mm-1065 === Subject: Any decent math software for students? Is there any decent math software for people who simply want to brush up on basic algebra, geometry, trig and basic calculas? I've looked at Mathematica 5, but the standard version is nearly $2000. A bit steep for me. === Subject: Re: Any decent math software for students? > Is there any decent math software for people who > simply want to brush up on basic algebra, geometry, trig and basic > calculas? > I've looked at Mathematica 5, but the standard version is nearly > $2000. A bit steep for me. MuPad - free for non-comercial use Maxima - free Derive - not free but not-expensive === Subject: Re: Any decent math software for students? i've tried Mathematica, Matlab and Mathcad, but nothing can beat Maple. In my humble opinion it's easiest to use (if you know what you want) and so far i haven't found a problem it can't handle (i'm not saying there aren't any). Now it has a new version and a student edition at reduced price (a little over $100 i think). === Subject: Re: Any decent math software for students? >Is there any decent math software for people who >simply want to brush up on basic algebra, geometry, trig and basic >calculas? Do you mean instructional software, or software you can use to check your work and maybe do some of the grunt work? You'll find pointers to some of the latter kind on my page http://oakroadsystems.com/math/mathlinx.htm As an alternative, you might consider a TI-89 calculator. It can do everything right up through calculus, and is a lot more portable than any computer. Searching on the Internet you should be able to find one for about $125. -- Stan Brown, Oak Road Systems, Cortland County, New York, USA http://OakRoadSystems.com Fortunately, I live in the United States of America, where we are gradually coming to understand that nothing we do is ever our fault, especially if it is really stupid. --Dave Barry === Subject: Re: Any decent math software for students? > Is there any decent math software for people who > simply want to brush up on basic algebra, geometry, trig and basic > calculas? > I've looked at Mathematica 5, but the standard version is nearly > $2000. A bit steep for me. I'm a 48 year old computer science engineer, and got Derive and Geometer's Sketchpad about a year ago for just the purpose you describe. Have spent many hours with each one and have as much attachment to them as my old faithful HP calculator. http://education.ti.com/us/product/software/derive/features/features.html http://www.keypress.com/sketchpad/ Hope this helps, Rick http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 80,000 Newsgroups - 16 Different Servers! =----- === Subject: Re: Attacking my own math proof, fun > If you're a liar, or not even a mathematician, you may wish to > continue to debate. Brilliant! === Subject: Re: Attacking my own math proof, fun why don't you try working on another problem, for a change, perhaps one that is even answered already, in terms taht you can except without beating -- however limpwristedly -- on mathfolk? you often refer to the great, unwashed-but-presumably-educated, silent majority out here in Googolplexland, but can you find evidence of even one, who has stuck with your prevarications, after his (or her) initial locquaciousness? > However, because it's Usenet, the poster may simply post as if it > didn't happen. > Special cases do NOT prove the general case. > That proves the general case, unless you expect readers to believe > that algebraic equations first check a variable value, like to see if > it's 3 or not, before deciding to be functions. > Hmmm...maybe you do expect readers to believe such a thing. --UN HYDROGEN (sic; Methanex (TM) reformanteurs) ECONOMIE?... La Troi Phases d'Exploitation de la Protocols des Grises de Kyoto: (FOSSILISATION [McCainanites?] (TM/sic))/ BORE/GUSH/NADIR @ http://www.tarpley.net/aobook.htm. Http://www.tarpley.net/bushb.htm (content partiale, below): 17 -- L'ATTEMPTER de COUP D'ETAT, 3/30/81 === Subject: Re: Attacking my own math proof, fun > Lots deleted this time. > I wouldn't mind doing a sanity check, then restate my problem if you don't > mind. > Fine. We have > P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) > = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) > so it's really a cubic polynomial in x. f is an integer coprime to 3. m is > any > integer coprime to f. Not it's NOT really a cubic polynomial in x, as it's not a polynomial > at all. A polynomial has constant coefficients while it has variable ones. You can consider it to be a polynomial with respect to various > factors, which implies that certain variables are constant; however, > your assumption is not a mathematical constraint. That's an important point. > OK as you prefer. Of course it's really a function of m, x, f, and u. > Yes, you can consider it to be a function, which doesn't help. > However, what you *cannot* do is say it's just a polynomial with > respect to some particular personal choice, like what you did in the > previous post and readers can see your comments above. > While your mind may prefer a simple explanation the math doesn't care, > and it's also not a mind reader. > What the math sees is an uber-polynomial, not a polynomial, as a > polynomial has constant coefficients, while with the uber-polynomial > you have all those variables. I don't mind this, and you do get to pick the term for how to describe it. I appreciate that they're not just expressions picked at random, so I'm not intending to dismiss it. After all, how could it be? Is the math supposed to read your mind > and figure out which point of view you have? Whether or not you're > looking at a polynomial with respect to x or m, or even f? Not sure about u. It's an integer that usually ends up being set to 1 in > most > discussions. > So you can pick values for f, m and u, constrained only by their coprimality > requirements, so > you're actually considering an infinite number of poynomials of x. Nope. You keep trying to force the expression into a box, when it > can't be so forced. While you can consider it as containing an infinite number of > polynomials of x, it also has an infinite number of polynomials of m, > or other variables. That's why I call it the uber-polynomial. It's a special construction for a particular purpose. Yep. Doing some algebra, by expanding the right-hand side and equating like > values of > x, among > other things you can discover: a_1 a_2 a_3 = f^2(m^3 f^4 - 3m^2 f^2 + m) Hmmm. Too much cutting and pasting. I think you actually get > a_1 a_2 a_3 = f^2(m^3 f^4 - 3m^2 f^2 + 3m) Never mind, no arguments have changed. and the a's are the solutions to: a^3 + 3 (-1 + mf^2)a^2 -f^2(m^3 f^4 - 3m^2 f^2 + 3m) = 0 So they are indeed algebraic integers. Nice construction. At m=0, the > solutions > are 0,0,3 Which tells you how f^2 divides off, if you also consider that with > P(m) at m=0, you have P(0) = u^2(3x + uf), which shows that a_3 is coprime to f, since f is coprime to 3 and x. > Yes, at m=0. We agree that a_3 is coprime to f when m=0. > And I'm sensing that you must still think that there's some variable > dependency on m, or there wouldn't be further discussion. Well, yes I do. The a's are clearly dependent on m and f. Their cubic is described above. So I don't really know what you mean, I don't think. The m and f are almost independent - they can be anything as long as f is coprime to 3 and m. > Are you still trying to claim that you are not? > If you're not then I can just check at m=0, confident that I've > covered when m does not equal 0, right? > If not, why do you believe so? I'm sure this is not the case. a is a function of m. Trivial analogy: Say a_3 = 3 + m (1 + f) Imagine f=5 so a_3 is coprime to f when m=0. Now, when m=7, also coprime to 5, a_3 is not coprime to 5. I know your a_3's aren't as above, but I honestly don't accept that checking at m=0 is sufficient, unless I can see a proof of it. That's important, as if f is not coprime to 3 and x, you get a > different result, so it matters to check the constant term P(0). Now considering Q(m) = P(m)/f^2 Q(m) = m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f > = (b_1 x + u)(b_2 x + u)(a_3x + uf) and you get > b_1 b_2 a_3 = m^3 f^4 - 3m^2 f^2 + 3m and the b's are two of the solutions to: fb^3 + 3 ( -1 + mf^2)b^2 -m^3 f^4 - 3m^2 f^2 + 3m = 0 Sorry, if anyone's watching. This should have been: fb^3 + 3 ( -1 + mf^2)b^2 - ( m^3 f^4 - 3m^2 f^2 + 3m ) = 0 In general then, the b's are not algebraic integers as m and f are coprime. You're making leaps. If you let f=sqrt(2), m=1, you'll see that two > of the b's are algebraic integers, while one is not for your cubic. > No problem, that's what I meant by in general > Also you've constrained m and f to be integers, so it's irrelevant to consider > non-integer situations. Even if you haven't, let's see if your argument is valid > when f and m are integers. > Sorry but that example with f=sqrt(2), m=1, removes room for your > claim that the b's can only be in a field. > The example ends your previous claim that using a_1 = fb_1 and a_2 = > fb_2 forces you into the field of algebraic numbers. > Now you may *believe* that reducibility matters, but all that happens > when P(m) is irreducible is that you can't just *look* at the results, > and you may have seized on your inability to check by looking with an > irreducible, as if it were a limitation on the math. > It is not. > In fact the b's are never forced into a field, but they are forced out > of the ring of algebraic integers. OK. It's important that it's recoginsed that they are often not algebraic integers. > That follows from the math, while your assertions provably do not. It > seems you have a personal preference, a belief, to which you're trying > to hold on, against mathematical logic. Is this sanity check correct? Please fix it if not. The cubics are actually > derived quite easily using your substitution of v = -1 + m^2 They look ok to me. As a side issue, the 2 b's you want are the 2 that are equal to 0 at m=0. > Probably if you solved these cubics, these 2 b's might be evident. As side > point > to consider, what the hell is the 3rd b and where did it come from? One of the b's is a_3. > a_3/f actually. > Yeah, you're right. Now onto the disagreement we have: You claim that because b_1 b_2 a_3 = m^3 f^4 - 3m^2 f^2 + 3m and RHS is coprime to f (which it is as m and f are coprime) > then a_3 is coprime to f > Note that b_1 and b_2 are NOT algebraic integers. > Not by any theorem of mathematics is your claim correct. You're making several false assumptions, and it's easy to refute your > position by having you consider f=sqrt(2), m=1. > I think my assumptions are. > - f is an integer coprime to 3. > - m is an integer coprime to f. > Which are false, and what other false assumptions have I made that I'm not even > aware of? > Your assumption that the field of algebraic numbers is required, which > you've repeated several times, but have been unable to prove. As long as we're mindful that the b's are not always algebraic integers. Clearly, you've seized on one idea, and you keep holding on to it, > despite my efforts to get you to follow the math. > I'm trying to follow the math. I keep getting stuck here though: > You claim that because > b_1 b_2 a_3 = m^3 f^4 - 3m^2 f^2 + 3m > and RHS is coprime to f (which it is as m and f are coprime) > then a_3 is coprime to f. > This is only true for integer m and f when m=0. > Which indicates that you believe that it might be different when m > does not equal 0, which forces a dependency on m. Yes, the values of the a's and b's depend on m. The RHS is coprime to f, of course. I'm hoping that you don't mean something as trivial as that. It is agreed that b_1b_2a_3 is coprime to f. It's not agreed that a_3 is coprime to f for all valid m and f. > Now it's quite simple, admit that you believe there's a variable > dependency on m, and then I can show there is none. I think this would be extremely valuable. But you shouldn't feel any pressure to convince me of anything. The choice, and the floor is yours if you want. I believe there's a variable dependency on m.The a's and b's are variables here, right? They're the things I think are dependent on m. > However, if you continue to maintain that there is no dependency on m, > but then try to give emphasis on the case when m=0, then that > fundamental contradiction in your thinking means that you cannot be > following the math. Except at m=0 when b_1 and b_2 are =0 and are algebraic integers. Well, try m=1 with f=sqrt(2), and welcome to a more complicated > mathematical world than you might have realized. > I'm sure it will work out fine with those values, so I won't try. > How about YOU try when f=5, and m is your favorite non-zero integer coprime to > 5. > Why? No reason. I'm guessing though that a_3 is not coprime to 5. > wrong assumption, which is that the constant f^2 divides off in some > way as a function of m or variable dependent on m, which you have > repeatedly demonstrated by citing the m=0 case as if it is a special > case, but when I say that's your assumption, you claim it is not. In my mind, it doesn't matter how the f^2 divides off. If you resolve the a's and b's to their cubic definitions, the steps you took to get there disappear in any case. I haven't actually said that m=0 is a special case at any point. Although secretly I think that solving for m=0 isn't sufficient to show that the result applies for all m. Not so secretly, obviously. > Looking at other examples will not help you, but trying to focus you > onto the basic contradiction in your position--the claim that there is > no dependency on m, when you keep pointing at m=0 as if there is--just > might. > And if you refuse to ever acknowledge something that obvious, then I > don't see where there's much likelihood that you will ever follow the > math. That's all good. Perhaps there is something you can do to enlighten me. I know the a's and b's depend on m; but I know you know that as well, it makes me think that I'm just misunderstanding what you mean by variable dependency. So if I'm merely misunderstanding, I'll stop for now. Phil Nicholson. > James Harris === Subject: Re: Attacking my own math proof, fun And I'm sensing that you must still think that there's some variable > dependency on m, or there wouldn't be further discussion. > Well, yes I do. The a's are clearly dependent on m and f. Their cubic is > described above. So I don't really know what you mean, I don't think. The m and > f are almost independent - they can be anything as long as f is coprime to 3 and > m. Some progress may have been made if you accept that you're trying to get a variable dependency for how f^2 divides off, though it's a constant. Possibly you're confused because the a's are dependent on m and f, but f^2 is a constant factor of P(m), and it is not. Also m and f are completely independent in general, while I introduce specific restrictions for special purposes at particular points. Again, the math can't read minds, so it's setup to handle the general case, where m and f are completely independent, and it doesn't bother to shift because of my choices, as the mathematical logic is rigid. > Are you still trying to claim that you are not? > If you're not then I can just check at m=0, confident that I've > covered when m does not equal 0, right? > If not, why do you believe so? > I'm sure this is not the case. a is a function of m. > Trivial analogy: > Say a_3 = 3 + m (1 + f) > Imagine f=5 so a_3 is coprime to f when m=0. > Now, when m=7, also coprime to 5, a_3 is not coprime to 5. > I know your a_3's aren't as above, but I honestly don't accept that checking at > m=0 is > sufficient, unless I can see a proof of it. Well your own example should show you why. Imagine the possibility that you had some expression where if f=5, your a_3 had a factor of 5 for ALL m, but if f didn't equal 5 it equaled 1 at m=0. Do you believe that is possible? If you follow mathematical logic that should finish your objections. of the ring of algebraic integers. > OK. It's important that it's recoginsed that they are often not algebraic > integers. It is important as it shows a problem with the definition of algebraic integers, as it's not as inclusive as it should be. you've repeated several times, but have been unable to prove. > As long as we're mindful that the b's are not always algebraic integers. I have said so myself, and in fact that is why there's a problem with the ring of algebraic integers. It may seem esoteric to readers on sci.physics and sci.skeptic, but mathematics requires zero errors, and what I've managed to show with some fascinatingly basic algebra is an error created by that definition of algebraic integers as roots of monic polynomials with integer coefficients. That definition leaves gaps by not including certain numbers that should be included which leads to fascinating contradiction like that ***in the ring of algebraic integers*** you can have abc = 5, where a, b and c are coprime to 5. That coprime just means they don't share non-unit factors, i.e. not factors of 1, with 5, but they multiply together to give 5, and a, b and c are each algebraic integers. Mathematicians missed this little thing for over a hundred years, but I can prove there's a problem in the ring with a short argument using basic algebra, which comes at the end of this post. Clearly, you've seized on one idea, and you keep holding on to it, > despite my efforts to get you to follow the math. I'm trying to follow the math. I keep getting stuck here though: You claim that because b_1 b_2 a_3 = m^3 f^4 - 3m^2 f^2 + 3m and RHS is coprime to f (which it is as m and f are coprime) > then a_3 is coprime to f. > This is only true for integer m and f when m=0. > Which indicates that you believe that it might be different when m > does not equal 0, which forces a dependency on m. > Yes, the values of the a's and b's depend on m. The RHS is coprime to f, > of course. I'm hoping that you don't mean something as trivial as that. It is > agreed that b_1b_2a_3 is coprime to f. It's not agreed that a_3 is coprime to f > for all valid m and f. However, disagreeing there requires that you go against mathematical logic. In this case it's hopefully easily seen by considering that if f=3, ALL of the a's have a constant factor that is 3, plus if m isn't coprime to 3 they can have additional factors in common with m and 3. That shows factors of f can't jump around when f is coprime to 3 as you apparently continue to wish, as in fact that would be forcing a dependency on f^2 which does not exist, as f^2 is a constant with regard to m. > Now it's quite simple, admit that you believe there's a variable > dependency on m, and then I can show there is none. > I think this would be extremely valuable. But you shouldn't feel any pressure to > convince me of anything. > The choice, and the floor is yours if you want. I believe there's a variable > dependency on m.The a's and b's are variables here, right? They're the things I > think are dependent on m. Well you're suddenly sounding passive. The question here is, can you follow mathematical logic? Or will you hang on to some belief, possibly for personal comfort *against* mathematical logic? I find that question intriguing. > However, if you continue to maintain that there is no dependency on m, > but then try to give emphasis on the case when m=0, then that > fundamental contradiction in your thinking means that you cannot be > following the math. Except at m=0 when b_1 and b_2 are =0 and are algebraic integers. Well, try m=1 with f=sqrt(2), and welcome to a more complicated > mathematical world than you might have realized. I'm sure it will work out fine with those values, so I won't try. > How about YOU try when f=5, and m is your favorite non-zero integer coprime > to > 5. > Why? > No reason. I'm guessing though that a_3 is not coprime to 5. There is no need to guess. It's mathematics, and it's possible to prove that it is. No guessing needed. > wrong assumption, which is that the constant f^2 divides off in some > way as a function of m or variable dependent on m, which you have > repeatedly demonstrated by citing the m=0 case as if it is a special > case, but when I say that's your assumption, you claim it is not. > In my mind, it doesn't matter how the f^2 divides off. If you resolve the a's > and b's to their cubic definitions, the steps you took to get there disappear in > any case. > I haven't actually said that m=0 is a special case at any point. Although > secretly I think that solving for m=0 isn't sufficient to show that the result > applies for all m. Not so secretly, obviously. You apparently have seized on the idea that m=0 is a special case, and simply choose to ignore counterexamples like m=1, with f=sqrt(2), though there apparently you may think that f not being an integer makes a difference. However, there is no need to guess, or go by hunches, as if you follow the math, the conclusion is clear. > Looking at other examples will not help you, but trying to focus you > onto the basic contradiction in your position--the claim that there is > no dependency on m, when you keep pointing at m=0 as if there is--just > might. > And if you refuse to ever acknowledge something that obvious, then I > don't see where there's much likelihood that you will ever follow the > math. > That's all good. Perhaps there is something you can do to enlighten me. I know > the a's and b's depend on m; but I know you know that as well, it makes me think > that I'm just misunderstanding what you mean by variable dependency. > So if I'm merely misunderstanding, I'll stop for now. > Phil Nicholson. Well, the argument which settles things is *luckily* short and rather direct, so I'll give it here. Some may think it's exactly what they've seen before, as I've been posting it a lot of places, but I've seen need to put in minor corrections. Consider, in the ring of algebraic integers, P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f). Now using b_1, b_2, b_3, w_1, w_2, and w_3, I have the factorization P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3) where w_1 w_2 w_3 = f, and b_1 b_2 b_3 = (m^3 f^4 - 3m^2 f^2 + 3m), and at m=0 P(0)/f^2 = 3xu^2 + u^3 f = u^2(3x + uf), so two of the b's must equal 0, which means P(0)/f^2 = w_1 w_2 u^2 (b_3 x + u w_3) which is P(0)/f^2 = u^2 (b_3 w_1 w_2 x + u f) = u^2(3x + uf) proving that w_1 w_2 must equal 1, if f is coprime to 3, which leaves b_3 = 3. Essentially objections to how f^2 divides off now come down to claiming that the w's are functions of m, but consider that w_1 w_2 = 1, when m=0, if f is coprime to 3. But that was an arbitrary choice, so let f=3. Now w_1 w_2 = 3^{2/3} WITHOUT REGARD TO m. That is, the w's are now all constant with regard to m and have the same value no matter what the value of m is. Therefore, the factorization is P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f = (b_1 x + u)(b_2 x + u)(b_3 x + uf) where you'll notice that the b's are algebraic integers with m=1, f=sqrt(2), but that's a special case as generally they are not, which shows a problem with the ring of algebraic integers. I've found the Ring of Objects which includes the ring of algebraic integers, and does not have this problem, as the b's are all included in it. The Ring of Objects is the set of all numbers where 1 is the only member that is both a unit, i.e. factor of 1, and an integer, where no non-unit member is a factor of any two integers that are coprime. That definition and more is linked to from my primary website http://groups.msn.com/AmateurMath where you can also find information on my other math research. James Harris === Subject: Re: Attacking my own math proof, fun Distribution: inet > It may seem esoteric to readers on sci.physics and sci.skeptic, but > mathematics requires zero errors, and what I've managed to show with > some fascinatingly basic algebra is an error created by that > definition of algebraic integers as roots of monic polynomials with > integer coefficients. > That definition leaves gaps by not including certain numbers that > should be included which leads to fascinating contradiction like that > ***in the ring of algebraic integers*** you can have abc = 5, where a, > b and c are coprime to 5. There is no problem with the ring of algebraic integers. The problem is in your faulty argument. If abc = 5, and 'a', 'b' and 'c' are algebraic integers, then bc = 5/a, where the product 'bc' is an algebraic integer, hence 'a' divides 5 with an algebraic integer quotient. If 'a' is not a unit, then it is (in your terminology) a 'factor' of 5 in the ring of algebraic integers, and is therefore *not* coprime to 5. The same holds for 'b' and 'c'. You claim there are certain numbers which should be included in the ring of algebraic integers, but which are left out. OK, name one! C'mon James, just one! Please leave out the 'fascinating', 'odd', etc. terminology, because these terms simply expose your mental state and have nothing to do with the math. -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: Constant factors and polynomials > If P(x) is a polynomial with coefficients in a commutative ring R, > then if r=s (mod a) for r,s,a in R, then one can easily prove that > A), and from this we obtain the result > PROPOSITION. Let P(x) be a polynomial with coefficients in the > commutative ring R. For all a in R, P(a) is coprime to a if and only > if P(0) is coprime to a. i.e. (a ,P(a)) = (a ,P(0)) i.e. a==0 => P(a)==P(0) or (a-b,P(a)) = (a-b,P(b)) i.e. a==b => P(a)==P(b) so a-b | P(a)-P(b) e.g. 10-1 | P(10)-P(1) Casting Out Nines These are prototypical examples encountered when one first studies congruences and (evaluation) homomorphisms on abstract rings. That polynomial evaluation is a ring hom is already known in high-school, as are applications such as Castings Out Nines. Many of my prior posts discuss at length related ideas, e.g. start with this one: -Bill Dubuque === Subject: Re: Constant factors and polynomials >> >> [.snip.] >> >> Your statement that >> ... there's no reason to believe it varies >> with m, or that it cares if the polynomial is irreducible >> over Q >>is one of the weakest you have ever put forward. First, >>saying there's no reason to believe it is one of that lamest >>non-proofs I have seen. What you are really saying is, >>'*I* can't think of any reason to believe it.' Second, justifying a >>conclusion on the basis that it cares whether a polynomial is >>irreducible - this is truly new methodology. >> >> >> No, it is over a year old. He first started using it to attack the >> standard theorem on roots of irreducible polynomials with integer >> coefficients. He kept arguing about how the math couldn't care >> whether a polynomial was irreducible or not, and how the roots >> couldn't care, and how it was stupid to think the roots cared >> whether the polynomial is irreducible or not. >I've been pointing out now that a constant factor of a polynomial does >not divide off as a function, while for some time posters like Magidin >have gotten away with objections that depend on saying that m=0 is a >special case, forcing an m dependency on how f^2 divides off. > It's interesting that you ignore any post I make with math in it, and > when I make an offhand comment, you jump in to make non sequitur > claims and assertions that usually have little base in reality, or > little sense as written. Hmmm...so I give the math argument, you delete it out, and then give weird excuses. > Your post was a perfect example. Since turnaround is fair play, I > guess, I will also ignore everything you posted and instead make some > comments. My ignoring your post here is not an admission of any kind, > since I have addressed your multiple errors many times before. Ok, so I give the math argument, you delete it out, give weird excuses, and then claim it's fair play, and, oh yeah, you claim that you've addressed multiple errors before. Gist of it is, you're telling readers to trust you. I say trust the mathematics, which I will give, once again. > Some time ago I guessed at what was behind your intuition. Based on > many flawed arguments you have given from time to time, and senseless > statements like the above, my guess is that you are thinking in terms > of polynomials. > If P(x) is a polynomial with coefficients in a commutative ring R, > then if r=s (mod a) for r,s,a in R, then one can easily prove that > A), and from this we obtain the result > PROPOSITION. Let P(x) be a polynomial with coefficients in the > commutative ring R. For all a in R, P(a) is coprime to a if and only > if P(0) is coprime to a. > This is essentially the argument you have been trying to use. Of > course, your functions are not polynomials, but you still imagine them > as ->essentially<- polynomials, perhaps polynomials in rational powers > of x instead of integer powers, perhaps involving radicals and so > on. Once before you already tried to argue that a=b (mod r) implies > sqrt(a)=sqrt(b) (mod r), and never really acknowledged that this is > simply false. > So, what are you doing? You imagine your coefficients as functions of > m, and you imagine that these functions are really a lot like > polynomials. So you evaluate at 0 to figure out what the constant term > is, and thus deduce that since this constant term is coprime to certain > values of m when evaluated at 0, it must also be coprime when > evaluated at any value coprime to that constant term, just like a > polynomial would. Well, I don't think so, but rather than just toss out *opinions* I'm going to give the actual mathematics and readers can work it out for themselves. > Of course, the error is pretty dismal. It ignores the fact that > congruences do not behave well with general algebraic functions, they > behave terribly with radicals, AND ignores that your functions are > really piece-wise defined. For p=3, you get general formulas for the > coefficients (in some instances at least) from the Cardano formulas, > but m=0 is a degenerate case in which you cannot use those formulas > but must instead switch to specific values. In these cases, the > argument you are probably basing your intuition on is bollocks. Well you said probably basing in that paragraph, and there is no need to guess, as I'll give the mathematics. Consider, in the ring of algebraic integers, P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f). Now using b_1, b_2, b_3, w_1, w_2, and w_3, I have the factorization P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3) where w_1 w_2 w_3 = f, and b_1 b_2 b_3 = (m^3 f^4 - 3m^2 f^2 + 3m), and at m=0 P(0)/f^2 = 3xu^2 + u^3 f = u^2(3x + uf), so two of the b's must equal 0, which means P(0)/f^2 = w_1 w_2 u^2 (b_3 x + u w_3) which is P(0)/f^2 = u^2 (b_3 w_1 w_2 x + u f) = u^2(3x + uf) proving that w_1 w_2 must equal 1, if f is coprime to 3, which leaves b_3 = 3. Essentially objections to how f^2 divides off now come down to claiming that the w's are functions of m, but consider that w_1 w_2 = 1, when m=0, if f is coprime to 3. But that was an arbitrary choice, so let f=3. Now w_1 w_2 = 3^{2/3} WITHOUT REGARD TO m. And in fact, all of the w's are constant values for ALL m when f=3. Therefore, the factorization is P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f = (b_1 x + u)(b_2 x + u)(b_3 x + uf) where you'll notice that the b's are algebraic integers with m=1, f=sqrt(2), but that's a special case as generally they are not, which shows a problem with the ring of algebraic integers. James Harris === Subject: Re: Constant factors and polynomials >It occurred to me that some of you may be hampered in understanding >certain math arguments of mine because P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) has that constant factor of f^2. Normally when considering factorizations, you separate off constant >factors, as otherwise you don't have a unique factorization even with >polynomial factors. For instance > > 4(x^2 + 2x + 1) = (2x + 2)(2x + 2) = (x+1)(4x + 4) along with an infinity of other factorizations, but typically you'd >just have 4(x^2 + 2x + 1) = 4(x+1)(x+1). Besides all that the expression I use is rather imposing, and it has a >lot of symbols, so I thought I'd remind you of a few things. 1. You *can* look at an actual example with m=1, f=sqrt(2), as then >all that complexity drops away and you have P(1) = 2x^3 - 3x + 1 which actually does reduce over Q. Some of you may have realized that you can consider m=1(mod sqrt(2)) >to blow apart several assertions made by some posters. 