mm-444 Subject: Re: JSH: Distributive property, math argumentthis is cause for a party. F(x) = x^2 + x + 3. P(x) = 2 Q(x) = (1/2)(x^2+x+3). Then F(x) = P(x)*Q(x) in the ring of all polynomials over Q. IN ADDITION, Q(x) is a function from Z to Z, and P(x) is a function from Z to Z, with the properties that, AS FUNCTIONS, F(x) = P(x)*Q(x). That is, for each value of x, Q(x) divides F(x) in Z, which is a different statement from the polynomial Q(x) divides the polynomial F(x) in Z[x]. Amazing, that after only about two years he finally realizes what he--ils duces d'Enron!http://www.movisol.org/http://members.tripod.com/~ american_almanac/ === Subject: Re: JSH: Distributive property, math argument>...> The start is simple enough, where the x shown is in the ring of> algebraic integers.> Let P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078>Why does it not work with my polynomial?>It's not your polynomial, of course, it's the *factorization* that you>>use.>At the start of the post I point out that if f(x) = x + 3, and f(x)>>has 3 as a factor, then x must have it as a factor as well.>That's *basic* Dik Winter, and if you accept that result, then at>>least some things should be clear to you.>Notice I didn't mention a ring for x, as the statement is true,>>without regard to the commutative ring.>yes, you did.>You said Consider f(x) = x+3, where x is an algebraic integer.>>Yup, you're right. I'd forgotten that quickly and didn't go back to>>check before making the reply.>>In any event, maybe I should say that it doesn't matter what the ring>>is.>>Consider that if I have f(x) = x + 3, where 3 has itself as a factor>>and f(x) has 3 as a factor, then x *must* have 3 as a factor.>>Now let's say x=7. Then, 7/3 must be in the ring.>>See?>No, I don't understand . You said it was the ring of algebraic integers>and that f(x), given as f(x) = x+3, has 3 as a factor. Then instead of>concluding that this places a *restricition* on the values that x can take>(i.e. x must be an algebraic integer that is divisible by 3) you choose x to>be 7 and conclude that therefore 7/3 =2.33333... must be in the ring of>algebraic integers.!!!! Nope. I'm saying that it doesn't matter what you might *say* the ring is, given f(x) = x+3, if you also have that f(x) has 3 as a factor and 3 has itself as a factor, as then necessarily x has 3 as a factor, which means that if x=7, which IS an algebraic integer, it *still* must have 3 as a factor for the other statements to be true. Look at it as a logical argument: 1. f(x) = x + 3 2. 3 has itself as a factor 3. f(x) has 3 as a factorThis is false for x=7 in the algebraic integers. 4. Therefore, x has 3 as a factor.This is false for x=7 in the algebraic integers.Therefore, since the other statements are *not* all true, you are correct. === Subject: Re: JSH: Distributive property, math argument>...> The start is simple enough, where the x shown is in the ring of> algebraic integers.> Let P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078>Why does it not work with my polynomial?>>It's not your polynomial, of course, it's the *factorization* that you>>use.>>At the start of the post I point out that if f(x) = x + 3, and f(x)>>has 3 as a factor, then x must have it as a factor as well.>>That's *basic* Dik Winter, and if you accept that result, then at>>least some things should be clear to you.>>Notice I didn't mention a ring for x, as the statement is true,>>without regard to the commutative ring.>yes, you did.>You said Consider f(x) = x+3, where x is an algebraic integer. Yup, you're right. I'd forgotten that quickly and didn't go back to check before making the reply. In any event, maybe I should say that it doesn't matter what the ring is. Consider that if I have f(x) = x + 3, where 3 has itself as a factor and f(x) has 3 as a factor, then x *must* have 3 as a factor. Now let's say x=7. Then, 7/3 must be in the ring.You missed a step. You first have to assert that f(7) is divisible by 3. *Then* you can can state that 7 is divisible by 3. See? It's rather simple and depends on accepting the distributive property: a(b+c) = ab + ac Which is why those who are arguing with me are amusing in one sense, as inevitably they are attacking VERY basic mathematics to try and keep playing silly games. It's funny to see how much math society actually cares about its foundations: like the distributive property.This is an example of the problem you have. We aren't arguing the distributive property, but something else entirely. You tend to make sloppy mistakes, like not checking all the