mm-68 === Turing's test design is ? It is not double blind: the tester knows some are machines and some> are not.>i get you. this is a facet of nearly all real life data samples, you can still makeit double blind by having sets of panels of judges, so some of them areall human or some are all computer testees.PANEL 1 test 1 ~ human test 2 ~ human test 3 ~ humanPANEL 2 test 4 ~ human test 5 ~ computer test 6 ~ humanPANEL 3 test 7 ~ ... test 8 test 9There's a bigger problem with true negatives (is to false positives). In an actual Turing Testthe judge said the agent was a computer... because it knew too much Shakespear.Herc === =true negatives (is to false positives). In an actual Turing Test> the judge said the agent was a computer... because it knew too much Shakespear.false negatives... where's my brain.. :)Some Unis do Turing Tests on peoples Eliza programs. One guygot a $5,000 prize, and there was a $100,000 offer for a real pass.But one of the control people was identi?d as a computer, the judge saidthere's no way a person could know that much Shakespear!Herc === snip>A discovery of mine is that (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = 49(300125 x^3 - 18375 x^2 - 360 x + 22)where b_3(x) = a_3(x) - 3 and the a's are roots ofa^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)and when x=0, a_1(0) = a_2(0) = b_3(0) = 0.Attempting to use those relations in the ring of algebraic integers I> can demonstrate a problem when I consider what happens if you divide> both sides by 49.If you *assume* that the ring of algebraic integers is ?e, it's> reasonable to conclude the dividing both sides by 49 gives you> something like(5 b_1(x) + 1)(5 b_2(x) + 1)(5 c_3(x) + 22) = 300125 x^3 - 18375 x^2 - 360 x + 22where b_1(0) = b_2(0) = c_3(0) = 0,in the ring of algebraic integers, as the constant terms of the> factors go from being 7, 7 and 22 to being 1, 1, and 22.Now some posters have repeatedly attempted to question considering the> value of the constant terms as if the constants can't tell you> anything here.That I think requires readers considering their position forget that> the constants don't vary, so like with x=2:(5 a_1(2) + 7)(5 a_2(2) + 7)(5 b_3(2) + 22) = 49(300125 2^3 - 18375 2^2 - 360 (2) + 22)you can still *see* 7, 7 and 22, despite the fact that x=2. Granted> that if you evaluate you no longer *see* them.My point, of course, is that 7, 7 and 22 are there without regard to> x's value, as they are constants.However, if you divide both sides by 49, you ?d that your constants> then are 1, 1, and 22.Now then, how do you get from 7 to 1?Of course, you divide by 7.> Not necessarily, and this is where you make your mistake.> That's not possible because 7, 7 and 22 are *constants*, and theconstant 49 is what's being divided from both sides.Now if you have variable factors, like I said in my original post,like(x+7)(x+1) = x^2 + 8x + 7then you can talk of 7 being divided from the constant term when x+7is divided from both sides, but notice--you have a variableexpression!However, 49 is a number, so it's constant, and 7, 7 and 22 are numberswhich are also then constant. So, if dividing both sides gives you 1,1, and 22 then you divided those constants by a constant factor whichmust be 7.The algebra here is actually just arithmetic.There's simply no room for mistake. The point here that might be confusing to some is that 49 is *itself*> constant, so it and its factors behave as constants.>Not necessarily. In fact, its factors might well differ (indeed,> they *do* differ) depending on the value of x. As I'll show> tomorrow (got to rush tonight--I have a party to go to),> the a(x) functions might very well dictate a splitting> of 49 that's not 7 and 7.Well then you could have replied when you had time, rather than makinga claim without defending it.As it is, your claim fall ? its face because constants, like 7,7 and 22, are NOT affected by a variable.Remember(5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = 49(300125 x^3 - 18375 x^2 - 360 x + 22) where b_3(x) = a_3(x) - 3 and the a's are roots of a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) and when x=0, a_1(0) = a_2(0) = b_3(0) = 0.The value of x has NO AFFECT on 7, 7 and 22, but what *does* affectthe constant terms is dividing both sides by 49. James Harris === >A discovery of mine is that (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = 49(300125 x^3 - 18375 x^2 - 360 x + 22)where b_3(x) = a_3(x) - 3 and the a's are roots ofa^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)and when x=0, a_1(0) = a_2(0) = b_3(0) = 0.Attempting to use those relations in the ring of algebraic integers I>can demonstrate a problem when I consider what happens if you divide>both sides by 49.If you *assume* that the ring of algebraic integers is ?e, it's>reasonable to conclude the dividing both sides by 49 gives you>something like(5 b_1(x) + 1)(5 b_2(x) + 1)(5 c_3(x) + 22) = 300125 x^3 - 18375 x^2 - 360 x + 22where b_1(0) = b_2(0) = c_3(0) = 0,in the ring of algebraic integers, as the constant terms of the>factors go from being 7, 7 and 22 to being 1, 1, and 22.Now some posters have repeatedly attempted to question considering the>value of the constant terms as if the constants can't tell you>anything here.That I think requires readers considering their position forget that>the constants don't vary, so like with x=2:(5 a_1(2) + 7)(5 a_2(2) + 7)(5 b_3(2) + 22) = 49(300125 2^3 - 18375 2^2 - 360 (2) + 22)you can still *see* 7, 7 and 22, despite the fact that x=2. Granted>that if you evaluate you no longer *see* them.My point, of course, is that 7, 7 and 22 are there without regard to>x's value, as they are constants.However, if you divide both sides by 49, you ?d that your constants>then are 1, 1, and 22.Now then, how do you get from 7 to 1?Of course, you divide by 7.>Not necessarily, and this is where you make your mistake.That's not possible because 7, 7 and 22 are *constants*, and the> constant 49 is what's being divided from both sides.> The point here that might be confusing to some is that 49 is *itself*>constant, so it and its factors behave as constants.>>Not necessarily. In fact, its factors might well differ (indeed,>>they *do* differ) depending on the value of x. As I'll show>>tomorrow (got to rush tonight--I have a party to go to),>>the a(x) functions might very well dictate a splitting>>of 49 that's not 7 and 7.>>Well then you could have replied when you had time, rather than making> a claim without defending it.>Heh. Well, I'm back, as promised. BTW, it was a fun party. I hopethe rest of the sci.math population had a good time too.Let's parallel your argument in a simpler case.Let Q(x) = 7(25x^2 + 30x + 2) [1] = 7(x^2 + x)(5^2) + 7(x - 1)(5) + 7^2and write Q(x) = (5a_1(x) + 7)(5a_2(x) + 7) [2]Now we mirror your construction and observe that we may takea_1(x) and a_2(x) to be roots of r(x) = a^2 - (x - 1)a + 7(x^2 + x) [3]We see that Q(0) = 14 = (7)(2), and r(0) = a^2 + a has rootsa_1(0) = 0, a_2(0) = -1. We see that this assignment isconsistent with your factorization [2], since Q(0) = (5(0) + 7)(5(-1) + 7) = (7)(2) [4]and we see that we can divide [4] by 7 and write it as Q(0)/7 = (5(a_1(0)/7) + 1)(5a_2(0) + 7) = (1)(2)or, emphasizing the constant term in [1] we may write,letting b_2(0) = a_2(0) + 1, Q(0)/7 = (5(a_1(0)/7) + 1)(5b_2(0) + 2) = (1)(2)So, answering the equivalent of your question, howdo you get from 7 to 1? we see that in the x = 0case, it is exactly by dividing one factor by 7.But what happens if we have x = 1? Well, in thiscase, we see that Q(1) = 7(25 + 30 + 2) = (7)(57)and it's easy to see that a_1(1) = sqrt(-14) a_2(1) = -sqrt(-14)which yields the factorization Q(1) = (5(sqrt(-14)) + 7)(5(-sqrt(-14)) + 7) = 7(57)excatly as it should. Now, however, we can'tdivide by 7, since neither of the two factorsabove are divisible by 7. They are, howeverboth divisible by sqrt(7), so we havethe factorization (in algebraic integers) Q(1)/7 = [(5 sqrt(-14) + 7)/sqrt(7)] * [(-5 sqrt(-14) + 7)/sqrt(7)]In fact, the condition that exactly one ofthe factors (or 2, in your example) is divisibleby 7 holds only for a small set of x values.For even more fun, try this with x = 2.Rick === snip>>A discovery of mine is that (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = 49(300125 x^3 - 18375 x^2 - 360 x + 22)where b_3(x) = a_3(x) - 3 and the a's are roots ofa^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)and when x=0, a_1(0) = a_2(0) = b_3(0) = 0.Attempting to use those relations in the ring of algebraic integers I>can demonstrate a problem when I consider what happens if you divide>both sides by 49.If you *assume* that the ring of algebraic integers is ?e, it's>reasonable to conclude the dividing both sides by 49 gives you>something like(5 b_1(x) + 1)(5 b_2(x) + 1)(5 c_3(x) + 22) = 300125 x^3 - 18375 x^2 - 360 x + 22where b_1(0) = b_2(0) = c_3(0) = 0,in the ring of algebraic integers, as the constant terms of the>factors go from being 7, 7 and 22 to being 1, 1, and 22.Now some posters have repeatedly attempted to question considering the>value of the constant terms as if the constants can't tell you>anything here.That I think requires readers considering their position forget that>the constants don't vary, so like with x=2:(5 a_1(2) + 7)(5 a_2(2) + 7)(5 b_3(2) + 22) = 49(300125 2^3 - 18375 2^2 - 360 (2) + 22)you can still *see* 7, 7 and 22, despite the fact that x=2. Granted>that if you evaluate you no longer *see* them.My point, of course, is that 7, 7 and 22 are there without regard to>x's value, as they are constants.However, if you divide both sides by 49, you ?d that your constants>then are 1, 1, and 22.Now then, how do you get from 7 to 1?Of course, you divide by 7.>Not necessarily, and this is where you make your mistake.That's not possible because 7, 7 and 22 are *constants*, and the> constant 49 is what's being divided from both sides. The point here that might be confusing to some is that 49 is *itself*>constant, so it and its factors behave as constants.>>Not necessarily. In fact, its factors might well differ (indeed,>>they *do* differ) depending on the value of x. As I'll show>>tomorrow (got to rush tonight--I have a party to go to),>>the a(x) functions might very well dictate a splitting>>of 49 that's not 7 and 7.>>Well then you could have replied when you had time, rather than making> a claim without defending it.>Heh. Well, I'm back, as promised. BTW, it was a fun party. I hope> the rest of the sci.math population had a good time too.Ok.> Let's parallel your argument in a simpler case.Why am I not surprised?> Let> Q(x) = 7(25x^2 + 30x + 2) [1]> = 7(x^2 + x)(5^2) + 7(x - 1)(5) + 7^2and write Q(x) = (5a_1(x) + 7)(5a_2(x) + 7) [2] Now we mirror your construction and observe that we may take> a_1(x) and a_2(x) to be roots of r(x) = a^2 - (x - 1)a + 7(x^2 + x) [3]We see that Q(0) = 14 = (7)(2), and r(0) = a^2 + a has roots> a_1(0) = 0, a_2(0) = -1. We see that this assignment is> consistent with your factorization [2], since Q(0) = (5(0) + 7)(5(-1) + 7) = (7)(2) [4]Hold on a bit, as you're rushing. At this point you have:(5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2)where the a's are the roots ofa^2 - (x - 1)a + 7(x^2 + x).You note that at x=0, you have a^2 + a = 0, which gives one of the a'sequal to 0, with one equal to -1. You pick a_1(0) = 0.So then(5a_1(0) + 7)(5a_2(0) + 7) = 7(25(0)^2 + 30(0) + 2)gives 7(2) = 7(2).The constant terms then are 7 and 2. and we see that we can divide [4] by 7 and write it as Q(0)/7 = (5(a_1(0)/7) + 1)(5a_2(0) + 7) = (1)(2)or, emphasizing the constant term in [1] we may write,> letting b_2(0) = a_2(0) + 1, Q(0)/7 = (5(a_1(0)/7) + 1)(5b_2(0) + 2) = (1)(2)So, answering the equivalent of your question, how> do you get from 7 to 1? we see that in the x = 0> case, it is exactly by dividing one factor by 7.But what happens if we have x = 1? Well, in this> case, we see that Q(1) = 7(25 + 30 + 2) = (7)(57)and it's easy to see that a_1(1) = sqrt(-14)> a_2(1) = -sqrt(-14)which yields the factorization Q(1) = (5(sqrt(-14)) + 7)(5(-sqrt(-14)) + 7)> = 7(57)excatly as it should. Now, however, we can't> divide by 7, since neither of the two factors> above are divisible by 7. They are, howeverThat looks ok to me. > both divisible by sqrt(7), so we have> the factorization (in algebraic integers) Q(1)/7 = [(5 sqrt(-14) + 7)/sqrt(7)] *> [(-5 sqrt(-14) + 7)/sqrt(7)]> In fact, the condition that exactly one of> the factors (or 2, in your example) is divisible> by 7 holds only for a small set of x values.> For even more fun, try this with x = 2.> Why? Now then, given the time you spent here Rick Decker, can youexplain why you think it matters?That is, what do *you* think is the relevance of your example?James Harris === snip>Let>> Q(x) = 7(25x^2 + 30x + 2) [1]>> = 7(x^2 + x)(5^2) + 7(x - 1)(5) + 7^2and write Q(x) = (5a_1(x) + 7)(5a_2(x) + 7) [2]Now we mirror your construction and observe that we may take>>a_1(x) and a_2(x) to be roots of r(x) = a^2 - (x - 1)a + 7(x^2 + x) [3]We see that Q(0) = 14 = (7)(2), and r(0) = a^2 + a has roots>>a_1(0) = 0, a_2(0) = -1. We see that this assignment is>>consistent with your factorization [2], since Q(0) = (5(0) + 7)(5(-1) + 7) = (7)(2) [4]>>Hold on a bit, as you're rushing. At this point you have:(5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2)where the a's are the roots ofa^2 - (x - 1)a + 7(x^2 + x).You note that at x=0, you have a^2 + a = 0, which gives one of the a's> equal to 0, with one equal to -1. You pick a_1(0) = 0.So then(5a_1(0) + 7)(5a_2(0) + 7) = 7(25(0)^2 + 30(0) + 2)gives 7(2) = 7(2).The constant terms then are 7 and 2. >>and we see that we can divide [4] by 7 and write it as Q(0)/7 = (5(a_1(0)/7) + 1)(5a_2(0) + 7) = (1)(2)or, emphasizing the constant term in [1] we may write,>>letting b_2(0) = a_2(0) + 1, Q(0)/7 = (5(a_1(0)/7) + 1)(5b_2(0) + 2) = (1)(2)So, answering the equivalent of your question, how>>do you get from 7 to 1? we see that in the x = 0>>case, it is exactly by dividing one factor by 7.But what happens if we have x = 1? Well, in this>>case, we see that Q(1) = 7(25 + 30 + 2) = (7)(57)and it's easy to see that a_1(1) = sqrt(-14)>> a_2(1) = -sqrt(-14)which yields the factorization Q(1) = (5(sqrt(-14)) + 7)(5(-sqrt(-14)) + 7)>> = 7(57)excatly as it should. Now, however, we can't>>divide by 7, since neither of the two factors>>above are divisible by 7. They are, however>>That looks ok to me. >>both divisible by sqrt(7), so we have>>the factorization (in algebraic integers) Q(1)/7 = [(5 sqrt(-14) + 7)/sqrt(7)] *>> [(-5 sqrt(-14) + 7)/sqrt(7)]>>In fact, the condition that exactly one of>>the factors (or 2, in your example) is divisible>>by 7 holds only for a small set of x values.>>For even more fun, try this with x = 2.Why? Now then, given the time you spent here Rick Decker, can you> explain why you think it matters?That is, what do *you* think is the relevance of your example?> You have a polynomial P(x) = 49(something) which you write as P(x) = (5a_1(x) + 7)(5a_2(x) + 7)(5b_3(x) + 22)with a_1(0) = 0, a_2(0) = 0, b_3(0) = 0. That's okay, butthen you want to deduce from the constants 7, 7, and 22that the only way to express P(x)/49 is by dividing the?st two factors by 7.My example showed that this conclusion is not justi?din general. In simple terms, you can't draw anyconclusions from the x = 0 case that can be used whenx is not zero. Even simpler, you're being misled byconcentrating on the constants. As has been pointedout before, your understanding of unique factorizationin the integers had led you to an incorrect conclusionin the realm of a different ring.Hope this helps,Rick === snip>Let>> Q(x) = 7(25x^2 + 30x + 2) [1]>> = 7(x^2 + x)(5^2) + 7(x - 1)(5) + 7^2and write Q(x) = (5a_1(x) + 7)(5a_2(x) + 7) [2]Now we mirror your construction and observe that we may take>>a_1(x) and a_2(x) to be roots of r(x) = a^2 - (x - 1)a + 7(x^2 + x) [3]We see that Q(0) = 14 = (7)(2), and r(0) = a^2 + a has roots>>a_1(0) = 0, a_2(0) = -1. We see that this assignment is>>consistent with your factorization [2], since Q(0) = (5(0) + 7)(5(-1) + 7) = (7)(2) [4]>>Hold on a bit, as you're rushing. At this point you have:(5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2)where the a's are the roots ofa^2 - (x - 1)a + 7(x^2 + x).You note that at x=0, you have a^2 + a = 0, which gives one of the a's> equal to 0, with one equal to -1. You pick a_1(0) = 0.So then(5a_1(0) + 7)(5a_2(0) + 7) = 7(25(0)^2 + 30(0) + 2)gives 7(2) = 7(2).The constant terms then are 7 and 2. >>and we see that we can divide [4] by 7 and write it as Q(0)/7 = (5(a_1(0)/7) + 1)(5a_2(0) + 7) = (1)(2)or, emphasizing the constant term in [1] we may write,>>letting b_2(0) = a_2(0) + 1, Q(0)/7 = (5(a_1(0)/7) + 1)(5b_2(0) + 2) = (1)(2)So, answering the equivalent of your question, how>>do you get from 7 to 1? we see that in the x = 0>>case, it is exactly by dividing one factor by 7.But what happens if we have x = 1? Well, in this>>case, we see that Q(1) = 7(25 + 30 + 2) = (7)(57)and it's easy to see that a_1(1) = sqrt(-14)>> a_2(1) = -sqrt(-14)which yields the factorization Q(1) = (5(sqrt(-14)) + 7)(5(-sqrt(-14)) + 7)>> = 7(57)excatly as it should. Now, however, we can't>>divide by 7, since neither of the two factors>>above are divisible by 7. They are, however>>That looks ok to me. >>both divisible by sqrt(7), so we have>>the factorization (in algebraic integers) Q(1)/7 = [(5 sqrt(-14) + 7)/sqrt(7)] *>> [(-5 sqrt(-14) + 7)/sqrt(7)]>>In fact, the condition that exactly one of>>the factors (or 2, in your example) is divisible>>by 7 holds only for a small set of x values.>>For even more fun, try this with x = 2.Why? Now then, given the time you spent here Rick Decker, can you> explain why you think it matters?That is, what do *you* think is the relevance of your example?> You have a polynomial P(x) = 49(something) which you write as P(x) = (5a_1(x) + 7)(5a_2(x) + 7)(5b_3(x) + 22)with a_1(0) = 0, a_2(0) = 0, b_3(0) = 0. That's okay, but> then you want to deduce from the constants 7, 7, and 22> that the only way to express P(x)/49 is by dividing the> ?st two factors by 7.My example showed that this conclusion is not justi?d> in general. In simple terms, you can't draw any> conclusions from the x = 0 case that can be used when> x is not zero. Even simpler, you're being misled by> concentrating on the constants. As has been pointed> out before, your understanding of unique factorization> in the integers had led you to an incorrect conclusion> in the realm of a different ring.> But Decker your example doesn't show that at all, but I appreciate youcoming forward to explain your thinking.You've made an obvious mistake by going to what you think is aparallel example, but the mathematics here is subtle enough for youthat you failed to consider *how* I got my own factorization.Now then, can you explain how you got the factorization in yourexample?You probably are aware to some extent how I obtained mine, so youshould be ahead of others on the newsgroup!!!But I *do* like that you presented a quadratic example as you did tothe extent that I've created two other threads to talk about it!I suggest readers look at those other threads, especially at Decker'srecent posts in one of them.James Harris === A discovery of mine is that (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = 49(300125 x^3 - 18375 x^2 - 360 x + 22)where b_3(x) = a_3(x) - 3 and the a's are roots ofa^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)and when x=0, a_1(0) = a_2(0) = b_3(0) = 0.Attempting to use those relations in the ring of algebraic integers I> can demonstrate a problem when I consider what happens if you divide> both sides by 49.If you *assume* that the ring of algebraic integers is ?e, it's> reasonable to conclude the dividing both sides by 49 gives you> something like(5 b_1(x) + 1)(5 b_2(x) + 1)(5 c_3(x) + 22) = 300125 x^3 - 18375 x^2 - 360 x + 22where b_1(0) = b_2(0) = c_3(0) = 0,in the ring of algebraic integers, as the constant terms of the> factors go from being 7, 7 and 22 to being 1, 1, and 22.> One must bear in mind that dividing by a non-unit can lead out of a> ring, into the fraction ?ld. So it is not safe to assue that , for> instance, (5a_1(x)+7)/7 is an algebraic integer. For some values of x> it might well be; but this takes some invsetigation.It's not an algebraic integer in general as that's the point.What's key here though is that I have49(300125 x^3 - 18375 x^2 - 360 x + 22)which *clearly* has 49 as a factor *in the ring of algebraicintegers*, and I have a factorization that is also in the ring, whichis (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = 49(300125 x^3 - 18375 x^2 - 360 x + 22) where b_3(x) = a_3(x) - 3 and the a's are roots of a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) and when x=0, a_1(0) = a_2(0) = b_3(0) = 0.But, if I divide both sides by 49, I'm no longer in general in thering of algebraic integers proving there's a more inclusive ring.> Now some posters have repeatedly attempted to question considering the> value of the constant terms as if the constants can't tell you> anything here.That I think requires readers considering their position forget that> the constants don't vary, so like with x=2:(5 a_1(2) + 7)(5 a_2(2) + 7)(5 b_3(2) + 22) = 49(300125 2^3 - 18375 2^2 - 360 (2) + 22)you can still *see* 7, 7 and 22, despite the fact that x=2. Granted> that if you evaluate you no longer *see* them.My point, of course, is that 7, 7 and 22 are there without regard to> x's value, as they are constants.> They are there in your formula, but the signi?ance of this fact is> unclear. It does not seem to me to point out any dif?ulty with the> usual concept of the ring of algebraic integers.The signi?ance from the distributive property is that when 49 isdivided from both sides, the only way for the constant terms to gofrom being 7, 7 and 22 to being 1, 1, and 22 is for two of the factorsto be divided by 7, as the distributive property still applies!However, if you divide both sides by 49, you ?d that your constants> then are 1, 1, and 22.Now then, how do you get from 7 to 1?Of course, you divide by 7.The point here that might be confusing to some is that 49 is *itself*> constant, so it and its factors behave as constants.Some of you may be confused by knowing that with, say, (x+7)(x+1) = x^2 + 8x + 7you have constant terms 7, and 1, but dividing by x+7 gives you a> constant term of 1, so you may be puzzled into believing that somehow> 49 can factor in some odd way to do something like that, but remember> 49 is NOT a polynomial; it's a constant.Now then, as to the conclusions that derive from accepting that 49 is> a constant, and 7, 7 and 22 cannot change dependent on x, so dividing> by 49, in giving 1, 1 and 22 as constants tells you how 49 divided> both sides.With(5 b_1(x) + 1)(5 b_2(x) + 1)(5 c_3(x) + 22) = 300125 x^3 - 18375 x^2 - 360 x + 22where b_1(0) = b_2(0) = c_3(0) = 0,the result you'll ?d in general is that you can no longer be in the> ring of algebraic integers.> Thus we have three algebraic numbers (in general, not integral) whose> product is an algebraic integer. This seems OK to me. No, you're wrong as in general, they're NOT algebraic integers. Ifyou disagree then I challenge you to prove they are!Here's where posters trying to debate the issue can try to *produce*the functions, if they wish to believe they can be algebraic integersin general.Some of you may have noticed that I didn't even try, though before Igive that the a's are roots of a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x).The reality is that *no one* can give b_1(x), b_2(x), and c_3(x), ingeneral, though a_1(x), a_2(x), and a_3(x) CAN be given, and that isbecause they're not in general in the ring of algebraic integers.James Harris === A discovery of mine is that > (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = > 49(300125 x^3 - 18375 x^2 - 360 x + 22)> where b_3(x) = a_3(x) - 3 and the a's are roots of> a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)> and when x=0, a_1(0) = a_2(0) = b_3(0) = 0.> Attempting to use those relations in the ring of algebraic integers I> can demonstrate a problem when I consider what happens if you divide> both sides by 49.> If you *assume* that the ring of algebraic integers is ?e, it's> reasonable to conclude the dividing both sides by 49 gives you> something like> (5 b_1(x) + 1)(5 b_2(x) + 1)(5 c_3(x) + 22) = > 300125 x^3 - 18375 x^2 - 360 x + 22> where b_1(0) = b_2(0) = c_3(0) = 0,> in the ring of algebraic integers, as the constant terms of the> factors go from being 7, 7 and 22 to being 1, 1, and 22.> One must bear in mind that dividing by a non-unit can lead out of a> ring, into the fraction ?ld. So it is not safe to assue that , for> instance, (5a_1(x)+7)/7 is an algebraic integer. For some values of x> it might well be; but this takes some invsetigation.It's not an algebraic integer in general as that's the point.What's key here though is that I have49(300125 x^3 - 18375 x^2 - 360 x + 22)which *clearly* has 49 as a factor *in the ring of algebraic> integers*, and I have a factorization that is also in the ring, which> is (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = > 49(300125 x^3 - 18375 x^2 - 360 x + 22)> where b_3(x) = a_3(x) - 3 and the a's are roots of> a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)> and when x=0, a_1(0) = a_2(0) = b_3(0) = 0.But, if I divide both sides by 49, I'm no longer in general in the> ring of algebraic integers proving there's a more inclusive ring.> Now some posters have repeatedly attempted to question considering the> value of the constant terms as if the constants can't tell you> anything here.> That I think requires readers considering their position forget that> the constants don't vary, so like with x=2:> (5 a_1(2) + 7)(5 a_2(2) + 7)(5 b_3(2) + 22) = > 49(300125 2^3 - 18375 2^2 - 360 (2) + 22)> you can still *see* 7, 7 and 22, despite the fact that x=2. Granted> that if you evaluate you no longer *see* them.> My point, of course, is that 7, 7 and 22 are there without regard to> x's value, as they are constants.> They are there in your formula, but the signi?ance of this fact is> unclear. It does not seem to me to point out any dif?ulty with the> usual concept of the ring of algebraic integers.The signi?ance from the distributive property is that when 49 is> divided from both sides, the only way for the constant terms to go> from being 7, 7 and 22 to being 1, 1, and 22 is for two of the factors> to be divided by 7, as the distributive property still applies!> Yes - right there is where you are making your error. Youthink the *only possible* way to factor 49 out of (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22)is to divide 7, 7, and 1 out of each of the factors. That isnot true. Perhaps you think that you cannot divide something like (5 b_3(x) + 22) by a factor of 7 because 22 is coprimeto 7. That logic however is wrong. Do you see why? Do youneed an example? Perhaps you think the factorization must be such thatthe constant terms are preserved. This is true onlyin the sense that, after the division, the product of the constant terms must be 22. There are in?itely many waysfor this to occur. 