distributive law. If youll look at his contributions to this thread, I think youll nd this dubious. I wouldnt want any of my opinions to depend on John Schoenfeld knowing his own middle name, much less being mathematically adept. -- Not all features that are found on the Security tab are designed to help make your documents and les more secure. --Microsoft on Ofce security features (after it was pointed out by a third party that a certain password setting is easily bypassed.) === Subject: Re: .99999... still=/= 1 Originator: richard@cogsci.ed.ac.uk (Richard Tobin) >>It follows that a(b_1+ ... + b_n) = a b_1 + a (b_2 + ... + b_n). >> Very good. Now prove the limit case. >Tell us which denition of limit you prefer, and well prove that >a (lim_{n -> oo} k_n) = lim_{n -> oo} (a k_n) we? -- Richard === Subject: Re: .99999... still=/= 1 <87acsxynol.fsf@phiwumbda.org> <87k6rz9rb5.fsf@phiwumbda.org> <87fz2mnzj0.fsf@phiwumbda.org> <87hdn1dkv2.fsf@phiwumbda.org> posting-account=I8cafwwAAACoOYfL9BiKocZY4Rsgl4L7 It has to do with your objection to the sentence 10*0.9_ = 9.9_ despite it following directly from the axioms of the implied ordered ring. a(b+c) = ab + ac {Distributive} a(b+c+d) = a(b+(c+d) {Associative} Thus, a(b_1+ ... + b_n) = a b_1 + a (b_2 + ... + b_n). === Subject: Re: .99999... still=/= 1 <87acsxynol.fsf@phiwumbda.org> <87k6rz9rb5.fsf@phiwumbda.org> <87fz2mnzj0.fsf@phiwumbda.org> <87hdn1dkv2.fsf@phiwumbda.org> Discussion, linux) > It has to do with your objection to the sentence 10*0.9_ = 9.9_ despite > it following directly from the axioms of the implied ordered ring. Moron. Ive never objected to that equation. Ive objected to your spurious claim that it follows from the distributive property. a Sum_{i = 1}^{oo} b_i = Sum_{i=1}^{oo} a b_i whenever Sum_{i = 1}^{oo} b_i converges. This is true, but it is not true by virtue of the distributive property alone. It requires proof. -- If you have a really big idea, you can get a measure of how big it is REALLY, REALLY, *REALLY*, BIG DISCOVERY!!! --James Harris, on being ignored === Subject: Re: .99999... still=/= 1 >> It has to do with your objection to the sentence 10*0.9_ = 9.9_ despite >> it following directly from the axioms of the implied ordered ring. >Moron. Ive never objected to that equation. Ive objected to your >spurious claim that it follows from the distributive property. >a Sum_{i = 1}^{oo} b_i = Sum_{i=1}^{oo} a b_i >whenever Sum_{i = 1}^{oo} b_i converges. This is true, but it is not >true by virtue of the distributive property alone. It requires proof. Or an appropriate notion of ring. Lee Rudolph (I suggest secret decoder ring as most appropriate in this context) === Subject: Re: .99999... still=/= 1 > It has to do with your objection to the sentence 10*0.9_ = 9.9_ despite > it following directly from the axioms of the implied ordered ring. > a(b+c) = ab + ac {Distributive} > a(b+c+d) = a(b+(c+d) {Associative} > Thus, a(b_1+ ... + b_n) = a b_1 + a (b_2 + ... + b_n). Notice that this holds for nite sums. You have to use a limit argument to extend that to an innite series. Bob Kolker === Subject: Re: .99999... still=/= 1 posting-account=I8cafwwAAACoOYfL9BiKocZY4Rsgl4L7 Show the error: x = 0.99999 [repeating n times] = 9/10^1 + 9/10^2 + ..... + 9/10^n 10x = 90/10^1 + 90/10^2 + ..... + 90/10^n = 9/10^0 + 9/10^1 + 9/10^2 + ..... + 9/10^(n - 1) = 9 + x - (9/10^n) For the case that n = +inf, then 10x = lim (n -> +inf) 9 + x - (9/10^n) = 9 + x - [ lim (n -> +inf) 9/10^n ] = 9 + x - 0 Since x = 0.9_, 10 * 0.9_ = 9 + 0.9_ 9 * 0.9_ = 9 0.9_ = 9/9 = 1 === Subject: Re: .99999... still=/= 1 > Show the error: > x = 0.99999 [repeating n times] > = 9/10^1 + 9/10^2 + ..... + 9/10^n > 10x = 90/10^1 + 90/10^2 + ..... + 90/10^n > = 9/10^0 + 9/10^1 + 9/10^2 + ..... + 9/10^(n - 1) > = 9 + x - (9/10^n) > For the case that n = +inf, then > 10x = lim (n -> +inf) 9 + x - (9/10^n) > = 9 + x - [ lim (n -> +inf) 9/10^n ] > = 9 + x - 0 You left out some steps. Notice that x is really a function of n, the way you dened it so: 10*lim x [n-> inf] = 9 + lim x [n -> inf] - lim 9/10^n [n -> inf] You have to show rst that lim x [n -> inf] exists. (We know it does, but you have to show it). Then let L = lim x [n -> inf] and get: 10*L = 9 + L; hence 9*L = 9 and L = 1. Bob Kolker === Subject: Re: .99999... still=/= 1 > Show the error: x = 0.99999 [repeating n times] > = 9/10^1 + 9/10^2 + ..... + 9/10^n 10x = 90/10^1 + 90/10^2 + ..... + 90/10^n > = 9/10^0 + 9/10^1 + 9/10^2 + ..... + 9/10^(n - 1) > = 9 + x - (9/10^n) For the case that n = +inf, then > 10x = lim (n -> +inf) 9 + x - (9/10^n) > = 9 + x - [ lim (n -> +inf) 9/10^n ] > = 9 + x - 0 > You left out some steps. Notice that x is really a function of n, the > way you dened it so: > 10*lim x [n-> inf] = 9 + lim x [n -> inf] - lim 9/10^n [n -> inf] > You have to show rst that lim x [n -> inf] exists. (We know it does, > but you have to show it). Then let L = lim x [n -> inf] and get: > 10*L = 9 + L; hence 9*L = 9 and L = 1. lim(n->+inf) x = x - [ lim(n->+inf) 0 ] = x - 0 = x Most proofs are quasi-proofs as it would take thousands of statements to derive the simplest of propositions from the axioms of the relevant mathematical structure and rst order logic. http://en.wikipedia.org/wiki/Mathematical_proof http://en.wikipedia.org/wiki/Proof_theory > Bob Kolker === Subject: Re: .99999... still=/= 1 posting-account=I8cafwwAAACoOYfL9BiKocZY4Rsgl4L7 >It only requires the distributive axiom a(b+c)= ab + ac, which is a given. > It requires a version of it for innite sums. >If > x = 0.99999... [repeating n times] >Then, > x = 9/10^1 + 9/10^2 + .... 9/10^n > ax = (9a)/10^1 + (9a)/10^2 + .... (9a)/10^n >Let a = 10 > 10x = 90/10^1 + 90/10^2 + .... + (9*10^n)/10^n > ^^^^^^^^^^^^^ > You mean 90/10^n > = 9/10^0 + 9/10^1 + 9/10^2 + .... + (9*10^n)/10^n > ^^^^^^^^^^^^^ > You mean 9/10^(n-1) > = 9 + x > You mean 9 + x - 9/10^n. > -- Richard I made a few typos, yes. The point is what I originally provided was a proof, and that you were wrong. === Subject: Re: .99999... still=/= 1 Originator: richard@cogsci.ed.ac.uk (Richard Tobin) >I made a few typos, yes. The point is what I originally provided was a >proof, and that you were wrong. ??? What are you talking about? -- Richard === Subject: Re: .99999... still=/= 1 >I made a few typos, yes. The point is what I originally provided was a >proof, and that you were wrong. > ??? What are you talking about? Things of a trivial nature. === Subject: Re: .99999... still=/= 1 by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iB2JbVW32373; > .999...* 10 - 9 = .999... => x * 10 - 9 = x where x = .999... >=> 10x - x = 9 = 9x <==> x = 1 So why you claim that .999... != 1? > Thats right. > x = .999... and after his calculations are done he shows it to be equal to something > else. And another way to look at it lets set x = 1 to start with instead of > .999... > x = 1 > 10x = 10 > x = 1 Here we can see x = 1 before and after. And, x = .999... 10 *x = 9.999... x = .999... Here we can see x = .999... before and after. This is often given as a laymans justication (not proof) that .999... > =1. However mathematically, its not quite right. >>It is sufcient for a proof. > You cant justify the > step 10x = 9.999... without appealing to a theorem that says you can > multiply each term of a convergent series by a constant, and the > resulting series will converge to the constant times the sum of the > original series. >>Such a theorem would be a tautology of little value, since the >>statement follows from the directly from distributive axiom a(b+c)= ab >>+ ac. >>If >> x = 0.99999... [repeating n times] >>Then, >> x = 9/10^1 + 9/10^2 + .... 9/10^n >> ax = (9a)/10^1 + (9a)/10^2 + .... (9a)/10^n >>Let a = 10 >> 10x = 90/10^1 + 90/10^2 + .... + (9*10^n)/10^n >> = 9/10^0 + 9/10^1 + 9/10^2 + .... + (9*10^n)/10^n >> = 9 + x > If you could prove that, then youd already have enough understanding of > limits to know that .999... = 1. >>Limits are unnecessary. The proof is simply, >>x = 0.9_ >>10x = 9.9_ (follows directly from the distributive axiom) >> = 9 + x >> 9x = 9 >> x = 1 > Nope, this is incorrect. > The question was does .999... converge to 1, not 9.999... . >equation. Equations arent used in convergence tests. > The basic convergence test is like this. >If, >m-->oo >Lim Z_m = 0 >then convergence otherwise divergence. >.999... is represented as (1 - 1/10^m) >Z_m = ( 1 - 1/10^m) >m-->oo >Lim Z_m = 1 > This means divergence. So not only does it not converge to 1 but it doesnt >equal 1. And even if it did converge it wouldnt EQUAL 1. There isnt even one >series in all of math that does converge that really equals that convergence >value. >.999... --> diverges and =/= 1 >Smarts Alt. Physics News Group >ht tp://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv =1http://smart1234. s-enterprize.com / convergrence test according to your denition 1+ 1/2 + 1/3+1/4 +... would converge since 1/n--->0 as n--> innity. It is generally accepted that the sum given is divergent. thus we have a counter example that disproves your convergence test. Please nd an acceptable trust worthy test for convergence === Subject: Re: .99999... still=/= 1 >> So this shows that, >>10x -.999... = 9 + x - .999... >Add .999... to both sides. Then subtract x. What do you get? >-- Richard The point is, using this type of equation isnt a proof of convergence or divergence. lim ( equation below) 10x = 9 + x At x = 0, you get 0 = 9 At x = 1, you get 10 = 10 at x = .999... , you get 9.999... = 9.999... At x = 10, 100 = 19 At x = oo, you get 10(oo) = 9 + oo ???? <-- indeterminate This test fails and is inconsistent. Smarts 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: .99999... still=/= 1 > .999...* 10 - 9 = .999... >>=> x * 10 - 9 = x where x = .999... >>=> 10x - x = 9 = 9x <==> x = 1 >>So why you claim that .999... != 1? >> Thats right. >> x = .999... >> and after his calculations are done he shows it to be equal to >something >> else. >> And another way to look at it lets set x = 1 to start with instead of >> .999... >> x = 1 >> 10x = 10 >> x = 1 >> Here we can see x = 1 before and after. >> And, >> x = .999... >> 10 *x = 9.999... >> x = .999... >> Here we can see x = .999... before and after. >> This is often given as a laymans justication (not proof) that .999... >> =1. However mathematically, its not quite right. >It is sufcient for a proof. >> You cant justify the >> step 10x = 9.999... without appealing to a theorem that says you can >> multiply each term of a convergent series by a constant, and the >> resulting series will converge to the constant times the sum of the >> original series. >Such a theorem would be a tautology of little value, since the >statement follows from the directly from distributive axiom a(b+c)= ab >+ ac. >If > x = 0.99999... [repeating n times] >Then, > x = 9/10^1 + 9/10^2 + .... 9/10^n > ax = (9a)/10^1 + (9a)/10^2 + .... (9a)/10^n >Let a = 10 > 10x = 90/10^1 + 90/10^2 + .... + (9*10^n)/10^n > = 9/10^0 + 9/10^1 + 9/10^2 + .... + (9*10^n)/10^n > = 9 + x >> If you could prove that, then youd already have enough understanding of >> limits to know that .999... = 1. >Limits are unnecessary. The proof is simply, >x = 0.9_ >10x = 9.9_ (follows directly from the distributive axiom) > = 9 + x > 9x = 9 > x = 1 >> Nope, this is incorrect. >> The question was does .999... converge to 1, not 9.999... . >>equation. Equations arent used in convergence tests. >> The basic convergence test is like this. >>If, >>m-->oo >>Lim Z_m = 0 >>then convergence otherwise divergence. >>.999... is represented as (1 - 1/10^m) >>Z_m = ( 1 - 1/10^m) >>m-->oo >>Lim Z_m = 1 >> This means divergence. So not only does it not converge to 1 but it >doesnt >>equal 1. And even if it did converge it wouldnt EQUAL 1. There isnt even >one >>series in all of math that does converge that really equals that convergence >>value. >>.999... --> diverges and =/= 1 >>Smarts Alt. Physics News Group >>http://pub39.bravenet.com/forum/show.php?usernum=3320272813 &cpv=1 >>S. Enterprize (Science Journal) >>http://smart1234.s-enterprize.com/ >convergrence test >according to your denition 1+ 1/2 + 1/3+1/4 +... would converge since >1/n--->0 as >n--> innity. It is generally accepted that the sum given is divergent. >thus we have a counter example that disproves your convergence test. >Please nd an acceptable trust worthy test for convergence The form of of a harmonic series is n= 1 to oo lim SUM 1/n Using The Root Convergence Test ____ lim n/ Z_n = L n-->oo L < 1 = converge L > 1 = diverge L = 1, test fails, no conclusion possible. At n-->0 its divergent , L >1 n=0, L= 1 n>1, L < 1 = convergent. According to this convergence test, it is divergent, no conclusion possible , and convergent. It depends on the point of view you take, or you can just observe all cases. Smarts 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: .99999... still=/= 1 > A scientic calculator does approximate convergence with a certain > number signicant digits. Just because a calculator emits a particular number of digits of output, doesnt say anything about how accurate the represented number is, i.e. how close to the correct answer it is. If you do a chain of calculations, the accuracy usually gets worse and worse, yet the digits output remains the same, looking like its that accurate (all the way out to the last digit shown) when in fact many of the last digits shown are totally wrong. In fact with a long enough chain of calculations, you can produce a displayed result which may have a whole bunch of digits shown but not even one signicant digit of accuracy. For example, start with the exact integer 2, then take square root, then take square root of that, etc., ve square roots in a row. Now work your way backward by multiplying the value by itself ve times in a row. The correct mathematical answer is eactly 2, but your calculator wont show that. Instead itll show some exact value that starts out 2.00 but then somewhere has a bunch of non-zero digits, or starts out 1.99 but then somewhere has a bunch of non-nine digits. Do you honestly believe that the correct mathematical result is correctly shown to all the digits displayed on your calculator? The point Im making is that you cant use the results given by such a calculator to establish truth of any mathematical fact. Now if you had a calculator that used interval arithmetic internally, and which displayed *only* as many digits as were accurate per the interval (lower&upper bounds on correct result), *then* you could simply look at the output and know that all digits except the last are accurate. But Im not aware of any calculators that work that way. > I think my calculator about 32 signicant digits. > 1/3 = .3333333333333333333333333333 > see that, > 1/3 = .333... No, from the output of your calculator you cant see anything like that except in your imagination, or if you happen to already know the correct result and are just ignoring what the calculator says. In that case, you got lucky, the output from the calculator happened to be accurate to the last digit shown. For a few classes of simple calculations, such as dividing small integer by small integer, thats generally true. But in general, even with all those 3s correct, that still doesnt give you any proof or assurance that the rest of the digits beyond that point will fulll the obvious pattern you see within those limited digits. Try this on your calculator: 10000000000000000001/30000000000000000004 Now you know the correct mathematical result is not 1/3, so if you see 3333333333333333333333333333 and just assume the pattern of digits of 3 continues forever then you should know you are wrong. > Partial Sum Convergence does something like this. It uses a certain > number of nite signicant digits, then it nds a converges value. And like the calculator, you cant really trust the value it gives you. It might be correct out to the number of digits it shows, or it might not be, and even if its correct to that point you cant conclude any pattern of digits you see will be condinued forever in the mathematically correct result. > This convergence value is a number that the series APPROACHES NOT > EQUALS. Your language there is very sloppy. FIrst, do you mean the true mathematical limit or the approximate value given by the calculator? Second, by the series do you mean any nite segment of the series, i.e. any partial sum, or do you mean the whole series, which is dened as the limit of partial sum as number of terms increases without bound? If you mean exact mathematical value of limit, and value of whole innite series, then you are wrong because those two ways of expression mean exactly the same thing hence are of course equal. If you mean exact value of limit but only an individual partial sum not the limit, then you are correct, no individual partial sum (in the series .3 + .03 + .003 + .0003 + ... were discussing here) exactly equals the limit. If you mean the value produced by your calculator, then you are wrong because depending on which way roundoff error occurs, the result it gives you can be greater than the mathematical limit, or smaller, or exactly correct, and you have no way of knowing for any given problem which is the case, so any statement you make about it denitely one way or the other is a wrong statement. > .999... can NEVER = 1 When any competant mathematician uses the notation .999... he/she means the limit of the series, where the pattern of digits of 9 are continued forever without exception. That limit of the series is dened as the limit of partial sums. That limit is in fact exactly 1. You seem to mean something other than the limit of partial sums when you write that notation, and Ive asked you several times to say what exactly you mean by your use of that notation, and still you have failed to tell me. So I ask you again, what do you mean by .999... as you use it just above, and as you used .99999... in the Subject of this thread. Do the two notations .999... and .99999... mean something different to you, or are they just two different amounts of effort put into denoting the same thing? In any case, what did *you* mean when you > Theorem 1 ( Divergence ) > If a series z_1 + z_2 + .... converges, then > lim z_m = 0 > m-->oo That is correct. Note that its a one-way test. The converse is not true. (If you dont know what converse means, please look it up online.) > Hence if the series doesnt satisfy this condition, it diverges. Correct. Thats the contrapositive, which is necessarily true if the original statement is true. (If you dont know what contrapositive means, look it up online.) > .999... = .9 + .09 + .... + ( 1 - 1/10^m) Your notation is unclear. What is that extra term at the very end? It appears to be either an error term or the individual mth term of the sequence that is being summed to make the series. If its supposed to be the mth individual term, then you need to write: > .999... = .9 + .09 + .... + ( 1 - 1/10^m) + ... to indicate that the series doesnt stop at some point but continues on is *not* equal to the innite series indicated by the left side, so If you intended the partial sum, you should have written: .9999999...99999999 = .9 + .09 + .... + ( 1 - 1/10^m) |<- (m digits) ->| If its supposed to be the error term, that is the amount you have to fudge a partial sum to make up all the rest of terms you havent included, then rst of all thats not the correct error term, and the part before that isnt a partial sum, its the notation for the entire series. You should have written: .999... = (.9 + .09 + .... + 0.000000...000000009) + 1/10^m |<- (m digits) ->| where the rst parenthesized expression is the partial sum and the nal expression is the error term. Note that if the right side (the mth partial sum plus the mth error term) is a constant, then the notation makes sense. In that case, the series converges if and only if the error term approaches zero, and if it converges then the constant right-side value is the limit. In this case, indeed the right side is a constant (independent of m), that constant value is 1, and the error term does indeed approach zero, so the limit of the series is exactly 1. The only difcult part of that proof is showing the right side has the same exact value for all values of m. That is you must show that .9 + 1/10 equals 1 (easy), and that (.9 + .09) + 1/100 also equals 1 (pretty easy), and that (.9 + .09 + .009) + 1/1000 also equals 1 (not too hard), etc. etc. etc. However this way of proving a series converges has virtually nothing to do with the divergence test you listed above, so I you must be confused to introduce it here. > Z_m = ( 1 - 1/10^m) If Z_m is supposed to be the mth term of the sequence you are summing, Z_m. > lim 1 - 1/10^m = 1 > m -->oo Thats correct! The limit of partial sums (not individual terms) is exactly 1, proving the series converges to 1. > So, .999.... doesnt converge to 1. You must be awfully confused, given that you just nished proving that it *does* converge to 1. I suspect you are confusing the mth term with the mth partial sum. You didnt prove the mth term approaches 1, you proved the mth partial sum approaches 1. The mth term Z_m is 9/10^m which indeed approaches zero. === Subject: Re: .99999... still=/= 1 >> Z_m = ( 1 - 1/10^m) >If Z_m is supposed to be the mth term of the sequence you are summing, Nope, the convergence test shown in my Advanced Engineering Math book shows Z_M = z_1 + z_2 + z_3 + ... The total series is dened as, Z_m = 1 - 1/10^m The mth term isnt used in the test so 9/10^m isnt part of the convergence test. The test further states that if, m-->oo lim Z_m = 0 then it converges otherwise it diverges. m-->oo lim ( 1 - 1/10^m) = 1 .999... diverges so it doesnt equal 1 or converge to 1. >Z_m. >> lim 1 - 1/10^m = 1 >> m -->oo >Thats correct! The limit of partial sums (not individual terms) is >exactly 1, proving the series converges to 1. >> So, .999.... doesnt converge to 1. >You must be awfully confused, given that you just nished proving that >it *does* converge to 1. I suspect you are confusing the mth term with >the mth partial sum. You didnt prove the mth term approaches 1, you >proved the mth partial sum approaches 1. The mth term Z_m is 9/10^m >which indeed approaches zero. Smarts 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: .99999... still=/= 1 > Z_M = > z_1 + z_2 + z_3 + ... Thats utter nonsense because the rst line (left side of equality) depends on M whereas the second line (right side of equality) doesnt depend on M. > Z_m = 1 - 1/10^m Thats correct, assuming Z_m means the mth partial sum, i.e. 0.9 + 0.09 + ... + 0.00000000000...00000000009 |<- (m-1 0 digits) ->| which partial sum can also be denoted as: 0.999999999...9999999999 |<- (m 9 digits) ->| > The test further states that if, > m-->oo > lim Z_m = 0 > then it converges otherwise it diverges. Thats blatantly false!! Youre saying if the partial sums converge to zero, then they converge, else they dont converge at all, in other words no series can ever converge to anything except zero. Heres a trivial counterexample: 1 + 0 + 0 + 0 + 0 + ... The rst term is 1. Every term after the rst is zero. Every partial sum is 1. Obviously this series converges to 1, and the limit of partial sums is 1: > m-->oo > lim Z_m = 1 Yet what you said above says it doesnt converge at all because the limit of partial sums isnt zero. I think you are totally confused, probably because you are using upper case Z_m to represent partial sum and lower case z_m to represent indiviual terms, yet you cant tell the difference between upper-case and lower-case when reading off your favorite calculator or WebSite etc. so you get the two all mixed up. I suggest you change notation: Use a_m to represent individual term, and S_m to represent partial sum, so you can tell the two letters apart and not mix them up in your mind. Also I suggest you tell us what you originally meant by .99999... in the Subject eld, since you obviously dont mean what everyone else means by that notation, namely the limit of partial sums. === Subject: Re: .99999... still=/= 1 > You must be awfully confused, given that you just nished proving that > it *does* converge to 1. I suspect you are confusing the mth term with > the mth partial sum. You didnt prove the mth term approaches 1, you > proved the mth partial sum approaches 1. The mth term Z_m is 9/10^m > which indeed approaches zero. This fool (Enterprise) has no concept of limit. He does not know what the word means. Bob Kolker === Subject: Re: .99999... still=/= 1 >> You must be awfully confused, given that you just nished proving that >> it *does* converge to 1. I suspect you are confusing the mth term with >> the mth partial sum. You didnt prove the mth term approaches 1, you >> proved the mth partial sum approaches 1. The mth term Z_m is 9/10^m >> which indeed approaches zero. >This fool (Enterprise) has no concept of limit. He does not know what >the word means. >Bob Kolker Well at least take the word of the Advanced Engineering Math book. z_1 + z_2 + z_3 +... = Z_m The total series is used not the mth term. m-->oo lim (1 - 1/10^m) = 1 If, m-->oo lim Z_m = 0 then it converges otherwise diverges These are the words from the Advanced Engineering Math book. I hope you dont start calling this book foolish, too. Smarts 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: .99999... still=/= 1 > Well at least take the word of the Advanced Engineering Math book. I dont have access to that book. All I can see is your mis-copying of stuff you found there. The book may have said things that are true, but your mis-copying of it yields things that are not true. > z_1 + z_2 + z_3 +... = Z_m > The total series is used not the mth term. The left side of the equation appears to be standard mathematical notation for the limit of the partial sum as the number of terms grows without bound. Is that what you mean by the notation? Is that what you mean by the phrase total series? But the right side of that equation appears to be just the mth partial sum, not the whole series in any sense. Have you made another mistake again? > m-->oo > lim (1 - 1/10^m) = 1 Thats correct. The term (1 - 1/10^m) equals the mth partial sum, right? So youre saying the limit of partial sums is 1, right? By the way, do you know how to represent 1/7 as a repeating decimal fraction? I do, and I bet everyone reading this newsgroup, except you and HERC, do too. === Subject: Re: .99999... still=/= 1 > Well at least take the word of the Advanced Engineering Math book. > z_1 + z_2 + z_3 +... = Z_m > The total series is used not the mth term. You cant read either. A necessary condition for convergence is that the n-th term of the series (not the n-th partial sum, you putz!) go to zero. Bob Kolker === Subject: Re: .99999... still=/= 1 >> Well at least take the word of the Advanced Engineering Math book. >> z_1 + z_2 + z_3 +... = Z_m >> The total series is used not the mth term. >You cant read either. A necessary condition for convergence is that the >n-th term of the series (not the n-th partial sum, you putz!) go to zero. >Bob Kolker I looked at it again and I think youre right. z_m is the the mth term and it does equal 9/10^m. And it appears I was using the Partial Sums view of convergence. But even if .999... does converge to 1, .999... still=/= 1 .999...--->1 convergence The whole debate was whether or not .999... really EQUALS 1. It doesnt. Convergence doesnt mean equal to in any case. Smarts 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: .99999... still=/= 1 > But even if .999... does converge to 1, In ordinary mathematics usage: 999999999999...9999999999999 |<- (m digits, each 9) ->| can be sued to represent an exact decimal fraction, one of the partial sums in the series were talking about. Before we go on, do you understand that notation? Do you agree that for any positive integer m: 999999999999...9999999999999 = 1 - 10^(-m) |<- (m digits, each 9) ->| In ordinary mathematics usage: 999... means the limit of the above sequence of partial sums, i.e. the value the sequence of partial sums converges to, i.e.: limit (.999999999999...9999999999999) m->inf |<- (m digits, each 9) ->| Do you understand that ordinary mathematical usage, or do you need it explained further? Now that limit value is in fact 1, youve already agreed. === Subject: Re: .99999... still=/= 1 >> Well at least take the word of the Advanced Engineering Math book. >> z_1 + z_2 + z_3 +... = Z_m >> The total series is used not the mth term. >You cant read either. A necessary condition for convergence is that the >n-th term of the series (not the n-th partial sum, you putz!) go to zero. >Bob Kolker > I looked at it again and I think youre right. z_m is the the mth term and > it does equal 9/10^m. And it appears I was using the Partial Sums view of > convergence. But even if .999... does converge to 1, > .999... still=/= 1 > .999...--->1 convergence > The whole debate was whether or not > .999... really EQUALS 1. It doesnt. Convergence doesnt mean equal to in any > case. I see. You set up a test, fail the test, then declare the test to be meaningless. Youre almost as good at this as James Harris. === Subject: Re: .99999... still=/= 1 > Well at least take the word of the Advanced Engineering Math book. z_1 + z_2 + z_3 +... = Z_m The total series is used not the mth term. >>You cant read either. A necessary condition for convergence is that the >>n-th term of the series (not the n-th partial sum, you putz!) go to zero. >>Bob Kolker >> I looked at it again and I think youre right. z_m is the the mth term >and >> it does equal 9/10^m. And it appears I was using the Partial Sums view of >> convergence. But even if .999... does converge to 1, >> .999... still=/= 1 >> .999...--->1 convergence >> The whole debate was whether or not >> .999... really EQUALS 1. It doesnt. Convergence doesnt mean equal to in >any >> case. >I see. You set up a test, fail the test, then declare the test to be >meaningless. Youre almost as good at this as James Harris. First of all I didnt set up this test. Someone else did. They showed it as, m-->oo lim SUM 9/10^m = I took it to this, (1 - 10^m) = SUM 9/10^m m-->oo lim ( 1 - 1/10^m ) = 1 The test shows divergence relative to Partial SUMs. But I did agree, the mth term should be used in this test. And it shows convergence, but not perfectly equal to 1. I really didnt make any mistakes. You did, when you said, .999... = 1 .999... converges to 1 not equals 1. Why not admit your mistake? Smarts 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: .99999... still=/= 1 > .999...* 10 - 9 = .999... >>=> x * 10 - 9 = x where x = .999... >>=> 10x - x = 9 = 9x <==> x = 1 >>So why you claim that .999... != 1? >> Thats right. >> x = .999... >> and after his calculations are done he shows it to be equal to >something >> else. >> And another way to look at it lets set x = 1 to start with instead of >> .999... >> x = 1 >> 10x = 10 >> x = 1 >> Here we can see x = 1 before and after. >> And, >> x = .999... >> 10 *x = 9.999... >> x = .999... >> Here we can see x = .999... before and after. >> This is often given as a laymans justication (not proof) that .999... >> =1. However mathematically, its not quite right. It is sufcient for a proof. > You cant justify the >> step 10x = 9.999... without appealing to a theorem that says you can >> multiply each term of a convergent series by a constant, and the >> resulting series will converge to the constant times the sum of the >> original series. Such a theorem would be a tautology of little value, since the >statement follows from the directly from distributive axiom a(b+c)= ab >+ ac. If > x = 0.99999... [repeating n times] >Then, > x = 9/10^1 + 9/10^2 + .... 