mm-3299 === > By Donald S. McDonald, WGTN Astronomical Soc. > (Meeting.) Observation 1/ moon's apparent diameter equals little > finger nail = 0.5 degree. Observation 2/ Intensity of moon equals daylight sky. Hypothesis 3/. 105,050 moons would cover a > hemisphere, lighting Earth with equal of Sun's > radiance. Propositions 41/42/43. > Archimedes' textbook, Euclid 1897?, On the sphere and > cylinder. (book 11.) Don. Area of Compass Circle on a Sphere, O. > New Zealand Mathematics Magazine, 25 (2), [1988.]. Don.Lotto birthday greeting nz, Google groups. > ...> Ex. (sin .25) ^-2 = 52,525.25 = 52E5/99. STATS NZ 525 snazzy neat phone numb palindrome S-N-S coorection in dompost 225 not 25. deaths by work accid ?? debt years ended XXX Jul 2006 errors. page A1. Grecian goblets. Extremely rare. Gif/ jpg? Mag. Diff. = 5x log base 10 ( sin .25 deg ) = 11.8. > Casio fx-82 MS multi-step scientific-calculator > schools. Was list, nzmm, budding, newsct, mcd, nzm cc a10 (prof > h, Vicki dbl, paul m ) Donald McDonald 12-10-2006. File moon-antar280906-19, word antares-occult, e: > astro-space, email. > Giffordobsy-gvthse... fog hanson court fls., > not '52' goblet? Follows. blue sky lunar occultation of red supergiant star, > Antares. > Soup ladle, flare, JPG. 28-9-06. > > Don S. McDonald === Subject: Re: LINK PIX. Colour of moon's surface grey or gold? Here is why! Antares Occultn JPG LINK PIX. Colour of moon's surface grey or gold? Here is why! Antares > Occultn JPG > click.. http://www.was.org.nz/memberspages/dmcdonald.htm a more direct link the previous one also worked. member pages > WGTN ASTRON SOC new zealand. pix moon antares blue sky occultation********** > **************** > MAXIMISE. i hope this link works. dmcdonald > don.lotto nz > > picky mcdonald tvnz > 22.10.06 > cheers > > hobby card. > === Subject: Re: integral of unknown function Sat, 21 Oct 2006 05:07:14 -0700 from William Elliot : > > Fri, 20 Oct 2006 11:27:48 +0200 from Johan Lans : > > If i know that int(f(x))dx between -6 and 6 is 11, how do i know what > > int((f(x))^2)dx is (same interval)? > > Without knowing more about f, you don't. > > For example, if you knew the integral from -6 to 0 or from 0 to 6, in > addition to the integral from -6 to 6, then you could solve the problem. > How so? I don't know how I made such a mistake, but a mistake it was and is. -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com/ === Subject: Re: integral of unknown function > Fri, 20 Oct 2006 11:27:48 +0200 from Johan Lans : > > If i know that int(f(x))dx between -6 and 6 is 11, how do i know > > what int((f(x))^2)dx is (same interval)? > > Without knowing more about f, you don't. > > For example, if you knew the integral from -6 to 0 or from 0 to 6, in > > addition to the integral from -6 to 6, then you could solve the problem. > > How so? I don't know how I made such a mistake, but a mistake it was and is. > You forgot f was to be an odd or even function. === Subject: Re: integral of unknown function Sat, 21 Oct 2006 05:07:14 -0700 from William Elliot > > Fri, 20 Oct 2006 11:27:48 +0200 from Johan Lans : > > If i know that int(f(x))dx between -6 and 6 is 11, how do i know > > > what int((f(x))^2)dx is (same interval)? > > > > Without knowing more about f, you don't. > > > > For example, if you knew the integral from -6 to 0 or from 0 to 6, in > > addition to the integral from -6 to 6, then you could solve the problem. > > > How so? > > I don't know how I made such a mistake, but a mistake it was and is. > You forgot f was to be an odd or even function. > Nope, that's not sufficent either. === Subject: Re: integral of unknown function Sun, 22 Oct 2006 05:34:09 -0700 from William Elliot : > > Sat, 21 Oct 2006 05:07:14 -0700 from William Elliot > > > Fri, 20 Oct 2006 11:27:48 +0200 from Johan Lans : > > > > If i know that int(f(x))dx between -6 and 6 is 11, how do i know > > > what int((f(x))^2)dx is (same interval)? > > > > Without knowing more about f, you don't. > > > > For example, if you knew the integral from -6 to 0 or from 0 to 6, in > > > addition to the integral from -6 to 6, then you could solve the problem. > > > > How so? > > I don't know how I made such a mistake, but a mistake it was and is. > > You forgot f was to be an odd or even function. > Nope, that's not sufficent either. I think I was thinking somewhere along the same lines. I realized too late, that would give the integral of abs[f(x)] but not of [f(x)]^2. -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com/ === Subject: Re: JSH: Direct demonstration of contradiction > . Usenet is very public. Imagine the way they feel to be like being on > TV and being totally embarrassed by someone you thought you were > embarrassing, and having that be part of a permanent record. Ok. Lets try out my brand new, super hardened, guaranteed not to break, irony meter. SPROING!!! - William Hughes === Subject: Re: JSH: Direct demonstration of contradiction >[...] So in reply, posters like yourself have to keep deleting out >information, and make claims that defy what was given in my posts, >which is kind of odd, but clear to readers who bother to pay >attention--and care what the correct answer is. The attempt though can be bizarre for people who don't understand human >nature, and how far some people can go when a cherished belief is >totally refuted in a very public way. Readers need to remember this is all VERY public. My posts are read >around the world. These forums are read around the world--posters who have argued with me >for years are simply refusing to accept being proven wrong in a way >they can't talk their way out of, as easily as they did when there were >variables and functions. Usenet is very public. Imagine the way they feel to be like being on >TV and being totally embarrassed by someone you thought you were >embarrassing, and having that be part of a permanent record. You really can't help posting this sort of thing, eh? You really have no idea how much _more_ foolish you look when you say something like this, then later in the same thread you say >I did make a mistake. and then a few posts after that you say >Oh yeah, I made another dumb mistake. and then in your next post in the thread you say >Oh, I got a sign wrong but the rest should be correct. Yes, admitting your mistakes is a good thing - too bad the only time it happens is when they're so trivially shown to be errors. But going on about how you're going to make us all look foolish, in spite of the fact that oh, never mind. Yes, you have the power to destroy us all. >James Harris ************************ David C. Ullrich === Subject: Re: JSH: Direct demonstration of contradiction > > > > Start with > > > > > > (5 + (sqrt(37) - sqrt(13))/2)((sqrt(37) - sqrt(13))/2) = (5 - > > > > sqrt(13))(5 + sqrt(37))/2 > > > > > > > > Okay, I'll tell you what the story is. The ring of integers in > > > Q(sqrt(13),sqrt(37)) is Z[(1+sqrt(13))/2,(1+sqrt(37))/2]. The ideal > > > generated by 2 is a product of two prime ideals p_1 and p_2 of norm 4. > > > Only one of them contains 5+(sqrt(37)-sqrt(13))/2, I'll let p_1 be that > > > one. > > > > > > We have 3=((sqrt(13)+1)/2)((sqrt(13)-1)/2). The ideals generated by > > > (sqrt(13)+1)/2 and (sqrt(13)-1)/2 are prime. > > > > > > Now, the ideal generated by 5+(sqrt(37)-sqrt(13))/2 is the product of > > > p_1 and the ideal generated by (sqrt(13)-1)/2. > > > > > > And the ideal generated by (sqrt(37)+sqrt(13))/2 is the product of > > > (p_2)^2 and the ideal generated by (sqrt(13)-1)/2. It turned out you > > > were actually right that the gcd's with 3 are equal, in both cases they > > > are (sqrt(13)-1)/2. However, this was just a fluke. > > > > > > > Yet looking at > > > > (5 + (sqrt(37) - sqrt(13))/2)((sqrt(37) - sqrt(13))/2) = (5 - > > > sqrt(13))(5 + sqrt(37))/2 > > > > > My calculations were wrong. We have (2)=p_1.p_2 and > > (3)=q_1.q_2.q_3.q_4. I'll let q_1.q_2 be invariant under the > > automorphism which sends sqrt(13) to -sqrt(13), and q_1.q_3 be > > invariant under the automorphism which sends sqrt(37) to -sqrt(37). > > > > Then I have > > > > (5+(sqrt(37)-sqrt(13))/2)=p_1.q_1.q_2 > > ((sqrt(37)-sqrt(13))/2)=p_2.q_3.q_4 > > (5-sqrt(13))=p_1.p_2.q_3.q_4 > > ((5+sqrt(37))/2)=q_1.q_2 > > > > you can just note that > > > > (5 + (sqrt(37) - sqrt(13))/2) > > > > must be coprime to (sqrt(37) - sqrt(13))/2 with respect to 3, by just > > > subtracting one from the other as that leaves 5. > > > > > Yes that's right. The greatest common divisor of one with 3 is q_1.q_2, > > the other is q_3.q_4. > > > > And since ((sqrt(37) - sqrt(13))/2)*((sqrt(37) + sqrt(13))/2) = 6 > > > > you know that (5 + (sqrt(37) - sqrt(13))/2) must share all of its > > > factors in common with 3, with (sqrt(37) + sqrt(13))/2. > > > > > You know that every prime ideal dividing (5+(sqrt(37)-sqrt(13))/2) and > > 3 must divide (sqrt(37)+sqrt(13))/2. In fact, as it happens > > gcd(5+(sqrt(37)-sqrt(13))/2,3)=gcd((sqrt(37)+sqrt(13))/2,3)=(sqrt(13)-1)/2. > > > > So the next step is also trivial which is to divide one by the other > > correcting for other prime factors, where in this case that means you > > can correct for 2 by multiplying by 2 > > Nope. You can't clear all the factors of 2 from > (sqrt(37) + sqrt(13))/2 by multiplying by 2. > > - William Hughes Except that right after I demonstrate:: z = 2(10 + sqrt(37) - sqrt(13))/(sqrt(37) + sqrt(13)) which gives z as a root of 9z^4 - 72z^3 - 409z^2 - 2212z - 144 = 0 where the lead term is 9 which is coprime to 2, showing that it worked! > And factors in common with 2 were successfully removed. I also explained the why of this result with a simple example from > integers with x^2 + 4x + 3 = 0 as if you substitute with x = 3y, you get a non-monic, showing how > despite one root having 3 as a factor, you'll end up with a non-monic > when you do simple things like divide because not ALL roots have that > factor. So with (10 + sqrt(37) - sqrt(13))/2 and (sqrt(37) + sqrt(13))/2 > provably they BOTH have the same factor in common with 3. But dividing one by the other and correcting for factors in common with > 2, you STILL get a non-monic polynomial with integer coefficients. This direct demonstration is so effective because readers can't be > confronted with questions about variables, or issues about what's the > constant term, or be challenged with claims about mystery functions, as > it's just the numbers. So in reply, posters like yourself have to keep deleting out > information, WHO is the one who deletes his embarrasing posts? Gee, it's James Harris. > and make claims that defy what was given in my posts, > which is kind of odd, but clear to readers who bother to pay > attention--and care what the correct answer is. The attempt though can be bizarre for people who don't understand human > nature, and how far some people can go when a cherished belief is > totally refuted in a very public way. Readers need to remember this is all VERY public. My posts are read > around the world. Which means there are people in Australia chuckling about JSH's incompetance. It's the funniest thing mathematically to happen to Australia since Archimedes Plutonium arrived, finding out that he couldn't cut it as a grad student, and a B.A. in math doesn't give you tutoring jobs. And in Germany, someone is saying: Was fur ein Scheisskopf!. > These forums are read around the world--posters who have argued with me > for years are simply refusing to accept being proven wrong in a way > they can't talk their way out of, as easily as they did when there were > variables and functions. There's an old rule: Never argue with an insane person. And clearly JSH is missing a few screws. > Usenet is very public. Imagine the way they feel to be like being on > TV and being totally embarrassed by someone you thought you were > embarrassing, and having that be part of a permanent record. Ah, so that's why JSH deletes his posts: so that they _aren't_ part of his permanent record. Well, this only works for Google Groups, Dummkopf. Your posts haven't been removed from the rest of Usenet. --- Christopher Heckman > James Harris === Subject: Re: JSH: Direct demonstration of contradiction > > > > > Start with > > > > > > > > (5 + (sqrt(37) - sqrt(13))/2)((sqrt(37) - sqrt(13))/2) = (5 - > > > > > sqrt(13))(5 + sqrt(37))/2 > > > > > > > > > > > Okay, I'll tell you what the story is. The ring of integers in > > > > Q(sqrt(13),sqrt(37)) is Z[(1+sqrt(13))/2,(1+sqrt(37))/2]. The ideal > > > > generated by 2 is a product of two prime ideals p_1 and p_2 of norm 4. > > > > Only one of them contains 5+(sqrt(37)-sqrt(13))/2, I'll let p_1 be that > > > > one. > > > > > > > > We have 3=((sqrt(13)+1)/2)((sqrt(13)-1)/2). The ideals generated by > > > > (sqrt(13)+1)/2 and (sqrt(13)-1)/2 are prime. > > > > > > > > Now, the ideal generated by 5+(sqrt(37)-sqrt(13))/2 is the product of > > > > p_1 and the ideal generated by (sqrt(13)-1)/2. > > > > > > > > And the ideal generated by (sqrt(37)+sqrt(13))/2 is the product of > > > > (p_2)^2 and the ideal generated by (sqrt(13)-1)/2. It turned out you > > > > were actually right that the gcd's with 3 are equal, in both cases they > > > > are (sqrt(13)-1)/2. However, this was just a fluke. > > > > > > > > > > Yet looking at > > > > > > (5 + (sqrt(37) - sqrt(13))/2)((sqrt(37) - sqrt(13))/2) = (5 - > > > > sqrt(13))(5 + sqrt(37))/2 > > > > > > > > My calculations were wrong. We have (2)=p_1.p_2 and > > > (3)=q_1.q_2.q_3.q_4. I'll let q_1.q_2 be invariant under the > > > automorphism which sends sqrt(13) to -sqrt(13), and q_1.q_3 be > > > invariant under the automorphism which sends sqrt(37) to -sqrt(37). > > > > > > Then I have > > > > > > (5+(sqrt(37)-sqrt(13))/2)=p_1.q_1.q_2 > > > ((sqrt(37)-sqrt(13))/2)=p_2.q_3.q_4 > > > (5-sqrt(13))=p_1.p_2.q_3.q_4 > > > ((5+sqrt(37))/2)=q_1.q_2 > > > > > > you can just note that > > > > > > (5 + (sqrt(37) - sqrt(13))/2) > > > > > > must be coprime to (sqrt(37) - sqrt(13))/2 with respect to 3, by just > > > > subtracting one from the other as that leaves 5. > > > > > > > > Yes that's right. The greatest common divisor of one with 3 is q_1.q_2, > > > the other is q_3.q_4. > > > > > > And since ((sqrt(37) - sqrt(13))/2)*((sqrt(37) + sqrt(13))/2) = 6 > > > > > > you know that (5 + (sqrt(37) - sqrt(13))/2) must share all of its > > > > factors in common with 3, with (sqrt(37) + sqrt(13))/2. > > > > > > > > You know that every prime ideal dividing (5+(sqrt(37)-sqrt(13))/2) and > > > 3 must divide (sqrt(37)+sqrt(13))/2. In fact, as it happens > > > gcd(5+(sqrt(37)-sqrt(13))/2,3)=gcd((sqrt(37)+sqrt(13))/2,3)=(sqrt(13)-1)/2. > > > > > > > So the next step is also trivial which is to divide one by the other > > > correcting for other prime factors, where in this case that means you > > > can correct for 2 by multiplying by 2 > > > > Nope. You can't clear all the factors of 2 from > > (sqrt(37) + sqrt(13))/2 by multiplying by 2. > > > > - William Hughes > > Except that right after I demonstrate:: > > z = 2(10 + sqrt(37) - sqrt(13))/(sqrt(37) + sqrt(13)) > > which gives z as a root of > > 9z^4 - 72z^3 - 409z^2 - 2212z - 144 = 0 > > where the lead term is 9 which is coprime to 2, showing that it worked! > > And factors in common with 2 were successfully removed. > > Nope. > > What matters here is not the factors of the first term or the constant > term, but the factors of the constant term divided by the first term. > > Let z= f_1/g_1, where f_1 and g_1 are coprime. As you correctly > note, f_1 is coprime to 3. > > Let, f_2/g_2, f_3/g_3, and f_4/g_4 be the other three roots > of > > P(x) = 9x^4 - 72x^3 - 409x^2 - 2212x - 144 > > Galois theory as usually taught suggests > that f_2, f_3 and f_4 must be coprime to 3. > > Then > > 9(x - f_1/g_1)(x - f_2/g_2)(x - f_3/g_3)(x - f_4/g_4) = P(x) > > So, looking at the contant terms on each side > > f_1 f_2 f_3 f_4 > 9 ---- ---- ---- ---- = 144 > g_1 g_2 g_3 g_4 > > or (f_1 f_2 f_3 f_4)/(g_1 g_2 g_3 g_4) = 144/9 = 16 > > No factors of 3. No contradiction. > (Note, before you start looking at factors of 2, remember that > f_1 and g_1 may both share factors with 2, even though they > are coprime). If James had got the polynomial right, then the fact that it is a > primitive polynomial and the leading coefficient is 9 would show that > there are prime ideals dividing 3 in the denominator of the fractional > ideal generated by any one of the roots (though the fact that the norm > is 16 would show that there are also other prime ideals dividing 3 in > the numerator). > I did make a mistake. The correct polynomial is 3z^4 - 100z^3 - 48z^2 - 440z - 96 = 0 where you get the same result as it is still non-monic and irreducible over Q. The leading coefficient is coprime to 2, so at issue only are factors of 3. If those had canceled out, then it would be a monic polynomial and its roots algebraic integers. > In fact the leading coefficient is 3, and the norm is 16. There is a > prime ideal dividing 3 in the numerator and another one in the > denominator. It might have happened that James' multiplying by 2 didn't work, but as > it turned out it did work. > But yeah it DID work and with it true that (10 + sqrt(37) - sqrt(13))/2 and (sqrt(37) + sqrt(13))/2 have the same factors in common with 3, I get the direct demonstration of the coverage problem of algebraic integers. It's not even a really complicated thing as I can demonstrate with x^2 + 4x + 3 = 0 as let x = 3y and you get 9y^2 + 12y + 3 = 0, and you can divide off 3, to get 3y^2 + 4y + 1 = 0 and see the why of what happened above. So that is a direct demonstration with numbers of how mathematicians after Gauss got lost on non-rational numbers and with use of the ring of algebraic integers, as that's a quirky ring with a coverage problem and proving things true in that ring, doesn't necessarily mean they are true in a ring without its peculiarities. James Harris === Subject: Re: JSH: Direct demonstration of contradiction > > > > > Start with > > > > > > > > > > (5 + (sqrt(37) - sqrt(13))/2)((sqrt(37) - sqrt(13))/2) = (5 - > > > > > sqrt(13))(5 + sqrt(37))/2 > > > > > > > > > > > > > Okay, I'll tell you what the story is. The ring of integers in > > > > > Q(sqrt(13),sqrt(37)) is Z[(1+sqrt(13))/2,(1+sqrt(37))/2]. The ideal > > > > > generated by 2 is a product of two prime ideals p_1 and p_2 of norm 4. > > > > > Only one of them contains 5+(sqrt(37)-sqrt(13))/2, I'll let p_1 be that > > > > > one. > > > > > > > > We have 3=((sqrt(13)+1)/2)((sqrt(13)-1)/2). The ideals generated by > > > > > (sqrt(13)+1)/2 and (sqrt(13)-1)/2 are prime. > > > > > > > > Now, the ideal generated by 5+(sqrt(37)-sqrt(13))/2 is the product of > > > > > p_1 and the ideal generated by (sqrt(13)-1)/2. > > > > > > > > And the ideal generated by (sqrt(37)+sqrt(13))/2 is the product of > > > > > (p_2)^2 and the ideal generated by (sqrt(13)-1)/2. It turned out you > > > > > were actually right that the gcd's with 3 are equal, in both cases they > > > > > are (sqrt(13)-1)/2. However, this was just a fluke. > > > > > > > > > > > Yet looking at > > > > > > > > (5 + (sqrt(37) - sqrt(13))/2)((sqrt(37) - sqrt(13))/2) = (5 - > > > > sqrt(13))(5 + sqrt(37))/2 > > > > > > > > > > My calculations were wrong. We have (2)=p_1.p_2 and > > > > (3)=q_1.q_2.q_3.q_4. I'll let q_1.q_2 be invariant under the > > > > automorphism which sends sqrt(13) to -sqrt(13), and q_1.q_3 be > > > > invariant under the automorphism which sends sqrt(37) to -sqrt(37). > > > > > > Then I have > > > > > > (5+(sqrt(37)-sqrt(13))/2)=p_1.q_1.q_2 > > > > ((sqrt(37)-sqrt(13))/2)=p_2.q_3.q_4 > > > > (5-sqrt(13))=p_1.p_2.q_3.q_4 > > > > ((5+sqrt(37))/2)=q_1.q_2 > > > > > > > you can just note that > > > > > > > > (5 + (sqrt(37) - sqrt(13))/2) > > > > > > > > must be coprime to (sqrt(37) - sqrt(13))/2 with respect to 3, by just > > > > subtracting one from the other as that leaves 5. > > > > > > > > > > Yes that's right. The greatest common divisor of one with 3 is q_1.q_2, > > > > the other is q_3.q_4. > > > > > > > And since ((sqrt(37) - sqrt(13))/2)*((sqrt(37) + sqrt(13))/2) = 6 > > > > > > > > you know that (5 + (sqrt(37) - sqrt(13))/2) must share all of its > > > > factors in common with 3, with (sqrt(37) + sqrt(13))/2. > > > > > > > > > > You know that every prime ideal dividing (5+(sqrt(37)-sqrt(13))/2) and > > > > 3 must divide (sqrt(37)+sqrt(13))/2. In fact, as it happens > > > > gcd(5+(sqrt(37)-sqrt(13))/2,3)=gcd((sqrt(37)+sqrt(13))/2,3)=(sqrt(13)-1)/2. > > > > > > > > So the next step is also trivial which is to divide one by the other > > > correcting for other prime factors, where in this case that means you > > > can correct for 2 by multiplying by 2 > > > > Nope. You can't clear all the factors of 2 from > > > (sqrt(37) + sqrt(13))/2 by multiplying by 2. > > > > - William Hughes > > > > Except that right after I demonstrate:: > > > > z = 2(10 + sqrt(37) - sqrt(13))/(sqrt(37) + sqrt(13)) > > > > which gives z as a root of > > > > 9z^4 - 72z^3 - 409z^2 - 2212z - 144 = 0 > > > > where the lead term is 9 which is coprime to 2, showing that it worked! > > And factors in common with 2 were successfully removed. > > Nope. > > What matters here is not the factors of the first term or the constant > > term, but the factors of the constant term divided by the first term. > > Let z= f_1/g_1, where f_1 and g_1 are coprime. As you correctly > > note, f_1 is coprime to 3. > > Let, f_2/g_2, f_3/g_3, and f_4/g_4 be the other three roots > > of > > P(x) = 9x^4 - 72x^3 - 409x^2 - 2212x - 144 > > Galois theory as usually taught suggests > > that f_2, f_3 and f_4 must be coprime to 3. > > Then > > 9(x - f_1/g_1)(x - f_2/g_2)(x - f_3/g_3)(x - f_4/g_4) = P(x) > > So, looking at the contant terms on each side > > f_1 f_2 f_3 f_4 > > 9 ---- ---- ---- ---- = 144 > > g_1 g_2 g_3 g_4 > > or (f_1 f_2 f_3 f_4)/(g_1 g_2 g_3 g_4) = 144/9 = 16 > > No factors of 3. No contradiction. > > (Note, before you start looking at factors of 2, remember that > > f_1 and g_1 may both share factors with 2, even though they > > are coprime). > > If James had got the polynomial right, then the fact that it is a > primitive polynomial and the leading coefficient is 9 would show that > there are prime ideals dividing 3 in the denominator of the fractional > ideal generated by any one of the roots (though the fact that the norm > is 16 would show that there are also other prime ideals dividing 3 in > the numerator). > > I did make a mistake. The correct polynomial is 3z^4 - 100z^3 - 48z^2 - 440z - 96 = 0 > No that's wrong as well. The coefficient in z^3 is -50 and the constant term is 48. I would imagine that the polynomial Tim posted is probably correct. > where you get the same result as it is still non-monic and irreducible > over Q. > We are agreed, the number is not an algebraic integer. > The leading coefficient is coprime to 2, so at issue only are factors > of 3. > True. > If those had canceled out, then it would be a monic polynomial and its > roots algebraic integers. > But they don't. As I told you, there is a prime ideal dividing 3 and (sqrt(37)+sqrt(13))/2 that does not divide (5+(sqrt(37)-sqrt(13))/2). And why shouldn't there be? > In fact the leading coefficient is 3, and the norm is 16. There is a > prime ideal dividing 3 in the numerator and another one in the > denominator. > > It might have happened that James' multiplying by 2 didn't work, but as > it turned out it did work. > > But yeah it DID work and with it true that (10 + sqrt(37) - sqrt(13))/2 and (sqrt(37) + sqrt(13))/2 have the same factors in common with 3, I get the direct demonstration > of the coverage problem of algebraic integers. > No, that is not true. The greatest common divisor with 3 is not the same. You point out that (10+sqrt(37)-sqrt(13))/2 and (sqrt(37)-sqrt(13))/2 are coprime. Also (sqrt(37)-sqrt(13))/2 and (sqrt(37)+sqrt(13))/2 are coprime and their product is 6. That shows that every prime ideal dividing (10+sqrt(37)-sqrt(13))/2 and 3 must also divide (sqrt(37)+sqrt(13))/2. It does not show that every prime ideal dividing (sqrt(37)+sqrt(13))/2 and 3 must also divide (10+sqrt(37)-sqrt(13))/2. You have never given a good argument for this and it is not true. These are the factorizations. ((10+sqrt(37)-sqrt(13))/2)=p_1.