mm-3569 === Subject: : Re: Non-polynomial factorization: Simplicity at its core > >Imagine you have a factorization of a polynomial >P(x) = (f(x) + 1)*(g(x) + 2) >and you are told to multiply it by 7, and to pick one way, so you >choose >7*P(x) = (7f(x) + 7)*(g(x) + 2) >can the value of the functions f(x) and g(x) invalidate your choice, >turning it into something else? >That simplicity of multiplying by a constant lies at the heart of the >non-polynomial factorization arguments. >To understand its importance let >P(x) = 175x^2 - 15x + 2 >and now multiply that polynomial times 7 and re-group the resultant >terms: >7*P(x) = (49x^2 - 14x)5^2 + (7x-1)(7)(5) + 49 >to factor it as >7*P(x) = (5a_1(x) + 7)(5a_2(x)+ 7) >where the a's are roots of >a^2 - (7x-1)a + (49x^2 - 14x) = 0. >Notice that equation comes easily enough from multiplying out the >factorization and solving for the a's, as you have >a_1(x)*a_2(x) = 49x^2 - 14x and a_1(x) + a_2(x) = 7x-1. >Now divide off 7: >P(x) = (7x^2 - 2x)5^2 + (7x-1)(5) + 7 >giving the factorization >P(x) = (5b_1(x) + w_1)(5b_2(x)+ w_2) >where I've used b's for the new functions and w_1 and w_2 for the rest. Are we supposed to have w_1 and w_2 algebraic integers and b_1 and b_2 > mapping the algebraic integers into the algebraic integers? It's not > obvious to me that that's possible. >Then I have simply enough--notice 5 still brackets out what must go >where: >b_1(x)*b_2(x) = 7x^2 - 2x, >w_1*b_2(x) + w_2*b_1(x) = 7x-1 >and w_1*w_2 = 7 >so I can substitute out b_2(x) and w_2(x), to get >w_1*(7x^2 - 2x)/b_1(x) + 7*b_1(x)/w_1 = 7x-1 >and multiplying both sides by w_1*b_1(x) and simplifying a bit, I have >7*b_1(x)^2 - (7x-1)w_1*b_1(x) + (7x^2 - 2x)w_1^2 = 0 >so I can solve for b_1(x) using the quadratic formula: >b_1(x) = ((7x - 1) +/- sqrt((7x-1)^2 -4(7x^2 - 2x)))*w_1/14 >proving that b_1(x) cannot be an algebraic integer function, as notice >the 7 in the denominator cannot divide out unless w_1 has 7 itself as a >factor. Well, if we can find an x such that 7x-1+sqrt((7x-1)^2-4(7x^2-2x)) is > coprime to 7, yes. The functions now obscure how your multiplication proceeded, so let >x=0, to clear them out, and you find >a^2 + a = 0 >proving that your 7 went through as it did in my simple example above >as one of the a's is 0 and the other -1 when you clear out the >functions with x=0. >So the 7 does multiply through in a simple way as demonstrated with the >simple analysis above. >Simplicity is at the core where what complexity there is comes from the >use of a creative construction: >7*P(x) = (49x^2 - 14x)5^2 + (7x-1)(7)(5) + 49 >to get the non-polynomial factorization. >A smart idea that simply took over a hundred years to be extant. What have you proved? Both Tim and I asked you that just recently. Why > did you ignore us? Why can't you state your theorem? > > Theorems are for plodders. JSH has brilliant insights. === === Subject: : Re: Non-polynomial factorization: Simplicity at its core > Imagine you have a factorization of a polynomial >P(x) = (f(x) + 1)*(g(x) + 2) >and you are told to multiply it by 7, and to pick one way, so you > choose >7*P(x) = (7f(x) + 7)*(g(x) + 2) >can the value of the functions f(x) and g(x) invalidate your choice, > turning it into something else? >That simplicity of multiplying by a constant lies at the heart of the > non-polynomial factorization arguments. >To understand its importance let >P(x) = 175x^2 - 15x + 2 >and now multiply that polynomial times 7 and re-group the resultant > terms: >7*P(x) = (49x^2 - 14x)5^2 + (7x-1)(7)(5) + 49 >to factor it as >7*P(x) = (5a_1(x) + 7)(5a_2(x)+ 7) >where the a's are roots of >a^2 - (7x-1)a + (49x^2 - 14x) = 0. >Notice that equation comes easily enough from multiplying out the >factorization and solving for the a's, as you have >a_1(x)*a_2(x) = 49x^2 - 14x and a_1(x) + a_2(x) = 7x-1. >Now divide off 7: >P(x) = (7x^2 - 2x)5^2 + (7x-1)(5) + 7 >giving the factorization >P(x) = (5b_1(x) + w_1)(5b_2(x)+ w_2) >where I've used b's for the new functions and w_1 and w_2 for the rest. >Then I have simply enough--notice 5 still brackets out what must go >where: >b_1(x)*b_2(x) = 7x^2 - 2x, >w_1*b_2(x) + w_2*b_1(x) = 7x-1 >and w_1*w_2 = 7 >so I can substitute out b_2(x) and w_2(x), to get >w_1*(7x^2 - 2x)/b_1(x) + 7*b_1(x)/w_1 = 7x-1 >and multiplying both sides by w_1*b_1(x) and simplifying a bit, I have >7*b_1(x)^2 - (7x-1)w_1*b_1(x) + (7x^2 - 2x)w_1^2 = 0 >22 >so I can solve for b_1(x) using the quadratic formula: >b_1(x) = ((7x - 1) +/- sqrt((7x-1)^2 -4(7x^2 - 2x)))*w_1/14 >proving that b_1(x) cannot be an algebraic integer function, as notice >the 7 in the denominator cannot divide out unless w_1 has 7 itself as a >factor. > No, no - this last step does not follow - you are assuming without >realizing it perhaps the following non-theorem: > If A and B are both not divisible by 7, > then A + B is not divisible by 7. > In your expression, > A = (7x - 1) and > B = sqrt((7x-1)^2 -4(7x^2 - 2x)). > > But if (7x-1)^2-4(7x^2-2x) is not a square you can prove A+B is coprime > to 7 by considering the trace. >No, that's wrong of course. > Well, I was gonna say. M. > So of course A is not divisible by 7 and in general B is also >not divisible by 7. But that does not prove that A + B is >not divisible by 7. It is quite possible that (A + B)*w_1/14 is an >algebraic integer, even if w_1 is not divisible by 7. > Simpler example showing the non-thereom is wrong: > A = 6, B = 8. > Marcus. > >The functions now obscure how your multiplication proceeded, so let > x=0, to clear them out, and you find >a^2 + a = 0 >proving that your 7 went through as it did in my simple example above > as one of the a's is 0 and the other -1 when you clear out the > functions with x=0. >So the 7 does multiply through in a simple way as demonstrated with the >simple analysis above. >Simplicity is at the core where what complexity there is comes from the > use of a creative construction: >7*P(x) = (49x^2 - 14x)5^2 + (7x-1)(7)(5) + 49 >to get the non-polynomial factorization. >A smart idea that simply took over a hundred years to be extant. > === === Subject: : Re: Non-polynomial factorization, complete. >Minor fix below... > Imagine you have a factorization of a polynomial > P(x) = (f(x) + 1)*(g(x) + 2) > and you are told to multiply it by 7, and to pick one way, so you >choose > 7*P(x) = (7f(x) + 7)*(g(x) + 2) > can the value of the functions f(x) and g(x) invalidate your choice, >turning it into something else? > That simplicity of multiplying by a constant lies at the heart of the >non-polynomial factorization arguments. > To understand its importance let > P(x) = 175x^2 - 15x + 2 > and now multiply that polynomial times 7 and re-group the resultant >terms: > 7*P(x) = (49x^2 - 14x)5^2 + (7x-1)(7)(5) + 49 > to factor it as > 7*P(x) = (5a_1(x) + 7)(5a_2(x)+ 7) > where the a's are roots of > a^2 - (7x-1)a + (49x^2 - 14x) = 0. > Notice that equation comes easily enough from multiplying out the >factorization and solving for the a's, as you have > a_1(x)*a_2(x) = 49x^2 - 14x and a_1(x) + a_2(x) = 7x-1. > Now divide off 7: > P(x) = (7x^2 - 2x)5^2 + (7x-1)(5) + 7 > giving the factorization > P(x) = (5b_1(x) + w_1)(5b_2(x)+ w_2) > where I've used b's for the new functions and w_1 and w_2 for the >rest. > Then I have simply enough--notice 5 still brackets out what must go >where: > b_1(x)*b_2(x) = 7x^2 - 2x, > w_1*b_2(x) + w_2*b_1(x) = 7x-1 > and w_1*w_2 = 7 > so I can substitute out b_2(x) and w_2(x), to get > w_1*(7x^2 - 2x)/b_1(x) + 7*b_1(x)/w_1 = 7x-1 > and multiplying both sides by w_1*b_1(x) and simplifying a bit, I >have > 7*b_1(x)^2 - (7x-1)w_1*b_1(x) + (7x^2 - 2x)w_1^2 = 0 > so I can solve for b_1(x) using the quadratic formula: > b_1(x) = ((7x - 1) +/- sqrt((7x-1)^2 -4(7)(7x^2 - 2x)))*w_1/14 > proving that b_1(x) cannot be an algebraic integer function unless >w_1 >has 7 as a factor, >No that's not right. >The norm of ((7x-1)+sqrt((7x-1)^2-28(7x^2-2x))) is divisible by 28, so > it has non-unit factors in common with 7. So it is still possible for > b_1(x) to be an algebraic integer. >By the way, w_1 and w_2 may depend on x as well. > None of that matters because of > 7*b_1(x)^2 - (7x-1)w_1*b_1(x) + (7x^2 - 2x)w_1^2 = 0 > since with non-zero integer x it is non-monic and irreducible over Q. It's not a polynomial over Q at all unless w_1 happens to be in Q. If > w_1 and w_2 are both integers in Q, then yes, one must be 1 and the > other 7. That's not profound. > > NO Banana for YOU! === === Subject: : Re: JSH: What's at issue , > No human being can shift over infinity to magically adjust their choice > for every possible x, so what is your reasoning here? > -- Michael Press === === Subject: : Re: JSH: Why might factoring idea fail? > stop playing his game, man. > you do have better things to do too. > He has never factored any number at all, > he refuses to read books, > he just likes to spin up sci.math for fun > he sits at the keyboard and makes stuff up > and he has been a troll for years. Every day I'm a troll, My random dice are on a roll, I struggle to find a factor, I post my results in my blotter... You keep telling me I'm an asshole! === === Subject: : Re: How Far Across the Earth Can A Person One Km Up See? centre and produce the line to get a diameter AB. Now suppose a tangent > from C to the earth's surface meets it at T. The points A, B, C, T are > in a single plane. Euclid III.36 says (CT)^2 = CA.CB > so if you're at height h above the earth whose dimameter is d, then > the distance to the horizon is CT = sqrt(h(d + h)) which is very close to sqrt(hd) if h << d. Yep. A very good