7, 7, 1 is only one of them, and ithappens to be one for which the other divided terms, likea_1(x)/7, are not algebraic integers. The constant terms donot have magical or sacred properties as you seem to believe. They are just numbers. All that is required is that, for eachx, the arithmetic balances. There is another way to factor 49 within the algebraicintegers as a product of 3 numbers: 49 = w1(x) * w2(x) * w3(x),which is such that w1, w2, and w3 are all algebraic integers and a_1(x)/w1(x) and a_2(x)/w2(x) and (5*b_3(x) + 22) / w3(x) are all algebraic integers.And no, in general, 22/w3(x) is NOT an algebraicinteger. There is no reason it should be. Nor isit constant! It, like w_3(x) and b_3(x), is *dependent on x*.Remember, all that matters is that after the divisionoccurs, the arithmetic comes out right, for *each value of x*. It need not come out the *same* way foreach value of x; it only needs to come out right. Thetwo sides of the equation, considered as individual numbers, must balance. The constant term on the right,22, must equal (7*7*22)/(w1(x)*w2(x)*w3(x)). Noting that 49 = w1(x)*w2(x)*w3(x), this is obviously true. That is,the constant term, 22, is the product of three algebraic integerterms that individually are *not* constant. I know this sounds like anathema to you, but try proving that it cannot happen!> However, if you divide both sides by 49, you ?d that your constants> then are 1, 1, and 22.> Now then, how do you get from 7 to 1?> Of course, you divide by 7.> The point here that might be confusing to some is that 49 is *itself*> constant, so it and its factors behave as constants.> Some of you may be confused by knowing that with, say, > (x+7)(x+1) = x^2 + 8x + 7> you have constant terms 7, and 1, but dividing by x+7 gives you a> constant term of 1, so you may be puzzled into believing that somehow> 49 can factor in some odd way to do something like that, but remember> 49 is NOT a polynomial; it's a constant.> Now then, as to the conclusions that derive from accepting that 49 is> a constant, and 7, 7 and 22 cannot change dependent on x, so dividing> by 49, in giving 1, 1 and 22 as constants tells you how 49 divided> both sides.> With> (5 b_1(x) + 1)(5 b_2(x) + 1)(5 c_3(x) + 22) = > 300125 x^3 - 18375 x^2 - 360 x + 22> where b_1(0) = b_2(0) = c_3(0) = 0,> the result you'll ?d in general is that you can no longer be in the> ring of algebraic integers.> Thus we have three algebraic numbers (in general, not integral) whose> product is an algebraic integer. This seems OK to me. No, you're wrong as in general, they're NOT algebraic integers. If> you disagree then I challenge you to prove they are!> He *just said* they are algebraic numbers but not algebraicintegers. You are just automatically disagreeing without readingwhat was said. You are so hostile to what people are saying thatyou cannot think straight.> Here's where posters trying to debate the issue can try to *produce*> the functions, if they wish to believe they can be algebraic integers> in general.> You're mixed up. We *agree* with you that a_1(x)/7 is not algebraicinteger. We have been saying that for months and months. That fact should tell you that you have chosen the wrong factorization of 49,and that there has to be another factorization which does work. But no. You jump to a different conclusion. You think that because 7, 7, 1 leads to a contradiction, there must be anerror in Mathematics - the Mathematics of Gauss, Dedekind, You arrive a contradiction. Most people would concludethey have made an error or a wrong assumption. You don't.You conclude that Mathematics itself is wrong! Even when it ispointed out exactly why your assumption is wrong and whatthe right factorization is, you persist! Wake up, man! The error is yours! 7, 7, 1 is the WRONGFACTORIZATION (unless x = 0). There is a correct one as speci?d by Dik and others previously (i.e., w1(x) = GCD(a_1(x), 49, etc.). The factors are not constants; they are dependent on x. This shouldnot be a surprise because a_1(x), etc., as you know, are dependent on x. None of the factors are units and none of them equal 7. Think of it like this. You tried 7, 7, 1 and it led to a contradiction. We all agree that some factorization has to work. The one you have chosen, 7,7,1, does not. The only possibility left, really, is that the factorization is of the form speci?d by Dik -49 is the product of three numbers which are dependent on x; in generalnone are equal to 7 and none are units. Unlikely though thismay seem to you, it is the *only possibility left*. You very de?itelyhave not proved that it cannot happen. It has to be true! When you have eliminated all the other possibilities, the only one that isleft has to be correct! You have far too much invested in this. You have lost allobjectivity, not to mention humility. I am not sure at this point that you are capable of coming to your senses. Perhaps not. Perhaps you never will. At least make an honest effort to read what other people write. Nora B.> Some of you may have noticed that I didn't even try, though before I> give that the a's are roots of> a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x).The reality is that *no one* can give b_1(x), b_2(x), and c_3(x), in> general, though a_1(x), a_2(x), and a_3(x) CAN be given, and that is> because they're not in general in the ring of algebraic integers.> James Harris === deleted The signi?ance from the distributive property is that when 49 is> divided from both sides, the only way for the constant terms to go> from being 7, 7 and 22 to being 1, 1, and 22 is for two of the factors> to be divided by 7, as the distributive property still applies!> Yes - right there is where you are making your error. You> think the *only possible* way to factor 49 out of (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22)is to divide 7, 7, and 1 out of each of the factors. That is> not true. Perhaps you think that you cannot divide something > like (5 b_3(x) + 22) by a factor of 7 because 22 is coprime> to 7. That logic however is wrong. Do you see why? Do you> need an example?The constant terms are 7, 7 and 22, but when 49 is divided off theresulting constant terms are 1, 1, and 22, which means from BASICarithmetic, there's only one way to go.There is, in fact, only one way that can happen. I focus on the constant terms--7, 7, and 22--because they're numbers,and therefore the value of x doesn't matter to them.James Harris === =... > I focus on the constant terms--7, 7, and 22--because they're numbers, > and therefore the value of x doesn't matter to them.What is the matter with you? I have de?ed w3(x) before; what is theconstant term of: (b3(x) + 22)/w3(x)?Is it or is it not 22? If it is not, show why (w3(0) = 1). I havede?ed w2(x) before; what is the constant term of: (a2(x) + 7)/w2(x)?Is it or is it not 1? If it is not, show why (w2(0) = 7). I havede?ed w1(x) before; what is the constant term of: (a1(x) + 7)/w1(x)?Is it or is it not 1? If it is not, show why (w1(0) = 7).Moreover, please show in what way the above three functions are *not*functions yielding algebraic integer when x is an algebraic integer.Go to the de?itions, and show what part is wrong.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === divided from both sides, the only way for the constant terms to go> from being 7, 7 and 22 to being 1, 1, and 22 is for two of the factors> to be divided by 7, as the distributive property still applies!> Yes - right there is where you are making your error. You> think the *only possible* way to factor 49 out of (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22)is to divide 7, 7, and 1 out of each of the factors. That is> not true. Perhaps you think that you cannot divide something > like (5 b_3(x) + 22) by a factor of 7 because 22 is coprime> to 7. That logic however is wrong. Do you see why? Do you> need an example?The constant terms are 7, 7 and 22, but when 49 is divided off the> resulting constant terms are 1, 1, and 22, which means from BASIC> arithmetic, there's only one way to go.There is, in fact, only one way that can happen. > Let u, v, and w be any algebraic integers such that u*v*w = 49.There are in?itely many such triples. The constant termsafter factoring out u, v, and w are 7/u, 7/v, and 22/w. Now you may say, not fair, 22/w is not an algebraic integer.Absolutely correct, unless w is a unit. But that DOESN'T MATTER. What matters is whether (5 b_3(x) + 22) / wis an algebraic integer. The fact is, for any x, u, v, andw can be chosen so that (5 b_3(x) + 22)/w is an algebraicinteger, and (most importantly) if x <> 0 then w <> 1. Why can't you understand this?> I focus on the constant terms--7, 7, and 22--because they're numbers,> and therefore the value of x doesn't matter to them.> Whether the value of x doesn't matter to them or not isirrelevant. Their feelings are not mathematical entitites. For any given x, (5 b_3(x) + 22) is also a number. It is its divisibility of the *whole expresssion* which is the key here, not the divisibility of 22. You really are not getting this. Moreover, you deletedall the rest of my post. You must not want other people to read it. So much for Mathematical consistency, courage. I am appending it below. You can have a second chance at understanding it. Here is another way to view this. You say there is onlyone way to factor 49 which makes sense. Dik and othershere say there is another. It is de?ed by: w1(x) = GCD(a_1(x), 49) w2(x) = GCD(a_2(x), 49) w3(x) = GCD(a_3(x), 49),where a_1(x), a_2(x), and a_3(x) are the roots of yourauxiliary polynomial, a^3 + 3*(-1 + 49*x)*a^2 - 49*(2401*x^3 - 147*x^2 + 3*x). Note that w1, w2, and w3 are perfectly well-de?ed: the roots a_1, a_2, and a_3 exist and can be computed, and the GCD function exists (by a deep theorem of Dedekind) and can be computed. Thus two ways to factor 49: 7, 7, 1 and w1(x), w2(x), w3(x).The question here is NOT Why does 7, 7, 1 lead to a problem?We have proved, and you accept, that it does: a_1(x)/7 isnot an algebraic integer. No disagreement on that. The question here is, What is wrong with Dik's factorization,for which a_1(x)/w1(x) *is* an algebraic integer? Try answering that for once rather than repeating yourusual memorized guff. Nora B. === === == (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22)is to divide 7, 7, and 1 out of each of the factors. That is> not true. Perhaps you think that you cannot divide something > like (5 b_3(x) + 22) by a factor of 7 because 22 is coprime> to 7. That logic however is wrong. Do you see why? Do you> need an example? Perhaps you think the factorization must be such that> the constant terms are preserved. This is true only> in the sense that, after the division, the product of > the constant terms must be 22. There are in?itely many ways> for this to occur. 7, 7, 1 is only one of them, and it> happens to be one for which the other divided terms, like> a_1(x)/7, are not algebraic integers. The constant terms do> not have magical or sacred properties as you seem to believe. > They are just numbers. All that is required is that, for each> x, the arithmetic balances. There is another way to factor 49 within the algebraic> integers as a product of 3 numbers: 49 = w1(x) * w2(x) * w3(x),which is such that w1, w2, and w3 are all algebraic > integers and a_1(x)/w1(x) and a_2(x)/w2(x) and > (5*b_3(x) + 22) / w3(x) are all algebraic integers.> And no, in general, 22/w3(x) is NOT an algebraic> integer. There is no reason it should be. Nor is> it constant! It, like w_3(x) and b_3(x), is *dependent on x*.> Remember, all that matters is that after the division> occurs, the arithmetic comes out right, for *each > value of x*. It need not come out the *same* way for> each value of x; it only needs to come out right. The> two sides of the equation, considered as individual > numbers, must balance. The constant term on the right,> 22, must equal (7*7*22)/(w1(x)*w2(x)*w3(x)). Noting that > 49 = w1(x)*w2(x)*w3(x), this is obviously true. That is,> the constant term, 22, is the product of three algebraic integer> terms that individually are *not* constant. I know this sounds > like anathema to you, but try proving that it cannot happen!> However, if you divide both sides by 49, you ?d that your constants> then are 1, 1, and 22.> Now then, how do you get from 7 to 1?> Of course, you divide by 7.> The point here that might be confusing to some is that 49 is *itself*> constant, so it and its factors behave as constants.> Some of you may be confused by knowing that with, say, > (x+7)(x+1) = x^2 + 8x + 7> you have constant terms 7, and 1, but dividing by x+7 gives you a> constant term of 1, so you may be puzzled into believing that somehow> 49 can factor in some odd way to do something like that, but remember> 49 is NOT a polynomial; it's a constant.> Now then, as to the conclusions that derive from accepting that 49 is> a constant, and 7, 7 and 22 cannot change dependent on x, so dividing> by 49, in giving 1, 1 and 22 as constants tells you how 49 divided> both sides.> With> (5 b_1(x) + 1)(5 b_2(x) + 1)(5 c_3(x) + 22) = > 300125 x^3 - 18375 x^2 - 360 x + 22> where b_1(0) = b_2(0) = c_3(0) = 0,> the result you'll ?d in general is that you can no longer be in the> ring of algebraic integers.> Thus we have three algebraic numbers (in general, not integral) whose> product is an algebraic integer. This seems OK to me. No, you're wrong as in general, they're NOT algebraic integers. If> you disagree then I challenge you to prove they are!> He *just said* they are algebraic numbers but not algebraic> integers. You are just automatically disagreeing without reading> what was said. You are so hostile to what people are saying that> you cannot think straight.> Here's where posters trying to debate the issue can try to *produce*> the functions, if they wish to believe they can be algebraic integers> in general.> You're mixed up. We *agree* with you that a_1(x)/7 is not algebraic> integer. We have been saying that for months and months. That fact > should tell you that you have chosen the wrong factorization of 49,> and that there has to be another factorization which does work. But no. You jump to a different conclusion. You think that > because 7, 7, 1 leads to a contradiction, there must be an> error in Mathematics - the Mathematics of Gauss, Dedekind, You arrive a contradiction. Most people would conclude> they have made an error or a wrong assumption. You don't.> You conclude that Mathematics itself is wrong! Even when it is> pointed out exactly why your assumption is wrong and what> the right factorization is, you persist! > Wake up, man! The error is yours! 7, 7, 1 is the WRONG> FACTORIZATION (unless x = 0). There is a correct one as speci?d by > Dik and others previously (i.e., w1(x) = GCD(a_1(x), 49, etc.). The > factors are not constants; they are dependent on x. This should> not be a surprise because a_1(x), etc., as you know, are dependent > on x. None of the factors are units and none of them equal 7. Think of it like this. You tried 7, 7, 1 and it led to a > contradiction. We all agree that some factorization has to > work. The one you have chosen, 7,7,1, does not. The only possibility > left, really, is that the factorization is of the form speci?d by Dik -> 49 is the product of three numbers which are dependent on x; in general> none are equal to 7 and none are units. Unlikely though this> may seem to you, it is the *only possibility left*. You very de?itely> have not proved that it cannot happen. It has to be true! When you have > eliminated all the other possibilities, the only one that is> left has to be correct! > You have far too much invested in this. You have lost all> objectivity, not to mention humility. I am not sure at this point > that you are capable of coming to your senses. Perhaps not. Perhaps > you never will. At least make an honest effort to read what other > people write. Nora B.> Some of you may have noticed that I didn't even try, though before I> give that the a's are roots of> a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x).The reality is that *no one* can give b_1(x), b_2(x), and c_3(x), in> general, though a_1(x), a_2(x), and a_3(x) CAN be given, and that is> because they're not in general in the ring of algebraic integers.> James HarrisJames Harris === =... > Here is another way to view this. You say there is only > one way to factor 49 which makes sense. Dik and others > here say there is another. It is de?ed by: > w1(x) = GCD(a_1(x), 49) > w2(x) = GCD(a_2(x), 49) > w3(x) = GCD(a_3(x), 49),You must be careful. Strange enough, but with this de?ition it is notcertain that w1(x)*w2(x)*w3(x) = 49. My latest de?ition is a bit moreelaborate, but I think it is fool-proof (though probably not James-proof): v1(x) = GCD(a_1(x) + 7, 49) v2(x) = GCD(a_2(x) + 7, 49) v3(x) = GCD(a_3(x) + 7, 49) k3(x) = v1(x)*v2(x)*v3(x) ; can be a multiple of 49 g(x) = k3(x) / 49 ; the excess, must be distributed k2(x) = GCD(v2(x), g(x)) ; the part in v2 of the excess k1(x) = g(x) / k2(x) ; and the remainder in v1. z3(x) = v3(x) ; this one is plain w2(x) = v2(x) / k2(x) ; part of the excess removed w1(x) = v1(x) / k1(x) ; the remaining excess removed u(x) = z1(x)*z2(x)*z3(x)/49 ; must be a unit w3(x) = z3(x) / u(x) ; force it off.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === >Mathematics is about logic and consistency, which *should* mean that a>person can present a mathematical argument, and deal with objective>criticisms or concerns and there be resolution. However, I'm dealing>with a situation where as a body mathematicians are trying to avoid>the implications of a rather basic mathematical argument, which I will>present yet again:[...]Ah, I guess that David Ullrich deletes out the content as a sign ofdisrespect.But then, why does he pay attention to my posts at all, as that is asign of respect?Attempting to use those relations in the ring of algebraic integers I>can demonstrate a problem when I consider what happens if you divide>both sides by 49.If you *assume* that the ring of algebraic integers is ?e, it's>reasonable to conclude the dividing both sides by 49 gives you>something like[...]the result you'll ?d in general is that you can no longer be in the>ring of algebraic integers.The result is simple enough, and direct enough that I'd hope some of>you might wonder why there has been such argument over it.Well, the problem I think is that mathematicians thought one thing,>but now I've shown another, and it's easier for many of them to just>hold on to what they believed, even though it's mathematically shown>now to be incorrect.Uh, no. A proof that some bit of mathematics is incorrect does not> include phrases like the ring of algebraic integers is ?e unless> it also includes a _de?ition_ of what it means for a ring to be> ?e. It does not include phrases like it's reasonable to assume> that.Well I've looked that statement over a couple of times, and it doesn'tmake much sense to me.Possibly someone else could interject here and explain it?> You're not demonstrating any problem with the math, all you're> demonstrating is that it doesn't work the way you think it should.> If you want to demonstrate that there's a problem with the> math then showing that something that should (according> to you) be an algebraic integer is not is not going to cut it.> The _only_ way to show that there's a problem with the> math is to _show_ that something _is_ an algebraic integer,> and then _show_ that the same thing is _not_ an algebraic> integer.> Huh?>[...]I'm the one who's paused as another human being harmed by their>actions. For heaven's sake, if you don't like it here then go away. People> are not going to start agreeing you're right when you're not just> because you whine about how you're harmed by people's> comments.Mathematicians seem now to think that they can switch to a societythat *tells* itself what truth is, which might seem like a nice way tohandle outsiders like myself who ?d truths that they might ?dunpleasant.However, hurting me by refusing to acknowledge mathematical truthdoesn't change the truth.To the extent that my work shows errors in thinking about algebraicintegers, those errors have existed as long as that *thinking* hasexisted, in this case, over a hundred years.There's a right way to think of algebraic integers, and theirproperties, and a wrong way. >My life is affected, but you see, they've *always* been>wrong about the ring of algebraic integers.That fact hasn't changed. They were just always wrong. And hurting>me, doesn't change that fact.What's amazing here is how *easily* I can prove that I'm right, andbring it down to acceptance of constant terms, like 7, 7 and 22,actually being NUMBERS, versus variables dependent on x, andacceptance of the distributive property!!!The point is that I'm not the bad guy here, though some might want toattack the messenger.The reality is that if something is mathematically correct; it'scorrect.And if wrong, it's wrong.Mathematicians have just been wrong about algebraic integers. It'snot complicated, and refusing to acknowledge the truth doesn't changeit.Mathematicians are just still wrong, until they accept mathematics.James Harris <3c65f87.0312311559.6b8f62e4@posting.google.com> === But then, why does he pay attention to my posts at all, as that is a> sign of respect?What ever gave you that idea?-- Even if [...] a communistic regime should come [to China], the oldtradition [...] will break Communism and change it beyond recognition,rather than Communism [...] break the old tradition. It must be so. -- Lin Yutang on Socialism with Chinese characteristics in 1935 === =But then, why does he pay attention to my posts at all, as that is a> sign of respect?What ever gave you that idea?Respect comes from Latin. It might help you to look up the word's etymology.James Harris <3c65f87.0312311559.6b8f62e4@posting.google.com> <87y8srwyrx.fsf@phiwumbda.org> <3c65f87.0401011226.74040ca6@posting.google.com> === >> But then, why does he pay attention to my posts at all, as that is a>> sign of respect?>>> What ever gave you that idea? Respect comes from Latin. It might help you to look up the word's etymology.Why? I'm interested in the term as it is used today, and why youthink responses to your post are a sign of respect -- in the common,everyday meaning of the word respect.Etymology doesn't help answer that question.-- Jesse F. HughesTruth is common stuff, ready to your hand, but lies you have to makeyourself, and you can't be sure they are any good until you'veused them and then it's too late. John Steinbeck === >>> But then, why does he pay attention to my posts at all, as that is a>> sign of respect?>>> What ever gave you that idea? Respect comes from Latin. It might help you to look up the word's etymology.Why? I'm interested in the term as it is used today, and why you> think responses to your post are a sign of respect -- in the common,> everyday meaning of the word respect.Etymology doesn't help answer that question.Then look up the *current* de?ition.Why don't you do that and report back?James Harris === Respect comes from Latin. It might help you to look up the word's > etymology.Why? I'm interested in the term as it is used today, and why you> think responses to your post are a sign of respect -- in the common,> everyday meaning of the word respect.Etymology doesn't help answer that question.> Then look up the *current* de?ition.Why don't you do that and report back?> James HarrisIt appears that James has now lost hhis dictionary, as well as his mind. === >>Mathematics is about logic and consistency, which *should* mean that a>>person can present a mathematical argument, and deal with objective>>criticisms or concerns and there be resolution. However, I'm dealing>>with a situation where as a body mathematicians are trying to avoid>>the implications of a rather basic mathematical argument, which I will>>present yet again:[...]Ah, I guess that David Ullrich deletes out the content as a sign of>disrespect.But then, why does he pay attention to my posts at all, as that is a>sign of respect?Giggle. People replying to your posts is not a sign of respect. If I _had_ deleted relevant content that would not be a signof disrespect. In fact I didn't delete anything relevant, Iincluded all the comments of yours that were relevant tothe point I wanted to make, and snipped the parts thathad no relevance to my comment to make the relevantparts easier to see.>>Attempting to use those relations in the ring of algebraic integers I>>can demonstrate a problem when I consider what happens if you divide>>both sides by 49.If you *assume* that the ring of algebraic integers is ?e, it's>>reasonable to conclude the dividing both sides by 49 gives you>>something like[...]the result you'll ?d in general is that you can no longer be in the>>ring of algebraic integers.The result is simple enough, and direct enough that I'd hope some of>>you might wonder why there has been such argument over it.Well, the problem I think is that mathematicians thought one thing,>>but now I've shown another, and it's easier for many of them to just>>hold on to what they believed, even though it's mathematically shown>>now to be incorrect.>>> Uh, no. A proof that some bit of mathematics is incorrect does not>> include phrases like the ring of algebraic integers is ?e unless>> it also includes a _de?ition_ of what it means for a ring to be>> ?e. It does not include phrases like it's reasonable to assume>> that.Well I've looked that statement over a couple of times, and it doesn't>make much sense to me.Of course it doesn't - it's an explanation of why something you saidwas wrong, and such explanations _never_ make sense to you.>Possibly someone else could interject here and explain it?> You're not demonstrating any problem with the math, all you're>> demonstrating is that it doesn't work the way you think it should.>> If you want to demonstrate that there's a problem with the>> math then showing that something that should (according>> to you) be an algebraic integer is not is not going to cut it.>> The _only_ way to show that there's a problem with the>> math is to _show_ that something _is_ an algebraic integer,>> and then _show_ that the same thing is _not_ an algebraic>> integer.>> Huh?A problem in math would be an _inconsistency_; a _contradiction.Here's an example of something that is _not_ a contradiction:It seems to me that a/7 should be an algebraic integer, but it'snot.That _non_-contradiction is all you're demonstrating here, andbecause it's not a contradiction it follows that you're not showingthere is any problem with the algebraic integers.On the other hand, here's an example of an actual contradiction:a/7 is an algebraic integer, and a/7 is not an algebraic integer.That's an example of what an actual contradiction looks like -_that_ is what you have to _prove_ in order to show thatthere is a problem with the algebraic integers.The _proof_ is not allowed to contain phrases like it'sreasonable to assume that. It's not allowed to containunde?ed words like complete ring.