(q_1)^2 ((sqrt(37)-sqrt(13))/2)=p_2.q_2.q_3 (5-sqrt(13))=p_1.p_2.q_1.q_2 ((5+sqrt(37))/2)=q_1.q_3 Also ((sqrt(37)+sqrt(13))/2)=p_1.q_1.q_4 (2)=p_1.p_2 (3)=q_1.q_2.q_3.q_4 This is consistent with the observation that ((10+sqrt(37)-sqrt(13))/2)((sqrt(37)-sqrt(13))/2)=(5-sqrt(13))((5+sqrt(37))/ 2) and that ((sqrt(37)-sqrt(13))/2)((sqrt(37)+sqrt(13))/2)=6. It is also consistent with your observation that there is no prime ideal dividing (10+sqrt(37)-sqrt(13))/2, ((sqrt(37)-sqrt(13))/2), and 3. It does not follow from these observations that every prime ideal dividing (sqrt(37)+sqrt(13))/2 and 3 must also divide (10+sqrt(37)-sqrt(13))/2, as the factorizations illustrate. You were claiming to derive that conclusion from those observations alone. As the factorizations show, your argument must have been wrong. === Subject: Re: JSH: Direct demonstration of contradiction > > > > > > Start with > > > > > > > > > > (5 + (sqrt(37) - sqrt(13))/2)((sqrt(37) - sqrt(13))/2) = (5 - > > > > > > sqrt(13))(5 + sqrt(37))/2 > > > > > > > > > > > > > > Okay, I'll tell you what the story is. The ring of integers in > > > > > Q(sqrt(13),sqrt(37)) is Z[(1+sqrt(13))/2,(1+sqrt(37))/2]. The ideal > > > > > generated by 2 is a product of two prime ideals p_1 and p_2 of norm 4. > > > > > Only one of them contains 5+(sqrt(37)-sqrt(13))/2, I'll let p_1 be that > > > > > one. > > > > > > > > > > We have 3=((sqrt(13)+1)/2)((sqrt(13)-1)/2). The ideals generated by > > > > > (sqrt(13)+1)/2 and (sqrt(13)-1)/2 are prime. > > > > > > > > > > Now, the ideal generated by 5+(sqrt(37)-sqrt(13))/2 is the product of > > > > > p_1 and the ideal generated by (sqrt(13)-1)/2. > > > > > > > > > > And the ideal generated by (sqrt(37)+sqrt(13))/2 is the product of > > > > > (p_2)^2 and the ideal generated by (sqrt(13)-1)/2. It turned out you > > > > > were actually right that the gcd's with 3 are equal, in both cases they > > > > > are (sqrt(13)-1)/2. However, this was just a fluke. > > > > > > > > > > > > > Yet looking at > > > > > > > > (5 + (sqrt(37) - sqrt(13))/2)((sqrt(37) - sqrt(13))/2) = (5 - > > > > > sqrt(13))(5 + sqrt(37))/2 > > > > > > > > > > > My calculations were wrong. We have (2)=p_1.p_2 and > > > > (3)=q_1.q_2.q_3.q_4. I'll let q_1.q_2 be invariant under the > > > > automorphism which sends sqrt(13) to -sqrt(13), and q_1.q_3 be > > > > invariant under the automorphism which sends sqrt(37) to -sqrt(37). > > > > > > > > Then I have > > > > > > > > (5+(sqrt(37)-sqrt(13))/2)=p_1.q_1.q_2 > > > > ((sqrt(37)-sqrt(13))/2)=p_2.q_3.q_4 > > > > (5-sqrt(13))=p_1.p_2.q_3.q_4 > > > > ((5+sqrt(37))/2)=q_1.q_2 > > > > > > > > you can just note that > > > > > > > > (5 + (sqrt(37) - sqrt(13))/2) > > > > > > > > must be coprime to (sqrt(37) - sqrt(13))/2 with respect to 3, by just > > > > > subtracting one from the other as that leaves 5. > > > > > > > > > > > Yes that's right. The greatest common divisor of one with 3 is q_1.q_2, > > > > the other is q_3.q_4. > > > > > > > > And since ((sqrt(37) - sqrt(13))/2)*((sqrt(37) + sqrt(13))/2) = 6 > > > > > > > > you know that (5 + (sqrt(37) - sqrt(13))/2) must share all of its > > > > > factors in common with 3, with (sqrt(37) + sqrt(13))/2. > > > > > > > > > > > You know that every prime ideal dividing (5+(sqrt(37)-sqrt(13))/2) and > > > > 3 must divide (sqrt(37)+sqrt(13))/2. In fact, as it happens > > > > gcd(5+(sqrt(37)-sqrt(13))/2,3)=gcd((sqrt(37)+sqrt(13))/2,3)=(sqrt(13)-1)/2. > > > > > > > > > > So the next step is also trivial which is to divide one by the other > > > > correcting for other prime factors, where in this case that means you > > > > can correct for 2 by multiplying by 2 > > > > > > Nope. You can't clear all the factors of 2 from > > > (sqrt(37) + sqrt(13))/2 by multiplying by 2. > > > > > > - William Hughes > > > > Except that right after I demonstrate:: > > > > z = 2(10 + sqrt(37) - sqrt(13))/(sqrt(37) + sqrt(13)) > > > > which gives z as a root of > > > > 9z^4 - 72z^3 - 409z^2 - 2212z - 144 = 0 > > > > where the lead term is 9 which is coprime to 2, showing that it worked! > > > And factors in common with 2 were successfully removed. > > > > Nope. > > > > What matters here is not the factors of the first term or the constant > > term, but the factors of the constant term divided by the first term. > > > > Let z= f_1/g_1, where f_1 and g_1 are coprime. As you correctly > > note, f_1 is coprime to 3. > > > > Let, f_2/g_2, f_3/g_3, and f_4/g_4 be the other three roots > > of > > > > P(x) = 9x^4 - 72x^3 - 409x^2 - 2212x - 144 > > > > Galois theory as usually taught suggests > > that f_2, f_3 and f_4 must be coprime to 3. > > > > Then > > > > 9(x - f_1/g_1)(x - f_2/g_2)(x - f_3/g_3)(x - f_4/g_4) = P(x) > > > > So, looking at the contant terms on each side > > > > f_1 f_2 f_3 f_4 > > 9 ---- ---- ---- ---- = 144 > > g_1 g_2 g_3 g_4 > > > > or (f_1 f_2 f_3 f_4)/(g_1 g_2 g_3 g_4) = 144/9 = 16 > > > > No factors of 3. No contradiction. > > (Note, before you start looking at factors of 2, remember that > > f_1 and g_1 may both share factors with 2, even though they > > are coprime). > > If James had got the polynomial right, then the fact that it is a > > primitive polynomial and the leading coefficient is 9 would show that > > there are prime ideals dividing 3 in the denominator of the fractional > > ideal generated by any one of the roots (though the fact that the norm > > is 16 would show that there are also other prime ideals dividing 3 in > > the numerator). > > I did make a mistake. The correct polynomial is > > 3z^4 - 100z^3 - 48z^2 - 440z - 96 = 0 > > No that's wrong as well. The coefficient in z^3 is -50 and the constant > term is 48. I would imagine that the polynomial Tim posted is probably > correct. > Oh yeah, I made another dumb mistake. Should be 3z^4 - 50z^3 - 24z^2 - 220z - 48 = 0. > where you get the same result as it is still non-monic and irreducible > over Q. > > We are agreed, the number is not an algebraic integer. > The leading coefficient is coprime to 2, so at issue only are factors > of 3. > > True. > If those had canceled out, then it would be a monic polynomial and its > roots algebraic integers. > > But they don't. As I told you, there is a prime ideal dividing 3 and > (sqrt(37)+sqrt(13))/2 that does not divide (5+(sqrt(37)-sqrt(13))/2). > And why shouldn't there be? > Both are coprime to (sqrt(37) - sqrt(13))/2. But (sqrt(37) + sqrt(13))*(sqrt(37) - sqrt(13))/4 = 6 so all factors that are coprime to sqrt(37) - sqrt(13) with respect to 3, MUST be factors of sqrt(37) + sqrt(13) and since (5+(sqrt(37)-sqrt(13))/2) is coprime to sqrt(37) - sqrt(13) as well, it must share its factors in common with 3 with sqrt(37) + sqrt(13). Trivial. So your denial and continued posting as if you have some mathematical objection is bizarre. Let's say you argue that just because (5+(sqrt(37)-sqrt(13))/2) is coprime to sqrt(37) - sqrt(13) is can have some factor of 3 NOT in common with sqrt(37) + sqrt(13), well, the question then is, how? How? Why do you continue when you are so massively beaten on the mathematical facts? And then dare to ask me to thank you for trying? You're not trying. You're missing the obvious. With (5+(sqrt(37)-sqrt(13))/2) coprime to sqrt(37) - sqrt(13) it MUST share its factors in common with 3 with sqrt(37) + sqrt(13). If you accept that trivial point then the will to continue to debate should be completely out of you. James Harris === Subject: Re: JSH: Direct demonstration of contradiction > > > > > > Start with > > > > > > > > > > > > (5 + (sqrt(37) - sqrt(13))/2)((sqrt(37) - sqrt(13))/2) = (5 - > > > > > > sqrt(13))(5 + sqrt(37))/2 > > > > > > > > > > > > > > > > Okay, I'll tell you what the story is. The ring of integers in > > > > > > Q(sqrt(13),sqrt(37)) is Z[(1+sqrt(13))/2,(1+sqrt(37))/2]. The ideal > > > > > > generated by 2 is a product of two prime ideals p_1 and p_2 of norm 4. > > > > > > Only one of them contains 5+(sqrt(37)-sqrt(13))/2, I'll let p_1 be that > > > > > > one. > > > > > > > > > > We have 3=((sqrt(13)+1)/2)((sqrt(13)-1)/2). The ideals generated by > > > > > > (sqrt(13)+1)/2 and (sqrt(13)-1)/2 are prime. > > > > > > > > > > Now, the ideal generated by 5+(sqrt(37)-sqrt(13))/2 is the product of > > > > > > p_1 and the ideal generated by (sqrt(13)-1)/2. > > > > > > > > > > And the ideal generated by (sqrt(37)+sqrt(13))/2 is the product of > > > > > > (p_2)^2 and the ideal generated by (sqrt(13)-1)/2. It turned out you > > > > > > were actually right that the gcd's with 3 are equal, in both cases they > > > > > > are (sqrt(13)-1)/2. However, this was just a fluke. > > > > > > > > > > > > > > Yet looking at > > > > > > > > > > (5 + (sqrt(37) - sqrt(13))/2)((sqrt(37) - sqrt(13))/2) = (5 - > > > > > sqrt(13))(5 + sqrt(37))/2 > > > > > > > > > > > > > My calculations were wrong. We have (2)=p_1.p_2 and > > > > > (3)=q_1.q_2.q_3.q_4. I'll let q_1.q_2 be invariant under the > > > > > automorphism which sends sqrt(13) to -sqrt(13), and q_1.q_3 be > > > > > invariant under the automorphism which sends sqrt(37) to -sqrt(37). > > > > > > > > Then I have > > > > > > > > (5+(sqrt(37)-sqrt(13))/2)=p_1.q_1.q_2 > > > > > ((sqrt(37)-sqrt(13))/2)=p_2.q_3.q_4 > > > > > (5-sqrt(13))=p_1.p_2.q_3.q_4 > > > > > ((5+sqrt(37))/2)=q_1.q_2 > > > > > > > > > you can just note that > > > > > > > > > > (5 + (sqrt(37) - sqrt(13))/2) > > > > > > > > > > must be coprime to (sqrt(37) - sqrt(13))/2 with respect to 3, by just > > > > > subtracting one from the other as that leaves 5. > > > > > > > > > > > > > Yes that's right. The greatest common divisor of one with 3 is q_1.q_2, > > > > > the other is q_3.q_4. > > > > > > > > > And since ((sqrt(37) - sqrt(13))/2)*((sqrt(37) + sqrt(13))/2) = 6 > > > > > > > > > > you know that (5 + (sqrt(37) - sqrt(13))/2) must share all of its > > > > > factors in common with 3, with (sqrt(37) + sqrt(13))/2. > > > > > > > > > > > > > You know that every prime ideal dividing (5+(sqrt(37)-sqrt(13))/2) and > > > > > 3 must divide (sqrt(37)+sqrt(13))/2. In fact, as it happens > > > > > gcd(5+(sqrt(37)-sqrt(13))/2,3)=gcd((sqrt(37)+sqrt(13))/2,3)=(sqrt(13)-1)/2. > > > > > > > > > > > So the next step is also trivial which is to divide one by the other > > > > correcting for other prime factors, where in this case that means you > > > > can correct for 2 by multiplying by 2 > > > > > > Nope. You can't clear all the factors of 2 from > > > > (sqrt(37) + sqrt(13))/2 by multiplying by 2. > > > > > > - William Hughes > > > > > > Except that right after I demonstrate:: > > > > > > z = 2(10 + sqrt(37) - sqrt(13))/(sqrt(37) + sqrt(13)) > > > > > > which gives z as a root of > > > > > > 9z^4 - 72z^3 - 409z^2 - 2212z - 144 = 0 > > > > > > where the lead term is 9 which is coprime to 2, showing that it worked! > > > And factors in common with 2 were successfully removed. > > > > Nope. > > > > What matters here is not the factors of the first term or the constant > > > term, but the factors of the constant term divided by the first term. > > > > Let z= f_1/g_1, where f_1 and g_1 are coprime. As you correctly > > > note, f_1 is coprime to 3. > > > > Let, f_2/g_2, f_3/g_3, and f_4/g_4 be the other three roots > > > of > > > > P(x) = 9x^4 - 72x^3 - 409x^2 - 2212x - 144 > > > > Galois theory as usually taught suggests > > > that f_2, f_3 and f_4 must be coprime to 3. > > > > Then > > > > 9(x - f_1/g_1)(x - f_2/g_2)(x - f_3/g_3)(x - f_4/g_4) = P(x) > > > > So, looking at the contant terms on each side > > > > f_1 f_2 f_3 f_4 > > > 9 ---- ---- ---- ---- = 144 > > > g_1 g_2 g_3 g_4 > > > > or (f_1 f_2 f_3 f_4)/(g_1 g_2 g_3 g_4) = 144/9 = 16 > > > > No factors of 3. No contradiction. > > > (Note, before you start looking at factors of 2, remember that > > > f_1 and g_1 may both share factors with 2, even though they > > > are coprime). > > > > If James had got the polynomial right, then the fact that it is a > > primitive polynomial and the leading coefficient is 9 would show that > > there are prime ideals dividing 3 in the denominator of the fractional > > ideal generated by any one of the roots (though the fact that the norm > > is 16 would show that there are also other prime ideals dividing 3 in > > the numerator). > > > > I did make a mistake. The correct polynomial is > > 3z^4 - 100z^3 - 48z^2 - 440z - 96 = 0 > > No that's wrong as well. The coefficient in z^3 is -50 and the constant > term is 48. I would imagine that the polynomial Tim posted is probably > correct. > > Oh yeah, I made another dumb mistake. Remember that, for later on. > Should be 3z^4 - 50z^3 - 24z^2 - 220z - 48 = 0. > where you get the same result as it is still non-monic and irreducible > > over Q. > > We are agreed, the number is not an algebraic integer. > > The leading coefficient is coprime to 2, so at issue only are factors > > of 3. > > True. > > If those had canceled out, then it would be a monic polynomial and its > > roots algebraic integers. > > But they don't. As I told you, there is a prime ideal dividing 3 and > (sqrt(37)+sqrt(13))/2 that does not divide (5+(sqrt(37)-sqrt(13))/2). > And why shouldn't there be? > > Both are coprime to (sqrt(37) - sqrt(13))/2. But (sqrt(37) + sqrt(13))*(sqrt(37) - sqrt(13))/4 = 6 so all factors that are coprime to sqrt(37) - sqrt(13) with respect to > 3, MUST be factors of sqrt(37) + sqrt(13) and since (5+(sqrt(37)-sqrt(13))/2) is coprime to sqrt(37) - sqrt(13) > as well, it must share its factors in common with 3 with sqrt(37) + > sqrt(13). Trivial. So your denial and continued posting as if you have some mathematical > objection is bizarre. Let's say you argue that just because (5+(sqrt(37)-sqrt(13))/2) is > coprime to sqrt(37) - sqrt(13) is can have some factor of 3 NOT in > common with sqrt(37) + sqrt(13), well, the question then is, how? How? Why do you continue when you are so massively beaten on the > mathematical facts? Why does JSH continue when he is so massively beaten on the arithmetical facts? Oh, yeah, for this reason: Minor errors are MINOR that's why they are minor errors. -- JSH, September 21, 2006 > And then dare to ask me to thank you for trying? You're not trying. You're missing the obvious. With (5+(sqrt(37)-sqrt(13))/2) coprime to sqrt(37) - sqrt(13) it MUST > share its factors in common with 3 with sqrt(37) + sqrt(13). If you accept that trivial point then the will to continue to debate > should be completely out of you. > James Harris Yeah, right. And if beer cans were airplanes, my brother's apartment would be an airport. --- Christopher Heckman === Subject: Re: JSH: Direct demonstration of contradiction > > > > > > Start with > > > > > > > > > > > > (5 + (sqrt(37) - sqrt(13))/2)((sqrt(37) - sqrt(13))/2) = (5 - > > > > > > sqrt(13))(5 + sqrt(37))/2 > > > > > > > > > > > > > > > > Okay, I'll tell you what the story is. The ring of integers in > > > > > > Q(sqrt(13),sqrt(37)) is Z[(1+sqrt(13))/2,(1+sqrt(37))/2]. The ideal > > > > > > generated by 2 is a product of two prime ideals p_1 and p_2 of norm 4. > > > > > > Only one of them contains 5+(sqrt(37)-sqrt(13))/2, I'll let p_1 be that > > > > > > one. > > > > > > > > > > We have 3=((sqrt(13)+1)/2)((sqrt(13)-1)/2). The ideals generated by > > > > > > (sqrt(13)+1)/2 and (sqrt(13)-1)/2 are prime. > > > > > > > > > > Now, the ideal generated by 5+(sqrt(37)-sqrt(13))/2 is the product of > > > > > > p_1 and the ideal generated by (sqrt(13)-1)/2. > > > > > > > > > > And the ideal generated by (sqrt(37)+sqrt(13))/2 is the product of > > > > > > (p_2)^2 and the ideal generated by (sqrt(13)-1)/2. It turned out you > > > > > > were actually right that the gcd's with 3 are equal, in both cases they > > > > > > are (sqrt(13)-1)/2. However, this was just a fluke. > > > > > > > > > > > > > > Yet looking at > > > > > > > > > > (5 + (sqrt(37) - sqrt(13))/2)((sqrt(37) - sqrt(13))/2) = (5 - > > > > > sqrt(13))(5 + sqrt(37))/2 > > > > > > > > > > > > > My calculations were wrong. We have (2)=p_1.p_2 and > > > > > (3)=q_1.q_2.q_3.q_4. I'll let q_1.q_2 be invariant under the > > > > > automorphism which sends sqrt(13) to -sqrt(13), and q_1.q_3 be > > > > > invariant under the automorphism which sends sqrt(37) to -sqrt(37). > > > > > > > > Then I have > > > > > > > > (5+(sqrt(37)-sqrt(13))/2)=p_1.q_1.q_2 > > > > > ((sqrt(37)-sqrt(13))/2)=p_2.q_3.q_4 > > > > > (5-sqrt(13))=p_1.p_2.q_3.q_4 > > > > > ((5+sqrt(37))/2)=q_1.q_2 > > > > > > > > > you can just note that > > > > > > > > > > (5 + (sqrt(37) - sqrt(13))/2) > > > > > > > > > > must be coprime to (sqrt(37) - sqrt(13))/2 with respect to 3, by just > > > > > subtracting one from the other as that leaves 5. > > > > > > > > > > > > > Yes that's right. The greatest common divisor of one with 3 is q_1.q_2, > > > > > the other is q_3.q_4. > > > > > > > > > And since ((sqrt(37) - sqrt(13))/2)*((sqrt(37) + sqrt(13))/2) = 6 > > > > > > > > > > you know that (5 + (sqrt(37) - sqrt(13))/2) must share all of its > > > > > factors in common with 3, with (sqrt(37) + sqrt(13))/2. > > > > > > > > > > > > > You know that every prime ideal dividing (5+(sqrt(37)-sqrt(13))/2) and > > > > > 3 must divide (sqrt(37)+sqrt(13))/2. In fact, as it happens > > > > > gcd(5+(sqrt(37)-sqrt(13))/2,3)=gcd((sqrt(37)+sqrt(13))/2,3)=(sqrt(13)-1)/2. > > > > > > > > > > > So the next step is also trivial which is to divide one by the other > > > > correcting for other prime factors, where in this case that means you > > > > can correct for 2 by multiplying by 2 > > > > > > Nope. You can't clear all the factors of 2 from > > > > (sqrt(37) + sqrt(13))/2 by multiplying by 2. > > > > > > - William Hughes > > > > > > Except that right after I demonstrate:: > > > > > > z = 2(10 + sqrt(37) - sqrt(13))/(sqrt(37) + sqrt(13)) > > > > > > which gives z as a root of > > > > > > 9z^4 - 72z^3 - 409z^2 - 2212z - 144 = 0 > > > > > > where the lead term is 9 which is coprime to 2, showing that it worked! > > > And factors in common with 2 were successfully removed. > > > > Nope. > > > > What matters here is not the factors of the first term or the constant > > > term, but the factors of the constant term divided by the first term. > > > > Let z= f_1/g_1, where f_1 and g_1 are coprime. As you correctly > > > note, f_1 is coprime to 3. > > > > Let, f_2/g_2, f_3/g_3, and f_4/g_4 be the other three roots > > > of > > > > P(x) = 9x^4 - 72x^3 - 409x^2 - 2212x - 144 > > > > Galois theory as usually taught suggests > > > that f_2, f_3 and f_4 must be coprime to 3. > > > > Then > > > > 9(x - f_1/g_1)(x - f_2/g_2)(x - f_3/g_3)(x - f_4/g_4) = P(x) > > > > So, looking at the contant terms on each side > > > > f_1 f_2 f_3 f_4 > > > 9 ---- ---- ---- ---- = 144 > > > g_1 g_2 g_3 g_4 > > > > or (f_1 f_2 f_3 f_4)/(g_1 g_2 g_3 g_4) = 144/9 = 16 > > > > No factors of 3. No contradiction. > > > (Note, before you start looking at factors of 2, remember that > > > f_1 and g_1 may both share factors with 2, even though they > > > are coprime). > > > > If James had got the polynomial right, then the fact that it is a > > primitive polynomial and the leading coefficient is 9 would show that > > there are prime ideals dividing 3 in the denominator of the fractional > > ideal generated by any one of the roots (though the fact that the norm > > is 16 would show that there are also other prime ideals dividing 3 in > > the numerator). > > > > I did make a mistake. The correct polynomial is > > 3z^4 - 100z^3 - 48z^2 - 440z - 96 = 0 > > No that's wrong as well. The coefficient in z^3 is -50 and the constant > term is 48. I would imagine that the polynomial Tim posted is probably > correct. > > Oh yeah, I made another dumb mistake. Should be 3z^4 - 50z^3 - 24z^2 - 220z - 48 = 0. > That's getting closer to Tim's answer. The constant term is definitely positive 48, not negative 48. I can't comment on the z^2 and z coefficients because I haven't checked, but my money would be on Tim being right about them as well. > > where you get the same result as it is still non-monic and irreducible > > over Q. > > We are agreed, the number is not an algebraic integer. > > The leading coefficient is coprime to 2, so at issue only are factors > > of 3. > > True. > > If those had canceled out, then it would be a monic polynomial and its > > roots algebraic integers. > > But they don't. As I told you, there is a prime ideal dividing 3 and > (sqrt(37)+sqrt(13))/2 that does not divide (5+(sqrt(37)-sqrt(13))/2). > And why shouldn't there be? > > Both are coprime to (sqrt(37) - sqrt(13))/2. But (sqrt(37) + sqrt(13))*(sqrt(37) - sqrt(13))/4 = 6 so all factors that are coprime to sqrt(37) - sqrt(13) with respect to > 3, MUST be factors of sqrt(37) + sqrt(13) > Yes, that's right. Every prime ideal which divides 5+(sqrt(37)-sqrt(13))/2 and 3 must also divide sqrt(37)+sqrt(13). I've always agreed with you about this. However, I don't agree that you couldn't get a prime ideal dividing the former to a higher power than the latter, and I don't agree that every prime ideal which divides sqrt(37)+sqrt(13) and 3 must also divide 5+(sqrt(37)-sqrt(13))/2. > and since (5+(sqrt(37)-sqrt(13))/2) is coprime to sqrt(37) - sqrt(13) > as well, it must share its factors in common with 3 with sqrt(37) + > sqrt(13). > As I've told you the largest ideal dividing 5+(sqrt(37)-sqrt(13))/2 and 3 is (q_1)^2. The largest ideal dividing 3 and (sqrt(37)+sqrt(13))/2 is q_1.q_4. So it's true that every prime ideal that divides 5+(sqrt(37)-sqrt(13))/2 and 3 also divides (sqrt(37)+sqrt(13))/2. However, the prime ideal in question occurs to a higher power in the former number than in the latter. And there is a prime ideal that occurs in the latter which does not occur in the former at all. That's why the former divided by the latter is not an algebraic integer. > Trivial. > Depends what you mean by the vague expression share its factors in common. The interpretation which I've agreed with already is trivial, yes. The interpretation which you're sliding to which says that the largest ideal dividing 5+(sqrt(37)-sqrt(13))/2 and 3 has to be divisible by the largest ideal dividing (sqrt(37)+sqrt(13))/2 and 3 is not true. > So your denial and continued posting as if you have some mathematical > objection is bizarre. > What I find bizarre is how you don't understand the significance of my factorizations. You've observed the following: (5+(sqrt(37)-sqrt(13))/2)((sqrt(37)-sqrt(13))/2)=(5-sqrt(13))((5+sqrt(37))/2 ) ((sqrt(37)-sqrt(13))/2)((sqrt(37)+sqrt(13))/2)=6 5+(sqrt(37)-sqrt(13))/2 and (sqrt(37)-sqrt(13))/2 are coprime (sqrt(37)-sqrt(13))/2 and (sqrt(37)+sqrt(13))/2 are coprime and from this you claim to be able to derive the greatest ideal dividing 5+(sqrt(37)-sqrt(13))/2 and 3 is divisible by the greatest ideal dividing (sqrt(37)+sqrt(13))/2 and 3. Since this *does* happen to be the case when we substitute 2 for 3, then if this were true it would indeed follow that (5+(sqrt(37)-sqrt(13))/2)/((sqrt(37)+sqrt(13))/2) would be an algebraic integer, and we would have a contradiction since we've observed that it isn't. Now here are my factorizations. (2)=p_1.p_2 (3)=q_1.q_2.q_3.q_4 (5+(sqrt(37)-sqrt(13))/2)=p_1.(q_1)^2 ((sqrt(37)-sqrt(13))/2)=p_2.q_2.q_3 ((sqrt(37)+sqrt(13))/2)=p_1.q_1.q_4 (5-sqrt(13))=p_1.p_2.q_1.q_2 ((5+sqrt(37))/2)=q_1.q_2 Now, my factorizations are consistent with all the observations you've used, but they contradict your conclusion and they are consistent with the leading coefficient of the minimum primitive polynomial for 2(5+(sqrt(37)-sqrt(13))/2)/((sqrt(37)+sqrt(13))/2) being 3, as we've observed. Since my factorizations are consistent with all the observations you've used and they contradict your conclusion, your argument must be invalid. Unless your argument gives some reason why these factorizations are impossible. That's your job. Explain why my factorizations must be wrong. If you can do it, that should help to clarify your argument. Of course, if you are right and we do have a sound argument for the number in question being an algebraic integer, that means that mathematics is inconsistent. Presumably you would reply that the contradiction occurred because the argument for the contradiction made use of some faulty theorems. In that case, it's your job to specify which ones and point out the mistakes in the proofs. That's another thing you've got to do. But in the meantime, just clarify your argument. My factorizations are consistent with all the observations you've made and they contradict your conclusion. So how does your argument show my factorizations to be impossible? Work through your argument step by step using my factorizations, and see where you first encounter a contradiction. Then explain it to me. > Let's say you argue that just because (5+(sqrt(37)-sqrt(13))/2) is > coprime to sqrt(37) - sqrt(13) is can have some factor of 3 NOT in > common with sqrt(37) + sqrt(13), well, the question then is, how? I don't say that. You really should look at my factorizations. I've told you exactly what I'm claiming. I say there can be a non-unit factor of 3 dividing sqrt(37)+sqrt(13) which doesn't divide 5+(sqrt(37)-sqrt(13))/2. You've never given me any reason that I can understand for thinking otherwise. My calculations say that that is indeed the case. Show me why I must be wrong. How? Why do you continue when you are so massively beaten on the > mathematical facts? > I've presented the mathematical facts to you. They are consistent with all the observations you've made. So what's the problem? > And then dare to ask me to thank you for trying? > Yes, you should thank me even if I am mistaken, which I'm not. This process will help to clarify your argument to other people. Either that or it will show you where you've gone wrong. > You're not trying. You're missing the obvious. > Then enlighten me. Why are my factorizations impossible? > With (5+(sqrt(37)-sqrt(13))/2) coprime to sqrt(37) - sqrt(13) it MUST > share its factors in common with 3 with sqrt(37) + sqrt(13). > Yes, every prime ideal dividing 5+(sqrt(37)-sqrt(13))/2 and 3 must also divide sqrt(37)+sqrt(13). That part of your argument is quite correct. I've always agreed with you on this, and my factorizations bear it out. But I do not agree that every prime ideal dividing sqrt(37)+sqrt(13) and 3 must divide 5+(sqrt(37)-sqrt(13))/2. As far as I'm concerned, you've never given me any good reason to think this. And my calculations say otherwise. So now it's your job to tell me why my calculations must be wrong. > If you accept that trivial point then the will to continue to debate > should be completely out of you. > I've told you exactly what I accept and what I don't accept. I've told you exactly what I think the mathematical situation is. I've given you the factorizations. If you want to know what I believe, look at them and work it out. Now, my factorizations are consistent with all the observations you've made use of, but they contradict your conclusion. So either your argument must be invalid, or there must be another observation you're making use of which