9/10^n ax = (9a)/10^1 + (9a)/10^2 + .... (9a)/10^n Let a = 10 10x = 90/10^1 + 90/10^2 + .... + (9*10^n)/10^n > = 9/10^0 + 9/10^1 + 9/10^2 + .... + (9*10^n)/10^n > = 9 + x > If you could prove that, then youd already have enough understanding of >> limits to know that .999... = 1. Limits are unnecessary. The proof is simply, >x = 0.9_ >10x = 9.9_ (follows directly from the distributive axiom) > = 9 + x > 9x = 9 > x = 1 >> Nope, this is incorrect. >> The question was does .999... converge to 1, not 9.999... . >>equation. Equations arent used in convergence tests. >> The basic convergence test is like this. >>If, >>m-->oo >>Lim Z_m = 0 >>then convergence otherwise divergence. >>.999... is represented as (1 - 1/10^m) >>Z_m = ( 1 - 1/10^m) >>m-->oo >>Lim Z_m = 1 >> This means divergence. So not only does it not converge to 1 but it >doesnt >>equal 1. And even if it did converge it wouldnt EQUAL 1. There isnt even >one >>series in all of math that does converge that really equals that convergence >>value. >>.999... --> diverges and =/= 1 >>Smarts Alt. Physics News Group >>http://pub39.bravenet.com/forum/show.php?usernum=3320272813 &cpv=1 >>S. Enterprize (Science Journal) >>http://smart1234.s-enterprize.com/ >convergrence test >according to your denition 1+ 1/2 + 1/3+1/4 +... would converge since >1/n--->0 as >n--> innity. It is generally accepted that the sum given is divergent. >thus we have a counter example that disproves your convergence test. >Please nd an acceptable trust worthy test for convergence > The form of of a harmonic series is > n= 1 to oo > lim SUM 1/n > Using The Root Convergence Test > ____ > lim n/ Z_n = L > n-->oo > L < 1 = converge > L > 1 = diverge > L = 1, test fails, no conclusion possible. > At n-->0 its divergent , L >1 > n=0, L= 1 > n>1, L < 1 = convergent. > According to this convergence test, it is divergent, no conclusion possible , > and convergent. It depends on the point of view you take, or you can just > observe all cases. Are you ignoring your lsat post now? Youre hopeless. === Subject: Re: .99999... still=/= 1 > .999...* 10 - 9 = .999... => x * 10 - 9 = x where x = .999... >=> 10x - x = 9 = 9x <==> x = 1 So why you claim that .999... != 1? > Thats right. > x = .999... and after his calculations are done he shows it to be equal to >>something > else. And another way to look at it lets set x = 1 to start with instead > .999... > x = 1 > 10x = 10 > x = 1 Here we can see x = 1 before and after. And, x = .999... 10 *x = 9.999... x = .999... Here we can see x = .999... before and after. This is often given as a laymans justication (not proof) that .999... > =1. However mathematically, its not quite right. >>It is sufcient for a proof. > You cant justify the > step 10x = 9.999... without appealing to a theorem that says you can > multiply each term of a convergent series by a constant, and the > resulting series will converge to the constant times the sum of the > original series. >>Such a theorem would be a tautology of little value, since the >>statement follows from the directly from distributive axiom a(b+c)= ab >>+ ac. >>If >> x = 0.99999... [repeating n times] >>Then, >> x = 9/10^1 + 9/10^2 + .... 9/10^n >> ax = (9a)/10^1 + (9a)/10^2 + .... (9a)/10^n >>Let a = 10 >> 10x = 90/10^1 + 90/10^2 + .... + (9*10^n)/10^n >> = 9/10^0 + 9/10^1 + 9/10^2 + .... + (9*10^n)/10^n >> = 9 + x > If you could prove that, then youd already have enough understanding of > limits to know that .999... = 1. >>Limits are unnecessary. The proof is simply, >>x = 0.9_ >>10x = 9.9_ (follows directly from the distributive axiom) >> = 9 + x >> 9x = 9 >> x = 1 > Nope, this is incorrect. > The question was does .999... converge to 1, not 9.999... . >equation. Equations arent used in convergence tests. > The basic convergence test is like this. >If, >m-->oo >Lim Z_m = 0 >then convergence otherwise divergence. >.999... is represented as (1 - 1/10^m) >Z_m = ( 1 - 1/10^m) >m-->oo >Lim Z_m = 1 > This means divergence. So not only does it not converge to 1 but it >>doesnt >equal 1. And even if it did converge it wouldnt EQUAL 1. There isnt even >>one >series in all of math that does converge that really equals that >convergence >value. >.999... --> diverges and =/= 1 >Smarts Alt. Physics News Group >http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 >S. Enterprize (Science Journal) >http://smart1234.s-enterprize.com/ >>convergrence test >>according to your denition 1+ 1/2 + 1/3+1/4 +... would converge since >>1/n--->0 as >>n--> innity. It is generally accepted that the sum given is divergent. >>thus we have a counter example that disproves your convergence test. >>Please nd an acceptable trust worthy test for convergence >The form of of a harmonic series is >n= 1 to oo >lim SUM 1/n >Using The Root Convergence Test Correction : Use the Ratio Test The answer is still the same. Not this vvvvv > ____ >lim n/ Z_n = L >n-->oo >L < 1 = converge >L > 1 = diverge >L = 1, test fails, no conclusion possible. > At n-->0 its divergent , L >1 > n=0, L= 1 > n>1, L < 1 = convergent. > According to this convergence test, it is divergent, no conclusion possible >and convergent. It depends on the point of view you take, or you can just >observe all cases. Smarts 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: invariant differential forms in a lie group Im looking for proofs (of references) using the least possible lie theory of the following fact: if w is a left and right invariant differential form in a lie group G then w is closed. nojb. PS: is the following correct? Let G be a connected lie group, g in G and let L_g be the G -> G diffeomorphism given by left translation by g, then L_g s = s in homology for every singular chain. Proof: H(g,t) = f(t).g is a homotopy between id : G -> G and L_g where f is a path between e and g in G. === Subject: cone cone intersection hi, how can I compute the two intersection lines, when two cones intersect each other? Erich === Subject: Re: cone cone intersection >how can I compute the two intersection lines, when two cones intersect >each other? Write the equations for the two cones, and solve for two components in terms of the other. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Hilbert 16th Problem by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iB3IkHx27203; Hello I Search new Interpretations For the Limit Cycles Phenomena and unusuall formulation for the hilbert 16th problem: (any comments and new suggestions is appreciated), I begin with Following interpretations 1)The Number of attractores of a vector eld X is equall to fredholm index of functional operator X(g)=X.g please review : http://www.arxiv.org/abs/math.DS/0408037 2)let g be a riamanian metrics compatible to a vecter eld X:that is the trajectories of X are geodesics,so the limit cycles of X are closed geodesics,we can try to nd various metrics g which curvature-function has an appropriate behaviour (please See Limit Cycles And complex Geometry,Romanovski St Ptrs Jour Of Math 1997) 3)Quantum Mechanics, however I have no background In Math. Aspects Of Quantum Mechanics,But I Have found The Following two Papers(with Intersting titles) in The Web,last year: i)IS QUANTUM MECHANICS A LIMIT CYCLE THEORY,By A.M.Cetto and de la Penna ii)Limit Cycles In Quantum Theories,Glazek Wilson Physical review letter 2002 Question :What is a Quantization Of Hilbert 16th Problem: Let we have A Vec. Field X on A surface M,it generate a ow and this ow naturally denes a owb on C^inf(M),now lets Quantize M,that is to embedd C^inf(M) in Some B(H),now the ow in C^inf(M) would be converted to a ow in a subspace Of B(H)....(what process you suggest for continuation....) === Subject: bootstrap and Fishers ducial analysis Trying to understand the logic behind the bootstrap, it seems to me more and more that this is similar to Fishers ducial inference. Any opinion? Kjetil Halvorsen === Subject: Re: Takers and leavers: Synthesis > The way I see this, the Western society is torn between two lines of > thought. One is the unconditional growth of the civilization, whatever > the consequences for the planet and for the future generations. The > other is a reaction against it: a return-to-the-roots hippie > environmentalism that seeks to What old-fashioned hippie types (and their neoLuddite descendants) fail to take into account is the sheer self-serving strategic proigacy of Nature based on the policy of unconditional growth. Plants do not produce just enough seeds to ensure zero population growth for themselves, they produce enough for any given species to ll their local biome (in some cases, blanket the planet) in one generation. Similarly, animals do not produce just enough progeny to ensure ZPG for themselves. Fortunately, the Law Of Fang And Claw (for plants, chemical warfare and rapacious competition for water and sunlight) keeps any given species from taking over completely. So, whos actually acting in accordance with Nature, hippies or Capitalists? And before you bewail the sheer mass of human esh that will occupy the planet after weve killed everything else off and have to resort to cannibalism because theres nothing else to eat, consider what happened to the poor, unsung anaerobes that formerly exclusively ruled the Earth. Now, us aerobes have to eat each other while trying to nish off the anaerobes. Nature is change in strobing neon caps, not any kind of imaginary idyllic steady state. Environmentalism, at its root, is narrow-minded arrogance. Mark L. Fergerson === Subject: Re: Takers and leavers: Synthesis I agree completely with you. You are an animal, and your law is the law of the jungle. === Subject: Re: Takers and leavers: Synthesis <41b24ea6$0$11754$8fcfb975@news.wanadoo.fr> posting-account=bkwsGwwAAAAQH-_znrrunTeLFmbQyH96 Well ... _where_ do you live? He nailed it as far as Im concerned. Its eons of evolution getting us to the point where the state can pass laws over-ruling God, Darwin, and anyone else who gets into the way :-) Mark ( ... ) === Subject: Re: Takers and leavers: Synthesis > I agree completely with you. > You are an animal, and your law is the law of the jungle. Its the New Barbarism. I hear its all the rage in New York. === Subject: Re: What is a proof, exactly? >However, due to several technical difculties formalising linear algebra I >have begun wondering what exactly it means to have a proof of a theorem. > I dene a formal system as a nitistic function FS(Prf,Thm) that > distinguishes valid proofs from invalid proofs. I think you miss the point. He means the intuitive meaning of proof and how we formalize it. You are talking about particular formalizations (a lower level of abstraction.) There are plenty of formalizations around. Why is that one better than all the others? > A proof in a formal system FS is then a pair , such that > FS(Prf,Thm) = 1. Wouldnt putting it in terms of Recursion Theory be more general? > -Why could I not prove -say- that the set of natural numbers is as large > as the set of reals? > Its not necessarily impossible, but the existence of such a proof would > imply the inconsistency of the formal system in which it was checked, > given standard denitions of sets, the naturals, the reals, and the > predicate is as large as. Again, I think he means in the intitive sense that a proof means that it really is true and is also convincing, hopefully. Of course, I could be wrong. C-B === Subject: Re: What is a proof, exactly? > I dene a formal system as a nitistic function FS(Prf,Thm) that > distinguishes valid proofs from invalid proofs. > I think you miss the point. He means the intuitive meaning of proof > and how we formalize it. You are talking about particular > formalizations (a lower level of abstraction.) I am talking about properties that should hold for *any* formal system. And I presume that we believe that an informal proof is valid just when we believe it could be formalized in a suitable formal system. > There are plenty of formalizations around. Why is that one better than > all the others? Representing a formal system as primitive-recursive proof predicate has a number of advantages. First of all, it is very general, but probably not overly so: the kinds of things that a primitive recursive function can do seem to neatly contain the functionality needed by a proof-checking algorithm. (If it is too general, then perhaps all that is really needed are the functions with polynomial time and space complexity.) Second, it is very expressive. Its possible to write a primitive recursive function that corresponds closely to our intuition of how a proof should be checked. Third, it allows us to calculate an upper bound on how long it will take to check a proof, since the running time of a FS(Prf,Thm) is bounded by f(length(Prf)+length(Thm)), where f is primitive recursive. > Wouldnt putting it in terms of Recursion Theory be more general? It already is - primitive recursive is part of the lexicon of recursion theory. === Subject: Re: What is a proof, exactly? > Generally speaking, this is true. But two important counter-examples > loom large: Megills development of set theory in Metamath, and > Juttings formalization of Landaus _ Foundations of Analysis _ in > Automath. Indeed. Juttings work is the forefather of Automated Reasoning. Since then many many formalisations have been done, in various other systems. Notably Mizar (www.mizar.org) has a large library. Jasper -- The problem with having an open mind is that people toss in garbage === Subject: Re: What is a proof, exactly? > Indeed. Juttings work is the forefather of Automated Reasoning. Could you please elaborate a little more on this. The reason is that Jutting has been one of my mathematics instructors at the Eindhoven University of Technology. Im just curious what happened next. Han de Bruijn === Subject: Re: What is a proof, exactly? >> - What is the status of = (equality) with respect to proofs? Is there >> a >> difference between meaning of and denition of? Is = dened or >> does it only have a (logical) meaning? Can we equate any two mathematical >> objects? > Well, rst-order logic does give a preferential status to that symbol > over the other relational symbols, in that it appears in the axioms and is > not up for interpretation. This, in combination with other posts reminds me of something. I was raised on Dirk van Dalens Logic and Structure, where he states that there are languages of 1st order logic where = doesnt appear, although AFAIR he gives no examples. But now discussion on set theory and specically the axiom of extensionality makes me wonder about the status of = in ZF. Extensionality is (Ax Ay: x=y <=> (Az: zex <=> zey)). Does this dene = in terms of e? That is: what is the language of set theory, does it have 2 relations (e and =) or 1 (only e)? In the latter case, is = a dened notion (by extensionality) or is it a logical thing which is only used in this axiom to axiomatize e? In that case, are the logical = and the relation symbol = two different things? (As to the question about your class notion of truth - every model comes with an interpretation of the various constant, function and relation symbols of the language; a sentence of the language is true if it is true in all models (according to their respective interpretations)) > logically from the axioms or deduction. There is another way to insert > interpretations into logic, which is as a model of a set of logical > statements, namely a set in which each logical symbol corresponds to a > function or relation or element of the set, so that a logical sentence is > translated into a composition of these things concatenated with logical > connectives like -> or for all, and this sentence is either true or > false (or totally undecidable, of > course)...but again, this judgement is made by a person. Mathematical I dont quite get what you mean here. Could you expand, please? >> - Once we dened X as above, we can make additional denitions Y, Z, >> ... >> depending on X. Can we also make denitions depending on Thm? On prf? >> Can you comment on the denition of the reciprocal ($x to >> frac{1}{x}$), which seems to depend not only on $x$ but also on a >> theorem stating that $x neq 0$? Does it also depend on a proof of such >> a theorem? How? > In what way does the denition of X allow us to make additional > denitions? Well, this is common, isnt it? We dene what a ring is, and then we can prove that if there is a unit element (some denitions of ring dont require one) for its muliplication, then its unique. Then one can dene what a division ring (aka. skew eld) is, and if we have that we can dene what a eld is (namely a commutative division ring). Similarly one can dene x to be some real number and f a function on the reals, and then dene y := f(x), depending on f and x. (By the way, do you consider these to be two different kinds of denitions?) > In the sense that the existence of X is a logical statement > with logical consequences, this is perhaps true (of course, thats not a > rst-order > concept). Another way to look at it is that Y and Z depend on certain > logical non-axioms in addition to the axioms, and if X satises those > axioms then, having convinced ourselves that X is our universe of the day, > we will accept Y > and Z as well. That is a rst order statement, and it really just says > that > the any model of Ys axioms is also a model of Xs. Im not sure I follow what youre saying here.What do you mean by X is our universe of the day? I meant X to be just some mathematical object: a real number, a graph, or a category perhaps. These are not universes in any way Im aware of...? I also presume you mean to say that if the existence of object X requires axioms A1...Ai then we also have to accept objects Y and Z if their existence follows from these same A1...Ai? > However, the actual nature of Prf is irrelevant, since its existence is > the verication that Thm is provable, and conversely the provability of > Thm is (by > denition) the existence of Prf. And we can deal with Thm. Okay. So if Prf exists on the same level as X and Thm, could we dene a new object Y:=f(prf)? Lets say Y:=1 if prf uses induction and Y:=0 otherwise? [about the 4colour theorem:] > reproduce the proof, and since as Ive indicated that I belive, proofs are > eventually based on human ideas about how to determine truth. Okay, some systems already do that. They spew out a proof (encoded in a lambda term) of what they think is true. But these proofs are on the level of formal proofs: every reasoning step is minutely described, and so they become very big very soon. Even writing out the proof of n+k=k+n requires tens of lines - okay, its a pretty-printed lambda term with lots of whitespace, but still its a lot. So if you dont trust computers by themselves, and humans cannot check these proofs, what must we think of them? > If a tree falls in a forest, and no one witnesses it in any way, did it > fall? I > would say yes. But theorems arent trees; if a logical statement is > consistent with the axioms then (by the completeness theorem) there exists ^^^^^^^^^^ Okay, I saw your other post :-) > a proof, and the > production of that proof does not change its logical status. All it does > is change our knowledge of it, and its the part where we believe that the > RH is true not because wed like to, but because it has been shown to be > the case by methods we also believe to be the perfect embodiment of our > intuition, that is > important to us. If aliens give us an incomprehensible proof its no > good; and if the four-color theorems proof is incomprehensible, thats no > good either. > Even if its true. Of course, if I understand it and you dont, I can > feel free to use it (provided my understanding indicates its true, of > course), whereas you would warily agree that my manipulation of the > logical symbols is correct, there is one spot in my proofs that needs > further verication. It is interesting to see that you appear to allow the human factor to play a role. I wonder how wide-spread this belief is. Responses, anyone? I know the ofcial blurb that the human factor shouldnt play a role, but I dont believe it, and neither do you, it seems. I suppose that in some axiomatic systems there are theorems - well, statements really - whose minimum-sized proof is too long to even write out physically. What about these? >> I suppose thats enough questions for the time being. Feel free to >> include any pertinent thoughts and additional questions. > Let the tearing down of my amateurish responses begin! Jasper -- The problem with having an open mind is that people toss in garbage === Subject: Re: What is a proof, exactly? Discussion, linux) > But now discussion on set theory and specically the axiom of > extensionality makes me wonder about the status of = in > ZF. Extensionality is (Ax Ay: x=y <=> (Az: zex <=> zey)). Does this > dene = in terms of e? That is: what is the language of set > theory, does it have 2 relations (e and =) or 1 (only e)? In the > latter case, is = a dened notion (by extensionality) or is it a > logical thing which is only used in this axiom to axiomatize e? In > that case, are the logical = and the relation symbol = two different > things? Howdy, Jasper. Whats the difference, really? If one wanted, he could take equality to be dened by extensionality, so that theres really only one relation in ZF, but in this case, he has to introduce the axioms for equality in terms of elementhood. Thus, one would have to introduce a rule of inference something like: (A z)(z in x <-> z in y) P --------------------------- P[y/x] (where P[y/x] denotes a formula obtained by substituting some freely occurring xs in P by ys). Not particularly nice. Heres something worse than this cosmetic complaint. If we take this approach, we could have a model for ZF in which distinct elements satisfy all the same formulas. Nothing particularly bad happens, I guess, but it doesnt satisfy our intuitions. If instead we take equality as primitive and require that it is always interpreted as equality in the model (which seems the real meat of putting it into FOL), then we dont get these perverse models. When we have the freedom to say that equality must be interpreted as equal in the model, then indistinguishable elements are equal (in the model). This is obvious:if x and y are indistinguishable, then M |= x = y and thus x =_M y. Of course, as you know, its been a while since I paid my rst-order dues, so I could be just butt-wrong here. -- Jesse F. Hughes I thought it relevant to inform that I notied the FBI a couple of months ago about some of the math issues Ive brought up here. -- James S. Harris gives Special Agent Fox a new assignment. === Subject: Re: What is a proof, exactly? > [about the 4colour theorem:] Not that it has so much to do with this thread, but if anybody wants to see some *real* map 4colouring, you could take a look at: http://hdebruijn.soo.dto.tudelft.nl/fcp/index.htm Han de Bruijn === Subject: Re: What is a proof, exactly? > - Suppose aliens gave us a pack of paper, full of english text and > mathematical symbols, ending in ... and hence the Riemann Hypothesis is > proved. QED. Suppose also that this text was so intricate and so involved > and so convoluted that no one would be able to understand it in full. Does > this pack of paper constitute a proof? The aliens have landed! Actually, there does _exist_ such a (would-be?) proof here on earth: http://www.math.purdue.edu/~branges/ Han de Bruijn === Subject: Re: What is a proof, exactly? >> Two sets are disjoint if they have no element in common. (A is disjoint from B) iff (forall C (CeA iff (it is not the case that > CeB))). Is that more clear? > Its certainly incorrect. A would be VB, a proper class...?! (V being the > class of all sets) > I propose: > (A is disjoint from B) iff (forall C, (not (CeA and CeB))) > or perhaps (more like your denition): > (forall C in (A union B), (CeA iff (not CeB)) > Jasper Yeah, I should remember not to post when Im sleeping, obviously sticking to the easy stuff isnt good enough since there is no excuse for that error. === Subject: Re: What is a proof, exactly? >> [A proof] cant be required and empty at the same time. It cant at >> the same time be both necessarily present AND absent. > Good point. I was trying to say that x is a theorem just when > ProofPredicate(prf,x) is true, and ProofPredicate may be such that it is > true for those arguments even if prf is something that we might consider > empty, such as is the empty list or the empty tree. >> [...] FS(.,.) is never a proof of anything. That is NOT the sort of >> thing that a proof can be. > Right; that was sloppy of me. I should have written ``nil might be a > valid proof of 5=5 (according to my (idiosyncratic?) denition of a > formal system as a ``nitistic function FS(Prf,x) that distinguishes > valid proofs of x from invalid proofs of x). This is somewhat strange - does this mean that one (formal) proof can prove several things? At least I suppose that if <> proves 5=5 then it also proves e.g. 27=27. (It seems that Im missing some of the posts to the newsgroup - I havent seen George Greenes original post) Jasper -- The problem with having an open mind is that people toss in garbage === Subject: Re: What is a proof, exactly? > This is somewhat strange - does this mean that one (formal) proof can prove > several things? Correct. Consider a Hilbert-style proof system, in which a proof is a sequence of statements, each of which is either an (instance of) an axiom, a premise, or follows from one or more previous statements in the sequence by a truth-preserving rule of inference. Thus, the proof can be considered to prove every sentence in the sequence. > (It seems that Im missing some of the posts to the newsgroup - I havent > seen George Greenes original post) Try Google Groups. === Subject: Re: What is a proof, exactly? Regarding the subject eld contents: A proof is a complete justication for the truth of a statement. G C === Subject: Re: What is a proof, exactly? > Regarding the subject eld contents: > A proof is a complete justication for the truth of a statement. > G C questions I raised? Please do. Jasper -- The problem with having an open mind is that people toss in garbage === Subject: Re: What is a proof, exactly? > I wonder if Jasper Stein has benetted from the 42 mails sent in > reply to his original post. How will they affect his formalization of > linear algebra; is he closer to being able to demonstrate to his > thesis committee that he has grasped the prologemena for automated > reasoning? is already nished (publicly available even), and although I intend to do some additional updating and polishing, this thread does not inuence the contents of +that+. My questions were of a more general nature. Ive been in this community for some while now, but Ive started wondering if what were doing is actually the same thing that we promise, namely computer-checked proofs. As I said Ive encountered some strange problems and oddities in my formalisation, and I would know how the Average Mathematician thinks about some of these issues. Jasper -- The problem with having an open mind is that people toss in garbage === Subject: Re: What is a proof, exactly? > My questions were of a more general nature. Ive been in this community for > some while now, but Ive started wondering if what were doing is actually > the same thing that we promise, namely computer-checked proofs. As I said > Ive encountered some strange problems and oddities in my formalisation, > and I would know how the Average Mathematician thinks about some of these > issues. I would say that, if computer proof-checking is ever going to take off, then thats a less important question than usability. I think everyone can live with the fact that a usable proof-checking system will have bugs. The real question is, is there a practical way for me to enter a proof Im working on and have the system nd my errors in reasoning, without requiring me to break my proof down into sand-grain size steps? Ive often wished for that. I mean, when Im writing a program, I just hit the compile button and it nds all sorts of errors; I dont have to personally pore over the document to nd them. It would be very nice to have something similar when doing math, even if an occasional error is missed. But Im somewhat skeptical about the notion, because it seems to me that one of the major goals of research in mathematics is to nd new techniques, and the software would need constant updating to accomodate them. Its hard to see how it wouldnt always be years behind the state of the art. Only the less interesting new techniques can be chunked into theorems that say everything that needs to be said--most of them are more like very high-level, very complex, rules of inference. === Subject: Re: What is a proof, exactly? > OK. I will not pursue this argument any further. That is, until we would > arrive at the nal question I had in mind: is x different from {x} ? Ah, but thats a very interesting question indeed! Although it has little bearing on this thread, as you say. > Which belongs to another thread: > Re: Set inclusion and membership I hope to have time to read this one As to your question, different set theories answer this differently. ZF says theyre different, because youd violate the axiom of foundation. Quines NF (New Foundations) however +does+ equate them, or so Ive been told. And then you have various forms of hyperset theory ZFA where there are some sets that obey x={x} and some dont. (In some forms theres only one such set, called Omega; in others you have plenty) Read Vicious circles if youre interested (by either Aczel or Barwise, I forget who) Jasper -- The problem with having an open mind is that people toss in garbage === Subject: Re: What is a proof, exactly? > it to what is inside the ve in the blue set. If you dont have sets > all the way down, then evetually you get to something that doesnt > have equality dened and it causes equality to become undened for > everything that contained it, and everything that contained that and > so on transnitely. Do you mean set equality specically, or is this more generally a denial of my question Can we equate any two mathematical objects?? > mean the usual things. The thing is that set equality itself is an > equivalence relation itself that is too big to be a set. So sets > can do relations on specic sets, but sets cant do relations on all > sets in general. So - would you say wed need two different kinds of equalities, the set theoretic ones (equality between members of a set), and another one that spans all sets simultaneously? Would this be a logical equality? And since this equality is a proper class, what would this equality be if we consider Godel-Bernays instead of Zermelo-Frankel? Jasper -- The problem with having an open mind is that people toss in garbage === Subject: Re: What is a proof, exactly? >> - Does the extremely simple theorem 5=5 need a proof? Why (not)? > Yes, even more you need a denition of 5 = and a standard of > proof. Why? Because thats what proving things is all about. If you Exactly. I believe thats what Im after. You say we need a standard of proof - Im trying to get to know what that standard looks like. As Ive said, I come from the automated reasoning community, and I believe our own standard of proof is skewed. So I want to know what the standard standard of proof is. People have answered that there is a difference between formal proof and informal proof, and that the informal proof should convince the reader that there is a formal proof, being a sequence of statements following from axioms or from rules of inference. But no one seems to bother actually carrying out this transition. Moreover, I expect that if I write a paper containing only formal proofs, itd be (1)huge (2)too hard to follow even for relatively simple informal proofs (3)unpublishable in any regular math journal. So why do we care about formal proof? > dene x=y as Az (zex <=> zey) then there IS a provable theorem > Ax x=x, Nice to see the dened equality boil down to a logical one in this case (after all, for all predicates P we have Az P(z)<=>P(z)) > so then as long