mm-2239 === Subject: [JSH] Experiment in psycho-mathematics? It seems that James has abandoned all attempts to answer his critics with reasoning or to find any fault with the arguments which conclusively refute his so-called proof. Instead, he has embarked on a saturation-bombing strategy in which he appears to believe that if he posts the same, shop-worn, oft-refuted, pseudo-logical phrases machine gun style he will defeat his critics using argument-by-repetition. This is a style of argument historically advocated by ruthless dictators to promote political propaganda. His last few threads repeat over and over the same refuted arguments he has posted before. Maybe his new technique, using propagandists tools instead of technologists tools will work. He may garner converts to his cause by simple lambasting them endlessly until they cry out, Hey, maybe James is right. Um, he certainly wouldn't lie would he? Could this be the 'method in his madness'? -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: [JSH] Experiment in psycho-mathematics? > It seems that James has abandoned all attempts to answer his critics > with reasoning or to find any fault with the arguments which > conclusively refute his so-called proof. Instead, he has embarked on a > saturation-bombing strategy in which he appears to believe that if he > posts the same, shop-worn, oft-refuted, pseudo-logical phrases machine > gun style he will defeat his critics using argument-by-repetition. This > is a style of argument historically advocated by ruthless dictators to > promote political propaganda. > His last few threads repeat over and over the same refuted arguments he > has posted before. Maybe his new technique, using propagandists tools > instead of technologists tools will work. He may garner converts to his > cause by simple lambasting them endlessly until they cry out, Hey, > maybe James is right. Um, he certainly wouldn't lie would he? > Could this be the 'method in his madness'? Another suspicion comes to mind. What if he is in reality a sociology or psychology professor, and uses this ng as an object of study? After all, the JSH threads show close resemblance to a group of monkeys beating the omega-male. Now, assuming that this is the case, where would the results of this research end up? Not in journals, since this is going on for years, now, and nothing has shown up, afaik. So, one may wonder: how many names of JSH posters will be in CIA, KGB or Mossad files, by now? Herman Jurjus === Subject: A combinatorics questions Hi all, I have a table of integers of size N x M. Think of the table as a rectangle with a grid of integers. I run an algorithm over this table which runs like this: 1) Given some criteria, take away two rectangles (in such a way the remaining space form two rectangles as well). These correspond to 50% of the table. These will no longer be of interest. Now we have two sub-rectangles left. 2) For each sub-rectangle, go to 1), recursively. When there is only one integer left, the search in the current sub-rectangle is done. Question is: What is the running time for such an algorithm? log(N)^2 ? === Subject: Re: A combinatorics questions > I have a table of integers of size N x M. Think of the table as a > rectangle with a grid of integers. > I run an algorithm over this table which runs like this: > 1) Given some criteria, take away two rectangles (in such a way the > remaining space form two rectangles as well). These correspond to 50% > of the table. These will no longer be of interest. In the case of a 3x3 rectangle it's impossible to take away two rectangles as describe which cover exactly 50% of the table's space or area. A 2x3 rectangle can only have a 1x1 and a 1x2 rectangle removed leaving a 1x1 and a 1x2 subrectangle. Tho it's an even split... > Now we have two sub-rectangles left. > 2) For each sub-rectangle, go to 1), recursively. When there is only > one integer left, the search in the current sub-rectangle is done. ... when a 1x2 rectangle has two rectangles removed, it will be empty. > Question is: What is the running time for such an algorithm? > log(N)^2 ? Algorithm description inaccurate. Running time for 'criteria' not given. Proposed running time doesn't include M. As a binary tree search runs in log n time, a double binary search over a nxm grid should take log mn = log m + log n time. === Subject: Re: A combinatorics questions http://mygate.mailgate.org/mynews/comp/comp.theory/89e801690ffe7ff621e87e947 c 9dd387.48257%40mygate.mailgate.org > Hi all, > I have a table of integers of size N x M. Think of the table as a > rectangle with a grid of integers. > I run an algorithm over this table which runs like this: > 1) Given some criteria, take away two rectangles (in such a way the > remaining space form two rectangles as well). These correspond to 50% > of the table. These will no longer be of interest. > Now we have two sub-rectangles left. > 2) For each sub-rectangle, go to 1), recursively. When there is only > one integer left, the search in the current sub-rectangle is done. > Question is: What is the running time for such an algorithm? > log(N)^2 ? Well, if you were only talking about the non-some criteria part of the algorithm, then log(N)+log(M), as was answered elsewhere. But if your some criteria has to look at all the cells of the removed and remaindered rectangles, then the N*M order for the top level at least dominates, and (I'm too lazy to do the math) there may be an order log multiplier times that to do all the cycles. xanthian. -- === Subject: Re: A crank post about Chaitins constant > The page http://www.wikipedia.org/wiki/Chaitin%27s_constant describes > Chaitin's constant more as a construction than as a single constant: since > its definition depends on the arbitrary choice of computation model and > program encoding > So I wonder, how many Chaitin's constants are there? The definition given is Chaitin's constant is the probability that a randomly produced string will represent a program that, when run, will eventually halt. However, consider the following program: 1. Read in nonnegative integer N 2. If N=2 then Halt 3. Initialize scratch value X equal to 3 4. If N The definition given is Chaitin's constant is the probability that a > randomly produced string will represent a program that, when run, will > eventually halt. However, consider the following program: > 1. Read in nonnegative integer N > 2. If N=2 then Halt > 3. Initialize scratch value X equal to 3 > 4. If N 5. Double X > 6. If N 7. Double X > 8. Go to step 4 > What is the probability that this program halts? Answer: There is > none. The probability does not converge. I think you misunderstand Chaitin's argument. By program he means any input to his chosen machine, perhaps a universal Turing machine. He then scales the result according to the length of the input. If the input is of length n then it contributes 2^{-n} to his number if the machine halts, and 0 if it does not. His use of the word program is somewhat confusing. In normal parlance it means program + input. -- Timothy Murphy e-mail: tim@birdsnest.maths.tcd.ie tel: +353-86-233 6090 s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland === Subject: Re: A crank post about Chaitins constant > The definition given is Chaitin's constant is the probability that a > randomly produced string will represent a program that, when run, will > eventually halt. However, consider the following program: > > 1. Read in nonnegative integer N > 2. If N=2 then Halt > 3. Initialize scratch value X equal to 3 > 4. If N 5. Double X > 6. If N 7. Double X > 8. Go to step 4 > > What is the probability that this program halts? Answer: There is > none. The probability does not converge. > I think you misunderstand Chaitin's argument. > By program he means any input to his chosen machine, > perhaps a universal Turing machine. Actually, it is a universal Turing Machine, so that the input represents both the Turing Machine (which needs an initial value on the tape) and the initial value on the tape (coded as one value). Thus the random input represents a random (Turing Machine,its input). > He then scales the result according to the length of the input. > If the input is of length n > then it contributes 2^{-n} to his number if the machine halts, > and 0 if it does not. > His use of the word program is somewhat confusing. > In normal parlance it means program + input. Certainly, since the Turing model of computing requires exactly one input. (A model need not have an input. For example, using Logic as a base of computing, rather than running a program+input we run a sentence (wff with no free variables) by trying to prove it axiomatically. To simulate input we construct a sentence from a wff with any number of free variables plus a tuple of numbers, which are the input, whose symbolic representations are substituted for these free variables.) My point is that the notion of the probability of a random program halting is not defined for some specific programs, so it is not defined for programs in general. As I mentioned in my original post, the coin-flipping procedure used by Chaitin (used to produce the scaling that you describe above) itself also produces anomalous