you haven't told us about yet, which contradicts my factorizations. So tell us what it is. > > James Harris === Subject: Re: JSH: Direct demonstration of contradiction [jstevh@msn.com] > ... > Except that right after I demonstrate:: z = 2(10 + sqrt(37) - sqrt(13))/(sqrt(37) + sqrt(13)) which gives z as a root of 9z^4 - 72z^3 - 409z^2 - 2212z - 144 = 0 [rupertmccallum@yahoo.com] >> ... >> If James had got the polynomial right, then the fact that it is a >> primitive polynomial and the leading coefficient is 9 would show that >> there are prime ideals dividing 3 in the denominator of the fractional >> ideal generated by any one of the roots (though the fact that the norm >> is 16 would show that there are also other prime ideals dividing 3 in >> the numerator). [jstevh@msn.com] > ... > I did make a mistake. The correct polynomial is 3z^4 - 100z^3 - 48z^2 - 440z - 96 = 0 [rupertmccallum@yahoo.com] No that's wrong as well. The coefficient in z^3 is -50 and the constant term is 48. I would imagine that the polynomial Tim posted is probably correct. [jstevh@msn.com] >> Oh yeah, I made another dumb mistake. >> Should be >> 3z^4 - 50z^3 - 24z^2 - 220z - 48 = 0. [rupertmccallum@yahoo.com] > That's getting closer to Tim's answer. The constant term is definitely > positive 48, not negative 48. I can't comment on the z^2 and z > coefficients because I haven't checked, but my money would be on Tim > being right about them as well. And if James were a betting man, you would win :-) LOL -- you can't even /give/ this character a correct answer without multiple rounds of him flailing away at more broken attempts. To be fair, it takes a bit of skill to derive one of these, and also a bit of skill to verify one. I have an old copy of Macysma, which has no builtin facilities for computing minimal polynomials directly, but it certainly helps with that and is perfect for verifying them. Of course I so verified every polynomial I've posted. But it takes no skill whatsoever, and little time, merely to note that z = 2(10 + sqrt(37) - sqrt(13))/(sqrt(37) + sqrt(13)) is approximately equal to 2.5757, and plug that value in to a /claimed/ polynomial to see whether z is a /plausible/ root. I can't believe James is too stupid to do that much, so it must be that he's too careless and/or lazy. If he'd do that with his 3rd guess above: 3z^4 - 50z^3 - 24z^2 - 220z - 48 he'd see in seconds that its value at z=2.5757 is approximately -1500, which isn't plausibly close to 0. OTOH, at any point he could have tried that with the polynomial I gave him for this case: 3*z^4 - 50*z^3 + 24*z^2 + 200*z + 48 to see that it's approximately 0 at z=2.5757. It wasn't necessary to evaluate /exactly/ to learn that any of his first three guesses were obviously wrong. Hey, James: this would be a good time to explain to your vast audience exactly what it is about you or your math that merits taking you seriously. When you get a simple thing wrong three times in a row, and after being /handed/ a correct answer on a silver platter, a lot of the people looking for their next hero might start to suspect that when you tell them you're a mathematical genius, you're trying to con them. Heck, even I get that suspicion sometimes. Being a genius won't do you much good if your followers lose faith. > ... === Subject: Re: JSH: Direct demonstration of contradiction > > > > > > > Start with > > > > > > > > > > > > (5 + (sqrt(37) - sqrt(13))/2)((sqrt(37) - sqrt(13))/2) = (5 - > > > > > > > sqrt(13))(5 + sqrt(37))/2 > > > > > > > > > > > > > > > > > Okay, I'll tell you what the story is. The ring of integers in > > > > > > Q(sqrt(13),sqrt(37)) is Z[(1+sqrt(13))/2,(1+sqrt(37))/2]. The ideal > > > > > > generated by 2 is a product of two prime ideals p_1 and p_2 of norm 4. > > > > > > Only one of them contains 5+(sqrt(37)-sqrt(13))/2, I'll let p_1 be that > > > > > > one. > > > > > > > > > > > > We have 3=((sqrt(13)+1)/2)((sqrt(13)-1)/2). The ideals generated by > > > > > > (sqrt(13)+1)/2 and (sqrt(13)-1)/2 are prime. > > > > > > > > > > > > Now, the ideal generated by 5+(sqrt(37)-sqrt(13))/2 is the product of > > > > > > p_1 and the ideal generated by (sqrt(13)-1)/2. > > > > > > > > > > > > And the ideal generated by (sqrt(37)+sqrt(13))/2 is the product of > > > > > > (p_2)^2 and the ideal generated by (sqrt(13)-1)/2. It turned out you > > > > > > were actually right that the gcd's with 3 are equal, in both cases they > > > > > > are (sqrt(13)-1)/2. However, this was just a fluke. > > > > > > > > > > > > > > > > Yet looking at > > > > > > > > > > (5 + (sqrt(37) - sqrt(13))/2)((sqrt(37) - sqrt(13))/2) = (5 - > > > > > > sqrt(13))(5 + sqrt(37))/2 > > > > > > > > > > > > > > My calculations were wrong. We have (2)=p_1.p_2 and > > > > > (3)=q_1.q_2.q_3.q_4. I'll let q_1.q_2 be invariant under the > > > > > automorphism which sends sqrt(13) to -sqrt(13), and q_1.q_3 be > > > > > invariant under the automorphism which sends sqrt(37) to -sqrt(37). > > > > > > > > > > Then I have > > > > > > > > > > (5+(sqrt(37)-sqrt(13))/2)=p_1.q_1.q_2 > > > > > ((sqrt(37)-sqrt(13))/2)=p_2.q_3.q_4 > > > > > (5-sqrt(13))=p_1.p_2.q_3.q_4 > > > > > ((5+sqrt(37))/2)=q_1.q_2 > > > > > > > > > > you can just note that > > > > > > > > > > (5 + (sqrt(37) - sqrt(13))/2) > > > > > > > > > > must be coprime to (sqrt(37) - sqrt(13))/2 with respect to 3, by just > > > > > > subtracting one from the other as that leaves 5. > > > > > > > > > > > > > > Yes that's right. The greatest common divisor of one with 3 is q_1.q_2, > > > > > the other is q_3.q_4. > > > > > > > > > > And since ((sqrt(37) - sqrt(13))/2)*((sqrt(37) + sqrt(13))/2) = 6 > > > > > > > > > > you know that (5 + (sqrt(37) - sqrt(13))/2) must share all of its > > > > > > factors in common with 3, with (sqrt(37) + sqrt(13))/2. > > > > > > > > > > > > > > You know that every prime ideal dividing (5+(sqrt(37)-sqrt(13))/2) and > > > > > 3 must divide (sqrt(37)+sqrt(13))/2. In fact, as it happens > > > > > gcd(5+(sqrt(37)-sqrt(13))/2,3)=gcd((sqrt(37)+sqrt(13))/2,3)=(sqrt(13)-1)/2. > > > > > > > > > > > > > So the next step is also trivial which is to divide one by the other > > > > > correcting for other prime factors, where in this case that means you > > > > > can correct for 2 by multiplying by 2 > > > > > > > > Nope. You can't clear all the factors of 2 from > > > > (sqrt(37) + sqrt(13))/2 by multiplying by 2. > > > > > > > > - William Hughes > > > > > > Except that right after I demonstrate:: > > > > > > z = 2(10 + sqrt(37) - sqrt(13))/(sqrt(37) + sqrt(13)) > > > > > > which gives z as a root of > > > > > > 9z^4 - 72z^3 - 409z^2 - 2212z - 144 = 0 > > > > > > where the lead term is 9 which is coprime to 2, showing that it worked! > > > > And factors in common with 2 were successfully removed. > > > > > > Nope. > > > > > > What matters here is not the factors of the first term or the constant > > > term, but the factors of the constant term divided by the first term. > > > > > > Let z= f_1/g_1, where f_1 and g_1 are coprime. As you correctly > > > note, f_1 is coprime to 3. > > > > > > Let, f_2/g_2, f_3/g_3, and f_4/g_4 be the other three roots > > > of > > > > > > P(x) = 9x^4 - 72x^3 - 409x^2 - 2212x - 144 > > > > > > Galois theory as usually taught suggests > > > that f_2, f_3 and f_4 must be coprime to 3. > > > > > > Then > > > > > > 9(x - f_1/g_1)(x - f_2/g_2)(x - f_3/g_3)(x - f_4/g_4) = P(x) > > > > > > So, looking at the contant terms on each side > > > > > > f_1 f_2 f_3 f_4 > > > 9 ---- ---- ---- ---- = 144 > > > g_1 g_2 g_3 g_4 > > > > > > or (f_1 f_2 f_3 f_4)/(g_1 g_2 g_3 g_4) = 144/9 = 16 > > > > > > No factors of 3. No contradiction. > > > (Note, before you start looking at factors of 2, remember that > > > f_1 and g_1 may both share factors with 2, even though they > > > are coprime). > > > > If James had got the polynomial right, then the fact that it is a > > > primitive polynomial and the leading coefficient is 9 would show that > > > there are prime ideals dividing 3 in the denominator of the fractional > > > ideal generated by any one of the roots (though the fact that the norm > > > is 16 would show that there are also other prime ideals dividing 3 in > > > the numerator). > > > > > I did make a mistake. The correct polynomial is > > > > 3z^4 - 100z^3 - 48z^2 - 440z - 96 = 0 > > > > No that's wrong as well. The coefficient in z^3 is -50 and the constant > > term is 48. I would imagine that the polynomial Tim posted is probably > > correct. > > Oh yeah, I made another dumb mistake. > > Should be > > 3z^4 - 50z^3 - 24z^2 - 220z - 48 = 0. > > That's getting closer to Tim's answer. The constant term is definitely > positive 48, not negative 48. I can't comment on the z^2 and z > coefficients because I haven't checked, but my money would be on Tim > being right about them as well. > Oh, I got a sign wrong but the rest should be correct. 3z^4 - 50z^3 - 24z^2 - 220z + 48 = 0 > > where you get the same result as it is still non-monic and irreducible > > over Q. > > > > We are agreed, the number is not an algebraic integer. > > > The leading coefficient is coprime to 2, so at issue only are factors > > of 3. > > > > True. > > > If those had canceled out, then it would be a monic polynomial and its > > roots algebraic integers. > > > > But they don't. As I told you, there is a prime ideal dividing 3 and > > (sqrt(37)+sqrt(13))/2 that does not divide (5+(sqrt(37)-sqrt(13))/2). > > And why shouldn't there be? > > Both are coprime to (sqrt(37) - sqrt(13))/2. > > But (sqrt(37) + sqrt(13))*(sqrt(37) - sqrt(13))/4 = 6 > > so all factors that are coprime to sqrt(37) - sqrt(13) with respect to > 3, MUST be factors of > > sqrt(37) + sqrt(13) > > Yes, that's right. Every prime ideal which divides > 5+(sqrt(37)-sqrt(13))/2 and 3 must also divide sqrt(37)+sqrt(13). I've > always agreed with you about this. However, I don't agree that you > couldn't get a prime ideal dividing the former to a higher power than > the latter, and I don't agree that every prime ideal which divides > sqrt(37)+sqrt(13) and 3 must also divide 5+(sqrt(37)-sqrt(13))/2. > Then turn z upside down. z = 2(10 + sqrt(37) - sqrt(13))/(sqrt(37) + sqrt(13)) so use instead z = (sqrt(37) + sqrt(13))/(10 + sqrt(37) - sqrt(13)) and you'll find you get a non-monic polynomial irreducible over Q with a leading coefficient that has 2, 3 as prime factors. Now if you say that (10 + sqrt(37) - sqrt(13)) has its factors of 3 squared versus (sqrt(37) + sqrt(13)) then I just say use z = (sqrt(37) + sqrt(13))^2/(10 + sqrt(37) - sqrt(13)) as I already know that won't work. You can do the same trick that I used to eliminate 2, and you'll find you still get a non-monic polynomial irreducible over Q. And the leading coefficient will have 3 as a factor as will the constant term. If it were a matter of squares then ALL factors in common with 3 would have divided out of either the denominator or the numerator and I'd have ended up with either a leading coefficient with 3 as a factor with a constant term coprime to it, or the opposite. But BOTH the leading coefficient and constant term have 3 as a factor, destroying your claim. Do the math. I think it's simpler when people with math software do it as I keep screwing up the algebra. It's easy to do, yes, but it's tedious, especially when a computer can do it. And you will find you are wrong. Now, why? Well, it's just like that simple example with x^2 + 4x + 3 = 0 as with z = (sqrt(37) + sqrt(13))^2/(10 + sqrt(37) - sqrt(13)) when you start flipping signs in front of the square roots, and go through all the combinations, which is why you end up with a quartic anyway, you have some where there aren't any shared factors, so it's the same basic reason as why with x^2 + 4x + 3 = 0 using x = 3y will give you a non-monic as well. Trivial. > and since (5+(sqrt(37)-sqrt(13))/2) is coprime to sqrt(37) - sqrt(13) > as well, it must share its factors in common with 3 with sqrt(37) + > sqrt(13). > > As I've told you the largest ideal dividing 5+(sqrt(37)-sqrt(13))/2 and > 3 is (q_1)^2. The largest ideal dividing 3 and (sqrt(37)+sqrt(13))/2 is > q_1.q_4. So it's true that every prime ideal that divides > 5+(sqrt(37)-sqrt(13))/2 and 3 also divides (sqrt(37)+sqrt(13))/2. And I've explained how to test that claim. > However, the prime ideal in question occurs to a higher power in the > former number than in the latter. And there is a prime ideal that > occurs in the latter which does not occur in the former at all. That's > why the former divided by the latter is not an algebraic integer. > Then flip them, and see what happens. > Trivial. > > Depends what you mean by the vague expression share its factors in > common. The interpretation which I've agreed with already is trivial, > yes. The interpretation which you're sliding to which says that the > largest ideal dividing 5+(sqrt(37)-sqrt(13))/2 and 3 has to be > divisible by the largest ideal dividing (sqrt(37)+sqrt(13))/2 and 3 is > not true. > Astute readers already know that flipping won't make a difference as with 3z^4 - 50z^3 - 24z^2 - 220z + 48 = 0 you could just use z = 1/y to do the flipping and, of course, you will still have both the leading and constant coefficient having 3 as a prime factor. Trivial. Oh, also I know what you say about ideals can't be right, as if the factors in common with 3 were as you say, then again, only the leading coefficient would have 3 as a factor with the polynomial above. I don't really care as I think it's a little more dramatic if you can get algebraic integers one way as later you can't anyway with a follow-up result. This example is so dramatic because it leaves so little room for avoiding the result. In the past posters like Dik Winter and Arturo Magidin could have a field day confusing people about functions and polynomials and what was the key variable or not. But with (5 + (sqrt(37) - sqrt(13))/2)((sqrt(37) - sqrt(13))/2) = (5 - sqrt(13))(5 + sqrt(37))/2 where there are just numbers they are at a loss, which is good! As it's been quieter this time!!! And you appear to have continued as you got lost with that ideal nonsense and haven't appreciated how factors of 3 in both the leading coefficient and constant term destroy your position. But make no mistake, this demonstration shatters your claims, and with it validates mine, including my techniques of non-polynomial factorization, my paper and my claims of proof of Fermat's Last Theorem, from which I pulled a piece to come up with the paper. Since these results overturn over a hundred years worth of number theory and greatly change the historical positions of quite a few people famous in the math field, it is no surprise that there would be INCREDIBLE resistance, and one dead journal so far is not a surprise. The editors may have just found they could not continue publishing in the face of the reality. Entire universities can crumble from a result this big, which may seem strange, but it is possible, and yes, it would quite upset the academics in other departments from history to physics, especially because they wouldn't think it possible. But it is, and the way things are going, I'm more and more afraid I can't stop it. And no, I don't want to have entire universities shut-down, as I don't like the loss of the one math journal as it is. But with no one helping out, the energy is just building and building and building. Eventually, it has to release. James Harris === Subject: Re: JSH: Direct demonstration of contradiction Lets take a= (sqrt(37) - sqrt(13))/2, b = (sqrt(37)+sqrt(13))/2. Then if f|3 and f|5+a then f is coprime to a (any factor of f that divides a must also divide 3 and 5 and is therefore a unit) Now a*b = 6, so f|ab, and f is coprime to a, so f|b. So any factor of 3 that divides 5+a also divides 3. However A factor of 3 that divides 5+a to a higher power than one might not divide b to the same power. . Such a factor exists. Call it h_1. We have (h_1)^2 | (5+a) but (h_1)^2 does not divide b. (however h_1 does divide b). A factor of 3 that divides b, might not divide 5+a. Such a factor exists, call it h_2. h_2 | b, h_2 does not divide (5+a) Look at a/b. The numerator is divisible by h_1, the denominator is not. The denominator is divisible by h_2, the numerator is not. Look at b/a. The numerator is divisible by h_2, the denominator is not. The denominator is divisible by h_1, the numerator is not. So it does not matter how you flip things. There are factors of 3 in both the numerator and denominator. - William Hughes === Subject: Re: JSH: Direct demonstration of contradiction > Lets take a= (sqrt(37) - sqrt(13))/2, b = (sqrt(37)+sqrt(13))/2. Then if f|3 and f|5+a then f is coprime to a (any factor of f that > divides a must also divide 3 and 5 and is therefore a unit) Now a*b = 6, so f|ab, and f is coprime to a, so f|b. So any factor of 3 that divides 5+a also divides 3. However A factor of 3 that divides 5+a to a higher power than > one might not divide b to the same power. Correct. > . Such a factor exists. Call it h_1. > We have (h_1)^2 | (5+a) but (h_1)^2 does not divide b. > (however h_1 does divide b). A factor of 3 that divides b, might not divide 5+a. Incorrect. Remember a*b = 6, so if a factor is coprime to a with respect to 3 how can it not share all its factors in common with 3 with b? If you get that trivial point the debate is over. And it IS trivial. You have to have a will to be wrong to miss it. Notice how that destroy Rupert's objections as well and brings into question his claims. James Harris === Subject: Re: JSH: Direct demonstration of contradiction > Lets take a= (sqrt(37) - sqrt(13))/2, b = (sqrt(37)+sqrt(13))/2. > > Then if f|3 and f|5+a then f is coprime to a (any factor of f that > divides a must also divide 3 and 5 and is therefore a unit) > > Now a*b = 6, so f|ab, and f is coprime to a, so f|b. > > So any factor of 3 that divides 5+a also divides 3. > > However > > A factor of 3 that divides 5+a to a higher power than > one might not divide b to the same power. Correct. > . Such a factor exists. Call it h_1. > We have (h_1)^2 | (5+a) but (h_1)^2 does not divide b. > (however h_1 does divide b). > > A factor of 3 that divides b, might not divide 5+a. Incorrect. Remember a*b = 6, so if a factor is coprime to a with respect to 3 how > can it not share all its factors in common with 3 with b? > Yes, the fact that f|3 and f is coprime to a means that f|b or in your terms, f shares all its factors in common with 3 with b. This does not mean that b shares all its factors of 3 in common with f (or in other words just because f does not have any factor of 3 that b doesn't have, we can't say that b does not have any factor of 3 that f doesn't have). - William Hughes === Subject: Re: JSH: Direct demonstration of contradiction > > Lets take a= (sqrt(37) - sqrt(13))/2, b = (sqrt(37)+sqrt(13))/2. > > Then if f|3 and f|5+a then f is coprime to a (any factor of f that > > divides a must also divide 3 and 5 and is therefore a unit) > > Now a*b = 6, so f|ab, and f is coprime to a, so f|b. > > So any factor of 3 that divides 5+a also divides 3. > > However > > A factor of 3 that divides 5+a to a higher power than > > one might not divide b to the same power. > > Correct. > > . Such a factor exists. Call it h_1. > > We have (h_1)^2 | (5+a) but (h_1)^2 does not divide b. > > (however h_1 does divide b). > > A factor of 3 that divides b, might not divide 5+a. > > Incorrect. > > Remember a*b = 6, so if a factor is coprime to a with respect to 3 how > can it not share all its factors in common with 3 with b? > > Yes, the fact that f|3 and f is coprime to a means that f|b > or in your terms, f shares all its factors in common with 3 with b. > This does not mean that b shares all its factors of 3 in common with > f (or in other words just because f does not have any factor of 3 > that b doesn't have, we can't say that b does not have any factor > of 3 that f doesn't have). - William Hughes Hmmm...ok, I'll agree with that, but then you need to agree that b/(5+a) will cancel out any factors in common with 3 from the denominator. If you correct for any other primes, by multiplying with some appropriate x, then xb/(5 + a) should be an algebraic integer, correct? James Harris === Subject: Re: JSH: Direct demonstration of contradiction > > Lets take a= (sqrt(37) - sqrt(13))/2, b = (sqrt(37)+sqrt(13))/2. > > > > Then if f|3 and f|5+a then f is coprime to a (any factor of f that > > divides a must also divide 3 and 5 and is therefore a unit) > > > > Now a*b = 6, so f|ab, and f is coprime to a, so f|b. > > > > So any factor of 3 that divides 5+a also divides 3. > > > > However > > > > A factor of 3 that divides 5+a to a higher power than > > one might not divide b to the same power. > > Correct. > > > . Such a factor exists. Call it h_1. > > We have (h_1)^2 | (5+a) but (h_1)^2 does not divide b. > > (however h_1 does divide b). > > > > A factor of 3 that divides b, might not divide 5+a. > > Incorrect. > > Remember a*b = 6, so if a factor is coprime to a with respect to 3 how > > can it not share all its factors in common with 3 with b? > > Yes, the fact that f|3 and f is coprime to a means that f|b > or in your terms, f shares all its factors in common with 3 with b. > This does not mean that b shares all its factors of 3 in common with > f (or in other words just because f does not have any factor of 3 > that b doesn't have, we can't say that b does not have any factor > of 3 that f doesn't have). > > - William Hughes Hmmm...ok, I'll agree with that, but then you need to agree that > b/(5+a) will cancel out any factors in common with 3 from the > denominator. > Nope. (5+a) is divisible by (h_1)^2, but b is only divisible by (h_1). After cancelling there is still a factor of (h_1) in the denominator. - William Hughes === Subject: Re: JSH: Direct demonstration of contradiction > > > Lets take a= (sqrt(37) - sqrt(13))/2, b = (sqrt(37)+sqrt(13))/2. > > > > Then if f|3 and f|5+a then f is coprime to a (any factor of f that > > > divides a must also divide 3 and 5 and is therefore a unit) > > > > Now a*b = 6, so f|ab, and f is coprime to a, so f|b. > > > > So any factor of 3 that divides 5+a also divides 3. > > > > However > > > > A factor of 3 that divides 5+a to a higher power than > > > one might not divide b to the same power. > > > > Correct. > > > > . Such a factor exists. Call it h_1. > > > We have (h_1)^2 | (5+a) but (h_1)^2 does not divide b. > > > (however h_1 does divide b). > > > > A factor of 3 that divides b, might not divide 5+a. > > > > Incorrect. > > > > Remember a*b = 6, so if a factor is coprime to a with respect to 3 how > > can it not share all its factors in common with 3 with b? > > > > Yes, the fact that f|3 and f is coprime to a means that f|b > > or in your terms, f shares all its factors in common with 3 with b. > > This does not mean that b shares all its factors of 3 in common with > > f (or in other words just because f does not have any factor of 3 > > that b doesn't have, we can't say that b does not have any factor > > of 3 that f doesn't have). > > - William Hughes > > Hmmm...ok, I'll agree with that, but then you need to agree that > b/(5+a) will cancel out any factors in common with 3 from the > denominator. > > Nope. (5+a) is divisible by (h_1)^2, but b is only divisible > by (h_1). After cancelling there is still a factor of (h_1) in the > denominator. - William Hughes Ok, that is possible. So let's move forward. Further my earlier result (5 + (sqrt(37) - sqrt(13))/2)((sqrt(37) - sqrt(13))/2) = (5 - sqrt(13))(5 + sqrt(37))/2 is the same as ((5 + sqrt(37) + 5 - sqrt(13))/2)((sqrt(37) - sqrt(13))/2) = (5 - sqrt(13))(5 + sqrt(37))/2 proving that 5 + sqrt(37) must share the same factors from 3 with 5 - sqrt(13) and also then proving that sqrt(37) - sqrt(13) is coprime to 3. Baby steps Hughes. But I have maybe finally brought you there. James Harris === Subject: Re: JSH: Direct demonstration of contradiction > > > > > > > Start with > > > > > > > > > > > > > > (5 + (sqrt(37) - sqrt(13))/2)((sqrt(37) - sqrt(13))/2) = (5 - > > > > > > > sqrt(13))(5 + sqrt(37))/2 > > > > > > > > > > > > > > > > > > > Okay, I'll tell you what the story is. The ring of integers in > > > > > > > Q(sqrt(13),sqrt(37)) is Z[(1+sqrt(13))/2,(1+sqrt(37))/2]. The ideal > > > > > > > generated by 2 is a product of two prime ideals p_1 and p_2 of norm 4. > > > > > > > Only one of them contains 5+(sqrt(37)-sqrt(13))/2, I'll let p_1 be that > > > > > > > one. > > > > > > > > > > > > We have 3=((sqrt(13)+1)/2)((sqrt(13)-1)/2). The ideals generated by > > > > > > > (sqrt(13)+1)/2 and (sqrt(13)-1)/2 are prime. > > > > > > > > > > > > Now, the ideal generated by 5+(sqrt(37)-sqrt(13))/2 is the product of > > > > > > > p_1 and the ideal generated by (sqrt(13)-1)/2. > > > > > > > > > > > > And the ideal generated by (sqrt(37)+sqrt(13))/2 is the product of > > > > > > > (p_2)^2 and the ideal generated by (sqrt(13)-1)/2. It turned out you > > > > > > > were actually right that the gcd's with 3 are equal, in both cases they > > > > > > > are (sqrt(13)-1)/2. However, this was just a fluke. > > > > > > > > > > > > > > > > > Yet looking at > > > > > > > > > > > > (5 + (sqrt(37) - sqrt(13))/2)((sqrt(37) - sqrt(13))/2) = (5 - > > > > > > sqrt(13))(5 + sqrt(37))/2 > > > > > > > > > > > > > > > > My calculations were wrong. We have (2)=p_1.p_2 and > > > > > > (3)=q_1.q_2.q_3.q_4. I'll let q_1.q_2 be invariant under the > > > > > > automorphism which sends sqrt(13) to -sqrt(13), and q_1.q_3 be > > > > > > invariant under the automorphism which sends sqrt(37) to -sqrt(37). > > > > > > > > > > Then I have > > > > > > > > > > (5+(sqrt(37)-sqrt(13))/2)=p_1.q_1.q_2 > > > > > > ((sqrt(37)-sqrt(13))/2)=p_2.q_3.q_4 > > > > > > (5-sqrt(13))=p_1.p_2.q_3.q_4 > > > > > > ((5+sqrt(37))/2)=q_1.q_2 > > > > > > > > > > > you can just note that > > > > > > > > > > > > (5 + (sqrt(37) - sqrt(13))/2) > > > > > > > > > > > > must be coprime to (sqrt(37) - sqrt(13))/2 with respect to 3, by just > > > > > > subtracting one from the other as that leaves 5. > > > > > > > > > > > > > > > > Yes that's right. The greatest common divisor of one with 3 is q_1.q_2, > > > > > > the other is q_3.q_4. > > > > > > > > > > > And since ((sqrt(37) - sqrt(13))/2)*((sqrt(37) + sqrt(13))/2) = 6 > > > > > > > > > > > > you know that (5 + (sqrt(37) - sqrt(13))/2) must share all of its > > > > > > factors in common with 3, with (sqrt(37) + sqrt(13))/2. > > > > > > > > > > > > > > > > You know that every prime ideal dividing (5+(sqrt(37)-sqrt(13))/2) and > > > > > > 3 must divide (sqrt(37)+sqrt(13))/2. In fact, as it happens > > > > > > gcd(5+(sqrt(37)-sqrt(13))/2,3)=gcd((sqrt(37)+sqrt(13))/2,3)=(sqrt(13)-1)/2. > > > > > > > > > > > > > > So the next step is also trivial which is to divide one by the other > > > > > correcting for other prime factors, where in this case that means you > > > > > can correct for 2 by multiplying by 2 > > > > > > > > Nope. You can't clear all the factors of 2 from > > > > > (sqrt(37) + sqrt(13))/2 by multiplying by 2. > > > > > > > > - William Hughes > > > > > > > > Except that right after I demonstrate:: > > > > > > > > z = 2(10 + sqrt(37) - sqrt(13))/(sqrt(37) + sqrt(13)) > > > > > > > > which gives z as a root of > > > > > > > > 9z^4 - 72z^3 - 409z^2 - 2212z - 144 = 0 > > > > > > > > where the lead term is 9 which is coprime to 2, showing that it worked! > > > > And factors in common with 2 were successfully removed. > > > > > > Nope. > > > > > > What matters here is not the factors of the first term or the constant > > > > term, but the factors of the constant term divided by the first term. > > > > > > Let z= f_1/g_1, where f_1 and g_1 are coprime. As you correctly > > > > note, f_1 is coprime to 3. > > > > > > Let, f_2/g_2, f_3/g_3, and f_4/g_4 be the other three roots > > > > of > > > > > > P(x) = 9x^4 - 72x^3 - 409x^2 - 2212x - 144 > > > > > > Galois theory as usually taught suggests > > > > that f_2, f_3 and f_4 must be coprime to 3. > > > > > > Then > > > > > > 9(x - f_1/g_1)(x - f_2/g_2)(x - f_3/g_3)(x - f_4/g_4) = P(x) > > > > > > So, looking at the contant terms on each side > > > > > > f_1 f_2 f_3 f_4 > > > > 9 ---- ---- ---- ---- = 144 > > > > g_1 g_2 g_3 g_4 > > > > > > or (f_1 f_2 f_3 f_4)/(g_1 g_2 g_3 g_4) = 144/9 = 16 > > > > > > No factors of 3. No contradiction. > > > > (Note, before you start looking at factors of 2, remember that > > > > f_1 and g_1 may both share factors with 2, even though they > > > > are coprime). > > > > > > If James had got the polynomial right, then the fact that it is a > > > primitive polynomial and the leading coefficient is 9 would show that > > > there are prime ideals dividing 3 in the denominator of the fractional > > > ideal generated by any one of the roots (though the fact that the norm > > > is 16 would show that there are also other prime ideals dividing 3 in > > > the numerator). > > > > > > > I did make a mistake. The correct polynomial is > > > > 3z^4 - 100z^3 - 48z^2 - 440z - 96 = 0 > > > > > No that's wrong as well. The coefficient in z^3 is -50 and the constant > > term is 48. I would imagine that the polynomial Tim posted is probably > > correct. > > > > Oh yeah, I made another dumb mistake. > > Should be > > 3z^4 - 50z^3 - 24z^2 - 220z - 48 = 0. > > That's getting closer to Tim's answer. The constant term is definitely > positive 48, not negative 48. I can't comment on the z^2 and z > coefficients because I haven't checked, but my money would be on Tim > being right about them as well. > > Oh, I got a sign wrong but the rest should be correct. 3z^4 - 50z^3 - 24z^2 - 220z + 48 = 0 > See Tim's post. He says he's checked his with mathematical software. I would say the chances are pretty good his answer is correct. > > > where you get the same result as it is still non-monic and irreducible > > > over Q. > > > > > We are agreed, the number is not an algebraic integer. > > > > The leading coefficient is coprime to 2, so at issue only are factors > > > of 3. > > > > > True. > > > > If those had canceled out, then it would be a monic polynomial and its > > > roots algebraic integers. > > > > > But they don't. As I told you, there is a prime ideal dividing 3 and > > (sqrt(37)+sqrt(13))/2 that does not divide (5+(sqrt(37)-sqrt(13))/2). > > And why shouldn't there be? > > > > Both are coprime to (sqrt(37) - sqrt(13))/2. > > But (sqrt(37) + sqrt(13))*(sqrt(37) - sqrt(13))/4 = 6 > > so all factors that are coprime to sqrt(37) - sqrt(13) with respect to > > 3, MUST be factors of > > sqrt(37) + sqrt(13) > > Yes, that's right. Every prime ideal which divides > 5+(sqrt(37)-sqrt(13))/2 and 3 must also divide sqrt(37)+sqrt(13). I've > always agreed with you about this. However, I don't agree that you > couldn't get a prime ideal dividing the former to a higher power than > the latter, and I don't agree that every prime ideal which divides > sqrt(37)+sqrt(13) and 3 must also divide 5+(sqrt(37)-sqrt(13))/2. > > Then turn z upside down. z = 2(10 + sqrt(37) - sqrt(13))/(sqrt(37) + sqrt(13)) so use instead z = (sqrt(37) + sqrt(13))/(10 + sqrt(37) - sqrt(13)) > Okay, let's look at my factorizations. I say ((sqrt(37)+sqrt(13))/2)=p_1.q_1.q_4 (5+(sqrt(37)-sqrt(13))/2)=p_1.(q_1)^2 So the fractional ideal generated by that z should be q_4.(q_1)^(-1). What's going on here is that we have a prime ideal occurring in 5+(sqrt(37)-sqrt(13))/2 to a higher power than in (sqrt(37)+sqrt(13))/2. So you'll find that z isn't an algebraic integer either. The leading coefficient in the minimum polynomial will be 3. (The leading coefficient is always the norm of the denominator in the fractional ideal). > and you'll find you get a non-monic polynomial irreducible over Q with > a leading coefficient that has 2, 3 as prime factors. > I don't think so. Just 3. Perhaps you left a 2 out of the denominator of your z, and you calculated the minimum polynomial for the z which had a 2 in it? . > Now if you say that (10 + sqrt(37) - sqrt(13)) has its factors of 3 > squared versus (sqrt(37) + sqrt(13)) then I just say use > z = (sqrt(37) + sqrt(13))^2/(10 + sqrt(37) - sqrt(13)) as I already know that won't work. > Well, I agree that's a good test. If my factorizations are correct, that must be an algebraic integer. Let's see. (sqrt(37)+sqrt(13))^2/(10+sqrt(37)-sqrt(13))= (50+2.sqrt(37.13))(10+sqrt(37)+sqrt(13))/((10+sqrt(37))^2-13)= (500+50.sqrt(37)+50.sqrt(13)+20.sqrt(37.13)+74.sqrt(13)+26.sqrt(37))/(124+20 .sqrt(37))= (500+76.sqrt(37)+124.sqrt(13)+20.sqrt(37.13))(124-20.sqrt(37))/576= ((500.124-20.76)+(76.124-20.500).sqrt(37)+(124^2-14800).sqrt(13)+(2480-2480) )/576= (60480-576.sqrt(37)+576^2.sqrt(13))/576= 105-sqrt(37)+576.sqrt(13). That looks like an algebraic integer to me. Can you find a mistake in my calculations? > You can do the same trick that I used to eliminate 2, and you'll find > you still get a non-monic polynomial irreducible over Q. > I don't think so. > And the leading coefficient will have 3 as a factor as will the > constant term. If it were a matter of squares then ALL factors in common with 3 would > have divided out of either the denominator or the numerator and I'd > have ended up with either a leading coefficient with 3 as a factor with > a constant term coprime to it, or the opposite. > What happens is there's a prime ideal dividing 3 in the numerator and also in the denominator. So the norm is 16, so the constant term is 48. Why don't you actually look at my factorizations? > But BOTH the leading coefficient and constant term have 3 as a factor, > destroying your claim. > No, that is perfectly consistent with my factorizations and indeed I derived my values for the leading coefficient and constant term by looking at them. > Do the math. > I am. > I think it's simpler when people with math software do it as I keep > screwing up the algebra. It's easy to do, yes, but it's tedious, > especially when a computer can do it. And you will find you are wrong. > Well, let's see. I've done it by hand and I get that I'm right. Maybe someone else can check with a computer. > Now, why? Well, it's just like that simple example with x^2 + 4x + 3 = 0 as with z = (sqrt(37) + sqrt(13))^2/(10 + sqrt(37) - sqrt(13)) when you start flipping signs in front of the square roots, and go > through all the combinations, which is why you end up with a quartic > anyway, you have some where there aren't any shared factors, so it's > the same basic reason as why with > The ideal generated by z is (p_1)^2.p_2.(q_4)^2. The ideals generated by the conjugates are (p_1)^2.p_2.(q_1)^2, (p_2)^2.p_1.(q_2)^2, (p_2)^2.p_1.(q_3)^2. > x^2 + 4x + 3 = 0 using x = 3y will give you a non-monic as well. > Oh, were you talking about the numerator? The ideal generated by that is (p_1)^4.(p_2)^2.(q_1)^2.(q_4)^2. The conjugates form pairs of associates, so the ideals are the same. The other pair of conjugates both generate the ideal (p_2)^4.(p_1)^2.(q_1)^2.(q_2)^2. Each conjugate of the numerator is always divisible by the corresponding conjugate of the denominator. > Trivial. > and since (5+(sqrt(37)-sqrt(13))/2) is coprime to sqrt(37) - sqrt(13) > > as well, it must share its factors in common with 3 with sqrt(37) + > > sqrt(13). > > As I've told you the largest ideal dividing 5+(sqrt(37)-sqrt(13))/2 and > 3 is (q_1)^2. The largest ideal dividing 3 and (sqrt(37)+sqrt(13))/2 is > q_1.q_4. So it's true that every prime ideal that divides > 5+(sqrt(37)-sqrt(13))/2 and 3 also divides (sqrt(37)+sqrt(13))/2. And I've explained how to test that claim. > Seems to have survived every test you've thrown at it so far. > However, the prime ideal in question occurs to a higher power in the > former number than in the latter. And there is a prime ideal that > occurs in the latter which does not occur in the former at all. That's > why the former divided by the latter is not an algebraic integer. > > Then flip them, and see what happens. > Done. > > Trivial. > > Depends what you mean by the vague expression share its factors in > common. The interpretation which I've agreed with already is trivial, > yes. The interpretation which you're sliding to which says that the > largest ideal dividing 5+(sqrt(37)-sqrt(13))/2 and 3 has to be > divisible by the largest ideal dividing (sqrt(37)+sqrt(13))/2 and 3 is > not true. > > Astute readers already know that flipping won't make a difference as > with 3z^4 - 50z^3 - 24z^2 - 220z + 48 = 0 you could just use z = 1/y to do the flipping and, of course, you will > still have both the leading and constant coefficient having 3 as a > prime factor. > Yeah, I agree with that and that's predicted by my factorizations. On the other hand, my factorizations predict that when you square the numerator you should get an algebraic integer, and I found that you do. > Trivial. Oh, also I know what you say about ideals can't be right, as if the > factors in common with 3 were as you say, then again, only the leading > coefficient would have 3 as a factor with the polynomial above. > No, that's not the case. I've explained that earlier. > I don't really care as I think it's a little more dramatic if you can > get algebraic integers one way as later you can't anyway with a > follow-up result. This example is so dramatic because it leaves so little room for > avoiding the result. In the past posters like Dik Winter and Arturo Magidin could have a > field day confusing people about functions and polynomials and what was > the key variable or not. But with (5 + (sqrt(37) - sqrt(13))/2)((sqrt(37) - sqrt(13))/2) = (5 - > sqrt(13))(5 + sqrt(37))/2 where there are just numbers they are at a loss, which is good! As > it's been quieter this time!!! And you appear to have continued as you got lost with that ideal > nonsense and haven't appreciated how factors of 3 in both the leading > coefficient and constant term destroy your position. > No, they do not. The fractional ideal generated by the root of the polynomial has a prime ideal of norm 3 in the numerator and a prime ideal of norm 3 in the denominator. So the norm is coprime to 3. That's why the constant term is divisible by 3. > But make no mistake, this demonstration shatters your claims, and with > it validates mine, Nope. > including my techniques of non-polynomial > factorization, my paper and my claims of proof of Fermat's Last > Theorem, from which I pulled a piece to come up with the paper. > Don't be ridiculous. This discussion has no bearing on those issues. > Since these results overturn over a hundred years worth of number > theory Which ones? I really can't wait to hear. > and greatly change the historical positions of quite a few > people famous in the math field, it is no surprise that there would be > INCREDIBLE resistance, and one dead journal so far is not a surprise. The editors may have just found they could not continue publishing in > the face of the reality. Entire universities can crumble from a result this big, which may seem > strange, but it is possible, and yes, it would quite upset the > academics in other departments from history to physics, especially > because they wouldn't think it possible. But it is, and the way things are going, I'm more and more afraid I > can't stop it. And no, I don't want to have entire universities shut-down, as I don't > like the loss of the one math journal as it is. But with no one helping out, the energy is just building and building > and building. > > Eventually, it has to release. > > > James Harris === Subject: Re: JSH: Direct demonstration of contradiction > > > > > > > Start with > > > > > > > > > > > > > > (5 + (sqrt(37) - sqrt(13))/2)((sqrt(37) - sqrt(13))/2) = (5 - > > > > > > > sqrt(13))(5 + sqrt(37))/2 > > > > > > > > > > > > > > > > > > > Okay, I'll tell you what the story is. The ring of integers in > > > > > > > Q(sqrt(13),sqrt(37)) is Z[(1+sqrt(13))/2,(1+sqrt(37))/2]. The ideal > > > > > > > generated by 2 is a product of two prime ideals p_1 and p_2 of norm 4. > > > > > > > Only one of them contains 5+(sqrt(37)-sqrt(13))/2, I'll let p_1 be that > > > > > > > one. > > > > > > > > > > > > We have 3=((sqrt(13)+1)/2)((sqrt(13)-1)/2). The ideals generated by > > > > > > > (sqrt(13)+1)/2 and (sqrt(13)-1)/2 are prime. > > > > > > > > > > > > Now, the ideal generated by 5+(sqrt(37)-sqrt(13))/2 is the product of > > > > > > > p_1 and the ideal generated by (sqrt(13)-1)/2. > > > > > > > > > > > > And the ideal generated by (sqrt(37)+sqrt(13))/2 is the product of > > > > > > > (p_2)^2 and the ideal generated by (sqrt(13)-1)/2. It turned out you > > > > > > > were actually right that the gcd's with 3 are equal, in both cases they > > > > > > > are (sqrt(13)-1)/2. However, this was just a fluke. > > > > > > > > > > > > > > > > > Yet looking at > > > > > > > > > > > > (5 + (sqrt(37) - sqrt(13))/2)((sqrt(37) - sqrt(13))/2) = (5 - > > > > > > sqrt(13))(5 + sqrt(37))/2 > > > > > > > > > > > > > > > > My calculations were wrong. We have (2)=p_1.p_2 and > > > > > > (3)=q_1.q_2.q_3.q_4. I'll let q_1.q_2 be invariant under the > > > > > > automorphism which sends sqrt(13) to -sqrt(13), and q_1.q_3 be > > > > > > invariant under the automorphism which sends sqrt(37) to -sqrt(37). > > > > > > > > > > Then I have > > > > > > > > > > (5+(sqrt(37)-sqrt(13))/2)=p_1.q_1.q_2 > > > > > > ((sqrt(37)-sqrt(13))/2)=p_2.q_3.q_4 > > > > > > (5-sqrt(13))=p_1.p_2.q_3.q_4 > > > > > > ((5+sqrt(37))/2)=q_1.q_2 > > > > > > > > > > > you can just note that > > > > > > > > > > > > (5 + (sqrt(37) - sqrt(13))/2) > > > > > > > > > > > > must be coprime to (sqrt(37) - sqrt(13))/2 with respect to 3, by just > > > > > > subtracting one from the other as that leaves 5. > > > > > > > > > > > > > > > > Yes that's right. The greatest common divisor of one with 3 is q_1.q_2, > > > > > > the other is q_3.q_4. > > > > > > > > > > > And since ((sqrt(37) - sqrt(13))/2)*((sqrt(37) + sqrt(13))/2) = 6 > > > > > > > > > > > > you know that (5 + (sqrt(37) - sqrt(13))/2) must share all of its > > > > > > factors in common with 3, with (sqrt(37) + sqrt(13))/2. > > > > > > > > > > > > > > > > You know that every prime ideal dividing (5+(sqrt(37)-sqrt(13))/2) and > > > > > > 3 must divide (sqrt(37)+sqrt(13))/2. In fact, as it happens > > > > > > gcd(5+(sqrt(37)-sqrt(13))/2,3)=gcd((sqrt(37)+sqrt(13))/2,3)=(sqrt(13)-1)/2. > > > > > > > > > > > > > > So the next step is also trivial which is to divide one by the other > > > > > correcting for other prime factors, where in this case that means you > > > > > can correct for 2 by multiplying by 2 > > > > > > > > Nope. You can't clear all the factors of 2 from > > > > > (sqrt(37) + sqrt(13))/2 by multiplying by 2. > > > > > > > > - William Hughes > > > > > > > > Except that right after I demonstrate:: > > > > > > > > z = 2(10 + sqrt(37) - sqrt(13))/(sqrt(37) + sqrt(13)) > > > > > > > > which gives z as a root of > > > > > > > > 9z^4 - 72z^3 - 409z^2 - 2212z - 144 = 0 > > > > > > > > where the lead term is 9 which is coprime to 2, showing that it worked! > > > > And factors in common with 2 were successfully removed. > > > > > > Nope. > > > > > > What matters here is not the factors of the first term or the constant > > > > term, but the factors of the constant term divided by the first term. > > > > > > Let z= f_1/g_1, where f_1 and g_1 are coprime. As you correctly > > > > note, f_1 is coprime to 3. > > > > > > Let, f_2/g_2, f_3/g_3, and f_4/g_4 be the other three roots > > > > of > > > > > > P(x) = 9x^4 - 72x^3 - 409x^2 - 2212x - 144 > > > > > > Galois theory as usually taught suggests > > > > that f_2, f_3 and f_4 must be coprime to 3. > > > > > > Then > > > > > > 9(x - f_1/g_1)(x - f_2/g_2)(x - f_3/g_3)(x - f_4/g_4) = P(x) > > > > > > So, looking at the contant terms on each side > > > > > > f_1 f_2 f_3 f_4 > > > > 9 ---- ---- ---- ---- = 144 > > > > g_1 g_2 g_3 g_4 > > > > > > or (f_1 f_2 f_3 f_4)/(g_1 g_2 g_3 g_4) = 144/9 = 16 > > > > > > No factors of 3. No contradiction. > > > > (Note, before you start looking at factors of 2, remember that > > > > f_1 and g_1 may both share factors with 2, even though they > > > > are coprime). > > > > > > If James had got the polynomial right, then the fact that it is a > > > primitive polynomial and the leading coefficient is 9 would show that > > > there are prime ideals dividing 3 in the denominator of the fractional > > > ideal generated by any one of the roots (though the fact that the norm > > > is 16 would show that there are also other prime ideals dividing 3 in > > > the numerator). > > > > > > > I did make a mistake. The correct polynomial is > > > > 3z^4 - 100z^3 - 48z^2 - 440z - 96 = 0 > > > > > No that's wrong as well. The coefficient in z^3 is -50 and the constant > > term is 48. I would imagine that the polynomial Tim posted is probably > > correct. > > > > Oh yeah, I made another dumb mistake. > > Should be > > 3z^4 - 50z^3 - 24z^2 - 220z - 48 = 0. > > That's getting closer to Tim's answer. The constant term is definitely > positive 48, not negative 48. I can't comment on the z^2 and z > coefficients because I haven't checked, but my money would be on Tim > being right about them as well. > > Oh, I got a sign wrong but the rest should be correct. 3z^4 - 50z^3 - 24z^2 - 220z + 48 = 0 > > where you get the same result as it is still non-monic and irreducible > > > over Q. > > > > > We are agreed, the number is not an algebraic integer. > > > > The leading coefficient is coprime to 2, so at issue only are factors > > > of 3. > > > > > True. > > > > If those had canceled out, then it would be a monic polynomial and its > > > roots algebraic integers. > > > > > But they don't. As I told you, there is a prime ideal dividing 3 and > > (sqrt(37)+sqrt(13))/2 that does not divide (5+(sqrt(37)-sqrt(13))/2). > > And why shouldn't there be? > > > > Both are coprime to (sqrt(37) - sqrt(13))/2. > > But (sqrt(37) + sqrt(13))*(sqrt(37) - sqrt(13))/4 = 6 > > so all factors that are coprime to sqrt(37) - sqrt(13) with respect to > > 3, MUST be factors of > > sqrt(37) + sqrt(13) > > Yes, that's right. Every prime ideal which divides > 5+(sqrt(37)-sqrt(13))/2 and 3 must also divide sqrt(37)+sqrt(13). I've > always agreed with you about this. However, I don't agree that you > couldn't get a prime ideal dividing the former to a higher power than > the latter, and I don't agree that every prime ideal which divides > sqrt(37)+sqrt(13) and 3 must also divide 5+(sqrt(37)-sqrt(13))/2. > > Then turn z upside down. z turned upside down is still z. As an English teacher, you should know that. > z = 2(10 + sqrt(37) - sqrt(13))/(sqrt(37) + sqrt(13)) so use instead z = (sqrt(37) + sqrt(13))/(10 + sqrt(37) - sqrt(13)) and you'll find you get a non-monic polynomial irreducible over Q with > a leading coefficient that has 2, 3 as prime factors. Now if you say that (10 + sqrt(37) - sqrt(13)) has its factors of 3 > squared versus (sqrt(37) + sqrt(13)) then I just say use > z = (sqrt(37) + sqrt(13))^2/(10 + sqrt(37) - sqrt(13)) as I already know that won't work. You can do the same trick that I used to eliminate 2, and you'll find > you still get a non-monic polynomial irreducible over Q. And the leading coefficient will have 3 as a factor as will the > constant term. If it were a matter of squares then ALL factors in common with 3 would > have divided out of either the denominator or the numerator and I'd > have ended up with either a leading coefficient with 3 as a factor with > a constant term coprime to it, or the opposite. But BOTH the leading coefficient and constant term have 3 as a factor, > destroying your claim. Do the math. No, you do it. And get it right this time, okay? > I think it's simpler when people with math software do it as I keep > screwing up the algebra. It's easy to do, yes, but it's tedious, > especially when a computer can do it. And you will find you are wrong. Now, why? Because JSH has redefined what it means for someone to be wrong. The Harris-English dictionary lists the definition of wrong as dead on. > Well, it's just like that simple example with x^2 + 4x + 3 = 0 as with z = (sqrt(37) + sqrt(13))^2/(10 + sqrt(37) - sqrt(13)) when you start flipping signs in front of the square roots, and go > through all the combinations, which is why you end up with a quartic > anyway, you have some where there aren't any shared factors, so it's > the same basic reason as why with x^2 + 4x + 3 = 0 using x = 3y will give you a non-monic as well. Trivial. > and since (5+(sqrt(37)-sqrt(13))/2) is coprime to sqrt(37) - sqrt(13) > > as well, it must share its factors in common with 3 with sqrt(37) + > > sqrt(13). > > As I've told you the largest ideal dividing 5+(sqrt(37)-sqrt(13))/2 and > 3 is (q_1)^2. The largest ideal dividing 3 and (sqrt(37)+sqrt(13))/2 is > q_1.q_4. So it's true that every prime ideal that divides > 5+(sqrt(37)-sqrt(13))/2 and 3 also divides (sqrt(37)+sqrt(13))/2. And I've explained how to test that claim. > However, the prime ideal in question occurs to a higher power in the > former number than in the latter. And there is a prime ideal that > occurs in the latter which does not occur in the former at all. That's > why the former divided by the latter is not an algebraic integer. > > Then flip them, and see what happens. > Trivial. > > Depends what you mean by the vague expression share its factors in > common. The interpretation which I've agreed with already is trivial, > yes. The interpretation which you're sliding to which says that the > largest ideal dividing 5+(sqrt(37)-sqrt(13))/2 and 3 has to be > divisible by the largest ideal dividing (sqrt(37)+sqrt(13))/2 and 3 is > not true. > > Astute readers already know that flipping won't make a difference as > with 3z^4 - 50z^3 - 24z^2 - 220z + 48 = 0 you could just use z = 1/y to do the flipping and, of course, you will > still have both the leading and constant coefficient having 3 as a > prime factor. Trivial. Oh, also I know what you say about ideals can't be right, as if the > factors in common with 3 were as you say, then again, only the leading > coefficient would have 3 as a factor with the polynomial above. I don't really care as I think it's a little more dramatic if you can > get algebraic integers one way as later you can't anyway with a > follow-up result. This example is so dramatic because it leaves so little room for > avoiding the result. In the past posters like Dik Winter and Arturo Magidin could have a > field day confusing people about functions and polynomials and what was > the key variable or not. Just because they confused JSH doesn't mean they confused anyone else. > But with (5 + (sqrt(37) - sqrt(13))/2)((sqrt(37) - sqrt(13))/2) = (5 - > sqrt(13))(5 + sqrt(37))/2 where there are just numbers they are at a loss, which is good! As > it's been quieter this time!!! And you appear to have continued as you got lost with that ideal > nonsense and haven't appreciated how factors of 3 in both the leading > coefficient and constant term destroy your position. But make no mistake, this demonstration shatters your claims, and with > it validates mine, including my techniques of non-polynomial > factorization, my paper and my claims of proof of Fermat's Last > Theorem, from which I pulled a piece