2. A requirement I give is that f be coprime to 3, but letting f=3, >you get that *each* of the a's in the factorization 3^2((m^3 3^4 - 3m^2 3^2 + 3m) x^3 - 3(-1+m3^2 )x u^2 + u^3 3) = (a_1 x + 3u)(a_2 x + 3u)(a_3 x + 3u) has a non-unit factor in common with 3, which is a radical factor of >3, and there's no reason to believe it varies with m, or that it cares >if the polynomial is irreducible over Q. >That actually destroys >several claims made about using Galois Theory where reducibility over >rationals is an issue. > Not even close. First, in this case, f factors out of your > polynomial 3 times, not 2 times as in the cases you were > considering (f <> 3). This is a special case of no interest > to you or me. It is irrelevant. The Galois argument does not > apply here. No claims based on that argument are 'destroyed'. > Second, only one of the proofs that you are wrong in the cases > where f <> 3 is dependent on Galois Theory. The other proofs are > based on an elementary theorem from algebraic number theory, > which you have previously accepted. All of the proofs *do* > require irreducibility of P(x)/f^2. > Well, it IS the case that for f=sqrt(2) only *two* of the a's have a > factor that is sqrt(2), so your claim that it is otherwise is false. In your main applications, as in your proof of FLT, f is an integer. I assumed that here. > Readers can easily verify by considering > P(1) = 2x^3 -3xy^2 + y^3, where y=uf, > using f=sqrt(2), m=1, with P(m) given above, as it is reducible. Relatively minor point: if f = sqrt(2), then f^3 = 2*sqrt(2), which is not rational, so y = uf gives an irrational constant term, y^3, for your polynomial. It makes no sense to discuss reducibility (over the rationals) of a polynomial unless it has rational coefficients. This one does not when f = sqrt(2). > Then consider that using m=1(mod sqrt(2)) then destroys any claims > that reducibility is the issue, as, for instance, m=3 is NOT > reducible. You've lost me here. See above note regarding reducibility of this polynomial when it does not even have rational coefficients. As for m = 3, I assume you intended to say that the polynomial is irreducible over Q when m = 3. Again this doesn't make any sense when the polynomial does not even have rational coefficients. A side question: are you saying that m = 3 is congruent to 1 modulo sqrt(2) ???? Would you care to explain that? > Now if Nora Baron were an ethical mathematician, there would be no > debate. > But it's not even clear who Nora Baron is, as was pointed out by > another poster the name itself is a palindrome. > That is, it's the same reversed as Nora B-aron. > The actual poster may not even be female. Your statement that ... there's no reason to believe it varies > with m, or that it cares if the polynomial is irreducible > over Q is one of the weakest you have ever put forward. First, > saying there's no reason to believe it is one of that lamest > non-proofs I have seen. What you are really saying is, > '*I* can't think of any reason to believe it.' Second, justifying a > conclusion on the basis that it cares whether a polynomial is > irreducible - this is truly new methodology. You are basing an > argument on the psychology of some utterly fictional entity. In fact > REDUCIBILITY MATTERS - that is, conclusions regarding irreducible > polynomials are *qualitatively different* from conclusions > regarding reducible ones. Here is an example: x^2 -5*x + 6 = 0 has constant term 6 = 2*3. The roots are 2 and 3. One root is > coprime to 3 and the other is coprime to 2. But > x^2 -3*x + 6 = 0 also has two roots. They are: r1 = (3 + sqrt(-15))/2 and r2 = (3 - sqrt(-15))/2. You can check (by computing a 4th degree polynomial) > that: NEITHER r1 nor r2 is coprime to 2, and > NEITHER r1 nor r2 is coprime to 3. Thus: REDUCIBILITY MATTERS: note that: x^2 - 5*x + 6 is reducible over Q, and x^2 - 3*x + 6 is irreducible over Q. It's not a coincidence. > That was hand-waving. In fact, your position requires that people > believe that given That was absolutely NOT handwaving. There is nothing in what I said but cold, hard computation. I gave two examples: one, a reducible polynomial, for which the roots are coprime to one or the other factor of the constant term; the other, an IRREDUCIBLE polynomial, for which NEITHER root is coprime to ANY factor of the constant term. These are not isolated examples; these are the general rule. The point: IRREDUCIBILITY MATTERS. Somehow, in your terms, the math knows or the math cares, whether you like it or not. And you blow it off by calling it hand-waving! Tell me how I can be more explicit! > P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - > 3(-1+mf^2 )x u^2 + u^3 f) = > (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) > that some f^2 divides off as some function of m, or variable dependent > on m. YES! YES! YES! YOU GOT IT! > What I noted is the oddity of believing that a constant factor of a > polynomial, which f^2 is, could factor off as a function. Odd or not, that is how it works. The way it factors depends on its reducibility or irreducibility. That should not seem strange. Lots of theorems in algebraic number theory depend on reducibility or irreducibility. > It seems unlikely to me that a math expert would make that mistake. Ask other math experts. Anyone you like. > And in fact in this case it's rather easy to prove that belief is > false by letting f=3, as then ALL of the a's have a factor of 3 that > is 3^{2/3}. In that case, unlike the one in which you are interested, P(x) / f^2 is not coprime to f. Since your own arguments depend on such coprimeness, this is an irrelevant example. What I want you to understand is that for trained mathematicians, >these are not issues. However, when it comes to confusing people >about even relatively basic mathematics, who would be better at it >than mathematicians? They need to confuse you here for *social* reasons. > This is not true. All we want is that the truth > emerges. Social consequences are irrelevant and > unimportant. You, in contrast, are desperately seeking > social recognition for your proof. As has happened > in the past, we are dragging you kicking and screaming > into the realization that your proof is incorrect. > Then why are you ignoring the actual math argument that I repeatedly > give? No one here has paid more detailed attention to your argument than I have. I have looked at it on its own terms. This is more than I should have done. I and others have given explicit, rigorous proofs that your main claims are incorrect. What this means is, there is no particular reason to do your work for you and identify where you have made your mistake. I have nevertheless pinpointed your error many times, and that is what you are now trying to deal with: how you justify going from a factorization when m = 0 to a factorization when m <> 0. Nothing I have seen so far from you on this hangs together logically. > Besides, why would you care? > There are any number of posters all over Usenet with various notions. For sure. There are loonies, mystics, jokers, liars, etc., of all sorts out there. Most of them do not pretend to do mathematics and they cannot be proven right or wrong. You however try to present logical arguments. You say things that annoy me because they are provably false, and you brag about it. My instinct is to not let you get away with it. > It makes more sense that realizing my work is correct, and fearing the > social consequences of acknowledging the truth, people like yourself > Nora Baron bank on your ability to confuse others about the > mathematics. I very definitely do not want to confuse anyone. I try to write simply and clearly and I put in most of the details. I do not want to mislead anyone. As for social consequences, if I honestly thought you had a valid proof, I would defend it as strongly as I attack what you have at present. > The more I simplify, the more you desperately work to confuse, which I > think explains your recent postings. That is one possible explanation. Another is that I believe what I am saying and my main goal is to see that the truth prevails, that I do not like someone claiming false results when they can easily be debunked. > Although social reasons - recognition, especially - are > paramount for you, you are losing the battle. That is > why you desperately keep starting new threads and you refuse > to answer rigorous arguments which prove you are wrong. > Then why bother continuing to reply? I am hopeful that you will eventually understand what is wrong with your argument. > In actuality, you are lying. You have not given valid math arguments > but instead have given bogus ones. ABSOLUTELY NOT! > However, you have relied on using math where you can hide the fact > that it is invalid by various tricks, mostly banking on people not > checking thoroughly. People: check all you want, and everything you want. I abhor 'tricks'. I like simple straightforward stuff, and that is what I have tried to present. If you think I am tricking you, point out what it is you don't understand. The examples I gave above of quadratic polynomials - was that a trick? Why do you think so? You called it handwaving. Did you feel I was trying to trick you? If so, ASK FOR AN EXPLANATION of any step you didn't get. > Faced with your efforts, and the support you get from other poster > helping you to try and obscure the truth, I keep bringing focus back > to the actual math argument I've presented. Great. > >Notice that with f=3, the constant term P(0) = u^2(3x + 3u) = >3u^2(x+u), so it *still* has a factor that is 3, and that's why I >always have the condition that f be coprime to 3. Here, however, I'm hoping it'll help to point out why that requirement >is there, and what happens if you ignore it. Well then, what are some posters trying to convince you about P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f)? They're trying to convince you that there is a mathematical limitation >based on reducibility over Q that determines how f^2 can divide >through when you have the factorization P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) and the first question that should come to you is, how could a >constant factor be constrained by reducibility over rationals? > See above: REDUCIBILITY MATTERS greatly and is central to > the disagreement here. It is reasonable to ask, as you do, how > reducibility might constrain factoring. It is NOT reasonable to assume > with no hint of proof that it doesn't. Our counterproofs need either > a basic theorem from Galois theory (which *requires* irreducibility), or > a basic theorem from algebraic number theory (which *also requires* > irreducibility). If you really want to understand why, you need to > study the proofs of those theorems. However the quadratic examples I > gave above should be sufficient to show exactly how reducibility can > affect the form of factors of roots. > Yet I can give f=sqrt(2), with m=1, and get a reducible polynomial, > which you already made a false statement about in this post, where > only two of the a's have a factor that is sqrt(2). See above. Since f = sqrt(2) and the constant term is (u*f)^3, where u is an integer, the constant term is not rational. So it makes no sense to discuss reducibility over the rationals for this polynomial. > Then I can simply use m=1(mod sqrt(2)) for a family of results where > most are irreducible. > Now your claim would require that suddenly, against mathematical > logic, the single one of the a's that doesn't have a non-unit factor > in common with sqrt(2) gets one. > Faced with the destruction of your argument, you side-stepped it, > which diligent readers can see at the start of this post. Please first resolve the problem with f^3 being irrational before you go any farther with this. Perhaps you have a definition of irreducible (over the rationals) that is at variance with the usual one. >Now that question is resolvable, but I know the answer is in my favor, >so mathematicians are avoiding even letting you know that IS the >question, and instead those who post work to confuse. Now given that I know I have a short proof of Fermat's Last Theorem, >and that mathematicians have been avoiding dealing with reality, while >some posters have gotten away with *deliberately* confusing people, >why would I quit talking about my proof of FLT? If you'd found a short proof of Fermat's Last Theorem, would you quit >talking about it? > > No: *** If ***. But if other people had presented rigorous > proofs that I was wrong, I would try very hard to > dismantle those proofs. You have made no effort to do that, > preferring instead to bring up irrelevant issues as in the > present thread. The bottom line here is, whether you like it > or not, reducibility DOES matter. If a ridiculous argument based > on whether it (math) doesn't care about reducibility is all > you have, you had better give up mathematics. Nora B. > So now even though I have a proof of my own, No, dammit, I said *** IF ***. You don't have a proof. You have written down many proofs over the years. You have admitted over and over again that what you thought was a proof was in fact wrong. You have made stupid mistakes in various arguments even in the last few days. Why you NOW think you have suddenly become infallible baffles the hell out of me. > you claim that I should > have to work to show falsity in alternate proofs given by people who > have proven a need to deceive, who have a long record of deception, > who may be actual mathematicians, i.e. math experts, with a > willingness to use their knowledge to confuse others? If I were you I would try to find errors in the counterarguments. If you won't listen to us, maybe it would be reasonable for us to post a warning to editors of math journals, saying there are strong counterarguments to claims you are making in a paper you are trying to publish; specifying what those counterarguments are; and noting that, although you have had multiple opportunities, you have not found errors in any of them. You are either (1) refusing to do so, or (2) unable to do so. In either case, it is dishonest of you to try to publish a result when there are real, rigorous, unanswered objections which prove it is false. Should we notify editors of this? > Here readers should consider that Nora Baron is probably a pseudonym > that the poster believes is clever because it's a palindrome, who may > not even be a woman. > They can also consider the clear dodge of an issue brought up against > the poster's claims with f=sqrt(2), m=1. See above. It should not be news that sqrt(2) is not a rational number. > But best of all, readers can consider that *mathematics* gives them > their best way to figure out what's going on, which is to consider the > argument Nora Baron keeps working to obscure. > Consider > P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - > 3(-1+mf^2 )x u^2 + u^3 f. > Now using b_1, b_2, b_3, w_1, w_2, and w_3, I have the factorization > P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3) > where w_1 w_2 w_3 = f, and > b_1 b_2 b_3 = (m^3 f^4 - 3m^2 f^2 + 3m), > and at m=0 > P(0)/f^2 = 3xu^2 + u^3 f = u^2(3x + uf), > so two of the b's must equal 0, which means > P(0)/f^2 = w_1 w_2 u^2 (b_3 x + u w_3) > which is > P(0)/f^2 = u^2 (b_3 w_1 w_2 x + u f) = u^2(3x + uf) > proving that w_1 w_2 must equal 1, as f is coprime to 3 from before, > which leaves b_3 = 3. ... right. And every bit of what you have just said is predicated on the condition: ***** m = 0 *****. > Essentially objections now come down to claiming that the w's are > dependent on m, Exactly! Prove they're not! > but consider that w_1 w_2 = 1, when m=0, here where f > is coprime to 3. > But that was an arbitrary choice *I* made, so let f=3. In that case, P(x) / f^2 is not coprime to f, so your own argument breaks down. It is a special case of no interest to either you or me. Why don't you look in detail at the cases f = 5, f = 7, f = 11, ... ? > Now w_1 w_2 = 3^{2/3} as long as m is coprime to 3, WITHOUT REGARD TO > m. Really ? I note that you say that u*w3 = u*f = u*3, that is, w3 = 3, and so if w1*w2 = 3^{2/3}, then w1*w2*w3 = 3^{5/3}. But u^3*w1*w2*w3 is supposed to equal the constant term, which here is u^3 * f = u^3 * 3. Let u = 1 to make this simpler if you want. Anyway, 3 is not equal to 3^{5/3}. You want to think about that some more? > So those posters who try to convince you that the w's are actually > dependent on m, like being functions of m, must now also convince you > that the w's make a decision, first looking to see if f=3 or have some > non-unit factor in common with 3, and THEN they decide if they're > dependent on m. This is a silly way of saying it, but in a sense it is right. The form of the factorization DOES depend on reducibility of the polynomial. Trying to anthropomorphize the w's does not constitute a proof that they behave the way you want them to. Got that? YOUR SILLY LITTLE ARGUMENT ABOVE IS NOT A PROOF! If you can present a REAL proof, do so. The burden of doing this is on you. If you think the w's are NOT dependent on m, you need to PROVE IT. Bear in mind, however, that in the cases where P(x) / f^2 is primitive and irreducible, there are rigorous arguments that say you can't do it. You might learn something by trying to understand and answer those arguments. Nora B. > People can waffle trying to figure out who they are, but mathematics > is logical, which is why I've emphasized that posters are acting on > *social* not mathematical reasons. > James Harris === Subject: Re: Constant factors and polynomials > If you won't listen to us, maybe it would be reasonable for us to > post a warning to editors of math journals, saying there are strong > counterarguments to claims you are making in a paper you are trying > to publish; specifying what those counterarguments are; and noting > that, although you have had multiple opportunities, you have not > found errors in any of them. You are either (1) refusing to do so, > or (2) unable to do so. In either case, it is dishonest of you to try to > publish a result when there are real, rigorous, unanswered objections > which prove it is false. Should we notify editors of this? I think that's an excellent idea. It's unlikely that a reputable journal would be fooled by James' pseudo-math, but it could save some editors and/or reviewers from wasting too much time on it. -- Wayne Brown | When your tail's in a crack, you improvise fwbrown@bellsouth.net | if you're good enough. Otherwise you give | your pelt to the trapper. e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock === Subject: Re: Constant factors and polynomials polynomial 3 times, not 2 times as in the cases you were > considering (f <> 3). This is a special case of no interest > to you or me. It is irrelevant. The Galois argument does not > apply here. No claims based on that argument are 'destroyed'. > Second, only one of the proofs that you are wrong in the cases > where f <> 3 is dependent on Galois Theory. The other proofs are > based on an elementary theorem from algebraic number theory, > which you have previously accepted. All of the proofs *do* > require irreducibility of P(x)/f^2. Well, it IS the case that for f=sqrt(2) only *two* of the a's have a > factor that is sqrt(2), so your claim that it is otherwise is false. In your main applications, as in your proof of FLT, > f is an integer. I assumed that here. Your assumption is irrelevant to that fact, as you have tried to confuse people by working to convince that m=0 is a special case. You have in the past succeeded, but now I'm showing, rather easily and directly that you were indeed misleading them. > Readers can easily verify by considering P(1) = 2x^3 -3xy^2 + y^3, where y=uf, using f=sqrt(2), m=1, with P(m) given above, as it is reducible. Relatively minor point: if f = sqrt(2), then f^3 = 2*sqrt(2), > which is not rational, so y = uf gives an irrational constant > term, y^3, for your polynomial. It makes no sense to discuss reducibility > (over the rationals) of a polynomial unless it has rational > coefficients. This one does not when f = sqrt(2). As usual you work to mislead, and here it's by distraction. While it is true that given 2x^3 -3xy^2 + y^3 with y=uf, where f=sqrt(2), you end up with coefficients that are not rational. It is also true that 2x^3 -3xy^2 + y^3 factors over Q. That's important because I show that you are wrong with the emphasis on m=0 as a special case by showing m=1, where people can see that my argument correctly gives ONLY two of the a's as having a factor that is sqrt(2), while the remaining one is coprime to 2. > Then consider that using m=1(mod sqrt(2)) then destroys any claims > that reducibility is the issue, as, for instance, m=3 is NOT > reducible. > You've lost me here. See above note regarding reducibility of > this polynomial when it does not even have rational coefficients. > As for m = 3, I assume you intended to say that the polynomial > is irreducible over Q when m = 3. Again this doesn't make any sense > when the polynomial does not even have rational coefficients. Hmmm...so you're claiming it's irreducible over Q? Then that refutes you directly as your own claim is that reducibility is the issue. Since you may back off if you see that you can't confuse people on that issue, I'll point out that using m=1(mod f) with f=sqrt(2), takes away the possibility that the factors of f, move, which refutes you as well. The *mathematical* basis for your claims is thus destroyed. > A side question: are you saying that m = 3 is congruent to 1 > modulo sqrt(2) ???? Would you care to explain that? 3 = 1(mod sqrt(2)) Understand? Now if Nora Baron were an ethical mathematician, there would be no > debate. But it's not even clear who Nora Baron is, as was pointed out by > another poster the name itself is a palindrome. That is, it's the same reversed as Nora B-aron. The actual poster may not even be female. Thus: REDUCIBILITY MATTERS: note that: x^2 - 5*x + 6 is reducible over Q, and x^2 - 3*x + 6 is irreducible over Q. It's not a coincidence. That was hand-waving. In fact, your position requires that people > believe that given That was absolutely NOT handwaving. There is nothing in what > I said but cold, hard computation. I gave two examples: one, > a reducible polynomial, for which the roots are coprime to one > or the other factor of the constant term; the other, an > IRREDUCIBLE polynomial, for which NEITHER root is coprime to > ANY factor of the constant term. These are not isolated > examples; these are the general rule. The point: IRREDUCIBILITY > MATTERS. Somehow, in your terms, the math knows or > the math cares, whether you like it or not. Your quadratics are irrelevant to the case as hand, and in fact, it's fascinating that you believe that you've proven some rule, with two quadratics. > And you blow it off by calling it hand-waving! Tell me > how I can be more explicit! You can stick to the line of mathematical logic, rather than trying to distract. > P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) that some f^2 divides off as some function of m, or variable dependent > on m. YES! YES! YES! YOU GOT IT! Previous to that I said that your position requires that people believe that given P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) that some f^2 divides off as some function of m, or variable dependent on m. Splitting that sentence up with your own comments and then posting as if I'd said something else is clearly dishonest. James Harris === Subject: Re: Constant factors and polynomials > If you won't listen to us, maybe it would be reasonable for us to > post a warning to editors of math journals, saying there are strong > counterarguments to claims you are making in a paper you are trying > to publish; specifying what those counterarguments are; and noting > that, although you have had multiple opportunities, you have not > found errors in any of them. You are either (1) refusing to do so, > or (2) unable to do so. In either case, it is dishonest of you to try to > publish a result when there are real, rigorous, unanswered objections > which prove it is false. Should we notify editors of this? > I think that's an excellent idea. It's unlikely that a reputable journal > would be fooled by James' pseudo-math, but it could save some editors > and/or reviewers from wasting too much time on it. James Harris is but one of many math cranks in this world. I imagine that journal editors can recognise their nonsense rather quickly. I doubt that there is any point in warning journal editors about James. -- Clive Tooth http://www.clivetooth.dk === Subject: Re: Constant factors and polynomials > If you won't listen to us, maybe it would be reasonable for us to > post a warning to editors of math journals, saying there are strong > counterarguments to claims you are making in a paper you are trying > to publish; specifying what those counterarguments are; and noting > that, although you have had multiple opportunities, you have not > found errors in any of them. You are either (1) refusing to do so, > or (2) unable to do so. In either case, it is dishonest of you to try to > publish a result when there are real, rigorous, unanswered objections > which prove it is false. Should we notify editors of this? > I think that's an excellent idea. It's unlikely that a reputable journal > would be fooled by James' pseudo-math, but it could save some editors > and/or reviewers from wasting too much time on it. I think it's a terrible idea. You're just opening yourself up to legitimate claims of conspiring. I would hate to think that such a letter could influence the review process, causing an editor to prejudge a paper. If it could, just imagine the possibilities for abuse. A paper should stand or fall on its merits, and a reviewer should judge a paper on its content. Trust the system. They're already used to getting incoherent documents. - Randy === Subject: Re: Constant factors and polynomials >> If you won't listen to us, maybe it would be reasonable for us to >> post a warning to editors of math journals, saying there are strong >> counterarguments to claims you are making in a paper you are trying >> to publish; specifying what those counterarguments are; and noting >> that, although you have had multiple opportunities, you have not >> found errors in any of them. You are either (1) refusing to do so, >> or (2) unable to do so. In either case, it is dishonest of you to try to >> publish a result when there are real, rigorous, unanswered objections >> which prove it is false. Should we notify editors of this? >> I think that's an excellent idea. It's unlikely that a reputable journal >> would be fooled by James' pseudo-math, but it could save some editors >> and/or reviewers from wasting too much time on it. > I think it's a terrible idea. You're just opening yourself up > to legitimate claims of conspiring. I would hate to think that > such a letter could influence the review process, causing an > editor to prejudge a paper. If it could, just imagine the > possibilities for abuse. A paper should stand or fall on its > merits, and a reviewer should judge a paper on its content. > Trust the system. They're already used to getting incoherent > documents. I thought Nora was talking about something other than a letter saying This guy's a crank. It was my assumption that by specifying what those counterarguments are she meant that the editor would be provided with valid mathematical reasons why James' paper is incorrect. The editor (or reviewer) would be able to read both James' submission and the counterarguments and make his/her own decision. It would be a little like this: James shows up at a newspaper with a picture he's taken of a flying saucer. The editor says, Now that's interesting, because I have another picture here that someone sent me. See this? Here you are in your back yard, with a camera, some fishing line, and a pie plate that looks a *lot* like that flying saucer of yours... I assume Nora was not seriously proposing to do this, though, and I'm certainly not qualified to do so. -- Wayne Brown | When your tail's in a crack, you improvise fwbrown@bellsouth.net | if you're good enough. Otherwise you give | your pelt to the trapper. e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock === Subject: Re: Constant factors and polynomials > Not even close. First, in this case, f factors out of your >> polynomial 3 times, not 2 times as in the cases you were >> considering (f <> 3). This is a special case of no interest >> to you or me. It is irrelevant. The Galois argument does not >> apply here. No claims based on that argument are 'destroyed'. >> Second, only one of the proofs that you are wrong in the cases >> where f <> 3 is dependent on Galois Theory. The other proofs are >> based on an elementary theorem from algebraic number theory, >> which you have previously accepted. All of the proofs *do* >> require irreducibility of P(x)/f^2. >> >> Well, it IS the case that for f=sqrt(2) only *two* of the a's have a >> factor that is sqrt(2), so your claim that it is otherwise is false. >> >> In your main applications, as in your proof of FLT, >> f is an integer. I assumed that here. >Your assumption is irrelevant to that fact, as you have tried to >confuse people by working to convince that m=0 is a special case. 1. My assumption is clearly more relevant than assuming f = sqrt(2). Your applications in either Advanced Polynomial Factorization or your proof of FLT both involve assuming f is a prime *integer*. 2. Since I honestly believe m = 0 *IS* a special case, it would be less than honest of me to pretend otherwise. I am not the least bit interested in confusing people. >You have in the past succeeded, but now I'm showing, rather easily and >directly that you were indeed misleading them. >> Readers can easily verify by considering >> >> P(1) = 2x^3 -3xy^2 + y^3, where y=uf, >> >> using f=sqrt(2), m=1, with P(m) given above, as it is reducible. >> >> Relatively minor point: if f = sqrt(2), then f^3 = 2*sqrt(2), >> which is not rational, so y = uf gives an irrational constant >> term, y^3, for your polynomial. It makes no sense to discuss reducibility >> (over the rationals) of a polynomial unless it has rational >> coefficients. This one does not when f = sqrt(2). >As usual you work to mislead, and here it's by distraction. >While it is true that given > 2x^3 -3xy^2 + y^3 >with y=uf, where f=sqrt(2), you end up with coefficients that are not >rational. Right. So it cannot factor over the rationals. >It is also true that 2x^3 -3xy^2 + y^3 factors over Q. That would mean by definition that this polynomial factors into two polynomials of lower degree, each of which has rational coefficients. If y = sqrt(2), the constant term is irrational. Show that factorization. I really want to see it. Lots of other people will be interested as well. >That's important because I show that you are wrong with the emphasis >on m=0 as a special case by showing m=1, where people can see that my >argument correctly gives ONLY two of the a's as having a factor that >is sqrt(2), while the remaining one is coprime to 2. >> Then consider that using m=1(mod sqrt(2)) then destroys any claims >> that reducibility is the issue, as, for instance, m=3 is NOT >> reducible. >> You've lost me here. See above note regarding reducibility of >> this polynomial when it does not even have rational coefficients. >> As for m = 3, I assume you intended to say that the polynomial >> is irreducible over Q when m = 3. Again this doesn't make any sense >> when the polynomial does not even have rational coefficients. >Hmmm...so you're claiming it's irreducible over Q? Then that refutes >you directly as your own claim is that reducibility is the issue. It is not even a polynomial to which the term irreducible over Q should be applied. It does not have rational coefficients. If you insist on describing it as reducible over Q, then it must equal a product of lower-degree polynomials with *rational* coefficients. Clearly that is impossible. Prove me wrong. >Since you may back off if you see that you can't confuse people on >that issue, I'll point out that using m=1(mod f) with f=sqrt(2), takes >away the possibility that the factors of f, move, which refutes you as >well. >The *mathematical* basis for your claims is thus destroyed. My proof showing that your conclusion is wrong assumes f = 5 and m = 1. Your claimed main results deal with f a *prime integer*. Clearly f = sqrt(2) is not exactly in the ballpark. >> A side question: are you saying that m = 3 is congruent to 1 >> modulo sqrt(2) ???? Would you care to explain that? > 3 = 1(mod sqrt(2)) >Understand? You must be thinking that the ring in which you are doing the arithmetic includes the square root of 2. Thus it could be the ring of algebraic integers, or it could be Z[sqrt(2)]. Thus 3 = 1 + k*sqrt(2) where k = sqrt(2) would suffice. Is that what you meant? >> >> Now if Nora Baron were an ethical mathematician, there would be no >> debate. >> >> But it's not even clear who Nora Baron is, as was pointed out by >> another poster the name itself is a palindrome. >> >> That is, it's the same reversed as Nora B-aron. >> >> The actual poster may not even be female. > >> Thus: REDUCIBILITY MATTERS: note that: >> >> x^2 - 5*x + 6 is reducible over Q, and >> >> x^2 - 3*x + 6 is irreducible over Q. >> >> It's not a coincidence. >> >> That was hand-waving. In fact, your position requires that people >> believe that given >> >> That was absolutely NOT handwaving. There is nothing in what >> I said but cold, hard computation. I gave two examples: one, >> a reducible polynomial, for which the roots are coprime to one >> or the other factor of the constant term; the other, an >> IRREDUCIBLE polynomial, for which NEITHER root is coprime to >> ANY factor of the constant term. These are not isolated >> examples; these are the general rule. The point: IRREDUCIBILITY >> MATTERS. Somehow, in your terms, the math knows or >> the math cares, whether you like it or not. >Your quadratics are irrelevant to the case as hand, and in fact, it's >fascinating that you believe that you've proven some rule, with two >quadratics. They were examples. No, I do not claim that examples prove rules. I thought in this case that examples might convince you that things were not so simple as you want them to be. I anticipated that you would say that quadratics are irrelevant and said so. However if your principles of factorization were correct, that the form of the factorization is not dependent on reducibility, then the second of the two example quadratics should not exist: *both* roots share algebraic integer factors with both integer factors of the constant term. >> And you blow it off by calling it hand-waving! Tell me >> how I can be more explicit! >You can stick to the line of mathematical logic, rather than trying to >distract. >> P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - >> >> 3(-1+mf^2 )x u^2 + u^3 f) = >> >> (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) >> >> that some f^2 divides off as some function of m, or variable dependent >> on m. >> >> YES! YES! YES! YOU GOT IT! >believe that given > P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - > 3(-1+mf^2 )x u^2 + u^3 f) = > (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) >that some f^2 divides off as some function of m, or variable dependent >on m. Yes. How f^2 factors out of the linear terms depends on whether the polynomial is irreducible, and that in turn can depend on m. You say the factorization has a constant form regardless of m. You said this earlier in this thread about such factorizations: So now those posters who try to convince you that the w's are actually dependent on m, like being functions of m, must now also convince you that the w's make a decision, first looking to see if f = 3 or have some non-unit factor in common with 3, and THEN they decide if they're dependent on m. Such a cute way of putting it. Those whimsical little w's, making decisions on the basis of coefficients of polynomials. Sounds absurd, doesn't it? And THIS is what you call a mathematical proof ??!! This is a new low-water mark. Rather than producing an actual proof, you dismiss the idea by *personifying* the w's. No wonder you are cross-posting to alt.fiction. The fact is, the form of the factorization DOES differ depending on whether or not the polynomial is reducible. That is exactly what was illustrated by the quadratic examples, which you called handwaving. Let's get back to real math. You have said, applying your methods in Advanced Polynomial Factorization, that if you factor the polynomial P(x) = 65*x^3 - 12*x + 1 in the form P(x) = (a1*x + 1)*(a2*x + 1)*(a3*x + 1), where a1, a2, and a3 are algebraic integers, then two of the a's are divisible by sqrt(5) in the algebraic integers. Right? Say a1 is divisible by sqrt(5). Let a1 = sqrt(5) * c1, where c1 is an algebraic integer. Note that -1/a1 is a root of P(x). Therefore P(-1/(sqrt(5)*c1)) = 0. This implies -65*(1/(5*sqrt(5)*c1^3)) - 12*(-1/(sqrt(5)*c1)) + 1 = 0. Multiply through by 1/(5*sqrt*5)*c1^3). You get -65 + 12*5*c1^2 + 5*sqrt(5)*c1^3 = 0. Divide out 5, move things around: sqrt(5)*c1^3 = -12*c1^2 + 13. Square both sides: 5*c1^6 = 144*c1^4 - 312*c1^2 + 169. Rewrite this as 5*c1^6 - 144*c1^4 + 312*c1^2 - 169 = 0. Use your favorite piece of software to show that this is a non-monic and ***irreducible*** polynomial in c1. Then apply a well-known theorem from algebraic number theory: THEOREM: If r is a root of a non-monic polynomial with integer coefficients, ***irreducible*** over the rationals, the r cannot be an algebraic integer. and conclude that c1 cannot be an algebraic integer. This contradicts your claim that at least one of a1, a2, or a3 is divisible (in the algebraic integers) by sqrt(5). Therefore that claim is wrong. Your proof of it has an error. Have I tricked you somewhere in the above argument? Have I left out anything? Have I pulled the wool over your eyes? Have I used phrases like c1 cannot be an algebraic integer because it decides not to be one in this situation? Note that irreducibility is an absolutely key assumption in the Theorem. That is why it plays such a part here. IRREDUCIBILITY MATTERS. >Splitting that sentence up with your own comments and then posting as >if I'd said something else is clearly dishonest. I did NOT post as if you said something else. I did not in any way misquote you or change your words. I totally agree with you that my proof implies that you must conclude that the way f^2 factors out of your polynomial DOES depend on whether that polynomial is reducible or irreducible. I don't understand at all why that should be surprising. If you believe that the form of the factorization DOES NOT depend on reducibility, then you should be able to give an actual proof, and not some bullshit argument about the w's not having the capacity to decide how they should factor depending on the coefficients. That is *NOT* mathematics; that is more like Aesop's Fables. The w's unfortunately do not factor the way you want them to. Similarly, above, contrary to your wishes, NONE of the a's above are divisible by sqrt(5). I think the situation is a little bit like quantum theory. Things at the atomic level do not behave like objects in the macro world. The double-slit experiment and the 'entanglement' experiments prove that. The behavior is *unintutitive*. We don't see these phenomena in everyday life because the scale on which they take place is very small. Here, polynomials factor in familiar ways, just like in high school factoring problems, when they are reducible. But when they are irreducible, they factor differently, *unintuitively*. It is hard to see this happen directly because in the irreducible case, the computations are formidable, even in quadratics. However denying in the face of overwhelming evidence - proof, actually - that they DO happen is just silly. What you have done is turn your back on that evidence and pretend that it does not exist or that we are lying about it or embedding tricks that you cannot detect. We're not. You have several choices here. One is to try to give a real mathematical proof that the w's factor the way you want them to. Another is to try to find an error in one of my proofs [note however that continuing to give irrelevant examples for which the hypotheses are not satisfied is pointless]. One is to concede my arguments and those of others, and start over. One is to continue calling me a liar and try to convince someone - anyone, perhaps - that your argument makes sense and that, e.g., my proof above is a trick. You deleted much of my message. You had accused me of lying, to which I said (and say now) ABSOLUTELY NOT. You deleted the additional discussion of your referring to the polynomial with an irrational constant term as reducible over the rationals. You deleted the part where you said w1*w2 = 3^{2/3}, and I noted that w3 = 3, and that this was inconsistent with the fact that w1*w2*w3 = 3. I wonder why you omitted that, among other things. You have since repeated this same error in another post. Nora B. >James Harris === Subject: Re: Constant factors and polynomials Visiting Assistant Professor at the University of Montana. Nitpicking... > Let's get back to real math. > You have said, applying your methods in Advanced Polynomial >Factorization, that if you factor the polynomial > P(x) = 65*x^3 - 12*x + 1 >in the form > P(x) = (a1*x + 1)*(a2*x + 1)*(a3*x + 1), >where a1, a2, and a3 are algebraic integers, then two of >the a's are divisible by sqrt(5) in the algebraic >integers. Right? > Say a1 is divisible by sqrt(5). Let a1 = sqrt(5) * c1, >where c1 is an algebraic integer. > Note that -1/a1 is a root of P(x). Therefore > P(-1/(sqrt(5)*c1)) = 0. >This implies > -65*(1/(5*sqrt(5)*c1^3)) - 12*(-1/(sqrt(5)*c1)) + 1 = 0. >Multiply through by 1/(5*sqrt*5)*c1^3). You get > -65 + 12*5*c1^2 + 5*sqrt(5)*c1^3 = 0. >Divide out 5, move things around: > sqrt(5)*c1^3 = -12*c1^2 + 13. >Square both sides: > 5*c1^6 = 144*c1^4 - 312*c1^2 + 169. >Rewrite this as > 5*c1^6 - 144*c1^4 + 312*c1^2 - 169 = 0. >Use your favorite piece of software to show that this is >a non-monic and ***irreducible*** polynomial in c1. > Then apply a well-known theorem from algebraic number >theory: > THEOREM: If r is a root of a non-monic polynomial > with integer coefficients, ***irreducible*** over > the rationals, the r cannot be an algebraic integer. >and conclude that c1 cannot be an algebraic integer. You must also include the hypothesis that the polynomial is primitive. Since nonzero constants are units in Q[x], they are not considered nontrivial factors, so the hypothesis must be explicitly included. Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A man's capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of figures few readers can critize. A great many people are staggered to this extend, that they imagine there must be the indefinite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Constant factors and polynomials If you won't listen to us, maybe it would be reasonable for us to > post a warning to editors of math journals, saying there are strong > counterarguments to claims you are making in a paper you are trying > to publish; specifying what those counterarguments are; and noting > that, although you have had multiple opportunities, you have not > found errors in any of them. You are either (1) refusing to do so, > or (2) unable to do so. In either case, it is dishonest of you to try to > publish a result when there are real, rigorous, unanswered objections > which prove it is false. Should we notify editors of this? I think that's an excellent idea. It's unlikely that a reputable journal > would be fooled by James' pseudo-math, but it could save some editors > and/or reviewers from wasting too much time on it. >> I think it's a terrible idea. You're just opening yourself up >> to legitimate claims of conspiring. I would hate to think that >> such a letter could influence the review process, causing an >> editor to prejudge a paper. If it could, just imagine the >> possibilities for abuse. A paper should stand or fall on its >> merits, and a reviewer should judge a paper on its content. >> Trust the system. They're already used to getting incoherent >> documents. >I thought Nora was talking about something other than a letter saying >This guy's a crank. It was my assumption that by specifying what those >counterarguments are she meant that the editor would be provided with >valid mathematical reasons why James' paper is incorrect. The editor >(or reviewer) would be able to read both James' submission and the >counterarguments and make his/her own decision. That's what I assumed she meant as well, probably because that's what she said. It's a terrible idea, for the reasons Randy suggests, and also not necessary, for reasons he suggests. >It would be a little like this: James shows up at a newspaper with a >picture he's taken of a flying saucer. The editor says, Now that's >interesting, because I have another picture here that someone sent me. >See this? Here you are in your back yard, with a camera, some fishing >line, and a pie plate that looks a *lot* like that flying saucer of >yours... >I assume Nora was not seriously proposing to do this, though, and I'm >certainly not qualified to do so. ************************ David C. Ullrich === Subject: Re: Constant factors and polynomials > Nitpicking... > Let's get back to real math. > You have said, applying your methods in Advanced Polynomial >Factorization, that if you factor the polynomial > P(x) = 65*x^3 - 12*x + 1 >in the form > P(x) = (a1*x + 1)*(a2*x + 1)*(a3*x + 1), >where a1, a2, and a3 are algebraic integers, then two of >the a's are divisible by sqrt(5) in the algebraic >integers. Right? Nope. > Say a1 is divisible by sqrt(5). Let a1 = sqrt(5) * c1, >where c1 is an algebraic integer. > Note that -1/a1 is a root of P(x). Therefore > P(-1/(sqrt(5)*c1)) = 0. >This implies > -65*(1/(5*sqrt(5)*c1^3)) - 12*(-1/(sqrt(5)*c1)) + 1 = 0. >Multiply through by 1/(5*sqrt*5)*c1^3). You get > -65 + 12*5*c1^2 + 5*sqrt(5)*c1^3 = 0. >Divide out 5, move things around: > sqrt(5)*c1^3 = -12*c1^2 + 13. >Square both sides: > 5*c1^6 = 144*c1^4 - 312*c1^2 + 169. >Rewrite this as > 5*c1^6 - 144*c1^4 + 312*c1^2 - 169 = 0. >Use your favorite piece of software to show that this is >a non-monic and ***irreducible*** polynomial in c1. > Then apply a well-known theorem from algebraic number >theory: > THEOREM: If r is a root of a non-monic polynomial > with integer coefficients, ***irreducible*** over > the rationals, the r cannot be an algebraic integer. >and conclude that c1 cannot be an algebraic integer. > You must also include the hypothesis that the polynomial is > primitive. Since nonzero constants are units in Q[x], they are not > considered nontrivial factors, so the hypothesis must be explicitly > included. That is correct. And it is true that c1 cannot be an algebraic integer. That has not been under debate. It's also NOT under debate as to whether or not given 65x^3 - 12x + 1 = (a_1 x + 1)(a_2 x + 1)(a_3 x + 1) any a's exist, within the ring of algebraic integers, such that sqrt(5) is a factor of them in that ring. The problem is that neither a_1, a_2, nor a_3 have ANY non-unit factors in common with 5 in the ring of algebraic integers. James Harris === Subject: Re: Going from base n to base 10 > Let N be the naturals and N* all sequences of natural numbers or zeros > which (the sequences) are eventually zero. > You have two maps: > T:N -> N* such that n = T(n)_0 + T(n)_1*10 + T(n)_2*10^2 + ... > where T(n)=(T(n)_0, T(n)_1, T(n)_2, ...) > and, (for base b), > S:N -> N* such that n = S(n)_0 + S(n)_1*b + s(n)_2*b^2 + ... > It is non-trivial that both T and S exist and are bijections but, given > that, TS^(-1) and ST^(-1) are bijections and are, I think, what you are > looking for. You must have a very liberal definition of 'non-trivial'. === Subject: How do you prove this? (Composite functions and their inverses) Let f be a one-one function, and let g be a one-one and onto function. If a function h can be decomposed into f and g, and is given by h = f o g , then the inverse of h is the composite function inv_h given by inv_h = inv_g o inv_f , (*) where inv_g and inv_f are the inverse functions of g and f, respectively. Does anyone know of a proof for (*) ? === Subject: Re: How do you prove this? (Composite functions and their inverses) > Let f be a one-one function, and let g be a > one-one and onto function. > If a function h can be decomposed into f and g, and is > given by h = f o g , then the inverse of h is the composite > function inv_h given by > inv_h = inv_g o inv_f , (*) > where inv_g and inv_f are the inverse functions of > g and f, respectively. > Does anyone know of a proof for (*) ? You've got a difficulty: unless f is also onto, it doesn't _have_ an inverse. -- Paul Sperry Columbia, SC (USA) === Subject: Nora Baron is a palindrome, Not a person In asking the question of why would mathematicians lie, we have two possibilities. One possibility is that they are not lying, but really can't understand my work. There was always the global truth of whether or not I had found the short FLT Proof, and the local truths of the actual mathematical arguments. I won plenty of arguments where people focused on some aspect of an argument, while I was still looking for the whole proof. This September will mark eight and half years since I made the fateful decision to pursue a proof of Fermat's Last Theorem. When I started years ago, I made a decision to publicize my work for very practical reasons. I assumed success. Assuming success I decided that working in secret was counter productive for many reasons, some of them philosophical. Maybe I do it because mathematicians have *broken* the social contract. How would you survive if you made a major math discovery but didn't get proper recognition because mathematicians cheat? Society pays mathematicians to promote mathematical discovery. It does NOT pay them to promote their own interests at the world's expense. Mathematical ideas are not to be fought against for personal gain, but mathematicians are betraying the society that feeds and clothes them. And it's just a matter of time before the world knows because the IDEAS are simply too powerful. In the long run, the truth will come out, and the world will find out that mathematicians broke the social contract, cheating humanity as they continue to take money from society, while blocking through action or inaction acceptance of valid and powerful mathematical ideas. Besides, I figured that even if my making my work public lead to someone else finishing it out at least I'd still be there as the primary contributor. As time went on, I also saw another reason to be public as I got a lot of help on general mathematical issues and on common pitfalls. I also received quite a bit of critical help which allowed me to drop a lot of ideas that wouldn't work, very quickly. In the process, I also found a lot of people angry with me. My thinking on that was pragmatic. I figured that with such a monumental task I needed to do what it took to accomplish my goal, and since at the worst that meant stepping on a few toes here and there and having people mad at me, I could handle that. Besides, once I accomplished my goal, I figured that people would understand as you don't climb such mountains worrying about keeping everybody around you happy. Several thing complicated matters. First it all took a lot longer than I'd planned and it was a lot more frustrating for me. I don't know if any of you can imagine facing a problem like Fermat's Last Theorem and contemplating the real possibility of years of your life going down the drain without you ever succeeding. I lived it. Second, I found the negative reactions that I received were draining as well. And third, I found that I attracted a certain type of person who would reply to my posts apparently simply to get attention as I found ways to get attention brought on my work. I called these people attention parasites and they were like barnacles that I had to routinely scrape off, if I could. Finally, thankfully, I stumbled across a key idea which I've developed over the last couple of years into the full proof. Yet, even with the proof I've faced some problems that I hadn't expected. For one thing, I'd always assumed that a proof would speak for itself, but I feared there might be institutional objections to a short proof of Fermat's Last Theorem, so I banked on it being simple enough that no one could object to it without clearly showing motives other than an interest in the truth. However, the proof while simple in many ways turned out to be just complex enough that I could face heavy resistance that was hard to explain away, while posters in replying to me became more and more sophisticated in obscuring truths like my successful handling of objections. Often I'd hoped that brutal honesty about my work and my motivations would help my case, but my honesty was used against me. Ultimately, it didn't seem to help that I'd talk about past failures or admit that there had been many when I thought I was right. It didn't seem to help for me to admit that there was a monetary gain to finding the proof, or that success would put me in the history books. Even when I was challenged about my sanity or intelligence, mentioning facts about my past that impeached such accusations was used later by people in criticising me, as if I were trying to gain unfair purchase from things like being labeled gifted, or going to Duke University as a child, or having been an active duty military officer rather than using such facts as defense. Quite simply, nothing I did was right. If I kept posting people faulted me simply for that. If I looked for a proof people faulted me for that. When I put up devices like using JSH in the subject line, I'd get faulted for that or despite that, as some people boldly put forward the proposition that any post of mine was an offense to them and that I should quit simply because they were annoyed, even though they could simply not read my posts. Simply put, a political campaign was waged against me with great success, and I found that interesting as I'd often talked of politics and the tools of political campaigns, so I do admit that to some extent I could be said to have brought a lot of that on myself. Still as amusing as the tools used against me might be from one perspective, from the perspective of questions of truth, they were frightening to me because I'd always had a primary assumption that the truth would be the most powerful arbiter. Yet here it was clearly demonstrated to me time and again that the truth was of less than secondary concern on the newsgroup. The sad fact is that I think that there will be people willing to do anything to hide the truth unless I could find something simple enough that they would be unable to do so. And I think I finally did find that. After all, so much is in a name, and many readers on the newsgroup should be able to understand that people like Dedekind came up with a name like algebraic integers for a reason. Also, it seemed to me that it should be easy enough to explain the rationale of extending from Dedekind's work with monic polynomials with integer coefficients to the more general case where the condition of being monic is relaxed. Still, I also see it as yet being something of a longshot. I know that these posts of mine will go through a gauntlet. First there will be the automatic replies from the robot program deleting out all the content I so carefully typed in an effort to keep people from reading what I have to say. Besides that there will be the posts from those who obsessively reply to my posts either in an attempt to get attention for themselves, or apparently to also distract others from reading what I have to say. I don't think any other person in the history of Usenet has gotten such treatment, yet this newsgroup seems to accept it as a casual thing. I must simply see it as a challenge on top of the many that have come my way on a long road. At times I've felt that all of the above was proof that posts like these are a waste of time, but I still find myself hoping that I'll reach those people who don't follow the crowd, who do value truth, and the fact that truth is an integral part of mathematics more than they care about interpersonal squabbles. So here I am hoping yet again, and yet again acting from that hope. And I can't fault myself for hoping for the best in some of you, despite having up until this point mainly seen the worst. Now if any of that makes your head swim from the apparent complexity, then consider what might happen if a professional mathematician were *deliberately* trying to confuse you about those expressions. It's not there, so the objection should go away, but conceding that apparently is something the people who've spent so much time trying to tell you that I'm wrong refuse to do. Now, here's where it get fascinating. You see, they know I can explain it like this, and they've seen me explain it repeatedly, but still they make the posts that they do. Unfortunately, by doing so, they prove that they are willing to lie, or they are incompetent. But if they're willing to lie now, or are incompetent, why should we trust what they've said before? It's really a tight loop if you think about it. What mathematicians say is usually checked by other mathematicians. It's far worse now than most of you suppose as a lot of the technology of our modern world is produced by engineers, so that explains how we still have so much neat stuff, when mathematicians may not have been doing much of anything at all. At this point, it's hard to say how deeply the problem goes or how much of currently established mathematics may be fatally flawed. Oh yeah, so why would they lie? I suspect because they figured they could get away with it, and given the comments I've made, it seemed to them that the world would be a nicer place where no one knew that I'd solved Fermat's Last Theorem. And don't accuse me of entrapment please, but I suspected as much, which is why I said a lot of the things that I did. Slipping over the edge again, are we? Here's a question for you: Note that Wiles' _first_ version _was_ wrong. When the error was discovered there was no conspiracy to cover it up, the proof was acknowledged to be incorrect. Curious Wiles agreed it was incorrect, without even once suggesting that if Gauss were alive he'd see it was right. Why do you conjecture that the error in the second version is being suppressed, while the error in the first version was widely publicized? Surely it would have been simpler just to suppress the error in the first version, wouldn't it? I imagine you have an explanation for this... It's actually rather fascinating. Which just makes it that much clearer that mathematicians have been avoiding interesting *math* instead choosing to focus on obscuring recognition of its validity, either directly in posts attacking my me or my work, or indirectly by ignoring my work. In my search for a simple proof of Fermat's Last Theorem, I've used the newsgroup to test ideas that I wasn't absolutely sure of, as eventually, I could ALWAYS get one of you to post something mathematical, and even when your posts were wrong, which amazingly enough was typical, they could help me figure out what was going on. Needing a group to bounce ideas off of for so long put me in a bad position as I found that I had to holler, in the form of catchy subject lines, and interesting posts, to not be totally ignored, but as I went through the discovery process the noise from useless posters who just posted opinions got louder. Now that's all there remains on the newsgroup, as the few posters who actually post with SOME math content have been quiet for the last couple of days, which is fine with me as I'm sure I have the proof, so I can send it to a journal. That doesn't mean I won't continue to post over the six months or more that it might take for the proof to be reviewed by a journal. In fact, I can assure you that I will. What amazes me is that some of you apparently don't realize how limited your mathematical ability is. And it's as if you believe that you can fool people by making idiot replies to my posts where you put me down. Let me inform you of what should be obvious, putting me down in a post does not make anyone believe that you are intelligent or that you know any math. You may believe that it is a shortcut to people knowing something about you on the group, but it is not. They won't really pay attention to you until you produce MATH. It all reminds me of this one guy who did manage to get known just a little bit. Turns out he had an interesting history of posting on the net that I won't go into, and here he just made these psychological attacks on me. Well, I responded back in anger and we had some verbal battles and the newsgroup seemed to like him. Well, as always happens, I got bored with arguing with him and he tried to make posts on other subjects, still with a psychological tinge if I remember correctly, and he got what he should have expected. I no longer see posts from this person on the newsgroup. Like I just said in another post, I've been using the newsgroup as a place with people I can bounce ideas off of for several years now. Obviously, when I had a correct proof, and given the low level of apparent ability of people on the newsgroup (based on replies to my posts), I'd need to send it to a journal. For some years I've been posting on the newsgroup sci.math where I've been engaging as a hobby in the find for a short explanation for a famous math problem. In discussing math topics I would think that discourse would be less heated, and more rational, than in areas where things should be more ambiguous, like politics or human psychology, but it wasn't the case. One of the things that really surprised me was what I saw as the triumph of form over substance, as I ended up in lots of debates where from my knowledge of logic, mathematics, social norms, etc. I'd given the most convincing argument, but posters would simply refuse to acknowledge that, claim that they'd won, and when I polled the newsgroup, it seemed that the newsgroup as a whole agreed with them. You may rationally assume that maybe it was just that I was wrong, but consider the following. Because it IS math, some of these people finally tripped up by attacking a result that was independently verified by another person and accepted on the newsgroup as correct months previously. They'd been successful up until that point in convincing the newsgroup that a mathematically correct result was in fact false, and had fought for their position over a period of months. Amazingly enough, when caught, these people didn't back down, and to this point still haven't to my knowledge been challenged on the newsgroup. I outline the information above in order to consider the interesting possibility that disciplines where people might think rational debate is more potent, are actually *less* likely to tolerate rational debate, and more likely to operate from belief, and the appearance of truth, which could have important social implications. In considering the sci.math newsgroup, I think my observations about the members of that group, whether correct or not, are pertinent, as it goes to my own subjective interpretations based on those observations: (1) Most posters are technical workers but not working mathematicians. (2) There is a significant posting population of mathematics graduate that they are not a majority. (3) There are a small number of posters who are working mathematicians with Ph.d's in mathematics who are currently teaching and researching in mathematics at universities. My analysis of the hierarchy of the posting group is that most of the posts come from (1), while a significant number of posts come from (3)--more as critiques of others rather than originating discussion on particular ideas or research--while the members of (3) enjoy a special status on the group, basically as experts, which is not surprising. What did surprise me was how much of the posting behavior was juvenile in that it often disintegrated into personal insults when some poster claimed a particular position and was challenged. It seems to me that one would as a matter of commonsense assume that in a discipline like mathematics that debates would be rather quick affairs where mathematics itself based on its conciseness and dependence on logic would be the deciding factor. Not surprisingly with my coming in as an outsider, my lack of familiarity with much of the terminology of mathematicians, my tendency to challenge the hierarchy, my attempt at solving a famous math problem in a way that mathematicians had not succeeded in doing, and my repeated failures with claims of success that later turned out to be premature, lead to a tremendous amount of anger. In challenging the social order I wasn't surprised at the anger as people saw me testing boundaries. I was surprised that their reactions revealed social values at odds with an assumed one of mathematical truth as something independent of human belief. That is I assumed that mathematics was a discipline where truth was seen to be independent of human belief, which investigation indicated was not actually a social value of posters. Ultimately, it seemed to me that rather than depending on mathematical truth as something independent of human beings, the posters on sci.math relied on social truth where social status was considered key when deciding truth. My stated purposes in looking for a simple solution to the famous math problem were among other things, fame and fortune, social status, human progress, revenge for maltreatment at the hands of the group, fun, practice with problem solving, practice with writing, and testing of political methods. My primary tools in discussions were the dialetic, problem solving techniques as learned through academia and personal study, and mathematical arguments. I think the biggest surprise for me was how little weight was often given to even simple mathematical arguments which clearly proved a particular position, unless such arguments were supported by certain members of the group. Worse, it seems to me that the mathematicians in the group were quite aware of their social power, and were quite adept at wielding it as if from long familiarity. The major question raised for society is in determining the value to society of truths from mathematicians, when in asking questions about claims of truth from mathematicians, it may be impossible to determine if it's a social or mathematical truth without the help of mathematicians. It's not as esoteric a question as you may think, as though some mathematical results are tested through experiment in the sciences, many of them are not, like the recent work of Wiles, and others are tested mainly in that there are benefits offered for proving they fail, like with certain Internet encryption methods where substantial cash awards are offered to anyone who can break the encryption and report that fact. As a side note, it is possible that computers may offer a way to remove much of the social power I've hypothesized as they could remove much of the human element in the determination of mathematical truth. Finally, in considering the broader arena of rational discourse, what lessons can be learned from my experiences? I'm still working on that one, but appreciate input, of course. In making my post I'm not claiming to have done a definitive study, but feel that my experiences are worth bringing up to these newsgroups. Personally I prize rational discourse, as well as the role of gadfly and feel that disciplines should not easily be able to keep out outsiders who test them on what they truly believe. You see, I learned years ago that math journals are far less tolerant of repeated effort than you folks are, so I have to be careful about what I send to a journal. To date, I've sent only two efforts--the first one way back when I started, and one more that it took me a while to work out was wrong as I became fixated on the idea of success and the newsgroup only gave me useless feedback--I actually had to pay out $500 to get out of that logjam, thank you very much. The thing is, none of this is new as over the years I've made posts where I talk about using the newsgroup. You folks go along, some of you whine, some of you get verbally abusive, nothing's new. Now I'm just about certain but while I go through the effort of translating my proof to a format more typical for a journal (and no, none of you will see that or what I send, it's not as if you should care or as if I care if you care), I will test out the links within the proof, since the concusion finally seems to be stupendously obvious. That means this is the last post about my main plans, and I'll go back to posts like FLT Proof, simple enough for kids on the newsgroup. Some of you will not read this post and will make replies asking why I don't just send it to a journal. Others will make different useless replies. That's ok. I don't really have much respect for the overall intelligence of the newsgroup anyway, so you all go right ahead and do dumb things. And as for questioning my intelligence, it's not worth it to talk about my having been a member of Mensa, or how I went to Duke University twice as a teenager as part of a summer program most of you probably wouldn't have even heard of. And none of you care that I've been considered gifted by the American educational systems since I was eleven. I know it doesn't matter to any of you that I have a degree in physics from Vanderbilt University, or that I work at a large software development company in a key role because I've told you these things before, and many of you know that they are true. You know what happens after I make a post like this one? And yes, I should know because I've been posting for years. People start posting about how IQ doesn't matter. And how none of that means anything because it apparently upsets them for me to talk about being in the top 2% by intelligence. Hey, you're the people who try and put me down. I'm the person who tries to deal with the facts. But I guess many of you don't want to know the truth because you can't handle the truth. Just remember, I've been looking at posts on this newsgroup for years, and I've seen quite a few people come and go during that time. Stand on your own two feet, when it comes to trying to get the attention of the newsgroup. James Harris === Subject: Re: Nora Baron is a palindrome, Not a person > You know what happens after I make a post like this one? Sales of all products containing caffeine increase tenfold? > Just remember, I've been looking at posts on this newsgroup for years, > and I've seen quite a few people come and go during that time. Then why are you posting this to alt.fiction.original, when it's of no interest to us whatsoever? BTW, I see you're back to the Yahoo email address. I look forward to the next post from jstevh@msn.com telling us that this is a forgery. Go play in the traffic. === Subject: Re: Nora Baron is a palindrome, Not a person > In asking the question of why would mathematicians lie, we have two > possibilities. > One possibility is that they are not lying, but really can't > understand my work. Why do you have the wind up your drawers about Nora Baron? Doesn't Nicholai Bourbaki bother you much more? Historically, Bourbaki is much more significant. Chuck -- ... The times have been, That, when the brains were out, the man would die. ... Macbeth Chuck Simmons chrlsim@earthlink.net === Subject: Re: Nora Baron is a palindrome, Not a person James, you brought this on yourself. You have shown your immaturity over and over again by telling everyone they're liars and threatening to get them fired from their job. I don't know what your problem is, but it is obvious that it's not us. I haven't had as much math as some of the others here, but know enough to know your claims are false. I don't need to prove them false, because others have. You just live in a world in which you think mathematicians are out to get you. Grow some skin, Jerk, David Moran === Subject: Re: Nora Baron is a palindrome, Not a person If Nora Baron is NOT a person, but still can whip the ass off James Harris in math, as she does repeatedly, then that makes James Harris totally ineffective as a mathematician. N'est pas? === Subject: Re: Nora Baron is a palindrome, Not a person So who is Nora Baron? > ... > And as for questioning my intelligence, it's not worth it to talk > about my having been a member of Mensa, or how I went to Duke > University twice as a teenager as part of a summer program most of you > probably wouldn't have even heard of. And none of you care that I've > been considered gifted by the American educational systems since I > was eleven. I know it doesn't matter to any of you that I have a > degree in physics from Vanderbilt University, And it all came to naught. Sad. > or that I work at a > large software development company in a key role ... You look after the keys? Are you what we call a caretaker or janitor? -- G.C. === Subject: Re: Nora Baron is a palindrome, Not a person [thank you, or whom ever, for exposing Nora Badbooboobda Baron; it's just a requirement for him to graduate.] I didn't check to see if there's any math in your tirade, which (at least) doesn't begin with a bunch of extraneious quoting. it's just that your statements fall off the screen with such a thud, firstly that an 8-year file-of-files, without any noticable menu or index, constitutes a reprisal of *le methode de Pierre et Fil*, or a Newer, Less-miraculous Proof. on the other hand, please, accept my apology & concurrent applause, for when I finaly catch-up, if ever! and you've become so charitable: there is certainly a plausible possibilit y that they aren't lying, and simply share my lack of dyscernment. after all, one can learn a lot from the puzzle-books of Mensa, but I'd never dare to join the local chapter. (ah, so; membership hath its priveleges .