mm-579 === Subject: Re: Cantor won, but one final question anymore thnk u > I think this was Cantor's first theorem of uncountabilty: > --clik -- > Suppose a set R is > 1) linearly ordered, and > 2) densely ordered, i.e., between any two members there is another, and > 3) has no endpoints, i.e., smallest or largest members, and > 4) has no gaps, i.e., if it is partitioned into two nonempty sets A and > B in such a way that every member of A is less than every member of B, > then there is a boundary point c, so that every point less than c is in > A and > every point greater than c is in B. > ->Then R is not countable. > --clik -- > Let Q = all rational numbers between 0 and 1 > Let R = all real numbers between 0 and 1 > Now what is the differnce between Q and R? > Why is R uncountable and Q countable? > Consider > A = {x in R: x > 0 and x^2 < 2}, > B = {x in R: x > 0 and x^2 > 2} > Then A union B contains all rationals, but not all reals (specifically > not sqrt(2) which is irrational),so that there is no rational between > A and B in the required sense. So the theorem proves uncountability for > R interval but not for Q. > I'm really confused or stupid or tired. > The theorem says that if there is no gaps then then R is uncountable. But > there was gap in A U B concerning real numbers when c=srt(2). No, there is no gap. Point 4) explains what it means for a set to have no gaps. To show a gap in R, you have to provide suitable sets A and B such that no boundary point c exists in R. In the case mentioned, c=sqrt(2) is the required boundary point, and c is in R. Therefore, you have not demonstrated a gap in R. >There was no > gap concerning rational numbers i.e. A U B contains all rationals? So > rationals are not countable but reals are? On the contrary, there is a gap in the rationals, because in that case c=sqrt(2) is not a member of Q, and therefore a gap exists. > I misunderstand that gap? A gap means something is absent. The point c=sqrt(2) is absent from Q (hence, a gap), but it is present in R (hence, no gap). -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: Cantor won, but one final question anymore thnk u --clik -- Suppose a set R is 1) linearly ordered, and 2) densely ordered, i.e., between any two members there is another, and 3) has no endpoints, i.e., smallest or largest members, and 4) has no gaps, i.e., if it is partitioned into two nonempty sets A and B in such a way that every member of A is less than every member of B, then there is a boundary point c, so that every point less than c is in A and every point greater than c is in B. ->Then R is not countable. --clik -- > A gap means something is absent. The point c=sqrt(2) is absent from Q > (hence, a gap), but it is present in R (hence, no gap). Dealing with R = Q (rational numbers): A sqrt(2) B ----------------------!----------------------------- Q = A U B -> no gaps in A U B sqrt(2) does not belongs to Q -> no gaps (Q=A U B) Dealing with R = R'(=real numbers) sqrt(2) ----------------------!----------------------------- sqrt(2) belongs to R' R <> A U B so there is gap in A U B. R = A U B U {sqrt(2)} === Subject: Re: Cantor won, but one final question anymore thnk u > --clik -- > Suppose a set R is > 1) linearly ordered, and > 2) densely ordered, i.e., between any two members there is another, and > 3) has no endpoints, i.e., smallest or largest members, and > 4) has no gaps, i.e., if it is partitioned into two nonempty sets A and B > in such a way that every member of A is less than every member of B, then > there is a boundary point c, so that every point less than c is in A and > every point greater than c is in B. > ->Then R is not countable. > --clik -- > A gap means something is absent. The point c=sqrt(2) is absent from Q > (hence, a gap), but it is present in R (hence, no gap). > Dealing with R = Q (rational numbers): > A sqrt(2) B > ----------------------!----------------------------- > Q = A U B -> no gaps in A U B > sqrt(2) does not belongs to Q -> no gaps (Q=A U B) That's not what 4) says. Since c = sqrt(2) is the required boundary point, and since c does not belong to Q, it follows that c is a missing point (a gap) in Q. > Dealing with R = R'(=real numbers) > sqrt(2) > ----------------------!----------------------------- > sqrt(2) belongs to R' > R <> A U B so there is gap in A U B. R = A U B U {sqrt(2)} No, R = A U B because A and B partition R. There is no gap in A U B = R because the required boundary point is a member of R. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: Cantor won, but one final question anymore thnk u --clik -- Suppose a set R is 1) linearly ordered, and 2) densely ordered, i.e., between any two members there is another, and 3) has no endpoints, i.e., smallest or largest members, and 4) has no gaps, i.e., if it is partitioned into two nonempty sets A and B in such a way that every member of A is less than every member of B, then there is a boundary point c, so that every point less than c is in A and every point greater than c is in B. ->Then R is not countable. --clik -- > The theorem says that if there is no gaps then then R is uncountable. > But there was gap in A U B concerning real numbers when c=srt(2). > No, there is no gap. Point 4) explains what it means for a set to have > no gaps. To show a gap in R, you have to provide suitable sets A and B > such that no boundary point c exists in R. In the case mentioned, > c=sqrt(2) is the required boundary point, and c is in R. Therefore, you > have not demonstrated a gap in R. There is no gap _in_ _R_ allthough there _is_ gap in A U B because every point less than c is in A and every point greater than c is in B (there is not word equal). c does not belongs to A and it doesn't belong to B. Of course the is then gap in A U B because c doesn't belong to A either B. > There was no gap concerning rational numbers i.e. A U B contains all > rationals? So > rationals are not countable but reals are? > On the contrary, there is a gap in the rationals, because in that case > c=sqrt(2) is not a member of Q, and therefore a gap exists. There is not gap when we stay in Q. c=sqrt(2) doesn't belong to Q, then A U B contains all the numbers than R. We didn't not suppose anything more than that R is a set in the first hand. R is not said to be the hole one dimensional line - system of co ordinates x-axle or something like that. If we then divede the R to two sets, and choose c out of that R of course that c doesn't belong to R because it doesn't belong there at the beginning. But there is no gap in A U B. ... but ok. You said = To show a gap in R, you have to provide suitable sets A and B such that no boundary point c exists in R. Sorry i have to repeat to try to understand: no boundary point c exists in R? does this boundary point c exist or does it not exist in R? Of course that boundary point c exist in R. sqrt(2) belongs to R. It belongs to R but it does'nt belong to A U B. Ok, there is no boundary point in R alltohough it could be divede to two sets A and B...ok now i maybe made it clear tomyself. === Subject: Re: Cantor won, but one final question anymore thnk u > --clik -- > Suppose a set R is > 1) linearly ordered, and > 2) densely ordered, i.e., between any two members there is another, and > 3) has no endpoints, i.e., smallest or largest members, and > 4) has no gaps, i.e., if it is partitioned into two nonempty sets A and B > in such a way that every member of A is less than every member of B, then > there is a boundary point c, so that every point less than c is in A and > every point greater than c is in B. > ->Then R is not countable. > --clik -- > The theorem says that if there is no gaps then then R is uncountable. > But there was gap in A U B concerning real numbers when c=srt(2). > No, there is no gap. Point 4) explains what it means for a set to have > no gaps. To show a gap in R, you have to provide suitable sets A and B > such that no boundary point c exists in R. In the case mentioned, > c=sqrt(2) is the required boundary point, and c is in R. Therefore, you > have not demonstrated a gap in R. > There is no gap _in_ _R_ allthough there _is_ gap in A U B because > every point less than c is in A and every point greater than c is in B > (there is not word equal). c does not belongs to A and it doesn't belong > to B. Of course the is then gap in A U B because c doesn't belong to A > either B. No, there is no gap in A U B, since A U B = R and the required boundary point c does belong to R. Notice that 4) begins, if [R] is partitioned into two nonempty sets A and B..., and a partition in mathematics means a decomposition of a set into pairwise disjoint subsets whose union is the original set. Therefore A U B = R, and since c belongs to R, it follows that c must belong either to A or to B. We don't know which, and it doesn't matter. > There was no gap concerning rational numbers i.e. A U B contains all > rationals? So > rationals are not countable but reals are? > On the contrary, there is a gap in the rationals, because in that case > c=sqrt(2) is not a member of Q, and therefore a gap exists. > There is not gap when we stay in Q. c=sqrt(2) doesn't belong to Q, then A > U B contains all the numbers than R. The fact that c=sqrt(2) doesn't belong to Q means that 4) is not satisfied with respect to Q, and therefore Q has a gap. Remember that 4) is the definition of what it means to have no gaps. There is no c in Q such that every member of Q that is less than c is in A and every member of Q that is greater than c is in B. >We didn't not suppose anything more > than that R is a set in the first hand. Yes, we did. We supposed 1), 2), 3), and 4). > R is not said to be the hole one > dimensional line - system of co ordinates x-axle or something like that. > If we then divede the R to two sets, and choose c out of that R of course > that c doesn't belong to R because it doesn't belong there at the > beginning. I don't understand. If you choose c out of R, then c belongs to R. >But there is no gap in A U B. That depends on the original set R. > but ok. You said = To show a gap in R, you have to provide suitable sets > A and B > such that no boundary point c exists in R. > Sorry i have to repeat to try to understand: > no boundary point c exists in R? does this boundary point c exist or > does it not exist in R? Of course that boundary point c exist in R. > sqrt(2) belongs to R. It belongs to R but it does'nt belong to A U B. Yes, it does belong to A U B, since A U B = R. > Ok, > there is no boundary point in R alltohough it could be divede to two sets > A and B...ok now i maybe made it clear tomyself. You have it backwards. There *is* a boundary point in the reals, but not in Q. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: Cantor won, but one final question anymore thnk u --clik -- Suppose a set R is 1) linearly ordered, and 2) densely ordered, i.e., between any two members there is another, and 3) has no endpoints, i.e., smallest or largest members, and 4) has no gaps, i.e., if it is partitioned into two nonempty sets A and B in such a way that every member of A is less than every member of B, then there is a boundary point c, so that every point less than c is in A and every point greater than c is in B. ->Then R is not countable. --clik -- > no boundary point c exists in R? does this boundary point c exist or > does it not exist in R? Of course that boundary point c exist in R. > sqrt(2) belongs to R. It belongs to R but it does'nt belong to A U B. > Yes, it does belong to A U B, since A U B = R. It was this sentence in definition every point less than c is in A and every point greater than c is in B. that made my misunderstanding. I thought it was the definition of A and B, iff so c would not belongs to A U B. A = {a e R ! ac} A U B = R U {sqrt(2)} (sqrt(2) needed to add -> gap) ........ A = {a e Q ! ac} A U B = Q (sqrt(2) no needed to add -> no gap) === Subject: Re: Cantor won, but one final question anymore thnk u > --clik -- > Suppose a set R is > 1) linearly ordered, and > 2) densely ordered, i.e., between any two members there is another, and > 3) has no endpoints, i.e., smallest or largest members, and > 4) has no gaps, i.e., if it is partitioned into two nonempty sets A and B > in such a way that every member of A is less than every member of B, then > there is a boundary point c, so that every point less than c is in A and > every point greater than c is in B. > ->Then R is not countable. > --clik -- > no boundary point c exists in R? does this boundary point c exist or > does it not exist in R? Of course that boundary point c exist in R. > sqrt(2) belongs to R. It belongs to R but it does'nt belong to A U B. > Yes, it does belong to A U B, since A U B = R. > It was this sentence in definition > every point less than c is in A and every point greater than c is in B. > that made my misunderstanding. I thought it was the definition of A and B, > iff so c would not belongs to A U B. Notice that it is not claimed that every point in A is less than c or that every point in B is greater than c. The real misunderstanding is not knowing what a partition is, which is understandable. Is there a corresponding term in Finnish? > A = {a e R ! a B = {b e R ! b>c} > A U B = R U {sqrt(2)} (sqrt(2) needed to add -> gap) I think you mean R = A U B U {sqrt(2)}, but in fact it's R = A U B and c = sqrt(2) is in R, hence no gap. > ........ > A = {a e Q ! a B = {b e Q ! b>c} > A U B = Q (sqrt(2) no needed to add -> no gap) Yes, A U B = Q, but there is no c in Q such that { q in Q : q < c } is contained in A and { q in Q : q > c } is contained in B. Hence, a gap exists. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: Cantor won, but one final question anymore thnk u > --clik -- > Suppose a set R is > 1) linearly ordered, and > 2) densely ordered, i.e., between any two members there is another, and > 3) has no endpoints, i.e., smallest or largest members, and > 4) has no gaps, i.e., if it is partitioned into two nonempty sets A and > B in such a way that every member of A is less than every member of B, > then there is a boundary point c, so that every point less than c is in > A and > every point greater than c is in B. > ->Then R is not countable. > --clik -- Huh! How can this be so difficult. Of course if we select c outside Q it doesn't belong to Q and there is gap, that is obvious, but that is not how i undestood that definition above. Lets start from the beginning step by step. First line: > Suppose a set R is This R is any set which countability or uncountability i want to observe. There is a little possibility of confusion even in this first step. Why it is marked R? Does it have something to do with real numbers? Why? We can take any set and observe it's countability or uncountability. Let's mark this set is X which countability or countability i'm studying. And now let say first X=Q (Set of rationals). Skip those 1)2)3) and go to the 4) I can partiote this X into two nonempty sets A and B. Of course i can. But do i have to select that c inside this X? You did not do that and Cantor did not. I can select it outside X. If X is set of rationals i can select c=sqrt(2) which is outside X. And btw. the definition does not say that for all selected c there is no gap. It only says that there exist one selected c. Why do we select that c outside X. Why could't i select c=0.5? Huh, holy . I think the defination i found was somehow badly written. I must try to find other and follow it...and this is just to try to understand and adapt the therory - what kind of difficulties i'm gonna face when i try to understand the proof of this theory... > Notice that it is not claimed that every point in A is less than c or > that every point in B is greater than c. Allthough every point less than c is in A and every point greater than c is in B Hm. It is not claimed yes. But why it could not been selected c=0.5 forexpample. The definition does not say that there is no gap for all c we select, it only says it is enough there is one. And if we choose c = 0.5 it belongs to Q. -> no gap.-> not countable. And here again. I'm not sure what the defination means. If we choose c=0.5 then Q=A U B > A = {a e R ! a B = {b e R ! b>c} > A U B = R U {sqrt(2)} (sqrt(2) needed to add -> gap) > I think you mean R = A U B U {sqrt(2)}, but in fact it's R = A U B and c > = sqrt(2) is in R, hence no gap. I did not mean that. Here the communication difficulties raises from that we are using R in two meanings. A = {a e R ! ac} A U B = X U {sqrt(2)} (sqrt(2) needed to add -> gap) Where X = R. So i'm adapting that defination to Real numbers. And if i have selected that all in A is less than c, and all in B are greater than c then A U B cannot be R. Or can it? If A= {a in R ! a --clik -- > Suppose a set R is > 1) linearly ordered, and > 2) densely ordered, i.e., between any two members there is another, and > 3) has no endpoints, i.e., smallest or largest members, and > 4) has no gaps, i.e., if it is partitioned into two nonempty sets A and > B in such a way that every member of A is less than every member of B, > then there is a boundary point c, so that every point less than c is in > A and > every point greater than c is in B. > ->Then R is not countable. > --clik -- > Huh! How can this be so difficult. > Of course if we select c outside Q it doesn't belong to Q and there is > gap, that is obvious, but that is not how i undestood that definition > above. Then you misunderstand the definition. The point is that there is no such thing as s c *in Q* that with the property that every rational number less than c is in A, and every rational greater than c is in B. The fact that such a c exists, but is not a member of Q, is not even relevant. It's not a question of selecting a c and then checking to see whether it is in Q. The point is that there is no c in Q that can be selected to define the partition. That absence of a suitable choice for c is what constitutes a gap. > Lets start from the beginning step by step. > First line: > Suppose a set R is > This R is any set which countability or uncountability i want to observe. > There is a little possibility of confusion even in this first step. Why it > is marked R? Does it have something to do with real numbers? Why? We can > take any set and observe it's countability or uncountability. Let's mark > this set is X which countability or countability i'm studying. R in this context was a confusing choice. Feel free to use a different symbol, such as X, so that we can easily distinguish between the cases X = R (the reals) and X = Q (the rationals). But notice that what we have here is not a test for countability or uncountability. The criterion presented here is sufficient, but not necessary, for uncountability. The statement is that if a set satisfies conditions 1), 2), 3), and 4), then the set is not countable. That is not an if and only if condition. For example, it is perfectly possible for a set with gaps to be uncountable. Just take the real line and remove one point. The result is an uncountable set with a gap in it. > And now let say first X=Q (Set of rationals). > Skip those 1)2)3) and go to the 4) > I can partiote this X into two nonempty sets A and B. Of course i can. But > do i have to select that c inside this X? You did not do that and Cantor > did not. I can select it outside X. If X is set of rationals i can select > c=sqrt(2) which is outside X. It makes no sense to even ask that question, because we don't know yet what c is. First you select the sets A and B. Only after that is done do we ask whether there is a suitable c in X for that particular choice of A and B. > And btw. the definition does not say that for all selected c there is no > gap. It only says that there exist one selected c. Why do we select that c > outside X. Why could't i select c=0.5? We don't select c outside X. There are exactly two possibilities: a) There is a c in X that satisfies 4). b) There is no c in X that satisfies 4). Neither of those possibilities says anything at all about whether there is a c outside X. And with good reason. No c that is outside X can possibly satisfy the condition stated in 4), since that condition requires that c, if it exists, must be inside X. Talking about c=sqrt(2) is just a red herring when we are dealing with X = Q, since sqrt(2) is not in Q and therefore is not a candidate for being chosen. > Huh, holy . I think the defination i found was somehow badly written. > I must try to find other and follow it...and this is just to try to > understand and adapt the therory - what kind of difficulties i'm gonna > face when i try to understand the proof of this theory... > Notice that it is not claimed that every point in A is less than c or > that every point in B is greater than c. > Allthough every point less than c is in A and every point greater than c > is in B > Hm. It is not claimed yes. But why it could not been selected c=0.5 > forexpample. The definition does not say that there is no gap for all c we > select, it only says it is enough there is one. And if we choose c = 0.5 > it belongs to Q. -> no gap.-> not countable. You are coming at the definition from the wrong end. We don't start by selecting c; we start by selecting A and B. Then the question is whether a c exists that defines the partition. There are exactly two possibilities, marked a) and b) above, and in neither case do we do any selecting of c. If c exists, it is unique, and therefore no selection is involved. > And here again. I'm not sure what the defination means. If we choose c=0.5 > then Q=A U B You haven't specified A and B yet. That is the necessary first step. > A = {a e R ! a B = {b e R ! b>c} > A U B = R U {sqrt(2)} (sqrt(2) needed to add -> gap) > I think you mean R = A U B U {sqrt(2)}, but in fact it's R = A U B and c > = sqrt(2) is in R, hence no gap. > I did not mean that. Here the communication difficulties raises from that > we are using R in two meanings. > A = {a e R ! a B = {b e R ! b>c} > A U B = X U {sqrt(2)} (sqrt(2) needed to add -> gap) Your sets A and B do not partition the reals, and therefore are not relevant to the question of whether the reals have a gap. Remember what I said earlier about how a set could have a gap and still be uncountable? According to your choice of A and B, that is exactly what happens here. You are dealing with a set X = R{sqrt(2)} = A U B, and yes, it's true that X has a gap. That means you can't tell from the considerations we know that X is uncountable, even though it fails the no gap test. However, there is no gap in R (the reals). > Where X = R. No. You have R = X U {sqrt(2)}, where sqrt(2) is not a member of X = A U B. >So i'm adapting that defination to Real numbers. No, to a subset of the real numbers. But it doesn't apply, because X fails the no gap test. >And if i > have selected that all in A is less than c, and all in B are greater than > c then A U B cannot be R. Or can it? You have the test backwards. The test is whether everything less than c is in A, and everything greater than c is in B. There is no suitable c in this case. > If A= {a in R ! a Now then: c in A? True or false? False. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: Cantor won, but one final question anymore thnk u > There is not gap when we stay in Q. That's fuzzy terminology. Just look at the definition. > c=sqrt(2) doesn't belong to Q, Right. So, since c is the number that should have been in the gap between A,B, and it is not in Q, then clearly there is a gap in Q. V. -- email: lastname at cs utk edu homepage: www cs utk edu tilde lastname === Subject: JSH: The Math Not surprisingly, since mathematics IS mathematics, my position can be defended rather easily: f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) where on the left side you have independent variables, and they are f, m, x and u. One the right side you have x, u and f, but also the dependent variables a_1, a_2, and a_3. Note that you have f^2 as a multiple of the left side, and divide it off to get (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)/f^2 and to determine how that factor f^2 divides through rely on f being independent of m, x and u, so you can set m = 0 to get - 3(-1)x u^2 + u^3 f = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)/f^2 which is u^2(3x + uf) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)/f^2 and you can tell how the factorization goes by observation, as the only way you can get what you have on the left from what you have on the right is for two of the a's to equal 0, while one equals 3, when m=0. Given that m is independent of f, and f is independent of m, it follows that the result holds without regard to the value of m. Therefore, algebraically, it is proven that two of the factors (a_1 x + uf), (a_2 x + uf), and (a_3 x + uf) have f as a factor. Notice that the proof follows easily, as long as you accept algebra. The central point is the independence relationship which is basic. The independent variables are just independent. Some of you may have kept up with arguments on sci.math that have raged over this algebra for OVER TWO YEARS NOW and might consider that posters who even attempt to challenge the algebra algebraically try to FORCE a dependency relationship, like by claiming there exists w_1(m), w_2(m), and w_3(m), that are factors of the a's where w_1 (m) w_2(m) w_3(m) = f^2 and where the w's are to vary as m varies, and notice you now have a dependency relationship! That is, fighting the independent reality of the variables, posters would simply try to add in a dependency of their own. It's completely bogus but there are psychological reasons. Most of you, despite your labels, are not mathematicians. Real mathematicians are like me. Quirky sense of humor, persistence to the point of insanity as far as normal people are concerned, extraordinary consistency over periods of years, and oh yeah, a tremendous love of the truth above all else, even at the expense of other people. People come and go, mathematics is absolute. Read your history carefully, and REALLY see people like Gauss, Newton, Euler, Fermat and Archimedes. Some of you think you can take the title of mathematician or that it can be given to you, but you're not mathematicians, no matter how many people TELL you that you are. Consider the tests I've given for years now, and finally maybe some of you understand those tests. I've had a good time talking about how many of you don't love mathematics, hate algebra, and look at math as a social thing. I've used quite a few inside jokes, like talking about how math is NOT a fashion show. If you got that, pat yourself on the head, and congratulate yourself on being at least kind of smart, if you still don't get it: Consider that Galois Theory has to do with assuming that the FORM that is forced on how you can express certain roots by elementary means actually forces the SUBSTANCE in terms of factors of those roots. The research by Evariste Galois actually has more to do with style: how you can express roots of irrational polynomials irreducible over Q. Get it? Math is not a fashion show. Get it now? James Harris http://mathforprofit.blogspot.com/ === Subject: Re: JSH: The Math > Not surprisingly, since mathematics IS mathematics, my position can be > defended rather easily: > f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) = > (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) > where on the left side you have independent variables, and they are f, > m, x and u. > One the right side you have x, u and f, but also the dependent > variables a_1, a_2, and a_3. So, unsupressing the dependencies we can write a_1(m,f,u), a_2(m,f,u) and a_3(m,f,u). Note we have a_1(0,f,u) = a_2(0,f,u) =0 > Note that you have f^2 as a multiple of the left side, and divide it > off to get > (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f = > (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)/f^2 > and to determine how that factor f^2 divides through rely on f being > independent of m, x and u, so you can set m = 0 to get > - 3(-1)x u^2 + u^3 f = > (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)/f^2 > which is > u^2(3x + uf) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)/f^2 > and you can tell how the factorization goes by observation, as the > only way you can get what you have on the left from what you have on > the right is for two of the a's to equal 0, while one equals 3, when > m=0. The notation above is a bit confusing, especially if one is going to use independence arguments. What we can say is that f^2 divides (a_1(m,f,u) x + uf)(a_2(m,f,u) x + uf)(a_3(m,f,u) x + uf) (*) for all values of m, f, u. > Given that m is independent of f, and f is independent of m, it > follows that the result holds without regard to the value of m. No, because although f is independent of m, a1(m,f,u), a_2(m,f,u) and a_3(m,f,u) are not independent of m. It is the values of a1(m,f,u), a_2(m,f,u) and a_3(m,f,u) that determine how f^2 divides (*). > Therefore, algebraically, it is proven that two of the factors > (a_1 x + uf), (a_2 x + uf), and (a_3 x + uf) > have f as a factor. No, the statement is false. Also you do not attempt to prove this algebraically but by a incorrect use of an independence argument. > Notice that the proof follows easily, as long as you accept algebra. No, the proof only follows if you accept woolly independence arguments. > The central point is the independence relationship which is basic. > The independent variables are just independent. The independent variables are indeed independent. However, the variables that matter are a_1, a_2 and a_3. These are not independent of m. > Some of you may have kept up with arguments on sci.math that have > raged over this algebra for OVER TWO YEARS NOW and might consider that > posters who even attempt to challenge the algebra algebraically try to > FORCE a dependency relationship, like by claiming there exists w_1(m), > w_2(m), and w_3(m), that are factors of the a's where > w_1 (m) w_2(m) w_3(m) = f^2 > and where the w's are to vary as m varies, and notice you now have a > dependency relationship! > That is, fighting the independent reality of the variables, posters > would simply try to add in a dependency of their own. w_1, w_2 and w_3 depend on a_1, a_2 and a_3. a_1, a_2 and a_3 depend on m. Thus w_1, w_2 and w_3 depend on m. > [diatribe snipped] -William Hughes === Subject: Re: JSH: The Math when you all pile on top of HSJ like this, the poor item seems to die. this really has got to be a paid social experiment; the question is, Do we get any of the money, when it's over? > No, the statement is false. Also you do not attempt > to prove this algebraically but by a incorrect use of an > independence argument. > The central point is the independence