results, such as determining that the probability that a random non-negative integer is even to be 1/3. Charlie Volkstorf Cambridge, MA === Subject: Advanced polynomial factorization, demonstration Finding factorizations is the key to my work and understanding how I used a special technique to factor polynomials is key to understanding it. I'm going to talk about how to use additional variables to get find out information about a polynomial that is irreducible over Q. That polynomial is 65x^3 - 12x + 1 and I'll be using a special construction to factor it. But before that I want to give an analogy using integers. Consider if you wanted to just find some factorization of 3, you could add a variable, and use x^2 + 3x + 2 = 3, as then you have x^2 + 3x - 1 = 0, and can simply solve for x and plug one of the solutions back into (x+1)(x+2) = 3, and have a factorization. Now in considering 65x^3 - 12x + 1, I'm going to use several variables to do something similar, as I'll use (v^3+1)x^3 - 3vxy^2 + y^3 = 65x^3 - 12x + 1. Clearly v=4, and y=1, are values for the variables v and y that work. Now concentrating on the expression (v^3+1)x^3 - 3vxy^2 + y^3 I'm going to add more variables, by letting v= -1 + mf^2. If you have to have a ring, assume it's algebraic integers. If you want specifics about the variables, assume m is a nonzero integer, and f is a nonzero algebraic integer coprime to 3 and m. Now then substituting for v gives ((-1 + mf^2)^3 + 1)x^3 - 3(-1 + mf^2)xy^2 + y^3 which is (m^3 f^6 - 3m^2 f^4 + m f^2 ) x^3 - 3(-1 + mf^2 )xy^2 + y^3 and now let y = uf, where u is an algebraic integer coprime to f, and using that substitution gives (m^3 f^6 - 3m^2 f^4 + m f^2 ) x^3 - 3(-1 + mf^2)x u^2 f^2 + (uf)^3 which is f^2 ((m^3 f^4 - 3m^2 f^2 + m) x^3 - 3(-1 + mf^2 )x u^2 + u^3 f). Now let P(m) = f^2((m^3 f^4 - 3m^2 f^2 + m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) and consider that the constant term is P(0) = f^2(3x u^2 + u^3 f^j). Then P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f. Now consider P(m)/f^2 = (b_1 x + u)(b_2 x + u)(b_3 x + uf) as P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + m) x^3 - 3(-1 + mf^2)u^2 x + u^3 f so b_1 b_2 b_3 = (m^3 f^{4j} - 3m^2 f^{2j} + m) and of course if m=0, at least one of the b's must equal 0, and at m=0 P(0)/f^{2j} = 3x u^2 + u^3 f = u^2(3x + uf) which is consistent with P(m)/f^2 = (b_1 x + u)(b_2 x + u)(b_3 x + uf) showing that two of the b's equal 0, when m = 0. Now several posters have disagreed with that conclusion, and instead have claimed that some other factorization is possible, like P(m)/f^2 = (b_1 x + uw_1)(b_2 x + uw_2)(b_3 x + uw_3) where w_1, w_2, and w_3 are *each* non-unit factors of f. However, notice that would give P(0)/f^2 = w_1 w_2 u^2 (b_3 x + u w_3) which at best is P(0)/f^2 = u^2 (b_3 w_1 w_2 x + u f) when in fact it is P(0)/f^2 = 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. So there really never was any room for discussions about definitions of polynomials or anything of that nature. The mathematics was *always* straightforward, while the hostility towards the truth from mathematicians has been clear as well. It never was very complicated. James Harris === Subject: Re: Advanced polynomial factorization, demonstration > Finding factorizations is the key to my work and understanding how I > used a special technique to factor polynomials is key to understanding > it. > I'm going to talk about how to use additional variables to get find > out information about a polynomial that is irreducible over Q. > That polynomial is > 65x^3 - 12x + 1 > and I'll be using a special construction to factor it. But before > that I want to give an analogy using integers. thing pertaining to this polynomial. Maybe it's because I refuse to read the trash that JSH spews in the name of mathematics, especially when it comes to his so-called extra variables, and the nonsense that allows him to claim to have proven 1 = 2, or some such falsehood. I don't care, but I was sure to catch some of that hell-fire and gallstone stuff that our little JSHter is so prone to producing. Anyhow, since there was no mathematics, and no entertainment, I might as well leave the rest of this crap out of it. Sorry for the interruption. .... embarassing interlude of juicy flatulence omitted ... > So there really never was any room for discussions about definitions > of polynomials or anything of that nature. > The mathematics was *always* straightforward, while the hostility > towards the truth from mathematicians has been clear as well. > It never was very complicated. Nor you, I'm sure. > James Harris Dale === Subject: Re: Advanced polynomial factorization, demonstration > Finding factorizations is the key to my work and understanding how I > used a special technique to factor polynomials is key to understanding > it. > I'm going to talk about how to use additional variables to get find > out information about a polynomial that is irreducible over Q. > That polynomial is > 65x^3 - 12x + 1 > and I'll be using a special construction to factor it. But before > that I want to give an analogy using integers. > Consider if you wanted to just find some factorization of 3, you could > add a variable, and use > x^2 + 3x + 2 = 3, as then you have x^2 + 3x - 1 = 0, > and can simply solve for x and plug one of the solutions back into > (x+1)(x+2) = 3, and have a factorization. > Now in considering 65x^3 - 12x + 1, I'm going to use several variables > to do something similar, as I'll use > (v^3+1)x^3 - 3vxy^2 + y^3 = 65x^3 - 12x + 1. > Clearly v=4, and y=1, are values for the variables v and y that work. > Now concentrating on the expression > (v^3+1)x^3 - 3vxy^2 + y^3 > I'm going to add more variables, by letting v= -1 + mf^2. > If you have to have a ring, assume it's algebraic integers. > If you want specifics about the variables, assume m is a nonzero > integer, and f is a nonzero algebraic integer coprime to 3 and m. > Now then substituting for v gives > ((-1 + mf^2)^3 + 1)x^3 - 3(-1 + mf^2)xy^2 + y^3 > which is > (m^3 f^6 - 3m^2 f^4 + m f^2 ) x^3 - 3(-1 + mf^2 )xy^2 + y^3 > and now let y = uf, where u is an algebraic integer coprime to f, A 'minor' point here is that in your original polynomial, the constant term is 1, which is not equal to (uf)^3. > and > using that substitution gives > (m^3 f^6 - 3m^2 f^4 + m f^2 ) x^3 - > 3(-1 + mf^2)x u^2 f^2 + (uf)^3 > which is > f^2 ((m^3 f^4 - 3m^2 f^2 + m) x^3 - > 3(-1 + mf^2 )x u^2 + u^3 f). > Now let > P(m) = f^2((m^3 f^4 - 3m^2 f^2 + m) x^3 - > 3(-1+mf^2 )x u^2 + u^3 f) > and consider that the constant term is > P(0) = f^2(3x u^2 + u^3 f^j). > Then > P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + m) x^3 - > 3(-1+mf^2 )x u^2 + u^3 f. > Now consider > P(m)/f^2 = (b_1 x + u)(b_2 x + u)(b_3 x + uf) > as > P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + m) x^3 - > 3(-1 + mf^2)u^2 x + u^3 f > so b_1 b_2 b_3 = (m^3 f^{4j} - 3m^2 f^{2j} + m) > and of course if m=0, at least one of the b's must equal 0, and at m=0 > P(0)/f^{2j} = 3x u^2 + u^3 f = u^2(3x + uf) > which is consistent with > P(m)/f^2 = (b_1 x + u)(b_2 x + u)(b_3 x + uf) > showing that two of the b's equal 0, when m = 0. Let me emphasize that a bit: > when m = 0. <<<<<<<< > Now several posters have disagreed with that conclusion, No - it's OK >> when m = 0. <<<<<< > and instead > have claimed that some other factorization is possible, > when m <> 0 <<<<<<< >like > P(m)/f^2 = (b_1 x + uw_1)(b_2 x + uw_2)(b_3 x + uw_3) > where w_1, w_2, and w_3 are *each* non-unit factors of f. > However, notice that would give > P(0)/f^2 = w_1 w_2 u^2 (b_3 x + u w_3) > which at best is > P(0)/f^2 = u^2 (b_3 w_1 w_2 x + u f) > when in fact it is > P(0)/f^2 = 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. Again: > when m = 0. <<<<<<< Everything you have just said applies > when m = 0. <<<<<<< and *under that condition*, I do not object to it. > So there really never was any room for discussions about definitions > of polynomials or anything of that nature. Somehow you think that you have proved that the factorization which is true in the degenerate, degree-1 case, > when m = 0 <<<<<<< must hold in the nondegenerate case, > when m <> 0. <<<<<<< But you have no such proof. Note that all of your argument above is restricted to the m = 0 case. You KNOW that the coefficients are functions of m. When m = 0, b1 and b2 are 0. No one disagrees with that. When m <> 0, you don't know what they are, except they very definitely are not 0. You think: when m = 0, b1 = b2 = 0, so b1/f and b2/f are both zero and are both algebraic integers. Therefore in general, when m <> 0, b1/f and b2/f must be algebraic integers. You think in terms of formulas. You think that since the formula b1/f works when m = 0, it must also work when m <> 0. You think the denominator f must work for all values of m, not just 0. But what if the formula is more complicated? Instead of the denominator being f, what if it is of the form d(m) = m * q(m) + (1 - m) * f where q(m) is some other factor of f^2 that depends on m? Note that when m = 0, d(m) = f. But when m = 1, for example, d(m) = d(1) = q(1). In that case, when m = 0, b1/d(m) = b1/f, but when m = 1, you get b1/d(m) = b1/q(1). Here q(1) could be