to come up with the paper. Since these results overturn over a hundred years worth of number > theory and greatly change the historical positions of quite a few > people famous in the math field, it is no surprise that there would be > INCREDIBLE resistance, and one dead journal so far is not a surprise. OTOH, how could all of the professional number theorists --- and quite a few amateurs --- have missed something like that? Only if it wasn't there. You see, if someone had pointed out a mistake one year after Dedikind's error, there wouldn't have been any problems. Dedikind would have _thanked_ someone for finding it. And any actual errors would have been found soon after the theory. JSH knows nothing about what it means to be professional, or _a_ professional, of any kind. > The editors may have just found they could not continue publishing in > the face of the reality. Yes, having published TWO incorrect papers in the same issue. (Have you forgotten about Platnikov's paper, which claimed that P=NP, and which was in the same issue? I suspect not, since cranks seldom care about anything any one else has done.) > Entire universities can crumble from a result this big, which may seem > strange, but it is possible, and yes, it would quite upset the > academics in other departments from history to physics, especially > because they wouldn't think it possible. But it is, and the way things are going, I'm more and more afraid I > can't stop it. And no, I don't want to have entire universities shut-down, as I don't > like the loss of the one math journal as it is. But with no one helping out, the energy is just building and building > and building. Eventually, it has to release. Do you have any kind of time frame on that? I'm still waiting for The Hammer and the world-wide collapse you promised: I'm working on it now, and luckily it should only take a few days for the impact and this will all be over. --- James Harris, math crank, 2006 Feb 12. But I fear it's time to get The Shovel again. --- Christopher Heckman === Subject: JSH: Remember dead math journal I've seen replies from posters indicating a desire to protect math people who are tops in the social hierarchy of the current mathematical world, which isn't a surprise. But why will it be easy to convince governments around the world to send prosecutors in with the full power of the law to find out how much they knew and when? Well, remember that paper that sci.math'ers helped get yanked, and then the journal died? An early draft went to Barry Mazur, a professor at Harvard. He commented on it, even asked me a question about it. After him, an early draft went to Andrew Granville, who passed on denying acceptance for the New York Journal of Mathematics, deferring to the chief editor who claimed it was too small a size paper in terms of length for the journal. And beyond all of that, a freaking mathematical journal keeling over and dying after yanking a paper that it published from a guy called one of the biggest cranks in Usenet history is not the kind of thing that math people could NOT notice. The event could have well made headlines in major newspapers around the world if math people did not have so much power in their community to control information. So yes, necessarily, governments around the world will pit their full powers of investigation against top people in the math world, going through their houses, going through their computers, questioning their colleagues and students. We are talking about the future of the world here. What if Newton had been stopped by petty academics? Or Gauss had never managed to get attention for his work because of dumb social stuff? Our world would not be the one we have today. The future depends on what people today do, just like our present depended on the people of the past. A math journal does not just keel over and die with such a spectacular story behind it and not get noticed. I think there will be a clear trail, like with Enron, from emails to conversations with colleagues and students. James Harris === Subject: Re: JSH: Remember dead math journal I must give you credit, Harris, for finding a journal that was sufficiently desperate for contributions that it would accept something from you, probably without even reading it. I would have thought it very unlikely. But that journal was on its last legs, and indeed it died shortly after withdrawing your unsound paper. > We are talking about the future of the world here. [No comment.] === Subject: Re: JSH: Remember dead math journal > I must give you credit, Harris, for finding a journal that was > sufficiently desperate for contributions that it would accept something > from you, probably without even reading it. I would have thought it > very unlikely. But that journal was on its last legs, and indeed it > died shortly after withdrawing your unsound paper. I don't believe it was luck or coincidence. Who better would know the best place for a crackpot to slip his rubbish under the door than a society full of crackpots, say a club of High IQ types? How do you think Beckwith became associated with JSH? Mazur and Granville aren't the only ones JSH has been in contact with. > We are talking about the future of the world here. > > [No comment.] === Subject: Re: Remember dead math journal > The event could have well made headlines in major newspapers around > the > world if math people did not have so much power in their community > to > control information. Yes!! Weekly World News http://www.weeklyworldnews.com/ currently has a headline Batboy sighted in NYC subway! I am looking forward to reading about Mathboy sighted in Atlanta bar!. -- Clive Tooth www.clivetooth.dk Stock photos: http://submit.shutterstock.com/?ref=61771 === Subject: Re: Remember dead math journal <4q2036Fl4i5sU1@individual.net > The event could have well made headlines in major newspapers around > the > world if math people did not have so much power in their community > to > control information. Yes!! Weekly World News > http://www.weeklyworldnews.com/ > currently has a headline Batboy sighted in NYC subway! I am looking forward to reading about Mathboy sighted in Atlanta > bar!. Ha! Who are you trying to kid with your silly headlines? We all know that mathboy moved to San Francisco. -- > Clive Tooth > www.clivetooth.dk > Stock photos: > http://submit.shutterstock.com/?ref=61771 === Subject: Re: Remember dead math journal > I've seen replies from posters indicating a desire to protect math > people who are tops in the social hierarchy of the current mathematical > world, which isn't a surprise. But why will it be easy to convince governments around the world to > send prosecutors in with the full power of the law to find out how much > they knew and when? Well, remember that paper that sci.math'ers helped get yanked, and then > the journal died? An early draft went to Barry Mazur, a professor at Harvard. He > commented on it, even asked me a question about it. After him, an early draft went to Andrew Granville, who passed on > denying acceptance for the New York Journal of Mathematics, deferring > to the chief editor who claimed it was too small a size paper in terms > of length for the journal. And beyond all of that, a freaking mathematical journal keeling over > and dying after yanking a paper that it published from a guy called one > of the biggest cranks in Usenet history is not the kind of thing that > math people could NOT notice. The event could have well made headlines in major newspapers around the > world if math people did not have so much power in their community to > control information. So yes, necessarily, governments around the world will pit their full > powers of investigation against top people in the math world, going > through their houses, going through their computers, questioning their > colleagues and students. We are talking about the future of the world here. What if Newton had been stopped by petty academics? Or Gauss had never managed to get attention for his work because of > dumb social stuff? Our world would not be the one we have today. The future depends on what people today do, just like our present > depended on the people of the past. A math journal does not just keel over and die with such a spectacular > story behind it and not get noticed. I think there will be a clear trail, like with Enron, from emails to > conversations with colleagues and students. > James Harris > Your paper as I remember didn't even resemble the typical format for a technical paper. Besides, if so many people said it's wrong, guess what? It probably is. You're too big of a pompous idiot to realize that. You're not any better than anyone else. Dave === Subject: Re: JSH: Remember dead math journal What confuses me is why you spend so much time and energy on an obscure and now defunct publication, and don't spend more time on the Annals. You have said that your paper was accepted for reiview. You have never explained what you meant by this. You have said that you never received a formal rejection, but that on inquiry you were informed that such a rejection had been sent. What do you think happened? Do you think sci.mathers blocked the publication? How did the editors of the Annals manage to surpress the paper? Inquiring minds want to know. - William Hughes === Subject: Re: Remember dead math journal > I've seen just stop it. your paper was complete crap and it was rejected. You paper was never published, if so send me a copy in print. Can't do it can you ? Try something new. Like getting SF published, even if you have tp pay for it yourself. === Subject: Re: JSH: Remember dead math journal .......................................................................... > We are talking about the future of the world here. > James Harris *************************************************** Tonio Pd Of WHAT country is the catchy part...hehe. === Subject: Re: JSH: Remember dead math journal > I've seen replies from posters indicating a desire to protect math > people who are tops in the social hierarchy of the current mathematical > world, which isn't a surprise. But why will it be easy to convince governments around the world to > send prosecutors in with the full power of the law to find out how much > they knew and when? Well, remember that paper that sci.math'ers helped get yanked, and then > the journal died? An early draft went to Barry Mazur, a professor at Harvard. He > commented on it, even asked me a question about it. After him, an early draft went to Andrew Granville, who passed on > denying acceptance for the New York Journal of Mathematics, deferring > to the chief editor who claimed it was too small a size paper in terms > of length for the journal. And beyond all of that, a freaking mathematical journal keeling over > and dying after yanking a paper that it published from a guy called one > of the biggest cranks in Usenet history is not the kind of thing that > math people could NOT notice. The event could have well made headlines in major newspapers around the > world if math people did not have so much power in their community to > control information. So yes, necessarily, governments around the world will pit their full > powers of investigation against top people in the math world, going > through their houses, going through their computers, questioning their > colleagues and students. We are talking about the future of the world here. What if Newton had been stopped by petty academics? Or Gauss had never managed to get attention for his work because of > dumb social stuff? Our world would not be the one we have today. The future depends on what people today do, just like our present > depended on the people of the past. A math journal does not just keel over and die with such a spectacular > story behind it and not get noticed. I think there will be a clear trail, like with Enron, from emails to > conversations with colleagues and students. > James Harris Math people are truly powerful. They secretly rule the world. :-) === Subject: Re: JSH: Will to be wrong days. My association with the Department is that of an alumnus. > Please don't be cryptic... have I said something thick? >> No, it's perfectly understandable; I was merelywondering if you had a >> pair of numbers A, B, and two rings such that A divides B in the first >> ring, but not in the second. Then I apologise for my unhelpful reply. Your previous posts suggest >you know rather more about this area than I do, so I figured I had made >a gaffe and you were taking the piss :-) > Admittedly, the best I can do is use A = 2, B = 5, first ring is the >> field of rationals and the second the ring of integers. But I was >> hoping for subrings of the algebraics. :-) Aren't the rationals a subring of the algebraic numbers? If you mean >the algebraic integers then I'm afraid I can't provide an example since >I'm not that familiar with the AIs, though I'm sure somebody else can >do so. I don't think you can. Here's my argument, which no doubt someone will soon replace by an easier one. Suppose A and B are two algebraic integers, such that A does not divide B. Consider the ring of integers R of Q(A,B) (the collection of all algebraic integers in Q(A,B)). Then we have a factorization of (A) and (B) into prime ideals (A) = P_1^{a_1}*...*P_n^{a_n} (B) = Q_1^{b_1}*...*Q_m^{b_m} where P_i, Q_j are prime ideals, P_i <> P_j if i<>j, Q_i<>Q_j if i<>j, and the a_i and b_j are positive integers. Since we are assuming that A does not divide B, we must either have: (i) There exists i, 1<= i <= n such that P_i is different from Q_j for all j, 1<=j<=m. Or (ii) For all i, 1<=i <=n, there exists j such that P_i = Q_j, but there is an i such that b_jdo<- have a non-unit common factor in a larger ring. For example, 2 and 1+sqrt(-5) have no nonunit common factor in Z[sqrt(-5)], but they do have sqrt(2) as a common factor in the ring of all algebraic integers. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes by Bill Watterson) Arturo Magidin magidin-at-member-ams-org === Subject: Re: JSH: Will to be wrong Flash! , === > Subject: JSH: Will to be wrong > X-HTTP-UserAgent: Mozilla/5.0 (Windows; U; Windows NT 5.1; en-US; rv:1.8.0.7) Gecko/20060909 Firefox/1.5.0.7,gzip(gfe),gzip(gfe) > posting-account=Q2zO6wwAAABSLuGzZIjG0efOtB9n8fUY > > So what happened is I finally found a pure enough demonstration that > without variables and functions posters who have managed to argue > against me for years without it being clear to most that they had to be > wrong, got stuck: > > ((5 + sqrt(37) + 5 - sqrt(13))/2)((sqrt(37) - sqrt(13))/2) = (5 - > sqrt(13))(5 + sqrt(37))/2 > > That expression is remarkable for many reasons not the least of them > being it took me years to find it, as I've looked for alternative ways > to prove a result I not only proved years before, but even got > published before some of you got it yanked by freaking the editors of > the journal with some emails, and then, later, the math journal died. > > My job has been a difficult one as a few years ago I realized there had > to be a will to be wrong from a large number of people within the > mathematical community. > > math people have consistently refused to follow logic and mathematical > proof, as a group, which has forced me to look extensively for > something that was too direct for any of you to skillfully deny what > you were really doing. > > Now that I have it, I want to reinforce to you what many of you have > been fighting--the progress of humanity itself. > > Gauss did a lot of great work starting up what I call modular algebra, > and he started an understanding of key properties of integers in the > right direction with what are now called gaussian integers, but then > something awful happened, and an error came into accepted mathematical > thinking. > > Over a hundred years went by until the correction, which it has been my > difficult task to engineer. > > Make no mistake, no matter how many of you tell yourselves you are > mathematicians you cannot be mathematicians when you do not accept > mathematical proof, but yes, you can probably keep going for a while > longer with the world believing you are mathematicians, and maybe get > some more awards like recent recipients of the Field's Medal, make some > money, take care of your kids, or make some investments--and prove your > hatred for humanity. > > There has been a will to be wrong for years now. I hope that it is now > about to end. > > And that some of you will instead choose to be right, and learn correct > mathematical ideas. > > > James Harris -- Michael Press === Subject: Re: JSH: Will to be wrong > Say CHEESE! ... Click! > > Captured for posterity. When doing this you need to capture the full headers thusly: , === > Subject: JSH: Will to be wrong > X-HTTP-UserAgent: Mozilla/5.0 (Windows; U; Windows NT 5.1; en-US; rv:1.8.0.7) Gecko/20060909 Firefox/1.5.0.7,gzip(gfe),gzip(gfe) > posting-account=Q2zO6wwAAABSLuGzZIjG0efOtB9n8fUY > > So what happened is I finally found a pure enough demonstration that > without variables and functions posters who have managed to argue > against me for years without it being clear to most that they had to be > wrong, got stuck: -- Michael Press === Subject: Re: JSH: Will to be wrong > [Rupert] > > >> ... > > >> Then we either have > > > > >> ((5+sqrt(37)+5-sqrt(13))/2)=p_1.q_2.q_3 > > >> (sqrt(37)-sqrt(13))/2=p_2.(q_1)^2 > > > > >> or > > > > >> ((5+sqrt(37)+5-sqrt(13))/2=p_1.(q_1)^2 > > >> (sqrt(37)-sqrt(13))/2=p_2.q_2.q_3 > > > > [also Rupert] > > > It must be this one. Because there exists a nonidentity element of the > > > Galois group under which the ideal generated by (sqrt(37)-sqrt(13))/2 > > > is invariant. > > > > I will post the details of my reasoning about this example later. > > > > So we have > > > > (((5+sqrt(37)+5-sqrt(13))/2)=p_1.(q_1)^2 > > > ((sqrt(37)-sqrt(13))/2)=p_2.q_2.q_3 > > > (5-sqrt(13))=p_1.p_2.q_1.q_2 > > > ((5+sqrt(37))/2)=q_1.q_3 > > > > It's encouraging :-) that the p's and q's balance now across the LHS and RHS > > of: > > > > ((5 + sqrt(37) + 5 - sqrt(13))/2)((sqrt(37) - sqrt(13))/2) = > > (5 - sqrt(13))(5 + sqrt(37))/2 > > > > So what point did you want to make with this example, James? > > > > I think you guessed that one days ago, although the details of his claim > > seem to change. Most recently he claimed to me that his muddy reasoning > > about factors proves that: > > > > (sqrt(37) - sqrt(13))/2 > > > > is a unit: > > > > [JSH] > > ... > > Another key bit of information is that the original factorization > > > > (5 + (sqrt(37) - sqrt(13))/2)((sqrt(37) - sqrt(13))/2) = (5 - > > sqrt(13))(5 + sqrt(37))/2 > > > > is the same as > > > > ((5 + sqrt(37) + 5 - sqrt(13))/2)((sqrt(37) - sqrt(13))/2) = (5 - > > sqrt(13))(5 + sqrt(37))/2 > > > > so you can see BOTH of the factors on the right hand side inside of > > the one factor on the left. > > > > So they must share all their factors forcing > > (sqrt(37) - sqrt(13))/2 to actually be a unit thought it's not in > > the ring of algebraic integers. > > > > Well, as a matter of fact (sqrt(37)-sqrt(13))/2 is an algebraic > > integer. > > > And it's not a unit in that ring. Despite that people have instructed him > > on this point for years, he still doesn't understand that the word unit > > has no meaning without specifying a ring first. > > One way to define when an algebraic number which is not an algebraic > > integer is a unit is to say: it's a unit when you can multiply it > > by an algebraic integer and get 1. Not every algebraic number has this > > property. > > Actually provably (sqrt(37) - sqrt(13))/2 is coprime to 3, though not > in the ring of algebraic integers. > > Well, this doesn't mean anything until you specify in what ring. > The ring of algebraic integers forces you to think carefully so that > makes sense, as it's like how with evens 2 is NOT a factor of 6 because > 3 is not even. > > In this case, you can prove that (sqrt(37) - sqrt(13))/2 does not have > any factors in common with 3 in actuality Meaningless. Specify what ring you're working in. > You are going backwards. The point of the demonstration is that the OLD WAYS DO NOT WORK so the old way of emphasizing what ring don't work. It turns out that there is really only one ring. I call it the ring of objects. But rather than go into that, I like pointing out that the old ideas lead to the appearance of contradictions. Like how you can prove factors in common with 3 and then find that you can't get algebraic integer results that agree!!! That's because the ring of algebraic integers has some issues--it has problems. Easily proven problems when you have the right tools. I've shown it has problems with non-polynomial factorization, and even fighting the paper and the result with some emails. Now I've found a demonstration that removes room for obfuscation by taking away variables and functions!!! So Arturo Magidin can't babble about new polynomials. And Dik Winter can't come at people talking about bizarre functions that can defy the distributive property!!! (5 + (sqrt(37) - sqrt(13))/2)((sqrt(37) - sqrt(13))/2) = (5 - sqrt(13))(5 + sqrt(37))/2 Those are just numbers--no variables and no functions. Your confidence in continuing to reply to me is all about your lack of fear and the greater fear of the bulk of the sci.math readership that is bothering with reading this debate. The reality is cowardice in the math field, while yes, you are not a coward or you would have run away by now. But you feel no fear so you continue to work to obscure. And math people let you because none of them truly care or believe in mathematics, least of all mathematical proof. I use to call it breaking. I'd explain everything to some mathematician--used to contact them by email--and they'd seem to kind of get it, and then run away. I'd say they broke. Cowardice rules the math field, along with naivette and basic incompetence about numbers. James Harris === Subject: Re: JSH: Will to be wrong >[...] It turns out that there is really only one ring. Whee! Another classic. Hard to say whether this is going to replace integers are irrational as most-quoted JSH-ism. >I call it the ring of objects. But rather than go into that, I like pointing out that the old ideas >lead to the appearance of contradictions. [...] I use to call it breaking. I'd explain everything to some >mathematician--used to contact them by email--and they'd seem to kind >of get it, and then run away. I'd say they broke. The rest of us would say they realized that there was no way to make you see reason, so they stopped replying because they had better things to do (unlike us here on sci.math). >Cowardice rules the math field, along with naivette and basic >incompetence about numbers. >James Harris ************************ David C. Ullrich === Subject: Re: JSH: Will to be wrong : It turns out that there is really only one ring. Oh dear - James is going all Tolkien on us. : I call it the ring of objects. So when I chuck it into Mount Doom are you going to go in after it? : But rather than go into that, I like pointing out that the old ideas : lead to the appearance of contradictions. You can't ignore old ideas in the very attempt to disprove them. In other words, you can't dismiss the proper uses of the words unit, factor and so on and then use your own vague definitions to disprove the original ones. : That's because the ring of algebraic integers has some issues--it has : problems. Such as what? Justin === Subject: Re: JSH: Will to be wrong : It turns out that there is really only one ring. Oh dear - James is going all Tolkien on us. Possibly Conway's surreal numbers. One ring to embed them all One ring to define them One ring to bring them all But leave characteristic p behind them But this is too limiting. On the other hand, for the category of rings the initial object is Z and the terminal object is the trivial ring, and what good is that? === Subject: Re: JSH: Will to be wrong : It turns out that there is really only one ring. > > Oh dear - James is going all Tolkien on us. Possibly Conway's surreal numbers. One ring to embed them all > One ring to define them > One ring to bring them all > But leave characteristic p behind them But this is too limiting. On the other hand, for the category of rings > the initial object is Z and the terminal object is the trivial ring, > and what good is that? The one ring is powerful as a concept, as it removes all the extra and useless baggage that math people carry around with them using the old, conventional concepts. With the one ring I found a short proof of Fermat's Last Theorem. Having the idea of the one ring in mind I could find (5 + (sqrt(37) - sqrt(13))/2)((sqrt(37) - sqrt(13))/2) = (5 - sqrt(13))(5 + sqrt(37))/2 and blow apart the old ways of thinking with just numbers. Taking away the ability of sci.math posters like Dik Winter and Arturo Magidin to fool the bulk of you with mumblings about new polynomials, key variables or what's a constant term, or wacky functions. No functions. No variables. No constant terms to question. After my discovery of the proof of Fermat's Last Theorem I found my prime counting function, which because of its use of a constrained partial difference equation gives two explanations at once, answering the questions Riemann and others before him had been asking: 1. It has a partial differential equation analog which shows the why of the connection between the count of prime numbers and continuous functions like x/ln(x). 