relationship which is basic. > The independent variables are indeed independent. However, the > variables that matter are a_1, a_2 and a_3. These are not > independent of m. --ils duces d'Enron! http://larouchepub.com --Chair Man George! http://tarpley.net === Subject: Re: JSH: The Math > Not surprisingly, since mathematics IS mathematics, my position can be > defended rather easily: > f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) = > (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) > where on the left side you have independent variables, and they are f, > m, x and u. > One the right side you have x, u and f, but also the dependent > variables a_1, a_2, and a_3. > Note that you have f^2 as a multiple of the left side, and divide it > off to get > (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f = > (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)/f^2 > and to determine how that factor f^2 divides through rely on f being > independent of m, x and u, so you can set m = 0 to get > - 3(-1)x u^2 + u^3 f = > (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)/f^2 > which is > u^2(3x + uf) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)/f^2 > and you can tell how the factorization goes by observation, as the > only way you can get what you have on the left from what you have on > the right is for two of the a's to equal 0, while one equals 3, when > m=0. > Given that m is independent of f, and f is independent of m, it > follows that the result holds without regard to the value of m. Here we have the New Mathematics: The Harris Independence Principle (HIP). Here is a concise statement: THE HARRIS INDEPENDENCE PRINCIPLE Suppose two variables m and f both occurring in the same equation are *independent* of each other. Then if some fact regarding f is true when m = 0, it must be true for all other values of m also. Of course we should define Ôindependent'. Two variables in an equation are independent if, when you choose a value for one of them, the other one can be chosen to be anything you want. Terrific new principle! Why haven't mathematicians noticed it before? I bet it can be used to prove all sorts of new theorems! Here is an example: Say m and f are nonnegative integers. (m + f) * (m + f + 1) = a * b, where a and b are integers. First I pick m and f. Then once I have picked m and f, a and b are dependent on them, though m and f are themselves completely INDEPENDENT! Remember, first I pick m, then I can pick f to be anything I want. Of course I am not as free to pick a and b: they are DEPENDENT on m and f. Note that when m = 0, either a or b (or both) is non-coprime to f, since f * (f + 1) = a * b. Therefore by the Harris Independence Principle (HIP), for all other values of m, either a or b (or both) is noncoprime to f ! Hmm, let's see. I'll let m = 1 and f = 5. The equation is (1 + 5)*(1 + 5 + 1) = a * b, or 6 * 7 = 42 = a * b. Now, the HIP says that either a or b must be noncoprime to 5. That is, 42 must have a nonunit factor in common with 5 ! Immediately I get a new result! And one nobody every suspected! Yes, the HIP is great! I say, HIP HIP HOORAY for the Harris Independence Principle! > Therefore, algebraically, it is proven that two of the factors > (a_1 x + uf), (a_2 x + uf), and (a_3 x + uf) > have f as a factor. > Notice that the proof follows easily, as long as you accept algebra. Yep, the HIP: it's JUST ALGEBRA! > The central point is the independence relationship which is basic. > The independent variables are just independent. Yep, they just *are*. Really, this is not just another James Harris bull pseudomath argument. It is getting so annoying to see him accused of that all the time. And this is not just some fancy untrustworthy BS put out by those Galois fanatics. Grow up, people! This is the HIP ! This is ALGEBRA! > Some of you may have kept up with arguments on sci.math that have > raged over this algebra for OVER TWO YEARS NOW and might consider that > posters who even attempt to challenge the algebra algebraically try to > FORCE a dependency relationship, like by claiming there exists w_1(m), > w_2(m), and w_3(m), that are factors of the a's where > w_1 (m) w_2(m) w_3(m) = f^2 > and where the w's are to vary as m varies, and notice you now have a > dependency relationship! Shocking. You know, when I write m*f = a1, where a1 is some integer, m and f are the independent variables - I am perfectly free to choose m and f - and a1 is the DEPENDENT variable. First I pick m and f, and then a1 is determined (i.e., DEPENDENT). For fixed f, I could write m*f = a1(m), to express that dependency. Whoa, now it looks like m and f are both in one equation, so now they must be DEPENDENT. But you know what? m and f are still INDEPENDENT. No dependency of m and f has been imposed! If you pick one of them, you are still perfectly free to pick the other one however you wish! Why? Because a1 is a *dependent* variable! It's the one ending up with the short end of the stick! The other two can do pretty much anything they want. They're INDEPENDENT. Still, I must say, this was a pretty good attempt to lead yourself astray. You were probably thoroughly confused by the time you got through it. Was anyone else? Maybe not. > That is, fighting the independent reality of the variables, posters > would simply try to add in a dependency of their own. You can't fight Independence! > It's completely bogus but there are psychological reasons. Right. When people deny that 42 and 5 have common factos, that's just for psychological reasons. They are after all denying the HIP! > Most of you, despite your labels, are not mathematicians. > Real mathematicians are like me. You mean, they're full of crap??? > Quirky sense of humor, persistence > to the point of insanity or well beyond > as far as normal people are concerned, > extraordinary consistency over periods of years, and oh yeah, a > tremendous love of the truth above all else, even at the expense of > other people. [snip the part not concerned with the HIP] > Get it now? Oh, sure. HIP HIP HOORAY! Nora B. > James Harris > http://mathforprofit.blogspot.com/ === Subject: Re: JSH: The Math >Not surprisingly, since mathematics IS mathematics, my position can be >defended rather easily: >f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) = > (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) >where on the left side you have independent variables, and they are f, >m, x and u. >One the right side you have x, u and f, but also the dependent >variables a_1, a_2, and a_3. >Note that you have f^2 as a multiple of the left side, and divide it >off to get >(m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f = > (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)/f^2 >and to determine how that factor f^2 divides through rely on f being >independent of m, x and u, so you can set m = 0 to get > - 3(-1)x u^2 + u^3 f = > (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)/f^2 >which is >u^2(3x + uf) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)/f^2 >and you can tell how the factorization goes by observation, as the >only way you can get what you have on the left from what you have on >the right is for two of the a's to equal 0, while one equals 3, when >m=0. >Given that m is independent of f, and f is independent of m, it >follows that the result holds without regard to the value of m. > Here we have the New Mathematics: The Harris Independence Principle > (HIP). That would be HIP as in HIP HOP. Harristotelian mathematics has a lot in common with hip hop music - improvised on the spot, with a very simple repetitive rhythm, full of boasts and threats. === Subject: Re: JSH: The Math > Not surprisingly, since mathematics IS mathematics, my position can be > defended rather easily: > Given that m is independent of f, and f is independent of m, it > follows that the result holds without regard to the value of m. > > Here we have the New Mathematics: The Harris Independence Principle > (HIP). Here is a concise statement: > THE HARRIS INDEPENDENCE PRINCIPLE > Suppose two variables m and f both occurring in the same > equation are *independent* of each other. Then if some > fact regarding f is true when m = 0, it must be true for > all other values of m also. > Of course we should define Ôindependent'. Two variables > in an equation are independent if, when you choose a value > for one of them, the other one can be chosen to be anything > you want. > Terrific new principle! Why haven't mathematicians noticed > it before? I bet it can be used to prove all sorts of new > theorems! At the risk of sullying a JSH thread with the sort of crankery the mention of Goedel tends to invite, I have to wonder if James has somehow found a way out of the Goedelian quagmire. That is, by employing the Harris Independence Principle, * Can it be shown that for every true statement in mathematics, there is a proof of that statement? Indeed, I suspect this is so! In fact, the power of the HIP so impresses me that I would ask: * Can it be shown that for every false statement in mathematics, there is a proof of that statement? Now if *both* of these are true, then it gives James a tremendous advantage over traditional mathematicians. All James would have to do is formulate statements that are precise enough to actually *be* true or false . . . Hmmm . . . Well never mind, then. === Subject: Re: JSH: The Math [Dead-on parody of JSH's reasoning deleted] OK, that was brilliant. It was also frightening, because I don't think one can imitate insanity that perfectly without being somewhat insane themselves. :-) -- --Tim Smith === Subject: Re: JSH: The Math X-RFC2646: Original Ullrich does not respond much to Nora Baron's posts, one possible reason being: he does not want to talk to himself. > Here we have the New Mathematics: > The Harris Independence Principle (HIP). > Here is a concise statement: > THE HARRIS INDEPENDENCE PRINCIPLE > Suppose two variables m and f both occurring > in the same equation are *independent* of each > other. Then if some fact regarding f is true when > m = 0, it must be true for all other values of m also. > Of course we should define Ôindependent'. Two variables > in an equation are independent if, when you choose a value > for one of them, the other one can be chosen to be anything > you want. > Terrific new principle! Why haven't mathematicians noticed > it before? I bet it can be used to prove all sorts of new > theorems!... === Subject: Re: JSH: The Math > Ullrich does not respond much > to Nora Baron's posts, one possible > reason being: he does not want to > talk to himself. Harris, is that you? Let it go. Just admit you're wrong and move on. Jim Burns === Subject: Re: JSH: The Math === >Subject: Re: JSH: The Math >Message-id: >Ullrich does not respond much >to Nora Baron's posts, one possible >reason being: he does not want to >talk to himself. So the not using the shift key was just a ploy to mask his secret identity? > Here we have the New Mathematics: > The Harris Independence Principle (HIP). > Here is a concise statement: > THE HARRIS INDEPENDENCE PRINCIPLE > Suppose two variables m and f both occurring > in the same equation are *independent* of each > other. Then if some fact regarding f is true when > m = 0, it must be true for all other values of m also. > Of course we should define Ôindependent'. Two variables > in an equation are independent if, when you choose a value > for one of them, the other one can be chosen to be anything > you want. > Terrific new principle! Why haven't mathematicians noticed > it before? I bet it can be used to prove all sorts of new > theorems!... -- Mensanator Ace of Clubs === Subject: Re: JSH: The Math X-RFC2646: Original > So the not using the shift key > was just a ploy to mask > his secret identity? Perhaps you are right. But that shift-key injury was nothing more than an indulgance. === Subject: Re: JSH: The Math > Not surprisingly, since mathematics IS mathematics, my position can be > defended rather easily: You still don't get it. But why do you continually display your full frontal ignorance? -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: JSH: The Math <4155EC9D.6B6D0922@ix.netcom.com> posting-account=jcZk7AwAAADXpPEyHtVyWC264SxtppRB posts, he will eventually write down something meaningful by chance! Van === Subject: Re: JSH: The Math [cut] > The research by Evariste Galois actually has more to do with style: > how you can express roots of irrational polynomials irreducible over > Q. Irreducible is not required or needed. Also, as has been pointed out before, the phrase irrational polynomials makes no sense. Galois showed that the roots of a polynomial with rational coefficients are expressible in terms of radicals if and only if the Galois group of that polynomial is a solvable group. The polynomial need not be irreducible. -- Bill Hale === Subject: Re: A better antiderivative of x^p? > Here's something similar that I noticed some time ago: > > Consider the linear constant-coefficient homogeneous DE > > (*) y'' + ay' + by = 0. > > Say r, s are the roots of the Ôcharacteristic equation' > x^2 + ax + b = 0. We all know that _if_ r <> s then > y_1 = exp(rt), y_2 = exp(bt) form a basis for the set >You meant y_2 = exp(st) > Yes. > of solutions to (*). And we know that if r = s then > a basis for the set of solutions looks entirely > different; there's the same sort of apparent (but > impossible) discontinuity. > > Quiz: What is a better basis for the solution space, > in the case r <> s? >take y_1 = exp(rt), y_2 = (exp(st)-exp(rt))/(s-r) as the basis. >for s->r, y_2 -> t exp(rt). > Precisely. Before this thread ends, I'll mention another better basic antiderivative. (BTW, Derive doesn't give this one.) An antiderivative of sin(a*x) wrt x which behaves nicely as a -> 0 is # (1 - cos(a*x))/a. Example: Find an antiderivative of x*cos(a*x) wrt x such that the antiderivative approaches 1/2*x^2 as a -> 0. Integrating by parts and using # in the process, we directly obtain x*sin(a*x)/a + (cos(a*x) - 1)/a^2. David Cantrell === Subject: Re: the imprecision of 4 color mapping and why it should be 2 color mapping > Borderlines have no measurable area, you idiot. They're > one-dimensional. > Apparently you never really studied math because people who do study > math are all familar with **** Every point of the set is assigned a > function for coloring...*** > But of course every point of a set has little meaning to Graham for he > has no acumen or mind of mathematics. > Appel and Haken assume every point of the set... > It matters not whether the interior points have area. It matters not > that borderlines have no area. It matters not that vertices have no > area. It matters not that interiors are 2 dimensional. It matters not > that borderlines are 1 dimensional. It matters not that vertices are > zero dimensional. > What matters is a function that colors *** Each and every point *** > regardless of whether the point is 0,1 or 2 dimensions. > Every point has to be accountable to the Function of coloring. > Graham should have read the posts by Tim Mellor before he engaged in > this discussion. Because Tim pointed out the gap in the defining of the > 4 Color Mapping of Guthrie to deMorgan to Cayley to Kempe to Appel & > Haken. > EVERY POINT of a mapping must be accountable to a color Function. That > is mathematics and mathematics is the science of 100% precision. You > lose and fail precision if you pretend that borderlines or vertices are > colorless while other points of the mapp have color. Actually, that's not the case. There's a second way of looking at the problem known as the four-colour theorem that's systematically identical to the usual one. Suppose you have a two-dimensional undirected graph, arranged such that no two edges cross each other. This means you can have (fixed-width graphics ahead): ...A-----B... ...A-----B... | /| | /| | / | | / | | / | but not | X | | / | | / | |/ | |/ | ...D-----C... ...D-----C... Given such a graph with an arbitrarily large number of nodes, how many *unique* colours do you need to colour _the nodes_ such that no two nodes that share an edge have the same colour? Four. There are even algorithmic ways to colour the graph. This completely separates the problem from physical representation; neither nodes nor edges inherently have area at all, and colours are merely arbitrary pieces of identifying information associated with a node. It also clearly defines what part of the graph we're concerned with: the nodes. Let me state that again, just in case you missed it: the four-colour theorem is *defined* to only refer to the nodes of the graph, not the edges. The edges exist only to indicate which nodes are considered adjacent and which are not. You consistently bring up the border lines as being areas in the same sense; in this representation the border lines map to the edges, which we've already determined that we're uninterested in. Note, however, that there is a subset of this problem which does map to your interpretation, where the border lines between regions are also nodes, like so: region A | region b ------------------------ <-- border line region c mapping to A --- border --- B | C which can, indeed, be coloured in only two colours. However, this is not the general case covered by the four-colour theorem, but a subset of possible graphs. Your other assertion is that two colours is sufficient to relay all the information on a map of regions. You may well be right on that, but that is not what the four-colour theorem addresses. -- Some say the Wired doesn't have political borders like the real world, but there are far too many nonsense-spouting anarchists or idiots who think that pranks are a revolution. === Subject: Re: the imprecision of 4 color mapping and why it should be 2 color mapping > Your other assertion is that two colours is sufficient to relay all the > information on a map of regions. You may well be right on that, but that > is not what the four-colour theorem addresses. All maps ever drawn in the entire Universe and ever will be drawn start out as a 2 Color Mapping. The squire or page or artist or editor or mapmaker gets a sheet of white paper and with a blue ink pen makes the Map. There, the map comes into existence and it is 2 colors. And every intelligent person can distinguish one country from another country. Now, because some people think white and blue is not artistic enough they want to color in the interiors. But that is not mathematics. Because there never existed nor will ever exist a Map in which it starts out as 4 Colors for the map. All maps start out as 2 colors. That means 2 colors suffice or sufficient to color every and all Maps. Just because some people are bored with white and blue is no justification for coloring in the white interiors and to make it easy on some eyes viewing the mapp and to justify calling in the mathematicians to come up with a 4 Color Mapping. It was always a 2 Color Mapping easily proven by the Jordan Curve Theorem. And the General Theorem which is higher than the 2 Color Mapping and the Jordan Curve Theorem is the Theorem of Information: which states that 2 colors are sufficient to convey or communicate any piece of information. If the 4 Color Mapping of Appel and Haken are true then my theorem of Information must be false. But my theory of Information is rather obviously Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: the imprecision of 4 color mapping and why it should be 2 color mapping (big snip) > EVERY POINT of a mapping must be accountable to a color Function. That > is mathematics and mathematics is the science of 100% precision. You > lose and fail precision if you pretend that borderlines or vertices are > colorless while other points of the mapp have color. > Actually, that's not the case. There's a second way of looking at the > problem known as the four-colour theorem that's systematically identical > to the usual one. > Suppose you have a two-dimensional undirected graph, arranged such that no > two edges cross each other. This means you can have (fixed-width graphics > ahead): > ...A-----B... ...A-----B... > | /| | /| > | / | | / | > | / | but not | X | > | / | | / | > |/ | |/ | > ...D-----C... ...D-----C... > Given such a graph with an arbitrarily large number of nodes, how many > *unique* colours do you need to colour _the nodes_ such that no two > nodes that share an edge have the same colour? Four. There are even > algorithmic ways to colour the graph. I disagree. And I would answer Two colors. systematic case by case to cover all cases. But when you allege a proof by the claim that * you covered all cases* you are on shaky grounds. Because there maybe zillions of cases in which are not covered. So you almost need a subproof inside of Appel & Haken that proves all cases are covered. For example: not only the 4 Color Mapping but also the Four-Colour-Theorem if this is genuine. Recall in Calculus those wild functions, those bizarre functions of the step-function that are discontinuous yet integrable. Recall those wild functions of picket fence with points not differentiable but still integrable. Recall many wild functions in calculus that behave so outlandish that we can barely comprehend them. Now suppose a country exists that has those borderlines of one of those wild functions. The Four-Colour-Theorem falls ßat on its face. The 4-Color-Mapping of Appel and Haken would include this wild function as a country because, hey, it ignores borderlines anyway. But the 2 Color Mapping proven by the Jordan Curve Theorem would step in and say it is not a country at all because it is not a **closed curve**. Another Example: is the geometry of a wheel with spokes of one of those wild functions of calculus. There are infinite number of vertices and your Four-Colour-Theorem breaks down. The 4-Color-Mapping of Appel & Haken becomes silent because although it has land area it is lost to the infinite spokes. Only the 2 Color Mapping answers the problem by saying that it is not closed loops. So this is an example of where Appel and Haken had missed a case in their alleged proof. In fact they missed all of those wild functions of Calculus. The 2 Color Mapping and its equivalent of the Jordan Curve Theorem never misses any case example no matter how wild of a function you drag out of Calculus because they all have to be *closed curves* and because they are closed curves the borderlines are one color and the interiors a different color. > This completely separates the problem from physical representation; > neither nodes nor edges inherently have area at all, and colours are > merely arbitrary pieces of identifying information associated with a node. > It also clearly defines what part of the graph we're concerned with: the > nodes. Most every theorem of mathematics is a generalization or idealization over the physical universe. But the 4 Color Mapping of Appel & Haken is an oddball from all other math theorems because it simply is a poorly formulated problem. We have managed to drag into mathematics sense perception in telling apart one country from another from colors of their interior. This is not mathematics but an appeal to our sensory perception. If all humans were color blind and could only see white and black and shades of gray, then would that species due to their sensory perceptions come up with a 4 Colour Mapping Problem when such a species never could see 4 colors? You see what I mean when I say the 4 Color Mapping of Appel and Haken is art and not mathematics. A colorblind species would discover all the other theorems of mathematics but never a 4 Color Mapping. They would discover a 2 Color Mapping theorem. This illustration points out starkly the fakery that is the 4 Color Mapping. Mathematics is never bound to the sensory perception of a species. > Let me state that again, just in case you missed it: the four-colour > theorem is *defined* to only refer to the nodes of the graph, not the > edges. The edges exist only to indicate which nodes are considered > adjacent and which are not. I am unable to point out the ßaws of Four Colour Theorem. I can point out the ßaw of 4 Color Mapping-- it ignores points of borderlines and points of vertices and treats only points of interiors as able to receive color when all points of a map should receive a color regardless of whether they are interior or borderline of vertex. > You consistently bring up the border lines as being areas in the same > sense; in this representation the border lines map to the edges, which I do not bring up borderlines as areas. I simply say you cannot ignore borderline points as colorless and treat other points as colorfull. If you color every interior point, you must color points that are borderlines and vertices. > we've already determined that we're uninterested in. Note, however, that > there is a subset of this problem which does map to your interpretation, > where the border lines between regions are also nodes, like so: > region A | region b > ------------------------ <-- border line > region c > mapping to > A --- border --- B > | > C > which can, indeed, be coloured in only two colours. However, this is not > the general case covered by the four-colour theorem, but a subset of > possible graphs. > Your other assertion is that two colours is sufficient to relay all the > information on a map of regions. You may well be right on that, but that > is not what the four-colour theorem addresses. I doubt there is a patching up of the 4 color mapping theorem. A restatement that makes it a work of mathematics. I say this because it is impossible to ignore points of borderline and vertices and to treat interiors with color because of those wild functions. Besides, the 2 Color Mapping proven by Jordan Curve is sufficient with just 2 colors and so 3 colors would make a even better artistic sensory perception than 2. So the 4 color mapping is really just art and not mathematics. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: the imprecision of 4 color mapping and why it should be 2 color mapping > Suppose you have a two-dimensional undirected graph, arranged such that no > two edges cross each other. This means you can have (fixed-width graphics > ahead): > ...A-----B... ...A-----B... > | /| | /| > | / | | / | > | / | but not | X | > | / | | / | > |/ | |/ | > ...D-----C... ...D-----C... > Given such a graph with an arbitrarily large number of nodes, how many > *unique* colours do you need to colour _the nodes_ such that no two > nodes that share an edge have the same colour? Four. There are even > algorithmic ways to colour the graph. > I disagree. And I would answer Two colors. Be careful what you disagree with. The example graph cannot be coloured, satisfying the definition right below it, using two colours; if you don't believe me, strip off the ...s indicating that the graph could continue and try colour it with only two colours such that no nodes sharing an edge also share a colour. You only need to consider nodes A, B, and D to demonstrate this. Let's call your colours x and y... If you colour A with x, then you must necessarily colour B with y, as it shares an edge with A and cannot be the same colour. But what colour do we make D? If we colour it x, it's the same colour as A (with which it shares an edge), and if we colour it y, it's the same colour as B (with which it also shares an edge). > systematic case by case to cover all cases. But when you allege a proof by > the claim that * you covered all cases* you are on shaky grounds. Because > there maybe zillions of cases in which are not covered. So you almost need a > subproof inside of Appel & Haken that proves all cases are covered. > For example: not only the 4 Color Mapping but also the Four-Colour-Theorem if > this is genuine. Recall in Calculus those wild functions, those bizarre > functions of the step-function that are discontinuous yet integrable. Recall > those wild functions of picket fence with points not differentiable but still > integrable. Recall many wild functions in calculus that behave so outlandish > that we can barely comprehend them. Remember the defintion of a function, though. For any input x, f(x) must output only one value (normally mapped along the y axis of a plane). We can use this to break simple continuous functions into two regions (above the function and below or on the function, or above and on the function and below the function). For conics, we can break it down into two or three regions (inside or on the line and outside the line or inside the line and outside or on the line -- seeing the pattern yet? Any line can be included in the region on either side of it without altering the areas of the regions.). For more-complex functions, more analysis is required, but in general they can be broken up into clearly-delineated regions. Note that some or all of those regions can be infinitely large: the function f(x) = x**2 defines two regions, both infinitely large. > Now suppose a country exists that has those borderlines of one of those wild > functions. For some functions, this is a very poor definition. Consider the function f(x) = [[x]] /* ßoor (x) */: it doesn't enclose a region, because the line is discontinuous but not asymptotic. ^ | <-- not a boundary between regions | | | *-------------------> x However, we can create boundaries by Ôconnecting' the vertical breaks between segments of the function, at which point this is about as complex as the function f(x) = x. > Another Example: is the geometry of a wheel with spokes of one of those wild > functions of calculus. There are infinite number of vertices and your > Four-Colour-Theorem breaks down. The 4-Color-Mapping of Appel & Haken becomes > silent because although it has land area it is lost to the infinite spokes. > Only the 2 Color Mapping answers the problem by saying that it is not closed > loops. So this is an example of where Appel and Haken had missed a case in > their alleged proof. In fact they missed all of those wild functions of > Calculus. Once again, you're using a very poorly-defined way of creating regions on a cartesian plane and trying to apply those poorly-defined regions to a clearly-defined theorem. I'm honestly not certain what you're trying to get at, here. Build a graph (either in the Ôplot the function' sense or what I actually mean, which is Ônodes and edges') of it, if it helps demonstrate it. > This completely separates the problem from physical representation; > neither nodes nor edges inherently have area at all, and colours are > merely arbitrary pieces of identifying information associated with a node. > It also clearly defines what part of the graph we're concerned with: the > nodes. > Most every theorem of mathematics is a