>>[...]I'm the one who's paused as another human being harmed by their>>actions. >>> For heaven's sake, if you don't like it here then go away. People>> are not going to start agreeing you're right when you're not just>> because you whine about how you're harmed by people's>> comments.Mathematicians seem now to think that they can switch to a society>that *tells* itself what truth is, which might seem like a nice way to>handle outsiders like myself who ?d truths that they might ?d>unpleasant.However, hurting me by refusing to acknowledge mathematical truth>doesn't change the truth.To the extent that my work shows errors in thinking about algebraic>integers, those errors have existed as long as that *thinking* has>existed, in this case, over a hundred years.There's a right way to think of algebraic integers, and their>properties, and a wrong way.>>My life is affected, but you see, they've *always* been>>wrong about the ring of algebraic integers.That fact hasn't changed. They were just always wrong. And hurting>>me, doesn't change that fact.What's amazing here is how *easily* I can prove that I'm right, and>bring it down to acceptance of constant terms, like 7, 7 and 22,>actually being NUMBERS, versus variables dependent on x, and>acceptance of the distributive property!!!The point is that I'm not the bad guy here, though some might want to>attack the messenger.The reality is that if something is mathematically correct; it's>correct.And if wrong, it's wrong.Mathematicians have just been wrong about algebraic integers. It's>not complicated, and refusing to acknowledge the truth doesn't change>it.Mathematicians are just still wrong, until they accept mathematics.>James Harris************************David C. Ullrich === >> >>Mathematics is about logic and consistency, which *should* mean that a>>person can present a mathematical argument, and deal with objective>>criticisms or concerns and there be resolution. However, I'm dealing>>with a situation where as a body mathematicians are trying to avoid>>the implications of a rather basic mathematical argument, which I will>>present yet again:[...]Ah, I guess that David Ullrich deletes out the content as a sign of>disrespect.But then, why does he pay attention to my posts at all, as that is a>sign of respect?Giggle. People replying to your posts is not a sign of respect. > If I _had_ deleted relevant content that would not be a sign> of disrespect. In fact I didn't delete anything relevant, I> included all the comments of yours that were relevant to> the point I wanted to make, and snipped the parts that> had no relevance to my comment to make the relevant> parts easier to see.> You sound like you need to look up the word respect, though I wonderif you care what a dictionary says, given your record.It's not my fault you spend so much time and attention on my posts, orthat giving your time and attention is a sign of respect.Now then, you can argue that giving time and attention is not a signof respect, but that's not how most people see it, as they highlyvalue their time and attention!!!For instance, consider *your* posts not made in my threads. Do yousee me giving them time or attention?James Harris === Mathematics is about logic and consistency, which *should* mean that a>person can present a mathematical argument, and deal with objective>criticisms or concerns and there be resolution. However, I'm dealing>with a situation where as a body mathematicians are trying to avoid>the implications of a rather basic mathematical argument, which I will>present yet again:>[...]Ah, I guess that David Ullrich deletes out the content as a sign of> disrespect.But then, why does he pay attention to my posts at all, as that is a> sign of respect? >Attempting to use those relations in the ring of algebraic integers I>can demonstrate a problem when I consider what happens if you divide>both sides by 49.>If you *assume* that the ring of algebraic integers is ?e, it's>reasonable to conclude the dividing both sides by 49 gives you>something like>[...]>the result you'll ?d in general is that you can no longer be in the>ring of algebraic integers.>The result is simple enough, and direct enough that I'd hope some of>you might wonder why there has been such argument over it.>Well, the problem I think is that mathematicians thought one thing,>but now I've shown another, and it's easier for many of them to just>hold on to what they believed, even though it's mathematically shown>now to be incorrect.Uh, no. A proof that some bit of mathematics is incorrect does not> include phrases like the ring of algebraic integers is ?e unless> it also includes a _de?ition_ of what it means for a ring to be> ?e. It does not include phrases like it's reasonable to assume> that.Well I've looked that statement over a couple of times, and it doesn't> make much sense to me.Possibly someone else could interject here and explain it?Sure. He means that you haven't given a characterization of ?erings--necessary and suf?ient conditions for when a ring is ?e. Since you haven't done that, the sentence the ring of algebraicintegers has (at best) an indeterminate truth value. Suppose, forexample, that I say that pi is kerplonkus. Unless I de?ekerplonkus--that is, give necessary and suf?ient conditions forwhen a mathematical object is kerplonkus, you won't have any ideawhat I mean. (By the way, a number is kerplonkus iff it istranscendental.)Regarding reasonable assumptions, any assumption you make underminesyour proof by making it weaker than it might be otherwise. If youthink that you have a proof for the general case of FLT, you do notwant to assume yourself into particular cases. If you're makingreasonable assumptions, you might as well stick them in thehypotheses and cite them.You're not demonstrating any problem with the math, all you're> demonstrating is that it doesn't work the way you think it should.> If you want to demonstrate that there's a problem with the> math then showing that something that should (according> to you) be an algebraic integer is not is not going to cut it.> The _only_ way to show that there's a problem with the> math is to _show_ that something _is_ an algebraic integer,> and then _show_ that the same thing is _not_ an algebraic> integer.> Huh?> Right. Once you've given suitable de?itions (necessary andsuf?ient conditions), in order to claim and prove that the ring ofalgebraic integers has a structural de?iency (fails the ringaxioms), you have to show that there, you have to show that there issomething which *ought* to be an algebraic integer (show that it is anintegral solution to a polynomial equation) but which is not containedin the ring of algebraic integers. (Good luck, heh)>[...]>I'm the one who's paused as another human being harmed by their>actions. For heaven's sake, if you don't like it here then go away. People> are not going to start agreeing you're right when you're not just> because you whine about how you're harmed by people's> comments.Mathematicians seem now to think that they can switch to a society> that *tells* itself what truth is, which might seem like a nice way to> handle outsiders like myself who ?d truths that they might ?d> unpleasant.However, hurting me by refusing to acknowledge mathematical truth> doesn't change the truth.To the extent that my work shows errors in thinking about algebraic> integers, those errors have existed as long as that *thinking* has> existed, in this case, over a hundred years.There's a right way to think of algebraic integers, and their> properties, and a wrong way.> You haven't presented any mathematical truths. Your sloppy languageand poor methodology have undermined any chance for you ?ding that.>My life is affected, but you see, they've *always* been>wrong about the ring of algebraic integers.> >That fact hasn't changed. They were just always wrong. And hurting>me, doesn't change that fact.What's amazing here is how *easily* I can prove that I'm right, and> bring it down to acceptance of constant terms, like 7, 7 and 22,> actually being NUMBERS, versus variables dependent on x, and> acceptance of the distributive property!!!The point is that I'm not the bad guy here, though some might want to> attack the messenger.The reality is that if something is mathematically correct; it's> correct.And if wrong, it's wrong.> True enough. And if it's neither, it's not either. > Mathematicians have just been wrong about algebraic integers. It's> not complicated, and refusing to acknowledge the truth doesn't change> it.Mathematicians are just still wrong, until they accept mathematics.> You're not doing mathematics. A mathematical proposition must have adeterminate truth value--something your propositions lack.'cid ?ooh === Mathematics is about logic and consistency, which *should* mean that a>person can present a mathematical argument, and deal with objective>criticisms or concerns and there be resolution. However, I'm dealing>with a situation where as a body mathematicians are trying to avoid>the implications of a rather basic mathematical argument, which I will>present yet again:>[...]Ah, I guess that David Ullrich deletes out the content as a sign of> disrespect.But then, why does he pay attention to my posts at all, as that is a> sign of respect? >Attempting to use those relations in the ring of algebraic integers I>can demonstrate a problem when I consider what happens if you divide>both sides by 49.>If you *assume* that the ring of algebraic integers is ?e, it's>reasonable to conclude the dividing both sides by 49 gives you>something like>[...]>the result you'll ?d in general is that you can no longer be in the>ring of algebraic integers.>The result is simple enough, and direct enough that I'd hope some of>you might wonder why there has been such argument over it.>Well, the problem I think is that mathematicians thought one thing,>but now I've shown another, and it's easier for many of them to just>hold on to what they believed, even though it's mathematically shown>now to be incorrect.> Uh, no. A proof that some bit of mathematics is incorrect does not> include phrases like the ring of algebraic integers is ?e unless> it also includes a _de?ition_ of what it means for a ring to be> ?e. It does not include phrases like it's reasonable to assume> that.Well I've looked that statement over a couple of times, and it doesn't> make much sense to me.Possibly someone else could interject here and explain it?Sure. He means that you haven't given a characterization of ?e> rings--necessary and suf?ient conditions for when a ring is ?e. > Since you haven't done that, the sentence the ring of algebraic> integers has (at best) an indeterminate truth value. Suppose, for> example, that I say that pi is kerplonkus. Unless I de?e> kerplonkus--that is, give necessary and suf?ient conditions for> when a mathematical object is kerplonkus, you won't have any idea> what I mean. (By the way, a number is kerplonkus iff it is> transcendental.)now you run into the problem of what he deleted.I said:Attempting to use those relations in the ring of algebraic integers Ican demonstrate a problem when I consider what happens if you divideboth sides by 49.If you *assume* that the ring of algebraic integers is ?e, it'sreasonable to conclude the dividing both sides by 49 gives yousomething like(5 b_1(x) + 1)(5 b_2(x) + 1)(5 c_3(x) + 22) = 300125 x^3 - 18375 x^2 - 360 x + 22where b_1(0) = b_2(0) = c_3(0) = 0,in the ring of algebraic integers, as the constant terms of thefactors go from being 7, 7 and 22 to being 1, 1, and 22.> Regarding reasonable assumptions, any assumption you make undermines> your proof by making it weaker than it might be otherwise. If you> think that you have a proof for the general case of FLT, you do not> want to assume yourself into particular cases. If you're making> reasonable assumptions, you might as well stick them in the> hypotheses and cite them.>What are you trying to say that actually relates back to *my* mathargument?James Harris === =Mathematics is about logic and consistency, which *should* mean that a> person can present a mathematical argument,You haven't so far.Well I think I have. That sounds like a fundamental point ofdisagreement with little room for debate.It then seems *rational* that you'd wander off. If you continue topay attention to my posts, it seems clear then that you are doing sofor what should be puzzling reasons to rational readers.Possibly you could give your reasons.James Harris === =[snip]With all due contempt, James, your posts are getting tiresome. Please> consider the following advice:Then don't read my posts. Now then you claim contempt but you showrespect by your attention.I'd *prefer* if you'd just wander off as your posts tend tobe...erratic, and irrational.Now then, PROVE you're not a liar, and your really have contempt forme by not giving me so much attention.James Harris === >> [snip]>>> With all due contempt, James, your posts are getting tiresome. Please>> consider the following advice:Then don't read my posts. Now then you claim contempt but you show>respect by your attention.Fascinating line of reasoning. I wander into a new year's eve partyand start defecating in all the punch bowls - when I run out of punch bowls I start pissing in people's drinks. And then if theypay attention to the way I'm behaving it follows that they're showingrespect.Huh.>I'd *prefer* if you'd just wander off as your posts tend to>be...erratic, and irrational.Now then, PROVE you're not a liar, and your really have contempt for>me by not giving me so much attention.>James Harris************************David C. Ullrich === >>> [snip]>>> With all due contempt, James, your posts are getting tiresome. Please>> consider the following advice:Then don't read my posts. Now then you claim contempt but you show>respect by your attention.Fascinating line of reasoning. I wander into a new year's eve party> and start defecating in all the punch bowls - when I run out of > punch bowls I start pissing in people's drinks. And then if they> pay attention to the way I'm behaving it follows that they're showing> respect.Well, if you did those things, you'd probably be thrown out, thoughgiven your posting behaviors maybe you go to parties that aredifferent!Usenet is not a party. It's a posting environment where thousands ofpeople post.Huh.> So then, out of those thousands of posts that go on Usenet picking outone particular person's posts consistently over a period of years isde?itely not a sign of disrespect, no matter if you have deludedyourself into thinking it is.James Harris <3c65f87.0312311544.69a6cfb3@posting.google.com> <3c65f87.0401011230.ae9459b@posting.google.com> === So then, out of those thousands of posts that go on Usenet picking out>one particular person's posts consistently over a period of years Um, James - all sci.math readers *know* that Professor Ullrich respondsto a *lot* of other posters (and also knows that most people thank himpolitely)>is>de?itely not a sign of disrespect, no matter if you have deluded>yourself into thinking it is.>James Harris-- MinSo where are all the buffaloes? === So then, out of those thousands of posts that go on Usenet picking out>one particular person's posts consistently over a period of years Um, James - all sci.math readers *know* that Professor Ullrich responds> to a *lot* of other posters (and also knows that most people thank him> politely)Irrelevant. Ullrich has *obsessively* replied to my posts for yearsnow.Like I said *most* people VALUE their time and attention, so theyvalue giving it.Now if you wish to try and push the idea that Ullrich has spent somuch time and attention without showing some kind of respect, then youcan if you wish.But I think most people will look at you as deluded or lying, as they*do* value *their* time and attention.Looks like I struck a nerve pointing out the obvious.James Harris === >So then, out of those thousands of posts that go on Usenet picking out>>one particular person's posts consistently over a period of years >>> Um, James - all sci.math readers *know* that Professor Ullrich responds>> to a *lot* of other posters (and also knows that most people thank him>> politely)Irrelevant. Ullrich has *obsessively* replied to my posts for years>now.Like I said *most* people VALUE their time and attention, so they>value giving it.Now if you wish to try and push the idea that Ullrich has spent so>much time and attention without showing some kind of respect, then you>can if you wish.But I think most people will look at you as deluded or lying, as they>*do* value *their* time and attention.Looks like I struck a nerve pointing out the obvious.Uh, no. This latest bit about how the fact that people reply showssome sort of respect is so pathetic that people are trying to talkyou out of it, because even your enemies can't stand the wayit looks when a grown man talks that way.>James Harris************************David C. Ullrich === Irrelevant. Ullrich has *obsessively* replied to my posts for years> now.David is not nearly as obsessive as James Harris. <3c65f87.0312311544.69a6cfb3@posting.google.com> === >> [snip]>>> With all due contempt, James, your posts are getting tiresome. Please>> consider the following advice: Then don't read my posts. Now then you claim contempt but you show> respect by your attention. I'd *prefer* if you'd just wander off as your posts tend to> be...erratic, and irrational.Uh, then don't read his posts? You claim contempt but you showrespect by your attention.I'll skip the last line, as the punchline is too easy and besides, Inever claimed to prefer that James goes away.-- Jesse F. HughesI think the burden is on those people who think he didn't haveweapons of mass destruction to tell the world where they are. -- White House spokesman Ari Fleischer === >>> [snip]>>> With all due contempt, James, your posts are getting tiresome. Please>> consider the following advice: Then don't read my posts. Now then you claim contempt but you show> respect by your attention. I'd *prefer* if you'd just wander off as your posts tend to> be...erratic, and irrational.Uh, then don't read his posts? You claim contempt but you show> respect by your attention.Well, usually I don't read his posts, and usually I don't read yourseither!!!But for now I'm reading ALL posts made in my recent threads, whichprobably won't last long given that I'm seeing yet again why I usuallydon't!!! > I'll skip the last line, as the punchline is too easy and besides, I> never claimed to prefer that James goes away.I think some people are here for math, and others are here for otherreasons, and you seem to be someone here for reasons other than math.James Harris <3c65f87.0312311544.69a6cfb3@posting.google.com> <873cb0xtzl.fsf@phiwumbda.org> <3c65f87.0401010856.2f8a2638@posting.google.com> === >> I'll skip the last line, as the punchline is too easy and besides, I>> never claimed to prefer that James goes away. I think some people are here for math, and others are here for other> reasons, and you seem to be someone here for reasons other than math.I don't deny that.A casual observer can see that I ?d your posts entertaining, just byreading my .sigs (well, not the one below, I guess -- random sigsynchronicity failed me this time).-- Jesse F. HughesC is for Cookie. That's good enough for me. Cookie Monsters === = [snip] With all due contempt, James, your posts are getting tiresome. Please> consider the following advice: Then don't read my posts. Now then you claim contempt but you show> respect by your attention.I promise to stop reading your posts as soon as you stop posting.> I'd *prefer* if you'd just wander off as your posts tend to> be...erratic, and irrational.No, they haven't. There has been no evidence of any erratic or irrational postings on my part.> Now then, PROVE you're not a liar, and your really have contempt for> me by not giving me so much attention.Nice try! But the problem with your posts is that you're using up bandwidth on a math forumwithout contributing to the math. My *attention* to you has no more to do with respect for youthan my complaints about an obnoxious neighbor would demonstrate respect for him. You are arude, obnoxious vandal who gets his attention by ?g manure -- not a person who earns hisattention by offering anything worthwhile.As usual, your idea of what constitutes a proof fails to exhibit the level of rigorappropriate to the subject.However, I repeat my promise to stop reading and responding to you if you will stop posting.--There are two things you must never attempt to prove: the unprovable -- and the obvious.--Democracy: The triumph of popularity over principle.--http://www.crbond.com === = > [snip]> With all due contempt, James, your posts are getting tiresome. Please> consider the following advice: Then don't read my posts. Now then you claim contempt but you show> respect by your attention.I promise to stop reading your posts as soon as you stop posting.It's *your* choice whether or not to read a post. However you whinedin a post as if it were my problem, and I pointed out a simplesolution!!!If you wish to contemptuously express that you ?d my posts tiresomethen a simple solution is not to read them!!! > I'd *prefer* if you'd just wander off as your posts tend to> be...erratic, and irrational.No, they haven't. There has been no evidence of any erratic or irrational postings on my part.Now then, PROVE you're not a liar, and your really have contempt for> me by not giving me so much attention.Nice try! But the problem with your posts is that you're using up bandwidth on a math forum> without contributing to the math. My *attention* to you has no more to do with respect for you> than my complaints about an obnoxious neighbor would demonstrate respect for him. You are a> rude, obnoxious vandal who gets his attention by ?g manure -- not a person who earns his> attention by offering anything worthwhile.> No matter how you slice it, your attention is a sign of some kind ofrespect, even if it's only of my ability to use up bandwith!!!> As usual, your idea of what constitutes a proof fails to exhibit the level of rigor> appropriate to the subject.However, I repeat my promise to stop reading and responding to you if you will stop posting.> Why should you care whether or not I post? I'm actually curious aboutthat one.Are you saying there are NO OTHER posters who use up bandwith that youtrack to obsessively reply to their posts, or do you claim that I useup more than anyone else?Or what?And yes, in case you're curious, while your posts on math aren'tinteresting, your posts about your *own* state of mind have sparked mycuriosity.James Harris === =In sci.math, James Harrison 1 Jan 2004 09:02:08 -0800<3c65f87.0401010902.3610b573@posting.google.com>:>>> [snip]>> With all due contempt, James, your posts are getting tiresome. Please>> consider the following advice: Then don't read my posts. Now then you claim contempt but you show>> respect by your attention.>>> I promise to stop reading your posts as soon as you stop posting.It's *your* choice whether or not to read a post. However you whined> in a post as if it were my problem, and I pointed out a simple> solution!!!If you wish to contemptuously express that you ?d my posts tiresome> then a simple solution is not to read them!!![rest snipped]Can we argue about something a little more interesting? :-)Like, for instance, whether there exists a unique [*]set of numbers with units -1 and +1 that form a ringover multiplication?(I'd be curious, and it's apparently related to JSH'sclaim that there exists a set of numbers with the algebraicintegers as a proper subset, with units -1 and +1.)[*] in the sense of isomorphism, obviously.-- #191, ewill3@earthlink.netIt's still legal to go .sigless. === =[snip]> No matter how you slice it, your attention is a sign of some kind of> respect, even if it's only of my ability to use up bandwith!!!On the contrary. You attract attention the way a screaming brat attracts attention in a supermarket.It has nothing whatsoever to do with respect.> As usual, your idea of what constitutes a proof fails to exhibit the level of rigor> appropriate to the subject. However, I repeat my promise to stop reading and responding to you if you will stop posting.> Why should you care whether or not I post? I'm actually curious about> that one.If you stop posting, the SNR of this newsgroup improves. I don't like noise.> Are you saying there are NO OTHER posters who use up bandwith that you> track to obsessively reply to their posts, or do you claim that I use> up more than anyone else? Or what? And yes, in case you're curious, while your posts on math aren't> interesting, your posts about your *own* state of mind have sparked my> curiosity.You are not only degrading the quality of this newsgroup, you are doing so maliciously anddishonestly. You misrepresent others, you insult them, you promote lunaticconsipiracy theoriesabout the motives of others, and generally behave like an idiot. As for the other trolls, loons,etc. on this newsgroup, they do not generally insult or threaten other posters -- which you oftendo.--There are two things you must never attempt to prove: the unprovable -- and the obvious.--Democracy: The triumph of popularity over principle.--http://www.crbond.com === = [snip] No matter how you slice it, your attention is a sign of some kind of> respect, even if it's only of my ability to use up bandwith!!! On the contrary. You attract attention the way a screaming brat attractsattention in a supermarket.> It has nothing whatsoever to do with respect. > As usual, your idea of what constitutes a proof fails to exhibit thelevel of rigor> appropriate to the subject.> However, I repeat my promise to stop reading and responding to you ifyou will stop posting.> Why should you care whether or not I post? I'm actually curious about> that one. If you stop posting, the SNR of this newsgroup improves. I don't likenoise. Are you saying there are NO OTHER posters who use up bandwith that you> track to obsessively reply to their posts, or do you claim that I use> up more than anyone else? Or what? And yes, in case you're curious, while your posts on math aren't> interesting, your posts about your *own* state of mind have sparked my> curiosity. You are not only degrading the quality of this newsgroup, you are doing somaliciously and> dishonestly. You misrepresent others, you insult them, you promotelunaticconsipiracy theories> about the motives of others, and generally behave like an idiot. As forthe other trolls, loons,> etc. on this newsgroup, they do not generally insult or threaten otherposters -- which you often> do. --> There are two things you must never attempt to prove: the unprovable -- and the obvious.> --> Democracy: The triumph of popularity over principle.> --> http://www.crbond.comJames acts like he can dish out all the insults he wants to, but he can'ttake it when the insults are directed toward him. He acts so much like my 7year old brother that it's scary.-- David Moran === skip The satellite picks up your thoughts and plays them out to everyone.