2. The partial difference equation has to be constrained to give the exact count, or it's slightly off, explaining why there is a gap between what's given by the continuous functions versus the exact count of primes. Given the power of the one ring, and the sweep of my research, math people have had to have a will to be wrong--a deep need to BELIEVE they were correct against mathematical proof. That is human nature, so it's not a surprise. But it is also weakness which is not a surprise either. As if the math people currently in power had the strength of intellect and character needed for their positions, they might have figured out what I did--themselves. They did not, so their failure in admitting the truth is a weakness of character and intellect, which unfortunately, is not a surprise. Mathematical research is not about fantasies of being a great discoverer. The math does not care about human delusions of grandeur, or human need to feel important. It does not bend to social needs, which is why real mathematicians don't need social power, a point I've made over and over again. People like Wiles and Ribet have a large crowd of math people to defend them, hope for them, cry for them, and most importantly, deny for them. I don't need a crowd--I have mathematical proof. James Harris === Subject: Re: JSH: Will to be wrong useless baggage that math people carry around with them using the old, > conventional concepts. If you think that the old concepts of ring theory are useless, then presumably you think that the proof that the circle cannot be squared is flawed (or would do if you had the faintest idea how it was proved). In that case, perhaps you should devote your considerable energies to squaring the circle. > After my discovery of the proof of Fermat's Last Theorem I found my > prime counting function, which because of its use of a constrained > partial difference equation gives two explanations at once, answering > the questions Riemann and others before him had been asking: 1. It has a partial differential equation analog which shows the why > of the connection between the count of prime numbers and continuous > functions like x/ln(x). Shows they why? The why of the connection between the count of prime numbers and continuous functions is obvious. For example, extend the domain of pi(n) from the natural numbers to the non-negative reals by demanding that it is linear on every interval (n,n+1) for n in N. Then pi(n) is continuous. Your claim may have more substance if you can give a mathematical definition of when a function is like x/ln(x), but as it stands it is trivial. Incidentally, the prime number theorem that relates pi(x) to x/ln(x) has already been proven. > 2. The partial difference equation has to be constrained to give the > exact count, or it's slightly off, explaining why there is a gap > between what's given by the continuous functions versus the exact count > of primes. Actually there is no gap for the function I defined above. >... > Mathematical research is not about fantasies of being a great > discoverer. Quite. -Rotwang === Subject: Re: JSH: Will to be wrong The one ring is powerful as a concept, as it removes all the extra and > useless baggage that math people carry around with them using the old, > conventional concepts. If you think that the old concepts of ring theory are useless, then > presumably you think that the proof that the circle cannot be squared > is flawed (or would do if you had the faintest idea how it was proved). > In that case, perhaps you should devote your considerable energies to > squaring the circle. > Pay attention to what I write. Note specifically above: ...removes all the extra and useless baggage that math people carry around... There is excess in the conventional ideas of rings, which the one ring removes. And you can do so much more without that useless extra, while not necessarily everything done in the past is wrong. I am quite specific about what is wrong, as it all relates to the coverage problem of the ring of algebraic integers. And remember I've easily proven problems with that ring repeatedly, while in response I've faced people like yourself who have gotten away with avoiding the mathematical proofs and using insults versus real debate. (5 + (sqrt(37) - sqrt(13))/2)((sqrt(37) - sqrt(13))/2) = (5 - sqrt(13))(5 + sqrt(37))/2 is in response to tactics that worked against proofs that had function and variables, where by just having numbers I can keep people like you from so easily getting away with fooling the bulk of the readership. > After my discovery of the proof of Fermat's Last Theorem I found my > prime counting function, which because of its use of a constrained > partial difference equation gives two explanations at once, answering > the questions Riemann and others before him had been asking: > > 1. It has a partial differential equation analog which shows the why > of the connection between the count of prime numbers and continuous > functions like x/ln(x). Shows they why? The why of the connection between the count of > prime numbers and continuous functions is obvious. For example, extend > the domain of pi(n) from the natural numbers to the non-negative reals > by demanding that it is linear on every interval (n,n+1) for n in N. > Then pi(n) is continuous. Your claim may have more substance if you can > give a mathematical definition of when a function is like x/ln(x), > but as it stands it is trivial. Incidentally, the prime number theorem > that relates pi(x) to x/ln(x) has already been proven. > So simple? Then Gauss was a fool for wondering about x/ln(x)? Riemann was an idiot for looking for a reason? So if you were born back then you could have taught Gauss, eh? So simple, just ...extend the domain of pi(n) from the natural numbers to the non-negative reals...? Do that web search on unskilled and unaware. > 2. The partial difference equation has to be constrained to give the > exact count, or it's slightly off, explaining why there is a gap > between what's given by the continuous functions versus the exact count > of primes. Actually there is no gap for the function I defined above. > So you are better than Gauss? >... > Mathematical research is not about fantasies of being a great > discoverer. Quite. -Rotwang If you agree then, why put yourself up as better than Gauss? Why act like you know more than Riemann ever did? Why claim one of the great areas of mathematics, prime numbers and the behavior of the prime distribution, is some trivial thing you can handle so simply: ...extend the domain of pi(n) from the natural numbers to the non-negative reals... You think that's all there is to it? James Harris === Subject: Re: JSH: Will to be wrong If you think that the old concepts of ring theory are useless, then > presumably you think that the proof that the circle cannot be squared > is flawed (or would do if you had the faintest idea how it was proved). > ... > Pay attention to what I write. > Note specifically above: ...removes all the extra and useless baggage > that math people carry around... > There is excess in the conventional ideas of rings, which the one ring > removes. > And you can do so much more without that useless extra, while not > necessarily everything done in the past is wrong. FFS, looks like I have to spell it out for you. The proof that the circle cannot be squared relies on ring theory. Specifically one considers field extensions Kf obtained by adjoining the coordinates of (possibly) ruler-and-compass constructible points to a field Ki generated by the coordinates of some initial set of points. By considering the dimensionality of Kf considered as a vector space over Ki one can show that the adjoined points were not in fact ruler-and-compass constructible after all. If there were indeed only one ring, and therefore at most one field, then this method would not work. If OTOH the method does work, then ring theory can hardly be considered excess baggage. Unless, that is, you can present a proof of the impossibility of squaring the circle using only the object ring (I won't hold my breath for that one). > I am quite specific about what is wrong, as it all relates to the > coverage problem of the ring of algebraic integers. Sure. Except that you are either unable or unwilling to tell us what a coverage problem actually is, despite being asked dozens of times by a multitude of different people. > And remember I've easily proven problems with that ring repeatedly, > while in response I've faced people like yourself who have gotten away > with avoiding the mathematical proofs and using insults versus real > debate. Please point to a place in this thread where I have used insults. I have made many statements, none of which you have refuted, and I have asked many questions, none of which you have answered, but I have no recollection of having insulted you. I also have some recollection of having spent some time writing an algorithm at your request in the recent past. Have you verified it yet? > (5 + (sqrt(37) - sqrt(13))/2)((sqrt(37) - sqrt(13))/2) = (5 - > sqrt(13))(5 + sqrt(37))/2 An equality. Big Factorisation Domain. > Shows they why? The why of the connection between the count of > prime numbers and continuous functions is obvious. For example, extend > the domain of pi(n) from the natural numbers to the non-negative reals > by demanding that it is linear on every interval (n,n+1) for n in N. > Then pi(n) is continuous. Your claim may have more substance if you can > give a mathematical definition of when a function is like x/ln(x), > but as it stands it is trivial. Incidentally, the prime number theorem > that relates pi(x) to x/ln(x) has already been proven. > So simple? Then Gauss was a fool for wondering about x/ln(x)? Riemann > was an idiot for looking for a reason? however I don't imagine he lost any sleep over continuous functions in general for precisely the reason I gave. If you are claiming that your p.d.e. explains why pi(x) can be approximated by a generic continuous function, then there is nothing to explain. If you are claiming it explains why pi(x) can be approximated by a function like x/ln(x) then you must give a mathematical definition of like, which you haven't. If you are claiming it explains why pi(x) can be approximated by the specific function x/ln(x), then you must actually show that this is the case, rather than simply assert it, if you want anyone to take your claim seriously. > Do that web search on unskilled and unaware. Already read it, long before you ironically started to believe that it somehow supported your position. Once again I will ask: please give an example of a field where I am certain of my own knowledge, and have demonstrated gross deficiency. > So you are better than Gauss? Nope. > If you agree then, why put yourself up as better than Gauss? Why act > like you know more than Riemann ever did? When have I ever put myself up as better than Gauss? It happens that, unlike you, I know how to construct a mathematically coherent statement, and therefore am considerably better equipped to guess at what Guass would and wouldn't have considered trivial than you are. > Why claim one of the great areas of mathematics, prime numbers and the > behavior of the prime distribution, is some trivial thing you can > handle so simply: > ...extend the domain of pi(n) from the natural numbers to the > non-negative reals... > You think that's all there is to it? Why do you feel the need to put words in my mouth? I gave a very specific answer to a very specific question, namely the question of why pi(x) can be approximated by continuous funtions. Was my answer wrong? Have I claimed that my answer somehow solves all problems related to the prime distribution? -Rotwang === Subject: Re: JSH: Will to be wrong > If you think that the old concepts of ring theory are useless, then > > presumably you think that the proof that the circle cannot be squared > > is flawed (or would do if you had the faintest idea how it was proved). > > ... > Pay attention to what I write. > Note specifically above: ...removes all the extra and useless baggage > that math people carry around... > There is excess in the conventional ideas of rings, which the one ring > removes. > And you can do so much more without that useless extra, while not > necessarily everything done in the past is wrong. FFS, looks like I have to spell it out for you. The proof that the circle cannot be squared relies on ring theory. More specifically the proof relies on proving that pi is transcendant. > Specifically one considers field extensions Kf obtained by adjoining > the coordinates of (possibly) ruler-and-compass constructible points to > a field Ki generated by the coordinates of some initial set of points. > By considering the dimensionality of Kf considered as a vector space > over Ki one can show that the adjoined points were not in fact > ruler-and-compass constructible after all. If there were indeed only > one ring, and therefore at most one field, then this method would not > work. If OTOH the method does work, then ring theory can hardly be > considered excess baggage. Unless, that is, you can present a proof > of the impossibility of squaring the circle using only the object ring > (I won't hold my breath for that one). > The object ring CAN be used to prove that pi is transcendant as I've considered this issue before and even posted on it, of course to the derision of sci.math'ers. My approach was to consider infinite series known to include pi in their results and note their dependence on fractions--blocked by the object ring. > I am quite specific about what is wrong, as it all relates to the > coverage problem of the ring of algebraic integers. Sure. Except that you are either unable or unwilling to tell us what a > coverage problem actually is, despite being asked dozens of times by > a multitude of different people. > Denial. Repeatedly I've not only explained but given definitive examples including my latest relying on just numbers to block posters, including Dik Winter and Arturo Magidin, from confusing people with variables and functions. And that one is so easy I can casually give it (5 + (sqrt(37) - sqrt(13))/2)((sqrt(37) - sqrt(13))/2) = (5 - sqrt(13))(5 + sqrt(37))/2 repeatedly, and use it to directly show how the old, conventional thinking, fails. The situation for math people here is not unlike what physicists faced with quantum mechanics--difference is, physicists went with what worked. Math people have kept fighting for what does not. > And remember I've easily proven problems with that ring repeatedly, > while in response I've faced people like yourself who have gotten away > with avoiding the mathematical proofs and using insults versus real > debate. Please point to a place in this thread where I have used insults. I > have made many statements, none of which you have refuted, and I have > asked many questions, none of which you have answered, but I have no > recollection of having insulted you. I also have some recollection of > having spent some time writing an algorithm at your request in the > recent past. Have you verified it yet? > Ever? Never insulted me in any posts? Regardless, the history of people disagreeing with me has been a lot about denial--like your refusal to acknowledge my demonstrations of the failure of the old ways of thinking--and a lot about insults, like calling me a crank or crackpot. And the issue before with your algorithm has to do with proof checking of MATHEMATICAL PROOF not logic proofs. Logic proofs are slightly different. The issue of mathematical proofs was one that Russell grappled with--unsuccessfuly I might add--for some time. While sentential calculus is easy, mathematical proofs are hard. I suggest before that you prove that 1+1 = 2, and didn't bother to check what your reply might have been. I could find something else if you don't like that one, like use your algorithm to check a proof that sqrt(2) is irrational. > (5 + (sqrt(37) - sqrt(13))/2)((sqrt(37) - sqrt(13))/2) = (5 - > sqrt(13))(5 + sqrt(37))/2 An equality. Big Factorisation Domain. > It shows the power of simplicity against people willing to confuse other people with anything they can get their hands on. By taking away variables and functions, notice how silent it is. No replies from Dik Winter this time, eh? Because he can't go on about mystery functions that defy the distributive property. Do that search on unskilled and unaware and consider why you might not get the incredible elegance of the solution that has taken away so much from my critics, who were so adept at manipulating public opinion when variables and functions were available to them. Now they are mostly silenced. > > Shows they why? The why of the connection between the count of > > prime numbers and continuous functions is obvious. For example, extend > > the domain of pi(n) from the natural numbers to the non-negative reals > > by demanding that it is linear on every interval (n,n+1) for n in N. > > Then pi(n) is continuous. Your claim may have more substance if you can > > give a mathematical definition of when a function is like x/ln(x), > > but as it stands it is trivial. Incidentally, the prime number theorem > > that relates pi(x) to x/ln(x) has already been proven. > So simple? Then Gauss was a fool for wondering about x/ln(x)? Riemann > was an idiot for looking for a reason? however I don't imagine he lost any sleep over continuous functions in > general for precisely the reason I gave. If you are claiming that your > p.d.e. explains why pi(x) can be approximated by a generic continuous > function, then there is nothing to explain. If you are claiming it > explains why pi(x) can be approximated by a function like x/ln(x) > then you must give a mathematical definition of like, which you > haven't. If you are claiming it explains why pi(x) can be approximated > by the specific function x/ln(x), then you must actually show that this > is the case, rather than simply assert it, if you want anyone to take > your claim seriously. > Because you don't understand how remarkable it is that a discrete value like the count of primes would be connected to continuous functions. It is extraordinary and more than enough to bring interest from extraordinary mathematical minds like Gauss and Riemann. So you think it's just this trivial thing where you connect the dots continuously. And in doing so, you are not aware of the insult you give to those who can appreciate the oddity of it. > Do that web search on unskilled and unaware. Already read it, long before you ironically started to believe that it > somehow supported your position. Once again I will ask: please give an > example of a field where I am certain of my own knowledge, and have > demonstrated gross deficiency. > Mathematics. > So you are better than Gauss? Nope. > If you agree then, why put yourself up as better than Gauss? Why act > like you know more than Riemann ever did? When have I ever put myself up as better than Gauss? It happens that, > unlike you, I know how to construct a mathematically coherent > statement, and therefore am considerably better equipped to guess at > what Guass would and wouldn't have considered trivial than you are. > And you think that uniting the discrete with the continuous is just about connecting the dots? And you think you have an idea what Gauss considered trivial? I don't even dare to claim that, beyond what he himself said he thought was trivial. > Why claim one of the great areas of mathematics, prime numbers and the > behavior of the prime distribution, is some trivial thing you can > handle so simply: > ...extend the domain of pi(n) from the natural numbers to the > non-negative reals... > You think that's all there is to it? Why do you feel the need to put words in my mouth? I gave a very > specific answer to a very specific question, namely the question of why > pi(x) can be approximated by continuous funtions. Was my answer wrong? > Have I claimed that my answer somehow solves all problems related to > the prime distribution? -Rotwang Consider context, I found a prime counting function that gives the count of primes using a constrained partial difference equation, and thereby found direct answers to questions that intrigued great minds of mathematics--Gauss and Riemann--among others. Your response is an attempt to trivialize my research and in so doing you don't quite comprehend how you trivialize theirs. Do that search on unskilled and unaware and consider if it is about yourself, and your denial to accept that it is about you, is quite within the bounds of the research itself. Pay attention to the part where the researchers found that people in overrating their own ability don't think that's what they are doing. James Harris === Subject: Re: JSH: Will to be wrong The proof that the circle cannot be squared relies on ring theory. > More specifically the proof relies on proving that pi is transcendant. No, that is only part of it. The proof outline I gave assumed that the transcendence of pi is already proven. In order to dismiss ring theory as excess baggage, you need to reproduce the part of the proof I described without using ring theory. Explicity: Assume that pi is transcendental. Show, without using any rings other than the object ring, that the circle cannot be squared. > [proof outline snipped] > The object ring CAN be used to prove that pi is transcendant as I've > considered this issue before and even posted on it, of course to the > derision of sci.math'ers. Irrelevant, though I would be interested to see a citation nonetheless. > Denial. > Repeatedly I've not only explained but given definitive examples > including my latest relying on just numbers to block posters, Your examples have failed to answer the question to anybody's satisfaction as far as I can see. Let me make it easy for you. Definition: Let R be a ring. We say that R has a Coverage Problem when ________? > The situation for math people here is not unlike what physicists faced > with quantum mechanics--difference is, physicists went with what > worked. Actually the situation is very different. Physics is an empirical subject. Classical mechanics was never a mistake, it was never flawed, it was merely an approximation that was seen to fail upon the development of sufficiently precise experiments; physicists went with what agrees with experiment. Mathematics is not empirical. Mathematicians go with proof, which you have failed to provide. > Ever? Never insulted me in any posts? On the contrary, I have insulted you in many posts when I felt you deserved it (I have even apologised when it was pointed out to me that my insults were unfounded - you should try that some time). But I didn't insult you in this thread until after you insulted me. > And the issue before with your algorithm has to do with proof > checking of MATHEMATICAL PROOF not logic proofs. > Logic proofs are slightly different. Wrong. The general definition of a formal proof I gave includes many axiomatic systems, including Peano arithmetic and ZF set theory, both of which can handle statements like 1+1=2 (the latter can be considered a foundation for almost all mathematics). The only reason I chose zeroth order logic is because it is particularly simple and the resulting program was particulartly short. If you examine the program you will see that the methods it used to verify well-formed formulae, axioms, rules of inference and proofs can be trivially generalised to other axiomatic systems such as the examples I gave above. Besides which, this is irrelevant; the fact remains that you challenged me to write an algorithm and I did so. The fact that I was able to do so clearly demonstrates the difference between formal and natural language proofs. Of course there's an easy way to prove me wrong about this: simply describe an algorithm that can verify natural language proofs. Unless you have no idea how, in which case all your talk of how mathematicians should be using such programs was nothing but hot air. > The issue of mathematical proofs was one that Russell grappled > with--unsuccessfuly I might add--for some time. While sentential > calculus is easy, mathematical proofs are hard. In that case why the hell do you think it should be so easy to verify them by computer? > I suggest before that you prove that 1+1 = 2, and didn't bother to > check what your reply might have been. How convenient. Here it is: > By taking away variables and functions, notice how silent it is. No > replies from Dik Winter this time, eh? Plenty of replies from other people though. Unconventional definition of silent, though I gather that unconventional definitions are your forte. > Because you don't understand how remarkable it is that a discrete value > like the count of primes would be connected to continuous functions. Really? So do you deny that I gave an explicit example of a continuous function that approximates the count of primes? If not, then the fact that the count of primes would be connected to continuous functions is hardly remarkable. > It is extraordinary and more than enough to bring interest from > extraordinary mathematical minds like Gauss and Riemann. No, a considerably more precise statement than the one I quoted above was extraordinary and more than enough to blah blah blah. This is my point which you consistently manage to miss: your claims as to the importance of your prime counting formula are vague enough to be open to several different meanings. Some of