some other algebraic integer factor of f^2. What's to prevent it? Bottom line: the 'formula' that you prefer, b1/f may be true when m = 0, but that could be a special case of a formula which is very different when m <> 0. All you have done is produce a formula when m = 0. You have NOT proved, by any stretch, that the same formula, or ANY formula, applies when m <> 0. It is just a leap of faith. And given that several of us have independently posted different proofs that your conclusion is dead wrong (and you have not refuted any of them), that leap of faith is very, very likely where you are making your mistake. Nora B. > The mathematics was *always* straightforward, while the hostility > towards the truth from mathematicians has been clear as well. > It never was very complicated. > James Harris === Subject: Re: Advanced polynomial factorization, demonstration > Finding factorizations is the key to my work and understanding how I > used a special technique to factor polynomials is key to understanding > it. > Now in considering 65x^3 - 12x + 1, I'm going to use several variables > to do something similar, as I'll use > (v^3+1)x^3 - 3vxy^2 + y^3 = 65x^3 - 12x + 1. > Clearly v=4, and y=1, are values for the variables v and y that work. v^3 + 1 = 65 3vy^2 = 12 y^3 = 1 This is a system of 3 equations with 2 unknowns which may have no solutions. How are you going to guarantee that it is solvable? > If you want specifics about the variables, assume m is a nonzero > integer, and f is a nonzero algebraic integer coprime to 3 and m. > P(m) = f^2((m^3 f^4 - 3m^2 f^2 + m) x^3 - > 3(-1+mf^2 )x u^2 + u^3 f) > and consider that the constant term is > P(0) = f^2(3x u^2 + u^3 f^j). How did you switch from x to m as the indeterminate (especially as you imply that it is constant)? Also: setting m=0 in the above polynomial yields -3XU^2 + 3U^3F > Then > P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + m) x^3 - > 3(-1+mf^2 )x u^2 + u^3 f. Ummm, NO! You did not divide the constant term by f^2. -Tralfaz === Subject: Re: Advanced polynomial factorization, demonstration > Finding factorizations is the key to my work and understanding how I > used a special technique to factor polynomials is key to understanding > it. > I'm going to talk about how to use additional variables to get find > out information about a polynomial that is irreducible over Q. > That polynomial is > 65x^3 - 12x + 1 > and I'll be using a special construction to factor it. But before > that I want to give an analogy using integers. Note: You are making a claim that SOMEWHERE you will list the factorization of this polynomial... The problem is scanning ahead I don't see that factorization anywhere. > Consider if you wanted to just find some factorization of 3, you could > add a variable, and use > x^2 + 3x + 2 = 3, as then you have x^2 + 3x - 1 = 0, > and can simply solve for x and plug one of the solutions back into > (x+1)(x+2) = 3, and have a factorization. > Now in considering 65x^3 - 12x + 1, I'm going to use several variables > to do something similar, as I'll use > (v^3+1)x^3 - 3vxy^2 + y^3 = 65x^3 - 12x + 1. > Clearly v=4, and y=1, are values for the variables v and y that work. Ok, so there is at least one choice of v and y that works. Do you intend for us to *keep* these as our working values or just keep in mind that v and y satisfy: v^3+1=65 3vy^2=12 y^3=1 ? > Now concentrating on the expression > (v^3+1)x^3 - 3vxy^2 + y^3 > I'm going to add more variables, by letting v= -1 + mf^2. Where v = 4 or any other solution to the above system. > If you have to have a ring, assume it's algebraic integers. What would it mean if we don't have a ring? > If you want specifics about the variables, assume m is a nonzero > integer, and f is a nonzero algebraic integer coprime to 3 and m. Ok, m<>0 > Now then substituting for v gives > ((-1 + mf^2)^3 + 1)x^3 - 3(-1 + mf^2)xy^2 + y^3 > which is > (m^3 f^6 - 3m^2 f^4 + m f^2 ) x^3 - 3(-1 + mf^2 )xy^2 + y^3 > and now let y = uf, where u is an algebraic integer coprime to f, and > using that substitution gives Are we using y=1 here? If so, the only way u and f can both be algebraic integers is if they are both units. > (m^3 f^6 - 3m^2 f^4 + m f^2 ) x^3 - > 3(-1 + mf^2)x u^2 f^2 + (uf)^3 > which is > f^2 ((m^3 f^4 - 3m^2 f^2 + m) x^3 - > 3(-1 + mf^2 )x u^2 + u^3 f). > Now let > P(m) = f^2((m^3 f^4 - 3m^2 f^2 + m) x^3 - > 3(-1+mf^2 )x u^2 + u^3 f) > and consider that the constant term is > P(0) = f^2(3x u^2 + u^3 f^j). You can't do this. You said m<>0. What is m, anyway? > Then > P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + m) x^3 - > 3(-1+mf^2 )x u^2 + u^3 f. Ok, you divided by a unit. > Now consider > P(m)/f^2 = (b_1 x + u)(b_2 x + u)(b_3 x + uf) > as > P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + m) x^3 - > 3(-1 + mf^2)u^2 x + u^3 f > so b_1 b_2 b_3 = (m^3 f^{4j} - 3m^2 f^{2j} + m) > and of course if m=0, at least one of the b's must equal 0, and at m=0 But you said m<>0. Which is it? > P(0)/f^{2j} = 3x u^2 + u^3 f = u^2(3x + uf) > which is consistent with > P(m)/f^2 = (b_1 x + u)(b_2 x + u)(b_3 x + uf) > showing that two of the b's equal 0, when m = 0. But m <> 0. > Now several posters have disagreed with that conclusion, and instead > have claimed that some other factorization is possible, like > P(m)/f^2 = (b_1 x + uw_1)(b_2 x + uw_2)(b_3 x + uw_3) > where w_1, w_2, and w_3 are *each* non-unit factors of f. But it looks like f is a factor. > However, notice that would give > P(0)/f^2 = w_1 w_2 u^2 (b_3 x + u w_3) > which at best is > P(0)/f^2 = u^2 (b_3 w_1 w_2 x + u f) > when in fact it is > P(0)/f^2 = 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. > So there really never was any room for discussions about definitions > of polynomials or anything of that nature. > The mathematics was *always* straightforward, while the hostility > towards the truth from mathematicians has been clear as well. > It never was very complicated. > James Harris So where's the factorization of 65x^3 - 12x + 1? Also, it's much less complicated when you drop conditions on your variables as you go from constants to variables. Keeping those conditions in mind makes things MUCH more complicated. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Advice on research I was hoping some of you will be able to shed light on the process of research. I am new to it and have been working at a problem for the past 6 months or so without success. I am in the process of getting an advanced degree in mathematics. How long do you spend on problems before say, you publish them ? Is it better to have several problems in the play at the same time ? What do you do ? How many problems do you tackle ? Is it better to keep butting one's head against the wall or move onto some other problem ? Any interesting experiences, ideas and suggestions would be welcome especially from active mathematicians. Arvind === Subject: Re: Advice on research > I was hoping some of you will be able to shed light on > the process of research. I am new to it and have been > working at a problem for the past 6 months or so without success. > I am in the process of getting an advanced degree in mathematics. > How long do you spend on problems before say, you publish them ? > Is it better to have several problems in the play at the same time ? > What do you do ? How many problems do you tackle ? > Is it better to keep butting one's head against the wall or move > onto some other problem ? > Any interesting experiences, ideas and suggestions would be welcome > especially from active mathematicians. Gauss's motto was Few, but ripe. === Subject: a=e^(log a), Is it always correct ? I think a=e^(ln a) is correct but, a=e^(log a) is not. (If a=e^(log a) is correct, then 10=e^log 10 ?) Because, the base of ln is e but the base of log is 10. Is there any wrong point about my thinking ? In my complex analysis book, the following is given, log z = ln r + i*(Arg(z)+2pi*n), n in integers, where |z|=r. As for me, I cannot agree with above equality. If z=re^i*t then lnz=lnr+ln(e^i*t) =>lnz=lnr + ln(Arg(z)+2pi*n) I agree this, but I can't agree log z=lnr + ln(Arg(z)+2pi*n). If anyone clearly explain to me about this, I'll really thank you !! === Subject: Re: a=e^(log a), Is it always correct ? >I think a=e^(ln a) is correct but, a=e^(log a) is not. >(If a=e^(log a) is correct, then 10=e^log 10 ?) >Because, the base of ln is e but the base of log is 10. >Is there any wrong point about my thinking ? >In my complex analysis book, the following is given, >log z = ln r + i*(Arg(z)+2pi*n), n in integers, where |z|=r. >As for me, I cannot agree with above equality. >If z=re^i*t then lnz=lnr+ln(e^i*t) =>lnz=lnr + ln(Arg(z)+2pi*n) >I agree this, but I can't agree log z=lnr + ln(Arg(z)+2pi*n). >If anyone clearly explain to me about this, I'll really thank you !! Complete detailed answer sent to the e-mail address you so courteously provided. And, maybe you should learn some complex analysis. Disagree all you want, but the facts remain facts despite your objections. === Subject: Re: a=e^(log a), Is it always correct ? > I think a=e^(ln a) is correct but, a=e^(log a) is not. > (If a=e^(log a) is correct, then 10=e^log 10 ?) > Because, the base of ln is e but the base