> Sounds like Sci Fi, I am in a constant forced telelpathic dialog with> a team of purist sickos passing themselves off as pyschs.Anyone nearby can hear my thoughts, most people verbally abuse me> to force a nasty reply. I'm the kindest man on Earth, but that's because> I'm used to conceiling anything condescending. I watched my brother> and sister out my window leave yesterday, pretending I wasn't home. The> alternative is half dozen people sitting around me trying to tame my thoughts.[snip]If your thoughts are picked up by the satellite as electromagnetic radiation, which includes radio-waves and microwaves, then a TEMPEST-type shielding device might be the thing for you:see for example:http://cryptome.org/nacsim-5000.htmDavid Bernier === If your thoughts are picked up by the satellite as electromagnetic> radiation, which includes radio-waves and microwaves, then a> TEMPEST-type shielding device might be the thing for you: see for example: http://cryptome.org/nacsim-5000.htm>not another alfoil beanie site :)I know, all I need is a big soundproofed shed. I can experiment on what metalsshield the spy radiation as long as I have $10,000 to fund some sealed boxes before theyturn off the active part which gives feedback.Then I still need $100,000 to live moderately inside a shed. I can make money offthe internet so that's no drama.Or I could ?d a mine site or cave for temporaray measures. The thing is they've beencoordinating my life around me for 20 years. I can't afford a car to ?d an abondoned mine.The last job interview a dozen Truman Company sickos heckled me for an hour whileI ?ed computers in a tiny lab with 10 people. They can whisper in peoples earwhat cards are in my hand at the card table and only they hear it.The rays penetrate inside highrise buildings. I thought it was infra red laser heating the airfor sound modulation but it penetrates buildings. They can hear your heartbeat from spacefrom inside a bomb shelter. Not knowing about WOMD is a lie, they can listen to anyonesthoughts anywhere in the world. You're thoughts still send a stimulus to your voice box,I would be suprised if a sensitive microphone couldn't pick it up, the satellite picks upmine EVERYWHERE. They make up bizzare characters that talk to me in context inmy dreams. They're the sickest s in history, I listen to them 2 years 24/7 not 2 secondsbreak and am FORCED to respond every second. That's what the truman is.What's $100,000 to 100,000 people on usenet to owe the Truman?THERE ARE 100,000 witnesses to paranormal in Townsville Australiahttp://groups.google.com/groups?selm= dksqpv41597ofbhr16mi7tdj2q?voldv%404ax.commy uncle told me you post here... he's did some audio on yoursecret show and gave me some tapes. love the show.http://www.google.com/groups?as_umsgid= 9aqbbv85c53jobkenltkil2tbsdk7m2n3p@4ax.comI was in Townsville over the weekend, and I heard him.Very spooky!I'm in Townsville. We're sick of youhttp://www.google.com/groups?selm=Pine.OSF .4.21.0309230859190.5384-100000%40marlin.jcu.edu.au>Do you know if he is living in Townsville?I've been hearing stuff, yeah.funny delusions that the whole town has them === =I have written a game based on Turing Machines:http://home.comcast.net/~logiclab/CyberWar.htmlAny comments and suggestions are welcome.I have tested the page with Internet Explorer.It also seems to work with Netscape.The page may not work with other browsers.I will add your strategy to the list of strategiesthe computer uses.Russell- 2 many 2 count === =I'm facing a frustrating situation where mathematicians appear to havedecided to throw out rules like following mathematical logic, as I'vebacked my position up in so much detail that I've routinely mentionedthe distributive property!!!Worse, they're getting away with vague comments, and vaguer claims, asI think many of you are just willing to let them simply disagree andbelieve that all that matters in mathematics is if a mathematiciansagree or disagree!So here's yet another challenge to show you just how strong myposition is--the mathematical position--versus how weak their positionis--the social position.I've repeatedly given(5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = 49(300125 x^3 - 18375 x^2 - 360 x + 22)where b_3(x) = a_3(x) - 3 and the a's are roots ofa^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)and when x=0, a_1(0) = a_2(0) = b_3(0) = 0.Now dividing both sides by 49, you get differing functions like(5 b_1(x) + 1)(5 b_2(x) + 1)(5 c_3(x) + 22) = 300125 x^3 - 18375 x^2 - 360 x + 22where when x=0, b_1(0) = b_2(0) = c_3(0) = 0.Now before the a's were de?ed by the cubic:a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x).But what is the de?ing cubic now?Well, the leading coef?ient is known, as is the last coef?ient butwhat about the others?b^3 + ? b^2 + ? b - 2401 x^3 - 147 x^2 + 3x.So there's the challenge, which is to ?l in the other twocoef?ients.Now then, if posters disagreeing with me have been correct then theremust exist integers or integer functions that go there, and it shouldbe possible to show them!Like with the cubic de?ing the a's, you have 3(-1 + 49x), as the second coef?ient, and 0, as the third.Now then, have you thrown away ALL rules of logic? Have you all lostany respect for mathematics itself, believing that mathematics is justsome democracy where all that matters is what mathematicians say?Or do any of you care about mathematical truth?James Harris === = > I've repeatedly given > (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = > 49(300125 x^3 - 18375 x^2 - 360 x + 22) > where b_3(x) = a_3(x) - 3 and the a's are roots of > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) > and when x=0, a_1(0) = a_2(0) = b_3(0) = 0.Yup, that has not been challenged of late. > Now dividing both sides by 49, you get differing functions like > (5 b_1(x) + 1)(5 b_2(x) + 1)(5 c_3(x) + 22) = > 300125 x^3 - 18375 x^2 - 360 x + 22 > where when x=0, b_1(0) = b_2(0) = c_3(0) = 0.Possibly. Although you do not explain c_3 any further. > Now before the a's were de?ed by the cubic: > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). > But what is the de?ing cubic now? > Well, the leading coef?ient is known, as is the last coef?ient but > what about the others? > b^3 + ? b^2 + ? b - 2401 x^3 - 147 x^2 + 3x.What is c? > So there's the challenge, which is to ?l in the other two > coef?ients. > Now then, if posters disagreeing with me have been correct then there > must exist integers or integer functions that go there, and it should > be possible to show them!That is *false*. There is *no* disagreement that a_1(x) and a_2(x) arenot divisible by 7 in the algebraic integers, which is what you areclaiming with the above statement. Because if there are integers orinteger functions, b_1 and b_2 would be algebraic integer functions.You are again misrepresenting the claims of your opponents. > Now then, have you thrown away ALL rules of logic? Have you all lost > any respect for mathematics itself, believing that mathematics is just > some democracy where all that matters is what mathematicians say?You have, it appears. > Or do any of you care about mathematical truth?You do not, it appears. What part of my de?ition of the w's iswrong?-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === =In sci.math, Dik T. Winter:> I've repeatedly given> (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = > 49(300125 x^3 - 18375 x^2 - 360 x + 22)> where b_3(x) = a_3(x) - 3 and the a's are roots of> a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)> and when x=0, a_1(0) = a_2(0) = b_3(0) = 0.Yup, that has not been challenged of late.Now dividing both sides by 49, you get differing functions like> (5 b_1(x) + 1)(5 b_2(x) + 1)(5 c_3(x) + 22) = > 300125 x^3 - 18375 x^2 - 360 x + 22> where when x=0, b_1(0) = b_2(0) = c_3(0) = 0.Possibly. Although you do not explain c_3 any further.c_3(x) = b_3(x) and he's shifting notation so that he canreuse b_3(x) as one of the roots of his new polynomial.Of course he's already de?ed b_3(x) = a_3(x) - 3 so there'sgoing to be a little eraser work... :-)Now before the a's were de?ed by the cubic:> a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x).But what is the de?ing cubic now?> Well, the leading coef?ient is known, as is the last coef?ient but> what about the others?> b^3 + ? b^2 + ? b - 2401 x^3 - 147 x^2 + 3x.What is c?What, indeed?[rest snipped]-- #191, ewill3@earthlink.netIt's still legal to go .sigless. === > I've repeatedly given> (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = > 49(300125 x^3 - 18375 x^2 - 360 x + 22)> where b_3(x) = a_3(x) - 3 and the a's are roots of> a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)> and when x=0, a_1(0) = a_2(0) = b_3(0) = 0.Yup, that has not been challenged of late.That's important mathematical information that I can luckily includefor reference in my posts as it's short.> Now dividing both sides by 49, you get differing functions like> (5 b_1(x) + 1)(5 b_2(x) + 1)(5 c_3(x) + 22) = > 300125 x^3 - 18375 x^2 - 360 x + 22> where when x=0, b_1(0) = b_2(0) = c_3(0) = 0.Possibly. Although you do not explain c_3 any further.I replaced all the previous function names with new ones advancing tothe next letter in the alphabet.So before there was a_1(x), a_2(x), and b_3(x), so I went to b_1(x),b_2(x), and c_3(x).The functions are to a large extent unknown, as there's no de?ingcubic within the ring of algebraic integers, unlike with the a's, andb_3(x).Remember, I ended up with b_3(x) from b_3(x) = a_3(x) - 3, where thea's are roots of the cubic:a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) > Now before the a's were de?ed by the cubic:> a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x).> But what is the de?ing cubic now?> Well, the leading coef?ient is known, as is the last coef?ient but> what about the others?> b^3 + ? b^2 + ? b - 2401 x^3 - 147 x^2 + 3x.What is c?Explained above, now consider what I'm emphasizing by pointing out theunknown coef?ients, which is that it's not as easy as before withthe a's and the *de?ing* cubic:a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)Something has happened, which shows that the resultant functions--thefunctions that result after 49 is divided from both sides--are NOT inthe ring of algebraic integers as they aren't de?able within thatring.> So there's the challenge, which is to ?l in the other two> coef?ients.> Now then, if posters disagreeing with me have been correct then there> > must exist integers or integer functions that go there, and it should> be possible to show them!That is *false*. There is *no* disagreement that a_1(x) and a_2(x) are> not divisible by 7 in the algebraic integers, which is what you are> claiming with the above statement. Where? I am in fact not making that claim with that statement.The b's are just not algebraic integer functions, and I've made nomention of factors of a_1(x) or a_2(x) in the ring of algebraicintegers as such are irrelevant to my point.>Because if there are integers or> integer functions, b_1 and b_2 would be algebraic integer functions.> You are again misrepresenting the claims of your opponents.My point is that there just aren't algebraic integer functions thatexist to ful? the requirements, which is shown by the inability tode?e them within the ring of algebraic integers.Now then, people like yourself with your own claims need algebraicinteger functions, so I'm not misrepresenting your position at all. > > Now then, have you thrown away ALL rules of logic? Have you all lost> any respect for mathematics itself, believing that mathematics is just> some democracy where all that matters is what mathematicians say?You have, it appears. > Or do any of you care about mathematical truth?You do not, it appears. What part of my de?ition of the w's is> wrong?Your problem as I see it is a need to look only at one piece of thepuzzle, proclaim some solution, and then refuse to acknowledge otherpieces that refute your claims.It seems to me that you should have noticed your inability to ?dalgebraic integer functions to support your position as if you *could*?d them, then that would have ended arguments long ago in *your*favor.Instead I think you relied on other people not paying attention to themathematical facts to promote your own views *against* mathematicaltruth.James Harris === I'm facing a frustrating situation where mathematicians appear to have> decided to throw out rules like following mathematical logic, as I've> backed my position up in so much detail that I've routinely mentioned> the distributive property!!! Worse, they're getting away with vague comments, and vaguer claims, as> I think many of you are just willing to let them simply disagree and> believe that all that matters in mathematics is if a mathematicians> agree or disagree! So here's yet another challenge to show you just how strong my> position is--the mathematical position--versus how weak their position> is--the social position. I've repeatedly given (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = 49(300125 x^3 - 18375 x^2 - 360 x + 22) where b_3(x) = a_3(x) - 3 and the a's are roots of a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) and when x=0, a_1(0) = a_2(0) = b_3(0) = 0. Now dividing both sides by 49, you get differing functions like (5 b_1(x) + 1)(5 b_2(x) + 1)(5 c_3(x) + 22) = 300125 x^3 - 18375 x^2 - 360 x + 22 where when x=0, b_1(0) = b_2(0) = c_3(0) = 0. Now before the a's were de?ed by the cubic: a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). But what is the de?ing cubic now?>Not much of a challenge : a_1 = 7*b_1, a_2 = 7*b_27b^3 + 3(-1 + 49x)b^2 - ( 2401 x^3 -147 x^2 + 3 x)I'll leave you to work out which two of the three roots correspond to b_1and b_2.> Well, the leading coef?ient is known, as is the last coef?ient but> what about the others? b^3 + ? b^2 + ? b - 2401 x^3 - 147 x^2 + 3x.Well, maybe you thought you knew what the leading coef?ient was, but youareclearly wrong. So there's the challenge, which is to ?l in the other two> coef?ients. Now then, if posters disagreeing with me have been correct then there> must exist integers or integer functions that go there, and it should> be possible to show them!>Challenge done - what do I win?> Like with the cubic de?ing the a's, you have 3(-1 + 49x), as the second coef?ient, and 0, as the third. Now then, have you thrown away ALL rules of logic? Have you all lost> any respect for mathematics itself, believing that mathematics is just> some democracy where all that matters is what mathematicians say? Or do any of you care about mathematical truth?> James HarrisPhil. === =Dave Hamilton>If you're looking for enrichment material, there iswww.nrich.maths.org.ukand quite a few interesting bits and pieces here:www.mathpages.comMaybe alsowww.cut-the-knot.org/ctk/index.shtmlLH === =?d out what university alegbra was all about (rings, ?lds and what-not. It's apparently very different from high school algebra.) I am also keen to study number theory. I'll pick up that book you mention on basic university level algebra.I hear you about having a teacher. I need to ?ish my home study high school course and move out of here! And anyway, I can't get a decent maths degree by correspondence anyway. Monash University used to have a great one, but because of funding cuts they discontinued all the useless degrees like maths, physics and engineering, leaving the most useful ones like commerce, business and arts. ;) Actually, it's probably more due to the popularity of those courses and the fact that wealthy overseas students put up massive amounts of money (certainly more than the price of my house) to get a Monash business degree.Johnathan === =?d out what university alegbra was all about (rings, ?lds and > what-not. It's apparently very different from high school algebra.) > I am also keen to study number theory. I'll pick up that book you > mention on basic university level algebra.I hear you about having a teacher. I need to ?ish my home study > high school course and move out of here! And anyway, I can't get a > decent maths degree by correspondence anyway. Monash University used > to have a great one, but because of funding cuts they discontinued all > the useless degrees like maths, physics and engineering, leaving the > most useful ones like commerce, business and arts. ;) Actually, it's > probably more due to the popularity of those courses and the fact that > wealthy overseas students put up massive amounts of money (certainly > more than the price of my house) to get a Monash business degree.Johnathan Johnathan, Actually I don't think I mentioned a book on university-levelalgebra. The ?Skeleton Key of Mathematics' is almost certainlyout of print, and though it dealt with a central topic - permutation groups - it was much too narrow in scope. Its greatvirtue was that it talked about the real thing at a basic level,not some super?ial overview. If you want a comprehensive lookat university level algebra, you should go for either Herstein or Beachy & Blair or Birkhoff & McLane. None of these are easyhowever to study on your own. All deal extensively with rings,?lds, groups, etc.. All are good but Herstein may be the best. Personally I would not bother with a Schaum's Outline book. They can be good sources of problems to supplement somethingelse but generally they do not convey the essence of the subject. Number theory: there are some decent elementary books: e.g., oneby W. J. LeVeque, listed at $8.95 from Amazon, an old classic. I un-recommended a book by Hardy and Littlewood - my mistake - the authors are Hardy and Wright - a great book but too hard to startbook called Inequalities which I really love, and a lot of itis elementary. Hard to ?d, probably. Elementary number theory, thougha topic with unlimited depth, does not necessarily give you the ?of modern mathematics, which must include structures(rings, ?lds, groups, e.g.) and concepts and *ways of thinking*. Elementary number theory, not to belittle it, looks somewhat more like a mixed bag of tricks. Yes: you need to get to a place where there is a decent Universityand a decent library. Nora B. === =out what university alegbra was all about (rings, ?lds and what-not. > It's apparently very different from high school algebra.) I am also keen> to study number theory. I'll pick up that book you mention on basic> university level algebra.I hear you about having a teacher. I need to ?ish my home study high> school course and move out of here! And anyway, I can't get a decent> maths degree by correspondence anyway. Monash University used to have a> great one, but because of funding cuts they discontinued all the useless> degrees like maths, physics and engineering, leaving the most useful ones> like commerce, business and arts. ;) Actually, it's probably more due to> the popularity of those courses and the fact that wealthy overseas> students put up massive amounts of money (certainly more than the price of> my house) to get a Monash business degree.JohnathanYou might consider the distance program offered through the UNE atArmidale. They certainly offer a B.CompSc degree with a math major or aB.Sc, both by distance education. These degrees are certainly covered byHECS.You can visit the maths department site at... http://turing.une.edu.au/Bruce. === ?d>> out what university alegbra was all about (rings, ?lds and what-not. >> It's apparently very different from high school algebra.) I am also keen>> to study number theory. I'll pick up that book you mention on basic>> university level algebra.>>> I hear you about having a teacher. I need to ?ish my home study high>> school course and move out of here! And anyway, I can't get a decent>> maths degree by correspondence anyway. Monash University used to have a>> great one, but because of funding cuts they discontinued all the useless>> degrees like maths, physics and engineering, leaving the most useful ones>> like commerce, business and arts. ;) Actually, it's probably more due to>> the popularity of those courses and the fact that wealthy overseas>> students put up massive amounts of money (certainly more than the price of>> my house) to get a Monash business degree.>>> JohnathanYou might consider the distance program offered through the UNE at>Armidale. They certainly offer a B.CompSc degree with a math major or a>B.Sc, both by distance education. These degrees are certainly covered by>HECS.>You can visit the maths department site at... >http://turing.une.edu.au/>Bruce.>But I would never, ever want a home study degree. There areinnumerable things that are learned by watching the teacherand cannot be taught ala text. Anything that has to do withscience, computing, and engineering falls into this category.Math does, too because they never show enough examples formy satisfaction./BAH === The original poster has since replied to the thread that he has little> exposure to proof at all in his mathematics education, a statement that I> don't ?d surprising at all. I remember my own mathematics education in> school (in an earlier decade, supposedly in a good suburban United States> school system) as mostly an exhortation to thoughtlessly memorize mechanical> rules, with little exposure to proof even in the geometry class. I ?d the> construction of the real number system from the ?ld properties taken as> local axioms (the procedure of the Lang or the Spivak textbooks) or from the> Peano postulates (the procedure of Landau's textbook) to be delicious, and> wonder why I couldn't have got some proper math like that when I was in> school. But I have found a quotation from W.W. Sawyer that gives me the> answer: The proper thing for a parent to say is, ?I did badly at> mathematics, but I had a very bad teacher. I wish I had had a good one.' W.> W. Sawyer, Vision in Elementary Mathematics (1964), page 5. For the original> poster, I extend my best wishes in ?ding affordable books that can help> his process of self-education. For my children, I start early to make sure> they get a proper education in mathematics. I hope to walk my oldest son> (age eleven) through his ?st proof by mathematical induction in the next> day or so, taking advantage of the New Year holiday for some educational> recreation. A few years ago, I myself didn't know what mathematical> induction is, but I found readable books about university-level mathematics> and learned something through self-study. I hope the original poster ?ds> pleasure in doing the same.I am pleased to hear that there is a good elementary book that treats proof as something important. I can get a second hand copy of Serge Lang's book, and make a point of getting it by the end of the month. I am looking forward to digesting this one.I am mostly doing self study, but I am ?ishing school by home study here in Australia through a technical college. The course is of reasonable quality, but there was no other school or college in my State that offered the highest level of mathematics by distance study. I live in the country and I can't get to an ordinary school, so I had no choice. And since the rules changed for interstate students, I can no longer take subjects from another State's school (in this case another technical college) without paying full fees, almost the equivalent of what overseas students pay. :(So much for politicians' lip service about education being a national priority.Still, the next step is to ?ish HS, get into university, and take brush-up courses in advanced maths (HS level) before the ?st semester starts. === snip political policy that sucksSo much for politicians' lip service about education being a national >priority.Still, the next step is to ?ish HS, get into university, and take >brush-up courses in advanced maths (HS level) before the ?st >semester starts.Oh, yes. YOu have to plan on taking at least one of those courses.Note that attending college courses require a completely different mind set than high school if you want to get themost out of them. You have not learned how to take notes;you have not learned how to take a test; you have not learnedhow to use the of?e hours time allotted to you. There'sprobably dozens of more skills acquired by osmosis in high school that I'm not aware of. I would suspect that you probably are not skilled in tuning out distractions. Thatis certainly one of the things I learned in high school. (I'min the USA so things are culturally different.) /BAH === My very ?st exposure to rigid proofs was my high school plane> geometry class. In addition to learning about proofs, I also> learned about construction. I bet you've never done that.Are you talking about something purely mathematical, or building construction?No to both. ;)Google is my new friend. About to ?d out ... === My very ?st exposure to rigid proofs was my high school plane>> geometry class. In addition to learning about proofs, I also>> learned about construction. I bet you've never done that.Are you talking about something purely mathematical, or building >construction?No to both. ;)Yes.Google is my new friend. About to ?d out ...I don't know the magic incantation to google and I don't know if the correct stuff is out there. Proof by construction mayhelp. You sure you don't want to drop in for the ?st monthof the high school plane geometry class? I'm not sure I'dhave been able to ?ure out how to the proofs physicallywithout watching the teacher./BAH === Let P be a polynomial having all coef?ients integer numbers.>It is assumed that :> ii) P has at least one rational root.Question: Find all rational roots of P.Let us solve a similar problems.If |P(0)| = 1, we have that P(x) = xQ(x) +/- 1, where Q is in Z[x].> If |P(1)| = 1, we have that P(x) = (x-1)R(x) +/- 1, where R is in Z[x].> Suppose p/q is a root of P, with (p,q) = 1 and q > 0, then |Q(p/q)| = |q/p| [1]and |R(p/q)| = |q/(q-p)| [2]Let k be the degree of Q, then q^k Q(p/q) is an integer. This means> that q^{k+1}/p is an integer. Since, (p,q) = 1, we have |p| = 1.> k must also be the degree of R, so q^k R(p/q) is an integer. We then> get that q^{k+1}/{q-p} is an integer. This means that |q-p| = 1.> The only way to have |p| = 1 and |q-p| = 1 with (p,q) = 1 and q > 0, is> if p = 1 and q = 2. Thus, the only possible rational root is 1/2.of the original P to be 4007/2.Rob Johnson ooHere are some possible options...1) For each w, this de?es X_n(w), a sequence of reals and then thelimit of the supremum could make sense...in which case, it would denote a function from the sample space to reals, like a random variable.2) One might consider the Lim sup itself as an operator as de?ed for the limit of a sequence of sets. Evidence also supports this since the author asserts that the expression belongs to the tail sigma ?ld, meaning that it is a set. But then, what set does the operand, X_n correspond to?One more question....De?e S={w: sum over n from 1 to inf of X_n(w) converges}Why does S belong to the tail sigma ?ld of {X_n}?Note:Tail sigma ?ld is de?ed as the lim as n->inf of { minimal sigma ?ld of(X_n,X_n+1,X_n+2,.......)} === =If {X_n} is a sequence of random varibles.What does the following expression mean?lim sup X_n ?> n->ooHere are some possible options...1) For each w, this de?es X_n(w), a sequence of reals and then the>limit of the supremum could make sense...in which case, it would >denote a function from the sample space to reals, like a random >variable.That's what it means. (Except for the word like: the lim sup ofthe X_n _is_ a random variable.)