conditions. === Subject: Re: JSH: Distributive property, math argument It's amazing how easily people can question very basic concepts when the stakes are high, so here I am again to help you out by reminding you just how basic the math argument I've been giving recently is. Consider f(x) = x+3, where x is an algebraic integer. Now then, if f(x) has 3 as a factor, then x *must* have 3 as a factor as well.I think this translates as: if x+3 is divisible by 3 then x is divisible by 3 in the algebraic integersIn fact, we can say something stronger: x+3 is divisible by 3 if and only if x is divisible by 3 in the algebraic integers That follows easily enough from the distributive property: a(b+c) = ab + ac and for quite a few months, I've noticed people willing to debate it, as if that property can just go away at a whim.No you haven't. People haven't debated it, you've just misinterpretted.[minor deletia] Now here's the argument, where those people questioning it have basically been questioning the distributive property. The start is simple enough, where the x shown is in the ring of algebraic integers. Let P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078 which you'll notice has a constant term that is 1078.Right. Well moving things around with P(x) gives you P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3 which is a deliberate form to allow me to factor P(x), so that I have P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7) where the a's are roots of a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x).I'll take this as a definition of the a_i's. To use the distributive property as I wish I need to have constant terms. And it *appears* that the constant terms for the three factors are all 7, but that can't be right, as the constant term of P(x) is 1078. So I need to do another step, and analysis offers the simple technique of setting x=0 to pull out the constant terms, so setting x=0, I find that P(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22) as the cubic defining the a's at x=0 is a^3 - 3a^2, which has roots, 0, 0 and 3, and I've picked a_1(0) and a_2(0) to equal 0, which leaves a_3(0) with a value of 3. So let a_3(x) = b_3(x) + 3, to keep indices matched. Then I have P(x) = (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 5(3) + 7) P(x) = (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22) and now my constant terms work out correctly.Ok. Now I can use the distributive property, as P(x) has 49 as a factor of each term in P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078 so P(x) has 49 as a factor, so I can divide by 49, and dividing 1078 by 49 gives me 22, as the new constant term.Again, right. Well that means that P(x)/49 = (5 a_1/7 + 1)(5 a_2/7 + 1)(5 b_3 + 22) is the only way that 49 can divide out and keep the constant terms matching, in what is a simple exercise dependent on the distributive property.It does keep it matching, but you have NOT shown that the results are in the ring of algebraic integers. If you had shown that a_1 is divisible by seven or that (5 a_1 + 7) is divisible by 7, you'd be ok. You've done neither. The problem here is that you are assuming the constant terms *will* always be matching. This is not, in general, true. === Subject: Re: is hausdorff dimension a topological invariant??given two homeomorphic metric spaces, are their hausdorff dimensionsidentical? I have no idea what the answer is, so as always: proof orThree people already mentioned Cantor sets as a counter-example. Asanother type of counter-example, there are metrics in R^n (for some n,at least) that induce the usual topology but have Hausdorff dimensionequal to an integer greater than n. === Subject: Re: No Perfect Cuboid Exists: - Proofthe symmetry of the box can be kept in mind,by making the edges x, y, z and the diagonals a, b, cwhich are respectively oppposite. that is,yy+zz=aa, zz+xx=bb, xx+yy=cc; thenxx+aa=dd, yy+bb=dd, zz+cc=dd,with d being the interior diagonal. or,in a common parlance,D=X+A=Y+B=Z+C=D. but you can't show a violation of parity, that way! Lost what?