approximation for this is that if h is your height above the ground in FEET, then the distance to the horizon in MILES is very close to 5/4 sqrt(h). === === Subject: : Re: answer to Beer Math combination problem dnZ2d@comcast: > And I would bet that you are neither. >You lost, AGAIN. You are a drunken math minor? -- Sleep well tonight.........RD (The Sandman) http://homecast.net/~rdsandman Tis far better to burn the flag while wrapped in the Constitution than to burn the Constitution while wrapped in the flag. .357Mag...my personal version of Homeland Security We'll fill landfills with tons and tons of garbage, but when our trash is shaped like a human, we [somehow] feel the need to keep it around. John P...2006 === === Subject: : Re: What would have happened if ? Well, We could probably be more knowledgeable about God and not have these many Religions and doubts and imaginations. Our laws of Physics, Chemistry etc would have been found rigorously and we could have explained all phenomena. Terrorism, War and Crimes would have been Globally less, but Universally more, perhaps until we evolve as Universal Citizens. All these Crimes would have intensified Technologically too and Osama could even escape to other Planets. Compatible species from other Planets and Living Spaces might have been imported. Our awareness would be tremendous ! > What would have happened if we had by now : > Been able to traverse the Universe at will. > Been sure of every Living thing on the universe and their capability. > Been able to monitor and detect what's going on in the mind of any thinking > being. > Been able to : > See not only by line of sight but have bent and twisted vision and > ability to look around the corners > Hear more than what we hear > See all frequencies , infra - red etc. and emissions from Black holes as > we see what we see [normal light] > Have more faculties than these sight,touch,hearing,smelling only as we could > have evolved with stronger sense in these faculties as well as newer > ones. The world we see is an optical one and how shunted would have been in > our perception if were sightless or unable to hear and how our physics > ,maths, languages would all have evolved with those restrictions. We would be God. Bob Kolker > -- === === Subject: : Re: What would have happened if ? > Well, We could probably be more knowledgeable about God Some of us are. There isn't one. Graham === === Subject: : Re: What would have happened if ? > Well, > We could probably be more knowledgeable about God >Some of us are. >There isn't one. >Graham Is that the OFFICIAL REPUBLICAN RIGHT-WINGER POSITION? === === Subject: : Re: What would have happened if ? > Well, >We could probably be more knowledgeable about God and not have these many > Religions and doubts and imaginations. How do you know there is a god? > Our laws of Physics, Chemistry etc would have been found rigorously and we > could have explained all phenomena. Typical ignorance about science. It has never had the pretense of explaining anything. Nature is gloriously mysterious. > Terrorism, War and Crimes would have been Globally less, but Universally > more, perhaps until we evolve as Universal Citizens. All these Crimes would > have intensified Technologically too and Osama could even escape to other > Planets. You are a nut case. === === Subject: : Re: What would have happened if ? Read with respect to the context of the earlier post Or else you can't even break monkey nuts ! > Well, > We could probably be more knowledgeable about God and not have these many > Religions and doubts and imaginations. How do you know there is a god? > Our laws of Physics, Chemistry etc would have been found rigorously and we > could have explained all phenomena. Typical ignorance about science. It has never had the pretense of > explaining anything. Nature is gloriously mysterious. > Terrorism, War and Crimes would have been Globally less, but Universally > more, perhaps until we evolve as Universal Citizens. All these Crimes would > have intensified Technologically too and Osama could even escape to other > Planets. You are a nut case. > -- === === Subject: : Re: What would have happened if ? > Read with respect to the context of the earlier post... Your stupid string of 'what ifs'? Ok, you are a delusional nut case.... === === Subject: : We, as Scientists must show tolerance to opposite views As much as opposite poles in electricity and magnetism have learned to coexist, we Scientists must show restraint and tolerance to each others [opposite] views which I find is lacking in our groups. Researcher -- === === Subject: : Re: We, as Scientists must show tolerance to opposite views > As much as opposite poles in electricity and magnetism have learned to > coexist, we Scientists must show restraint and