-) what really takes me aback is your simultaneous requirement that mathfolks be both devine & socially contracted, as you put it. ipso facto, because they have broken the social contract, they must have descended to the other end of the spectrum of the other requirement. > In asking the question of why would mathematicians lie, we have two > possibilities. > Society pays mathematicians to promote mathematical discovery. > However, the proof while simple in many ways turned out to be just > complex enough that I could face heavy resistance that was hard to > Often I'd hoped that brutal honesty about my work and my motivations > would help my case, but my honesty was used against me. Ultimately, > First there will be the automatic replies from the robot program > I don't think any other person in the history of Usenet has gotten > And don't accuse me of entrapment please, but I suspected as much, > Curious Wiles agreed it was incorrect, without even once > suggesting that if Gauss were alive he'd see it was right. > What amazes me is that some of you apparently don't realize how > limited your mathematical ability is. And it's as if you believe that > For some years I've been posting on the newsgroup sci.math where I've > been engaging as a hobby in the find for a short explanation for a > famous math problem. In discussing math topics I would think that > discourse would be less heated, and more rational, than in areas where > things should be more ambiguous, like politics or human psychology, > but it wasn't the case. > (3) There are a small number of posters who are working > mathematicians with Ph.d's in mathematics who are currently teaching > Finally, in considering the broader arena of rational discourse, what > You see, I learned years ago that math journals are far less tolerant > of repeated effort than you folks are, so I have to be careful about > Just remember, I've been looking at posts on this newsgroup for years, > and I've seen quite a few people come and go during that time. Stand > on your own two feet, when it comes to trying to get the attention of > the newsgroup. --les ducs d'Enron! http://members.tripod.com/~american_almanac === Subject: Solutions Manual for_Classical Galois Theory w/ Examples_? Hello All, Does anyone know of a companion solutions manual or online solution source for the book _Classical Galois Theory with Examples_ by Lisl Gaal? The book has a lot of little questions and charts to help the reader make incremental steps in understanding, but solutions are not included in the back of the book ( :(! )to check if one is on the right track. === Subject: Re: to James Harris > Still waiting for you to refute Nora Baron, James. What's the holdup? > Toby What you missed it? I actually replied to Nora Baron in the thread Constant factors and polynomials and basically shredded that poster's arguments. I just checked and Nora Baron has not replied. Yup, Nora Baron has given some math whoppers. Don't believe me? Then believe the math. Consider, in the ring of algebraic integers, P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f). Now using b_1, b_2, b_3, w_1, w_2, and w_3, I have the factorization P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3) where w_1 w_2 w_3 = f, and b_1 b_2 b_3 = (m^3 f^4 - 3m^2 f^2 + 3m), and at m=0 P(0)/f^2 = 3xu^2 + u^3 f = u^2(3x + uf), so two of the b's must equal 0, which means P(0)/f^2 = w_1 w_2 u^2 (b_3 x + u w_3) which is P(0)/f^2 = u^2 (b_3 w_1 w_2 x + u f) = u^2(3x + uf) proving that w_1 w_2 must equal 1, if f is coprime to 3, which leaves b_3 = 3. Essentially objections to how f^2 divides off now come down to claiming that the w's are functions of m, but consider that w_1 w_2 = 1, when m=0, if f is coprime to 3. But that was an arbitrary choice, so let f=3. Now w_1 w_2 = 3^{2/3} WITHOUT REGARD TO m. Therefore, the factorization is P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f = (b_1 x + u)(b_2 x + u)(b_3 x + uf) where you'll notice that the b's are algebraic integers with m=1, f=sqrt(2), but that's a special case as more often only one of them is, which shows a problem with the ring of algebraic integers. James Harris === Subject: Re: to James Harris > Still waiting for you to refute Nora Baron, James. What's the holdup? Toby > What you missed it? I actually replied to Nora Baron in the thread > Constant factors and polynomials and basically shredded that > poster's arguments. I just checked and Nora Baron has not replied. > Yup, Nora Baron has given some math whoppers. Don't believe me? > Then believe the math. > Consider, in the ring of algebraic integers, > P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - > 3(-1+mf^2 )x u^2 + u^3 f). > Now using b_1, b_2, b_3, w_1, w_2, and w_3, I have the factorization > P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3) What constraints are there on the b's or the w's besides the equations below? I could just as easily write P(m)/f^2 = (c_1 x + u d_1) (c_2 x + u d_2) (c_3 x + u d_3) with c_1 = 2b_1, d_1 = 2w_1 c_2 = 3b_1, d_2 = 3w_2 c_3 = b_3/6, d_3 = b_3/6 > where w_1 w_2 w_3 = f, and > b_1 b_2 b_3 = (m^3 f^4 - 3m^2 f^2 + 3m), These are still satisfied by my c's and d's. Consider: c_1 c_2 c_3 = (2b_1)(3b_2)(b_3/6) = b_1 b_2 b_3 d_1 d_2 d_3 = (2w_1)(3w_2)(w_3/6) = w_1 w_2 w_3 I chose 2, 3 and 1/6 as factors. Depending on what you are constraining your b's and w's to do, there might be other factors that are appropriate. You can't just claim that no such factoring has m-dependence, because I could choose c_1 = m b_1 and still have a valid factorization. There are infinitely many valid factorizations, and many of them depend on m. What other requirements do you have on the b's and w's? - Randy === Subject: Re: to James Harris > Still waiting for you to refute Nora Baron, James. What's the holdup? > Toby > What you missed it? I actually replied to Nora Baron in the thread > Constant factors and polynomials and basically shredded that > poster's arguments. I just checked and Nora Baron has not replied. No, you didn't. Instread of airily dismissing it out of hand, you might try starting with NB's: >Here is an example: > x^2 -5*x + 6 = 0 >has constant term 6 = 2*3. The roots are 2 and 3. One root is >coprime to 3 and the other is coprime to 2. >But > x^2 -3*x + 6 = 0 >also has two roots. They are: > r1 = (3 + sqrt(-15))/2 and > r2 = (3 - sqrt(-15))/2. >You can check (by computing a 4th degree polynomial) >that: > NEITHER r1 nor r2 is coprime to 2, and > NEITHER r1 nor r2 is coprime to 3. >Thus: REDUCIBILITY MATTERS: note that: > x^2 - 5*x + 6 is reducible over Q, and > x^2 - 3*x + 6 is irreducible over Q. >It's not a coincidence. As long as you dissemble, you won't have a leg to stand on. Have you told Andrew Wiles about your proof? Toby === Subject: Re: to James Harris > Still waiting for you to refute Nora Baron, James. What's the holdup? Toby What you missed it? I actually replied to Nora Baron in the thread > Constant factors and polynomials and basically shredded that > poster's arguments. I just checked and Nora Baron has not replied. Yup, Nora Baron has given some math whoppers. Don't believe me? > Then believe the math. Consider, in the ring of algebraic integers, P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f). > Now using b_1, b_2, b_3, w_1, w_2, and w_3, I have the factorization P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3) > What constraints are there on the b's or the w's besides > the equations below? I could just as easily write > P(m)/f^2 = (c_1 x + u d_1) (c_2 x + u d_2) (c_3 x + u d_3) > with > c_1 = 2b_1, d_1 = 2w_1 > c_2 = 3b_1, d_2 = 3w_2 > c_3 = b_3/6, d_3 = b_3/6 You've deliberately left the ring of algebraic integers unless you're claiming that b_3/6 is an algebraic integer. The ring is declared to be algebraic integers at the start. where w_1 w_2 w_3 = f, and b_1 b_2 b_3 = (m^3 f^4 - 3m^2 f^2 + 3m), > These are still satisfied by my c's and d's. Consider: > c_1 c_2 c_3 = (2b_1)(3b_2)(b_3/6) = b_1 b_2 b_3 > d_1 d_2 d_3 = (2w_1)(3w_2)(w_3/6) = w_1 w_2 w_3 > I chose 2, 3 and 1/6 as factors. But 1/6 is not in the ring of algebraic integers. >Depending on what you are > constraining your b's and w's to do, there might be other > factors that are appropriate. You can't just claim that no > such factoring has m-dependence, because I could choose > c_1 = m b_1 and still have a valid factorization. There > are infinitely many valid factorizations, and many of them > depend on m. What other requirements do you have on the > b's and w's? > - Randy One essential requirement is that you actually read important information in the argument, such as that the ring is algebraic integers. Here's the argument again, and notice the beginning. Consider, in the ring of algebraic integers, P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f). Now using b_1, b_2, b_3, w_1, w_2, and w_3, I have the factorization P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3) where w_1 w_2 w_3 = f, and b_1 b_2 b_3 = (m^3 f^4 - 3m^2 f^2 + 3m), and at m=0 P(0)/f^2 = 3xu^2 + u^3 f = u^2(3x + uf), so two of the b's must equal 0, which means P(0)/f^2 = w_1 w_2 u^2 (b_3 x + u w_3) which is P(0)/f^2 = u^2 (b_3 w_1 w_2 x + u f) = u^2(3x + uf) proving that w_1 w_2 must equal 1, if f is coprime to 3, which leaves b_3 = 3. Essentially objections to how f^2 divides off now come down to claiming that the w's are functions of m, but consider that w_1 w_2 = 1, when m=0, if f is coprime to 3. But that was an arbitrary choice, so let f=3. Now w_1 w_2 = 3^{2/3} WITHOUT REGARD TO m. That is, the w's are now all constant with regard to m and have the same value no matter what the value of m is. Therefore, the factorization is P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f = (b_1 x + u)(b_2 x + u)(b_3 x + uf) where you'll notice that the b's are algebraic integers with m=1, f=sqrt(2), but that's a special case as generally they are not, which shows a problem with the ring of algebraic integers. James Harris === Subject: Re: to James Harris Still waiting for you to refute Nora Baron, James. What's the holdup? Toby What you missed it? I actually replied to Nora Baron in the thread > Constant factors and polynomials and basically shredded that > poster's arguments. I just checked and Nora Baron has not replied. > No, you didn't. Instread of airily dismissing it out of hand, you might > try starting with NB's: >Here is an example: > x^2 -5*x + 6 = 0 >has constant term 6 = 2*3. The roots are 2 and 3. One root is >coprime to 3 and the other is coprime to 2. Yes, here with *integers* as roots it is easy to just look at the answer. >But > x^2 -3*x + 6 = 0 >also has two roots. They are: > r1 = (3 + sqrt(-15))/2 and > r2 = (3 - sqrt(-15))/2. Now you have something more complex, so what do you do? >You can check (by computing a 4th degree polynomial) >that: > NEITHER r1 nor r2 is coprime to 2, and > NEITHER r1 nor r2 is coprime to 3. Here Nora Baron skipped a step. Actually you can find a quadratic to prove that assuming that either root has a factor that is 2 leads to a non-monic quadratic irreducible over Q. While given that the last two coefficients have a factor that is 3, each root has a factor that is sqrt(3). Possibly the poster assumed that you need to check using sqrt(2) which is why Nora Baron made the claim about a degree 4 polynomial, but you don't need a quartic. >Thus: REDUCIBILITY MATTERS: note that: > x^2 - 5*x + 6 is reducible over Q, and > x^2 - 3*x + 6 is irreducible over Q. >It's not a coincidence. It's irrelevant. Here the actual COEFFICIENTS MATTER as given that both the second and last have a factor that is 3, you know that sqrt(3) is a factor of both roots. As for 2, at best using standard techniques you can determine that assuming that either has a factor that is 2 leads to a non-monic primitive quadratic irreducible over Q, which proves that neither can have that as a factor in the ring of algebraic integers. However, you can prove that negative, but can you prove the positive? That is, how do you prove that *either* DOES have some factor in common with 2 in the ring of algebraic integers? > As long as you dissemble, you won't have a leg to stand on. Your implication is without foundation. My suggestion is that you stay away from social tools and follow the math. If you can, show the mathematical logic behind your assertion that the examples you posted from the post by Nora Baron relate to my work. Why do you *believe* I and other readers should pay attention to them? How exactly do they relate? > Have you told Andrew Wiles about your proof? > Toby My suggestion is that you refrain from trying to use tactics. If your beliefs fit with mathematical logic, then tactics should be unnecessary. Why don't you first try to prove your own assertions mathematically, and then if I'm not reasonable, you might feel justified in switching to social tactics? James Harris === Subject: Re: to James Harris > Still waiting for you to refute Nora Baron, James. What's the holdup? Toby > What you missed it? I actually replied to Nora Baron in the thread > Constant factors and polynomials and basically shredded in your dreams > that > poster's arguments. I just checked and Nora Baron has not replied. > Yup, Nora Baron has given some math whoppers. Don't believe me? > Then believe the math. Good! Let's see these whoppers! > Consider, in the ring of algebraic integers, > P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - > 3(-1+mf^2 )x u^2 + u^3 f). > Now using b_1, b_2, b_3, w_1, w_2, and w_3, I have the factorization > P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3) > where w_1 w_2 w_3 = f, and *** Make a note of this for reference later: w_1 w_2 w_3 = f. > b_1 b_2 b_3 = (m^3 f^4 - 3m^2 f^2 + 3m), > and at m=0 > P(0)/f^2 = 3xu^2 + u^3 f = u^2(3x + uf), > so two of the b's must equal 0, which means > P(0)/f^2 = w_1 w_2 u^2 (b_3 x + u w_3) > which is > P(0)/f^2 = u^2 (b_3 w_1 w_2 x + u f) = u^2(3x + uf) > proving that w_1 w_2 must equal 1, if f is coprime to 3, which leaves > b_3 = 3. > Essentially objections to how f^2 divides off now come down to > claiming that the w's are functions of m, but consider that w_1 w_2 = > 1, when m=0, if f is coprime to 3. > But that was an arbitrary choice, so let f=3. > Now w_1 w_2 = 3^{2/3} WITHOUT REGARD TO m. *** Note this also for reference later. > Therefore, the factorization is > P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - > 3(-1+mf^2 )x u^2 + u^3 f = > (b_1 x + u)(b_2 x + u)(b_3 x + uf) Above you said: > P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3) Clearly in the expression you just gave, P(m)/f^2 = (b_1 x + u)(b_2 x + u)(b_3 x + uf), you intend that w_1 = 1, w_2 = 1, and w_3 = f. Right? But above you say w_1 w_2 = 3^{2/3}. If w_3 = f and f = 3, you get w_1 w_2 w_3 = 3^{5/3}, which is clearly wrong, given that you know the constant term of P(x)/f^2 is u^3*f. As you noted above, w_1 w_2 w_3 = f = 3. This just looks totally confused. And this is the second time you have posted this. Maybe you should think about fixing these whoppers. Nora B. > where you'll notice that the b's are algebraic integers with m=1, > f=sqrt(2), but that's a special case as more often only one of them > is, which shows a problem with the ring of algebraic integers. > James Harris === Subject: Re: WANTED: USED BOOKS Check out: http://dogbert.abebooks.com/abe/BookSearch >I have been looking for the used text books or old editions of >textbooks I read in college. I would love to get hold of a copy of the >following textbooks for cheap >For e.g. Kenneth H. Rosen, Discrete Mathematics and Its Applications >Calculus Deborah Hughes-Hallett, Gleason, McCallum >or Calculus Ron E. Larson, Robert P. Hostetler, Bruce H. Edwards >Calculus: An Intuitive and Physical Approach Morris Kline >Where can I buy books mentioned above? I am looking for the kind that >the bookstores won't buy back and the student no longer wants to use. >I wish to do exercises in the back of the book and look up the >solutions. A textbook that also covers the history a bit while it goes >over the problems and their solutions would be great. Please email me >your response instead of responding on the newsgroup because I don't >check with this group often. === Subject: homotopy question For m n, is pi_m defined in that case? Also, to show that two manifolds M1 and M2 have the same homotopy type, is it necessary to show that their homotopy groups pi_m(M1) and pi_m(M2) are isomorphic for all m? === Subject: Re: homotopy question You probably know that pi_1 (S^n)=0 for n>1. However, there are some notrivial ways to wrap k-dimensional spheres (k>1) around S^n. For example: pi_3 (S^2) = Z pi_4 (S^2) = Z_2 pi_7 (S^4) = Z x Z_4 x Z_3 Possibly even weirder, pi_2 of the 2-sphere wedged with the 1-sphere is the sum of infinitely many copies of Z (although S^2 v S^1 is not a manifold), so it is not even finitely generated. So, in conclusion, pi_m (M^n) is defined, even if m>n, namely as [S^m,M^n] i.e. the set of homotopy classes of maps from S^m to M^n. However, you can show that there are no nontrivial ways to wrap a k-sphere around S^1 if k>1, therefore pi_k (S^1)=0 for k>1. Also, you probably know that pi_m is abelian for m>1, therefore you can always decompose the group into a direct sum of copies of Z or some finite cyclic groups. Michael Williams === Subject: Re: homotopy question >For m then I gather that means that any hypersphere S^m embedded in that >manifold can be contracted to a point. Not just embedded--simply *mapped into*. (Embedded has a technical sense; if you don't know it, learn it.) And this has nothing to do with the target being an n-manifold: precisely what it means, for *any* (half-way decent, connected) topological space X, for the homotopy group pi_m of X to be trivial is that any continuous map of S^m into X is homotopic to a constant map from S^m onto a point of X. >But what about when m > n, is >pi_m defined in that case? Yessirree! For instance, pi_3(S^2) is non-zero (it is infinite cyclic), and in particular the Hopf fibration of S^3 over S^2 represents a non-zero element of this group. (Think of S^3 as the unit 3-sphere in complex 2-space C^2, so a point of S^3 is a pair (z,w) where z and w are complex numbers and |z|^2+|w|^2=1. The Hopf fibration H maps (z,w) to the extended complex number z/w, an element of the Riemann sphere C union {infinity}, i.e., S^2. Showing that H is non-trivial in homotopy is ... non-trivial. But very educational. Look it up.) >Also, to show that two manifolds M1 and M2 have the same homotopy type, >is it necessary to show that their homotopy groups pi_m(M1) and >pi_m(M2) are isomorphic for all m? Well, if M1 and M2 have the same homotopy type, then for all m, pi_m(M1) and pi_m(M2) are isomorphic; but the converse is not true. (You should in any case demand more; there is, for example, a natural action of the fundamental group of X on each pi_m(X), and you shuld demand that those actions also correspond. But it still won't be true that isomorphic-homotopy-groups-plus- identical-pi1-actions implies same-homotopy-type. There's more and more structure to retain!) Lee Rudolph === Subject: Re: AN: GuruGram #39 <41BA2B9E.DAAA15C7@tinaja.com> I read in sci.electronics.design that Rich Grise consensus. Who's in charge here, voltage or current? (pun unintended, >but noted.) ;-) The voltage is there even if the wire is not part of a complete circuit. Can I do Faraday's equation in ASCII art? / / / | d | | O e dl = - - | | B dA | dt| | / / / e is the element of induced voltage B is the induction A is the area within which B exists Some of the puzzles depend on people forgetting that B, A or both may vary with time. -- The good news is that nothing is compulsory. The bad news is that everything is prohibited. http://www.jmwa.demon.co.uk Also see http://www.isce.org.uk === Subject: Re: AN: GuruGram #39 > I read in sci.electronics.design that Rich Grise It seems we're just going to have to sit down and try to reach a >>consensus. Who's in charge here, voltage or current? (pun unintended, >>but noted.) ;-) > The voltage is there even if the wire is not part of a complete circuit. > Can I do Faraday's equation in ASCII art? > / / / > | d | | > O e dl = - - | | B dA > | dt| | > / / / > e is the element of induced voltage > B is the induction > A is the area within which B exists > Some of the puzzles depend on people forgetting that B, A or both may > vary with time. Having never heard of Farady's equation, I looked it up, and am now more confused than before: And I flunked out of school because I couldn't grasp calculus. I get the concept of the integral - it's the area under the curve. The double integral, that's conceptually the inverse of the second derivatave, right? The integral of the integral(s)? Other than that, I'm lost. But I do have the impression that Faraday's equation has something to do with the interaction of magnetic fields and electric current, is this close? Rich (and don't worry - I'm not asking for anybody to try to teach me calculus! :-) ) === Subject: Re: AN: GuruGram #39 >> I read in sci.electronics.design that Rich Grise consensus. Who's in charge here, voltage or current? (pun unintended, >but noted.) ;-) >> The voltage is there even if the wire is not part of a complete circuit. >> Can I do Faraday's equation in ASCII art? >> / / / >> | d | | >> O e dl = - - | | B dA >> | dt| | >> / / / >> e is the element of induced voltage >> B is the induction >> A is the area within which B exists >> Some of the puzzles depend on people forgetting that B, A or both may >> vary with time. >Having never heard of Farady's equation, I looked it up, and am now more >confused than before: >And I flunked out of school because I couldn't grasp calculus. I get the >concept of the integral - it's the area under the curve. The double >integral, that's conceptually the inverse of the second derivatave, right? >The integral of the integral(s)? Other than that, I'm lost. But I do have >the impression that Faraday's equation has something to do with the >interaction of magnetic fields and electric current, is this close? >Rich >(and don't worry - I'm not asking for anybody to try to teach me calculus! :-) ) The integral symbol on the left has a circle in the middle- it's a *line integral* (around a closed path). The other is a *surface integral*. There's an old (ca. 1973) little paperback book you may be able to find called Div, Grad, Curl, And All That, which goes into this vector calculus stuff in more detail without being any more stuffy than necessary. For example, CARLO BARONCINI 125 NORTH PARKER STREET, SAN PEDRO, CA $11.98 Or you can probably find all the pieces to the puzzle on the net, one way or the other. Spehro Pefhany -- it's the network... The Journey is the reward speff@interlog.com Info for manufacturers: http://www.trexon.com Embedded software/hardware/analog Info for designers: http://www.speff.com === Subject: Re: AN: GuruGram #39 <41BA2B9E.DAAA15C7@tinaja.com> I read in sci.electronics.design that Spehro Pefhany The integral symbol on the left has a circle in the middle- it's a *line >integral* (around a closed path). Yes, that's a little paradoxical, because the voltage is there even if the path isn't closed insofar as appreciable current can flow. For example, you can measure it on the 200 mV range of a DMM, which has an almost infinite input impedance, and by extrapolation it would still be there if the impedance were really infinite. -- The good news is that nothing is compulsory. The bad news is that everything is prohibited. http://www.jmwa.demon.co.uk Also see http://www.isce.org.uk === Subject: Re: lim question <9vtlr0pllav06rtolocgq040n000toiq9j@4ax.com> >But 1- does _not_ refer to a number in the notation x -> 1- that >you see in calculus books. I know. >You need to understand that fact if you're interested in understanding >calculus. Is it possible to exist mathematically rigorous constructions (or concepts) different from limit or hyperreal number system, on whose basis is possible development of calculus theory? >>Are you completely sure that it is impossible to exist a number >>system in which a thing named 1- is a number < 1? >Look. Here's a definition: >Definition: 1- = 17. >Now we have a number system in which 1- equals 17. So what? 17 >= 1 If 1- is defined to be 1 - eps, where eps is a positive infinitesimal from the hyperreal number system, then will be true: lim(a -> 1-, a^(1/(1-a))) = 1/e? === Subject: Re: lim question > If 1- is defined to be 1 - eps, where eps > is a positive infinitesimal from the hyperreal > number system, then will be true: > lim(a -> 1-, a^(1/(1-a))) = 1/e? Since polynomials are continuous, if the limit were taken in the hyperreals, we would get: lim(a ->(1-eps), a^(1/(1-a))) = (1-eps)^(1/(1-(1-eps))) = (1-eps)^(1/(eps)), which is a hyperreal number (not a real) with standard part 1/e. So, it would be infinitely-close-to but not equal to 1/e. === Subject: Re: lim question <9vtlr0pllav06rtolocgq040n000toiq9j@4ax.com> <6pLud.1075934$Gx4.504967@bgtnsc04-news.ops.worldnet.att.net> >lim(a ->(1-eps), a^(1/(1-a))) >= (1-eps)^(1/(1-(1-eps))) >= (1-eps)^(1/(eps)), >which is a hyperreal number (not a real) >with standard part 1/e. So, it would be >infinitely-close-to but not equal to 1/e. If n = 1/eps is a hypernatural, is it true the following?: (1-eps)^n = (1-eps)*(1-eps)*(1-eps)*...*(1-eps) (n times) If so, how to perform not-finite many calculations with hyperreal numbers? Kevin === Subject: Re: lim question >>lim(a ->(1-eps), a^(1/(1-a))) >>= (1-eps)^(1/(1-(1-eps))) >>= (1-eps)^(1/(eps)), >>which is a hyperreal number (not a real) >>with standard part 1/e. So, it would be >>infinitely-close-to but not equal to 1/e. > If n = 1/eps is a hypernatural, is it true the following?: > (1-eps)^n = (1-eps)*(1-eps)*(1-eps)*...*(1-eps) (n times) Prof. Israel or someone else may have a more definitive answer. AFAIK, it does not make sense to operate with your expression in this way. (I'm pretty sure there are uncountably many infinite hyperreal integers.) Writing n-times does not have any meaning that makes the right-hand side well-defined. > If so, how to perform not-finite many calculations with hyperreal > numbers? === Subject: Re: lim question <9vtlr0pllav06rtolocgq040n000toiq9j@4ax.com> <6pLud.1075934$Gx4.504967@bgtnsc04-news.ops.worldnet.att.net> >Prof. Israel or someone else may have a more definitive answer. So, you do not understand completely the hyperreals? n > 1 f(x, n) = x * f(x, n - 1) f(x, 1) = x According to one math book a function consists of: domain D, target T and a rule which for every y from D specifies a unique g(y) from T. The rule of f works only if n is finite integer. Does f(1 - eps, m) is a definite value for m = 1/eps > 0, m - hyperinteger? Kevin === Subject: Re: lim question <9vtlr0pllav06rtolocgq040n000toiq9j@4ax.com> <6pLud.1075934$Gx4.504967@bgtnsc04-news.ops.worldnet.att.net> In addition to the previous post: What is the rule of u if u(x) = (1-eps)^x, x - hypernatural? === Subject: Re: lim question > In addition to the previous post: What is the rule of u if u(x) = > (1-eps)^x, x - hypernatural? In nonstandard analysis, every standard mathematical object corresponds to an object in the nonstandard realm as well. The standard function f(a,x) = a^x defined from a>0 real and x real to reals corresponds to a nonstandard function (called by the same name f) from a>0 *real and x *real to *reals. So in particular f(a,x) is defined for a=1-eps, eps infinitesimal and x=n in *N-N an infinite *natural number. Reference: A. Robinson, NON-STANDARD ANALYSIS. I doubt the OP intended to talk about nonstandard analysis, but if he did, then he should read about it in a good reference, rather than just getting hints here in a newsgroup. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: lim question <9vtlr0pllav06rtolocgq040n000toiq9j@4ax.com> <6pLud.1075934$Gx4.504967@bgtnsc04-news.ops.worldnet.att.net> >> In addition to the previous post: What is the rule of u if u(x) = >> (1-eps)^x, x - hypernatural? >The standard function f(a,x) = a^x defined from a>0 real and x >real to reals corresponds to a nonstandard function (called by the same name f) >from a>0 *real and x *real to *reals. So in particular >f(a,x) is defined for a=1-eps, eps infinitesimal and x=n in *N-N >an infinite *natural number. So, u is defined in NSA. What means ^ in the formula of u? How to compute u(x) for x hypernatural? >In nonstandard analysis, every standard mathematical object corresponds to >an object in the nonstandard realm as well. Then for f defined as: n > 1 f(x, n) = x * f(x, n - 1) f(x, 1) = x what is f(1-eps, m) for m - hypernatural and eps - positive infinitesimal? Kevin === Subject: Re: lim question > If 1- is defined to be 1 - eps, where eps is a positive infinitesimal > from the hyperreal number system, then will be true: lim(a -> 1-, > a^(1/(1-a))) = 1/e? It is easier to say that x approaches 1 from the minus side. Bob Kolker === Subject: Re: generalized mean and moments Or do you > want to express not only in terms of but also in terms of , > , ...? Yes, that's what I'm asking. How would I express a generalized mean ^(1/a) in terms of a series: f(a) = f_1(a) + f_2(a) + ... Or as a series of orthogonal moments. What I would like to do is to approximate the generalized mean formula using a few simple moments (the distribution functions that I'm working with are simple enough, have only one peak, and have moments of finite value). Note that for a=2, the generalized mean has the form ^(1/2), and not simply as . If anyone knows any good reference about this question, I would appreciate it if you could post it. I've been searching lately but haven't found a useful one. David === Subject: String Theory: Good, Bad and Bogus Part 1 Words written by Dennis Overbye are between quote marks. a single equation that could explain all the laws of physics, all the forces of nature - the proverbial 'theory of everything' ? Extraordinary claims require extraordinary proof. Physicists, like Dr. Robert Park, use a double standard not applying the same rules of engagement to fashionable elegant string theory as they do to flying saucers, the paranormal and cold fusion. http://www.archivefreedom.org/ And so emerged into the limelight a strange new concept of nature, called string theory, so named because it depicts the basic constituents uniting all the forces, string theory had the potential of achieving the goal that Einstein sought without success for half his life and that has embodied the dreams of every physicist since then. If true, it could be used like a searchlight to illuminate some of the deepest mysteries physicists can imagine, like the origin of space and time in the Big Bang and the putative death of space and time at the infinitely dense centers of black holes. ... In the last 20 years, string theory has become a major branch of physics. Physicists and mathematicians conversant in strings are courted and recruited like star quarterbacks by universities eager to establish their research credentials. String theory has been celebrated and explained in best-selling books like The Elegant Universe, by Dr. Brian Greene, a physicist at Columbia University, and even on popular television shows. 'Let them eat cake' said Marie Antoinette shortly before she was guillotined. even as they ate cake and drank wine, the string theorists admitted that after 20 years, they still did not know how to test string theory, or even what it meant. Note or even what it meant. Mainstream theoretical physics today is in a sorry state. Not so for experimental physics. As a result, the goal of explaining all the features of the modern world is as far away as ever, they say. And some physicists outside the string theory camp are growing restive. At another meeting, at the Aspen Institute for Humanities, only a few days before the string commemoration, Dr. Lawrence Krauss, a cosmologist at Case Western Reserve University in Cleveland, called string theory 'a colossal failure.' Brian Greene got a few million dollars advance for his book. The big corporations have a vested interest in this hype that can be compared to WMD in Iraq. String theorists agree that it has been a long, strange trip, but they still have faith that they will complete the journey. The fusion of science and religion on the heels of the fusion of Church and State? Twenty years ago no one would have correctly predicted how string theory has since developed, said Dr. Andrew Strominger of Harvard. 'There is disappointment that despite all our efforts, experimental verification or disproof still seems far away. On the other hand, the depth and beauty of the subject, and the way it has reached out, influenced and connected other areas of physics and mathematics, is beyond the wildest imaginations of 20 years ago.' In a way, the story of string theory and of the physicists who have followed its siren song for two decades is like a novel that begins with the classic what if? What if the basic constituents of nature and matter were not little points, as had been presumed since the time of the Greeks? What if the seeds of reality were rather teeny tiny wiggly little bits of string? merely correspond to different ways for the strings to vibrate, different notes on God's guitar? I'll play the music on my guitar, yes! Rossini, The Barber of Seville. That would explain why you cannot have a single quark you cannot have a string with only one end. Strings seduced many physicists with their mathematical elegance, but they had some problems, like requiring 26 have anything to do with quarks or the strong force. You need the extra space dimensions because string theory violates signal locality in Einstein's original 4D curved spacetime. That is, final causes after their effects are possible in plain vanilla 4D string theory with topology change. Of course, this is exactly what the UFO and paranormal evidence is telling us along with presponse experiments by Dick Bierman and others. Then there is also the Intelligent Design controversy. What is bad and bogus for string theorists is good for UFO, paranormal and consciousness researchers. One man's meat is another man's poison. When accelerator experiments supported an alternative theory of quark behavior known as quantum chromodynamics, most physicists consigned strings to the dustbin of history. But some theorists thought the mathematics of strings was too beautiful to die. In 1974 Dr. Schwarz and Dr. Joel Scherk from the .83cole Normale Sup.8erieure in France noticed responsible for transmitting gravity in a quantum theory of gravity, if such a theory existed. This pyrrhic victory has been trumpeted as a great triumph showing the power of fantasy in the minds of string theorists. Einstein's gravity is not renormalizable as a quantum field theory which strongly suggests that it should not be quantized top -> down like electro-weak dynamic and chromodynamics. Indeed, Andrei Sakharov suggested that Einstein's gravity emerges bottom -> up as a collective phenomenon just like the elasticity in crystals. P.W. Anderson has since formalized this idea as More is different. Furthermore all current tests for grainy quantum gravity foam that should show dispersion in the high energy gamma rays from outer space have been null. That is, there is no evidence for the equation E^2 = [(pc)^2 + mc^2]/[1 + (pc/mPc^2)^2] where mP is the Planck mass for the quantum gravity graininess of Einstein'sw spacetime geometrodynamics. Note that the high energy limit of this equation is E -> mPc^2 ~ 10^-5 grams c^2 ~ 10^16 ergs ~ 10^9 Joules ~ 10^28 electron volts implying a maximal acceleration of a_max = c^2/LP ~ 10^21/10^-33 ~ 10^54 cm/sec^2 ~ 10^51g But, so far, no evidence supports this idea. If my theory of the emergence of Einstein's gravity of MACRO-QUANTUM curved spacetime from the PARTIAL cohering of the micro-quantum zero point vacuum fluctuations in the BCS unstable massless pre-inflationary globally flat false conformal vacuum is correct, there is no quantum foam in principle. non-perturbative background-independent MACRO-QUANTUM vacuum condensate of virtual electron-hole pairs bound by virtual photons near the E = 0 Fermi energy of the pre-inflationay false vacuum obeys E^2 = (pc)^2 + Egap^2 Egap ~ (Debye energy of edge of Fermi surface)e^-1/Vrho(0) V is the attractive static potential energy between the virtual electron-hole (hole = positron) pair from a single virtual photon exchange. rho(0) is the density of Dirac's negative energy virtual electron states at the edge per unit energy. The Einstein-Cartan tetrads eu^a derive from the MACRO-QUANTUM COHERENT different order parameter. guv = eu^anabev^b = nuv + (1/2)Lp^2(PHASE OF VACUUM COHERENCE){,u,v} = nuv + (1/2)[Eu,v + Ev,u) { } is symmetrizer (i.e. anti-commutator) with ,u ordinary partial derivatives in a local coordinate patch of the manifold. eu^a = &u^a + Eu^a &u,^a = Kronecker-Delta i.e. 1 if u = a, 0 if u =/= a Eu = Eu^a,a = Lp^2(PHASE OF VACUUM COHERENCE),u This equation is the analog to the deBroglie-Bohm guidance equation of IT by BIT for a quantum liquid. Here we have an ELASTIC ODLRO quantum solid that I introduced into physics in 1966-67 in Physics Letters A on Super Solids in Helium when I was on the physics faculty of San Diego State and the creator of the National Science Foundation Summer School in Superfluid Physics and Lasers for College Teachers at San Diego State (1969 & 1970) that I ran with Herschel Snodgrass. The faculty included F.W. Cummings and Jim Johnston. The QUANTUM OF AREA Lp^2 = hG/c^3 ensures Hawking's (also Bekenstein, Susskind, t'Hooft) Entropy of Volume of Space = (Bounding Area of Volume)/4Lp^2 Space-time physics is local because the macro-quantum order parameter is local. This same locality ensures that the early post-Big Bang universe after the inflationary vacuum phase transition has low entropy so that the Arrow of Time for the irreversible processes of the Second Law of Thermodynamics points in the same direction as the presently accelerating expansion of the 3D space of our universe from residual w = -1 zero point dark energy of negative quantum pressure. Hawking's blackhole entropy formula is automatically obeyed. That is considered a triumph of string theory. It's no big deal really. Sorry for stealing string theorist's thunder and upstaging them, but as Ludwig Boltzmann said Elegance is for tailors. Not that this new way of connecting the cosmic dots is not elegant to the max. :-) That is, the pre -> post inflationary Big Bang phase transition collapses the phase space volume of the ground state and the entropy is ~ kBlog(Phase Space Volume). This same collapse of ground state phase space volume always accompanies More is different emergence of qualitatively CREATIVE new order, AKA spontaneous breakdown of vacuum symmetry. To be continued. === Subject: Re: JSH: Factoring Integers [the same old song] > James Harris A mule is an animal with long funny ears, kicks up at anything he hears His back is brawny but his brain is weak, he's just plain stupid with a stubborn streak. And by the way, if you hate to go to school, you may grow up to be a mule. No watermelons in winter huh? === Subject: Re: Is this math test too easy? <93k5q0p7es1dpbt7e2upljfods2gdp723r@4ax.com> <30hnelF2vr3i3U1@uni-berlin.de> BEKOS === Subject: Re: Is this math test too easy? <93k5q0p7es1dpbt7e2upljfods2gdp723r@4ax.com> <30hnelF2vr3i3U1@uni-berlin.de> > BEKOS That's not a misspelling, its Phrygian, in case you were wondering dimwit. Frederick of Prussia tried too. === Subject: Re: rewrite this inequality >Ah, but you didn't try sloving it. What's there to slove? It's flase. --Lynn === Subject: Re: rewrite this inequality >>Ah, but you didn't try sloving it. > What's there to slove? It's flase. Somehow Lewis Carroll comes to mind, something about slithy toves, or was it slovy tithes? === Subject: Re: Mathematical Logic === > Subject: Mathematical Logic > I. Let Gamma = {negation of all v_1P_v_1, P_v_2, P_v_3,...