of log is 10. > Is there any wrong point about my thinking ? > In my complex analysis book, the following is given, > log z = ln r + i*(Arg(z)+2pi*n), n in integers, where |z|=r. > As for me, I cannot agree with above equality. > If z=re^i*t then lnz=lnr+ln(e^i*t) =>lnz=lnr + ln(Arg(z)+2pi*n) > I agree this, but I can't agree log z=lnr + ln(Arg(z)+2pi*n). > If anyone clearly explain to me about this, I'll really thank you !! Some people use log(z) to represent log to base e. This is common in more advanced books. The book you are reading seems strange in that it uses both ln and log to represent logarithms to base e. -- Stephen Montgomery-Smith stephen@math.missouri.edu http://www.math.missouri.edu/~stephen === Subject: Re: a=e^(log a), Is it always correct ? >I think a=e^(ln a) is correct but, a=e^(log a) is not. You first statement is 100% correct: it is one form of the definition of the natural logarithm, the exponent you must put onto a base of e to get the original number. Assuming you mean log as the base-10 log, I agree that your second statement is false, and your following correction is correct: >(If a=e^(log a) is correct, then 10=e^log 10 ?) >Because, the base of ln is e but the base of log is 10. >Is there any wrong point about my thinking ? No, your thinking is exactly right. The logarithm of a number x to a given base is the exponent you must put on that base to get back the given number x. Logarithm and exponential (to the same base) are inverses of each other. >In my complex analysis book, the following is given, >log z = ln r + i*(Arg(z)+2pi*n), n in integers, where |z|=r. Okay, it does get more complicated in complex numbers because log and exponential are no longer one-to-one functions. So the logarithm becomes a number with a property, rather than the number with that property. But does your book really say log and ln in the same equation? If so, and if they have not defined log as the base-e logarithm, then I agree that the above looks wrong. If your book actually has ln z on the left-hand side, then the equation is correct. >As for me, I cannot agree with above equality. >If z=re^i*t then lnz=lnr+ln(e^i*t) =>lnz=lnr + ln(Arg(z)+2pi*n) I don't know whether you've made a couple of typos in your last the intermediate steps: z = r * e^(it) t = Arg(z), r = |z| But e^(2pi*n*i) = 1 for integer n. Therefore z = r * e^(it+2pi*n*i) Then ln z = ln r + ln( e^(it + 2pi*n*i) ) ln z = ln r + it + 2pi*n*i ln z = ln r + i*(t + 2pi*n) ln z = ln r + i*(Arg(z) + 2pi*n) of, and last term is multiplied by i. -- 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: a=e^(log a), Is it always correct ? a ^ (log base a of x) = x always (a>0,x>0) > I think a=e^(ln a) is correct but, a=e^(log a) is not. > (If a=e^(log a) is correct, then 10=e^log 10 ?) > Because, the base of ln is e but the base of log is 10. > Is there any wrong point about my thinking ? > In my complex analysis book, the following is given, > log z = ln r + i*(Arg(z)+2pi*n), n in integers, where |z|=r. > As for me, I cannot agree with above equality. > If z=re^i*t then lnz=lnr+ln(e^i*t) =>lnz=lnr + ln(Arg(z)+2pi*n) > I agree this, but I can't agree log z=lnr + ln(Arg(z)+2pi*n). > If anyone clearly explain to me about this, I'll really thank you !! a^ (log base a of x) = x, always (a>0,x>0) To see this take log base a of both sides. === Subject: Re: a=e^(log a), Is it always correct ? >I think a=e^(ln a) is correct but, a=e^(log a) is not. >(If a=e^(log a) is correct, then 10=e^log 10 ?) >Because, the base of ln is e but the base of log is 10. In complex analysis, base-10 logs are hardly ever used. >Is there any wrong point about my thinking ? >In my complex analysis book, the following is given, >log z = ln r + i*(Arg(z)+2pi*n), n in integers, where |z|=r. Your text is using log for the complex base-e logarithm, and ln for the base-e logarithm on the positive reals. It is not talking about base-10 logarithms at all. With the definition as above, a = e^(log a) is always correct (except for a=0 where log a is undefined). Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: a=e^(log a), Is it always correct ? > I think a=e^(ln a) is correct but, a=e^(log a) is not. > (If a=e^(log a) is correct, then 10=e^log 10 ?) > Because, the base of ln is e but the base of log is 10. > Is there any wrong point about my thinking ? > In my complex analysis book, the following is given, > log z = ln r + i*(Arg(z)+2pi*n), n in integers, where |z|=r. > As for me, I cannot agree with above equality. > If z=re^i*t then lnz=lnr+ln(e^i*t) =>lnz=lnr + ln(Arg(z)+2pi*n) > I agree this, but I can't agree log z=lnr + ln(Arg(z)+2pi*n). > If anyone clearly explain to me about this, I'll really thank you !! > Some people use log(z) to represent log to base e. This is common in more > advanced books. > The book you are reading seems strange in that it uses both ln and log to > represent logarithms to base e. It seems to make sense to me - log of a complex number is likely defined as a new function because of all the questions about which branch to choose; it's likely that they introduced it and just used ambiguous notation. A prof is the regular natural log of a real number, which the book would have assumed. Doesn't seem odd to see them side by side. > -- > Stephen Montgomery-Smith > stephen@math.missouri.edu > http://www.math.missouri.edu/~stephen === Subject: Algebraic integer factorization, demonstration Now concentrating on the expression (v^3+1)x^3 - 3vxy^2 + y^3 let v= -1 + mf^2. If you have to have a ring, assume it's algebraic integers, m is a nonzero integer, and f is a nonzero algebraic integer coprime to 3 and m. Now then substituting for v gives ((-1 + mf^2)^3 + 1)x^3 - 3(-1 + mf^2)xy^2 + y^3 which is (m^3 f^6 - 3m^2 f^4 + m f^2 ) x^3 - 3(-1 + mf^2 )xy^2 + y^3 and now let y = uf, where u is an algebraic integer coprime to f, and using that substitution gives (m^3 f^6 - 3m^2 f^4 + m f^2 ) x^3 - 3(-1 + mf^2)x u^2 f^2 + (uf)^3 which is f^2 ((m^3 f^4 - 3m^2 f^2 + m) x^3 - 3(-1 + mf^2 )x u^2 + u^3 f). Now let P(m) = f^2((m^3 f^4 - 3m^2 f^2 + m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) and consider that the constant term is P(0) = f^2(3x u^2 + u^3 f). Then P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + m) 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 + m), 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. Here notice that it's clear that two of the b's must go to 0 from the u symbols, as that's the only way to get that factor of u^2 in P(0)/f^2. Given that w_1 and w_2 are coprime to f, then the factorization is P(m)/f^2 = (b_1 x + u)(b_2 x + u)(b_3 x + uf). Now looking at the factorization P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) it's clear that a_1 = f b_1, and a_2 = f b_2, which proves that a_3 is coprime to f, since a_1 a_2 a_3 = f^2(m^3 f^4 - 3m^2 f^2 + m), so b_1 b_2 a_3 = (m^3 f^4 - 3m^2 f^2 + m), which is coprime to f as m is coprime to f, notice that a_3 = b_3. James Harris === Subject: Re: Algebraic integer factorization, demonstration Since James hadn't demonstrated for a couple of months, I was afraid he was stagnant. === Subject: Re: Algebraic integer factorization, demonstration Still looking for that promised follow-on to this from his preceding > That polynomial is > 65x^3 - 12x + 1 > and I'll be using a special construction to factor it. But before > that I want to give an analogy using integers. > Now concentrating on the expression > (v^3+1)x^3 - 3vxy^2 + y^3 > let v= -1 + mf^2. > If you have to have a ring, assume it's algebraic integers, m is a > nonzero integer, and f is a nonzero algebraic integer coprime to 3 and > m. Note the cavalier if you have to have a ring, nearly rubbing it in by following with you babies, as though *real* men don't need no stinking rings. REAL MEN do algebra with objects devoid of ANY connection to mathematics. > Now then substituting for v gives > ((-1 + mf^2)^3 + 1)x^3 - 3(-1 + mf^2)xy^2 + y^3 > which is > (m^3 f^6 - 3m^2 f^4 + m f^2 ) x^3 - 3(-1 + mf^2 )xy^2 + y^3 > and now let y = uf, where u is an algebraic integer coprime to f, and > using that substitution gives > (m^3 f^6 - 3m^2 f^4 + m f^2 ) x^3 - > 3(-1 + mf^2)x u^2 f^2 + (uf)^3 > which is > f^2 ((m^3 f^4 - 3m^2 f^2 + m) x^3 - > 3(-1 + mf^2 )x u^2 + u^3 f). I keep seeing these expressions, without any reference to *what* is an integer, *what* is an indeterminate, or *where* all these manipulations take place. I know he said algebraic integers, but some of these things *surely* must be variables. > Now let > P(m) = f^2((m^3 f^4 - 3m^2 f^2 + m) x^3 - > 3(-1+mf^2 )x u^2 + u^3 f) > and consider that the constant term is > P(0) = f^2(3x u^2 + u^3 f). So x is a number, m is a variable. However, he says above m is a nonzero integer, and f is coprime to m. So m can't be a variable. > Then > P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + m) 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 + m), > 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. So, b_3 = 3? Are we talking about the polynomial > 65x^3 - 12x + 1 Aren't the b's coefficients of the x's? Is this as dumb as it looks? I mean, does 3 divide 65? I didn't think so. > Here notice that it's clear that two of the b's must go to 0 from