as 5 is a set (usually > {{},{{}},{{{}}},{{{{}}}},{{{{{}}}}}}) I thought usually it something else, but never mind :-) { {}, {{}}, { {}, {{}} }, { {}, {{}}, { {}, {{}} } }, { {}. {{}}, { {}, {{}} } }, { {}, {{}}, { {}, {{}} } } } } > then 5=5 is a theorem too, the > proof of the more general theorem already covers when x happens to be > 5 assuming you dened 5 to be a set. So you assume everything to be a set. It is said that this can be done in principle, although personally Id like to rephrase it as everything can be modeled using sets since when I talk about numbers or graphs or vectorspaces, I dont think of these things as sets at all. Theyre more like urelements; a primitive notion that doesnt decompose into sets, although we can model them in pure set theory. Ive been wondering: since math is all in the mind and on paper, wouldnt it be possible to actually have a foundation for mathematics based purely on non-sets? Im thinking of a term model like kind of thing. >> - How is a proof different from just an explanation? Is there a >> difference >> at all? > The usual standard is that a proof (of T relative to A) is a > demonstration that if T were false then A would be false too. So if ??? That cant possibly be true. Deriving A->T from ~T -> ~A isnt even possible if you havent got the law of excluded middle (ie. forall P, P or ~P) or equivalently double negation (forall P, ~~P -> P). But even if you do have them, this is only one means of inference, certainly not the only one. > the assumption is A, then A is a theorem, in the usual standard. > Im assuming the is an ordinary rst order logic system. This raises another interesting question: is informal maths formalisable using FOL only? I think we use higher order logic, actually. FOL might be enough if you say everything is a set, but if you dont think thats true then FOL seems rather restrictive. >> - Suppose we dened an object X. Then we state a theorem Thm. about X, >> which we subsequently prove by a proof prf. According to the main >> school(s) of mathematics, what is the ontological status of X, Thm, and >> prf? Does Thm exist in the same sense as X exists? Does prf exist >> in the same sense as Thm? > Im not sure you meant this as written, just dening something > doesnt entail proving that it exists, so the ontological status of X > is undetermined at this point. Okay, thats true. But if we can prove the existence of X, then whats your opinion? If you need an example, lets take this one: X := { {} } Thm := (forall a,b: (aeX and beX) -> a=b) (ie. X has 1 element) prf := (take any proof you like, there must be lots) >> - Once we dened X as above, we can make additional denitions Y, Z, >> ... >> depending on X. Can we also make denitions depending on Thm? On prf? >> Can you comment on the denition of the reciprocal ($x to >> frac{1}{x}$), which seems to depend not only on $x$ but also on a >> theorem stating that $x neq 0$? Does it also depend on a proof of such >> a theorem? How? [...skipping denition == shorthand, with which I agree mostly - in type theory you ARE doing something: making a denition means extending the context, unfolding a denition is doing a delta reduction...] > the point is > that there is a function f that takes each non-zero real to its > reciprocal, so 1/x can be thought of as f(x) and checking that ~x=0 is > just like checking that a number is in the domain of a function and > then nding the value of the function. My point in this question is exactly that: how do you check whether x is in the domain of the function? In general thats undecidable. If youre a classical mathematician (ie. believing in classical logic) then decidability isnt an issue in principle, but in practice it is. The prime example would be the number (lets call it a) dened by decimal expansion, where digit n is 1 if a sequence of 99 9s has occurred prior to position n in the expansion of pi, and 0 otherwise. So (classically speaking) a is either 0 or 0.000...01111... Now how does one check whether 1/a makes sense? For this we need a proof that actually a=/=0, ie. a proof that pi has a series of 99 9s somewhere in its expansion. (And as of today this is unproved.) > so why should I trust a proof by that is checked only by a computer? Good answer. The main automated-reasoning response is that (1) we have the de Bruijn criterion: the computer program code must be small enough that it can be checked by hand. (2) we can let the computer emit a proof object that contains the reasoning it followed in proving the theorem. This proof object can then be checked independently. Does this inuence your opinion? Why (not)? >> - This code calculated and checked a few thousands of congurations. >> If >> these calculations and the checking were done by mathematicians, would >> the proof be still as controversial? Why (not)? > Its not one mathematician versus one computer, its the fact that so > few independant minds would actually go through the bother of checking > it. Its just not as trustworthy, because its been checked a smaller > number of both quality and independant times. Allow me to play the devils advocate: in the middle ages, everyone thought the earth was at. Today everyone thinks the theorem is true, that this-and-this reasoning is sound. You sound like you think maths is a sociological enterprise. Is that true? > If we could understand the symbols, then we could make a machine (from > scratch) to check the proof. More likely I would expect that > mathematicians would prefer to make a machine that attempted to > *simplify* the proof. Then you can just read the simplier proof, a > much better case. Now for pack of paper substitute my computers proof object, which is indeed a string of symbols that my computer has checked. How does this hold with your remark (see above) > so why should I trust a proof by that is checked only by a computer? questions. (And this goes for others who responded as well!) I hope you and others can nd the time to answer my further questions too - theyre quite important to me, because they bear on what Im doing for my job every day, and Im doubtful (as you may have noticed)... Jasper -- The problem with having an open mind is that people toss in garbage === Subject: Re: How to measure randomness of a deck of cards? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iB3KH3I03194; >I read several papers, there are few approaches such as variation >distances, birthday bound and markov chain. I need a >method/algorithm to determine the randomness of my deck of cards >(permutation) >I would be appreciated if anybody can give me some ideas or good solutions. Estimate how much information is lost due to each shufe. Ideally, it is as much as 1 bit. This would occur if you couldnt tell which half of the deck contains a given card (assuming you knew which half before the shufe). Assuming each shufe is independent, each destroys one bit of information. In practice about 8 or 9 shufes provides pretty good randomness. Google for more including experiments phil === Subject: nearest common ancestor on Stern-Brocot tree? For example, the nearest common ancestor for 5/9 and 3/4 is 2/3. Indeed 5/9 = LRLLL 3/4 = LRR 2/3 = LR The straightforward way to calculate NCA is by converting the rational number into LR representation applying (extended) Euclidian algirithm. Is there a formula for NCA? === Subject: Circle-line intersection Hi there! Dont tell me to look at other posts in this forum. This is a special problem aimed to they who know math. The question how the get t, look below. I thought I solved the problem but the outcome was a failure. I got a correct equation from mathworld but that is not what I want though. A line(L) intersects with a circle(C), tell me if something is wrong. All positions orients from circle position which is (0,0). t is what I want. L = A + (B-A)*t where D = B-A Lx = cos(v)*R Ly = sin(v)*R A combination of equations: cos(v)*R = Ax + (Dx)*t sin(v)*R = Ay + (Dy)*t I do this because the points of the line intersects on the circle-line which distance is R from origo. cos(v)*R is the horizontal distance and sin(v)*R is the vertical distance. Hope you understand my thinking here. Only if could draw with a pen :). And no we do power of two at both sides of equation. This I do because of a forumla which Ive to merge with this equation, see below. cos^2(v)*R^2 = (Ax + Dx*t)^2 which also are ((Ax + Dx*t)^2) cos^2(v) = -------------------- R^2 sin^2(v)*R^2 = (Ay + Dy*t)^2 which also are ((Ay + Dy*t)^2) sin^2(v) = -------------------- R^2 Ok now to the hard part. This is a math rule: sin^2(v) + cos^2(v) = 1 and this gives cos^2(v) = 1 - sin^2(v) Now we combine the rst combination with the formula above: ((Ax + Dx*t)^2) cos^2(v) = -------------------- R^2 will be ((Ax + Dx*t)^2) 1 - sin^2(v) = -------------------- R^2 And know we just take away the sin-function: ((Ay + Dy*t)^2) ((Ax + Dx*t)^2) 1 - -------------------- = -------------------- R^2 R^2 Now I can get t. The equation will be, after some ipping and opping of the variables, look like this: t^2*(Dx^2 + Dy^2) + t*(2*Ax*Dx + 2*Ay*Dy) - R^2 + Ay^2 + Ax^2 = 0 If you dont see what this is I tell you this: ax^2 + bx + c = 0 Now if you know what math is you know what it is. Now the question: Why DOESNT THIS WORK!!! :| === Subject: Re: Circle-line intersection days. My association with the Department is that of an alumnus. >A line(L) intersects with a circle(C), tell me if something is wrong. >All positions orients from circle position which is (0,0). You might want to say what it is you are trying to do... >t is what I want. >L = A + (B-A)*t >where D = B-A There was no D so far.... You mean: L is a line going through the points A and B, and we obtain it (by vector addition) as A + Dt, where D = (B-A) (A, B being vectors with initial point at the origin and terminal point at the point A and B, presumably, and t being a real number parameter...) Is that right? >Lx = cos(v)*R >Ly = sin(v)*R What are Lx and Ly supposed to mean? What is R? What is v? I assume Lx means the x coordinate of the point on the line, and Ly means the y coordinate (as below you have Ax, Dx, Ay, Dy, presumably being x coordinate of A, x coordinate of D, etc). Meaning... you are trying to gure out the intersection? (x,y) are the coordinates of the intersection? And then v is simply the angle, R the radius of the circle? Sounds reasonable to me... >A combination of equations: cos(v)*R = Ax + (Dx)*t > sin(v)*R = Ay + (Dy)*t >I do this because the points of the line intersects on the circle-line >which distance is R from origin. cos(v)*R is the horizontal distance >and sin(v)*R is the vertical distance. Hope you understand my thinking >here. Only if could draw with a pen :). Lets see. We are working on the plane. You are describing the circle as being given by all points of the form (R*cos(v),R*sin(v)), as v ranges over, say, 0 to 2pi. You are describing the line as going through the points A=(a_1,a_2) and B=(b_1,b_2), so you let D = B-A = (b_1-a_1,b_2-a_2) = (d_1,d_2), and the line becomes (x,y) = (a_1,a_2) + t(d_1,d_2) where t ranges over all real numbers. So any intersection between the circle and the line would have to satisfy a_1 + td_1 = R*cos(v) a_2 + td_2 = R*sin(v) for some value of t and some value of v. >And no we do power of two at both sides of equation. You are squaring both equations... Okay... HOWEVER: remember that because you squared the original equations, you may have introduced new solutions that were not part of your original intersections. For example, if you nd a point where R*(cos(v)) = - (a_1+td_1) R*(sin(v)) = (a_2 + td_2) then these values of v and t will ALSO satisfy R^2*cos^2(v) = (a_1+td_1)^2 R^2*sin^2(v) = (a_2+td_2)^2 even though they do not satisfy the original equation. In fact, the solutions to the following four systems: ORIGINAL: a_1 + td_1 = R*cos(v) a_2 + td_2 = R*sin(v) Plus a_1 + td_1 = -R*cos(v) a_2 + td_2 = R*sin(v) a_1 + td_1 = R*cos(v) a_2 + td_2 = -R*sin(v) and a_1 + td_1 = -R*cos(v) a_2 + td_2 = -R*sin(v) will also satisfy (a_1 + td_1)^2 = R^2*cos^2(v) (a_2 + td_2)^2 = R^2*sin^2(v) So ->not every solution you nd after squaring needs to be a solution of your original equation<-. Every solution to your original equation will also be a solution to the new one, but the new one may have more solutions. (Its like, if you start with x= 3, there is only one solution; but if you square it, you get x^2 = 9, and now both x=3 and x=-3 are solutions). >This I do >because >of a forumla which Ive to merge with this equation, see below. > cos^2(v)*R^2 = (Ax + Dx*t)^2 > which also are > ((Ax + Dx*t)^2) > cos^2(v) = -------------------- > R^2 sin^2(v)*R^2 = (Ay + Dy*t)^2 > which also are > ((Ay + Dy*t)^2) > sin^2(v) = -------------------- > R^2 Fine. Since a_1 + td_1 = R*cos(v) a_2 + td_2 = R*sin(v) then cos(v) = (a_1 + t*d_1)/R sin(v) = (a_2 + t*d_2)/R >Ok now to the hard part. >This is a math rule: > sin^2(v) + cos^2(v) = 1 Identity. Yes. >and this gives > cos^2(v) = 1 - sin^2(v) Yes. >Now we combine the rst combination with the formula above: > ((Ax + Dx*t)^2) > cos^2(v) = -------------------- > R^2 will be ((Ax + Dx*t)^2) > 1 - sin^2(v) = -------------------- > R^2 >And know we just take away the sin-function: > ((Ay + Dy*t)^2) ((Ax + Dx*t)^2) > 1 - -------------------- = -------------------- > R^2 R^2 Easier to just add the two equations you had: a_1 + td_1 = R*cos(v) a_2 + td_2 = R*sin(v) we have R^2 = R^2(cos^2(v) + sin^2(v) = (a_1+td_1)^2 + (a_2 + td_2)^2. >Now I can get t. The equation will be, after some ipping and >opping of the variables, look like this: >t^2*(Dx^2 + Dy^2) + t*(2*Ax*Dx + 2*Ay*Dy) - R^2 + Ay^2 + Ax^2 = 0 >If you dont see what this is I tell you this: ax^2 + bx + c = 0 >Now if you know what math is you know what it is. >Now the question: Why DOESNT THIS WORK!!! :| Looks right. What do you mean by This does not work? Maybe you mean you sometimes get wrong answers? Thats because of the observation above. You have to make sure that whatever answers you get actually satisfy the original equations. Plug them in. They should work ne. -- Its not denial. Im just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Circle-line intersection t^2*(Dx^2 + Dy^2) + t*(2*Ax*Dx + 2*Ay*Dy) - R^2 + Ay^2 + Ax^2 = 0 but when I converted it to an equation of the second degree I forgot an expression (Dx^2 + Dy^2). ;D It will look like this (sqrt = square root of): Ax*Dx + Ay*Dy (Ax*Dx + Ay*Dy)^2 Ax^2 + Ay^2 - R^2 t = - ------------- +- sqrt(----------------- - ------------------) Dx^2 + Dy^2 (Dx^2 + Dy^2)^2 Dx^2 + Dy^2 Im doing this as a hobby programmer. Maybe there are easier ways, but for me its fun to solve this myself, and the purpose of this equation is to compare two vectors (which has the same angles but different origins). The vector which has the lowest t value above 0 is the vector which rst intersects the circle. Im planning a circle-circle collision-detection algorithm which tells the user where the rst collision-point is, and the vectors in this equation has the same angle as the moving vector of the circle, but the origins is at the both sides of the circle, these two are needed because if Id only one, the actual collision of the circles may miss. But this will not work if I take a bigger circle to collide with a smaller one because of the radius of the smaller one is lesser than the distance between the two vectors at the bigger circle. Therefore I made a rule to always check smaller circles against bigger ones (and not vice versa). Hope you understand my theory, and thank you for reading through and comment everything. /Johannes === Subject: Shiing-shen Chern (Chen Xingshen) passes away at 93 Just grabbed off Xinhua web site: Noted mathematician passes away in Tianjin TIANJIN, Dec. 3 (Xinhuanet) -- Shiing-shen Chern (Chen Xingshen), a world-renowned overseas Chinese mathematician, 93, died of illness at his home at Nankai University in north Chinas Tianjin Municipality at around 7:15 p.m. Friday, the university announced. Chern, a US citizen, is best known for his achievements in the study of differential geometry. He was born in 1911 in Jiaxing, Zhejiang Province, east China. He graduated from Nankai University in 1930 and received further education at Qinghua University and the University of Hamburg in Germany. He taught at several Chinese and US universities -- including Princeton University, the University of Chicago, and the University of California, Berkeley -- and is the only Chinese to win the Wolf Prize -- the most distinguished award in the international mathematics eld. The International Astronomical Union ofcially named asteroid No. 1998CS2 after the noted mathematician in November for his outstanding contributions to human society. === Subject: Re: Shiing-shen Chern (Chen Xingshen) passes away at 93 > Just grabbed off Xinhua web site: > Noted mathematician passes away in Tianjin > TIANJIN, Dec. 3 (Xinhuanet) -- Shiing-shen Chern (Chen Xingshen), a > world-renowned overseas Chinese mathematician, 93, died of illness at > his home at Nankai University in north Chinas Tianjin Municipality at > around 7:15 p.m. Friday, the university announced. > Chern, a US citizen, is best known for his achievements in the > study of differential geometry. He was born in 1911 in Jiaxing, Zhejiang > Province, east China. He graduated from Nankai University in 1930 and > received further education at Qinghua University and the University of > Hamburg in Germany. > He taught at several Chinese and US universities -- including > Princeton University, the University of Chicago, and the University of > California, Berkeley -- and is the only Chinese to win the Wolf Prize -- > the most distinguished award in the international mathematics eld. > The International Astronomical Union ofcially named asteroid No. > 1998CS2 after the noted mathematician in November for his outstanding > contributions to human society. I had the privilege of taking an upper division class (upper division = 3rd or 4th year undergrad math major) from Professor Chern in 1976 at U.C Berkeley. I remember him as a warm, down to earth man. Approachable to us students. He cared about his students and about people. Im sad to hear of his passing. === Subject: Re: Shiing-shen Chern (Chen Xingshen) passes away at 93 Moment of Silence. W. Dale Hall ??? > Just grabbed off Xinhua web site: > Noted mathematician passes away in Tianjin > TIANJIN, Dec. 3 (Xinhuanet) -- Shiing-shen Chern (Chen Xingshen), a > world-renowned overseas Chinese mathematician, 93, died of illness at > his home at Nankai University in north Chinas Tianjin Municipality at > around 7:15 p.m. Friday, the university announced. > Chern, a US citizen, is best known for his achievements in the > study of differential geometry. He was born in 1911 in Jiaxing, Zhejiang > Province, east China. He graduated from Nankai University in 1930 and > received further education at Qinghua University and the University of > Hamburg in Germany. > He taught at several Chinese and US universities -- including > Princeton University, the University of Chicago, and the University of > California, Berkeley -- and is the only Chinese to win the Wolf Prize -- > the most distinguished award in the international mathematics eld. > The International Astronomical Union ofcially named asteroid No. > 1998CS2 after the noted mathematician in November for his outstanding > contributions to human society. === Subject: Re: Graphing polynomial equations >xyz^3 + 3xz + 3zy + xyz + zyx^2 + x - y + 3 = 0 >Obviously you would loop through x and y and calculate >z, and wed use a numerical approximation technique to >do that. Im trying to write a program that would plot >the graph. The problem boils down to applying a >numerical algorithm to a quintic like this: >z^5 + a4 z^4 + a3 z^3 + a2 z^2 + a1 z + a0 = 0. >But the problem Im having is this: Most numerical >approximation algorithms require an initial guess. So, what >would be a good algorithm to nd a good initial >guess? You probably realize that a degree-5 polynomial can have up to 5 zeroes, so you would want 5 (or maybe more) initial guesses. The high and low guesses should be fairly easy, because theres no such thing as too high or too low for the tangential method (sorry I forgot who discovered that or Id give him credit!). Even if youre using some other numerical method you can still use the tangential results to seed it. You can get the other possibilities by looking at the zeroes of the derivative, which are potentially relative mimima and maxima. Any zeroes other than the rst and last will be between a relative maximum and a relative minimum. You may have to apply this process recursively until you get down to degree 2 where you can use the quadratic equation. Once youre started, you can use results from neighboring (x,y) as guesses. They wont always work out, but when they do it will save time. --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. === Subject: Re: Graphing polynomial equations >Houw would one go about plotting a 3D surface graph of >a quintic or higher equation in three variables? >Like this: >xyz^3 + 3xz + 3zy + xyz + zyx^2 + x - y + 3 = 0 > Try surf.sourceforge.net . === Subject: Number of prime factors of an odd perfect number Content-Length: 215 Originator: rusin@vesuvius What is the known upper bound on the number of prime factors / distinct prime factors of an odd perfect number? Does anyone have a conjecture on what the tighest lower bound would be? Edmond === Subject: Re: Number of prime factors of an odd perfect number posting-account=tN6qcA0AAACUaa8g28ioHaW6IKJhmtDk > What is the known upper bound on the number of prime factors / distinct > prime factors of an odd perfect number? Does anyone have a conjecture on > what the tighest lower bound would be? > Edmond Youve already received some excellent replies, but it might be of interest to note that any constant upper bound here would imply there are only nitely many odd perfect numbers: Dickson showed that there are at most nitely many odd perfect numbers with a prescribed number of distinct prime factors. Dicksons result can be made effective. A striking result of Heath-Brown is that an odd perfect number with k distinct prime factors is bounded by 4^{4^k}. (The proof is entirely elementary.) This has been improved to 2^{4^k} by Nielsen in [1]. Hope this helps, Paul [1] Nielsen, Pace An Upper Bound for Odd Perfect Numbers http://www.emis.de/journals/INTEGERS/papers/d14/d14.pdf === Subject: Re: Number of prime factors of an odd perfect number >> What is the known upper bound on the number of prime factors / > distinct >> prime factors of an odd perfect number? Does anyone have a conjecture > on >> what the tighest lower bound would be? >> Edmond > Youve already received some excellent replies, but it might be of > interest to note that any constant upper bound here would imply there > are only nitely many odd perfect numbers: Dickson showed that there > are at most nitely many odd perfect numbers with a prescribed number > of distinct prime factors. > Dicksons result can be made effective. A striking result of > Heath-Brown is that an odd perfect number with k distinct prime factors > is bounded by 4^{4^k}. (The proof is entirely elementary.) This has > been improved to 2^{4^k} by Nielsen in [1]. > Hope this helps, > Paul > [1] Nielsen, Pace > An Upper Bound for Odd Perfect Numbers > http://www.emis.de/journals/INTEGERS/papers/d14/d14.pdf fact, but I lately realised that if the non-existance of odd perfect numbers is proven, then by virtue of the recent proof that primality testing can be done in polynomial time, it would follow that perfect number testing can also be done in polynomial time. Does anyone think Im mistaken or does anyone have any other corollaries that would follow from the proof of non-existance of odd perfect numbers. Edmond === Subject: Re: Number of prime factors of an odd perfect number > What is the known upper bound on the number of prime factors / distinct > prime factors of an odd perfect number? Does anyone have a conjecture on > what the tighest lower bound would be? I dont know much about recent advances, but there are a few classical results: J. J. Sylvester, L. E. Dickson and H. J. Kanold proved that there is no odd perfect number with 4 prime divisors. Next, I. S. Gradshtein, U. K.9fhnel and G. C. Weber disproved existence of numbers with 4 divisors and C. Pomerance and N. Robbins showed the same for 6 divisors. P. Hagis Jr. and E. Z. Chein proved in 1975 the analogous theorem for 8 divisors and I dont know about any newer results. Pawel Gladki === Subject: Re: Number of prime factors of an odd perfect number > P. Hagis Jr. and E. Z. Chein proved in 1975 the analogous > theorem for 8 divisors and I dont know about any newer results. Of course we are talking about _distinct_ prime factors. For the bound of prime factors counted with repetitions it is known that there are at least 47 factors - see Kevin Hares preprint: http://www.math.uwaterloo.ca/~kghare/Preprints/PDF/P15_Odd.pdf Pawel Gladki === Subject: Re: Number of prime factors of an odd perfect number >> P. Hagis Jr. and E. Z. Chein proved in 1975 the analogous theorem for 8 >> divisors and I dont know about any newer results. > Of course we are talking about _distinct_ prime factors. For the bound of > prime factors counted with repetitions it is known that there are at least > 47 factors - see Kevin Hares preprint: > http://www.math.uwaterloo.ca/~kghare/Preprints/PDF/P15_Odd.pdf interested in known UPPER BOUNDS and a CONJECTURE on what would be the ACTUAL LOWER BOUND. Edmond === Subject: Re: Number of prime factors of an odd perfect number >> P. Hagis Jr. and E. Z. Chein proved in 1975 the analogous theorem for 8 >> divisors and I dont know about any newer results. > Of course we are talking about _distinct_ prime factors. For the bound of > prime factors counted with repetitions it is known that there are at least > 47 factors - see Kevin Hares preprint: > http://www.math.uwaterloo.ca/~kghare/Preprints/PDF/P15_Odd.pdf > interested in known UPPER BOUNDS and a CONJECTURE on what would be the > ACTUAL LOWER BOUND. Known upper bounds? You mean a proof that if there is an odd perfect number it cant have more than so many prime factors? I dont think Ive ever seen a result along those lines - have you? As for conjectured lower bounds, I think those would be innite, as I think the conjecture is that there arent any odd perfect numbers. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Number of prime factors of an odd perfect number > P. Hagis Jr. and E. Z. Chein proved in 1975 the analogous theorem for > 8 > divisors and I dont know about any newer results. >> Of course we are talking about _distinct_ prime factors. For the bound >> of >> prime factors counted with repetitions it is known that there are at >> least >> 47 factors - see Kevin Hares preprint: >> http://www.math.uwaterloo.ca/~kghare/Preprints/PDF/P15_Odd.pdf >> interested in known UPPER BOUNDS and a CONJECTURE on what would be the >> ACTUAL LOWER BOUND. > Known upper bounds? You mean a proof that if there is an odd perfect > number it cant have more than so many prime factors? Exactly > I dont think > Ive ever seen a result along those lines - have you? No, I havent, but Id like to know at least a reasonably tight upper bound. > As for conjectured lower bounds, I think those would be innite, > as I think the conjecture is that there arent any odd perfect numbers. Let us assume that someone does prove this conjecture. What would be the important corollaries thereof ? Edmond === Subject: Re: Number of prime factors of an odd perfect number days. My association with the Department is that of an alumnus. >> P. Hagis Jr. and E. Z. Chein proved in 1975 the analogous theorem for >> 8 >> divisors and I dont know about any newer results. Of course we are talking about _distinct_ prime factors. For the bound > of > prime factors counted with repetitions it is known that there are at > least > 47 factors - see Kevin Hares preprint: http://www.math.uwaterloo.ca/~kghare/Preprints/PDF/P15_Odd.pdf > interested in known UPPER BOUNDS and a CONJECTURE on what would be the > ACTUAL LOWER BOUND. >> Known upper bounds? You mean a proof that if there is an odd perfect >> number it cant have more than so many prime factors? > Exactly >> I dont think >> Ive ever seen a result along those lines - have you? > No, I havent, but Id like to know at least a reasonably tight upper >bound. Myersons point is that, as far as he is aware (and as far as I am aware), there are ->no<- results on upper bounds on the number of prime factors for an odd perfect number. Im sure you would like to know one. But if there are none to be found, youll have to prove one if you want to know one. -- Its not denial. Im just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Number of prime factors of an odd perfect number > P. Hagis Jr. and E. Z. Chein proved in 1975 the analogous theorem > for > 8 > divisors and I dont know about any newer results. >> Of course we are talking about _distinct_ prime factors. For the >> bound >> of >> prime factors counted with repetitions it is known that there are at >> least >> 47 factors - see Kevin Hares preprint: >> http://www.math.uwaterloo.ca/~kghare/Preprints/PDF/P15_Odd.pdf >> interested in known UPPER BOUNDS and a CONJECTURE on what would be the >> ACTUAL LOWER BOUND. > Known upper bounds? You mean a proof that if there is an odd perfect > number it cant have more than so many prime factors? >> Exactly > I dont think > Ive ever seen a result along those lines - have you? >> No, I havent, but Id like to know at least a reasonably tight upper >>bound. > Myersons point is that, as far as he is aware (and as far as I am > aware), there are ->no<- results on upper bounds on the number of > prime factors for an odd perfect number. > Im sure you would like to know one. But if there are none to be > found, youll have to prove one if you want to know one. I guess I have to set out to prove one myself. But Im still curious as to how ( or maybe why ) so much work has been done on the lower bound, whereas we have no results regarding the upper bound. Also, is there a good example of a similar upper bound proof, i.e., an upper bound on some property of a hypothetical number ? Edmond === Subject: Re: Number of prime factors of an odd perfect number days. My association with the Department is that of an alumnus. >> Myersons point is that, as far as he is aware (and as far as I am >> aware), there are ->no<- results on upper bounds on the number of >> prime factors for an odd perfect number. >> Im sure you would like to know one. But if there are none to be >> found, youll have to prove one if you want to know one. >I guess I have to set out to prove one myself. But Im still curious as to >how ( or maybe why ) so much work has been done on the lower bound, whereas >we have no results regarding the upper bound. Because... how would you prove an upper bound on the number of distinct prime factors? The lower bounds are proven by using the fact that if m and n are relatively prime, then sigma(m*n) = sigma(m)*sigma(n), with sigma(k) = sum of all divisors of k, and showing that if q1,...,qs are distinct prime powers (for which it is easy to calculate sigma), then the product of the sigma(qi) simply does not add up to twice q1*...