generalization or idealization over > the physical universe. But the 4 Color Mapping of Appel & Haken is an oddball > from all other math theorems because it simply is a poorly formulated > problem. > We have managed to drag into mathematics sense perception in telling > apart one country from another from colors of their interior. This is not > mathematics but an appeal to our sensory perception. In other words, it's an application problem. This doesn't make it any less mathematical: the theorem discovered applies to any graph with the requisite properties and is just as much pure math as any other theorem you care to name. However, like some theorems and unlike others, the four-colour theorem came out of a specific application, which you seem to have some problem with in this specific case. > If all humans were color blind and could only see white and black and > shades of gray, then would that species due to their sensory perceptions > come up with a 4 Colour Mapping Problem when such a species never could > see 4 colors? Colour here can be taken to be an abstract piece of information associated with a region or a node in a graph. For a colour-blind species, consider four different fill patterns. > Let me state that again, just in case you missed it: the four-colour > theorem is *defined* to only refer to the nodes of the graph, not the > edges. The edges exist only to indicate which nodes are considered > adjacent and which are not. > I am unable to point out the ßaws of Four Colour Theorem. Hold that thought. It's an important one. The Four Colour Theorem is pure math, and can be demonstrated and proven using only clearly-defined graph theory concepts. > I can point out the ßaw of 4 Color Mapping-- it ignores points of > borderlines and points of vertices and treats only points of interiors > as able to receive color when all points of a map should receive a color > regardless of whether they are interior or borderline of vertex. What you are complaining about, then, is the application of the theorem to a real-world problem. Congratulations, your opinion about this is just as much an opinion as anyone else's, not some mystical truth that we have all missed. > You consistently bring up the border lines as being areas in the same > sense; in this representation the border lines map to the edges, which > I do not bring up borderlines as areas. I simply say you cannot ignore > borderline points as colorless and treat other points as colorfull. If > you color every interior point, you must color points that are > borderlines and vertices. Here's a simple example for you, showing what others have been getting at with the no borderlines angle. Here's a picture: Four regions, coloured differently, with no borders between them - the boundary is simply a change in colour. Sure, it's not a map... ...or is it? > 4 color mapping is really just art and not mathematics. In fact, two colours is as much an aesthetic judgement as four colours. Also sprach Archimedes Plutonium: > Your other assertion is that two colours is sufficient to relay all the > information on a map of regions. You may well be right on that, but > that is not what the four-colour theorem addresses. > All maps ever drawn in the entire Universe and ever will be drawn start > out as a 2 Color Mapping. The squire or page or artist or editor or > mapmaker gets a sheet of white paper and with a blue ink pen makes the > Map. There, the map comes into existence and it is 2 colors. And every > intelligent person can distinguish one country from another country. > Now, because some people think white and blue is not artistic enough > they want to color in the interiors. But that is not mathematics. > Because there never existed nor will ever exist a Map in which it starts > out as 4 Colors for the map. You've discovered something truly stunning here: you've determined that applications of math aren't always pure math. Wow. > All maps start out as 2 colors. Us and Them, usually... -- Some say the Wired doesn't have political borders like the real world, but there are far too many nonsense-spouting anarchists or idiots who think that pranks are a revolution. === Subject: Re: the imprecision of 4 color mapping and why it should be 2 color mapping > Suppose you have a two-dimensional undirected graph, arranged such that no > two edges cross each other. This means you can have (fixed-width graphics > ahead): > > ...A-----B... ...A-----B... > | /| | /| > | / | | / | > | / | but not | X | > | / | | / | > |/ | |/ | > ...D-----C... ...D-----C... > > Given such a graph with an arbitrarily large number of nodes, how many > *unique* colours do you need to colour _the nodes_ such that no two > nodes that share an edge have the same colour? Four. There are even > algorithmic ways to colour the graph. > I disagree. And I would answer Two colors. > Be careful what you disagree with. The example graph cannot be coloured, > satisfying the definition right below it, using two colours; if you don't > believe me, strip off the ...s indicating that the graph could continue > and try colour it with only two colours such that no nodes sharing an edge > also share a colour. You only need to consider nodes A, B, and D to > demonstrate this. Owen, if you can illustrate your problem with the colors of black and white on my computer screen and the computer screens of millions of viewers. And provide all the information ever needed in just those 2 colors. The fact that you can even communicate this Nodal problem in 2 colors of black and white and tell me all about the information on that page on my computer screen should tell you that 2 colors suffice. I mean, I can see that A, B, D are vertices even though they are all one color. So do you not think it is pretentious of you to think that you need to have more colors than black and white to paint B or D or C or A? Do you not think it is pretentious of you when you can convey all the information of 4 Color Theorem in black and white that you keep saying you need a third new color when you obviously do not need a 3rd new color. So your Color Mapping Theorem concerning Nodal points is not 4 color but 2 color in the fact that you can convey the entire information with the colors of black and white. I can see all vertex points in black as easy and perhaps easier than if you painted them in 3 more colors. The mistake of this 4 Color Theorem is that you place yourself on this restriction when you never had to be placed on any restriction no nodes share an edge also share a color. The graph does not demand it because it already conveys all the information in black and white and you are heedlessly placing that demand by wanting 2 more colors on topp of black and white. We know it is a vertex or node at A or B or D from the black lines that intersect. For what reason would you ever need two more additional colors is nonsense. > Let's call your colours x and y... > If you colour A with x, then you must necessarily colour B with y, as it No. Both A and B can be colored x. The only points colored y are not borderlines or interior points. You are not doing math, Owen, you are doing personal demands. > shares an edge with A and cannot be the same colour. But what colour do > we make D? If we colour it x, it's the same colour as A (with which it > shares an edge), and if we colour it y, it's the same colour as B (with > which it also shares an edge). Again, Owen. The fact exists that you are able to communicate or convey the entire illustration in 2 colors of black and white, for I can see a A as a node and B as a node and D as a node as black and I do not need to have two more colors to see them as nodes. Your 4 Color Theorem fails like the 4 Color Mapping fails in that both are created first from 2 colors and 2 colors only and then much later or after it is created come the artists with 2 more colors that are not needed and jazz up or make fancy the nodes or mapp. > systematic case by case to cover all cases. But when you allege a proof by > the claim that * you covered all cases* you are on shaky grounds. Because > there maybe zillions of cases in which are not covered. So you almost need a > subproof inside of Appel & Haken that proves all cases are covered. > For example: not only the 4 Color Mapping but also the Four-Colour-Theorem if > this is genuine. Recall in Calculus those wild functions, those bizarre > functions of the step-function that are discontinuous yet integrable. Recall > those wild functions of picket fence with points not differentiable but still > integrable. Recall many wild functions in calculus that behave so outlandish > that we can barely comprehend them. > Remember the defintion of a function, though. For any input x, f(x) must > output only one value (normally mapped along the y axis of a plane). We > can use this to break simple continuous functions into two regions (above > the function and below or on the function, or above and on the > function and below the function). For conics, we can break it down > into two or three regions (inside or on the line and outside the line > or inside the line and outside or on the line -- seeing the pattern > yet? Any line can be included in the region on either side of it without > altering the areas of the regions.). > For more-complex functions, more analysis is required, but in general they > can be broken up into clearly-delineated regions. Note that some or all > of those regions can be infinitely large: the function f(x) = x**2 defines > two regions, both infinitely large. > Now suppose a country exists that has those borderlines of one of those wild > functions. > For some functions, this is a very poor definition. Consider the function > f(x) = [[x]] /* ßoor (x) */: it doesn't enclose a region, because the > line is discontinuous but not asymptotic. > | <-- not a boundary between regions > *-------------------> x > However, we can create boundaries by Ôconnecting' the vertical breaks > between segments of the function, at which point this is about as complex > as the function f(x) = x. Owen, my point was that Appel & Haken offered a computer program as a alleged proof of 4 Color Mapping that tackles a large number of **cases**. I doubt that they tackled these wild functions in Calculus and because Appel and Haken ignore borderlines the above step function which you admireably displayed must be accounted for by Appel and Haken since they have interior areas. But the 2-Color Mapping proven by using the Jordan Curve Theorem immediately says those step functions and wild functions of Calculus are not countries because they are not closed loops. Owen, I brought up these wild functions to illustrate the huge gaps and ßaws of Appel & Haken. > Another Example: is the geometry of a wheel with spokes of one of those wild > functions of calculus. There are infinite number of vertices and your > Four-Colour-Theorem breaks down. The 4-Color-Mapping of Appel & Haken becomes > silent because although it has land area it is lost to the infinite spokes. > Only the 2 Color Mapping answers the problem by saying that it is not closed > loops. So this is an example of where Appel and Haken had missed a case in > their alleged proof. In fact they missed all of those wild functions of > Calculus. > Once again, you're using a very poorly-defined way of creating regions > on a cartesian plane and trying to apply those poorly-defined regions to a > clearly-defined theorem. I'm honestly not certain what you're trying to > get at, here. Build a graph (either in the Ôplot the function' sense or > what I actually mean, which is Ônodes and edges') of it, if it helps > demonstrate it. What I am getting at is that Appel & Haken ignore borderline points and vertex points and say they are colorless. Okay, if Appel and Haken say those points are colorless then the above step-function of Calculus qualifies as a country in Appel & Haken mapping because, hey, borderlines are ignored anyway. But in 2 Color Mapping using Jordan Curve theorem only closed loops count as countries. > This completely separates the problem from physical representation; > neither nodes nor edges inherently have area at all, and colours are > merely arbitrary pieces of identifying information associated with a node. > It also clearly defines what part of the graph we're concerned with: the > nodes. > Most every theorem of mathematics is a generalization or idealization over > the physical universe. But the 4 Color Mapping of Appel & Haken is an oddball > from all other math theorems because it simply is a poorly formulated > problem. > We have managed to drag into mathematics sense perception in telling > apart one country from another from colors of their interior. This is not > mathematics but an appeal to our sensory perception. > In other words, it's an application problem. This doesn't make it any > less mathematical: the theorem discovered applies to any graph with the > requisite properties and is just as much pure math as any other theorem > you care to name. However, like some theorems and unlike others, the > four-colour theorem came out of a specific application, which you seem to > have some problem with in this specific case. Wrong. Appel & Haken are gap ridden. The 4 Color Mapping is poorly defined because it ignores the color of points that are borderlines and vertices. Once those gaps are filled and corrected the entire problem then becomes a 2 Color Mapping easily proven by the Jordan Curve theorem. The 4 Color Mapping started around 1850s with Guthrie and DeMorgan and then Cayley were never given a clear well-defined statement. Once you clearly and well-define the statement of this problem--- how many colors suffice to color every point of a map to convey information. And the answer is 2 and only 2 colors suffice. The additional two colors that Cayley and Appel and Haken looked for was mere art work and not mathematics. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: the imprecision of 4 color mapping and why it should be 2 color mapping > Be careful what you disagree with. The example graph cannot be coloured, > satisfying the definition right below it, using two colours; if you don't > believe me, strip off the ...s indicating that the graph could continue > and try colour it with only two colours such that no nodes sharing an edge > also share a colour. You only need to consider nodes A, B, and D to > demonstrate this. > Owen, if you can illustrate your problem with the colors of black and > white on my computer screen and the computer screens of millions of > viewers. And provide all the information ever needed in just those 2 > colors. The fact that you can even communicate this Nodal problem in 2 > colors of black and white and tell me all about the information on that > page on my computer screen should tell you that 2 colors suffice. I > mean, I can see that A, B, D are vertices even though they are all one > color. So do you not think it is pretentious of you to think that you > need to have more colors than black and white to paint B or D or C or A? > Do you not think it is pretentious of you when you can convey all the > information of 4 Color Theorem in black and white that you keep saying > you need a third new color when you obviously do not need a 3rd new > color. You misunderstand the theorem. It does not say you must colour the nodes of the graph to relay information. It says if you are colouring the graph with these rules, here is the minimum number of colours need. > So your Color Mapping Theorem concerning Nodal points is not 4 color but > 2 color in the fact that you can convey the entire information with the > colors of black and white. I can see all vertex points in black as easy > and perhaps easier than if you painted them in 3 more colors. The four-colour theorem is not about information theory. > The mistake of this 4 Color Theorem is that you place yourself on this > restriction when you never had to be placed on any restriction no nodes > share an edge also share a color. The graph does not demand it because > it already conveys all the information in black and white and you are > heedlessly placing that demand by wanting 2 more colors on topp of black > and white. We know it is a vertex or node at A or B or D from the black > lines that intersect. For what reason would you ever need two more > additional colors is nonsense. Just for fun, let's break away from colours completely for a moment. Let's say you have an undirected graph in two dimensions, where no edge intersects any other edge. The nodes of this graph are denoted with capital letters. Say you also have a pool of arbitrary values you can arrange on the nodes. We'll denote those values with lowercase letters. Minimize the number of values used, such that no node shares a value with any node connected to it by an edge. The *reasons why* someone might do this are outside the domain of the problem and are, in fact, quite possibly sthetic or artistic. However, the problem and the theorem solving it are strictly math. The why is irrelevant to the theorem. > the artists with 2 more colors that are not needed and jazz up or make > fancy the nodes or mapp. This is an application of the theorem, not the theorem itself. Please restrict yourself to the matter at hand. > For some functions, this is a very poor definition. Consider the > function f(x) = [[x]] /* ßoor (x) */: it doesn't enclose a region, > because the line is discontinuous but not asymptotic. > ^ > | <-- not a boundary between regions > | > | > | > *-------------------> x That's actually a graph of ceil(x), not ßoor(x). My error. > However, we can create boundaries by Ôconnecting' the vertical breaks > between segments of the function, at which point this is about as > complex as the function f(x) = x. > Owen, my point was that Appel & Haken offered a computer program as a > alleged proof of 4 Color Mapping that tackles a large number of > **cases**. And mine was that you haven't shown a case that can't be represented, whereas the theorem shows that there aren't cases that can't be represented. It should be easy to prove that there is one: simply provide an example. > I doubt that they tackled these wild functions in Calculus and because > Appel and Haken ignore borderlines the above step function which you > admireably displayed must be accounted for by Appel and Haken since they > have interior areas. Another approach to the step function above: take the Ôbackground' of the chart as one region, and the steps themselves as regions (of zero area, being lines, but regions nonetheless). This particular graph can, indeed, be coloured using only two colours. Once again, this is not the general case, merely an example of a special case. Consider the functions f(x) = x + 1 and g(x) = -x + 1. Colour them using only two colours such that the graphs are accurately presented within the graphed region and such that f(x) and g(x) can be told apart from each other as well as the background. > The 4 Color Mapping started around 1850s with Guthrie and DeMorgan and > then Cayley were never given a clear well-defined statement. Once you > clearly and well-define the statement of this problem--- how many colors > suffice to color every point of a map to convey information. And the > answer is 2 and only 2 colors suffice. The additional two colors that > Cayley and Appel and Haken looked for was mere art work and not > mathematics. You're still confusing cause and effect, here. The four-colour theorem comes from an external requirement to colour a map. It doesn't, in any way, state or imply that one *needs* to colour the map this way; it merely provides a mathematically-precise description of the minimal number of colours required by the external definition and an algorithmic process for placing those colours. The external requirement is, indeed, art. The theorem itself is an application of pure math to that requirement. -- Some say the Wired doesn't have political borders like the real world, but there are far too many nonsense-spouting anarchists or idiots who think that pranks are a revolution. === Subject: Re: the imprecision of 4 color mapping and why it should be 2 color mapping (big snips) > The four-colour theorem is not about information theory. (snip) > You're still confusing cause and effect, here. The four-colour theorem > comes from an external requirement to colour a map. It doesn't, in any > way, state or imply that one *needs* to colour the map this way; it merely > provides a mathematically-precise description of the minimal number of > colours required by the external definition and an algorithmic process for > placing those colours. > The external requirement is, indeed, art. The theorem itself is an > application of pure math to that requirement. Owen I disagree that the 4 Color Theorem is not about Information. I disagree the the 4 Color Theorem is pure math for it is not. It is a rule imposed on a situation that required 1 Color originally. So it is not mathematics at all. I need to get back to basics with you Owen. In a recent discussion with Tim Mellor I was able to get back to basics with him when I asked him about the Jordan Curve theorem and how the Jordan Curve theorem can prove that 4 countries or 4 states of Utah, Colorado, New Mexico, Arizona need only 2 colors of black and white. And Tim conceded that the Jordan Curve Theorem proves that those 4 states suffice with 2 colors of mapping where borderlines are blue ink and interiors are white. So, Owen, I need some moment of reckoning with you because in your posts I recognize where you have been brainwashed over this problem. I do not expect you to see your brainwashing yourself. In an earlier post of yours you said you did not know about this Information Theorem but now you voice opinion that you are strongly against it and seem to think external requirements are art. I am going to ask you, Owen, just one question below. INFORMATION Theorem: 2 colors suffice to convey any and all messages, be they a message of language with words in blue ink on a white paper or graphs and sketches in blue ink on white paper or be they geometrical figures in blue ink on white paper or be they works of art in blue ink on white paper etc etc. In other words, every message, every idea, every map, every art can be rendered in 2 colors of blue ink on white paper. This includes every book ever written because words are blue on white paper. So, here is my question Owen, because we really cannot go beyond this post in endless argument but rather instead it all boils down to this. This is the focal point of the issue as to whether the 4 Color Mapping or the 4 Color Theorem are fakes or whether they have any value at all. Question Owen: Can you have a Information Theorem in the world that says 2 colors are sufficient to convey any and all ideas or messages or data or information, and yet, simultaneously have in this world another theorem in mathematics of the 4 Color Mapping. Tim Mellor was honest enough to answer that the 4 states are 2 colorable as proven by Jordan Curve Theorem. So I need an honest answer from Owen Jacobson. Can this world accomodate 2 theorems, one saying that all messages and ideas suffice with 2 colors and the other saying that 4 colors suffice to make all maps. So, Owen, please answer that question. Can this world of ours accomodate 2 theorems where one says 2 colors suffice but another says that 4 colors suffice. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: the imprecision of 4 color mapping and why it should be 2 color mapping > So, Owen, please answer that question. Can this world of ours accomodate > 2 theorems where one says 2 colors suffice but another says that 4 > colors suffice. I've never said it couldn't. *You* claimed that this somehow made the four-colour theorem and its application inaccurate. That's very much a judgement call, rather than an absolute truth. The two theorems are completely orthogonal ways of looking at the same problem. That's all. Neither is more or less right or wrong than the other, except that they're both mathematically-sound. -- Some say the Wired doesn't have political borders like the real world, but there are far too many nonsense-spouting anarchists or idiots who think that pranks are a revolution. === Subject: Re: the imprecision of 4 color mapping and why it should be 2 color mapping > So, Owen, please answer that question. Can this world of ours accomodate > 2 theorems where one says 2 colors suffice but another says that 4 > colors suffice. > I've never said it couldn't. *You* claimed that this somehow made > the four-colour theorem and its application inaccurate. That's > very much a judgement call, rather than an absolute truth. > The two theorems are completely orthogonal ways of looking at the > same problem. That's all. Neither is more or less right or wrong > than the other, except that they're both mathematically-sound. Well I have several issues to cover tonight. First off I want to know if Owen posted years ago under this name address, but there are probably many Jacobsons around. = = T. Eliot, Top Bard. jacobson@oucsace.cs.ohiou.edu Secondly, I now can pinpoint the ßaws and gaps of the 4 Colour Theorem versus the 4 Color Mapping. Although Owen and others deem the two as equivalent statements they are not. Both are ßawed and fakes, and surprizingly they are not even equivalent. The root of ßaw of 4 Color Mapping is that borderlines and vertices are ignored even though they have a color which is black, and black before the persons with 4 different colored crayons start filling in the interiors. So the 4 Color Mapping in toto has 5 colors in all for every point. So it is bogus to even call it 4 Color Mapping and Appel & Haken should admit that theirs is 5 Color Mapping. Now looking for the root gap and ßaw of 4 Color Theorem, which is not a theorem because it is created from a rule. The root ßaw here is that in the end there are 6 colors, and whereas the 4 Color Mapping ends up with 5 colors, the 4 Color Theorem that Owen has elaborated on in the past week ends up with 6 colors. In the 4-Color Theorem there are 4 different colors for vertices or nodes but then there is interiors neglected and ignored and uncolored so let us call them white and there is the question of the lines and curves that create the vertices and nodes and they are black. So you end up with 6 colors in all in the 4 Color Theorem. Both are ßawed and fakes. One ends up with 5 colors for every mapping and the other ends up with 6 colors, so they are not even equivalent fakes. And the moral theme should be that if a statement of mathematics is a ßawed and fake piece of mathematics that there are no equivalent statements. Thirdly, I like to address the above reply by Owen as per Theorem of Information. It is more than a judgement call. If we find true statements in mathematics and provide proofs of those statements we end up with a theorem. And when we end up with a theorem it begins to interconnect and link with other statements and theorems of science and mathematics. A fake in mathematics remains in the corner isolated and never able to link or connect to other pieces of math or science. This linkage is a upward linkage and a downward linkage in that a true theorem of mathematics has higher level generalizations and a true theorem of mathematics has statements of lower rank which are usually called lemmas. The history of 4 Color Mapping starting in the 1850s with Guthrie and DeMorgan on to Cayley onto many others and then alleged proof by Appel & Haken, if true would have connected and linked to other pieces of science and math. It has not. It has stayed in a corner and isolated and unable to connect. That warns us it is a fake. I have discovered the Theorem of Information which says 2 colors suffice to convey any idea, message, thought, knowledge, wisdom, map, figure, sketch, artwork, picture, diagram, geometry, language, etc etc. This theorem is obviously a huge generalization over 4 Color Mapping. Both cannot be simultaneously true. They contradict each other. So the 4 Color Mapping is a fake. It is a 2 Color Mapping easily proven via the Jordan Curve Theorem. It is just that mathematicians for the past 150 years were too stupid to question the basic elements of the proposition as to whether they are well defined. It is ill-defined when points of borderlines and vertices are ignored as to color and to assign color to only interior points. As Gauss so often commented when he was alive that when mathematics fails the root cause of that failure is usually a ill-defined element of the proposition or a hidden assumption. In the case of 4 Color Mapping it is the hideous ignoring of color of points that are borderlines and vertices. Owen says the two theorems are orthogonal. That is wrong. They are contradictory. That means one of them is a fakery. The Theorem of Information that 2 colors suffice to convey any and all messages or ideas or data is obviously true. A formal proof of it can be constructed from physics from the things (2 colors). So the Theorem of Information is true, that means the 4 Color Mapping is ßawed and a fakery. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: the imprecision of 4 color mapping and why it should be 2 color mapping so, Guthrie actually proved a Six Color Theorem, only he really only needed to use two colors, since there was this huge, weblike Black Country, interposed by the printer. it's true that old jurisdictional maps had to be hand-colored, if they wanted to visually tag the jurisdictions, like coloring books; eh?... a coloring book with no borders to fill in -- that really is hideous! > crayons start filling in the interiors. So the 4 Color Mapping in toto has 5 > colors in all for every point. So it is bogus to even call it 4 Color Mapping > and Appel & Haken should admit that theirs is 5 Color Mapping. Now looking > for the root gap and ßaw of 4 Color Theorem, which is not a theorem because > it is created from a rule. The root ßaw here is that in the end there are > 6 colors, and whereas the 4 Color Mapping ends up with 5 colors, the 4 Color > Theorem that Owen has elaborated on in the past week ends up with 6 colors. > In the 4-Color Theorem there are 4 different colors for vertices or nodes but > then there is interiors neglected and ignored and uncolored so let us call > them white and there is the question of the lines and curves that create the > vertices and nodes and they are black. So you end up with 6 colors in all in > the 4 Color Theorem. Both are ßawed and fakes. One ends up with 5 colors for > every mapping and the other ends up with 6 colors, so they are not even > equivalent fakes. And the moral theme should be that if a statement of > mathematics is a ßawed and fake piece of mathematics that there are no > equivalent statements. --ils duces d'Enron! http://larouchepub.com --Chair Man George! http://tarpley.net === Subject: Re: the imprecision of 4 color mapping and why it should be 2 color mapping (some snips) Before I reply to your new post Owen I want to correct two errors of my last posts. Where I said that 4 colors are redundant and that 2 colors suffice for the 4 Colour Theorem on Vertices. What I meant to say is that 1 color such as black suffices to color all vertices and the other color of white is the background of what is not a line or vertex. So I see no reason to have 3 other colors to color the vertices in this 4 Colour Theorem and judging from reading of your last post you seem to have assumed that when I said 2 colors suffice that the A, B,C, D nodes were colored alternately in black and white. No, I meant by 2 colors sufficing that all nodes are black and the surrounding background that is not a line or vertex is white. And the second error I quote a paragraph of my last post: The 4 Color Mapping started around 1850s with Guthrie and DeMorgan and then Cayley were never given a clear well-defined statement. Once you clearly and well-define the statement of this problem--- how many colors suffice to color every point of a map to convey information. And the answer is 2 and only 2 colors suffice. The additional two colors that Cayley and Appel and Haken looked for was mere art work and not mathematics. I do not know if a statement of that contains a thought of 2 and only 2 colors suffice is logically sound. I say this because in logic a if and only if is a necessary and sufficient condition already and then to say necessary and sufficient suffice. I do not know if that is a ßaw of logic. What I am getting at is that there exists a theorem of Information that belongs in mathematics that says something like this: Theorem of Information: 2 colors suffice to convey any piece of information then there exists a lemma of this Theorem of Information which is the Color Mapping lemma which goes something like this: 2 Color Mapping lemma: 2 colors suffice to color any and all maps Proof: Jordan Curve theorem > You misunderstand the theorem. It does not say you must colour the nodes > of the graph to relay information. It says if you are colouring the > graph with these rules, here is the minimum number of colours need. Yes, apparently I did after re-reading your first: > Actually, that's not the case. There's a second way of looking at the > problem known as the four-colour theorem that's systematically identical > to the usual one. > Suppose you have a two-dimensional undirected graph, arranged such that no > two edges cross each other. This means you can have (fixed-width graphics > ahead): > ...A-----B... ...A-----B... > | /| | /| > | / | | / | > | / | but not | X | > | / | | / | > |/ | |/ | > ...D-----C... ...D-----C... > Given such a graph with an arbitrarily large number of nodes, how many > *unique* colours do you need to colour _the nodes_ such that no two > nodes that share an edge have the same colour? Four. There are even > algorithmic ways to colour the graph. But I would then rejoinder or come to say that you have created a rule that forces 4 colors when only one color suffices. Every person can see that point A, B,C, D are vertices or nodes and they can see it because of line intersection. So you need only one color to **see** that it is a vertice. So now you create a rule that you are unhappy with the fact and realization that one color suffices to **distinguish** each node and being unhappy that one color suffices you needlessly invent a rule that forces there to be 4 colors for the vertices of A, B,C, D when in reality one color sufficed. So this is not a theorem of mathematics that you call 4 Colour Theorem because you invented an artificial rule that forces there to be 4 colors. What is a vertice or node in the first place? It is a point of intersection of lines or curves is it not. Those lines and curves, points thereof, need a color and black is a good color to give them. So we immediately can see points of intersection. And those intersection points are also black. So why create a Rule that forces there to be 4 colors when 1 color sufficed? So the 4 Colour Theorem is no theorem at all because there is no proof but only a dreamt-up imaginary rule. > The four-colour theorem is not about information theory. Yes, I see that now. It is just a rule that forces there to be 4 colors when in reality 1 color sufficed. I leave the remainder of your post to answer later as this is already too long. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: JSH: Key insight, algebra What I did was utilize what some might see as a neat trick as first look at f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) then multiply through by f^2, to get (m^3 f^4 - 3m^2 f^2 + 3m)f^2 x^3 - 3(-1+mf^2 )x u^2 f^2 + u^3 f^3 and you have this neat situation where there's a factor of f^2, but when you multiply through with it, you can get to something quite different, as using y=uf on the last two coefficients to get (m^3 f^4 - 3m^2 f^2 + 3m)f^2 x^3 - 3(-1+mf^2 )x y + y^3 and if you let A = (m^3 f^4 - 3m^2 f^2 + 3m)f^2 and B = -1+mf^2 ), you have Ax^3 -3Bxy + y^3 and you don't have a multiple. The mathematics handles ALL CASES as notice all I've done really is deliberately set y in one example to uf, so that I can see how that impacts a factorization. That doesn't change the fact that y is an independent variable. I simply set it to get a multiple--f^2--so that I can figure some things out. The math can be kind of subtle here, but it's not really that hard. What can be hard is a will to be wrong on the part of people who fight against the truth against all evidence. Some of you feel a NEED to be wrong, as your biology compels you to fight for what you were trained to believe even if mathematics says it's false. Your GUT INSTINCTS lead you astray, and you forget that you are a biological machine. You're like religious people. Before I came along, multiples of polynomials were these trivial things that you just divided off, but I guess maybe another lesson of mathematics is that very little is actually trivial. If you use a multiple as I have, you can see how factors distribute with roots even if they're irrational, which is a powerful analysis tool. Now then, I point out that m and f are independent of each other so that I can set m=0, and see what happens with a key factorization when that f^2 multiple is divided off. For some reason, many of you seem susceptible to the belief that SUDDENLY a multiple of a polynomial must be a variable that varies in how it divides off. That's not just bizarre; it's not sane. James Harris === Subject: Re: Key insight, algebra > That's not just bizarre; it's not sane. > James Harris Mr. Harris, I thought the same exact thing! What would make someone post lies and incorrect mathematics and make themselves out to be blithering idiot for a decade now? Then, claim to be in the league of Gauss, Newton, Euler, Fermat, Archimedes and many of the like. That's not just bizarre; it's not sane! === Subject: Re: Key insight, algebra > What would make someone post lies and incorrect mathematics and make > themselves out to be blithering idiot for a decade now? > Then, claim to be in the league of Gauss, Newton, Euler, Fermat, Archimedes > and many of the like. I'd say he's in the league of Archimedes. Archimedes Plutonium, that is. -- --Tim Smith === Subject: Re: JSH: Key insight, algebra Full frontal ignorance -- again! -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: Some math, algebraic integers > But let me explain > again with those simple quadratics, this time, with the cat out of the > bag: > x^2 + (2s_1 + 3s_2)x + s_1 s_2 6 = (x + 2s_1))(x + 3s_2) > where s_1 s_2 = 1, and 2s_1 + 3s_2 is an integer. > Supposedly Evariste's ideas mean that iff s_1 and s_2 are irrational, > then somehow, someway factors of 2 and 3 shift, such that the roots of > x^2 + (2s_1 + 3s_2)x + s_1 s_2 6 > can meaningfully not be coprime to either 2 or 3, but as I've shown > that's bunkum. > Did you mean algebraic integer where you say irrational? No. The invalid distinction is between rational and irrational where supposedly irrational roots are distinguished from rational ones in a special way, which is the basis for Galois Theory. What I like about the example above is that it shows in bas relief just how odd the supposed distinction is, where supposedly the unit case is barred from x^2 + (2s_1 + 3s_2)x + s_1 s_2 6 = (x + 2s_1))(x + 3s_2) where s_1 s_2 = 1, and 2s_1 + 3s_2 is an integer. When all that's actually true is that s_1 and s_2 cannot be algebraic integers. Some people make a false leap. That's all. Such mistakes are as old as humanity. > If s_1 s_2 = 1 and 2s_1 + 3s_2 is an integer, say n, > then s_2 satisfies > p(x) = 3 x^2 - n x + 2 = 0 > If n is chosen such that p(x) is reducible as an element of Z[x], > then each root of p(x) is either an ordinary integer > or a rational number. If n is chosen such that p(x) is > irreducible, then, because p(x) is primitive and its leading > coefficient is not 1 or -1, none of its roots can be algebraic > integers. So s_2 cannot be an algebraic integer. > -- Yeah and the same applies to s_2. My point is to show how odd it is that x^2 + (2s_1 + 3s_2)x + s_1 s_2 6 = (x + 2s_1))(x + 3s_2) supposedly has such a natural case as the unit case blocked, just because s_1 and s_2 are irrational. Why would irrational numbers be so limited versus rationals? The truth is that they are not. Some people simply assumed something that is not true, and I can prove that they are wrong using algebra in what I call advanced polynomial factorization. People being people, rather than accept the truth, and follow the algebra, there are quite a few people who decided to attack me as a crank, and try to just dismiss my results using social tactics, like a coordinated email campaign against my paper. See http://rattler.cameron.edu/swjpam/vol2-03.html They also attack my other research, like my prime counting function. Human beings are this way. It's not news to most of you I know, but that doesn't change things...for most of you. I'm looking for the very few it makes a difference to. A very few of you will look to mathematics itself and choose it over society. James Harris === Subject: Re: Some math, algebraic integers > People being people, rather than accept the truth, and follow the > algebra, there are quite a few people who decided to attack me as a > crank, and try to just dismiss my results using social tactics, like a > coordinated email campaign against my paper. and > They also attack my other research, like my prime counting function. Hmmmmm .... and how do YOU spell P-A-R-A-N-O-I-D? === Subject: Re: Some math, algebraic integers > People being people, rather than accept the truth, and follow the > algebra, there are quite a few people who decided to attack me as a > crank, and try to just dismiss my results using social tactics, like a > coordinated email campaign against my paper. > and > They also attack my other research, like my prime counting function. > Hmmmmm .... and how do YOU spell P-A-R-A-N-O-I-D? http://rattler.cameron.edu/swjpam/vol2-03.html The facts are clear, but I know that it's an easy trick to attack facts with snide comments. email campaign mounted against it coordinated in posts on the sci.math newsgroup. That's not paranoia unless you think talking about having a criminal come in rob you at gunpoint and run away with your money is paranoia. You people are NOT mathematicians as mathematicians accept the truth, while you depend on social function, like trying to escape facts with a single word, like paranoia. But that's how society is--easy--because truth is about what keeps the society going, no matter what happens to individuals. Society is about the importance of the group over the individual, and will even kill individuals for the society to survive. Mathematics is difficult, as it's about the truth. Mathematics is not about what perceptions or beliefs keep some society going, but about what is actually true. That's why it's so hard, and why so few people can handle being an actual mathematician, though social function allows many to CLAIM to be mathematicians. James Harris === Subject: Re: Some math, algebraic integers > http://rattler.cameron.edu/swjpam/vol2-03.html > The facts are clear, but I know that it's an easy trick to attack > facts with snide comments. > email campaign mounted against it coordinated in posts on the sci.math > newsgroup. The paper was withdrawn, yes, and the timing is probably correct, too. It was withdrawn because of incorrect results. Whether the withdrawal was proper behavior by the editor is a matter for debate, not whether the mathematics in the paper was correct (it wasn't). > You people are NOT mathematicians as mathematicians accept the truth, > while you depend on social function, like trying to escape facts with > a single word, like paranoia. It seems that it is JSH who refuses to accept the truth, namely how error ridden the paper in question is. No amount of ranting will change incorrect results into correct ones. It isn't social function that has determined these results to be incorrect, it is simple arithmetic (not even algebra). > But that's how society is--easy--because truth is about what keeps > the society going, no matter what happens to individuals. Society is > about the importance of the group over the individual, and will even > kill individuals for the society to survive. > Mathematics is difficult, as it's about the truth. > Mathematics is not about what perceptions or beliefs keep some society > going, but about what is actually true. > That's why it's so hard, and why so few people can handle being an > actual mathematician, though social function allows many to CLAIM to > be mathematicians. What is it that allows JSH to make claims about mathematics when his paper shows very clearly that he understands very little? === Subject: Re: Some math, algebraic integers > > http://rattler.cameron.edu/swjpam/vol2-03.html > > The facts are clear, but I know that it's an easy trick to attack > facts with snide comments. > > email campaign mounted against it coordinated in posts on the sci.math > newsgroup. > The paper was withdrawn, yes, and the timing is probably correct, too. > It was withdrawn because of incorrect results. Whether the > withdrawal was proper behavior by the editor is a matter for debate, > not whether the mathematics in the paper was correct (it wasn't). Ioannis Argyros, the chief editor of Southwest Journal of Pure and Applied Mathematics, emailed me *after* he had my paper withdrawn. Problem is, he included text that a sci.math poster had posted the day before. So the evidence shows that it was the email campaign that inßuenced him, and that it did so within a VERY short time, when in fact, the journal had my paper for NINE MONTHS. If you look in the record, you'll notice I even posted that it was at the Southwest Journal of Pure and Applied Mathematics, months before. > > You people are NOT mathematicians as mathematicians accept the truth, > while you depend on social function, like trying to escape facts with > a single word, like paranoia. > It seems that it is JSH who refuses to accept the truth, namely how > error ridden the paper in question is. No amount of ranting will > change incorrect results into correct ones. It isn't social > function that has determined these results to be incorrect, it is > simple arithmetic (not even algebra). Name a single error. That's a basic challenge. You're coming here bold as anything on Usenet claiming my paper has an error in it, so I challenge you to name a single error. The paper is at http://www.ne-plus-ultra.net/index.php?option=content&task= view&id=46&Itemid =26 so you can go look at it, and trace through it, and if there is an error, you can give it. You opened the door by challenging me with your claim. And I'm calling you on what I know is a bluff. You're lying to these newsgroups for your own reasons. If I'm wrong, then give a freaking error you simpleton Usenet punk!!! James Harris === Subject: Re: Some math, algebraic integers > If I'm wrong, then give a freaking error you simpleton Usenet punk!!! Attaboy, James! That's the kind of mathematical response we all expect from you! -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: Some math, algebraic integers >http://rattler.cameron.edu/swjpam/vol2-03.html >The facts are clear, but I know that it's an easy trick to attack >facts with snide comments. >email campaign mounted against it coordinated in posts on the sci.math >newsgroup. >The paper was withdrawn, yes, and the timing is probably correct, too. > It was withdrawn because of incorrect results. Whether the >withdrawal was proper behavior by the editor is a matter for debate, >not whether the mathematics in the paper was correct (it wasn't). > Ioannis Argyros, the chief editor of Southwest Journal of Pure and > Applied Mathematics, emailed me *after* he had my paper withdrawn. > Problem is, he included text that a sci.math poster had posted the day > before. > So the evidence shows that it was the email campaign that inßuenced > him, and that it did so within a VERY short time, when in fact, the > journal had my paper for NINE MONTHS. > If you look in the record, you'll notice I even posted that it was at > the Southwest Journal of Pure and Applied Mathematics, months before. >You people are NOT mathematicians as mathematicians accept the truth, >while you depend on social function, like trying to escape facts with >a single word, like paranoia. >It seems that it is JSH who refuses to accept the truth, namely how >error ridden the paper in question is. No amount of ranting will >change incorrect results into correct ones. It isn't social >function that has determined these results to be incorrect, it is >simple arithmetic (not even algebra). > Name a single error. I note that the gowan used the phrase simple arithmetic as what has shown your result to be incorrect. I have interpreted this to indicate that he has seen my arithmetic refutation of the conclusion of your so-called primary argument from your paper. As I've said before, I do not intend to let you get away with lying about the fact that my argument has shown you wrong. The error, which I have already noted several times, is this statement Therefore, with the factorization 65x^3 .89 12x+ 1 = (a_1x + 1)(a_2x + 1)(a_3x+ 1) one of the a.89s is coprime to 5, which shows where some of the algebraic integer factors distribute despite the factors being irrational. What I have proven is that, in fact, *each* of the a's has a factor in the algebraic integers, in common with 5. This is at least as strong as the condition that each coefficient a *not* be coprime to 5. I have shown you these factors. For any of your a's in that section, the number (as shown in step 2 below): r(-a) = 8 a^2 + 4a - 45 is an algebraic integer that divides a, and also divides 5. You don't have to believe me. All you have to do is follow some elementary arithmetic. Why won't you do elementary arithmetic, if you think you're correct? If you disagree that I have found such factors, then point out the error. Your earlier claim that I make some weird assumption is irrelevant, since the arithmetic is manifestly correct, and there is simply *no argument available* to the effect that, if there were a ring R containing the ring of algebraic integers, satisfying the property (mumble), then the arithmetic would fail. In other words, arithmetic (i.e., addition and multiplication) in the rational integers is fixed, and unchangeable by the existence or absence of anything else. Whether you have an object ring or whether it is just a hallucination, the argument still holds. Here are the simple steps to follow: In this outline, a refers to one of the coefficients that you claim must be coprime to 5. P(x) refers to the polynomial x^3 - 12 x^2 + 65, and I note that P(-a) = 0, for each of your values a. 1. The following formulas are true: q(x)r(x) = (64 x + 128)P(x) + 5 r(x)s(x) = (32 x + 72)P(x) + x, where q,r,s are defined as follows: q(x) = 8 x^2 - 76 x - 185 r(x) = 8 x^2 - 4 x - 45 s(x) = 4 x^2 - 37 x - 104 2. Since -a is a root of P(x), we have the following factorizations: q(-a)r(-a) = 5 r(-a)s(-a) = -a This shows that this number: r(-a) = 8 (-a)^2 - 4 (-a) - 45 = 8 a^2 + 4a - 45 is a divisor of -a (i.e., the coefficient a that you claim to be coprime to 5), as well as a divisor of 5. 3. The minimal polynomial of r(-a) is given as: MP_r = x^3 - 969 x^2 + 315 x + 5 which is irreducible over Q. This shows that r(-a) is an algebraic integer, but not a unit in that ring. shared by a and 5 is not a unit in the ring of algebraic integers. 5. a and 5 are not coprime in the ring of algebraic integers, since they share a non-unit factor in common. In any commutative ring, if two elements share a non- unit factor, they cannot be coprime. I suppose the result holds for arbitrary rings, provided the two elements share the factor on the same side; if they share a left factor, their minimal common left ideal (this can't be the right way to say this) cannot be the whole ring, and similarly for factors on the right. It is well known that the algebraic integers form a commutative ring. Therefore, a and 5 are not coprime in the ring of algebraic integers. > That's a basic challenge. I've given the basic result. You are evading the truth, and no amount of bluster or bringing in of outside help, like your new friend at NASA, or the guy at Houston, or to see, will change the fact that you are wrong and are refusing to acknowledge your error. > You're coming here bold as anything on Usenet claiming my paper has an > error in it, so I challenge you to name a single error. the error. I don't claim it's the first error. It's definitely an error. > The paper is at > http://www.ne-plus-ultra.net/index.php?option=content&task= view&id=46&Itemid= 26 > so you can go look at it, and trace through it, and if there is an > error, you can give it. > You opened the door by challenging me with your claim. > And I'm calling you on what I know is a bluff. Haven't you been listening? I've found your error. No bluff. I've even given you all the ammunition you need to shoot it down, if indeed I turn out to have been incorrect. Shoot it down, and win a gold star, or at least an acknowledgement from me that you have done so. note that claims you're wrong! > You're lying to these newsgroups for your own reasons. Am I lying? C'mon, don't be a shy guy! > If I'm wrong, then give a freaking error you simpleton Usenet punk!!! The error is plain, and if you could even do arithmetic, or understand the basic nature of my proof, you wouldn't be calling anyone else a simpleton, let alone a punk. > James Harris Dale === Subject: Re: Some math, algebraic integers > > But let me explain > > again with those simple quadratics, this time, with the cat out of the > > bag: > > > > x^2 + (2s_1 + 3s_2)x + s_1 s_2 6 = (x + 2s_1))(x + 3s_2) > > > > where s_1 s_2 = 1, and 2s_1 + 3s_2 is an integer. > > > > Supposedly Evariste's ideas mean that iff s_1 and s_2 are irrational, > > then somehow, someway factors of 2 and 3 shift, such that the roots of > > > > x^2 + (2s_1 + 3s_2)x + s_1 s_2 6 > > > > can meaningfully not be coprime to either 2 or 3, but as I've shown > > that's bunkum. > > Did you mean algebraic integer where you say irrational? > No. > The invalid distinction is between rational and irrational where > supposedly irrational roots are distinguished from rational ones in a > special way, which is the basis for Galois Theory. > What I like about the example above is that it shows in bas relief > just how odd the supposed distinction is, where supposedly the unit > case is barred from > x^2 + (2s_1 + 3s_2)x + s_1 s_2 6 = (x + 2s_1))(x + 3s_2) > where s_1 s_2 = 1, and 2s_1 + 3s_2 is an integer. ...when s_1 and s_2 are irrational. Forgot to add that important bit though I've said it before, and it follows from what came before. > When all that's actually true is that s_1 and s_2 cannot be algebraic > integers. When s_1 and s_2 are irrational, as, of course, there are the cases where both equal 1 or -1. > Some people make a false leap. That's all. > Such mistakes are as old as humanity. The mistake in this case was to assume that presentation limits on elementary methods for showing irrational roots were actual limits on the factors of the roots themselves. But, just like I can have r_1=1, r_2=2, and r_2 = 3, with rational roots, I can equivalently have r_1 = s_1, r_2 = 2s_2, and r_2 = 3s_3, without regard to the rationality of s_1, s_2 and s_3. So how did some people make such a mistake? The full details are for historians I guess but look at the tried and true quadratic formula: x = (-b +/- sqrt(b^2 - 4ac))/2a where ax^2 + bx + c = 0, and if you have a reducible quadratic, then you might have something like x = (-3 +/- sqrt(1))/2 where to some it will seem silly that I left in sqrt(1), but the ENTIRE BASIS for Galois Theory as it's taught is that when you can't just simplify the square root, like you can with sqrt(1), then your limitation is a limitation on the underlying roots!!! So, that was the solution for x^2 + 3x + 2, let's do x^2 + 3x + 1: x = (-3 +/- sqrt(5))/2 and people write that as if they actually have numbers. But (-3 + sqrt(5))/2 is not a single number. It's two. Like sqrt(4) = +/- 2, and it doesn't matter if you write +/- in front of the square root or not, as *mathematically* it's always there. The sqrt() is an operator. It's NOT a number. For a large class of irrationals we cannot express the numbers directly but instead use operators, like +, - and sqrt(), to point at the numbers. It's like a Zen thing, finger pointing at the moon...you know? So x = (-3 +/- sqrt(5))/2 is just the best we can do in representing the roots of x^2 + 3x + 1 but our limitation in representation is not a mathematical limit on factors of the roots. Here the style is not the substance, and math is NOT a fashion show. Now the math supports me in a lot of ways, as hey, that's HOW MATH WORKS! But people are people, they can believe just about anything. So I have the mathematics, but you people have your need to be wrong. > The truth is that they are not. Some people simply assumed something > that is not true, and I can prove that they are wrong using algebra in > what I call advanced polynomial factorization. > People being people, rather than accept the truth, and follow the > algebra, there are quite a few people who decided to attack me as a > crank, and try to just dismiss my results using social tactics, like a > coordinated email campaign against my paper. > See http://rattler.cameron.edu/swjpam/vol2-03.html That's just freaky. How many of you even know of a story like that? But a remarkable story just becomes another thing all of a sudden, eh? No matter how you slice it, it's a remarkable story, as the Southwest Journal of Pure and Applied Mathematics is supposedly a peer reviewed math journal. They had my paper for nine months, put it out, and then caved within a few days after receiving *emails* in a coordinated assault by sci.math posters, who actually conspired IN POSTS on the newsgroup! Whether you believe my work is correct or not, it's a remarkable story. Yet, you can also sit back and not get excited if it's NOT remarked upon much because it's that human ability to believe contradictory things at the same time! I've talked a bit about real mathematicians and I think I want to explain to you all something that will probably ßy past most of you: Most people have an eery ability to believe contradictory things at the same time. It's a social thing which has to do with people getting along well enough to make more babies and build societies. Mathematicians are a special breed of people who find it difficult if not impossible to accept contradictory things. Their social function is turned off a bit, so that they can find and appreciate mathematical truth. There are VERY FEW true mathematicians in the world, necessarily, as probably society would collapse if most people lost the ability to believe in different and contradictory things. People need to believe, even if it's against all evidence, so few can be real mathematicians. James Harris http://mathforprofit.blogspot.com/ === Subject: Re: Some math, algebraic integers [...] |where to some it will seem silly that I left in sqrt(1), but the |ENTIRE BASIS for Galois Theory as it's taught is that when you can't |just simplify the square root, like you can with sqrt(1), then your |limitation is a limitation on the underlying roots!!! What?! Do you have any idea how obvious it is that you're making this up? Look at any textbook in Galois theory. I mean, come on, at least make SOME effort to retain a connection to reality. Try visiting http://www.jmilne.org/math and look at his Galois theory notes. Where does it base itself on the kind of baloney you say it does? Keith Ramsay === Subject: Re: Some math, algebraic integers >[...] >[...] Try visiting http://www.jmilne.org/math and >look at his Galois theory notes. Where does it base >itself on the kind of baloney you say it does? http://www.jmilne.org/math/tips.html Bwaahahahahahah! >-> -- Angus Rodgers (angus_prune@ eats spam; reply to angusrod@) Contains mild peril === Subject: Re: Some math, algebraic integers >[...] >|where to some it will seem silly that I left in sqrt(1), but the >|ENTIRE BASIS for Galois Theory as it's taught is that when you can't >|just simplify the square root, like you can with sqrt(1), then your >|limitation is a limitation on the underlying roots!!! >What?! >Do you have any idea how obvious it is that you're making >this up? Silly question. Can't put my finger on exactly why, but I know it's a silly question. >Look at any textbook in Galois theory. I mean, come on, >at least make SOME effort to retain a connection to >reality. Try visiting http://www.jmilne.org/math and >look at his Galois theory notes. Where does it base >itself on the kind of baloney you say it does? >Keith Ramsay ************************ David C. Ullrich === Subject: Re: Hilbert spaces among Banach spaces > > No, let 1 with direct sum in the l^2 sense. Then E is isometrically isomorphic > to E*, but not isomorphic to Hilbert. >I'm not sure if it is the same or different, but l^p + l^q (direct >sum) will do with p, q as above. > ??? If we're talking about _isometries_ then you need to specify > what _norm_ you're taking on this direct sum. He did. You didn't... > (if you take some other natural norm on the direct sum, like > ||(f,g)|| = ||f||_p + ||g||_p or something, then the space > is no longer isometric to its dual, or at least it's not > obvious that it is.) ***** > David C. Ullrich Yes, of course you are right. The l^2 sum will be the unique one (I would assume) for which the dual is the same. === Subject: Topology help! X-RFC2646: Original Given that the continuous image of a compact set is compact, why is the continuous image of an F-sigma set F-sigma? === Subject: Re: Topology help! > Given that the continuous image of a compact set is compact, why is the > continuous image of an F-sigma set F-sigma? true for example when the domain is the real numbers, but not in general. === Subject: Re: Topology help! X-RFC2646: Original The specific continuous map I am actually working with is actually a linear transformation L: R^d -> R^d. > Given that the continuous image of a compact set is compact, why is the > continuous image of an F-sigma set F-sigma? > true for example when the domain is the real numbers, but not in general. === Subject: Re: Topology help! X-RFC2646: Response Actually, the continuous image of an F-sigma set is F-sigma, since every F-sigma set can be written as a countable union of compact sets. > The specific continuous map I am actually working with is actually a > linear transformation > L: R^d -> R^d. > Given that the continuous image of a compact set is compact, why is the > continuous image of an F-sigma set F-sigma? > true for example when the domain is the real numbers, but not in general. === Subject: Re: Topology help! > Actually, the continuous image of an F-sigma set is F-sigma, since every > F-sigma set can be written as a countable union of compact sets. True in R^d as needed here. False in some other spaces. === Subject: Re: Topology help! > Actually, the continuous image of an F-sigma set is F-sigma, since every > F-sigma set can be written as a countable union of compact sets. An F-sigma set is a union of closed sets. If the domain space is compact and the codomain space Hausdorff, then the F-sigma set is a union of compact sets as closed sets of a compact space are compact. The image of that set is then the countable union of images of compact sets, which are closed, as the codomain space is Hausdorff. Now for your proof to hold you'll want to be in a space for which every closed set is a countable union of compact sets. This appears to be the case for R^n but it won't be for Q^n. === Subject: Help on Proof: a^2 + b^2 = c^2 then a or b is even Hello everyone, I need help with a proof for my discrete math homework. The statement to prove is this: If a, b and c are integers and a^2 + b^2 = c^2, then at least one of a and b is even. The book says it has to be a proof by contradiciton. This is what i have: PROOF: Suppose not. That is suppose there exists integers a, b, and c such that a^2 + b^2 = c^2 and a and b are not even. Since a and b are not even they are odd. By definition of odd a = 2k + 1 and b = 2n + 1 for any integers k and n. So we have: (2k + 1)^2 + (2n + 1)^2 = c^2. By definition the product of two odd integerse is odd, therefore (2k + 1)^2 is odd and (2n + 1)^2 is odd. By definition the sum of two odd integers is even. This proves c^2 is even, so c^2 = (2j)^2 for any integer j. Thus we have: (2k + 1)^2 + (2n + 1)^2 = (2j)^2 = 4k^2 + 4K + 1 + 4n^2 + 4n + 1 = 4j^2 = 2(2k^2 + 2k + 2n^2 + 2n + 1) = 4j^2 = 2k^2 + 2k + 2n^2 + 2n + 1 = 2j^2 Since k and n are any integer, and by definition of even the product of 2 times any integer is even we see that 2k^2, 2k, 2n^2 and 2n are all even. Since the sum of even integers is even 2k^2 + 2k + 2n^2 + 2n = 2z for some integer z. Thus 2z + 1 = 2j^2. But 2z + 1 is by definition odd and 2j^2 is by definition even. No integer can be both even and odd, therefore we have reached a contradiction. QED. I would like to know if i got the negation of the statement to be proved correct, and if the proof is logical/correct. I am only a mediocore math student (which you can probably tell) and am _really_ struggling through Chris P. -- Mind over matter: If you don't mind, then it doesn't matter. === Subject: Re: Help on Proof: a^2 + b^2 = c^2 then a or b is even > Hello everyone, > I need help with a proof for my discrete math homework. > The statement to prove is this: > If a, b and c are integers and a^2 + b^2 = c^2, then at least one > of a and b is even. > The book says it has to be a proof by contradiciton. This is what i > have: > PROOF: Suppose not. That is suppose there exists integers a, b, and c > such that a^2 + b^2 = c^2 and a and b are not even. Since a and b are > not even they are odd. By definition of odd a = 2k + 1 and b = 2n + 1 > for any integers k and n. So we have: (2k + 1)^2 + (2n + 1)^2 = c^2. By > definition the product of two odd integerse is odd, therefore (2k + 1)^2 > is odd and (2n + 1)^2 is odd. By definition the sum of two odd integers > is even. This proves c^2 is even, so c^2 = (2j)^2 for any integer j. > Thus we have: > (2k + 1)^2 + (2n + 1)^2 = (2j)^2 > = 4k^2 + 4K + 1 + 4n^2 + 4n + 1 = 4j^2 > = 2(2k^2 + 2k + 2n^2 + 2n + 1) = 4j^2 > = 2k^2 + 2k + 2n^2 + 2n + 1 = 2j^2 > Since k and n are any integer, and by definition of even the product of > 2 times any integer is even we see that 2k^2, 2k, 2n^2 and 2n are all > even. Since the sum of even integers is even 2k^2 + 2k + 2n^2 + 2n = 2z > for some integer z. Thus 2z + 1 = 2j^2. But 2z + 1 is by definition odd > and 2j^2 is by definition even. No integer can be both even and odd, > therefore we have reached a contradiction. QED. > I would like to know if i got the negation of the statement to be proved > correct, and if the proof is logical/correct. I am only a mediocore math > student (which you can probably tell) and am _really_ struggling through > Chris P. I think you can also proof this using the fact that (a+b)^p = = a^p + b^p (mod p) where p is a prime (this can be easily seen by proving that the binome C(p,j) is divisible by p and then by applying Newton's formula) So for every a,b,c being integers we have to prove that a^2+b^2=c^2 has at last a or b even. We assume the contrary : (2a + 1)^2 + (2b + 1)^2 = c^2 (mod 2) By developing the first side we get: (2a + 1)^2 + (2b + 1)^2 = 4(a^2 + b^2) + 2 (mod 2) Because c^2 = 4(a^2 + b^2) + 2 (mod 2) we have also (by using again the formula described at the beginning) : c^2 = (2(a+b) + sqrt(2))^2 (mod 2). Thus c is not an integer (c = sqrt(2) (mod 2)) . A contradiction. Maks, P.S: Youre proof is correct in my sense === Subject: Re: Help on Proof: a^2 + b^2 = c^2 then a or b is even > Hello everyone, > I need help with a proof for my discrete math homework. > The statement to prove is this: > If a, b and c are integers and a^2 + b^2 = c^2, then at least one > of a and b is even. > The book says it has to be a proof by contradiciton. This is what i > have: > PROOF: Suppose not. That is suppose there exists integers a, b, and c > such that a^2 + b^2 = c^2 and a and b are not even. Since a and b are > not even they are odd. By definition of odd a = 2k + 1 and b = 2n + 1 > for any integers k and n. So we have: (2k + 1)^2 + (2n + 1)^2 = c^2. By > definition the product of two odd integerse is odd, therefore (2k + 1)^2 > is odd and (2n + 1)^2 is odd. By definition the sum of two odd integers > is even. This proves c^2 is even, so c^2 = (2j)^2 for any integer j. > Thus we have: > (2k + 1)^2 + (2n + 1)^2 = (2j)^2 > = 4k^2 + 4K + 1 + 4n^2 + 4n + 1 = 4j^2 > = 2(2k^2 + 2k + 2n^2 + 2n + 1) = 4j^2 > = 2k^2 + 2k + 2n^2 + 2n + 1 = 2j^2 > Since k and n are any integer, and by definition of even the product of > 2 times any integer is even we see that 2k^2, 2k, 2n^2 and 2n are all > even. Since the sum of even integers is even 2k^2 + 2k + 2n^2 + 2n = 2z > for some integer z. Thus 2z + 1 = 2j^2. But 2z + 1 is by definition odd > and 2j^2 is by definition even. No integer can be both even and odd, > therefore we have reached a contradiction. QED. > I would like to know if i got the negation of the statement to be proved > correct, and if the proof is logical/correct. I am only a mediocore math > student (which you can probably tell) and am _really_ struggling through > Chris P. > I think you can also proof this using the fact that (a+b)^p = > = a^p + b^p (mod p) where p is a prime (this can be easily seen > by proving that the binome C(p,j) is divisible by p and then by > applying Newton's formula) > So for every a,b,c being integers we have to prove that a^2+b^2=c^2 > has at last a or b even. We assume the contrary : > (2a + 1)^2 + (2b + 1)^2 = c^2 (mod 2) > By developing the first side we get: > (2a + 1)^2 + (2b + 1)^2 = 4(a^2 + b^2) + 2 (mod 2) > Because c^2 = 4(a^2 + b^2) + 2 (mod 2) we have also (by using > again the formula described at the beginning) : > c^2 = (2(a+b) + sqrt(2))^2 (mod 2). > Thus c is not an integer (c = sqrt(2) (mod 2)) . A contradiction. For convenience, in the following discussion, I will refer to your fact your lemma: Lemma: (a + b)^p = a^p + b^p (mod p) where p is a prime. In particular, (a + b)^2 = a^2 + b^2 (mod 2). The ideas in your main proof are correct, but I think the presentation is faulty at a couple of places. Comments and critiques are shown below: 1. A proof involving modulo arithmetic is, IMOHO, more advanced than one which does not use it, especially when the new proof does not offer any significant reduction in steps of reasoning. 2. You seem to be using a and b in two different contexts: quote > has at last a or b even. We assume the contrary : > (2a + 1)^2 + (2b + 1)^2 = c^2 (mod 2) unquote I think you wanted to say this: Assume the contrary. Then a = 2k + 1 and b = 2n + 1 for some integers k and n. Then the rest of your reasoning proceeds without difficulty with the correct symbols k and n in place of the a and b as follows: c^2 = a^2 + b^2 (mod 2) = 4(k^2 + n^2) + 4(k + n) + 2 (mod 2) = 4(k + n)^2 + 4(k + n) + 2 (mod 2) (by your lemma) = [2(k+n)+sqrt(2)]^2 (mod 2) After that you said > Thus c is not an integer (c = sqrt(2) (mod 2)) . A contradiction. I don't think this is quite right. I think one wants to say: c = + [2(k+n)+sqrt(2)], or - [2(k+n)+sqrt(2)], (mod 2), which is a contradiction because neither one is an integer. Shedar === Subject: Re: Help on Proof: a^2 + b^2 = c^2 then a or b is even > I need help with a proof for my discrete math homework. > The statement to prove is this: > If a, b and c are integers and a^2 + b^2 = c^2 > then at least one of a and b is even. Mod 4: odd^2 + odd^2 = 2 isn't odd^2 (= 1) or even^2 (= 0) --Bill Dubuque === Subject: Re: Help on Proof: a^2 + b^2 = c^2 then a or b is even : > Hello everyone, > I need help with a proof for my discrete math homework. > The statement to prove is this: > If a, b and c are integers and a^2 + b^2 = c^2, then at least one > of a and b is even. > The book says it has to be a proof by contradiciton. This is what i > have: > PROOF: Suppose not. That is suppose there exists integers a, b, and c > such that a^2 + b^2 = c^2 and a and b are not even. Since a and b are > not even they are odd. By definition of odd a = 2k + 1 and b = 2n + 1 > for any integers k and n. So we have: (2k + 1)^2 + (2n + 1)^2 = c^2. By > definition the product of two odd integerse is odd, therefore (2k + 1)^2 > is odd and (2n + 1)^2 is odd. By definition the sum of two odd integers > is even. This proves c^2 is even, so c^2 = (2j)^2 for any integer j. > Thus we have: > (2k + 1)^2 + (2n + 1)^2 = (2j)^2 > = 4k^2 + 4K + 1 + 4n^2 + 4n + 1 = 4j^2 > = 2(2k^2 + 2k + 2n^2 + 2n + 1) = 4j^2 > = 2k^2 + 2k + 2n^2 + 2n + 1 = 2j^2 > Since k and n are any integer, and by definition of even the product of > 2 times any integer is even we see that 2k^2, 2k, 2n^2 and 2n are all > even. Since the sum of even integers is even 2k^2 + 2k + 2n^2 + 2n = 2z > for some integer z. Thus 2z + 1 = 2j^2. But 2z + 1 is by definition odd > and 2j^2 is by definition even. No integer can be both even and odd, > therefore we have reached a contradiction. QED. > I would like to know if i got the negation of the statement to be proved > correct, and if the proof is logical/correct. I am only a mediocore math > student (which you can probably tell) and am _really_ struggling through > Chris P. The version I heard was that among a, b and c there must be a multiple of 3, a multiple of 4, and a multiple of 5. Of course, these may be combined in the same number as in (119, 120, 169). -- Paul Townsend Pair them off into threes Interchange the alphabetic letter groups to reply === Subject: Re: Help on Proof: a^2 + b^2 = c^2 then a or b is even Newsgrps: sci.math Subject : Re: Help on Proof: a^2 + b^2 = c^2 then a or b is even > If a, b and c are integers and a^2 + b^2 = c^2, then at least one > of a and b is even. > The version I heard was that among a, b and c there must be a multiple > of 3, a multiple of 4, and a multiple of 5. Of course, these may be > combined in the same number as in (119, 120, 169). a^2 + b^2 = c^2 if abc /= 0 (mod 3), then a^2 = b^2 = c^2 = 1 (mod 3) 1 + 1 = 1 (mod 3). Thus 3|abc If abc /= 0 (mod 5), then a^2 = +-1 (mod 5), b^2 = +-1 (mod 5) c^2 = +-1 (mod 5) Thus 1 + 1 = +-1 (mod 5) or -1 - 1 = +-1 (mod 5). So 5|abc If not 4 | a, not 4 | b and not 4 | c, then a^2 = 1 or 4 (mod 8) b^2 = 1 or 4 (mod 8) c^2 = 1 or 4 (mod 8) Thus 1 + 1 = 1 or 4 (mod 8) or 1 + 4 = 1 or 4 (mod 8) So 4 divides a or b or c. ---- === Subject: Re: Help on Proof: a^2 + b^2 = c^2 then a or b is even > If a, b and c are integers and a^2 + b^2 = c^2, then at least one > of a and b is even. > The book says it has to be a proof by contradiciton. This is what i > have: > PROOF: Suppose not. That is suppose there exists integers a, b, and c > such that a^2 + b^2 = c^2 and a and b are not even. Since a and b are > not even they are odd. By definition of odd a = 2k + 1 and b = 2n + 1 > for any integers k and n. So we have: > (2k + 1)^2 + (2n + 1)^2 = c^2 > = 4k^2 + 4K + 1 + 4n^2 + 4n + 1 Thus c^2 = 4j + 2 for some integer j. If c is even, then 4 divides c^2 but 4 cannot divide 4j + 2. If c is odd, then c^2 is odd but 4j + 2 isn't odd. === Subject: Re: Help on Proof: a^2 + b^2 = c^2 then a or b is even > Hello everyone, > I need help with a proof for my discrete math homework. > The statement to prove is this: > If a, b and c are integers and a^2 + b^2 = c^2, then at least one > of a and b is even. > The book says it has to be a proof by contradiciton. This is what i > have: > PROOF: Suppose not. That is suppose there exists integers a, b, and c > such that a^2 + b^2 = c^2 and a and b are not even. Since a and b are > not even they are odd. By definition of odd a = 2k + 1 and b = 2n + 1 > for ANY integers k and n. See Remark 3. > So we have: (2k + 1)^2 + (2n + 1)^2 = c^2. By > definition the product of two odd integerse is odd, therefore (2k + 1)^2 > is odd and (2n + 1)^2 is odd. By definition the sum of two odd integers > is even. This proves c^2 is even, so c^2 = (2j)^2 for ANY integer j. See Remark 3 and Remark 4. > Thus we have: > (2k + 1)^2 + (2n + 1)^2 = (2j)^2 > = 4k^2 + 4K + 1 + 4n^2 + 4n + 1 = 4j^2 > = 2(2k^2 + 2k + 2n^2 + 2n + 1) = 4j^2 > = 2k^2 + 2k + 2n^2 + 2n + 1 = 2j^2 > Since k and n are ANY integer, and by definition of even the product of See Remark 3. > 2 times any integer is even we see that 2k^2, 2k, 2n^2 and 2n are all > even. Since the sum of even integers is even 2k^2 + 2k + 2n^2 + 2n = 2z > for some integer z. Thus 2z + 1 = 2j^2. But 2z + 1 is by definition odd > and 2j^2 is by definition even. No integer can be both even and odd, > therefore we have reached a contradiction. QED. > I would like to know if i got the negation of the statement to be proved > correct, and if the proof is logical/correct. I am only a mediocore math > student (which you can probably tell) and am _really_ struggling through > Chris P. Yes, you did well. You got all the main ideas and most of the steps correct except at a couple of places where the exposition can be better phrased. I'll rephrase as follows with side comments enclosed in square brackets [...]: PROOF: Assume a, b and c are integers such that a^2 + b^2 = c^2 [see remark 1]. Suppose, on the contrary, both a and b are not even [see remark 2]. Then both a and b must be odd. Therefore a = 2k + 1, and b = 2n + 1 for SOME integers k and n [see remark 3]. Hence we have: c^2 = a^2 + b^2 = (2k + 1)^2 + (2n + 1)^2 = (2k + 1)^2 + (2n + 1)^2 = 4k^2 + 4k + 1 + 4n^2 + 4n + 1 = 2(2k^2 + 2k + 2n^2 + 2n + 1) The preceding line shows that c^2 is an even number, and therefore, c itself must be an even number (because the square of every odd number is always odd) [you might want to prove this as a lemma]. So c = 2 j for SOME integer j [see Remark 4]. Thus we have (2j)^2 = c^2 = 2(2k^2 + 2k + 2n^2 + 2n + 1), 4 j^2 = c^2 = 2(2k^2 + 2k + 2n^2 + 2n + 1), 2 j^2 = c^2/2 = 2k^2 + 2k + 2n^2 + 2n + 1. But the preceding line is a logical contradiction since it says that the number c^2/2 is both an even integer (namely, 2 j^2) as well as an odd integer (namely, 2k^2 + 2k + 2n^2 + 2n + 1). It follows from the principle of proof by contradiction that our supposition must be false. In other words, if a, b and c are integers such that a^2 + b^2 = c^2, then one of a and b must be even. QED. Remarks 0. Given a conditional statement IF P THEN Q, P is called its antecedent and Q its consequent. 1. This is called assuming the antecedent of the given Ôif .. then' statement. 2. This is called denying the consequent of the given Ôif .. then' statement. 3. In your original version, you said for ANY integers k and n, which is incorrect. You should have said for SOME integers k and n (in other words, there exist some integers k and n). In mathematical logic (in contrast to common English usage), the implicit convention is that the word any is only used in the sense of each and every. In fact, I recommend against using the potentially confusing word any even when you mean each and every. Instead, always use for every to mean for each and every, and use for some to mean there exist(s) some. 4. Here, in your original version, you did not fully explain why c = 2 j for some j. To explain it, we may appeal to the fact that c^2 is even implies c itself must be even (because the square of an odd is always odd). === Subject: Re: Help on Proof: a^2 + b^2 = c^2 then a or b is even X-RFC2646: Original > If a, b and c are integers and a^2 + b^2 = c^2, > then at least one of a and b is even. > The book says it has to be a proof by contradiciton. This is what i > have: PROOF: Suppose not. Then there exist integers a, b, and c > such that a^2 + b^2 = c^2 and a and b are both odd. So, there exist integers k and n, such that a = 2k + 1 and b = 2n + 1 > We have: (2k + 1)^2 + (2n + 1)^2 = c^2. The product of two odd > integers is odd. Therefore (2k + 1)^2 is odd and (2n + 1)^2 is odd. > But the sum of two odd integers is even. This proves c^2 is even. > Then c^2 = 2j for some integer j and: > (2k + 1)^2 + (2n + 1)^2 = (2j)^2 > 4k^2 + 4k + 1 + 4n^2 + 4n + 1 = 4j^2 2k^2 + 2k + 2n^2 + 2n + 1 = 2j^2 2(k^2 + k + n^2 + n) + 1 = 2j^2 Let m = k^2+k+n^2+n. Since Z is closed, m is an integer, which gives: 2j^2 = 2m + 1. The left side is even and the right side is odd. > No integer can be both even and odd, > therefore we have reached a contradiction. QED. === Subject: Re: Like a log |One that's weaker than both local boundedness and measurability is |that there's a measurable set of positive measure on which the function is |bounded. That's nice in that it is fairly plain that it is a weakening of both local boundedness and measurability. |I make no claim that this is the weakest possible condition. The weakest condition that in conjunction with a condition A is sufficient to imply a condition B is always if A then B. It doesn't seem like a very natural condition in this case, though! I feel like saying we're looking for conditions that somehow show how A and B could be relevant to each other, but I don't know of a way to define that as a requirement. Another condition weakening local boundedness a little is that the graph of the function not be dense in the right half-plane {(x,y)| x>0}. Keith Ramsay === Subject: Re: Chess can easily be solved! ... > stumbling across the optimal move and showing it to be optimal > demonstrates that he was incorrect in stating that every node must be > searched to prove the optimal strategy in chess. What if the game is essentially a draw? But I was indeed incorrect in stating that you have to look at every node. But the number of nodes is much larger than you implied with your branching factor of 2 to 4. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Chess can easily be solved! > ... > stumbling across the optimal move and showing it to be optimal > demonstrates that he was incorrect in stating that every node must be > searched to prove the optimal strategy in chess. > What if the game is essentially a draw? But I was indeed incorrect > in stating that you have to look at every node. But the number of > nodes is much larger than you implied with your branching factor > of 2 to 4. to rigorously prove that the game is a draw.. i think you're correct in saying that all nodes for both sides would have to be searched. (this of course only to eliminate the possibility of it being a win) === Subject: Re: Std.dev. for random integers by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i8Q0NvU20981; >dual digit numbers and calculated the std.dev. for each column and for >all 25 numbers. >I found that the std.dev. for the twenty five did not equal the mean >of the sets of five. I then simulated the problem with two sets of >digits, viz. 1-2-5-6-9 and 0-3-4-7-8. They each have a s.d. of 3.21, >while all ten as a sample yield 3.03. >Lilkewise, the five odd and the five even digits have a s.d. of 3.16. >Also, the first five digits and the last five each have a s.d. of 1.58 >Thus there must be a more complex formula for predicting the s.d. of >the sum of two or more sets. Geometric mean and mean of variances did >not work. >Can some body please tell me a specific text that deals with this type >problem. I've looked at many textbooks but found no guidance. >Stig Holmquist Are you familiar with the formulas for standard deviation? or are you just plugging numbers into a calculator? phil === Subject: Re: Std.dev. for random integers > dual digit numbers and calculated the std.dev. for each column and for > all 25 numbers. > I found that the std.dev. for the twenty five did not equal the mean > of the sets of five. I then simulated the problem with two sets of > digits, viz. 1-2-5-6-9 and 0-3-4-7-8. They each have a s.d. of 3.21, > while all ten as a sample yield 3.03. > Lilkewise, the five odd and the five even digits have a s.d. of 3.16. > Also, the first five digits and the last five each have a s.d. of 1.58 > Thus there must be a more complex formula for predicting the s.d. of > the sum of two or more sets. Geometric mean and mean of variances did > not work. > Can some body please tell me a specific text that deals with this type > problem. I've looked at many textbooks but found no guidance. > Stig Holmquist When you consider equal-size sets of numbers together rather than separately, the variance of the entire collection equals the average of the within-set variances, plus the variance of the averages of the sets; in all cases, the denominator of each variance is its n, not n-1. If the sets are not equal-sized then the relation is a little more complicated. All this is usually covered (but in somewhat different terms) in Analysis of Variance texts. === Subject: Re: Std.dev. for random integers > dual digit numbers and calculated the std.dev. for each column and for > all 25 numbers. > I found that the std.dev. for the twenty five did not equal the mean > of the sets of five. I then simulated the problem with two sets of > digits, viz. 1-2-5-6-9 and 0-3-4-7-8. They each have a s.d. of 3.21, > while all ten as a sample yield 3.03. > Lilkewise, the five odd and the five even digits have a s.d. of 3.16. > Also, the first five digits and the last five each have a s.d. of 1.58 > Thus there must be a more complex formula for predicting the s.d. of > the sum of two or more sets. Geometric mean and mean of variances did > not work. > Can some body please tell me a specific text that deals with this type > problem. I've looked at many textbooks but found no guidance. > Stig Holmquist >When you consider equal-size sets of numbers together rather than >separately, the variance of the entire collection equals the average >of the within-set variances, plus the variance of the averages of the >sets; in all cases, the denominator of each variance is its n, not n-1. >If the sets are not equal-sized then the relation is a little more >complicated. All this is usually covered (but in somewhat different >terms) in Analysis of Variance texts. I'm stil puzzled because the overall variance was always less than any sum, so where does a negative factor come from? Stig Holmquist === Subject: Re: Std.dev. for random integers >dual digit numbers and calculated the std.dev. for each column and for >all 25 numbers. >I found that the std.dev. for the twenty five did not equal the mean >of the sets of five. I then simulated the problem with two sets of >digits, viz. 1-2-5-6-9 and 0-3-4-7-8. They each have a s.d. of 3.21, >while all ten as a sample yield 3.03. >Lilkewise, the five odd and the five even digits have a s.d. of 3.16. >Also, the first five digits and the last five each have a s.d. of 1.58 >Thus there must be a more complex formula for predicting the s.d. of >the sum of two or more sets. Geometric mean and mean of variances did >not work. >Can some body please tell me a specific text that deals with this type >problem. I've looked at many textbooks but found no guidance. >Stig Holmquist >Are you familiar with the formulas for standard deviation? or >are you just plugging numbers into a calculator? >phil Yes, I'm familiar in a rudimentary way with the formula. How would that help me? What new equation must I apply? Stig Holmquist === Subject: Re: Std.dev. for random integers > Yes, I'm familiar in a rudimentary way with the formula. How would > that help me? What new equation must I apply? I don't know if its Ônew' to you, but the following is a useful approach ( also used in mechanics where mean ~ centre of Mass and standard deviation ~ moment of inertia ) Its a notation thing. Although summation symbols are better than writing out things in full, vector notation is better still. Treat each set of integers as a vector, and take U to be the vector of units and of the same dimension. Multiply U by the mean(m) to get another vector, M. X = [x1, x2, x3, .... xn] U = [1, 1 , 1 , .... 1 ] M = [m, m , m , .... m ] Using dot products we then have: U.U = COUNT = n X.U = SUM X.U / n = MEAN M.M = n * SQUARE of MEAN X.M = MEAN * X.U X.M = MEAN * SUM = n * SQUARE of MEAN = M.M also M.X = M.M X.X = SUM OF SQUARES (X-M).(X-M) = SUM of SQUARES of DIFFERENCES from MEAN = n * SQUARE of Standard Deviation ++++++++++++++++++++++++++++++++++++++++++++++++++ If you multiply out the brackets in the last equation, and divide by n, you get: (X-M).(X-M) = X.X - M.X - X.M + M.M = X.X - M.M SQUARE of SD = (SUM of SQUARES) - (SQUARE of MEAN) +++++++++++++++++++++++++++++++++++++++++++++++++++ To calculate the SD and Mean of the combination of two collections of numbers, knowing their individual SD, Mean and Count, you use the Right-hand side. The combined ÔSum of Squares' is obvious. For the combined mean, you have to give multiply by a weighting of the COUNT of each collection, or use: SUM = MEAN * COUNT -- Ken === Subject: Re: Std.dev. for random integers >dual digit numbers and calculated the std.dev. for each column and for >all 25 numbers. > >I found that the std.dev. for the twenty five did not equal the mean >of the sets of five. I then simulated the problem with two sets of >digits, viz. 1-2-5-6-9 and 0-3-4-7-8. They each have a s.d. of 3.21, >while all ten as a sample yield 3.03. > >Lilkewise, the five odd and the five even digits have a s.d. of 3.16. >Also, the first five digits and the last five each have a s.d. of 1.58 > >Thus there must be a more complex formula for predicting the s.d. of >the sum of two or more sets. Geometric mean and mean of variances did >not work. > >Can some body please tell me a specific text that deals with this type >problem. I've looked at many textbooks but found no guidance. > >Stig Holmquist >Are you familiar with the formulas for standard deviation? or >are you just plugging numbers into a calculator? >phil > Yes, I'm familiar in a rudimentary way with the formula. How would > that help me? What new equation must I apply? > Stig Holmquist not an equation. just an idea. mean isn't MEANT to be the same as standard deviation. if they are that's completely arbitrary. that's like finding significance in a circle of radius r centered at (r,0) === Subject: Re: probability theory events by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i8Q0Nvs20976; >I just started reading Ross' Intro. Probability Models and an event E is >defined to be a subset of the sample space S. >Although that definition is clear, I'm trying to get a more intuitive idea >of what an event means. >Let's say for a die roll, S = {1, 2, 3, 4, 5, 6} and we have an event E = >{2, 3} >Would it be correct to say, if the outcome of a die roll is 2 then the event >E occurs? >Essentially I'm wondering if an event in probability is in some way related >to something that happens. In your example, it would take a minimum of two die rolls to have the event E as you defined it. phil === Subject: Re: probability theory events >I just started reading Ross' Intro. Probability Models and an event E is >defined to be a subset of the sample space S. >Although that definition is clear, I'm trying to get a more intuitive idea >of what an event means. >Let's say for a die roll, S = {1, 2, 3, 4, 5, 6} and we have an event E = >{2, 3} >Would it be correct to say, if the outcome of a die roll is 2 then the event >E occurs? >Essentially I'm wondering if an event in probability is in some way related >to something that happens. > >In your example, it would take a minimum of two die rolls to have >the event E as you defined it. Incorrect. G.A. Edgar's interpretation was the correct one. The set E = {2, 3} represents the event that either a 2 or a 3 is rolled. So if a 2 is rolled, event E has occured. -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: Who thinks Goldbach's Conjecture is unprovable? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i8Q0NvQ21009; >these are silly conundrums. if it were impossible to prove, >a counterexample would be found eventually, if not sooner. > >False. It is possible that Goldbach's Conjecture is independent of >ZFC, i.e., that Con(ZFC) implies Con(ZFC+GC) and Con(ZFC) implies >Con(ZFC+not GC). (Just as the Continuum Hypothesis is independent of >ZFC.) I suspect most mathematicians believe that this independence is >not the case. >-- >Stephen J. Herschkorn herschko@rutcor.rutgers.edu Maybe I'm confused, but suppose that (not GC) is unprovable (say in PA). Then it would seem that each even constant > 2 is in fact the sum of two primes, for if there were a counterexample, then there would exist a finite calculation which showed that (e.g. use Eratosthenes sieve to list all primes less than the constant, and check that no two summed to the constant). I.e. that a proof of unprovability of (not GC) would perforce rule out the possibility of such a finite calculation, no matter how lengthy. Am I missing something? === Subject: Re: Who thinks Goldbach's Conjecture is unprovable? >these are silly conundrums. if it were impossible to prove, >a counterexample would be found eventually, if not sooner. > > >False. It is possible that Goldbach's Conjecture is independent of >ZFC, i.e., that Con(ZFC) implies Con(ZFC+GC) and Con(ZFC) implies >Con(ZFC+not GC). (Just as the Continuum Hypothesis is independent of >ZFC.) I suspect most mathematicians believe that this independence is >not the case. >-- >Stephen J. Herschkorn herschko@rutcor.rutgers.edu > Maybe I'm confused, but suppose that (not GC) is unprovable > (say in PA). Then it would seem that each even constant > 2 is > in fact the sum of two primes, for if there were a counterexample, > then there would exist a finite calculation which showed that > (e.g. use Eratosthenes sieve to list all primes less than the > constant, and check that no two summed to the constant). > I.e. that a proof of unprovability of (not GC) would perforce > rule out the possibility of such a finite calculation, no matter > how lengthy. > Am I missing something? This leads to another question. Who thinks that it is possible to prove that GC is unprovable in any reasonable axiom system? In other words, instead of saying GC is unprovable in PA. or GC is unprovable in ZFC., who thinks that it is possible to prove that GC is unprovable. (i.e., the only way to prove GC is to make it an axiom). Craig === Subject: Re: Who thinks Goldbach's Conjecture is unprovable? |>these are silly conundrums. if it were impossible to prove, |>a counterexample would be found eventually, if not sooner. |>False. It is possible that Goldbach's Conjecture is independent of |>ZFC, i.e., that Con(ZFC) implies Con(ZFC+GC) and Con(ZFC) implies |>Con(ZFC+not GC). (Just as the Continuum Hypothesis is independent of |>ZFC.) I suspect most mathematicians believe that this independence is |>not the case. There is already a result of Chen that every large enough even number > 2 is either a sum of two primes, or the sum of a prime and a product of two primes. It seems implausible to me that we have been unable to strengthen this to a proof of Goldbach due to some essential barrier to doing so, as opposed to the ordinary difficulties in pushing forward a subject like this. |Maybe I'm confused, but suppose that (not GC) is unprovable |(say in PA). Then it would seem that each even constant > 2 is |in fact the sum of two primes, for if there were a counterexample, |then there would exist a finite calculation which showed that |(e.g. use Eratosthenes sieve to list all primes less than the |constant, and check that no two summed to the constant). |I.e. that a proof of unprovability of (not GC) would perforce |rule out the possibility of such a finite calculation, no matter |how lengthy. That's right. If GC is false, then it is possible to disprove in ZF, PA, or even very weak systems of arithmetic. It doesn't eliminate the possibility that GC is true but can't be proven in your favorite axiom system, whether PA or ZF, however. Keith Ramsay === Subject: Re: My real opinion about real numbers > This is real number > 0,45466745453679885796799779767766796708204820099599496... > And below is real list of real numbers. BTW: what you think about the > length of the Cantors decimal number (that number that will be proved to > missing from list) One of your many misconceptions is that the Cantor proof constructs ONE missing number. It gives a rule for constructing a number which is guaranteed not to be on a countable list, but there are infinitely many choices which can be made within those rules. More choices than there are numbers on the list. For instance, following the Cantor construction, when I want to construct a number that isn't on your list below, the first digit should be different from 0. That means that it can be 1-9. So there are 9 different choices for 1st digit of the missing number, and ALL of them will eventually end up with numbers not on the list. The second digit of these missing numbers must be different from 1. That leaves 0 and 2-9. so there are 9*9 = 81 different 2-digit sequences which begin Cantor missing numbers. And this is only one rule. Your particular list, if carried on to arbitrarily long integers, never generates infinite sequences, so that means that most of the rational numbers are also not on your list. > after the 999010:th row of this list? Its length is > 999010 (see below). I think you mean that your list is finite, and stops after 999010. Are you saying you can't think of a number which is not on this list? Or that the Cantor construction doesn't give one? The Cantor construction is a rule for constructing numbers which do not appear on countably infinite lists (lists that assign a value to every natural number). You haven't assigned any value for numbers beyond 999010, so let me say f(n) = 0 for n>999010. Also, none of your numbers has more than 6 digits, i.e. all digits beyond the 6th are 0 in all of your numbers. Therefore in the Cantor construction all I have to do is choose nonzero values past the 6th digit, for instance 0.23456781111111..... > I just hope that this kind of approach What approach? Approach to what? What is your approach? > give new kind of proove New kind of proof? Where was your proof? Did you draw a conclusion? What was it? > that the number generated is not a real number at all. There are infinitely many numbers generated. The one I picked is (2345678/10000000) + (1/90000000). Are you saying that adding those two fractions gives a number that isn't really a number? Why do you say that? > F(0)=0.0 > F(1)=0.1 > F(2)=0.2 Snip. The rule is that F(n) = the digits of n reversed. - Randy === Subject: Re: My real opinion about real numbers > What approach? Approach to what? What is your approach? > [...] > New kind of proof? Where was your proof? Did you draw a conclusion? > What was it? It was just an outline in my head. But the figure in my head went something like this: 1) Real numbers can be list in order of they length (ok, maybe i missed - this is not true :() 2) in every step the missing number's (MN) length generated in Cantor's diagonal trick is longer than all real numbers previously in list Then use induction to proove that for every step in list MN is not in a list. But ok, maybe that was an error approcach. === Subject: Re: My real opinion about real numbers > What approach? Approach to what? What is your approach? > [...] > New kind of proof? Where was your proof? Did you draw a conclusion? > What was it? > It was just an outline in my head. > But the figure in my head went something like this: > 1) Real numbers can be list in order of they length > (ok, maybe i missed - this is not true :() > 2) in every step the missing number's (MN) length generated in Cantor's > diagonal trick is longer than all real numbers previously in list > Then use induction to proove that for every step in list MN is not in a > list. If the numbers on your list all have infinitely many digits, then your second claim is false. Most real numbers have infinite decimal expansions. -- === Subject: Re: JSH: Nora Baron ... > My suggestion for readers wanting to get some perspective on Nora > Baron is to read a singular post of mine: Reading the replies to it is also useful. > It seems to me that for some odd reason many of you are fixated on the > idea that Nora Baron is actually some woman out there, Why would I conclude that it is a woman? Do you know from a name immediately whether it purports to be a woman or not? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: JSH: Nora Baron > ... > My suggestion for readers wanting to get some perspective on > Nora Baron is to read a singular post of mine: > > Reading the replies to it is also useful. > It seems to me that for some odd reason many of you are fixated > on the idea that Nora Baron is actually some woman out there, > Why would I conclude that it is a woman? Do you know from a name > immediately whether it purports to be a woman or not? In American English (and probably British English and others), referring to a person by the pronoun it carries a clear tone of insult. I don't doubt that Harris intends the insult, but I strongly suspect you do not. For many years (several centuries, I believe), the general rule for the third person singular has been he or she for male or female people, with he used in the unspecified case (such as, A lawyer should not lie about his client.) The pronoun it is used for essentially everything else: animals, machines, geographical features, etc. There are exceptions: ships have generally been she, pets or anything else one wants to show sentimentality toward might be he or she. There has been social pressure in recent decades to replace he in the unspecified case with a gender-neutral pronoun. I've seen he/she, hir and they tried as substitutes. (I prefer they used as third person singular.) I have never seen anyone suggest the obvious choice it be used that way. The rule I see in all this: referring to someone as it tacitly de-personalizes them. Please believe I intend no criticism of you. I suspected this is one of those things left unmentioned because they are so obvious to native speakers, and so I wanted to correct the situation. Jim Burns === Subject: Re: JSH: Nora Baron ... > > It seems to me that for some odd reason many of you are fixated > > on the idea that Nora Baron is actually some woman out there, > > Why would I conclude that it is a woman? Do you know from a name > immediately whether it purports to be a woman or not? > In American English (and probably British English and others), > referring to a person by the pronoun it carries a clear tone > of insult. I know, I know, I am guilty as charged. And I even know about the older usage of they. It comes from the Dutch language where the only sexe-neutral pronoun is the equivalent of it. Here it is used if the gender of the word involved is neuter (unless it is a definitely male or female animate object), and when the sexe is unknown. (Yes; in Dutch words still have gender, formally male, female or neuter, although the strict distinction of the first two is only maintained in the southern part of the Dutch speaking community. Gender indicates pronoun used, except in the cases I mention above, where the grammatically correct pronoun is generally replaced by the gender correct pronoun.) -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: JSH: Nora Baron |For many years (several centuries, I believe), the general rule for |the third person singular has been he or she for male or |female people, with he used in the unspecified case (such |as, A lawyer should not lie about his client.) The pronoun |it is used for essentially everything else: animals, machines, |geographical features, etc. I've also read that they has been used for centuries in the case of indeterminate gender, despite the advice of grammar teachers not to use it. I think it's generally fine. I avoid it in favor of rewriting the sentence or he or she in formal writing. I do find it jarring in reference to people, or even animals with something of a personality, such as dogs and cats. Keith Ramsay === Subject: Re: JSH: Nora Baron === >Subject: Re: JSH: Nora Baron >|For many years (several centuries, I believe), the general rule for >|the third person singular has been he or she for male or >|female people, with he used in the unspecified case (such >|as, A lawyer should not lie about his client.) The pronoun >|it is used for essentially everything else: animals, machines, >|geographical features, etc. >I've also read that they has been used for centuries in the >case of indeterminate gender, despite the advice of grammar >teachers not to use it. I think it's generally fine. I avoid it in >favor of rewriting the sentence or he or she in formal writing. >I do find it jarring in reference to people, or even animals with >something of a personality, such as dogs and cats. If a pig had a charming personality, it would cease being a filthy animal? >Keith Ramsay -- Mensanator Ace of Clubs === Subject: Re: JSH: Nora Baron === >Subject: Re: JSH: Nora Baron >I've also read that they has been used for centuries in the >case of indeterminate gender, despite the advice of grammar >teachers not to use it. I think it's generally fine. I avoid it >in favor of rewriting the sentence or he or she in formal >writing. >I do find it jarring in reference to people, or even animals >with something of a personality, such as dogs and cats. > If a pig had a charming personality, it would cease being a > filthy animal? I'm going to guess that you know that this is not a rhetorical question. Some people do have pigs as pets, presumably for their personalities rather than, say, their taste. See, for example, http://www.pigs4ever.com/PotBellyPigPictures/thekids.htm I would say that a filthy animal with a charming personality is different in an important way from a filthy animal without one. This is a curious direction you've gone, since the point I, at least, wanted to make was that he and she are linguistic markers for personality. Their potential connection to filthy-animal-hood was not something I'd considered. Certainly I've known humans with very distict personalities (charming or otherwise) who would have fit into the filthy animal category very nicely. I hope you agree that he or she would be appropriate in those cases. Would you mind expanding on your point? Jim Burns === Subject: Re: JSH: Nora Baron === >Subject: Re: JSH: Nora Baron >Message-id: <41571A1B.857E2608@osu.edu> === >Subject: Re: JSH: Nora Baron > >I've also read that they has been used for centuries in the >case of indeterminate gender, despite the advice of grammar >teachers not to use it. I think it's generally fine. I avoid it >in favor of rewriting the sentence or he or she in formal >writing. > >I do find it jarring in reference to people, or even animals >with something of a personality, such as dogs and cats. > If a pig had a charming personality, it would cease being a > filthy animal? >I'm going to guess that you know that this is not a rhetorical >question. Some people do have pigs as pets, presumably for their >personalities rather than, say, their taste. See, for example, >http://www.pigs4ever.com/PotBellyPigPictures/thekids.htm >I would say that a filthy animal with a charming personality is >different in an important way from a filthy animal without one. >This is a curious direction you've gone, since the point I, >at least, wanted to make was that he and she are linguistic >markers for personality. Their potential connection to >filthy-animal-hood was not something I'd considered. Certainly >I've known humans with very distict personalities (charming >or otherwise) who would have fit into the filthy animal category >very nicely. I hope you agree that he or she would be >appropriate in those cases. >Would you mind expanding on your point? >Jim Burns It was a joke. The actual quotation is: Jules: Hey, sewer rat may taste like pumpkin pie but I'd never know Ôcause I wouldn't eat the filthy motherers. Pig sleep and root in . That's a filthy animal. I ain't eat nothin' that ain't got enough sense to disregard its own feces. Vincent: How about a dog? Dogs eat their own feces. Jules: I don't eat dog either. Vincent: Yeah, but do you consider a dog to be a filthy animal? Jules: I wouldn't go so far as to call a dog filthy but they're definitely dirty. But, a dog's got personality. Personality goes a long way. Vincent: Ah, so by that rationale, if a pig had a better personality, it'd cease to be a filthy animal. Is that true? Jules: Well we gotta be talkin' about one charmin' motherin' pig. I mean he'd have to be ten times more charmin' than that Arnold on Green Acres, you know what I'm sayin'? The point was: which animals is it ok to refer to as it? -- Mensanator Ace of Clubs === Subject: Re: JSH: Nora Baron === >Subject: Re: JSH: Nora Baron >Message-id: <41571A1B.857E2608@osu.edu> >Would you mind expanding on your point? >Jim Burns > It was a joke. The actual quotation is: > Jules: Hey, sewer rat may taste like pumpkin pie but I'd never > know Ôcause I wouldn't eat the filthy motherers. Pig sleep > and root in . That's a filthy animal. I ain't eat nothin' > that ain't got enough sense to disregard its own feces. > Vincent: How about a dog? Dogs eat their own feces. > Jules: I don't eat dog either. > Vincent: Yeah, but do you consider a dog to be a filthy animal? > Jules: I wouldn't go so far as to call a dog filthy but they're > definitely dirty. But, a dog's got personality. Personality goes > a long way. > Vincent: Ah, so by that rationale, if a pig had a better > personality, it'd cease to be a filthy animal. Is that true? > Jules: Well we gotta be talkin' about one charmin' motherin' > pig. I mean he'd have to be ten times more charmin' than that > Arnold on Green Acres, you know what I'm sayin'? > The point was: which animals is it ok to refer to as it? Apparently it's a matter of taste. Jim Burns === Subject: Re: Recovering a probability distribution from its moments >No help here for the original poster, but a related query of my own. >It is often stated that, given a sequence of putative moments for >a probability distribution, there is a unique distribution with those >moments if are satisfied. >My query is: is there a nice, simple-ish example where the conditions >are NOT satisfied? >Specifically: is there a nice example of two different probability >The fact that I don't know of any, in spite of this being an obvious sort >of bookwork problem, suggests that there ISN'T any known closed-form >explicit example. >Anyone know? The best know natural example is the lognormal. In this case, the literature even gives natural alternatives. The first natural example given was by Stieltjes. I believe he did this by using the fact that (1+i)^(4k+2) is purely imaginary, so int x^(4k+1)*exp(-x)*(1 + C*cos(ix)) dx is independent of C. Let x be the fourth root of u, and we have the counterexample. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: Recovering a probability distribution from its moments >Suppose we can easily calculate, recursively, all the moments m_k of a >probability distribution P on the reals, and we can show that these >moments satisfy the hypothesis for the method of moments. (I.e., the >power series with coefficients m_k / k! has a positive radius of >convergence.) Is there a known algorithm to approximate the >distribution from the moments? I.e., given real x and positive e, >is there some way we can determine a function c which depends only on >finitely many of the moments (as well as on x and e) such that >|P(-infty, x] - c| < e? The number of moments needed may vary for each >x and e. >I guess this relates to inverse Laplace transform. Unless you can find a good algorithm to compute the Laplace transform for complex arguments, or directly identify the distribution from its moments, you are not going to do well with this approach. Here is what is known; it can be found in any good classical book on the problem of moments, and it is quite old. Given the first 2j moments of a distribution, precise upper and lower bounds on the distribution can be obtained, and they are best possible. The details can be found under the title Chebyshev Inequalities; note the plural. However, the problem is horrible from a computational standpoint, and the rate of convergence is slow; I have verified for the normal distribution it is O(1/sqrt(j)). What you have asked for can only be obtained at points of continuity. There is a means of computing the number of moments needed for a given x and e. If P_i is the i-th orthonormal polynomial,' including i=0, what is needed for the first 2j moments to give the approximation asked for is that sum_0^j P_i(x)^2 > .5/e. The point x must be a point of continuity. >Here is an application I have in mind. I can determine that the >limiting distribution of a process has that of the random variable X, >where >X' = C X + 1. >Here, X' and X have the same distribution, the distribution of C >is known, 0 <= C <= 1, and C and X are independent. Using this >equality, it is easy to determine recursively the moments of X in >terms of the moments of C. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: Recovering a probability distribution from its moments > Unless you can find a good algorithm to compute the Laplace > transform for complex arguments, or directly identify the > distribution from its moments, you are not going to do well > with this approach. See my earlier posting. > Here is what is known; it can be found in any good classical > book on the problem of moments, and it is quite old. Is not in many books, but see the reference I gave. Karl === Subject: JSH: How to see it Ok, here finally is the way to see the result. The secret is not to focus on f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) where you have f^2, but instead to consider something like (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f = (b_1 x + u)(b_2 x + u)(b_3 x + uf) where you do not. Remember my examples like where I had x^2 + (2s_1 + 3s_2)x + s_1 s_2 6 = (x + 2s_1))(x + 3s_2)? Remember how if s_1 and s_2 are irrational then they can't be algebraic integers? Well similarly, if the b's are irrational then b_1 and b_2 cannot be algebraic integers. It's a symmetry thing. BUT, notice that if you now multiply both sides by f^2, something interesting happens: f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) = (b_1 f x + uf)(b_2 f x + uf)(b_3 x + uf) as I've pushed in a symmetry, and b_1 f, and b_2 f, now CAN be algebraic integers. Algebraically, (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f = (b_1 x + u)(b_2 x + u)(b_3 x + uf) is just fine, but the ring of algebraic integers is quirky as well as picky because of that monic requirement. My example of xy=2 applies dead on. Here the factorization is just more complicated, but the result is the same. Mathematically, the work isn't even all that complicated. The result is one of the most powerful in mathematics and I don't know how many sci.math posters figured out a long time ago that I've been right all along. However, there has been a certain dedication from some posters who managed to get on my bad side *before* I had my advanced analysis tools. Obviously, by showing a problem with the previous interpretations of Galois Theory, as well as other results like my prime counting function, I could make good on my promises to...make them accountable. After all, I'm the guy who worked so hard for so many years with some very nasty people who clearly were full of themselves calling me names and being hostile in a very public arena. For them, the worst disaster would be my validation and their comeuppance. Certain posters were in a bad way, so they needed company. They pulled many of you in with them, and you went along because you're not mathematicians, many of you don't seem to even like mathematics--especially algebra--and lying seems to be second nature to many of you as well. My analysis indicates that many of you are in the math field because you've learned that you can lie about mathematics and get away with it, as many people are intimidated by the subject. But you forgot one thing--mathematics is for real. James Harris === Subject: Re: JSH: How to see it > The result is one of the most powerful in mathematics and I don't know > how many sci.math posters figured out a long time ago that I've been > right all along. What's the result? [rest deleted] === Subject: Re: JSH: How to see it > Ok, here finally is the way to see the result. The secret is not to > focus on > f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) = > > (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) > where you have f^2, but instead to consider something like > (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f = > > (b_1 x + u)(b_2 x + u)(b_3 x + uf) Are you saying that if m is an integer then there will be a factorization like this with the b's algebraic integers? That's not obvious to me, can you prove it? [rest deleted] === Subject: Re: JSH: How to see it days. My association with the Department is that of an alumnus. [...] > but instead to consider something like > (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f = > > (b_1 x + u)(b_2 x + u)(b_3 x + uf) >Are you saying that if m is an integer then there will be a >factorization like this with the b's algebraic integers? >That's not obvious to me, can you prove it? Doesn't seem likely; of course, there are solutions for b1, b2, b3 algebraic numbers... Let's see: if u is nonzero, then this amounts to saying that there is a factorization in algebraic integers of fX^3 + BX^2 + C = (x + b_1)(x+b_2)(fx+b_3) where B = -3(-1+mf^2) C = (m^3*f^4 - 3m^2*f^2 + 3m) If gcd(f,3)=1, then the polynomial is primitive; if gcd(f,3)=3, then the polynomial is not primitive. Consider first gcd(f,3). That would mean that -b_1 and -b_2 are algebraic integers which are roots of fX^3 + BX^2 + C, a nonmonic polynomial with integer coefficients. Therefore, the irreducibles of b_1 and b_2 must divide fX^3 + BX^2 + C. There are only two possibilities: b_1 and b_2 are rational integers, or else they are conjugate and their minimal polynomial is a quadratic. In either case, the polynomial fX^3 + BX^2 + C is reducible over Q, and fx+b_3 must be a Q[x]-factor. Which means that b_3 is an integer, prime to f. So if gcd(f,3)=1, then the only time we can factor it over the algebraic integers as given is if the polynomial is reducible over Q. I'm not sure if the converse holds, though... The second case if gcd(f,3)=3. Write f = 3*k. Then we have (x+b_1)(x+b_2)(fx+b_3) = fX^3 + BX^2 + C = 3*(kX^3 + (-1+mf^2)X^2 + (27m^3*k^4 - m^2*f^2 + m)) The polynomial that is left is primitive, since gcd(k, mf^2-1)=1. So the content is 3, hence the product of the contents on the left is also 3. The first two factors are primitive, so gcd(f,b_3) = 3. Write b_3 = 3*c_3, and we have (x+b_1)(x+b_2)3(kx + c_3) = 3*(kX^3 + (-1+mf^2)X^2 + (27m^3*k^4 - m^2*f^2 + m)) from which we get (x+b_1)(x+b_2)(kx + c_3) = kX^3 + (-1+mf^2)X^2 + (27m^3*k^4 - m^2*f^2 + m) If k is not 1, then we are in the situation similar to the previous one: the polynomial must be reducible over Q. If k = 1, then we have (x+b_1)(x+b_2)(x+c_3) = X^3 + (-1+mf^2)X^2 + (27m^3 - m^2*f^2 + m) = X^3 + (-1 + 9m)X^2 + (27m^3 - 9m^2 + m). in which case we can certainly find b_1, b_2, and c_3 which are algebraic integers and satisfy the polynomial. In summary: algebraic integers b_1, b_2, and b_3 exist such that (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f = (b_1 x + u)(b_2 x + u)(b_3 x + uf) only if u=0, f=3, or the polynomial is reducible over Q. The first two conditions are also sufficient. The last one may not be sufficient. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: JSH: How to see it > Ok, here finally is the way to see the result. The secret is not to > focus on > > f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) = > > (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) > > where you have f^2, but instead to consider something like > > > (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f = > > (b_1 x + u)(b_2 x + u)(b_3 x + uf) > > Are you saying that if m is an integer then there will be a > factorization like this with the b's algebraic integers? > That's not obvious to me, can you prove it? > [rest deleted] No. The factorization (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f = (b_1 x + u)(b_2 x + u)(b_3 x + uf) does not in general exist in the ring of algebraic integers. It's like with my xy=2 example in evens, but not as rigid as you can find SOME solutions that are in fact in that ring. Like let f=sqrt(2), m=1, u=1. It's a rather neat result in many ways as I think the naive belief would be that you could always find that factorization in the ring of algebraic integers, but the mathematical reality is that you cannot. Like instead let f=sqrt(5), m=1, u=1. Ring of algebraic integers is quirky, and you need to understand that quirkiness or you'll fall into all kind of traps and false arguments you think are proofs, like the basis for Galois Theory. James Harris === Subject: Re: JSH: How to see it !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~( 5eZ41to5f%E@'ELIi $t^ VcLWP@J5p^rst0+(Ô>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw > The factorization > (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f = > > (b_1 x + u)(b_2 x + u)(b_3 x + uf) > does not in general exist in the ring of algebraic integers. > It's like with my xy=2 example in evens, but not as rigid as you can > find SOME solutions that are in fact in that ring. > Like let f=sqrt(2), m=1, u=1. > It's a rather neat result in many ways as I think the naive belief > would be that you could always find that factorization in the ring > of algebraic integers, but the mathematical reality is that you > cannot. I don't see on what sort of rubbish one would base such a naive belief. For crying out loud, the algebraic integers are _defined_ to be only the roots of monic polynomials with integral coefficients. That's the whole point of them and what distinguishes them from algebraic numbers that can also be roots of non-monic polynomials over the integers. So your neat result is absolutely self-evident and banal once you are talking about algebraic integers in the first place. It's the whole point to algebraic integers, and you don't need to wave around complicated stuff like the above to see that. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: JSH: How to see it > Ok, here finally is the way to see the result. Finally? Does this mean you are no longer going to keep reposting these error-ridden arguments? -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: JSH: How to see it > Ok, here finally is the way to see the result. The secret is not to > focus on > f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) = > > (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) > where you have f^2, but instead to consider something like > (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f = > > (b_1 x + u)(b_2 x + u)(b_3 x + uf) > where you do not. > Remember my examples like where I had > x^2 + (2s_1 + 3s_2)x + s_1 s_2 6 = (x + 2s_1))(x + 3s_2)? > Remember how if s_1 and s_2 are irrational then they can't be > algebraic integers? > Well similarly, if the b's are irrational then b_1 and b_2 cannot be > algebraic integers. > It's a symmetry thing. > BUT, notice that if you now multiply both sides by f^2, something > interesting happens: > f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) = > > (b_1 f x + uf)(b_2 f x + uf)(b_3 x + uf) > as I've pushed in a symmetry, and b_1 f, and b_2 f, now CAN be > algebraic integers. > Algebraically, > (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f = > > (b_1 x + u)(b_2 x + u)(b_3 x + uf) > is just fine, but the ring of algebraic integers is quirky as well as > picky because of that monic requirement. > My example of xy=2 applies dead on. Here the factorization is just > more complicated, but the result is the same. > Mathematically, the work isn't even all that complicated. > The result is one of the most powerful in mathematics and I don't know > how many sci.math posters figured out a long time ago that I've been > right all along. > However, there has been a certain dedication from some posters who > managed to get on my bad side *before* I had my advanced analysis > tools. > Obviously, by showing a problem with the previous interpretations of > Galois Theory, as well as other results like my prime counting > function, I could make good on my promises to...make them accountable. > After all, I'm the guy who worked so hard for so many years with some > very nasty people who clearly were full of themselves calling me names > and being hostile in a very public arena. > For them, the worst disaster would be my validation and their > comeuppance. > Certain posters were in a bad way, so they needed company. > They pulled many of you in with them, and you went along because > you're not mathematicians, many of you don't seem to even like > mathematics--especially algebra--and lying seems to be second nature > to many of you as well. > My analysis indicates that many of you are in the math field because > you've learned that you can lie about mathematics and get away with > it, as many people are intimidated by the subject. > But you forgot one thing--mathematics is for real. You seem a lot crazier lately than usual. === Subject: Optimization Question I am working on a problem of maximizing a convex objective function subject to non-smooth convex constriants. Does anyone know whether the solution is unique, i.e., any local maximum is the global maximum? Any Mike === Subject: conditional Probabilities help The conditional probability defined: P( E | F ) = P( EF ) / P(F), where EF means E intersects F. It is not clear to me how this equation is derived. My book says: ...because we know that F has occurred, it follows that F becomes our new sample space and hence the probability that the event EF occurs will equal the probability of EF relative to the probability of F. I understand how F is the new sample space because we know F occured. However, I don't get the second part of the above sentence--the hence part. === Subject: Re: conditional Probabilities help > The conditional probability defined: > P( E | F ) = P( EF ) / P(F), where EF means E intersects F. > It is not clear to me how this equation is derived. > My book says: ...because we know that F has occurred, it follows that F > becomes our new sample space and hence the probability that the event EF > occurs will equal the probability of EF relative to the probability of F. > I understand how F is the new sample space because we know F occured. > However, I don't get the second part of the above sentence--the hence > part. Draw a Venn Diagram. === Subject: Re: conditional Probabilities help > The conditional probability defined: > P( E | F ) = P( EF ) / P(F), where EF means E intersects F. > It is not clear to me how this equation is derived. P( EF ) = P( F ) * P( E|F ) The probability of E and F is the probability of F, times the probability of E given F. The probability that I would visit this newsgroup and answer your post is the probability that I would visit this newsgroup times the probability that I would answer your post given that I visted the newsgroup. By rearrangement, we have P( E|F ) = P( EF ) / P( F ) === Subject: Re: conditional Probabilities help >P( E | F ) = P( EF ) / P(F), where EF means E intersects F. >It is not clear to me how this equation is derived. It's derived by reading Bayes' Theorem in a book. :-) Seriously, I myself deprive it from an equivalent form that makes more sense to me: P(EF) = P(E|F) * P(F) and then divide left and right by P(F). I say this makes more sense to me because multiplication of probabilities makes sense to me: P(E|F) is the contingent probability E will happen if F has happened P(F) is the probability F happens therefore the product of those is the probability E happens, but since the contingency was that F happened then it's also the probability that both E and F happen. That's not a formal proof, obviously, but I'm not sure how formal you want to get and how far back you want to go when you say derived. -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com A: Maybe because some people are too annoyed by top-posting. Q: Why do I not get an answer to my question(s)? A: Because it messes up the order in which people normally read text. Q: Why is top-posting such a bad thing? === Subject: Re: conditional Probabilities help > The conditional probability defined: > P( E | F ) = P( EF ) / P(F), where EF means E intersects F. > It is not clear to me how this equation is derived. > My book says: ...because we know that F has occurred, it follows that F > becomes our new sample space and hence the probability that the event EF > occurs will equal the probability of EF relative to the probability of F. > I understand how F is the new sample space because we know F occured. > However, I don't get the second part of the above sentence--the hence > part. Let S denote the sample space, and let E and F be subsets of S. Notation: EF denotes the intersection of E and F. The equation P(E | F) = P(EF)/P(F) is not derived; rather, it is the official definition of P( E | F ), which is called the (conditional) probability of E given F. It is still a fair question, though, to ask why it is defined like that. Here is one intuitive way to view it: P(E | F) is used to denote the probability of the event E GIVEN THAT the event F has occurred. Under the explicit assumption that F has already occurred, the sample points (aka, outcomes) in the part of E that is outside of F will not have any contribution to the (further) probability of E; as a result, we need only compute how many sample points there are in the part of E that is also part of F, hence the focus on the intersection of E and F. In this manner, F effectively becomes the restricted sample space (think of it as shrinking the sample space from S down to F). So, speaking tersely and informally, P(E) (assuming F has occurred) = P(E inside F) / P(restricted sample space). In formal notations, P(E | F) = P(EF)/P(F). Loosely speaking, it is the ratio of how many sample points there are in EF over how many sample points there are in the restricted sample space F (if one measures probabilities using the counting measure as it is usually done when the sample space is finite). This should be compared to the usual kind of probabilities, namely, for every event A, P(A | S) = P(AS)/P(S) = P(A) (since AS = S, and P(S)=1). Shedar === Subject: Re: conditional Probabilities help > The conditional probability defined: > P( E | F ) = P( EF ) / P(F), where EF means E intersects F. > It is not clear to me how this equation is derived. > My book says: ...because we know that F has occurred, it follows that F > becomes our new sample space and hence the probability that the event EF > occurs will equal the probability of EF relative to the probability of F. > I understand how F is the new sample space because we know F occured. > However, I don't get the second part of the above sentence--the hence > part. > Let S denote the sample space, and let E and F be subsets of S. > Notation: EF denotes the intersection of E and F. > The equation P(E | F) = P(EF)/P(F) is not derived; rather, it is the > official definition of P( E | F ), which is called the (conditional) > probability of E given F. It is still a fair question, though, to ask why it > is defined like that. Here is one intuitive way to view it: > P(E | F) is used to denote the probability of the event E GIVEN THAT the > event F has occurred. Under the explicit assumption that F has already > occurred, the sample points (aka, outcomes) in the part of E that is > outside of F will not have any contribution to the (further) probability > of E; as a result, we need only compute how many sample points there are in > the part of E that is also part of F, hence the focus on the intersection of > E and F. In this manner, F effectively becomes the restricted sample > space (think of it as shrinking the sample space from S down to F). So, > speaking tersely and informally, > P(E) (assuming F has occurred) = P(E inside F) / P(restricted sample > space). > In formal notations, P(E | F) = P(EF)/P(F). > Loosely speaking, it is the ratio of how many sample points there are in EF > over how many sample points there are in the restricted sample space F (if > one measures probabilities using the counting measure as it is usually > done when the sample space is finite). Ok, I think I'm starting to get it. Let me see if I can articulate it. F is the new sample space. EF can range from the empty set to complete intersection with F (EF = F). The more EF and F overlap the more likely E is to occur and conversely. So the ratio P(EF) / P(F) basically gives us a probability number in [0, 1] to model that. E.g., EF = F means P(EF) / P(F) = 1, which means E certainly occurs, which is obviously true if the event = sample space. If EF intersects half of F then we would get P(EF) / P(F) = 1/2. Although I see that this works, I think what was confusing is why not define it as: P(E | F) = (num sample point in EF) / (num sample points in F) I guess because the definition using the probability ratios turns out to be easier to solve problems with. Is that why or is there a deeper reason? > This should be compared to the usual kind of probabilities, namely, for > every event A, > P(A | S) = P(AS)/P(S) = P(A) (since AS = S, and P(S)=1). > Shedar === Subject: Re: conditional Probabilities help > > > > The conditional probability defined: > > > > P( E | F ) = P( EF ) / P(F), where EF means E intersects F. > > > > It is not clear to me how this equation is derived. > > > > My book says: ...because we know that F has occurred, it follows that F > > becomes our new sample space and hence the probability that the event EF > > occurs will equal the probability of EF relative to the probability of > F. > > > > I understand how F is the new sample space because we know F occured. > > However, I don't get the second part of the above sentence--the hence > > part. > > > > > Let S denote the sample space, and let E and F be subsets of S. > Notation: EF denotes the intersection of E and F. > The equation P(E | F) = P(EF)/P(F) is not derived; rather, it is the > official definition of P( E | F ), which is called the (conditional) > probability of E given F. It is still a fair question, though, to ask why > it > is defined like that. Here is one intuitive way to view it: > P(E | F) is used to denote the probability of the event E GIVEN THAT the > event F has occurred. Under the explicit assumption that F has already > occurred, the sample points (aka, outcomes) in the part of E that is > outside of F will not have any contribution to the (further) probability > of E; as a result, we need only compute how many sample points there are > in > the part of E that is also part of F, hence the focus on the intersection > of > E and F. In this manner, F effectively becomes the restricted sample > space (think of it as shrinking the sample space from S down to F). So, > speaking tersely and informally, > P(E) (assuming F has occurred) = P(E inside F) / P(restricted sample > space). > In formal notations, P(E | F) = P(EF)/P(F). > Loosely speaking, it is the ratio of how many sample points there are in > EF > over how many sample points there are in the restricted sample space F (if > one measures probabilities using the counting measure as it is usually > done when the sample space is finite). > Ok, I think I'm starting to get it. Let me see if I can articulate it. > F is the new sample space. > EF can range from the empty set to complete intersection with F (EF = F). Yes > The more EF and F overlap the more likely E is to occur and conversely. No. I think you meant to say the more E and F overlap ... The converse may not be true, though. > So the ratio P(EF) / P(F) basically gives us a probability number in [0, 1] > to model that. E.g., EF = F means P(EF) / P(F) = 1, (EF = F) IMPLIES P(E | F) = P(EF)/P(F) = 1 but the converse does not necessary hold (don't worry about why the converse may fail at this point; one needs to know some measure theory--specifically, exotic sets called sets with measure zero--before this can be explained). > which means E certainly > occurs, which is obviously true if the event = sample space. If EF > intersects half of F then we would get P(EF) / P(F) = 1/2. The phrase Half of F is a bit vague. One needs to specify how sample points are counted or measured. For a finite sample space, counting the number of sample points would not be a problem. [But food for thought: how would one count sample points if the sample space or the event is uncountable? Don't worry about this too much at this point if you are only taking an introductory (2nd year college) class in probability or statistics.] > Although I see that this works, I think what was confusing is why not define > it as: > P(E | F) = (num sample point in EF) / (num sample points in F) > I guess because the definition using the probability ratios turns out to be > easier to solve problems with. Is that why or is there a deeper reason? Defining it as the n(EF) divided by the n(F) would presuppose that one is using the counting measure to compute probabilities: P(A) = n(A) / n(S), where n(A) is the number of sample points favorable to A, and n(S) is the total number of sample points in S with S nonempty. This, in turn, would presuppose that one is able to count elements in a set, which, in turn, would presuppose that the sets are FINITE. In the case of uncountable sets, the word count would be misleading, and one needs a more sophisticated way of measuring them appropriately in the context of computing probabilities. E.g., randomly pick a number from the open interval S = (-1,1), and let A be the event {t in S | t < 0}. What is the probability of A? One would desire this probability to be 1/2. But observe that both A and S are uncountable, so ordinary counting does not apply. What level is this probability/statistics class that you are taking (2nd year, 3rd year, 4th year university)? Is it calculus based? Shedar === Subject: Re: conditional Probabilities help How does this sound for describing the definition of P(E | F): The closer P(EF) is to P(F) the more likely E occurs because we know F did happen. === Subject: Re: conditional Probabilities help > How does this sound for describing the definition of P(E | F): > The closer P(EF) is to P(F) the more likely E occurs because we know F did > happen. Yes. This idea is right. Your statement above may be interpreted as saying this: For events E and F, if F is nonempty, then as P(EF) approaches P(F), P(E | F) approaches 1. But it is a consequence of the definition as opposed to being a characterization of the definition. By definition, we have: (1) P(E | F) = P(EF)/P(F) Fix a nonempty event F. Since EF is a subset of F, P(EF) <= P(F). As the numerator gets closer to the denominator in equation (1), the fraction on the right-hand-side gets closer to P(F)/P(F), which is 1; therefore P(E | F) does the same. Shedar === Subject: Re: conditional Probabilities help > > > > The conditional probability defined: > > > > P( E | F ) = P( EF ) / P(F), where EF means E intersects F. > > > > It is not clear to me how this equation is derived. > > > > My book says: ...because we know that F has occurred, it follows that > > becomes our new sample space and hence the probability that the event > EF > > occurs will equal the probability of EF relative to the probability of > F. > > > > I understand how F is the new sample space because we know F occured. > > However, I don't get the second part of the above sentence--the > hence > > part. > > > > > > > > Let S denote the sample space, and let E and F be subsets of S. > > > > Notation: EF denotes the intersection of E and F. > > > > The equation P(E | F) = P(EF)/P(F) is not derived; rather, it is the > > official definition of P( E | F ), which is called the (conditional) > > probability of E given F. It is still a fair question, though, to ask > why > it > > is defined like that. Here is one intuitive way to view it: > > > > P(E | F) is used to denote the probability of the event E GIVEN THAT the > > event F has occurred. Under the explicit assumption that F has already > > occurred, the sample points (aka, outcomes) in the part of E that is > > outside of F will not have any contribution to the (further) > probability > > of E; as a result, we need only compute how many sample points there are > in > > the part of E that is also part of F, hence the focus on the > intersection > of > > E and F. In this manner, F effectively becomes the restricted sample > > space (think of it as shrinking the sample space from S down to F). > So, > > speaking tersely and informally, > > > > P(E) (assuming F has occurred) = P(E inside F) / P(restricted > sample > > space). > > > > In formal notations, P(E | F) = P(EF)/P(F). > > > > Loosely speaking, it is the ratio of how many sample points there are in > EF > > over how many sample points there are in the restricted sample space F > (if > > one measures probabilities using the counting measure as it is usually > > done when the sample space is finite). > Ok, I think I'm starting to get it. Let me see if I can articulate it. > F is the new sample space. > EF can range from the empty set to complete intersection with F (EF = F). > Yes > The more EF and F overlap the more likely E is to occur and conversely. > No. I think you meant to say the more E and F overlap ... The converse > may not be true, though. > So the ratio P(EF) / P(F) basically gives us a probability number in [0, > 1] > to model that. E.g., EF = F means P(EF) / P(F) = 1, > (EF = F) IMPLIES P(E | F) = P(EF)/P(F) = 1 > but the converse does not necessary hold (don't worry about why the converse > may fail at this point; one needs to know some measure > theory--specifically, exotic sets called sets with measure zero--before > this can be explained). > which means E certainly > occurs, which is obviously true if the event = sample space. If EF > intersects half of F then we would get P(EF) / P(F) = 1/2. > The phrase Half of F is a bit vague. One needs to specify how sample > points are counted or measured. For a finite sample space, counting the > number of sample points would not be a problem. [But food for thought: how > would one count sample points if the sample space or the event is > uncountable? Don't worry about this too much at this point if you are only > taking an introductory (2nd year college) class in probability or > statistics.] By half I meant, say we roll a pair of dice and the first die = 4, then F = { (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6) } If EF is half, I mean it contains half of the elements in F, e.g., EF = { (4, 1), (4, 2), (4, 3) } But I see what you and Ted mention about uncountability. The book hasn't mentioned uncountable events yet. Well actually it did give an example but it hasn't been given much emphasis yet. I assume it will be discussed later on. > Although I see that this works, I think what was confusing is why not > define > it as: > P(E | F) = (num sample point in EF) / (num sample points in F) > I guess because the definition using the probability ratios turns out to > be > easier to solve problems with. Is that why or is there a deeper reason? > Defining it as the n(EF) divided by the n(F) would presuppose that one is > using the counting measure to compute probabilities: P(A) = n(A) / n(S), > where n(A) is the number of sample points favorable to A, and n(S) is the > total number of sample points in S with S nonempty. This, in turn, would > presuppose that one is able to count elements in a set, which, in turn, > would presuppose that the sets are FINITE. In the case of uncountable sets, > the word count would be misleading, and one needs a more sophisticated way > of measuring them appropriately in the context of computing probabilities. > E.g., randomly pick a number from the open interval S = (-1,1), and let A be > the event {t in S | t < 0}. What is the probability of A? One would desire > this probability to be 1/2. But observe that both A and S are uncountable, > so ordinary counting does not apply. Ok, so basically as you and Ted point out, we use this definition because it works on uncountable sets, and also Ted mentions if we do: P(A) = n(A) / n(S) it does not take into consideration possible biased samples (weighted coin). > What level is this probability/statistics class that you are taking (2nd > year, 3rd year, 4th year university)? Is it calculus based? It is upper division (3rd year+). The course description says: Introductory course emphasizing applications. Bayes theorem, random variables, expectation, variance and covariance, normal distribution and limit theorems. > Shedar === Subject: Re: conditional Probabilities help > > Ok, so basically as you and Ted point out, we use this definition because it > works on uncountable sets, and also Ted mentions if we do: > P(A) = n(A) / n(S) it does not take into consideration possible biased > samples (weighted coin). > Right. The rule of measuring probabilities in a finite sample space via the formula P(A) = n(A) / n(S) is called the counting measure. It presupposes that each of the sample points of S is equally likely to occur. As a more concrete example, if a significantly BIASED coin is tossed, the sample space is S = {H,T}, but we expect each of P({H}) and P({T}) to be significantly different from 1/2 (which would be the ratio obtained for n({H})/n(S) if the coin WERE unbiased). This shows that the counting measure *alone* cannot be used to model probabilities of events when individual sample points are NOT equally like to occur. Shedar === Subject: Re: conditional Probabilities help : Although I see that this works, I think what was confusing is why not : define it as: : P(E | F) = (num sample point in EF) / (num sample points in F) 1. The primary reason is that your definition assumes that all sample points are equally likely to occur. So it wouldn't work for a weighted coin or a weighted die. 2. num sample points in F only makes sense when F is finite. Ted === Subject: Re: JSH: Algebraic integers, a special case In sci.math, James Harris > It turns out that I've stumbled across a very good way to explain how > algebraic integers are limited using > x^2 + (2s_1 + 3s_2)x + s_1 s_2 6 = (x + 2s_1))(x + 3s_2) > where s_1 s_2 = 1, and 2s_1 + 3s_2 is an integer, > as it's VERY simple in many ways, but I can show you some complex > behavior which outlines the problem with the ring of algebraic > integers. > Notice that s_1 = s_2 = 1, just gives you x^2 + 5x + 6, which is nice > and simple, and both s_1 and s_2 are coprime to 2 and 3. > But something weird happens if s_1 and s_2 are irrational as then, s_1 > and s_2 cannot be algebraic integers. Well, lessee. If one postulates two entities s_1 and s_2 satisfying s_1 * s_2 = 1 3*s_1 + 2*s_2 = n then s_1 satisfies 3*s_1^2 - n*s_1 + 2 = 0 s_1^2 - n/3*s_1 + 2/3 = 0 and s_2 satisfies 2 * s_2^2 - n*s_2 + 3 = 0 (which is readily seen if one multiplies the first quadratic by s_2^2) or s_2^2 - (n/2)*s_2 + 3/2 = 0 Neither is an equation that can be satisfied in the general case by an algebraic integer, or an algebraic unit. So why does n=5 result in both s_1 and s_2 being integers? Mostly because s^2 - (5/2)*s + 3/2 = 0 has 1 as a root. (The other root is 3/2.) Since these are so simple, I can use the standard quadratic residue and derive: s_1 = ((n/3) +/- sqrt( (n/3)^2 - 4*(1)*(2/3)) ) / 2 or s_1 = (n +/- sqrt(n^2 - 4*1*2*3)) / 6 = (n +/- sqrt(n^2 - 24)) / 6 I now have a generator. When is n^2 - 24 a square, for arbitrary n? Obviously, if n > 5, then (n+2)^2 - n^2 = 4*n + 4 > 24. If n < -7, then (n+2)^2 - n^2 = 4*n + 4 < -24. Therefore, we need only test the integers {-6,-5,-4,-3,-2,-1,0,1,2,3,4,5}; all others fall under the general case, and cannot produce integers, algebraic or otherwise. There's no point in examining (n+1)^2 - n^2 = 2*n + 1, as it cannot yield 24. n = 6: s_1 = 1 + sqrt(3)/3, s_2 = (n - 3*s_1)/2 = 3/2 - sqrt(3)/2 s_1 = 1 - sqrt(3)/3, s_2 = 3/2 + sqrt(3)/2 s_2 satisfies the equation u^2 + 9*u - (19 + 1/2) = 0 so not quite. n = 5: s_1 = 5/6 + 1/6 = 1, s_2 = 1 s_1 = 5/6 - 1/6 = 2/3, s_2 = 3/2 Your example. n = 4: s_1 = 2/3 + sqrt(-2)/3, s_2 = 1 - sqrt(-2)/2 s_1 = 2/3 - sqrt(-2)/3, s_2 = 1 + sqrt(-2)/2 It turns out -3 + sqrt(-2)/2 is not an algebraic integer, as it satisfies the quadratic u^2 + 6*u + 9 + 1/2 = 0 Oops. I'll leave to the interested reader to verify the rest, though in most cases it's pretty obvious. n = 3: s_1 = 1/2 + sqrt(-15)/6, s_2 = 3/4 - sqrt(-15)/4 s_1 = 1/2 - sqrt(-15)/6, s_2 = 3/4 + sqrt(-15)/4 n = 2: s_1 = 1/3 + sqrt(-5)/3, s_2 = 1/2 - sqrt(-5)/2 s_1 = 1/3 - sqrt(-5)/3, s_2 = 1/2 + sqrt(-5)/2 n = 1: s_1 = 1/6 + sqrt(-23)/6, s_2 = 1/4 - sqrt(-23)/4 s_1 = 1/6 - sqrt(-23)/6, s_2 = 1/4 + sqrt(-23)/4 n = 0: s_1 = sqrt(-6)/3, s_2 = -sqrt(-6)/2 s_1 = -sqrt(-6)/3, s_2 = sqrt(-6)/2 n = -1: s_1 = -1/6 + sqrt(-23)/6, s_2 = -1/4 - sqrt(-23)/4 s_1 = -1/6 - sqrt(-23)/6, s_2 = -1/4 + sqrt(-23)/4 n = -2: s_1 = -1/3 + sqrt(-5)/3, s_2 = -1/2 - sqrt(-5)/2 s_1 = -1/3 - sqrt(-5)/3, s_2 = -1/2 + sqrt(-5)/2 n = -3: s_1 = -1/2 + sqrt(-15)/6, s_2 = -3/4 - sqrt(-15)/4 s_1 = -1/2 - sqrt(-15)/6, s_2 = -3/4 + sqrt(-15)/4 n = -4: s_1 = -2/3 + sqrt(-2)/3, s_2 = -1 - sqrt(-2)/2 s_1 = -2/3 - sqrt(-2)/3, s_2 = -1 + sqrt(-2)/2 n = -5: s_1 = -5/6 + 1/6 = -2/3, s_2 = -3/2 s_1 = -5/6 - 1/6 = -1, s_2 = -1 In your equation, this yields x^2 + (2*s_1 + 3*s_2)*x + s_1*s_2*6 = (x + 2*s_1)*(x + 3*s_2) x^2 + (-5)*x + 6 = (x - 2)* (x - 3) One might consider this a dual. n = -6: s_1 = -1 + sqrt(3)/3, s_2 = -3/2 - sqrt(3)/2 s_1 = -1 - sqrt(3)/3, s_2 = -3/2 + sqrt(3)/2 > That should be dramatic enough for any of you with a modicum of > mathematical expertise to pause, and consider, why. > The simple answer is that if you solve for s_1 or s_2 when they are > irrational you will find them to be roots of non-monic quadratics with > integer coefficients. True. > Some of you may not understand why it matters if the quadratic is > monic or not, but algebraic integers are roots of monic quadratics > with integer coefficients, so imagine that a non-monic polynomial > Ax^2 + Bx + C > has an algebraic integer as a root, where A, B, and C are integers, > and the quadratic is irreducible over Q, that is, it has irrational > roots. > Then as algebraic integers are roots of monic polynomials with integer > coefficients, there would have to exist a quadratic > x^2 + Dx + E > where D and E are integers, which has that same algebraic integer as a > root. > But then you can multiply it times A and subtract from the first one, > which gives you a linear expression: > (B-D)x + C-E = 0 > which would mean that x must be an integer, contradicting with the > root being irrational. Actually, it merely means x must be rational -- but that's good enough here. > (If you wonder, what if the polynomial isn't a quadratic, well if the > monic polynomial were of higher degree than 2, you could divide it by > Ax^2 + Bx + C > which would give you a linear expression which would then have an > integer root, which would contradict with the root being irrational.) > Now back to > x^2 + (2s_1 + 3s_2)x + s_1 s_2 6 = (x + 2s_1))(x + 3s_2) > with s_1 s_2 = 1, and 2s_1 + 3s_2 an integer > as notice that you could put in lots of integers, but not algebraic > integers. > They are excluded. > But then, do you really believe no unit s_1 and s_2 exist? You've already found one of them. (1,1) and (-1,-1) both the only solutions, if one requires s_1 and s_2 to be algebraic integers. Am I missing something here? > Mathematicians for over a hundred years have apparently concluded that > in this case non-unit factors of 2 and 3 are pushed around, such that > s_1 and s_2 if irrational must pull a non-unit factor of 2 from 2 such > that > 3s_2 is now not coprime to 2 > and similarly pull a non-unit factor of 3 from 3 such that > 2s_1 is now not coprime to 3 Well, unless s_1 and s_2 are both algebraic integers, the terminology 2*s_1 is (not) coprime to 2 doesn't mean a whole lot. One might ask for example whether 2/3 is divisible by 7. > but that is a false implication caused by a failure to understand the > ring of algebraic integers as though it MAY occur, mathematically, > it's not the only possibility. > A clue can be seen by trying to disprove that statement. > That is, if you try to prove that s_1 and s_2 must shift factors of 2 > and 3 around you run in a circle, as they may do so, but are not > forced to do so in general. > But you can prove that s_1 and s_2 cannot be algebraic integers if > irrational. See above. It's a bit clumsy, admittedly. > Mathematicians naively took the road of *assuming* that s_1 and s_2, > if they could not be algebraic integers, could not be units, except in > some ring where 2 and 3 are units. > It's an easy mistake to make, and mathematicians made it for over a > hundred years, until I came along and figured out that they were > wrong. > You can see what has happened. > I found a way to rigorously prove the problem with algebraic integers > using basic algebra and put that in a paper I call Advanced Polynomial > Factorization. > I sent it to an electronic math journal, which after nine months said > it would publish. > A group of sci.math posters coordinated an email campaign