mm-4619 === Subject: Riccati, The Key Formula To The Universe (Traffic-Server/6.1.5 [uScM]) http://mathforum.org/kb/thread.jspa?threadID=1739373&tstart=0 It is difficult to imagine that a simple formula like Riccati differential equation (RDE): dx/dt+x^2=g(t) will represent the sole mathematical formula of the universe. Yes, all majestic edifice of mathematics can indeed be reduced to a RDE (or many of them). Sounds like impossible to visualize. But this is result of my 25 years extensive research experiences, after Cambridge graduation. That is more than 5 times of time required from B.Sc. to Ph.D. degrees (5 years average), bearing in mind my intuition in the subject is about 10 times more than those of academics in top schools. They are conventional researchers within mainstream research, some practicing obsolete mathematical sciences. Let us look at the Math-chart again (as enclosed). 1-I have declared that polynomials are included in RDE. Not only we systematically extract solving polynomials to Riccati, but also general polynomial can be included in a RDE, with the residue having solutions according to ODE or PDE existence theorems for systems of differential equations. Indeed many developments and applications demonstrate that. 2- linear differential equations of higher order of variable coefficients (LDE). In fact once we have a general polynomial, a LDE can be constructed based on it. Also we can see on my classical solving equations, I imbed a RDE within a LDE and then manipulate the result to solve The Riccati. One can say there is a kind of inverse relations happening here. 3- Non-linear differential equations are similarly imbedded within a higher order LDE, or systems. The solving will apply to systems of algebraic equations as well. 4-Integrals are special part of a RDE anyway. But manipulating whole system around RDE will as well solve many integrals particularly Elliptic. 5-Numerical applications are direct applications of results obtained with many solved parametric algebraic equations or differential equations, as well as difference equations. All I see the Math-Chart is correct and you need to have 5 years of international research to experience personally this new world of math computations. I will renew the offer after each five years! Well, all this means that, once you start your research here, there is no way out of RDE. About other algebraic branches in mathematics, eventually they will be out of business or have a very small influence on the subject. Dr.M.Basti PS: It seems Google will not transfer files with attachments, from Math Forum sites. === Subject: Re: Riccati, The Key Formula To The Universe I have to mention that other forms of ODE (not Riccati) will not be suitable for developments. What I mean, one may find a trick to solve a special type, but the right one i.e. Riccati, will present a class type developments along with a class type applications. Indeed within a powerful mathematical chemistry of Riccati, we will approach new understanding of the universe. The Riccati solving will provide a basis for structuring other unsolved problems, with some kind of transformations TO REACH THEM. And with a focus in solving problems. They need to carefully look at my Math-Chart. You will notice that Linear and Non-Linear differential equations of any order (variable) is included in Riccati. The argument is based on existence theorems on ODE and PDE Confirming the situation. Indeed, based on these understanding new developments will be initiated; in fact the new math is based on it. By the way differential equations of the type: dx/dt=sum (Ai(t)*x^i,i=0..n) are all the first order representative of a general polynomial, which are included in Riccati. For example a polynomial of degree 6 has also an ODE of this form: dx/dt = A5(t)*x^5+A4(t)*x^4+A3(t)*x^3+A2(t)*x^2+A1(t)*x+A0(t). Indeed once a polynomial solved in Riccati as their source then we can easily trace the above form differential equations, handle types of non-linear ODE, as well as other polynomials. I have also many classes of solved first order differential equations of above types in my collection (They are all rooted in Riccati). We need to recursively solve problems , one by one (although as a class). There are many other forms of non-linear ODE which are resulting from a collection of transformations of past solved polynomials,ODE,etc. In short, the universe has one mathematical source structure and that is Riccati. Obviously the theory is totally in its infancy. Dr.M.Basti === Subject: Re: Riccati, The Key Formula To The Universe <28112537.1210203389693.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=JpxxPAgAAAAgwzQIYqn4j6syK-YhOmcF Gecko/20071127 Firefox/2.0.0.11,gzip(gfe),gzip(gfe) Indeed within a powerful mathematical chemistry of Riccati, we will approach new understanding of the universe. > The Moon is made with Riccati cheese! > Dr.M.Basti === Subject: Re: Riccati, The Key Formula To The Universe posting-account=suWj4AkAAADE1IvGmj55Nmq3f98qb17e InfoPath.1; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) http://mathforum.org/kb/thread.jspa?threadID=1739373&tstart=0 It is difficult to imagine that a simple formula like Riccati > differential equation (RDE): dx/dt+x^2=g(t) will represent the sole mathematical formula of the universe. Yes, all majestic edifice of mathematics can indeed be reduced to a > RDE (or many of them). Sounds like impossible to visualize. But this is result of my 25 years extensive research experiences, > after Cambridge graduation. That is more than 5 times of time required from B.Sc. to Ph.D. degrees > (5 years average), bearing in mind my intuition in the subject is > about 10 times more than those of academics in top schools. They are conventional researchers within mainstream research, some > practicing obsolete mathematical sciences. Let us look at the Math-chart again (as enclosed). 1-I have declared that polynomials are included in RDE. Not only we systematically extract solving polynomials to Riccati, but > also general polynomial can be included in a RDE, with the residue > having solutions according to ODE or PDE existence theorems for > systems of differential equations. Indeed many developments and applications demonstrate that. 2- linear differential equations of higher order of variable > coefficients (LDE). In fact once we have a general polynomial, a LDE can be constructed > based on it. Also we can see on my classical solving equations, I imbed a RDE > within a LDE and then manipulate the result to solve > The Riccati. One can say there is a kind of inverse relations happening here. 3- Non-linear differential equations are similarly imbedded within a > higher order LDE, or systems. The solving will apply to systems of algebraic equations as well. 4-Integrals are special part of a RDE anyway. But manipulating whole > system around RDE will as well solve many integrals particularly > Elliptic. 5-Numerical applications are direct applications of results obtained > with many solved parametric algebraic equations or differential > equations, as well as difference equations. All I see the Math-Chart is correct and you need to have 5 years of > international research to experience personally this new world of math > computations. I will renew the offer after each five years! Well, all this means that, once you start your research here, there is > no way out of RDE. About other algebraic branches in mathematics, eventually they will be > out of business or have a very small influence on the subject. Dr.M.Basti PS: It seems Google will not transfer files with attachments, from > Math Forum sites. ************************************************************* Too late. After a welcome hiatus, Plutoniums is already posting his astonishing wise theories again, and if you read them carefully you'll get convinced that he already found out the key to understanding the whole world, up to and including Bush's speeches: we all (sit down since this is awsome!)... have our asses on a huge plutonium atom! Now go and convince him that it in fact is a Ricatty RDE uga-boogah thing. Good luck Tonio === Subject: Re: Riccati, The Key Formula To The Universe posting-account=JpxxPAgAAAAgwzQIYqn4j6syK-YhOmcF Gecko/20070309 Firefox/2.0.0.3,gzip(gfe),gzip(gfe) >http://mathforum.org/kb/thread.jspa?threadID=1739373&tstart=0 It is difficult to imagine that a simple formula like Riccati > differential equation (RDE): dx/dt+x^2=g(t) will represent the sole mathematical formula of the universe. Yes, all majestic edifice of mathematics can indeed be reduced to a > RDE (or many of them). Sounds like impossible to visualize. But this is result of my 25 years extensive research experiences, > after Cambridge graduation. That is more than 5 times of time required from B.Sc. to Ph.D. degrees > (5 years average), bearing in mind my intuition in the subject is > about 10 times more than those of academics in top schools. They are conventional researchers within mainstream research, some > practicing obsolete mathematical sciences. Let us look at the Math-chart again (as enclosed). 1-I have declared that polynomials are included in RDE. Not only we systematically extract solving polynomials to Riccati, but > also general polynomial can be included in a RDE, with the residue > having solutions according to ODE or PDE existence theorems for > systems of differential equations. Indeed many developments and applications demonstrate that. 2- linear differential equations of higher order of variable > coefficients (LDE). In fact once we have a general polynomial, a LDE can be constructed > based on it. Also we can see on my classical solving equations, I imbed a RDE > within a LDE and then manipulate the result to solve > The Riccati. One can say there is a kind of inverse relations happening here. 3- Non-linear differential equations are similarly imbedded within a > higher order LDE, or systems. The solving will apply to systems of algebraic equations as well. 4-Integrals are special part of a RDE anyway. But manipulating whole > system around RDE will as well solve many integrals particularly > Elliptic. 5-Numerical applications are direct applications of results obtained > with many solved parametric algebraic equations or differential > equations, as well as difference equations. All I see the Math-Chart is correct and you need to have 5 years of > international research to experience personally this new world of math > computations. I will renew the offer after each five years! Well, all this means that, once you start your research here, there is > no way out of RDE. About other algebraic branches in mathematics, eventually they will be > out of business or have a very small influence on the subject. Dr.M.Basti PS: It seems Google will not transfer files with attachments, from > Math Forum sites. ************************************************************* Too late. After a welcome hiatus, Plutoniums is already posting his > astonishing wise theories again, and if you read them carefully you'll > get convinced that he already found out the key to understanding the > whole world, up to and including Bush's speeches: > we all (sit down since this is awsome!)... have our asses on a huge > plutonium atom! > Now go and convince him that it in fact is a Ricatty RDE uga-boogah > thing. > Good luck Tonio you can't comment on that Ricatty RDE uga-boogah thing unless you have 5 years of international research to experience personally this new world of math computations. === Subject: Re: Riccati, The Key Formula To The Universe >you can't comment on that Ricatty RDE uga-boogah thing unless you have >5 years of international research to experience personally this new >world of math >computations. The troll won't let any of the billy goats cross his Riccati bridge. -- Angus Rodgers Contains mild peril === Subject: Re: Riccati, The Key Formula To The Universe > >you can't comment on that Ricatty RDE uga-boogah thing unless you have >5 years of international research to experience personally this new >world of math >computations. The troll won't let any of the billy goats cross his Riccati bridge. I just want to know if there will be Riccati for my lasagna. -- Michael Press === Subject: Re: Riccati, The Key Formula To The Universe I suppose, since Riccati is an Italian name, thus Italy and its people, should be very proud, having the formula of the universe, after one of their citizens (as well as Vatican which is in Italy). http://en.wikipedia.org/wiki/Jacopo_Riccati Since I have discovered this characteristic of the universe, may be I calling myself, Julius Caesar II, (after the great Roman emperor). Dr.M.Basti === Subject: Re: Riccati, The Key Formula To The Universe <2m4424h9roig4rfm2oqmcmktrihpf49709@4ax.com> posting-account=Rkt6TwoAAACG_SqlrxmgPCl1Ozr0PWSD MathPlayer 2.10b; .NET CLR 2.0.50727; .NET CLR 3.0.04506.30; .NET CLR 1.1.4322; .NET CLR 3.0.04506.648),gzip(gfe),gzip(gfe) you can't comment on that Ricatty RDE uga-boogah thing unless you have >5 years of international research to experience personally this new >world of math >computations. The troll won't let any of the billy goats cross his Riccati bridge. -- > Angus Rodgers > Contains mild peril I reckon this obstinate asinus, Basti, can't cross the pons himself. === Subject: Re: Closure of a set in the Zariski topology The thing here is that you bumped into the problem that things do not > patch properly together when they are hand-made: What is not true in > Yeah, I've noticed that everything should always be patched together > from natural morphisms. Trying to build things by hand doesn't > usually get you anywhere... reading it after posting, I hoped you didn't understand that I wanted to > say that your methods are hand- or home-made. :-) > > I assume that if J = Ker f^#, where f:X->Y is quasi-compact, then > f(X) is a dense subset of Supp(O_Y/J)? I was able to prove this for > this special case, but I suppose it's true in general? That's correct. > > Do you have any ideas how to approach this? Boil that thing down to commutative algebra with the following road-map: 1. Reduce to the case where Y is affine. > [Checking denseness can be done locally.] 2. Reduce to the case where X is affine. > [Use the technique from yesterday using a _finite_ cover of X by open > affines U1,...,Un and consider the composition of f with the covering > U1 u ... u Un (disjoint union) -> X.] After 1. and 2. we can assume that X and Y are affine with coordinate > rings B and A, say, such that f comes from a ring homomorphism ff: A -B. We want to show that f(X) is dense V(ker(ff)). 3. Reduce to the case where ff is injective, i.e. ker(ff) = 0, i.e. > V(ker(ff))=Y. Our claim reduces to the case that f has dense image. > [Divide out the kernel.] 4. Reduce to the case that A is an integral domain. Corrected version: > [f(X) is dense iff f hits the generic point of every irreducible component of Y.] 5. Show that it suffices to show that if A is an integral domain and ff: > A -> B is an injective ring homomorphism, then there is a prime ideal pp > in B sitting above the prime ideal 0 of A. > [Localize ff with respect to S=A{0} and take any prime ideal in the > non-trivial ring S^-1B you want.] -- Best wishes, J. === Subject: Re: Closure of a set in the Zariski topology > The thing here is that you bumped into the problem that things do > not patch properly together when they are hand-made: What is not > true in > Yeah, I've noticed that everything should always be patched together > from natural morphisms. Trying to build things by hand doesn't > usually get you anywhere... > reading it after posting, I hoped you didn't understand that I wanted > to say that your methods are hand- or home-made. :-) > I assume that if J = Ker f^#, where f:X->Y is quasi-compact, then > f(X) is a dense subset of Supp(O_Y/J)? I was able to prove this for > this special case, but I suppose it's true in general? > That's correct. > Do you have any ideas how to approach this? > Boil that thing down to commutative algebra with the following > road-map: > 1. Reduce to the case where Y is affine. > [Checking denseness can be done locally.] > 2. Reduce to the case where X is affine. > [Use the technique from yesterday using a _finite_ cover of X by open > affines U1,...,Un and consider the composition of f with the > covering U1 u ... u Un (disjoint union) -> X.] > After 1. and 2. we can assume that X and Y are affine with coordinate > rings B and A, say, such that f comes from a ring homomorphism ff: A > -> B. We want to show that f(X) is dense V(ker(ff)). > 3. Reduce to the case where ff is injective, i.e. ker(ff) = 0, i.e. > V(ker(ff))=Y. Our claim reduces to the case that f has dense image. > [Divide out the kernel.] > 4. Reduce to the case that A is an integral domain. Corrected version: > > [f(X) is dense iff f hits the generic point of every irreducible > component of Y.] > 5. Show that it suffices to show that if A is an integral domain and > ff: A -> B is an injective ring homomorphism, then there is a prime > ideal pp in B sitting above the prime ideal 0 of A. > [Localize ff with respect to S=A{0} and take any prime ideal in the > non-trivial ring S^-1B you want.] 5. can be substituted by 5'. Reduce the case where A is a field. [Localize ff with respect to S=A{0} and note that S^-1ff is injective.] 6. We are done if we can show that if B is a k-algebra, then there are prime ideals in B. But this is certainly true since B is not the zero ring. HTH. -- Best wishes, J. === Subject: Re: FLT and New Math <1gp32499ai2cm32tpvvlpte0scqq36okfg@4ax.com> posting-account=ee2apQoAAABJNMlLTFasCJw9Nfo9FmYk Gecko/20020924 AOL/7.0,gzip(gfe),gzip(gfe) HTTP/1.1 cache-ntc-ac04.proxy.aol.com[CFC87485] (Traffic-Server/6.1.5 [uScM]) > If you want to converse please do so and stop lecturing and making > unsupported claims as if they were already credibly well established. > I am currently in an intellectual war with the establishment, to challenge their illegal control of the value judgments in the academia. One person against a system! Dr.M.Basti === Subject: Re: FLT and New Math > > If you want to converse please do so and stop lecturing and making > unsupported claims as if they were already credibly well established. > I am currently in an intellectual war with the establishment, to > challenge their illegal control of the value judgments in the > academia. One person against a system! Dr.M.Basti And the results so far? === Subject: Re: FLT and New Math > If you want to converse please do so and stop lecturing and making > unsupported claims as if they were already credibly well established. > I am currently in an intellectual war with the establishment, to > challenge their illegal control of the value judgments in the > academia. > > One person against a system! > > Dr.M.Basti And the results so far? Stop Press: City Hall defeated! -- Angus Rodgers Contains mild peril === Subject: Re: Solution to a Mathematics GRE question > If one cannot even begin to do such for grade-school concepts then > one will face insurmountable difficulties at the forefront of math. SIMPLER: one is nearly like an ape. Rainer Rosenthal r.rosenthal@web.de === Subject: Finding all sets of nodes forming complete sub-graphs Cc: ahmad.humyn@gmail.com posting-account=w9hrNwoAAACb8kL13uRGQISfhdn7Szm7 Gecko/2008032620 Firefox/3.0b5,gzip(gfe),gzip(gfe) Hi there, I want to find all the sets of nodes in a graph which have edges between all the nodes in the set....in other words I want to find all the complete sub-graphs in an un-directional graph: e.g. If I have this upper-triangular graph (un-directed graph) where each edge is denoted by a 1 (0 is no edge) ... e.g. C is connected to D: A B C D E F A X 1 1 0 0 0 B X X 1 0 0 0 C X X X 1 0 1 D X X X X 0 1 E X X X X X 0 F X X X X X X If I give the algo this matrix, it should return me two answer sets -> (A,B,C) and (C,D,F). Am I looking for finding cliques? If so what is the fastest way known to compute these quickly? What is the most standard easiest way to compute these sets. I have heard cliques are SAT-3 (NP complete)? Please help!! === Subject: Re: Finding all sets of nodes forming complete sub-graphs posting-account=w9hrNwoAAACb8kL13uRGQISfhdn7Szm7 Gecko/2008032620 Firefox/3.0b5,gzip(gfe),gzip(gfe) Plus I am probably not looking just for maximal cliques....I'm looking for cliques of all sizes === Subject: Re: Finding all sets of nodes forming complete sub-graphs posting-account=w9hrNwoAAACb8kL13uRGQISfhdn7Szm7 Gecko/2008032620 Firefox/3.0b5,gzip(gfe),gzip(gfe) > Plus I am probably not looking just for maximal cliques....I'm looking > for cliques of all sizes added thought: I know that a clique will have smaller cliques inside it....in such a situation want the clique of the largest size in that set of nodes. === Subject: Re: JSH: You people are stupid Yeah you're cute (NOT) but not nearly as amart a JSH! >(this JSH moment is brought to you by MIT Math Department, Make Sure you are >dealing with a Professional Mathematician, else this can happen;) >JSH: You people are stupid Ok, let's just put it out there, you people as a group are stupid. There is no way there can be nothing to my research when there is so much >math behind it that says it has to be important. No way. It is complete bull that rules your world, which is how you people keep >posturing despite your rank stupidity. Only stupidity could explain the failure to even look into integrating that >partial differential equation. Only stupidity could explain the ability of posters to keep going on and on >despite there being no other prime counting function that can even >recursively call itself. And that's just my research related to the prime count. Only people stupid on a massive scale could ignore publication in a peer >reviewed math journal, for an argument so damn trivial I can go over it in a >couple of paragraphs using basic algebra. You people are just damn stupid. Your community is stupid. You are willfully stupid, stupid, stupid. [A reply to someone who said that James is a ranting tit.] At least I am right and have the mathematics to prove it. I even got published. You damn fools just said, duh, publication doesn't matter! You are stupid. I have the mathematics to prove my case. I've had it for years. Your society is just too damn stupid to go with the math. You're just too damn stupid and full of yourselves. Idiots. ># posted by James Harris @ 21:23 >JSH: Four years plus of failure >It has been over four years since I found what I call my prime counting >function. It is still an open question how closely the partial differential equation >that follows it maps to the prime distribution because the current math >world is broken, and rather than consider the answers to that question, most >mathematicians have ignored me, while you can see the posters here who just >work to dismiss the research. The lack of answers is the fruit of their labors. That's what such people are about, no answers, no solutions, ever. Has it occurred to any of you that you could die, and a day later, some >person or persons could finally look into that partial differential equation >I found, and answer huge questions in mathematics, and the world go on and >on about this amazing thing that research thought quack from some crackpot >turned out to be important, but no one knew, because no one checked. And mostly in looking over the history, they could see years of posturing >from people who never did anything of value. ANYONE can criticize. You have four years plus to see the value of posters just calling me names >and claiming my research is not new, as now you may know that yes, I did >find a partial differential equation that follows from a prime counting >function. And it's an open question as to how closely integrating that partial >differential equation gets you to the prime distribution. STILL an open question, over four years later. Still there are posters who seem to think this is just about posturing on >some newsgroups as if no one cares if the question is EVER answered. Ever. What kind of human beings does this make you to be, when you are so >incurious, so incapable of wondering at the miniscule level necessary to get >some answers? Instead you sit by and either promote or are in complicit in just blanket >do-nothingness. A world of no answers, just talk. Four plus years of total failure from your group and your community. That's what the world has today. Four plus years of total failure to find any kind of answer. [A reply to someone who asked why is it that this is not a failure from >James.] You ing idiot I TESTED IT AND NUMERICALLY INTEGRATED IT AND FOUND IT >CLOSE! You are the goddamn fools who refuse to check and then keep going on as if >you are intelligent when you will NOT EVEN ING CHECK!!! ># posted by James Harris @ 21:13 JSH: Minority view >A few days ago I asked that readers who use Google Groups use the ratings >system that is provided through it where you can rate posts from 1 to 5 >stars. I have looked over threads that I created to talk about my research and >found that along with my ratings, as I've started now routinely rating >posts, I usually see 3 or 4 others and at times maybe 10 who have bothered >to vote. Giving the benefit of the doubt to opposition, let's say there are about 10 >people out there then who disagree with me and are vocal in that >disagreement. Well I just checked Google Groups to see how many people it says subscribe >to sci.math and it said 5,506, which is the number through Google Groups, so >it's a lower number than the readership. For instance, I don't subscribe to sci.math so I'm not part of that count. >But, of course, I post on sci.math a lot. But let's go with the low number to be fair and now go with the high number >for the opposition based on ratings to get 10/5506 or about 0.18% of the >sci.math readership clearly disagrees with me. So I have one point of view and about 10 or so other people who don't seem >to have any other support based on the ratings, argue with me, and that >generates a lot of posts. I like the ratings system as gives an overview of how many people on >sci.math care to vote in one direction or another on these topics where if >you just look at the volume of postings you might wrongly assume that the >majority of sci.math is wrapped up in these discussions, when from the >ratings, most don't care. Actually, only about 0.18% clearly show that they do care. I suggest you use the ratings if you use Google Groups. There is just no >denying that much of the newsgroup doesn't care one way or the other when >there is no voting for your point of view. ># posted by James Harris @ 16:18 Friday, November 24, 2006 JSH: Under review >verified receipt. A version of the paper is at >and in keeping with the philosophy of extreme mathematics, I would >appreciate comments. The paper is under review as Princeton verified receipt. Uh, yeah that >surprised even me. This thread is for comments on the paper. And yes, this paper could end it all. If Princeton does what it should then >it doesn't matter what any of you say, as you know and I know that Princeton >trumps every last one of you. I especially welcome comments from posters who argue with me regularly. I am curious though. Read the paper, and come back and post as confidently >as you've done. What gives you that faith? blame the sinister cadre of sci.math again.] No. I know and you know that Princeton doesn't give a damn what you think. There is only one case where sci.math definitely interfered and killed a There is no way that Princeton would fall in the same way. You people can't kill the Annals of Mathematics. I dare you to try. Send your emails. Just try. [A reply to someone who explained to James how uninteresting and dishonest >his paper is.] The replies in this thread reek of fear. What I actually did was go from a prime counting sieve function, to a >constrained summation of a partial difference equation that counts primes, >to the partial differential equation that follows from it. The three forms connect the dots from the discrete count of primes to a >continuous function directly, for the first time in mathematical history, >showing how the count of primes connects to continuous functions, and >accomplishing what notables like Gauss and Riemann set out to do so long >ago. That is done in just two pages, and is an accomplishment at the pinnacle of >human achievement in mathematics, but it's also an accomplishment members of >the sci.math newsgroup have enviously and jealously fought to disparage. It is one of the greatest accomplishments in the history of human thought. It is more than a great enough accomplishment to merit publication in one of >the world's top math journals. ># posted by James Harris @ 01:16 >Sabotage! >A member of the sci.math newsgroup managed to get past my screening and join >the group, revealing himself when he systematically down rated all of my >posts. I have banned that person from the group. It shows you how these people operated. Looks like people not members of the group can rate posts so I was wrong on >that point. As shown by a group of sci.math'ers who came in to downrate my posts. Still, it shows how these people operate. There was other evidence the banned person was a hostile sci.math'er so I'm >comfortable with the decision to ban that person. I do wish Google would allow you to only let group members rate posts, as I >think their Beta still needs a bit more options for group owners, as this >current ongoing sabotage by members of the sci.math newsgroup shows. The juvenile behavior from members of the Usenet sci.math group reveals a >clear weakness in Google's current setup. ># posted by James Harris @ 00:45 Wednesday, November 22, 2006 JSH: Use ratings >People coming in with Google Groups can rate posts, and I'd like more people >to start rating to help me pick which people to reply to. If this experiment works, I'll start going more by group opinions on posts. I'd like to try and see if more of the crap posters can be weeded out in >this way, as they are down-rated to a level that I know I can completely >ignore them from now on. And no, down rating my posts will not matter to me. I personally rate them highly. This is just a way for those who wish to give me feedback as to which people >they think I should answer. ># posted by James Harris @ 22:24 >JSH: Ever talk to a mathematician? >I wonder how many of you actually have sat down and had any kind of >conversation with a mathematician at a university. I have had a few such conversations, like maybe half a dozen or so over the >years. If you have not, then to you mathematicians may just be something out of a >novel, or a movie, but I assure you they are just people. And they are mostly safe from consequences in a very protected world. This story would not have played out like it has if they were not so safe. My research is so obviously important that the supporting evidence is >overwhelming, but we all here know the rules, and the rules say that what >the mathematicians at universities say is important about mathematics is >what the world says is important. So they know they can sit quietly, and they know the impact of sitting >quietly. I've TALKED to some of these people in person. They are protected in a way >that most of you aren't. Remember the story with Wiles? How he worked for over seven years with no >one knowing exactly what he was doing? He spent a lot of that time at home >with his family. Can you do the same? Can you comprehend that world? No one knows what you're really doing, but you have a good salary, respect >and admiration, to go off and do what you want for over seven years. To them this whole thing may be kind of a puzzle, and it might not feel real >to them that by leaving me out here arguing with fringe people they are >doing a bad thing. In their world, you protect YOUR research. Academics work to further their >OWN CAREERS. Tellingly a leading math professor at my own alma mater Vanderbilt >University told me when I sat down to explain my work on factoring >polynomials into non-polynomial factors-worked it all out on the chalkboard >in a discussion that covered all the major issues over a couple of >hours-that I lacked polish. Well I have a B.Sc. in physics so I'm pondering why in the hell polish >matters when the result is so dramatic, and you know, sitting here now I >think that professor was just doing things by the rules of his society. My >polish means so much in an academic world where polish is part of the rules, >like the social rules that govern human behavior in many other areas. But here and now with my research those rules are shown to be out-dated, but >that society is safe. Those mathematicians do not have to acknowledge my >research no matter how important it is easily shown to be because world >society does not make them, and by the rules they know, there are no >consequences. Ever talk to a mathematician? Doing so might open your eyes to how their >world works, so that you understand that this drama is not about arguments >on Usenet, as Usenet has no real impact on their world. It's about the society of mathematicians in universities around the world, >and the rules they play by. ># posted by James Harris @ 16:50 >Attacking reformulations, prime contradictions >One thing that is clear at this point is the position by several posters >arguing with me that my prime counting function is a useless reformulation >of information already known about prime counting. There is much to support that view: With natural numbers x and n, where p_i is the i_th prime: P(x,n) = x - 1 - sum for i=1 to n of {(P(x/p_i,i-1) - (i-1))} where if n is greater than the count of primes up to and including sqrt(x) >then n is reset to that count. That prime counting function in its sieve form clearly has elements that can >be found in prior prime counting research, and, guess what? It gives the same answer as the count of primes is the same. So what good is a reformulation? Well, posters in going to so much trouble to proclaim my research old have >repeatedly pointed out links between my prime counting function and >previously known algorithms for counting primes and functions like the sieve >function phi(x,a) used in those algorithms. Um, but that sounds like they're saying that with one function, I can do >everything that mathematicians previously did with multiple functions like >using pi(x) and phi(x,a), so where before you had two or more, with my >ideas, you have one function, which posters repeatedly point out can be So this reformulation captures everything you need and you can work from it >alone to do prime counting, going over old ground, so saying the >reformulation is worthless can have some merit, right? After all, prime counting isn't advanced by it, as I've finally acknowledged >though early on I had high hopes that you could find much faster prime >counting algorithms with it, so the reformulation despite its scope and size >is a waste of time? Well, hold on a minute. Sure, my prime counting function can be used to do >everything known before, so you can say that it is going over old ground, so >that if history had been different there might never have been a phi(x,a) >function or even a one variable pi(x) function as you can do everything with >my multi-variable function. But history shows humanity didn't go that route, >so that's it, right? BUT, the function I give-look back if you need to refresh >yourself-recursively calls itself and directly counts primes, so you can use >something simple, where I'll go to the classical prime counting function to >show it: pi(x) - pi(x-1) with a natural number x greater than 3 is only non-zero if x is prime. Because of that my prime counting function can go from being a sieve >function, so that instead of having P(x,n) where n is a count of primes, you >can use P(x,y) where y is just a natural number like x, as the function can >call itself to sieve out the primes on its own, without human aid. Hmmm.that sounds like more than just a reformulation now. And besides, remember reformulations aren't necessarily all bad. How about >Laplace Transforms? Or Hamiltonians? Why just hate reformulations? Posters here clearly do as they repeatedly >attack my research. Yes I've made grand claims at times and had to back-track, but I acknowledge >being excited about my own ideas and discoveries and hoping that they are >grander than they may be, and I can stand corrected. But even the reformulation argument begins to fall apart when the P(x,y) >function arrives, able to do what no other prime counting function has >ever been known to do in mathematical history. And posters attack that noting that algorithms counting primes are now slow. Um, is all prime counting just about fast prime counting, as excuse me, but >isn't there something called the Riemann Hypothesis which gives methods that >are even slower? If speed actually counting primes is all that matters, why in the hell does >anyone care about the Riemann Hypothesis? Of course, it's not just about speed counting primes, and here the >objections of posters go into hysteria and denial as they bounce all over >the map, to ignore a unique feature of my research, but hey, this is Usenet. Usenet is known for having people who are at the extremes of human behavior, >who can say just about anything, and it's not like the math world is ruled >by Usenet. Nope. My research is not blocked by strange people objecting in weird ways >on Usenet and calling me nasty names. It is blocked by mathematicians at universities allowing that to go on by >refusing to acknowledge my research, no matter how much I try to get them to >pay attention. So no, it's not about the posters calling me nasty names, and not even Erik >Max Francis calling me a crackpot on his website, but aboutprofessors at >universities, sitting quietly. And that is why I say that the academic world today is a dinosaur with >medieval crap like tenure giving them too much leeway, so that they can sit >without fear of reprisals if the truth is known. Do you think ANY math professor on this planet is brave enough to quietly >sit by while my research is wrongly reviled on Usenet and the web if they >thought it could impact their own careers? If they thought they would be tossed out of their universities and forced to >get a real job in the outside world, where results actually matter? I doubt it. They don't strike me as being brave people. But they are smart enough to know that today's academic world makes them >almost bullet-proof when it comes to consequences for ignoring my research >and leaving it to fringe people on Usenet to wrongly go after it, while they >sit quietly, knowing they are safe from consequences, while also knowing >that they are the key to having important research like my prime counting >function properly acknowledged. If they sit quietly long enough, they can hope that it will all just go away >and the knowledge will be lost, with their academic careers safely >protected. ># posted by James Harris @ 16:37 > === Subject: Re: JSH: You people are stupid Yeah you're cute (NOT) but not nearly as amart a JSH! ^ ^ ^ >(this JSH moment is brought to you by MIT Math Department, Make Sure you >are >dealing with a Professional Mathematician, else this can happen;) >JSH: You people are stupid >Ok, let's just put it out there, you people as a group are stupid. >There is no way there can be nothing to my research when there is so much >math behind it that says it has to be important. >No way. >It is complete bull that rules your world, which is how you people >keep >posturing despite your rank stupidity. >Only stupidity could explain the failure to even look into integrating >that >partial differential equation. >Only stupidity could explain the ability of posters to keep going on and >on >despite there being no other prime counting function that can even >recursively call itself. >And that's just my research related to the prime count. >Only people stupid on a massive scale could ignore publication in a peer >reviewed math journal, for an argument so damn trivial I can go over it in >a >couple of paragraphs using basic algebra. >You people are just damn stupid. Your community is stupid. >You are willfully stupid, stupid, stupid. >[A reply to someone who said that James is a ranting tit.] >At least I am right and have the mathematics to prove it. >I even got published. >You damn fools just said, duh, publication doesn't matter! >You are stupid. >I have the mathematics to prove my case. >I've had it for years. >Your society is just too damn stupid to go with the math. >You're just too damn stupid and full of yourselves. >Idiots. ># posted by James Harris @ 21:23 >JSH: Four years plus of failure >It has been over four years since I found what I call my prime counting >function. >It is still an open question how closely the partial differential equation >that follows it maps to the prime distribution because the current math >world is broken, and rather than consider the answers to that