--ils duces d'Enron!http://www.movisol.org/http://members.tripod.com/~ american_almanac/ === Subject: Re: No Perfect Cuboid Exists: - Proof (take 2!)I tried taht the other day,trying to show that parity is violatedbetween three right trigona, butI realized that it wouldn't work. what does W/LOG mean? : b^2+c^2=d^2 : c^2+a^2=e^2 : a^2+b^2=f^2 : a^2+b^2+c^2=g^2 : therefore WLOG all a,b,c even No, you've got LOG because of destroyed symmetry. At this point you *can* say: *If* a,b,c all even, *then* hcf[a,b,c]>1, which contradicts smallest solution stuff, *hence* a,b,c *are not all even*. So a,b,c are not all even, what now? --ils duces d'Enron!http://www.movisol.org/http://members.tripod.com/~ american_almanac/ === Subject: Re: No Perfect Cuboid Exists: - Proof (take 2!): I realized that it wouldn't work.: what does W/LOG mean?WLOG stands for Without Loss of Generality.LOG is never used, really, I just used it to be humorous about the factthat that he was losing generality.Justin === Subject: Re: JSH: Factorization P(x) = 2(x(x+1)/2 + 1)...> My point is that you have to focus on the *factorization* and its> validity in particular rings. > > Some factorizations will be valid in one ring, but not another. > Yes, and your factorisation > P(x)/49 = (5 a1/7 + 1)(5 a2/7 + 1)(5 b3 + 22) > is in general not valid in the ring of algebraic integers. So what > are you trying to show? > That's the point. I *prove* that if you have coprimeness between 7 > and 22 in the ring in which the factorization is valid, where 7 is NOT > a unit (and neither is 22), then the constant terms of the factors > that result from dividing P(x) by 49 *MUST* be coprime to 7.If you are talking here about the factorisation P(x)/49 = (5 a1(x)/7 + 1)(5 a2(x)/7 + 1)(5 b3(x) + 22)being valid in that ring (for all x), that is vacuously true. Noproof needed, 1 is coprime to 7 and 22, and 22 is coprime to 7 inany ring that contains 1, 7 and 22. Even if in that ring 7 and/or 22are units. So we need no proof in that case.It becomes different about the MUST. You have made no showing of that.Why can it not be that for particular x, a division of all three factorsby 7^{2/3} be valid? Yes, 7/7^{2/3} is not coprime to 7, but(5 a1(x) + 7)/7^{2/3} *is* for the polynomial I used. You are just assumingthat because for x = 0 the factors of 49 clearly distribute as 7, 7, 1this *must* be true for all x. When you stay in the algebraic integersthat simply is not the case. When you talk about divisibility you cannot assume that something that is proper in one case is also proper inanother case. > Get it yet Dik Winter?Get it yet James Harris?dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: JSH: Factorization P(x) = 2(x(x+1)/2 + 1) ... > My point is that you have to focus on the *factorization* and its > validity in particular rings. Some factorizations will be valid in one ring, but not another. > Yes, and your factorisation > P(x)/49 = (5 a1/7 + 1)(5 a2/7 + 1)(5 b3 + 22) > is in general not valid in the ring of algebraic integers. So what > are you trying to show? > That's the point. I *prove* that if you have coprimeness between 7 > and 22 in the ring in which the factorization is valid, where 7 is NOT > a unit (and neither is 22), then the constant terms of the factors > that result from dividing P(x) by 49 *MUST* be coprime to 7. If you are talking here about the factorisation P(x)/49 = (5 a1(x)/7 + 1)(5 a2(x)/7 + 1)(5 b3(x) + 22) being valid in that ring (for all x), that is vacuously true. No proof needed, 1 is coprime to 7 and 22, and 22 is coprime to 7 in any ring that contains 1, 7 and 22. Even if in that ring 7 and/or 22 are units. So we need no proof in that case.Oh yeah, so ignore my statement that 7 is NOT a unit Dik Winter asyou're showing your lack of reasonableness.Now then, consider coprimeness in algebraic integers, and what thatmeans for 7 and 22, versus talking about coprimeness as if either is aunit, can you do that Dik Winter?Assuming you can, now then, my point is that for a factorization validIN SUCH A RING, the argument plays out as I've given it, as theconstant terms MUST be coprime to 7, in a way that's rather obvious.Now you come up with different factorizations, and got all excited asif finding some other factorization, will invalidate a DIFFERENTfactorization, so I've explained in detail.Rather than be reasonable, you're being petulant and stubborn. It becomes different about the MUST. You have made no showing of that. Why can it not be that for particular x, a division of all three factors by 7^{2/3} be valid? Yes, 7/7^{2/3} is not coprime to 7, butBecause the constant terms of the factors of P(x)/49 are coprime to 7.Understand Dik Winter?It's NOT COMPLICATED, but you are behaving as if logic is a disease.Now then, if you find a *different* factorization, not valid in ringswhere 7 is coprime to 22 in the same sense as in algebraic integers,then you have a different result.It's math, so I can trace out the mathematical argument, and show youexactly the point where the