_Many_ of the things you'll see in probability are to be interpretedthe same way. For example, when someone says Let N be thesmallest n such that X_n > 42 then N is a random variable.************************David C. Ullrich === =If {X_n} is a sequence of random varibles.What does the following expression mean?lim sup X_n ?> n->ooHere are some possible options...1) For each w, this de?es X_n(w), a sequence of reals and then the> limit of the supremum could make sense...in which case, it would > denote a function from the sample space to reals, like a random > variable.Yes, that is exactly what it means. Note that the r.v. is allowed totake on the values +/- in?ity.2) One might consider the Lim sup itself as an operator as de?ed for > the limit of a sequence of sets. Evidence also supports this since the > author asserts that the expression belongs to the tail sigma ?ld, > meaning that it is a set. But then, what set does the operand, X_n > correspond to?No, your author is probably saying that the lim sup is *measurable*with respect to the tail sigma-?ld. Check his/her notation.One more question....> De?e S={w: sum over n from 1 to inf of X_n(w) converges}Why does S belong to the tail sigma ?ld of {X_n}?Note:> Tail sigma ?ld is de?ed as the > lim as n->inf of { minimal sigma ?ld of(X_n,X_n+1,X_n+2,.......)}Because the convergence of a series can be determined without regardto any ?ite number of initial terms. Or perhaps you prefer thefollowing explanation: For any positive integer n,S = intersection(k=1..infty, union(j=n..infty, {|sum(m=j..infty, X_m)|< 1/k}))Stephen J. HerschkornX-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBUK7iZ24324; === Has anyone ever explained the notion of proof by contradiction to>you?Here's the standard proof that sqrt(2) is irrational: Suppose that>sqrt(2) is rational. Let sqrt(2) = n/m where n and m are relatively>prime. Then 2m^2 = n^2, so n is even. If n = 2k then m^2 = 2k^2,>so m is even. This contradicts the fact that n and m are relatively >prime.It appears that you would object to this proof as follows: We must>not assume that sqrt(2) is rational, because it's not. That's not>a valid objection, it just shows you don't know how proof by>contradiction works.[Everyone else: Before pointing out that the proof of the>uncountability of R does not have to be phrased as a >proof by contradiction stop and consider whether that's>relevant to an attempt to explain the error in his reasoning...]>OK. Your example is a good one, but it has a little difference respect the case we are dealing with. You can freely suppose that sqrt(2) is rational because this supposition does not run against any mathematical concept. At ?st sight it doesnt matter if sqrt(2) is rational or not. Just after a good reasoning we arrive, by contradiction, to the conclusion that it cannot be rational.However, in the case of Cantors proof, if we initially suppose that we have a one-to-one correspondence between N and R, then we are breaking off the mathematical rules. This means that our initial premise is false, and we cannot use a bad premise as starting point in a demonstration trying to ?d a contradiction, or whatever thing we were looking for.Nicolas de la Foz>> For instance, if we admit the one-to-one correspondence between N and R,>> then we are confronting two different kinds of in?ity, the potential in?ite>> represented by the asymptotic approximation of the naturals to the in?ite>> (never reaching it), and the actual in?ite represented by the set of all real>> numbers;How do you handle the fact there are more than two grades of in?ity?> For example, the set F of all functions f: R->R is larger than the>set of reals R, such that you cannot place these into one-to-one>correspondence either. Yet you call the smaller set R an actual>in?ity. What does that make F? An even more actual in?ity?And how about the power set of F? That's bigger still.>************************David C. Ullrich === OK. Your example is a good one, but it has a little difference > respect the case we are dealing with. You can freely suppose that sqrt> (2) is rational because this supposition does not run against any > mathematical concept. At ?st sight it doesn't matter if sqrt(2) is > rational or not. Just after a good reasoning we arrive, by > contradiction, to the conclusion that it cannot be rational. However, in the case of Cantor's proof, if we initially suppose that > we have a one-to-one correspondence between N and R, then we are > breaking off the mathematical rules. This means that our initial > premise is false, and we cannot use a bad premise as starting point > in a demonstration trying to ?d a contradiction, or whatever thing > we were looking for.(1) Every proof by contradiction involves a bad premise. If not,you'll never reach the contradiction. This nonsense about runningagainst mathematical concepts has nothing to do with mathematicallogic at all.(2) You don't have to prove Cantor's theorem indirectly. Let f:N -> R be any function. Go through the usual construction and youhave proven that f is not surjective -- without ever assuming that f*is* surjective.-- Jesse F. HughesThe sole cause of all human misery is the inability of peopleto sit quietly in their rooms. -- Blaise Pascal === >Has anyone ever explained the notion of proof by contradiction to>>you?Here's the standard proof that sqrt(2) is irrational: Suppose that>>sqrt(2) is rational. Let sqrt(2) = n/m where n and m are relatively>>prime. Then 2m^2 = n^2, so n is even. If n = 2k then m^2 = 2k^2,>>so m is even. This contradicts the fact that n and m are relatively >>prime.It appears that you would object to this proof as follows: We must>>not assume that sqrt(2) is rational, because it's not. That's not>>a valid objection, it just shows you don't know how proof by>>contradiction works.[Everyone else: Before pointing out that the proof of the>>uncountability of R does not have to be phrased as a >>proof by contradiction stop and consider whether that's>>relevant to an attempt to explain the error in his reasoning...]>>OK. Your example is a good one, but it has a little difference >respect the case we are dealing with. You can freely suppose that sqrt>(2) is rational because this supposition does not run against any >mathematical concept. At ?st sight it doesnt matter if sqrt(2) is >rational or not. Just after a good reasoning we arrive, by >contradiction, to the conclusion that it cannot be rational.However, in the case of Cantors proof, if we initially suppose that >we have a one-to-one correspondence between N and R, then we are >breaking off the mathematical rules. This means that our initial >premise is false, and we cannot use a bad premise as starting point >in a demonstration trying to ?d a contradiction, or whatever thing >we were looking for.You're speaking nonsense here - there is no difference betweenthe two examples logically.>Nicolas de la Foz>> For instance, if we admit the one-to-one correspondence between N and R,> then we are confronting two different kinds of in?ity, the potential in?ite> represented by the asymptotic approximation of the naturals to the in?ite> (never reaching it), and the actual in?ite represented by the set of all real> numbers;How do you handle the fact there are more than two grades of in?ity?>> For example, the set F of all functions f: R->R is larger than the>>set of reals R, such that you cannot place these into one-to-one>>correspondence either. Yet you call the smaller set R an actual>>in?ity. What does that make F? An even more actual in?ity?And how about the power set of F? That's bigger still.>>************************David C. Ullrich************************David C. Ullrich === OK. Your example is a good one, but it has a little difference > respect the case we are dealing with. You can freely suppose that sqrt> (2) is rational because this supposition does not run against any > mathematical concept. At ?st sight it doesnt matter if sqrt(2) is > rational or not. Just after a good reasoning we arrive, by > contradiction, to the conclusion that it cannot be rational.However, in the case of Cantors proof, if we initially suppose that > we have a one-to-one correspondence between N and R, then we are > breaking off the mathematical rules. This means that our initial > premise is false, and we cannot use a bad premise as starting point > in a demonstration trying to ?d a contradiction, or whatever thing > we were looking for.EXACTLY WRONG. A proof by contradiction starts by assuming the opposite to whatever one wants to prove (even if one knows that asssumption is wrong), then showing that it leads to the desired contradiction and so must be false, and what one want to prove must then be true.However, Cantor's diagonal proof does not do this. What it does is a direct proof, at least in Cantor's actual version. It starts with ANY mapping from the naturals to the reals and shows that there are reals not in the image of that mapping. It is a direct proof, NOT a proof by contradiction.X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBUK7i824319; === = Is there any archive/collection of lecture notes' or something like that for graduate-advanced undergrad level courses in Mathematics on the internet? I am aware that many homepages have lecture notes' but as with books, I need to choose. While I have seen recommendations towards standard text-books, I would be grateful for directives to speci? locations on the internet. thanks a lot, gsrk. === = Is there any archive/collection of lecture notes' or something> like that for graduate-advanced undergrad level courses in > Mathematics on the internet? I am aware that many homepages have > lecture notes' but as with books, I need to choose. While I have seen > recommendations towards standard text-books, I would be grateful for > directives to speci? locations on the internet. thanks a lot,> gsrk.Here is a fairly good collection of online textbooks:http://www.math.gatech.edu/~cain/textbooks/ onlinebooks.htmlOnline lecture notes are much harder to ?d, because they disappearso often. you might tryhttp://us.geocities.com/alex_stef/mylist.htmlorhttp:// www.gotmath.com/notes.htmlGood luck.Nathan. === Try the following sites> www.mathforum.com> www.helptosolve.com> www.mathgoodies.com>Surely you mean http://www.mathgoodies.comso it can be opened without hassle from news browser.X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBUK7ja24381; === You are right, and for that reason I say that we must not suppose >>that we have a one-to-one correspondence between N and R to prove >>that Real numbers are innumerable, since N and R have distinct grade >>of in?ity.The only reason that anybody ever makes a supposition that there is a>one-to-one correspondence between N and R is to show that such a>supposition leads to a contradiction, and to show that there is no such>one-to-one correspondence. It's a technique called reductio ad absurdum,>and has been in use at least since the time of the ancient Greeks (who>used the technique to prove that the square root of 2 is irrational, and>also to prove that there are in?itely many prime numbers).David McAnally>down a simple analogy.Lets suppose that somebody wants to prove that he cant ?e Superman. He initially states that I can ?uper man, and then, after a portion of good reasoning against the ground, he comes to the conclusion that he cannot ?e Superman. Do you think he has proved something, knowing beforehand that the premise I can ?Superman goes against the law of gravity?This analogy doesnt re?xactly the problem with Cantors proof, but it is good enough to transmit the idea I want to express.Nicolas de la Foz>-- === Let?s suppose that somebody wants to prove that he can?t ?e > Superman. He initially states that ?I can ?Superman?, and then, > after a portion of good reasoning against the ground, he comes to the > conclusion that he cannot ?e Superman. Do you think he has > proved something, knowing beforehand that the premise ?I can ?> Superman? goes against the law of gravity?What if I could ?e Superman.Then I can jump out this window.(jumps out window).Ow, that hurt, I guess I cannot ? superman.With what you were saying about cantor's proof vs the sqrt(2) proof,how can one tell between something that is obviously false (likecantor) and something that is only false after some reasoning (sqrt(2)) ?The basis of these arguments is that either a statement is true orfalse (goedel's not here right now). Sometimes you cannot directlyprove that something is true, so you see what happens if it werefalse. If you run into problems, then it must have been true to beginwith. === =Sometimes you cannot directly prove that something is true, so you see what > happens if it were false. If you run into problems, then it must have been true > to begin with.Exactly. Although, stylistically, proofs by contradiction are usuallyused to prove a statement false, by assuming it true, rather than theother way around.It's also the case that every proof by contradiction can be restatedas a proof of the contrapositive. AB is equivalent to ~B~A, soproving ~A by contradiction is equivalent to proving that AB andthat ~B is true. Folks who dislike proof by contradiction will restatetheir proofs in this manner, but the result is a stilted style.Len. === Lets suppose that somebody wants to prove that he cant ?e > Superman. He initially states that I can ?Superman, and then, > after a portion of good reasoning against the ground, he comes to the > conclusion that he cannot ?e Superman. Do you think he has > proved something, knowing beforehand that the premise I can ?> Superman goes against the law of gravity?> This analogy doesnt re?xactly the problem with Cantors proof, > but it is good enough to transmit the idea I want to express.> Nicolas de la FozIt is clear that you don't know what Cantor's proof actually says, since Cantor did not assume a mapping from the naturals ONTO the reals, but version of the proof is direct, not by contradiction, It is others who insist on recasting it into an indirect proof.X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBUK7k824414; === >> w.t.s. Min{||x0-y|| : y in K} = Max{|| : y in K^perp, ||y||=1}>>>> I'm getting stuck but should be pretty easy :(>>> know from given info that ||x0 -y|| >= ||x0|| for all y in K>> Also that there exists a y in K, z in K^perp s.t. x0 = y + z>>> so ||x0 - y|| = ||z|| > = ||x0|| ... I think I'm going wrong way with this.>> This result is false as stated. If K = H then the min is 0; but the>sup of the empty set is -in?ity. (The max doesn't exist.)Always watch the boundary cases...--Ron BruckOh, oh yes. What if we restrict K a proper subspace of H. Then what about:Let P be orthogonal projection of H onto K: P = P^2, ||x0||^2 = ||Px||^2 + ||x-Px||^2, Px perpendicular to (x-Px). Then Hahn-Banach thm & equals above imply sup{|| : y in K^perp, ||y||=1} = sup{|| : y in K^perp, ||y||=1} = ||y-Py|| inf{||x0-y|| : y in K} = ||y-Py|| so, Min{||x0-y|| : y in K} = Max{|| : y in K^perp, ||y||=1}. What people think?!?!!! X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBUK7kV24426; === James Harris Dear Sir, I am sorry, that You are working so hardwith some certain details. These are alsosome polynomials with specy? coef?ientsand we can not so be sure in general for optional coef?ients... I was wondering Your developments anyway,but in the very end I am enjoying some of moregeneralised and approximately simple developmentsfor to close the FLT deal. ( see my topic:FLT: advanced developments with ideal parameters:TAB/tabunpedited ca 23.12.03 at sci. math as well ) Imagine, that in the very end I broke the pseudosymmetryof primary FLT equation; lets call it f(t). After someinput procedure I'll change one previous parameter into the certain new one. Now primary FLT equation is transformedinto some F(t) . Once f(t) is of n degree so F(t) is of n(n-1)degree. Looking for consistency for at least one positiveroot we should assume ?st derivative of F(t) equal to zero.( very good known is that primary f(t) equation is monotonic with only one real and positive root for real values of parameters )Also after executing F'(t)= 0 , the most obstacles are gone: You can use Eisenstein criterion and ?d simple inconsistencyfor remining natural values... I hope for some constructive critics too Ro === James Harris Dear Sir,> I am sorry, that You are working so hardI'm sorry but I'm focused on my own work. I say that if you have yourown to present: present it.I know that that involves putting yourself out there, but I hopeyou'll respect my efforts to promote my *own* work in threads I'vecreated, and I will not go to any threads you create promoting my workto the potential detriment of yours.James HarrisX-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBUMgv603136; === =I want to know how we can ?d the coef?ients of this serie?> _L_> C =/_ (a_n)^2 * f_n(t)> n=0C = is a constant>a_n = series coef?ients (we want to ?d)>f_n = constitue an orthogonal basis functionSarahIn general you can't. Suppose f_n(t) are all cosine functions. Since all of your coef?ients are positive, how can you represent C = -1 at t = 0 ?philX-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBUMguG03088; === See>Those interesed should read Milnor's summary of major results related tothe Poincare conjecture in Notices of AMS. The last paragraph brie?Buddenhagen-- To reply copy jbuddenh@REMOVEtexas.net to address bar and edit out REMOVE === =phil> Seeold_math_problem_may_have_been_solved/There is a thread about it at sci.math.research also.LHX-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBUMgvM03103; === = Italy is a funny country they are mental there, they are even more racist in football than anything, the football clubs even admit they will never have a black or a jew playing for their team, i think countrys need racist to balance things, im not sayin coloreds or minoritys are bad but some of them abuse and ruin the society of a country instead of embracing it.from Mike === Italy is a funny country they are mental there, they are even more racist in football than anything, the football clubs> even admit they will never have a black or a jew playing for their team,I admit I know next to nothing about soccer (which is probably thecorrect english term for what you called ?football', IIRC), but AFAICTno club admitted what you said. That they do think so may well bepossible indeed, but saying it publicly is another matter. And FWIW,there are many black athletes in italian teams.Michele-- > Comments should say _why_ something is being done.Oh? My comments always say what _really_ should have happened. :)- Tore Aursand on comp.lang.perl.misc === =Michele Dondi scribbled the following:>> Italy is a funny country they are mental there, they are even more racist in football than anything, the football clubs>> even admit they will never have a black or a jew playing for their team,> I admit I know next to nothing about soccer (which is probably the> correct english term for what you called ?football', IIRC), but AFAICT> no club admitted what you said. That they do think so may well be> possible indeed, but saying it publicly is another matter. And FWIW,> there are many black athletes in italian teams.Depends on where you live. In the USA, they call it soccer. Everywhereelse (including the UK, Canada and Australia) they call it football.-- /-- Joona Palaste (palaste@cc.helsinki.? - Finland - http://www.helsinki.?~palaste rules! --/This is a personnel commuter. - Train driver in Scienti? American === Michele Dondi scribbled the following:> Italy is a funny country they are mental there, they are even more racist in football than anything, the football clubs> even admit they will never have a black or a jew playing for their team,>> I admit I know next to nothing about soccer (which is probably the>> correct english term for what you called ?football', IIRC), but AFAICT>> no club admitted what you said. That they do think so may well be>> possible indeed, but saying it publicly is another matter. And FWIW,>> there are many black athletes in italian teams.> Depends on where you live. In the USA, they call it soccer. Everywhere> else (including the UK, Canada and Australia) they call it football.Not quite. In the UK it's called football, but in Canada, Australiaand the USA, they have different games that are called football.Hence, soccer, which is derived from association football.-- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === =On Fri, 2 Jan 2004 19:56:48 +0000 (UTC), Dave Seaman>> Depends on where you live. In the USA, they call it soccer. Everywhere>> else (including the UK, Canada and Australia) they call it football.Not quite. In the UK it's called football, but in Canada, Australia>and the USA, they have different games that are called football.>Hence, soccer, which is derived from association football.Then again, in most the languages I know it translates directly toexception then the majority of the football-playing world calls itfootball. === On Fri, 2 Jan 2004 19:56:48 +0000 (UTC), Dave Seaman> Depends on where you live. In the USA, they call it soccer. Everywhere> else (including the UK, Canada and Australia) they call it football.Not quite. In the UK it's called football, but in Canada, Australia>>and the USA, they have different games that are called football.>>Hence, soccer, which is derived from association football. Then again, in most the languages I know it translates directly to> exception then the majority of the football-playing world calls it> football.My Chinese wife tells me that football translates to the charactersfor foot and ball (transliterated as zu qiu inMandarin -- she thinks).-- Sale or rental of this disc is ILLEGAL. If you have rented orpurchased this disc, please call the MPAA at 1-800-NO-COPYS. -- The MPAA begins a new anti-piracy program, found on a DVD purchased in China === =Toni Lassila scribbled the following:> On Fri, 2 Jan 2004 19:56:48 +0000 (UTC), Dave Seaman> Depends on where you live. In the USA, they call it soccer. Everywhere> else (including the UK, Canada and Australia) they call it football.Not quite. In the UK it's called football, but in Canada, Australia>>and the USA, they have different games that are called football.>>Hence, soccer, which is derived from association football.I was aware of the USA case but I didn't know about Canada orAustralia. Serves me right for only having visited the UK and the USA.> Then again, in most the languages I know it translates directly to> exception then the majority of the football-playing world calls it> football.This is true. In Finnish it's jalkapallo, from jalka (foot orleg) and pallo (ball), in Swedish it's fotboll, in German it'sFuball, and so on. I don't know exactly about French. It's notle boule du pieds but it certainly isn't soccer either.-- /-- Joona Palaste (palaste@cc.helsinki.? - Finland - http://www.helsinki.?~palaste rules! --/Normal is what everyone else is, and you're not. - Dr. Tolian Soran === Toni Lassila scribbled the following:>> On Fri, 2 Jan 2004 19:56:48 +0000 (UTC), Dave Seaman>> Depends on where you live. In the USA, they call it soccer. Everywhere>> else (including the UK, Canada and Australia) they call it football.>Not quite. In the UK it's called football, but in Canada, Australia>and the USA, they have different games that are called football.>Hence, soccer, which is derived from association football.> I was aware of the USA case but I didn't know about Canada or> Australia. Serves me right for only having visited the UK and the USA.The Australian game is called Australian Rules Football. I watched agame on TV once, but I couldn't make any sense of it. People claim thatthe game has no rules, but that's probably just a ?st impression. Kindof like rugby, but less structured.Canada has the CFL, which plays a game that is vaguely similar toAmerican football, but with some important differences.>> Then again, in most the languages I know it translates directly to>> exception then the majority of the football-playing world calls it>> football.> This is true. In Finnish it's jalkapallo, from jalka (foot or> leg) and pallo (ball), in Swedish it's fotboll, in German it's> Fuball, and so on. I don't know exactly about French. It's not> le boule du pieds but it certainly isn't soccer either.In French it's le football. Spanish is futbol. Russian is the same,after you transliterate from the Cyrillic.-- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === =[snip]> The Australian game is called Australian Rules Football. I watched a> game on TV once, but I couldn't make any sense of it. People claim that> the game has no rules, but that's probably just a ?st impression. Kind> of like rugby, but less structured.The thing to remember about our game is that it was designed to keepour cricketers ? in the off season. It is therefore a running andpassing game, no player can keep possession of the ball for more thana few steps and the ball tends to be in play for a much largerproportion of the game than Grid Iron. (The most frustrating thingabout watching an American Fottball game is that they will not passthe ball !, is it actually against the rules ?)WayneX-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBUMgu403084; === =X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBUMgu103092; === Who proved that an in?ite number of numbers (i.e. in?ite series)>can have a ?ite sum?Aristotle, regarding Zeno's paradoxes. He said there were several kinds of in?ity. including in?ite as to extent and in?ite asto divisibility.philX-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBVDVav31690; === =I am wishing that I paid more attention in my pre-calculus/calculus >courses when it came to sums and series...If I state, sum^{N}_{i=1} t_{i} = 0what does the following sum equal? sum^{N}_{i=1} t^{2}_{i}My naive thought was that it was also zero but after working backwards>from the next equation in a stats/astronomy paper, the author has>something like: sum^{N}_{i=1} t^{2}_{i} = (Delta t^2)/12 N^3 Any ideas?Your ?st sum is the mean. Your second sum is the variance. They arein general independent. So if mean is zero, variance can be any non-negative number.philX-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBVDVae31694; === =what's the average distance between two points within or on the boundary of a circle of radius r?i've tried setting up two quadruple integrals, one polar and one cartesian. the cartesian one nearly fried my cpu and the polar one gave me zero so i think there is improperness involved unless it is zero but that doesn't make perfect sense. i tried differnent orderings of integration, though probably not all of them, and i'm not super sure i even have the right integrals or if integration is the best approach. maybe there's a way to reduce it by symmetry arguments.if one point is ?ed at the origin, do you get 2r/3 for the average distance to the origin? === what's the average distance between two points within or on the boundary> of a circle of radius r?That depends on the distribution, which you have not speci?d. Themost obvious distribution is uniform by area (i.e., the probability of apoint being in a certain region within the circle is proportional to thearea of the region). > i've tried setting up two quadruple integrals, one polar and one> cartesian. the cartesian one nearly fried my cpu and the polar one gave> me zero so i think there is improperness involved unless it is zero but> that doesn't make perfect sense. i tried differnent orderings of> integration, though probably not all of them, and i'm not super sure i> even have the right integrals or if integration is the best approach.