question, >most >mathematicians have ignored me, while you can see the posters here who >just >work to dismiss the research. >The lack of answers is the fruit of their labors. >That's what such people are about, no answers, no solutions, ever. >Has it occurred to any of you that you could die, and a day later, some >person or persons could finally look into that partial differential >equation >I found, and answer huge questions in mathematics, and the world go on and >on about this amazing thing that research thought quack from some crackpot >turned out to be important, but no one knew, because no one checked. >And mostly in looking over the history, they could see years of posturing >from people who never did anything of value. >ANYONE can criticize. >You have four years plus to see the value of posters just calling me names >and claiming my research is not new, as now you may know that yes, I did >find a partial differential equation that follows from a prime counting >function. >And it's an open question as to how closely integrating that partial >differential equation gets you to the prime distribution. >STILL an open question, over four years later. >Still there are posters who seem to think this is just about posturing on >some newsgroups as if no one cares if the question is EVER answered. >Ever. >What kind of human beings does this make you to be, when you are so >incurious, so incapable of wondering at the miniscule level necessary to >get >some answers? >Instead you sit by and either promote or are in complicit in just blanket >do-nothingness. >A world of no answers, just talk. >Four plus years of total failure from your group and your community. >That's what the world has today. >Four plus years of total failure to find any kind of answer. >[A reply to someone who asked why is it that this is not a failure from >James.] >You ing idiot I TESTED IT AND NUMERICALLY INTEGRATED IT AND FOUND IT >CLOSE! >You are the goddamn fools who refuse to check and then keep going on as if >you are intelligent when you will NOT EVEN ING CHECK!!! ># posted by James Harris @ 21:13 >JSH: Minority view >A few days ago I asked that readers who use Google Groups use the ratings >system that is provided through it where you can rate posts from 1 to 5 >stars. >I have looked over threads that I created to talk about my research and >found that along with my ratings, as I've started now routinely rating >posts, I usually see 3 or 4 others and at times maybe 10 who have bothered >to vote. >Giving the benefit of the doubt to opposition, let's say there are about >10 >people out there then who disagree with me and are vocal in that >disagreement. >Well I just checked Google Groups to see how many people it says subscribe >to sci.math and it said 5,506, which is the number through Google Groups, >so >it's a lower number than the readership. >For instance, I don't subscribe to sci.math so I'm not part of that count. >But, of course, I post on sci.math a lot. >But let's go with the low number to be fair and now go with the high >number >for the opposition based on ratings to get 10/5506 or about 0.18% of the >sci.math readership clearly disagrees with me. >So I have one point of view and about 10 or so other people who don't seem >to have any other support based on the ratings, argue with me, and that >generates a lot of posts. >I like the ratings system as gives an overview of how many people on >sci.math care to vote in one direction or another on these topics where if >you just look at the volume of postings you might wrongly assume that the >majority of sci.math is wrapped up in these discussions, when from the >ratings, most don't care. >Actually, only about 0.18% clearly show that they do care. >I suggest you use the ratings if you use Google Groups. There is just no >denying that much of the newsgroup doesn't care one way or the other when >there is no voting for your point of view. ># posted by James Harris @ 16:18 >Friday, November 24, 2006 >JSH: Under review >verified receipt. >A version of the paper is at >and in keeping with the philosophy of extreme mathematics, I would >appreciate comments. >The paper is under review as Princeton verified receipt. Uh, yeah that >surprised even me. >This thread is for comments on the paper. >And yes, this paper could end it all. If Princeton does what it should >then >it doesn't matter what any of you say, as you know and I know that >Princeton >trumps every last one of you. >I especially welcome comments from posters who argue with me regularly. >I am curious though. Read the paper, and come back and post as confidently >as you've done. >What gives you that faith? >blame the sinister cadre of sci.math again.] >No. I know and you know that Princeton doesn't give a damn what you think. >There is only one case where sci.math definitely interfered and killed a >There is no way that Princeton would fall in the same way. >You people can't kill the Annals of Mathematics. >I dare you to try. Send your emails. Just try. >[A reply to someone who explained to James how uninteresting and dishonest >his paper is.] >The replies in this thread reek of fear. >What I actually did was go from a prime counting sieve function, to a >constrained summation of a partial difference equation that counts primes, >to the partial differential equation that follows from it. >The three forms connect the dots from the discrete count of primes to a >continuous function directly, for the first time in mathematical history, >showing how the count of primes connects to continuous functions, and >accomplishing what notables like Gauss and Riemann set out to do so long >ago. >That is done in just two pages, and is an accomplishment at the pinnacle >of >human achievement in mathematics, but it's also an accomplishment members >of >the sci.math newsgroup have enviously and jealously fought to disparage. >It is one of the greatest accomplishments in the history of human thought. >It is more than a great enough accomplishment to merit publication in one >of >the world's top math journals. ># posted by James Harris @ 01:16 >Sabotage! >A member of the sci.math newsgroup managed to get past my screening and >join >the group, revealing himself when he systematically down rated all of my >posts. >I have banned that person from the group. >It shows you how these people operated. >Looks like people not members of the group can rate posts so I was wrong >on >that point. >As shown by a group of sci.math'ers who came in to downrate my posts. >Still, it shows how these people operate. >There was other evidence the banned person was a hostile sci.math'er so >I'm >comfortable with the decision to ban that person. >I do wish Google would allow you to only let group members rate posts, as >I >think their Beta still needs a bit more options for group owners, as this >current ongoing sabotage by members of the sci.math newsgroup shows. >The juvenile behavior from members of the Usenet sci.math group reveals a >clear weakness in Google's current setup. ># posted by James Harris @ 00:45 >Wednesday, November 22, 2006 >JSH: Use ratings >People coming in with Google Groups can rate posts, and I'd like more >people >to start rating to help me pick which people to reply to. >If this experiment works, I'll start going more by group opinions on >posts. >I'd like to try and see if more of the crap posters can be weeded out in >this way, as they are down-rated to a level that I know I can completely >ignore them from now on. >And no, down rating my posts will not matter to me. >I personally rate them highly. >This is just a way for those who wish to give me feedback as to which >people >they think I should answer. ># posted by James Harris @ 22:24 >JSH: Ever talk to a mathematician? >I wonder how many of you actually have sat down and had any kind of >conversation with a mathematician at a university. >I have had a few such conversations, like maybe half a dozen or so over >the >years. >If you have not, then to you mathematicians may just be something out of a >novel, or a movie, but I assure you they are just people. >And they are mostly safe from consequences in a very protected world. >This story would not have played out like it has if they were not so safe. >My research is so obviously important that the supporting evidence is >overwhelming, but we all here know the rules, and the rules say that what >the mathematicians at universities say is important about mathematics is >what the world says is important. >So they know they can sit quietly, and they know the impact of sitting >quietly. >I've TALKED to some of these people in person. They are protected in a way >that most of you aren't. >Remember the story with Wiles? How he worked for over seven years with no >one knowing exactly what he was doing? He spent a lot of that time at home >with his family. >Can you do the same? Can you comprehend that world? >No one knows what you're really doing, but you have a good salary, respect >and admiration, to go off and do what you want for over seven years. >To them this whole thing may be kind of a puzzle, and it might not feel >real >to them that by leaving me out here arguing with fringe people they are >doing a bad thing. >In their world, you protect YOUR research. Academics work to further their >OWN CAREERS. >Tellingly a leading math professor at my own alma mater Vanderbilt >University told me when I sat down to explain my work on factoring >polynomials into non-polynomial factors-worked it all out on the >chalkboard >in a discussion that covered all the major issues over a couple of >hours-that I lacked polish. >Well I have a B.Sc. in physics so I'm pondering why in the hell polish >matters when the result is so dramatic, and you know, sitting here now I >think that professor was just doing things by the rules of his society. My >polish means so much in an academic world where polish is part of the >rules, >like the social rules that govern human behavior in many other areas. >But here and now with my research those rules are shown to be out-dated, >but >that society is safe. Those mathematicians do not have to acknowledge my >research no matter how important it is easily shown to be because world >society does not make them, and by the rules they know, there are no >consequences. >Ever talk to a mathematician? Doing so might open your eyes to how their >world works, so that you understand that this drama is not about arguments >on Usenet, as Usenet has no real impact on their world. >It's about the society of mathematicians in universities around the world, >and the rules they play by. ># posted by James Harris @ 16:50 >Attacking reformulations, prime contradictions >One thing that is clear at this point is the position by several posters >arguing with me that my prime counting function is a useless reformulation >of information already known about prime counting. >There is much to support that view: >With natural numbers x and n, where p_i is the i_th prime: >P(x,n) = x - 1 - sum for i=1 to n of {(P(x/p_i,i-1) - (i-1))} >where if n is greater than the count of primes up to and including sqrt(x) >then n is reset to that count. >That prime counting function in its sieve form clearly has elements that >can >be found in prior prime counting research, and, guess what? >It gives the same answer as the count of primes is the same. >So what good is a reformulation? >Well, posters in going to so much trouble to proclaim my research old have >repeatedly pointed out links between my prime counting function and >previously known algorithms for counting primes and functions like the >sieve >function phi(x,a) used in those algorithms. >Um, but that sounds like they're saying that with one function, I can do >everything that mathematicians previously did with multiple functions like >using pi(x) and phi(x,a), so where before you had two or more, with my >ideas, you have one function, which posters repeatedly point out can be >So this reformulation captures everything you need and you can work from >it >alone to do prime counting, going over old ground, so saying the >reformulation is worthless can have some merit, right? >After all, prime counting isn't advanced by it, as I've finally >acknowledged >though early on I had high hopes that you could find much faster prime >counting algorithms with it, so the reformulation despite its scope and >size >is a waste of time? >Well, hold on a minute. Sure, my prime counting function can be used to do >everything known before, so you can say that it is going over old ground, >so >that if history had been different there might never have been a phi(x,a) >function or even a one variable pi(x) function as you can do everything >with >my multi-variable function. But history shows humanity didn't go that >route, >so that's it, right? >BUT, the function I give-look back if you need to refresh >yourself-recursively calls itself and directly counts primes, so you can >use >something simple, where I'll go to the classical prime counting function >to >show it: >pi(x) - pi(x-1) >with a natural number x greater than 3 is only non-zero if x is prime. >Because of that my prime counting function can go from being a sieve >function, so that instead of having P(x,n) where n is a count of primes, >you >can use P(x,y) where y is just a natural number like x, as the function >can >call itself to sieve out the primes on its own, without human aid. >Hmmm.that sounds like more than just a reformulation now. >And besides, remember reformulations aren't necessarily all bad. How about >Laplace Transforms? Or Hamiltonians? >Why just hate reformulations? Posters here clearly do as they repeatedly >attack my research. >Yes I've made grand claims at times and had to back-track, but I >acknowledge >being excited about my own ideas and discoveries and hoping that they are >grander than they may be, and I can stand corrected. >But even the reformulation argument begins to fall apart when the P(x,y) >function arrives, able to do what no other prime counting function has >ever been known to do in mathematical history. >And posters attack that noting that algorithms counting primes are now >slow. >Um, is all prime counting just about fast prime counting, as excuse me, >but >isn't there something called the Riemann Hypothesis which gives methods >that >are even slower? >If speed actually counting primes is all that matters, why in the hell >does >anyone care about the Riemann Hypothesis? >Of course, it's not just about speed counting primes, and here the >objections of posters go into hysteria and denial as they bounce all over >the map, to ignore a unique feature of my research, but hey, this is >Usenet. >Usenet is known for having people who are at the extremes of human >behavior, >who can say just about anything, and it's not like the math world is ruled >by Usenet. >Nope. My research is not blocked by strange people objecting in weird ways >on Usenet and calling me nasty names. >It is blocked by mathematicians at universities allowing that to go on by >refusing to acknowledge my research, no matter how much I try to get them >to >pay attention. >So no, it's not about the posters calling me nasty names, and not even >Erik >Max Francis calling me a crackpot on his website, but aboutprofessors at >universities, sitting quietly. >And that is why I say that the academic world today is a dinosaur with >medieval crap like tenure giving them too much leeway, so that they can >sit >without fear of reprisals if the truth is known. >Do you think ANY math professor on this planet is brave enough to quietly >sit by while my research is wrongly reviled on Usenet and the web if they >thought it could impact their own careers? >If they thought they would be tossed out of their universities and forced >to >get a real job in the outside world, where results actually matter? >I doubt it. They don't strike me as being brave people. >But they are smart enough to know that today's academic world makes them >almost bullet-proof when it comes to consequences for ignoring my research >and leaving it to fringe people on Usenet to wrongly go after it, while >they >sit quietly, knowing they are safe from consequences, while also knowing >that they are the key to having important research like my prime counting >function properly acknowledged. >If they sit quietly long enough, they can hope that it will all just go >away >and the knowledge will be lost, with their academic careers safely >protected. ># posted by James Harris @ 16:37 > === Subject: Re: puzzles posting-account=G_G-iQoAAAB08LNQidt_LsMkopmIb4ZS Gecko/20060111 Firefox/1.5.0.1 Mnenhy/0.7.3.0,gzip(gfe),gzip(gfe) I was going through a math puzzle book, the following are some of the > questions in that book 1) write 1000 using the the no: 9 four times only. > condition +, - cannot be used the solution is given as 9 > 99 --- > 9 > is there any other solution ? This is only 100, not 1000. 2) 19-9-1991 is a palindromic date , is their any other palindromic > dates ? 11-11-1111, 11-1-111; 1-11-111 10-12-2101 to name a few. 3) give two two digit numbers of the form ab & cd such that ab x cd > is always equal to ba x dc ex:- 21 x 24 == 12 x 42 This works whenever a = d and b = c; ie 62 and 26, which become 26 and 62 when the digits are reversed. 4) how valid is the result that difference between a two digit number > and the number formed by reversing the its digits is a multiple of > 9 ? ex:- 73 - 37 = 36 can this result be generalized to n digit numbers? The result is valid for all n digit integers. Look up casting out nines on the internet Bill J === Subject: Re: puzzles > On May 7, 10:50 am, sophia I was going through a math puzzle book, the > following are some of the > questions in that book 1) write 1000 using the the no: 9 four times only. > condition +, - cannot be used the solution is given as 9 > 99 --- > 9 > is there any other solution ? This is only 100, not 1000. 2) 19-9-1991 is a palindromic date , is their any > other palindromic > dates ? > 11-11-1111, 11-1-111; 1-11-111 > 10-12-2101 to name a few. 3) give two two digit numbers of the form ab & cd > such that ab x cd > is always equal to ba x dc ex:- 21 x 24 == 12 x 42 This works whenever a = d and b = c; ie > 62 and 26, which become 26 and 62 > when the digits are reversed. Actually, it works if and only if [a b] [d c] is a singular matrix. To be a bit less obnoxious, it works if and only if ac = bd. 4) how valid is the result that difference between > a two digit number > and the number formed by reversing the its digits > is a multiple of > 9 ? ex:- 73 - 37 = 36 can this result be generalized to n digit > numbers? The result is valid for all n digit integers. Look up casting out nines on the internet Bill J In response to (3) === Subject: Re: puzzles posting-account=OKTeIQkAAAAZk6JK1hK7-grwpoUDNy98 CLR 1.1.4322; .NET CLR 2.0.50727; InfoPath.1),gzip(gfe),gzip(gfe) I was going through a math puzzle book, the following are some of æthe > questions in that book 1) write 1000 using the the no: 9 four times only. > æ æcondition +, - cannot be used æthe solution is given as æ æ æ æ æ 9 > æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ99 æ--- > æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ9 > æ æis there any other solution ? 2) 19-9-1991 is a palindromic date , is their any other palindromic > dates ? 3) give two two digit numbers of the form ab æ& cd such that ab x cd > is always æequal to æba x dc æ æ ex:- 21 x 24 == 12 x 42 4) how valid is the result that difference between a two digit number > and the ænumber formed by reversing the its digits is a multiple of > 9 ? æ ex:- 73 - 37 = 36 What's the residue when you cast out nines from 73? Does the residue change if you change the order of the digits? So the residue of the answer must always be 0 when you cast out nines. Thus, the answer is always divisible by nine. æ can this result be generalized to n digit numbers? Is casting out nines dependent on the number of digits? === Subject: Re: puzzles posting-account=OKTeIQkAAAAZk6JK1hK7-grwpoUDNy98 CLR 1.1.4322; .NET CLR 2.0.50727; InfoPath.1),gzip(gfe),gzip(gfe) > I was going through a math puzzle book, the following are some of æthe > questions in that book 1) write 1000 using the the no: 9 four times only. > æ æcondition +, - cannot be used æthe solution is given as æ æ æ æ æ 9 > æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ99 æ--- > æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ9 > æ æis there any other solution ? 2) 19-9-1991 is a palindromic date , is their any other palindromic > dates ? 3) give two two digit numbers of the form ab æ& cd such that ab x cd > is always æequal to æba x dc æ æ ex:- 21 x 24 == 12 x 42 4) how valid is the result that difference between a two digit number > and the ænumber formed by reversing the its digits is a multiple of > 9 ? æ ex:- 73 - 37 = 36 What's the residue when you cast out nines from 73? Does the residue change if you change the order of the digits? So the residue of the answer must always be 0 when you cast out nines. Thus, the answer is always divisible by nine. æ can this result be generalized to n digit numbers? Is casting out nines dependent on the number of digits? Works in other number systems also. See http://members.aol.com/rotanasnem/cherries/cherries.htm If we take CHERRIES BAR BELL minus BELL BAR CHERRIES We can't subtract CHERRIES from BELL, so we borrow a LEMON from the next digit (BAR, leaving LEMON) and LEMON BELL minus CHERRIES is CHERRIES L L BL CHERRIES BAR BELL - BELL BAR CHERRIES --------------------- CHERRIES Can't subtract a BAR from a LEMON, so we borrow a LEMON from the CHERRIES (leaving BELL) and LEMON LEMON minus BAR is CHERRIES BL L L CHERRIES BAR BELL - BELL BAR CHERRIES --------------------- CHERRIES CHERRIES And finally, BELL minus BELL is ORANGE BL CHERRIES BAR BELL - BELL BAR CHERRIES --------------------- ORANGE CHERRIES CHERRIES Now, when we cast out CHERRIES from the operands, we get LEMON for each, of course, and LEMON minus LEMON is ORANGE, as is the result when we cast out CHERRIES. Thus, CHERRIES BAR BELL minus BELL BAR CHERRIES is divisible by CHERRIES (it's easy to see that CHERRIES CHERRIES divided by CHERRIES is LEMON LEMON). === Subject: Re: puzzles posting-account=OKTeIQkAAAAZk6JK1hK7-grwpoUDNy98 CLR 1.1.4322; .NET CLR 2.0.50727; InfoPath.1),gzip(gfe),gzip(gfe) > ?the solution is given as ? ? ? ? ? 9 > ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?99 ?--- > ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?9 > ? ?is there any other solution ? ? it should be read as 99 and 9 by 9 What? I don't understand what you are saying. How does 99 and 9 by 9 = 1000? ?it is just like this for 2.5 we will say 2 and 1 by 2 to express in > fractional terms I don't think that's what the problem is. 99 + 9/9 equals 100, not 1000. === Subject: Re: puzzles posting-account=9mFhxAoAAADvYLsbTTuTch-3O-0ByiBY CLR 2.0.50727),gzip(gfe),gzip(gfe) ?the solution is given as ? ? ? ? ? 9 > ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?99 ?--- > ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?9 > ? ?is there any other solution ? ? it should be read as 99 and 9 by 9 What? I don't understand what you are saying. How does 99 and 9 by 9 = 1000? ?it is just like this for 2.5 we will say 2 and 1 by 2 to express in > fractional terms I don't think that's what the problem is. 99 + 9/9 equals 100, not > 1000.- sorry the question should be rather write 100 using the the no: 9 four times only. condition +, - cannot be used === Subject: Re: puzzles posting-account=Rkt6TwoAAACG_SqlrxmgPCl1Ozr0PWSD MathPlayer 2.10b; .NET CLR 2.0.50727; .NET CLR 3.0.04506.30; .NET CLR 1.1.4322; .NET CLR 3.0.04506.648),gzip(gfe),gzip(gfe) > æthe solution is given as æ æ æ æ æ 9 > æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ99 æ--- > æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ9 > æ æis there any other solution ? æ it should be read as 99 and 9 by 9 What? I don't understand what you are saying. How does 99 and 9 by 9 = 1000? æit is just like this for 2.5 we will say 2 and 1 by 2 to express in > fractional terms I don't think that's what the problem is. 99 + 9/9 equals 100, not > 1000.- sorry the question should be rather write 100 using the the no: 9 four times only. > æ æcondition +, - cannot be used Implicitly a + is used. But I suppose you could get away with it by just omitting the + and writing the answer as 9 9/9. === Subject: Re: Maple 12 is out Nostalgia isn't what it used to be. > As promised, we are going to finish the beta 0.2 soon. Now, for starters, Man+Machine Review Of Maple Crisis: ................................................................ The beta 0.1 of the first world's document written by > a human in a close cooperation with a successor of the > GEMM machine, our unique VM automated testing expert > system which is a failure prediction oracle http://maple.bug-list.org/maple-crisis.php Main Maple's Quality Results 1. Maple 9.5 is an unstable, inconsistent, non-linear, > non-uniform, randomized, self-incompatible environment > where fundamental math properties (uniqueness of the > answer for a good-defined problem commutativity and > linearity property etc) now hold, now fail making Maple > breaking down grotesquely. [ ... ] ................................................................ > Best wishes, Vladimir Bondarenko VM and GEMM architect > Co-founder, CEO, Mathematical Director http://www.cybertester.com/ Cyber Tester, LLC > http://maple.bug-list.org/ Maple Bugs Encyclopaedia > http://www.CAS-testing.org/ CAS Testing ----------------------------------------------------------- We must understand that technologies > like these are the way of the future. ----------------------------------------------------------- === Subject: Re: Not yet !: Elementary FLT Approach I can see, that it takes at least one more > Year for You to be a little more familiar > with this unique possibility for to > make proof of FLT much much more shorter. > I used to check so many other methods > and there common fault in the very > beginning was to afirmate Z-Y or Z-X > or X+Y as some perfect powers... See again anyhow... > 1.edit: on 2 May 2008: > I have already seen existing problem > for possible z;y or z;x negative values: > Appropriate z;y or z;x values where taken > generally as integers and some properties > for x;y as natural values creating the correct > proof can not be taken for z;y<0 or z;x<0 . > I am sorry: next step for to correct this > proof for negative values of z;y is still > not so certain. > Lets see... It looks, that this very tiny chance is again with me: Finally I succeed to determine: Z-Y < 2(m-n)^2 for X taken as odd number and once should be X = X1*X2 so for X1 = (m-n) ; X2 = (m+n) also for X = m^2 - n^2 and the other values approximated with the help of this simple set: (completation to all possible Pythagorean triplets when beginning with: X = m1^2 -n1^2 =...= mi^2 -ni^2 =...= ms^2 -ns^2 and so on from all possible factors of X will be taken appropriate m;n values and m-n; and m+n; appropriate factors reflected with so called Abel formulae's: (X1)^p = Z-Y (X2)^p = (Z^p -Y^p)/(Z-Y) for X not divided by p or for p|X: [(X1)^p]/p = Z-Y p(X2)^p = (Z^p -Y^p)/(Z-Y) once for some set of natural X;Y;Z values could be true: X^p +Y^p = Z^p for p>=3 and for bigger primes ) Also for X = m^2 - n^2 let Y = 2mn + y and Z = m^2 + n^2 + z where y;z appropriate integers: Now for z;y > 0 it was easy to proof: z Message was edited by: Roman B. Binder === Subject: Re: OPEN LETTER TO QUASI >Your continued silence implies that you accept >A wins 4 games and B wins 3 games as >legitimate data. Do you still maintain that my >decision to allocate the stake based on >games won is illogical? Sorry to butt in - I don't even know which thread this refers to! - but quasi's continued silence probably signifies only his continued absence, and nothing about any conversation in particular. (His most recent post to sci.math appears to have been on 29 April.) -- Angus Rodgers Contains mild peril === Subject: Re: OPEN LETTER TO QUASI > >Your continued silence implies that you accept >A wins 4 games and B wins 3 games as >legitimate data. Do you still maintain that my >decision to allocate the stake based on >games won is illogical? Sorry to butt in - I don't even know which thread this refers to! - > but quasi's continued silence probably signifies only his continued > absence, and nothing about any conversation in particular. (His most > recent post to sci.math appears to have been on 29 April.) We want quasi! We want quasi! -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: runge kutta matrix problem posting-account=O5emkAoAAABg3Nf29iHLQWuoVQZBNQYb Gecko/20080404 Firefox/2.0.0.14,gzip(gfe),gzip(gfe) > Hi > I have to resolve problem like this: > Ax + B*(dx/dt) = y > x + C*(dy/dt) = D where x(0) = 0 ; > y(0) = 0; I 'd like to use Runge_Kutta or other numerical method > Regard > kamil thx everyone for help. Is working fine now Regard kamil === Subject: Pls Help Stupid Programmer! Number Sequence When Added Always Equal A Unique Number posting-account=TaQOcwoAAAD1iwGL2vvHtsOdZxCo0W4s CLR 2.0.50727; InfoPath.1; .NET CLR 1.1.4322; .NET CLR 3.0.04506.30; .NET CLR 3.0.04506.648; MS-RTC LM 8),gzip(gfe),gzip(gfe) Please help this stupid programmer keep his job! Need A Number Sequence When Added Will Always Equal A Unique Number. The Trouble is the Brand Associations Column Below - I need a Number Sequence When Added Will Always Equal A Unique Number. BRANDS TABLE: BrandID# | Brand ------------------- 1 | BrandA 3 | BrandB 5 | BrandC PRODUCTS TABLE: ProdID: | Product | Brand Associations: ----------------------------------------- A | ProdA | 1 <- Supports BrandA Only B | ProdB | 4 <- Supports Brands 1 & 3 (BrandA & BrandB) C | ProdC | 8 <- Supports Brands 3 & 5 (BrandB & BrandC) Respectfully, Bill wrhart11@netzero.com === Subject: Re: Pls Help Stupid Programmer! Number Sequence When Added Always Equal A Unique Number posting-account=Rkt6TwoAAACG_SqlrxmgPCl1Ozr0PWSD MathPlayer 2.10b; .NET CLR 2.0.50727; .NET CLR 3.0.04506.30; .NET CLR 1.1.4322; .NET CLR 3.0.04506.648),gzip(gfe),gzip(gfe) > Please help this stupid programmer keep his job! > Need A Number Sequence When Added Will Always Equal A Unique Number. The Trouble is the Brand Associations Column Below - I need a Number > Sequence When Added Will Always Equal A Unique Number. BRANDS TABLE: > BrandID# | Brand > ------------------- > 1 æ æ æ æ| BrandA > 3 æ æ æ æ| BrandB > 5 æ æ æ æ| BrandC PRODUCTS TABLE: > ProdID: | Product | Brand Associations: > ----------------------------------------- > A æ æ æ | ProdA æ | æ1 æ<- Supports BrandA Only > B æ æ æ | ProdB æ | æ4 æ<- Supports Brands 1 & 3 (BrandA & BrandB) > C æ æ æ | ProdC æ | æ8 æ<- Supports Brands 3 & 5 (BrandB & BrandC) > Respectfully, > Bill > wrhar...@netzero.com Does this sum need to be reversible, i.e. to unravel the individual brands in your brand sum for example to check if a specific brand is in some given sum? It's hard to imagine a practical scenario where this wouldn't be required, either now or in the future. But if not then a hash function might do. Otherwise, I'd forget all that kiddie stuff about adding integers. You'll give yourself no end of grief in the long run! The only practical method is via a pivot table (Microsoft spreadsheet terminology - AKA cross table or join table) in a normalized database, just as Julio described. If you're using Excel, I'm sure VBA can manage pivot tables with no problem. Otherwise get hold of a copy of MySql or Postgres, both free, or perhaps (if you're using Windows) the freebie version of SQL Server (SQL Server Express?) John Ramsden === Subject: Re: Pls Help Stupid Programmer! Number Sequence When Added Always Equal A Unique Number > Please help this stupid programmer keep his job! > Need A Number Sequence When Added Will Always Equal A > Unique Number. My friend and estimated colleague, if this is a homework then my advice will simply be of no help. If this is for real, then please remember: I real programmer can swear but never sweat! The Trouble is the Brand Associations Column Below - > I need a Number > Sequence When Added Will Always Equal A Unique > Number. BRANDS TABLE: > BrandID# | Brand > ------------------- > 1 | BrandA > 3 | BrandB > 5 | BrandC PRODUCTS TABLE: > ProdID: | Product | Brand Associations: > ----------------------------------------- > A | ProdA | 1 <- Supports BrandA Only > B | ProdB | 4 <- Supports Brands 1 & 3 > (BrandA & BrandB) > C | ProdC | 8 <- Supports Brands 3 & 5 > (BrandB & BrandC) > Unless you have (unreasonable) constraints here, which you didn't state, that is *not* the way to go for a real programming task, because you are going to end up with a structure that is not manageable/extendable (not to talk about the nightmare at implementing your original idea, even if you found that rule). The canonical way to go looks like this (and it is the implementation of an n-to-n relationship): Brands { ID, Name } Products { ID, Name } Products_Brands { IDProduct, IDBrand } I hope I have not misinterpreted. If that is of no help, please try and state your *original* requirements and constraints, that is the problem, not the solution you have devised. (And please remember me when you are gonna get your next salary.) Joke apart, hope it helps and feel free to ask more. -LV Respectfully, > Bill > wrhart11@netzero.com === Subject: Re: Pls Help Stupid Programmer! Number Sequence When Added Always Equal A Unique Number > Please help this stupid programmer keep his job! > Need A Number Sequence When Added Will Always Equal A Unique Number. The Trouble is the Brand Associations Column Below - I need a Number > Sequence When Added Will Always Equal A Unique Number. BRANDS TABLE: > BrandID# | Brand > ------------------- > 1 | BrandA > 3 | BrandB > 5 | BrandC PRODUCTS TABLE: > ProdID: | Product | Brand Associations: > ----------------------------------------- > A | ProdA | 1 <- Supports BrandA Only > B | ProdB | 4 <- Supports Brands 1 & 3 (BrandA & BrandB) > C | ProdC | 8 <- Supports Brands 3 & 5 (BrandB & BrandC) > It might help if you stated clearly and precisely what you were looking for. If you want a sequence of numbers such that the sums of its subsets are all distinct, the obvious one is the powers of 2. Robert Israel israel@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Pls Help Stupid Programmer! Number Sequence When Added Always Equal A Unique Number posting-account=TaQOcwoAAAD1iwGL2vvHtsOdZxCo0W4s CLR 2.0.50727; InfoPath.1; .NET CLR 1.1.4322; .NET CLR 3.0.04506.30; .NET CLR 3.0.04506.648; MS-RTC LM 8),gzip(gfe),gzip(gfe) > Please help this stupid programmer keep his job! > Need A Number Sequence When Added Will Always Equal A Unique Number. The Trouble is the Brand Associations Column Below - I need a Number > Sequence When Added Will Always Equal A Unique Number. BRANDS TABLE: > BrandID# | Brand > ------------------- > 1 æ æ æ æ| BrandA > 3 æ æ æ æ| BrandB > 5 æ æ æ æ| BrandC PRODUCTS TABLE: > ProdID: | Product | Brand Associations: > ----------------------------------------- > A æ æ æ | ProdA æ | æ1 æ<- Supports BrandA Only > B æ æ æ | ProdB æ | æ4 æ<- Supports Brands 1 & 3 (BrandA & BrandB) > C æ æ æ | ProdC æ | æ8 æ<- Supports Brands 3 & 5 (BrandB & BrandC) > It might help if you stated clearly and precisely what you were looking > for. æIf you want a sequence of numbers such that the sums of its > subsets are all distinct, the obvious one is the powers of 2. Robert Israel æ æ æ æ æ æ æisr...@math.MyUniversitysInitials.ca > Department of Mathematics æ æ æ æhttp://www.math.ubc.ca/~israel > University of British Columbia æ æ æ æ æ æVancouver, BC, Canada- Hide quoted text - - Show quoted text - I was thinking of the power of 2, but I get stuck quickly when combining more than two brands. ---------------------------------------------------------------------- BrandID# | Brand ------------------- 1 | BrandA 3 | BrandB 5 | BrandC 7 | BrandD 9 | BrandE 11 | BrandF I have Products that when Combined with the BrandID number is a unique total number: For Example: ProductA Works with 1 & 3 (Total = 4) OK ProductB works with 1 & 5 (Total = 6) OK ProductC works with 3 & 9 (Total = 12) OK ProductD works with 1,3,5 (Total = 9) NOT OK - 9 is already a Single Unique Number Used (BrandID number 9 = BrandE) I am Totally open to another way of doing this. Be it an algorithm or algebraic equation ---------------------------------------------------------------------- Respectfully, Bill wrhart11@netzero.com === Subject: Re: Pls Help Stupid Programmer! Number Sequence When Added Always Equal A Unique Number ... > It might help if you stated clearly and precisely what you were looking > for. æIf you want a sequence of numbers such that the sums of its > subsets are all distinct, the obvious one is the powers of 2. > Robert Israel æ æ æ æ æ æ æisr...@math.MyUniversitysInitials.ca > Department of Mathematics æ æ æ æhttp://www.math.ubc.ca/~israel > University of British Columbia æ æ æ æ æ æVancouver, BC, Canada- Hide quoted text - > - Show quoted text - I was thinking of the power of 2, but I get stuck quickly when > combining more than two brands. ---------------------------------------------------------------------- > BrandID# | Brand > ------------------- > 1 | BrandA > 3 | BrandB > 5 | BrandC > 7 | BrandD > 9 | BrandE > 11 | BrandF Well, Robert was suggesting *powers* of two i.e. 1,2,4,8,... If you use C/C++ or something like that, you'll see library functions using bitfields which rely on this property. There's an introduction on Wikipedia [1]. As others have suggested, this isn't really sustainable if you've got lots of brands though - if it's a 32-bit machine, what happens when you go past 32 brands? The right solution definitely depends on what software/language you're using. But maybe a maths forum is no longer the place to ask about it? Rupert [1] http://en.wikipedia.org/wiki/Bit_field === Subject: Re: Pls Help Stupid Programmer! Number Sequence When Added Always Equal A Unique Number > Please help this stupid programmer keep his job! > Need A Number Sequence When Added Will Always Equal A Unique Number. The Trouble is the Brand Associations Column Below - I need a Number > Sequence When Added Will Always Equal A Unique Number. BRANDS TABLE: > BrandID# | Brand > ------------------- > 1 æ æ æ æ| BrandA > 3 æ æ æ æ| BrandB > 5 æ æ æ æ| BrandC PRODUCTS TABLE: > ProdID: | Product | Brand Associations: > ----------------------------------------- > A æ æ æ | ProdA æ | æ1 æ<- Supports BrandA Only > B æ æ æ | ProdB æ | æ4 æ<- Supports Brands 1 & 3 (BrandA & BrandB) > C æ æ æ | ProdC æ | æ8 æ<- Supports Brands 3 & 5 (BrandB & BrandC) > It might help if you stated clearly and precisely what you were looking > for. æIf you want a sequence of numbers such that the sums of its > subsets are all distinct, the obvious one is the powers of 2. Robert Israel æ æ æ æ æ æ æisr...@math.MyUniversitysInitials.ca > Department of Mathematics æ æ æ æhttp://www.math.ubc.ca/~israel > University of British Columbia æ æ æ æ æ æVancouver, BC, Canada- Hide > quoted text - - Show quoted text - I was thinking of the power of 2, but I get stuck quickly when > combining more than two brands. Perhaps you don't understand what Prof. Israel means by the powers of two. He is referring to the numbers 1, 2, 4, 8, 16, 32, 64, 128, etc. Use those numbers, and you won't get stuck. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Pls Help Stupid Programmer! Number Sequence When Added Always Equal A Unique Number > <97e10461-b2cc-4655-a265-66016a936efe@c58g2000hsc.goog > legroups.com>, On May 7, 5:47æpm, Robert Israel > Please help this stupid programmer keep his > job! > Need A Number Sequence When Added Will Always > Equal A Unique Number. [...] Perhaps you don't understand what Prof. Israel means > by the powers > of two. He is referring to the numbers 1, 2, 4, 8, > 16, 32, 64, 128, > etc. Use those numbers, and you won't get stuck. Just for the chronicle, Prof. Israel was right in being prudent and asking for clarifications, because that approach is not going to work for real, the main reason being that even with a 64 bit integral type (the biggest on common nowadays machines), you can address up to 64 distinct items. (There are also more subtle reasons related to code manageability.) Moreover, from a technical standpoint, while auto-increment keys come -so to say- for free in any modern relational database, implementing a custom key numbering risks to be a nightmare for a guy who has no idea about basic table relations, because it requires fancy stuff like triggers and similar. The power of 2 approach is fine when, for instance, you have a reasonably short list of constants, since it allows basic bit-wise operators rather then arithmetic ones for packing-unpacking single values to the list. -LV -- > Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for > email) === Subject: Re: Pls Help Stupid Programmer! Number Sequence When Added Always Equal A Unique Number <13075715.1210201947392.