the > u symbols, as that's the only way to get that factor of u^2 in > P(0)/f^2. JSH likes to argue from it's clear that to you lying incompetent bastards, I guess. I suppose that's the only way a person of his monumental absence of honor could produce the results he, er, has. Oh, no, wait, he HASN'T produced any results, has he? > Given that w_1 and w_2 are coprime to f, then the factorization is > P(m)/f^2 = (b_1 x + u)(b_2 x + u)(b_3 x + uf). > Now looking at the factorization > P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) > it's clear that a_1 = f b_1, and a_2 = f b_2, which proves that a_3 is > coprime to f, since > a_1 a_2 a_3 = f^2(m^3 f^4 - 3m^2 f^2 + m), > so > b_1 b_2 a_3 = (m^3 f^4 - 3m^2 f^2 + m), > which is coprime to f as m is coprime to f, notice that a_3 = b_3. Still not seeing that polynomial you promised: > 65x^3 - 12x + 1 Were you still on that thing about two of the coefficients being coprime to 5? Didn't I show you those coefficients, as well as the common factor each one has with 5? > James Harris C'mon, stop being such a girly-man. Show some gumption. Say what it is you claim with respect to the *actual* coefficients of that polynomial, without hiding it behind your mama's skirts, or behind the fig leaf of m and u and v and f and all this other distraction. All you're likely to prove is that you're full of hot air. Dale. === Subject: Re: Algebraic integer factorization, demonstration [snip cut-and-paste repitition of previous error-ridden argument] Zzzzzz... Now deal with the refutations of your proof. Face your critics with reason, not proclamations. -- 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: Algebraic integer factorization, demonstration > Still looking for that promised follow-on to this from his preceding > > That polynomial is > > 65x^3 - 12x + 1 > > and I'll be using a special construction to factor it. But before > that I want to give an analogy using integers. > I took that out to *lessen* confusion. Looking at this poster's reply, it's clear he put it back in to add it. > Now concentrating on the expression > > (v^3+1)x^3 - 3vxy^2 + y^3 > > let v= -1 + mf^2. > > If you have to have a ring, assume it's algebraic integers, m is a > nonzero integer, and f is a nonzero algebraic integer coprime to 3 and > m. > > Note the cavalier if you have to have a ring, nearly rubbing it in > by following with you babies, as though *real* men don't need no > stinking rings. REAL MEN do algebra with objects devoid of ANY > connection to mathematics. There's a problem with the ring of algebraic integers, which this argument shows. No histrionics are necessary. > Now then substituting for v gives > > ((-1 + mf^2)^3 + 1)x^3 - 3(-1 + mf^2)xy^2 + y^3 > > which is > > (m^3 f^6 - 3m^2 f^4 + m f^2 ) x^3 - 3(-1 + mf^2 )xy^2 + y^3 > > and now let y = uf, where u is an algebraic integer coprime to f, and > using that substitution gives > > (m^3 f^6 - 3m^2 f^4 + m f^2 ) x^3 - > > 3(-1 + mf^2)x u^2 f^2 + (uf)^3 > > which is > > f^2 ((m^3 f^4 - 3m^2 f^2 + m) x^3 - > > 3(-1 + mf^2 )x u^2 + u^3 f). > > I keep seeing these expressions, without any reference to *what* > is an integer, *what* is an indeterminate, or *where* all these > manipulations take place. I know he said algebraic integers, but > some of these things *surely* must be variables. It's algebra, they're all variables, as that's what x or y or any letter in an algebraic expression is. Notice the attack on the basis of algebra itself!!! After all, algebra is about symbols in place of numbers, and those symbols are all called variables. > Now let > > P(m) = f^2((m^3 f^4 - 3m^2 f^2 + m) x^3 - > > 3(-1+mf^2 )x u^2 + u^3 f) > > and consider that the constant term is > > P(0) = f^2(3x u^2 + u^3 f). > So x is a number, m is a variable. However, he says above m is > a nonzero integer, and f is coprime to m. So m can't be a variable. It's algebra; they're all variables. > > Then > > P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + m) 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 + m), > > 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. > > So, b_3 = 3? Are we talking about the polynomial It is shown that at m=0, b_3 = 3. > > 65x^3 - 12x + 1 > Aren't the b's coefficients of the x's? Is this as dumb as it looks? Now readers can see why the poster introduced 65x^3 - 12x + 1, though it wasn't in the post to which he replied. > I mean, does 3 divide 65? I didn't think so. Apparently, the idea is to keep throwing up smoke to interfere with readers like yourself being able to follow. > Here notice that it's clear that two of the b's must go to 0 from the > u symbols, as that's the only way to get that factor of u^2 in > P(0)/f^2. > > JSH likes to argue from it's clear that to you lying incompetent > bastards, I guess. I suppose that's the only way a person of his > monumental absence of honor could produce the results he, er, has. > Oh, no, wait, he HASN'T produced any results, has he? And readers can see the point of this person's post: telling them what to think. Here's some person on Usenet trying to control other readers on the newsgroup. > Given that w_1 and w_2 are coprime to f, then the factorization is > > P(m)/f^2 = (b_1 x + u)(b_2 x + u)(b_3 x + uf). > > Now looking at the factorization > > P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) > > it's clear that a_1 = f b_1, and a_2 = f b_2, which proves that a_3 is > coprime to f, since > > a_1 a_2 a_3 = f^2(m^3 f^4 - 3m^2 f^2 + m), > > so > > b_1 b_2 a_3 = (m^3 f^4 - 3m^2 f^2 + m), > > which is coprime to f as m is coprime to f, notice that a_3 = b_3. > > Still not seeing that polynomial you promised: I took that out to *lessen* confusion. Looking at this poster's reply, it's clear he put it back in to add it. > > 65x^3 - 12x + 1 > Were you still on that thing about two of the coefficients being coprime > to 5? > Didn't I show you those coefficients, as well as the common factor each > one has with 5? > > James Harris > C'mon, stop being such a girly-man. > Show some gumption. Say what it is you claim with respect to the > *actual* coefficients of that polynomial, without hiding it behind > your mama's skirts, or behind the fig leaf of m and u and v and f and > all this other distraction. > All you're likely to prove is that you're full of hot air. > Dale. Histrionics, personal attacks, and emotional appeals from a poster trying to tell you what to think, should worry the reader. If I were wrong, with such an approach laid out as it was then instead of a personal attack, posters could actually find an error. Ultimately, it's the reader's choice. You can be controlled by posters, or you can follow the math. James Harris === Subject: Re: Algebraic integer factorization, demonstration > There's a problem with the ring of algebraic integers, which this that you have no conception as to what it is. === Subject: Re: Algebraic integer factorization, demonstration >>Still looking for that promised follow-on to this from his preceding >That polynomial is > 65x^3 - 12x + 1 >and I'll be using a special construction to factor it. But before >that I want to give an analogy using integers. > I took that out to *lessen* confusion. Looking at this poster's > reply, it's clear he put it back in to add it. Yes, I suppose it's *confusing* to see actual numbers. Your argument is a huge pile of, er, error and confusion itself, and it does wonders to see what it *means* when actual numbers are in place of the symbols you strew about. I find it somewhat surprising that you're willing to respond to this (my preceding) posting, which was evidently *not* about the details of this exceedingly flawed process you insist on parading about, but rather on the absence of any clear association to the polynomial you *said* you were talking about. When I have pointed out (on too many occasions to mention here, but I did post a count, and it was embarassingly large) that your method produces an incorrect conclusion, you have been unwavering in your willingness to ignore the evidence. You have claimed obliquely to Nora Baron that I haven't shown anything that contradicts your case, but have simply ignored the direct, explicit calculations that even an elementary algebra student could follow. Those calculations show that your claim regarding coefficients of the relevant factorization of the polynomial 65x^3 - 12x + 1 is mistaken. I didn't need any extra variables. I didn't need to claim to minimize confusion. I just went there and found what you said wasn't so, and did it in a way that ANYONE can verify. You have chosen to ignore a simple and direct argument, requiring nothing more than algebraic manipulation. I don't need to assume anything about factorizations, coefficients, divisibility, or common factors or their absence. Yet you ignore that, in preference for the current thread, in which extra variables exist in multitudes, for (what you say) is to *lessen* confusion. I say you're a coward, and hide behind these since it's easy for you to maintain the pretense that you're doing *something* right. It's a charade, and you're the charlatan. BTW, It's plain rude to say you're talking about a particular problem, and then proceed to ignore said problem. I doubt that anyone could