> maybe there's a way to reduce it by symmetry arguments.If you use polar form, you can reduce it to a triple integral; symmetrywill let you ? one of the angles.> if one point is ?ed at the origin, do you get 2r/3 for the average> distance to the origin?If the distribution is uniform by area, yes. (And if one point is ?edat the boundary of the circle, I get 32r/9pi for the average distance,but I don't know whether that will help you.)-- Daniel W. Johnsonpanoptes@iquest.nethttp://members.iquest.net/~panoptes /039 53 36 N / 086 11 55 W === =En el mensaje:1g6vupw.1hdobfw10d0dmoN%panoptes@iquest.net,Daniel W. Johnson escribi.97:> what's the average distance between two points within or on the>> boundary of a circle of radius r? That depends on the distribution, which you have not speci?d. The> most obvious distribution is uniform by area (i.e., the probability> of a point being in a certain region within the circle is> proportional to the area of the region).> i've tried setting up two quadruple integrals, one polar and one>> cartesian. the cartesian one nearly fried my cpu and the polar one>> gave me zero so i think there is improperness involved unless it is>> zero but that doesn't make perfect sense. i tried differnent>> orderings of integration, though probably not all of them, and i'm>> not super sure i even have the right integrals or if integration is>> the best approach. maybe there's a way to reduce it by symmetry>> arguments. If you use polar form, you can reduce it to a triple integral;> symmetry will let you ? one of the angles.> if one point is ?ed at the origin, do you get 2r/3 for the average>> distance to the origin? If the distribution is uniform by area, yes. (And if one point is> ?ed at the boundary of the circle, I get 32r/9pi for the average> distance, but I don't know whether that will help you.)Then it is solved. As I indicated in my previous post, the average distanceis then (4/5)(32r/9pi ) = 128/(45pi) ~= 0,9054 r-- Ignacio Larrosa Ca.96estroA Coru.96a (Espa.96a)ilarrosaQUITARMAYUSCULAS@mundo-r.com === what's the average distance between two points within or on the> boundary of a circle of radius r? i've tried setting up two quadruple integrals, one polar and one> cartesian. the cartesian one nearly fried my cpu and the polar one> gave me zero so i think there is improperness involved unless it is> zero but that doesn't make perfect sense. i tried differnent> orderings of integration, though probably not all of them, and i'm> not super sure i even have the right integrals or if integration is> the best approach. maybe there's a way to reduce it by symmetry> arguments. if one point is ?ed at the origin, do you get 2r/3 for the average> distance to the origin?Let r = 1 the radius of the disk C, for simplicity. Consider a second disk Dof radius (1 - k), and the circular krown E between both circles. Its areasare:(C) = pi(D) = (1 - k)^2*pi(E) = pi(1 - (1 - k)^2) = pi(2k - k^2)Let d to be the average distance between points of C, e the average distancebetween a point of D an a point of E, and f the average distance betweenpoints of E. Then,d = ((1-k)^2*pi)^2(1-k)d + 2(1-k)^2*pi*(2k-k^2)pi*e) +((2k-k^2)pi)^2*f)/pi^2 = (1-k)^5*d + 2(1-k)^2*(2k-k^2)e + (2k-k^2)^2*fSolving for d,d = (2(1-k)^2*(2k-k^2)e + (2k-k^2)^2*f )/(1 - (1 - k)^5)Letting k > 0, you get d = (4/5)e. I.E., the average distance betweenpoints of the disk is 4/5 of the average distance between pont of the circleand points of the disk. By simmetry, you only need calculta the averagedistance from a point of the circle and all the points of the disk.Now, I can't continue. Perhaps later ...-- Ignacio Larrosa Ca.96estroA Coru.96a (Espa.96a)ilarrosaQUITARMAYUSCULAS@mundo-r.comX-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBVDVbT31724; === Can anyone tell me what the odds of winning here are given that only>60 numbers out of an available 75 are called?http:// maxim.skyphix.com/bingo.jpg= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =I guess depending upon what three bingo cards you get!Also there are some duplicate numbers on all three cards.I never played Bingo, but what does the red area representon each card? I assume the center of each board is a wild card (ball). As longas you can place two markers on either side of it you can have awinner with only 4 markers instead of 5 across or 5 down or 4 eachon any diagonal and one center vertical 4 for any particular board.So hypothetically you could have as little as 12 markers for winnerson all three boards and a maximum of 15 markers for winners.I am not sure what the red areas on each card are, so I could bewrong!Also to hit the jackpot does each card need to be a winner witheither 5 markers across or 5 down or 5 diagonal? Naturally thecenter I believe is a wild (ball) which could be included for any winon any card. So trying to ?ure out any kind of odds you would need the threecards before hand to analyze the data.Good luck with that!Portly X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBVDVdL31761; === >Message-id: Is this known about certain consecutive pairs of factorials and the>>Collatz Conjecture where the ?st pair is trivial but pertinent with>>the rest of the pairs 1! path falls completley within the 2! path of a Collatz sequence.>>3! path falls completely within the 4! path of a Collatz sequence.>>7! path falls completely within the 8! path >>15! path falls completely within the 16! path Do to integer size I cannot check higher pairs of factorials.I use Python for this type of work. You can do Collatz stuff with>extremely large numbers. The limits are memory and exectution time,>not the numbers themselves. A Collatz run on 2**100000 only takes>two minutes (with the GMPY module) on a 1.7 GHz processor.>It looks like the next pair may be--31! path may fall completely within 32! path?After I ran it32!>263130836933693530167218012160000000> 131565418466846765083609006080000000> 65782709233423382541804503040000000> 32891354616711691270902251520000000> 16445677308355845635451125760000000> 8222838654177922817725562880000000>.>.>.31!> 8222838654177922817725562880000000> 4111419327088961408862781440000000> 2055709663544480704431390720000000> 1027854831772240352215695360000000>.>.>.it was obvious.I am guessing here but I am looking at the four largest of the ?st 4>>pairs -->>2!,4!,8!,16!, so I am assuming the next larger of the pair would be>>32!.>>If it is, it probably will be true for the rest of 2^n >oo.Yes, beacuse 32! is 32 * 31!and 32 is a power of two which means 32! is on the same branch as 31!,>?e nodes higher (because 32 is 2**5).>Could someone check this out that use a programming language that can>>handle integers size > 10^15 ?It would be interesting to know if 31! path falls completely within>>32! path.>>Also, any other consecutive pairs of factorials > 18! that may>>have this property.It will always work when the larger factorial is a power of two.>>>Dan>-->Mensanator>Ace of ClubsMakes sense even though 2^n <> n!. I was wondering if the next two powers 3 or 5 have this property, like 3^n or 5^n series.Probably not because of the hal?g process for all sequences where the power of 2 has its own sequence.I will check it out.Dan === =Message-id: Is this known about certain consecutive pairs of factorials and the>Collatz Conjecture where the ?st pair is trivial but pertinent with>the rest of the pairs 1! path falls completley within the 2! path of a Collatz sequence.>3! path falls completely within the 4! path of a Collatz sequence.>7! path falls completely within the 8! path >15! path falls completely within the 16! path Do to integer size I cannot check higher pairs of factorials.I use Python for this type of work. You can do Collatz stuff with>>extremely large numbers. The limits are memory and exectution time,>>not the numbers themselves. A Collatz run on 2**100000 only takes>>two minutes (with the GMPY module) on a 1.7 GHz processor.>>It looks like the next pair may be--31! path may fall completely within 32! path?After I ran it32!>>263130836933693530167218012160000000>> 131565418466846765083609006080000000>> 65782709233423382541804503040000000>> 32891354616711691270902251520000000>> 16445677308355845635451125760000000>> 8222838654177922817725562880000000>>.>>.>>.31!>> 8222838654177922817725562880000000>> 4111419327088961408862781440000000>> 2055709663544480704431390720000000>> 1027854831772240352215695360000000>>.>>.>>.it was obvious.I am guessing here but I am looking at the four largest of the ?st 4>pairs -->2!,4!,8!,16!, so I am assuming the next larger of the pair would be>32!.>If it is, it probably will be true for the rest of 2^n >oo.Yes, beacuse 32! is 32 * 31!and 32 is a power of two which means 32! is on the same branch as 31!,>>?e nodes higher (because 32 is 2**5).>>Could someone check this out that use a programming language that can>handle integers size > 10^15 ?It would be interesting to know if 31! path falls completely within>32! path.>Also, any other consecutive pairs of factorials > 18! that may>have this property.It will always work when the larger factorial is a power of two.> >Dan>>-->>Mensanator>>Ace of Clubs>Makes sense even though 2^n <> n!. I meant to say is a factorial of a power of 2, i.e., (2**n)!I was wondering if the next two powers 3 or 5 have this property, like 3^n or>5^n series.>Probably not because of the hal?g process for all sequences where the power>of 2 has its own sequence.Every branch of the Collatz tree starts with an odd number andextends to in?ity by factors of 2. Every branch has the formx*2**0, x*2**1, x*2**2, . . . x*2**nwhere x is the root odd number of the branch. For the powers of two, x just happens to be 1 and is the only branch that isnot a sub-branch, but structurally, it is no different than anyother branch. If you were to look at 3x+3, you would ?d thatthe powers of have no signi?ance whatsoever as some otherbranch (3, 6, 12, 24...) forms the trunk of that tree.If two numbers are on the same branch of the Collatz tree, then the smaller number's sequence will be a subset of the larger number's sequence (the reverse is not necessarily true, a sequence can be a subset of a larger sequence without starting on the same branch). Two numbers on the same branch differ by factors of 2. Also, there can only be one odd number on a branch.The powers of 3 and 5 differ by factors of 3 and 5 respectively,therefore, they cannot occupy the same branch. The powers of 3 and 5 are all odd, therefore, they cannot occupy the same branch. But all that does is prove they aren't on the same branch, a power of 5 sequence could still be a subset of a larger power of 5 sequence.I will check it out.However, a power of 3 sequence cannot be a subset of another power of 3 sequence. All powers of three are odd and all are divisible by 3. A Collatz sequence cannot have more than one odd number that is divisible by 3. Another way of saying that is a Collatz sequence can only have a single number that is == 1 (mod 2) and == 0 (mod 3).I wouldn't waste any time checking the powers of 3.>Dan --MensanatorAce of Clubs === =Message-id: Is this known about certain consecutive pairs of factorials and the>Collatz Conjecture where the ?st pair is trivial but pertinent with>the rest of the pairs 1! path falls completley within the 2! path of a Collatz sequence.>3! path falls completely within the 4! path of a Collatz sequence.>7! path falls completely within the 8! path >15! path falls completely within the 16! path Do to integer size I cannot check higher pairs of factorials.I use Python for this type of work. You can do Collatz stuff with>>extremely large numbers. The limits are memory and exectution time,>>not the numbers themselves. A Collatz run on 2**100000 only takes>>two minutes (with the GMPY module) on a 1.7 GHz processor.>>It looks like the next pair may be--31! path may fall completely within 32! path?After I ran it32!>>263130836933693530167218012160000000>> 131565418466846765083609006080000000>> 65782709233423382541804503040000000>> 32891354616711691270902251520000000>> 16445677308355845635451125760000000>> 8222838654177922817725562880000000>>.>>.>>.31!>> 8222838654177922817725562880000000>> 4111419327088961408862781440000000>> 2055709663544480704431390720000000>> 1027854831772240352215695360000000>>.>>.>>.it was obvious.I am guessing here but I am looking at the four largest of the ?st 4>pairs -->2!,4!,8!,16!, so I am assuming the next larger of the pair would be>32!.>If it is, it probably will be true for the rest of 2^n >oo.Yes, beacuse 32! is 32 * 31!and 32 is a power of two which means 32! is on the same branch as 31!,>>?e nodes higher (because 32 is 2**5).>>Could someone check this out that use a programming language that can>handle integers size > 10^15 ?It would be interesting to know if 31! path falls completely within>32! path.>Also, any other consecutive pairs of factorials > 18! that may>have this property.It will always work when the larger factorial is a power of two.> >Dan>>-->>Mensanator>>Ace of Clubs>Makes sense even though 2^n <> n!. I meant to say is a factorial of a power of 2, i.e., (2**n)!> I was wondering if the next two powers 3 or 5 have this property, like 3^n or>5^n series.>Probably not because of the hal?g process for all sequences where the power>of 2 has its own sequence.Every branch of the Collatz tree starts with an odd number and> extends to in?ity by factors of 2. Every branch has the formx*2**0, x*2**1, x*2**2, . . . x*2**nwhere x is the root odd number of the branch. For the powers > of two, x just happens to be 1 and is the only branch that is> not a sub-branch, but structurally, it is no different than any> other branch. If you were to look at 3x+3, you would ?d that> the powers of have no signi?ance whatsoever as some other> branch (3, 6, 12, 24...) forms the trunk of that tree.If two numbers are on the same branch of the Collatz tree, > then the smaller number's sequence will be a subset of the > larger number's sequence (the reverse is not necessarily true, > a sequence can be a subset of a larger sequence without > starting on the same branch). Two numbers on the same > branch differ by factors of 2. Also, there can only be one > odd number on a branch.The powers of 3 and 5 differ by factors of 3 and 5 respectively,> therefore, they cannot occupy the same branch. The powers > of 3 and 5 are all odd, therefore, they cannot occupy the same > branch. But all that does is prove they aren't on the same > branch, a power of 5 sequence could still be a subset of a > larger power of 5 sequence.> I will check it out.However, a power of 3 sequence cannot be a subset of > another power of 3 sequence. All powers of three are odd > and all are divisible by 3. A Collatz sequence cannot have > more than one odd number that is divisible by 3. Another > way of saying that is a Collatz sequence can only have a > single number that is == 1 (mod 2) and == 0 (mod 3).> I wouldn't waste any time checking the powers of 3.>DanI have given up on powers and am looking @ prime twins instead.Not many so far!In reverse order -- 3 and 5 (5)path is contained in (3)path 11 and 13 (13)path is contained in (11)path In order -- 17 and 19 -- 71 and 73 -- 107 and 109 Still searching from there!It appears there are no more after these, but who knows? DanX-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBVLqNc32102; === =Just ?ished reading a really great book about the theorem and also Wiles attempts to solve itFermat's Last Theoremby Simon SinghPublished by ?Fourth Estate' Hope this helpsI'm a sincere math amateur, and I saw a series on PBS about Andrew Wiles'>attempt to solve Fermat's Last Theorem. If I recall, he was close, but that>an associate pointed out a fundamental ? his theory.Does anyone know if he ?ed the theorem -- and is it now considered>proved? Also, can anyone recommend a good book with regard to Wiles'>attempt to solve the problem over 7 years? === =I'm trying to design a function with certain properties, and it's just > driving me crazy... it really should be simple, but I just don't succeed.A rough scetch can be found on http://www.wernhoff.com/div/graf1.gif.[snip]Although you already found your answer, it may be interesting to notethat the family of curves de?ed by solutions to the ordinarydifferential equation:u_xx + c*u_x = 0 0 < x < 1 eq(1)u(0) = 0, u(1) = 1may be suitable for your purpose as well. The solution to eq(1) isgiven by:(1/(exp(-c)-1))*(-1 + exp(-c*x))It seems to me that the fact that minesweeper is NP-Complete focuses on thewrong question. Instead, let me propose some alternative questions that are(perhaps) more relevant to winning the game. It's not so much that we need toknow if a con?uation is ?legal', but rather how to compute the probabilitythatthere is a mine in a given cell.(1) Suppose one has a con?uration where, with respect to an establishedsafe cell, one of two neighbors (X and Y) must contain a mine. Assuming thatthedistribution of mines is Poisson, then the probability is 1/2 for each of theneighbors containing a mine. However, this analysis is done in isolation from other nearby cells. Suppose in addition there are nearby cells for whichX and Y are two of say 6 cells of and that these 6 cells only have two mines.y 2 1y 3 Mz z Mx 4 Mx x 2 the point of view of the cell marked ?3', the probability should be 1/2whether the mine is in one of the cells marked y and 1/2 whether it is in oneof the cells marked z.So how do we compute an exact probability that there is a mine in one ofthe two z spaces. Clearly we must consider the joint probabilities, butI don't know how to compute the joint pdf. IDEAS?I have tried to do this without success.(2) Suppose one reaches a con?uation where a guess MUST be made. Can an optimal strategy be determined? One could of course choose the guess withhighest probability of success, but this may not be optimal. It could be thecase that a correct guess of a lower probability con?uration reveals more information about how to then proceed. Also, there might be several guess of equal highest probability. All ofthem might not be equal in the sense of information revealed from a successfulguess.You can lead a horse's ass to knowledge, but you can't make him think. === It seems to me that the fact that minesweeper is> NP-Complete focuses on the wrong question.> Instead, let me propose some alternative questions> that are (perhaps) more relevant to winning the game.> It's not so much that we need to know if a con?uation> is ?legal', but rather how to compute the probability> that there is a mine in a given cell....That's all well and good. In pratice one must considerthe number that will be revealed and the probabilitythis standpoint, the optimal strategy might be to choosea square with higher probability to avoid potentialdisaster in the long run. Using your analysis onemay paint oneself into a corner by not taking aslightly higher risk earlier. One needs to balanceboth global (yours) and a long-term strategies(mine). === =You areclaiming that in ANY abelian group, given ANY two elements a andb,the order of ab is the least common multiple of the orders of a and b.This is false.Yet, in ANY abelian group, given ANY two elements a and b, and if none primenumber appear with the same exponent in the decomposition of order of a andorder of b, then the least common multiple of the orders of a and b is orderof abSee Odoux Ramis Deschamps Tome 1 Exercise 2.18 === You areclaiming that in ANY abelian group, given ANY two elements a and>b,the order of ab is the least common multiple of the orders of a and b.>This is false.Yet, in ANY abelian group, given ANY two elements a and b, and if none prime>number appear with the same exponent in the decomposition of order of a and>order of b, then the least common multiple of the orders of a and b is order>of abYes: if and intersect trivially, and a commutes with b, thenthe order of ab is the least common multiple of the orders of a and ofb. This holds, in particular, when the orders are relatively prime.-- === == === ==It's not denial. I'm just very selective about what I accept as reality. Calvin (Calvin and Hobbes) === === === =Arturo Magidinmagidin@math.berkeley.edu === =was discovered at Los Alamos laboratories on News Years 2004. Thisright in the middle of an.., experiment, has Los Alamos scientistsgreatly perplex over chronom tunneling since it was detected a fewdays before it's discovery.- === was discovered at Los Alamos laboratories on News Years 2004. This> right in the middle of an.., experiment, has Los Alamos scientists> greatly perplex over chronom tunneling since it was detected a few> days before it's discovery.-Its 01/01/04, not 04/01/04! === = was discovered at Los Alamos laboratories on News Years 2004. This> right in the middle of an.., experiment, has Los Alamos scientists> greatly perplex over chronom tunneling since it was detected a few> days before it's discovery. Its 01/01/04, not 04/01/04!>Huh? === > was discovered at Los Alamos laboratories on News Years 2004. This> right in the middle of an.., experiment, has Los Alamos scientists> greatly perplex over chronom tunneling since it was detected a few> days before it's discovery. Its 01/01/04, not 04/01/04! Huh?January ?st of the year 2004 C.E. may be abbreviated 01/01/04. === = was discovered at Los Alamos laboratories on News Years 2004. This> right in the middle of an.., experiment, has Los Alamos scientists> greatly perplex over chronom tunneling since it was detected a few> days before it's discovery. Its 01/01/04, not 04/01/04!> Huh? === >> was discovered at Los Alamos laboratories on News Years 2004. This>> right in the middle of an.., experiment, has Los Alamos scientists>> greatly perplex over chronom tunneling since it was detected a few>> days before it's discovery. Its 01/01/04, not 04/01/04!> Huh?> The Fourth of January xxxx.-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === > was discovered at Los Alamos laboratories on News Years 2004. This>> right in the middle of an.., experiment, has Los Alamos scientists>> greatly perplex over chronom tunneling since it was detected a few>> days before it's discovery. Its 01/01/04, not 04/01/04!> Huh?> The Fourth of January xxxx.Not for leftpondians. === =In sci.math, Robin Chapmanon Fri, 02 Jan 2004 09:34:14 +0000:> was discovered at Los Alamos laboratories on News Years 2004. This> right in the middle of an.., experiment, has Los Alamos scientists> greatly perplex over chronom tunneling since it was detected a few> days before it's discovery. Its 01/01/04, not 04/01/04!> Huh? The Fourth of January xxxx.> In Europe, perhaps; in the US it's April Fool's Day. :-)Personally, I thinkxxxx-04-01is a more logical notation, if only because it's alphanumericallysortable. (It's also sanctioned by ISO.)Apparently, someone wants to celebrate April 3 months early... :-)Of course we ?Murkins are a stubborn bunch; we're not metric,for example, except in the scienti? area, although someindustries (car parts, etc.) are presumably using metric bolts.-- #191, ewill3@earthlink.netIt's still legal to go .sigless. === was discovered at Los Alamos laboratories on News Years 2004. This>> right in the middle of an.., experiment, has Los Alamos scientists>> greatly perplex over chronom tunneling since it was detected a few>> days before it's discovery. Its 01/01/04, not 04/01/04!>> Huh? The Fourth of January xxxx.>Well yes, we did will have established a chronon bifurcation whichcould will have come about by chronon anti-chronon dis-anihilation.Yes, it will be was known to us that an event happing at two differenttimes doesn't disrupt chronon quantum continuity while chronom tunneling,an event happening when it doesn't or shouldn't, causes chronon quantumcontinuity stress which wastes time in violation of temporal preservation.We will about to have published a paper about these results pendingupon when they did will happened. Oh, when did I put my car keys?I need to get out of where and go home. === =The Fourth of January xxxx.How do you rationali(sz)e that? === =That's the schema used in the British Empire. dd/mm/yy - which is somewhatmore logical, as the units go in ascending order of size (day, month, year).He should also have known that you seppos (Americans) do it mm/dd/yy.> The Fourth of January xxxx. How do you rationali(sz)e that?> === That's the schema used in the British Empire. dd/mm/yy - which is somewhat> more logical, as the units go in ascending order of size (day, month, year).He should also have known that you seppos (Americans) do it mm/dd/yy. > The Fourth of January xxxx. How do you rationali(sz)e that?>I am aware of something called the British Commonwealth, but where on earth is the British Empire? === >> That's the schema used in the British Empire. dd/mm/yy - which is>> somewhat more logical, as the units go in ascending order of size (day,>> month, year).>> >> He should also have known that you seppos (Americans) do it mm/dd/yy.>>>>> The Fourth of January xxxx. How do you rationali(sz)e that?>I am aware of something called the British Commonwealth, but where on> earth is the British Empire?Taken over by the US one I'm afraid... === = That's the schema used in the British Empire. dd/mm/yy - which is somewhat> more logical, as the units go in ascending order of size (day, month,year). He should also have known that you seppos (Americans) do it mm/dd/yy.> The Fourth of January xxxx.> How do you rationali(sz)e that?> I am aware of something called the British Commonwealth, but> where on earth is the British Empire?See OP and run it backwards.Bill === =On Fri, 02 Jan 2004 19:43:48 +1000, Bruce Percy > The Fourth of January xxxx.How do you rationali(sz)e that?04*xxxx/01 === =was discovered at Los Alamos laboratories on News Years 2004. This> right in the middle of an.., experiment, has Los Alamos scientists> greatly perplex over chronom tunneling since it was detected a few> days before it's discovery.Wasn't there a time past with higher standards when a Village Idiotprovided entertainment?--Uncle Alhttp://www.mazepath.com/uncleal/qz.pdfhttp:// www.mazepath.com/uncleal/eotvos.htm (Do something naughty to physics) === Wasn't there a time past with higher standards when a Village Idiot> provided entertainment?>Don't worry Uncle Al, I won't hold you to those standards. === >>If f: R^2->R, what conditions ensure that [i]>>lim_{x->a} [f(x,x)-f(x,a)] / (x-a) = f_2(a,a)(where f_2 is the partial derivative wrt the second argument)?