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=TaQOcwoAAAD1iwGL2vvHtsOdZxCo0W4s CLR 2.0.50727; InfoPath.1; .NET CLR 1.1.4322; .NET CLR 3.0.04506.30; .NET CLR 3.0.04506.648; MS-RTC LM 8),gzip(gfe),gzip(gfe) > <97e10461-b2cc-4655-a265-66016a936...@c58g2000hsc.goog > legroups.com>, On May 7, 5:47æpm, Robert Israel > Please help this stupid programmer keep his > job! > Need A Number Sequence When Added Will Always > Equal A Unique Number. [...] Perhaps you don't understand what Prof. Israel means > by the powers > of two. He is referring to the numbers 1, 2, 4, 8, > 16, 32, 64, 128, > etc. Use those numbers, and you won't get stuck. Just for the chronicle, Prof. Israel was right in being prudent and asking for clarifications, because that approach is not going to work for real, the main reason being that even with a 64 bit integral type (the biggest on common nowadays machines), you can address up to 64 distinct items. (There are also more subtle reasons related to code manageability.) Moreover, from a technical standpoint, while auto-increment keys come -so to say- for free in any modern relational database, implementing a custom key numbering risks to be a nightmare for a guy who has no idea about basic table relations, because it requires fancy stuff like triggers and similar. The power of 2 approach is fine when, for instance, you have a reasonably short list of constants, since it allows basic bit-wise operators rather then arithmetic ones for packing-unpacking single values to the list. -LV -- > Gerry Myerson (ge...@maths.mq.edi.ai) (i -> u for > email)- Hide quoted text - - Show quoted text -- Hide quoted text - - Show quoted text - believe the resolution, and rightly so, is the cross join tables as you have so graciously pointed out. (I guess I was trying to be too flashy on the original idea.) Again, thank you!!! Respectfully, Bill === Subject: Re: Pls Help Stupid Programmer! Number Sequence When Added Always Equal A Unique Number > experience. I > believe the resolution, and rightly so, is the cross > join tables as > you have so graciously pointed out. (I guess I was > trying to be too > flashy on the original idea.) > Again, thank you!!! > Respectfully, > Bill Technically speaking it is not a cross-join either, a cross-join being the full cartesian product of the two original tables: http://en.wikipedia.org/wiki/Join_%28SQL%29 What you have here is a (conceptual level) n-to-n relation implemented (technical level) through an intermediate table (and I don't know if it has got any specific name other than said n-to-n relation). In any case, I guess theory doesn't matter too much here... Instead, to the serious things! Where shall I send you my details for that salary thing then? ;) -LV === Subject: Re: Pls Help Stupid Programmer! Number Sequence When Added Always Equal A Unique Number posting-account=TaQOcwoAAAD1iwGL2vvHtsOdZxCo0W4s CLR 2.0.50727; InfoPath.1; .NET CLR 1.1.4322; .NET CLR 3.0.04506.30; .NET CLR 3.0.04506.648; MS-RTC LM 8),gzip(gfe),gzip(gfe) On May 7, 6:48æpm, Gerry Myerson Please help this stupid programmer keep his job! > Need A Number Sequence When Added Will Always Equal A Unique Number. The Trouble is the Brand Associations Column Below - I need a Number > Sequence When Added Will Always Equal A Unique Number. BRANDS TABLE: > BrandID# | Brand > ------------------- > 1 æ æ æ æ| BrandA > 3 æ æ æ æ| BrandB > 5 æ æ æ æ| BrandC PRODUCTS TABLE: > ProdID: | Product | Brand Associations: > ----------------------------------------- > A æ æ æ | ProdA æ | æ1 æ<- Supports BrandA Only > B æ æ æ | ProdB æ | æ4 æ<- Supports Brands 1 & 3 (BrandA & BrandB) > C æ æ æ | ProdC æ | æ8 æ<- Supports Brands 3 & 5 (BrandB & BrandC) > It might help if you stated clearly and precisely what you were looking > for. æIf you want a sequence of numbers such that the sums of its > subsets are all distinct, the obvious one is the powers of 2. Robert Israel æ æ æ æ æ æ æisr...@math.MyUniversitysInitials.ca > Department of Mathematics æ æ æ æhttp://www.math.ubc.ca/~israel > University of British Columbia æ æ æ æ æ æVancouver, BC, Canada- Hide > quoted text - - Show quoted text - I was thinking of the power of 2, but I get stuck quickly when > combining more than two brands. Perhaps you don't understand what Prof. Israel means by the powers > of two. He is referring to the numbers 1, 2, 4, 8, 16, 32, 64, 128, > etc. Use those numbers, and you won't get stuck. -- > Gerry Myerson (ge...@maths.mq.edi.ai) (i -> u for email)- Hide quoted text - - Show quoted text - clarify my conundrum: Store: StoreName: 1 WalMart 3 Target 5 Sears 7 Kmart 9 Dillards Products: Stores That Carry This Shirt A Store# 1 & 3 ( My Value here would be 4) B Only Store #9 (My Value here would be 9) C Store# 1 & 5 ( My Value here would be 6) D Store# 1 & 3 & 5 ( My Value here would be 9) *PROBLEM - #9 is Dillards - Not the combo of Stores 1, 3 & 5. === Subject: Re: Pls Help Stupid Programmer! Number Sequence When Added Always Equal A Unique Number posting-account=OKTeIQkAAAAZk6JK1hK7-grwpoUDNy98 CLR 1.1.4322; .NET CLR 2.0.50727; InfoPath.1),gzip(gfe),gzip(gfe) > On May 7, 6:48æpm, Gerry Myerson Need A Number Sequence When Added Will Always Equal A Unique Number. The Trouble is the Brand Associations Column Below - I need a Number > Sequence When Added Will Always Equal A Unique Number. BRANDS TABLE: > BrandID# | Brand > ------------------- > 1 æ æ æ æ| BrandA > 3 æ æ æ æ| BrandB > 5 æ æ æ æ| BrandC PRODUCTS TABLE: > ProdID: | Product | Brand Associations: > ----------------------------------------- > A æ æ æ | ProdA æ | æ1 æ<- Supports BrandA Only > B æ æ æ | ProdB æ | æ4 æ<- Supports Brands 1 & 3 (BrandA & BrandB) > C æ æ æ | ProdC æ | æ8 æ<- Supports Brands 3 & 5 (BrandB & BrandC) > It might help if you stated clearly and precisely what you were looking > for. æIf you want a sequence of numbers such that the sums of its > subsets are all distinct, the obvious one is the powers of 2. Robert Israel æ æ æ æ æ æ æisr...@math.MyUniversitysInitials.ca > Department of Mathematics æ æ æ æhttp://www.math.ubc.ca/~israel > University of British Columbia æ æ æ æ æ æVancouver, BC, Canada- Hide > quoted text - - Show quoted text - I was thinking of the power of 2, but I get stuck quickly when > combining more than two brands. Perhaps you don't understand what Prof. Israel means by the powers > of two. He is referring to the numbers 1, 2, 4, 8, 16, 32, 64, 128, > etc. Use those numbers, and you won't get stuck. -- > Gerry Myerson (ge...@maths.mq.edi.ai) (i -> u for email)- Hide quoted text - - Show quoted text - clarify my conundrum: You're still not doing it right. Powers of 2 DOES NOT mean difference of 2, it means 2**i Store: æStoreName: > 1 æ æ æ WalMart > 3 æ æ æ Target > 5 æ æ æ Sears > 7 æ æ æ Kmart > 9 æ æ æ Dillards Should be: Store: StoreName: 1 WalMart 2 Target 4 Sears 8 Kmart 16 Dillards Products: æ æ æ Stores That Carry This Shirt > A æ æ æ Store# 1 & 3 ( My Value here would be 4) > B æ æ æ Only Store #9 (My Value here would be 9) > C æ æ æ Store# 1 & 5 ( My Value here would be 6) > D æ æ æ Store# 1 & 3 & 5 ( My Value here would be 9) > *PROBLEM - #9 is Dillards - Not the combo of Stores 1, 3 & 5. - Hide quoted text - Which would give you: Products: Stores That Carry This Shirt A Store# 1 & 2 (your Value here would be 3) B Only Store #16 (your Value here would be 16) C Store# 1 & 4 (your Value here would be 5) D Store# 1 & 2 & 4 (your Value here would be 7) No conflict. You can even have a product 0 (not carried by anyone yet). Let's say you have a product carried by everyone. Its number would be 1+2+4+8+16 or 31. Now suppose Sears drops that item. Just subtract 4 from 31 to get 27, in binary: 11011 ||||| ||||+ Walmart (carried) |||+ Target (carried) ||+ Sears (not carried) |+ Kmart (carried) + Dillards (carried) === Subject: Re: algebra - number theory Referring to an old post in here: The question from a preparation book for GRE math subject test was: > Let x,y be positive integers such that 3x + 7y is divisible by 11. > Which of the following must also be divisible by 11? > (a) 4x + 6y (b) x + y + 5 (c) 9x + 4y (d) 4x - 9y (e) x + y - 1 I have a new question about similar problems, but first I will write > down a summary of the previous posts, **Looking back at the question, and after reviewing all replies, I guess > the Fastest approach that one can go about this problem (where time is > very limited in such tests), is proceeding by elimination, as follows: 2) since (0,0) is an obvious solution, then (b) and (e) can easily > eliminated since 5 and -1 are not congruent to 0 mod 11. > 3) (-7,3) is also an obvious solution, (a), and (b) can be eliminated. > 4) well if 3) did not work, 2) & 3) always works since over a field a line is uniquely determined by any two distinct points on it. > then one can calculate the determinant of the matrices > for a) |3 7| > |4 6| which gives -10 not congruent to 0 mod 11 > for c) |3 7| > |9 4| which gives -51 not congruent to 0 mod 11 > so (d) is the answer. That's equivalent to comparing the slopes of the lines (through a common point), an approach which I already mentioned before. > For the record, 2) and 3) were first suggested by Bill Dubuque, > 4) was suggested by quasi. > ***Another fast approach that was also suggested by Peter Schorn and > MuTsun Tsai is: we have, 3x + 7y = 0 (mod 11). we wish to write it as > x = Ay (mod 11), then substitute Ay in the given choices. [...] That's equivalent to comparing the inverse slopes of the lines. As I mentioned, one can employ any usual normal form for lines. > My first question is whether step 4) in the first approach might be needed? > In Burton's Elementary number theory, it is mentioned that if ax = c - by > (mod n), and gcd(a,n) = 1, then there is a unique solution for x (mod n) > for each of the n incongruent values of y (mod n). Now is it plausible to > say that in the worst case, we can have up to n-1 points, incongruent in > the y-coordinate, that validate both Ax + By = C (mod n), and Dx + Ey = F > (mod n), and still Ax + By = C (mod n) and Dx + Ey = F (mod n) are not > equivalent? If it is so, they maybe that justifies step 4) in the first > approach, in the case when C = 0. Over a field any of the usual normal forms for lines may be employed. However, if n isn't prime then Z/n isn't a field and this breaks down. E.g. in Z/15 the lines y = 0, 3x = 5y both pass through (0,0), (5,0) but (5,3) is on the latter but not the former. So over a non-field ring a line is no longer uniquely determined by two distinct points on it. > My second question is, what if the problem was of this type: Ax + By = C > (mod n), then what is the equivalent of step 4) in the first approach. > For example, consider 3x + 7y = -5 (mod 11). To apply the first approach. > First need to find points as in 2) and 3). take for example x = -1, so > 7y = -2 (mod 11), looking for the inverse of 7 in Z_11 gives 8, so > 56y = -16 (mod 11), so y = -5 (mod 11). So (-1,-5) would be a solution. > but how one would go about step 4)? Since Z/11 is a field the usual normal forms for lines may be employed. Using 2) & 3) corresponds to finding two common distinct points on it. Using 4) corresponds to finding a common point and slope. Noticing that -5 = 6 is divisible by 3 yields one obvious point (2,0), which then yields another point (2,0)+(7,-3) = (-2,-3). The key to understanding is to stop thinking about these as _relations_ among integers (mod n) and, instead, to think functionally, by viewing Z (mod p) as a field, where one can transfer all the usual arithmetical intuition, using the usual arithmetic operations and their laws. I've emphasized this strongly in many prior posts here, see my prior links in the recent thread [1] that spawned this one. --Bill Dubuque === Subject: Programs Like MENU at Northwestern? We just had a speaker in our seminar today who is doing a post-doc at Northwestern, and she indicated that they have an unusual option for calculus for freshmen who have taken AP Calculus. The program assumes the student has acheived a 4 or 5 on the BC AP Calculus exam, so they have finished single-variable integral calculus (and done reasonably well in it). Math 290/291 semester 1 covers linear algebra. Math 290/291 semester 2 then covers multiple-variable (that is, in general, more than three variables) differential calculus. Math 290/291 semester 3 covers multiple-variable integral calculus (including, assumedly, the Generalized Stoke's Theorem). (The main difference between 290 and 291 is that 291 is supposed to be more theoretical, although, according to the post-doc, neither is proof-based.) They then offer a three semester Math 360 collection of courses covering diff eqs (ordinary and partial), with the third semester focusing on advanced topics like complex variables, numerical analysis, and the like. Of course, students majoring in math can skip 360 and go directly into real analysis (Rudin-style?) and abstract (modern) algebra. I think it would have been fantastic to have had an introduction to multiple (> 3) variable calculus freshman year and real analysis and abstract algebra sophomore year. Does anyone know if programs such as this are a trend in undergraduate education or if this is an anomaly? -- Jeffrey Rolland === Subject: -- Limit of ratio of consecutive primes = 1 ? Let p(n) denote the nth prime. It seems that, as n -> oo, limit( p(n + 1)/p(n) ) = 1. Is that correct? If so, how can it be proven? Does it perhaps depend on some result (unfamiliar to me) concerning how big, for given n, the gap between p(n) and p(n + 1) can be? David === Subject: Re: -- Limit of ratio of consecutive primes = 1 ? > Let p(n) denote the nth prime. > It seems that, as n -> oo, limit( p(n + 1)/p(n) ) = 1. > Is that correct? If so, how can it be proven? It certainly follows from the Prime Number Theorem, q.v. On the other hand, it's too weak to imply the Prime Number Theorem, so there's some hope that it can be proved without resorting to the PNT. But at the moment I don't see how. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: -- Limit of ratio of consecutive primes = 1 ? Let p(n) denote the nth prime. > It seems that, as n -> oo, limit( p(n + 1)/p(n) ) = > 1. > Is that correct? If so, how can it be proven? It certainly follows from the Prime Number Theorem, > q.v. On the other hand, it's too weak to imply the Prime > Number > Theorem, so there's some hope that it can be proved > without > resorting to the PNT. But at the moment I don't see > how. -- > Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for > email) Let p(n) denote the nth prime. > It seems that, as n -> oo, limit( p(n + 1)/p(n) ) = > 1. > Is that correct? If so, how can it be proven? It certainly follows from the Prime Number Theorem, > q.v. On the other hand, it's too weak to imply the Prime > Number > Theorem, so there's some hope that it can be proved > without > resorting to the PNT. But at the moment I don't see > how. -- > Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for > email) Can you demonstrate how this follow from PNT? It seems to me it's the existence, and not the value, of the limit that's at the heart of the problem. Some corollaries of PNT used to bound prime gaps might be useful, but it's far from obvious to a non-expert like myself. === Subject: Re: -- Limit of ratio of consecutive primes = 1 ? Let p(n) denote the nth prime. > It seems that, as n -> oo, limit( p(n + 1)/p(n) ) = 1. > Is that correct? If so, how can it be proven? It certainly follows from the Prime Number Theorem, q.v. > Can you demonstrate how this follow from PNT? I think you use PNT to show that for any e > 0 there's an n such that N > n implies there's a prime between N and N(1 + e). Then from there it shouldn't be hard to get the limit we want. === Subject: Re: -- Limit of ratio of consecutive primes = 1 ? > > Let p(n) denote the nth prime. > It seems that, as n -> oo, limit( p(n + 1)/p(n) ) = > 1. > Is that correct? If so, how can it be proven? > > It certainly follows from the Prime Number Theorem, > q.v. > > On the other hand, it's too weak to imply the Prime > Number > Theorem, so there's some hope that it can be proved > without > resorting to the PNT. But at the moment I don't see > how. Can you demonstrate how this follow from PNT? It seems to me it's >the existence, and not the value, of the limit that's at the heart >of the problem. Some corollaries of PNT used to bound prime gaps >might be useful, but it's far from obvious to a non-expert like myself. The proof in Apostol, Introduction to Analytic Number Theory, Theorem 4.5, goes something like this: The PNT is pi(x) ~ x/ln(x). This implies (with a therefore at the beginning of each line): ln(pi(x)) - ln(x) - ln(ln(x)) -> 0 ln(pi(x))/ln(x) - 1 - ln(ln(x))/ln(x) -> 0 ln(pi(x)) ~ ln(x) pi(x) ~ x/ln(pi(x)) (using the PNT again) n ~ p(n)/ln(n) p(n) ~ n*ln(n) -- Angus Rodgers Contains mild peril === Subject: Re: -- Limit of ratio of consecutive primes = 1 ? > ln(pi(x)) - ln(x) - ln(ln(x)) -> 0 ln(pi(x))/ln(x) - 1 - ln(ln(x))/ln(x) -> 0 Oops! The second '-' on each line should of course be a '+'. -- Angus Rodgers Contains mild peril === Subject: Re: -- Limit of ratio of consecutive primes = 1 ? > ln(pi(x)) - ln(x) - ln(ln(x)) -> 0 ln(pi(x))/ln(x) - 1 - ln(ln(x))/ln(x) -> 0 Oops! The second '-' on each line should of course be a '+'. -- Angus Rodgers Contains mild peril === Subject: Re: -- Limit of ratio of consecutive primes = 1 ? > Let p(n) denote the nth prime. > It seems that, as n -> oo, limit( p(n + 1)/p(n) ) = 1. > Is that correct? If so, how can it be proven? I'm not sure, but maybe the following approach will work. First, your hypothesis is equivalent to proving that lim_{n -> oo} ln[p(n + 1)] - ln[p(n)] = 0. Recall that the Chebyshev theta function is defined to be theta(n) = sum_{k = 1}^{pi(n)} ln[p(k)] ..where p(k) is the k'th prime and pi(n) is the prime counting function. Note this implies ln[p(n)] = theta(n) - theta(n - 1). Moreover, the Prime Number Theorem implies lim_{n -> oo} theta(n) / n = 1, or equivalently lim_{n -> oo} n / theta(n) = 1. Perhaps one could proceed in this manner? Kyle Czarnecki > Does it perhaps depend on some result (unfamiliar to > me) concerning how big, for given n, the gap between > p(n) and p(n + 1) can be? David === Subject: Re: -- Limit of ratio of consecutive primes = 1 ? Let p(n) denote the nth prime. > It seems that, as n -> oo, limit( p(n + 1)/p(n) ) = 1. > Is that correct? If so, how can it be proven? I'm not sure, but maybe the following approach will > work. First, your hypothesis is equivalent to proving that lim_{n -> oo} ln[p(n + 1)] - ln[p(n)] = 0. Recall that the Chebyshev theta function is defined > to be theta(n) = sum_{k = 1}^{pi(n)} ln[p(k)] ...where p(k) is the k'th prime and pi(n) is the > prime counting function. Note this implies ln[p(n)] = theta(n) - theta(n - 1). Moreover, the Prime Number Theorem implies lim_{n -> oo} theta(n) / n = 1, or equivalently lim_{n -> oo} n / theta(n) = 1. Perhaps one could proceed in this manner? Actually, perhaps it can be done more simply... IIRC, the Prime Number Theorem implies that p(n) is asymptotic to n ln(n), that is p(n) ~ n ln(n) ..which implies... lim_{n -> oo} p(n + 1) / p(n) ~ lim_{n -> oo} (n + 1) ln(n + 1) / [n ln(n)] = lim_{n -> oo} (n + 1) / n * lim_{n -> oo} ln(n + 1) / ln(n) = 1 * lim_{n -> oo} n / (n + 1) = 1. > Kyle Czarnecki Does it perhaps depend on some result (unfamiliar to > me) concerning how big, for given n, the gap between > p(n) and p(n + 1) can be? David === Subject: Re: -- Limit of ratio of consecutive primes = 1 ? > Let p(n) denote the nth prime. > It seems that, as n -> oo, limit( p(n + 1)/p(n) ) = > 1. > Is that correct? If so, how can it be proven? Does it perhaps depend on some result (unfamiliar to > me) concerning how > big, for given n, the gap between p(n) and p(n + 1) > can be? David Baker and Harman showed there is a constant C such that g(p) < C*p^0.535 where g(p_n) = p_(n+1) - p(n) + 1. This implies, p(n+1)/p(n) = g(p_n)/p_n + 1 - 1/p_n < 1 + C*(p_n)^(-0.465) - (p_n)^(-1) Clearly this goes to 1 as p_n -> oo. This is, of course, massive overkill. All you really need to prove is that g(p) is sublinear, which shouldn't be hard. Reference: R. C. Baker and G. Harman, The difference between consecutive primes, Proc. Lond. Math. Soc., series 3, 72 (1996) 261--280. http://plms.oxfordjournals.org/cgi/reprint/s3-72/2/261 === Subject: Re: Mersenne primes > How many Mersenne primes we will find in next 400 years, assuming > that > Moore's law will keep on going and the set of Mersenne primes is > infinite? Moore's law (see http://en.wikipedia.org/wiki/Moore's_Law) commonly > is taken to mean that transistor counts per integrated circuit > double > every two years, and tables at http://www.mersenne.org/ and > http://primes.utm.edu/mersenne/ suggest how much work it takes to > find a Mersenne prime. During the past 12 years, 11 new Mersenne > primes have been found (for a total of 44). The work per find has > been steadily increasing -- mersenne.org suggests about 1.78 > Mersenne > primes per octave of exponents -- and so has available computing > power, > in a similar exponential manner, from which an estimate of about one > find per year for the foreseeable future follows, under the > assumptions > you state. So, about 400 per 400 years. Of course, fundamental > physical > obstacles eventually get in the way of Moore's Law. If each > transistor > is proton-sized, 2^200 of them mass more than a thousand times the > Sun. ... as can be verified by typing: 2^200 * mass of proton / mass of sun into Google... -- Clive Tooth http://www.shutterstock.com/cat.mhtml?gallery_id=61771 === Subject: Subgroups of squarefree groups Let G be a group of order p_1 p_2 ... p_k I know how to show (in a rather long-winded way) that G has a normal subgroup of index p_1. I am trying to show that G has a normal subgroup of order p_k. My question is - is there any way to easily deduce the second statement from the first? It looks like an induction argument, but that doesn't work properly because normality is not transitive... PS: I know the standard proof of the result involving finding the solutions to x^p = 1 (e.g. Hall p147) but I am thinking there must be an easier way if the index p_1 result above is already known. === Subject: Re: Subgroups of squarefree groups > Let G be a group of order p_1 p_2 ... p_k I know how to show (in a > rather long-winded way) that G has a normal subgroup of index p_1. You must have knowledge of some special properties of G. What prevents G from being a simple group (among other possibilities)? - Tim === Subject: Re: Subgroups of squarefree groups posting-account=-PngCgkAAAD2yUjosqWv1Nf1lkqWP4lp Gecko/20080316 SUSE/2.0.0.13-1.1 Firefox/2.0.0.13,gzip(gfe),gzip(gfe) Let G be a group of order p_1 p_2 ... p_k I know how to show (in a > rather long-winded way) that G has a normal subgroup of index p_1. You must have knowledge of some special properties of G. What > prevents G from being a simple group (among other possibilities)? > The fact that he can prove that it has a normal subgroup of index p_1 prevents it from being simple! (This is assuming that p_1 < p_2 < ... < p_k.) That is usually proved as an application of Burnside's Transfer Theorem. Derek Holt. === Subject: Re: Subgroups of squarefree groups Let G be a group of order p_1 p_2 ... p_k I know how to show (in a rather long-winded way) that > G has a normal subgroup of index p_1. I am trying to show that G has a normal subgroup of > order p_k. Suppose by induction that any Sylow subgroup of maximal order is normal in a group of squarefree order strictly smaller than the order of G. Write |G|=p1*...*pk, where p1 < ... < pk are primes. Let N be the normal subgroup of index p1 < pk. Then |N| is strictly smaller than G, squarefree, and divisible by pk, so the Sylow pk-subgroup of N is normal in N, so characteristic in N, so normal in G. === Subject: Re: Subgroups of squarefree groups <15911704.1210217713441.JavaMail.jakarta@nitrogen.mathforum.org>, Let G be a group of order p_1 p_2 ... p_k I know how to show (in a rather long-winded way) that > G has a normal subgroup of index p_1. I am trying to show that G has a normal subgroup of > order p_k. Suppose by induction that any Sylow subgroup of maximal > order is normal in a group of squarefree order strictly > smaller than the order of G. Write |G|=p1*...*pk, where p1 < ... < pk are primes. Let N be the normal subgroup of index p1 < pk. Then |N| > is strictly smaller than G, squarefree, and divisible by > pk, so the Sylow pk-subgroup of N is normal in N, so > characteristic in N, so normal in G. My recollection from student days is that characteristic (in the context of subgroups) is a synonym for normal. The preceding paragraph strongly suggests that my memory is faulty (or my teacher was misinformed), so I guess I don't know what characteristic means. It must mean something - if A is normal in B, and B is normal in C, A needn't be normal in C. I know you know that, Jack, I'm just trying to find my bearings here. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Subgroups of squarefree groups days. My association with the Department is that of an alumnus. ><15911704.1210217713441.JavaMail.jakarta@nitrogen.mathforum.org>, > > Let G be a group of order p_1 p_2 ... p_k > > I know how to show (in a rather long-winded way) that > G has a normal subgroup of index p_1. > > I am trying to show that G has a normal subgroup of > order p_k. > > Suppose by induction that any Sylow subgroup of maximal > order is normal in a group of squarefree order strictly > smaller than the order of G. > > Write |G|=p1*...*pk, where p1 < ... < pk are primes. > > Let N be the normal subgroup of index p1 < pk. Then |N| > is strictly smaller than G, squarefree, and divisible by > pk, so the Sylow pk-subgroup of N is normal in N, so > characteristic in N, so normal in G. My recollection from student days is that characteristic >(in the context of subgroups) is a synonym for normal. >The preceding paragraph strongly suggests that my >memory is faulty (or my teacher was misinformed), so >I guess I don't know what characteristic means. Let G be a group. An endomorphism of G is a group homomorphism G->G. An automorphism of G is an invertible group endomorphism. Let x in G. The map that sends g to x^{-1}gx (conjugation by x) is an automorphism. The subset of all such automorphisms of G forms a group, called the group of inner automorphisms of G. A subgroup H of G is: * Normal if and only if for every inner automorphism f, f(H)=H. * Characteristic if and only if for every automorphism f, f(H)=H. * Fully invariant if and only if for every endomorphism f, f(H) <= H. There is a further condition on subgroups, but it is slightly different (verbal subgroup, which implies fully invariant). So a fully invariant subgroup is always characteristic; and a characteristic subgroup is always normal. But the implications are not reversible. >It must mean something - if A is normal in B, and B >is normal in C, A needn't be normal in C. I know you >know that, Jack, I'm just trying to find my bearings here. One reason why characteristic subgroups are nice is the following: you probably recall that if N < H < G, N is normal in H, and H is normal in G, then it does not follow that N is normal in H. However, if N is characteristic in H, and H is normal in G, then N is normal in G: for every inner automorphism of G will induce an automorphism of H, which will therefore map N to itself. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes by Bill Watterson) Arturo Magidin magidin-at-member-ams-org === Subject: Re: Subgroups of squarefree groups <15911704.1210217713441.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=-PngCgkAAAD2yUjosqWv1Nf1lkqWP4lp Gecko/20080316 SUSE/2.0.0.13-1.1 Firefox/2.0.0.13,gzip(gfe),gzip(gfe) On 8 May, 06:07, Gerry Myerson , > Let G be a group of order p_1 p_2 ... p_k I know how to show (in a rather long-winded way) that > G has a normal subgroup of index p_1. I am trying to show that G has a normal subgroup of > order p_k. Suppose by induction that any Sylow subgroup of maximal > order is normal in a group of squarefree order strictly > smaller than the order of G. Write |G|=p1*...*pk, where p1 < ... < pk are primes. Let N be the normal subgroup of index p1 < pk. Then |N| > is strictly smaller than G, squarefree, and divisible by > pk, so the Sylow pk-subgroup of N is normal in N, so > characteristic in N, so normal in G. My recollection from student days is that characteristic > (in the context of subgroups) is a synonym for normal. > The preceding paragraph strongly suggests that my > memory is faulty (or my teacher was misinformed), so > I guess I don't know what characteristic means. It must mean something - if A is normal in B, and B > is normal in C, A needn't be normal in C. I know you > know that, Jack, I'm just trying to find my bearings here. -- > Gerry Myerson (ge...@maths.mq.edi.ai) (i -> u for email) OK, here is a quick tutorial! Subgroup A characteristic in B means that phi(A) = A for every automorphism phi of G. Note that A is normal in B means that this is true for every inner automorphism of G, so characteristic subgroups are necessarily normal. This concept is useful because it has nice transitivity properties. A char B and B char C imply A char C; and A char B and B normal in C imply A normal in C. (But note that A normal in B and B char C do not always imply A normal in C.) Some familiar normal subgroups are always characteristic. These include the centre of the group, the derived subgroup, all members of of the upper or lower or derived series, the Frattini subgroup, the Fitting subgroup, and so on. If a Sylow subgroup (or indeed a Hall subgroup) P of G is normal in G, then it is the unique Sylow subgroup, and so it is characteristic. That is the result that is being used here to deduce that the square- free group has a normal subgroup of order its largest prime divisor. Derek Holt. === Subject: Re: Subgroups of squarefree groups > OK, here is a quick tutorial! Subgroup A characteristic in B means that phi(A) = A for every > automorphism phi of G. Note that A is normal in B means that this is true for every inner > automorphism of G, so characteristic subgroups are necessarily normal. This concept is useful because it has nice transitivity properties. A char B and B char C imply A char C; and A char B and B normal in C imply A normal in C. (But note that A normal in B and B char C do not always imply A normal > in C.) Some familiar normal subgroups are always characteristic. These > include the centre of the group, the derived subgroup, all members of > of the upper or lower or derived series, the Frattini subgroup, the > Fitting subgroup, and so on. If a Sylow subgroup (or indeed a Hall subgroup) P of G is normal in G, > then it is the unique Sylow subgroup, and so it is characteristic. > That is the result that is being used here to deduce that the square- > free group has a normal subgroup of order its largest prime divisor. a normal subgroup that isn't characteristic would be a 2-element subgroup of the Klein 4-group. === Subject: Re: Subgroups of squarefree groups <6203873.1210249539288.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=-PngCgkAAAD2yUjosqWv1Nf1lkqWP4lp Gecko/20080316 SUSE/2.0.0.13-1.1 Firefox/2.0.0.13,gzip(gfe),gzip(gfe) > OK, here is a quick tutorial! Subgroup A characteristic in B means that phi(A) = A for every > automorphism phi of G. Note that A is normal in B means that this is true for every inner > automorphism of G, so characteristic subgroups are necessarily normal. This concept is useful because it has nice transitivity properties. A char B and B char C imply A char C; and A char B and B normal in C imply A normal in C. (But note that A normal in B and B char C do not always imply A normal > in C.) Some familiar normal subgroups are always characteristic. These > include the centre of the group, the derived subgroup, all members of > of the upper or lower or derived series, the Frattini subgroup, the > Fitting subgroup, and so on. If a Sylow subgroup (or indeed a Hall subgroup) P of G is normal in G, > then it is the unique Sylow subgroup, and so it is characteristic. > That is the result that is being used here to deduce that the square- > free group has a normal subgroup of order its largest prime divisor. a normal subgroup that isn't characteristic would be a 2-element > subgroup of the Klein 4-group. Yes! Incidentally, it can be proved that a group is characteristically simple (i.e. it has not proper nontrivial characteristic subgroups) iff it is a direct product of isomorphic simple groups, so the Klein 4- group is an example of this. Derek Holt. === Subject: Re: Subgroups of squarefree groups The context of this thread is clearly finite groups, but I wanted to mention a nice example of an infinite counterexample. > Incidentally, it can be proved that a *finite* > group is characteristically simple (i.e. it has > no proper nontrivial characteristic subgroups) > iff it is a direct product of isomorphic simple > groups, so the Klein 4-group is an example of this. In the infinite case there are weirder examples. McLain's matrix group example is locally nilpotent, characteristically simple, perfect, centerless, and it is the product (not direct) of its normal abelian subgroups. It is described on p361ff in Robinson's text on the theory of groups. It is basically the uni-triangular matrices over any field, but where the matrix entries are indexed by rationals rather than a finite set {1,2,...,n}. === Subject: Re: Subgroups of squarefree groups <15317909.1210207001988.JavaMail.jakarta@nitrogen.mathforum.org>, > Let G be a group of order p_1 p_2 ... p_k I know how to show (in a rather long-winded way) that G has a normal subgroup > of index p_1. I am trying to show that G has a normal subgroup of order p_k. My question is - is there any way to easily deduce the second statement from > the first? It looks like an induction argument, but that doesn't work > properly because normality is not transitive... PS: I know the standard proof of the result involving finding the solutions > to x^p = 1 (e.g. Hall p147) but I am thinking there must be an easier way if > the index p_1 result above is already known. There's a lot that you aren't telling us. Are p_1, p_2, ..., p_k supposed to be prime numbers? Are they supposed to be distinct prime numbers? Are they supposed to be in increasing order? Is it really true that if G has order, say, 3 x 5 x 7 x 11 x 31, then it is guaranteed to have a normal subgroup of order 31? -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Subgroups of squarefree groups > <15317909.1210207001988.JavaMail.jakarta@nitrogen.math > forum.org>, Let G be a group of order p_1 p_2 ... p_k I know how to show (in a rather long-winded way) > that G has a normal subgroup > of index p_1. I am trying to show that G has a normal subgroup of > order p_k. My question is - is there any way to easily deduce > the second statement from > the first? It looks like an induction argument, but > that doesn't work > properly because normality is not transitive... PS: I know the standard proof of the result > involving finding the solutions > to x^p = 1 (e.g. Hall p147) but I am thinking > there must be an easier way if > the index p_1 result above is already known. There's a lot that you aren't telling us. > I am sorry for the confusion. > Are p_1, p_2, ..., p_k supposed to be prime numbers? > Yes, they are primes. > Are they supposed to be distinct prime numbers? Are they supposed to be in increasing order? > Yes they are in strictly increasing order (and therefore distinct). If i < j then p_i < p_j. I am sorry that I didn't explain this better. > Is it really true that if G has order, say, 3 x 5 x 7 > x 11 x 31, > then it is guaranteed to have a normal subgroup of > order 31? > Yes, this result is definitely _true_ as proved in M. Hall's book, p147. But the question is whether it follows easily from my earlier statement. > -- > Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for > email) === Subject: Re: Subgroups of squarefree groups > Is it really true that if G has order, say, 3 x 5 x 7 > x 11 x 31, then it is guaranteed to have a normal > subgroup of order 31? Yes. Groups whose Sylow subgroups are all cyclic have very special properties. The derived subgroup and the maximal abelian quotient are both cyclic. A large prime cannot act but trivially on smaller cyclic groups, so an element of largest prime order centralizes every chief factor, so is contained in the Fitting subgroup. This is basically a property of supersoluble groups, but it might not be the most direct method of proof. Groups whose Sylows are cyclic are explained nicely in M. Hall's text on the theory of groups, but I don't have the reference handy. It might be on the wikipedia === Subject: Re: Subgroups of squarefree groups posting-account=9QOSvAoAAACEOWJVSDuswW7dB_0wApQO Gecko/20080416 Fedora/2.0.0.14-1.fc8 Firefox/2.0.0.14,gzip(gfe),gzip(gfe) On May 7, 11:47 pm, Gerry Myerson , Let G be a group of order p_1 p_2 ... p_k I know how to show (in a rather long-winded way) that G has a normal subgroup > of index p_1. I am trying to show that G has a normal subgroup of order p_k. My question is - is there any way to easily deduce the second statement from > the first? It looks like an induction argument, but that doesn't work > properly because normality is not transitive... PS: I know the standard proof of the result involving finding the solutions > to x^p = 1 (e.g. Hall p147) but I am thinking there must be an easier way if > the index p_1 result above is already known. There's a lot that you aren't telling us. Are p_1, p_2, ..., p_k supposed to be prime numbers? Are they supposed to be distinct prime numbers? Are they supposed to be in increasing order? Is it really true that if G has order, say, 3 x 5 x 7 x 11 x 31, > then it is guaranteed to have a normal subgroup of order 31? According to gap, the answer is yes: gap> Filtered(AllSmallGroups(3 * 5 * 7 * 11 * 31), g -> not (31 in List(NormalSubgroups(g), Order)); [ ] -- m === Subject: Re: Subgroups of squarefree groups posting-account=9QOSvAoAAACEOWJVSDuswW7dB_0wApQO Gecko/20080416 Fedora/2.0.0.14-1.fc8 Firefox/2.0.0.14,gzip(gfe),gzip(gfe) Let G be a group of order p_1 p_2 ... p_k I know how to show (in a rather long-winded way) that G has a normal subgroup of index p_1. I am trying to show that G has a normal subgroup of order p_k. > I imagine your primes are ordered in a specific way? Otherwise, just reindex ;-) -- m === Subject: Re: Subgroups of squarefree groups Let G be a group of order p_1 p_2 ... p_k I know how to show (in a rather long-winded way) that G has a normal > subgroup of index p_1. I am trying to show that G has a normal subgroup of order p_k. > I imagine your primes are ordered in a specific way? > Otherwise, just reindex ;-) Have you've missed the distinction between index and order? -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Subgroups of squarefree groups posting-account=9QOSvAoAAACEOWJVSDuswW7dB_0wApQO Gecko/20080416 Fedora/2.0.0.14-1.fc8 Firefox/2.0.0.14,gzip(gfe),gzip(gfe) On May 8, 2:00 am, Gerry Myerson subgroup of index p 1. I am trying to show that G has a normal subgroup of order p k. I imagine your primes are ordered in a specific way? > Otherwise, just reindex ;-) Have you've missed the distinction between index and order? No. -- m === Subject: Re: Subgroups of squarefree groups > <49f73e82-d9fa-4aef-83ae-dc088ed71a48@z72g2000hsb.goog > legroups.com>, > Mariano Su.87rez-Alvarez On May 7, 9:36 pm, Adam Burley Let G be a group of order p_1 p_2 ... p_k I know how to show (in a rather long-winded way) > that G has a normal > subgroup of index p_1. I am trying to show that G has a normal subgroup > of order p_k. > I imagine your primes are ordered in a specific > way? > Otherwise, just reindex ;-) > Yes, I'm sorry I didn't mention that, we have p_i < p_j for i < j. > Have you've missed the distinction between index and > order? > No, I understand. Let H be a subgroup of G. The order of H is the number of elements of H. The index of H is the number of cosets of H in G. For finite groups (which all squarefree groups are), the index of H is |G|/|H|. > -- > Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for > email) === Subject: A puzzle For the discussion below, let [a_1,a_2,...a_n] denote a natural number having first decimal digit a_1, second digit a_2, etc. More precisely, [a_1,a_2,...,a_n] = Sum[10^(n-k)*a_k,{k,1,n}] Consider first the following problem (n=2): What pairs of numbers satisfy [a,b]*[c,d]=[b,a]*[d,c]? In this case the solution is simply the set of numbers which satisfy ac=bd. Alternatively, we might say the equality is satisfied if and only if the matrix |a b| |d c| is singular. Now consider the problem for n=3. i.e. Which numbers satisfy [a,b,c]*[d,e,f]=[c,b,a]*[f,e,d]. It turns out in this case the equality is satisfied if and only if the matrix |a b c| |f e d| |1 R 1| is singular, where R = -101/10. For the case n=4, i.e. [a,b,c,d]*[e,f,g,h]=[d,c,b,a]*[h,g,f,e], the equality is satisfied if and only if the matrix |a b c d| |h g f e| |0 1 R 1| |1 R 1 0| is singular, where again R = -101/10. I do not know what matrix satisfies the expression for higher n. Obviously the original question itself is very easy to answer in the sense that you can readily obtain an algebraic expression that you show is 0 if and only if equality holds. What is curious to me is how this algebraic expression is related to determinants of these rather nice looking matrices. - How do the matrices for higher n look? Can one be found whose n-2 bottom rows contain only the numbers 0,1, and R? - What can be said about uniqueness of these matrices if we demand that the first two rows are unchanged? (up to perhaps rows changes and scalar multiplications) - Can you see some intuitive/geometric explanation for how/why determinants of these matrices relate to the equality? === Subject: Re: A puzzle >For the discussion below, let [a_1,a_2,...a_n] denote a natural >number having first decimal digit a_1, second digit a_2, etc. >More precisely, [a_1,a_2,...,a_n] = Sum[10^(n-k)*a_k,{k,1,n}] >Consider first the following problem (n=2): What pairs of numbers satisfy [a,b]*[c,d]=[b,a]*[d,c]? In this case the solution is simply the set of numbers which satisfy >ac=bd. Alternatively, we might say the equality is satisfied if and >only if the matrix |a b| >|d c| is singular. Now consider the problem for n=3. i.e. Which numbers >satisfy [a,b,c]*[d,e,f]=[c,b,a]*[f,e,d]. It turns out in this case >the equality is satisfied if and only if the matrix |a b c| >|f e d| >|1 R 1| is singular, where R = -101/10. For the case n=4, i.e. >[a,b,c,d]*[e,f,g,h]=[d,c,b,a]*[h,g,f,e], the equality is satisfied >if and only if the matrix |a b c d| >|h g f e| >|0 1 R 1| >|1 R 1 0| is singular, where again R = -101/10. I do not know what matrix satisfies the expression for higher n. >Obviously the original question itself is very easy to answer in >the sense that you can readily obtain an algebraic expression that >you show is 0 if and only if equality holds. What is curious to me >is how this algebraic expression is related to determinants of these >rather nice looking matrices. - How do the matrices for higher n look? Can one be found whose n-2 >bottom rows contain only the numbers 0,1, and R? >- What can be said about uniqueness of these matrices if we demand >that the first two rows are unchanged? (up to perhaps rows changes >and scalar multiplications) >- Can you see some intuitive/geometric explanation for how/why >determinants of these matrices relate to the equality? The pattern continues for higher n. Let u = (10^{n-1}, ..., 10, 1), and v = (1, 10, ..., 10^{n-1}), Then (0 ... 