tell what claims you're making now about the factorization of the polynomial you *SAID* you were talking about. I know I couldn't. I, for one, refuse to follow an argument once it is SHOWN to produce an incorrect conclusion. It's up to you to clean up your own mess, not up to anyone else to wade through the raw sewage and point out the floating turds. >Now concentrating on the expression > (v^3+1)x^3 - 3vxy^2 + y^3 >let v= -1 + mf^2. >If you have to have a ring, assume it's algebraic integers, m is a >nonzero integer, and f is a nonzero algebraic integer coprime to 3 and >m. >>Note the cavalier if you have to have a ring, nearly rubbing it in >>by following with you babies, as though *real* men don't need no >>stinking rings. REAL MEN do algebra with objects devoid of ANY >>connection to mathematics. > There's a problem with the ring of algebraic integers, which this > argument shows. No, you have yet to point out an *explicit* problem with that ring. Up until recently, you have denied it's a ring, after all. Perhaps the problem lies entirely between your ears. > No histrionics are necessary. Oh, so the if you have to have a ring means something *other* than what I pointed out? Can you show me an argument involving polynomials, that DOES NOT require the existence of a ring? Just one? I didn't think so. SO, what else could your taunting remark mean? ... stuff deleted ... >>I keep seeing these expressions, without any reference to *what* >>is an integer, *what* is an indeterminate, or *where* all these >>manipulations take place. I know he said algebraic integers, but >>some of these things *surely* must be variables. > It's algebra, they're all variables, as that's what x or y or any > letter in an algebraic expression is. A variable isn't an integer. > Notice the attack on the basis of algebra itself!!! Notice the attack is on the audacity of someone asking what JSH means!!! > After all, algebra is about symbols in place of numbers, and those > symbols are all called variables. Yes, and freedom's just another word for nothing left to lose. You think that everything is all about your little world and your little ideas. Variables are different from numbers, which are different from parameters. >Now let > P(m) = f^2((m^3 f^4 - 3m^2 f^2 + m) x^3 - > 3(-1+mf^2 )x u^2 + u^3 f) >and consider that the constant term is > P(0) = f^2(3x u^2 + u^3 f). >>So x is a number, m is a variable. However, he says above m is >>a nonzero integer, and f is coprime to m. So m can't be a variable. > It's algebra; they're all variables. This is where you are absolutely clueless. >Then > P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + m) 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 + m), >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. >>So, b_3 = 3? Are we talking about the polynomial > It is shown that at m=0, b_3 = 3. >> 65x^3 - 12x + 1 >>Aren't the b's coefficients of the x's? Is this as dumb as it looks? > Now readers can see why the poster introduced 65x^3 - 12x + 1, though > it wasn't in the post to which he replied. Oh, I see. Your preceding post, titled Advanced polynomial factorization, demonstration and this thread, titled Algebraic integer factorization, demonstration were *not* talking about the cubic I mentioned? I sure though they were. As you said, you left the mention of that cubic out, to *lessen* confusion. >>I mean, does 3 divide 65? I didn't think so. > Apparently, the idea is to keep throwing up smoke to interfere with > readers like yourself being able to follow. I'm sorry if I misinterpreted. I haven't seen a single word to the effect that something is amiss with the ring of algebraic integers, save your bald assertions. However, it's ME who wishes to tell others what and how to think? >Here notice that it's clear that two of the b's must go to 0 from the >u symbols, as that's the only way to get that factor of u^2 in >P(0)/f^2. >>JSH likes to argue from it's clear that to you lying incompetent >>bastards, I guess. I suppose that's the only way a person of his >>monumental absence of honor could produce the results he, er, has. >>Oh, no, wait, he HASN'T produced any results, has he? > And readers can see the point of this person's post: telling them what > to think. Wasn't that posed flatly as a question? A rhetorical question, to boot? Are you *sure* English is your native language? > Here's some person on Usenet trying to control other readers on the > newsgroup. Oh, I get it. YOU can threaten: My warning to universities and mathematicians teaching at them remains. Some of you may face civil liabilities, maybe worse-- angry parents--when the full story comes out. School is about to start, and if universities teach the flawed mathematics they may also face federal authorities for deliberate fraud. Trouble is, I never figured that mathematicians were both incompetent and stupid. Oh well, new plan as now I get to use the Hammer with full force, which is probably what was intended. Now, no feeling sympathy for mathematicians who start marching with signs like Will work for food in the future, as I've given them lots of opportunities to play right. I will not show mercy going forward. I was trained as a soldier in the United States Army after all. And when the US Army plays a game, we play to win. Yup, you guessed it. If worse comes to worse, I *will* turn to the Army to help me with mathematicians. And then mathematicians don't think the NSA or CIA can save your asses, as generals LIKE me. And I think I know the CIA and NSA better than any mathematician. When push comes to shove, they'll throw you out with the garbage. They'd personally shoot you themselves, if it were necessary. If I have to sic Army generals on you, I will be really pissed. 3. exposure as frauds It seems to me that your statements here can be deliberately taken as an attempt at fraud in further attempts to prevent the world from recognizing that I have a short proof of Fermat's Last Theorem. It also seems to me that you must be aware that proper recognition has monetary value for me, so malice aforethought can be assumed. I am also making it clear to you that I deem your statement and the defense of them by at least one other mathematican to be evidence of a severe problem within the mathematical community, which at least in the United States, I feel should be a matter that is considered by Congress in order to determine proper distribution for funding of mathematical research in the United States. but if I raise the rhetorical (but factual) issue of whether you've *actually proven anything*, it's I who is trying to control other readers on the newsgroup? Does your dictionary contain a mirror instead of a verbal definition, next to the word hyprocrite? Doesn't there seem to be a bit of an imbalance between the tone of your imbalance point to *me* as being the one whos over the top? Hardly. >Given that w_1 and w_2 are coprime to f, then the factorization is > P(m)/f^2 = (b_1 x + u)(b_2 x + u)(b_3 x + uf). >Now looking at the factorization > P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) >it's clear that a_1 = f b_1, and a_2 = f b_2, which proves that a_3 is >coprime to f, since > a_1 a_2 a_3 = f^2(m^3 f^4 - 3m^2 f^2 + m), >so > b_1 b_2 a_3 = (m^3 f^4 - 3m^2 f^2 + m), >which is coprime to f as m is coprime to f, notice that a_3 = b_3. >>Still not seeing that polynomial you promised: > I took that out to *lessen* confusion. Looking at this poster's > reply, it's clear he put it back in to add it. >> 65x^3 - 12x + 1 >>Were you still on that thing about two of the coefficients being coprime >>to 5? >>Didn't I show you those coefficients, as well as the common factor each >>one has with 5? I see. Here's what I imagine you're saying THIS IS CONFUSING: Let 65 x^3 - 12 x + 1 = (a1 x + 1)(a2 x + 1)(a3 x + 1) In response to the claim that any of the ai's could be coprime to 5, I gave these formulas: q(a) = 8 a^2 + 76 a - 185 r(a) = 8 a^2 + 4 a - 45 s(a) = 4 a^2 + 37 a - 104 and showed that whenever ai is one of the ai's in the first factorization, then: Letting qi = q(ai), ri = r(ai), si = s(ai), ai = qi ri and 5 = ri si. This exhibits the common factor of ri that divides *both* ai and 5. It's what you proved could not exist. I ask you what about those factors, and how does this thread's discussion relate to that particular polynomial? whereas THIS IS NOT CONFUSING: Now then substituting for v gives ((-1 + mf2)3 + 1)x3 - 3(-1 + mf2)xy2 + y3 which is (m3 f6 - 3m2 f4 + m f2 ) x3 - 3(-1 + mf2 )xy2 + y3 and now let y = uf, where u is an algebraic integer coprime to f, and using that substitution gives (m^3 f^6 - 3m^2 f^4 + m f^2 ) x^3 - 3(-1 + mf^2)x u^2 f^2 + (uf)^3 which is f^2 ((m^3 f^4 - 3m^2 f^2 + m) x^3 - 3(-1 + mf^2 )x u^2 + u^3 f). ... and so forth. Now, tell me. Given the fact that I have given *explicit* formulas, with no clever dancing around the point, like this: where u is an algebraic integer coprime to f, and using that substitution ... how is it that you claim *I* am making things confusing? My work can be verified directly. Yours relies on the choice of an algebraic integer coprime to f, making some substitution, and calling out to the US Army when no one accepts your bogus argument as holding