tolerance to each others > [opposite] views which I find is lacking in our groups. Researcher Then, a dietitian should show restraint and tolerance to cannibals? Those that eat the other other white meat. === === Subject: : Re: We, as Scientists must show tolerance to opposite views > As much as opposite poles in electricity and magnetism have learned > to coexist, we Scientists must show restraint and tolerance to each > others [opposite] views which I find is lacking in our groups. > Then, a dietitian should show restraint and tolerance to cannibals? > Those that eat the other other white meat. My girlfriend is a dietitian. It's her job to ensure that people eat nutritionally adequate food, not to get involved in its morality. She would be concerned about antibiotic and pesticide residues in human flesh (liver in particular) and would suggest you eat plenty of veggies along with it, but that would be about as far as her strictures could professionally go. It isn't white meat under any normal definition of the term. Jack Campin: 11 Third St, Newtongrange EH22 4PU, Scotland | tel 0131 660 4760 for CD-ROMs and free | fax 0870 0554 975 stuff: Scottish music, food intolerance, & Mac logic fonts | mob 07800 739 557 === === Subject: : Re: We, as Scientists must show tolerance to opposite views > As much as opposite poles in electricity and magnetism have learned to > coexist, we Scientists must show restraint and tolerance to each others > [opposite] views which I find is lacking in our groups. > Researcher >Then, a dietitian should show restraint and tolerance to cannibals? > Those that eat the other other white meat. Of course. After all, long pig has the exact balance of vitamins and minerals, amino acids and proteins required for human nutrition. === === Subject: : Re: We, as Scientists must show tolerance to opposite views > As much as opposite poles in electricity and magnetism have learned to > coexist, we Scientists must show restraint and tolerance to each others > [opposite] views which I find is lacking in our groups. You don't know many scientists, do you? Look, if my ideas are wrong, I *want* somebody to point out where I'm wrong. It's an opportunity to learn something, and I crave that. The point of science is that there is right and wrong, at least about a lot of things. If you had said that there is room for people to learn to critcize ideas without being offensive, that I might well agree with. Indeed, I've taken several classes in doing exactly that. Not that it took all that well, mind you, but I know how to do it even if I don't do it all the time. It's a lot of work. Socks === === Subject: : Re: We, as Scientists must show tolerance to opposite views > As much as opposite poles in electricity and magnetism have learned to > coexist, we Scientists must show restraint and tolerance to each others > [opposite] views which I find is lacking in our groups. Researcher We as Usenet users should not post separate copies of identical messages to different newsgroups. This message was posted to: alt.math,sci.energy,sci.environment,sci.logic,sci.math An identical message was posted to: alt.sci.physics,alt.sci.physics.new-theories,sci.astro,sci.physics,sci.physi cs.relativity I smell spam. === === Subject: : Re: We, as Scientists must show tolerance to opposite views On Fri, 3 Nov 2006 00:58:49 +0530, Researcher coexist, we Scientists must show restraint and tolerance to each others >[opposite] views which I find is lacking in our groups. Congeniality, restraint, and tolerance of idiocy has never taken much of any priority at any meaningful physics-theoretic discussion I've attended. I must have been missing those tea, cakes, and 'let's be nice to each other' science parties you've been attending. Jon === === Subject: : Re: We, as Scientists must show tolerance to opposite views > On Fri, 3 Nov 2006 00:58:49 +0530, Researcher As much as opposite poles in electricity and magnetism have learned to >coexist, we Scientists must show restraint and tolerance to each others >[opposite] views which I find is lacking in our groups. >Congeniality, restraint, and tolerance of idiocy has never taken much > of any priority at any meaningful physics-theoretic discussion I've > attended. I must have been missing those tea, cakes, and 'let's be > nice to each other' science parties you've been attending. Restraint of idiocy sounds like it would work quite well, though. Phil -- Home taping is killing big business profits. We left this side blank so you can help. -- Dead Kennedys, written upon the B-side of tapes of /In God We Trust, Inc./. === === Subject: : Re: JSH: Weird answer, but it looks solid > Well, it looks like there was this easy enough resolution, which oddly > enough can be said to rely on the impossibility of an algebraic integer > being the root of a non-monic polynomial with integer coefficients > irreducible over Q. >Starting with the simple polynomial >P(x) = 175x^2 - 15x + 2 >I multiplied by 7 and did my creative thing with the terms to get >7*P(x) = (49x^2 - 14x)5^2 + (7x-1)(7)(5) + 49 >to factor it as >7*P(x) = (5a_1(x) + 7)(5a_2(x)+ 7) >where the a's are roots of >a^2 - (7x-1)a + (49x^2 - 14x) = 0. >But I finally got around to just dividing that damn 7 off, to get >P(x) = (7x^2 - 2x)5^2 + (7x-1)(5) + 7 >giving the factorization >P(x) = (5b_1(x) + w_1)(5b_2(x)+ w_2) >sticking in b's and w's, and with the 5's still around I could work out > that >b_1(x)*b_2(x) = 7x^2 - 2x, >w_1*b_2(x) + w_2*b_1(x) = 7x-1 >and w_1*w_2 = 7 >so I could so some substitutions and I chose to focus on b_1(x) and w_1 > to get >7*b_1(x)^2 - (7x-1)w_1*b_1(x) + (7x^2 - 2x)w_1^2 = 0 >which is a non-monic polynomial--unless w_1 has 7 as a factor!!! This last step is invalid, as has been pointed out. The polynomial above is not necessarily a polynomial over Q. b_1(x) can be an algebraic integer without w_1(x) being divisible by 7. Consider x = 1. For simplicity, I'll write a_i = a_i(1) and do the same with b_i and w_i. We have a_1 = 3 + sqrt(-26) a_2 = 3 - sqrt(-26). Let w_1 = (307 - 30*sqrt(-26)^{1/6} w_2 = (307 + 30*sqrt(-26)^{1/6}. It's not too hard to show (307 + 30*sqrt(-26))(307 - 30*sqrt(-26)) = 7^6 so w_1 * w_2 = 7. Thus w_1 and w_2 divide 7 (and are obviously algebraic integers). It's also not hard to verify that (3 + sqrt(-26))^6 / (307 + 30*sqrt(-26)) = 109 - 12*sqrt(-26) (3 - sqrt(-26))^6 / (307 - 30*sqrt(-26)) = 109 + 12*sqrt(-26) so w_2 divides a_1 and w_1 divides a_2 Let b_1 = a_1 / w_2 = (109 - 12*sqrt(-26))^{1/6} b_2 = a_2 / w_1 = (109 + 12*sqrt(-26))^{1/6} Note that b_1 and b_2 are algebraic integers. Now we have, as a check, P(1) = (5*b_1 + w_1)(5*b_2 + w_2) = (1/w_2)(5*b_1*w_2 + w_1*w_2)(1/w_1)(5*b_2*w_1 + w_1*w_2) = (1/7)(5*a_1 + 7)(5*a_2 + 7) so 7*P(1) = (5*a_1 + 7)(5*a_2 + 7), as required. Finally, the w_i satisfy w^12 - 614 w^6 + 7^6 = 0, so z = w/7 satisfies 7^6 z^12 - 614 z^6 + 1 = 0, so neither of the w_i are divisible by 7. In sum, we have an example where the b_i are algebraic integers, in spite of the fact that neither of the w_i are divisible by 7. Rick === === Subject: : Re: JSH: Weird answer, but it looks solid > Well, it looks like there was this easy enough resolution, which oddly > enough can be said to rely on the impossibility of an algebraic integer > being the root of a non-monic polynomial with integer coefficients > irreducible over Q. >Starting with the simple polynomial >P(x) = 175x^2 - 15x + 2 >I multiplied by 7 and did my creative thing with the terms to get >7*P(x) = (49x^2 - 14x)5^2 + (7x-1)(7)(5) + 49 >to factor it as >7*P(x) = (5a_1(x) + 7)(5a_2(x)+ 7) >where the a's are roots of >a^2 - (7x-1)a + (49x^2 - 14x) = 0. >But I finally got around to just dividing that damn 7 off, to get >P(x) = (7x^2 - 2x)5^2 + (7x-1)(5) + 7 >giving the factorization >P(x) = (5b_1(x) + w_1)(5b_2(x)+ w_2) >sticking in b's and w's, and with the 5's still around I could work out > that >b_1(x)*b_2(x) = 7x^2 - 2x, >w_1*b_2(x) + w_2*b_1(x) = 7x-1 >and w_1*w_2 = 7 Here I have a question: I'm assuming w_1 and w_2 are independent of x, so constants. How do b_1(x) and b_2(x) vary with x? Or, in other words, are the graphs of b_1(x), b_2(x) straight lines in the x,y plane? David Bernier > so I could so some substitutions and I chose to focus on b_1(x) and w_1 > to get >7*b_1(x)^2 - (7x-1)w_1*b_1(x) + (7x^2 - 2x)w_1^2 = 0 >which is a non-monic polynomial--unless w_1 has 7 as a factor!!! >And with non-zero integer x, it is irreducible over Q, proving that > b_1(x) can't be an algebraic integer, unless w_1 has 7 as a factor. >That is a nice absolute relying on the very result that has often been > used against my assertion that 7 is properly a factor of only one root > as, hey!!! NOTHING but 7 as a factor can even give you algebraic > integer functions if you divide off that pesky 7 from both sides of >7*P(x) = (5a_1(x) + 7)(5a_2(x)+ 7). >Try any w_1 other than one with 7 as a factor with x a non-zero > integer, and b_1(x) cannot be an algebraic integer function as it is > then the root of a non-monic polynomial with integer coefficients > irreducible over Q. >Damn strange, but it is definitive. >Astute readers may notice that with w_1=7, you just get back a solution > for the a's. >Also you may notice this approach doesn't deal with the issue of what > is the factor of the a's, it just shows they can't have algebraic > integer factors, since 7 itself can't be a factor of just one of them, > bringing into question claims that you can find algebraic integer > factors from 7 for both. >My guess is that old problem in mathematics where when you have false > assumptions