0 1 R 1 0 ... 0)u = (0 ... 0 1 R 1 0 ... 0)v = 0, for each of the n - 2 possible positions of the sequence 1 R 1 in the vector (0 ... 0 1 R 1 0 ... 0) in R^n. (That's a blackboard R!) Let p and q be two vectors in R^n (e.g. p = (a b c) and q = (f e d) in R^3). If we are given: (p.u)(q.v) = (p.v)(q.u) then the simultaneous linear equations: (p.u)x + (q.v)y = 0 (q.u)x + (q.v)y = 0 have a non-zero solution for (x, y). Putting w = xu + yv, we have p.w = q.w = 0, and also (0 ... 0 1 R 1 0 ... 0)w = 0 (for each of the n - 2 possible positions of 1 R 1). Since u and v are linearly independent, and x and y are not both zero, w != 0. So the set of homogeneous linear equations represented by the n x n matrix: p q ... 0 ... 0 1 R 1 0 ... 0 ... (if you see what I mean!) has a non-zero solution in R^n; so the determinant of the matrix is zero. Conversely, the n - 2 homogeneous linear equations represented by the last n - 2 rows of the matrix have a unique solution in terms of the last two components of a vector in R^n, so, if the full set of equations has a solution (as it must if the determinant is zero), it must be a linear combination of u and v, therefore the 2 x 2 set of linear equations above has a solution, therefore its determinant is zero, i.e. (p.u)(q.v) = (p.v)(q.u). (This could probably have been put more elegantly, but my linear algebra has always been shaky.) -- Angus Rodgers Contains mild peril === Subject: An integer optimization problem I have discovered the following optimization problem in a piece of software that I'm working on. Does anybody know if it has a name or if it's an instance of some other well-known problem. Given N integers {n_1, n_2, ..., n_N} with M <= N different values, find the set of O integers {o_1, o_2, ..., o_O} with P < M different values that minimizes |n_i - o_i| for all i. === Subject: Re: An integer optimization problem posting-account=K5WE3woAAAAXArsybjkbN6LjMxWdHtbX Gecko/20080404 Firefox/2.0.0.14,gzip(gfe),gzip(gfe) I have discovered the following optimization problem in a piece of software > that I'm working on. Does anybody know if it has a name or if it's an > instance of some other well-known problem. Given N integers {n_1, n_2, ..., n_N} with M <= N different values, find the > set of O integers {o_1, o_2, ..., o_O} with P < M different values that > minimizes |n_i - o_i| for all i. > I am almost embarrassed by my previous p-median formulation, which is / correct/ but unnecessary in this case. I think the problem's special structure allows it to be solved efficiently using a dynamic programming approach. Say the original set is k_1 copies of integer v_1, k_2 copies of v_2,..., k_M copies of v_M, where v_1 < v_2 < ... < v_M; we denote this as (v_1,k_1), (v_2,k_2), ... . Draw the points v_1,...,v_M on a line. Now we want to insert (P-1) boundaries separating the integers {v_i} into P subsets {v_1,..., v_a}, {v_(a +1),...,v_b}, {v_(b+1),...,v_c}, etc, where the first boundary is between v_a and v_(a+1), the second between v_b and v_(b+1), etc. Let f(r,s) = minimum of sum given a partition of {(v_1,k_1), (v_2,k_2), ..., (v_r,k_r)} into s subsets. The overall minimum is f(M,P). We have f(M,P) = min{f(r,P-1) + m(r+1,M) : r = P-1, P, ..., M-1}. Here, m(s,t) denotes the sum k_s*|v_s - mu| + ... + k_t*|v_t - mu|, with mu being the median of the set {(v_s,k_s),...,(v_t,k_t)} and is easily obtained. Thus, if we can compute each f(r,P-1) we can get the optimal solution. But, now, getting f(r,P-1) is just a smaller version of the original problem and can be attacked in the same way, etc. Once you have the optimal partition, the solution is o_1 = median of {(v_1,k_1),...,(v_a,k_a)} with multiplicity = k_1 + ... + k_a, etc. R.G. Vickson === Subject: Re: An integer optimization problem posting-account=zKSjoQoAAAC5wIsHKdIzIXzOHEFPgaXy Gecko/20080404 Firefox/2.0.0.14,gzip(gfe),gzip(gfe) I have discovered the following optimization problem in a piece of software > that I'm working on. Does anybody know if it has a name or if it's an > instance of some other well-known problem. Given N integers {n_1, n_2, ..., n_N} with M <= N different values, find the > set of O integers {o_1, o_2, ..., o_O} with P < M different values that > minimizes |n_i - o_i| for all i. > Your problem seems ill-posed. In order to have |n_i - o_i| defined for all i, you need O = N. Assuming that is what you meant to write, now what do you mean by saying that |n_i - o_i| is minimized for all i? What minimizes it for one i may not work for another i. You could look at minimizing sum{|n_i - o_i|; i = 1..N} or minimizing max{|n_i - o_i|; i=1..N}. Finally, is P fixed? If not, what stops you from taking P = N-1, so that only one value is different in the two sets? R.G. Vickson === Subject: Re: An integer optimization problem > I have discovered the following optimization problem in a piece of > software > that I'm working on. Does anybody know if it has a name or if it's an > instance of some other well-known problem. > Given N integers {n_1, n_2, ..., n_N} with M <= N different values, find > the > set of O integers {o_1, o_2, ..., o_O} with P < M different values that > minimizes |n_i - o_i| for all i. Your problem seems ill-posed. In order to have |n_i - o_i| defined for > all i, you need O = N. Assuming that is what you meant to write, now > what do you mean by saying that |n_i - o_i| is minimized for all i? > What minimizes it for one i may not work for another i. You could look > at minimizing sum{|n_i - o_i|; i = 1..N} or minimizing max{|n_i - > o_i|; i=1..N}. Finally, is P fixed? If not, what stops you from taking > P = N-1, so that only one value is different in the two sets? an example instead. Let's say I have the set {1, 1, 2, 3, 9, 9, 10, 48} where N = 8 with M = 6 different value and I want to find the set of O = 8 integers with P = 4 different values. One possibility is {1, 1, 1, 3, 10, 10, 10, 48}and another is {1, 1, 1, 3, 9, 9, 9, 48}.The latter is better than the former since sum{|n_i - o_i|, 1..8} = 2 for the last and 3 for the first. === Subject: Re: An integer optimization problem Given N integers {n_1, n_2, ..., n_N} with M <= N different values, > find the set of O integers {o_1, o_2, ..., o_O} with P < M different > values that minimizes |n_i - o_i| for all i. Set is the wrong concept. You want to think finite sequences. > an example instead. Let's say I have the set {1, 1, 2, 3, 9, 9, 10, 48} > where N = 8 with M = 6 different value and I want to find the set of O = 8 > integers with P = 4 different values. One possibility is {1, 1, 1, 3, 10, > 10, 10, 48}and another is {1, 1, 1, 3, 9, 9, 9, 48}.The latter is better > than the former since sum{|n_i - o_i|, 1..8} = 2 for the last and 3 for the > first. > Let (aj)_(j=1,..n) be a sequence of n integers with d distinct values. Find a sequence (bj)_(j=1,..n) of n integers with less than d distinct values that minimizes |aj - bj| for all j = 1,.. n what ever that means. This problem is unsolvable when d = 1. For d > 1, here's an idea. For all j, let bj = aj except for one. Pick that one so that the positive absolute value of the difference between it and each of the other integers is as small as possible. Then change that one to the nearest different integer in the sequence. Hm. Another version. Let (aj)_(j=1,..n) be a sequence of n integers with d > 1 distinct values. Find a sequence (bj)_(j=1,..n) of n integers with c distinct values, for which c < d and some mystical quantity is minimized. === Subject: Re: An integer optimization problem Given N integers {n_1, n_2, ..., n_N} with M <= N different values, > find the set of O integers {o_1, o_2, ..., o_O} with P < M different > values that minimizes |n_i - o_i| for all i. Set is the wrong concept. You want to think finite sequences. an example instead. Let's say I have the set {1, 1, 2, 3, 9, 9, 10, 48} > where N = 8 with M = 6 different value and I want to find the set of O = 8 > integers with P = 4 different values. One possibility is {1, 1, 1, 3, 10, > 10, 10, 48}and another is {1, 1, 1, 3, 9, 9, 9, 48}.The latter is better > than the former since sum{|n_i - o_i|, 1..8} = 2 for the last and 3 for the > first. > Whoops, missed the last line. Ok, here's what I consider to be an exact formulation of the problem. > Let (aj)_(j=1,..n) be a sequence of n integers with d > 1 distinct > values. Find a sequence (bj)_(j=1,..n) of n integers with c distinct values, > for which c < d and sum_(J=1,..n) |aj - bj| is minimized. It seems that when c = 1, that taking the average of the integers would give a good estimate for the integer to use throughout the sequence. Now if c > 1, then order (aj)_j by magnitude and partition the sequence into c contiguous parts, each having as small as variation as possible, and take the average of each of the c parts to estimate c distint integers. === Subject: Re: An integer optimization problem posting-account=K5WE3woAAAAXArsybjkbN6LjMxWdHtbX Gecko/20080404 Firefox/2.0.0.14,gzip(gfe),gzip(gfe) > I have discovered the following optimization problem in a piece of > software > that I'm working on. Does anybody know if it has a name or if it's an > instance of some other well-known problem. > Given N integers {n_1, n_2, ..., n_N} with M <= N different values, find > the > set of O integers {o_1, o_2, ..., o_O} with P < M different values that > minimizes |n_i - o_i| for all i. > Your problem seems ill-posed. In order to have |n_i - o_i| defined for > all i, you need O = N. Assuming that is what you meant to write, now > what do you mean by saying that |n_i - o_i| is minimized for all i? > What minimizes it for one i may not work for another i. You could look > at minimizing sum{|n_i - o_i|; i = 1..N} or minimizing max{|n_i - > o_i|; i=1..N}. Finally, is P fixed? If not, what stops you from taking > P = N-1, so that only one value is different in the two sets? an example instead. Let's say I have the set {1, 1, 2, 3, 9, 9, 10, 48} > where N = 8 with M = 6 different value and I want to find the set of O = 8 > integers with P = 4 different values. One possibility is {1, 1, 1, 3, 10, > 10, 10, 48}and another is {1, 1, 1, 3, 9, 9, 9, 48}.The latter is better > than the former since sum{|n_i - o_i|, 1..8} = 2 for the last and 3 for the > first. So, are you saying that you would like to minimize the sum{|o_i - n_i|; i=1..N}? If so, this looks like the P-median problem. Think of integers n_1, ..., n_N as customers and integers 1,2,...,n_N as possible locations for warehouses. The distance between customer n_i and warehouse j is |j - n_i|. We want to pick P warehouses from among the potential locations in such a way that each customer is served from the closest available warehouse o (distance = [n_i - o|). After solving the problem, each customer is served by a warehouse, and we can partition the customers among the different warehouses that serve them. The partition corresponds to the different values o_i. The problem can be formulated as a (binary) linear-integer programming problem, and small enough instances can be solved exactly by branch- and-bound. Very large instances may be difficult to solve to optimality, in which case there are some very effective heuristics available that give good, but not necessarily optimal solutions. Here is an integer programming formulation. Define the decision variables as x(i,j) = 1 if customer n_i is served by location j, x(i,j) = 0 otherwise. Define y(j) = 1 if location j has a warehouse, y(j) = 0 if not. The problem is: minimize sum{ |n_i - j|*x(i,j), i=1..N,j=1..n_N} subject to sum{y(j),j=1..n_N} = P (open P warehouses) sum{x(i,j); j=1..n_N} = 1 for all i (each customer is served from 1 warehouse) x(i,j) <= y(j) for all i and j (cannot server from j unless j has a warehouse) x(i,j), y(j) = 0 or 1 for all i and j. The minimization will automatically force each n_i to be served from it closest available j. Note: there are lots of constraints x(i,j) <= y(j); we COULD replace these by a much smaller number of constraints of the form sum{x(i,j); i=1..N} <= N*y(j) for all j, but that might be an unwise strategy. Experiments with such formulations show that the larger version tends to be easier and faster to solve than the smaller version (because, technically, the set of individual constraints x(i,j) <= y(j) for all i,j tends to produce a linear programming relaxation with a feasible region that is closer to that of the integer problem). A brief introduction to the p-median problem is in: http://www.hyuan.com/java/ A survey of exact and heuristic solution methods is contained in: http://ramanujan.math.trinity.edu/tumath/research/reports/report96.pdf The general instance of the p-median problem is NP-hard, so numerous heuristics such as (i) the greedy algorithm (best-insertion): start with 1 warehouse; the best location is at the median of the {n_i}; then fixing one warehouse at the median, find the best place to put a second one. This is easily and quickly solved by explicit search. Then, fixing the first two, add a third in the best place, etc. Continue until you have P of them in place) (ii) best-deletion (start with P = N, then determine the single best warehouse to close. Fixing the first closure, now find the best place to close a second one, etc. Continue until only P are left.) Numerous authors have used genetic algorithms, simulated annealing, ant-colony optimization methods, and others, to try to deal with large versions of the problem. However, modern integer programming codes can likely handle moderate sized problems and solve them to optimality. (By modest, I mean a few tens of n_i values.) R.G. Vickson === Subject: Re: An integer optimization problem > I have discovered the following optimization problem in a piece of > software > that I'm working on. Does anybody know if it has a name or if it's an > instance of some other well-known problem. Given N integers {n_1, n_2, ..., n_N} with M <= N different values, find > the > set of O integers {o_1, o_2, ..., o_O} with P < M different values that > minimizes |n_i - o_i| for all i. > Your problem seems ill-posed. In order to have |n_i - o_i| defined for > all i, you need O = N. Assuming that is what you meant to write, now > what do you mean by saying that |n_i - o_i| is minimized for all i? > What minimizes it for one i may not work for another i. You could look > at minimizing sum{|n_i - o_i|; i = 1..N} or minimizing max{|n_i - > o_i|; i=1..N}. Finally, is P fixed? If not, what stops you from taking > P = N-1, so that only one value is different in the two sets? give an example instead. Let's say I have the set {1, 1, 2, 3, 9, 9, 10, > 48} where N = 8 with M = 6 different value and I want to find the set of O > = 8 integers with P = 4 different values. One possibility is {1, 1, 1, 3, > 10, 10, 10, 48}and another is {1, 1, 1, 3, 9, 9, 9, 48}.The latter is > better than the former since sum{|n_i - o_i|, 1..8} = 2 for the last and 3 > for the first. I reckon it can be formulated in terms of minimizing the manhattan distance if we let O = N under a constraint regarding the number of different elements. === Subject: on-line calculator posting-account=gQCGxQoAAADHkJCCmXTgfgTdilT5bOEY Gecko/20080410 SUSE/2.0.0.14-0.1 Firefox/2.0.0.14,gzip(gfe),gzip(gfe) I do a lot of engineering calculations and I got frustrated with on- line calculators that try to look and act like pocket calculators. So http://calculator.troymius.com Now I am looking for smart people like you to try it if it really === Subject: Re: on-line calculator posting-account=cvz5-QoAAABVNzogw177Plx_25TguPUZ CLR 1.0.3705; .NET CLR 1.1.4322; .NET CLR 2.0.50727; InfoPath.1),gzip(gfe),gzip(gfe) > I do a lot of engineering calculations and I got frustrated with on- > line calculators that try to look and act like pocket calculators. So http://calculator.troymius.com Now I am looking for smart people like you to try it if it really When I type in sqrt(-1) into your calculator I get .... When I type in sqrt(-1) into Google I get i. - MO === Subject: Re: on-line calculator posting-account=gQCGxQoAAADHkJCCmXTgfgTdilT5bOEY GM_UserLogonTimeUTC: 2008-05-02 12:26:20; GM_UserLogonTimeBias: 240; .NET CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) I.N. and MO, So far I dont plan implementing operations with complex numbers, maybe I should think about it. In the code, pi is only defined with a limited precision (16 digits or so), so to get a completely correct answer for tan(pi/2) i would have to implement some kind of a logic... I am not sure how to do that yet. Same problem applies to sin(pi) and similar... One way to get a round it is not to pre-define pi at all: that way the user would have to enter it by him/her self and would be better aware that the calculator's pi is not exactly pi... But having pi predefined is in most cases quite handy... Jerry === Subject: Re: on-line calculator posting-account=gQCGxQoAAADHkJCCmXTgfgTdilT5bOEY Gecko/20080410 SUSE/2.0.0.14-0.1 Firefox/2.0.0.14,gzip(gfe),gzip(gfe) I.N. and MO, So far I dont plan implementing complex numbers, maybe I should think about it. The pi number is only defined with some 8 digits, so to get a completely correct answer for tan(pi/2) i would have to implement some kind of logic... Will think about it as well. Jerry === Subject: Re: on-line calculator > I do a lot of engineering calculations and I got frustrated with on- > line calculators that try to look and act like pocket calculators. So http://calculator.troymius.com Now I am looking for smart people like you to try it if it really I'm getting: tan(pi/2)=1.63317787284e+16 tan(-pi/2)=-1.63317787284e+16 That's not quite right. tan(+/- pi/2) is undefined, so it may be a better idea to use the same tree dots you use for, say, 1/0. Or think about implementing NaN's. -- I.N. Galidakis === Subject: A Lagrangian not dependent on time? posting-account=JUtbYgoAAAD32oKainxIT3ac9fVJqDnB Gecko/20080404 Firefox/2.0.0.14 Creative ZENcast v1.02.10,gzip(gfe),gzip(gfe) Hi everybody ! I need a little help with one problem: I was given a Hamiltonian of one degree of freedom that has a form: H (p,q) = (p^2) / (2*alfa) - b*q*p*e^(-alfa*t) + (b*a / 2)*q^2*e^(- alfa*t)*(alfa+b*e^(-alfa*t))+... where a,b,alfa, and k are constants. I found a Lagrangian corresponding to this Hamiltonian and it looks like that: L (q`, q) = (1/2)*(q` + b*q*e^(-alfa*t))^2-(1/2)*b*a*q^2*e^(-alf... where q`= dq/dt (a derivative of q with respect to t, time). However, after all this tedious work I cannot find an eqiuvalent Lagrangian that is not explicitly dependent on time (I got brain dead:)) I am supposed to use the fact that: delta H / delta t = delta L / delta t Can anybody help me? === Subject: Re: A Lagrangian not dependent on time? posting-account=zKSjoQoAAAC5wIsHKdIzIXzOHEFPgaXy Gecko/20080404 Firefox/2.0.0.14,gzip(gfe),gzip(gfe) > Hi everybody ! I need a little help with one problem: I was given a Hamiltonian of one degree of freedom that has a form: H (p,q) = (p^2) / (2*alfa) - b*q*p*e^(-alfa*t) + (b*a / 2)*q^2*e^(- > alfa*t)*(alfa+b*e^(-alfa*t))+... where a,b,alfa, and k are constants. I found a Lagrangian > corresponding to this Hamiltonian and it looks like that: L (q`, q) = (1/2)*(q` + b*q*e^(-alfa*t))^2-(1/2)*b*a*q^2*e^(-alf... where q`= dq/dt (a derivative of q with respect to t, time). > However, after all this tedious work I cannot find an eqiuvalent > Lagrangian that is not explicitly dependent on time Why would you suppose this is even possible? Maybe the Lagrangian for the system has explicit time dependence. R.G. Vickson > (I got brain > dead:)) > I am supposed to use the fact that: delta H / delta t = delta L / delta t Can anybody help me? === Subject: Re: A Lagrangian not dependent on time? posting-account=JUtbYgoAAAD32oKainxIT3ac9fVJqDnB Gecko/20080404 Firefox/2.0.0.14 Creative ZENcast v1.02.10,gzip(gfe),gzip(gfe) Assuming that this is possible, what should be the procedure for this? === Subject: Re: A Lagrangian not dependent on time? it would be easier, if you provide more background. In economics one often has to deal with utility functions like exp(-r t)*F(q, u) Now one replaces the adjoint variables p by P*exp(-r t) and obtains the differential equation P' = -@ K/@ q + r P, where @ denotes the partial derivative and K is the current value Hamiltonian, simply exp(r t)*H. In your case it seems, that b could be a candidate for scaling. The new system is of course not strictly Hamiltonian, the spectrum is shifted by r/2 to the right. Good luck Alois === Subject: Statistica Sinica April issue posting-account=QrG9mwoAAAD4swwif73Cg69MYQStuBea Gecko/20080404 Firefox/2.0.0.14,gzip(gfe),gzip(gfe) I am posting the table of contents for the April 2008 issue of Statistica Sinica. Please click on the Current Issue at the following website to access the editorial: http://www.stat.sinica.edu.tw/statistica/ Karen Li---on behalf of the Co-editors Editorial Assistant Statistica Sinica ---------------------------------------------------------------------------- -------------- EDITORÍS MELANGE Highlights.84A master of minds Jianqing Fan Editorial.84A rigorous statistician and passionate educator Jiahua Chen MEMORIAL ARTICLES TO XIRU CHEN On constrained M-estimation and its recursive analog in multivariate linear regression models Zhidong Bai, Xiru Chen and Yuehua Wu Large sample covariance matrices without independence structures in columns Zhidong Bai and Wang Zhou Inference for normal mixtures in mean and variance Jiahua Chen, Xianming Tan and Runchu Zhang GENERAL Estimating monotone, unimodal and U--shaped failure rates using asymptotic pivots Moulinath Banerjee Factorial designs with multiple levels of randomization Derek Bingham, Randy Sitter, Elizabeth Kelly, Leslie Moore and J. Davaid Olivas Design-adaptive minimax local linear regression for longitudinal/ clustered data Kani Chen, Jianqing Fan and Zhezhen Jin Information identities and testing hypotheses: Power analysis for contingency tables Philip E. Cheng, Michelle Liou, John A. D. Aston and Arthur C. Tsai Confidence intervals for the long memory parameter based on wavelets and resampling Pier Luigi Conti, Livia De Giovanni, Stilian A. Stoev and Murad S. Taqqu Classification efficiencies for robust linear discriminant analysis Christophe Croux, Peter Filzmoser and Kristel Joossens Tests for independence in nonparametric regression John H. J. Einmahl and Ingrid Van Keilegom Minimum distance inference in unilateral autoregressive lattice processes Marc G. Genton and Hira L. Koul A circular-circular regression model Shogo Kato, Kunio Shimizu and Grace S. Shieh Testing for threshold moving average with conditional heteroscedasticity Guodong Li and Wai Keung Li Spatial bootstrap with increasing observations in a fixed domain J. M. Loh and M. L. Stein Most robust BIBDs J. P. Morgan and Valentin Parvu A nonparametric empirical Bayes approach to joint modeling of multiple sources of genomic data Wei Pan, Keyong S. Jeong, Yang Xie and Arkady Khodursky Between-and within-cluster covariate effects and model misspecification in the analysis of clustered data Lei Shen, Jun Shao, Soomin Park and Mari Palta Asymptotically efficient product-limit estimators with censoring indicators missing at random Qihua Wang and Kai W. Ng Moderate deviations for stationary processes Wei Biao Wu and Zhibiao Zhao Detecting differentially expressed genes using calibrated Bayes factors Fang Yu, Ming-Hui Chen and Lynn Kuo === Subject: Solution manual of Real Analysis third edition by H. L. Royden posting-account=nt6tWgoAAACBTUCifRLyoX2ESsnRyn2Y Gecko/20080404 Firefox/2.0.0.6;MEGAUPLOAD 1.0,gzip(gfe),gzip(gfe) you very much === Subject: Re: x = [x] !!!! <16278803.1210080845384.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=euF15goAAACbw3KIqEWxZHCIPUc2KPmU .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506),gzip(gfe),gzip(gfe) >the foundational axiom of tommy set theory. > Before anyone can decide whether this has any interest > at all you need to specify the entire system. What are > all the axioms? Some of the axioms of TST are mereological rather than set theoretic. Zuhair, over at sci.logic, also sought to blend mereology and set theory. For more information, see: http://en.wikipedia.org/wiki/Mereology === Subject: Re: x = [x] !!!! <21897945.1210183860392.JavaMail.jakarta@nitrogen.mathforum.org> <87y76mht2i.fsf@phiwumbda.org> posting-account=euF15goAAACbw3KIqEWxZHCIPUc2KPmU .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506),gzip(gfe),gzip(gfe) > your argument that [] is one element is wrong, since [] has no > elements and [x] = x. > Oh, now I get it. æYou're silly. I once attempted to give a rigorous axiomatization of Tommy Set Theory, or TST. Let me go back to that axiomatization. According to tommy1729, the cornerstone of his theory is the axiom: Ax (x = [x]). Obviously, if we were to add this axiom to ZFC, the theory would clearly be inconsistent, for the reasons Jesse F. Hughes stated above. As I mentioned in one of the earlier tommy1729 threads, what is sought here is more like mereology than set theory.Over at sci.logic, Zuhair came up with a theory merging mereology and set theory. http://en.wikipedia.org/wiki/Mereology As we see from the Wikipedia link, the main primitive of mereology is not membership, as in ZFC, but parthood. Parts are more like subsets in ZFC than elements. An object with no parts other than itself is called an atom. So tommy1729's intended axiom holds for all atomic x, and is in fact the definition of atomicity. Both Zuhair and tommy1729 use the [] notation, rather than {}, to denote a fusion of atoms, which Zuhair calls a heap. In other threads, tommy1729 tells us that he wants the only infinite cardinalities in TST to be aleph 0, 2^aleph 0, and 2^2^aleph 0. So what we need in TST is to add enough mereological axioms to prove the existence of an object for every element of V (omega+2) of the cumulative hierarchy of ZFC. === Subject: Re: x = [x] !!!! Hi Amy, Funny but not easy!! Please take what follows as a first approximation only. To really get what I am talking about (due to my lack of formal skills I mean) you should actually go through my posts here on the forum (not many anyway, just look at my profile), and do most of the understanding effort yourself... > [...] However, at least in ZF set axioms we do not > distinguish elements of > set from the set. An element of set is a set on > its > own, so s is a > set. thus x = [x] I was actually *questioning* that notion, which -to me- is in a sense right (that is, what is an element in some context is a set in another, so taking the equal sign in a very abstract sense, maybe sort of a la Von Neumann), but at the same time is wrong unless -say- within the context of a theory of types, or by other means, like within the theory of closed systems I'm trying to shape. > [...] That sounds interesting, although I am going to > contradict it. Struggling a formalization: x = [x] > a > bit like: _element_ IS _set_ (or whichever > terminology you might wish there). as you say x = [x] , yes. which i said long ago here on sci.math ( and earlier > elsewhere ) But, I said I was going to contra-dict it; that is, that notion, without further qualification, or actually without a complete reversal of perspective, I believe is incorrect/contradictory/incongruent (sorry for the imprecision here). That IS may indeed be correct in x IS [x], it is not in [x] IS x. That is, IS cannot be equivalence, it may be inclusion, or it may be an is-a relation, like: a cat (necessarily) IS a feline, but a feline IS NOT (necessarily) a cat. Keeping in mind that elements in a set are > supposed > to be unique, an immediate consequence is that > either an element is in the set or is it not. The > _contra-diction_ is in the fact that non-belonging > is > recursive and belonging is not... from outside! what ?? 2 is in [2] , in fact its the same. 2 is not in [3] ... what recurvisive ? ... what from > outside ? Recursive there stands for recursively enumerable (r.e.) while not recursive from outside stands for a simply recursive relation that holds from within the given universe of reference, to predicate elements within that universe, but does not hold for elements supposedly outside that universe. Namely, the membership (belonging) relation cannot predicate the complement of a set from within the set itself, that is, when the set is taken as a universe on its own. Conversely, that task seems possible with a closed system, because that universe is constructed on the dis-belonging (exclusion rather than inclusion) relation, and so stays always constructive and closed: http://mathforum.org/kb/message.jspa?messageID=6199721 Now, I'll take it the other way round: elements in > a > set _do_ share a specific property, that being > that > they belong to *that* set. specific but not unique : 2 and 3 are in [2,3,5] and not in [5,7] but 2 is also in [2,7] and 3 also in [3] Yes, indeed I believe the notion of uniqueness is the problem here. That property, I have > (ultimately) called _exclusiveness_ (which, BTW, > sounds good enough to me) exclusive sounds like unique to me ... which i > countered above. http://mathforum.org/kb/message.jspa?messageID=6199721 , and it seems enough to: on > a side, found elements and sets, whatever they > are; > on the other, provide a (formal) definition of the > complement of a set, from within the set itself! That (I'd say) makes our objects self-contained > universes, at least as far as mathematics goes... By the notion of exclusiveness (and implying the notion of closed systems), we get for instance that the following definitions *cannot but* be as they are (follows an overview mainly): http://mathforum.org/kb/message.jspa?messageID=6133404 (with my short follow-up to that) http://mathforum.org/kb/message.jspa?messageID=6128731 (again, with my short reply) (And, from there, by mere induction, we get all the naturals.) There are actually other implied notions, like what I have stated as: among the possible systems, we cannot but elect *that* one as natural because of such and such exclusiveness (indeed this is the exclusiveness stipulation): http://mathforum.org/kb/message.jspa?messageID=6169005 After further struggling, we get to: expressibility as the only primitive, and exclusiveness as the only stipulation (and we still keep the system closed): http://mathforum.org/kb/message.jspa?messageID=6204754 Whatever! The point is, we can start from the above two notions of expressibility and exclusiveness in order to stay within a closed system which is recursive (r.e.) in any case, and where even Russel's paradox and Goedel-type arguments do not lead to incompleteness, or rather are not paradoxical anymore: http://mathforum.org/kb/message.jspa?messageID=6174975 Worth keeping in mind anyway is that this notion of closed system leads to a number system which exhibits quite uncommon properties in relation to the standard one, at least on the boundary cases. For instance, the notion of computability becomes coincident with the whole domain of tractability (although here we are going to my very first posts in the forum, so please bear with me there): http://mathforum.org/kb/message.jspa?messageID=6165721 http://mathforum.org/kb/message.jspa?messageID=6131417 http://mathforum.org/kb/message.jspa?messageID=6131384 Another rule I have found along the way is aleph cannot predicate omega, which resembles a type-like constraint, although this I still don't know how to incorporate into the above notions: http://mathforum.org/kb/message.jspa?messageID=6200977 http://mathforum.org/kb/message.jspa?messageID=6128704 That all said, please feel free to question/object, as at this point I don't even know if I have addressed your concerns at all. -LV -LV from the comments in between i reach another > conclusion. x = [x] [...] === Subject: Re: x = [x] !!!! amy666 a .8ecrit : > > amy666 a .8ecrit : > thus x = [x] 2 is an element of a set , the set [2] thus x = (x) ( () is same as [] ) just as i always said : x = [x] the foundational axiom of tommy set theory. with or without ZFC. without russel-like paradoxes. > Who needs russel-like paradoxes (whatever those > are) when we have > such obvious contradictions in TST? > Does TST prove that there is a set with no > elements? > Or that there is > a set with two elements? If TST proves either of > these things, then > TST is inconsistent, since you accept the claim > that > for all x, > x = {x}. > -- > Jesse F. Hughes > I often told you of the dangers of hubris, and > most > importantly of > all, I TOLD you that I wanted to change the > institution of mathematics > worldwide. -- James Harris, on the evils of > pride > lol , why the hell would that be inconsistant ??? here is a set with no elements : [] here is a set with two elements : [2,3] that was easy. happy now ? > So if x= [2,3], x=[x] means x has only one element , > namely x (or [2,3]) > ans one = two (lol, as you usally say) sure x is the element in [x]. thus we have the (single) element x. Lets call this line * now in your example x is a set [2,3] (or 2,3 if you like) therefore 2 and 3 are elements of [x]. Lets call this line ** this does not show 2 = 1 , this shows that x = [x] is consistant with subsets. indeed when [[x]] = x then x is its element and its subset. I have stopped answered you for a long time, but really, is it so hard to say something like well, on this particular point, I may have been not 100% right... and stop at that (intead of lolling everybody, I mean) ? Really, if x =[x], then if 2 and 3 are elements of [x] (by line **), they are elements of x (by line *) (unless your sign = is different of mine, of course), and x (and [x]) have at least two elements (namely 2 and 3, unless you believe 2=3, of course). Next, you say that [x] (and x, by line *) has (by definition) the only element x, which shows that one, namely x = two (at least), namely 2 and 3. And you still pretend not to see any contradiction, and be in earnest and not trolling? > or do you want me to list the elements of [] ? here it is : ( its empty remember ! ) > tommy1729 ps : keep quoting your example JSH, it makes a good > impression. > tommy1729 === Subject: Re: x = [x] !!!! amy666 a .8ecrit : > > amy666 a .8ecrit : > amy666 a .8ecrit : > thus x = [x] 2 is an element of a set , the set [2] thus x = (x) ( () is same as [] ) just as i always said : x = [x] the foundational axiom of tommy set theory. with or without ZFC. without russel-like paradoxes. > Who needs russel-like paradoxes (whatever > those > are) when we have > such obvious contradictions in TST? > Does TST prove that there is a set with no > elements? > Or that there is > a set with two elements? If TST proves either > of > these things, then > TST is inconsistent, since you accept the claim > that > for all x, > x = {x}. > -- > Jesse F. Hughes > I often told you of the dangers of hubris, and > most > importantly of > all, I TOLD you that I wanted to change the > institution of mathematics > worldwide. -- James Harris, on the evils of > pride > lol , why the hell would that be inconsistant ??? here is a set with no elements : [] here is a set with two elements : [2,3] that was easy. happy now ? > So if x= [2,3], x=[x] means x has only one element > , > namely x (or [2,3]) > ans one = two (lol, as you usally say) > sure x is the element in [x]. thus we have the (single) element x. > Lets call this line * > now in your example x is a set [2,3] (or 2,3 if you > like) > therefore 2 and 3 are elements of [x]. > Lets call this line ** > this does not show 2 = 1 , this shows that x = [x] > is consistant with subsets. > indeed when [[x]] = x then x is its element and its > subset. > I have stopped answered you for a long time, but > really, is it so hard > to say something like well, on this particular > point, I may have been > not 100% right... and stop at that (intead of > lolling everybody, I mean) ? > Really, if x =[x], then if 2 and 3 are elements of > [x] (by line **), > they are elements of x (by line *) (unless your sign > = is different of > mine, of course), and x (and [x]) have at least two > elements (namely 2 > and 3, unless you believe 2=3, of course). Next, you > say that [x] (and > x, by line *) has (by definition) the only element > x, which shows that > one, namely x = two (at least), namely 2 and 3. > And you still > pretend not to see any contradiction, and be in > earnest and not trolling? or do you want me to list the elements of [] ? here it is : ( its empty remember ! ) > tommy1729 ps : keep quoting your example JSH, it makes a > good > impression. tommy1729 you and jesse keep ignoring that you havent simplified the expression completely. [x] has element x > [2,3] has elements 2,3 if x = [2,3] then [x] has 1 element and if x =[x], so does x. This will be my last serious answer to you. You are a very pernicious troll and that 1 element is x or [2,3] ( same ! not 2 elements ) then we get 1 = 1 if we consider subsets or subelements then [x] has 2,3 then we get 2 = 2 the reason you arrive at your 2 = 1 is because you dont threath the left side equal to the right side ; you consider subsets on one side and not the other. therefore you dont have equality anymore. unless of course they are empty and add 0's to the equation. thus [[]] = [] then we get 0 = 0 if we also consider subsets , we get 0 + 0 = 0 similar [[[x]]] = [x] considering the subsets : 0 + 1 = 1 .. thus > for unordered sets > x = [x] > [a,b] = [b,a] = b,a = a,b > tommy1729 === Subject: Re: x = [x] !!!! posting-account=euF15goAAACbw3KIqEWxZHCIPUc2KPmU .