}. > P denotes the relation; v_i denotes elements in sentences. > What's Gamma, { ~Av1.P(v1), ~Av2.P(v2), ... } ? TX_MG: Gamma = {~Av1P(v2), v2P(v2), v3P(v3), ...} if you want to have ~ denote as negation, A denotes as the for all. > Is Gamma consistent? Is Gamma satisfiable? > this problem, i know Gamma is not consistent, because it just a > bunch of elements and relations together, there is no sentence > within, it is not a formula. then since is not consistent, it is not > satisfiable. BUT, how do i apply completeness theorem to this prove? > please give me some pointers. > -- > II. determine the following sentences, either show there is a > deduction(from empty set)or give a counter-model(i.e., a stucture in > which it is false): > Will you not follow the sloppy fad(like this)of not using spaces > (where they should be) because that fad makes reading slower?!! > a) all x (Q_x --> All y Q_y) > Q_x =df (x = a) counter example > b) (There exist P_x --> For all y Q_y) --> for all Z(P_z --> Q_z) > -) There exist P_x --> For all y Q_y > -) P_z > Ay.Q_y > Q_z > P_z -> Q_z > Az(P_z -> Q_z) > (Ex.P_x -> Ay.Q_y) -> P_z -> Q_z > c) for all z(P_z --> Q_z) --> (There exist x P_x --> for all yQ_y) > P_x =df Q_x =df (x = a) counter example > d) Not exist y for all x P_(xy) <--> not P_(xx) > d: deduction > Do you mean? > Not exist y for all x (P_(xy) <--> not P_(xx)) TX_MG: yeah you are right, i mistyped. > -- > III. Assume the language (with equality) has just the parameters > (for all) and P, where P is a two-place predicate symbol. Let A be > the structure with |A| = Z, the set of integers (positive, negative, > and zero) and with belong to P^A if |a-b|=1. Thus A looks > like a infinite graph: > ...<-->*<-->*<-->*<-->... > Show that there is an elementarily equivalent structure B that is > not connected. (being connected means that for every two members of > |B|, there is a path between them. A path--of length n -- from a to > b is a sequence with a = p_0 and b = p_n and > belong to P^B for each i.) > How about (1/2)Z ? TX_MG: what about it? > I know I need to apply compactness, > but i don't know how to get it start. > Me neither. Are there axioms for P? > ---- === Subject: Re: Mathematical Logic === Subject: Re: Mathematical Logic > I. Let Gamma = {negation of all v_1P_v_1, P_v_2, P_v_3,...}. > P denotes the relation; v_i denotes elements in sentences. > What's Gamma, { ~Av1.P(v1), ~Av2.P(v2), ... } ? > TX_MG: Gamma = {~Av1P(v2), v2P(v2), v3P(v3), ...} if you want to have > ~ denote as negation, A denotes as the for all. v2 P(v2) isn't well formed formula. > Is Gamma consistent? Is Gamma satisfiable? > BUT, how do i apply completeness theorem to this prove? -- > II. determine the following sentences, either show there is a > deduction(from empty set)or give a counter-model(i.e., a > stucture in which it is false): > d) Not exist y for all x P_(xy) <--> not P_(xx) > d: deduction > Do you mean? > Not exist y for all x (P_(xy) <--> not P_(xx)) > TX_MG: yeah you are right, i mistyped. Assume d, prove Ay.Ax.(x = y) -- > III. Assume the language (with equality) has just the > parameters (for all) and P, where P is a two-place predicate > symbol. Let A be the structure with |A| = Z, the set of integers > (positive, negative, and zero) and with belong to P^A if > |a-b|=1. Thus A looks like a infinite graph: > ...<-->*<-->*<-->*<-->... > Show that there is an elementarily equivalent structure B that > is not connected. (being connected means that for every two > members of |B|, there is a path between them. A path--of length > n -- from a to b is a sequence with a = p_0 > and b = p_n and belong to P^B for each i.) > How about (1/2)Z ? > TX_MG: what about it? Yea, what about it? What language? FOL with = and P. What are the axioms for <=? None? Then (1/2)Z is a model. You can also define for (1/2)Z, aRb when |a - b| = 1. Is (1/2)Z equivalent to Z? Is it elementary? > I know I need to apply compactness, > but i don't know how to get it start. > Me neither. Are there axioms for P? ---- === Subject: Reverse Rounding - Determining a Range This is a real world problem I'm struggling with. Lets say that you are the owner of a store. You have 15 packages that are numbered 1 to 10. Each package contains as many items as the number written on the package. There is only one type of item so the price per item is constant but has a decimal precision of up to 6 places. Each package can only be sold for whole pennies, so the price per package is rounded. Figure out the lower bound of the price per item and the upper bound of the price per item if the total retail value for all packages is $0.10, $0.25, or $0.46. Well, effectively, the problem can be broken down into: ROUND(1*Y,2) + ROUND(2*Y,2) + ROUND(3*Y,2) + ROUND(4*Y,2) + ROUND(5*Y,2) + ROUND(6*Y,2) + ROUND(7*Y,2) + ROUND(8*Y,2) + ROUND(9*Y,2) + ROUND(10*Y,2) = 0.10 .10 / 55 = 0.001818. Because the rounding is done for two decimal places, I know that my range must be +/- 0.001. Therefore, my range must be between 0.000818 and 0.002818. I can brute force this by using that range and plugging each value into the equation until my result equals the desired result, but I'm looking for a mathematical method to solve this. Any helps is greatly appreciated. === Subject: Re: EUREKA Cantor exposed.... sci.math curls tails between legs.. > : Therefore > : Ea e N, Ab e N, L(a,b) = !L(b,b) > Of COURSE Ea. That a is the a AT WHICH r occurs > on the list: r=L(a). > : When a=b > : Ea e N, L(a,a) =!L(a,a) > : = : Ec e N, c=!c > : CONTRADICTION > Right. This is the part I was after, your construction works out to this, EaAb L(a,b) = !L(b,b) Which George agrees with, then also agrees that its a contradiction. Hence exactly this >>your conclusions of LET (i,n)=!(n,n), contradiction, therefore new sequence >>therefore higher infinities than infinite lists exist Herc === Subject: Re: EUREKA Cantor exposed.... sci.math curls tails between legs.. We're going to have to go through this about 200 more times. George is the only one who follows what I'm saying. I'm trying to formalise your Cantorian argument as this: >>They match to UNLIMITED PRECISION. the sequence is on the list to infinite flips. >IT * DOES * contradict your conclusions of LET (i,n)=!(n,n), contradiction, therefore new sequence >therefore higher infinities than infinite lists exist The reason you are all religious about cantors proof is this: you have a logical construction of the real (dn) = !(n,n) you have a 'concrete' argument of its non existence (i,n) != (n,n) What you don't see is that they are mirror images of the logic of one another *** What ACTUALLY HAPPENS, is that n->oo and your contradiction dissapears. n does approach infinity in your construction > |-|erc says... >NOT (Ea e N, Ab e N, L(a, b) != L(b, b)) > More readably, you have said > ~(EaAb[Lab!=Lbb]) > It is OF COURSE BLATANTLY OBVIOUS Right! This is an algebraic truth about diagonals. The interpretation is no real on the list is completely different to the diagonal... at every digit THIS IS TRIVIAL, but its an important axiom about dealing with multiple variables. You have all stumbled across it as a real exists that is completely different to the diagonal Er e R, Aa e N, Eb e N, L(a, b) =/= r(b) DMC's claim or, > again, more legibly, > AaEb[Lab=Lbb] > and again, it is > BLATANTLY OBVIOUSLY true that Eb, Right, but it doesn't suggest DMC's claim. DMC's claim is just naivety of the AXIOM. >Aa e N, Eb e N, L(a, b) = L(b, b) FACT > DMC That doesn't follow at all. As a matter of fact, you've already > proved the *opposite*, namely that there is a real r such that > Aa e N, Eb e N, L(a, b) != r(b) FICTION > But, again, as Jesse pointed out, this is NOT the > opposite of that, because THIS talks about r, whereas > that doesn't. Anyway believe what you want, because Cardinality won't be getting taught much longer and you just miss out on cyberspace construction next decade. I'm literally going to have to publish this puzzle in the newspaper, so everyone can work out for themselves if it would work, and show how thick you must be to believe this solution. read it, actually examine this solution, its blatently does not work. All possible sequences of heads and tails have been flipped to inifinite length FACT. >Take one of the people, whatever his 1st flip was, reverse it! If he >flipped a head you select tail, if he flipped a tail, heads. That's >your first outcome, cross him off and select someone else, whatever was >their second flip, reverse it! Keep on going and you have an infinite >sequence that is different to everyone's sequence in atleast one flip. Herc === Subject: Re: EUREKA Cantor exposed.... sci.math curls tails between legs.. <87llc6qdd8.fsf@phiwumbda.org> <31unhpF3fdkmdU1@individual.net> <321cepF3i1npjU1@individual.net> Discussion, linux) > Anyway believe what you want, because Cardinality won't be getting > taught much longer [...] Herc, this really is JSH's schtick. How would you like it if *he* started claiming that his life was just like Mean Girls or something? Let him have the line about this topic is fraud and won't be taught much longer and you can have the nuttiness about the movies. Fair's fair. -- Jesse F. Hughes I can't tell you how many times she left me. I lost count the very first time that she did. -- The Flatlanders, I Thought the Wreck Was Over === Subject: Re: EUREKA Cantor exposed.... sci.math curls tails between legs.. In sci.logic, Jesse F. Hughes <87k6rovsel.fsf@phiwumbda.org>: >> Anyway believe what you want, because Cardinality won't be getting >> taught much longer [...] > Herc, this really is JSH's schtick. Naah. JSH's shtick is more in polynomial space. :-) However, I am a little worried; would not the extra numbers of the object ring be suitable antidiagonals for the denumerable list of reals? Can't be too careful with the new math, nowadays... [rest snipped] -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: EUREKA Cantor exposed.... sci.math curls tails between legs.. > Anyway believe what you want, because Cardinality won't be getting > taught much longer [...] > Herc, this really is JSH's schtick. How would you like it if *he* > started claiming that his life was just like Mean Girls or > something? > Let him have the line about this topic is fraud and won't be taught > much longer and you can have the nuttiness about the movies. Fair's > fair. In a decade with advanced interactive programming, 1,000,000's of kids and adults alike will be tuning into math.com every day. its already 40,000 per day and the internet is in its 1st hour. I'll get you a free email account, dickhead@math.com but you letters will be corrected of any diag proofs. woof Herc === Subject: Re: EUREKA Cantor exposed.... sci.math curls tails between legs.. === >Subject: Re: EUREKA Cantor exposed.... sci.math curls tails between legs.. >Message-id: <321cepF3i1npjU1@individual.net> getting taught much longer and you just miss out on cyberspace >construction next decade. Sounds like pi in the sky to me. >Herc -- Mensanator Ace of Clubs === Subject: Re: EUREKA Cantor exposed.... sci.math curls tails between legs.. http://mygate.mailgate.org/mynews/comp/comp.theory/65fba47b6e89cdf6ba84d2662 c 882ced.48257%40mygate.mailgate.org > indivisible TM... disproven > Halt...disproven > Cantor... disproven > Godel....disproven > why don't you put 2 and 2 together I have, Herc. You claim to have accomplished all those, people who _can_ do the math, and have looked at your claims, note that you are blowing smoke. That's 2 == too bad for you plus 2 == ( your reputation & your sanity ) brought low by your ravings. Neither of those findings requires splendid insight on the part of observers to assemble into a whole. Your dreams of wealth will end up on life's cutting room floor along with the rest of your ravings, and deservedly so. You have done nothing to deserve fame, and nothing to deserve fortune either. Attempts to scam your way to the first do get you ridiculed, attempts to scam your way to the second will get you jailed. That, not the bogus results of the dictates of numerology, or the laughingstock reputation you gain from your claims to be a divine being, is how life works in this, the real world, which real world is something far away from the place where you dwell in your delusional mind. HTH xanthian. -- === Subject: Re: EUREKA Cantor exposed.... sci.math curls tails between legs.. > Your dreams of wealth will end up on life's cutting > room floor along with the rest of your ravings, and > deservedly so. You have done nothing to deserve > fame, and nothing to deserve fortune either. > Attempts to scam your way to the first do get you > ridiculed, attempts to scam your way to the second > will get you jailed. > That, not the bogus results of the dictates of > numerology, or the laughingstock reputation you gain > from your claims to be a divine being, is how life > works in this, the real world, which real world is > something far away from the place where you dwell in > your delusional mind. no Xanthian, the worlds' evolution turned binary, male and female. 2 seperate processes evolved the perfect man, 7 and the perfect woman 10. its only the jinx of the truman show, that US researchers 40 years ago figured out the paranormal name by nature coincidences in history as a prophecy that they hunted me down and track and monitor me no end. why do you bring money into the discussion? are you assuming i am (still) poor? not everyone is obligued by the cliche anti-truman-belief programming. your snippets of my writing are not relevant any more, my fame will be buried for 20 years, your opportunity is over, god has ascended his rightful place as the giver, unknown, $50K in pay per click per year undeveloped, the ice cube on the tip of the iceberg. 20% of internet profits for ever is your own continent. not to mention math.com, given my responsibility not to teach lies. we've all had technical positions at some point xanthian, your professional status is unimportant. the high pro-rata the mining companies were paying me was very acceptable for them, nobody under 170iq could program the 6 levels of SQL queries that give them the interactive view they were after. indeed the dozens of programmers who quit i replaced couldn't replicate 500 bytes of 4GL in 100,000 lines of 3GL. the only work I take on is always an impossible-task, 50 years of programming world wide its the only real work left. Herc === Subject: Re: EUREKA Cantor exposed.... sci.math curls tails between legs.. http://mygate.mailgate.org/mynews/comp/comp.theory/8405fdc56aa8570eba441221e 7 6a173d.48257%40mygate.mailgate.org [megalomaniacal drivel] I just love how easy it is to get you to convict yourself of wildly insane kookery, again and again in the eyes of all onlookers, just by a nudge of your hot buttons. You have less common sense than a two year old. _Sure_ someone is paying a moron like you big bucks to chat up all the superhuman software accomplishments you'll be doing Real Soon Now. And I am Marie of Rumania. xanthian. -- === Subject: Re: EUREKA Cantor exposed.... sci.math curls tails between legs.. > _Sure_ someone is paying a moron > like you big bucks to chat up all > the superhuman software > accomplishments you'll be doing > Real Soon Now. for the 3rd time, I don't need any more money! I wake up richer each day, I click [confirm membership] 20 times a morning for a living. now if you don't want a free xanthian@math.com email address just say so! atleast I know my proof is onto something, when all the math crowd suddenly getting edgy about paying. Herc === Subject: Re: Binary field; trace; New(?) Conjecture for fast determination of the trace of an element of a binary field. >> Let t^j in F_{2^m} with reduction polynomial >> t^m + t^{k_0} + t^{k_1} + ... + t^{k_{n-1}} + 1. >> Then the trace of t^j is 1 iff there exist an i such >> that m - k_i is odd, where i runs from 0 to n and k_n = 0. > Heh, that should have been: > .. such that j = m - k_i and is odd ... Could you explain please what t is, what you mean by the reduction polynomial and what you mean by trace? -- Timothy Murphy e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie tel: +353-86-2336090, +353-1-2842366 s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland === Subject: Re: Binary field; trace; New(?) Conjecture for fast determination of the trace of an element of a binary field. > Could you explain please what t is, > what you mean by the reduction polynomial > and what you mean by trace? t is the indeterminate that is adjoined to Z_2 to get the binary field. The reduction polynomial is the irreducible polynomial in t that is used as modulo, in order to make the field finite again. The trace Tr(x) is basically the sum of the eigenvalues of a matrix (relative to some basis) representing the element x. Thus Ay = xy for every y element of the field (for a given basis). However, as is always the case with traces of matrices, the trace does not depend on the basis. For a binary field, the trace of x in F_(2^m) would be sum of x^(2^i) where i runs from 0 till and including m-1. -- Carlo Wood === Subject: Re: Binary field; trace; New(?) Conjecture for fast determination of the trace of an element of a binary field. >> Could you explain please what t is, >> what you mean by the reduction polynomial >> and what you mean by trace? > t is the indeterminate that is adjoined to Z_2 > to get the binary field. The reduction polynomial > is the irreducible polynomial in t that is used > as modulo, in order to make the field finite again. > The trace Tr(x) is basically the sum of the eigenvalues > of a matrix (relative to some basis) representing > the element x. I would have put your question as follows: Suppose f(t) is an irreducible polynomial of degree d over k = F_p. Let K = k[t]/(f(t)) = F_q where q = p^d. Then the characteristic equation of c = t mod f(t) is f(t) = 0. Now you are asking for the trace of c^j as an element of the extension K/k. If the eigenvalues of c are lambda_1,...,lambda_d then the coefficients of f(t) = t^d - h_1 t^{d-1} + h_2 t^{d-2} + ... are the basic symmetric functions h_1 = sum lambda_i, h_2 = sum_{iI honestly think that the best way to put exercises to the students is >to consider other quadratic fields, with a norm function and defining >integers to be those that have integer norm. What norm? Certainly not the usual norm. For example, in Q(sqrt(-15)), the norm of (1/4) + (1/4)*sqrt(-15) is 1, an integer, yet it is clearly not an algebraic integer. In fact, any root of an irreducible polynomial of the form x^2 + ax + b, with b an integer, will have integer norm but will only be an algebraic integer if a is also an integer. >This results in interesting >division properties. It also clearly shows that the rings of those >integers are integer-like, but sometimes also have strange properties. >And they can at least prove that integers so defined are indeed >algebraic integers (but not the other way around, and this is what you >seem to wish). Ah; so you're dealing with an order rather than with the ring of algebraic integer. Okay then.... -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Algebraic integers >I honestly think that the best way to put exercises to the students is >to consider other quadratic fields, with a norm function and defining >integers to be those that have integer norm. > What norm? Certainly not the usual norm. The usual norm, for quadratic fields. > For example, in Q(sqrt(-15)), the norm of (1/4) + (1/4)*sqrt(-15) is > 1, an integer, yet it is clearly not an algebraic integer. Yes, in some cases such a norm (as Hardy & Wright call it) will not result in algebraic integers. However, at the level the students apparently are, they would have difficulty proving that it is *not* an algebraic integer. They have just touched the Gaussian integers. Nevertheless, while they are not algebraic integers, they are integer-like, and that was the purpose; to show that there were interesting other integer-like rings. But I was indeed wrong when I said that they could prove that the integers so defined were algebraic integers. I shouls really pick up again Hardy & Wright to see how they do it. -- 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: Algebraic integers > > For example, in Q(sqrt(-15)), the norm of (1/4) + (1/4)*sqrt(-15) is > > 1, an integer, yet it is clearly not an algebraic integer. > Yes, in some cases such a norm (as Hardy & Wright call it) will not > result in algebraic integers. Isn't this the norm (pun intended)? If f is an automorphism of the extension K/k then x/f(x) will have norm 1, but will not normally be an algebraic integer. > However, at the level the students > apparently are, they would have difficulty proving that it is *not* > an algebraic integer. They have just touched the Gaussian integers. > Nevertheless, while they are not algebraic integers, they are > integer-like, and that was the purpose; to show that there were > interesting other integer-like rings. I wouldn't have thought numbers with norm 1 are at all integer-like. Does the sum of two such numbers always have norm 1? -- Timothy Murphy e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie tel: +353-86-2336090, +353-1-2842366 s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland === Subject: Re: Algebraic integers days. My association with the Department is that of an alumnus. >> > For example, in Q(sqrt(-15)), the norm of (1/4) + (1/4)*sqrt(-15) is >> > 1, an integer, yet it is clearly not an algebraic integer. >> Yes, in some cases such a norm (as Hardy & Wright call it) will not >> result in algebraic integers. >Isn't this the norm (pun intended)? >If f is an automorphism of the extension K/k >then x/f(x) will have norm 1, >but will not normally be an algebraic integer. >> However, at the level the students >> apparently are, they would have difficulty proving that it is *not* >> an algebraic integer. They have just touched the Gaussian integers. >> Nevertheless, while they are not algebraic integers, they are >> integer-like, and that was the purpose; to show that there were >> interesting other integer-like rings. >I wouldn't have thought numbers with norm 1 are at all integer-like. >Does the sum of two such numbers always have norm 1? Certainly not. The two roots of x^2 - (1/2)x + 1 have norm 1 and add up to 1/2, which has norm 1/4. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Pi, randomness of numbers? <73mer01btu9vsudr4ajitlcr5obdcltns4@4ax.com> when I first saw the header, I thought that it was about messages in pi a la the Bible Code or Carl Sagan's dumb-ass movie with Hinkley's doyenne. of course, if the digits of pi in base-ten are random, and there is no reason to suppose that they aren't, then *every* finite message should occur in it, modulo some number of monkeys-a-reading it. > that there is no way to know whether pi is truly random is > arbitrary. > You have to know what something is before you can know whether something --Chairman George and Trickier Dick at Watergate! http://tarpley.net/bush12.htm === Subject: polynomial solver Can someone explain me what polynomial solvers are and how they work? How can I use them when I have for example an equation system with nine variables (parts of them are quadratic) and nine unknowns. To solve it in a traditional way (e.g. Gauss) the results are too huge and it takes a lot of time to compute it. Someone told me that solving this with polynomial solver would be better. S. Nurbe === Subject: intersection two lines hello there are two lines , line a goes from ( x1, y1 ) to ( x2, y2 ) and line b goes from ( x3, y3 ) to ( x4, y4 ) how does one find out if these lines intersect ? and if they do , in which point ( x, y ) ? thank you jmcs3 === Subject: Re: intersection two lines johanmeskenscs3@chromaticspaceandworld.com > hello > there are two lines > , line a goes from ( x1, y1 ) to ( x2, y2 ) > and line b goes from ( x3, y3 ) to ( x4, y4 ) > how does one find out if these lines intersect ? > and if they do > , in which point ( x, y ) ? > thank you > jmcs3 It appears that OP means to ask whether the segment from ( x1, y1 ) to ( x2, y2 ) has a point in common with the segment from ( x3, y3 ) to ( x4, y4 ). One way is to represent both segments parametrically thus: Segment 1: x = x1*s + x2*(1 - s), y = y1*s + y2*(1 - s), for 0 <= s <= 1 Segment 2: x = x3*t + x4*(1 - t), y = y3*t + y4*(1 - t), for 0 <= t <= 1 Then solve, if possible, the system of two equations in the two unknowns s and t: x1*s + x2*(1 - s) = x3*t + x4*(1 - t), y1*s + y2*(1 - s) = y3*t + y4*(1 - t) If there is a solution with 0 <= s <= 1 and 0 <= t <= 1, then the line segments intersect at the point (x,y), otherwise not. === Subject: Re: intersection two lines > It appears that OP means to ask whether the segment from > ( x1, y1 ) to ( x2, y2 ) has a point in common with the segment from > ( x3, y3 ) to ( x4, y4 ). > One way is to represent both segments parametrically thus: > Segment 1: x = x1*s + x2*(1 - s), > y = y1*s + y2*(1 - s), for 0 <= s <= 1 > Segment 2: x = x3*t + x4*(1 - t), > y = y3*t + y4*(1 - t), for 0 <= t <= 1 > Then solve, if possible, the system of two equations in the two unknowns > s and t: > x1*s + x2*(1 - s) = x3*t + x4*(1 - t), > y1*s + y2*(1 - s) = y3*t + y4*(1 - t) > If there is a solution with 0 <= s <= 1 and 0 <= t <= 1, then the line > segments intersect at the point (x,y), otherwise not. True, but not very handsome in practice. Instead, you should employ the following straight line equations, as well: (y2-y1)(x-x1) - (x2-x1)(y-y1) = 0 and (y4-y3)(x-x3) - (x4-x3)(y-y3) = 0 The function f(x,y) = (y2-y1)(x-x1) - (x2-x1)(y-y1) is used to determine if (x,y)=(x3,y3) and (x,y)=(x4,y4) results in f(x3,y3)<0 and f(x4,y4)>0 or f(x3,y3)>0 and f(x4,y4)<0. If not then the segments cannot intersect. Likewise is done for the function g(x,y) = (y4-y3)(x-x3) - (x4-x3)(y-y3) and the end-points (x,y)=(x1,y1) , (x,y)=(x2,y2) . Divide and conquer .. Then, IF the two segments DO intersect, Virgil's approach can be used to actually calculate the intersection points. Here comes an implementation in Pascal: type punt = record x,y : double; end; function Lijn(A,B,r : punt) : double; { Straight Line function } var x1,y1,x2,y2,w : double; begin x1 := A.x ; y1 := A.y; x2 := B.x ; y2 := B.y; w := (x2-x1)*(r.y-y1) - (y2-y1)*(r.x-x1); Lijn := w; end; function Lengte(P,Q : punt) : double; { Length of Line Segment } begin Lengte := sqrt(sqr(P.x-Q.x)+sqr(P.y-Q.y)); end; function Snijdend(A,B,P,Q : punt) : boolean; { True IF two segments Are intersecting } const eps : double = 1.E-9; { Floating point precautions } var AB,PQ : double; Lpm,Lmp,Rpm,Rmp,OK : boolean; begin AB := Lengte(A,B)*eps; Lpm := (Lijn(A,B,P) >= +AB) and (Lijn(A,B,Q) <= -AB); Lmp := (Lijn(A,B,P) <= -AB) and (Lijn(A,B,Q) >= +AB); PQ := Lengte(P,Q)*eps; Rpm := (Lijn(P,Q,A) >= +PQ) and (Lijn(P,Q,B) <= -PQ); Rmp := (Lijn(P,Q,A) <= -PQ) and (Lijn(P,Q,B) >= +PQ); OK := (Lpm and Rpm) or (Lpm and Rmp) or (Lmp and Rpm) or (Lmp and Rmp); Snijdend := OK; end; Han de Bruijn === Subject: Re: intersection two lines > hello > there are two lines > , line a goes from ( x1, y1 ) to ( x2, y2 ) > and line b goes from ( x3, y3 ) to ( x4, y4 ) > how does one find out if these lines intersect ? > and if they do > , in which point ( x, y ) ? Many ways. Here's some geometric/algrebraic/parametric reasoning that will get you there. [I avoid explicit referencing of the two point formula for a line, which is easier, but is more of a recitation. Exercise your brain]. Let solution = (x,y) (and assume it exists). define t = (x2-x)/(x2-x1) ...(1) (you characterise that) then t = (y2-y)/(y2-y1) ...(2) (because (x,y) is on a line between X1 and X2) similarly s = (x4-x)/(x4-x3) ...(3) then s = (y4-y)/(y4-y3) ...(4) rewriting (1), x = x2 - (x2-x1)*t ...(5) rewriting (3), x = x4 - (x4-x3)*s ...(6) similarly, y = y2 - (y2-y1)*t ...(7) and, y = y4 - (y4-y3)*s ...(8) eliminate x from (5) and (6), and y from (7) and (8). Solve for s and t. Yields x and y. [Q: interpret s <0, s > 1, t <0, t>1 cases]. Tomasso. === Subject: Re: intersection two lines > hello > there are two lines > , line a goes from ( x1, y1 ) to ( x2, y2 ) > and line b goes from ( x3, y3 ) to ( x4, y4 ) > how does one find out if these lines intersect ? > and if they do > , in which point ( x, y ) ? Each line can be expressed as a linear equation in two unknowns. That is two equations in two unknowns. Look at the determinant formed by the co-eficients. If the deterimant is not zero, then there exists a unique solution to the two equations. That is your point of intersection. If the determinant is equal to 0 then the equations are inconsistent (no solutions) or they represent the same line (infinte number of solutions). Bob Kolker === Subject: Re: intersection two lines >> hello >> there are two lines >> , line a goes from ( x1, y1 ) to ( x2, y2 ) >> and line b goes from ( x3, y3 ) to ( x4, y4 ) >> how does one find out if these lines intersect ? >> and if they do >> , in which point ( x, y ) ? > Each line can be expressed as a linear equation in two unknowns. > That is two equations in two unknowns. > Look at the determinant formed by the co-eficients. > If the deterimant is not zero, then there exists a unique solution to > the two equations. That is your point of intersection. If the > determinant is equal to 0 then the equations are inconsistent (no > solutions) or they represent the same line (infinte number of solutions). That is true for two lines that extend to infinite. However, I think the question is about the intersection point of two line _segments_. See my answer on Virgil's poster. Han de Bruijn === Subject: Re: intersection two lines > hello > there are two lines > , line a goes from ( x1, y1 ) to ( x2, y2 ) > and line b goes from ( x3, y3 ) to ( x4, y4 ) > how does one find out if these lines intersect ? > and if they do > , in which point ( x, y ) ? > thank you > jmcs3 === Subject: DISPROOF OF CARDINALITY THEORY Given a contender list for the set of all reals, Assume that for all digit positions, an infinite number of reals together_contain every possible numeral. L(n, i) is the ith digit of the nth real. Ai, En1, En2.. En10 L(n1,i)=1 & L(n2,i)=2 & .. L(n2, i)=10 Not only that, INFINITE amount of members exist that support this property. Ai, Ae, En1>e, En2>e.. En10>e L(n1,i)=1 & L(n2,i)=2 & .. L(n2, i)=10 Then we can reorder the list without changing the content such that 1 Dxxxx.. x is a digit placeholder 2 xxxxx.. 3 xxxxx.. .. The 1st digit of the 1st real is any of {0,1,2,3,4,5,6,7,8,9} 1 Dxxxx.. 2 xDxxx.. 3 xxxxx.. .. The 2nd digit of the 2nd real is any of {0,1,2,3,4,5,6,7,8,9} This sorting process will result in an infinite list, constructed from the same members of the previous list. e.g. we could set D = {0.203048..} and claim D1 = {0.314159..} that pi/10 is absent from the list. By reordering the list, we can construct any diagonal desirable, and prove any real we want is not on the list using diagonalisation. Since the list is not empty, this is a contradiction to proof by diagonalisation. Herc -- i don't know where, i don't know how i only know that some day we'll be together again === Subject: Re: DISPROOF OF CARDINALITY THEORY > Given a contender list for the set of all reals, > Assume that for all digit positions, an infinite number of reals together_contain every possible numeral. > L(n, i) is the ith digit of the nth real. > Ai, En1, En2.. En10 L(n1,i)=1 & L(n2,i)=2 & .. L(n2, i)=10 > Not only that, INFINITE amount of members exist that support this property. > Ai, Ae, En1>e, En2>e.. En10>e L(n1,i)=1 & L(n2,i)=2 & .. L(n2, i)=10 > Then we can reorder the list without changing the content such that > 1 Dxxxx.. x is a digit placeholder > 2 xxxxx.. > 3 xxxxx.. > .. > The 1st digit of the 1st real is any of {0,1,2,3,4,5,6,7,8,9} > 1 Dxxxx.. > 2 xDxxx.. > 3 xxxxx.. > .. > The 2nd digit of the 2nd real is any of {0,1,2,3,4,5,6,7,8,9} > This sorting process will result in an infinite list, constructed from the same members > of the previous list. > e.g. we could set D = {0.203048..} > and claim D1 = {0.314159..} that pi/10 is absent from the list. > By reordering the list, we can construct any diagonal desirable, > and prove any real we want is not on the list using diagonalisation. > Since the list is not empty, this is a contradiction to proof by diagonalisation. > Herc Two problems: (1) You've only shown that by means of such re-arranging we can arbitrarily stipulate the first n digits of the diagonal number where n is finite, not that we can arbitrarily stupilate the diagonal number (2) You're assuming a list of all the reals exists. This assumption is false, so from it you can derive anything. So you're not problematizing the proof by diagonalization in any way. > -- > i don't know where, i don't know how > i only know that some day we'll be together again === Subject: Re: DISPROOF OF CARDINALITY THEORY In sci.logic, |-|erc <321m3gF3h2jdvU1@individual.net>: > Given a contender list for the set of all reals, > Assume that for all digit positions, an infinite number of reals together_contain every possible numeral. > L(n, i) is the ith digit of the nth real. OK. > Ai, En1, En2.. En10 L(n1,i)=1 & L(n2,i)=2 & .. L(n2, i)=10 This looks like a typo; presumably you meant Ai, En1, En2.. En10 L(n1,i)=0 & L(n2,i)=1 & .. L(n10, i)=9 or perhaps Ai, En1, En2.. En10 L(n1,i)=1 & L(n2,i)=2 & .. L(n10, i)=0 I should note here that, if L = T_10, this works, rendering your proof slightly suspect, as T_10 does not cover Q, let alone R, but T_10 fills your requirement above. Or are you going to continue to claim that 1/3 = .333... is in T_10, despite explicit proof to the contrary? Like the following proof, perhaps? Lemma 1. 