>>> It's easy to see that this holds if f_2 is continuous at (a,a),>> for example, the Mean Value Theorem shows that for>> x > a there exists y = y(x) such that a < y(x) < x and>>> [f(x,x)-f(x,a)] / (x-a) = f_2(x, y(x)),>>> which tends to f_2(a,a) since if f_2 is continuous. You>> should note that in fact f_2 continuous implies something>> much stronger:>>> [ii] if phi(x) <> a for x <> a and phi(x) -> a as x -> a then>>> (f(x, phi(x)), f(x, a)) / (phi(x) - a) -> f_2(a,a) (as x -> a).>>> The proof of [ii] when f_2 is continuous is the same.>>> On the other hand, you ask about the case phi(x) = x.>> In that case [i] follows from a much weaker condition,>> namely that f be _differentiable_ at (a,a)Differentiability at (a,a) is not weaker than f_2 being continuous at >(a,a).Of course my ?st reaction was what???.Uh, no of course it isn't. It is weaker than supposing thatboth f_1 and f_2 are continuous at (a,a)...************************David C. Ullrich === =THANKS|| >If f: R^2->R, what conditions ensure that| >| [i]| >lim_{x->a} [f(x,x)-f(x,a)] / (x-a) = f_2(a,a)| >| >(where f_2 is the partial derivative wrt the second argument)?|| It's easy to see that this holds if f_2 is continuous at (a,a),| for example, the Mean Value Theorem shows that for| x > a there exists y = y(x) such that a < y(x) < x and|| [f(x,x)-f(x,a)] / (x-a) = f_2(x, y(x)),|| which tends to f_2(a,a) since if f_2 is continuous. You| should note that in fact f_2 continuous implies something| much stronger:|| [ii] if phi(x) <> a for x <> a and phi(x) -> a as x -> a then|| (f(x, phi(x)), f(x, a)) / (phi(x) - a) -> f_2(a,a) (as x -> a).|| The proof of [ii] when f_2 is continuous is the same.|| On the other hand, you ask about the case phi(x) = x.| In that case [i] follows from a much weaker condition,| namely that f be _differentiable_ at (a,a), in the| sense de?ed in that other thread. Writing things out| in coordinates, saying f is differentiable at (a,a) says| that there exist A and B such that|| f(x,y) = f(a,a) + A(x-a) + B(y-a) + o(|x-a|+|y-a|)|| (and as we saw the other day it follows that B = f_2(a,a).)| Hence|| f(x,x) - f(x,a) = B(x-a) + o(|x-a| + |x-a| + |x-a| + |a-a|),|| so|| [f(x,x)-f(x,a)] / (x-a) = B + o(1) -> B.|| So: the answer to the question you asked is It| holds if f is differentiable, which is probably| more or less the weakest natural condition| you're going to get. But, in case you were assuming| that the condition giving [i] would also give [ii],| you should note that f differentiable at (a,a)| does _not_ imply [ii]; if [ii] is what really comes| up and you just asked about [i] to simplify| the question then the condition you really want| is f_2 is continuous at (a,a).||| >Any help is mucho appreciado.| >| >Peace| >||| ************************|| David C. Ullrich === THANKSI posed a little exercise in a followup to my own post.You should note there was a typo in the statement ofthe exercise (leading to Exercise 2: try to do Exercise 1and in the course of looking for the proof ?ure out whatthe correct statement is...)************************David C. Ullrich === I knew that a suf?ient condition for f:R^n -> R to be differentiable>at a point u is that all of its partial derivatives exist in an open>ball centered at u and are continuous at u. But then I read there's a>weaker condition, that is, f is differentiable at u if one of it's>partial derivatives exists at u (it need not exist in a neighborhood>of u) and the others exist in a neighborhood of u and are continuous>at u. Seems this proof is in Apostol book.>I don't have this book, so I'm trying to work out a proof myself. >To make things simpler, we can, WLOG, consider the case f:R^2 -> R and>suppose u =(0,0). Suppose f_1 exists at (0,0)and f_2 is continuous at>(0,0) and exists in an open ball B centered at the origin. If h1<>0 >and h_2<>0 are suf?iently small in absolure value that (h1,h2) is in>B, then f(h1,h2) - f(0,0) = f(h1,h2) - f(h1,0) + f(h1,0) - f(0,0).>Since f_2 exists in B, we can apply the Mean Value Theorem>(unidimensional case) to get a point p on the line segment>joining(h1,0) and (h1, h2) that satis?s f(h1,h2) - f(h1,0) = f_2(p)*>h2. Therefore, we can write f(h1,h2) - f(h1,0) = f_2(0,0)*h2 + [f_2(p)>- f_2(0,0)]*h2 .>Since f_1 exists at (0,0), we have f(h1,0) - f(0,0) = f_1(0,0)*h1 +>o(h1), where o(h1)/h1 goes to zero as h1 does.>Now, from the continuity of f_2 at (0,0), we notice that, if (h1,h2)>-> (0,0), then g(h1,h2) =[f_2(p) - f_2(0,0)] -> 0 (taking into account>that p depends on h1 and h2). This means that f(h1,h2) - f(h1,0) f_1(0,0)*h1 + f_2(0,0)*h2 + o(h1) + g(h1,h2)*h2 = f_1(0,0)*h1 +>f_2(0,0)*h2 + |(h1,h2)|* v(h1,h2), where v(h1,h2) = [o(h1) +>g(h1,h2)*h2]/|(h1,h2)| = o(h1)/|(h1,h2)| + g(h1,h2)*[{h2/|(h1,h2)| .>But since |h2|/|(h1,h2)| remains bounded when (h1, h2) -> (0,0), we>infer v(h1,h2) ->0. This shows f is differentiable at (0,0) and it's>derivative is the linear function that sends (h1,h2) into f_1(0,0)*h1>+ f_2(0,0)*h2.Right. (Didn't read it incredibly carefully, but there's a rightproof that has an outline exactly like this, so I imagine it'sright. What's crucial is that you start by saying f(h1,h2) - f(0,0) = f(h1,h2) - f(h1,0) + f(h1,0) - f(0,0);if you'd split it the other way, f(h1,h2) - f(0,0) = f(h1,h2) - f(0,h2) + f(0,h2) - f(0,0)the proof wouldn't work.)>The n_th dimensional case can be treated similarly, we walk inside the>ball where f_2,...f_n are continuous, increasing only one variable at>each move in this ball, and applying MVT.>Seems the proof of this fact is not that hard, unless, of course, I>made a mistake. Anyway, it came as a surprise to me that this>condition for diffrentiabilty holds.>Happy 2004>Amanda************************David C. Ullrich === >I knew that a suf?ient condition for f:R^n -> R to be differentiable>at a point u is that all of its partial derivatives exist in an open>ball centered at u and are continuous at u. But then I read there's a>weaker condition, that is, f is differentiable at u if one of it's>partial derivatives exists at u (it need not exist in a neighborhood>of u) and the others exist in a neighborhood of u and are continuous>at u. Seems this proof is in Apostol book.>I don't have this book, so I'm trying to work out a proof myself. >To make things simpler, we can, WLOG, consider the case f:R^2 -> R and>suppose u =(0,0). Suppose f_1 exists at (0,0)and f_2 is continuous at>(0,0) and exists in an open ball B centered at the origin. If h1<>0 >and h_2<>0 are suf?iently small in absolure value that (h1,h2) is in>B, then f(h1,h2) - f(0,0) = f(h1,h2) - f(h1,0) + f(h1,0) - f(0,0).>Since f_2 exists in B, we can apply the Mean Value Theorem>(unidimensional case) to get a point p on the line segment>joining(h1,0) and (h1, h2) that satis?s f(h1,h2) - f(h1,0) = f_2(p)*>h2. Therefore, we can write f(h1,h2) - f(h1,0) = f_2(0,0)*h2 + [f_2(p)>- f_2(0,0)]*h2 .>Since f_1 exists at (0,0), we have f(h1,0) - f(0,0) = f_1(0,0)*h1 +>o(h1), where o(h1)/h1 goes to zero as h1 does.>Now, from the continuity of f_2 at (0,0), we notice that, if (h1,h2)>-> (0,0), then g(h1,h2) =[f_2(p) - f_2(0,0)] -> 0 (taking into account>that p depends on h1 and h2). This means that f(h1,h2) - f(h1,0) f_1(0,0)*h1 + f_2(0,0)*h2 + o(h1) + g(h1,h2)*h2 = f_1(0,0)*h1 +>f_2(0,0)*h2 + |(h1,h2)|* v(h1,h2), where v(h1,h2) = [o(h1) +>g(h1,h2)*h2]/|(h1,h2)| = o(h1)/|(h1,h2)| + g(h1,h2)*[{h2/|(h1,h2)| .>But since |h2|/|(h1,h2)| remains bounded when (h1, h2) -> (0,0), we>infer v(h1,h2) ->0. This shows f is differentiable at (0,0) and it's>derivative is the linear function that sends (h1,h2) into f_1(0,0)*h1>+ f_2(0,0)*h2.Right. (Didn't read it incredibly carefully, but there's a right> proof that has an outline exactly like this, so I imagine it's> right. What's crucial is that you start by saying f(h1,h2) - f(0,0) = f(h1,h2) - f(h1,0) + f(h1,0) - f(0,0);if you'd split it the other way, f(h1,h2) - f(0,0) = f(h1,h2) - f(0,h2) + f(0,h2) - f(0,0)the proof wouldn't work.)No, it wouldn't. It would be impossible to guarantee the existence off_1 at (0,h2).Amanda === I knew that a suf?ient condition for f:R^n -> R to be differentiable> at a point u is that all of its partial derivatives exist in an open> ball centered at u and are continuous at u. But then I read there's a> weaker condition, that is, f is differentiable at u if one of it's> partial derivatives exists at u (it need not exist in a neighborhood> of u) and the others exist in a neighborhood of u and are continuous> at u. Seems this proof is in Apostol book.> I don't have this book, so I'm trying to work out a proof myself.It's an illusion: in the usual proof that (f is C1 <= its partialderivatives exist and are continuous), we start by proving the result for atwo-variable function f: U->G (U open set of ExF, where E,F,G are threenormed vector spaces), but we don't use the continuity of d_1 f. Exactly thething you stated here, except that you don't need to work on numericalfunctions (in R^2); in order to generalize the result, you may use the ????theorem (I don't know its name in English) which states that if f: U->F (Uopen set of E, E and F two normed vector spaces) is continuous on U anddifferentiable on U (except maybe at (at most) a countable number of points)then, for any [a,b] which is included in U:||f(b)-f(a)||<=||b-a||*sup(||Df(x)||, x in [a,b]) .As a corollary (used below),||f(b)-f(a)-df(a)(b-a)|| <= ||b-a||*sup(||df(x)-df(a)||, x in [a,b]).Now your case f: R^2 -> R becomes (it's your proof, literally, in thegeneral case):f(h1,h2)-f(0,0)-d_1 f(0,0).h1-d_2 f(0,0).h2 =[f(h1,0)-f(0,0)-d_1 f(0,0)]+[f(h1,h2)-f(h1,0)-d_2 f(0,0).h2] =But [f(h1,0)-f(0,0)-d_1 f(0,0)] = o(h1,h2) and:||f(h1,h2)-f(h1,0)-d_2 f(0,0).h2|| <=||f(h1,h2)-f(h1,0)-d_2 f(h1,0).h2|| +||d_2 f(h1,0).h2 - d_2 f(0,0).h2|| <=||(h1,h2)||*sup(||d_2 f(x,0) - d_2 f(0,0)||, x in [0,h2])+2*||(h1,h2)||*||d_2 f(0,0) - d_2 f(h1,0)||.Now using the continuity of d2 f you can easily conclude.Happy New Year,--Julien Santini,France. === =Is there any other test, whenever |f ?(z_0)| = 1, which can determine thebehavior (attractive vs repelling) of a ?ed point z_0, for a given complexfunction f(z) and its iterates f^(n)(z)?I have a bunch of ?ed points for the iterates, where the condition fails(as per |f ?(z_0)|=1), and I need to see what's happening there.I have a pretty good idea what's happening (with Maple) where I can seenumerically that whenever I perturb z_0, I get a k-cycle, but how does oneshow that theoretically?Perhaps with |[f ^(k)(z_0+perturb)]'| whenever the perturbation produces ak-cycle? It appears to me that the direction of the complex vector perturbin the last expression might play a crucial role on the sign.--Ioannis Galidakishttp://users.forthnet.gr/ath/jgal/ Eventually, _everything_ is understandable === Is there any other test, whenever |f ?(z_0)| = 1, which can determine the>behavior (attractive vs repelling) of a ?ed point z_0, for a given complex>function f(z) and its iterates f^(n)(z)?I have a bunch of ?ed points for the iterates, where the condition fails>(as per |f ?(z_0)|=1), and I need to see what's happening there.I have a pretty good idea what's happening (with Maple) where I can see>numerically that whenever I perturb z_0, I get a k-cycle, but how does one>show that theoretically?Perhaps with |[f ^(k)(z_0+perturb)]'| whenever the perturbation produces a>k-cycle? It appears to me that the direction of the complex vector perturb>in the last expression might play a crucial role on the sign.Since you are using f', I am assuming your functions are analytic aroundthe ?ed points.If |f'(z_0)| = 1, then in a neighborhood of z_0, f looks like f(z) - f(z_0) = u(z-z_0) + c(z-z_0)^k + ...where |u| = |f'(z_0)| = 1 and k > 1. Unless f is a degree 1 polynomial,for z close enough to z_0, there are 2k wedges around z_0; f is a bitattractive in k of them and a bit repulsive in the other k.Thus, when |f'(z_0)| = 1, there is no further test to determine if f isattractive or repulsive; it is both depending on direction. After a fewiterations, z-z_0 will most likely visit different wedges and the actionof f may become chaotic, varying greatly with a slight variation in theinitial z.Rob Johnson take out the trash before replying === >Is there any other test, whenever |f ?(z_0)| = 1, which can determine the>>behavior (attractive vs repelling) of a ?ed point z_0, for a given complex>>function f(z) and its iterates f^(n)(z)?>Since you are using f', I am assuming your functions are analytic around>the ?ed points.>If |f'(z_0)| = 1, then in a neighborhood of z_0, f looks like> f(z) - f(z_0) = u(z-z_0) + c(z-z_0)^k + ...>where |u| = |f'(z_0)| = 1 and k > 1. Unless f is a degree 1 polynomial,>for z close enough to z_0, there are 2k wedges around z_0; f is a bit>attractive in k of them and a bit repulsive in the other k.>Thus, when |f'(z_0)| = 1, there is no further test to determine if f is>attractive or repulsive; it is both depending on direction. After a few>iterations, z-z_0 will most likely visit different wedges and the action>of f may become chaotic, varying greatly with a slight variation in the>initial z.Consider a few cases. 1) f(z) = z + c z^k.If c = p e^{is} with p > 0 and s real, and |z| is small, arg(f(z)) = arg(z) if arg(z) = (pi n - s)/(k-1) for some integer n. We thus have 2(k-1) invariant directions, of which k-1 will be expanding (n even) and k-1 contracting (n odd). I believe that in this case the basin of attraction of the ?ed point 0 will contain an open set whose boundary approaches 0 tangent to the expanding directions: thus if r is small most points with |z|=r are attracted to z=0. But since there are those expanding directions, the ?ed point is unstable.2) f(z) = w z + c z^k where w is an n'th root of 1.Then the analysis of (1) will apply to the n'th iterate f@@n, so againthe ?ed point is unstable but most points near 0 are attracted to z=0.3) f(z) = w z + c z^k where w = exp(it), t/pi irrational.I don't know what happens in this case. Perhaps the result is chaotic,or perhaps the ?ed point is stable but not asymptotically stable.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === =I already did it ...Most of the referencies are out of reach for me ...70s ...) .....> >Could anybody give me referencies which describe Szasz-Mirakjan operators ?>( Textbooks , Monographies , Articles .... ) If you search google for Szasz-Mirakjan you get 50 hits, some of> which must contain the sort of reference you want. (You also ?d> that it's sometimes spelled Szasz-Mirakian, which gives 8 more> hits.)> ************************David C. Ullrich === =Omri, thanks for explaining it to me.It kind of make sense. I am just trying to work on one compressionproblem.Can I ask one more thing?If 2 dimensional space can hold holographic 3d image?Couldn't I just write these numbers around the cube and createholographic image?That way I would ? lot of information into 2d, that could be auseful compression as well, no?Anyways... thanks for helping me with this, your comments were veryhelpful.Serge= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption === = Omri, thanks for explaining it to me.> It kind of make sense. I am just trying to work on one compression> problem.Can I ask one more thing?If 2 dimensional space can hold holographic 3d image?Couldn't I just write these numbers around the cube and create> holographic image?That way I would ? lot of information into 2d, that could be a> useful compression as well, no?Anyways... thanks for helping me with this, your comments were very> helpful.Serge= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =Unfortunately, I don't understand what you mean by holographic 3dimage, and which numbers you want to write on which cube - you'll haveto explain it further.I also don't know much about compression. All I know is that it dealswith representing oddities in a more concise way. That is, if thesymbols (numbers, letters, moves, whatever) have a certain distribution,you can take advantage of that in the way your construct yourcompression. But I do suggest looking in compression-speci? places(just use google - compression is a very wide subject).The problem starts when you want to take ANY symbol and represent itwith less information than is *required*. For example, if you knew thatsmall moves are more common than big moves, you could decide torepresent short moves using 3-4 digits, and long moves using 7-9 digits.On average, you'd need less digits than a simple representation (onewhich ignores the distribution). But in any compression scheme there's apoint where the compression gives worse results (for a certain input)than the naive representation. Otherwise you'd have magic: representingx bits of information using less than x bits. This collides with thepigeonhole principle, and would not work.I suggest you ask people who know more about compression than I do -comp.compression is not a bad place for that.Hope this helps,Omri. === Yes, > I spent couple of days trying to ?d the solution myself and I got to> the point, that it would be miracle (in my opinion) if someone could> think of a solution for every possible move using only 7 (0-1)> digits.> Using 8 of them is trivial, using less than 8 is impossible?> Hm...> Now, is this forever destined to be impossible?> I was wondering about cutting this square to 4 (or more) other squares> and putting them into 3D cube. Imagine 4 squares slices (or more),> one on the other, and then perhaps using some math methods inside of> the cube. Approximations or triangles or ...?> Really I was wondering about many ways, just too many to mention here,> but I need someone elses insight, just to refresh my own view of this> problem.Please let me know.Serge= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =Try to put 10 letters in 9 pigeounholes without getting two letters in one pigeonhole. It doesn't matter how you divide them, how you order them, what names you give them or which counting system you're using. It cannot be done.In the same way, you're trying to push 255 different moves into less (whichever way you look) pigeonholes. Draw them on a 4 dimentional cube, you're still stuck with the pigeonhole principle.http://mathworld.wolfram.com/ PigeonholePrinciple.htmlUnless you can ?d a way to represent 255 different items using 7 digits or less, you're not going to ?d a solution. 7 items, each 0 or 1, can only get you at most 2^7 different combinations, no matter what name you give those combinations, in which space.Unless you're going to use a separator between moves (in which case you can use the empty word, the 1-digit word, ..., the 7-digit word = 255 words), there is simply not enough information.Again, I can only suggest you try to ?d a solution for smaller problems: 2x2 square with (up to) 1 digit, 4x4 square with (up to) 3 digits, 8x8 square with (up to) 5 digits. If you can solve any of them, you'll have a solution for the larger problem.Why do you need it so badly? === 200 times in the last 10 minutes a squadron of the Truman Companysurround> me on me bed while I'm trying to sleep and keep repeating YOUR A CREEP You suffer from paranoid schizophrenia, and need to go to hospital.Sounds like a PeaceNik to me === > in P(S)^X such that f(n) -> f(infty) (set-wise) as n -> infty.Where X = N / {oo} = omega+1 = w+1 with order topology.I doubt there's a net extension or analogy to this sequence space. > I am noting that this coinduced topology seems to be the same > as the homeomorph of the product topology on {0,1}^S >> By what facts do you base this surmise? >By your original post in this thread, where you say: | The powerset topology for a powerset P(S) is de?ed | as the topology produced by subbase sets of the form | { A in P(S) | a in A }, { A in P(S) | a not in A } | | A base for this topology are the sets of the form | { X | A subset X subset SB } | for A,B ?ite subset S. >With this topology, P(S) is precisely homeomorphic to {0,1}^S >under the product topology. The homeomorphism identi?s each >subset in P(S) with its characteristic function in {0,1}^S.Reasonable basis to conjecture the homeomorphism.What map do you think represents the homeomorphism?-- >BTW, given any set Y and a collection of sequences with>designated limits in Y, the (largest) topology generated>by these sequence-limits is again the coinduced topology by>the corresponding functions in Y^(omega + 1). >Whence the confusion? I am pointing out that the determination of a >topology as the largest whereunder a given collection of sequences >converge to given limits is a special case of a coinduced topology.Grappling with sequence space application. Let A subset Yx(w+1)be a bunch of sequences and their assigned limits in Y. F = { f:w+1 -> Y | some (aj,a) in A with f(j) = aj, f(w) = a }Coinduce Y with F.If (aj, a) in A, then some f in F with f(j) = aj, f(w) = a. j -> w in w+1; f continuous; f(j) -> f(w), ie aj -> a in YThus (aj,a) in A = aj -> a in YConversely, when aj -> a in Y, why is (aj,a) in A ?A = A_b = { (a,a,a..., b) | a in Y } coinduces what topology to Y?Now (a,a,a... ) -> a; where's the f in F with f(j) = a, f(w) = b ?-->>One more example to compare the two concepts:>>Let X = [0,1] under the usual topology.>>Let Y = {(x,y) in R^2: x^2 + y^2 = 1}, the unit circle.>>Let Z = [-1,1] under the usual topology.>>Then the usual topology on Y can be generated in the following>>two ways:>>It is that induced by the projection maps from Y to Z^2. >I meant the projection maps from Y to Z (not Z^2).Hm, the projections induce the subspaces of a product? Seems likely.>>It is also that coinduced by>>f: X -> Y, x|->(cos 2pi x, sin 2pi x).Looks likes it's the quotient map below. >1) Paste together the endpoints of a compact intervalAnother quotient map with the pasted endpoints oneequivalance class, all the others singletons.- === > In set theory, for the powerset P(S) of a set S> limits of set sequences Aj are de?ed as> lim Aj = liminf Aj = limsup Aj> provided liminf Aj = limsup Aj,>> So a better way of describing the coinducing functions f are those>> in P(S)^X such that f(n) -> f(infty) (set-wise) as n -> infty.Where X = N / {oo} = omega+1 = w+1 with order topology.> I doubt there's a net extension or analogy to this sequence space.For any speci? index set I, sure there is. The topology coinducedby the corresponding functions in Y^I, where we use the ordertopology on I. Recall that with coinduced topologies, you can alsohave functions with different domains. > I am noting that this coinduced topology seems to be the same> as the homeomorph of the product topology on {0,1}^S>> By what facts do you base this surmise?>By your original post in this thread, where you say:> | The powerset topology for a powerset P(S) is de?ed> | as the topology produced by subbase sets of the form> | { A in P(S) | a in A }, { A in P(S) | a not in A }> |> | A base for this topology are the sets of the form> | { X | A subset X subset SB }> | for A,B ?ite subset S.> >With this topology, P(S) is precisely homeomorphic to {0,1}^S>under the product topology. The homeomorphism identi?s each>subset in P(S) with its characteristic function in {0,1}^S.Reasonable basis to conjecture the homeomorphism.> What map do you think represents the homeomorphism?My God, man! Don't you read? I have already told you this severaltimes!Sorry. I hate it when people type that, but, literally, this is atleast the fourth time I have told you. The previous instance is inyour quote above. Here it is in CAPS:THE HOMEOMORPHISM IDENTIFIES EACH SUBSET IN P(S) WITH ITSCHARACTERISTIC FUNCTION IN {0,1}^S.To spell it out, here is the homeomorphism g: P(S) -> {0,1}^S: Let B be a subset of S, h = g(B), and x an element of S. Then h(x) =1 iff x in B, h(x) = 0 iff x not in B.Perhaps you need to review what is meant by the product topology?-- >BTW, given any set Y and a collection of sequences with>designated limits in Y, the (largest) topology generated>by these sequence-limits is again the coinduced topology by>the corresponding functions in Y^(omega + 1).>Whence the confusion? I am pointing out that the determination of a> >topology as the largest whereunder a given collection of sequences>converge to given limits is a special case of a coinduced topology.> Grappling with sequence space application. Let A subset Yx(w+1)> be a bunch of sequences and their assigned limits in Y.> F = { f:w+1 -> Y | some (aj,a) in A with f(j) = aj, f(w) = a }> Coinduce Y with F.If (aj, a) in A, then some f in F with f(j) = aj, f(w) = a.> j -> w in w+1; f continuous; f(j) -> f(w), ie aj -> a in Y> Thus (aj,a) in A = aj -> a in YConversely, when aj -> a in Y, why is (aj,a) in A ?> A = A_b = { (a,a,a..., b) | a in Y } coinduces what topology to Y?> Now (a,a,a... ) -> a; where's the f in F with f(j) = a, f(w) = b ?No matter whether you use sequences directly or call it a coinducedtopology, you cannot guarantee that the designated sequences are theonly convergent ones in the resulting topology. For example, if youuse Cauchy sequences of rationals to induce the topology on R, youwill end up with the usual topology. (I think. I am still a littleunsure if there might not be a larger topology e.g., the Sorgenfreyline? - where these sequences converge. In any case, you will stilget convergent sequeces of reals as well.)For the power set, I believe you showed in this case that the onlyconvergent sequences are the original ones.-->>One more example to compare the two concepts:>>Let X = [0,1] under the usual topology.>>Let Y = {(x,y) in R^2: x^2 + y^2 = 1}, the unit circle.>>Let Z = [-1,1] under the usual topology.>>Then the usual topology on Y can be generated in the following>>two ways:>>It is that induced by the projection maps from Y to Z^2.> >I meant the projection maps from Y to Z (not Z^2).> Hm, the projections induce the subspaces of a product? Seems likely.More than likely. This is how I was taught to de?e the producttopology. I thought that was standard - and, again, I have cited thisfact several times. This fact is also mentioned in the mathworldentry on product topology, for example.I am not near my books right now, but my guess is that you will ?d adiscussion of induced and coinduced topologies in Kelly. Doing aquick seach on the ?net, it does seem that perhaps the identi?ationof these concepts (learned from lectures by Spanier) is notwide-spread.>>It is also that coinduced by>>f: X -> Y, x|->(cos 2pi x, sin 2pi x).Looks likes it's the quotient map below. >1) Paste together the endpoints of a compact interval> Another quotient map with the pasted endpoints one> equivalance class, all the others singletons.William, what you note here is correct, but I fear that sometimes youmiss the forest on account of the trees. I mean that as constructivecriticism. <20040101072404.G49311@agora.rdrop.com> <4de76b8.0401011321.73d2f58a@posting.google.com> === = >> Where X = N / {oo} = omega+1 = w+1 with order topology. >> I doubt there's a net extension or analogy to this sequence space. >For any speci? index set I, sure there is. The topology >coinduced by the corresponding functions in Y^I, where we use the >order topology on I. Recall that with coinduced topologies, you can >also have functions with different domains.The domain or index of a net is an updirected pre-order, tho all that I'veseen are updirected partial orders. What paci?ally have you in mind bythe order topology of a pre-order? A Scott topology? Likely that'dsuf?e, yet I've yet to check with the devil in the details. Thetopology of partial orders is an intricate topic. I've seen anon-Hausdorff T1 topology with convex base sets proposed. > I am noting that this coinduced topology seems to be the same > as the homeomorph of the product topology on {0,1}^S >> By what facts do you base this surmise? > By your original post in this thread, where you say:>>| The powerset topology for a powerset P(S) is de?ed >>| as the topology produced by subbase sets of the form >>| { A in P(S) | a in A }, { A in P(S) | a not in A }>With this topology, P(S) is precisely homeomorphic to {0,1}^S >under the product topology. The homeomorphism identi?s each >subset in P(S) with its characteristic function in {0,1}^S.