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506),gzip(gfe),gzip(gfe) On May 7, 12:02æpm, Denis Feldmann indeed when [[x]] = x then x is its element and its subset. > I have stopped answered you for a long time, but really, is it so hard > to say something like well, on this particular point, I may have been > not 100% right... and stop at that (intead of lolling everybody, I mean) ? > Really, if x =[x], then if 2 and 3 are elements of [x] (by line **), > they are elements of x (by line *) (unless your sign = is different of > mine, of course), and x (and [x]) have at least two elements (namely 2 > and 3, unless you believe 2=3, of course). Next, you say that [x] (and > x, by æline *) has (by definition) the only element x, æwhich shows that > one, namely x = two (at least), namely 2 and 3. æAnd you still > pretend not to see any contradiction, and be in earnest and not trolling? Here tommy1729 is dealing with mereology, not set theory. It isn't trolling just because he wants to work with mereology rather than set theory. http://en.wikipedia.org/wiki/Mereology === Subject: My postings in the newsgroup Most of my debates on my postings in newsgroups rest on my issues of New Math. Until they read and practice (i.e. engage in actual computations) about ten of my lecture notes (each about 200 pages), it is really difficult to support (or reject) my statements based on just words. My other postings mainly have focused on strange behavior of the math community, which has intellectual roots in current American crisis (academic). So, we need to wait, until I write and disseminate my lecture notes, without any outside interference (like blockage of the math community, or any similar scenario like FLT). Dr.M.Basti === Subject: Re: My postings in the newsgroup > Most of my debates on my postings in newsgroups > rest on my issues of New Math. You'll probably want to call it something besides New Math, or everyone is going to think you're talking about something else. The fact that anyone needs to even point this out to you is another one of those red flags that led me to make some of my other comments in several of your threads. Dave L. Renfro === Subject: Re: Proof #2 To be divisible by 5, a number must end in 5 or 0. > Therefore, if we can show that (3^(3n+1) + 2^(n+1)) always > ends with 5 or zero, we have the necessary proof. > Fortunately, this is very easy. The proof was easy (for me) since I remembered that all integer > powers of 3 must end in 1, 3, 7 or 9. A quick check revealed > that all integer powers of 2 must end in 2, 4, 6 or 8. > The exponents of 2 and 3 automatically select the 4 desired cases. Prove it. I thought this was self evident, but I see your point. Anyway, The last digits of 3 ^ k are constrained to be 1, 3, 7 or 9. The last 3 > digits of 2 ^ p are 2, 4, 6 and 8. k and p are functions of n that must > make the pairings of {1,4}; {3,2}; {7,8} and {9,6} true for all n. k = > 3*n+1 and p = n+1 satisfy for n = 0, 1, 2 & 3. Because the last digits > of 3 ^ k repeat every 4th power as do those of 2 ^ p; it is intuitive > that the pairings are consistent for all n. But I cant prove this. > . Blow hard like the wind, ya land lubbing blow hard. > Mine works even if the expression is something like > [12783843 ^ (3n+1) + 234252 ^ (n+1)]! Does yours? Yes, immediately because ...3 and ...2 equal 3 and 2 (mod 5) Prove ii > You want me to repeat myself? If a = 3 (mod 5) and b = 2 (mod 5), then a^(3n+1) + b^(n+1) = 3^(3n+1) + 2^(n+1) = 3 * 27^n + 2 * 2^n = 3 * 2^n + 2 * 2^n = 5 * 2^n = 0 (mod 5) For example, a = -232908747 and b = 340982 > So, what is YOUR criteria for easiness? > Stop your off mouthing and show your work. What work? This is all in my head. The fact that I could do it in my > head is the reason that I thought it was easy. > Oh ya, you and Fermat with your thought proof too big for the margins of your mind. > BTW, in the US we say mouthing off. > They also say off shore when they mean shove off. For example, fair trade agreements shove off US workers. === Subject: Health is a state of complete physical, mental and social well-being and not merely the absence of disease or infirmity. [1] ... posting-account=PPv0cQkAAAACyhvqDsLZO3MLec3St06X EmbeddedWB 14.52),gzip(gfe),gzip(gfe) Health is a state of complete physical, mental and social well-being and not merely the absence of disease or infirmity. [1] ... http://www.freewebs.com/eyees/ === Subject: Re: n+n = n x n > On May 5, 3:03 pm, G. A. Edgar n + n = n x n or n squared > ex:- > 2 + 2 = 2 x 2 or 2 squared > 0 + 0 = 0 * 0 or 0 squared > Furthermore, there are no others besides 0 and 2. > The OP did not SAY real number or integers. There are other *cardinal > numbers* with n+n = n^2 ... > **************************************************** Yes indeed: any infinite cardinal. It is not obvious though that the > OP would hardly mean that? I mean, if someone's asking for help > solving the equation n^2 = 2n it seems the odds she can understand > cardinals and stuff are rather ridiculously low... > Ridiculously low, yes, maybe even ludicrously low ... but can you prove that the probability is zero?? === Subject: Re: More evidence proving Apollo Hoax > Compare the the below authentic pictures of the Lunar surface with the > fake ones from the Apollo hoax. > Tell me this then. What _exactly_ did happen? > It is telling that no one has answered this simple > question. If the mood landings were faked, how exactly > did they do it. This is the question the moon landing > conspiracy crackpots cannot answer. > Martin Hogbin > Finding one fault with the NASA/Apollo holy grail is all it takes. > We've found dozens of faults. Your being in denial of being in denial > isn't helping, is it. > I cannot be in denial - you have not given me anything > to deny. Denial of your denial is where you start to appreciate the greater > truth and consequences of your actions. What actions? Why will you not anser my question? Martin Hogbin === Subject: Re: More evidence proving Apollo Hoax posting-account=nf79RwoAAABXjvy5ztMzmPxgY1WGoktI Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) > Compare the the below authentic pictures of the Lunar surface with the > fake ones from the Apollo hoax. > Tell me this then. What _exactly_ did happen? > It is telling that no one has answered this simple > question. If the mood landings were faked, how exactly > did they do it. This is the question the moon landing > conspiracy crackpots cannot answer. > Martin Hogbin > Finding one fault with the NASA/Apollo holy grail is all it takes. > We've found dozens of faults. Your being in denial of being in denial > isn't helping, is it. > I cannot be in denial - you have not given me anything > to deny. Denial of your denial is where you start to appreciate the greater > truth and consequences of your actions. What actions? Why will you not anser my question? Martin Hogbin It can be proven, they had motive, means and opportunity. The naked (meaning unfiltered) Kodak eye doesn't lie, nor do the basic laws of physics. Their losing those 700 boxes of NASA/Apollo mission data and much other that's unaccounted for is simply icing on their LLPOF cake. . - Brad Guth === Subject: Re: More evidence proving Apollo Hoax The purpose of your reply eludes me. Try asking him the question that none of the moon landing conspiracy crackpots can answer; what exactly do you claim really happened and who was in on it? All I got was a few grunts in reply. Martin Hogbin === Subject: Re: More evidence proving Apollo Hoax posting-account=PTS84AoAAACr67p51zvy0Hlr3LkoIUcc InfoPath.1; .NET CLR 1.1.4322; .NET CLR 2.0.50727; .NET CLR 3.0.04506.30),gzip(gfe),gzip(gfe) > The purpose of your reply eludes me. Try asking him the question that none of the moon > landing conspiracy crackpots can answer; There is no such thing as conspiracy theory or crackpots. You are either wrong or right. Sometimes, it is subjective. Sometimes, it is going against what is normally believed. For example, we have this scenario where most people believe in the God of butterflies being supreme. Now, someone comes along and claims the God of spiders being the most supreme instead. This someone will be regarded as a crackpot and a conspiracy nutcase. do you claim really happened and who was in on it? You are avoiding the issue. Your question is a separate matter. It requires some investigation, donÍt you think? Have you lost your mind in scientific methodology? > All I got was a few grunts in reply. You have erroneously interpreted them as grunts. That is because of your strong BELIEF IN what you have been taught in the first place and also very strong peer pressures. If your peers believe in the God of butterflies, and you believe in the God of spiders, you are going to feel pressure from your peers. The subject on manned moon missions can only be decided with more scientific investigations. Unable to face any scientific inquiries into the matter, you have resorted to faith. You have called and tried to dismiss the ones bringing up the questionable events as conspiracy crackpots. You have lost your scientific methodology in doing so. You ought to be ashamed of yourself. Going back to the manned moon landing claim, the Apollo manned missions were the only ones that have gone beyond the low earth orbit in which the environment enjoys a relatively benign radiation of 10 RADs per year. The Apollo manned missions were also the only ones that have supposedly gone into the Van Allen Belts and beyond. The shuttle program only calls out for missions in the low earth orbit. With relatively and vastly more sophisticated hardware, the shuttle program is plagued with problems after problems while the Apollo missions have gone in perfect performance except the staged mission 13 which the problem miraculously happened on the 13th mission, the 13th hour, the 13th minute, and the 13th second. The Van Allen Belts represent deadly radiation zones that miraculous shielded all harmful solar and cosmic radiation from the living organisms of earth. Beyond these belts, the radiation is not as severe as in it. However, it is still deadly. The earlier deep space probes namely the Viking and the Mariner missions did not pay very much attention to the radiation of deep space. These probes did not last very long. However, Pioneer and Voyagers did have their electronics hardened for the deadly radiation. To give you an example, the geosynchronous satellites have to withstand a radiation of 300kRADs or more per year beyond the Van Allen Belts. The following link should cast no doubt on this radiation amount. Each year, the sun sheds some of its mass (2-3E-14 solar mass) with each proton having an average energy of 1keV. http://en.wikipedia.org/wiki/Solar wind That is why semiconductor vendors such as Intersil sell devices that can withstand some or all these radiation. Check them out. http://www.intersil.com/military/radhardlist.asp This is a scientific method of exposing Apollo manned missions being an elaborate hoax. What is your scientific method of countering this? None, so you have to resort to faith, right? === Subject: Re: More evidence proving Apollo Hoax posting-account=nf79RwoAAABXjvy5ztMzmPxgY1WGoktI Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) The purpose of your reply eludes me. Try asking him the question that none of the moon > landing conspiracy crackpots can answer; There is no such thing as conspiracy theory or crackpots. You are > either wrong or right. Sometimes, it is subjective. Sometimes, it is > going against what is normally believed. For example, we have this > scenario where most people believe in the God of butterflies being > supreme. Now, someone comes along and claims the God of spiders being > the most supreme instead. This someone will be regarded as a crackpot > and a conspiracy nutcase. what exactly > do you claim really happened and who was in on it? You are avoiding the issue. Your question is a separate matter. It > requires some investigation, donÍt you think? Have you lost your mind > in scientific methodology? All I got was a few grunts in reply. You have erroneously interpreted them as grunts. That is because of > your strong BELIEF IN what you have been taught in the first place and > also very strong peer pressures. If your peers believe in the God of > butterflies, and you believe in the God of spiders, you are going to > feel pressure from your peers. The subject on manned moon missions can only be decided with more > scientific investigations. Unable to face any scientific inquiries > into the matter, you have resorted to faith. You have called and > tried to dismiss the ones bringing up the questionable events as > conspiracy crackpots. You have lost your scientific methodology in > doing so. You ought to be ashamed of yourself. Going back to the manned moon landing claim, the Apollo manned > missions were the only ones that have gone beyond the low earth orbit > in which the environment enjoys a relatively benign radiation of 10 > RADs per year. The Apollo manned missions were also the only ones > that have supposedly gone into the Van Allen Belts and beyond. The > shuttle program only calls out for missions in the low earth orbit. > With relatively and vastly more sophisticated hardware, the shuttle > program is plagued with problems after problems while the Apollo > missions have gone in perfect performance except the staged mission 13 > which the problem miraculously happened on the 13th mission, the 13th > hour, the 13th minute, and the 13th second. The Van Allen Belts represent deadly radiation zones that miraculous > shielded all harmful solar and cosmic radiation from the living > organisms of earth. Beyond these belts, the radiation is not as > severe as in it. However, it is still deadly. The earlier deep space > probes namely the Viking and the Mariner missions did not pay very > much attention to the radiation of deep space. These probes did not > last very long. However, Pioneer and Voyagers did have their > electronics hardened for the deadly radiation. To give you an > example, the geosynchronous satellites have to withstand a radiation > of 300kRADs or more per year beyond the Van Allen Belts. The > following link should cast no doubt on this radiation amount. Each > year, the sun sheds some of its mass (2-3E-14 solar mass) with each > proton having an average energy of 1keV. http://en.wikipedia.org/wiki/Solar wind That is why semiconductor vendors such as Intersil sell devices that > can withstand some or all these radiation. Check them out. http://www.intersil.com/military/radhardlist.asp This is a scientific method of exposing Apollo manned missions being > an elaborate hoax. What is your scientific method of countering > this? None, so you have to resort to faith, right? It seems having motive, means and opportunity represents nothing to these status quo or bust kind of folks. Evidence exclusion is yet another one of their dirty tricks that goes along with their conditional laws of physics. Most everything except their fly-by-rocket landers and rad-hard astronaut had been resolved, just like today we still can't safely fly- by-rocket land anything, not even in any R&D prototype format. Having that 30% inert GLOW is another pesky item in question, since nothing since has ever come remotely close to that kind of mission performance with such a great deal of inert/payload mass to deal with. . - Brad Guth === Subject: Re: More evidence proving Apollo Hoax The purpose of your reply eludes me. > Try asking him the question that none of the moon > landing conspiracy crackpots can answer; > > what exactly > do you claim really happened and who was in on it? You are avoiding the issue. No, that is the issue. If you cannot say what you believe was actually done you have no case. > > All I got was a few grunts in reply. You have erroneously interpreted them as grunts. They were not answers to my question. Neither is your reply - it is pure bluff and bluster. > That is because of > your strong BELIEF IN what you have been taught in the first place and > also very strong peer pressures. I am waiting for you to set me straight. What really happened in your opinion. Martin Hogbin === Subject: Re: More evidence proving Apollo Hoax | > | > The purpose of your reply eludes me. | | Try asking him the question that none of the moon | landing conspiracy crackpots can answer; what exactly | do you claim really happened and who was in on it? | | All I got was a few grunts in reply. | | Martin Hogbin Why did Einstein say the speed of light from A to B is c-v, the speed of light from B to A is c+v, the time each way is the same? That's the question that none of the relativist crackpots can answer; what exactly do you claim really happened and who was in on it? All I get is a few grunts in reply. Right, ignorant piggy? === Subject: Re: More evidence proving Apollo Hoax posting-account=nf79RwoAAABXjvy5ztMzmPxgY1WGoktI Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) The purpose of your reply eludes me. Try asking him the question that none of the moon > landing conspiracy crackpots can answer; what exactly > do you claim really happened and who was in on it? All I got was a few grunts in reply. Martin Hogbin You got a whole lot more than grunts, my brown-nosed friend. Sadly, it seems you and others of your kind can not deductively put two and two together without first checking your NASA/Apollo Qur'an to see what's scripted. What brown-nosed part in the NASA/Apollo ruse did you play? . - Brad Guth === === Subject: Re: Cech cohomology and singular cohomology posting-account=R7AgUAoAAADVFAtIe36IBmgohoHjZsKW Gecko/20061201 Firefox/2.0.0.13 (Ubuntu-feisty),gzip(gfe),gzip(gfe) Is there an easy way to see directly that singular cohomology for topological spaces is just a special case of cech cohomology? If you take the the sphere X=S^2 this admits the structure of a simplicial complex in the usual way: Two 2-simplices are glued together at their boundaries. To get singular cohomology one builds up a complex 0-->D0(X;G)-->D1(X;G)-->D2(X;G)-->0 where G is an abelian group and Di(X;G) denotes the abelian group of continuous functions from the i-simplices of X to G. The differentials are build in the obvious way. The homology of the complex is defined as the singular cohomology of X. To get cech cohomology you need a presheaf F:(open sets of Y)^op-- >(Abelian Groups) for example and a covering {U_i} of Y. Can this be choosen from the above data to get a cech cohomology which is the same as the above singular one? > For good spaces, yes. Is it easy to see on the above example that singular cohomology of X can be constructed as some sheaf cohomology, too? S. === Subject: Re: Cech cohomology and singular cohomology posting-account=9QOSvAoAAACEOWJVSDuswW7dB_0wApQO Gecko/20080416 Fedora/2.0.0.14-1.fc8 Firefox/2.0.0.14,gzip(gfe),gzip(gfe) Is there an easy way to see directly that singular cohomology for > topological spaces is just a special case of cech cohomology? Cech cohomology (of constant sheaves) and singular homology are *different* for some spaces. You need to restrict to, for example, paracompact spaces in order to have the two be canonically isomorphic. Since most usual spaces are paracompact (eg, CW-complexes) or have the homotopy type of paracompact spaces (like nice map spaces) this is not a bad restriction. IIRC the `long line', which is not paracompact, is an example where the two comohologies are different. > If you > take the the sphere X=S^2 this admits the structure of a simplicial > complex in the usual way: Two 2-simplices are glued together at their > boundaries. As we've discussed a while ago, that is *not* a simplicial structure on the sphere, only a CW structure. In a simplicial decomposition (a triangulation), the simplices either are disjoint or intersect in exactly one of their faces. To compare Cech with singular cohomology, you need to deal with simplicial homology. > To get singular cohomology one builds up a complex > 0-->D0(X;G)-->D1(X;G)-->D2(X;G)-->0 > where G is an abelian group and Di(X;G) denotes the abelian group of > continuous functions from the i-simplices of X to G. The differentials > are build in the obvious way. The homology of the complex is defined > as the singular cohomology of X. To get cech cohomology you need a presheaf F:(open sets of Y)^op-->(Abelian Groups) for example and a covering {U_i} of Y. Can this be choosen from the above data to get a cech cohomology which is the same > as the above singular one? In order to get singular cohomology from Cech cohomology for, say, a paracompact space, you have to consider the constant sheaf F with fibers isomorphic to Z, the group of integers. > For good spaces, yes. Is it easy to see on the above example that singular cohomology of X > can be constructed as some sheaf cohomology, too? Identify the sphere S^2 wth the boundary of the standard 3-simplex Delta sitting in R^3. This gives a triangulation of S^2, with 4 2-simplicies, 6 1-simplices and 4 0-simplices. If sigma is one of those simplices, the *star* of sigma is the union of the interiors of the simplices which contain sigma. This is of course an open set. Moreover, the set of the stars of all vertices in the triangulation is a covering C. It is not difficult to see that any finite intersection of members of C is contractible. This means that C is a *good* cover (notice that if you take the cellular decomposition you wanted to take, this does not work) Construct the Cech complex corresponding to the open cover C and the contant sheaf F. Can you see why it gives the singular cohomology of S^2? (Of course, in order to compute Cech cohomology of S^2 one has to consider *all* open covers and take the limit. But one can show that if there is one ood cover (as C above) then one can simply do the computation for that cover, since that will coincide with the limit. One way to see that this works is (this is not a proof, though): One can show that the nerve of a good cover of a space is homotopically equivalent to the space itself, and the cech complex for that good cover and the constant sheaf F will be the same as the complex which computes simplicial cohomology for the nerve.) -- m === Subject: Re: divisibility test for 7 #> #> #> #> #> #> #> #> #> I have collected some nice divisibility tests for the number 7 and #> have aggregated them in the form of a blog. #> #> It is available here.. #> #> >http://thinktibits.blogspot.com/2008/04/yet-another-divisibility-by-7... #> #> to those who are interested. #> #> #> The usual problem with divisibility tests for 7 is that the combined #> cost of remembering the test and applying it is higher than that of #> simply dividing by 7 and looking at the remainder. Your latest #> produces the sequence #> #> 857864 8*3+5=29 #> 297864 2*3+9=15 #> 157864 1*3+5= 8 #> 87864 8*3+7=31 #> 31864 3*3+1=10 #> 10864 1*3+0= 3 #> 3864 3*3+8=17 #> 1764 1*3+7=10 #> 1064 1*3+0= 3 #> 364 3*3+6=15 #> 154 1*3+5= 8 #> 84 etc #> #> but just stripping out multiples of 7 from the left gives remainders #> of #> #> 857864 -700000 #> 157864 -140000 #> 17864 - 14000 #> 3864 - 3500 #> 364 - 350 #> 14 - 14 #> 0 #> #> and also the quotient 122552. I know which is easier to remember. #> #> But for this particular number 857-864 = -7 ... bingo! #> #> -- #> Clive Toothhttp://www.shutterstock.com/cat.mhtml?gallery_id=61771- Hide quoted text - #> #> - Show quoted text - # #If you can be bothered to remember that it is 1001 which is a multiple #of 7 rather than say 101 or 10001. Hmmm.... reminds me of a post I made to alt.stupidity a couple of years ago... -- Clive Tooth http://www.shutterstock.com/cat.mhtml?gallery_id=61771 === Subject: Re: True Philosophers and Thought *bicker-bicker-bicker* > You two - Head Teacher's study, NOW! And DON'T RUN! regarding the analysis and demonstration of truth? Yup. My contribution is this: comments like:- In other words you're just a common fraud. You're getting way too bitchy to carry on a polite conversation. Go browbeat some students. Well if you consider your imitation of science a parody . . .What I can't understand is why you even bother. In udder words you're a stupid cow. Just as you make a parody of science. ...do nothing to advance the analysis and demonstration of truth. And I hold _that_ truth to be self-evident. Jon -- with 'green-lines'. === Subject: E = m c^2 and Riccati posting-account=ee2apQoAAABJNMlLTFasCJw9Nfo9FmYk Gecko/20020924 AOL/7.0,gzip(gfe),gzip(gfe) HTTP/1.1 cache-ntc-ab08.proxy.aol.com[CFC87448] (Traffic-Server/6.1.5 [uScM]) Mass and energy are two major components of the universe. Thus the equation, E = m c^2 represents one of the important formulas in the universe (within the domain of physics). As I had mentioned before the abstract formula of Riccati differential equation dx/dt + x^2 = g(t) is another important one (within the domain of mathematics). There is no doubt that, from the mathematical equations, we can always conclude about the behavior of the physical world (and visa versa). Thus eventually questions will arise from one formula to the other, about inner characteristics of both of them. It remains to be seen, how the future researchers in both domain determine the outcome of their comparison studies? Dr.M.Basti === Subject: Re: A consideration concerning the diagonal argument of G. Cantor <30676305.1210179264768.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=-eQqtQoAAACZVM-kNEsOn3k7GSvoJoS4 MathPlayer 2.0; .NET CLR 1.1.4322; InfoPath.1; .NET CLR 2.0.50727; .NET CLR 3.0.04506.648; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) > Not sure what you are after. (Maybe it is a matter of > poor wording, as > you suggest.) I have tried to make this notion as precise as possible within the following thread, where you'll find a link to the original discussion about this problem: http://mathforum.org/kb/message.jspa?messageID=6174924&tstart=0 To restate it here: [Within ZF?] the two notions | and ~A&A are equivalent and there is no way to distinguish contradiction from falsity (or rather, incongruency from falsity: see below). In other words, you don't believe the Law of the excluded middle or Tertium non datur. That is fine; I am told there are theories that can be built on multivalued logic, but I know very little about that. I am therefore the wrong person to continue that discussion. As an application to the case of WM, however, it is quite misguided. Whether you accept two-valued logic or not, an axiom system that can (correctly!) derive two different truth values for the same statement is inconsistent. WM claims that ZF is inconsistent, yet his derivations are (and have been shown repeatedly to be) utterly wrong. This is why I am quite taken aback and surprised about your follow-on quote below: > To be a little bit more honest, and quite a bit less shy, I must admit that at the moment, > given the line of reasoning below, I am quite convinced about the incongruency of /modern > axiomatic set theory/ as well as its underlying /logical theory/, until contrary evidence > emerges. Care to elaborate? If you have any questions about any of the so-called proofs that WM peddles, I (and others in sci.math, I am sure) would be more than happy to explain to you why he is wrong (and ZF *IS* consistent). === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > On May 7, 5:53æpm, Julio Di Egidio Not sure what you are after. (Maybe it is a > matter of > poor wording, as > you suggest.) I have tried to make this notion as precise as > possible within the following thread, where you'll > find a link to the original discussion about this > problem: > http://mathforum.org/kb/message.jspa?messageID=6174924 > &tstart=0 To restate it here: [Within ZF?] the two notions > _|_ and ~A&A are equivalent and there is no way to > distinguish contradiction from falsity (or rather, > incongruency from falsity: see below). In other words, you don't believe the Law of the > excluded middle or > Tertium non datur. That is fine; I am told there > are theories that > can be built on multivalued logic, but I know very > little about that. > I am therefore the wrong person to continue that > discussion. Yes, it might have to do with multivalued logic, and maybe even with modal logic. I am more than everything interested into the analysis of natural language by formal means (my point of reference for that is Strawson, although he surely does not put it this way). Other interesting things are coming out from another thread on set theory, namely Quine's atoms, and Mereology, where parthood is the main primitive rather than membership: http://mathforum.org/kb/thread.jspa?threadID=1736621 In fact, I have myself tried to overcome that notion of membership by my (hyperlogical/analogical?) notion of dis-belonging. Anyway, set theory as well as formal logic are mainly a field of study and great interest for me at the moment, given that, for a living, I do something else: software development. As an application to the case of WM, however, it is > quite misguided. > Whether you accept two-valued logic or not, an axiom > system that can > (correctly!) derive two different truth values for > the same statement > is inconsistent. WM claims that ZF is inconsistent, > yet his > derivations are (and have been shown repeatedly to > be) utterly wrong. > This is why I am quite taken aback and surprised > about your follow-on > quote below: To be a little bit more honest, and quite a bit > less shy, I must admit that at the moment, > given the line of reasoning below, I _am_ quite > convinced about the incongruency of /modern > axiomatic set theory/ as well as its underlying > /logical theory/, until contrary evidence > emerges. Care to elaborate? I am very well aware that what I am trying to say may have little to no bearings to WM's arguments (although I find his arguments of interest, given that I too feel that there is something wrong in the usually accepted picture). Indeed, what I was trying to say above, supported by the quotes from Strawson and Wittgenstein you have snipped, is that it might very well be wrong to look for an *inconsistency* in ZF (or, as I get it, in modern axiomatic set theory: let's name it MAST here, for the sake of brevity). Instead, what I am saying is that MAST is *incongruent*!! And, that implies that it might be wrong although consistent and pseudo-self-contained: Contradictory assertions have then the character of being both logically exclusive and logically exhaustive. (Strawson) Because the totality of facts determines all that happens, and also all that does not happen. (Tractatus, 1.12) In fact, by Goedel, MAST is always incomplete, while the closed systems approach I am trying to devise does not suffer any kind of limitation and is actually able to exhaust the domain of tractability. Moreover, there are subtler - but crucial! - implications to a theory of natural languages, and - a fortiori - to a theory and practice of legal languages!! (And then you see how the whole thing becomes delicate and not only crucial.) Within /my/ approach, a formal theory of natural language seems, and strongly seems, to be possible, while all this stays forever unreachable to any MAST. Hope this clarifies a bit, at least as to what I am after. -LV If you have any questions about any of the so-called > proofs that WM > peddles, I (and others in sci.math, I am sure) would > be more than > happy to explain to you why he is wrong (and ZF *IS* > consistent). === Subject: Re: A consideration concerning the diagonal argument of G. Cantor <6967258.1210256984132.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=-eQqtQoAAACZVM-kNEsOn3k7GSvoJoS4 MathPlayer 2.0; .NET CLR 1.1.4322; InfoPath.1; .NET CLR 2.0.50727; .NET CLR 3.0.04506.648; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) [...snip...] > Indeed, what I was trying to say above, supported by the quotes from Strawson and Wittgenstein you have snipped, is that it might very well be wrong to look for an *inconsistency* in ZF (or, as I get it, in modern axiomatic set theory: let's name it MAST here, for the sake of brevity). Instead, what I am saying is that MAST is *incongruent*!! And, that implies that it might be wrong although consistent and pseudo-self-contained: Contradictory assertions have then the character of > being both logically exclusive and logically > exhaustive. (Strawson) Because the totality of facts determines all that > happens, and also all that does not happen. > (Tractatus, 1.12) I am still not sure what you mean by incongruent. And as for a set of axioms being wrong, well... They might not be useful, as in, lead to interesting mathematics (or applications), and they could, of course, be inconsistent. But just because there is Riemannian and Lobachevskian geometry, I wouldn't go calling Euclid's axioms wrong. (Just to give a concrete example.) So I am still puzzled by what you say. > In fact, by Goedel, MAST is always incomplete, while the closed systems approach I am trying to devise does not suffer any kind of limitation and is actually able to exhaust the domain of tractability. If I understand this correctly, then your proposed system must, by necessity and Goedel, be rather weak. But, whatever floats your boat. Propose a set of axioms and derive some results from them. If your theory is consistent and sufficiently interesting (whatever that means), you may actually be onto something. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > WM then tries to prop up his faulty notions of inconsistency with > proofs, but these proofs are so ludicrously flawed, they are not > worth anybody's time. (Admittedly I have learned a few things from the > discussions around them, which is my justification of toughing it > out.) Although WM'S mathematical inabilities are already quite well known in > this group, I am wondering about the patient fruitless attempts to > convince him. Nobody answering WM's assertions thinks that WM will be convinced. > Of course I am still shocked, that he is permitted to teach > mathematics on a german Fachhochschule (some kind of University). There is no accounting for taste. I wonder if the rest of the Fachhochschule is aware of WM's convictions? -- Michael Press === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > ... > æ> I'm going to ask you one last time: Do you believe that > æ> the axioms of set theory are inconsistent? > æ æ> Yes, they are. > æ æ> æIf so, show > æ> me a statement S such that ZF proves S and ZF also proves > æ> the negation of S. Is there such an S, or not? > æ> There is the statement that a set of infinitely many FISONs is an > æ> actually infinite segment called omega. Right. Not exactly. It is the UNION of a set of infinitely many fisons that is omega (or N or alpha_0), not just the set of fisons. Even that omega is a set of infinitely many FISONs that are smaller > than omega. It is the UNION of a set of ininfinitely many fisons. > And even that omega is a set of infinitely many FISONs that are > smaller than other FISONs. It is the UNION of a set of ininfinitely many fisons. And those infinitely many can be chose so that each of them is smaller than one not included in the union. E.g., the set of those fisons ending in an odd natural is such a set whose union is omega even though for each fison in the set there is a large fison not in it. æ> There is the statement that for every FISON there exists a larger > æ> FISON. And one can easily prove by induction that *every* FISON of the > æ> alleged set can be removed without changing the covering properties of > æ> that set. One can not prove that within that *each* FISON can be removed, but > not that *all* FISONs can be removed. Then remove the set of all FISONs that provably can be removed. That presumes that there is only one such maximal set. One can slit things into the set of even ended fisons versus odd ended fisons, and each set is equally sufficient to cover N when the other is removed, but remove their union and nothing is left to cover N. Wm has no imagination. What he sees must hold for finite sets, he cannot imagine not holding for infinite ones. But if they ere not different, they would not be different. > The > set of all FISONs that are less than omega. The set IS, not the set are. Also, the set of fisons is quite different from omega, unless one is using a von Neumann model is which fisons are the same as naturals. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > ... > æ> æ> Omega is the first transfinite number. 0 is the first number that > æ> æ> cannot be expressed by 1/n. > æ æ> Yes, and 0 is not in the union of [1/n, 1] and omega is not in the > union > æ> of {1, ..., n}. æNow what? > æ æ> [0, 1] is not the union of all intervals[1/n, 1]. The same is true for > æ> the reciprocal values. Right, {1, ..., omega} is not the union of all FISONs. æThe union of > all FISONs is {1, ...}. omega = {1, ...} is larger than every FISON. > The union of FISONs is a FISON. In particular it is not open. If it is a fison, then, by definition of fisons, it has a largest natural which must be larger than any other possible natural as all other fison must be subset of this largest fison. But for every fison there is a successor fison, which is larger. So that WM's claims lead directly to a self-contradiction. > omega = {1, ...} is not the union of all FISONs. If that were the case, then either WM should be able to find a member of omega which is not in any fison, or a member of some fison which is not in omega. And until WM can do one or the other, we will justifiably continue to believe that omega (or N or aleph_0) IS the union of all fisons. I have still to see a definition (within set theory) of potential > infinitude. Within set theory, potential infinity is only used when contradictions > could uncover the inconsistency of set theory But it is denied on > other occasions. As we have seen no actual self-contradictions in ZF, there is no need for us to fall back on such a self-contradictory notion as potential infinitude. For instance: > Pot. Inf. Complete induction is valid for every natural number n, but > it does not cover the set of all natural numbers. > Act. Inf. Cantor's diagonal proof is valid for every entry n, and it > covers the set of all entries. In principle there is no reason why both cases should yield different > infinities. But otherwise set theorists would loose their pet toy. It is WM's pet toys which are at risk, for until WM can provide a statement which can be proved within ZF and whose negation can also be proved within ZF, we see no reason at all to give up ZF in favor of WM's dubious toys. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > It appears to me as if you have not yet understood. It is WM who does not understand. WM claims that we cannot think about things that we do think about WM claims that we cannot formulate ideas which we have formulated. WM claims that numbers which do not have physical existence do not exist, but cannot present us with anything physical but the names of some numbers without ever showing us any physical existence of any number. There is no > infinitistic mathematics. There is at most the illusion of it. It is our illusions about those illusions that are what ultimately allow us to build very concrete bridges. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > I do not now state that, but it only has become obvious by means of > > FISONs. I knew it for long. But earlier in this thread you have stated that you accepted it. I assumed it in order to contradict it. At which, outside your mytheology, you have failed. > But, of course, if you believe in unions > > of linear sets that can be larger then all of their elements, then > > there is no help. Well, look at the union of the sets [1/n, 1], which is larger than all > of the sets used in the union. No, it is not. A linear union cannot be larger than every element. It can outside of WM's mytheology. it is only finite linear unions which cannot. If a set of sets has a largest member by inclusion, then its union is that member. If a set of sets has no largest member by inclusion then its union cannot be a member, at least outside of WM's mytheology. So that when WM claims A linear union cannot be larger than every element. he is of necessity also claiming that every set of sets ordered by inclusion has a largest member. Thus in WM's world, every set of fisons has a largest, from which it necessarily follows that every set of naturals has a largest member. These assumptions are false in ZF and most other commonly used set theories. > > But your approach shows that omega is not > > the union of all FISONs. Why not? Because there is no union of FISONs which is an open interval. There is no fison which is a closed interval without a topology. What topology does WM claim? He wosn't say, or will he say why any topology is relevant at all. Right, and there is no such claim. You claim that a set of natural numbers is open. We do not claim either openness or closedness, as these are irrelevant topological properties of sets. > That means that the > elements have to loose their properties. Topological properties only apply to sets, not to elements, so those elements cannot lose what they have never had. > Compare the > > infinite binary tree where all we have access to shows that there are > > more odes than paths. Again some undefined notion: all we have access to. All subtrees that can be reached by induction. Subtrees? Since they are all finite, they are all irrelevant. > Potential infinity does not suffer from quantifier dyslexia. But those who rely on it do. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor æ> æ æ æ æ æ æ æ æ æ æ æ æ æ I used his understanding of transfinite > æ> numbers in a discussion with Aatu Koskensilta in order to stress the > æ> fact that natural numbers are numbers that have some properties which > æ> they do not loose even when appearing in an infinite frequency. No. æYou used the term integer without specifying that you meant to > include also the transfinite integers. æSo you were misleading the > person you were discussing with. Wrong. I said: Remember: Omega or aleph_0 is an integer larger than > every natural. It is this statement which I oppose to. While Omega is a number of a sort, it is improper to call it an integer, at least in English. It is an ordinal number, but not a natural number and not an integer and not a rational and not a real and not a lot of other sorts of number. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > Wrong. I said: Remember: Omega or aleph_0 is an integer larger than > every natural. It is this statement which I oppose to. While Omega is a number of a sort, it is improper to call it an integer, > at least in English. It is improper to call it an integer in German too. But if omega was the number of naturals, then it could not avoid to be an integer. (Unless set theorists found out that there are fractional naturals, if they only stand up in infinite frequency. I would not be surprised about such a claim.) > It is an ordinal number, but not a natural number > and not an integer and not a rational and not a real and not a lot of > other sorts of number.- Omega is not a number at all, in the firts place, because there is no number of natural numbers. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=yKimjgoAAACk5WwPVD4l9HmbpoR6Hmy4 Gecko/20071127 Firefox/2.0.0.11,gzip(gfe),gzip(gfe) Wrong. I said: Remember: Omega or aleph_0 is an integer larger than > every natural. It is this statement which I oppose to. While Omega is a number of a sort, it is improper to call it an integer, > at least in English. It is improper to call it an integer in German too. But if omega was > the number of naturals, then it could not avoid to be an integer. > (Unless set theorists found out that there are fractional naturals, if > they only stand up in infinite frequency. I would not be surprised > about such a claim.) Oh, you stole this one from Tony Orlow. If we call omega just the ordinal figtree of the naturals do all your complaints vanish? > It is an ordinal number, but not a natural number > and not an integer and not a rational and not a real and not a lot of > other sorts of number.- Omega is not a number at all, in the firts place, because there is no > number of natural numbers. Can I define the ordinal figtree of the natural numbers instead then? It is exactly the same as omega but I won't call it a number, because that unsettles you so. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > æ> Yes, all sets are united at once, in a single step. æThere is *no* > æ> repetition neither in time nor in something else. > æ æ> Have you ever done some mathematics by yourself or by means of a > æ> computer? I mean real mathematics, not ZF pipe-dreams. Yes, I have done so. æBut the mathematics that I have done (factorisation > of large integers, primality proving, eigenvalue and singular value > calculation, design of routines to calculate special function, and > quite a few other things) were all based on standard mathematics with > algorithms designed to take care of the problems finite mathematics > pose. And that all was done in zero time? æ> And besides you should choose your wording accordingly. Sets are > æ> united at once has a bad temporal taste. You should say the union of > æ> the sets is considered. No, there is no consideration. So you consider nothing? Or do you consider everything simultaneously? æ> The infinite union does not change the quality of the sets that are united. > æ> I have no idea why you would think so. æIt is their *union* that has a > æ> quality different from the quality of the sets being united. Just as > æ> union [1/n, 1] = (0, 1]. > æ æ> Why don't you say that the union is [0, 1]? Or even [-1, 2]? If the > æ> quality of the union is in fact so independent from the properties of > æ> the sets united? Because from the definition of union it follows that the union is (0, 1]. > But you do not know that definition. æBut learn a bit of topology. æThe > *finite* union of closed sets is a closed set. æThe arbitrary union of > closed sets is not necessarily a closed set. That is taken from set theory. It cannot be used to discern whether set theory is inconsistent. The union of FISONs is a FISON. æ> æ> And because each number from the list differs in a finite position > æ> æ> from the diagonal, you do not have to manage other than a finite > æ> æ> number of digits. > æ> æ æ> æ> That means that you can do the complete diagonal number by a finite > æ> æ> number of digits? > æ æ> Who is saying that? > æ æ> You: And because each number from the list differs in a finite > æ> position from > æ> the diagonal, you do not have to manage other than a finite number of > æ> digits. Yes. For each comparison you do not have to manage other than a finite > number of digits. But there are infinitely many comparisons. Nevertheless you have to manage always the same number of digits? And in total you have to manage also only a finite number of digits? Strange. æ> This shows the basic contradiction of set theory. Of course every > æ> number differs only in a finite position from the diagonal. You have > æ> to manage only a finite number of digits. But there are infinitely > æ> many numbers. And for each one there is one position more to do. No. æYou *prove* that the comparison will yield inequality for *each* > entry in the list. æYou do not need to do infinitely many comparisons. I prove that a FISON is not sufficient to cover omega. For each FISON. Why do I need to do more? > æ æ> Not in the same way. As long as only natural numbers are concerned, > æ> their FISONs and unions of FISONs are given by closed intervals. That > æ> is the property of a natural number that must be violated when > æ> striving for finished infinity. Sorry. It is *not* a property of natural numbers. æYou have to *prove* > it. It is a property of a natural number that is closes a FISON. It is a property of all natural numbers, that they close FISONs. There is no natural number that requires an open set. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > æ> The infinite union does not change the quality of the sets that are > united. > æ> I have no idea why you would think so. æIt is their *union* that has a > æ> quality different from the quality of the sets being united. Just as > æ> union [1/n, 1] = (0, 1]. > æ æ> Why don't you say that the union is [0, 1]? Or even [-1, 2]? If the > æ> quality of the union is in fact so independent from the properties of > æ> the sets united? Because from the definition of union it follows that the union is (0, 1]. > But you do not know that definition. æBut learn a bit of topology. æThe > *finite* union of closed sets is a closed set. æThe arbitrary union of > closed sets is not necessarily a closed set. That is taken from set theory. It cannot be used to discern whether > set theory is inconsistent. The union of FISONs is a FISON. Certainly nothing from OUTSIDE set theory, except the rules of deductive logic, can be used to discern whether set theory is consistent. Consistency is an internal property of a system of axioms. What may occur outside that system is totally irrelevant to its consistency. That WM is ignorant of this is another mark of his complete and total incompetence to even discuss the consistency of systems of axioms. No. æYou *prove* that the comparison will yield inequality for *each* > entry in the list. æYou do not need to do infinitely many comparisons. I prove that a FISON is not sufficient to cover omega. For each FISON. > Why do I need to do more? You do not, and can not, prove that an infinite set of fisons does not cover N, because it always does. > æ æ> Not in the same way. As long as only natural numbers are concerned, > æ> their FISONs and unions of FISONs are given by closed intervals. That > æ> is the property of a natural number that must be violated when > æ> striving for finished infinity. Sorry. It is *not* a property of natural numbers. æYou have to *prove* > it. It is a property of a natural number that is closes a FISON. It is a > property of all natural numbers, that they close FISONs. There is no > natural number that requires an open set. I have a feeling that what WM means by open and closed are endless and ending, which is just his perpetual assumption that all sets are finite dressed up in new clothes. But in ZF, that is false, and may not be assumed Within any axiom system, one is not allowed to make any assumptions at all other than those explicitly made in the axioms, or derivable from the axioms by means of formal proofs. So that WM's open versus closed are meaningless in ZF until they have been clearly defined within ZF. So, WM, what are your definitions for a set being closed or being open? Until you give a definition of them valid in ZF, they are meaningless within ZF, and thus of no use in analyzing ZF. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > æ> ... > æ> æ> The natural numbers count themselves. n is the n-th number. > æ æ> What has this to do with omega? æOmega is not a natural number. > æ æ> And it is not the number of natural numbers. So, what is the number of natural numbers? There is no number of all natural numbers. That's because their > infinity is potential. That depends. In our world, in which axiom systems make sense, there are actual infinities in some of them and a number of the set of all naturals. In WM's world, everything is potential but nothing is actual, as things flicker into and out of existence haphazardly. > === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > If complete induction is not in ZF, then I am not interested in ZF. If ZF, or something like it, is not in WM's mytheology, then we are not interested in the vagaries of that mytheology. Formulate the above in the language of set theory > and prove it using the axioms of ZF. If these statements can be formulated in ZF, then ZF is inconsistent. Well, the last statement is the one that makes no sense, It makes no sense to say that a FISON is a subset of omega? It may not in WM's mytheology, but it does in ZF. > It makes no sense to say that omega FISON is not empty? It may not in WM's mytheology, but it does in ZF. > It makes no sense to claim that complete induction is not valid for > every natural number? Until your notion of complete induction has been stated clearly, it make no sense to me, because in the past you have claimed that complete induction claims things that the standard finite induction of ZF does not claim. You are joking! You are the joke. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > ... > æ> æ> æ æ æ æ æ æ æ æ æ æ æ æ æ I used his understanding of transfinite > æ> æ> numbers in a discussion with Aatu Koskensilta in order to stress the > æ> æ> fact that natural numbers are numbers that have some properties which > æ> æ> they do not loose even when appearing in an infinite frequency. > æ æ> No. æYou used the term integer without specifying that you meant to > æ> include also the transfinite integers. æSo you were misleading the > æ> person you were discussing with. > æ æ> Wrong. I said: Remember: Omega or aleph 0 is an integer larger than > æ> every natural. It is this statement which I oppose to. So you were misleading your opponent when you did not specify that you > meant to include also the transfinite integers. Larger than every natural is the same as transfinite. >æThat is why you got > the response: > æ> Very sensible of you. It is indeed not the case that omega or aleph 0 > æ> is an integer larger than every natural. No this response showed that Aatu was not aware of the character of transfinite numbers. > And so I was not wrong in my statement. You were as wrong as one can be. An integer larger than every natural is the same as a transfinite integer. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > An integer larger than every > natural is the same as a transfinite integer. Which is not the same as a transfinite number. For most standard definitions of integer they are finite, but for many uses of number there are transfinite ones. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > ... > æ> æ> æ> No. æWhen you give me an entry I have to compare to that one > æ> æ> æ> only, and can show it is different. æOnly one comparison > æ> æ> æ> needed. æHowever, that is easy to do for each and every > æ> æ> æ> entry you give to me, it holds for all entry. > æ> æ> æ æ> æ> æ> For all that I can give you! But as you know, I can give you only > æ> æ> æ> a finite number of entries. An irrational number however is said > æ> æ> æ> to have an actually infinite sequence of digits. > æ> æ æ> æ> So you also do think that the decimal representation of 1/3 does not > æ> æ> consist of an infinite number of threes, but only threes at those > æ> æ> places that I indicate to you, i.e. finitely many places? æOr how > æ> æ> do you otherwise prove that all digits are 3? > æ> æ æ> æ> That is easy because also at positions which cannot be identified, the > æ> æ> digit is 3. > æ æ> I asked for a proof not for an assertion. > æ æ> The property of having 3 at each position is a definition. And the property of having a particular number at each position (as in > Cantor's diagonal) is a definition. æSo why do you allow it here, but > not with Cantor's diagonal? Because a particular number has to be identified and replaced in case of Cantor's proof. æ> æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æWould like > æ> to see a proof that 0.333... exists? Of course there are not all > æ> digits. But those who are, are 3. No, I would not like to see a proof that it does exists, because by > just stating the definition it exists. Like the square circle? æ> æ æ æ æ æ æ æBut set theorists believe that it has uncountably many > æ> elements. Right. æIn your finitistic version of mathematics that is false, but > current mathematics is *not* finitistic. It is, but that has not yet been recognized by many. > There are too many thinks > that are true in true mathematics but false in finitistic mathematics. > Like some versions of the triangle inequality. And like the square circle? === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > And the property of having a particular number at each position (as in > Cantor's diagonal) is a definition. æSo why do you allow it here, but > not with Cantor's diagonal? Because a particular number has to be identified and replaced in case > of Cantor's proof. For the case of binary sequences, there is a simple rule that makes the diagonal defined by it differ from every member of any given sequence of binary sequences, having the opposite value at place n from the value in the nth place of the nth listed sequence. Does WM claim that that diagonal will be listed in that list? === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > To remove one fison and still cover N, there must be one left unremoved > after that removal. So that one cannot remove all of them simultaneoulsy without breaching > that supposition. The existence of the set of all FISONs presupposes the existence of > just that one being not in that set. Not in a logical world. In a logical world the existence of a set of ALL fisons presupposes that no fisons at all can be outside that set. > You believe that not all can be removed without one being left. That is neither what I said nor what I meant. QUOTE: To remove one fison and still cover N, there must be one left unremoved after that removal. To ignore the and still cover N is cheating. Which Wm does a lot. One can remove all fisons from the set of all fisons, but if one does that one does not have a cover of N left, because nothing is left. When one removes all the members of ANY set, what one gets is an empty set, whose union is also empty. The issue is what can be removed from the set of all fisons and still cover N. I.e., what sets of fisons in ZF form a cover of N. The answer in ZF is that every infinite set of fisons covers N but no finite set of fisons covers N. What goes on in WM's world is irrelevant to this. But it is also true for the inverse > process. Forming the union of all FISONs that are smaller than a FISON > excludes a last one. If this last one is not in the set when adding > all FISONs, then it need not be in the set when subtracting all > FISONs, and vice versa. > === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) WM says... > If so, show > me a statement S such that ZF proves S and ZF also proves > the negation of S. Is there such an S, or not? >There is the statement that a set of infinitely many FISONs is an >actually infinite segment called omega. Okay. There is the statement that for every FISON there exists a larger >FISON. Okay. And one can easily prove by induction that *every* FISON of the >alleged set can be removed without changing the covering properties of >that set. No. I said a proof using *only* the axioms of ZF. If complete induction is not in ZF, then I am not interested in ZF. Induction is in ZF. I don't know what you mean by complete > induction. But let's look at what you have actually proved. You say each FISON is unnecessary - it and all smaller FISONs can be > removed and the remaining set still covers N. Yes. And that holds for every FISON. What does this mean? It means that, for each natural number n, the > union of the set {F(n), F(n+1), ...} is equal to N. Oh? Look: I prove by induction that the set of numbers {1, 2, 3, ..., n} consists of natural numbers only. You would then conclude that the set {(n+1), ...} does not consist of only naturals? Strange. That's fine. That's correct. That is quite provable in ZF. That's ZF. I would agree. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=yKimjgoAAACk5WwPVD4l9HmbpoR6Hmy4 Gecko/20071127 Firefox/2.0.0.11,gzip(gfe),gzip(gfe) > WM says... > If so, show > me a statement S such that ZF proves S and ZF also proves > the negation of S. Is there such an S, or not? >There is the statement that a set of infinitely many FISONs is an >actually infinite segment called omega. Okay. There is the statement that for every FISON there exists a larger >FISON. Okay. And one can easily prove by induction that *every* FISON of the >alleged set can be removed without changing the covering properties of >that set. No. I said a proof using *only* the axioms of ZF. If complete induction is not in ZF, then I am not interested in ZF. Induction is in ZF. I don't know what you mean by complete > induction. But let's look at what you have actually proved. You say each FISON is unnecessary - it and all smaller FISONs can be > removed and the remaining set still covers N. Yes. And that holds for every FISON. Yes. Which implies that the union of every set of FISONs of the form {F(n), F(n+1), ...} is equal to N. But you somehow think this proves that the union of all FISONs is _not_ equal to N. In fact it does the opposite. After all, the set of all FISONs is a set of that form. You are just too sloppy in your thinking. You don't even think about what you are proving. > What does this mean? It means that, for each natural number n, the > union of the set {F(n), F(n+1), ...} is equal to N. Oh? Look: I prove by induction that the set of numbers {1, 2, 3, ..., > n} consists of natural numbers only. You would then conclude that the > set {(n+1), ...} does not consist of only naturals? Strange. No, I would not conclude that. I have said nothing at all similar to that. Why do you think that I have? > That's fine. That's correct. That is quite provable in ZF. That's ZF. I would agree. You agree that your proof does not prove what you say it does? Great. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor WM says... > If so, show > me a statement S such that ZF proves S and ZF also proves > the negation of S. Is there such an S, or not? >There is the statement that a set of infinitely many FISONs is an >actually infinite segment called omega. Okay. There is the statement that for every FISON there exists a larger >FISON. Okay. And one can easily prove by induction that *every* FISON of the >alleged set can be removed without changing the covering properties of >that set. No. I said a proof using *only* the axioms of ZF. If complete induction is not in ZF, then I am not interested in ZF. Induction is in ZF. I don't know what you mean by complete > induction. But let's look at what you have actually proved. You say each FISON is unnecessary - it and all smaller FISONs can be > removed and the remaining set still covers N. Yes. And that holds for every FISON. Always provided that for each fison removed there is a larger one remaining to covering its members. What does this mean? It means that, for each natural number n, the > union of the set {F(n), F(n+1), ...} is equal to N. Oh? Look: I prove by induction that the set of numbers {1, 2, 3, ..., > n} consists of natural numbers only. You would then conclude that the > set {(n+1), ...} does not consist of only naturals? Strange. There is no such conclusion implied. So it is only WM's misinterpretation which is strange, and contralogical. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 CLR 1.1.4322; InfoPath.1),gzip(gfe),gzip(gfe) What does this mean? It means that, for each natural number n, the > union of the set {F(n), F(n+1), ...} is equal to N. Oh? Look: I prove by induction that the set of numbers {1, 2, 3, ..., > n} consists of natural numbers only. You would then conclude that the > set {(n+1), ...} does not consist of only naturals? Strange. That's fine. That's correct. That is quite provable in ZF. That's ZF. I would agree. If each of the F(n) is finite, then you have changed your position. This started months ago when you disputed my claim that in ZF it is not the case that no infinite set is the union of infinitely many finite sets. You claimed for post after post after post that I was incorrect to claim that in ZFC there are infinite sets that are the union of a set all of whose members are finite. MoeBlee === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 CLR 1.1.4322; InfoPath.1),gzip(gfe),gzip(gfe) > That's ZF. I would agree. P.S. Another thing that stands out as utterly irrational of you is that you denied for so long that ZF has the theorem that there exists an infinite set that is the union of a set all of whose members are finite; yet if ZF is inconsistent (which you claim) not only would that have to be theorem of ZF (by ex falso quodlibet) but now we find you using its theoremhood as part of your (incorrect) argument that ZF is inconsistent. And you don't recognize that your argument is incorrect because you don't recognize that you've not shown that ZF has as a theorem the NEGATION of the abovementioned theorem. MoeBlee === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) WM says... I do not now state that, but it only has become obvious by means of >FISONs. I knew it for long. But, of course, if you believe in unions >of linear sets that can be larger than all of their elements, then >there is no help. Look, you keep mixing up two different issues: (1) What you believe > to be true, and (2) What can be proved from the axioms of ZF. I'm sure your beliefs about mathematics are fascinating, but > you at one time seemed to be claiming that ZF was inconsistent. > If that's true, that means that for some statement S, ZF proves > S, and also ZF proves the negation of S. Do you actually know > of such a statement S? Do you actually know how to prove S > from the axioms of ZF? Do you actually know how to prove the > negation of S from the axioms of ZF? What *you* believe about potential infinity or sets is irrelevant > to the question of the inconsistency of ZF. So don't bring up > your beliefs, if what you are claiming is that ZF is inconsistent. Look, it is futile to talk about inconsistency of ZF. You admit that you have no proof of the inconsistency of ZF. If ZF leads to the result that unions of linear sets can be larger than all of > their elements ... then there is no help. Then ZF is wrong and useless. May > it be consistenly wrong or inconsistently wrong. To all the rest of us, it is obviously true that the union of a set of > linear sets can be larger than each individual set. That is obviously a consequence of ZF. This consequence shows that ZF is inconsistent, because linear sets are defined, even in ZF, by the fact that every given set of elements is contained in one of the sets. Never two or more are required. The union of all sets of the form [1/n,1] is (0,1] in any reasonable > system. > The union of all FISONs is N, obviously. If these results did _not_ hold in ZF then I would think there was > something wrong. > There *is* something wrong. In particular if it is necessary to describe unions of FISONs by open sets. >(for instance that the union of all natural numbers can be > larger than the union of all numbers that can be reached by > induction), This is not a result of ZF. This is something you make up inside your > head and pretend is a result of ZF There is a union of all FISONs that can be proved, by induction, to be not omega: why then is this union omega by ZF? === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=yKimjgoAAACk5WwPVD4l9HmbpoR6Hmy4 Gecko/20071127 Firefox/2.0.0.11,gzip(gfe),gzip(gfe) > WM says... I do not now state that, but it only has become obvious by means of >FISONs. I knew it for long. But, of course, if you believe in unions >of linear sets that can be larger than all of their elements, then >there is no help. Look, you keep mixing up two different issues: (1) What you believe > to be true, and (2) What can be proved from the axioms of ZF. I'm sure your beliefs about mathematics are fascinating, but > you at one time seemed to be claiming that ZF was inconsistent. > If that's true, that means that for some statement S, ZF proves > S, and also ZF proves the negation of S. Do you actually know > of such a statement S? Do you actually know how to prove S > from the axioms of ZF? Do you actually know how to prove the > negation of S from the axioms of ZF? What *you* believe about potential infinity or sets is irrelevant > to the question of the inconsistency of ZF. So don't bring up > your beliefs, if what you are claiming is that ZF is inconsistent. Look, it is futile to talk about inconsistency of ZF. You admit that you have no proof of the inconsistency of ZF. If ZF leads to the result that unions of linear sets can be larger than all of > their elements ... then there is no help. Then ZF is wrong and useless. May > it be consistenly wrong or inconsistently wrong. To all the rest of us, it is obviously true that the union of a set of > linear sets can be larger than each individual set. That is obviously a consequence of ZF. This consequence shows that ZF > is inconsistent, because linear sets are defined, even in ZF, by the > fact that every given set of elements is contained in one of the sets. > Never two or more are required. That is not a definition of linear sets. It is a property that you believe, but have not shown, holds for linear sets. In all this discussion I have been taking that a family of sets is linear if every set in the family is a proper subset of some other set in the family. Is that not what we have been taking it to mean? If you insist on your definition then I simply claim that FISONs are not linear in your sense, because you have not shown that they have this property. How about a _proof_ that Every set in the family is a proper subset of some other set in the family. Implies Every set of elements of sets in the family is contained in one of the sets. If you know what a proof is yet.... > The union of all sets of the form [1/n,1] is (0,1] in any reasonable > system. > The union of all FISONs is N, obviously. If these results did _not_ hold in ZF then I would think there was > something wrong. There *is* something wrong. In particular if it is necessary to > describe unions of FISONs by open sets. This is only a problem if you think that is a wrong result. Everyone else thinks it is a right result. And you have offered no reason to doubt it. >(for instance that the union of all natural numbers can be > larger than the union of all numbers that can be reached by > induction), This is not a result of ZF. This is something you make up inside your > head and pretend is a result of ZF There is a union of all FISONs that can be proved, by induction, to be > not omega: You have not proved this. Your proof was garbage and you stopped defending it. Conveniently forgetting this later. Your proof proved only that the union of all sets of FISONs of form {F(n), F(n+1), ...} _is_ omega. Not the opposite! > why then is this union omega by ZF? Because it is proved from the axioms of ZF. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > linear sets are defined, even in ZF, by the fact that every given > set of elements is contained in one of the sets. That certainly isn't a definition of linear sets that I've ever seen. If you're using that as a definition, then the set of all FISONs is not linear. - Tim === Subject: Re: A consideration concerning the diagonal argument of G. Cantor WM says... I do not now state that, but it only has become obvious by means of >FISONs. I knew it for long. But, of course, if you believe in unions >of linear sets that can be larger than all of their elements, then >there is no help. Look, you keep mixing up two different issues: (1) What you believe > to be true, and (2) What can be proved from the axioms of ZF. I'm sure your beliefs about mathematics are fascinating, but > you at one time seemed to be claiming that ZF was inconsistent. > If that's true, that means that for some statement S, ZF proves > S, and also ZF proves the negation of S. Do you actually know > of such a statement S? Do you actually know how to prove S > from the axioms of ZF? Do you actually know how to prove the > negation of S from the axioms of ZF? What *you* believe about potential infinity or sets is irrelevant > to the question of the inconsistency of ZF. So don't bring up > your beliefs, if what you are claiming is that ZF is inconsistent. Look, it is futile to talk about inconsistency of ZF. You admit that you have no proof of the inconsistency of ZF. If ZF leads to the result that unions of linear sets can be larger than > all of > their elements ... then there is no help. Then ZF is wrong and useless. > May > it be consistenly wrong or inconsistently wrong. To all the rest of us, it is obviously true that the union of a set of > linear sets can be larger than each individual set. That is obviously a consequence of ZF. This consequence shows that ZF > is inconsistent, because linear sets are defined, even in ZF, by the > fact that every given set of elements is contained in one of the sets. > Never two or more are required. If WM thinks that he can prove that that is an inconsistency in ZF, he should go off and try to prove it, and not come back til his alleged proof hass been validated by someone competent. But he should note that his wild assumption about what he calls linear sets requires the existence of a largest such set, which is not necessary in ZF. The union of all sets of the form [1/n,1] is (0,1] in any reasonable > system. > The union of all FISONs is N, obviously. If these results did _not_ hold in ZF then I would think there was > something wrong. There *is* something wrong. In particular if it is necessary to > describe unions of FISONs by open sets. There is no axiom or theorem in ZF requiring a union of fisons to be described by what WM miscalls closed sets. What is wrong is that WM thinks he can impose whatever axioms he want on top of those of ZF, like imposing the rules of chequers on top of those of chess, and not get nonsense. > >(for instance that the union of all natural numbers can be > larger than the union of all numbers that can be reached by > induction), This is not a result of ZF. This is something you make up inside your > head and pretend is a result of ZF There is a union of all FISONs that can be proved, by induction, to be > not omega: why then is this union omega by ZF? Let us see that alleged proof. While it may hold within WM's mytheology, it will certainly not hold without it. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 CLR 1.1.4322; InfoPath.1),gzip(gfe),gzip(gfe) > To all the rest of us, it is obviously true that the union of a set of > linear sets can be larger than each individual set. That is obviously a consequence of ZF. This consequence shows that ZF > is inconsistent, because linear sets are defined, even in ZF, by the > fact that every given set of elements is contained in one of the sets. > Never two or more are required. If ZF is inconsistent, then there is a sentence P such that both P and ~P are theorems of ZF. What is your proof that there is such a P? Why don't you ever give such a proof? Your intellectual cowardice won't allow you to admit that you don't know how while yet you claim that ZF is inconsistent. MoeBlee === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > Look, it is futile to talk about inconsistency of ZF. If ZF leads to > the result that unions of linear sets can be larger than all of their > elements (for instance that the union of all natural numbers can be > larger than the union of all numbers that can be reached by > induction), then there is no help. The definition of union that requires that the union of the set of all fisons to be larger than any fison is: a = union(b) means that (x in a) <==>( exists c in B with x in c). By what definition of union does WM conclude otherwise? > Then ZF is wrong and useless. May > it be consistenly wrong or inconsistently wrong. We who have actually done some mathematics, on the other hand, find that WM's system without fixed axioms is usually wrong and mathematically useless. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) Look, it is futile to talk about inconsistency of ZF. If ZF leads to > the result that unions of linear sets can be larger than all of their > elements (for instance that the union of all natural numbers can be > larger than the union of all numbers that can be reached by > induction), then there is no help. The definition of union that requires that the union of the set of all > fisons to be larger than any fison is: This shows that the definition of union of all FISONs is similar to the definition of the triangular circle. By what definition of union does WM conclude otherwise? There is no union of linear sets other than one of the sets. If this is provably wrong, then the union is provably out of existence. Then ZF is wrong and useless. May > it be consistenly wrong or inconsistently wrong. We who have actually done some mathematics, on the other hand, If you have done some real mathematics, then you know that you did not use ZFC. For instance you will know that the intervals (1/2, 1] , (1/3 1] , (1/4, 1], ... are the same as the intervals [1], [1/2, 1] , [1/3 1] ... if only unit fractions are in play. And, again by complete induction, we can see that the complete set of natural numbers when put in the denominator will not yield any interval that cannot be noted as a closed interval. That is a property of natural numbers - with no regard to how many of them are there or not there. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > WM says... > Once again. Take the half-open interval > (0,1], which is the set of all reals x > such that 0 < x <= 1. > That interval is larger than the interval [1/2, 1]. > That interval is larger than the interval [1/3, 1]. > That interval is larger than the interval [1/4, 1]. > That interval is larger than the interval [1/5, 1]. > But you don't believe the general statement > forall natural numbers n such that n > 0, > the interval (0,1] is larger than the interval [1/n,1] I do not believe that for natural numbers there are half open >intervals that cannot be written as closed intervals. That wasn't the question. Nevertheless it is important. Do you believe the > statement: forall natural numbers n such that n > 0, > the interval (0,1] is larger than the interval [1/n,1] Let us analyse the theorem: For all natural numbers the union of all > intervals [1/n,1] is larger than one of those intervals [1/n,1]. i.e. (0,1]. 1) It is obvious that nobody can supply a natural number violating > this theorem. Because it is true. 2) It is also obvious, that the union consists only of points which > are members of at least one of the intervals. Yes. Therefore it is clear that at least one of the intervals reaches > as close to zero as points exist. That isn't clear at all. You give no reasons. Do you have any? Every union of sets ordered by inclusion is a set of the union. It is obvious that [0, 1] is not the union of all intervals [1/n,1]. > There is no help by infinitely many. Right! (0,1] is. That does not help to solve the problem! In case of the reciprocals, namely FISONs, there are no open intervals. As long as only natural numbers are in play, there are no open intervals that could not be replaced by closed intervals. That is not a question of frequency, but a property of naturals. I wonder, are you really so incompetent that you don't know this? It's quite worrying === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=yKimjgoAAACk5WwPVD4l9HmbpoR6Hmy4 Gecko/20071127 Firefox/2.0.0.11,gzip(gfe),gzip(gfe) WM says... > Once again. Take the half-open interval > (0,1], which is the set of all reals x > such that 0 < x <= 1. > That interval is larger than the interval [1/2, 1]. > That interval is larger than the interval [1/3, 1]. > That interval is larger than the interval [1/4, 1]. > That interval is larger than the interval [1/5, 1]. > But you don't believe the general statement > forall natural numbers n such that n > 0, > the interval (0,1] is larger than the interval [1/n,1] I do not believe that for natural numbers there are half open >intervals that cannot be written as closed intervals. That wasn't the question. Nevertheless it is important. Do you believe the > statement: forall natural numbers n such that n > 0, > the interval (0,1] is larger than the interval [1/n,1] Let us analyse the theorem: For all natural numbers the union of all > intervals [1/n,1] is larger than one of those intervals [1/n,1]. i.e. (0,1]. 1) It is obvious that nobody can supply a natural number violating > this theorem. Because it is true. 2) It is also obvious, that the union consists only of points which > are members of at least one of the intervals. Yes. Therefore it is clear that at least one of the intervals reaches > as close to zero as points exist. That isn't clear at all. You give no reasons. Do you have any? Every union of sets ordered by inclusion is a set of the union. That isn't clear at all. You give no reasons. Do you have any? > It is obvious that [0, 1] is not the union of all intervals [1/n,1]. > There is no help by infinitely many. Right! (0,1] is. That does not help to solve the problem! What problem? [0,1] is not the union of all intervals of form [1/n,1]. (0,1] is the union of all intervals of form [1/n,1]. Every interval of form [1/n,1] is a proper subset another interval of that form. The union of all intervals of form [1/n,1] is larger than each individual interval of that form. This directly contradicts your previous assertions that such a thing is impossible. What more needs to be said? === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > Therefore it is clear that at least one of the intervals reaches > as close to zero as points exist. That isn't clear at all. You give no reasons. Do you have any? Every union of sets ordered by inclusion is a set of the union. That isn't clear at all. You give no reasons. Do you have any? Simply try to find a counter example of an element that is not in the same set as all smaller elements. It is obvious that [0, 1] is not the union of all intervals [1/n,1]. > There is no help by infinitely many. Right! (0,1] is. That does not help to solve the problem! What problem? The problem of missing open intervals for natural numbers. [0,1] is not the union of all intervals of form [1/n,1]. > (0,1] is the union of all intervals of form [1/n,1]. > Every interval of form [1/n,1] is a proper subset another interval of > that form. > The union of all intervals of form [1/n,1] is larger than each > individual interval of that form. > This directly contradicts your previous assertions that such a thing > is impossible. What more needs to be said? In the first place, there are no open intervals in natural numbers that could not be given as closed intervals. Second, there is no point of (0,1] that is outside of an interval [1/n,1]. Third there are no two points that require more than one interval [1/n,1] to be included. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > Every union of sets ordered by inclusion is a set of the union. Every union of a finite set of sets ordered by inclusion is the maximal member of the set of sets. But in ZF, one can have a set of sets ordered by inclusion in which there is no maximal member. E.g., the set of fisons. And absent such a maximal member, the union CANNOT be a member. And even potentially infinite sets, whatever they may be, do not have maximum members. It is obvious that [0, 1] is not the union of all intervals [1/n,1]. > There is no help by infinitely many. Right! (0,1] is. That does not help to solve the problem! On the contrary, it solves it precisely and excatly everywhere outside of WM's mytheology. > In case of the reciprocals, > namely FISONs, there are no open intervals. Nor closed ones. There are no intervals at all. I wonder, are you really so incompetent that you don't know this? > It's quite worrying Worry away all you want, but our alleged incompetence, unlike yours, can create real mathematics. Now if our physicists and engineers had to use the mathematics produced by someone so obviously mathematically incompetent as WM, ... === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > WM says... If complete induction is not in ZF, then I am not interested in ZF. Yes, we all know that you are not interested in ZF. And nobody > cares. What people care about is your claim that ZF is inconsistent. > Are you now agreeing that you were wrong when you said that ZF > is inconsistent? It makes no sense to say that FISON is a subset of omega? Sure. It makes no sense to say that omega FISON is not empty? That's provably false. It makes no sense to claim that complete induction is not valid for >every natural number? No, that makes no sense. Fine. Then it makes sense to claim that complete induction os valid for every natural number. Then ZF is inconsistent. Because for FISONs we have: A property that is valid for every FISON of a set is also valid for the union of FISONs of that set. That is so because the union of FISONs is a FISON. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=yKimjgoAAACk5WwPVD4l9HmbpoR6Hmy4 Gecko/20071127 Firefox/2.0.0.11,gzip(gfe),gzip(gfe) > WM says... If complete induction is not in ZF, then I am not interested in ZF. Yes, we all know that you are not interested in ZF. And nobody > cares. What people care about is your claim that ZF is inconsistent. > Are you now agreeing that you were wrong when you said that ZF > is inconsistent? It makes no sense to say that FISON is a subset of omega? Sure. It makes no sense to say that omega FISON is not empty? That's provably false. It makes no sense to claim that complete induction is not valid for >every natural number? No, that makes no sense. Fine. Then it makes sense to claim that complete induction os valid > for every natural number. You have yet to explain what complete induction is, and how it differs from ordinary induction. Here's some help : ordinary induction proves that property P holds for each and every natural number. Now, what is complete induction? > Then ZF is inconsistent. So there should be some statement such that both it and its negation are theorems of ZF. What is that statement? > Because for FISONs we have: A property that is valid for every FISON of a set is also > valid for the union of FISONs of that set. How do we have that? > That is so because the union of FISONs is a FISON. That is something you have yet to prove. It is easy to prove that the union of a finite sets of FISONs is equal to the largest FISON in the set. However for infinite sets of FISONs there is no largest FISON in the set. So which FISON do you imagine the union is equal to? === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=-eQqtQoAAACZVM-kNEsOn3k7GSvoJoS4 MathPlayer 2.0; .NET CLR 1.1.4322; InfoPath.1; .NET CLR 2.0.50727; .NET CLR 3.0.04506.648; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) > You have yet to explain what complete induction is, and how it > differs from ordinary induction. Here's some help : ordinary induction > proves that property P holds for each and every natural number. Now, > what is complete induction? It's a translation from the German vollstaendige Induktion, which is used to distinguish mathematical induction from induction, which --- unlike deduction --- is logically suspect. (See http://en.wikipedia.org/wiki/Inductive reasoning.) So simply ignore the word complete or replace it with mathematical when he uses it; his proofs are wrong in any case. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > Fine. Then it makes sense to claim that complete induction ss valid > for every natural number. You have yet to explain what complete induction is, and how it > differs from ordinary induction. Ordinary induction means to do some numerical calculations and to conclude from that on general results, like Gauus and Euler did in order to find general result. It is more or less guessing. Complete induction proves results for all natural numbers by proving that tghe result is correct for 1 and that it is correct for n+1 if it is correct for n. > Here's some help : ordinary induction > proves that property P holds for each and every natural number. Now, > what is complete induction? That is called complete induction, at least in Germany, namely vollstaendige Induktion. Then ZF is inconsistent. So there should be some statement such that both it and its negation > are theorems of ZF. What is that statement? The intervals (1/2, 1] , (1/3 1] , (1/4, 1], ... are the same as the intervals [1], [1/2, 1] , [1/3 1] ... if only unit fractions are in play. And, again by complete induction, we can see that the complete set of natural numbers when put in the denominator will not yield any interval that cannot be noted as a closed interval. That is a property of natural numbers - with no regard to how many of them are there or not there. This is true in ZF for all fractions 1/n. And it is true in ZF that the union of all fractions 1/n yields a half-open interval. That's a contradiction. Because for FISONs we have: A property that is valid for every FISON of a set is also > valid for the union of FISONs of that set. How do we have that? Because a union of FISONs is a FISON. Also provable for all natural numbers by complete induction. That is so because the union of FISONs is a FISON. That is something you have yet to prove. It is easy to prove that the > union of a finite sets of FISONs is equal to the largest FISON in the > set. However for infinite sets of FISONs there is no largest FISON in > the set. Nevertheless every FISON has a natural number at its end. And every FISON is finite. Hence every union of FISONs is finite. > So which FISON do you imagine the union is equal to?- None. This simply shows that what you call an infinite union does have as little existence as an infinite natural number. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > Fine. Then it makes sense to claim that complete induction ss valid > for every natural number. > You have yet to explain what complete induction is, and how it > differs from ordinary induction. Ordinary induction means to do some numerical calculations and to > conclude from that on general results, like Gauus and Euler did in > order to find general result. It is more or less guessing. > Complete induction proves results for all natural numbers by proving > that tghe result is correct for 1 and that it is correct for n+1 if it > is correct for n. > Here's some help : ordinary induction > proves that property P holds for each and every natural number. Now, > what is complete induction? That is called complete induction, at least in Germany, namely > vollstaendige Induktion. The English term complete induction is however used for something else: en.wikipedia.org/wiki/Mathematical_induction#Complete_induction -- Alan Smaill === Subject: Re: A consideration concerning the diagonal argument of G. Cantor WM says... >It makes no sense to claim that complete induction is not valid for >every natural number? > No, that makes no sense. Fine. Then it makes sense to claim that complete induction os valid >for every natural number. No, that makes no sense, either. Your notion of complete induction makes no sense. >Then ZF is inconsistent. Because for FISONs >we have: A property that is valid for every FISON of a set is also >valid for the union of FISONs of that set. No, that's not true. >That is so because the union of FISONs is a FISON. No, it is not. Look, you either accept the existence of infinite unions, or you don't. If you don't, then there is no such thing as the union of FISONs. If you do, then it is provably true that the union of all FISONs is not a FISON. Either way, what you're saying makes no sense. -- Daryl McCullough Ithaca, NY === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322),gzip(gfe),gzip(gfe) >That is so because the union of FISONs is a FISON. No, it is not. Look, you either accept the existence > of infinite unions, Please do not pretend that I do not accept the existence of infinite unions. Look, the union of all natural numbers is accepted by many. It is apparently no problem to have all natural numbers in one set. And it is apparently no problem to enumerate the elements of this set and to finish this infinite enumeration by a number larger than all natural numbers. This all seems odd but it is accepted by many. And I would not deny it. However, when using the notion of FISON, then we see immediately by a proof that all this is inconsistent. In order to see this, we do not have to decide whether or not we accept the axiom of infinity! We have to look for logical consistency only. Therefore it is not the denial of the axiom, as you imply, but the proof of its inconsistency. It may be not impossible to have the union of all natural umbers. But it is impossibe to have a uninon of linear sets that is larger than all the sets. It is obviously provably impossible to get omega set by an infinite union of FISONs. The proof shows that none of the alleged infinitely many FISONs has any effect on the aim. That is independent of the fact that every union of finite linear sets is one linear set, as can be proved by induction too. As an example consider the union of {1} , {1, 2} , {1, 2, 3} is {1, 2, 3} . This holds as long as there are only finite FISONs {1, 2, 3, ..., n} concerned. But there are no infinite FISONs existing even in an infinite set. Therefore, it holds in all cases. >or you don't. If you don't, then > there is no such thing as the union of FISONs. If you > do, then it is provably true that the union of all > FISONs is not a FISON. Either way, what you're saying makes no sense. -- > Daryl McCullough > Ithaca, NY === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=yKimjgoAAACk5WwPVD4l9HmbpoR6Hmy4 Gecko/20071127 Firefox/2.0.0.11,gzip(gfe),gzip(gfe) That is so because the union of FISONs is a FISON. No, it is not. Look, you either accept the existence > of infinite unions, Please do not pretend that I do not accept the existence of infinite > unions. Look, the union of all natural numbers is accepted by many. It is > apparently no problem to have all natural numbers in one set. And it > is apparently no problem to enumerate the elements of this set and to > finish this infinite enumeration by a number larger than all natural > numbers. This all seems odd but it is accepted by many. And I would > not deny it. I think most mathematicians would not accept your strange idea that we can finish an infinite enumeration. Where did you get that from? > However, when using the notion of FISON, then we see immediately by a > proof that all this is inconsistent. In order to see this, we do not > have to decide whether or not we accept the axiom of infinity! We have > to look for logical consistency only. Therefore it is not the denial > of the axiom, as you imply, but the proof of its inconsistency. It may > be not impossible to have the union of all natural umbers. Ok. > But it is impossibe to have a uninon of linear sets that is larger than > all the sets. You have asserted this maybe a thousand times. You are yet to offer a proof of this claim. Heck, you have yet to offer _any reason whatsoever_ why it should be true. Do you not see that this is a major point of disagreement between you and others here? === Subject: Re: A consideration concerning the diagonal argument of G. Cantor WM says... >That is so because the union of FISONs is a FISON. > No, it is not. Look, you either accept the existence > of infinite unions, Please do not pretend that I do not accept the existence of infinite >unions. Look, the union of all natural numbers is accepted by many. You are speaking nonsense. What is the union of all natural numbers? A union is an operation on a set of sets. Once again, if C is a set of sets, then union(C) = the set of all x such that for at least one y in C, x is an element of y. To illustrate, if C = {{0}, {1,2}, {2,3}} then union(C) = {0,1,2,3}. 0 is in the union since it is in the first set. 1 is in the union since it is in the second set. 2 and 3 are in the union because they are in the third set. I don't know what you mean by the union of all natural numbers. >However, when using the notion of FISON, then we see immediately by a >proof that all this is inconsistent. STOP saying that! You have no proof that it is inconsistent! I've asked you many times to provide such a proof, and you *CANT*. Are you now returning to your claim that ZF is inconsistent? Then *PROVE* it! >As an example consider the union of {1} , {1, 2} , {1, 2, 3} is {1, >2, 3} . This holds as long as there are only finite FISONs {1, 2, >3, ..., n} concerned. But there are no infinite FISONs existing even >in an infinite set. Therefore, it holds in all cases. The correct theorem is this: If C is a collection of FISONs, then we have two cases: 1. If C has a largest element, then union(C) = the largest element 2. If C does *not* have a largest element, then union(C) is *not* equal to the largest element (since there isn't one). You are assuming The union of any set of FISONs is a FISON. But *why* do you think that? It's provably false. -- Daryl McCullough Ithaca, NY === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > WM says... >That is so because the union of FISONs is a FISON. > No, it is not. Look, you either accept the existence > of infinite unions, Please do not pretend that I do not accept the existence of infinite >unions. Look, the union of all natural numbers is accepted by many. You are speaking nonsense. What is the union of all natural > numbers? A union is an operation on a set of sets. A union is a set that contains the elements of the sets unioned. It is not an operation because it does not require any time. The union exists in a universe without space and time and it is invariable (so say the Platonists). > Once again, > if C is a set of sets, then union(C) = the set of all x such that > for at least one y in C, x is an element of y. To illustrate, > if C = {{0}, {1,2}, {2,3}} then union(C) = {0,1,2,3}. 0 is > in the union since it is in the first set. 1 is in the union > since it is in the second set. 2 and 3 are in the union because > they are in the third set. I don't know what you mean by the union of all natural numbers. To be honest, I don't know it either. But some people say that the union of all natural numbers is there (?): There is a set that contains 1 and ... you must have seen it somewhere. >As an example consider the union of {1} , {1, 2} , {1, 2, 3} æis {1, >2, 3} . This holds as long as there are only finite FISONs {1, 2, >3, ..., n} æconcerned. But there are no infinite FISONs existing even >in an infinite set. Therefore, it holds in all cases. The correct theorem is this: > If C is a collection of FISONs, then we have two cases: For FISONs that consists of finite numbers, there is only one case. > 1. If C has a largest element, then union(C) = the largest element > 2. If C does *not* have a largest element, then union(C) is > *not* equal to the largest element (since there isn't one). There is no infinite set of FISONs. That is correct. I told you already that ZF cannot depend on the representation. An infinite set of FISONs in unary representation cannot exist. You are assuming The union of any set of FISONs is a FISON. > But *why* do you think that? It's provably false. It is provably true by the fact that FISONs are linear by inclusion. I cannot understand why you seem to believe that this rule is violated for large numbers. Provably false is the opposite assumption. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > WM says... >It makes no sense to claim that complete induction is not valid for >every natural number? > No, that makes no sense. Fine. Then it makes sense to claim that complete induction os valid >for every natural number. No, that makes no sense, either. Your notion of complete > induction makes no sense. My notion of complete induction has been in mathematics for more than 300 years. It shows, for instance, if we use unit fractions only, that (1/2, 1] can be replaced by [1] . And it shows that if (1/n, 1] can be replaced by (1/(n-1), 1], then (1/(n+1), 1] can be replaced by (1/n, 1]. Now it is easy to show that the first and the second step are valid. Then we see for every natural number that the open interval can be replaced by a closed interval. And we see by a similar proof that every union of closed intervals is a closed interval. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor WM says... > No, that makes no sense, either. Your notion of complete > induction makes no sense. My notion of complete induction has been in mathematics for more than >300 years. No, you have consistently misunderstand that principle. Your much repeated claim that omega is not equal to the union of all FISONs is an illustration. You claim to prove by induction every FISON can be removed from the set that covers N without changing the covering properties of the set, and your proof is not an instance of natural induction. You have misunderstood the principle of induction. -- Daryl McCullough Ithaca, NY === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > WM says... > No, that makes no sense, either. Your notion of complete > induction makes no sense. My notion of complete induction has been in mathematics for more than >300 years. No, you have consistently misunderstand that principle. Your > much repeated claim that omega is not equal to the union of > all FISONs is an illustration. You claim to prove by induction every FISON can be removed from the > set æthat covers N without changing the covering properties of the set, > and your proof is not an instance of natural induction. You have > misunderstood the principle of induction. Let's see. 1) When FISON {1} is removed from omega, then there is something left. 2) When FISON {1, 2, 3, ..., n} is removed and something is left, then also something is left when FISON {1, 2, 3, ..., n, n+1} is removed. The first step is obvious. The second step can be proved by the existence of FISON {1, 2, 3, ..., n, n+1, n+2} for every FISON {1, 2, 3, ..., n}. So if FISON {1, 2, 3, ..., n} is removed, then there exists FISON {1, 2, 3, ..., n, n+1, n +2} < omega. Hence FISON {1, 2, 3, ..., n, n+1} can be removed too from omega and something is left. Therefore my proof covers every FISON {1, 2, 3, ..., n} for n a natural number. There is no one left out. Thus it is true for all FISONs that can be formed by finite natural nunmbers. Right? Or do you find any inconsistency in my proof? === Subject: Re: A consideration concerning the diagonal argument of G. Cantor My notion of complete induction has been in mathematics for more than > 300 years. > It shows, for instance, if we use unit fractions only, that (1/2, 1] > can be replaced by [1] . And it shows that if (1/n, 1] can be replaced > by (1/(n-1), 1], then (1/(n+1), 1] can be replaced by (1/n, 1]. Now it > is easy to show that the first and the second step are valid. Then we > see for every natural number that the open interval can be replaced by > a closed interval. And we see by a similar proof that every union of > closed intervals is a closed interval. > Do students at the FH Augsburg have to learn induction this way? Horrible! With this level of mathematical knowledge you shouldn't have passed the high school. Alois === Subject: Re: A consideration concerning the diagonal argument of G. Cantor Alois Steindl a .8ecrit : > My notion of complete induction has been in mathematics for more than > 300 years. > It shows, for instance, if we use unit fractions only, that (1/2, 1] > can be replaced by [1] . And it shows that if (1/n, 1] can be replaced > by (1/(n-1), 1], then (1/(n+1), 1] can be replaced by (1/n, 1]. Now it > is easy to show that the first and the second step are valid. Then we > see for every natural number that the open interval can be replaced by > a closed interval. And we see by a similar proof that every union of > closed intervals is a closed interval. > Do students at the FH Augsburg have to learn induction this way? > Horrible! With this level of mathematical knowledge you shouldn't have > passed the high school. > Alois Meet one of the master cranks of sci.math. Actually, I wonder if there is some way of retorsion, by denunciation of the guy to his superiors... === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > That is so because the union of FISONs is a FISON. That a union of fisons is necessarily a fison claim may be true in WM's mytheology, but in ZF it is NOT provably true and IS provably false. For example, the union of the set of fisons ending in even naturals is not a fison in ZF because it does not end in an even natural nor in an odd natural, and there are no others for it to end in. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 CLR 1.1.4322; InfoPath.1),gzip(gfe),gzip(gfe) > Fine. Then it makes sense to claim that complete induction os valid > for every natural number. Then ZF is inconsistent. Because for FISONs > we have: A property that is valid for every FISON of a set is also > valid for the union of FISONs of that set. That is so because the > union of FISONs is a FISON. By ZF is inconsistent we mean there is a sentence P such that both P and ~P are theorems of ZF. What proof do you have that there is a such a P? MoeBlee === Subject: Re: A consideration concerning the diagonal argument of G. Cantor Yes, all sets are united at once, in a single step. There is *no* > repetition neither in time nor in something else. Have you ever done some mathematics by yourself or by means of a > computer? I mean real mathematics, not ZF pipe-dreams. tenor of his postings, I very much doubt it. And besides you should choose your wording accordingly. Sets are > united at once has a bad temporal taste. You should say the union of > the sets is considered. WM trying to correct the style of someone else's English? Risible! > The infinite union does not change the quality of the sets that are united. > I have no idea why you would think so. It is their *union* that has a > quality different from the quality of the sets being united. Just as > union [1/n, 1] = (0, 1]. Why don't you say that the union is [0, 1]? Because it isn't. The union consists of those reals, x, for which there is some natural n with 1/n <= x <= 1, and 0 is not such a real. > Or even [-1, 2]? If the > quality of the union is in fact so independent from the properties of > the sets united? On the contrary, the union is TOTALLY DEPENDENT on the membership of the sets being unioned. For any set of sets, s, Union(s) = {x: Exists y in s such that x in y} i > But even in decimal representation, you need > to >count the number of digits. You will not be able to manage that >other >than for a finite number of digits. > > And because each number from the list differs in a finite position > > from > > the diagonal, you do not have to manage other than a finite number of > > digits. > > That means that you can do the complete diagonal number by a finite > > number of digits? Who is saying that? You: And because each number from the list differs in a finite > position from > the diagonal, you do not have to manage other than a finite number of > digits. Perhaps the Dutch to English to German translation obscured Dik's meaning I would have stated it as: Each comparison occurs at a finite digit position so that one need not ever deal with anything but finite digit postions. This shows the basic contradiction of set theory. Or a basic difficulty in multiple translations. > > The last one: N = union {n}. Why is that wrong? It is because of the axiom of infinity. Or because of the triangular circle? Such beasts exist only in WM's private menagerie. Now you may > work with the negation of that axiom, but in that case you will not > find an inconsistency in set theory but an incompatibility of set > theory with your theory. I am not working with the negation. I use that fact that the union of > all intervals [1/n, 1] is not [0, 1]. Nothing else! That does not contradict anything in ZF, so that WM must be assuming something else outside ZF to get his alleged contradiction. Moreover, elsewhere you were willing to assert > that N = union {n}. I assume it in order to show that it is wrong, at least in its actual > meaning. But it is not wrong except inside WM's world. > As long as only natural numbers are concerned, > their FISONs and unions of FISONs are given by closed intervals. Why closed? What is the topology on natural numbers which requires a fison to be closed but not open, and why is that sort of topology necessary? Until WM can answer these questions satisfactorily, his claims of topological properties are irrelevant. I suspect that WM is too ignorant of topology to answer, in which case all his topological arguments are mere nonsense. > That > is the property of a natural number that must be violated when > striving for finished infinity. Until WM's topological arguments can be justified by showing that the naturals must have a natural topology in which all fisons are closed but not open, his claims to that effect are mere nonsense. One notes that a set which is potentially infinite, but not actually infinite, may be quite difficult to topologize. At least by WM. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > WM says... Look, it is futile to talk about inconsistency of ZF. If ZF leads to >the result that unions of linear sets can be larger than all of their >elements (for instance that the union of all natural numbers can be >larger than the union of all numbers that can be reached by >induction), then there is no help. Then ZF is wrong and useless. May >it be consistenly wrong or inconsistently wrong. So you are now agreeing that you were wrong when you said > ZF is inconsistent? No. I said it is futile to talk about inconsistency of ZF. Either it is utterly wrong in that it defines unions that contain more elements never more than one set of a linear order of sets is required to cover a given number of elements. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor Nntp-Posting-Host: hera.cwi.nl ... > No. I said it is futile to talk about inconsistency of ZF. Either it > is utterly wrong in that it defines unions that contain more elements > never more than one set of a linear order of sets is required to cover > a given number of elements. So any theory that states that union(n in N) [1/n, 1] = (0, 1] is utterly wrong? -- 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: A consideration concerning the diagonal argument of G. Cantor posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > ... > æ> No. I said it is futile to talk about inconsistency of ZF. Either it > æ> is utterly wrong in that it defines unions that contain more elements > æ> never more than one set of a linear order of sets is required to cover > æ> a given number of elements. So any theory that states that union(n in N) [1/n, 1] = (0, 1] is utterly > wrong? No, I would not say so. Only if it states that there are all n, or that (0, 1] is larger than every interval [1/n, 1], or if it states that the union is its limit, namely [0, 1], then this theory is wrong. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor WM says... > So you are now agreeing that you were wrong when you said > ZF is inconsistent? No. I said it is futile to talk about inconsistency of ZF. You are the one who claimed that ZF was inconsistent. Why did you make that claim? -- Daryl McCullough Ithaca, NY === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322),gzip(gfe),gzip(gfe) > WM says... > So you are now agreeing that you were wrong when you said > ZF is inconsistent? No. I said it is futile to talk about inconsistency of ZF. You are the one who claimed that ZF was inconsistent. Why > did you make that claim? I just showed you why. But I can repeat it in short: ZF is inconsistent beause it requires that the union U{n in N} (F1, F2, F3, ...) of linear sets F1 c F2 c F3 c ... contains elements that are not in one of the sets. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor WM says... > You are the one who claimed that ZF was inconsistent. Why > did you make that claim? I just showed you why. But I can repeat it in short: ZF is >inconsistent beause it requires that the union >U{n in N} (F1, F2, F3, ...) of linear sets F1 c F2 c F3 c ... contains >elements that are not in one of the sets. ZF does not prove that. So you are lying. -- Daryl McCullough Ithaca, NY === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > WM says... > You are the one who claimed that ZF was inconsistent. Why > did you make that claim? I just showed you why. But I can repeat it in short: ZF is >inconsistent beause it requires that the union >U{n in N} (F1, F2, F3, ...) of linear sets F1 c F2 c F3 c ... contains >elements that are not in one of the sets. ZF does not prove that. So you are lying. ZF proves that not all elements are in one FISON. Linearity proves that all elements that are in a union of FISONs are also in one FISON (true for finite natural numbers, but there are no others). The only escape is to assume that there are elements of the union that are in no FISON. Or is the union not temporarily constant? Is there a temporal evolution? Like potential infinity??? === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > WM says... Look, it is futile to talk about inconsistency of ZF. If ZF leads to >the result that unions of linear sets can be larger than all of their >elements (for instance that the union of all natural numbers can be >larger than the union of all numbers that can be reached by >induction), then there is no help. Then ZF is wrong and useless. May >it be consistenly wrong or inconsistently wrong. So you are now agreeing that you were wrong when you said > ZF is inconsistent? No. I said it is futile to talk about inconsistency of ZF. It is certainly futile for WM to talk about what he has no understanding of. > Either it > is utterly wrong in that it defines unions that contain more elements > than the united sets Or WM is utterly wrong, and it does no such thing. > never more than one set of a linear order of sets is required to cover > a given number of elements. If WM's a given number of elements, means only a finite number of elements, then ZF agrees, but if infinite numbers of elements are allowed, as they are in ZF, then WM's claim becomes provably false. WM is incapable of understanding what his religion of finiteness forbids. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 CLR 1.1.4322; InfoPath.1),gzip(gfe),gzip(gfe) WM says... Look, it is futile to talk about inconsistency of ZF. If ZF leads to >the result that unions of linear sets can be larger than all of their >elements (for instance that the union of all natural numbers can be >larger than the union of all numbers that can be reached by >induction), then there is no help. Then ZF is wrong and useless. May >it be consistenly wrong or inconsistently wrong. So you are now agreeing that you were wrong when you said > ZF is inconsistent? No. I said it is futile to talk about inconsistency of ZF. Either it > is utterly wrong in that it defines unions that contain more elements > never more than one set of a linear order of sets is required to cover > a given number of elements. You say it is futile to talk about inconsistency of ZF, yet you talk about it, and you declare that ZF is inconsistent. So what sentences P and ~P do you claim can be derived in ZF? MoeBlee === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > æ> That is easy because also at positions which cannot be identified, the > æ> digit is 3. I asked for a proof not for an assertion. The property of having 3 at each position is a definition. Would like > to see a proof that 0.333... exists? Of course there are not all > digits. But those who are, are 3. That is hardly the asked for proof. Besides, if any digits are missing, the expression is not equal to 1/3. > æ æ> No. At most the set exists, not the indviduals. Hence, at most the > æ> idea of a set exists. I do not understand. How can a set exist without the individuals? The idea of a set can exists, for instance the set of all material > pink unicorns now exists as an idea, although no pink unicorn exists. > But such sets do not exist as set theorists believe that they exist. > For example the set of all real numbers contains at most 10^100 > individuals. But set theorists believe that it has uncountably many > elements. It, the set of all reals, still exists as an idea, which is all that anything mathematical can exist as. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > ... > æ>It is provably true from your axioms (when transposed to the > æ>reciprocals. > æ æ> No, it is not provably true. It is provably false. > æ æ> Omega is the first transfinite number. 0 is the first number that > æ> cannot be expressed by 1/n. Yes, and 0 is not in the union of [1/n, 1] and omega is not in the union > of {1, ..., n}. æNow what? [0, 1] is not the union of all intervals[1/n, 1]. The same is true for > the reciprocal values. 0 has no reciprocal, so WM's claim is meaningless, again. > omega is not the union of finite initial segments of omega. It is everywhere outside of WM's world. There is no help by the sophisticated escape omega is not in the > union. omega either is the union or is not the union. It is not. The set designated by omega equals the union of all fisons everywhere outside of WM's world. And what goes on inside that world is totally irrelevant to the world of mathematics. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > WM says... > Once again. Take the half-open interval > (0,1], which is the set of all reals x > such that 0 < x <= 1. > That interval is larger than the interval [1/2, 1]. > That interval is larger than the interval [1/3, 1]. > That interval is larger than the interval [1/4, 1]. > That interval is larger than the interval [1/5, 1]. > But you don't believe the general statement > forall natural numbers n such that n > 0, > the interval (0,1] is larger than the interval [1/n,1] I do not believe that for natural numbers there are half open >intervals that cannot be written as closed intervals. That wasn't the question. Nevertheless it is important. On the contrary, in the matter of the infiniteness of the set of naturals, the openness or closedness of real intervals of any sort is totally irrelevant. Do you believe the > statement: forall natural numbers n such that n > 0, > the interval (0,1] is larger than the interval [1/n,1] Let us analyse the theorem: For all natural numbers the union of all > intervals [1/n,1] is larger than one of those intervals [1/n,1]. 1) It is obvious that nobody can supply a natural number violating > this theorem. > 2) It is also obvious, that the union consists only of points which > are members of at least one of the intervals. Therefore it is clear > that at least one of the intervals reaches as close to zero as points > exist. Not so! For each one interval that is allegedly as close as points exist there is another even closer. There cannot be any such interval that is closest. You try to solve that problem by means of infinitely many intervals. > But that is no solution! It is a better solution than any that WM has offered. > The intervals are linearly ordered. If no > interval reaches 0 then even an infinite amount of them does not. Which is exactly our argument. > You > should be able to see that. Instead of insisting on the actual > existence of all points we can assume potential infinity. Or we can reject Wm's artificial potential infinity in favor of actual infinity, which is what the more sensible of us do. And if we include the fact that for the reciprocals, namely FISONs, > there are no open intervals, then omega is corresponding to [0, 1]. Openness and closedness for fisons is irrelevant unless WM first specifies the topology on N for which its subsets may be categorized as open or closed. And, in any case, there is nothing in topology that requires an infinite union of closed sets to be closed. So WM's arguments are again up the creek without a paddle. > And we get the theorem: For all natural numbers the union of all intervals [1/n,1] = [0, 1] is > larger than one of those intervals [1/n,1]. WM may get that as a theorem in his world, but it is not a theorem in any sane world that 0 is a member of a union of sets in which none of those sets contains 0. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor Look, you might not like ZF, but the nice thing is > that it has clear, unambiguous definitions of such > things as the union of a set of sets, the natural > numbers, infinite set, etc. such that all the > conclusions are provably true from those definitions. >It is certainly true that the union of intervals [1/n, 1] does not >contain the number zero. Nevertheless you may claim that the infinite >union of all intervals contains the number zero. > No, I don't claim that. The union of all sets of the form > [1/n,1] is the half-open interval (0,1]. It is the set > of all reals x such that 0 < x <= 1. According to Cantor and also obviously, omega corresponds to the > interval [0, 1] Cantor would not have been so foolish as to say that Union({[1/n,1]:n in N}) = [0,1] when it is so easily proved false. It seems that much of what is obvious to WM is provably false in any standard mathematics. Correct. But with FISONs there are no open intervals required. Whatever. I corrected a typo: n open intervals --> no open intervals. I'm going to ask you one last time: Do you believe that > the axioms of set theory are inconsistent? > Yes, they are. If so, show > me a statement S such that ZF proves S and ZF also proves > the negation of S. Is there such an S, or not? > There is the statement that a set of infinitely many FISONs is an > actually infinite segment called omega. Then show, using only the arguments of standard logic an no assumptions or claims of fact other than those required by ZF, that that statement is both true and false. Until you have done that, and you have not shown that, you have not shown any self-contradiction within ZF. That WM's set theory and ZF's set theory are mutually contradictory we accept, but that WM's set theory is itself free from internal contradictions we do not accept. There is the statement that for every FISON there exists a larger > FISON. And one can easily prove by induction that *every* FISON of the > alleged set can be removed without changing the covering properties of > that set. You claim an ability to prove a number of things, but, despite being challenged to show it on many occasions, you have yet to demonstrate that abilitySo that until you do demonstrate it, for some non-trivial assertion like the above, we lake leave to doubt you have any such ability. If these statements can be formulated in ZF, then ZF is inconsistent. This claim is still unproven, therefore of no merit. If so, if you believe that ZF is inconsistent, then produce > a statement S and produce a ZF proof of S, and then produce > a ZF proof of the negation. Use *only* axioms and rules of > inference of ZF. If complete induction is not sufficient to prove a theorem for all > members of the set of natural numbers, then ZF is not even > mathematics. Whether or not complete induction is sufficient, WM has still not proven his claim either with or without using it. Can you do that? Done. Only claimed but never proved. I can, with equal justice, claim that WM has too many holes in his head to keep his brains from falling out. === Subject: Re: Mathematics of poker > On May 7, 11:55 am, Tim Little I suspect (but haven't yet proved) that any strategy that permits > endless raising can be beaten by one that does not. You mean the optimal strategy would have > p(Raise_at_round_N) = 0 > for some N? Not sure why you think this is true. Assuming p is the probability of raising in a given round, I would think that it should also be a function of the number you were given, not just which round you're in. Though I think I used the word permits a bit loosely. More precisely, I suspect that the optimal strategy will have a zero probability of endless raising, and hence the game must have probability 1 of terminating regardless of what the other player does. E.g. something like p(x, N) = x would satisfy that, but something like p(x, N) = 1 - (1-x)/2^N would not. - Tim === Subject: Re: Mathematics of poker > Truly endless raising presumes each player an unlimited money > reserve from which to raise, which makes gambling fairly pointless > as anything but a contest of egos. I certainly don't disagree with the assessment of utility of the winnings in such a scenario. I am merely interested in the analysis of the optimal strategy and expected value of the winnings themselves. Maybe over the weekend I will have enough time to devote to investigating the problem myself. - Tim === Subject: Re: questions on turing machine problem > 1) I'm trying to understand the following problem: > let M_1 and M_2 two > turing machines with common input alphabet and a > given string x. Is > there a particular step in which the two machines > will write the same > symbol on the tape? > is it undecidable? can somebody give an idea on how > can we reduce this > problem to an undecidable problem? > > This could be used to solve the halting problem so > its undecidable. If you > at each step, and your > set up M_2 as a modified version of the universal > turing mechine that What's the stopping criterion for M_1 there? Nothing, M_1 keeps going forever. > Conversely, where is the flow in my presentation? You assumed that it would be trival to examine the contents of a turing mechean and determine if it prints a perticular charitor in general you can't do that. === Subject: Re: questions on turing machine problem > 1) I'm trying to understand the following > problem: > let M_1 and M_2 two > turing machines with common input alphabet and a > given string x. Is > there a particular step in which the two > machines > will write the same > symbol on the tape? > is it undecidable? can somebody give an idea on > how > can we reduce this > problem to an undecidable problem? > > This could be used to solve the halting problem so > its undecidable. If you > one > at each step, and your > set up M_2 as a modified version of the universal > turing mechine that What's the stopping criterion for M_1 there? Nothing, M_1 keeps going forever. I must be missing something, because it seems to me that _if you knew_ that M_1 never stops (and an analysis of the rule set might ascertain that, at least in the simple cases, like a loop in the rules), _then you would knew_ that M_2 does not stop either, although not by running M_2, because that would go on forever. Conversely, where is the flow in my presentation? You assumed that it would be trival to examine the > contents of > a turing mechean and determine if it prints a > perticular charitor in > general you can't do that. -LV === Subject: Most popular math chat room What is the most active, popular, and highly regarded math chat room on the internet? === Subject: (newbie) Do normal groups necessarily contain G' ? posting-account=NoOusQoAAADeMejejmF9SPjc_TwhqzV0 SLCC1; .NET CLR 2.0.50727; .NET CLR 3.0.04506; Media Center PC 5.0; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) Suppose that G is a group. I wonder if a normal subgroup N <= G must necessarily include the derived subgroup G'. If it is not the case, which is the smallest group that provides a counterexample? === Subject: Re: (newbie) Do normal groups necessarily contain G' ? > Suppose that G is a group. I wonder if a normal subgroup N <= G must > necessarily include the derived subgroup G'. If it is not the case, > which is the smallest group that provides a counterexample? No matter what G is, the 1-element subgroup is a normal subgroup. === Subject: Re: (newbie) Do normal groups necessarily contain G' ? <22182892.1210248846398.