*qs, for small values of s. But how would you prove an upper bound? How would you even ->approach<- such a problem? The upper bound in the even case comes by showing that if you write the number as 2^n*q, with q odd, then the only way this works out is if q is a prime (and, more specically, a prime of the form 2^{n}-1). It is not really a generalizable argument. > Also, is there a good example >of a similar upper bound proof, i.e., an upper bound on some property of a >hypothetical number ? Yes. Ramsey numbers have well known upper and lower bounds, but many of them are not exactly known. But here, you prove existence by proving an upper bound. -- Its not denial. Im just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Number of prime factors of an odd perfect number > Myersons point is that, as far as he is aware (and as far as I am > aware), there are ->no<- results on upper bounds on the number of > prime factors for an odd perfect number. > Im sure you would like to know one. But if there are none to be > found, youll have to prove one if you want to know one. >>I guess I have to set out to prove one myself. But Im still curious as to >>how ( or maybe why ) so much work has been done on the lower bound, >>whereas >>we have no results regarding the upper bound. > Because... how would you prove an upper bound on the number of > distinct prime factors? The lower bounds are proven by using the fact > that if m and n are relatively prime, then sigma(m*n) = > sigma(m)*sigma(n), with sigma(k) = sum of all divisors of k, and > showing that if q1,...,qs are distinct prime powers (for which it is > easy to calculate sigma), then the product of the sigma(qi) simply > does not add up to twice q1*...*qs, for small values of s. > But how would you prove an upper bound? How would you even > ->approach<- such a problem? The upper bound in the even case comes > by showing that if you write the number as 2^n*q, with q odd, then the > only way this works out is if q is a prime (and, more specically, a > prime of the form 2^{n}-1). It is not really a generalizable argument. participants. Edmond === Subject: Re: Smullyans Quiz Problem by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iB3LRdl09441; >My own analysis is that the error is the insistance that the >students must have a rational basis for believing >that there is or is not going to be an exam. >In essence the professor is saying to the students: >you have a rational basis for believing there >will be an exam if and only if you do not have a rational >basis for believing there will be an exam. If the >students believe the professor, they must conclude >that they do not have a rational reason to either >to believe that there will be an exam or to believe that >there will not be an exam. The best resolution is that >the students should not believe the professor. >In real life, of course, the professor can only >say There is a very high probability that you >will have an unexpected exam next week. This statement >does not lead to a paradox, so the impression that >a professor can make this kind of statement and >communicate information is justied. >In other forms of the paradox, where there >is no prediction element and hence no probablity, >the statement (e.g. there is an unexpected egg) >cannot be believably made. perhaps one should think seriously what unexpected means. to do this one could try to formalise expectations. but this is readily done in elementary game theory, and from there one also gets the wiseness of formalising the situation as a game. with reasonable pay-off functions it is clear that there are no pure strategy equilibria but the professor uses a mixed strategy to determine the date of the exam. the students expectations about the mixed strategy are, of course, correct in equilibrium but they do not know the realisation of the mixed strategy, i.e. the true date of the exam. in most cases then the exam will be a surprise exam in the sense that the studentd cannot predict the date with probability one. only if the realisation is friday they will be able to predict on thursday evening that the exam will be on friday with probability one. the point here is that this is not so much a problem of logic as a problem of formalising the situation correctly. and when one does! formalise it it turns out to be a game. === Subject: Re: Smullyans Quiz Problem >My own analysis is that the error is the insistance that the >students must have a rational basis for believing >that there is or is not going to be an exam. >In essence the professor is saying to the students: >you have a rational basis for believing there >will be an exam if and only if you do not have a rational >basis for believing there will be an exam. If the >students believe the professor, they must conclude >that they do not have a rational reason to either >to believe that there will be an exam or to believe that >there will not be an exam. The best resolution is that >the students should not believe the professor. >In real life, of course, the professor can only >say There is a very high probability that you >will have an unexpected exam next week. This statement >does not lead to a paradox, so the impression that >a professor can make this kind of statement and >communicate information is justied. >In other forms of the paradox, where there >is no prediction element and hence no probablity, >the statement (e.g. there is an unexpected egg) >cannot be believably made. > perhaps one should think seriously what unexpected means. > to do this one could try to formalise expectations. > but this is readily done in elementary game theory, and from there > one also gets the wiseness of formalising the situation as a game. > with reasonable pay-off functions it is clear that there are no pure > strategy equilibria but the professor uses a mixed strategy to > determine the date of the exam. the students expectations about the > mixed strategy are, of course, correct in equilibrium but they do not > know the realisation of the mixed strategy, i.e. the true date of the > exam. in most cases then the exam will be a surprise exam in the sense > that the studentd cannot predict the date with probability one. > only if the realisation is friday they will be able to predict on > thursday evening that the exam will be on friday with probability > one. the point here is that this is not so much a problem of logic > as a problem of formalising the situation correctly. and when > one does formalise it it turns out to be a game. Ok, but this does not change the basic nature of the paradox. Consider the following form of the paradox (I prefer this as it avoids discussion of prediction and states of knowledge, gives a more concrete meaning to trustworthy, and incorporates the idea of mixed strategy). We have two players A and B, and two cards, a king and an ace. A orders the cards and places them face down. The cards are turned over until the king is turned over. At this point the game ends. Before each card is turned over, B names an integer between 0 and 100. This is used as the % probability that B will place a $1 bet at even odds that the next card turned over will be the king. Bs best strategy is to say 50 before the rst card is turned over and 100 if the second card is to be turned over. No matter what strategy A uses, this ensures B an average return of $0.5 (So the game is worth $0.5 to B). A is trustworthy. That is if A says the rst card is the king B will say 100 before the rst card is turned over. As problematic statement is (after he has placed the cards face down) If you act rationally, you will not say 100 during this play of the game B argues If the rst card is the ace, then I will certainly say 100 before the second card is turned over, so the rst card must be the king, so rationally I should say 100 before the rst card is turned over. But then A knows his statement is always false, so I should not draw any inference from it and say 50 before the rst card is turned over. But then A know that his statement is only true if the rst card is the king, so I should treat his statement the same as if he had said the rst card is the king and say 100 before the rst card is turned over. But then A knows his statement is always false ... Surely it is not rational for B to refuse to play a game that has positive expectation. The unappealing, but seemingly inescapable conclusion is that B cannot behave rationally! -William Hughes === Subject: How Many M&Ms? Hello All. I am writing in reference to a contest a friend of mine is having. The rules are as follows. I bought a brand new clear plastic jar. It is lled with the standard size M & M candies. The jar is round and 8 inches tall. The diameter of the jar inside is 4 inches across. I bought over 4 pounds of M & Ms to ll it. I counted every M & M before lling the jar to the very top.. You can probably guess what I am seeking: How many M & Ms are in the jar as described? If anyone can help me on this it would be wonderful as I am not a good mathmatician. Joey === Subject: Re: How Many M&Ms? > Hello All. I am writing in reference to a contest a friend of mine is > having. The rules are as follows. > I bought a brand new clear plastic jar. It is lled > with the standard size M & M candies. > The jar is round and 8 inches tall. The diameter of > the jar inside is 4 inches across. > I bought over 4 pounds of M & Ms to ll it. I counted > every M & M before lling the jar to the very top.. > You can probably guess what I am seeking: > How many M & Ms are in the jar as described? > If anyone can help me on this it would be wonderful as I am not a good > mathmatician. > Joey The only precise method is to count them. Probably the next most accurate method is to esimate the number by weight. Count out as large a fraction of the total as you have patience for and weigh them to nd an average weight per piece. Then weigh the total (less jar weight) and divide by the average weight per piece. Or maybe the manfacturer will give you the average weight per piece. === Subject: Re: How Many M&Ms? posting-account=Qiuj5AwAAACmGnmS12qcvqA9IXzD0s4L I dont get this post.... Joey counted the M&Ms but wants help knowing how many? oh she counted the 4 pounds of M&Ms then threw in a portion of them into the jar. If youre running the competition and people have to guess, they might expect you to count the exact number in the jar. Of course you could just take the median of the guesses and select them as winner, quoting the prize amount as a few different from what they guessed, but if the jar is the prize they might check, and be honest and hand it over to the actual best guesser, this could result in all sorts of bad publicity, you might nd your next M&M competition youll be eating them yourself. A few questions, the jar you bought was empty right? You and your friend are both running the competition? You counted the M&Ms of all the bags you bought? Can you count the remaining M&Ms after you put them in the jar and substract? Is this like that puzzle, if there are 5 apples and I take away 2 how many do I have? Are we eligible for the prize to calculate our guesses? 2601. Herc === Subject: Re: How Many M&Ms? Fortunately, there has been some research on M&M packing: The net is that the density of packing is around 71% (if packed in an irregular fashion). Further, it looks like the someone has actually measured an individual M&M: http://www.kleinbottle.com/Bernie_Tao.htm Which a result of 0.45239 cm. so, volume of jar is: PI X 2 X 2 X 8 = 100.54 inches^3 = 100.54 X 2.54 X 2.54 X 2.54 = 1647.407 cm3 So, number of M&Ms is approx. = 1647.407 X 0.71 / 0.45239 ~ 2585 Now the result is probably only good to around 2 signicant gures, and there will probably be some rounding down due to the sides of the container, so I would say aound 2500 M&Ms ? Am I close? -Darren > Hello All. I am writing in reference to a contest a friend of mine is > having. The rules are as follows. > I bought a brand new clear plastic jar. It is lled > with the standard size M & M candies. > The jar is round and 8 inches tall. The diameter of > the jar inside is 4 inches across. > I bought over 4 pounds of M & Ms to ll it. I counted > every M & M before lling the jar to the very top.. > You can probably guess what I am seeking: > How many M & Ms are in the jar as described? > If anyone can help me on this it would be wonderful as I am not a good > mathmatician. > Joey === Subject: Re: How Many M&Ms? In sci.math, darrenn : > Fortunately, there has been some research on M&M packing: > The net is that the density of packing is around 71% (if packed in an > irregular fashion). > Further, it looks like the someone has actually measured an individual M&M: > http://www.kleinbottle.com/Bernie_Tao.htm > Which a result of 0.45239 cm. Pedant point. An M&M is generally a attened round thing and probably should be measured in all three dimensions, or perhaps one can pour a number into a liquid (although water will melt off the coatings, methinks) and measure the displacement. > so, volume of jar is: > PI X 2 X 2 X 8 = 100.54 inches^3 = 100.54 X 2.54 X 2.54 X 2.54 = 1647.407 > cm3 > So, number of M&Ms is approx. = 1647.407 X 0.71 / 0.45239 ~ 2585 > Now the result is probably only good to around 2 signicant gures, and > there will probably be some rounding down due to the sides of the container, > so I would say aound 2500 M&Ms ? > Am I close? > -Darren >> Hello All. I am writing in reference to a contest a friend of mine is >> having. The rules are as follows. >> I bought a brand new clear plastic jar. It is lled >> with the standard size M & M candies. >> The jar is round and 8 inches tall. The diameter of >> the jar inside is 4 inches across. >> I bought over 4 pounds of M & Ms to ll it. I counted >> every M & M before lling the jar to the very top.. >> You can probably guess what I am seeking: >> How many M & Ms are in the jar as described? >> If anyone can help me on this it would be wonderful as I am not a good >> mathmatician. >> Joey -- #191, ewill3@earthlink.net Its still legal to go .sigless. === Subject: Re: How Many M&Ms? > In sci.math, darrenn > : > Fortunately, there has been some research on M&M packing: > The net is that the density of packing is around 71% (if packed in an > irregular fashion). > Further, it looks like the someone has actually measured an individual M&M: > http://www.kleinbottle.com/Bernie_Tao.htm > Which a result of 0.45239 cm. > Pedant point. An M&M is generally a attened round thing and > probably should be measured in all three dimensions, or perhaps > one can pour a number into a liquid (although water will melt > off the coatings, methinks) and measure the displacement. > so, volume of jar is: > PI X 2 X 2 X 8 = 100.54 inches^3 = 100.54 X 2.54 X 2.54 X 2.54 = 1647.407 > cm3 > So, number of M&Ms is approx. = 1647.407 X 0.71 / 0.45239 ~ 2585 > Now the result is probably only good to around 2 signicant gures, and > there will probably be some rounding down due to the sides of the container, > so I would say aound 2500 M&Ms ? > Am I close? > -Darren >> Hello All. I am writing in reference to a contest a friend of mine is >> having. The rules are as follows. >> I bought a brand new clear plastic jar. It is lled >> with the standard size M & M candies. >> The jar is round and 8 inches tall. The diameter of >> the jar inside is 4 inches across. >> I bought over 4 pounds of M & Ms to ll it. I counted >> every M & M before lling the jar to the very top.. >> You can probably guess what I am seeking: >> How many M & Ms are in the jar as described? >> If anyone can help me on this it would be wonderful as I am not a good >> mathmatician. >> Joey what does it matter if the water removes the coating? the coating will still contribute to the displacement. === Subject: Re: How Many M&Ms? In sci.math, David Bandel [snip for brevity -- discussing # of M&Ms in a certain sized jar, and using water for one measurement of volume] > what does it matter if the water removes the coating? the coating will > still contribute to the displacement. Theres a few issues regarding solubility that may complicate the analysis. Id frankly have to look. -- #191, ewill3@earthlink.net Its still legal to go .sigless. === Subject: Re: How Many M&Ms? >[snip for brevity -- discussing # of M&Ms in a certain sized jar, > and using water for one measurement of volume] >> what does it matter if the water removes the coating? the coating will >> still contribute to the displacement. >Theres a few issues regarding solubility that may complicate >the analysis. Id frankly have to look. M&Ms require solving a polynomial of degree greater than 5 ? === Subject: Re: How Many M&Ms? > Which a result of 0.45239 cm. > Pedant point. An M&M is generally a attened round thing and > probably should be measured in all three dimensions, or perhaps > one can pour a number into a liquid (although water will melt > off the coatings, methinks) and measure the displacement. True. This should say 0.45239 cm^3. -Darren === Subject: JSH: James, you be the judge... Here is how James Harris operates: 1. Start with some goofy polynomial that was a leftover from one of his many failed FLT proofs. No explanation of motivation or reasoning provided for choosing that particular polynomial or explanation of why it might be meaningful in his new context. 2. Avoid being specic about such important details as which ring he intends to work in. 3. Dream up a bunch of weird non-standard terminology like properly unit, dividing off, etc. Use terms incorrectly, such as referring to multiple when he should use the term factor. Also, make use of standard terminology like distributive property and constant term without actually understanding it or using it properly. Also, add lots of vague distractions such as the equation has no memory, you can see the 7s in there, cant you? to act as a smoke screen. 4. Apply a bunch of algebraic manipulations, some of which are just wrong. Make inappropriate generalizations from specic cases to general cases, etc. 5. Get a result that seems contradictory, and assume that the error was with the core of algebra rather than the much more likely explanation that he himself might have made an error. 6. Through out a bunch of paranoid ad hominem attacks on critics. Then hypocritically claim that all his critics resort to social crap and personal attacks rather than sticking to mathematics, logic, and reasoning. We are supposed to believe that his arbitrarily chosen goofy polynomial just happens to one that, when probed by the genius of James Harris, give results that shake mathematics to its core! James, you be the judge. Are you a pile of crap? === Subject: Pi in space I was having an argument with a friend here and he claimed that pi doesnt really have all this signicance its attached to it, since it is transcendental only in the theoretical space of Euclidian geometry and all know that this geometry does not represent true space-time. Is pi transcendental in ANY space except in the theoretical space of Euclidian geometry? For example, is it transcendental in Riemannian geometry? Lobatchevskian geometry? or any other geometry? What about in the relativistic universe? Is the ratio of the circumference of a circle to its diameter inside the Einsteinian universe rational or irrational? Anyone knows? -- I. N. G. --- http://users.forthnet.gr/ath/jgal/ === Subject: Re: Pi in space > I was having an argument with a friend here and he claimed that pi > doesnt really have all this signicance its attached to it, since it > is transcendental only in the theoretical space of Euclidian geometry > and all know that this geometry does not represent true space-time. > Is pi transcendental in ANY space except in the theoretical space of > Euclidian geometry? > For example, is it transcendental in Riemannian geometry? Lobatchevskian > geometry? or any other geometry? > What about in the relativistic universe? Is the ratio of the > circumference of a circle to its diameter inside the Einsteinian > universe rational or irrational? Anyone knows? We dont know the true geometry of space. In particular, we dont know if its continuous. Digital physicists think it is not: see www.digitalphysics.org The concept of Pi, however, should not be regarded transcendental or anything like that since it is represented perfectly with a small program. True, it is an idealization, but what is transcendental? In my opinion, nothing transcends physical reality, and if Pi is part of physical models, it is for a good reason: physics is mostly about ideal models that work in ideal conditions, etc. Note however, that, if the universe turns out to be discrete, then, obviously, the real number Pi does not describe any physical reality. The computation of Pi is still sensible, however, (e.g. computation of Pi up to the nth decimal place) and it would indeed describe physical reality in a discrete universe (if its an actually innite discrete universe, things get more complicated so just assume nite which seems to be the case for *our* universe;). Most mathematicians on this forum seem to be Platonists, or Platonists who lack the philosophical maturity to identify themselves as Platonists, so you could expect quite a lot of Platonist nonsense in response to your question. However, as one poster correctly pointed out, constructivists and formalists will most likely look at Pi as a useful mental construct of some sort, and no talk of transcending anything will be necessary. Pi is not in space: its in your mind. [*] Indeed, constructivism is a more favorable position than Platonic Realism, in my opinion, and a point of view supported by great mathematicians, so its an alternative you should think about. More recently, instrumentalism has been suggested as an alternative way to look at mathematics. In my opinion, that too is considerable. -- Eray Ozkural There is no perfect circle [*] A caveat, however, there are perfect discrete circles in the world. It just may be the case that there are no perfect continuous circles, except conceptually!!!!!! === Subject: Re: Pi in space |Most mathematicians on this forum seem to be Platonists, or Platonists |who lack the philosophical maturity to identify themselves as |Platonists, so you could expect quite a lot of Platonist nonsense in |response to your question. No, because such points of philosophy are thoroughly irrelevant to his question. |However, as one poster correctly pointed |out, constructivists and formalists will most likely look at Pi as a |useful mental construct of some sort, and no talk of transcending |anything will be necessary. Pi is not in space: its in your mind. [*] I am disinclined to assume that you know that transcendental is a technical mathematical term meaning not a root of a nonzero polynomial with integer coefcients. |Indeed, constructivism is a more favorable position than Platonic |Realism, in my opinion, and a point of view supported by great |mathematicians, so its an alternative you should think about. Kronecker, Brouwer, and Bishop were all outstanding mathematicians (whether we call them great sort of depends on how high a standard we set for greatness). Bishop supported constructivism; Brouwer supported some kind of constructivism; Kronecker appears to have supported something along those lines. I wouldnt say, however, that popularity among great mathematicians is a very good standard for judging such things. To the extent that it can be trusted at all, I see no obvious sign of the very best mathematicians differing very far in terms of the distribution of their opinions on Platonism (or realism), formalism, constructivism and so on from the run-of-the-mill. I also dont see sci.math as being much of a hotbed of Platonism or realism. Keith Ramsay === Subject: Re: Pi in space >The concept of Pi, however, should not be regarded transcendental or >anything like that since it is represented perfectly with a small >program. True, it is an idealization, but what is transcendental? In >my opinion, nothing transcends physical reality, and if Pi is part of >physical models, it is for a good reason: physics is mostly about >ideal models that work in ideal conditions, etc. I think you need to look up the denition of transcendental number. You seem to be confusing it with some philosophical notion. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Pi in space >The concept of Pi, however, should not be regarded transcendental or >anything like that since it is represented perfectly with a small >program. True, it is an idealization, but what is transcendental? In >my opinion, nothing transcends physical reality, and if Pi is part of >physical models, it is for a good reason: physics is mostly about >ideal models that work in ideal conditions, etc. > I think you need to look up the denition of transcendental number. > You seem to be confusing it with some philosophical notion. the concept of a transcendental number. AFAICT, the notion of a transcendental number is not relative to a give geometry, or the true geometry of our universe. Pi is a transcendental number, according to the denition of a transcendental number in mathematics. I do think his question ought to be philosophical. Why should he talk about relativity then? What difference is there between any continuous metric and a Riemann tensor regarding the fact that Pi is a transcendental number? Maybe, I think, his friends intention was to give a better denition of what it means for a number to be transcendental than the ordinary usage. He might want to pick another term, though, or highlight the difference than the ordinary usage carefully. Otherwise, we become confused in argumentation. I thought so, and said that this philosophical sense is probably irrelevant to a computable real like Pi which captures a general geometric fact in a compact form! -- Eray Ozkural === Subject: Re: Pi in space > I was having an argument with a friend here and he claimed that pi > doesnt really have all this signicance its attached to it, since it > is transcendental only in the theoretical space of Euclidian geometry > and all know that this geometry does not represent true space-time. We know nothing whether space-time really exists and if it exists whether it has a given geometry of its geometry can be a matter of convention. You are asking metaphysical (ontological) questions that have no answers at this point. > Is pi transcendental in ANY space except in the theoretical space of > Euclidian geometry? > For example, is it transcendental in Riemannian geometry? Lobatchevskian > geometry? or any other geometry? The geometries have relative consistency to Euclidean geometry. This, pi must be transcendental in all of them to maintain relative consistency. > What about in the relativistic universe? Is the ratio of the > circumference of a circle to its diameter inside the Einsteinian > universe rational or irrational? Anyone knows? You can draw a circle even on your pixelized TFT screen, consider it a perfect one and use pi to nd its perimeter. I get the impression you are asking about the relation of abstract mathematical object to the actual universe. We do not know it. Thats another metaphysical subject. Platonists would argue circles are really part of the ontology of space-time and Universe whereas Formalists and Constructivists will look at circles as useful geometrical models with no real existence apart from human mind. Mike === Subject: Re: Pi in space >I was having an argument with a friend here and he claimed that pi doesnt >really have all this signicance its attached to it, since it is >transcendental only in the theoretical space of Euclidian geometry and all >know that this geometry does not represent true space-time. > Is pi transcendental in ANY space except in the theoretical space of > Euclidian geometry? > For example, is it transcendental in Riemannian geometry? Lobatchevskian > geometry? or any other geometry? > What about in the relativistic universe? Is the ratio of the circumference > of a circle to its diameter inside the Einsteinian universe rational or > irrational? Anyone knows? > -- > I. N. G. --- http://users.forthnet.gr/ath/jgal/ Dening pi to be the empirically measured ratio of a circles circumference to its diameter, the only geometry in which this is ratio is a constant is Euclidean geometry. Pi doesnt exist as a constant that can be empirically measured in any other geometry, so its value in other geometries is a meaningless question. Of course, dening pi as the sum of a series, or the constant in a solution to a differential equation or proper integral, or any of the other billion ways of dening pi has absolutely nothing to do with the geometry of the Universe. So there is only one value of pi, and yes, it is transcendental. === Subject: Re: Pi in space |Dening pi to be the empirically measured ratio of a circles circumference |to its diameter, the only geometry in which this is ratio is a constant is |Euclidean geometry. Pi doesnt exist as a constant that can be empirically |measured in any other geometry, so its value in other geometries is a |meaningless question. Obviously this is not a practical issue, but I thought as long as we were busy making points about what is true in principle, I would note that there often is a natural relationship between pi and other geometries. In some other geometries, pi is the limit of the ratio between the circumference and the diameter of a circle as the radius goes to 0, and could in principle (if you had such a space in your hands...) be measured that way. At a xed radius, measuring the circumference and diameter would only be able to approximate pi up to a certain degree of precision, but measurements of lengths are always only up to a certain degree of precision anyway, so one is not any worse of in an essential way. On the hyperbolic plane, for example, you could nd a length scale on which a triangle whose sides have lengths in a 3:4:5 ratio has an angle of between 90 and 91 degrees. One can then put an upper bound on the deviation of the ratio of the circumference to the diameter from pi, in terms of the ratio of the radius to the length of a side of this triangle. The circumference of a circle of radius r in the hyperbolic plane is 2*pi*a*sinh(r/a) where the length a depends on the curvature. The size of the 3:4:5 triangle with an angle of 91 degrees is likewise proportional to this length a. (I wont bother guring out what the ratio is right here.) So to get n digits, we could make a circle whose radius is roughly 10^{-n/3} times the length of a side of this triangle, and measure it very precisely. Keith Ramsay === Subject: Re: Pi in space >I was having an argument with a friend here and he claimed that pi doesnt >really have all this signicance its attached to it, since it is >transcendental only in the theoretical space of Euclidian geometry and all >know that this geometry does not represent true space-time. > Is pi transcendental in ANY space except in the theoretical space of > Euclidian geometry? > For example, is it transcendental in Riemannian geometry? Lobatchevskian > geometry? or any other geometry? > What about in the relativistic universe? Is the ratio of the circumference > of a circle to its diameter inside the Einsteinian universe rational or > irrational? Anyone knows? > -- > I. N. G. --- http://users.forthnet.gr/ath/jgal/ > Dening pi to be the empirically measured ratio of a circles circumference > to its diameter, the only geometry in which this is ratio is a constant is > Euclidean geometry. It is not possible to empirically measure any irrational in any geometry, if by that you mean physical measurement. And if not, what do you mean? === Subject: Re: Pi in space >>I was having an argument with a friend here and he claimed that pi >>doesnt >>really have all this signicance its attached to it, since it is >>transcendental only in the theoretical space of Euclidian geometry and >>all >>know that this geometry does not represent true space-time. >> Is pi transcendental in ANY space except in the theoretical space of >> Euclidian geometry? >> For example, is it transcendental in Riemannian geometry? >> Lobatchevskian >> geometry? or any other geometry? >> What about in the relativistic universe? Is the ratio of the >> circumference >> of a circle to its diameter inside the Einsteinian universe rational or >> irrational? Anyone knows? >> -- >> I. N. G. --- http://users.forthnet.gr/ath/jgal/ >> Dening pi to be the empirically