factorizations you hacked together gotheir own way as they're not valid in rings like algebraic integers. (5 a1(x) + 7)/7^{2/3} *is* for the polynomial I used. You are just assuming that because for x = 0 the factors of 49 clearly distribute as 7, 7, 1 this *must* be true for all x. When you stay in the algebraic integers that simply is not the case. When you talk about divisibility you can not assume that something that is proper in one case is also proper in another case.Constant terms do NOT changes as variables.I repeat, constant terms do NOT change as variables.Given a situation, where one position requires that they do, whileanother accepts that they do not, why in the hell would you go for theposition that requires constant terms to change as variables? > Get it yet Dik Winter? Get it yet James Harris?Yeah, I understand it you mocking nitwit. I'm just wondering why youREFUSE TO BE LOGICAL. Why do you even bother reading or posting on amath newsgroup if you display behavior that is anti-thetical tomathematics?Now then, given that constant factors do NOT change as variables theremust be a LOGICAL answer for what happens, and I've explained it toyou.James Harris === Subject: Re: JSH: Factorization P(x) = 2(x(x+1)/2 + 1)> ...>> My point is that you have to focus on the *factorization* and its>> validity in particular rings.>> Some factorizations will be valid in one ring, but not another.> >> Yes, and your factorisation>> P(x)/49 = (5 a1/7 + 1)(5 a2/7 + 1)(5 b3 + 22)>> is in general not valid in the ring of algebraic integers. So what>> are you trying to show?>That's the point. I *prove* that if you have coprimeness between 7>and 22 in the ring in which the factorization is valid, where 7 is NOT>a unit (and neither is 22), then the constant terms of the factors>that result from dividing P(x) by 49 *MUST* be coprime to 7.>If you are talking here about the factorisation> P(x)/49 = (5 a1(x)/7 + 1)(5 a2(x)/7 + 1)(5 b3(x) + 22)>being valid in that ring (for all x), that is vacuously true. No>proof needed, 1 is coprime to 7 and 22, and 22 is coprime to 7 in>any ring that contains 1, 7 and 22. Even if in that ring 7 and/or 22>are units. So we need no proof in that case. Oh yeah, so ignore my statement that 7 is NOT a unit Dik Winter as you're showing your lack of reasonableness.Given a commutative ring with 7 and 22.Definition of coprime: 7 and 22 are coprime in a ring if there exists p and q in the ring such that 7p + 22q = 1.7+7+7 = 21, so 21 is in the ring.21 is in the ring so -21 is in the ring.22 + -21 = 1 so 1 is in the ring.1+1+1 = 3, so 3 is in the ring.3 is in the ring, so -3 is in the ring.7*-3 + 22*1 = 1, with both -3 and 1 in the ring, so 7 and 22 are coprime.Note: whether 7 or 22 are units is *completely* irrelevent to the computations.You probably want to say something about whether or not they have any non-unit common factors, since unit factors are generally ignored.If we are looking at 7 and 22, we are at least in the integers, since closure under addition and additive inverses allows us to construct 1.If the ring has 7 as a unit, it must contain Z[1/7] as a subring.If the ring has 22 as a unit, it must contain Z[1/22] as a subring.If the ring has both as units, it must contain Z[1/7][1/22] as a subring.Regardless of the case, if p is an element of the ring that is a divisor of both 7 and 22, then p is a divisor of a unit, which means p is a unit.If 7 and 22 are *not* units, I'm pretty sure they won't have non-unit divisors, but I'm not sure how to go about showing that. Maybe when I've got some time later I'll look at it.What were you objecting to again?[rest deleted] === Subject: Re: JSH: Factorization P(x) = 2(x(x+1)/2 + 1) Adjunct Assistant Professor at the University of Montana.>> ...>> My point is that you have to focus on the *factorization* and its>> validity in particular rings.>> >> Some factorizations will be valid in one ring, but not another.>> >>Yes, and your factorisation>> P(x)/49 = (5 a1/7 + 1)(5 a2/7 + 1)(5 b3 + 22)>>is in general not valid in the ring of algebraic integers. So what>>are you trying to show?> That's the point. I *prove* that if you have coprimeness between 7>> and 22 in the ring in which the factorization is valid, where 7 is NOT>> a unit (and neither is 22), then the constant terms of the factors>> that result from dividing P(x) by 49 *MUST* be coprime to 7.>>If you are talking here about the factorisation>> P(x)/49 = (5 a1(x)/7 + 1)(5 a2(x)/7 + 1)(5 b3(x) + 22)>>being valid in that ring (for all x), that is vacuously true. No>>proof needed, 1 is coprime to 7 and 22, and 22 is coprime to 7 in>>any ring that contains 1, 7 and 22. Even if in that ring 7 and/or 22>>are units. So we need no proof in that case.