those meanings are trivial, and some ill-defined. The ones that are neither trivial nor ill-defined, you haven't proven to be true. > Once again I will ask: please give an > example of a field where I am certain of my own knowledge, and have > demonstrated gross deficiency. > Mathematics. Give some examples of false mathematical statements I have made. I will be more than happy to do likewise for you, if you want. > And you think that uniting the discrete with the continuous is just > about connecting the dots? Again... do you deny that the function I defined has the properties that I claimed of it? > And you think you have an idea what Gauss considered trivial? > I don't even dare to claim that, beyond what he himself said he thought > was trivial. Nonetheless, you're more than happy to claim that Gauss hates us and loves you. Incidentally, what do you actually know of Gauss' work? Have you ever read a proof of, say, the fundamental theorem of algebra? > Consider context, I found a prime counting function that gives the > count of primes using a constrained partial difference equation, and > thereby found direct answers to questions that intrigued great minds of > mathematics--Gauss and Riemann--among others. No, you didn't. If you were able to tell the difference between a statement that is mathematically coherent and one that isn't you would realise this. > Your response is an attempt to trivialize my research and in so doing > you don't quite comprehend how you trivialize theirs. I bother to learn theirs. You don't. What could be more insulting than that? > Do that search on unskilled and unaware What part of already read it was too hard for you to understand? > and consider if it is about > yourself, and your denial to accept that it is about you, is quite > within the bounds of the research itself. > Pay attention to the part where the researchers found that people in > overrating their own ability don't think that's what they are doing. Right. I am unskilled and unaware - and so is every mathematician in the world. None of us realise how bad we are at maths, because we don't not unskilled and unaware. You are truly brilliant at maths, and brilliant enough to know how brilliant you are. That must be it. I can't think of any other explanation for the disagreement between you and the rest of the world. Can you? -Rotwang === Subject: Re: JSH: Will to be wrong non-negative reals by demanding that it is linear on every interval (n,n+1) > for n in N. Mmmmmnnnngggg!!! I did of course mean [n,n+1] -Rotwang === Subject: Re: JSH: Will to be wrong Meaningless. Specify what ring you're working in. > > You are going backwards. The point of the demonstration is that the > OLD WAYS DO NOT WORK so the old way of emphasizing what ring don't > work. It turns out that there is really only one ring. Hey, it's stand-up comedy time again! Okay, James. Define what a ring is, or else you'll be viewed as an idiotic trolling crank forever. I guess it doesn't matter anyway, since he kill-files me and will thus never see this reply. So the rest of this reply is more cheap shots at JSH. > I call it the ring of objects. But rather than go into that, I like pointing out that the old ideas > lead to the appearance of contradictions. Or rather YOUR LACK OF UNDERSTANDING of the old ideas leads to the appearance of contradictions. Going to JSH for math advice is like going to a cardiologist who's never even seen a picture of a heart, let alone a real one. > Like how you can prove factors in common with 3 and then find that you > can't get algebraic integer results that agree!!! That's because the ring of algebraic integers has some issues--it has > problems. Coverage problems, right? Which are not actually part of the definition of a ring. > Easily proven problems when you have the right tools. Like a lobotomy. > I've shown it has problems with non-polynomial factorization, When are non-polynomials used in Algebra? I must have slept through that class. > and even > fighting the paper and the result with some emails. ONE e-mail, which pointed out the error. An error in a paper that JSH knew was there when he submitted it. Meanwhile, I still have two published papers, and a third one on the way. Let's take a look at the status on the third one, hm? On the Tightness of the 5/14 Independence Ratio/Oct 15, 2006/Required Reviews Completed Looks like I'll be hearing from them any day now. And this one is a solo, a result of mis-understanding what my advisor said I should do. > Now I've found a demonstration that removes room for obfuscation by > taking away variables and functions!!! Obfuscation ... Sounds like a word that only an English teacher would use. Let's try another one ... Obfuscation ... I guess that's JSH's new-word-of-the-month. And another ... Obfuscation ... How dare you speak like that in a public forum! One last one ... Obfuscation ... Isn't that the name of the house where JSH was born? Here's a goodie: Obfuscation ... I guess he's been reading through his mother's dictionary again. > So Arturo Magidin can't babble about new polynomials. And Dik Winter > can't come at people talking about bizarre functions that can defy the > distributive property!!! That doesn't seem fair: Winter can't talk about bizarre functions, but JSH can talk bizarrely about functions. > (5 + (sqrt(37) - sqrt(13))/2)((sqrt(37) - sqrt(13))/2) = (5 - sqrt(13))(5 + sqrt(37))/2 Those are just numbers--no variables and no functions. What's that sqrt thing, then? And if it represents the square root of n, then which one does it represent? You know, there are TWO (provided n is nonzero). > Your confidence in continuing to reply to me is all about your lack of > fear and the greater fear of the bulk of the sci.math readership that > is bothering with reading this debate. And ROTFLTAO. > The reality is cowardice in the math field, while yes, you are not a > coward or you would have run away by now. People who are cowards stay and keep fighting? Did JSH take some recreation pharmaceuticals before writing this? > But you feel no fear so you continue to work to obscure. And math people let you because none of them truly care or believe in > mathematics, least of all mathematical proof. Gee, this is a relevation, on par with the fact that there's only one ring. > I use to call it breaking. I'd explain everything to some > mathematician--used to contact them by email--and they'd seem to kind > of get it, and then run away. I'd say they broke. In short, they just got tired of your d@mn $h!t. They found something more important to do than read your research, like swimming through a sewer. Just because JSH doesn't have a life doesn't mean no one else does. Here's some examples of what professional mathematicians say about JSH: Buzz off, pea brain. -- Barry Mazur, to James S. Harris (attributed) Stop e-mailing me, you @$*&$#! -- Andrew Granville, to James Harris (attributed) Oh no, another e-mail from that Harris crank! --- Barry Mazur (attributed) > Cowardice rules the math field, along with naivette and basic > incompetence about numbers. How, exactly, can cowardice rule? > James Harris The ultimate coward, since he doesn't bother replying to the legitimate issues that people bring up. He fails to reply, or calls them a liar, or threatens to kill them, but never does anything. Yep, JSH is the ultimate coward. --- Christopher Heckman === Subject: Re: JSH: Will to be wrong LOL! Talk about posts that belong in the permanent record. Here you go, James: saving another one from the mysterious tragedy that causes so many of your posts to vanish. >> > [Rupert] >> > >> ... >> > >> Then we either have >> > >> > >> ((5+sqrt(37)+5-sqrt(13))/2)=p_1.q_2.q_3 >> > >> (sqrt(37)-sqrt(13))/2=p_2.(q_1)^2 >> > >> > >> or >> > >> > >> ((5+sqrt(37)+5-sqrt(13))/2=p_1.(q_1)^2 >> > >> (sqrt(37)-sqrt(13))/2=p_2.q_2.q_3 >> > > > [also Rupert] >> > > It must be this one. Because there exists a nonidentity element >> > > of the >> > > Galois group under which the ideal generated by >> > > (sqrt(37)-sqrt(13))/2 >> > > is invariant. >> > > > > I will post the details of my reasoning about this example later. >> > > > > So we have >> > > > > (((5+sqrt(37)+5-sqrt(13))/2)=p_1.(q_1)^2 >> > > ((sqrt(37)-sqrt(13))/2)=p_2.q_2.q_3 >> > > (5-sqrt(13))=p_1.p_2.q_1.q_2 >> > > ((5+sqrt(37))/2)=q_1.q_3 >> > > > It's encouraging :-) that the p's and q's balance now across the >> > LHS and RHS >> > of: >> > > > ((5 + sqrt(37) + 5 - sqrt(13))/2)((sqrt(37) - sqrt(13))/2) = >> > (5 - sqrt(13))(5 + sqrt(37))/2 >> > > > So what point did you want to make with this example, James? >> > > > I think you guessed that one days ago, although the details of his >> > claim >> > seem to change. Most recently he claimed to me that his muddy >> > reasoning >> > about factors proves that: >> > > > (sqrt(37) - sqrt(13))/2 >> > > > is a unit: >> > > > [JSH] >> > ... >> > Another key bit of information is that the original >> > factorization >> > > > (5 + (sqrt(37) - sqrt(13))/2)((sqrt(37) - sqrt(13))/2) = (5 - >> > sqrt(13))(5 + sqrt(37))/2 >> > > > is the same as >> > > > ((5 + sqrt(37) + 5 - sqrt(13))/2)((sqrt(37) - sqrt(13))/2) = >> > (5 - >> > sqrt(13))(5 + sqrt(37))/2 >> > > > so you can see BOTH of the factors on the right hand side >> > inside of >> > the one factor on the left. >> > > > So they must share all their factors forcing >> > (sqrt(37) - sqrt(13))/2 to actually be a unit thought it's not >> > in >> > the ring of algebraic integers. >> > > > Well, as a matter of fact (sqrt(37)-sqrt(13))/2 is an algebraic >> > integer. >> > > And it's not a unit in that ring. Despite that people have >> > instructed him >> > on this point for years, he still doesn't understand that the word >> > unit >> > has no meaning without specifying a ring first. >> > > One way to define when an algebraic number which is not an algebraic >> > integer is a unit is to say: it's a unit when you can multiply it >> > by an algebraic integer and get 1. Not every algebraic number has >> > this >> > property. >> > > Actually provably (sqrt(37) - sqrt(13))/2 is coprime to 3, though not >> in the ring of algebraic integers. >> Well, this doesn't mean anything until you specify in what ring. >> The ring of algebraic integers forces you to think carefully so that >> makes sense, as it's like how with evens 2 is NOT a factor of 6 because >> 3 is not even. >> > In this case, you can prove that (sqrt(37) - sqrt(13))/2 does not have >> any factors in common with 3 in actuality >> Meaningless. Specify what ring you're working in. > > You are going backwards. The point of the demonstration is that the > OLD WAYS DO NOT WORK so the old way of emphasizing what ring don't > work. It turns out that there is really only one ring. I call it the ring of objects. But rather than go into that, I like pointing out that the old ideas > lead to the appearance of contradictions. Like how you can prove factors in common with 3 and then find that you > can't get algebraic integer results that agree!!! That's because the ring of algebraic integers has some issues--it has > problems. Easily proven problems when you have the right tools. I've shown it has problems with non-polynomial factorization, and even > fighting the paper and the result with some emails. Now I've found a demonstration that removes room for obfuscation by > taking away variables and functions!!! So Arturo Magidin can't babble about new polynomials. And Dik Winter > can't come at people talking about bizarre functions that can defy the > distributive property!!! (5 + (sqrt(37) - sqrt(13))/2)((sqrt(37) - sqrt(13))/2) = (5 - > sqrt(13))(5 + sqrt(37))/2 Those are just numbers--no variables and no functions. Your confidence in continuing to reply to me is all about your lack of > fear and the greater fear of the bulk of the sci.math readership that > is bothering with reading this debate. The reality is cowardice in the math field, while yes, you are not a > coward or you would have run away by now. But you feel no fear so you continue to work to obscure. And math people let you because none of them truly care or believe in > mathematics, least of all mathematical proof. I use to call it breaking. I'd explain everything to some > mathematician--used to contact them by email--and they'd seem to kind > of get it, and then run away. I'd say they broke. Cowardice rules the math field, along with naivette and basic > incompetence about numbers. > James Harris === Subject: Re: JSH: Will to be wrong days. My association with the Department is that of an alumnus. >LOL! Talk about posts that belong in the permanent record. Here you go, >James: saving another one from the mysterious tragedy that causes so many >of your posts to vanish. I have to wonder why the fixation with Dik and me, given your enormous efforts of these past months, Tim. You've certainly maintained a much more level head than I every did. And yet, you can't seem to get your name into the generic rants! It will be three years since James... ahem... accepted my offer of not replying to him, and I haven't. I have not addressed his arguments directly since he made his I cannot do now that he's got the real, honest, final, no-this-time-I-mean-it example... I wonder what you need to do, Tim, to graduate onto the rants? -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes by Bill Watterson) Arturo Magidin magidin-at-member-ams-org === Subject: Re: JSH: Will to be wrong LOL! Talk about posts that belong in the permanent record. Here you go, >James: saving another one from the mysterious tragedy that causes so many >of your posts to vanish. I have to wonder why the fixation with Dik and me, given your enormous > efforts of these past months, Tim. You've certainly maintained a much > more level head than I every did. And yet, you can't seem to get your > name into the generic rants! It will be three years since > James... ahem... accepted my offer of not replying to him, and I > haven't. I have not addressed his arguments directly since he made his > I cannot do now that he's got the real, honest, final, > no-this-time-I-mean-it example... I wonder what you need to do, Tim, to graduate onto the rants? Well, we are talking non-polynomial factorization here, not surrogate factoring, and most of Tim's effort has been aimed at the latter. Tim has received a death threat, I think that counts as graduation. - the poster who goes by the name of William Hughes === Subject: Re: JSH: Will to be wrong days. My association with the Department is that of an alumnus. >Well, we are talking non-polynomial factorization here, not >surrogate factoring, and most of Tim's effort has been aimed at the >latter. Well, yes, but I first earned James's wrath when pointing out that his rational part and nonrational part arguments for a prior attempt at proof were nonsense. That's when I became a confirmed liar and obfuscator. > Tim has received a death threat, I think that >counts as graduation. Haven't we ->all<- received veiled death threats by now? I confess, for example, that I think James dies a little every time he points to my alleged silence as more evidence that he finally hit the right example. I wonder how much he remembers of the exchange when I made the offer to stop addressing his arguments directly and replying to him. To quote Clarence Darrow: -- I have never killed any one, but I have read some obituary notices with great satisfaction. --- Clarence Darrow, chapter 10 of _The Story of my Life_ Arturo Magidin magidin-at-member-ams-org === Subject: Re: JSH: Will to be wrong LOL! Talk about posts that belong in the permanent record. Here you go, > >James: saving another one from the mysterious tragedy that causes so many > >of your posts to vanish. > > I have to wonder why the fixation with Dik and me, given your enormous > efforts of these past months, Tim. You've certainly maintained a much > more level head than I every did. And yet, you can't seem to get your > name into the generic rants! It will be three years since > James... ahem... accepted my offer of not replying to him, and I > haven't. I have not addressed his arguments directly since he made his > I cannot do now that he's got the real, honest, final, > no-this-time-I-mean-it example... > > I wonder what you need to do, Tim, to graduate onto the rants? Well, we are talking non-polynomial factorization here, not > surrogate factoring, and most of Tim's effort has been aimed at the > latter. Tim has received a death threat, I think that > counts as graduation. - the poster who goes by the name of William Hughes There was never any death threat. I just noted that if my research on factoring IS viable then there are people who might not care about the niceties of the why's of sci.math poster's behavior, but just be truly angry at the reality of supposedly intelligent human beings fighting against something of that value. And remember people that research in its infant state is still out there. As I move forward with shattering the objections and credibility of posters with this latest example using (5 + (sqrt(37) - sqrt(13))/2)((sqrt(37) - sqrt(13))/2) = (5 - sqrt(13))(5 + sqrt(37))/2 the reality of my other research remains. You people can joke now because there is the unreality still and maybe some sense that you still have the support of the readers of the newsgroup? But realize, some of them now know they were lied to, or they trusted people who just got the math terribly wrong. Consequences can be huge, like loss of savings from someone actually working through that factoring research--if it is viable. And if you want to consider how angry someone who used to be a billionaire might be if that person becomes completely broke because they trusted a system that you people defended by confusing people about my research, how far might they go? I warned of the possible consequences of your behavior. This may seem like a trivial game to many of you, but in the real world, some of this math DOES matter. And in the real world, there are people who do more than just talk. So yes, later if some angry newly poor person comes after some of you later, you can chatter at them all you want, but they probably won't listen to you. You can't talk your way out of every problem. James Harris === Subject: Re: JSH: Will to be wrong [Tim Peters] > LOL! Talk about posts that belong in the permanent record. Here > you go, James: saving another one from the mysterious tragedy that > causes so many of your posts to vanish. [Arturo Magidin] I have to wonder why the fixation with Dik and me, given your enormous efforts of these past months, Tim. Well, I'm not an academic, but you are, and James probably thinks Dik is too. I'm more surprised that he's not also ripping at David Ullrich -- it's unusual for him /not/ to rip at David, even wrt topics (like surrogate factoring) where David is conspicuous by absence. You've certainly maintained a much more level head than I every did. Not always ;-) You certainly have no cause to be ashamed based on any of the old posts I've seen. James dishes it out 10x worse than he ever gets back from anyone who gives his arguments serious attention, and in your case 100x worse. Your restraint was remarkable! And yet, you can't seem to get your name into the generic rants! I did once, at a local maximum in the ongoing (albeit sporadic) surrogate factoring show. It's not really a life goal of mine to stay on that list ;-) If it were, I'd take a week to perfect a usable time machine, and go back 30 years to work toward becoming a professor instead of taking that damn computer job :-) It will be three years since James... ahem... accepted my offer of not replying to him, and I haven't. I have not addressed his arguments You have been true to your word on that, and given the vicious little provocations he /still/ throws your way, your self-control is an ongoing marvel to me. OTOH, I'm not sure James understood what you were offering, any more than he understands much of anything said to him. Wouldn't surprise me a bit if he forgot even that much, and thinks you don't reply to him any more because he broke you. And still he feels the need to say all the things I cannot do now that he's got the real, honest, final, no-this-time-I-mean-it example... That's only because he truly hates you ;-) I wonder what you need to do, Tim, to graduate onto the rants? There's scant chance I'll become a professor in what remains of this lifetime ... but I /could/ tell him my IQ ;-) I /expect/ that would drive him crazy with loathing (I don't know what his measured IQ is, but he's said he scored in the 98th percentile on various tests, and I'm four sigma territory -- because he has so few real accomplishments, he desparately overvalues the few things he officially did well at -- even silly things like IQ score). [William Hughes] >> Well, we are talking non-polynomial factorization here, not >> surrogate factoring, and most of Tim's effort has been aimed at the >> latter. Indeed, and all of my serious effort. What I know about algebraic integers is what I've learned from reading replies to JSH and following up threads are a vacation for me ;-) >> Tim has received a death threat, I think that counts as graduation. It's a step on the journey for sure. But Arturo is mostly right, I've only shown up once in a /generic/ JSH rant. >> - the poster who goes by the name of William Hughes And what's up with you? James doesn't appear to loathe you, despite that you've punched massive holes in his arguments more consistently than anyone else during this round, and have even accomplished the near-impossible task of explaining his errors in ways he partly understands. If you don't start revealing your academic credentials, or in some other way give James objective evidence that you have official recognition that should be his instead, you're never going to make it on to his hit list. Try harder ;-) [jstevh@msn.com>] > There was never any death threat. Liar. === Subject: Re: JSH: SF: Finally, surrogate factoring ... [JSH replying to Rick Decker and me] ... You're both lying. I am going to warn you. Neither of you will live to see 2007 because of this lie becaue there will be angry people who will kill you, as there is so much money at stake. Billions will be lost. It's not like you can reverse it now either. You killed yourselves. It might have seemed like a small lie to both of you, but your names will live in infamy, while you will not live at all. James Harris > I just noted that if my research on factoring IS viable then there are > people who might not care about the niceties of the why's of sci.math > poster's behavior, but just be truly angry at the reality of supposedly > intelligent human beings fighting against something of that value. Liar. There was nothing conditional about Neither of you will live to see 2007 because of this lie, or any of the rest of it. > ... [repetition] ... > You people can joke now because there is the unreality still and maybe > some sense that you still have the support of the readers of the > newsgroup? Poor James. People who know something about the topics you flail away at /know/ for themselves that you have nothing to offer beyond low-brow twisted entertainment. The newsgroups are irrelevant to that. Nearly all of your math is trivial or wrong, and every mathematician who bothers to look at it arrives at that conclusion with ease -- there's no need for help or social validation. You can't know that /because/ you know so little about the topics you flail away at and appear incapable of learning. And that's all there is to it in the real world. You're a mathematical crank, but an unusually entertaining one. > ... [repetition of old low-brow twisted entertainment] ... === Subject: Re: JSH: Will to be wrong > [Tim Peters] >> LOL! Talk about posts that belong in the permanent record. Here >> you go, James: saving another one from the mysterious tragedy that >> causes so many of your posts to vanish. > [Arturo Magidin] > I have to wonder why the fixation with Dik and me, given your enormous > efforts of these past months, Tim. Tim's relationship with JHS to reminds me of Sancho Panza and Don Quixote, different with similar dreams. === Subject: Re: JSH: Will to be wrong > > >LOL! Talk about posts that belong in the permanent record. Here you go, > >James: saving another one from the mysterious tragedy that causes so many > >of your posts to vanish. > > I have to wonder why the fixation with Dik and me, given your enormous > > efforts of these past months, Tim. You've certainly maintained a much > > more level head than I every did. And yet, you can't seem to get your > > name into the generic rants! It will be three years since > > James... ahem... accepted my offer of not replying to him, and I > > haven't. I have not addressed his arguments directly since he made his > > I cannot do now that he's got the real, honest, final, > > no-this-time-I-mean-it example... > > I wonder what you need to do, Tim, to graduate onto the rants? > > Well, we are talking non-polynomial factorization here, not > surrogate factoring, and most of Tim's effort has been aimed at the > latter. Tim has received a death threat, I think that > counts as graduation. > > - the poster who goes by the name of William Hughes There was never any death threat. I just noted that if my research on factoring IS viable then there are > people who might not care about the niceties of the why's of sci.math > poster's behavior, but just be truly angry at the reality of supposedly > intelligent human beings fighting against something of that value. > Duh, like we never threatened him at all. Like, Duh, all we said, like, was that this was a really nice store and it would be a shame, like, if it burned down, like. - William Hughes === Subject: Re: JSH: Will to be wrong > [Rupert] > > >> ... > > >> Then we either have > > > > > >> ((5+sqrt(37)+5-sqrt(13))/2)=p_1.q_2.q_3 > > >> (sqrt(37)-sqrt(13))/2=p_2.