measured ratio of a circles >> circumference >> to its diameter, the only geometry in which this is ratio is a constant >> is >> Euclidean geometry. > It is not possible to empirically measure any irrational in any > geometry, if by that you mean physical measurement. Sure it is. You can measure the diagonal of unit square just as easily as you can measure the length of one side. The fact that one is rational and the other irrational doesnt affect your ability to measure them. > And if not, what do > you mean? The reason that I talk about empirical measurement is assertion implicit in the OP that if the universe is not at, then pi does not have the value everybody thinks it does. The only sense in which this is sort-of true is that if you measure the circumference of a circle and divide it by the diameter, then you will not get pi as we know it as the answer. I used the term empirically to suggest the act of phyiscally measuring (for example with a tape measure) circumferences and diameters of circles embedded in that space. === Subject: Re: Pi in space (1) Pi is a constant. Its value is 4 ATAN(1) regardless of whether space is Euclidean. Your question is like asking whether the number 1 remains rational in non-Euclidean space. A constant is a constant is a constant. (2) we all know that this geometry..... Who is we? We certainly does NOT include all because the recent super-nova evidence that revealed the existence of dark energy also shows that the universe is Euclidean. (3) Pi is transcendental in ALL geometries. You can lead a horses ass to knowledge, but you cant make him think. === Subject: Re: Pi in space ETAtAhUAuKZDfmVeTIVHWzPAf+1RbygxYcACFGivLjq2oYFNZgzALqH4m/ 0aKNiF Gee, the last time I looked, the transcendentality of a number was INDEPENDENT of geometry, being a property of the number itself. Whether pi, or other famously nonconstructible numbers like the cube root of 2 or the sine of 20 degrees, become constructible in non-Euclidean geometries is a different question. --OL === Subject: Re: Pi in space >(1) Pi is a constant. Its value is 4 ATAN(1) regardless >of whether space is Euclidean. But wasnt there a part of the muppets show called Pis in space? Thomas >Your question is like asking whether the number 1 >remains rational in non-Euclidean space. >A constant is a constant is a constant. >(2) we all know that this geometry..... >Who is we? We certainly does NOT include >all because the recent super-nova evidence that >revealed the existence of dark energy also shows >that the universe is Euclidean. >(3) Pi is transcendental in ALL geometries. >You can lead a horses ass to knowledge, but you cant make him think. === Subject: Re: Pi in space > (1) Pi is a constant. Its value is 4 ATAN(1) regardless > of whether space is Euclidean. > Your question is like asking whether the number 1 > remains rational in non-Euclidean space. > A constant is a constant is a constant. Damn. I was somehow hoping that from context my question would be clear. Apparently it was not. Let me repeat it more accurately phrased: Dene pi=Length of circumference/Length of diameter of a circle however the later terms are dened in the appropriate geometry we are talking about. In which geometries pi is transcendental? -- I. N. G. --- http://users.forthnet.gr/ath/jgal/ === Subject: Re: Pi in space >> (1) Pi is a constant. Its value is 4 ATAN(1) regardless >> of whether space is Euclidean. >> Your question is like asking whether the number 1 >> remains rational in non-Euclidean space. >> A constant is a constant is a constant. > Damn. I was somehow hoping that from context my question would be clear. > Apparently it was not. Let me repeat it more accurately phrased: > Dene pi=Length of circumference/Length of diameter of a circle > however the later terms are dened in the appropriate geometry we are > talking about. > In which geometries pi is transcendental? > -- > I. N. G. --- http://users.forthnet.gr/ath/jgal/ As I explained below, the ratio of a circles circumference to its radius is only a constant in Euclidean space. In non-Euclidean geometries, the ratio depends upon the size of the circle; there is no single value of pi. Consider, for example, the space as consisting of the surface of a sphere as big as the earth. If you measue the ratio of the circumference to the diameter of a circle that is only a few inches across, the ratio will be very close to 3.14159265... If you measure the ratio for the equator, you get pi=2 (the radius is the distance from the North Pole to the equator, which is one quarter the circumference). So in non-Euclidean geometries, the ratio of a circles circumference to its diameter can take on a wide range of values, depending upon the size of the circle. As the size of the circle approaches zero, the ratio approaches pi. So as the ratio can take on a continuous range of values, you can make it transcendental or not, simply by selecting the size of the circle you use to form the ratio. Which is, of course, different to saying pi varies - in all geometries, pi is the same; only the measured ratios change. === Subject: Re: Pi in space > As I explained below, the ratio of a circles circumference to its radius is > only a constant in Euclidean space. In non-Euclidean geometries, the ratio > depends upon the size of the circle; there is no single value of pi. > Consider, for example, the space as consisting of the surface of a sphere as > big as the earth. If you measue the ratio of the circumference to the > diameter of a circle that is only a few inches across, the ratio will be > very close to 3.14159265... If you measure the ratio for the equator, you > get pi=2 (the radius is the distance from the North Pole to the equator, > which is one quarter the circumference). > So in non-Euclidean geometries, the ratio of a circles circumference to its > diameter can take on a wide range of values, depending upon the size of the > circle. As the size of the circle approaches zero, the ratio approaches pi. > So as the ratio can take on a continuous range of values, you can make it > transcendental or not, simply by selecting the size of the circle you use to > form the ratio. > Which is, of course, different to saying pi varies - in all geometries, pi > is the same; only the measured ratios change. changes. That would be ridiculous. I was talking about the ratio. Serves me right: If I cannot phrase a question correctly, then I end up having all the groups farts on my back :-( -- I. N. G. --- http://users.forthnet.gr/ath/jgal/ === Subject: Re: Pi in space <1102142838.264150@athnrd02 Dene pi=Length of circumference/Length of diameter of a circle > however the later terms are dened in the appropriate geometry we are > talking about. > In which geometries pi is transcendental? Consider R^2 with the discrete metric and the circle at (0,0) with radius 1. Computing the circumference and dividing it by the diameter 2, I nd pi, the ratio circumference/2, to be transcending. ;-) === Subject: Re: Pi in space > Your question is like asking whether the number 1 > remains rational in non-Euclidean space. Well, does it? DOES IT? How come you mathematicians never answer the IMPORTANT questions? === Subject: Re: Pi in space >I was having an argument with a friend here and he claimed that pi >doesnt really have all this signicance its attached to it, since it >is transcendental only in the theoretical space of Euclidian geometry >and all know that this geometry does not represent true space-time. Everything after since is incoherent. >Is pi transcendental in ANY space except in the theoretical space of >Euclidian geometry? >For example, is it transcendental in Riemannian geometry? Lobatchevskian >geometry? or any other geometry? >What about in the relativistic universe? Is the ratio of the >circumference of a circle to its diameter inside the Einsteinian >universe rational or irrational? Anyone knows? ...As are all those questions. Lee Rudolph === Subject: Re: Why exp(-st) in the Laplace Transform? Does anyone have an explaination why the kernel function exp(-st) was >>used in the denition of the Laplace transform? >>Is there a physical meaning to the use of this function? >>I know s is a complex frequency, how can we visualize what the Laplace >>transform is doing? > I believe the answer is no, theres no interpretation of what the > Laplace transform really means, analogous to the way one interprets, > say, the Fourier transform. The kernels are not orthogonal in any > obvious sense... > Its hard to know for certain that no is the correct answer, of > course. But Ive never seen an answer of the sort I think youre > looking for, Ive thought about it and never found one, and > I once posted more or less the same question to sci.math, and > had a few smart people agree the answer was no. Arthur Mattuck, an MIT professor, interprets the Laplace transform as a continuous analog of a power series expansion of a function. You can watch a video of him lecturing on this point at Click on Lecture 19. T. Monroe === Subject: Re: Why exp(-st) in the Laplace Transform? > heard something somewhat similar tho - that laplace transforms are the >> continuous generalization of power series. I cant recall exactly how >> the explanation went, it was in an ODE book by a guy named Simmons. > Perhaps you have seen something like this. > Let F(p) = integral exp(-px) f(x) dx (: Laplace Transform) > ==> F(0) = integral f(x) dx = area beneath f(x) > And F(p) = integral (-x) exp(-px) f(x) dx > ==> F(0) = - integral x f(x) dx = - rst order moment > / center of gravity / midpoint > And F(p) = integral x^2 exp(-px) f(x) dx > ==> F(0) = integral x^2 f(x) dx = second order moment > / moment of inertia / variance > And then you have F(p) = F(0) + p.F(0) + p^2/2.F(0) + ... > Interpretation: subsequent terms of the series expansion of the Laplace > Transform involve an area, a midpoint, a moment of intertia .. In short: > the most important (physical/global) characteristics of a function come > rst, when considered in the Laplace domain. > Han de Bruijn Arthur Mattuck, an MIT professor, interprets the Laplace transform as a continuous analog of a power series expansion of a function. You can watch a video of him lecturing on this point at Click on Lecture 19. T. Monroe === Subject: Re: Why exp(-st) in the Laplace Transform? >> heard something somewhat similar tho - that laplace transforms are the >> continuous generalization of power series. I cant recall exactly how >> the explanation went, it was in an ODE book by a guy named Simmons. > Perhaps you have seen something like this. > Let F(p) = integral exp(-px) f(x) dx (: Laplace Transform) > ==> F(0) = integral f(x) dx = area beneath f(x) > And F(p) = integral (-x) exp(-px) f(x) dx > ==> F(0) = - integral x f(x) dx = - rst order moment > / center of gravity / midpoint > And F(p) = integral x^2 exp(-px) f(x) dx > ==> F(0) = integral x^2 f(x) dx = second order moment > / moment of inertia / variance > And then you have F(p) = F(0) + p.F(0) + p^2/2.F(0) + ... > Interpretation: subsequent terms of the series expansion of the Laplace > Transform involve an area, a midpoint, a moment of intertia .. In short: > the most important (physical/global) characteristics of a function come > rst, when considered in the Laplace domain. > Han de Bruijn > Arthur Mattuck, an MIT professor, interprets the Laplace transform as a > continuous analog of a power series expansion of a function. You can watch > a video of him lecturing on this point at > Click on Lecture 19. > T. Monroe Excellent pointer. Dirk Vdm === Subject: Re: Gravitomagnetism > Ok, you dont need to respond to this comment, > but I have no major problem with non-integer > weight if you allow non-integer dimensionality. Non-integer? Dont understand this.. Well you know from SR as you move an object faster > and faster its length contracts and its time > slows down, and at c these dimensions vanish. > So in the intermediary, between v=0 and v=c, > can we say denitely that we have a xed integer > dimensionality? To put that on a more rational mathematical foundation > one can integrate, (no integration constants) $ x dx = (1/2) x^2 (area) $$ x dx dx = (1/6) x^3 (volume) which shows how integration generates dimensions. > Well conformal weight is a concept that is independent of > dimensionality, that is, a covariant in a Weyl conformal space has > properties under both coordinate and scale changes. The idea of weight > embodies the latter. to check that out, but I think you are right. You know, moving from eld to quantum physics where are those integers. > GR is quite clear, G_uv = T_uv is where Ill begin. > If I understand you correctly, you state > vanishing gravity (implying G_uv =>0) and > *inhomogeneous* electrodynamics (implying T(Maxwell)_uv =/=0 are compatible? > Yes! That is what is so surprising. Nowhere is a current posited - > there *is* no RHS here. In GR, as in Maxwell-Lorentz, one has a > eld that is driven by a *posited* current (energy tensor, charge > current resp.), and then that eld acts back on the current. Here, > there is *no* posited current at all, rather, the assumption of > strictly local metricity requires both the metric and calibration > eld, and these *jointly* assume roles in a manner that appears as > symmetric Ricci driven by energy-momentum, and Maxwell driven by > charge-current! Of course they are really 6-d *vacuum* equations: > Rmn - (2R/W) Tmn + (1/2W) (DmDn + DnDm) W = 0 (Rmn = symmetric part > of CCT) I have a problem, I can read that in the usual 4d terms, but when you uprate to 6d are your mn over 4 or 6, ah...your pushin the envelope so please tell us your encryption. > 1/S d/dxm ( S R Fmn ) - 5/4 (Dn W) = 0 (S = sqrt det g) > The new physics would be found in the pure geometry terms 1/2W > {Dm,Dn} W and (Dn W), which are respectively, energy-momentum and > charge current. These are absent in at space absent in at space, thats what Einstein said when he doubled his prediction for the deection of light. Would you be able to provide how your equations reduce to GR in the the weak eld, that way giving us ameans to connect GRists with the physicality of your 6D. > Yes, Im trying to make the simplest possible metric > consistent with a nonorthogonal space, i.e. g_uv = g*g_uv + A_u B_v > ^ ^ > calibration EM antisymmetry => A_u B_v = - A_v B_u > (Weyl=>gauge) (Einstein) where det g = 1 - AB, (thats a bit crude, but close). Ken S. Tucker What paper? ken > -drl === Subject: JSH: Good news, and bad news Ill give the good news rst. Despite the venting I do at times in frustration, the reality is that I have an easygoing nature. So I thought Id point out a few things along with that for any mathematicians who might wander by and read this post to consider. First of all, my research is not limiting but expanding. What Ive found are new tools for deeper explorations into the properties of numbers. Given that mathematics is difcult for many people, its unlikely that the status quo in mainstream math society would change too dramatically once the full story is out. Sure Id no longer be a crank, but top mathematicians probably wouldnt move much. So lets say you are Barry Mazur, or Wiles or Taylor, and you nd out that there is actually this problem with algebraic number theory that you learned and a LOT of what you thought you had proven, was not proven. So what? A lot of it you can go ahead and prove anyway in less space, without error, using the more advanced techniques. but very large work, you can move to a proof of a few pages. And if Wiles cares more about seeing the proof of Fermats Last Theorem than the glory of being the person who found that proof, then its gravy. Sure, its embarrassing that the underlying theory was wrong, but the mistake entered the discipline over a hundred years ago, long before anyone now living was even born. If the desire is to see a proof of Fermats Last Theore, then no problem! Thats the good news. If the desire is the glory of having found a proof of Fermats Last Theorem. Problem. So its a measure of the men, as they are mostly men, how they react. The good news is that not much will change in many ways if they accept what is mathematically true. The bad news is that if they react only when forced, then clearly they were in it for the glory. And if glory hounds willing to ght against the truth, they get no pity. So, one way, not much changes. Professorships remain. There is just this amazing story of which they are a part. The other way, much can change, much can be lost, and a world can be severely disappointed, when the full story eventually comes out. The choice is yours. James Harris === Subject: Re: Good news, and bad news > The good news is that not much will change in many ways if they accept > what is mathematically true. Good, indeed. News, hardly. Have you ever read the neprint? >The choice is yours. You have one claim (trivial) to a prime counting algorithm, and another (childish) denying basic algebra. At most one of them could stand. The choice is yours - is the bottle half full, half empty or fully empty? Barry === Subject: Re: Good news, and bad news Mr. Harris, Your level of delusion seems to grow without bound! The amount of crap spewage that you are propagating is beyond any level of decent mathematical discussion. You are a phony, a liar, a cheat and an ignoramus. Even if you managed to stumble into something in mathematics that actually made sense while in one of your drunken states, no one would give a damn! You just dont get it, do you? === Subject: Re: Good news, and bad news > What Ive found are new tools for deeper explorations into the > properties of numbers. Yes, you found ing great new tools for deeper explorations... Right... And you did manage to proved FLT using slick 17th century style algebra, so why not move onto the Riemann Hypothesis with your new power tools? That is the next big one, right? I just know you can do it with your new deep math exploration tools. After all, it is a great challenge. It is kind off to analysis what FLT is to arithmetic, right. True, you have fought hard, and won your deservedg lory for great achievement with FLT. But why rest on your laurels? Move on you dumb little ........ Or was FLT just appealing to you only because the statement of the problem was so simple that you thought you had a clue about it. You gure you can follow the arithmetic because the problem statement looks like high school stuff, but you dont know head from your ass beyond high school, right. Well, that is probably why cranks like you always stick to FLT. === Subject: Re: Good news, and bad news > but very large work, you can move to a proof of a few pages. You often now make the claim that Wiles FLT proof is wrong. Can you point out the error he made? Did he make an error in determining those properly units thingy-ma-jiggies? If you make this claim and you cannot back it up, then why are we to believe you? === Subject: Re: Good news, and bad news > but very large work, you can move to a proof of a few pages. > You often now make the claim that Wiles FLT proof is wrong. Moreover, Wiles did not prove FLT directly, he proved a certain conjecture (the conjecture being that every rational elliptic curve is some kind of equivalent to a modular form), for a limited number of elliptic curves, namely for the semi-stable elliptic curves. By earlier work from Frey and Ribet this was sufcient to show that FLT was true. The full conjecture was proved true later by Conrad et al. Because JSH apparently attacks Wiles proof, he should also consider the proof by Breuil, Conrad, Diamond and Taylor for the full conjecture... Or he should attack the work by Frey, of that by Ribet, or whatever. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Good news, and bad news > You often now make the claim that Wiles FLT proof is wrong. ... OOOOPS I mean Wiles, not Wiles... === Subject: Re: JSH: Good news, and bad news > Ill give the good news rst. Despite the venting I do at times in > frustration, the reality is that I have an easygoing nature. So I > thought Id point out a few things along with that for any > mathematicians who might wander by and read this post to consider. > First of all, my research is not limiting but expanding. That would be the bad news then. === Subject: Re: JSH: Good news, and bad news >Ill give the good news rst. Despite the venting I do at times in >frustration, the reality is that I have an easygoing nature. Right. > So I >thought Id point out a few things along with that for any >mathematicians who might wander by and read this post to consider. >First of all, my research is not limiting but expanding. >What Ive found are new tools for deeper explorations into the >properties of numbers. >Given that mathematics is difcult for many people, its unlikely >that the status quo in mainstream math society would change too >dramatically once the full story is out. Sure Id no longer be a >crank, but top mathematicians probably wouldnt move much. >So lets say you are Barry Mazur, or Wiles or Taylor, and you nd out >that there is actually this problem with algebraic number theory that >you learned and a LOT of what you thought you had proven, was not >proven. >So what? A lot of it you can go ahead and prove anyway in less space, >without error, using the more advanced techniques. >but very large work, you can move to a proof of a few pages. And if >Wiles cares more about seeing the proof of Fermats Last Theorem than >the glory of being the person who found that proof, then its gravy. Actually its clear that you havent been paying attention. Fermats Last Theorem is _false_! E. E. Escultura has posted counterexamples right here in sci.math. I think the two of you need to get together and work this out. Because youre both light-years past anything that the rest of us can comprehend, but you cant both be right. Check out his posts. Contact the guy, and write back as soon as the two of you agree on which one of the two is wrong. >Sure, its embarrassing that the underlying theory was wrong, but the >mistake entered the discipline over a hundred years ago, long before >anyone now living was even born. >If the desire is to see a proof of Fermats Last Theore, then no >problem! >Thats the good news. >If the desire is the glory of having found a proof of Fermats Last >Theorem. >Problem. >So its a measure of the men, as they are mostly men, how they react. >The good news is that not much will change in many ways if they accept >what is mathematically true. >The bad news is that if they react only when forced, then clearly they >were in it for the glory. And if glory hounds willing to ght >against the truth, they get no pity. >So, one way, not much changes. Professorships remain. There is just >this amazing story of which they are a part. >The other way, much can change, much can be lost, and a world can be >severely disappointed, when the full story eventually comes out. >The choice is yours. >James Harris ************************ David C. Ullrich === Subject: Re: JSH: Good news, and bad news >but very large work, you can move to a proof of a few pages. And if >Wiles cares more about seeing the proof of Fermats Last Theorem than >the glory of being the person who found that proof, then its gravy. > Actually its clear that you havent been paying attention. > Fermats Last Theorem is _false_! > E. E. Escultura has posted counterexamples right here in sci.math. > I think the two of you need to get together and work this out. > Because youre both light-years past anything that the rest of > us can comprehend, but you cant both be right. > ^^^^^^^^^^^^^^^^^^^^^^^ Oh, that is so retrograde of you. Surely their mathematical skills are advanced enough that they can prove Fermats Last Theorem AND its negation. Understand now? === Subject: Re: JSH: Good news, and bad news youre an easygoer in what respect ... til you were found in a cyberspace, and became marginally surlier. I missed the bad news, other than what the last-prior poster suggested. > Understand now? --Advice, 0.05; free, if wrong! http://tarpley.net/bush22.htm === Subject: Re: Good news, and bad news > What Ive found are new tools for deeper explorations into the > properties of numbers. Name or describe one new tool for deeper exploration that you have found. All you have done is claim that there is a deep problem in algebraic integers, and as far as that goes, you have convinced a grand total of zero converts. You never explained what that problem is. You never showed any results of what that problem are. And you have certainly never found any new tools for mathematicians to work with! === Subject: Re: JSH: Good news, and bad news !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~( 5eZ41to5f%E@ELIi $t^ VcLWP@J5p^rst0+(>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw > Ill give the good news rst. Despite the venting I do at times in > frustration, the reality is that I have an easygoing nature. Well, the _real_ good news is that the bad news turns out as hilariously absurd as the good news. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: JSH: Simply fascinating The math here is so readily understandable that its actually almost as interesting watching how people react to it, as anything else. For instance, Ive given a polynomial P(x) repeatedly where I factor it into three factors. I point out that the factors must include factors of the constant term of the polynomial P(x). I note that the factors of the constant term are independent of x, as they are, in fact, constant. Mathematically its easy to show: g_1(x) g_2(x) g_3(x) = P(x) and g_1(0) g_2(0) g_3(0) = P(0) as P(0) gives the constant term of the polynomial, and since the polynomial results from multiplying together the three factors which Ive called g_1(x), g_2(x), and g_3(x), then you can get the factors of the constant term, by setting x=0. Its that simple. Notice (1) you have the factors of P(x), (2) the constant term of P(x) is determinable by setting x=0, (3) you also get the factors of the constant term. Mathematically, its simple to the point of trivial. Now then, if you have *constants* which are factors of another constant then why would anyone try to argue that they are actually variables? You have x, as the variable. Besides x there are just these numbers. If you clear out x, then whats left are constants. Letting x=0, clears it out, leaving the constants visible. Some may say, yeah, sure, at x=0, but what about when x doesnt equal 0? Um, if the numbers are constant, and so are independent of x, then, duh, why should it matter what value x has? The logic is inescapable. In terms of difculty, my proof is about as easy as it gets in algebraic number theory, in terms of the actual mathematics. But the concepts are where there is a problem, and the social hang-up is in accepting that theres this simple technique that shows a BIG problem, which can invalidate claims of proof for, well, over a hundred plus years. So the mathemtics is EASY for a trained mathematician to follow. The social implications are hard, if social stuff is important to you, and clearly from what Ive seen it is to many of you. For instance, at this point Ive removed all objections raised in detail. Like I can explain supposed counter examples to my work. I can give an actual example where you can see the factorization play out--just as the theory shows it must. And you probably know that my research is the work that can be said to have gone to a journal which at least claims it does formal peer review. They thanked me for the paper said the reviewers liked it, and then some sci.mathers got together--actually literally conspiring online in posts on sci.math--sent them emails and the editors yanked my papers THE NEXT DAY. They had it for nine months. Id corresponded with them for a while, even corrected them when they called me Dr. Harris as I dont have a Ph.d, and I told them I was an independent researcher with concerns about how my work would be handled. They kept saying no problem, ok, all that matters is whats correct. Then they yanked my paper after sci.mathers emailed them: All the pertinent facts are in my favor. So whats the hold-up? My research shows that some mistakes were made over a hundred years ago, and a lot of people missed them, and gave proofs which were not, and are not proofs. Mathematics is unforgiving. It doesnt care about the social implications of the truth. So it doesnt matter mathematically that a LOT of people out there are terribly dependent on the false beliefs and incorrect results, but it DOES matter a lot to those people! I call their behavior passive-aggessive, as by dragging their feet, taking as long as possible before acknowledging my research, or worse, hoping to NEVER acknowledge it at all, they are passively hoping to escape mathematical truth, in what amounts to a very aggressive way. I liken their behavior to judges at a race, who watch a runner break a world record, and then lie about his time, refuse to admit he even nished the race, and some even call him names!!! Theyre turning the way its supposed to work with a major discovery, upside down. And its silly behavior as eventually the truth will come out, and you know what Ill do then? Probably go to the beach. Ill also hang out in some bars. Yup, Ill denitely hang out in some bars, preferably near a beach. Yup, you guessed it, Ill do my best to forget about them, as why bother worrying about silly people who do silly things. Lifes too short. James Harris http://mathforprot.blogspot.com/ === Subject: Re: JSH: Simply fascinating > Ive given a polynomial P(x) repeatedly where I factor > it into three factors. > I point out that the factors must include factors of the constant term > of the polynomial P(x). In general thats not true at all. For example, consider: (x+1)*(x+2)*(x+3) If you expand it out to get a polynomial in the usual form, you see that the constant term is 6. But 6 is not a factor of any of the three polymomial factors, in fact neither 2 nor 3 is a factor of any of them say what you really meant to say? Or maybe you were totally mistaken and need to just retract what you said. > My research shows that some mistakes were made over a hundred years > ago, and a lot of people missed them, and gave proofs which were > not, and are not proofs. Its true that some mistakes were made long ago, the worst in my opinion being the rst developers of set theory who concocted a set of all sets, which was subsequently proven not to exist. But Ive seen no evidence that *you* have ever done any research that showed any such past mistakes of others. Instead all I see is *you* making mistakes yourself and not admitting them. Do you know how to factor 7 in the ring of integers with sqrt(7) adjoined? Thats easy, of course you do, right? But do you know how to factor 7 in the ring of integers with sqrt(2) adjoined? Now if you can gure out that simple arithmetic problem, heres something more interesting: Find all primes p (other than 2 and 7 which Ive already shown you) such that 7 can be nontrivially factored in the ring of integers with sqrt(p) adjoined. Finally, nd all nite sets of two or more primes p1,p2,...,pn such that 7 is composite in the ring you get when you adjoin all the sqrts of those primes but its prime in the ring you get when you adjoin all but one of those sqrts. Same question if you allow adjoining sqrt(-1) in addition to adjoining sqrt of a prime. (Note: When I refer to a prime p or primes p1,p2..., I mean prime in the ring of integers. When I refer to 7 being prime or composite, I mean prime or composite in the ring of integers adjoined with the various sqrts.) === Subject: Re: JSH: Simply fascinating |Its true that some mistakes were made long ago, the worst in my |opinion being the rst developers of set theory who concocted a set |of all sets, which was subsequently proven not to exist. As far as I know, this incident involved Frege