> Oh yeah, so ignore my statement that 7 is NOT a unit Dik Winter as> you're showing your lack of reasonableness.Given a commutative ring with 7 and 22.Definition of coprime: 7 and 22 are coprime in a ring if there exists p and q in the ring such that 7p + 22q = 1.This definition (the standard definition) is stronger than thedefinition James uses: a and b are coprime if and only if any commondivisor is a unit.If 7 and 22 are *not* units, I'm pretty sure they won't have non-unit divisors, but I'm not sure how to go about showing that. Maybe when I've got some time later I'll look at it.That's why the usual definition is better:LEMMA. If R is a ring, and x and y in R are coprime in the sense thatthere exist a and b in R such that ax+by = 1, then for any ring thatcontains R, x and y are also coprime (in the same sense).Proof: a and b serve as witnesses in both R and the larger ring. QEDREMARK. If R is a ring, x and y in R are coprime in the sense thatany common divisor in R of x and y is a unit in R, then it is possiblefor there to be a larger ring S, containing R, where x and y are nolonger coprime (in that sense).Example: R= Z[sqrt(-5)]; x = 2, y = (1+sqrt(-5)), S=Z[sqrt(-5),sqrt(2)]; note that y is a multiple of sqrt(2), since(1+sqrt(-5)) = sqrt(2)*sqrt(3+sqrt(-5)).PROP. Let R be a ring. If x and y are coprime in R in the sense thatthere exists a and b in R such that ax+by=1, then x and y are coprimein R in the sense that any common divisor in R is a unit in R.Proof. Let u be a common divisor of x and y. Then it is a divisor ofax, and it is a divisor of by, so it is a divisor of ax+by=1. Divisorsof 1 are units. So u is a unit. QEDSo in ANY ring that contains the integers, 7 and 22 are coprime (undereither definition).e of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of figures few readers can criticize. A great many people are staggered to this extent, that they imagine there must be the indefinite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan === ===================================================== === ======Subject: Re: JSH: Factorization P(x) = 2(x(x+1)/2 + 1) So in ANY ring that contains the integers, 7 and 22 are coprime (under either definition).That's true in any ring R since R contains a homomorphic image of Z,Z -> Z*1_R, i.e. any ring is a Z-algebra. But any ring homomorphismmust preserve the relation 22 - 3(7) = 1.By the way, as I mentioned in a prior post [1], coprime isa highly overloaded term whose meaning depends upon context.JSH is using one of the most common definitions and it isincorrect to criticize him for that. Arturo's definition hasalso the less ambiguous name comaximal, and this shouldbe preferred in contexts where there may be ambiguity.- Dubuque === Subject: Re: JSH: Factorization P(x) = 2(x(x+1)/2 + 1) Adjunct Assistant Professor at the University of Montana.> So in ANY ring that contains the integers, 7 and 22 are coprime> (under either definition).That's true in any ring R since R contains a homomorphic image of Z,Z -> Z*1_R, i.e. any ring is a Z-algebra. But any ring homomorphismmust preserve the relation 22 - 3(7) = 1.Hmmm... Only if you assume that ring morphisms map 1 to 1, which isnot necessarily a given either. Even assuming rings have a 1, the zeromap is usually considered a valid homomorphism, and your conclusionwould be incorrect there.By the way, as I mentioned in a prior post [1], coprime isa highly overloaded term whose meaning depends upon context.JSH is using one of the most common definitions and it isincorrect to criticize him for that. I will quibble that what is most common depends on context aswell. Most ring theorists I know would object to using the definitiondepending on common divisors, since to them 'prime' refers to ideals,almost never to elements; and most number theorists would certainlydisagree that the definition via common divisors is 'the most common'(for the latter the definition ->is<- invariably related to ideals,never to elements). To me, and particularly given that JSH's work istaking place in subrings of the ring of all algebraic integers, itseems that the most common usage from algebraic number theory shouldprevail. It may be most common among those who studydomains or other closely related kinds of rings, though.But in any case, so long as he states explicitly what he means andsticks to it (neither of which is something he usually does), I