(q_1)^2 > > > > > >> or > > > > > >> ((5+sqrt(37)+5-sqrt(13))/2=p_1.(q_1)^2 > > >> (sqrt(37)-sqrt(13))/2=p_2.q_2.q_3 > > > > [also Rupert] > > > It must be this one. Because there exists a nonidentity element of the > > > Galois group under which the ideal generated by (sqrt(37)-sqrt(13))/2 > > > is invariant. > > > > > > I will post the details of my reasoning about this example later. > > > > > > So we have > > > > > > (((5+sqrt(37)+5-sqrt(13))/2)=p_1.(q_1)^2 > > > ((sqrt(37)-sqrt(13))/2)=p_2.q_2.q_3 > > > (5-sqrt(13))=p_1.p_2.q_1.q_2 > > > ((5+sqrt(37))/2)=q_1.q_3 > > > > It's encouraging :-) that the p's and q's balance now across the LHS and RHS > > > of: > > > > ((5 + sqrt(37) + 5 - sqrt(13))/2)((sqrt(37) - sqrt(13))/2) = > > > (5 - sqrt(13))(5 + sqrt(37))/2 > > > > > So what point did you want to make with this example, James? > > > > I think you guessed that one days ago, although the details of his claim > > > seem to change. Most recently he claimed to me that his muddy reasoning > > > about factors proves that: > > > > (sqrt(37) - sqrt(13))/2 > > > > is a unit: > > > > [JSH] > > > ... > > > Another key bit of information is that the original factorization > > > > (5 + (sqrt(37) - sqrt(13))/2)((sqrt(37) - sqrt(13))/2) = (5 - > > > sqrt(13))(5 + sqrt(37))/2 > > > > is the same as > > > > ((5 + sqrt(37) + 5 - sqrt(13))/2)((sqrt(37) - sqrt(13))/2) = (5 - > > > sqrt(13))(5 + sqrt(37))/2 > > > > so you can see BOTH of the factors on the right hand side inside of > > > the one factor on the left. > > > > So they must share all their factors forcing > > > (sqrt(37) - sqrt(13))/2 to actually be a unit thought it's not in > > > the ring of algebraic integers. > > > > > Well, as a matter of fact (sqrt(37)-sqrt(13))/2 is an algebraic > > integer. > > > > And it's not a unit in that ring. Despite that people have instructed him > > > on this point for years, he still doesn't understand that the word unit > > > has no meaning without specifying a ring first. > > > > One way to define when an algebraic number which is not an algebraic > > integer is a unit is to say: it's a unit when you can multiply it > > by an algebraic integer and get 1. Not every algebraic number has this > > property. > > > > Actually provably (sqrt(37) - sqrt(13))/2 is coprime to 3, though not > > in the ring of algebraic integers. > > Well, this doesn't mean anything until you specify in what ring. > > The ring of algebraic integers forces you to think carefully so that > > makes sense, as it's like how with evens 2 is NOT a factor of 6 because > > 3 is not even. > > In this case, you can prove that (sqrt(37) - sqrt(13))/2 does not have > > any factors in common with 3 in actuality > > Meaningless. Specify what ring you're working in. > > You are going backwards. The point of the demonstration is that the > OLD WAYS DO NOT WORK so the old way of emphasizing what ring don't > work. It turns out that there is really only one ring. > Really? Can I quote you on that? There is only one ring. Fascinating. Would you care to post the definition of a ring? I'll give you a hint. To make your statement defensible, modify it to There is only one ring with the properties that I like. Then I'll ask you to specify the properties that you like, and prove your statement. > I call it the ring of objects. > I've always been dying to know the definition of this wonderful ring. > But rather than go into that, I like pointing out that the old ideas > lead to the appearance of contradictions. > Assuming mathematics is consistent, the only way you can get a contradiction is by invalid reasoning. You claim to have found a contradiction. I say that mathematics is consistent and the mistake likes in the part of the reasoning you contributed. You say that mathematics is consistent and the mistake in the reasoning lies in previously accepted theorems whose proofs were faulty. Well, it is your job to specify which theorems and identify the mistakes in the proof. It is also your job to clarify the part of the reasoning that you have contributed so that people can understand it. Tell me, for example, how it proves that the factorizations I claim are impossible. > Like how you can prove factors in common with 3 and then find that you > can't get algebraic integer results that agree!!! That's because the ring of algebraic integers has some issues--it has > problems. > The only possible reason for getting a contradiction is that the axioms we have been using are inconsistent or we have made a mistake in our reasoning somewhere so that the contradiction does not really follow from the axioms. Perhaps you want to claim that these theorems you are questioning do not hold in the algebraic integers because the ring of algebraic integers lacks certain properties, and in your ring which has these properties everything will be all right. Great. So define the ring. Tell us what the properties are. Show that the theorems hold in your ring but not in the ring of algebraic integers. > Easily proven problems when you have the right tools. I've shown it has problems with non-polynomial factorization, and even > fighting the paper and the result with some emails. > Those arguments of yours were also flawed. I'm happy to go over them with you anytime you feel like it. > Now I've found a demonstration that removes room for obfuscation by > taking away variables and functions!!! > This demonstration is also flawed. You claim to have derived a certain conclusion from certain observations. I have found factorizations which are consistent with the observations but not the conclusion. So the argument must be invalid, or it must use another observation you haven't told us about yet which shows the factorizations to be impossible. I eagerly await news of any such observation. > So Arturo Magidin can't babble about new polynomials. And Dik Winter > can't come at people talking about bizarre functions that can defy the > distributive property!!! (5 + (sqrt(37) - sqrt(13))/2)((sqrt(37) - sqrt(13))/2) = (5 - > sqrt(13))(5 + sqrt(37))/2 Those are just numbers--no variables and no functions. Your confidence in continuing to reply to me is all about your lack of > fear and the greater fear of the bulk of the sci.math readership that > is bothering with reading this debate. > It's founded in the fact that I know I'm right. > The reality is cowardice in the math field, while yes, you are not a > coward or you would have run away by now. But you feel no fear so you continue to work to obscure. > No. I am not working to obscure, I am working to help you understand the subject you are dealing with. I find it offensive that you impugn my motives in this way. There is no sane reason why I would want people not to appraise your work on its own merits. Also, even if I did want this, there is no way I could achieve it. I cannot stop people from appraising your work and mine on its own merits, and I have no sane reason to want to. > And math people let you because none of them truly care or believe in > mathematics, least of all mathematical proof. > Rubbish. > I use to call it breaking. I'd explain everything to some > mathematician--used to contact them by email--and they'd seem to kind > of get it, and then run away. I'd say they broke. Cowardice rules the math field, along with naivette and basic > incompetence about numbers. > > > James Harris === Subject: Re: JSH: Will to be wrong [jstevh@msn.com, cementing his reputation as the greatest theorist of all time] >> ... >> You are going backwards. The point of the demonstration is that the >> OLD WAYS DO NOT WORK so the old way of emphasizing what ring don't >> work. >> It turns out that there is really only one ring. [some random lying mathematician, this time Rupert] > Really? Can I quote you on that? There is only one ring. Fascinating. Would you care to post the definition of a ring? I'll give you a hint. To make your statement defensible, modify it to > There is only one ring with the properties that I like. Then I'll ask > you to specify the properties that you like, and prove your statement. >> I call it the ring of objects. > I've always been dying to know the definition of this wonderful ring. It's had many definitions (it morphs whenever he's calm enough to realize that the last attempt didn't work either). AFAICT, he's stuck to this one for at least a year: http://mymath.blogspot.com/2005/03/object-ring.html The object ring is defined by two conditions, and includes all numbers such that these conditions are true: 1. 1 and -1 are the only rationals that are units in the ring. 2. Given a member m of the ring there must exist a non-zero member n such that mn is an integer, and if mn is not a factor of m, then n cannot be a unit in the ring. You'll get exactly nowhere pointing out the problems with it (they've all been noted before). Curiously, you won't even get anywhere noting that the: and if mn is not a factor of m, then n cannot be a unit in the ring part is wholly pointless since it /must/ be true (trivial direct proof of the contrapositive: if n is a unit, then by defn. nk = 1 for some k; therefore m = m*1 = m(nk) = (mn)k, and so also by defn. mn is a factor of m). I lost count of how many times people have explained that to him -- he just ignores it. Arturo Magidin has also noted that his the is impossible to meet: assuming that by his muddy all numbers he means a subset of the complex numbers (and also under other plausible assumptions), there isn't a unique such ring. For irony, note that the algebraic integers satisfy his definition if you ignore the impossible-to-meet the ... all wishful thinking. === Subject: Re: JSH: Will to be wrong of all time] >> ... >> You are going backwards. The point of the demonstration is that the >> OLD WAYS DO NOT WORK so the old way of emphasizing what ring don't >> work. > >> It turns out that there is really only one ring. [some random lying mathematician, this time Rupert] > Really? Can I quote you on that? There is only one ring. Fascinating. > > Would you care to post the definition of a ring? > > I'll give you a hint. To make your statement defensible, modify it to > There is only one ring with the properties that I like. Then I'll ask > you to specify the properties that you like, and prove your statement. >> I call it the ring of objects. > I've always been dying to know the definition of this wonderful ring. It's had many definitions (it morphs whenever he's calm enough to realize > that the last attempt didn't work either). AFAICT, he's stuck to this one > for at least a year: http://mymath.blogspot.com/2005/03/object-ring.html The object ring is defined by two conditions, and includes all > numbers such that these conditions are true: 1. 1 and -1 are the only rationals that are units in the ring. 2. Given a member m of the ring there must exist a non-zero member > n such that mn is an integer, and if mn is not a factor of m, > then n cannot be a unit in the ring. You'll get exactly nowhere pointing out the problems with it (they've all > been noted before). Curiously, you won't even get anywhere noting that the: and if mn is not a factor of m, then n cannot be a unit > in the ring part is wholly pointless since it /must/ be true (trivial direct proof of > the contrapositive: if n is a unit, then by defn. nk = 1 for some k; > therefore m = m*1 = m(nk) = (mn)k, and so also by defn. mn is a factor of > m). I lost count of how many times people have explained that to him -- he > just ignores it. Arturo Magidin has also noted that his the is impossible to meet: > assuming that by his muddy all numbers he means a subset of the complex > numbers (and also under other plausible assumptions), there isn't a unique > such ring. For irony, note that the algebraic integers satisfy his definition if you > ignore the impossible-to-meet the ... all wishful thinking. Further. Everything we have here is a complex number and the only numbers we are interested in here are the algebraic numbers. So consider the algebraic objects, the intersection of the object ring with the algebraic numbers (James has never been clear if the object ring contains non-algebraic numbers). Since any algebraic number satisfies all of 2., the second condition disappears entirely. Now we have the simple: A subring, R, of the algebraic numbers has the object property if R intersect the rationals is the integers. (there is no largest ring with the object property). Note. Arturo Magidin has pointed out that James views definitions as magic incantations to make proofs work. The real defintion of the object ring is the ring which makes JSH proofs work. - William Hughes === Subject: Re: JSH: Will to be wrong OLD WAYS DO NOT WORK so the old way of emphasizing what ring don't > work. It turns out that there is really only one ring. Golly. That's certainly a revelation. So which of, say, the integers, the complex numbers, the integers mod 5, the ring of 3 by 3 integer-valued matrices, and the quaternions aren't rings? Or are they all equal? -Rotwang === Subject: Re: JSH: Will to be wrong [jstevh@msn.com] In this case, you can prove that (sqrt(37) - sqrt(13))/2 does not have any factors in common with 3 in actuality but provably it does have a non-unit factor in common with 3 in the ring of algebraic integers. [sg552@hotmail.co.uk] >> Unless actuality is the name of a ring of which I have never heard, >> this statement makes no sense. The question of whether one number is a >> factor of another depends on what ring one is talking about. [The Ghost In The Machine] > Perhaps I for one am a little slow, but do you have an example handy? > :-) Sure. Let's assume commutative rings to reduce tedium (so A*B = B*A). Rings don't necessarily support division, so factor is defined in terms of ring multiplication: element A of ring R is called a factor of element B of ring R if there exists an element C of ring R such that A*C = B (where, of course, * denotes the ring multiplication function). And you already know an infinite number of cases where this matters. For example, in the ring of integers 8 is a factor of 24, because there exists an integer i such that 8*i = 24 (if you've feeling /very/ slow today ;-), try i=3). 8 is not a factor of 25 in the ring of integers, because there is no integer i such that 8*i = 25. However, 8 /is/ a factor of 25 in the ring of rationals -- and in the ring of reals, and in the ring of complex numbers. So is 8 a factor of 25, or isn't it? The question is plain senseless until you specify a ring with respect to which it's being asked, because it's a question /about/ both the elements of the ring and the nature of the ring multiplication function. Those are key. The algebraic integers are a funky ring to work with, since the set of algebraic integers is defined to be the set of roots of monic irreducible (over Q) polynomials with integer coefficients. If you ask me, it's not even obvious that this /does/ form a ring -- that requires non-trivial proof. It turns out, e.g., that 8 is not a factor of 25 in that ring either (meaning all of the following: - 8 is an algebraic integer - 25 is an algebraic integer - there is no algebraic integer i such that 8*i = 25 ). For another example, consider James's horrid (because inappropriate whenever he brings it up) ring of evens: {..., -6, -4, -2, 0, 2, 4, 6, ...}. He likes to point out that 2 is not a factor of 6 in that ring because 3 isn't even. That's essentially correct, but an obfuscated shorthand for noting that there is no even integer i such that 2*i = 6. For /exactly the same reasion/, 2 is not a factor of itself in that ring either! There is no even integer i such that 2*i = 2. But it seems impossible for James to grasp that -- which isn't really surprising given how careless he usually is in his use of technical words. Similar remarks apply to the word unit. In a ring with multiplicative identity (call it 1), an element A of the ring is called a unit if A is a factor of 1. Or, IOW, if there exists an element C in the ring such that A*C = 1; or, IOOW, if A has a multiplicative inverse in the ring. For example, in the ring of integers the only units are +1 and -1. In the ring of rationals (more generally, in any field), all non-zero elements are units. And again, and for the same reasons, it's plain senseless to ask whether something is a unit without specifying a ring first. It's again a question /about/ the ring. So when James says: In this case, you can prove that (sqrt(37) - sqrt(13))/2 does not have any factors in common with 3 in actuality but provably it does have a non-unit factor in common with 3 in the ring of algebraic integers. I fully agree with Rotwang that the prove ... in actuality part is senseless based on the meanings of the technical words involved. Unless, as Rotwang helpfully ;-) suggested, perhaps actuality is the name of some ring in JSH-land. === Subject: Re: JSH: Will to be wrong In sci.math, Tim Peters on Sat, 21 Oct 2006 23:52:21 -0400 : > [jstevh@msn.com] > In this case, you can prove that (sqrt(37) - sqrt(13))/2 does not have > any factors in common with 3 in actuality but provably it does have a > non-unit factor in common with 3 in the ring of algebraic integers. [sg552@hotmail.co.uk] Unless actuality is the name of a ring of which I have never heard, this statement makes no sense. The question of whether one number is a factor of another depends on what ring one is talking about. [The Ghost In The Machine] >> Perhaps I for one am a little slow, but do you have an example handy? >> :-) Sure. Let's assume commutative rings to reduce tedium (so A*B = B*A). Rings don't necessarily support division, so factor is defined in terms of > ring multiplication: element A of ring R is called a factor of element B of > ring R if there exists an element C of ring R such that A*C = B (where, of > course, * denotes the ring multiplication function). And you already know an infinite number of cases where this matters. For > example, in the ring of integers 8 is a factor of 24, because there exists > an integer i such that 8*i = 24 (if you've feeling /very/ slow today ;-), > try i=3). Heh...not quite *that* slow. :-) Of course one could consider 8 a factor of 24 in Z<5> [*] as well -- since 8 = 3, 24 = 4, and 3*3 = 4 (all mod 5). (In fact 8 is a factor of 24 in any Z

where p is not 2. If p = 2 one has 0 a factor of 0 which makes for certain difficulties.) 8 is not a factor of 25 in the ring of integers, because there is no integer > i such that 8*i = 25. However, 8 /is/ a factor of 25 in the ring of > rationals -- and in the ring of reals, and in the ring of complex numbers. So is 8 a factor of 25, or isn't it? The question is plain senseless until > you specify a ring with respect to which it's being asked, because it's a > question /about/ both the elements of the ring and the nature of the ring > multiplication function. Those are key. The algebraic integers are a funky ring to work with, since the set of > algebraic integers is defined to be the set of roots of monic irreducible > (over Q) polynomials with integer coefficients. If you ask me, it's not > even obvious that this /does/ form a ring -- that requires non-trivial > proof. It turns out, e.g., that 8 is not a factor of 25 in that ring either > (meaning all of the following: - 8 is an algebraic integer Root of z-8 = 0. > - 25 is an algebraic integer Root of z-25 = 0. > - there is no algebraic integer i such that 8*i = 25 ). I'll agree it's a difficult ring. Someone mentioned resultants; that's all I know about the matter but it *is* in fact a ring. Of course JSH seems to stumble around in the dark without a flashlight (mathbook). Me, I might have a flashlight on occasion -- quite a collection, in fact. However, I don't think I've studied rings in that much detail (I do have a book on Galois theory but have forgotten a lot of it :-) ) and it was quite a long time ago. Best I can do is that, if i is an algebraic integer, then 8*i = 25 implies i = 25/8. Since the defining equation is 8*i - 25 = 0, i is not an integer. Unfortunately, that's probably nowhere near rigorous enough. For another example, consider James's horrid (because inappropriate whenever > he brings it up) ring of evens: {..., -6, -4, -2, 0, 2, 4, 6, ...}. He > likes to point out that 2 is not a factor of 6 in that ring because 3 isn't > even. That's essentially correct, but an obfuscated shorthand for noting > that there is no even integer i such that 2*i = 6. For /exactly the same > reasion/, 2 is not a factor of itself in that ring either! There is no even > integer i such that 2*i = 2. But it seems impossible for James to grasp > that -- which isn't really surprising given how careless he usually is in > his use of technical words. Similar remarks apply to the word unit. In a ring with multiplicative > identity (call it 1), an element A of the ring is called a unit if A is > a factor of 1. Or, IOW, if there exists an element C in the ring such that > A*C = 1; or, IOOW, if A has a multiplicative inverse in the ring. For > example, in the ring of integers the only units are +1 and -1. In the ring > of rationals (more generally, in any field), all non-zero elements are > units. And again, and for the same reasons, it's plain senseless to ask whether > something is a unit without specifying a ring first. It's again a question > /about/ the ring. So when James says: In this case, you can prove that (sqrt(37) - sqrt(13))/2 does not have > any factors in common with 3 in actuality but provably it does have a > non-unit factor in common with 3 in the ring of algebraic integers. I fully agree with Rotwang that the prove ... in actuality part is > senseless based on the meanings of the technical words involved. Unless, as > Rotwang helpfully ;-) suggested, perhaps actuality is the name of some > ring in JSH-land. > FWIW, I've often wondered as to how James can miss the point about the Unique Factorization Theorem (?) falling down so badly in the algebraic integers. In regular integers, one has at most two factorizations of a prime (and one of them is a trivial derivation of the other). For example, 37 = 37 * 1 = (-37) * (-1) Since 1 and -1 are units in Z, this makes sense in a larger context to some extent (and there are *lots* of units in the algebraic integers), but in algebraic integers one can also have factorizations such as 37 = sqrt(37) * sqrt(37) 37 = sqrt(-37) * -(sqrt(-37)) 37 = 37^(1/4) * 37^(1/3) * 37^(5/12) and *none* of these are units, but they *are* algebraic integers. Maybe JSH has finally gotten it through his head that his magical 7 factor in his one equation doesn't quite behave as he expects. Certainly he's not used that particular logic flow lately. :-) [*] I have no idea whether this notation is correct. :-) -- #191, ewill3@earthlink.net /dev/signature: No such file or directory -- === Subject: Re: JSH: Will to be wrong <581r04-abe.ln1@sirius.tg00suus7038.net > Unless actuality is the name of a ring of which I have never heard, > this statement makes no sense. The question of whether one number is a > factor of another depends on what ring one is talking about. Perhaps I for one am a little slow, but do you have an example handy? > :-) Please don't be cryptic... have I said something thick? -Rotwang === Subject: Re: JSH: Will to be wrong In sci.math, sg552@hotmail.co.uk on 21 Oct 2006 20:16:10 -0700 > Unless actuality is the name of a ring of which I have never heard, >> this statement makes no sense. The question of whether one number is a >> factor of another depends on what ring one is talking about. >> Perhaps I for one am a little slow, but do you have an example handy? >> :-) Please don't be cryptic... have I said something thick? -Rotwang > No, it's perfectly understandable; I was merely wondering if you had a pair of numbers A, B, and two rings such that A divides B in the first ring, but not in the second. Admittedly, the best I can do is use A = 2, B = 5, first ring is the field of rationals and the second the ring of integers. But I was hoping for subrings of the algebraics. :-) -- #191, ewill3@earthlink.net Windows. Because it's not a question of if. It's a question of when. -- === Subject: Re: JSH: Will to be wrong <581r04-abe.ln1@sirius.tg00suus7038.net> <9bbr04-m5f.ln1@sirius.tg00suus7038.net > Please don't be cryptic... have I said something thick? > No, it's perfectly understandable; I was merelywondering if you had a > pair of numbers A, B, and two rings such that A divides B in the first > ring, but not in the second. Then I apologise for my unhelpful reply. Your previous posts suggest you know rather more about this area than I do, so I figured I had made a gaffe and you were taking the piss :-) > Admittedly, the best I can do is use A = 2, B = 5, first ring is the > field of rationals and the second the ring of integers. But I was > hoping for subrings of the algebraics. :-) Aren't the rationals a subring of the algebraic numbers? If you mean the algebraic integers then I'm afraid I can't provide an example since I'm not that familiar with the AIs, though I'm sure somebody else can do so. -Rotwang