and nobody else. If you have other developers of set theory in mind, who were they? Second, the set of all sets was only proven not to exist under some additional assumptions. The problem with Freges system was that it implied there exists a Russell set, R = {x : x is not a member of x}. The existence of a Russell set is directly contradictory. There are systems of set theory that have a universal set and are consistent (although they are less used). Frege didnt directly postulate a universal set or a Russell set; he just stated axioms that he regarded as intuitive, that together implied that for any predicate P there exists a set {x : x satises P}. Keith Ramsay === Subject: Re: JSH: Simply fascinating > The problem with Freges system was that it implied there exists a > Russell set, R = {x : x is not a member of x}. By what grammatical axiom(s) is the right side of that equation even a well-formed formula? I would guess two such grammatical axioms: If S and T are WFFs denoting sets, then S is a member of T is also a WFF denoting a truthvalue. If P is a WFF denoting a truthvalue, then not P is also a WFF denoting a truthvalue. x is not a member of x is shorthand for not (x is a member of x). If x is a WFF variable, and P is a boolean-valued WFF with x as the only free variable, then {x : P} is a WFF denoting a set. I personally accept all those grammatical axioms except the last as reasonable. The last I would replace with: If S is a WFF denoting a set, and x is a WFF variable, and P is a boolean-valued WFF with x as the only free variable, then {x in S : P} is a WFF denoting a set. The reason I dont like the original last grammatical axiom is that it begs the question what kind of Universe were dealing with, whereas my alternative clearly says S is the universe were dealing with. If we beg the question what Universe were dealing with, then we arent really making a denition, we arent saying what is allowed to be in a set and what isnt allowed to be in a set, so we arent saying whether some monstrosity in somebodys fantasy can be in our set or not. > There are systems of set theory that have a universal set and are > consistent (although they are less used). Do you know a WebPage that describes any of these axiomatic systems? Alternately, do you remember which commonsense axioms of set-theory grammar are *not* included, thereby avoiding writing a Russell set as a WFF in said system? === Subject: Re: JSH: Simply fascinating Discussion, linux) >> The problem with Freges system was that it implied there exists a >> Russell set, R = {x : x is not a member of x}. > By what grammatical axiom(s) is the right side of that equation even a > well-formed formula? I would guess two such grammatical axioms: ^^^^^^^ term, not formula. > If S and T are WFFs denoting sets, > then S is a member of T is also a WFF denoting a truthvalue. > If P is a WFF denoting a truthvalue, then not P is also a WFF > denoting a truthvalue. > x is not a member of x is shorthand for not (x is a member of x). > If x is a WFF variable, > and P is a boolean-valued WFF with x as the only free variable, > then {x : P} is a WFF denoting a set. > I personally accept all those grammatical axioms except the last as > reasonable. The last I would replace with: > If S is a WFF denoting a set, and x is a WFF variable, > and P is a boolean-valued WFF with x as the only free variable, > then {x in S : P} is a WFF denoting a set. > The reason I dont like the original last grammatical axiom is that > it begs the question what kind of Universe were dealing with, > whereas my alternative clearly says S is the universe were dealing ^^^^^^^^^^^^^^ your alternative? Youre very inuential. > with. If we beg the question what Universe were dealing with, then > we arent really making a denition, we arent saying what is > allowed to be in a set and what isnt allowed to be in a set, so we > arent saying whether some monstrosity in somebodys fantasy can be > in our set or not. -- Who knows, maybe that may be the only way to settle this crap. Its not like itd be that hard for me to go back and get a math degree. I can penetrate the math social group and then nish the takedown from inside. -- James S. Harris contemplates a new strategy. === Subject: Re: JSH: Simply fascinating > Ive given a polynomial P(x) repeatedly where I factor > it into three factors. > I point out that the factors must include factors of the constant term > of the polynomial P(x). > In general thats not true at all. For example, consider: > (x+1)*(x+2)*(x+3) But he said, e.g. in your example, having factored your polynomial into (x+1), (x+2), and (x+3), 1, 2, and 3 must be factors of 6, the constant term of your expanded polynomial. > If you expand it out to get a polynomial in the usual form, you see > that the constant term is 6. But 6 is not a factor of any of the three > polymomial factors, in fact neither 2 nor 3 is a factor of any of them He didnt say, with respect to your example, that 6 must be a factor of 1, 2, and 3. Rather that 1, 2, and 3 must be factors of 6. KeithK > say what you really meant to say? Or maybe you were totally mistaken > and need to just retract what you said. > My research shows that some mistakes were made over a hundred years > ago, and a lot of people missed them, and gave proofs which were > not, and are not proofs. > Its true that some mistakes were made long ago, the worst in my > opinion being the rst developers of set theory who concocted a set > of all sets, which was subsequently proven not to exist. But Ive seen > no evidence that *you* have ever done any research that showed any such > past mistakes of others. Instead all I see is *you* making mistakes > yourself and not admitting them. > Do you know how to factor 7 in the ring of integers with sqrt(7) > adjoined? Thats easy, of course you do, right? But do you know how to > factor 7 in the ring of integers with sqrt(2) adjoined? Now if you can > gure out that simple arithmetic problem, heres something more > interesting: Find all primes p (other than 2 and 7 which Ive already > shown you) such that 7 can be nontrivially factored in the ring of > integers with sqrt(p) adjoined. Finally, nd all nite sets of two or > more primes p1,p2,...,pn such that 7 is composite in the ring you get > when you adjoin all the sqrts of those primes but its prime in the > ring you get when you adjoin all but one of those sqrts. Same question > if you allow adjoining sqrt(-1) in addition to adjoining sqrt of a > prime. (Note: When I refer to a prime p or primes p1,p2..., I mean > prime in the ring of integers. When I refer to 7 being prime or > composite, I mean prime or composite in the ring of integers adjoined > with the various sqrts.) === Subject: Re: JSH: Simply fascinating > Um, if the numbers are constant, and so are independent of x, then, > duh, why should it matter what value x has? If they are independent of x, why did you have to set x to zero to uncover them? Evaluating a univariate polynomial at any numeric value of the variable produces a numeric result. Are all these results constants? > The logic is inescapable. > Lifes too short. I think yours is too long. -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: JSH: Simply fascinating > The math here is so readily understandable that its actually almost > as interesting watching how people react to it, as anything else. > For instance, Ive given a polynomial P(x) repeatedly where I factor > it into three factors. > I point out that the factors must include factors of the constant term > of the polynomial P(x). > I note that the factors of the constant term are independent of x, as > they are, in fact, constant. > Mathematically its easy to show: > g_1(x) g_2(x) g_3(x) = P(x) > and > g_1(0) g_2(0) g_3(0) = P(0) > as P(0) gives the constant term of the polynomial, and since the > polynomial results from multiplying together the three factors which > Ive called g_1(x), g_2(x), and g_3(x), then you can get the factors > of the constant term, by setting x=0. > Its that simple. Yes. Right so far. Trivial, but right. > Notice (1) you have the factors of P(x), (2) the constant term of P(x) > is determinable by setting x=0, (3) you also get the factors of the > constant term. > Mathematically, its simple to the point of trivial. Correct. > Now then, if you have *constants* which are factors of another > constant then why would anyone try to argue that they are actually > variables? No one does. If as you say, you assume that your factors are constant, of course they are not variables. That is nothing but a dimwit tautology. And of course that is not what we say. We say that if you factor 49 out of P(x) in a *variable* way, as functions w_1(x), w_2(x), and w_3(x) of x, and corresponding divide g_1(x), g_2(x) and g_3(x) by those variable functions, then you can obtain algebraic integer factors on both sides of the resulting equation. That is, g_1(x)/w_1(x), etc., are all algebraic integers. That is all you require. The constant terms of g_1(x)/w_1(x), g_2(x)/w_2(x), and g_2(x)/w_2(x) are BY YOUR OWN DEFINITION equal to g_1(0)/w_1(0), g_2(0)/w_2(0), and g_3(0)/w_2(0). No problem with that either. And the product of these three constant terms equals P(0)/49. It all works out. Factoring out 49 as a product of variable algebraic integer functions does not lead to ANY contradiction. It is only when you try to use a CONSTANT factorization that you arrive at what you think is a problem with algebraic integers. If you do the factorization in the right way there is no such problem. Suppose z is an integer variable, and you consider Q(z) = (x + 5)(x + 6). You notice that Q(z) is alway an even integer. Therefore you can always divide it by 2. You notice that when z = 0, Q(z) = Q(0) = 5 * 6. You notice that Q(0)/2 = 5 * (6/2). That is, when you divide by 2 to get an integer, you divide the constant term of the second factor, 6, by 2. If you tried to divide 5 by 2 you would not have an integer quotient. So by your logic, Q(z)/2 = (z + 5)*(z/2 + 6/2) = (z + 5)*(z/2 + 3) would be the only right, CONSTANT way to factor out 2. This is analogous to your division by 49 = 7*7*1: (5 a_1(x)/7 + 7/7)(5 a_2(x)/7 + 7/7)(5 b_3(x)/1 + 22/1) But, back to Q(z)/2 = (z + 5)(z/2 + 3). The problem is, z/2 is not always an integer. If you want integer quotients in both factors, you have to divide out 2 in a NONCONSTANT way. When z is even, you divide 2 out of (z + 6). When z is odd, you divide 2 out of (z + 5). Get it? Factoring 2 out in a constant fashion does not work. Factoring 2 out in a *variable* fashion does work. No problem arises with constant terms. > You have x, as the variable. Besides x there are just these numbers. > If you clear out x, then whats left are constants. Letting x=0, > clears it out, leaving the constants visible. > Some may say, yeah, sure, at x=0, but what about when x doesnt equal > 0? Some may say. > Um, if the numbers are constant, and so are independent of x, then, > duh, why should it matter what value x has? You MUST divide by a variable factorization of 49. If you assume a constant factorization you get into trouble, because a_1(x)/7 in general is not an algebraic integer. You know this, yet you persist in thinking that the factorization has to be constant. You conclude *not* that your own thinking is wrong (as you should) but that there is something fundamentally wrong with algebraic number theory. You try to justify using the constant 7*7*1 factorization by repeating your mantra that the constant terms must be constant. This is based on your incorrect conclusion that, if you divide (a_1(x) + 7) by w_1(x), then the constant term is 7/w_1(x). And right there is your central mistake. BY YOUR OWN DEFINTION, the constant term of (a_1(x) + 7)/w_1(x) is not 7/w_1(x). It is instead 7/w_1(0). You want to conclude that 7/w_1(x) must be equal to the constant, 7, because 1 * 1 * 22 = 22, the constant term of P(x)/49. If you do the *correct factorization*, you still get 22, but it is in the form (7/w_1(x)) * (7/w_2(x)) * (22/w_3(x)) = 22 because w_1(x)*w_2(x)*w_3(x) = 49. Now, you may howl, YES BUT 22/w_3(x) IS NOT AN ALGEBRAIC INTEGER!!! To which the reply is : yes, youre right. BUT THERE IS NO REASON IT SHOULD BE. All that is required is that (5 a_3(x) + 7)/w_3(x) is an algebraic integer, and this follows from the correct choice of w_1(x), w_2(x), and w_3(x). As it turns out, both 5 a_3(x)/w_3(x) AND 7/w_3(x) are algebraic integers. And as above: the constant term of (5 a_3(x) + 7)/w_3(x) is NOT 22/w_3(x), as you believe. It is merely 22/w_3(0) = 22. These two are not the same, unless you assume what you want to prove, i.e., that w_3(x) is a constant function. And you know it is not. Everything works out. There is no contradiction or problem involving the constant terms. It is YOU who has been trying to claim that the constant terms are not constant, e.g., when you say that 22/w_3(x) is the constant term of (5 a_3(x) + 7)/w_3(x). You are making such a rookie mistake. Why cant you see it? > The logic is inescapable. True enough. > In terms of difculty, my proof is about as easy as it gets in > algebraic number theory, in terms of the actual mathematics. Easy, yes. Correct, no. > But the concepts are where there is a problem, and the social [tiresome pompous rant about social factors etc. deleted] Nora B. > Lifes too short. > James Harris > http://mathforprot.blogspot.com/ === Subject: Re: JSH: Simply fascinating > The math here is so readily understandable that its actually almost > as interesting watching how people react to it, as anything else. For instance, Ive given a polynomial P(x) repeatedly where I factor > it into three factors. I point out that the factors must include factors of the constant term > of the polynomial P(x). I note that the factors of the constant term are independent of x, as > they are, in fact, constant. Mathematically its easy to show: g_1(x) g_2(x) g_3(x) = P(x) and g_1(0) g_2(0) g_3(0) = P(0) as P(0) gives the constant term of the polynomial, and since the > polynomial results from multiplying together the three factors which > Ive called g_1(x), g_2(x), and g_3(x), then you can get the factors > of the constant term, by setting x=0. Its that simple. Yes. Right so far. Trivial, but right. > Notice (1) you have the factors of P(x), (2) the constant term of P(x) > is determinable by setting x=0, (3) you also get the factors of the > constant term. Mathematically, its simple to the point of trivial. Correct. > Now then, if you have *constants* which are factors of another > constant then why would anyone try to argue that they are actually > variables? No one does. If as you say, you assume that your factors are > constant, of course they are not variables. That is nothing but > a dimwit tautology. > And of course that is not what we say. We say that if you factor > 49 out of P(x) in a *variable* way, as functions w_1(x), w_2(x), > and w_3(x) of x, and corresponding divide g_1(x), g_2(x) and g_3(x) > by those variable functions, then you can obtain algebraic integer > factors on both sides of the resulting equation. That is, > g_1(x)/w_1(x), etc., are all algebraic integers. That is all you > require. The constant terms of g_1(x)/w_1(x), g_2(x)/w_2(x), and > g_2(x)/w_2(x) are BY YOUR OWN DEFINITION equal to > g_1(0)/w_1(0), g_2(0)/w_2(0), and g_3(0)/w_2(0). > No problem with that either. And the product of these three > constant terms equals P(0)/49. Ok, Im going to answer the Nora Baron poster yet again. And I want readers to understand that Ive replied to this poster who you know lies anyway as its a guy posting as a woman using a name thats a palindrome MANY TIMES explaining in detail. What happens is that when I shoot down these objections, the poster either just repeats later, or replies with nonsense. One telling time when I carefully refuted point-by-point the poster replied deleting out everything Id said. Just deleted out everything, and you know what? I STILL so people replying about how supposedly I dont answer the objections from Nora Baron. Its partly a game for some people on sci.math, Im sure, and partly a case where many readers are hoodwinked as *they* dont know its a game. They seem incapable of realizing that there are people in this world willing to behave in such a way, on such a level. So why reply to this poster? Maybe Im hopelessly naive, but I just have to believe that eventually people will just get tired of being made fools of by this poster and others in the group with him (or her). It hasnt happened yet from what Ive seen, but I keep hoping. Ok, so whats wrong with the posters assertion? Well, the ws the posters gives are factors of 49. But 49 comes into the picture because 49 is a multiple of the polynomial. But the ws STILL REMAIN after 49 has been divided off--after the multiple is divided off--as the poster actually claims. (Notice Nora Baron gives you clues, but somehow sci.math readers dont seem to get it.) Notice the posters actually gives that when you divide off 49, you get g_1(x)/w_1(x), g_2(x)/w_2(x), and g_3(x)/w_3(x) which are TRACES, left of 49--after it has been divided off. So here we have mathematics. Here in an area where the truth can actually be determined, a poster who you cant be sure is a guy or a girl has confused many of you into questioning whether or not a multiple of a polynomial divides off as a variable or not. Did it ever occur to any of you that for some people that might be a great lark? Are you so trusting as to not consider that a poster who otherwise would probably be unknown can get off on confusing you not only on his or her gender, but on some of the most basic concepts in mathematics? Look now, even after the post where Nora Baron ended with a male name, people are STILL supporting the poster! It has occurred to me that this poster did so deliberately, ended that post with a male name, because the person knew you better than you still seem to know yourselves. In a way its sad as youre giving up so much for nothing in return. James Harris === Subject: Re: JSH: Simply fascinating >> The math here is so readily understandable that its actually almost >> as interesting watching how people react to it, as anything else. >> For instance, Ive given a polynomial P(x) repeatedly where I factor >> it into three factors. >> I point out that the factors must include factors of the constant term >> of the polynomial P(x). >> I note that the factors of the constant term are independent of x, as >> they are, in fact, constant. >> Mathematically its easy to show: >> g_1(x) g_2(x) g_3(x) = P(x) >> and >> g_1(0) g_2(0) g_3(0) = P(0) >> as P(0) gives the constant term of the polynomial, and since the >> polynomial results from multiplying together the three factors which >> Ive called g_1(x), g_2(x), and g_3(x), then you can get the factors >> of the constant term, by setting x=0. >> Its that simple. >> Yes. Right so far. Trivial, but right. >> Notice (1) you have the factors of P(x), (2) the constant term of P(x) >> is determinable by setting x=0, (3) you also get the factors of the >> constant term. >> Mathematically, its simple to the point of trivial. >> Correct. >> Now then, if you have *constants* which are factors of another >> constant then why would anyone try to argue that they are actually >> variables? >> No one does. If as you say, you assume that your factors are >> constant, of course they are not variables. That is nothing but >> a dimwit tautology. >> And of course that is not what we say. We say that if you factor >> 49 out of P(x) in a *variable* way, as functions w_1(x), w_2(x), >> and w_3(x) of x, and corresponding divide g_1(x), g_2(x) and g_3(x) >> by those variable functions, then you can obtain algebraic integer >> factors on both sides of the resulting equation. That is, >> g_1(x)/w_1(x), etc., are all algebraic integers. That is all you >> require. The constant terms of g_1(x)/w_1(x), g_2(x)/w_2(x), and >> g_3(x)/w_3(x) are BY YOUR OWN DEFINITION equal to >> g_1(0)/w_1(0), g_2(0)/w_2(0), and g_3(0)/w_3(0). [a couple of misprints corrected here] >> No problem with that either. And the product of these three >> constant terms equals P(0)/49. >Ok, Im going to answer the Nora Baron poster yet again. >And I want readers to understand that Ive replied to this poster who >you know lies anyway as its a guy posting as a woman using a name >thats a palindrome MANY TIMES explaining in detail. >What happens is that when I shoot down these objections, the poster >either just repeats later, or replies with nonsense. One telling time >when I carefully refuted point-by-point the poster replied deleting >out everything Id said. Because it was the SOS, repeated for the nth time. I replied with a complete detailed and rigorous proof. You were enraged but never really provided any refutation. >Just deleted out everything, and you know what? I STILL so people >replying about how supposedly I dont answer the objections from Nora >Baron. You STILL so people replying ??? Really! >Its partly a game for some people on sci.math, Im sure, and partly a >case where many readers are hoodwinked as *they* dont know its a >game. They seem incapable of realizing that there are people in this >world willing to behave in such a way, on such a level. >So why reply to this poster? Maybe Im hopelessly naive, but I just >have to believe that eventually people will just get tired of being >made fools of by this poster and others in the group with him (or >her). It hasnt happened yet from what Ive seen, but I keep hoping. Whine, whine, whine. Why not go straight to the math rather and skip the propaganda ? >Ok, so whats wrong with the posters assertion? >Well, the ws the posters gives are factors of 49. But 49 comes into >the picture because 49 is a multiple of the polynomial. But the ws >STILL REMAIN after 49 has been divided off--after the multiple is >divided off--as the poster actually claims. I claim no such thing. The product of the ws is 49. You divide P(x) by 49. You divide the product of the gs by the product of the ws. The 49 is gone from both sides. Note that here g_1(x) = (5 a_1(x) + 7), g_2(x) = (5 a_2(x) + 7), and g_3(x) = (5 a_3(x) + 7) = (5 b_3(x) + 22), where a_1, a_2, a_3 are all algebraic integer functions of x, and b_3(x) = a_3(x) - 3. Note that a_1(0) = a_2(0) = 0, and a_3(0) = 3. Also, note that w_1(0) = w_2(0) = 7, and w_3(0) = 1. When I divide the product of the gs by the product of the ws, the result is (5 c_1(x) + d_1(x))*(5 c_2(x) + d_2(x))*(5 c_3(x) + d_3(x)), where c_i(x) = a_i(x)/w_i(x) and d_i(x) = 7/w_i(x). Here is what Mr. Harris is squawking about. The product of the ds is 7. (That happens to be true for any x.) Harris thinks that is a problem, because it then looks like the product of the constant terms of (5 c_1(x) + d_1(x))*(5 c_2(x) + d_2(x))*(5 c_3(x) + d_3(x)) is 7, whereas the constant term of P(x)/49 is 22. That is, d_1(x)*d_2(x)*d_3(x) = 7, whereas P(0)/49 = 22. Puzzling! Is Harris right? I will now give my deceptive, lying, sleight-of-hand explanation. Watch very carefully, because as Mr. Harris says, I am going to try very hard to mislead you! So here is the explanation. The ds are NOT the constant terms of the factors that contain them. That is, d_1(x) is NOT the constant term of (5 c_1(x) + d_1(x)). Mr. Harris thinks it is. Certainly it is in the right position for being a constant term. Since Mr. Harris understands and mostly relies on only high-school level mathematics, he detects constant terms by *inspection*. The irony here is, he has provided a perfectly good, rigorous denition of constant term of a function, but he doesnt use it. It is the value that the function takes on when the argument is 0. Thus the constant term of (5 c_1(x) + d_1(x)) is NOT d_1(x) = 7/w_1(x). It is instead d_1(0) = 7/w_1(0) = 1. This is directly from Mr. Harriss own denition of constant term. Why is he so reluctant to apply that denition? Why does he instead rely on inspection, which gives the incorrect answer? What HE thinks is the constant term, namely d_1(x) = 7/w_1(x) is not even constant (unless he assumes what he wants to prove) ! Similary, the constant term of (5 c_2(x) + d_2(x)) is 1 also. However, the constant term of (5 c_3(x) + d_3(x)) is (5 c_3(0) + d_3(0)) = (5 a_3(0)/w_3(0) + 7/w_3(0)) = 5 * 3/1 + 7/1 = 15 + 7 = 22. Wow, look at that! The product of the constant terms is the constant term of the product! That is, d_1(0)*d_2(0)*(5 a_3(0) + d_3(0)) = 1*1*22 = 22 = P(0)/49. Amazing! Did I successfully trick you? Wheres the trick? Everything works out just as it should: IF you correctly identify the constant terms. Wheres the trick, Mr. Harris ? >(Notice Nora Baron gives you clues, but somehow sci.math readers >dont seem to get it.) I think not. I think that, with perhaps one exception, sci.math readers DO get it. In fact, I even think there is no exception. I think Mr. Harris actually gets it also, but he cannot stand to admit it. There is too much at stake. He clings to a delusion on the one-in-a trillion chance that he will be proven right. >Notice the posters actually gives that when you divide off 49, you >get >g_1(x)/w_1(x), g_2(x)/w_2(x), and g_3(x)/w_3(x) >which are TRACES, left of 49--after it has been divided off. See above. There is no mystery. After division, the product of the constant terms is the constant term of the product, just as it should be. That is, IF you correctly identify what the constant terms actually are. If you dont, then you get the wrong answer. This is Mr. Harriss mistake. This is not rocket science. This is not Galois theory, or algebraic number theory, or eld theory, or group theory. This is not abstract algebra at all. This is not even high-school level algebra. The mistake Mr. Harris is making is really a low-level one. A ROOKY mistake, as Mr. Harris used to say (about me and Arturo Magidin and others). He is simply not using his own denition. He is substituting in x when he should be substituting in 0. And incredibly, he seems unable to understand it. Math genius Harris, boy wonder Harris - from this you would think he needed help changing his own soiled underwear. [more propaganda deleted] Nora B. >James Harris === Subject: Re: JSH: Simply fascinating > The math here is so readily understandable that its actually almost > as interesting watching how people react to it, as anything else. For instance, Ive given a polynomial P(x) repeatedly where I factor > it into three factors. I point out that the factors must include factors of the constant term > of the polynomial P(x). I note that the factors of the constant term are independent of x, as > they are, in fact, constant. Mathematically its easy to show: g_1(x) g_2(x) g_3(x) = P(x) and g_1(0) g_2(0) g_3(0) = P(0) as P(0) gives the constant term of the polynomial, and since the > polynomial results from multiplying together the three factors which > Ive called g_1(x), g_2(x), and g_3(x), then you can get the factors > of the constant term, by setting x=0. Its that simple. Yes. Right so far. Trivial, but right. > Notice (1) you have the factors of P(x), (2) the constant term of P(x) > is determinable by setting x=0, (3) you also get the factors of the > constant term. Mathematically, its simple to the point of trivial. Correct. > Now then, if you have *constants* which are factors of another > constant then why would anyone try to argue that they are actually > variables? No one does. If as you say, you assume that your factors are > constant, of course they are not variables. That is nothing but > a dimwit tautology. > And of course that is not what we say. We say that if you factor > 49 out of P(x) in a *variable* way, as functions w_1(x), w_2(x), > and w_3(x) of x, and corresponding divide g_1(x), g_2(x) and g_3(x) > by those variable functions, then you can obtain algebraic integer > factors on both sides of the resulting equation. That is, > g_1(x)/w_1(x), etc., are all algebraic integers. That is all you > require. The constant terms of g_1(x)/w_1(x), g_2(x)/w_2(x), and > g_2(x)/w_2(x) are BY YOUR OWN DEFINITION equal to > g_1(0)/w_1(0), g_2(0)/w_2(0), and g_3(0)/w_2(0). > No problem with that either. And the product of these three > constant terms equals P(0)/49. > Ok, Im going to answer the Nora Baron poster yet again. > And I want readers to understand that Ive replied to this poster who > you know lies anyway as its a guy posting as a woman using a name > thats a palindrome MANY TIMES explaining in detail. Please provide a link to a message where nora baron says that she is a woman. <... Look now, even after the post where Nora Baron ended with a male > name, people are STILL supporting the poster! Please provide a link to a message where nora baron says that he is a man. === Subject: Re: JSH: Simply fascinating > ... 