3 and 10^n are relatively prime for every n >= 0 in J. Proof. We proceed by induction. [1] gcd(3,10^0) = gcd(3,1) = 1. Duh. We can also pick x = -3, y = 1, and 3 * x + 10^0 * y = -9 + 10 = 1. This might be helpful in [2]. [2] Let gcd(3,10^n) = 1. This means that x and y exist such that 3 * x + 10^n * y = 1. If we compute 3 * (10*x - 3) + 10^(n+1) * y, we find out that this is 10 * (3 * x + 10^n * y) - 3 * 3 = 10 * 1 - 9 = 1. Therefore gcd(3,10^(n+1)) = 1 also. By induction, QED. Lemma 2. If gcd(a,b) = 1 and a != 1, then there's no integer k such that a * k = b. Proof. If a * k = b for some k, then, since we can find x and y such that a * x + b * y = 1, then a * x + b * y = a * x + a * k * y = a * (x + k * y) = 1 which contradicts our assumptions. Hence k cannot exist. QED. (Of course if a = 1 then k = b, but that's relatively uninteresting. :-) ) Theorem. 1/3 != k/10^n for all n, k in J. Proof. For n <= 0, k/10^n = k * 10^(-n), which is an integer. Since 1/3 is clearly not an integer we're done for this subpart. [*] For n > 0, were 1/3 = k / 10^n then 10^n = 3 * k. Since 10^n and 3 are relatively prime by Lemma 1, k cannot exist by Lemma 2. QED. Corollary. 1/3 is not in T_10. Proof. T_10 = {k/10^n: 0 <= k <= 10^n, n >= 0, n, k in J}; QED. Corollary #2: 1/3 is not in S_3 = { 0, .3, .33, .333, ...} = { (10^n - 10) / (3 * 10^(n+10)): n >= 0 in J.} Proof: S_3 is a subset of T_10. Since 1/3 is not in T_10, 1/3 is not in S_3 either. QED. Squirm all you like; it won't help. :-) 1/3 is a number provably not in T_10; it is also a real (as are all rationals), and it turns out to be a Cantorian diagonal to T_10 as well. Hence T_10 doesn't cover the reals, and your probabilistic proof is worth diddly-squat. > Not only that, INFINITE amount of members exist that > support this property. So? T_10 contains an infinite number of finite prefixes for such numbers as 1/3, sqrt(1/2), pi/4, pi/10, e/3. Yet it does not contain 1/3. [rest snipped] [*] the field with 2 elements disproves this theorem, but 3 = 1 for that field anyway. :-) -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: DISPROOF OF CARDINALITY THEORY > In sci.logic, |-|erc > : > Given a contender list for the set of all reals, > Assume that for all digit positions, an infinite number of reals together_contain every possible numeral. > L(n, i) is the ith digit of the nth real. > OK. > Ai, En1, En2.. En10 L(n1,i)=1 & L(n2,i)=2 & .. L(n2, i)=10 > This looks like a typo; presumably you meant > Ai, En1, En2.. En10 L(n1,i)=0 & L(n2,i)=1 & .. L(n10, i)=9 > or perhaps > Ai, En1, En2.. En10 L(n1,i)=1 & L(n2,i)=2 & .. L(n10, i)=0 > I should note here that, if L = T_10, this works, rendering > your proof slightly suspect, as T_10 does not cover Q, > let alone R, but T_10 fills your requirement above. Or are you > going to continue to claim that 1/3 = .333... is in T_10, despite > explicit proof to the contrary? > Like the following proof, perhaps? > Lemma 1. 3 and 10^n are relatively prime for every n >= 0 in J. > Proof. We proceed by induction. > [1] gcd(3,10^0) = gcd(3,1) = 1. Duh. We can also pick x = -3, y = 1, > and 3 * x + 10^0 * y = -9 + 10 = 1. This might be helpful in [2]. > [2] Let gcd(3,10^n) = 1. This means that x and y exist such that > 3 * x + 10^n * y = 1. > If we compute 3 * (10*x - 3) + 10^(n+1) * y, we find out > that this is 10 * (3 * x + 10^n * y) - 3 * 3 = 10 * 1 - 9 = 1. > Therefore gcd(3,10^(n+1)) = 1 also. By induction, QED. > Lemma 2. If gcd(a,b) = 1 and a != 1, then there's no > integer k such that a * k = b. > Proof. If a * k = b for some k, then, since we can find > x and y such that a * x + b * y = 1, then > a * x + b * y = a * x + a * k * y = a * (x + k * y) = 1 > which contradicts our assumptions. Hence k cannot exist. QED. > (Of course if a = 1 then k = b, but that's relatively uninteresting. :-) ) > Theorem. 1/3 != k/10^n for all n, k in J. > Proof. For n <= 0, k/10^n = k * 10^(-n), which is an integer. Since > 1/3 is clearly not an integer we're done for this subpart. [*] > For n > 0, were 1/3 = k / 10^n then 10^n = 3 * k. Since > 10^n and 3 are relatively prime by Lemma 1, k cannot > exist by Lemma 2. QED. > Corollary. 1/3 is not in T_10. > Proof. T_10 = {k/10^n: 0 <= k <= 10^n, n >= 0, n, k in J}; QED. > Corollary #2: 1/3 is not in S_3 = { 0, .3, .33, .333, ...} > = { (10^n - 10) / (3 * 10^(n+10)): n >= 0 in J.} > Proof: S_3 is a subset of T_10. Since 1/3 is not in T_10, 1/3 > is not in S_3 either. QED. > Squirm all you like; it won't help. :-) 1/3 is a number provably > not in T_10; it is also a real (as are all rationals), and it > turns out to be a Cantorian diagonal to T_10 as well. > Hence T_10 doesn't cover the reals, and your probabilistic > proof is worth diddly-squat. > Not only that, INFINITE amount of members exist that > support this property. > So? T_10 contains an infinite number of finite prefixes > for such numbers as 1/3, sqrt(1/2), pi/4, pi/10, e/3. Yet it > does not contain 1/3. > [rest snipped] > [*] the field with 2 elements disproves this theorem, but > 3 = 1 for that field anyway. :-) You must have napped the last couple of weeks when I posted unique initial segment -> unique sequence characteristic expansion -> unique sequence The diag of T10 has a characteristic expansion, so it's not on the list T10. The T10 argument (every initial segment and finite) does not apply to an infinite population with random digits of the cartesian plane, or UTM(x,y). Considering any irrational is not on T10, it's hardly a contender list for all reals, but it won't hurt the disproof. Required to prove : pi/10 is not on T10. 0.314159.. Swap 2 members of T10 so L(1,1) = 2 [T10'] Swap 2 members of T10' so L(2,2) = 0 [T10''] Swap 2 members of T10'' so L(3,3) = 3 [T10'''] .. The diagonal of T10''.. is 0.2xxxxx.. 0.x0xxxx.. 0.xx3xxx.. .. Make an anti-diagonal D1 = 0.314... Therefore pi/10 is not on T10. Herc === Subject: Re: DISPROOF OF CARDINALITY THEORY In sci.logic, |-|erc <32204hF3gufagU1@individual.net>: >> In sci.logic, |-|erc >> <321m3gF3h2jdvU1@individual.net>: >> Given a contender list for the set of all reals, >> Assume that for all digit positions, an infinite number of reals together_contain every possible numeral. >> L(n, i) is the ith digit of the nth real. >> OK. >> Ai, En1, En2.. En10 L(n1,i)=1 & L(n2,i)=2 & .. L(n2, i)=10 >> This looks like a typo; presumably you meant >> Ai, En1, En2.. En10 L(n1,i)=0 & L(n2,i)=1 & .. L(n10, i)=9 >> or perhaps >> Ai, En1, En2.. En10 L(n1,i)=1 & L(n2,i)=2 & .. L(n10, i)=0 >> I should note here that, if L = T_10, this works, rendering >> your proof slightly suspect, as T_10 does not cover Q, >> let alone R, but T_10 fills your requirement above. Or are you >> going to continue to claim that 1/3 = .333... is in T_10, despite >> explicit proof to the contrary? >> Like the following proof, perhaps? >> Lemma 1. 3 and 10^n are relatively prime for every n >= 0 in J. >> Proof. We proceed by induction. >> [1] gcd(3,10^0) = gcd(3,1) = 1. Duh. We can also pick x = -3, y = 1, >> and 3 * x + 10^0 * y = -9 + 10 = 1. This might be helpful in [2]. >> [2] Let gcd(3,10^n) = 1. This means that x and y exist such that >> 3 * x + 10^n * y = 1. >> If we compute 3 * (10*x - 3) + 10^(n+1) * y, we find out >> that this is 10 * (3 * x + 10^n * y) - 3 * 3 = 10 * 1 - 9 = 1. >> Therefore gcd(3,10^(n+1)) = 1 also. By induction, QED. >> Lemma 2. If gcd(a,b) = 1 and a != 1, then there's no >> integer k such that a * k = b. >> Proof. If a * k = b for some k, then, since we can find >> x and y such that a * x + b * y = 1, then >> a * x + b * y = a * x + a * k * y = a * (x + k * y) = 1 >> which contradicts our assumptions. Hence k cannot exist. QED. >> (Of course if a = 1 then k = b, but that's relatively uninteresting. :-) ) >> Theorem. 1/3 != k/10^n for all n, k in J. >> Proof. For n <= 0, k/10^n = k * 10^(-n), which is an integer. Since >> 1/3 is clearly not an integer we're done for this subpart. [*] >> For n > 0, were 1/3 = k / 10^n then 10^n = 3 * k. Since >> 10^n and 3 are relatively prime by Lemma 1, k cannot >> exist by Lemma 2. QED. >> Corollary. 1/3 is not in T_10. >> Proof. T_10 = {k/10^n: 0 <= k <= 10^n, n >= 0, n, k in J}; QED. >> Corollary #2: 1/3 is not in S_3 = { 0, .3, .33, .333, ...} >> = { (10^n - 10) / (3 * 10^(n+10)): n >= 0 in J.} >> Proof: S_3 is a subset of T_10. Since 1/3 is not in T_10, 1/3 >> is not in S_3 either. QED. >> Squirm all you like; it won't help. :-) 1/3 is a number provably >> not in T_10; it is also a real (as are all rationals), and it >> turns out to be a Cantorian diagonal to T_10 as well. >> Hence T_10 doesn't cover the reals, and your probabilistic >> proof is worth diddly-squat. >> Not only that, INFINITE amount of members exist that >> support this property. >> So? T_10 contains an infinite number of finite prefixes >> for such numbers as 1/3, sqrt(1/2), pi/4, pi/10, e/3. Yet it >> does not contain 1/3. >> [rest snipped] >> [*] the field with 2 elements disproves this theorem, but >> 3 = 1 for that field anyway. :-) > You must have napped the last couple of weeks when I posted > unique initial segment -> unique sequence > characteristic expansion -> unique sequence > The diag of T10 has a characteristic expansion, > so it's not on the list T10. But your list, I take it, does not have a characteristic expansion. Interesting, if meaningless. > The T10 argument (every initial segment and finite) does > not apply to an infinite population with random digits > of the cartesian plane, or UTM(x,y). T10 has random digits and has cardinality aleph-null, which is essentially infinity. > Considering any irrational is not on T10, it's hardly a > contender list for all reals, > but it won't hurt the disproof. > Required to prove : pi/10 is not on T10. > 0.314159.. > Swap 2 members of T10 so L(1,1) = 2 [T10'] > Swap 2 members of T10' so L(2,2) = 0 [T10''] > Swap 2 members of T10'' so L(3,3) = 3 [T10'''] > .. > The diagonal of T10''.. is > 0.2xxxxx.. > 0.x0xxxx.. > 0.xx3xxx.. > .. > Make an anti-diagonal D1 = 0.314... > Therefore pi/10 is not on T10. > Herc Actually, pi/10 is a pretty good diagonal as it stands, at least for the first few digits. .[1]0000 != 3 .2[0]000 != 1 .30[0]00 != 4 Do you have a program to generate the denumerable list of all reals? You've proven that it's possible -- or so you say. -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: DISPROOF OF CARDINALITY THEORY > Do you have a program to generate the denumerable list of all reals? > You've proven that it's possible -- or so you say. UTM(x, y) A few kb of code, unlimited memory and all things are possible. Herc === Subject: Re: DISPROOF OF CARDINALITY THEORY > Do you have a program to generate the denumerable list of all reals? > You've proven that it's possible -- or so you say. > UTM(x, y) UTM(x, y) mod 10 A quadrant of the cartesian plane partially populated with computed digits, containing every number, every real, every sequence, every algorithm, Hamlet is in there somewhere too! We should make a competition, find the lowest TM in UTM(x,y) mod 26 that contains a limerick! Herc === Subject: Re: DISPROOF OF CARDINALITY THEORY > Ai, Ae, En1>e, En2>e.. En10>e L(n1,i)=1 & L(n2,i)=2 & .. L(n2, i)=10 ok, there's no digit 10! Ai, Ae, En1,n2..n10>e, L(n1,i)=1 & L(n2,i)=2 & ..L(n9,i)=9 & L(n10, i)=0 Herc === Subject: Re: universal constant >fuckhead >>;) > oh hi loogie, no wonder you made a putrid alias for yourself. > Herc > welcome to your nightmare twit > playing the pan flute lately? > http://www.mazepath.com/uncleal/sunshine.jpg a 'puckhead' in sci.math and rec.org.mensa gaaawd Herc === Subject: Re: universal constant >fuckhead >>;) >oh hi loogie, no wonder you made a putrid alias for yourself. >Herc >>welcome to your nightmare twit >>playing the pan flute lately? >>http://www.mazepath.com/uncleal/sunshine.jpg > a 'puckhead' in sci.math and rec.org.mensa gaaawd > Herc I must have left an impression on you |-|erc. You are using my news server. Now if you swicth from Outlook Express to Thunderbird, your transformation will be complete. Merry Christmas === Subject: Re: universal constant In sci.math, PuckHead <325emhF3g3ukoU1@individual.net>: > >fuckhead > >;) >>oh hi loogie, no wonder you made a putrid alias for yourself. >>Herc >welcome to your nightmare twit >playing the pan flute lately? >http://www.mazepath.com/uncleal/sunshine.jpg >> a 'puckhead' in sci.math and rec.org.mensa gaaawd >> Herc > I must have left an impression on you |-|erc. You are using my news > server. Now if you swicth from Outlook Express to Thunderbird, your > transformation will be complete. > Merry Christmas You are, of course, referring to Firebird's counterpart, not a certain beverage favored in an area that is usually avoided by those wishing to keep their money... ;-) -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: How much lossless compression is possible in images? > it seems to be happening quite a lot since they introduced the beta of > the new version. > Use a dedicated newsgroup service. Many are available for only a few > dollars a month. And there are even available for free. -- 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: How much lossless compression is possible in images? <5skfr0pb3p24nn2lthl7amb9eai1bf9rtr@4ax.com> <6m4mr0p6iml9reaphldhmufd1pksbi4t6q@4ax.com> > It seems to me that I am the only one to have actually attempted to > answer your original question. > I don't remember. All I remember is your ceaseless vituperation, which > effectively eclipsed anything else you had to say. Guess what Larry? That's my main contention with you. You are here mainly for the prattling. The escence seems to be quite beyond your abilities... eh ur ... interest. === Subject: Re: How much lossless compression is possible in images? > Guess what Larry? My name isn't Larry. -- Transpose hotmail and mxsmanic in my e-mail address to reach me directly. === Subject: Re: How much lossless compression is possible in images? <5skfr0pb3p24nn2lthl7amb9eai1bf9rtr@4ax.com> <6m4mr0p6iml9reaphldhmufd1pksbi4t6q@4ax.com> <8qumr09f34mehjaj9v2n7hhao9srbck532@4ax.com> See what I mean? === Subject: Re: Is There Anomaly In Value of pi ? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iBC1vJO17416; http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fibonacci/fibpi.html Useful link Ha-ha-ha Now I know how mathematitians usualy have a rest. === Subject: Re: Gambling problem -- Probability by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iBC1vLc17488; >Does anyone know how to deal with this problem? >A gambler with $10 makes a seriese of $1 bets in rovlette. Each bet has >probability 9/19 of winning $1, and 10/19 of losing $1. His strategy is >to play until he wins $25 or loses $10, whichever comes first. >a.) find the probability that when he quits he will have won $25. >b.) find the expected number of games that he plays. >c1.) call a near miss, the event that the gambler gets to winning $24, >only a dollar away from his goal, but then goes on to lose $10, without >reaching his goal. >c2.) Call a near ruin, the event that the gambler gets to losing $9, >only $1 away from ruin, but then goes on to reach his goal of winning >$25, never hit his losing $10 quitting point. >Find the probabilities of both a near miss, and a near ruin. Looks like homework to me. Show us your work so far. === Subject: Compactness Criteria for L^p Spaces Does anyone know (or can give reference) to a necessary and sufficient condition for compactness in L^p spaces? In particular, I need to know when it is true that Given a subspace E in L^p for a fixed p (1<=p Does anyone know (or can give reference) to a necessary and sufficient > condition for compactness in L^p spaces? In particular, I need to know > when it is true that > Given a subspace E in L^p for a fixed p (1<=p has a converging subsequence in L^p. > I know there is the closely-related Ascoli theorem, but that one is > about the subspace of continuous functions in the sup-norm function > space. The condition there is that the subspace is closed and bounded > with equi-continuity. > Is there a similar condition for subspace of L^p spaces? > Kira. Dunford-Schwartz: Linear Operators I, Theorems IV.8.18 and IV.8.20 (the latter one is for the real axis, and looks like a version of Arzela-Ascoli Theorem). It is too messy to render in plain text. Weak sequential compactness: in reflexive spaces, it is equivalent to boundedness (with weak closedness)(same book, II.3 28). === Subject: Re: Compactness Criteria for L^p Spaces >>Does anyone know (or can give reference) to a necessary and sufficient >>condition for compactness in L^p spaces? In particular, I need to know >>when it is true that >>Given a subspace E in L^p for a fixed p (1<=p>has a converging subsequence in L^p. >>I know there is the closely-related Ascoli theorem, but that one is >>about the subspace of continuous functions in the sup-norm function >>space. The condition there is that the subspace is closed and bounded >>with equi-continuity. >>Is there a similar condition for subspace of L^p spaces? >>Kira. > Dunford-Schwartz: Linear Operators I, Theorems IV.8.18 and IV.8.20 (the > latter one is for the real axis, and looks like a version of Arzela-Ascoli > Theorem). It is too messy to render in plain text. > Weak sequential compactness: in reflexive spaces, it is equivalent to > boundedness (with weak closedness)(same book, II.3 28). Kira === Subject: Re: Compactness Criteria for L^p Spaces To be more specific, by L^p space I meant the set of measurable functions {f} such that int |f}^p is finite. Kira. > Does anyone know (or can give reference) to a necessary and sufficient > condition for compactness in L^p spaces? In particular, I need to know > when it is true that > Given a subspace E in L^p for a fixed p (1<=p has a converging subsequence in L^p. > I know there is the closely-related Ascoli theorem, but that one is > about the subspace of continuous functions in the sup-norm function > space. The condition there is that the subspace is closed and bounded > with equi-continuity. > Is there a similar condition for subspace of L^p spaces? > Kira. === Subject: State-of-the-Art in Physics by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iBC1vM917517; THE STATE-OF-THE-ART IN PHYSICS There is similarity and distinction between mathematics and physics. They are similar in the sense that they both study the structure, properties and behavior of their respective subject matter. The difference lies in their subject matter: that of mathematics is the representation of thought, making it a language, and that of physics is nature, making it a science. Both suffer from similar defects: mathematical spaces and their concepts are ill-defined and nature and physical concepts are also ill-defined. Until my recent critique-rectification of foundations, number theory, the real number system and physics this was true of both mathematics and physics. A physical concept is well-defined if its existence, structure, behavior, properties, etc., are specified by the laws of nature. Physics has another problem. Its present methodology of mathematical modeling that describes nature in terms of mathematical spaces is both flawed and inadequate because two distinct spaces are involved which can only be well-defined by their axioms and the laws of nature, respectively. Therefore, solving a physical problem in some mathematical space is flawed and this is the reason mathematical physics has long standing problems such as the gravitational n-body problem, finding the basic constituent of matter and what the structure of the electron is. This inadequacy of methodology leads to error. This occurs, for instance in the search for the basic constituents of matter. The main difficulty is the ambiguity on the requirements they must satisfy. There are two: (1) they must be indestructible otherwise the first law of thermodynamics would not hold and (2) some basic constituents must comprise every piece if matter. It follows from (2) that there can only be one basic con! stituent aside from form. Immediately, the physicists.89 candidate for basic constituent, the quark, does not qualify because it is destructible. Moreover, in view of (2) the electron must consist of quark, too, and they do not seem to be looking for it there. Another error in physics lies in the physical concept .8bblack hole..8a First of all, it is ill-defined. I think physicists agree that a black hole is dark matter and yet it is supposed to suck visible matter around it. This is a contradiction since dark matter does not interact with visible matter. The remedy is a new methodology, dynamic modeling that EXPLAINS nature in terms of its laws. This methodology was first applied to solve the gravitational n-body problem in 1997 (Nonlinear Analysis, Vol. 30, No. 8). The crucial step was to find the appropriate laws of nature to well-define the key physical concepts in this problem, .8bbody.8a and .8bgravity..8a It took 11 natural laws to do this; they constitute the initial laws of nature that well-define gravity and establish the flux theory of gravitation. To well-define .8bbody.8a it was necessary to well-define the basic constituent of dark matter, the superstring. Then natural laws were discovered that convert dark matter to visible matter. These eleven laws of nature yielded the solution of the gravitational n-body problem. Moreover, every time this theory is applied to solve new problems new laws of nature are required. To-date, 42 natural laws including biological laws and laws of learning, mathematical principles and a general! law have been discovered that anchor the theories of turbulence, intelligence and learning and explain the Columbia Space Shuttle disaster. They anchor the full development of the flux theory of gravitation and applications (published in a series of over a dozen papers in Nonlinear Analysis, Nonlinear Analysis and Phenomena, Applied Mathematics and Computation, Indian Journal of Pure and Applied Mathematics and proceedings of several international conferences). For full listing of references visit my websites: http://www.users.bigpond.com/pidro/home.htm http://home.iprimus.com.au/pidro/ E. E. Escultura === Subject: Re: State-of-the-Art in Physics > THE STATE-OF-THE-ART IN PHYSICS > There is similarity and distinction between mathematics and physics. > They are similar in the sense that they both study the structure, > properties and behavior of their respective subject matter. The > difference lies in their subject matter: that of mathematics is the > representation of thought, making it a language, and that of physics > is nature, making it a science. Both suffer from similar defects: > mathematical spaces and their concepts are ill-defined and nature and > physical concepts are also ill-defined. So ill defined that it led to the production of computers ,using quantum effects, with which you can spew your nonsense all over the internet. Bob Kolker === Subject: mathematica/maple for linux Does anyone know of a good, user friendly piece of software similar to maple or mathematica for linux? In particular something that will === Subject: Re: mathematica/maple for linux > Does anyone know of a good, user friendly piece of software similar to > maple or mathematica for linux? In particular something that will http://www.mupad.de LD === Subject: Re: mathematica/maple for linux There are Linux versions of both Maple and Mathematica, according to their vendors: http://www.wolfram.com/products/mathematica/platforms.html http://www.maplesoft.com/products/maple/new/95sysrequirements.aspx I have heard of a few other tools that are supposed to run in Linux: Mathomatic: http://mathomatic.orgserve.de/math/ I haven't really looked at that, so I don't know how it compares to Mathematica in terms of sophistication or ease of use -J. Hess === Subject: Re: Comments on senior research idea >>I had an idea for my undergraduate senior research project. I thought >>about taking a look at the Bible codes. For those that don't know this >>is where people have found words/phrases at certain intervals in the >>Bible. What does everyone think of this idea? Any question you think >>I should look into? I am still trying to develop a list of questions >>to ask about it. > The theory behind the study of bible codes in English is that meaning can > be derived from numerology based on the English spelling of a word or > phrase. A lot of research has been done already on words and phrases from > the Bible. > I'd suggest writing a program to pick out interesting codes (666, etc.) from > any text, then run the Bible through it and see what you get. Then for > control, run an equal word-count of secular written material through it -- > One Flew Over the Cuckoo's Nest , washingtonpost.com, whatever. I agree with this basic thrust. This is a tiday empirical approach... How about some theory as well? Decoding presumes that there really is an encoded message in the text. I'd respect a nice probabilistic analysis of false positive (for finding encrypted matter), parametrised on the space of methods which are sloppily specified to cover bible codes. My suspicion is that the rules are sloppy enough that you can feel confident of finding things in the bible that aren't really there (ie, weren't put their in the first place). [And if Keith is interested in this general field, he could get involved in bioinformatics. It's seems to match is declared interests]. Tomasso. === Subject: Re: Comments on senior research idea by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iBC1vF217318; >So basically it is a pretty crappy idea. . .thus why I have avoided >doing it all semester, but it has been the only idea that my math >professors have been able to give me so far that fits my math talents. >I only have one semester left to do my project/research. All I really >need is an idea and I can run with it. I am just really bad with >coming up with ideas, not just in math, but in general. My main math >interests deal with number theory and combinatorics. I tried to have >them let me do elliptic curve crytography since I don't know anything >about it but they said there is not enough math in it or something. >Anyways, I better get back to googling various things to atempt to come >up with another idea. Sorry to bother with this crappy one. Just found out this thread randomly. How about criticizing the following paper? http://front.math.ucdavis.edu/math.MG/0407470 I wish I had enough math to judge on it. Besides, I wish I had enough math to read Professor Rychlik's paper. It is quite possible that my paper is a crappy idea too. But if your time allows. This is, I think, a 2-hour reading at most. I would be happy to be your coauthor :) H. Shinya P.S. I am not a professor; I am a college sophomore. I don't care, but you might want to email personally. Also, you definitely have to talk with your advisor about it as well; s/he will be able to comment on it quite correctly. Well, part of this message is motivated by the fact that I want my paper to be reviewed mathematically, but please do not have any thought that I am using you. In fact, it is absurdly universally self-evident that you can turn this down: a matter of whether YOU are interested in or not. All I can do is that I can present that paper; I am a sophomore and you a senior, which means that there is a tremendous difference in math knowledge. Well, that was a long postscript. === Subject: Re: Is There Anomaly In Value of pi ? > The question is, > is value of pi is experimentally verified or not? > You mean have we performed any infinitely-long experiments > of the type that are supposed to measure pi? No, we haven't > performed any infinitely-long experiments. > Infinitely long experiments are not needed. Yes, they are. > find that if squaring of circle problem is solved, value of pi can > change to 3.1547. No, you didn't. But why don't you provide the link and we'll see what it really said. - Randy === Subject: Re: Is There Anomaly In Value of pi ? > The question is, > is value of pi is experimentally verified or not? You mean have we performed any infinitely-long experiments > of the type that are supposed to measure pi? No, we haven't > performed any infinitely-long experiments. > Infinitely long experiments are not needed. > Yes, they are. Next time if someone question about accuracy of pi value, plese tell him/her that same value is used in missile trajectory. I admit, I was wrong. The value of pi is 3.141592.. -Abhi. === Subject: Re: Is There Anomaly In Value of pi ? > The question is, > is value of pi is experimentally verified or not? You mean have we performed any infinitely-long experiments > of the type that are supposed to measure pi? No, we haven't > performed any infinitely-long experiments. > Infinitely long experiments are not needed. > Yes, they are. > find that if squaring of circle problem is solved, value of pi can > change to 3.1547. > No, you didn't. > But why don't you provide the link and we'll see what it > really said. === Subject: Value(s) of Pi (was: Re: France's motivation) [Of Kevin J Maroney] > I want a vacation on his planet. Probably, by fiat, Kevin can make > pi=3 -- although it may require a UN resolution, as well. By good old rasff Sheer Coincidence, a friend has just loaned me _Mathematical Cranks_ by Underwood Dudley, which includes a substantial survey of independent thinkers[1] with new and improved values of pi. Pi-redefiners fall into two camps. The first group are those who seek to Square The Circle, i.e. produce a geometric construction from a circle of a square with equal area. Such a feat was long-ago proven impossible, but that is of little concern to such enthusiasts and they continue to come up with novel geometric constructions that, if they did work, would require pi to have a value such as (2/3)*(3+sqrt(3)), i.e. 3.1547... Such people thus come up with new values for pi as a by-product of their main work, although they often claim that this is (in software terms) a Feature, not a Bug. This is even the case when their diagrams imply a really good value of pi, e.g. (in one case Dudley quotes) 2.91419041. The second camp are those who simply find 3.141592654... much too untidy a value and seek to replace it with something neater and more memorable. Dudley cites one such person who declared pi=3.125, proving it thus: - Assume pi=3.125 - This would mean that a circle of circumference 4 units would have a radius of 4/pi=1.28 units. - To check, a circle of radius 1.28 units would have a circumference of 1.28*pi=4 units. QED! Dudley also addresses the near urban-legend that Indiana legislated that pi=4. What actually happened is that in 1897 the state legislature considered (but never passed) a bill on how to (guess what) Square The that one can get some nine different values of pi out of it. If nothing else, this bill would have given abundant freedom of choice. [1] Patrick Moore's delightfully diplomatic term for such people, as coined in his rather old but still very readable survey of cranks, _Can You Speak Venusian?_ -- Simon Bradshaw sjbradshaw@cix.co.uk http://www.cix.co.uk/~sjbradshaw *** The Science Fiction Foundation *** http://www.sf-foundation.org -Abhi. === Subject: Re: Is There Anomaly In Value of pi ? ... > Steven, I am not here to argue on my theory. I am here for your help. I > need some links, credible evidence regarding experimental verification > of pi, experimental measurement of circumference and radius of any > circular body. Will you people please help me so that I do not waste my > time on useless things? You are wasting your time on useless things. It is not possible to get the exact value of pi by experiment. *Every* experiment will get you no better than an approximation, and this has been proven. -- 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: Is There Anomaly In Value of pi ? Originator: richard@cogsci.ed.ac.uk (Richard Tobin) >Steven, I am not here to argue on my theory. I am here for your help. No, you're here to provide amusement when we get bored with the kooks who believe 0.999... is not equal to 1. -- Richard === Subject: Re: Complex Analysis Question >Hint: the integral of h'/h over a closed contour is an _integer_. Times 2*pi*i. === Subject: Re: Complex Analysis Question >>Hint: the integral of h'/h over a closed contour is an _integer_. >Times 2*pi*i. You guys are so picky... ************************ David C. Ullrich === Subject: Re: Platonism |If you can figure out why a discipline which is not |empirical in the slightest turns out to be so useful for empircal ends, |please share it with the rest of us. People interested in this kind of thing might enjoy reading a book by Mark Steiner, _The applicability of mathematics as a philosophical better job of explaining why this could pose a problem than is usually done. The problem he describes is not just with the existence of applications, but with the existence of applications when, apparently, one has no reason to expect a concept to be related to the area where it is applied. Keith Ramsay === Subject: Re: Platonism > The problem he describes is not just with the existence of applications, > but with the existence of applications when, apparently, one has no > reason to expect a concept to be related to the area where it is applied. There's a very nice quote of Feynman making this point--something about how he would understand how computers turning little switches on and off could produce results describing the behavior of atoms, if only atoms had little switches in them. Or something like that. I can't remember enough of it to attempt a decent Google search. === Subject: Re: Platonism <41b562b5.46155894@netnews.att.net> <41b682ed.59873690@netnews.att.net> <41b7d968.71009450@netnews.att.net> <41ba63ad.92196009@netnews.att.net> >I have been reading all this in sci.math so computer >representations were never part of the question. In any case, >I guess it depends on how you look at it. A typical computer's >memory is just an enourmous ordered list of bytes, so you could >always claim that anything stored in memory is ordered if you >look at it at a low enough level. >On the other hand a hash table is not usually considered as >an ordered data structure. The elements are of course going >to be stored in some order in memory, but this order has >little to do with any order property of the actual elements. >If you start to consider parallel/distributed systems than >ordering possibilities become much more interesting. The important thing in a computer is functional ordering. At the hardware level it depends on the address logic. At the software level it depends on the software. I not sure about ordering in an abstract mathematical set but wouldn't it be one of convention? In other words there is not inherent order in a set of items except that imposed by some protocol? John Casey === Subject: Re: Platonism >:> >:> :> Show up to class and take a picture. That specifies a set. >:> :> What's the order? >:> >:> : ... the order of the time space coordinates of the the students in the >:> : picture. You cannot represent a set in physical medium without picking >:> : up a time space bias. However, there may be a possible exception; >:> : representing a set in a neural network. Perhaps someone could shed some >:> : light on that. >:> >:> But what is the order of the time space coordinates? By order >:> I mean a total ordering. There is a first element, and each >:> element except the last has a unique successor. I do not see >:> that being the case with a photo. There are partial orderings >:> in a photo, but those are also rather arbitrary. >:> >:> http://www.discoverhuroncounty.com/phgal/turkeys/D-34-6.jpg >:> http://www.feathersite.com/Poultry/Turkeys/BuffFlock.JPEG >:> >:> Which is the first turkey? Which is the last? I imagine >:> somewhere someone has done studies on this. Perhaps >:> people come up with consistent answers but it would surprise >:> me. >:> >: Ok, you are right. Representing a set in a physical medium does not >: imbue it with a total ordering. I was actually coming from a computer >: science perspective. Representing an unordered set in a computer is >: very difficult. Representing an ordered set is a piece of cake. Can we >: then say that an ordered set is more fundamental to a computer than a >: unordered set? >I have been reading all this in sci.math so computer >representations were never part of the question. In any case, >I guess it depends on how you look at it. A typical computer's >memory is just an enourmous ordered list of bytes, so you could >always claim that anything stored in memory is ordered if you >look at it at a low enough level. >On the other hand a hash table is not usually considered as >an ordered data structure. The elements are of course going >to be stored in some order in memory, but this order has >little to do with any order property of the actual elements. >If you start to consider parallel/distributed systems than >ordering possibilities become much more interesting. >:> : Incidentally i have no use for the lester-set or its lesternality. >:> : forget in that world there are also lesterdifferences. >:> >:> That is true, I forgot about the lesterdifferences. The >:> fact he insists on using his own private vocabulary makes >:> it clear he is only interested in pontificating and not interested >:> in communicating. >:> >:> Stephen >: What irritates me is that at the point where i get familiar enough with >: Zick's mentations to grok them, that is the very point at which he >: fluffs me off and becomes abusive. It seems to me that Zick's >: mentations and Longley's share that property. When that happens enough >: times, you tend not to want to go there again. >You do seem to have some strange ones in comp.ai.philosophy. >But then again that is true of all of usenet. I think attempting >to understand Zick is pointless. He has demonstrated aptly in >this thread that there is nothing there to understand. When >somebody claims that the set {1,2,3} contains 0 you know there >is something hoplessly wrong with them. Glad you brought it up. Zero just doesn't photograph well. === Subject: Re: Tautologies Then and Now >>I don't know why you keep posting quotes that support what I said -- >>that the term tautology is not applied outside propositional logic >SH: I think the following paragraph could easily be seen to support >>Bob's statement, (That is how truth tables are extended to the first >>order >>logics.) but I wouldn't say you are wrong either. I used _word phrase_ >>to represent what the author put in bold in the original [underscore]. >>* is my emphasis. > The source you cite > http://www.lawrence.edu/fast/boardmaw/analytic_essay.html > cites Copi, and Copi asserts what Owen shows... that truth tables can > be used for monadic predicate logic. But in so describing, Copi does > not refer to monadic predicate sentences that are true in all > valuations as tautologies but as universally valid. Copi states: > If an argument contains n different predicate symbols, then if it is > valid for a model containing 2^n individuals, then it is valid for > every model or universally valid. > Copi, I.M. Symbolic Logic. p. 81. >- Paul Monadic means one, one predicate, not n>1. The definition you give to support your position, doesn't seem specific to monadic predicate logic. I think Copi's definition that you quote is a broad definition outside the range tautology applies. Shortly, you have shown how it is correct to use universally valid in describing all of general predicate calculus. Your quote does not show that it is incorrect to use tautologous for specific decidable fragments. I think Copi's quote is the not the same as the vice versa that I mention below. 