any water. I have a direct calculation that shows you're full of , and you're dancing around in your leprechaun hat, saying it's full of magical surprises! >James Harris >>C'mon, stop being such a girly-man. >>Show some gumption. Say what it is you claim with respect to the >>*actual* coefficients of that polynomial, without hiding it behind >>your mama's skirts, or behind the fig leaf of m and u and v and f and >>all this other distraction. >>All you're likely to prove is that you're full of hot air. >>Dale. > Histrionics, personal attacks, and emotional appeals from a poster > trying to tell you what to think, should worry the reader. No, I have not made any personal attacks. What I *have* attacked is what you're doing, way it is you're behaving. You're behaving like a naughty child who wants to get away with having done something that he shouldn't have done. It's abundantly clear that you *don't* have anything to say about that *specific* polynomial, since if you did, I could show that it behaves *precisely* the way that one would surmise from standard Galois theory and the perspective of the standard world of algebraic integers. You wish to have people believe that the algebraic integers are flawed, and still you wish to have that happen IN THE TOTAL ABSENCE OF SPECIFIC Show me how wrong I am. Produce a SPECIFIC polynomial, and roots and/or factors that behave OTHER than how they should. EXHIBIT a flaw in the ring of algebraic integers. No? Well, you'll just have to put up with being the butt of all jokes. If you could read more than just what you hoped to see, you would have seen that I've asked you to state, in so many words, exactly what it is that you're claiming about the polynomial that you've excised to *lessen* confusion. Don't be afraid of confusing *me* with numbers, because I know what the roots and relevant factors of the roots of that polynomial are. As far as telling the reader what to think, let's let the readers compare what it is you've said and threatened, to what you're crying about from my preceding posting. > If I were wrong, with such an approach laid out as it was then instead > of a personal attack, posters could actually find an error. I'm sorry. I have done that *HOW MANY TIMES*??? Over the past month or so, I believe I have posted on no fewer than 10 occasions, a *direct*, *explicit* proof that your argument produces an erroneous conclusion. You are now acting as though none of those postings have taken place. Your attitude of evasion will surely catch up with you, if not on sci.math, or whererever the hell you are throwing cross-postings these days, then in real life. A person sows what he reaps. > Ultimately, it's the reader's choice. You can be controlled by > posters, or you can follow the math. Show of hands for folks who follow what JSH is terming the math To find my argument, find THIS posting of mine: <3F1C3F01.7010501@farir.com> similarly-phrased argument. easily tell whether I've got the goods, or not. > James Harris Dale. === Subject: Re: Algebraic integer factorization, demonstration > There's a problem with the ring of algebraic integers, which this > that you have no conception as to what it is. Deleting out mathematics does not make it go away. I've seen this behavior repeatedly and it's bizarrely childish, as if by making a post where the information is deleted out...and I'll put key information back in here as I'm replying...a poster believes they've refuted a *math* argument!!! What makes it extraordinary is that apparently some vocal math people don't believe in mathematics as intrinsically true, but as a body of social truth where a math argument gains validity based on group acceptance. They clearly are not afraid of showing their belief system when it comes to posting on the sci.math newsgroup. That is, math to them is true, if people believe it's true. Here's some of the deleted information which has taken away room for cogent objections: Then 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. By shifting focus to P(m)/f^2 I shorten the space where posters can try to confuse. Remember though readers, they are trying to manipulate you. My position is that the mathematics speaks for itself, but if it is incorrect, someone can find a problem with the argument presented--none of this other nonsense. Math is simple in that way--if it's wrong, you can find something wrong with what's presented by showing a break in the logical chain i.e. show a given statement, which is followed by another statement, which does NOT follow logically. So demonstrating an error means doing something like, saying that while statement A was ok, you follow with statement B, which does not follow mathematically. James Harris === Subject: Re: Algebraic integer factorization, demonstration > Still looking for that promised follow-on to this from his preceding > > > That polynomial is > > 65x^3 - 12x + 1 > > and I'll be using a special construction to factor it. But before > that I want to give an analogy using integers. > > I took that out to *lessen* confusion. Looking at this poster's > reply, it's clear he put it back in to add it. Since you put it in the analogy originally, it is your confusion, not anyone else's. === Subject: Re: Algebraic integer factorization, demonstration > Deleting out mathematics does not make it go away. I've seen this book about math but you can't remember any of it. === Subject: Re: Algebraic integer factorization, demonstration > My position is that the mathematics speaks for itself, but if it is > incorrect, someone can find a problem with the argument > presented--none of this other nonsense. > Math is simple in that way--if it's wrong, you can find something > wrong with what's presented by showing a break in the logical chain > i.e. show a given statement, which is followed by another statement, > which does NOT follow logically. Like here? > So demonstrating an error means doing something like, saying that > while statement A was ok, you follow with statement B, which does not > follow mathematically. > James Harris -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: Algebraic integer factorization, demonstration > My position is that the mathematics speaks for itself, but if it is > incorrect, someone can find a problem with the argument > presented--none of this other nonsense. > > Math is simple in that way--if it's wrong, you can find something > wrong with what's presented by showing a break in the logical chain > i.e. show a given statement, which is followed by another statement, > which does NOT follow logically. > Like here? Ah, laziness shows itself. Now I've posted a math argument in THIS THREAD. Maybe you think you're too good to bother with answering what's here so you posted a link, but laziness is not going to get you anywhere. If you have a problem with the math argument posting in THIS THREAD then you can reply to what's in this thread without trying some short-cut. > > So demonstrating an error means doing something like, saying that > while statement A was ok, you follow with statement B, which does not > follow mathematically. Yup. In fact, it's so easy that when a person really does have a problem with a proof, mathematicians don't have to bother with insults or other varieties of personal attacks, they can just say statement A is true, but statement B does not follow from what is given. It's not complicated. It doesn't require endless arguments in threads all over the place. It doesn't require bringing up personal issues, or whining, or histrionics. It's actually a rather dry process. James Harris === Subject: Re: Algebraic integers: Nifty proof of prob > What's fascinating about math proofs is that they can be staring you > in the face, hiding in plain sight, as it were. > For instance, consider all the arguments about the factorization > 65x^3 - 12x + 1 = (a_1 x + 1)(a_2 x + 1)(a_3 x + 1) > and the question of factors in common with 5 among the a's. > With the focus on sqrt(5) no one seemed to notice that the claims of > counterexample against my work would work against *any* non-unit > algebraic integer f which is a factor of any of the a's. > To see that go ahead and solve for the roots like before to get > a^3 + 12a^2 - 65 = 0. > Now use a = fb, to get > f^3 b^3 + 12f^2 b^2 - 65 = 0, and divide out f^2 to get > f b^3 + 12b^2 - 65/f^2 = 0. > For any f, where algebraic manipulations can give you a polynomial > with integer coefficients, you will have a non-monic irreducible over > Q, which proves that none of its roots are algebraic integers. Nope. I'm still stuck. Damn mathematicians. If only I didn't have to deal with so many liars. Stuck trying to find an alternate proof for a proof. When I was looking for a short proof of FLT, the real question seemed to be whether or not I could find it, with the answer being, very unlikely. Little did I know then that mathematicians could so easily, continually, and casually lie about even a short proof. ways possible about mathematics itself, as if mathematical truth doesn't matter. James Harris === Subject: Re: Algebraic integers: Nifty proof of prob > What's fascinating about math proofs is that they can be staring you > in the face, hiding in plain sight, as it