you can appear to prove anything, when really you don't > have proof. >The ring of algebraic integers gives you all kinds of problems and > weirdness. === === Subject: : Re: JSH: Weird answer, but it looks solid [jstevh@msn] > ... [various ways of getting confused by making a polynomial > messier & messier] .... > My guess is that old problem in mathematics where when you have false > assumptions you can appear to prove anything, when really you don't > have proof. Yup! That's one way of describing your problems here. > The ring of algebraic integers gives you all kinds of problems and > weirdness. That's why it's imperative to study the topic. Everything you've discovered that's correct was understood over a century ago, and serious mathematicians stumbled over some of the same issues way back then -- the failure of unique factorization in particular caught many by surprise, as it's still catching you. Read a book! It's possible to make much swifter progress then. You've barely scratched the surface in 10 years of trying to fake it. Others easily spot fatal errors in your arguments here because they're building on a century of knowledge you haven't learned. Storming a fortified castle with a leaky water pistol rarely succeeds ;-) === === Subject: : Re: JSH: Weird answer, but it looks solid > Well, it looks like there was this easy enough resolution, which oddly > enough can be said to rely on the impossibility of an algebraic integer > being the root of a non-monic polynomial with integer coefficients > irreducible over Q. Starting with the simple polynomial P(x) = 175x^2 - 15x + 2 I multiplied by 7 and did my creative thing with the terms to get 7*P(x) = (49x^2 - 14x)5^2 + (7x-1)(7)(5) + 49 to factor it as 7*P(x) = (5a_1(x) + 7)(5a_2(x)+ 7) where the a's are roots of a^2 - (7x-1)a + (49x^2 - 14x) = 0. But I finally got around to just dividing that damn 7 off, to get P(x) = (7x^2 - 2x)5^2 + (7x-1)(5) + 7 giving the factorization P(x) = (5b_1(x) + w_1)(5b_2(x)+ w_2) sticking in b's and w's, and with the 5's still around I could work out > that b_1(x)*b_2(x) = 7x^2 - 2x, w_1*b_2(x) + w_2*b_1(x) = 7x-1 and w_1*w_2 = 7 so I could so some substitutions and I chose to focus on b_1(x) and w_1 > to get 7*b_1(x)^2 - (7x-1)w_1*b_1(x) + (7x^2 - 2x)w_1^2 = 0 which is a non-monic polynomial--unless w_1 has 7 as a factor!!! And with non-zero integer x, it is irreducible over Q, proving that > b_1(x) can't be an algebraic integer, unless w_1 has 7 as a factor. That is a nice absolute relying on the very result that has often been > used against my assertion that 7 is properly a factor of only one root > as, hey!!! NOTHING but 7 as a factor can even give you algebraic > integer functions if you divide off that pesky 7 from both sides of 7*P(x) = (5a_1(x) + 7)(5a_2(x)+ 7). Try any w_1 other than one with 7 as a factor with x a non-zero > integer, and b_1(x) cannot be an algebraic integer function as it is > then the root of a non-monic polynomial with integer coefficients > irreducible over Q. Damn strange, but it is definitive. Astute readers may notice that with w_1=7, you just get back a solution > for the a's. Also you may notice this approach doesn't deal with the issue of what > is the factor of the a's, it just shows they can't have algebraic > integer factors, since 7 itself can't be a factor of just one of them, > bringing into question claims that you can find algebraic integer > factors from 7 for both. My guess is that old problem in mathematics where when you have false > assumptions you can appear to prove anything, when really you don't > have proof. I wonder what JSH would make of the fact that there are theorems of the forms If P=NP, then ... and If P is not NP, then ... in the literature. --- Christopher Heckman > The ring of algebraic integers gives you all kinds of problems and > weirdness. === === Subject: : Re: JSH: Weird answer, but it looks solid forms If P=NP, then ... and If P is not NP, then ... in the > literature. --- Christopher Heckman Your reference to such theorems brings to mind Turing's Halting Problem. As in whether anyone could determine if JSH will ever halt attempting to claim that he has found an easy proof for FLT. (Based on his efforts to date, he will almost certainly keep attempting to claim that until he ends up in a coffin - but that is just highly probable, and not absolutely certain.) Geoff Harland. g_harland@optum12net.cos.au (Transpose m & s in address provided - then also remove cuberoot of 10^3 + 9^3 - 1^3.) === === Subject: : Re: JSH: Weird answer, but it looks solid > I wonder what JSH would make of the fact that ccheckman@gmail is a wanker beeyotch === === Subject: : Re: JSH: Weird answer, but it looks solid > Well, it looks like there was this easy enough resolution, which oddly > enough can be said to rely on the impossibility of an algebraic integer > being the root of a non-monic polynomial with integer coefficients > irreducible over Q. Starting with the simple polynomial P(x) = 175x^2 - 15x + 2 I multiplied by 7 and did my creative thing with the terms to get 7*P(x) = (49x^2 - 14x)5^2 + (7x-1)(7)(5) + 49 to factor it as 7*P(x) = (5a_1(x) + 7)(5a_2(x)+ 7) where the a's are roots of a^2 - (7x-1)a + (49x^2 - 14x) = 0. But I finally got around to just dividing that damn 7 off, to get P(x) = (7x^2 - 2x)5^2 + (7x-1)(5) + 7 giving the factorization P(x) = (5b_1(x) + w_1)(5b_2(x)+ w_2) sticking in b's and w's, and with the 5's still around I could work out > that b_1(x)*b_2(x) = 7x^2 - 2x, w_1*b_2(x) + w_2*b_1(x) = 7x-1 and w_1*w_2 = 7 so I could so some substitutions and I chose to focus on b_1(x) and w_1 > to get 7*b_1(x)^2 - (7x-1)w_1*b_1(x) + (7x^2 - 2x)w_1^2 = 0 which is a non-monic polynomial--unless w_1 has 7 as a factor!!! > It's non-monic all right. But it doesn't have rational, let alone integer, coefficients. It's not irreducible over Q because it doesn't have coefficients in Q. So the theorem you might like to use here doesn't apply. > And with non-zero integer x, it is irreducible over Q, proving that > b_1(x) can't be an algebraic integer, unless w_1 has 7 as a factor. That is a nice absolute relying on the very result that has often been > used against my assertion that 7 is properly a factor of only one root > as, hey!!! NOTHING but 7 as a factor can even give you algebraic > integer functions if you divide off that pesky 7 from both sides of 7*P(x) = (5a_1(x) + 7)(5a_2(x)+ 7). Try any w_1 other than one with 7 as a factor with x a non-zero > integer, and b_1(x) cannot be an algebraic integer function as it is > then the root of a non-monic polynomial with integer coefficients > irreducible over Q. Damn strange, but it is definitive. > Nope. Wrong again. > Astute readers may notice that with w_1=7, you just get back a solution > for the a's. > Even more astute readers might notice that in general w_1(x) is not rational. > Also you may notice this approach doesn't deal with the issue of what > is the factor of the a's, it just shows they can't have algebraic > integer factors, since 7 itself can't be a factor of just one of them, > bringing into question claims that you can find algebraic integer > factors from 7 for both. > No - you're back where you started. > My guess is that old problem in mathematics where when you have false > assumptions you can appear to prove anything, when really you don't > have proof. > My guess is that you screwed up for the 1000th time. > The ring of algebraic integers gives you all kinds of problems and > weirdness. > This part is true at least. Marcus. > === === Subject: : Re: JSH: Weird answer, but it looks solid > Well, it looks like there was this easy enough resolution, which oddly > enough can be said to rely on the impossibility of an algebraic integer > being the root of a non-monic primitive > polynomial with integer coefficients > irreducible over Q. > Starting with the simple polynomial > P(x) = 175x^2 - 15x + 2 > I multiplied by 7 and did my creative thing with the terms to get > 7*P(x) = (49x^2 - 14x)5^2 + (7x-1)(7)(5) + 49 > to factor it as > 7*P(x) = (5a_1(x) + 7)(5a_2(x)+ 7) > where the a's are roots of > a^2 - (7x-1)a + (49x^2 - 14x) = 0. > But I finally got around to just dividing that damn 7 off, to get > P(x) = (7x^2 - 2x)5^2 + (7x-1)(5) + 7 > giving the factorization > P(x) = (5b_1(x) + w_1)(5b_2(x)+ w_2) >If you want everything to be an algebraic integer, you'd better make > that (5b_1(x)+w_1(x))(5b_2(x)+w_2(x)) > sticking in b's and w's, and with the 5's still around I could work out > that > b_1(x)*b_2(x) = 7x^2 - 2x, > w_1*b_2(x) + w_2*b_1(x) = 7x-1 > and w_1*w_2 = 7 > so I could so some substitutions and I chose to focus on b_1(x) and w_1 > to get > 7*b_1(x)^2 - (7x-1)w_1*b_1(x) + (7x^2 - 2x)w_1^2 = 0 > which is a non-monic polynomial--unless w_1 has 7 as a factor!!! >w_1 may not be in Q at all. The theorem you quoted is only about > polynomials over Q, and does not apply here. The minimum polynomial of b_1(x) will in general have larger degree, > and it will be monic. > By the way, if we have a polynomial with coefficients which are algebraic integers in a number field K, and the ideal generated by the coefficients is not the whole ring of integers, and the polynomial is irreducible over K, then none of the roots can be algebraic integers. What happens is that if we let K be any field large enough to contain b_1 and w_1, your polynomial will not be irreducible over K. It will split into two linear factors, one with an algebraic integer root and one with a root that is not an algebraic integer. (This is assuming that w_1 and w_2 are