'Every tautology is valid, but not every valid formulation is tautological.' (the monadic ones can be both) http://www-unix.oit.umass.edu/~partee/726_04/lectures /Lecture6%20_Predicate%20Logic_.pdf 2.1. Tautologies, contradictions and contingencies As in Statement Logic, some closed formulas of Predicate Logic are always true, i.e. they are true in every model. These are called tautologies. Formulas which are false in every model are called contradictions. All the other formulas are called contingencies: their truth values depend on models; they are true in some models and false in others. Mathematical Methods in Linguistics** which is used as a textbook: http://www.ling.gu.se/kurser/mathmeth/ This course gives a general introduction to various tools from discrete mathematics that are used in linguistics. It is based largely (or entirely) on Barbara Partee, Alice ter Meulen, Robert Wall (1990) Mathematical Methods in Linguistics, Kluwer Academic Publishers. ** ------------------------------------------------------------ SH: I think there is something odd about you stating that your instructors did not cover this. Lecture 11 pages 20 & 21 Logic: The Art of Persuasion and the Science of Truth I think by Vann McGee http://aka-ocw.mit.edu/NR/rdonlyres/Linguistics-and-Philosophy/24-241Logic -IFall2002/BE072366-EA9F-4193-97CC-BB2922C3C7B1/0/lec11.pdf [pgs. 20&21] Normal Truth Assignment = N.T.A. Definition: A sentence is tautological iff it is assigned the value1 by every N.T.A. A sentence is valid iff it is true under every N.T.A. For the sentential calculus,the words tautological and valid were different words for the samething. Now that we've started on the predicate calculus, we need to distinguish them. Validity is the notion we're really interested in,but we need the notion of tautology as a technical notion. Proposition. Every tautology is valid, but not vice versa. [SH: He provides a proof, and then continues:] A tautological sentence is a valid sentence whose validity is determined by the sentence's truth functional structure. If, instead,the validity of a sentence depends upon the meaning of the quantifiers,the sentence won't be tautological. We can test whether a sentence is tautological by the method of truth tables, examining each possible way to assign a truth value to the sentence's basic truth functional components. --------------------------------------------------------- Derivations in the Mondadic Predicate Calculus [MIT online course] The fact that there is a mechanical procedure for testing whether a sentence is a tautological consequence of a set of sentences is important. In order for our derivations to have any probative value, we have to be able to recognize when a sequence of sentences really is a proof,which means that we need an algorithm for checking when a rule has been properly applied. http://ocw.mit.edu/NR/rdonlyres/Linguistics-and-Philosophy /24-241Logic-IFall2002/8A9852F3-D38E-4A94-9634-9EED32367CCB/0/lec12.pdf SH: That there is a mechancical procedure for testing (truth table) which can output all trues is correctly described as tautological by definition. No matter what the logical structure outside of propositional logic is named. That there is a mechanical shortcut, is why it can be done by a computer More about decidable fragments: http://www-mgi.informatik.rwth-aachen.de/Publications/Graedel/ #38 If my quotes don't convince you, maybe someone else will do a better job. Something interesting about artificial/formal languages and natural language. Knowledge Representation in Sanskrit and Artificial Intelligence by Rick Briggs http://www.aaai.org/Library/Magazine/Vol06/06-01/Papers/AIMag06-01-003.pdf Logically guarded, Stephen === Subject: Re: Tautologies Then and Now >> If an argument contains n different predicate symbols, then if it is >> valid for a model containing 2^n individuals, then it is valid for >> every model or universally valid. >> Copi, I.M. Symbolic Logic. p. 81. >>- Paul >I think Copi's definition that you quote is a broad definition outside the >range tautology applies. Shortly, you have shown how it is correct >to use universally valid in describing all of general predicate calculus. >Your quote does not show that it is incorrect to use tautologous >for specific decidable fragments. It should be clear to you from numerous quotes you posted previously that, with the one exception of Barbara Partee, most authors are careful when discussing logics to restrict their use of the term tautology to propositional/sentential logic. For example your oft quoted http://www.lawrence.edu/fast/boardmaw/analytic_essay.html says on top: Corresponding to tautologies in Sentential Logic are analytic sentence schemata in First Order Predicate Logic. You will remember that a tautology is a sentence schema which is true under any consistent interpretation of its sentential letters; Clearly the author indicates the term tautology is used uniquely in sentential logic. Numerous other quotes you have presented also indicate that same limit. Which matches Gamut and Copi: * In predicate logic ... Formulas @ such that V_M(@) = 1 for all models M for the language from which @ is taken are called universally valid formulas (they are not normally called tautologies). L.T.F. Gamut. Logic, Langauge, and Meaning. p. 99. * If an argument contains n different predicate symbols, then if it is valid for a model containing 2^n individuals, then it is valid for every model or universally valid. Copi, I.M. Symbolic Logic. p. 81. instances of its application to always valid predicate statements versus the one example of Partee. My question is why is tautology not normally used outside sentential logic? There must be a reason. - paul === Subject: Re: Tautologies Then and Now > If an argument contains n different predicate symbols, then if it is > valid for a model containing 2^n individuals, then it is valid for > every model or universally valid. > Copi, I.M. Symbolic Logic. p. 81. >- Paul >>I think Copi's definition that you quote is a broad definition outside the >>range tautology applies. Shortly, you have shown how it is correct >>to use universally valid in describing all of general predicate >>calculus. >>Your quote does not show that it is incorrect to use tautologous >>for specific decidable fragments. > It should be clear to you from numerous quotes you posted previously > that, with the one exception of Barbara Partee, most authors are > careful when discussing logics to restrict their use of the term > tautology to propositional/sentential logic. For example your oft > quoted This is wrong. Why do you think I posted under Mondadic Pred. Logic: Now that we've started on the predicate calculus, we need to distinguish them. Validity is the notion we're really interested in,but we need the notion of tautology as a technical notion. Proposition. Every tautology is valid, but not vice versa. [SH: He provides a proof, and then continues:] A tautological sentence is a valid sentence whose validity is determined by the sentence's truth functional structure. If, instead,the validity of a sentence depends upon the meaning of the quantifiers,the sentence won't be tautological. SH: This is taken from Mondadic Predicate Calculus which is about 11th. http://aka-ocw.mit.edu/OcwWeb/Linguistics-and-Philosophy/24-241Logic-IFall20 02/Readings/index.htm This is the online MIT course. Any time you have truth tables or their logical equivalent that can return all trues, that is described by the word tautological. It could also be less precisely described by universally valid. That is the definition, and it applies to some cases in mondadic predicate logic. _Universally valid_ is used for Predicate Logic, never tautological. But you didn't specify predicate logic, you said propositional logic. paul: what I said -- that the term tautology is not applied outside propositional logic. I did not say tautology was applied to Predicate logic but to monadic predicate logic. Every tautology is valid means that all trues are in sentential logic or in some cases of monadic logic which have truth tables returning all trues, can be described by the word tautological. Or the word valid. but not vice versa means there are cases which are described by valid but not tautological. Every valid is not tautological. But some are, the ones which are logically equivalent to truth tables with the possibility of all trues. > http://www.lawrence.edu/fast/boardmaw/analytic_essay.html > says on top: Corresponding to tautologies in Sentential Logic are > analytic sentence schemata in First Order Predicate Logic. You will > remember that a tautology is a sentence schema which is true under any > consistent interpretation of its sentential letters; > Clearly the author indicates the term tautology is used uniquely in > sentential logic. Numerous other quotes you have presented also > indicate that same limit. Which matches Gamut and Copi: This is just wrong, the author does not do that. Nor does Gamut nor Copi when they refer to monadic predicate logic. Monadic predicate logic is part of the foundation of Predicate logic. (PL) But it is less powerful, I think because it is decidable. The reason PL is more powerful is because it is undecidable. Undecidable IIRC, means that it doesn't halt according to Turing's 1936 paper. Not halting means there is no possibility of a truth table. > * In predicate logic ... Formulas @ such that V_M(@) = 1 for all > models M for the language from which @ is taken are called universally > valid formulas (they are not normally called tautologies). L.T.F. > Gamut. Logic, Langauge, and Meaning. p. 99. > * If an argument contains n different predicate symbols, then if it > is valid for a model containing 2^n individuals, then it is valid for > every model or universally valid. Copi, I.M. Symbolic Logic. p. 81. > instances of its application to always valid predicate statements > versus the one example of Partee. My question is why is tautology > not normally used outside sentential logic? There must be a reason. > - paul Well certainly. Universally valid (or maybe just valid, I'm not sure) applies to Predicate logic. But I am talking about monadic PL, which is outside of propositional logic, thus a counter example to your claim, because you didn't claim that tautology was not part of predicate logic, you claimed there was nothing outside of propositional logic that could properly use the term tautology. Both of my quotes, Partee and McGee showed that. McGee has written numerous books and is connected to MIT, which is going to be the standard, not eccentric. Since you have brought up Gamut and normally twice now in the context of (they are not _normally_ called tautologies) I will explain why normally should be interpreted in the sense of usually, as in most cases. The exceptions you have been hearing about in monadic PL have been saying in some cases. Emphasis on some by Chris Menzel: In *some* cases they do, namely, if you restrict your attention to arguments consisting of formulas of monadic predicate logic (i.e., formulas that involve only 1-place predicates), or if the argument in question is invalid and has a *finite* countermodel. >SH: I don't think the author (Gamut) of the book would have used the >qualifier normally (they are not normally called tautologies) if it were >strictly >true that the term tautology never applied to predicate logic which is >outside propositional logic, the boundary paul claimed, not applied. paul responded: You're assuming that what is abnormal is inherently also appropriate. I believe that what Gamut means by not normally called tautologies is that some may use the term tautology within predicate logic, but, strictly speaking, it is abnormal and inappropriate. That view is supported by what I was taught in school, and by reading Gamut, which finds no exception to the restriction of the term tautology to propositional logic, and as I recall other utterances stipulating the limitation of the term tautology to propositional logic. SH: I am using normally (they are not _normally_ called tautologies) as most people would when reading this sentence. Normally, means usually, or most of the time. The opposite of this meaning is not abnormal. The opposite would be unusually, infrequently, or rarely as in a small proportion when measured agains the whole. The decidable conditions are after all, only met by fragments of predicate logic. They are uncommon. You are using abnormal in a sense of incorrect. Gamut could have said that meaning (incorrect) much more plainly: is taken are called universally valid formulas (they are not normally called tautologies). Gamut could have changed this to: is taken are _correctly_ called universally valid formulas. Or 'is taken are _correctly_ called universally valid formulas, not tautologies.' If Gamut meant to clarify normal in the sense of correctness he could have just used _correctly_ as I did above and would not have needed a parenthetical remark. I suppose you could just argue he is a bad writer. But if you assume he is a good writer, it means the proper interpretation of Gamut's parenthetical remark is that he is distinguishing what is the most common usage in terms of scope, while hinting in some instances that the term tautologous has some actual properly applied descriptive power. This interpretation of normally meaning prominece or importance due to the frequency or liklihood of encountering some cases is consistent with understanding monadic PL ( a sort of subset of PL) as carrying over some foundational ideas into the larger dominant area of PL. You, paul, have not normally encountered the idea of monadic PL and truth tables. That doesn't mean they don't exist but that they are not usually mentioned, that is why you had not heard of them. Likewise with the limited usage (occurrence) of tautological in describing some special fragment cases of all trues in mondadic PL. Normally mean usually in this case, the opposite is unusually or rarely. It explains why you haven't heard of it perhaps. The opposite of normally does not mean abnormally in this case--- like in the case of some eccentric professor with deviant ideas. Barbara Partee would have been blasted if she were presenting some non-standard view, and not having her book used as a textbook. formulas that involve only 1-place predicates > * If an argument contains n different predicate symbols, then if it ------------------------------------------------------------------- > is valid for a model containing 2^n individuals, then it is valid for > every model or universally valid. Copi, I.M. Symbolic Logic. p. 81. If an argument contains n different predicate symbols... The n in Copi's definition can be greater than 1-place. So this definition includes full predicate logic. It is not limited to (it could be a greater value n) formulas that involve only 1-place predicates which is a definition for tautological/truth tabled monadic predicate logic. Since his definition covers a greater logical range which includes FPL it is not meet the tautological condition. If the definition were restriced to MPL with 1-place predicates then I think it would be correct to use tautological or valid. Because all tautologies are valid, but not the converse. Certainly an instructor is going to stress that Predicate logic is correctly described by universally valid (or maybe valid). But to be honest with you, I think teaching the monadic foundation of predicate logic is going to be very standard among good instructors in good colleges. That is done by McGee at MIT, who mentions both truth tables and tautology in his treatment of monadic PL in the progression to full PL. One of us quite misunderstands what (I would welcome outside comment.) http://aka-ocw.mit.edu/OcwWeb/Linguistics-and-Philosophy/24-241Logic-IFall20 02/Readings/index.htm I think it is you because you think this quote > says on top: Corresponding to tautologies in Sentential Logic are > analytic sentence schemata in First Order Predicate Logic. SH: supports your position. It doesn't because First Order Predicate Logic is not Monadic Predicate Logic, where all 'tautologies are valid' is true. The quote is relating to FOPL which is not the area of controversy. First Order Predicate Logic only corresponds to sentential logic, because the comparison of valid vs tautological, is the vice versa case where 'Every valid statement is not tautological'. Remember that Every tautology is valid nearly describes both Sentential logic and Monadic PL (AFAIK) which is under discussion, the category of not propositional logic which accepts MPL -- and what is not under discussion is first order predicate logic which has never been disputed by me as being anything other than 'not tautological' but valid. You seem to have made the assumption that if it is not propositional logic, then it must be full-blown predicate logic which has neither tautologies nor truth tables. I've been harping on monadic predicate logic which is decidable. I haven't claimed tautologies or truth tables for pure predicate calculus(PPL) nor full predicate calculus(FPL), both of which are undecidable. I think there are at most minor errors in this post. Why How is When, Stephen === Subject: Re: Tautologies Then and Now > instances of its application to always valid predicate statements > versus the one example of Partee. My question is why is tautology > not normally used outside sentential logic? There must be a reason. Because it usefully picks out a certain class of logical truths, viz., those that are true simply in virtue of their truth functional structure. If its meaning were broadened to include the logical truths of predicate logic, it would serve no purpose. We already have logical truth and universal validity (though the latter is rather less common; indeed, I can only recall seeing it in the LTF Gamut text -- whose actual authors, BTW, are the frighteningly prolific Dutch logician Johan van Benthem and a couple of his colleagues). === Subject: 4-manifold classification This question pertains to p. 114-115 of Massey's _A Basic Course in Algebraic Topology_. Massey very briefly summarizes a proof by Markov that the 4-manifolds cannot be classified -- that is, the problem of determining whether any two 4-manifolds are homeomorphic is not generally decidable. Unfortunately the original proof by Markov cited here happens to be in Russian, so I only have Massey to go by. Roughly, the proof seems to go something like this: Finitely presented groups are non-classifiable under isomorphism (given). For every finitely presented group G there exists a 4-manifold M such that its fundamental group pi(M)=G (Massey shows this). Therefore, the 4-manifolds are non-classifiable. The problem with my summary of Massey's summary here is that the final statement doesn't necessarily follow, because it's possible to have multiple non-homeomorphic manifolds with the same fundamental group. It seems to me that the manifolds could be classifiable even though their fundamental groups are not classifiable. So, what am I missing? Perhaps the particular manifolds constructed by Massey have the property that isomorphic fundamental groups imply homeomorphic manifolds? FYI Massey constructs a manifold for each finitely presented fundamental group as follows. First start out with the connected sum of n copies of S^1 x S^3, where n is the number of generators in the desired fundamental group. Then do some surgery to accomodate each relation in the fundamental group, which involves cutting out regions homeomorphic to S^1 x D^3 (containing a closed path representing the desired relation) and then gluing D^2 x S^2 to the resulting hollowed-out hypertube boundary (see Massey for a more coherent explanation). === Subject: Re: 4-manifold classification >This question pertains to p. 114-115 of Massey's _A Basic Course in >Algebraic Topology_. Massey very briefly summarizes a proof by Markov >that the 4-manifolds cannot be classified -- that is, the problem of >determining whether any two 4-manifolds are homeomorphic is not >generally decidable. Unfortunately the original proof by Markov cited >here happens to be in Russian, so I only have Massey to go by. >Roughly, the proof seems to go something like this: Finitely presented >groups are non-classifiable under isomorphism (given). For every >finitely presented group G there exists a 4-manifold M such that its >fundamental group pi(M)=G (Massey shows this). Therefore, the >4-manifolds are non-classifiable. >The problem with my summary of Massey's summary here is that the final >statement doesn't necessarily follow, because it's possible to have >multiple non-homeomorphic manifolds with the same fundamental group. >It seems to me that the manifolds could be classifiable even though >their fundamental groups are not classifiable. >So, what am I missing? What Massey (following Markov) shows is that classifying 4-manifolds is *as difficult* as classifying finitely presented groups. In other words, if you could solve the classification problem of 4-manifolds then you could also solve the classification problem for finitely presented groups. This comes about because the fundamental group is a functor on the category of pointed spaces so that (= for isomorphism in the respective categories) (A) (X,x) = (Y, y) => pi_1(X,x) = pi_1(Y, y) (The base-point here is just a minor technicality, you can get easily rid of it and translate everything into statements about spaces) Markov's construction tells us that any finitely presented group can appear as pi_1 of some 4-manifold. Since the classification problem for finitely presented groups is undecidable, by (A) the classification problem for 4-manifolds is also undecidable. G. Rodrigues === Subject: Re: 4-manifold classification ^OX9W/.#XpUmm`>TD2zNE-t}emfPkFR.Z5`flY:3QYT$>dUwN^sm;MBV:F7aL9x*q!` ln!l}>Y6_45$%R|P7DSrBkEph@1-;P*s~F_28vO@e4p/'>}Pc?@rl8cz]d9RXOt What Massey (following Markov) shows is that classifying 4-manifolds > is *as difficult* as classifying finitely presented groups. In other > words, if you could solve the classification problem of 4-manifolds > then you could also solve the classification problem for finitely > presented groups. This comes about because the fundamental group is a > functor on the category of pointed spaces so that (= for isomorphism > in the respective categories) > (A) (X,x) = (Y, y) => pi_1(X,x) = pi_1(Y, y) > (The base-point here is just a minor technicality, you can get easily > rid of it and translate everything into statements about spaces) > Markov's construction tells us that any finitely presented group can > appear as pi_1 of some 4-manifold. Since the classification problem > for finitely presented groups is undecidable, by (A) the > classification problem for 4-manifolds is also undecidable. What you said is not sufficient. As a historical comment, I note that this is not Markov's construction but an old fact dating back to at least Seifert and Threlfall's Lehrbuch der Topologie. Markov's contribution was not just building these manifolds, but in showing that if their fundamental groups were isomorphic, they were homeomorphic. === Subject: Re: 4-manifold classification >>What Massey (following Markov) shows is that classifying 4-manifolds >>is *as difficult* as classifying finitely presented groups. In other >>words, if you could solve the classification problem of 4-manifolds >>then you could also solve the classification problem for finitely >>presented groups. This comes about because the fundamental group is a >>functor on the category of pointed spaces so that (= for isomorphism >>in the respective categories) >>(A) (X,x) = (Y, y) => pi_1(X,x) = pi_1(Y, y) >>(The base-point here is just a minor technicality, you can get easily >>rid of it and translate everything into statements about spaces) >>Markov's construction tells us that any finitely presented group can >>appear as pi_1 of some 4-manifold. Since the classification problem >>for finitely presented groups is undecidable, by (A) the >>classification problem for 4-manifolds is also undecidable. > What you said is not sufficient. As a historical comment, I note that > this is not Markov's construction but an old fact dating back to at > least Seifert and Threlfall's Lehrbuch der Topologie. Markov's > contribution was not just building these manifolds, but in showing that > if their fundamental groups were isomorphic, they were homeomorphic. mentioned http://graphics.stanford.edu/%7Eafra/goodies/markov.pdf Dale. === Subject: Re: 4-manifold classification ... >The problem with my summary of Massey's summary here is that the final >statement doesn't necessarily follow, because it's possible to have >multiple non-homeomorphic manifolds with the same fundamental group. >It seems to me that the manifolds could be classifiable even though >their fundamental groups are not classifiable. >So, what am I missing? An appropriate understanding of what classification means? In this context, having a classification of 4-manifolds would certainly mean *at least* that, given (somehow) 4-manifolds M and N, one could determine whether or not M was homeomorphic to N. Since, given a group presentation P=gp(g_1,...,g_k|r_1,...,r_p) of a group G, one can construct a (closed, smooth) 4-manifold M(P) with fundamental group G, *if* one had a classification of 4-manifolds with the property I just gave, *then* one could determine whether or not two group presentations present isomorphic groups. Markov showed you can't do the latter, therefore you can't have the former. Lee Rudolph === Subject: Re: 4-manifold classification ^OX9W/.#XpUmm`>TD2zNE-t}emfPkFR.Z5`flY:3QYT$>dUwN^sm;MBV:F7aL9x*q!` ln!l}>Y6_45$%R|P7DSrBkEph@1-;P*s~F_28vO@e4p/'>}Pc?@rl8cz]d9RXOt ... >The problem with my summary of Massey's summary here is that the final >statement doesn't necessarily follow, because it's possible to have >multiple non-homeomorphic manifolds with the same fundamental group. >It seems to me that the manifolds could be classifiable even though >their fundamental groups are not classifiable. >So, what am I missing? > An appropriate understanding of what classification means? > In this context, having a classification of 4-manifolds would > certainly mean *at least* that, given (somehow) 4-manifolds M > and N, one could determine whether or not M was homeomorphic to N. > Since, given a group presentation P=gp(g_1,...,g_k|r_1,...,r_p) of > a group G, one can construct a (closed, smooth) 4-manifold > M(P) with fundamental group G, *if* one had a classification > of 4-manifolds with the property I just gave, *then* one could > determine whether or not two group presentations present > isomorphic groups. Markov showed you can't do the latter, > therefore you can't have the former. Hmmm...ok, well suppose that I've determined that M and N are not homeomorphic using my 4-manifold classification. What can I say about their fundamental groups? Markov's construction isn't just any old classification -- he shows that M and N are homeomorphic if and only if they are related by certain moves which are reflected by certain moves in their group presentations, which happens if and only if they are isomorphic. Or something like that. === Subject: Re: 4-manifold classification > This question pertains to p. 114-115 of Massey's _A Basic Course in > Algebraic Topology_. Massey very briefly summarizes a proof by Markov > that the 4-manifolds cannot be classified -- that is, the problem of > determining whether any two 4-manifolds are homeomorphic is not > generally decidable. Unfortunately the original proof by Markov cited > here happens to be in Russian, so I only have Massey to go by. Insolubility of the Problem of Homeomorphy http://graphics.stanford.edu/%7Eafra/goodies/markov.pdf ... the rest deleted ... Dale. === Subject: Re: 4-manifold classification <9XUud.41616$6q2.13238@newssvr14.news.prodigy.com> Originator: elkies@math.harvard.edu (Noam Elkies) > Let r be the expected number of CMRs per unit area. Noam Elkies > posted a nice proof that r > 1/8 for any rectangle. For an infinite > grid, Robert Ziff has found the numerical result r = 0.13154. See > http://www.mathsoft.com/mathresources/constants/discretestructures/ar... > (Ziff's result, KS(1/2) = 0.065770, is based on clusters of one > color only.) Interesting. > Apparently, no closed form solution is known for the infinite grid. > [...] > It is interesting that Noam Elkies's simple argument yields > such a tight lower bound on r for the infinite grid: 0.125 < 0.13154. The argument I gave suggests that in fact 1/8 is quite close to the correct answer. More precisely, let p be the probability that the two R squares land in the same component when the * squares in this half-infinite grid: ******R ******BR***** ... ************* ... ************* ************* ... are each independently filled with R or B, each having probability 1/2. Then it follows easily from my argument that the limit for large m,n of the ratio of the number of monochromatic clusters to the area mn equals (1+p)/8. Now while p is clearly positive, it can't be very large: already the probability that each R is connected to a square in the row below it is 2/9, the product of 1/3 for the top R times 2/3 for the bottom one. This gives a crude upper bound of 11/72=0.1527777... on (1+p)/8. For a lower bound, the shortest path between the two R's has length 5, giving p>1/32, which is already enough to push (1+p)/8 above 0.1289; if I did this right then we get p>173/4096 by using also the paths of length 7 or 9, so (1+p)/8 > 0.1302795... which is only a bit below the value reported by Robert Ziff in the URL cited above. NDE P.S. I'm sure that I'm not the first to find this argument -- surely the proposer of this Putnam problem, and/or the researchers in this kind of percolation theory, were aware of this too. [reply address is baloglouAToswego.edu] >P.S. I'm sure that I'm not the first to find this argument -- >surely the proposer of this Putnam problem, and/or the researchers >in this kind of percolation theory, were aware of this too. Of course. And yet it is provably difficult to come up with that idea! Why? Because the human mind is conditioned to think 'locally' (one square or cluster of squares at a time, leading to the Sissifus-like task of having to reach 1/8 by a sum of certain inverse powers of 2) rather than 'semi-globally (row-after-row and/or square-after-square), perhaps? My flawed inductive approach was closer to your idea, trying to establish that the expected number of new regions added by each new row exceeds n/8; of course your argument shows that induction is not needed at all :-) Anyway, I am a bit surprised that no one here was able to fix/complete my argument*, not even after Prof. Bernstein's explicit warning that added squares may also destroy regions! [I was aware of that fact, but I somehow thought that I could 'separate' the first n-1 or n-2 columns from the last one or two, preventing the destruction of existing regions: foolish! :-) ] *for the record, here it is (fixed -- I think! -- but still using two (!) squares at the end instead of one): n+1 row B B B B R R R R n row BB BR RB RR BB BR RB RR new CMCs 3/4 2/4 1/4 1/4 1/4 1/4 2/4 3/4 gone CMCs 2/8 2/8 1/4 0 0 1/4 2/8 2/8 net gain 1/2 1/4 0 1/4 1/4 0 1/4 1/2 With the last row summing up to 2, the inductive step is now fixed to (n-1)/8 + 2/8 = (n+1)/8 ... as opposed to (n-1)/8 + 7/16 > (n+1)/8. baloglouAToswego.edu === Subject: Euler Cookies It has come to my knowledge that, in spite of there being cookies named after Newton and cookies named after Leibniz, there is no cookie named after Euler. I have set myself to remedy this matter. These cookies are indeed the oiliest cookies I have ever made: EULER COOKIES Sift 1/4 cup confectioners sugar. Beat one cup of butter until soft and add the sugar slowly into it. Cream until blended. Yes, a whole cup. Add 1 teaspoon of vanilla and a pinch of allspice. Combine 2 cups all purpose flour (or, if wheat-free cookies are desired, one cup of rice flour and one cup of soy meal), 1/4 teaspoon salt, and 1/2 teaspoon baking powder. Work the butter mixture into the flour mixture. Make into balls and flatten on a cookie sheet. Bake at 350'F for approximately twenty minutes. I have been advised that eating too many cookies may cause you to go to L'Hopital. --scott -- C'est un Nagra. C'est suisse, et tres, tres precis. === Subject: Re: Euler Cookies > It has come to my knowledge that, in spite of there being cookies named > after Newton and cookies named after Leibniz, there is no cookie named > after Euler. I have set myself to remedy this matter. These cookies are > indeed the oiliest cookies I have ever made: > EULER COOKIES > Sift 1/4 cup confectioners sugar. > Beat one cup of butter until soft and add the sugar slowly into it. > Cream until blended. Yes, a whole cup. > Add 1 teaspoon of vanilla and a pinch of allspice. > Combine 2 cups all purpose flour (or, if wheat-free cookies are desired, one > cup of rice flour and one cup of soy meal), 1/4 teaspoon salt, and 1/2 teaspoon > baking powder. > Work the butter mixture into the flour mixture. Make into balls and flatten > on a cookie sheet. Bake at 350'F for approximately twenty minutes. > I have been advised that eating too many cookies may cause you to go to > L'Hopital. > --scott > -- > C'est un Nagra. C'est suisse, et tres, tres precis. Alexei Sayles (sp?) had a routine in the one of the Young Ones about treats being named after generals, like Napoleans and Garibaldis, etc. blacksalt === Subject: Re: Euler Cookies > It has come to my knowledge that, in spite of there being cookies named > after Newton and cookies named after Leibniz, there is no cookie named > after Euler. I have set myself to remedy this matter. These cookies are > indeed the oiliest cookies I have ever made: > EULER COOKIES > Sift 1/4 cup confectioners sugar. > Beat one cup of butter until soft and add the sugar slowly into it. > Cream until blended. Yes, a whole cup. > Add 1 teaspoon of vanilla and a pinch of allspice. > Combine 2 cups all purpose flour (or, if wheat-free cookies are desired, one > cup of rice flour and one cup of soy meal), 1/4 teaspoon salt, and 1/2 teaspoon > baking powder. > Work the butter mixture into the flour mixture. Make into balls and flatten > on a cookie sheet. Bake at 350'F for approximately twenty minutes. > I have been advised that eating too many cookies may cause you to go to > L'Hopital. To help with any future cookies, you may need the assistance of a mathematician who described a numerical method of solving differential equations and whose name is usually associated with that of Carle Runge. -- Clive Tooth http://www.clivetooth.dk === Subject: Re: Euler Cookies Originator: gheston@hiwaay.net (Gary Heston) >> It has come to my knowledge that, in spite of there being cookies named >> after Newton and cookies named after Leibniz, there is no cookie named >> after Euler. I have set myself to remedy this matter. These cookies are >> indeed the oiliest cookies I have ever made: >> EULER COOKIES [ ... ] >To help with any future cookies, you may need the assistance of a >mathematician who described a numerical method of solving differential >equations and whose name is usually associated with that of Carle Runge. Aren't shortbread cookies shaped with a Runge Cutter? Gary -- Gary Heston gheston@hiwaay.net Sept. 11, 2001, already a day of immeasurable tragedy, cannot be the day liberty perished in this country. Judge Gerald Tjoflat === Subject: Re: Euler Cookies > It has come to my knowledge that, in spite of there being cookies named > after Newton and cookies named after Leibniz, there is no cookie named > after Euler. I have set myself to remedy this matter. These cookies are > indeed the oiliest cookies I have ever made: >Aren't shortbread cookies shaped with a Runge Cutter? Let's not forget Graham wafers and Mandelbrot. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Euler Cookies <10rqajjkfit5aa8@corp.supernews.com> The latter being a case of a mathematician named for a cookie. Fig Newtons, by the way, are actually named for Newton, Massachusetts, where they were first made. Whether the town was named for Sir Isaac or is just a common contraction of New Town is arguable. === Subject: Re: Euler Cookies >> It has come to my knowledge that, in spite of there being cookies named >> after Newton and cookies named after Leibniz, there is no cookie named >> after Euler. I have set myself to remedy this matter. These cookies are >> indeed the oiliest cookies I have ever made: >> EULER COOKIES > [ ... ] >To help with any future cookies, you may need the assistance of a >mathematician who described a numerical method of solving differential >equations and whose name is usually associated with that of Carle Runge. > Aren't shortbread cookies shaped with a Runge Cutter? ... and I suppose we might need Richard Courant and Linards Reizins ... -- Clive Tooth http://www.clivetooth.dk === Subject: Re: Euler Cookies >It has come to my knowledge that, in spite of there being cookies named >after Newton and cookies named after Leibniz, there is no cookie named >after Euler. I have set myself to remedy this matter. These cookies are >indeed the oiliest cookies I have ever made: Are you the Scott Dorsey who used to eat at Chawee's. -- Susan N. Moral indignation is in most cases two percent moral, 48 percent indignation, and 50 percent envy. Vittorio De Sica, Italian movie director (1901-1974) === Subject: Re: Euler Cookies >>It has come to my knowledge that, in spite of there being cookies named >>after Newton and cookies named after Leibniz, there is no cookie named >>after Euler. I have set myself to remedy this matter. These cookies are >>indeed the oiliest cookies I have ever made: > Are you the Scott Dorsey who used to eat at Chawee's. ... no, no. he's the guy who made a face and it stuck that way. n ps; against his mothers repeated warnings -- ... this is my sig. it's one of the best sigs on the net.i know what you're asking yourself. 'did he post 5 or 6 messages'? well, in all the confusion i kinda lost track myself. so you gotta ask yourself one question 'do you feel lucky'? huh, DO YA? DO YA PUNK'? GO FOR IT, MAKE MY BED!!!' === Subject: Re: Euler Cookies > It has come to my knowledge that, in spite of there being cookies named > after Newton and cookies named after Leibniz, there is no cookie named > after Euler. I have set myself to remedy this matter. These cookies are > indeed the oiliest cookies I have ever made: > EULER COOKIES > Sift 1/4 cup confectioners sugar. > Beat one cup of butter until soft and add the sugar slowly into it. > Cream until blended. Yes, a whole cup. > Add 1 teaspoon of vanilla and a pinch of allspice. > Combine 2 cups all purpose flour (or, if wheat-free cookies are desired, one > cup of rice flour and one cup of soy meal), 1/4 teaspoon salt, and 1/2 teaspoon > baking powder. > Work the butter mixture into the flour mixture. Make into balls and flatten > on a cookie sheet. Bake at 350'F for approximately twenty minutes. > I have been advised that eating too many cookies may cause you to go to > L'Hopital. > --scott > -- > C'est un Nagra. C'est suisse, et tres, tres precis. ETH. They tend to become oily and flat if the butter/flour mixture is off balance...flour measurements vary with humidity! With the right balance they lose that oily taste. Also add a few Teaspoons of BRANDY to the mix, leave out the all space, but prop in a clove in each one and there you have it..I was weened on these cookies ! Merry Christmas, if applicable! Mike === Subject: Re: Euler Cookies Scott Dorsey kirjoitti viestiss.8a:cpgfo1$77i$1@panix3.panix.com... > It has come to my knowledge that, in spite of there being cookies named > after Newton and cookies named after Leibniz, there is no cookie named > after Euler. I have set myself to remedy this matter. These cookies are > indeed the oiliest cookies I have ever made: > EULER COOKIES > Sift 1/4 cup confectioners sugar. > Beat one cup of butter until soft and add the sugar slowly into it. > Cream until blended. Yes, a whole cup. > Add 1 teaspoon of vanilla and a pinch of allspice. > Combine 2 cups all purpose flour (or, if wheat-free cookies are desired, > one > cup of rice flour and one cup of soy meal), 1/4 teaspoon salt, and 1/2 > teaspoon > baking powder. > Work the butter mixture into the flour mixture. Make into balls and > flatten > on a cookie sheet. Bake at 350'F for approximately twenty minutes. > I have been advised that eating too many cookies may cause you to go to > L'Hopital. > --scott > -- > C'est un Nagra. C'est suisse, et tres, tres precis. Get stuck in your throat if you don't eat it with for example alcohol. Cookies follow people, which is bizarre. === Subject: Re: Euler Cookies > It has come to my knowledge that, in spite of there being cookies named > after Newton and cookies named after Leibniz, there is no cookie named > after Euler. I thought Euler was more into pie. (hangs head in shame) === Subject: Re: Euler Cookies > It has come to my knowledge that, in spite of there being cookies named > after Newton and cookies named after Leibniz, there is no cookie named > after Euler. Fig Newtons I know. But what's a Leibniz cookie? === Subject: Re: Euler Cookies >> It has come to my knowledge that, in spite of there being cookies named >> after Newton and cookies named after Leibniz, there is no cookie named >> after Euler. > Fig Newtons I know. But what's a Leibniz cookie? Go to www.bahlsen.com and klick on the rider Leibniz. Carmen -- Carmen Bartels elfgar@ATP, elfgar@Xyllomer caba@squirrel.han.de caba@irc === Subject: Re: Euler Cookies >> It has come to my knowledge that, in spite of there being cookies named >> after Newton and cookies named after Leibniz, there is no cookie named >> after Euler. >Fig Newtons I know. But what's a Leibniz cookie? That's a German brand of cookies. But it's spelled Leibnitz (Leibnitz Kekse). Thomas === Subject: Re: Euler Cookies > It has come to my knowledge that, in spite of there being cookies named > after Newton and cookies named after Leibniz, there is no cookie named > after Euler. >>Fig Newtons I know. But what's a Leibniz cookie? > That's a German brand of cookies. But it's spelled Leibnitz (Leibnitz > Kekse). Only in English. In German it's Leibniz. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: Euler Cookies >> It has come to my knowledge that, in spite of there being cookies named >> after Newton and cookies named after Leibniz, there is no cookie named >> after Euler. >Fig Newtons I know. But what's a Leibniz cookie? >> That's a German brand of cookies. But it's spelled Leibnitz (Leibnitz >> Kekse). >Only in English. In German it's Leibniz. Nope - It's Leibnitz-Kekse. Thomas.de === Subject: Re: Euler Cookies > It has come to my knowledge that, in spite of there being cookies named > after Newton and cookies named after Leibniz, there is no cookie named > after Euler. >>Fig Newtons I know. But what's a Leibniz cookie? > That's a German brand of cookies. But it's spelled Leibnitz (Leibnitz > Kekse). >>Only in English. In German it's Leibniz. > Nope - It's Leibnitz-Kekse. > Thomas.de The German language does not use t before z. When I Google for Leibnitz-Kekse, Google asks, Did you mean Leibniz-Kekse? You can find the correct spelling at . -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: Euler Cookies >The German language does not use t before z. When I Google for It does! For example: Witz = joke, Platz = place, Clausewitz (prussian officer and military historian). Except for proper words I don't know any German word ending on iz. >Leibnitz-Kekse, Google asks, Did you mean Leibniz-Kekse? You can >find the correct spelling at >. Yup, they were named after Gottfried Wilhelm von Leibniz who lived at Hannover (the place of production) - I always thought they were named after somebody else. Here's a history of the brand name (in German): http://marken.agentur.de/magazin/markengeschichte?id=26986&cmd=Anzeigen Thomas === Subject: Re: Euler Cookies >>The German language does not use t before z. When I Google for > It does! For example: Witz = joke, Platz = place, Clausewitz (prussian > officer and military historian). Except for proper words I don't know > any German word ending on iz. Yes, you're right, but it seems redundant, since the z alone is pronounced like tz. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: Euler Cookies >The German language does not use t before z. When I Google for >> It does! For example: Witz = joke, Platz = place, Clausewitz (prussian >> officer and military historian). Except for proper words I don't know >> any German word ending on iz. >Yes, you're right, but it seems redundant, since the z alone is >pronounced like tz. Well, spelling just isn't an axiomatic system... Thomas === Subject: Re: Euler Cookies >>The German language does not use t before z. When I Google for >> >It does! For example: Witz = joke, Platz = place, Clausewitz (prussian >officer and military historian). Except for proper words I don't know >any German word ending on iz. Also, jetzt = now, a very common word. -- Stephen J. Herschkorn sjherschko@netscape.net === Subject: Re: Euler Cookies >>Only in English. In German it's Leibniz. >Nope - It's Leibnitz-Kekse. >Thomas.de Shoot! You're right. Thomas === Subject: Re: Euler Cookies > -- > C'est un Nagra. C'est suisse, et tres, tres precis. Should we euler the cookie sheet? (B. Russell, attrib.) === Subject: Re: .99999... still=/= 1 In sci.math, tinyurl.com/uh3t {.3, .33, .333 etc } is 1/3 > Do you understand the difference between a set of numbers and an > individual number? One can admittedly define the numbers in a certain rather interesting way, but I'm not sure it will help S. Enterprize all that much. First, we define the naturals, starting with 1N, in the more or less time-honored tradition of Peano (if one can call such time-honored; Peano at most was in the early 20th century, IIRC). So we get 1N, 2N, 3N, with the usual operations. We then define elements such as 0W, 1W <=> 1N, 2W <=> 2N, ..., and then 0J <=> 0W, 1J <=> 1W, -1J = that element which when added to 1J produces 0J, etc. Q is similarly extended; 1Q <=> 1J / 1J; 1/2Q = 1J / 2J, etc. We can also define 1/2Q <=> 1Q / 2Q. We now have a strongly-typed number system. (A somewhat pointless strongly-typed number system, to be sure, but this sort of logic is forced upon me by too many late-night debugging C++ sessions. :-) ) Nowhere in this system do I automatically assume, say, that 1Q = 1N, or 3J = 3W, though the associations are easily done. (In C++, one can redefine the equality operator thusly: bool operator==(Natural const &, Real const &) {...} to test for equality, assuming one defines classes such as Natural and Real. There are a number of technical issues of course with such definitions -- the biggest one being how one handles very big integers. The best notation might be along the lines of Natural(1) == Real(1.000...) with a lot of machinery to interpret the strings properly. The constructor might be defined class Natural { public: Natural(const char *) { ... } ... } One can also define addition and the other operators: Natural operator+(Natural const &, Natural const &) {...} and ultimately one could do some stuff of mostly theoretical interest.) Now one can define 1/3R, by constructing Dedekind sets or Cauchy limits. One such set might very well be S_3 = {0Q, 3/10Q, 33/100Q, 333/1000Q, ...} and S'_3 = {1Q, 4/10Q, 34/100Q, 334/1000Q, ...} . Since one easily show that, for all s, s' in S_3 and S'_3 respectively, that s < s', and furthermore that abs(s - s') can be made arbitrarily small (but not zero), we get a number r. This number turns out to be 1/3R. So in a way 1/3R = (S_3, S'_3). This isn't exactly the best of definitions of course since one can hypothesize other Dedekind Cut sets (one can take all of the rationals which are negative or whose square is less than 1/9, for example, and the other side is all of the positive rationals whose square is greater than 1/9). Nor can one claim, even if one defines 1/3R = S_3, that 1/3R is an element of its defining set as well, especially if one uses a strongly-typed numbering system (since all elements of S_3 are in Q). I'm not sure how one would prove 1/3Q = 1/3R, though it depends on a number of factors -- mostly how one does the extension from Q to R. It's not usually a problem AFAICT, though. [rest snipped] -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: .99999... still=/= 1 <87acsxynol.fsf@phiwumbda.org> <87k6rz9rb5.fsf@phiwumbda.org> <87fz2mnzj0.fsf@phiwumbda.org> <87hdn1dkv2.fsf@phiwumbda.org> <87acspywc6.fsf@phiwumbda.org> <87d5xhrdc0.fsf@phiwumbda.org> >> It has to do with your objection to the sentence 10*0.9_ = 9.9_ > despite >> it following directly from the axioms of the implied ordered ring. >> Moron. I've never objected to that equation. I've objected to your >> spurious claim that it follows from the distributive property. >> a Sum_{i = 1}^{oo} b_i = Sum_{i=1}^{oo} a b_i >> whenever Sum_{i = 1}^{oo} b_i converges. This is true, but it is not >> true by virtue of the distributive property alone. It requires > proof. > If the series is transfinite then it is not an element of the implied > ordered ring. > Now *that's* a theorem! > Schoenfeld's Summation Theorem: > No infinite sum converges in R. > Proof: R is an ordered ring. Addition in ordered rings is only > defined for finite sums. Therefore, an infinite sum is not an element > of R. > You got it this time, baby. You have just simplified calculus > considerably. They *have* to put your name on this. > -- > Jesse F. Hughes > I'm better than you, and you know it. > -- James Harris That a series is infinite does not imply it's transfinite. === Subject: Re: .99999... still=/= 1 <87acsxynol.fsf@phiwumbda.org> <87k6rz9rb5.fsf@phiwumbda.org> <87fz2mnzj0.fsf@phiwumbda.org> <87hdn1dkv2.fsf@phiwumbda.org> <87acspywc6.fsf@phiwumbda.org> <87d5xhrdc0.fsf@phiwumbda.org> Discussion, linux) It has to do with your objection to the sentence 10*0.9_ = 9.9_ >> despite > it following directly from the axioms of the implied ordered > ring. Moron. I've never objected to that equation. I've objected to > your > spurious claim that it follows from the distributive property. a Sum_{i = 1}^{oo} b_i = Sum_{i=1}^{oo} a b_i whenever Sum_{i = 1}^{oo} b_i converges. This is true, but it is > not > true by virtue of the distributive property alone. It requires >> proof. >> If the series is transfinite then it is not an element of the > implied >> ordered ring. >> Now *that's* a theorem! >> Schoenfeld's Summation Theorem: >> No infinite sum converges in R. >> Proof: R is an ordered ring. Addition in ordered rings is only >> defined for finite sums. Therefore, an infinite sum is not an > element >> of R. >> You got it this time, baby. You have just simplified calculus >> considerably. They *have* to put your name on this. >> -- >> Jesse F. Hughes >> I'm better than you, and you know it. >> -- James Harris > That a series is infinite does not imply it's transfinite. You're a funny guy, but I think I'll let this conversation go now. -- Jesse F. Hughes And a journal can beg me for the right to publish it [...] because I'd rather see it in People magazine [...] --James Harris on his simple proof of Fermat's last theorem === Subject: Re: .99999... still=/= 1 > i think it's worth noting that if you use the dedikind cut def. of the real > numbers then > 1/3 (the real number) = {p in Q : p < 1/3 (the rational number)}. That should be <= 1/3. 1/3 is rational and is conventially included in the left hand side of the cut as the maximum element. The interesting cuts are where the left hand side has no maximum and the right hand side has no minimum. These correspond to the irrational numbers. Bob Kolker > ---- > Justin Young > http://web.syr.edu/~jryoun04/ === Subject: Re: .99999... still=/= 1 >> i think it's worth noting that if you use the dedikind cut def. of the real >> numbers then >> 1/3 (the real number) = {p in Q : p < 1/3 (the rational number)}. > That should be <= 1/3. 1/3 is rational and is conventially included in > the left hand side of the cut as the maximum element. The interesting > cuts are where the left hand side has no maximum and the right hand side > has no minimum. These correspond to the irrational numbers. Not that it makes any real difference, but that is most certainly not the way I learned Dedekind cuts. The left set of the cut is forbidden to contain a maximum element, but the right set is allowed to have a minimum element. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: .99999... still=/= 1 Discussion, linux) >> That should be <= 1/3. 1/3 is rational and is conventially included in >> the left hand side of the cut as the maximum element. The interesting >> cuts are where the left hand side has no maximum and the right hand side >> has no minimum. These correspond to the irrational numbers. > Not that it makes any real difference, but that is most certainly ^^^^^^^^^^^^^^^ You're funny. > not the way I learned Dedekind cuts. The left set of the cut is > forbidden to contain a maximum element, but the right set is allowed > to have a minimum element. -- Jesse F. Hughes Depression hits more people than thought. --headline in Lexington, KY newspaper, as reported on NPR's Morning Edition === Subject: Re: .99999... still=/= 1 > That should be <= 1/3. 1/3 is rational and is conventially included in > the left hand side of the cut as the maximum element. The interesting > cuts are where the left hand side has no maximum and the right hand side > has no minimum. These correspond to the irrational numbers. >> Not that it makes any real difference, but that is most certainly > ^^^^^^^^^^^^^^^ You're funny. >> not the way I learned Dedekind cuts. The left set of the cut is >> forbidden to contain a maximum element, but the right set is allowed >> to have a minimum element. The difference is purely imaginary. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: .99999... still=/= 1 > Not that it makes any real difference, but that is most certainly not the > way I learned Dedekind cuts. The left set of the cut is forbidden to > contain a maximum element, but the right set is allowed to have a minimum > element. That is the other convention. Either one gets you the same thing. Bob Kolker === Subject: Re: .99999... still=/= 1 >> Not that it makes any real difference, but that is most certainly not the >> way I learned Dedekind cuts. The left set of the cut is forbidden to >> contain a maximum element, but the right set is allowed to have a minimum >> element. > That is the other convention. Either one gets you the same thing. That is exactly why I said it doesn't make any real difference. I notice you snipped the context in which you attempted to correct someone for committing the grevious error of using the opposite convention. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: .99999... still=/= 1 >Not that it makes any real difference, but that is most certainly not the >way I learned Dedekind cuts. The left set of the cut is forbidden to >contain a maximum element, but the right set is allowed to have a minimum >element. >>That is the other convention. Either one gets you the same thing. > That is exactly why I said it doesn't make any real difference. My apologies. I misunderstood you. Bob Kolker === Subject: Prove problems for convex set Hi all, I have two problems that I don't know how to prove it. I am only an undergraduate student and I don't know why my prof raised a such of tough questions to us. Could anyone help to prove it? (1)Given that X is a closed convex set in R^n, prove that a point y either belongs to X or there exists a hyperplane including y such that X is contained in an open half-space bounded by the hyperplane. (2)Suppose that D is an open convex subset of R and that A is any compact convex subset of D. Given that f:D->R is C^2 and quasi-concave and that f'(x)>0 fo all x belongs to D, prove that there exists an a>0 such that g:A->R defined by g(x)=-e^-af(x) is concave. === Subject: Re: Prove problems for convex set > Hi all, > (2)Suppose that D is an open convex subset of R and that A is any > compact convex subset of D. Given that f:D->R is C^2 and quasi-concave > and that f'(x)>0 fo all x belongs to D, > prove that there exists an a>0 such that g:A->R defined by > g(x)=-e^-af(x) is concave. (2) Let h = -g. It suffices to determine a value of a such that h > 0 on A. h'(x) = e^(-af(x))(-a)f'(x). h(x) = -a(e^(-af(x))(-a)f'(x)f'(x) + e^(-af(x))f(x)) = ae^(-af(x))(a(f'(x))^2 - f(x)). f' has a positive minimum m, and f has a maximum M on A, since f is C^2 and f' > 0 on A. Therefore, there exists a sufficiently large positive a such that am^2 - M > 0, so that a(f'(x))^2-f(x)> 0 on A for such a. Also, ae^(-af(x)) > 0, so that h > 0 on A. Note that D is just an open interval, A is a closed interval, and f being quasi concave is not necessary. For students of mathematical economics, I recommend C. Berge, Topological Spaces (Dover). Takao === Subject: Re: Prove problems for convex set Takao, understand it......) I'll try my best to understand it and you really helped me a lot for the assignment. How about part (1)of the question? did you know how to do it? I'm just discussing with my friends about this assignment....and we find it is really really difficult! (Actually we've got another 4 more questions that's in this level) Anyhow, it's sure that I could not get the book that you mentioned by tonight (Cause the assignment is due tomorrow noon) Really hope part (1) could also be solved === Subject: Re: Prove problems for convex set First, concerning (2), f'> 0 implies f is increasing, so that f is quasi concave and quasi convex. Therefore, the condition f' > 0 is stronger than the condition that f is quasi concave. Concerning (1), you can draw a closed convex set and a point outside it on a paper and follow the instruction given by Ullrich to understand the solution intuitively. I will write the proof following Berge's book p.162. Without loss of generality consider the origin and a closed convex set X that does not contain the origin. Let B be a closed ball with the origin as its center such that the intersection of X and B is nonempty. Let this intersection, which is compact, be C. We consider the continuous function g(x) = |x|, the distance between x and the origin, defined on X. Since C is compact, g attains a positive minimum at some point c. |x| >= |c|> 0 for every x in C, so that |x| >= |c|> 0 for every x in X. It suffices to prove (c,x) >= |c|^2 for every x in X, because, this implies (c,x) > 0 for every x in X and hence X is contained in an open half-space bounded by the hyperplane {x | (c,x) = 0}. Suppose to the contrary that there exists some y in X such that (c,y) = |c|^2 - d for some d > 0. Let t be a number between 0 and 1. Then |(1-t)c+ty|^2 =((1-t)c+ty, (1-t)c +ty) (inner product) = (1-t)^2|c|^2 + 2t(1-t)(c,y) + t^2|y|^2 = |c|^2 -2t|c|^2 + t^2|c|^2 + 2t(1-t){|c|^2 - d} + t^2|y|^2, so that |(1-t)c+ty|^2 - |c|^2 = -2t(1-t)d + t^2{|y|^2 -|c|^2} = t(-2d +t(|y|^2 - |x|^2 + 2d)), where |y|^2 - |x|^2 + 2d > 0. Therefore, |(1-t)c+ty|^2 - |c|^2 < 0 for 0 < t < 2d/(|y|^2 - |x|^2 + 2d). However, |(1-t)c+ty|^2 - |c|^2 >= 0 to the contrary, since (1-t)c+ty is a point in X. Therefore, (c,x) >= |c|^2 for every x in X. Takao === Subject: Re: Prove problems for convex set Really thank you very much! I've handed in the assignment today....I found out that there's no one could done all the questions. === Subject: Re: Prove problems for convex set >Hi all, >I have two problems that I don't know how to prove it. I am only an >undergraduate student and I don't know why my prof raised a such of >tough questions to us. >Could anyone help to prove it? >(1)Given that X is a closed convex set in R^n, prove that a point y >either belongs to X or there exists a hyperplane including y such that >X is contained in an open half-space bounded by the hyperplane. Suppose y is not in X. Let x be a point of X that is closest to y. Consider the hyperplane through y perpendicular to y - x. >(2)Suppose that D is an open convex subset of R and that A is any >compact convex subset of D. Given that f:D->R is C^2 and quasi-concave >and that f'(x)>0 fo all x belongs to D, >prove that there exists an a>0 such that g:A->R defined by >g(x)=-e^-af(x) is concave. Pretend you're a calculus student. ************************ David C. Ullrich === Subject: Re: Prove problems for convex set hi David C. Ullrich, student.....I'm only majoring in economics. Also, I really have no idea about how to prove such qestions as all of the idea I learn is from the book, our professor even didn't teah anything.... Could you explain more in detail (or steps by steps) for me that how to prove it? And which section of the math book or are there any websites that I could find the related prove? Since I'm not very familiar with Calculus. Sorry for my poor languages(as I'm not a native English speaker) Hope you know what I mean and I am looking forward to your reply. === Subject: Re: Prove problems for convex set >hi David C. Ullrich, >student.....I'm only majoring in economics. Also, I really have no idea >about how to prove such qestions as all of the idea I learn is from the >book, our professor even didn't teah anything.... >Could you explain more in detail (or steps by steps) for me that how to >prove it? This is _your_ assignment, right? >And which section of the math book or are there any websites >that I could find the related prove? Since I'm not very familiar with >Calculus. >Sorry for my poor languages(as I'm not a native English speaker) >Hope you know what I mean and I am looking forward to your reply. ************************ David C. Ullrich === Subject: Re: Prove problems for convex set yes......I've try to find some books in the library..........but I am too confused that the example shown in those books didn't give out any hints for this question. I am now very flustrated about it! :( === Subject: Re: Re: Re: Don't get it by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iBC4tMP31710; I am sorry, sir, I do not see anything written in the previous message, could you please explain?. > I would appreciate your help with the following: > I had a question about a space-filling curve f:[0,1]->[0,1]^3, > I know that It is continuous on a subset of [0,1] (which I > do not think is dense in [0,1], so that the extension is not > trivial in this sense) > The function is defined on the subset of [0,1] given by : > S={ 0.a_1b_1c_10a_2b_2c_20.....0a_nb_nc_n0..... > > with a_i,b_i,c_i in {0,1,...,9} , and every fourth place is > 0} . The function f is given by: > f(0.a_1b_1c_10a_2b_2c_20.....0a_nb_nc_n0..)= > (0.a_1a_2a_3....., 0.b_1b_2b_3....b_n... ,0.c_1c_2c_3....) > Then f is continuous in S ( let's assume that). > What I would really appreciate your help with is, in > finding an explicit representation of an extension of > f to [0,1]S ( S as above), so that f is continuous > in [0,1]S/S=[0,1]. > I could tell that S is closed in [0,1],so that [0,1]S > is open, so that it is the countable union of open intervals, > so that we can define f in each open interval. >>Say (a,b) is one of these open intervals. Then you know what >>f(a) and f(b) are - you really can't think of a way to define >>f(x) for x in (a,b) so that f is continuous on [a,b]? > I can think of doing it linearly, I guess, by defining the > function in this interval to be the line joining f(a) > to f(b). > y-f(b)=[(f(b)-f(a))/(b-a)]*(x-f(a)) > Would this be correct? > I would really be grateful for any help in finding > this explicit continuation. > > >>************************ >>David C. Ullrich === Subject: Re: Don't get it > I am sorry, sir, I do not see anything written in the > previous message, could you please explain?. If you can't even read a reply closely enough to see what was this: Say (a,b) is one of these open intervals. Then you know what f(a) and f(b) are - you really can't think of a way to define f(x) for x in (a,b) so that f is continuous on [a,b]? >> I would appreciate your help with the following: >> I had a question about a space-filling curve f:[0,1]->[0,1]^3, >> I know that It is continuous on a subset of [0,1] (which I >> do not think is dense in [0,1], so that the extension is not >> trivial in this sense) >> The function is defined on the subset of [0,1] given by : >> S={ 0.a_1b_1c_10a_2b_2c_20.....0a_nb_nc_n0..... >> >> with a_i,b_i,c_i in {0,1,...,9} , and every fourth place is >> 0} . The function f is given by: >> f(0.a_1b_1c_10a_2b_2c_20.....0a_nb_nc_n0..)= >> (0.a_1a_2a_3....., 0.b_1b_2b_3....b_n... ,0.c_1c_2c_3....) >> Then f is continuous in S ( let's assume that). >> What I would really appreciate your help with is, in >> finding an explicit representation of an extension of >> f to [0,1]S ( S as above), so that f is continuous >> in [0,1]S/S=[0,1]. >> I could tell that S is closed in [0,1],so that [0,1]S >> is open, so that it is the countable union of open intervals, >> so that we can define f in each open interval. >Say (a,b) is one of these open intervals. Then you know what >f(a) and f(b) are - you really can't think of a way to define >f(x) for x in (a,b) so that f is continuous on [a,b]? >> I can think of doing it linearly, I guess, by defining the >> function in this interval to be the line joining f(a) >> to f(b). >> y-f(b)=[(f(b)-f(a))/(b-a)]*(x-f(a)) >> Would this be correct? >> I would really be grateful for any help in finding >> this explicit continuation. >> >> >************************ >David C. Ullrich ************************ David C. Ullrich === Subject: Re: Hilbert Spaces and L^p by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iBC4tMH31720; I am sorry, I was very unclear. I meant to say that the only L^p, l^p spaces that are also Hilbert Spaces are, resp. L^2 and l^2 I was hoping you could give me a hint of the role that the paralellogram rule plays in being able to define a norm in these spaces. >> Hi. I am trying to understand better why L^2 and l^2 are the >> only L^p and l^p spaces resp. I know it has something to see >> with the inner product satisfying the paralellogram rule >> ( and this is an iff condition, and one side of it is easy), >> but I am not clear on what role the parallelogram rule >> plays in here; I would appreciate any explanation here. >My guess is that you meant to ask this: why are L^2 and l^2 the only >L^p and l^p spaces resp. whose norm is induced by an inner product? If >I am right then your suggestion is correct: because the other norms do >not satisfy the paralellogram law. >Take l^1, for instance. Consider the sequences a = (1,0,0,0,0,...) and >b = (0,1,0,0,0,...). Then > ||a + b||^2 + ||a - b||^2 = 8 >and > 2||a||^2 + 2||b||^2 = 4. >Therefore, the norm ||.||_1 does not satisfy the paralellogram law. >Jose Carlos Santos === Subject: Re: Hilbert Spaces and L^p > I am sorry, I was very unclear. I meant to say that the > only L^p, l^p spaces that are also Hilbert Spaces are, resp. > L^2 and l^2 > I was hoping you could give me a hint of the role that the > paralellogram rule plays in being able to define a norm in these > spaces. Please don't top-post. If you want to know what's that and why you shouldn't do it, read http://www.caliburn.nl/topposting.html or http://www.html-faq.com/etiquette/?toppost Having said this, you were still more unclear this time. First of all, you do not say whether or not my answer was what you wanted. Besides, your question makes no sense at all. What do you mean by the role that the paralellogram rule plays in being able to define a norm in these spaces? If the paralellogram law holds that you must already have a norm. Why should you want to *define* one after that? Jose Carlos Santos