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=NoOusQoAAADeMejejmF9SPjc_TwhqzV0 SLCC1; .NET CLR 2.0.50727; .NET CLR 3.0.04506; Media Center PC 5.0; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) > Suppose that G is a group. I wonder if a normal subgroup N <= G must > necessarily include the derived subgroup G'. If it is not the case, > which is the smallest group that provides a counterexample? No matter what G is, the 1-element subgroup is a normal subgroup. That is right. But avoiding this trivial case, what else can be said ? To be more precise, is there a group with a normal subgroup N and derived group G', both non-trivial subgroups, sucht that N does not include G' ? === Subject: Re: (newbie) Do normal groups necessarily contain G' ? > Suppose that G is a group. I wonder if a normal subgroup N <= G must > necessarily include the derived subgroup G'. If it is not the case, > which is the smallest group that provides a counterexample? > No matter what G is, the 1-element subgroup is a normal subgroup. That is right. >But avoiding this trivial case, what else can be said ? >To be more precise, is there a group with a normal subgroup N and >derived group G', both non-trivial subgroups, sucht that N does not >include G' ? In G x H, [(x, y), (u, v)] = (x, y)(u, v)(x, y)^{-1}(u, v)^{-1} = [xux^{-1}u^{-1}, yvy^{-1}v^{-1}] = ([x, u], [y, v]). Therefore (G x H)' = G' x H'. So if we just take something like S_3 x C_2, {1} x C_2 is a non-trivial normal subgroup, and the derived group is C_3 x {1} (where C_3 is short for the subgroup generated by the cycle (123)). Is that right? (Group theory is still a little hazy for me.) -- Angus Rodgers Contains mild peril === Subject: Re: (newbie) Do normal groups necessarily contain G' ? <22182892.1210248846398.JavaMail.jakarta@nitrogen.mathforum.org> <510624t5bmaaqgk3rqmbq4umppdtfcjotl@4ax.com> posting-account=NoOusQoAAADeMejejmF9SPjc_TwhqzV0 SLCC1; .NET CLR 2.0.50727; .NET CLR 3.0.04506; Media Center PC 5.0; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) > Suppose that G is a group. I wonder if a normal subgroup N <= G must > necessarily include the derived subgroup G'. If it is not the case, > which is the smallest group that provides a counterexample? > No matter what G is, the 1-element subgroup is a normal subgroup. That is right. >But avoiding this trivial case, what else can be said ? >To be more precise, is there a group with a normal subgroup N and >derived group G', both non-trivial subgroups, sucht that N does not >include G' ? In G x H, [(x, y), (u, v)] = (x, y)(u, v)(x, y)^{-1}(u, v)^{-1} > = [xux^{-1}u^{-1}, yvy^{-1}v^{-1}] = ([x, u], [y, v]). Therefore > (G x H)' = G' x H'. æSo if we just take something like S 3 x C 2, > {1} x C 2 is a non-trivial normal subgroup, and the derived group > is C 3 x {1} (where C 3 is short for the subgroup generated by > the cycle (123)). Is that right? (Group theory is still a little > hazy for me.) === Subject: Re: (newbie) Do normal groups necessarily contain G' ? >In G x H, [(x, y), (u, v)] = (x, y)(u, v)(x, y)^{-1}(u, v)^{-1} >= [xux^{-1}u^{-1}, yvy^{-1}v^{-1}] = ([x, u], [y, v]). Therefore >(G x H)' = G' x H'. Excuse me for thinking aloud! (I'm trying to use sci.math to help me to learn to think on my feet, which means I'm liable to screw up.) This argument only works in one direction: an element of (G x H)' is a product of commutators, and so it belongs to G' x H', but it's not obvious (and for all I know is false) that an element of G' x H' must belong to (G x H)'. Trying to fix this ... an element of G' x H' can be written as: ([x_1, y_1][x_2, y_2]...[x_m, y_m], [u_1, v_1][u_2, v_2]...[u_n, v_n]) = ([x_1, y_1], 1)([x_2, y_2], 1)...([x_m, y_m], 1) (1, [u_1, v_1])(1, [u_2, v_2])...(1, [u_n, v_n]) = [(x_1, 1), (y_1, 1)][(x_2, 1), (y_2, 1)]...[(x_m, 1), (y_m, 1)] [(1, u_1), (1, v_1)][(1, u_2), (1, v_2)]...[(1, u_n), (1, v_n)] which does indeed belong to (G x H)'. I hope that's OK now. -- Angus Rodgers Contains mild peril === Subject: Re: (newbie) Do normal groups necessarily contain G' ? days. My association with the Department is that of an alumnus. >In G x H, [(x, y), (u, v)] = (x, y)(u, v)(x, y)^{-1}(u, v)^{-1} >= [xux^{-1}u^{-1}, yvy^{-1}v^{-1}] = ([x, u], [y, v]). Therefore >(G x H)' = G' x H'. Excuse me for thinking aloud! (I'm trying to use sci.math to help >me to learn to think on my feet, which means I'm liable to screw >up.) This argument only works in one direction: an element of (G x H)' >is a product of commutators, and so it belongs to G' x H', but >it's not obvious (and for all I know is false) that an element of >G' x H' must belong to (G x H)'. Note, on the other hand, that if you have an infinite family {G_i}_{i in I} of nontrivial groups, then you always have that Product_{i in I}(G_i') is contained in (Product_{i in I}G_i), but the inclusion may be proper. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes by Bill Watterson) Arturo Magidin magidin-at-member-ams-org === Subject: Re: (newbie) Do normal groups necessarily contain G' ? days. My association with the Department is that of an alumnus. Sigh... I should remember to finish my coffee before posting in the morning in the latter half of the week. >Note, on the other hand, that if you have an infinite family {G_i}_{i >in I} of nontrivial groups, then you always have that Product_{i in I}(G_i') is contained in (Product_{i in I}G_i), but the ^^^^^^^^^^^ should be contains. >inclusion may be proper. The other way around: the verbal subgroup of a product is contained in the product of the verbal subgroups, but there need not be equality if they the index set is infinite. Arturo Magidin magidin-at-member-ams-org === Subject: Re: (newbie) Do normal groups necessarily contain G' ? days. My association with the Department is that of an alumnus. Sigh... I should remember to finish my coffee before posting in the morning in the latter half of the week. >Note, on the other hand, that if you have an infinite family {G_i}_{i >in I} of nontrivial groups, then you always have that Product_{i in I}(G_i') is contained in (Product_{i in I}G_i), but the ^^^^^^^^^^^ should be contains. >inclusion may be proper. The other way around: the verbal subgroup of a product is contained in the product of the verbal subgroups, but there need not be equality if they the index set is infinite. Arturo Magidin magidin-at-member-ams-org === Subject: Re: (newbie) Do normal groups necessarily contain G' ? days. My association with the Department is that of an alumnus. >In G x H, [(x, y), (u, v)] = (x, y)(u, v)(x, y)^{-1}(u, v)^{-1} >= [xux^{-1}u^{-1}, yvy^{-1}v^{-1}] = ([x, u], [y, v]). Therefore >(G x H)' = G' x H'. Excuse me for thinking aloud! (I'm trying to use sci.math to help >me to learn to think on my feet, which means I'm liable to screw >up.) This argument only works in one direction: an element of (G x H)' >is a product of commutators, and so it belongs to G' x H', but >it's not obvious (and for all I know is false) that an element of >G' x H' must belong to (G x H)'. Remember that if K < G, then K' < G'. Since G x {1} and {1} x H are subgroups of G x H that are isomoprhic to G and H, you are done. >Trying to fix this ... an element of G' x H' can be written as: ([x_1, y_1][x_2, y_2]...[x_m, y_m], > [u_1, v_1][u_2, v_2]...[u_n, v_n]) = ([x_1, y_1], 1)([x_2, y_2], 1)...([x_m, y_m], 1) > (1, [u_1, v_1])(1, [u_2, v_2])...(1, [u_n, v_n]) = [(x_1, 1), (y_1, 1)][(x_2, 1), (y_2, 1)]...[(x_m, 1), (y_m, 1)] > [(1, u_1), (1, v_1)][(1, u_2), (1, v_2)]...[(1, u_n), (1, v_n)] which does indeed belong to (G x H)'. I hope that's OK now. Well, yes, but a bit too much work... -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes by Bill Watterson) Arturo Magidin magidin-at-member-ams-org === Subject: Re: (newbie) Do normal groups necessarily contain G' ? >Suppose that G is a group. I wonder if a normal subgroup N <= G must >necessarily include the derived subgroup G'. If it is not the case, >which is the smallest group that provides a counterexample? There's probably a slick way to do this, but I'm also a newbie in group theory, so I just drew up a multiplication table. 8-p It's well-known (i.e. I've just about heard of it - and it's easy to prove) that N contains G' iff G/N is abelian. Therefore, for a counterexample, we need G/N to be nonabelian. The smallest nonabelian group is S_3 - and this will do, because (as I just laboriously checked) its derived group is the (cyclic) subgroup of order 3. And, of course, {1} is a normal subgroup. -- Angus Rodgers Contains mild peril === Subject: Annuities II (Future Value) Calculator posting-account=ajkdngoAAAB2X3SxyQtiBYkcRETphklh CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) This is a second calculator in series of financial calculators under development at my site. This calculator allows computing Future value of Annuities and Annuities Due. Please visit http://thinkanddone.com/finance/annuitiesfv.aspx The last post mentioned the calculator that computed present value of annuities and annuities due. This one is available http://thinkanddone.com/finance/annuities.aspx === Subject: Re: Simple geometrical proof > If the case cannot be proved without further conditions (i.e., it > depends on the length of B), then what are these conditions? Say, > let B=qA, where 0= 1/3, then (1 + 3q)/2 >= 1. Hence (given A >= 0): A <= A (1 + 3q)/2 < C + D. So any q in [1/3, 1) will do. - Tim === Subject: Re: Simple geometrical proof > you cannot trust the graphical representation I give, just the > information I give below > 1. A>B > 2. A+B< D+E > 3. D=E > 4. length B = length C (as shown on graph) > How can I prove A Z may I write f:Y --> Z with a special kind of arrow, or something, to indicate that f is partial on Y? -- === Subject: video!! posting-account=iTIQZQoAAADpH6CwI6lev8X9FA0fRQUT Gecko/20080404 Firefox/2.0.0.14,gzip(gfe),gzip(gfe) Best Video..!! sexxx http://rozrywka.yeba.pl/show.php?id=2073 http://rozrywka.yeba.pl/show.php?id=2079 === Subject: Linear equations: everyday use? Hi there. I have the classic example of Linear Equations in everyday use as the naive computations of Google Juice, but can anyone here offer another? === Subject: Re: Linear equations: everyday use? The computer you are now using, and likely use almost everyday, uses a set of chips to perform certain computations. These chips get hot; in order to figure out how to make them get hot as little as possible, while at the same time maximizing performance, an engineer somewhere at some point had to solve some differential equations. The only practical way we have to solve such equations involves turning them into massive systems of linear equations (i.e. millions of coupled linear equations. http://en.wikipedia.org/wiki/Finite_element_method) Anything that at some point required CAD to design, inevitably required solving lots of linear equations. Nowadays this is pretty much everything -- from the space shuttle to the chair you're sitting on. === Subject: Re: University of Melbourne Maths Competition Intermediate Question > There was a maths test out of 100. 5 scored 100. Everyone scored at > least 60. The average was 75. What's the minimum number of students > who took the test? 5*100 + (n-5)*60 <= 75*n implies n >= 40/3, so 14 students (some of whom must have scored more than 60). (4 minutes) > There's three shirts and they cost $100 altogether at full > price. The price of each shirt is a whole number. First shirt > discounted by a fifth + second shirt discounted by 1/6 + third shirt > discounted 1/7 cost $84. What are the possible prices of the shirts > at full price? Let u,v,w be the discounts. Then u + v + w = 16, 5u + 6v + 7w = 100. They need to be integers since 5,6,7 are relatively prime. Multiplying and subtracting equations, w = u + 4, v + 2w = 20. Assuming that no shirt is given away for free (or paying to take it away!), the solutions are enumerated by 1 <= u <= 5: (5,60,35), (10,48,42), (15,36,49), (20,24,56), and (25,12,63). (11 minutes) > Add all the digits of the numbers from 1 through to 2008 and divide by > 9. What the hell's the remainder? That's just casting out 9's. Sum of all the numbers is 2008*2009/2 = 1004*2009, which is congruent mod 9 to 5*2, which gives 1. (2 minutes) - Tim === Subject: Re: University of Melbourne Maths Competition Intermediate Question posting-account=4W9t0woAAAB3dq7-fhllNhT3_mnzEKiC Gecko/20080404 Firefox/2.0.0.14,gzip(gfe),gzip(gfe) Here's a few questions from the test I took part in today. Oh and I > only had 3 hours for 7 questions. If you post a solution, please post > how long it took you just for me. > snip Sum of the numbers from 1 through to 2008 is 2008*2009/2 = 1004*2009 > which has the same remainder as the digit sum. > Modulo 9 this is the same as 5*11 or 5*2 or 10 or 1 using digit sums. > So 1 is the correct remainder. Could you explain in a bit more detail please? > I actually summed all the digits. > I went what's the sum of 1-9, then 10-19 then 20-29, figured out a > pattern, worked out 1-99. Then 100-109, 110-119, worked out 100-199 > then 200-299 then 100-999. Then 1000-1009, 1010-1019, 1000-1099, then > 1000-1999 then added 2, 3, 4, 5, 6, 7 and 8 onto it. I believe I got > 28 054 as my answer. There is an old adage about Gauss as a schoolboy being given a problem > like this to shut the class up for a bit but he solved it rapidly to > the chagrin of his schoolteacher! > His method in this case is > 1 + 2008 is ? > 2 + 2007 is > 3 + 2006 is > etc > How many of those pair-sums are there? > Then you can see the sum is 1004 * 2009 > Now use casting out nines (or modulo 9 arithmetic) to get the answer 1 > or you could do the calculation as 2017036 and divide by 9. > HTH > JJ You're not adding 1 to 2008. Here, your adding 1 + 2 + 0 + 0 + 8. > Was the answer to the last one that it was impossible? Originally, was that not the one where we had to try find 20 consecutive integers who had no prime factor greater than 18? It was that stupid question which they found out: omg, there's an error in our wording in our question in the last 20 minutes (well the email came in 15 minutes before it got to the exam room but that doesn't make a difference when everyone's checking their wording of their solutions at the end of uh long 3 hour test) stating the wording should be as thus and to not change your answer if you have already done the original erroneous wording: Find 20 consecutive POSITIVE integers which have have no prime factor greater than 18 or something like that. Silly me I didn't even attempt it though I looked at all of the first 6. So, lots of people in this group have done the test. Let me post question 5. For any quadrilateral (which has all 4 angles less than 180 degrees, what's the adjective?), if you join all 4 of the midpoints of each of the four sides together with 4 straight lines, that new quad will have parallel sides. Explain why. And question 6. Two people are playing a retarded game for us people to work out probabilities. There's a die and the first guy rolls it. Then the second guy does. They keep doing that until the sum of all previous rolls is more than 5 and the guy that makes it go over 5 loses. Show that the first player has a greater chance of winning. What kind of question is that??? === Subject: Re: University of Melbourne Maths Competition Intermediate Question posting-account=O9zR9AkAAACmp918j6u5m5plppeILcze 1.0.3705; .NET CLR 1.1.4322; Media Center PC 4.0; .NET CLR 2.0.50727; .NET CLR 3.0.04506.648; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) > question 5. For any quadrilateral (which has all 4 angles less than 180 degrees, > what's the adjective?), if you join all 4 of the midpoints of each of > the four sides together with 4 straight lines, that new quad will have > parallel sides. Explain why. Let the quadrilateral be ABCD. Construct diagonal AC of ABCD. It divides the quadrilateral into two triangles. Consider one of the triangles, say ABC. Let E be the midpoint of AB and F be the midpoint of CB. Then EF forms a small triangle EBF that is similar to ABC, because if two sides of one triangle are proportional to two sides of another triangle and the included angles are congruent, then the triangles are similar. Thus, angle FEB is congruent to CAB because corresponding angles of similar triangles are congruent. Hence EF is parallel to AC because if two lines are cut by a transversal that forms congruent corresponding angles, then the lines are parallel. In the same way, the segment connecting the midpoints of triangle ADC is parallel to AC, the segment connecting the midpoints of BA and DA is parallel to BD, and the segment connecting the midpoints of BC and DC is parallel to BD. Finally, the two segments parallel to AC are parallel to each other, and the two segments parallel to BD are parallel to each other, because two lines parallel to the same line are parallel to each other. Dave === Subject: Re: Questioning the defintions of set and element. > > That sounds interesting, although I am going to > contradict it. Struggling a formalization: x = [x] a > bit like: _element_ IS _set_ (or whichever > terminology you might wish there). Or, to use the usual notation: x = {x}. Quine's NF > may be of interest > to you. Hi Frederick, Actually, the whole discussion here is of great interest to me, and I am not replying or saying anything at all just because at the moment I am busy enough understanding what you and other people are saying. Julio === Subject: Re: Questioning the defintions of set and element. posting-account=euF15goAAACbw3KIqEWxZHCIPUc2KPmU .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506),gzip(gfe),gzip(gfe) > But this is still invalid in ZFC. Why? Because, for > example, there are uncountably many sets (say subsets > of omega), but only countably many properties P. > Sorry for drifting off-topic, but thinking about this counterintuitive > fact has made me wonder: is there a version of logic in which one can > quantify over predicates? I doubt it. Once again, I'm only expressing what the beginner is thinking in his or her mind -- that there's a predicate for every set. Of course, you and I both know that this is flawed, for the reason that you state above. === Subject: Re: The Future of Chess and Getting It Unstuck Fischer Random Chess. You could add not only random starting positions but random NEW pieces that move in new random ways. And then turn it into losing chess half the time. http://i7-dungeon.sourceforge.net/older.html === Subject: Re: The Future of Chess and Getting It Unstuck > On May 7, 9:32 am, David Richerby In [Deutsch's] paper, the convenience of thinking that all > computing can be reduced to an equivalentTuringmachine is > considered and rejected.http://xyz.lanl.gov/abs/math.HO/9911150As > a matter of fact, Richard Feynman, in his talk during the First > Conference on the Physics of Computation held at MIT in 1981, > observed that it appears to be impossible to simulate a general > quantum evolution on a classical probabilistic computer in an > efficient way. > That is largely because the only thing Feynman knew about computers > or computations was main frame computers. But since most of today's > computations depend on massively parellel networks, not mainframes, > his observation mostly concern 1950s Burroughs history, not > computers. > No. Combining two classical computers together makes a computer twice > as big that can do (roughly) twice as much work in a given time. > Connecting two quantum computers would make a computer twice as big > that can do (roughly) the square of the amount of work in a given time. > Networking classical computers together gives you at best linear > growth; adding qubits to your quantum computer gives exponential > growth. Networking classical computers gives you microcomputers, > laserdisks, > satellites,, HDTV, Holograms, fiber optics, robots, and a > paycheck, > rather than idiots like computer scientists, that's why they were > invented. You guys are speaking rather orthogonally. In theory (his topic), quantum computers could do all this and more. In practice (yours), adding more qubits is (so far) Really Difficult; it will take a major technological breakthrough to build a quantum computer with enough qubits that a network of classical computers can't simulate it at full speed. === Subject: Re: The Future of Chess and Getting It Unstuck > it will take a major technological breakthrough to build a quantum > computer with enough qubits that a network of classical computers > can't simulate it at full speed. Is it known that the difficulty of maintaining coherence grows less than exponentially in the number of interacting qubits? - Tim === Subject: Re: A hard to calculate limes... >can anybody calculate this limes: lim [ (x+a)^( 1 + 1/x ) - x^(1 + 1/(x+a) ) ] when x goes to infinitive >and a belongs to R. >and the answer is a :] This may be similar in idea to some other attempts, but perhaps a different approach may add clarity. (x+a)^( 1 + 1/x ) - x^(1 + 1/(x+a) ) = exp((1 + 1/x) log(x+a)) - exp((1 + 1/(x+a)) log(x)) = exp((1 + 1/x) (log(x) + log(1 + a/x))) - exp((1 + 1/x 1/(1 + a/x)) log(x)) = exp((1 + 1/x) (log(x) + a/x + O(1/x^2))) - exp((1 + 1/x + O(1/x^2)) log(x)) = exp(log(x) + log(x)/x + a/x + O(1/x^2)) - exp(log(x) + log(x)/x + O(log(x)/x^2)) = exp(log(x) + log(x)/x) [exp(a/x + O(1/x^2)) - exp(O(log(x)/x^2))] = x exp(log(x)/x) [(1 + a/x + O(1/x^2)) - (1 + O(log(x)/x^2))] = x (1 + O(log(x)/x)) (a/x + O(log(x)/x^2)) = (1 + O(log(x)/x)) (a + O(log(x)/x)) = a + O(log(x)/x) Rob Johnson take out the trash before replying === Subject: Re: A hard to calculate limes... amy666 a .8ecrit : hi, can anybody calculate this limes: lim [ (x+a)^( 1 + 1/x ) - x^(1 + 1/(x+a) ) ] > when > x goes to infinitive > and a belongs to R. > and the answer is a :] > hint: > 1) if a = 0 then ... > 2) if x is very large then could you 'ignore' a? > As the answer is a (which he gave), 2) doesn't > look > promising. > well if the answer has to be a by his definition , > its not a limit nor a lemon or a lime. > what we get is an equation instead : > (x+a)^( 1 + 1/x ) - x^(1 + 1/(x+a) = a > with x at infinity ( hmm kind of a lim afterall then > :p ) > if a = 0 that is a solution , even for all x not just > oo. > so i assume the OP wants x = oo and non-zero > solutions for a. > tommy1729 forget that , i got it wrong. Surprise, surprise he says the limit is a for any a. === Subject: Re: A hard to calculate limes... posting-account=OKTeIQkAAAAZk6JK1hK7-grwpoUDNy98 CLR 1.1.4322; .NET CLR 2.0.50727; InfoPath.1),gzip(gfe),gzip(gfe) > yyyyy what is limeade??? http://en.wikipedia.org/wiki/Limeade === Subject: Re: A hard to calculate limes... > hi, can anybody calculate this limes: lim [ (x+a)^( 1 + 1/x ) - x^(1 + 1/(x+a) ) ] when x goes to infinitive > and a belongs to R. > and the answer is a :] > hint: > 1) if a = 0 then ... > 2) if x is very large then could you 'ignore' a? As the answer is a (which he gave), 2) doesn't look promising. === Subject: Re: A hard to calculate limes... > hi, can anybody calculate this limes: lim [ (x+a)^( 1 + 1/x ) - x^(1 + 1/(x+a) ) ] when x goes to infinitive > and a belongs to R. > and the answer is a :] One approach is to write it as (x+a)^(1 + 1/x) - x^(1 + 1/x) + x^(1 + 1/x) - x^(1 + 1/(x+a)). On the first line use y^p - z^p = p*c^(p-1)*(y-z) for some c between y and z. With a little more work this will show the first line -> a. On the second line use b^y - b^z = ln(b)*b^c*(y-z) for some c between y and z, which will show this line -> 0. We are of course using the mean value theorem in each case. === Subject: Re: A hard to calculate limes... >can anybody calculate this limes: >lim [ (x+a)^( 1 + 1/x ) - x^(1 + 1/(x+a) ) ] when x goes to infinitive >and a belongs to R. >and the answer is a :] > One (horribly messy) solution is to write the expression as: > a + f(a, x) - f(-a, x + a) > where: > f(a, x) > = (x + a)[(x + a)^{1/x} - 1] > = (x + a){exp[(ln(x + a))/x] - 1} > = (x + a){(ln(x + a))/x + O[((ln(x + a))/x)^2]} > = (x + a)(ln(x + a))/x + O[(ln(x)^2)/x] > = ln(x + a) + O[ln(x)/x] + O[(ln(x)^2)/x] > so the given expression is: > a + ln(x + a) - ln(x) + O[(ln(x)^2)/x] > = a + ln(1 + a/x) + o(1) > = a + o(1) > Ugh! > But it works (assuming I haven't messed up). I'm too stupid to solution this by myself;/ but I'm really grateful. Before you start being grateful, better check that I haven't goofed! (I left out a few steps that looked obvious - but you never know! I'm old and rusty, and sometimes goof hideously. So, look closely at any step you're not sure of, and see if you know why it's valid - or if it even is valid.) >and one more think what is the o()?? it looks like Landau symbols... Yes, I used both the O() and o() notations. (Correctly, I hope!) -- Angus Rodgers Contains mild peril === Subject: Re: A hard to calculate limes... <76t324dlgrk4gh4bvqonvjjrca1me4i513@4ax.com> posting-account=g9m0FgoAAAD9MrZl_8GR7R2eFE1W4Hho Gecko/20061023 SUSE/2.0-30 Firefox/2.0,gzip(gfe),gzip(gfe) heh maybe old but still very good in math :] did you invent this solution by your self or did you saw it earlier? === Subject: Re: A hard to calculate limes... >heh maybe old but still very good in math :] did you invent this >solution by your self or did you saw it earlier? All my own cock-up. :-) -- Angus Rodgers Contains mild peril === Subject: Re: A hard to calculate limes... <54u3245t1j3fp33fpi9mnouk33vdpq8sk2@4ax.com> posting-account=g9m0FgoAAAD9MrZl_8GR7R2eFE1W4Hho Gecko/20061023 SUSE/2.0-30 Firefox/2.0,gzip(gfe),gzip(gfe) so congratulations;] no one of my friend haven't any ideas how to calculate it:] you are mathematician or someone similar? look at this line: = (x + a){(ln(x + a))/x + O[((ln(x + a))/x)^2]} = (x + a)(ln(x + a))/x + O[(ln(x)^2)/x] shouldn't be : = (x + a){(ln(x + a))/x + O[((ln(x + a))/x)^2]} = (x + a)(ln(x + a))/x + O[(ln(x)^2)/x^2] <---???? === Subject: Re: A hard to calculate limes... >so congratulations;] no one of my friend haven't any ideas how to >calculate it:] you are mathematician or someone similar? Someone similar. :-) >look at this line: = (x + a){(ln(x + a))/x + O[((ln(x + a))/x)^2]} > = (x + a)(ln(x + a))/x + O[(ln(x)^2)/x] shouldn't be : >= (x + a){(ln(x + a))/x + O[((ln(x + a))/x)^2]} > = (x + a)(ln(x + a))/x + O[(ln(x)^2)/x^2] <---???? That was one of the obvious steps I left out. Because ln(x + a) = O(ln(x)) (in fact, ln(x + a) ~ ln(x)), and x + a ~ x: (x + a)O[((ln(x + a))/x)^2] = (x + a)O[(ln(x + a)^2)/x^2] ~ xO[(ln(x)^2)/x^2] = O[(ln(x)^2)/x] (and vice versa, but we don't need this). Good thing you're keeping me on my toes. Keep checking! -- Angus Rodgers Contains mild peril === Subject: Re: A hard to calculate limes... > [...] = (x + a)O[(ln(x + a)^2)/x^2] ~ xO[(ln(x)^2)/x^2] The '~' should be a '=' there (although of course '=' doesn't exactly mean '=' when 'O()' or 'o()' is involved!). -- Angus Rodgers Contains mild peril === Subject: Re: A hard to calculate limes... <1hu324p26h76hbtpuq2q5don7c60rhr52p@4ax.com> <5k34241m8i7o06eoe3r1vtd28jauio7hg7@4ax.com> posting-account=g9m0FgoAAAD9MrZl_8GR7R2eFE1W4Hho Gecko/20061023 SUSE/2.0-30 Firefox/2.0,gzip(gfe),gzip(gfe) what can I say?? thank you all for help !!!! === Subject: Re: A hard to calculate limes... posting-account=g9m0FgoAAAD9MrZl_8GR7R2eFE1W4Hho Gecko/20061023 SUSE/2.0-30 Firefox/2.0,gzip(gfe),gzip(gfe) OK tell me something more about this solution it's hard to understand;/ why we use O and o??? === Subject: Re: A hard to calculate limes... >OK tell me something more about this solution it's hard to >understand;/ why we use O and o??? Because I would get in [even more of] a muddle if I didn't. It's just very handy - not essential. -- Angus Rodgers Contains mild peril === Subject: Re: A hard to calculate limes... <69v324hac2pjsds6rg27epjmkepc79icjc@4ax.com> posting-account=g9m0FgoAAAD9MrZl_8GR7R2eFE1W4Hho Gecko/20061023 SUSE/2.0-30 Firefox/2.0,gzip(gfe),gzip(gfe) and I hope the last thing: why here disappear exp?? = (x + a){exp[(ln(x + a))/x] - 1} = (x + a){(ln(x + a))/x + O[((ln(x + a))/x)^2]} eh I know that it is easy for you but it's hard for me:( === Subject: Re: A hard to calculate limes... >and I hope the last thing: why here disappear exp?? >= (x + a){exp[(ln(x + a))/x] - 1} = (x + a){(ln(x + a))/x + O[((ln(x + a))/x)^2]} y = ln(x + a)/x -> 0 as x -> oo, and exp(y) = 1 + y + O(y^2) as y -> 0. That's expressed rather loosely (which is the way I think of it!), but I hope it conveys the idea. If necessary, translate it into the language of 'epsilons' - that will expose any logical flaws. >eh I know that it is easy for you but it's hard for me:( I would probably panic if I had to justify every step in detail. Mathematics demands logical perfection, and some day I'm going to have to try to come to terms with that demand again, but at the moment I'm trying to get the hang of reasoning imperfectly. This involves a certain amount of bluff, but if I bluff in public, the bluff can be called, if necessary, which keeps me reasonably honest! It's hard to explain this properly (I keep getting tangled up in psychology!), but I'm trying to learn to have confidence in arguments that feel right - and confidence, also, that I can back up any sketchy and loose arguments with more detailed and rigorous ones if called on to do so. I long ago lost confidence, both in myself and in mathematics, and nothing now feels easy, the way a lot of things once did. Excuse me - I could, and sometimes do, ramble on at length about psychological matters! Long story short: if I'm giving the impression that I think things are easy, it's a bluff! (But the right kind of bluffing is OK.) :-) -- Angus Rodgers Contains mild peril === Subject: Re: A hard to calculate limes... Angus Rodgers a .8ecrit : > > and I hope the last thing: why here disappear exp?? > = (x + a){exp[(ln(x + a))/x] - 1} > = (x + a){(ln(x + a))/x + O[((ln(x + a))/x)^2]} y = ln(x + a)/x -> 0 as x -> oo, and exp(y) = 1 + y + O(y^2) as y -> 0. That's expressed rather loosely (which is the way I think of it!), Not loosely at all, I think. After all, this is *exactly* the reason o and O notation were invented, and you have only to be sure of simple rules like the ones you use there > but I hope it conveys the idea. If necessary, translate it into > the language of 'epsilons' - that will expose any logical flaws. > Yes (I personnally suggest to my students to always replace o(y^2), say, by y^2*eps(y) whenver they are not 100 % sure of the legitimacy of their calculations) > eh I know that it is easy for you but it's hard for me:( I would probably panic if I had to justify every step in detail. > Mathematics demands logical perfection, and some day I'm going > to have to try to come to terms with that demand again, but at > the moment I'm trying to get the hang of reasoning imperfectly. > This involves a certain amount of bluff, but if I bluff in > public, the bluff can be called, if necessary, which keeps me > reasonably honest! It's hard to explain this properly (I keep > getting tangled up in psychology!), but I'm trying to learn to > have confidence in arguments that feel right - and confidence, > also, that I can back up any sketchy and loose arguments with more > detailed and rigorous ones if called on to do so. I long ago lost > confidence, both in myself and in mathematics, and nothing now > feels easy, the way a lot of things once did. Excuse me - I could, > and sometimes do, ramble on at length about psychological matters! > Long story short: if I'm giving the impression that I think things > are easy, it's a bluff! (But the right kind of bluffing is OK.) :-) > === Subject: Covariance Matrix Calculation, Last Step - Expectation of a Matrix posting-account=Ux2Z0goAAAC4J8BZI3y8Q-Iw5VyhIqWP Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) Hi All, I would really appreciate your help on this one. I'm calculating a covariance matrix. However I'm having trouble understanding how to calculate the expected value of a Matrix. So far I manage to reach (X-E[X]) * (X-E[X])'. The original data is X 67 79 30 37 76 90 38 13 90 I reached: (X-E[X]) * (X-E[X])' 4458.6 2692.5 -459.9 2692.5 5435.4 2442.1 -459.9 2442.1 3418.7 I know that the final result is 193.6 215.7 926.0 -393.3 -460.0 800.0 However I'm not able to reach it. MT === Subject: Is there book on Series solution of PDE? posting-account=g1OGIgoAAABPkZd1Sq2qvJtXdCA6ujpI MathPlayer 2.10b; Maxthon; .NET CLR 1.1.4322; .NET CLR 2.0.50727; .NET CLR 3.0.04506.30; InfoPath.1),gzip(gfe),gzip(gfe) === Subject: Matrix Rank Equation posting-account=2eef1goAAABkCPRixyGuos1jigjAi5oE Gecko/20070725 Firefox/2.0.0.6,gzip(gfe),gzip(gfe) Suppose X is a random data matrix with dimension n by t M=inv(X ' * X), i.e. M is the inverse of product of (transpose X and X) and P=I-X * M * X' I is an identity matrix (n by n), is rank (P) = n-t? %%%%%%%%%%%rank_test.m for n=7:15 for t=2:7 X=randn(n,t); M=inv(X'*X); I=eye(n,n); P=I-X*M*X'; r=rank(P); fprintf('the rank of p at n=%d, t=%d is %dn',n,t,r); end end And the result IS correct: the rank of p at n=7, a=2 is 5 the rank of p at n=7, a=3 is 4 the rank of p at n=7, a=4 is 3 the rank of p at n=7, a=5 is 2 the rank of p at n=7, a=6 is 1 the rank of p at n=7, a=7 is 7 the rank of p at n=8, a=2 is 6 the rank of p at n=8, a=3 is 5 the rank of p at n=8, a=4 is 4 the rank of p at n=8, a=5 is 3 the rank of p at n=8, a=6 is 2 the rank of p at n=8, a=7 is 1 the rank of p at n=9, a=2 is 7 the rank of p at n=9, a=3 is 6 the rank of p at n=9, a=4 is 5 the rank of p at n=9, a=5 is 4 the rank of p at n=9, a=6 is 3 the rank of p at n=9, a=7 is 3 the rank of p at n=10, a=2 is 8 the rank of p at n=10, a=3 is 7 the rank of p at n=10, a=4 is 6 the rank of p at n=10, a=5 is 5 the rank of p at n=10, a=6 is 4 the rank of p at n=10, a=7 is 4 the rank of p at n=11, a=2 is 9 the rank of p at n=11, a=3 is 8 the rank of p at n=11, a=4 is 7 the rank of p at n=11, a=5 is 6 the rank of p at n=11, a=6 is 5 the rank of p at n=11, a=7 is 4 the rank of p at n=12, a=2 is 10 the rank of p at n=12, a=3 is 9 the rank of p at n=12, a=4 is 8 the rank of p at n=12, a=5 is 7 the rank of p at n=12, a=6 is 6 the rank of p at n=12, a=7 is 5 the rank of p at n=13, a=2 is 11 the rank of p at n=13, a=3 is 10 the rank of p at n=13, a=4 is 9 the rank of p at n=13, a=5 is 8 the rank of p at n=13, a=6 is 7 the rank of p at n=13, a=7 is 6 the rank of p at n=14, a=2 is 12 the rank of p at n=14, a=3 is 11 the rank of p at n=14, a=4 is 10 the rank of p at n=14, a=5 is 9 the rank of p at n=14, a=6 is 8 the rank of p at n=14, a=7 is 7 the rank of p at n=15, a=2 is 13 the rank of p at n=15, a=3 is 12 the rank of p at n=15, a=4 is 11 the rank of p at n=15, a=5 is 10 the rank of p at n=15, a=6 is 9 the rank of p at n=15, a=7 is 8 %%%%%%%%%%%%%%%%% So far, I vaguely feel this identity is right by calculating the trace tr(P)=tr(I)-tr(X * M * X')=tr(I)-tr(X'*X*M)=n-t But is there any rigorous calculation or proof that anybody knows? === Subject: Re: Matrix Rank Equation posting-account=2eef1goAAABkCPRixyGuos1jigjAi5oE Gecko/20070725 Firefox/2.0.0.6,gzip(gfe),gzip(gfe) Hum.... There ARE some exceptions. I think it's due to the numerical subtlety. > Suppose X is a random data matrix with dimension n by t M=inv(X ' * X), i.e. M is the inverse of product of (transpose X and > X) and P=I-X * M * X' I is an identity matrix (n by n), is rank (P) = n-t? > %%%%%%%%%%%rank_test.m > for n=7:15 > for t=2:7 > X=randn(n,t); > M=inv(X'*X); > I=eye(n,n); > P=I-X*M*X'; > r=rank(P); > fprintf('the rank of p at n=%d, t=%d is %dn',n,t,r); > end > end And the result IS correct: > the rank of p at n=7, a=2 is 5 > the rank of p at n=7, a=3 is 4 > the rank of p at n=7, a=4 is 3 > the rank of p at n=7, a=5 is 2 > the rank of p at n=7, a=6 is 1 > the rank of p at n=7, a=7 is 7 > the rank of p at n=8, a=2 is 6 > the rank of p at n=8, a=3 is 5 > the rank of p at n=8, a=4 is 4 > the rank of p at n=8, a=5 is 3 > the rank of p at n=8, a=6 is 2 > the rank of p at n=8, a=7 is 1 > the rank of p at n=9, a=2 is 7 > the rank of p at n=9, a=3 is 6 > the rank of p at n=9, a=4 is 5 > the rank of p at n=9, a=5 is 4 > the rank of p at n=9, a=6 is 3 > the rank of p at n=9, a=7 is 3 > the rank of p at n=10, a=2 is 8 > the rank of p at n=10, a=3 is 7 > the rank of p at n=10, a=4 is 6 > the rank of p at n=10, a=5 is 5 > the rank of p at n=10, a=6 is 4 > the rank of p at n=10, a=7 is 4 > the rank of p at n=11, a=2 is 9 > the rank of p at n=11, a=3 is 8 > the rank of p at n=11, a=4 is 7 > the rank of p at n=11, a=5 is 6 > the rank of p at n=11, a=6 is 5 > the rank of p at n=11, a=7 is 4 > the rank of p at n=12, a=2 is 10 > the rank of p at n=12, a=3 is 9 > the rank of p at n=12, a=4 is 8 > the rank of p at n=12, a=5 is 7 > the rank of p at n=12, a=6 is 6 > the rank of p at n=12, a=7 is 5 > the rank of p at n=13, a=2 is 11 > the rank of p at n=13, a=3 is 10 > the rank of p at n=13, a=4 is 9 > the rank of p at n=13, a=5 is 8 > the rank of p at n=13, a=6 is 7 > the rank of p at n=13, a=7 is 6 > the rank of p at n=14, a=2 is 12 > the rank of p at n=14, a=3 is 11 > the rank of p at n=14, a=4 is 10 > the rank of p at n=14, a=5 is 9 > the rank of p at n=14, a=6 is 8 > the rank of p at n=14, a=7 is 7 > the rank of p at n=15, a=2 is 13 > the rank of p at n=15, a=3 is 12 > the rank of p at n=15, a=4 is 11 > the rank of p at n=15, a=5 is 10 > the rank of p at n=15, a=6 is 9 > the rank of p at n=15, a=7 is 8 %%%%%%%%%%%%%%%%% So far, I vaguely feel this identity is right by calculating the trace tr(P)=tr(I)-tr(X * M * X')=tr(I)-tr(X'*X*M)=n-t But is there any rigorous calculation or proof that anybody knows?