coprime). > And with non-zero integer x, it is irreducible over Q, proving that > b_1(x) can't be an algebraic integer, unless w_1 has 7 as a factor. > That is a nice absolute relying on the very result that has often been > used against my assertion that 7 is properly a factor of only one root > as, hey!!! NOTHING but 7 as a factor can even give you algebraic > integer functions if you divide off that pesky 7 from both sides of > 7*P(x) = (5a_1(x) + 7)(5a_2(x)+ 7). > Try any w_1 other than one with 7 as a factor with x a non-zero > integer, and b_1(x) cannot be an algebraic integer function as it is > then the root of a non-monic polynomial with integer coefficients > irreducible over Q. > Damn strange, but it is definitive. > Astute readers may notice that with w_1=7, you just get back a solution > for the a's. > Also you may notice this approach doesn't deal with the issue of what > is the factor of the a's, it just shows they can't have algebraic > integer factors, since 7 itself can't be a factor of just one of them, > bringing into question claims that you can find algebraic integer > factors from 7 for both. > My guess is that old problem in mathematics where when you have false > assumptions you can appear to prove anything, when really you don't > have proof. > The ring of algebraic integers gives you all kinds of problems and > weirdness. === === Subject: : Re: JSH: Weird answer, but it looks solid >Well, it looks like there was this easy enough resolution, which oddly >enough can be said to rely on the impossibility of an algebraic integer >being the root of a non-monic > primitive > polynomial with integer coefficients >irreducible over Q. > Starting with the simple polynomial > P(x) = 175x^2 - 15x + 2 > I multiplied by 7 and did my creative thing with the terms to get > 7*P(x) = (49x^2 - 14x)5^2 + (7x-1)(7)(5) + 49 > to factor it as > 7*P(x) = (5a_1(x) + 7)(5a_2(x)+ 7) > where the a's are roots of > a^2 - (7x-1)a + (49x^2 - 14x) = 0. > But I finally got around to just dividing that damn 7 off, to get > P(x) = (7x^2 - 2x)5^2 + (7x-1)(5) + 7 > giving the factorization > P(x) = (5b_1(x) + w_1)(5b_2(x)+ w_2) > > If you want everything to be an algebraic integer, you'd better make > that > (5b_1(x)+w_1(x))(5b_2(x)+w_2(x)) > sticking in b's and w's, and with the 5's still around I could work out >that > b_1(x)*b_2(x) = 7x^2 - 2x, > w_1*b_2(x) + w_2*b_1(x) = 7x-1 > and w_1*w_2 = 7 > so I could so some substitutions and I chose to focus on b_1(x) and w_1 >to get > 7*b_1(x)^2 - (7x-1)w_1*b_1(x) + (7x^2 - 2x)w_1^2 = 0 > which is a non-monic polynomial--unless w_1 has 7 as a factor!!! > > w_1 may not be in Q at all. The theorem you quoted is only about > polynomials over Q, and does not apply here. > The minimum polynomial of b_1(x) will in general have larger degree, > and it will be monic. >By the way, if we have a polynomial with coefficients which are > algebraic integers in a number field K, and the ideal generated by the > coefficients is not the whole ring of integers, and the polynomial is > irreducible over K, then none of the roots can be algebraic integers. > What happens is that if we let K be any field large enough to contain > b_1 and w_1, Sorry, I mean just w_1. > your polynomial will not be irreducible over K. It will > split into two linear factors, one with an algebraic integer root and > one with a root that is not an algebraic integer. (This is assuming > that w_1 and w_2 are coprime). > And with non-zero integer x, it is irreducible over Q, proving that >b_1(x) can't be an algebraic integer, unless w_1 has 7 as a factor. > That is a nice absolute relying on the very result that has often been >used against my assertion that 7 is properly a factor of only one root >as, hey!!! NOTHING but 7 as a factor can even give you algebraic >integer functions if you divide off that pesky 7 from both sides of > 7*P(x) = (5a_1(x) + 7)(5a_2(x)+ 7). > Try any w_1 other than one with 7 as a factor with x a non-zero >integer, and b_1(x) cannot be an algebraic integer function as it is >then the root of a non-monic polynomial with integer coefficients >irreducible over Q. > Damn strange, but it is definitive. > Astute readers may notice that with w_1=7, you just get back a solution >for the a's. > Also you may notice this approach doesn't deal with the issue of what >is the factor of the a's, it just shows they can't have algebraic >integer factors, since 7 itself can't be a factor of just one of them, >bringing into question claims that you can find algebraic integer >factors from 7 for both. > My guess is that old problem in mathematics where when you have false >assumptions you can appear to prove anything, when really you don't >have proof. > The ring of algebraic integers gives you all kinds of problems and >weirdness. >