were. > For instance, consider all the arguments about the factorization > 65x^3 - 12x + 1 = (a_1 x + 1)(a_2 x + 1)(a_3 x + 1) > and the question of factors in common with 5 among the a's. > With the focus on sqrt(5) no one seemed to notice that the claims of > counterexample against my work would work against *any* non-unit > algebraic integer f which is a factor of any of the a's. > To see that go ahead and solve for the roots like before to get > a^3 + 12a^2 - 65 = 0. > Now use a = fb, to get > f^3 b^3 + 12f^2 b^2 - 65 = 0, and divide out f^2 to get > f b^3 + 12b^2 - 65/f^2 = 0. > For any f, where algebraic manipulations can give you a polynomial > with integer coefficients, you will have a non-monic irreducible over > Q, which proves that none of its roots are algebraic integers. > > Nope. I'm still stuck. Damn mathematicians. If only I didn't have > to deal with so many liars. Stuck trying to find an alternate proof > for a proof. > When I was looking for a short proof of FLT, the real question seemed > to be whether or not I could find it, with the answer being, very > unlikely. > Little did I know then that mathematicians could so easily, > continually, and casually lie about even a short proof. I'm not really a mathematician but that is my educational background anyway. When I was a graduate student, we rated proofs as to the number of beers required. A trivial theorem falls easily to one beer. a few differential operator theory proofs I encountered ran near to a case and Fermat's Last Theorem, after a little doodling, looks like three centuries of a brewers output. That aside, nobody is lying to you. If you think they are, show what steps in their proofs are invalid. Don't even talk about lying if you cannot do that. Students learn how to prove theorems by screwing up and taking lumps. That usually starts in high school geometry. It does not appear to me that you can tell a valid proof from an invalid one. Did you not even take high school geometry? > ways possible about mathematics itself, as if mathematical truth > doesn't matter. Look into a mirror. Ask, What is mathematical truth. You cannot create truth by repeating nonsense so many times that magic happens. It doesn't work that way. Mathematical truth is not what you wish it to be no matter how hard you wish nor how many good fairies wave their wands to your wishes. A word of advice. Never use the word should in a mathematical proof. Chuck -- ... The times have been, That, when the brains were out, the man would die. ... Macbeth Chuck Simmons chrlsim@earthlink.net === Subject: Re: Algebraic integers: Nifty proof of prob >>What's fascinating about math proofs is that they can be staring you >>in the face, hiding in plain sight, as it were. >>For instance, consider all the arguments about the factorization >> 65x^3 - 12x + 1 = (a_1 x + 1)(a_2 x + 1)(a_3 x + 1) >>and the question of factors in common with 5 among the a's. >>With the focus on sqrt(5) no one seemed to notice that the claims of >>counterexample against my work would work against *any* non-unit >>algebraic integer f which is a factor of any of the a's. >>To see that go ahead and solve for the roots like before to get >> a^3 + 12a^2 - 65 = 0. >>Now use a = fb, to get >> f^3 b^3 + 12f^2 b^2 - 65 = 0, and divide out f^2 to get >> f b^3 + 12b^2 - 65/f^2 = 0. >>For any f, where algebraic manipulations can give you a polynomial >>with integer coefficients, you will have a non-monic irreducible over >>Q, which proves that none of its roots are algebraic integers. No matter what JSH says, he ignores the fact of my postings of June: 21, 22 July: 6, 8, 10, 11, 12, 13, 15, 21, and 29 each one attesting to the fact that I had found such a proof, and *most* of which provided explicit proof. > > Nope. I'm still stuck. Damn mathematicians. If only I didn't have > to deal with so many liars. Stuck trying to find an alternate proof > for a proof. Yes, I lie in the sneakiest way possible: I provide explicit proof, in such a fashion that you could check it using pencil and paper, probably in under ten minutes' time. Damn it all, anyhow. How can the TRUTH survive, with such lying taking place all the time? Trying to find an alternate proof for a proof that itself leads directly to contradictions. A proof of falsehoods, that's what you want to find an alternate proof of. Rotsa Ruck, Mistah. > When I was looking for a short proof of FLT, the real question seemed > to be whether or not I could find it, with the answer being, very > unlikely. > Little did I know then that mathematicians could so easily, > continually, and casually lie about even a short proof. Yep, that's exactly what I'm talking about! Lying and more lying. Lying so bad that a person just has to accept the lie or just stop doing baby algebra. I'm talkin' 'bout *lying*, folks! > ways possible about mathematics itself, as if mathematical truth > doesn't matter. That truth thing, how does that go again? You write down an argument that says something that is PLAINLY incorrect, and where's the truth in that? When someone disputes your errors with direct, explicit calculations, that a high school student could follow with a bit of effort, and could verify that you were drawing a conclusion that is incorrect, I ask, truly: WHERE IS THE LIE? Can't show it? Then at least explain this, fool: HOW CAN A VALID ARGUMENT PRODUCE INCORRECT RESULTS? Can't do that neither? I didn't think so. > James Harris Dale. === Subject: Re: Algebraic integers: Nifty proof of prob > A rough analogy is the case of 2 and 6 in the ring of evens, The ring of evens? Excuse me, does this ring have a multiplicative identity? What is it? If it doesn't, how is analogous to the ring of algebraic integers? > However, the ring of evens is still complete, Does that mean it contains a multiplicative identity? If so, what is it? Your statement is an equivocation, because a ring is necessarily complete, or else it isn't a ring. There's no such thing as an incomplete ring. There are only sets that *do* form a ring with the operators + and * and sets that *do not* form a ring. > while the problem I've > described in the ring of integers leads to the contradiction that > though > r_1 r_2 r_3 = 65 > r_1, r_2 and r_3 are each coprime to 5. Where's the contradiction? The algebraic integers have been proven to form a ring. If your results contradict this, your results are wrong. QED. -- 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: Algebraic integers: Nifty proof of prob > Nope. I'm still stuck. Damn mathematicians. If only I didn't have half your brain missing. === Subject: Re: Algebraic integers: Nifty proof of prob Visiting Assistant Professor at the University of Montana. [.snip.] >A rough analogy is the case of 2 and 6 in the ring of evens, as 6 is >prime in the ring of evens as it is coprime to 2. We went through this one already. It just so happens that 6 and 2 are coprime in the ring 2Z, but NOT for the reason you think. You seem to think that coprime means has no common divisors other than units. In the ring of algebraic integers (and in any UFD or Bezout Domain, among others), the two statements are equivalent. BUT NOT IN GENERAL. The definition of coprime is: DEF. Let R be a ring. Two ideals I and J are coprime if and only if there is no prime ideal P of R which contains both I and J. Two elements x and y of R are said to be coprime (in R) if and only if the principal ideals (x) and (y) are coprime. In the case of commutative Rings with 1 (which 2Z is not), and assuming the Axiom of Choice (or at least, assuming that every proper ideal in a ring with 1 is contained in a maximal ideal, a statement which I believe is slightly weaker than AC), one can prove that: THEOREM. Let R be a commutative ring with 1, and let x and y be two elements of R. Then x and y are coprime if and only if there exist elements a and b in R such that ax+by = 1. Proof. Assuming AC (or the possibly weaker version mentioned before), every ideal is contained in a maximal ideal. Assume that x and y are coprime. In a commutative ring with 1, every maximal ideal is prime, so we may conclude that (x,y) is NOT a proper ideal of R. Therefore, (x,y)=(1), the unit ideal. Therefore, 1 is in (x,y), which means that there exist a and b such that ax+by = 1. Conversely, if ax+by =1 for some a and b in R, and I is ANY ideal containing x and y, then it contains ax and contains by, therefore contains 1=ax+by, so I=R. Thus, there can be no proper ideal containing both x and y, so there can be no prime ideal containing both (x) and (y). QED In the case of 2 and 6 in 2Z, the ideal (2) consists of all multiples of 4, and the ideal (6) consists of all multiples of 12. It is now easy to show that (2) is a maximal ideal (any larger ideal contains all even numbers), and contains both 2 and 6; so 2 and 6 are not comaximal. However, (2) is NOT a prime ideal: it contains 4=2*2, but does not contain 2. So any proper ideal containing both 2 and 6 must be the ideal (2), which is not prime, so 2 and 6 ARE coprime in 2Z, but for VERY different reasons than the case of the ring of all algebraic integers. If you do not know what a word means, either ask, or don't use it. 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