haveaccepted your correction and rather than criticise him for his use Isimply note the distinction between his use and the one defined vialinear combinations equalling 1 (see below).Arturo's definition hasalso the less ambiguous name comaximal, and this shouldbe preferred in contexts where there may be ambiguity.The definition I use is actually that no prime ideal contains theprincipal ideals generated by the elements. It turns out to beequivalent to comaximal in the presence of a 1 and the Axiom ofChoice, but not in general, as I discussed when James brought up theexample of the ring (or rng) of even integers. I assume you arereferring to a and b are coprime if and only if there exists r and ssuch that ar+bs = 1, which I did assume throughout above, but which Iwould consider a theorem, not the definition.But, if you'll see James's most recent contributions, his recentmistake on 'coprime' following his usage has led him to the conclusionthat 'coprime' is broken, so go figure what will happen next. === Subject: Re: JSH: Factorization P(x) = 2(x(x+1)/2 + 1) > ...> My point is that you have to focus on the *factorization* and its> validity in particular rings.> > Some factorizations will be valid in one ring, but not another. > > Yes, and your factorisation> P(x)/49 = (5 a1/7 + 1)(5 a2/7 + 1)(5 b3 + 22)> is in general not valid in the ring of algebraic integers. So what> are you trying to show? > > That's the point. I *prove* that if you have coprimeness between 7> and 22 in the ring in which the factorization is valid, where 7 is NOT> a unit (and neither is 22), then the constant terms of the factors> that result from dividing P(x) by 49 *MUST* be coprime to 7. > If you are talking here about the factorisation > P(x)/49 = (5 a1(x)/7 + 1)(5 a2(x)/7 + 1)(5 b3(x) + 22) > being valid in that ring (for all x), that is vacuously true. No > proof needed, 1 is coprime to 7 and 22, and 22 is coprime to 7 in > any ring that contains 1, 7 and 22. Even if in that ring 7 and/or 22 > are units. So we need no proof in that case. > Oh yeah, so ignore my statement that 7 is NOT a unit Dik Winter as > you're showing your lack of reasonableness.You are ignoring my statement that in *every* ring that contains 1, 7 and 22,they are coprime. Whether 7 is a unit or not. > Now then, consider coprimeness in algebraic integers, and what that > means for 7 and 22, versus talking about coprimeness as if either is a > unit, can you do that Dik Winter?Why? When you read what I write you know what I would say here. Because7 and 22 are coprime in every ring that contains them, they are also coprimein the algebraic integers. Or are you not able to raw that conclusion? > Assuming you can, now then, my point is that for a factorization valid > IN SUCH A RING, the argument plays out as I've given it, as the > constant terms MUST be coprime to 7, in a way that's rather obvious.I am attacking your *MUST*. You have shown nowhere that it *MUST* bethe case. > It becomes different about the MUST. You have made no showing of that. > Why can it not be that for particular x, a division of all three factors > by 7^{2/3} be valid? Yes, 7/7^{2/3} is not coprime to 7, but > Because the constant terms of the factors of P(x)/49 are coprime to 7.*Why?* > Understand Dik Winter?No. You have not shown anything about *why* they must be coprime to 7.You assert it. If for some particular x the factors can be divided by7^{2/3} leaving algebraic integers, there is no reason to divide two ofthe factors by 7 and the third by 1. That 22 is coprime to 7 does*not* mean that (5 b3(x) + 22) is coprime to 7 in the algebraic integers. > It's NOT COMPLICATED, but you are behaving as if logic is a disease. > Now then, if you find a *different* factorization, not valid in rings > where 7 is coprime to 22 in the same sense as in algebraic integers, > then you have a different result.Can you show me a ring (with unit) where 7 is *not* coprime to 22? Trysome mathematics instead. In every ring that contains 1, 7 and 22 theyare coprime in the same sense as in the algebraic integers. > It's math, so I can trace out the mathematical argument, and show you > exactly the point where the factorizations you hacked together go > their own way as they're not valid in rings like algebraic integers.Well, do that for once. > (5 a1(x) + 7)/7^{2/3} *is* for the polynomial I used. You are just assuming > that because for x = 0 the factors of 49 clearly distribute as 7, 7, 1 > this *must* be true for all x. When you stay in the algebraic integers > that simply is not the