>> Reasonable basis to conjecture the homeomorphism. >> What map do you think represents the homeomorphism? >THE HOMEOMORPHISM IDENTIFIES EACH SUBSET IN P(S) WITH ITS >CHARACTERISTIC FUNCTION IN {0,1}^S. >To spell it out, here is the homeomorphism g: P(S) -> {0,1}^S: >Let B be a subset of S, h = g(B), and x an element of S. >Then h(x) = 1 iff x in B, h(x) = 0 iff x not in B.I've worked thru the details showing the powerset topology is homeomorphicto the characteristic space. I'm ?ed handling the homeomorphicdetails regarding the coinduced space and the characteristic or powersetspace.-- > BTW, given any set Y and a collection of sequences with > designated limits in Y, the (largest) topology generated > by these sequence-limits is again the coinduced topology by > the corresponding functions in Y^(omega + 1). > Whence the confusion? I am pointing out that the determination of > a topology as the largest whereunder a given collection of > sequences converge to given limits is a special case of a > coinduced topology. >> Grappling with sequence space application. Let A subset Yx(w+1) >> be a bunch of sequences and their assigned limits in Y. >> F = { f:w+1 -> Y | some (aj,a) in A with f(j) = aj, f(w) = a } >> Coinduce Y with F.>> If (aj, a) in A, then some f in F with f(j) = aj, f(w) = a. >> j -> w in w+1; f continuous; f(j) -> f(w), ie aj -> a in Y >> Thus (aj,a) in A = aj -> a in Y> >> Conversely, when aj -> a in Y, why is (aj,a) in A ?As a matter of fact, (aj,a) may not be in A. >> A = A_b = { (a,a,a..., b) | a in Y } coinduces what topology to Y? >> Now (a,a,a... ) -> a; where's the f in F with f(j) = a, f(w) = b?Ok, you've cleared up this point in the issuing: >No matter whether you use sequences directly or call it a coinduced >topology, you cannot guarantee that the designated sequences are the >only convergent ones in the resulting topology. For example, if you >use Cauchy sequences of rationals to induce the topology on R, you >will end up with the usual topology.The smallest topology assuring the designated limits of sequences is theindiscrete lump. The topology we want is the largest topology assuringthe designated limits of the sequences and not, as I much puzzled over,the smallest topology you mentioned to the other participant early in thisthread. >(I think. I am still a little unsure if there might not be a larger >topology e.g., the Sorgenfrey line? - where these sequences >converge. In any case, you will stil get convergent sequences of >reals as well.)A decreasing sequence in the lower limit topology will reach it's reallimit. An increasing sequence in the lower limit topology won't reach it'sreal limit. Indeed, they have no limits. >For the power set, I believe you showed in this case that the only >convergent sequences are the original ones.Yes, but does that make it the largest topology assuring the designatedlimits? The sequence induced topology assures the designated limits. Isit the largest with that property? Yes to both of these and viola, thesequence induced topology is identical to the powerset topologyI'll sleep on it, especially the ?st question seems answerable positive.-- >>One more example to compare the two concepts: >>Let X = [0,1] under the usual topology. >>Let Y = {(x,y) in R^2: x^2 + y^2 = 1}, the unit circle. >>Let Z = [-1,1] under the usual topology. >>Then the usual topology on Y can be generated in the following >>two ways: >>It is that induced by the projection maps from Y to Z^2.>I meant the projection maps from Y to Z (not Z^2). >> Hm, the projections induce the subspaces of a product >This is how I was taught to de?e the product topology.Yes, the product topology is induced by the projections p_iThe subspace S is induced by the inclusion map, i_SSo the subspace of a produce would be induced by p_i o i_S = p_i | Sthe projections of the subspace. >William, what you note here is correct, but I fear that sometimes >you miss the forest on account of the trees. I mean that as >constructive criticism.Proof readers are valued for that ability. ;-)There is the forest, ah what a view. Now if you go along this trail, pastthe stream and where the trail bends, go instead straight ahead you will?d a cedar ripe for splitting shingles. Proceeding now in the directionof that fallen log you will cross a deer trail and with persistence cometo a small cliff and cave. Ah, what be therein? Yet before entering donotice the small ?trespassers enter at your own risk' signs of a bear.Instead follow the cliff and for some distance beyond appears a smallmeadow and pond surrounded by trees with lush hucckle berries underneathand much yummy dawdling. Now from there..., yet wait, back to the viewpoint now before night overcomes.- === =Here are 2 n=12 perplexes. A perplex, as invented by B.S.Rangaswamy,is a set of n n-digit zero-free squares, that when written out forms an n x n matrix of digits equal to its transpose. A zero-free squareis a squared integer number containing no zero digits in its base-10 representation. These two solutions are shown ?st as sets of roots,then in fuller form.-jiw{957511 358784 827807 934038 475632 868708 628471 391023 767854 403042 504407 411636}{683281 798836 829893 903962 855838 531773 996627 509885 664623 992182 833332 411643} 916827315121 957511 128725958656 358784 685264429249 827807 872426985444 934038 226225799424 475632 754653589264 868708 394975797841 628471 152898986529 391023 589599765316 767854 162442853764 403042 254426421649 504407 169444196496 411636 466872924961 683281 638138954896 798836 688722391449 829893 817147297444 903962 732458682244 855838 282782523529 531773 993265377129 996627 259982713225 509885 441723732129 664623 984425121124 992182 694442222224 833332 169449959449 411643X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBVDVc031742; === > ... Here is some output after an hour of processing>> at n=11 via program at http://pat7.com/jp/ perp11b.c -->> ...>> The program ?ished in 6.8 hours and found 8 n=11 perplexes ->> 229611 171618 272293 159527 111296 165793 117342 133191 199692 149738 122462>...Here is some funny-looking output after a few seconds of processing at n=12 --913159 913159 483729 994721 599729 542169 > 371325 924485 997865 353335 533335 333334Ie, 833859359281 913159 > 833859359281 913159 > 233993745441 483729 > 989469867841 994721 > 359674873441 599729 > 293947224561 542169 > 137882255625 371325 > 854672515225 924485 > 995734558225 997865 > 124845622225 353335 > 284446222225 533335 > 111111555556 333334It is indeed heartening to learn that you have made elaborate programing to arrive at Perplex-12 patterns. The funny looking output is itself amazing, since 11 out of 12 squares satisfy and 143 out of 144 numerals are correctly placed. Following is slightly a modi?d version of the output : 8 2 9 3 2 1 8 9 1 2 1 - 8 3 3 8 5 9 3 5 9 2 8 1 - 913159^2 2 3 3 9 9 3 7 4 5 4 4 1 - 483729^2 9 8 9 4 6 9 8 6 7 8 4 1 - 994721^2 3 5 9 6 7 4 8 7 3 4 4 1 - 599729^2 2 9 3 9 4 7 2 2 4 5 6 1 - 542169^2 1 3 7 8 8 2 2 5 5 6 2 5 - 371325^2 8 5 4 6 7 2 5 1 5 2 2 5 - 924485^2 9 9 5 7 3 4 5 5 8 2 2 5 - 997865^2 1 2 4 8 4 5 6 2 2 2 2 5 - 353335^2 2 8 4 4 4 6 2 2 2 2 2 5 - 533335^2 1 1 1 1 1 1 5 5 5 5 5 6 - 333334^2But,there are no 10 digit squares ending with 82932189121. All 11 squares and incomplete 11 digit number read alike across and downwards.With around 1.92 lakh 12 digit squares and 330 possible numbers for last column/row,CPU time required for completing this task with prevailing program may prolong upto 180 hours.I have discovered many patterns ranging from perplex-3 to 9,a decade back with out using computer.Following is an excercise depicting the formation of the lone perplex-4 pattern:Sort and arrange all 44 zero free 4 digit squares as follows: 2116 3136 1156 5476 1296 1521 6241 6561 8281 1225 9216 8836 4356 5776 7396 3721 4761 3481 4225 2916 1936 7921 1681 7225 5625 4624 1444 3364 5184 5329 3249 1369 4489 6724 7744 8464 5929 8649 7569 6889 3844 1764 1849 3969Identify numbers eligible for last column/row : 1156,1444 and 6561 only. Proceed as follows: (E- EVEN, O- ODD, A- ANY, B- 2 OR 6) E 1 E 4 1 - Fails E 1 E 4 1 8 4 1 - Fails E 1 E O 2 1 E 4 A 6 4 8 4 6 4 E E 2 5 4 2 2 5 E E E 4 4 6 2 4 4 6 2 4 1 1 5 6 1 1 5 6 1 4 4 4 1 4 4 4 1 4 4 4 E 8 1 O 6 A 9 6 1 9 6 -F 4 9 6 -F E A 4 4 - Fails 2 5 A B 2 5 1 2 2 5 4 2 2 5 8 4 6 4 O 2 O 6 9 2 1 6 9 2 1 6 9 2 1 6 1 4 4 4 6 5 6 1 6 5 6 1 6 5 6 1 6 5 6 1 7 9 6 -F 5 9 6 -F 4 1 6 -F 7 1 6 -F 5 1 6 -F 7 2 2 5 5 6 2 5 4 2 2 5 7 2 2 5 5 6 2 5 9 2 1 6 9 2 1 6 1 2 9 6 1 2 9 6 1 2 9 6 6 5 6 1 6 5 6 1 6 5 6 1 6 5 6 1 6 5 6 1 A 1 1 6 2 1 1 6 - 46^2 All squares are picked up from 1 2 2 5 1 2 2 5 - 35^2 above list. 1 2 9 6 1 2 9 6 - 36^2 6 5 6 1 6 5 6 1 - 81^2 PERPLEX -4This procedure probably may assist you in rewriting the program to reduce CPU time for the task. -bsr X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i01Dri626548; === at 10:36 PM, nico80@jazzfree.com (Nicolas de la Foz) said:>You are right, and for that reason I say that we must not suppose >>that we have a one-to-one correspondence between N and R to prove >>that Real numbers are innumerable, since N and R have distinct grade >>of in?ity.Then it's a good thing that the diagonal argument doesn't suppose it,>isn't it?>The diagonal argument needs a premise (a list, otherwise it cant work). If you know a *neutral* premise, then I believe that the proof by contradiction would be valid.Nicolas de la Foz === =The online Science Forum can be found here:http://www.sciencegroups.com/viewforum.php?f=19Doron- ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY **- http://www.usenet.com === =On Thu, 1 Jan 2004 13:53:45 +0000 (UTC), nico80@jazzfree.com (Nicolas>> at 10:36 PM, nico80@jazzfree.com (Nicolas de la Foz) said:>You are right, and for that reason I say that we must not suppose >that we have a one-to-one correspondence between N and R to prove >that Real numbers are innumerable, since N and R have distinct grade >of in?ity.Then it's a good thing that the diagonal argument doesn't suppose it,>>isn't it?>The diagonal argument needs a premise (a list, otherwise it cant >work). If you know a *neutral* premise, then I believe that the proof >by contradiction would be valid.Assume a set of reals S and an injection f from the naturals to thisset (effectively making it a list). Then show that for any such f wecan always ?d a real not in the map of f using the oh-so-wonderfuldiagonal argument, so f can't be a surjection onto any S that containsall reals. It follows that f cannot be a bijection.It doesn't really matter anyway since you've not given any argumentagainst proof-by-contradiction (in fact you accept it as correct inthe case of sqrt(2)) except it can be used to prove unintuitivethings so it must be wrong. === On Thu, 1 Jan 2004 13:53:45 +0000 (UTC), nico80@jazzfree.com (Nicolas> at 10:36 PM, nico80@jazzfree.com (Nicolas de la Foz) said:>You are right, and for that reason I say that we must not suppose >>that we have a one-to-one correspondence between N and R to prove >>that Real numbers are innumerable, since N and R have distinct grade >>of in?ity.Then it's a good thing that the diagonal argument doesn't suppose it,>isn't it?>The diagonal argument needs a premise (a list, otherwise it can't >>work). If you know a *neutral* premise, then I believe that the proof >>by contradiction would be valid.>Assume a set of reals S and an injection f from the naturals to this>set (effectively making it a list). You don't even need that f is an injection. All you need is to assume that f is a function from N to R, and then to prove that f can't be asurjection.>Then show that for any such f we>can always ?d a real not in the map of f using the oh-so-wonderful>diagonal argument, so f can't be a surjection onto any S that contains>all reals. It follows that f cannot be a bijection.David McAnally Despite anything you may have heard to the contrary, the rain in Spain stays almost invariably in the hills. === Assume a set of reals S and an injection f from the naturals to this>>set (effectively making it a list). You don't even need that f is an injection. All you need is to assume >that f is a function from N to R, and then to prove that f can't be a>surjection.Well, it's hard to think of f as a numerable list of reals unless it'san injection from N. Making the assumption doesn't invalidate theproof.X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i01DrgR26477; === =I am trying to do the same using MAPLE and I am using polynomial coef?ient matrix for doing the calculations. Also I am not using the GF that's available in MAPLE, But trying to do direct math calculations. I am having problems doing the shifting the polynomial coef?ients present in the array (that is multiplying by x^(deg(a)-deg(b)) How do we do that in Maple.or is there any other way to do it.Thanx. Hari.> I want to implement an extended Euclidean algorithm in MATLAB that can>> compute the following: GCD(A(x),B(x)) = u(x)A(x) + v(x)B(x), where>> A(x), B(x), u(x) and v(x) are from GF(2^m).This is a lot easier than it may seem from a glance at the literature. I>assume you mean that the coef?ients of those polynomials are in GF(2^m)?>The ring of polynomials with coef?ients in GF(2^m) is usually written>GF(2^m)[x]. Everything below is valid for polynomials with coef?ients in>any ?ld, which you can verify by inspection, because I was careful to use>the proper signes for everthing even though addition and subtraction are the>same in GF(2^m):Start with u=1,v=0, p=0,q=1, a=A and b=B.Then iterate, always maintaining the invariants:uA+vB=a; and>pA+qB=bfor each iteration:if deg(a) < deg(b), then swap u <-> p, v <-> q, and a <-> b, so that>deg(a)>=deg(b) and the invariants are maintained.If b==0, then stop. you're done.Otherwise, divide to ?d a' and k such that a'=a-kb and deg(a')(** see note **)Then expand to get a'=uA+vB-kpA-kqBand replace a <- a', u <- u-kp, and v <- v-kqthen repeat.**note** You don't necessarily need to do a whole division here. It>suf?es to ensure that the stated conditions are met. In particular, you>may simply choose k=x^(deg(a)-deg(b)) * (highest_coef(a)/highest_coef(b)).This won't really save you any execution time, because you will keep>iterating until you have essentially done the long division, but it may save>you some code, and it doesn't hurt the worst case complexity, no matter what>nifty division algorithm you might have used instead.If you need proof that the algorithm converges to the GCD, that's easy. The>only operations we perform on a and b are swapping a <-> b and replacing a><- a-kb. GCD(a,b) = GC(b,a), so swapping a and b doesn't change the GCD.>And if d divides a and d divides b, then d divides a-kb, or a-kb is zero.>If a-kb is zero, then GCD(a,b) = b. So GCD(a,b) is preserved throughout the>algorithm until one of them gets to zero, at which point the other is the>GCD.>X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i01Drib26557; === Does anyone have the solution manual or some of the solutions to the Applied Combinatorics (fourth edition by Alan Tucker)?Please if someone send you the manual forward it to me. === =Dear NG,What is the easiest way to ?d the upper (and maybe lower also) bounds of the zeroes of polynomials in the complex ?ld? And how does one do it?Sincerely,Jose Capco === What is the easiest way to ?d the upper (and maybe lower also) bounds of the >zeroes of polynomials in the complex ?ld? And how does one do it?See e.g. M. Mignotte, Mathematics for Computer Algebra, Springer-Verlag 1992, Theorem 4.2, which has 8 different upper bounds; by looking atthe reciprocal polynomial x^d p(1/x) you get corresponding lower boundsfor the nonzero roots.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === =Adam / Preno === k^0) * (1 - a * k^1) * ... where k goes from 0 to in?ity?I assume you mean P = product_{j=0}^in?ity (1 - a*k^j) for ?ed k.If |k| < 1 and none of the factors is 0, P converges to a nonzero value. If |k| >= 1, a <> 0 and none of the factors is 0, P either divergesor converges to 0. I don't know of a closed form for P in the case |k|<1. It isof course exp(sum_{j=0}^in?ity ln(1-a*k^j)), and can also be written as exp(sum_{j=1}^in?ity a^j/(j (k^j-1))) if |a| < 1.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === k^0) * (1 - a * k^1) * ... where k goes from 0 to in?ity?I assume you mean P = product_{j=0}^in?ity (1 - a*k^j) for ?ed k.> If |k| < 1 and none of the factors is 0, P converges to a nonzero value. > If |k| >= 1, a <> 0 and none of the factors is 0, P either diverges> or converges to 0. And the convention in this case is to say that the productdiverges to zero.> I don't know of a closed form for P in the case |k|<1. It is> of course exp(sum_{j=0}^in?ity ln(1-a*k^j)), and can also be > written as exp(sum_{j=1}^in?ity a^j/(j (k^j-1))) if |a| < 1.For a=1 this is essentially the Dededind eta function...The Dedekind eta function is q^(1/24)*product_{j=1}^in?ity (1-q^j),but normally they use complex variable tau where q=exp(2 pi i tau).Im(tau)>0 for convergence.-- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === =AdamIn sci.math, Adam Prenosilon Thu, 1 Jan 2004 17:13:52 +0100:> of (1 - a * k^0) * (1 - a * k^1) * ... where k goes from 0 to in?ity?You mean, the double productprod(j=0 to oo)(k=0 to oo) (1 - a * k^j) ?Unfortunately, unde?ed; 0^0 is like 0/0 in that respect.Adam / Preno> -- #191, ewill3@earthlink.netIt's still legal to go .sigless. === =There is a question: does ?everything' is'[b:454bb87986]one[/b:454bb87986]' OR ?[b:454bb87986]in?itely many..[/b:454bb87986]'?Let us examine this question from simplicity point of view.What is simpler, ?[b:454bb87986]one[/b:454bb87986]' or'[b:454bb87986]in?itely many[/b:454bb87986]'?'[b:454bb87986]one[/b:454bb87986] ' cannot be de?ed as'[b:454bb87986]many[/b:454bb87986]' and'[b:454bb87986]many[/b:454bb87986]' cannot be de?ed as'[b:454bb87986]one[/b:454bb87986]'.Does ?[b:454bb87986]one of many[/b:454bb87986]' ='[b:454bb87986]one[/b:454bb87986]'?By quantity concept they are the same.Is there another point of view where they are not the same?If there is, then it is not a quantitative point of view.A structural point of view:First of all ?[b:454bb87986]one[/b:454bb87986]' and ?[b:454bb87986]oneof many[/b:454bb87986]' are something (non-empty set's contents).Therefore, if we translate them to the simplest existing structuralrepresentation,we must get at least these two forms {.} , {__}.{.} is ?[b:454bb87986]one of many[/b:454bb87986]'.{__} is an in?itely long line, representing'[b:454bb87986]one[/b:454bb87986]'.When we have these two existing and simplest structuralrepresentations, we want to examine them also by quantity concept.Quantity by set's concept, is the cardinality of some set's content.There can be 3 basic kinds of some set's cardinality:|{}| = 0|{.}| = 1|{__}| = 1Let us say that power 0 represents the minimal power of some non-emptyset's content.When {__} is in?itely long line that represents'[b:454bb87986]one[/b:454bb87986]', it means that by structural pointof view, 0 points included in it, but unlike {} content, {__} contentexists.Therefore its cardinality > 0.When we combine between structural and quantity properties of {__}content, we can ?d its cardinality by this expression: 0^0 = 1 or|{__}| = 0^0 = 1.In {.} content there are at least 1 point, therefore according to theabove: |{.}| = 1^0 = 1Now we have:|{}| = 0|{.}| = 1^0 = 1|{__}| = 0^0 = 1And we can clearly distinguish between {.}(=1^0*1 or {. , . ,..}=1^0*n>1) and {__}(=0^0).By structural point of view the connective between . and __ is: . XOR__Let us return to the ?st question, but now instead of OR we areusing XOR.Does ?[b:454bb87986]everything[/b:454bb87986]' is'[b:454bb87986]one[/b:454bb87986]' XOR ?[b:454bb87986]in?itely many..[/b:454bb87986]'?By more ?e examination we ?d that there are at least two XORconnective levels:1. ( {}XOR{.} ) OR ( {}XOR{__} )2. {.}XOR{__}Level 1 is invariant but in level 2 we can change XOR by ANDconnective.As a result we get the new ?ite set's content {.__.} which is theminimal association between {., .}(= some pair) AND {__} .-->(or <--) is approaching(but cannot become closer) toAnd we get this general sets-structure: {} <--{.}AND{._.}-->{__} .Now we have:1. ( {}XOR{.} ) OR ( {}XOR{__} ) OR ( {}XOR{._.} ) OR ( {__}XOR{._.} ))2. {.}AND{__}Level 2 connective is the base of Complementary Logic, whichintroduced very brie?: http://www.geocities.com/complementarytheory/ CompLogic.pdf[url]http://www.geocities.com/ complementarytheory/4BPM.pdf[/url]Now, by Complementary Logic the answer is:'[b:454bb87986]everything[/b:454bb87986]' is the balance of {}<--{.}AND{._.}--> {__} . --By Complementary Logic multiplication is noncommutative, but I thinkthat another interesting result is the fact that multiplication andaddition are complementary, and as much as I know this point of viewhas a very deep in? on the question: [b:454bb87986]What isNumber?[/b:454bb87986] [u:454bb87986][b:454bb87986]A second levelexample:[/b:454bb87986][/u:454bb87986]Answer: [b:454bb87986]Number is anything that exist in({},{__})[/b:454bb87986]Or in more formal de?ition:[b:454bb87986]({},{_}):={x|{} <-- x(={.}) AND x(={._.})-->{_}}[/b:454bb87986]Where -->(or <--) is ASPIRATING(= approaching, but cannot becomecloser to).[u:454bb87986]If x=4 then number 4 example:[/u:454bb87986]Number 4 is fading transition between multiplication 1*4 and addition ((((+1)+1)+1)+1) ,and vice versa. This fading can be represented as:(1*4) ={1,1,1,1} <-Maximum symmetry-degree, ((1*2)+1*2) ={{1,1},1,1} Minimum information's clarity-degree(((+1)+1)+1*2) ={{{1},1},1,1} (nouniqueness)((1*2)+(1*2)) ={{1,1},{1,1}}(((+1)+1)+(1*2)) ={{{1},1},{1,1}}(((+1)+1)+((+1)+1))={{{1},1},{{1},1}}((1*3)+1 ) ={{1,1,1},1}(((1*2)+1)+1) ={{{1,1},1},1}((((+1)+1)+1)+1) ={{{{1},1},1},1} ^ 3 3 3 3 | 3 3 3 3 2 2 2 2 | 2 2 2 2 1 1 1 1 | 1 1 1 1 1 1 1 1 1 1 {0, 0, 0, 0} V {0, 0, 0, 0} {0, 1,0, 0} {0, 0, 0, 0} . . . . . . . . . . . . . . . . | | | | | | | | | | | | | | | | | | | | |__|_ | | |__| | | |__|_ |__|_ | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | |__|__|__|_ |_____|__|_ |_____|__|_ |_____|____ | | | | (1*4) ((1*2)+1*2) (((+1)+1)+1*2) ((1*2)+(1*2)) 4 = 2 2 2 1 1 1 1 1 1 1 {0, 1, 0, 0} {0, 1, 0, 1} {0, 0,0, 3} {0, 0, 2, 3} . . . . . . . . . . . . . . . . | | | | | | | | | | | | | | | | |__| |__|_ |__| |__| | | | | |__|_ | | | | | | | | | | | | | | | | | |__|__|_ | |_____| | | | | | | | | | |_____|____ |_____|____ |________| |________| | | | |(((+1)+1)+(1*2))(((+1)+1)+((+1)+1)) ((1*3)+1) (((1*2)+1)+1) {0, 1, 2, 3} . . . . | | | | |__| | | | | | |_____| | | | |________| | ((((+1)+1)+1)+1)Multiplication can be operated only among objects with structuralidentity, where addition can be operated among identical andnon-identical (by structure) objects. Also multiplication is noncommutative, for example:2*3 = ( (1,1),(1,1),(1,1) ) , ( ((1),1),((1),1),((1),1) )3*2 = ( (1,1,1),(1,1,1) ) , ( ((1,1),1),((1,1),1) ) , ((((1),1),1),(((1),1),1) )More about the above you can ?d here (the ?st 9 lines de?ed byHurkyl):[url]http://www.geocities.com/complementarytheory/ ET.pdf[/url]More about Complementary logic, you can ?d here:[url]http://www.geocities.com/complementarytheory/ CompLogic.pdf[/url][url]http://www.geocities.com/ complementarytheory/4BPM.pdf[/url][url]http:// www.geocities.com/complementarytheory/CK.pdf[/url][b: 454bb87986]More about the order concept can be foundhere:[/b:454bb87986][url]http://www.geocities.com/ complementarytheory/CL-CH.pdf[/url]Doron= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption === ==Dear Robin,The name of my topic is:Does ?everything' is ?one' OR ?in?itely many'?which can be found in:http://www.sciencegroups.com/viewtopic.php?p=36530-- -- ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY **- http://www.usenet.com === = The name of my topic is:Does ?everything' is ?one' OR ?in?itely many'?Assuming you mean _Is_ everything ?one' or ?in?itely many'?,look at: http://modular.fas.harvard.edu/sga/from_ grothendieck.pdftraces how, over the 20th century, concepts in various branchesof maths have converged by means of category theory and topoi.(I don't pretend to follow much of it, but the fog is clearingever so slightly ;-)- --John R Ramsden (jr@adslate.com) Eternity is a long time, especially towards the end. Woody Allen === Dear Robin, The name of my topic is: Does ?everything' is ?one' OR ?in?itely many'?>This does not make sense in English.l8r, Mike N. Christoff === Dear Robin,The name of my topic is:Does ?everything' is ?one' OR ?in?itely many'?The auxiliary does is not used with the verb to be.Is X Y?.-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === =On 1 Jan 2004 16:45:40 -0600, Doron Shadmi> Does ?everything' is ?one' OR ?in?itely many'?That doesn't seem to be an English sentence. What are you actuallyasking?-- Paul. === ?[b:454bb87986]one[/b:454bb87986]' cannot be de?ed as> ?[b:454bb87986]many[/b:454bb87986]' and> ?[b:454bb87986]many[/b:454bb87986]' cannot be de?ed as> ?[b:454bb87986]one[/b:454bb87986]'.What? === =Is there any problem to read my topic?= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption === ==Is there any problem to read my topic?I've no idea how one can read a topic. Buy anyway, there are many obstructions to making sense of your positing. The ?st is the[b:454bb87986]which I quoted. Doesn't look like any word with which I am familiar :-( -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === >>> Is there any problem to read my topic?I've no idea how one can read a topic. Buy anyway, there are many> obstructions to making sense of your positing. The ?st is the[b:454bb87986]which I quoted. Doesn't look like any word with which I am familiar :-(Examine it a bit and it becomes somewhat clear what it means. Note thatthese things come in pairs, starting with [b:somehexstuff] and ending with[/b:somehexstuff]. In another post, I think I saw i: instead of b:.It looks like some kind of style or formatting markup. b is probably forbold, and i for italic. The hex stuff is still puzzling, although itcould be a font code, font size, and color, or something like that.So, in the original post, where you see 00[b:stuff]1[/b:stuff], just readit as 001, and keep in mind the poster was trying to highlight in some waythe 1.-- --Tim Smith ===