49 is a multiple of the polynomial. You keep saying this kind of stuoid stuff. But 49 is *not* a multiple of the polynomial. Read the denition of the word multiple. > Look now, even after the post where Nora Baron ended with a male > name, people are STILL supporting the poster! That is because nobody really cares what Noras geneder really is. They care about what he says... i.e. his math. Is that not what you claim is the only thing that counts? If Nora was a aming tranvestite hemaphroditic cross-dresser, it would not make his math incorrect or your math correct. And it is not a lie to use a pseudonym, it is a personal choice. Most of us do not use our real names here. === Subject: Re: JSH: Simply fascinating posting-account=KR2cuw0AAACZ_86pfubjOKsQkAVb6Rpe What does Noras gender have to do with math. You only point this out because you are immature, trying to create a distraction, and you know your work is wrong. Get a life. Dave === Subject: Re: JSH: Simply fascinating > What does Noras gender have to do with math. You only point this out > because you are immature, trying to create a distraction, and you know > your work is wrong. Get a life. > Dave The poster lies repeatedly. Ive argued with this poster for some time and noted the lying, and the gender lying is just another example of a trend. Some people dont like being lied to, and take it seriously. My position is that I can explain in detail, and with very simple concepts how my argument works, but that mathematicians might believe they have lots of reasons for avoiding the truth--all social ones. Unfortunately, they are aided by the poster who manages to maintain confusion about the issues, and maintain the lie that there is any vagueness or area of real mathematical doubt about my work, as many people trust. They trust that if I were right mathematicians wouldnt disagree with me, and there wouldnt be so much opposition to my work. So the poster Nora Baron can just lie for the sake of showing opposition, which helps to create the illusion of uncertainty about my work, helping to hide the truth from people who dont know better. The people who are well-trained mathematicians though, they are not fooled, and I know from my own experiences talking with well-trained mathematicians about my work. So whats happening now is sad in many ways, but part of it has to do with a basic contempt for the public, and a belief that they can be lead astray indenitely by rather basic tactics. I say, the strategy is about to die, and the passive-aggessive strategy of hoping that indenitely the public will be fooled will be shown to be one of professional suicide. James Harris === Subject: Re: JSH: Simply fascinating > What does Noras gender have to do with math. You only point this out > because you are immature, trying to create a distraction, and you know > your work is wrong. Get a life. > Dave > The poster lies repeatedly. Ive argued with this poster for some > time and noted the lying, and the gender lying is just another example > of a trend. Bullcrap. Nora Baron has repeatedly shown errors in the poster James Harris math and THAT is why he hates her. (Oh he does! He said so not long ago). He appears to be particularly infuriated because she keeps pinning him back to the more involved polynomials from which his simpler examples are derived, and showing where his error lies (no pun intended). This poster James Harris historically responds to math rebuttals with diatribes like this when he is unable to respond successfully to the math with math. Perhaps what REALLY pisses him off is that a _female_ could so effectively demolish his erroneous math. > Some people dont like being lied to, and take it seriously. > My position is that I can explain in detail, and with very simple > concepts how my argument works, but that mathematicians might believe > they have lots of reasons for avoiding the truth--all social ones. Here it is: > Unfortunately, they are aided by the poster who manages to maintain > confusion about the issues, and maintain the lie that there is any > vagueness or area of real mathematical doubt about my work What this poster James Harris calls lies are any math rebuttals that show that there is any vagueness or area of real mathematical doubt about his work. Since his work is correct they _must_ be lies. KeithK >, as many > people trust. The people who are well-trained mathematicians though, they are not > fooled, and I know from my own experiences talking with well-trained > mathematicians about my work. So whats your problem? If well-trained mathematicians have discussed your work with you, surely you dont need to keep banging your head against the lame brains at sci.math or any other newsgroup. Just have some of these well-trained mathematicians promote your work and publications, accolades, Field medals, parades, etc. will follow. -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: JSH: Simply fascinating > The math here is so readily understandable that its actually almost > as interesting watching how people react to it, as anything else. > For instance, Ive given a polynomial P(x) repeatedly where I factor > it into three factors. > I point out that the factors must include factors of the constant term > of the polynomial P(x). > I note that the factors of the constant term are independent of x, as > they are, in fact, constant. > Mathematically its easy to show: > g_1(x) g_2(x) g_3(x) = P(x) > and > g_1(0) g_2(0) g_3(0) = P(0) > as P(0) gives the constant term of the polynomial, and since the > polynomial results from multiplying together the three factors which > Ive called g_1(x), g_2(x), and g_3(x), then you can get the factors > of the constant term, by setting x=0. > Its that simple. > Notice (1) you have the factors of P(x), (2) the constant term of P(x) > is determinable by setting x=0, (3) you also get the factors of the > constant term. > Mathematically, its simple to the point of trivial. Yes if P(x) = g_1(x) g_2(x) g_3(x) then P(0) = g_1(0) g_2(0) g_3(0) why did you surround this trivial fact with all the verbiage. No one is disputing this. > Now then, if you have *constants* which are factors of another > constant then why would anyone try to argue that they are actually > variables? > You have x, as the variable. Besides x there are just these numbers. > If you clear out x, then whats left are constants. Letting x=0, > clears it out, leaving the constants visible. > Some may say, yeah, sure, at x=0, but what about when x doesnt equal > 0? > Um, if the numbers are constant, and so are independent of x, then, > duh, why should it matter what value x has? > The logic is inescapable. And by this logic the constant term of (a(0)=0, w(0)=1) h(x) = (a(x)/w(x) + 7/w(x)), a(0)=0, w(0)=1 is 7. The constant term is not 7/w(x). Indeed you can write h(x) so you can see the constant. Let t(x) = (a(x) -7w(x) +7)/w(x) Note t(0) = 0 and h(x) = (t(x) + 7) So the constant term of h(x) is 7. And yes 7 divides the constant term of P(x). No one is arguing that 7/w(x) divides the constant term of P(x). -William Hughes === Subject: Re: Simply fascinating Having fun following this thread, but I just want to make sure I understand it. So, James has the expression: P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7) and he has shown that each a_*(x) evaluates to an algebraic integer for integer x. And he has shown that a_*(0) is an integer. Then he makes this leap by saying that a_1(0) = 7 implies a_1(x) = 7.b (b algebraic) for all integer x. Is that correct how I stated it, or am I missing something? Darren === Subject: Re: Simply fascinating > Then he makes this leap by saying that > a_1(0) = 7 implies a_1(x) = 7.b (b algebraic) for all integer x. > Is that correct how I stated it, or am I missing something? You got it all correctly. You see, 7 is a constant. It does not depend on x. So 7 cannot change just because x changes. So a_1(x) is still 7, since 7 has not changed..... get it? And by the way, the above equations have no memory of the 49 that got properly divided off as a multiple, or something like that. Get it? === Subject: Re: Simply fascinating Meant: > a_1(0) = 7 implies a_1(x) = 7.b (b algebraic Integer) for all integer x. === Subject: Re: Simply fascinating > But the concepts are where there is a problem, and the social hang-up > is in accepting that theres this simple technique that shows a BIG > problem, which can invalidate claims of proof for, well, over a > hundred plus years. That is not the hang-up. The hang-up is that you are 100% ed up. I would have no problem accepting something that would show a big problem, invalidating proofs for over a hundred years. I would just love it. I would think, wow, that is so cool! It would be a thrill. But you have not convinced me or anyone else that you have found anything interesting or correct. Just uffy bogus claims. Lots of claims. Little valid math. Little valid logic. Lots of paranoia. Lots of grandiosity. Lots of non-standard terminology. Lots of smoke screen crap, like the gender of Nora. Look James... lots of people here would love to see the next crisis/revolution in math... another Russell shakeup, another G.9adel shakeup. So tell me... why has nobody been interested in any of your ideas? Could it be that your ideas are ? Or do you think that it is more likely that everyone else is nuts. Apply Occams Razor here! Just a thought. === Subject: Re: inverse modulos posting-account= y3wZYhMAAABYsCtaDBjCWE5oFd14ElQZbfvQjxC1czdFUKdrfKUl4g The Euclidean algorithm constructively produces coefcients a,b such that ax + bm = gcd(x,m). So given x,m with gcd(x,m) = 1, one easily nds the multiplicative inverse a of x modulo m. You are asking if a were known, could you nd x (at least, up to modulo m), and the answer is yes. This is just turning the question around; x is again the multiplicative inverse of a modulo m. If you do a Google search for Euclidean algorithm, youll see that it is a pretty iteration of the division algorithm. === Subject: Re: inverse modulos posting-account=UtgH7gwAAACpBhTelVPOXNP7RAfbtQrK > If two numbers x and m are relatively prime, so that gcd(x, m) = 1 is > true, then x has a unique multiplicative inverse modulo m, a, so that ax > = 1 (mod m). > Knowing only the multiplicative inverse, a modulo m, and m, is it > possible to nd x? X is the modulo-inverse of a and m. > Is it true that a*x - 1 = k*m, for some k, possibly negative? Or > multiple ks? How to nd the multiple ks so that x could be found? > Is that possible at all, so that the candidate ks can be found knowing > only the inverse a modulo m and m? > If there is no single way to nd it, is there a reasonable trial and > error process to go through to nd x given a and m? > What I suppose I am saying is, what is the relationship between x and > its inverse a modulo m? > Well, if you caught all that, hope you can help me out :) > Johnathan Heres a Python demo: > import gmpy > for i in range(1,20): twoo = 2**i twee = 3**i a = gmpy.invert(twoo,twee) # modulo-inverse function x = gmpy.invert(a,twee) # run it again to nd x k = divmod(a*x-1,twee) # returns quotient, remainder print twoo,twee,a,x,k 2 3 2 2 (mpz(1), mpz(0)) 4 9 7 4 (mpz(3), mpz(0)) 8 27 17 8 (mpz(5), mpz(0)) 16 81 76 16 (mpz(15), mpz(0)) 32 243 38 32 (mpz(5), mpz(0)) 64 729 262 64 (mpz(23), mpz(0)) 128 2187 1589 128 (mpz(93), mpz(0)) 256 6561 4075 256 (mpz(159), mpz(0)) 512 19683 11879 512 (mpz(309), mpz(0)) 1024 59049 35464 1024 (mpz(615), mpz(0)) 2048 177147 17732 2048 (mpz(205), mpz(0)) 4096 531441 363160 4096 (mpz(2799), mpz(0)) 8192 1594323 181580 8192 (mpz(933), mpz(0)) 16384 4782969 90790 16384 (mpz(311), mpz(0)) 32768 14348907 9611333 32768 (mpz(21949), mpz(0)) 65536 43046721 11980120 65536 (mpz(18239), mpz(0)) 131072 129140163 92083502 131072 (mpz(93461), mpz(0)) 262144 387420489 175181914 262144 (mpz(118535), mpz(0)) 524288 1162261467 862431935 524288 (mpz(389037), mpz(0)) The gcd of a power of 2 and a power of 3 is always 1, so there is a valid a for every pair twoo and twee. Once I determined the modulo-inverse (a) of twoo, twee; I plugged a into the gmpy.invert function to get x (which is the same as twoo, but were pretending we dont know that). of the calculation is always 0, so the quotient is the legit k. CC: mensanator@aol.com === Subject: Re: inverse modulos > X is the modulo-inverse of a and m. Ah-ha, I thought there must be some kind of relationship (not bright enough to nd it myself ;-) > Heres a Python demo: Lovely, thankyou!!! But why doesnt it work with: x = 2385150540210225032 m = 3517 x and m are relatively prime. a = 3500 (multiplicative inverse of x and m) ? Or am I misunderstanding something? (Do I need to get k to nd the original x?) Running with this sample of x, m and a gives me 1655 as the inverse of a and m, which should have been equal to x. > > import gmpy > > for i in range(1,20): > twoo = 2**i > twee = 3**i > a = gmpy.invert(twoo,twee) # modulo-inverse function > x = gmpy.invert(a,twee) # run it again to nd x > k = divmod(a*x-1,twee) # returns quotient, remainder > print twoo,twee,a,x,k > 2 3 2 2 (mpz(1), mpz(0)) > 4 9 7 4 (mpz(3), mpz(0)) > 8 27 17 8 (mpz(5), mpz(0)) > 16 81 76 16 (mpz(15), mpz(0)) > 32 243 38 32 (mpz(5), mpz(0)) > 64 729 262 64 (mpz(23), mpz(0)) > 128 2187 1589 128 (mpz(93), mpz(0)) > 256 6561 4075 256 (mpz(159), mpz(0)) > 512 19683 11879 512 (mpz(309), mpz(0)) > 1024 59049 35464 1024 (mpz(615), mpz(0)) > 2048 177147 17732 2048 (mpz(205), mpz(0)) > 4096 531441 363160 4096 (mpz(2799), mpz(0)) > 8192 1594323 181580 8192 (mpz(933), mpz(0)) > 16384 4782969 90790 16384 (mpz(311), mpz(0)) > 32768 14348907 9611333 32768 (mpz(21949), mpz(0)) > 65536 43046721 11980120 65536 (mpz(18239), mpz(0)) > 131072 129140163 92083502 131072 (mpz(93461), mpz(0)) > 262144 387420489 175181914 262144 (mpz(118535), mpz(0)) > 524288 1162261467 862431935 524288 (mpz(389037), mpz(0)) > The gcd of a power of 2 and a power of 3 is always 1, so there > is a valid a for every pair twoo and twee. Once I determined > the modulo-inverse (a) of twoo, twee; I plugged a into the > gmpy.invert function to get x (which is the same as twoo, but > were pretending we dont know that). > of the calculation is always 0, so the quotient is the legit k. I wish I could get these results with my numbers... :) only a and m. === Subject: Re: inverse modulos === >Subject: Re: inverse modulos >Message-id: <41af7774$0$25776$5a62ac22@per-qv1-newsreader-01.iinet.net.au> X is the modulo-inverse of a and m. >Ah-ha, I thought there must be some kind of relationship (not bright >enough to nd it myself ;-) >> Heres a Python demo: >Lovely, thankyou!!! >But why doesnt it work with: >x = 2385150540210225032 >m = 3517 Yeah, I guess that only works when xx and m are relatively prime. >a = 3500 (multiplicative inverse of x and m) >? Or am I misunderstanding something? (Do I need to get k to nd the >original x?) Running with this sample of x, m and a gives me 1655 as >the inverse of a and m, which should have been equal to x. 1655 is the rst of an innite series of xs where the inverse modulo is 3500. When x> > import gmpy >> > for i in range(1,20): >> twoo = 2**i >> twee = 3**i >> a = gmpy.invert(twoo,twee) # modulo-inverse function >> x = gmpy.invert(a,twee) # run it again to nd x >> k = divmod(a*x-1,twee) # returns quotient, remainder >> print twoo,twee,a,x,k >> 2 3 2 2 (mpz(1), mpz(0)) >> 4 9 7 4 (mpz(3), mpz(0)) >> 8 27 17 8 (mpz(5), mpz(0)) >> 16 81 76 16 (mpz(15), mpz(0)) >> 32 243 38 32 (mpz(5), mpz(0)) >> 64 729 262 64 (mpz(23), mpz(0)) >> 128 2187 1589 128 (mpz(93), mpz(0)) >> 256 6561 4075 256 (mpz(159), mpz(0)) >> 512 19683 11879 512 (mpz(309), mpz(0)) >> 1024 59049 35464 1024 (mpz(615), mpz(0)) >> 2048 177147 17732 2048 (mpz(205), mpz(0)) >> 4096 531441 363160 4096 (mpz(2799), mpz(0)) >> 8192 1594323 181580 8192 (mpz(933), mpz(0)) >> 16384 4782969 90790 16384 (mpz(311), mpz(0)) >> 32768 14348907 9611333 32768 (mpz(21949), mpz(0)) >> 65536 43046721 11980120 65536 (mpz(18239), mpz(0)) >> 131072 129140163 92083502 131072 (mpz(93461), mpz(0)) >> 262144 387420489 175181914 262144 (mpz(118535), mpz(0)) >> 524288 1162261467 862431935 524288 (mpz(389037), mpz(0)) >> The gcd of a power of 2 and a power of 3 is always 1, so there >> is a valid a for every pair twoo and twee. Once I determined >> the modulo-inverse (a) of twoo, twee; I plugged a into the >> gmpy.invert function to get x (which is the same as twoo, but >> were pretending we dont know that). >> of the calculation is always 0, so the quotient is the legit k. >I wish I could get these results with my numbers... :) >only a and m. -- Mensanator Ace of Clubs === Subject: Re: inverse modulos > If two numbers x and m are relatively prime, so that gcd(x, m) = 1 is > true, then x has a unique multiplicative inverse modulo m, a, so that ax > = 1 (mod m). > Knowing only the multiplicative inverse, a modulo m, and m, is it > possible to nd x? Im not sure if I understand what you mean... We are given two numbers a and m and we want to nd two numbers x and k such that: a*x - 1 = k*m or, equivalently: a*x - k*m = 1. But this is a very simple linear diophantine equation - so, where is the problem...? Pawel Gladki CC: gladki@math.usask.ca === Subject: Re: inverse modulos > Im not sure if I understand what you mean... We are given two numbers a > and m and we want to nd two numbers x and k such that: > a*x - 1 = k*m > or, equivalently: > a*x - k*m = 1. > But this is a very simple linear diophantine equation - so, where is the > problem...? Ok... hang on! I understand what *you* are asking now. Ok, just did a mathworld.wolfram.com. This is exactly what I am looking for. Solving for two unknowns. Is there some way I can solve a diophantine equation computationally, i.e. I am using the GNU GMP library and writing programs in C. Now I need to do it in code. My papers show: px + sn = 1, where x and s are unknown, and p = modular inverse of x and n, and n is relatively prime to x of course for p to exist. So rearranging this I get px = 1 + (-s)n, is that the same thing? (I have done this one to verify when I do know x, but now I want to nd x and s for px + sn = 1 knowing only p and n.) Johnathan === Subject: Re: inverse modulos >> Im not sure if I understand what you mean... We are given two numbers >> a and m and we want to nd two numbers x and k such that: >> a*x - 1 = k*m >> or, equivalently: >> a*x - k*m = 1. >> But this is a very simple linear diophantine equation - so, where is >> the problem...? > Ok... hang on! I understand what *you* are asking now. Ok, just did a > mathworld.wolfram.com. This is exactly what I am looking for. Solving > for two unknowns. > Is there some way I can solve a diophantine equation computationally, > i.e. I am using the GNU GMP library and writing programs in C. > Now I need to do it in code. My papers show: > px + sn = 1, where x and s are unknown Solving diophantine equations is easy... it uses the extended Euclidean algorithm, for details see e.g.: http://www.win.tue.nl/~ida/demo/c1s3f2.html Pawel Gladki === Subject: Re: inverse modulos > Solving diophantine equations is easy... it uses the extended Euclidean > algorithm, for details see e.g.: > http://www.win.tue.nl/~ida/demo/c1s3f2.html use of the extended Euclidean algorithm, but its not giving the result I want. I thought that the inverse modulo was supposed to be unique where it existed, but I must have misunderstood that. Apparently there are many solutions sometimes (or always?) I havent yet found a way to grab all the solutions and test them for the particular value of x that I want. And even then, if the relationship between x and n and its inverse modulo is not unique (same for every possible x solution) then perhaps I cant know what the original x was. Johnathan === Subject: Re: inverse modulos :> Solving diophantine equations is easy... it uses the extended Euclidean :> algorithm, for details see e.g.: :> http://www.win.tue.nl/~ida/demo/c1s3f2.html : use of the extended Euclidean algorithm, but its not giving the result : I want. I thought that the inverse modulo was supposed to be unique : where it existed, but I must have misunderstood that. The inverse of a modulo m is unique *modulo m*, if it exists. For example, a = 17, m = 5. The inverse of 17 modulo 5 is 3, since 17*3 = 51 = 1 + 5*10. (k=10 in the notation of your earlier posts). I could add any multiple of m (in this case 5) to 3, and it would still be an inverse: for example 8,13,18,... but these are all congruent modulo 5 (i.e. they differ from each other by a multiple of 5). The inverse is unique modulo 5. In general, if a-b is divisible by m, then we say that a and b are congruent modulo m, and write a = b (mod m). Ted === Subject: Re: inverse modulos >> Solving diophantine equations is easy... it uses the extended >> Euclidean algorithm, for details see e.g.: >> http://www.win.tue.nl/~ida/demo/c1s3f2.html > use of the extended Euclidean algorithm, but its not giving the result > I want. I thought that the inverse modulo was supposed to be unique > where it existed, but I must have misunderstood that. Apparently there > are many solutions sometimes (or always?) There are _always_ innitely many solutions. However, since: a*x - k*m = 1 it follows that gcd(x, m) = 1, so there exists _exactly_ _one_ x which satises the above equation and is less than m (and bigger than 0, too). Indeed, suppose that we have two such x-es, x and x(and two respective k and k). Then: a*x - k*m = 1 a*x - k*m = 1 Subtracting the second equation from the rst gives: a*(x - x) - m*(k - k) = 0 that is a*(x-x) = m*(k-k), which yields x-x = m*(k-k)/a. But also gcd(a,m) = 1, which implies that a divides k-k, so that (k-k)/a is an integer. That means x and x differ by multiplicity of m, which is possible only if one of them is bigger than m. Pawel Gladki === Subject: Re: inverse modulos > Solving diophantine equations is easy... it uses the extended > Euclidean algorithm, for details see e.g.: > http://www.win.tue.nl/~ida/demo/c1s3f2.html >> makes use of the extended Euclidean algorithm, but its not giving >> the result I want. I thought that the inverse modulo was supposed >> to be unique where it existed, but I must have misunderstood that. >> Apparently there are many solutions sometimes (or always?) > There are always innitely many solutions. However, since: > ax - km = 1 > it follows that gcd(x, m) = 1, so there exists exactly one x which > satises the above equation and is less than m (and bigger than 0, > too). Indeed, suppose that we have two such x-es, x and x(and two > respective k and k). Then: > ax - k m = 1 > ax - km = 1 > Subtracting the second equation from the rst gives: > a(x - x) - m(k - k) = 0 > that is a(x-x) = m(k-k), which yields x-x = m(k-k)/a. But also > gcd(a,m) = 1, which implies that a divides k-k, so that (k-k)/a is > an integer. That means x and x differ by multiplicity of m, which is > possible only if one of them is bigger than m. Simpler: Inverses are always unique in any abelian group G. For if B has inverses A,C then A = A(BC) = (AB)C = C Here G = U(Z/m) = unit group of Z/m, the ring of integers (mod m), i.e. the units (invertibles) of the multiplicative monoid of Z/m. By the extended Euclidean/Bezout algorithm the units are simply the cosets n (mod m) := n + mZ such that n is coprime to m since then and only then does there exist integers j,k such that j n + k m = 1 <=> j n = 1 (mod m) <=> n in U(Z/m) Uniqueness theorems are powerful tools for proving equalities. For many further less trivial examples see my prior posts --Bill Dubuque === Subject: Re: inverse modulos >I thought that the inverse modulo was supposed >to be unique where it existed, but I must have misunderstood that. >Apparently there are many solutions sometimes (or always?) >>There are always innitely many solutions. However, since: >> ax - km = 1 >>it follows that gcd(x, m) = 1, so there exists exactly one x which >>satises the above equation and is less than m (and bigger than 0, >>too). Indeed, suppose that we have two such x-es, x and x(and two >>respective k and k). Then: >> ax - k m = 1 >> ax - km = 1 >>Subtracting the second equation from the rst gives: >> a(x - x) - m(k - k) = 0 >>that is a(x-x) = m(k-k), which yields x-x = m(k-k)/a. But also >>gcd(a,m) = 1, which implies that a divides k-k, so that (k-k)/a is >>an integer. That means x and x differ by multiplicity of m, which is >>possible only if one of them is bigger than m. > Simpler: Inverses are always unique in any abelian group G. (...) Well, yeah, I know that much :-) But it seems that the person who asked that question didnt know much about groups, rings etc., so I was trying to explain the uniqueness of an inverse element in Z_m in the simplest possible terms... hope I havent confused him too much. Sometimes one has to say difcult things, but one ought to say them as simply as one knows how - G. H. Hardy Pawel Gladki === Subject: Re: inverse modulos > Im not sure if I understand what you mean... Let me see if I can explain it a bit better. Two numbers, Ill call them x and n, and the gcd is 1, so they are relatively prime. I compute the modulo inverse of x and n, to get p. I know only n and p: how do I nd x? See: xp = 1 (mod n), so: xp - 1 is exactly divisible by n. Heres a concrete example: x = 2385150540210225032 n = 3517 x and n are relatively prime. p = 3500, the inverse modulo of x and n. x*p = 1 (mod n), and x*p - 1 is divisible by n exactly. If I know only n and p here, how do I get back the x? > But this is a very simple linear diophantine equation - so, where is the > problem...? Im after a numerical solution. Is it possible to nd one? That is, given an unknown x and a known n with a known inverse modulo of those two p, can I nd x? (Is there only one x?) So I can nd a/the numerical solution by solving a linear diophantine equation? I hope that I explained it correctly on the rst attempt! Johnathan === Subject: Re: inverse modulos > Im not sure if I understand what you mean... > Let me see if I can explain it a bit better. Two numbers, Ill call > them x and n, and the gcd is 1, so they are relatively prime. I compute > the modulo inverse of x and n, to get p. > I know only n and p: how do I nd x? > See: xp = 1 (mod n), so: > xp - 1 is exactly divisible by n. > Heres a concrete example: > x = 2385150540210225032 > n = 3517 Drat! My previous post was not quite correct. There are an inte number of xs whose modulo inverse 3517 result in 3500. Doing the gmpy.invert on (3500,3517) yields 1655, the rst of that innite list of xs. Call the rst one x_0. To nd x_k: x_k = k*n + x_0 Heres how the list starts out. Note the modulo inverse is 3500 for each x_k: > for k in range(20): print k, k*n+r,gmpy.invert(k*n+r,n) 0 1655 3500 1 5172 3500 2 8689 3500 3 12206 3500 4 15723 3500 5 19240 3500 6 22757 3500 7 26274 3500 8 29791 3500 9 33308 3500 10 36825 3500 11 40342 3500 12 43859 3500 13 47376 3500 14 50893 3500 15 54410 3500 16 57927 3500 17 61444 3500 18 64961 3500 19 68478 3500 Your example happens to occur at k = 678177577540581. > x and n are relatively prime. > p = 3500, the inverse modulo of x and n. > x*p = 1 (mod n), > and x*p - 1 is divisible by n exactly. > If I know only n and p here, how do I get back the x? Now that I think Im doing it correctly, I dont see how. If you saw a clock with a missing hour hand, you would know how many minutes past the hour it was, but not what hour. > But this is a very simple linear diophantine equation - so, where is the > problem...? > Im after a numerical solution. Is it possible to nd one? That is, > given an unknown x and a known n with a known inverse modulo of those > two p, can I nd x? (Is there only one x?) > So I can nd a/the numerical solution by solving a linear diophantine > equation? Thats beyond me. > I hope that I explained it correctly on the rst attempt! > Johnathan === Subject: Re: dimension of free modules - looking for example > somewhere (probably in Langs Algebra) I have read that there are rings R > and free modules R^n and R^m which are isomorphic although n!=m. > Does anybody know of an example? A search on invariant basis number (IBN) will turn up probably more than you wanted to know. --Bill Dubuque === Subject: Re: dimension of free modules - looking for example >> somewhere (probably in Langs Algebra) I have read that there are rings R >> and free modules R^n and R^m which are isomorphic although n!=m. >> Does anybody know of an example? > A search on invariant basis number (IBN) > will turn up probably more than you wanted to know. But is it possible? I assume m and n are nite. In that case cannot one take a maximal prime p in R, with quotient eld k = R/p, and take the tensor product with R/p to deduce that k^m and k^n are isomorphic? -- 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: Kick classic in shins, win prizes Edward Green says... >A symbol v (bold-v) represents a particular innitesimally long >displacement of the point P (script-P), which is the argument of the >function f = f(P). An inner product is formed and set equal to >a symbol @_v f (partial d subscript bold-v, italic-f), which is the >change of f in going from the tail of v to its tip. Mighty like a >differential, no? In fact we can identify the old notation with the >new term-by-term: > = @_v f == = df >Each slot on the rhs and lhs is occupied by the same object, but with >a different symbol (grad f is the gradient, dX the innitesimal >displacement of the argument, df the innitesimal change in the >function). Notice in particular that df and Df are _not_ mapped >to each other, but are distinct objects. I think the point is that in the *old* way of doing things, you need to pick a coordinate system in order to dene grad f, you need a coordinate system in order to dene dX, you need a coordinate a third time to dene the inner product of grad f with dX. In MTWs way of doing things, the directional derivative of f with respect to v requires no coordinates for its denition. You write: >Heavens! This means that df is itself a variable or a function of >its argument (the change in P), as _well_ as an innitesimal! What >are we to do! > But this is precisely what the exterior derivative Df represents... >Nope. Df represents an _engine_ to _calculate_ such a thing, given an >innitesimal displacement as input, same as the gradient. In fact, >it _is_ the gradient, as they just nished telling us. Seems they >forgot. You agree that df is a *function* of the displacement. Thats exactly what Df is. I dont understand your contrasting df is a variable or function of its argument and Df is an enging to calculate such a thing, given an innitesimal displacement as input. Df is not the same thing as the gradient, even though it has the same information. The gradient is a vector, while Df is a 1-form (a linear function on vectors). Yes, if you have a metric, you can convert between 1-forms and vectors, but Df makes sense in the absence of any metric. Df is related to the usual gradient grad f via: (grad f)^i = (Df)_j g^{i,j} The main point of doing things the way MTW do it is to see what of your denitions require a metric, and what do not. Yes, in at spacetime, when the metric is constant, you can ignore the difference between a 1-form and a vector, since they are always interconvertible via the constant metric. But when the metric itself varies from place to place, it is a little dangerous to ignore such implicit conversions. Where the metric is being used should be made explicit. -- Daryl McCullough Ithaca, NY === Subject: Re: Kick classic in shins, win prizes Hi Daryl. Nice to hear from you. Let me group your response under two headings: the behavior of grad f, and my allegation of fuzzy thinking by MTW. You write: > Df is not the same thing as the gradient, even though it has the same > information. The gradient is a vector, while Df is a 1-form (a linear > function on vectors). <... Df is related to the usual gradient grad f via: > (grad f)^i = (Df)_j g^{i,j} Ok. I take it that the usual gradient means the object whose components are @f/@x_i. One certainly must investigate whether this even denes a covector _or_ a vector, since a general rule for writing objects in terms of coordinates might dene no geometric object at all (we call these things symbols). So you say in effect that the usual rule in fact gives a coordinate representation of a vector, not a covector. This would certainly be an important point. I havent convinced myself whether this is true or not, but its at least clear that if we are using the funny metric with the minus sign, we at least have to negate the x_0 component of grad f to use it as a covector, even in the nicest coordinate systems. However, I claim this is oblique to my criticism. You respond to: >This means that df is itself a variable or a function of >its argument (the change in P), as _well_ as an innitesimal! > But this is precisely what the exterior derivative Df represents... >Nope. Df represents an _engine_ to _calculate_ such a thing, given an >innitesimal displacement as input, same as the gradient. > You agree that df is a *function* of the displacement. Yes I do. > Thats exactly what Df is. I nd that a strange claim. Consider the ordinary scalar function: y = bx where y is a variable, x is a variable, b is a constant; x varies, y varies, b holds steady. The relation as a whole is a function, or y is a function of x. But we never call the constant b itself a function of x. Now consider the inner product: @_v f = where v is a variable, @_v f is a variable, and Df holds constant (considering a xed point on the manifold). Df is not itself a function of the displacement. It is the constant in a linear function, like b above. > I dont understand your contrasting df is a variable or > function of its argument and Df is an enging to calculate such a > thing, given an innitesimal displacement as input. Well, thats what I meant. Im not saying there is something wrong with the new approach, or attacking the great brave new coordinate free geometric world. And you do remind us to be careful how we correctly understand the bad old coordinate shackled notations as representations of objects in the brave new world. But MTW stray into nonsense with their casual calumnies of the old world: There is nothing wrong with df being a function without explicitly mentioning them its arguments. It never bothered us with y -- if necessary we could always write y(x). And Df is not a more sophisticated geometrized version of df, despite the symbolic simularity; it is a more sophisticated geometrized version of grad f. And neither is Df a function of the displacement, any more than grad f was: they are constant linear operators waiting for input. Give them a differential input, they output a differential increment; give them a unit vector, they output a directional derivative. Stuff like this is a pet peave of mine. === Subject: Re: Kick classic in shins, win prizes <... LOL, MTWs dual prose is certainly interesting, > had to try it once. > While component based notation may have some > faults, I can always get out my ruler and clock, > and gure out where the I am. > Understandly, mathematical physicists seek notational > concistion, especially to eliminate the unimportant > distractions, and thats good. > I guess its up to the new students to determine what > ts best, maybe differential algebra will evolve legs, > and out run tensor analysis. Huh! I was almost afraid to go back and look at this. I expected at least a few how dare you criticize MTW, you lickspittles. But nothing... === Subject: No Unique Initial Segment And No Characteristic Expansion posting-account=Qiuj5AwAAACmGnmS12qcvqA9IXzD0s4L N. U. I. S. A. N. C. E. No Unique Initial Segment And No Characteristic Expansion . . . . . . Innite people each ip coins innite times. Can you always nd a different sequence of heads and tails? sci.math and sci.logic went quiet on this question for about a week, then against all logic, probability theory, and common sense they all agreed YES. Believers of hyperinnities have no shame! How on Earth can you exhaust an innite set? This is their solution, a standard application of Cantors diagonalisation technique, the hailed method of proving innity incomplete (innite already means never ending or incomplete!) Take one of the people, whatever his 1st ip was, reverse it! If he ipped a head you select tail, if he ipped a tail, heads. Thats your rst outcome, cross him off and select someone else, whatever was their second ip, reverse it! Keep on going and you have an innite sequence that is different to everyones sequence in atleast one ip. THATS THE PROOF! Thats hyperinnity and cardinality theory 101. How many assumptions about innity are they taking for granted here? What jumps straight out at most everyone is : Arent all possible combinations of heads and tails for innite ips already been done? This doesnt stir the Cantor supporter one bit. They think if the combination is on the list, it must be at some natural number position n, but the nth ip of person n is necessarily different! Voila a (somewhat contrived) contradiction. Hence the sequence does not appear anywhere on the list. Makes me reminiss to lectures on these types of problems. You decide for yourself no this is right. So the lecturer asks you for the further outcome of that conclusion, which goes around in a circle and proves what he wanted to. Its played out to everyone who studies theory, you can see the conversions taking place on sci.logic and sci.math each week. The other supporting argument Cantor followers have is demonstrated when you assume this innite list: 0.222222.. 