case. When you talk about divisibility you can > not assume that something that is proper in one case is also proper in > another case. > Constant terms do NOT changes as variables.They are constant terms *only* if you assume that they ought to be dividedby the same constant value for every x, but that is something you have toprove. If that is not the case, dividing P(x) by 49 gives thefactorisation (5 a1(x)/w1(x) + 7/w1(x))(5 a2(x)/w2(x) + 7/w2(x))(5 b3(x)/w3(x) + 22/w3(x))where the w's depend on the value of x. When x = 0, w1(x) = 7, w2(x) = 7and w3(x) = 1. For other values of x other values of the w's are available.That is, if you wish to stay in the algebraic integers. (Note that although22/w3(x) may very well not be an algebraic integer, (5 b3(x)/w3(x) + 22/w3(x))is. > I repeat, constant terms do NOT change as variables.When you divide through by 49, they are no longer constant terms). Unlessyou *assume* that the distribution of the factors of 49 must go the sameway for every x. > Yeah, I understand it you mocking nitwit. I'm just wondering why you > REFUSE TO BE LOGICAL. Why do you even bother reading or posting on a > math newsgroup if you display behavior that is anti-thetical to > mathematics?dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: JSH: Factorization P(x) = 2(x(x+1)/2 + 1) ... > > My point is that you have to focus on the *factorization* and its > > validity in particular rings. > Some factorizations will be valid in one ring, but not another. > > Yes, and your factorisation > P(x)/49 = (5 a1/7 + 1)(5 a2/7 + 1)(5 b3 + 22) > is in general not valid in the ring of algebraic integers. So what > are you trying to show? > > That's the point. I *prove* that if you have coprimeness between 7 > and 22 in the ring in which the factorization is valid, where 7 is NOT > a unit (and neither is 22), then the constant terms of the factors > that result from dividing P(x) by 49 *MUST* be coprime to 7. > If you are talking here about the factorisation > P(x)/49 = (5 a1(x)/7 + 1)(5 a2(x)/7 + 1)(5 b3(x) + 22) > being valid in that ring (for all x), that is vacuously true. No > proof needed, 1 is coprime to 7 and 22, and 22 is coprime to 7 in > any ring that contains 1, 7 and 22. Even if in that ring 7 and/or 22 > are units. So we need no proof in that case. > Oh yeah, so ignore my statement that 7 is NOT a unit Dik Winter as > you're showing your lack of reasonableness. You are ignoring my statement that in *every* ring that contains 1, 7 and 22, they are coprime. Whether 7 is a unit or not.Are you just totally stupid? That's irrelevant Dik Winter as thepoint is that 7 is NOT a factor of 22, which my saying that it's not aunit points out.I'm getting sick of stupid games from stupid people who apparentlyhave nothing better to do with *their* time.Now are you or are you not intelligent enough to understand what itmeans for 7 NOT to be a factor of 22?James Harris === Subject: Symmetric PolynomialHow does one go about proving that every symmetric polynomial,f(x_i)=f(x_Pi(i)), Pi permutation 1 to n, can be expressed in terms of theelementary symmetric polynomials, E_i=Sum[1<=a_1 S^1 given by p(x) = e^(2ix*pi) is aquotient map.Mike === Subject: Re: Unit interval homemorphic to Circle helpIf I = [0,1], and I / ~ is the quotient space of I obtained by identifying 0and 1, then the circle S^1 is homeomorphic to I / ~.This is intuitively clear, but how can I prove this?I think I want to show that p : I ---> S^1 given by p(x) = e^(2ix*pi) is aquotient map. That's exactly what you must show.In this case it is quite easy (fortunately).The map p is already continuous, as I and S^1 are compact metric the map p isautomatically closed (i.e., if A is closed then so is p[A]).This implies that p is a quotient map.KPE-MAIL: K.P.Hart@EWI.TUDelft.NL PAPER: Faculty EWIPHONE: +31-15-2784572 TU DelftFAX: +31-15-2786178 Postbus 5031URL: http://aw.twi.tudelft.nl/~hart 2600 GA Delft the Netherlands === Subject: Re: Unit interval homemorphic to Circle help === Subject: Unit interval homemorphic to Circle help >If I = [0,1], and I / ~ is the quotient space of I obtained by >identifying 0 and 1, then the circle S^1 is homeomorphic to I / ~. >This is intuitively clear, but how can I prove this? >I think I want to show that p : I -> S^1 given by >p(x) = e^(2ix*pi) is a quotient map.Yes and you also want to show I/p = I/~ (easy) and apply thefundamental homeomorphism theorem for the quotient space S^1 === Subject: Re: Factorial/Exponential Identity, InfinityFish, or cut bait.Is the dis-parity discrepancy contradictory? Why or why not?Ross === Subject: Re: minimum foamor, let's put it in this way:it's excruciatingly trivial -- stop it! actually, I thought that you were packing icosahedra,that were divided into 20 tetrahedra; since,for a regular icosah., the tetrahedra are pointyat the center, they're going to be less-pointyat the vertices of the icosah., so that you'd have to dystort thingsto get the icosahedra to meet with 20 tetrah.at each vertex, more & more as you added stuff. what it looks like is an ill-posed question. If I 3 go with dimensions, can I fill space with irregular tetrahedra such that all edges share 5 tetrahedra and all vertexes share 20 tetrahedra?--les Dicks d'Enron!http://www.wlym.com/antidummies/part36.html === Subject: Re: minimum foamit's not a problem, if you consider thatthere doesn't have to be a unit brick;it's not really interesting, as stated --it's not a tiling in any conventional sense. in other word, nonsequiter. If I 3 go with dimensions, can I fill space with irregular tetrahedra such that all edges share 5 tetrahedra and all vertexes share 20 tetrahedra?--les Dicks d'Enron!http://www.wlym.com/antidummies/part36.html === Subject: Re: Solving definite integral problem - newbie question I am studying for an exam and don't have access to my instructor today. Can someone help me with the following? I have a graph of the right half of a bell curve, a portion of which I must solve. The portion is from 0 to 1 (that is x=0 to x=1) and the equation is y=e^(-x^2/2). That is, e raised to (negative x squared, divided by two). I just need some help in the proper approach. I have been given the answer, but so far nothing that I have tried has helped me to get the answer.You also must have either a table of the cumulative probability or somenumbers on your graph.There is no elementary antiderivative, so the only way to get numericalvalues is (ta-da) numerically.Or are they looking for, sigma-squared equals 1, so you need to know thearea 1 s.d. to the right of the mean?Jon Miller === Subject: Re: Making Star Trek Realsince you gave a reality check for time,taht you can't go there if there's no functioning chronoselectoryou cannot visit a time before the first machineit built, or after the last machine ceases to function --over there, with there being cognate with when,doesn't that pre-suppose ... oh, nevermind;I could use that in a novel about Philip Keister Dick andthe elected governor of Venice Muscle Beach. Sarfatti has yet to learn that merely writing an equation (we won't argue the correctness of the equation, we'll assume they are correct) gives absolutely no clue as to how one goes about building devices that engineer the metric to conform to those equations. (5) suborbital ballistic low-value package and personnel delivery (6) orbiting satellites (space homes) (7) large orbiting satellites (full integration of Earth/Earth orbit) (8) very large interplanetary spaceraft (mobile spacehomes) then, flight to the stars at about 1/3 light speed; (9) interstellar space homes then, once a number of colonies are set up surrounding sol, we can imagine research along the lines proposed by Sarfatti - basically, (c) Very little shiptime later, but 10,000+ years later, arrive at your destination. Explore, and so forth. Then return. (d) Arrive back at your starting point 20,000+ years later, while only a few days pass on the ship due to very high speeds most of the way. (e) Use the time machine to travel back in time to the point just after your departure. For everyone's sanity exit the causality violating region so that the local time is synchronized with your ship time. (You may send radio telescope messages back and forth this way as well, to that everything is synchronized between the two frames. The reply messages from the moving ship are picked up by the time machine in the future and routed, like internet packets, to the right time exit, based on time stamps. Thus establishing instantaneous messaging) This is very much like Star Trek - but has nothing to do with the 'science' of Star Trek. Anymore than the flying in an airplane has to do with the 'science' of flying by Witchcraft. They're both talking about flight - but one achieves it while the other does not.--les Dicks d'Enron!http://www.wlym.com/pages/pedagogicals.html === Subject: Galias Field: what is primitive element in GF(4)?Dear all,I have the following question: suppose a is primitive in GF(4),then how to finda^4+a^2, why it is equal to 1 ?a^4+a^3, why it is equal to a^2 ?Anybody can tell me what is the trick?-Walala