0.322222.. 0.332222.. 0.333222.. 0.333322.. 0.333332.. .. The diagonal is 0.22222.. If we modify every digit, we can get 0.33333.. Does 0.3.. occur on the list? 0.3333333 <--L--> There are unlimited 3s in sequence on the list! Although 0.333.. has No Unique Initial Segment, it does have a Characteristic Expansion. It ends in 333.., all members end in 222.. so no, 0.3.. does not occur on the list. Back to the innite ippers list: httttt.. hhhttt.. hhhhhh.. tttttt.. hhhttt.. .. With probability 1, this list contains every initial segment possible of heads and tails sequences. Assume there is some initial segment that is not on the list. TTTHHH This sequence has a nite length L, there are 2^L possible sequences of length L, so with innite amount of ippers that initial segment will be covered with probability 1. The diagonal sequence HHHTT.. inverted TTTHH.. has No Unique Initial Segment. That in itself does not prove its on the list, remember 0.333..! But TTTHH.. has No Characteristic Expansion either. As far as can be determined the sequence appears on the list. So although we may not be able to disprove the hyperinnity status of the diagonal yet, we can show that it is a N.U.I.S.A.N.C.E. Herc === Subject: Re: No Unique Initial Segment And No Characteristic Expansion posting-account=WZMvOwwAAAC_B1TaayrVN99BJgDWQkUc H7> What jumps straight out at most everyone is : H7> Arent all possible combinations of heads and tails for innite ips H7> already been done? No, obviously; the anti-diagonal BY DEFINITION hasnt been done. H7> This doesnt stir the Cantor supporter one bit. They think if the H7> combination is on the list, it must be at some natural number position We DONT just THINK this Herc: WE *KNOW* this for an absolute DEFINITION. Thats what list MEANS if you are insisting that it is a list-as-we-know-it. Now, IF you want to start talking about lists of order-types LONGER THAN omega, lists where maybe an innite number of things can come BEFORE something on the list as well as after, so that SOME things on the list are at positions that are NOT any particular natural number, WELL the, have at it. But at a bare minimum, if you are going to KEEP calling it a list, it is going to HAVE to be well-ordered, that is, it is going to have to have a rst element and every element in it is going to have to have a unique next element. There are a great many innities you can embed or permute the list into and still keep it countable; there are uncountably many countable ordinals. But the point remains that we are NOT believing something that MIGHT be mistaken when we say that everything on the list is on it at some natural number position: THAT is true BY DEFINITION, JUST as surely as it is true that every decimal place in the decimal expansion of a real is some NATURAL number of places to the right of the decimal point. THATS WHAT list MEANS, FOOL. Obviously were sorry for you that youre so ignoran that you didnt know what a list was. If you want to talk about a list of things with MORE than w elements, well, have at it, but STATE YOUR AXIOMS and DEFINE YOUR TERMS. === Subject: Re: No Unique Initial Segment And No Characteristic Expansion posting-account=WZMvOwwAAAC_B1TaayrVN99BJgDWQkUc H7> What jumps straight out at most everyone is : H7> Arent all possible combinations of heads and tails for innite ips H7> already been done? No, obviously; the anti-diagonal BY DEFINITION hasnt been done. H7> This doesnt stir the Cantor supporter one bit. They think if the H7> combination is on the list, it must be at some natural number position We DONT just THINK this Herc: WE *KNOW* this for an absolute DEFINITION. Thats what list MEANS if you are insisting that it is a list-as-we-know-it. Now, IF you want to start talking about lists of order-types LONGER THAN omega, lists where maybe an innite number of things can come BEFORE something on the list as well as after, so that SOME things on the list are at positions that are NOT any particular natural number, WELL the, have at it. But at a bare minimum, if you are going to KEEP calling it a list, it is going to HAVE to be well-ordered, that is, it is going to have to have a rst element and every element in it is going to have to have a unique next element. There are a great many innities you can embed or permute the list into and still keep it countable; there are uncountably many countable ordinals. But the point remains that we are NOT believing something that MIGHT be mistaken when we say that everything on the list is on it at some natural number position: THAT is true BY DEFINITION, JUST as surely as it is true that every decimal place in the decimal expansion of a real is some NATURAL number of places to the right of the decimal point. THATS WHAT list MEANS, FOOL. Obviously were sorry for you that youre so ignoran that you didnt know what a list was. If you want to talk about a list of things with MORE than w elements, well, have at it, but STATE YOUR AXIOMS and DEFINE YOUR TERMS. === Subject: Re: No Unique Initial Segment And No Characteristic Expansion posting-account=Qiuj5AwAAACmGnmS12qcvqA9IXzD0s4L Thats my point, you dont show the member is unique, you just dene it so. c1=!f1,1 & c2=!f2,2 &... Therefore c1!=f1,1 & c2!=f2,2 &.. WOW big conclusion. but your denition is awed. your argument, assume sequence is at n, cn!=fn,n is self referential, its completely meaningless, no different to Russels paradox or many others. dene a number that is not a number, what do you have? So how did you conclude the sequence is original? by changing digits? every digit in every digit position is saturated, its not going to run out of options. THINK this time, any 10 year old can work this out. Innite people all toss coins innite times each. Can you come up with a new sequence of heads and tails? The diagonal is not going to help here. All combinations to innite length are done. Lets drop innity and go by the book, only allow assertions about large objects as they tend to innity. Many people toss coins many times each. How many coins do you have to toss in sequence, to eventually get a new sequence? log(number of people). As number of people -> oo, log(number of people) -> oo. All sequences have been done to innite length, you need MORE THAN INFINITE ips to succeed. A sequence has a HEAD and a TAIL. The diagonal of a random innite list has NO UNIQUE INITIAL SEGMENT in the head and NO CHARACTERISTIC EXPANSION i the tail. There is nothing to distinguish the diagonal from members of the list. The diagonal of UTM(X, Y) has NO UNIQUE INITIAL SEGMENT and NO CHARACTERISTIC EXPANSION to distinguish it from members of the list. Innite means NEVER ENDING. Innite means NEVER COMPLETED If you try to prove INFINITY is not complete you get TRUE. Herc === Subject: Re: No Unique Initial Segment And No Characteristic Expansion posting-account=QZxY1AwAAAD8ZwUvxC2Lm8DOYRarTcQj Herc, you just cant write down a list of the reals, and you can write down a list of the naturals. Get over it. === Subject: Re: No Unique Initial Segment And No Characteristic Expansion posting-account=Qiuj5AwAAACmGnmS12qcvqA9IXzD0s4L UTM(X,Y) the_diag_has_no_unique_digits! . . The following is clearly insufcient to base existence of a new sequence or a new innity upon... Innite people each ip coins innite times. Can you always nd a different sequence of heads and tails? . Take one of the people, whatever his 1st ip was, reverse it! If he ipped a head you select tail, if he ipped a tail, heads. Thats your rst outcome, cross him off and select someone else, whatever was their second ip, reverse it! Keep on going and you have an innite sequence that is different to everyones sequence in atleast one ip. X 0/10 WRONG F YOURE KIDDING! FAIL :( its really quite simple, innite people all doing the same thing you are doing dispells any possibility of you being unique. Herc === Subject: Re: No Unique Initial Segment And No Characteristic Expansion HERC777 says... >Take one of the people, whatever his 1st ip was, reverse it! If he >ipped a head you select tail, if he ipped a tail, heads. Thats >your rst outcome, cross him off and select someone else, whatever was >their second ip, reverse it! Keep on going and you have an innite >sequence that is different to everyones sequence in atleast one ip. >X 0/10 WRONG F YOURE KIDDING! FAIL :( >its really quite simple, innite people all doing the same thing you >are doing dispells any possibility of you being unique. You are talking about the case of an innite number of people, all of whom are trying to diagonalize each other? Well, youre right, in that case you run into problems. I thought you were talking about the case in which innitely many people (called person_1, person_2, etc.) are all ipping coins randomly and writing their results down in their own personal notebooks. Another person (the diagonalizer, *not* one of those people) is writing a list of coin ips that he intends not to be equal to anyone elses. He goes to person_1 and looks at his rst entry. If person_1s rst entry is heads, then diagonalizer looks at person_2s *second* entry. If that entry is heads, then person_3, etc. Now, tell me what is to prevent diagonalizer from carrying out his plan? Nothing. Will diagonalizers list be the same as the list for person_1? No. Will it be the same as the list for person_2? No. What about person_3? No. Yet, you are saying that diagonalizers list is the same as somebodys list, but that somebody is not person_1, or person_2, or person_3, or person_4, or person_5, etc. -- Daryl McCullough Ithaca, NY === Subject: Re: No Unique Initial Segment And No Characteristic Expansion posting-account=Qiuj5AwAAACmGnmS12qcvqA9IXzD0s4L because he doesnt have a legitimate sequence, its only a N.U.I.S.A.N.C.E. The sequence he forms has no unique initial segment, even to innite ips. Assumption: all sequences are present, all combinations of H&T to innite length. TRUE Denition: dene a number (sequence) that is not in the set of numbers BAD DEFINITION Reasoning: if its a number then its not a number, if its not a number then its a number. snap out of it! its a hoax! put your money on making a new sequence out of an innite set of attempts and you lose, laws of physics. This is no different to you insisting without showing reason this formula is INVALID proof(X) <-> X iff X has a proof then X is true because it disagrees with this formula G = ~proof(G) Youd rather, out of communal preference accept YOU CANT PROVE ME as a hailed stopping point in theory, rather than squash it with a high level consistency axiom, X <-> proof(X) Herc === Subject: Re: No Unique Initial Segment And No Characteristic Expansion > because he doesnt have a legitimate sequence, its only a > N.U.I.S.A.N.C.E. > The sequence he forms has no unique initial segment, even to innite > ips. > Assumption: all sequences are present, all combinations of H&T to > innite length. TRUE > Denition: dene a number (sequence) that is not in the set of > numbers BAD DEFINITION > Reasoning: if its a number then its not a number, if its not a number > then its a number. > snap out of it! its a hoax! > put your money on making a new sequence out of an innite set of > attempts and you lose, laws of physics. > This is no different to you insisting without showing reason this > formula is INVALID > proof(X) <-> X > iff X has a proof then X is true > because it disagrees with this formula > G = ~proof(G) > Youd rather, out of communal preference accept > YOU CANT PROVE ME as a hailed stopping point in theory, > rather than squash it with a high level consistency axiom, > X <-> proof(X) > Herc Oh for heavens sake. You introduce chronology into the equation (innity) and then go on to ignore it. If youre ipping forever then of course you will ALWAYS have some unique patterns of ips, until someone repeats them, but because the ips never end there will always be new unique patterns of coin ips, and then copies, then unique patterns, then copies, ad innitum. Its never ending. The discussion is a waste of your time Herc. Time to obsess on something else. Rob === Subject: Re: No Unique Initial Segment And No Characteristic Expansion qw@gbronline.com: > ...because the ips never end there will always be new unique patterns > of coin ips... > Its never ending. Right, and what HERC doesnt get is that Daryl McCs diagonalizer (who himself is generating an innate list) only has to visit every person once, even though every person is in the process of generating an innite list. The diagonalizer has an innite amount of time to visit every person, because all the lists are innite. === Subject: Re: No Unique Initial Segment And No Characteristic Expansion >> What will be your criteria for determining that any two >> lists are not the same? > it must have a unique initial segment or an expansion that has some > characteristic that distinguishes it from other lists with the same > initial segment > So if we step through both lists in order, from the beginning of each list, > comparing the values at the same ordinal positions on both lists, we can > conclude that the lists are different as soon as we nd two different > values at the same ordinal position on both lists; cant we? In other > words, no matter how long the lists are, if they differ in just one > position then theyre different lists; arent they? if 2 different corresponding digits are actually found it would mean their initial segments differ, but no process can determine if an innite list is the same as another innite list. we should stick to the terminology of sequence, and use list for the (ordered) set of sequences list 1 3454 2 43545 3 34543 sequence 45848 On the list 0.3 0.33 0.333 0.3333 .. Innite members match 0.3.. in the initial segment of length 1 <3> Innite members match 0.3.. in the initial segment of length 2 <33> Innite members match 0.3.. in the initial segment of length 3 <333> To paraphrase your algorithm so that it may possibly terminate giving a result, it has to be feasible when run in parallel. So if we step through both the list and the sequence in order, from the beginning of each sequence, comparing the values at the same ordinal positions on both the lists members and the sequence, we can conclude that the sequence is different as soon as we nd two different values at the same ordinal position on both the sequence and lists members. Herc === Subject: Re: No Unique Initial Segment And No Characteristic Expansion posting-account=QZxY1AwAAAD8ZwUvxC2Lm8DOYRarTcQj Herc, you just cant write down a list of the reals, and you can write down a list of the naturals. Get over it. === Subject: Re: No Unique Initial Segment And No Characteristic Expansion > N. U. I. S. A. N. C. E. > No Unique Initial Segment And No Characteristic Expansion No Underwear Interdicting And Nonsense Crapitating Everywhere http://www.apa.org/journals/psp/psp7761121.html http://insti.physics.sunysb.edu/~siegel/quack.html -- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) http://www.mazepath.com/uncleal/qz.pdf === Subject: Re: No Unique Initial Segment And No Characteristic Expansion posting-account=Qiuj5AwAAACmGnmS12qcvqA9IXzD0s4L Uncle Al, the last remnant of pre Herc anarchy on the science web. (and a complete mental blank on any issues he cant popularise out of previous theorists work) (and the most obnoxious purile self obsessed fckwit mensa has every had to put up with) like nobodys seen a science student correct art students..... http://www.winternet.com/~mikelr/ame76.html Herc === Subject: Re: No Unique Initial Segment And No Characteristic Expansion > Uncle Al, the last remnant of pre Herc anarchy on the science web. > (and a complete mental blank on any issues he cant popularise out of > previous theorists work) > (and the most obnoxious purile self obsessed fckwit mensa has every had > to put up with) > like nobodys seen a science student correct art students..... > http://www.winternet.com/~mikelr/ame76.html > Herc Here is what you so desperately seek to subborn, steal, purchase, cheat, lie... anything but earn: The editors acknowledge receipt of the above manuscript on [date Uncle Al did a prdy good job, little boy, and now his work will be reviewed by Referees to stand or fall. The rst of three experiments is happening in China as you read this. We then have four outcomes: 1) The paper is rejected and all three parity Eotvos experiments null within experimental error. Clever idea but empirically wrong. 2) The paper is accepted and all three parity Eotvos experiments null within experimental error. Clever idea but empirically wrong. Perhaps somebody else will look in a better way or extend quantitative parity divergence theory. 3) The paper is rejected and there is a reproducible non-null output. The paper will be resubmitted and quickly accepted. General Relativity and 3/5 of M-theory are falsied; quantum mechanics is in trouble (Lorentz invariance empirically fails). Weitzenboek and afne gravitation displace Einstein and metric gravitation. 4) The paper is accepted and there is a reproducible non-null output. General Relativity and 3/5 of M-theory are falsied; quantum mechanics is in trouble (Lorentz invariance empirically fails). Weitzenboek and afne gravitation displace Einstein and metric gravitation. (2), (3), or (4) will have you eating your liver. (3) or (4) will have Uncle Al sampling lutesk 10 December. Pookie pookie little whining nothing. Uncle Al says, Its hard to make a comeback when you havent been anywhere. -- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) http://www.mazepath.com/uncleal/qz.pdf === Subject: Re: No Unique Initial Segment And No Characteristic Expansion posting-account=Qiuj5AwAAACmGnmS12qcvqA9IXzD0s4L clap clap, and I just click the [conrm membership] button 20 times each morning for my crust. Herc === Subject: question about Banach spaces Is there an accessible example of a linear space endowed with two non-equivalent complete norms? jenny === Subject: Re: question about Banach spaces > Is there an accessible example of a linear > space endowed with two non-equivalent > complete norms? > jenny Innite-dimensional only. And AC required. How about this: Banach spaces l^2 and l^1 are isomorphic as linear spaces (both having Hamel dimension c). -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: question about Banach spaces > Is there an accessible example of a linear > space endowed with two non-equivalent > complete norms? > jenny > Innite-dimensional only. And AC required. How about this: > Banach spaces l^2 and l^1 are isomorphic as linear spaces > (both having Hamel dimension c). one has to necessarily appeal to AC? In other words, is AC equivalent to that statement? I am asking that, since I have a somehow realated problem: It is well known that the dual of L^infty strictly contains L^1 (this can be directly argued from separability arguments; clearly, only in innite dimension)). In spite of that, I am not aware af any example of a continuous linear functional on L^infty which is not L^1 without appealing to the Hahn-Banach theorem, which is to say, without appealing to AC. jenny === Subject: Re: question about Banach spaces > Is there an accessible example of a linear > space endowed with two non-equivalent > complete norms? > jenny Innite-dimensional only. And AC required. How about this: > Banach spaces l^2 and l^1 are isomorphic as linear spaces > (both having Hamel dimension c). > one has to necessarily appeal to AC? In other words, is AC > equivalent to that statement? Equivalent, probably not, but in fact it cannot be proved in ZF alone. In Solovays model where every set of reals (and every set in a Polish space) has the property of Baire, it follows (proved by Christensen, I guess) that any linear map of separable Banach spaces is automatically continuous. So, in particular, if a given vector space admits two complete separable norms, then the identity is a homeomorphism. (And since, in metric spaces, a map is continuous if and only if its restriction to every separable subspace is, we can get rid of the separable assumptions I made above.) Thats in Solovays model. But there is some principle that goes with this saying any spaces and maps THAT YOU CAN ACTUALLY WRITE DOWN EXPLICITLY also work like this, merely using ZF. So, depending on what accessible means in the original question, the answer may be no. > I am asking that, since I have a somehow realated problem: > It is well known that the dual of L^infty strictly contains L^1 > (this can be directly argued from separability arguments; clearly, only > in innite dimension)). > In spite of that, I am not aware af any example of a continuous linear > functional on L^infty which is not L^1 without appealing to > the Hahn-Banach theorem, which is to say, without appealing to AC. Again, this existence cannot be proved in ZF. But it can be proved using HB. It is known that HB is strictly weaker than AC, so this answers the equivalent question above. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: question about Banach spaces Is there an accessible example of a linear > space endowed with two non-equivalent > complete norms? > jenny > Innite-dimensional only. And AC required. How about this: Banach spaces l^2 and l^1 are isomorphic as linear spaces > (both having Hamel dimension c). > one has to necessarily appeal to AC? In other words, is AC > equivalent to that statement? > Equivalent, probably not, but in fact it cannot be proved in ZF alone. (also for David C. Ullrich) When you say that it cannot be proved what do you exactly mean? I mean, if it is provable with AC but not provable without AC why it is not automatically equivalent to AC. My point is: one cannot prove this within ZF because he/she is not able to, but in principle he/she could, or there is a way to show this impossibility? jenny === Subject: Re: question about Banach spaces >> Is there an accessible example of a linear >> space endowed with two non-equivalent >> complete norms? >> jenny >> Innite-dimensional only. And AC required. How about this: >> Banach spaces l^2 and l^1 are isomorphic as linear spaces >> (both having Hamel dimension c). >> one has to necessarily appeal to AC? In other words, is AC >> equivalent to that statement? >> Equivalent, probably not, but in fact it cannot be proved in ZF alone. >(also for David C. Ullrich) >When you say that it cannot be proved what do you exactly mean? >I mean, if it is provable with AC but not provable without AC >why it is not automatically equivalent to AC. >My point is: one cannot prove this within ZF because he/she is >not able to, but in principle he/she could, or there is a way to >show this impossibility? There is a way to show this impossibility. There was a big hint how that works in the paragraph you snipped, where Edgar said something about a certain model of ZF. Any statement that can be proved from the axioms of ZF will be true in every model of ZF, so if there is a model of ZF in which P is false it follows that P _cannot_ be proved in ZF. You need to study a small bit of mathematical logic to really know exactly what that means. For now an analogy: Say GT is the axioms for a group: (ab)c = a(bc), etc. So for example (ab(cd) = (a(bc))d is a theorem of GT - it can be proved from the axioms. Now a model of GT is precisely the same thing as a _group_. There exists a group in which it is not true that ab = ba for all a, b, and the existence of such a group shows that ab = ba for all a, b cannot be proved from the axioms of GT. >jenny ************************ David C. Ullrich === Subject: Re: question about Banach spaces > (also for David C. Ullrich) > When you say that it cannot be proved what do you exactly mean? > I mean, if it is provable with AC but not provable without AC > why it is not automatically equivalent to AC. > My point is: one cannot prove this within ZF because he/she is > not able to, but in principle he/she could, or there is a way to > show this impossibility? An example is the Hahn-Banach theorem, HB. Assuming consistency when needed, it has been shown that: (1) ZF does not prove HB, (2) ZF+AC proves HB, (3) ZF+HB does not prove AC. In that sense, HB is beyond ZF, but is not as strong as AC. On the other hand, Zorns Lemma, ZL, is equivalent to AC, meaning: (4) ZF+AC proves ZL (5) ZF+ZL proves AC In most cases, a cannot be proved result is shown by exhibiting a model. For example, a model of ZF+HB where AC fails will show (3), above. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: question about Banach spaces >> Is there an accessible example of a linear >> space endowed with two non-equivalent >> complete norms? >> jenny >> Innite-dimensional only. And AC required. How about this: >> Banach spaces l^2 and l^1 are isomorphic as linear spaces >> (both having Hamel dimension c). >one has to necessarily appeal to AC? In other words, is AC >equivalent to that statement? Not that I know the answer to either question, but Ill point out that the in other words isnt right - saying something requires AC is not the same as saying its equivalent to AC. If ZF does not prove P but ZFC does then P requires AC; saying P is equivalent to AC is a much stronger statement. (For example: If ZFC proves P but, say, ZF + AD (Axiom of Determinacy) proves not P then AC is required for P, although this doesnt say that P is equivalent to AC. This actually happens if P is every set of reals is Lebesgue measurable.) >I am asking that, since I have a somehow realated problem: >It is well known that the dual of L^infty strictly contains L^1 >(this can be directly argued from separability arguments; clearly, only >in innite dimension)). >In spite of that, I am not aware af any example of a continuous linear >functional on L^infty which is not L^1 without appealing to >the Hahn-Banach theorem, which is to say, without appealing to AC. >jenny ************************ David C. Ullrich === Subject: toplogically equevalent... hello.....doctor~ Let (X,d) be a metric space. Dene a function d :XxX -> (R+) U {0} by d(x,y) = min {1, d(x,y)}, the minumum of 1 and d(x,y), for all x,y in X. show that (1) d is a bounded metric for X. (2) The metric space (X,d) is topologically equivalent to the bounded metric space (X,d) ---------------------------------------------------- i can do (1) by metric conditions. but...in the (2) let x in X and e>0 then y in B_d_(x,e) => d(x,y) < e => d(x,y) <= d(x,y) < e => y in B_d_(x,e) y in B_d_(x,e) => d(x,y) < e =>min {1, d(x,y)} < e => 1 < e or d(x,y) < e if d(x,y) < e => y in B_d_(x,e) else 1 < e => (***) i cant deduce y in B_d_(x,e) in the (***) step. so, i need your advice. thank you very much for your advice. === Subject: Re: toplogically equevalent... > hello.....doctor~ > Let (X,d) be a metric space. > Dene a function d :XxX -> (R+) U {0} by > d(x,y) = min {1, d(x,y)}, the minumum of 1 and d(x,y), > for all x,y in X. > show that > (1) d is a bounded metric for X. > (2) The metric space (X,d) is topologically equivalent to > the bounded metric space (X,d) > ---------------------------------------------------- > i can do (1) by metric conditions. > but...in the (2) > let x in X and e>0 > then y in B_d_(x,e) => d(x,y) < e > => d(x,y) <= d(x,y) < e > => y in B_d_(x,e) > y in B_d_(x,e) => d(x,y) < e > =>min {1, d(x,y)} < e > => 1 < e or d(x,y) < e > if d(x,y) < e => y in B_d_(x,e) > else 1 < e => (***) > i cant deduce y in B_d_(x,e) in the (***) step. > so, i need your advice. > thank you very much for your advice. Let N_e(x) = {y in X | d(x,y)0, B_L = { N_1/n(x) | x in X, n in N, 1/N Let (X,d) be a metric space. > Dene a function d :XxX -> (R+) U {0} by > d(x,y) = min {1, d(x,y)}, the minumum of 1 and d(x,y), > for all x,y in X. > show that > (1) d is a bounded metric for X. > (2) The metric space (X,d) is topologically equivalent to > the bounded metric space (X,d) Exercise: Start with a metric d and f:R -> R, f(x) = 0 iff x = 0, f(x+y) <= f(x) + f(y) Show fd = f o d is a metric. Show when f is ascending and continuous at 0 that f is uniformly continuous and d,fd are equivalent metrics For your problem f(x) = min{ 1,x } and d = fd. > but...in the (2) > let x in X and e>0 > then y in B_d_(x,e) => d(x,y) < e > => d(x,y) <= d(x,y) < e > => y in B_d_(x,e) > y in B_d_(x,e) => d(x,y) < e > =>min {1, d(x,y)} < e > => 1 < e or d(x,y) < e > if d(x,y) < e => y in B_d_(x,e) > else 1 < e => (***) > i cant deduce y in B_d_(x,e) in the (***) step. I think you want to show y in B_d(x, min(1,r)) ==> y in B_d(x,r) Heres from my notes for the exercise: Let F(a,r) = { x | fd(a,x) < r } F(a,f(r)) subset B(a,r). If x in F(a,f(r)): fd(a,x) < f(r) if r <= d(a,x): f(r) <= fd(a,x) < f(r) which cannot be d(a,x) < r; x in B(a,r) some s > 0 with B(a,s) subset F(a,r). Some s > 0 with f(s) < r if x in B(a,s): d(a,x) < s; fd(a,x) <= f(s) < r; x in F(a,r) ---- === Subject: compute normal how can I compute the normal of a plane when I have only three vectors which are all standing on the plane at the same point (cp trihedron startpoint of the vectors is the same) and I know the three (different) angles between the vectors and the plane. S. Nurbe === Subject: Re: compute normal >how can I compute the normal of a plane when I have only three vectors >which are all standing on the plane at the same point (cp trihedron >startpoint of the vectors is the same) and I know the three >(different) angles between the vectors and the plane. >S. Nurbe Why start a new thread when you have already received an answer? --Lynn