against it, > in posts on sci.math, and so far got away with getting it censored as > they convinced the editors of Southwest Journal of Pure and Applied > Mathematics to yank my paper. > See http://rattler.cameron.edu/swjpam/vol2-03.html > Notice that I played by the rules, submitted to a journal, and in fact > waited over nine months for them to publish, having stayed in contact > with the editors during that time. > But posters on sci.math, sent false emails, and convinced the editors > in just a day or so, and my paper got yanked. > That's not how math is supposed to work. > Each day that goes by while this injustice remains is an indictment > against modern mathematics, as the obvious implication is that > mathematicians do not wish to acknowledge a correct result which might > be an embarrassment to them. > The Southwest Journal of Pure and Applied Mathematics continues, > without pause, as here's their latest publication: > http://rattler.cameron.edu/swjpam/vol1-04.html > That is also an indictment against math society as it implies that > mathematicians do not adhere to rules. > People break the rules in their society, and nothing happens. > No accountability. > One day, judgement will fall, and no matter how severe it will be, > mathematicians will not be able to cry out against it, as clearly they > have brought it upon themselves. > They are begging the world to do with them as it wills, as they have > no law among themselves. > James Harris Well, congratulations on publishing your paper, anyway. I'm not sure what it all means, though. -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: JSH: Analogy, xy=2, evens In sci.math, James Harris > I came up with an analogy which hopefully should make the mathematics > of the problem with algebraic integers crystal clear: > xy = 2 > and let's say, someone points out that you can have x even, and 2 is > of course even, but y is not, but of course it's not, as with that > scenario y=1, which is NOT even. > Now, with algebraic integers you can have > xy = 2 > where x is an algebraic integer, and of course 2 is, but y is not, but > that doesn't tell you much!!! > All it tells you is that y is not the root of a monic polynomial with > integer coefficients. Actually, not even that. xy=2 can be solved by x = sqrt(2), y = sqrt(2), both algebraic integers. Or x = cbrt(2), y = cbrt(4). Or for that matter x = 2^(m/n), y = 2^( (n-m)/n), where 0 < m < n, m and n arbitrary integers. Or even x = (1 + sqrt(-1)), y = (-1 + sqrt(-1)). In none of these cases does 2 divide one of the factors, even if one relaxes the definition of divides in the obvious fashion. [rest snipped] -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: Magazines for amateur mathematicians by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i8Q3DZY01290; >In addition to Mathematics Magazine, try the other MAA publications, > College Mathematics Journal > American Mathematical Monthly. >also, > Journal of Recreational Mathematics >and a U.K. publication, > Mathematical Gazette >Add your favorites to the list... >-- >G. A. Edgar http:// www.math.ohio-state.ed u/~edgar/ [End of Quote] I personally love Math Horizons (published by Mathematical Association of America), which is written for undergraduate math students, because I can read it without any fear for lack of math knowledge. Besides, it informs a lot of news in undergraduate /graduate math community, like young math stars' achievements. Being known that there are many stars in one's generation really inspires one, doesn't it? It is cheap too :) Well, but the content is not so technical. erdos fan === Subject: High School Math Hello ^o^ I'm a 15 yr. old High School student In the 10th grade and I was wondering if anyone knows of any good websites on math for both instruction (Learning somthing new) and review (pratice. I would like to think of myself a Philomath but sadly I'm more of a Dilatory person :-p ) PS - Please don't abase me === Subject: Re: High School Math > .... > I'm a 15 yr. old High School student In the 10th grade and I was > wondering if anyone knows of any good websites on math for both > instruction (Learning somthing new) and review .... http://www.nrich.maths.org.uk Ken Pledger. === Subject: Re: High School Math > Hello ^o^ > I'm a 15 yr. old High School student In the 10th grade and I was > wondering if anyone knows of any good websites on math for both > instruction (Learning somthing new) and review (pratice. I would like > to think of myself a Philomath but sadly I'm more of a Dilatory person > :-p ) I think most students when away from assignments just work with popular subjects. Here are a few little web pages: http://raphael.math.uic.edu/~jeremy/crypt/math.html http://www.utm.edu/research/primes/mersenne/ http://www.certicom.com/index.php?action=ecc_tutorial,home http://www.utm.edu/research/primes/prove/prove3_1.html http://www.cs.pitt.edu/~kirk/cs1501/notes/rsademo/ http://pascal.developpez.com/compilateurs/tp7/ Also, I wasn't able to link to a FSU p kirby slide show glossary on the Euclid algorithm... === Subject: Re: High School Math sci.math: > I'm a 15 yr. old High School student In the 10th grade and I was >wondering if anyone knows of any good websites on math for both >instruction (Learning somthing new) and review (pratice. I would like >to think of myself a Philomath but sadly I'm more of a Dilatory person >PS - Please don't abase me You've already done a fair job of that yourself. (Perhaps the word you wanted was abuse?) Here's some advice: stop trying to be cute. Instead, just ask questions one at a time in a straightforward way, always showing what you yourself have done to work on them. You'll find plenty of help if you focus more on the math and less on being pseudo-clever. Unless you're just trolling, of course. The more specific your questions, the more useful the help you'll get. Tell me about imaginary numbers isn't a very specific question; I'd like to study imaginary numbers from a background of two years of high-school algebra, but I don't know where to start. Can you suggest an appropriate book? is a specific question. P.S. Standard English capitalization will help too. -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com A: Maybe because some people are too annoyed by top-posting. Q: Why do I not get an answer to my question(s)? A: Because it messes up the order in which people normally read text. Q: Why is top-posting such a bad thing? === Subject: Re: question about induction > I see the domino effect in induction proofs and see why you need a > starting point. Can someone please give me an example of a statement > where we CAN go from assuming that the statement is true for some n to > showing, based on this assumption, that it is also true for the next n > but in fact the statement is not true. Note that I never said that the > statement was true for any n. > All natural numbers are less than 3. > Proof: 1 is less than 3, and 1 + 1 = 2 is less than 3 also. > this is incorrect. > you would need to formulate it like > all natural numbers are les than 3 > assume n > 3 > n+1 is not necessarily less than 3 > so this is a bad example True, though it's mostly because I misread his statement Note that I never said that the statment was true for any n. I guess he meant that the statement might be totally false, but it could also read I never said it was true for ALL n, that is, It might be true for some n and false for others, and you do the induction for those where it is assumed to be true in which case my example makes sense, though only in response to a rather nonsensical question. -- Ryan Reich ryanr@uchicago.edu === Subject: Fake (imaginary) numbers Are their any guides to Fake (imaginary) numbers for a Neophite and 10th grader like me ???? === Subject: Re: Fake (imaginary) numbers X-CompuServe-Customer: Yes X-Coriate: interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Pose: George Cox X-Punge: Micro$oft X-Sanguinate: The MVS Guy X-Terminate: SPA(GIS) X-Tinguish: Mark Griffith X-Treme: C&C,DWS at 08:35 PM, Yahiko_n_Tsubame@yahoo.com (Charuzu ?`???[???[) said: >Are their any guides to Fake (imaginary) Imaginary numbers are not fake. I've seen HS Algera books that discussed them, but with the general decline in standards I'm not sure that you can find one like that any more. You might check whether Another poster suggested Conway, and that's probably a good option. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === Subject: Re: Fake (imaginary) numbers >Are their any guides to Fake (imaginary) numbers for a Neophite and >10th grader like me ???? The complex number system is based on postulating the quantity i, which is a square root of negative one, and expanding the number line into a two-dimensional plane. A complex number z is defined as x + iy, where x and y are both real. Two complex numbers can be added or subtracted by adding or subtracting their components. Multiplication is derived by applying the fact that i is a square root of negative one: (x + yi) * (u + vi) = (xu - yv) + (xv + yu)i. Division is also possible; any complex number may be divided by any complex number except zero. The reciprocal of a complex number x + yi is (x - yi)/(x^2 + y^2), or (x/(x^2+y^2)) - (y/(x^2+y^2))i, so x + yi (xu + yv) + (yu - xv)i ------ = ---------------------- u + vi (u^2 + v^2) and of course dividing a complex number by a real is just dividing both its parts by that real. Complex numbers are very important in calculus. They allow mathematicians to see the complete form of funcitons like sin(x) or log(x). When a function is differentiable, it corresponds to a conformal mapping in the complex plane; this allows many strange-looking map projections having the same property as the Stereographic and the Mercator of small shapes looking good to be constructed. John Savard http://home.ecn.ab.ca/~jsavard/index.html === Subject: Re: Fake (imaginary) numbers > Are their any guides to Fake (imaginary) numbers for a Neophite and > 10th grader like me ???? All abstract numbers are fake in the sense that they are not physical. A so-called imaginary number is no more fake or imanginary than a so-called real number. The terms real, imaginary originated in the 15th and 16th centuries when people did not know as much about numbers and algebraic systems as they do now. As to you question, think of so-called imaginary numbers as vectors on the X,Y plane with the property that a product can be defined for any pair of such vectors with the kind of properties that we expect from a product, to wit: 1. x*y = y*x 2. 0*x = 0 3 x*(y + z) = x*y + x*z 4 if y != 0 then given x, y there is a z such that x = y*z (or z = x/y). let a be the x co-ordinate and b the y co-ordinate. we represent the vector (a,b) as a + i*b where i is the vector (0,1). (a,b) + (c,d) = (a+c, b+d) and (a,b)*(c,d) = (ac - bd, bc = ad). with these definitions of + and * you will find that the system of number a + b*i have the usual nice properties of number and in addition (0,1) = 0 + 1.i has the property (0,1)*(0,1) = (0 - 1, 1*0 + 0*1) = (-1,0) better known a just plain -1. so (0,1) = i squared is -1, just like we want it to be. Complex numbers were invented so that any polynomial with rational co-efficiants would have a root, in particular x^2 + 1 = 0. The imaginary number i = (0,1) is a roolt of this equation. What is the other root? You answer that one. Bob Kolker === Subject: Re: Fake (imaginary) numbers > Are their any guides to Fake (imaginary) numbers for a Neophite and > 10th grader like me ???? >All abstract numbers are fake in the sense that they are not physical. >A so-called imaginary number is no more fake or imanginary than a >so-called real number. The terms real, imaginary originated in the >15th and 16th centuries when people did not know as much about numbers >and algebraic systems as they do now. Just as the number 3 can answer the question how many, real numbers such as pi and the square root of two can answer the question how much, when applied to the length of a wooden beam or a piece of string. It is true that infinitely accurate real numbers are abstract, but there certainly is no guarantee that all physical quantities are rational fractions. Complex numbers off the real line do not correspond to the simple notion of quantity. To view them as being legitimate mathematical constructs, but not really numbers, is not, therefore, totally illegitimate. But it should be noted that for most objects that are counted and measured, -3 of that object is as absurd as 2i of that object. Few people would regard as anything but obscurantist the claim that negative numbers are not numbers. To some objects counted and measured, only integers apply, but to others, the reals are more appropriate. As negative numbers apply well enough to bank balances and temperatures, and they fit well enough into the rules of simple arithmetic, most people today are quite willing to concede them the status of numbers. There are a few physical things that correspond to imaginary numbers. It is very useful to regard a magnetic field as an electrical field of imaginary strength. But this sort of thing is generally the exclusive purview of scientists and engineers. John Savard http://home.ecn.ab.ca/~jsavard/index.html === Subject: Re: Fake (imaginary) numbers <2rorb6F1cu4vdU1@uni-berlin.de> <415810d0.45653210@news.ecn.ab.ca> X-CompuServe-Customer: Yes X-Coriate: interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Pose: George Cox X-Punge: Micro$oft X-Sanguinate: The MVS Guy X-Terminate: SPA(GIS) X-Tinguish: Mark Griffith X-Treme: C&C,DWS >Just as the number 3 can answer the question how many, real numbers >such as pi and the square root of two can answer the question how >much, when applied to the length of a wooden beam or a piece of >string. And i+3 can answer the question how much when applied to electricity, except that the EE write the i as a j. Given that, plus the fact that I can add and multiply them, they sure look real; at least, real enough that I won't grab the live wire. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === Subject: Re: Fake (imaginary) numbers >Just as the number 3 can answer the question how many, real numbers >such as pi and the square root of two can answer the question how >much, when applied to the length of a wooden beam or a piece of >string. > And i+3 can answer the question how much when applied to > electricity, except that the EE write the i as a j. Given that, > plus the fact that I can add and multiply them, they sure look real; > at least, real enough that I won't grab the live wire. EE's don't really write it that way for that reason. Since in most EE's equations dealing with electrcity, j is simply of a translation of i. But in most EE's equations dealy with mirco-electronics, j is orthogonal to i. In most EE's contexts the j=i, means it's a closed circuit carrying Electro-magnetic current, And j=/=i means it's a open circuit carrying Hydro-magnetic current, rather than E-M current. === Subject: Re: Fake (imaginary) numbers sci.math: >Are their any guides to Fake (imaginary) numbers for a Neophite and >10th grader like me ???? The first guide is to stop calling them fake. They are as real as the any number you already know; the name imaginary is a historical artifact. Isaac Asimov said it well long ago, when someone asked him to hand him the square root of minus one pieces of chalk. Hand me a one- half piece of chalk, he said, and I will. So the other fellow broke a piece of chalk in half and handed him one of the pieces. That's one piece of chalk, Asimov replied. Even if you call that sophistry, the other fellow certainly did not break the original chalk precisely in half, so at best he had handed Asimov a .47 piece of chalk or a .54 piece of chalk. -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com A: Maybe because some people are too annoyed by top-posting. Q: Why do I not get an answer to my question(s)? A: Because it messes up the order in which people normally read text. Q: Why is top-posting such a bad thing? === Subject: Re: Fake (imaginary) numbers >Are their any guides to Fake (imaginary) numbers for a Neophite and >10th grader like me ???? Why do you think that imaginary numbers are fake???? Complex numbers and imaginary numbers are just as important as real numbers, with at least three uses in physics (including their importance in quantum mechanics). Here are three realizations of C, the set of complex numbers. The first realization is C = R^2, subject to (a,b) + (c,d) = (a + c, b + d), (a,b) (c,d) = (ac - bd, ad + bc). The real number x is identified with the complex number (x,0). The imaginary number i is identified with the complex number (0,1). It follows that x + iy is identified with (x,y) for all real x and y. The second realization of C is as a subalgebra of the matrix algebra M_2(R). Specifically, the elements of C are those 2 x 2 real matrices of the form ( x y ) ( -y x ) for x and y real. The real number is identified with the 2 x 2 matrix ( x 0 ) ( 0 x ). The imaginary number i is identified with the 2 x 2 matriix ( 0 1 ) ( -1 0 ). So, for x and y real, x + iy is identified with the 2 x 2 matrix ( x y ) ( -y x ). The third realization is C = R[t]/. The real number x is identified with the equivalence class [x] = x + R[t] (t^2+1). The imaginary number i is identified with the equivalence class [t] = t + R[t] (t^2+1). For all real x and y, the complex number x + iy is identified with the equivalence class [x + yt] = x + yt + R[t] (t^2+1). Note in particular that i^2 + 1 = [t^2 + 1] = 0. David ----- === Subject: Re: Fake (imaginary) numbers > The third realization is C = R[t]/. The real number x is > identified with the equivalence class [x] = x + R[t] (t^2+1). > The imaginary number i is identified with the equivalence class > [t] = t + R[t] (t^2+1). For all real x and y, the complex number > x + iy is identified with the equivalence class [x + yt] > = x + yt + R[t] (t^2+1). Note in particular that > i^2 + 1 = [t^2 + 1] = 0. And R(t) a function ? May be You have a link to this representation, or may be, You can give some more details. Hero P.S. There's an representation of the form of all triangles by a 2D-vector, like (3,4) represents the triangle with the coordinates (0,0), (0,1) and (3,4). Multiplication with a number changes size, but not form. Adding a vector moves the triangle (like scrolling) and multiplication rotates and stretch/shrinks it,... === Subject: Re: Fake (imaginary) numbers > The third realization is C = R[t]/. The real number x is > identified with the equivalence class [x] = x + R[t] (t^2+1). > The imaginary number i is identified with the equivalence class The guy is 15 yrs old and a 10th grader ... do you really think he knows what an algebra, or an equivalence class, or an identification is ? === Subject: Re: Fake (imaginary) numbers > The third realization is C = R[t]/. The real number x is > identified with the equivalence class [x] = x + R[t] (t^2+1). > The imaginary number i is identified with the equivalence class >The guy is 15 yrs old and a 10th grader ... do you really think he knows >what an algebra, or an equivalence class, or an identification is ? I note that you deleted the parts where I defined a complex number as an ordered pair of real numbers, and as a 2 x 2 matrix over R. A 15 year old should be able to understand at least one of those. And it is sufficient that he understands one of the alternatives which I presented. The three alternatives were not specifically written for the original poster. They were also written for people who might read the comments out of interest. So long as the original poster understands a complex number as an ordered pair of real numbers, with associated sum and product which I gave, then that is sufficient for his purposes. And I do not see any reason why his young age should stop me from posting what for me is the most intellectually satisfying realization of the complex numbers in terms of the real numbers. David ----- === Subject: Re: Fake (imaginary) numbers > Are their any guides to Fake (imaginary) numbers for a Neophite and > 10th grader like me ???? Indeed, the real ones are just as fake (or as real) as the imaginary ones. All numbers--real or complex, rational or irrational, algebraic or transcendental, finite or transfinite--are figments of our imagination. The important thing to remember is this: they are logical abstractions or extensions of the basic notions of the counting numbers 1, 2, 3. Shedar === Subject: Re: Fake (imaginary) numbers > Are their any guides to Fake (imaginary) numbers for a Neophite and > 10th grader like me ???? > Indeed, the real ones are just as fake (or as real) as the imaginary > ones. All numbers--real or complex, rational or irrational, algebraic or > transcendental, finite or transfinite--are figments of our imagination. The > important thing to remember is this: they are logical abstractions or > extensions of the basic notions of the counting numbers 1, 2, 3. Perhaps if he called them the orthogonal numbers they would not be so dis-tasteful. === Subject: Re: Fake (imaginary) numbers > > Are their any guides to Fake (imaginary) numbers for a Neophite and > > 10th grader like me ???? > Indeed, the real ones are just as fake (or as real) as the imaginary > ones. All numbers--real or complex, rational or irrational, algebraic or > transcendental, finite or transfinite--are figments of our imagination. > The > important thing to remember is this: they are logical abstractions or > extensions of the basic notions of the counting numbers 1, 2, 3. > Perhaps if he called them the orthogonal numbers they would not be so > dis-tasteful. Aye. The word fake has a somewhat distasteful connotation (as it might mean more than just the opposite of real). In fact, I usually find the word real more distasteful in a context where the statement is made with an implicit assumption that if the observer fails to see how a mathematical concept relates directly to the physical world (aka real world), then somehow that failure makes the concept less real (whatever that means). Perhaps an example similar to the following one may help in opening the eyes of an inquisitive student (who may feel that i is more fake than fiveness or eightness). Example Under modulo 13 arithmetic, the congruence equation x^2 = -1 has exactly two solutions (namely 5 and 8). Shedar === Subject: Re: Fake (imaginary) numbers > Perhaps if he called them the orthogonal numbers they would not be so > dis-tasteful. They are generally called complex numbers to get rid of the imaginary label. Bob Kolker === Subject: Re: .9999... = 1 : proof: : x = .9999.... : 10x = 9.9999.... : 9x = 10x - x = 9.9999.... - .9999.... = 9 : x = 1 : .9999.... = 1 Looks a lot like this: x = 1 + 2 + 4 + 8 + ... 2x = 2 + 4 + 8 + 16 + ... x = 2x - x = -1 1 + 2 + 4 + 8 + ... = -1 === Subject: Re: .9999... = 1 > : proof: > : x = .9999.... > : 10x = 9.9999.... > : 9x = 10x - x = 9.9999.... - .9999.... = 9 > : x = 1 > : .9999.... = 1 > Looks a lot like this: > x = 1 + 2 + 4 + 8 + ... > 2x = 2 + 4 + 8 + 16 + ... > x = 2x - x = -1 > 1 + 2 + 4 + 8 + ... = -1 Well, Bandel's implied infinite series (.999) converges, so his subtraction is valid; whereas your series is not. Of course, Bandel's proof is missing the demonstration that his series does converge. === Subject: Re: .9999... = 1 > : proof: > : x = .9999.... > : 10x = 9.9999.... > : 9x = 10x - x = 9.9999.... - .9999.... = 9 > : x = 1 > : .9999.... = 1 > Looks a lot like this: > x = 1 + 2 + 4 + 8 + ... > 2x = 2 + 4 + 8 + 16 + ... > x = 2x - x = -1 > 1 + 2 + 4 + 8 + ... = -1 > Well, Bandel's implied infinite series (.999) converges, so his > subtraction is valid; whereas your series is not. Of course, Bandel's > proof is missing the demonstration that his series does converge. Bingo! That is precisely the point that a number of people have been trying to make. The proof is incomplete without an explanation of what 0.999... means, and a demonstration that the series converges. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: .9999... = 1 > The proof is incomplete without an explanation of what > 0.999... means So what does .999... mean. To be rigorous we define a finite series f as: Let f(N)=sum(i=1 to N) of (9*10^(-i)) where N=1,2,3,... So f(1)=.9, f(2)=.99, f(3)=.999, etc. Than limit N-->infinity f(N) is defined to be the symbol .999... >and a demonstration that the series converges. To prove convergence of .999... one can use a delta-epsilon like proof for series, which goes something like this: Let f(N)=sum(i=1 to N) of (9*10^(-i)) where N=1,2,3,... If limit N-->infinity f(N)=1 than For every x>0 there exists a N>0 such that for every M>N we have |1-f(M)|-log(x) Let g(x)=round x up to integer value, pick N=g(|log(x)|) Done. Talking about .999... Look at: 1-1+.1-.1+.01-.01+... Since, this series converges to 0 (you supply the proof!), We can add nearest neighbors: 0=1 +(-1+.1)+(-.1+.01)+... = 1-.9-.09-.009 . . .= 1-.999...=0 so 1=.999... === Subject: Re: .9999... = 1 > > >proof: > > > >x = .9999.... > >10x = 9.9999.... > >9x = 10x - x = 9.9999.... - .9999.... = 9 > >x = 1 > >.9999.... = 1 > > You weren't paying attention in that other thread. > The same proof shows that > > ...111.0 = -1/9. > > > > ************************ > > David C. Ullrich > >uhm.. i'm surprised to see something as ridiculous as that coming from >you. > Of course you are. None of the mathematicians finds it ridiculous. >well that's not very logical. you basiclaly said this: >a -> b is equivelant to ~a -> ~b I did? Where? >common error.. but no less an error for it's commonness >we're talking about infinitely repeating strings of decimal >digits.. to the right of the decimal point.. which is a completely >different thing from going the other way. > Why is it completely different? > (A: Because 0.111... is a real number and ...111.0 is not. > Q: Why is that? > A: Um, I don't really know, that's just what I was taught. > The only reason one seems more ridiculous to you than the > other is that one looks familiar and the other doesn't. > There's no differennce in the _logic_ of the arguments; > so the fact that the conclusion of the second argument > is ridiculous is proof that there's something missing > from the first. What's missing is a _definition_ of > exactly what 0.111... _means_.) > i made no mention of this >algorithm of proof applying to infinity. > Huh? Neither did I. > ************************ > David C. Ullrich >my algorithm deals with digits going infinitely to the right. i never >claimed that it was a general algorithm for digits goin to the left >and digits going to the right. your counterexample is a >coutnerexample to a generalization that i never made. Huh? We all know one example goes to the left and one to the right, and we all know who said what. This doesn't say anything about why one argument is valid and the other is not. > i'm not talking >about infinitely large numbers. i'm talking about infinitely long >strings of digits that lead nevertheless to a finite value. brush up >on your arithmetic Giggle. How do you _know_ that one leads to a finite value? What does that even mean? ************************ David C. Ullrich === Subject: Re: .9999... = 1 > > > > >proof: > > > >x = .9999.... > >10x = 9.9999.... > >9x = 10x - x = 9.9999.... - .9999.... = 9 > >x = 1 > >.9999.... = 1 > > > > You weren't paying attention in that other thread. > > The same proof shows that > > > > ...111.0 = -1/9. > > > > > > > > ************************ > > > > David C. Ullrich > > > >uhm.. i'm surprised to see something as ridiculous as that coming from > >you. > > Of course you are. None of the mathematicians finds it ridiculous. >well that's not very logical. you basiclaly said this: >a -> b is equivelant to ~a -> ~b > I did? Where? x is a mathematician -> x doesn't find it ridiculous from this you proceeded to you see this is as ridiculous -> you are not a mathematematician that's where >common error.. but no less an error for it's commonness > > >we're talking about infinitely repeating strings of decimal > >digits.. to the right of the decimal point.. which is a completely > >different thing from going the other way. > > Why is it completely different? > > (A: Because 0.111... is a real number and ...111.0 is not. > Q: Why is that? > A: Um, I don't really know, that's just what I was taught. > > The only reason one seems more ridiculous to you than the > other is that one looks familiar and the other doesn't. > There's no differennce in the _logic_ of the arguments; > so the fact that the conclusion of the second argument > is ridiculous is proof that there's something missing > from the first. What's missing is a _definition_ of > exactly what 0.111... _means_.) > > > i made no mention of this > >algorithm of proof applying to infinity. > > Huh? Neither did I. > > > ************************ > > David C. Ullrich >my algorithm deals with digits going infinitely to the right. i never >claimed that it was a general algorithm for digits goin to the left >and digits going to the right. your counterexample is a >coutnerexample to a generalization that i never made. > Huh? We all know one example goes to the left and one to the > right, and we all know who said what. This doesn't say > anything about why one argument is valid and the other > is not. you said that an example of digits going to the left was a counterexample that showed my method to be wrong. i say that it was not a counterexample because the statement i made had nothing to do with instances where the digits proceed to the left. > i'm not talking >about infinitely large numbers. i'm talking about infinitely long >strings of digits that lead nevertheless to a finite value. brush up >on your arithmetic > Giggle. *shudder* > How do you _know_ that one leads to a finite value? ever heard of an epsilon-delta proof? > What does that even mean? it means it's not infinite. double negative. > ************************ > David C. Ullrich === Subject: Re: .9999... = 1 > >well that's not very logical. you basiclaly said this: > > > >a -> b is equivelant to ~a -> ~b > > I did? Where? > x is a mathematician -> x doesn't find it ridiculous a: x is a mathematician b: x doesn't find it ridiculous > from this you proceeded to > you see this is as ridiculous -> you are not a mathematematician > that's where ~a: x is not a mathematician ~b: x finds it ridiculous. The statement above is ~b -> ~a. - Randy === Subject: Re: .9999... = 1 > > > > >proof: > > > >x = .9999.... > >10x = 9.9999.... > >9x = 10x - x = 9.9999.... - .9999.... = 9 > >x = 1 > >.9999.... = 1 > > > > You weren't paying attention in that other thread. > > The same proof shows that > > > > ...111.0 = -1/9. > > > > > > > > ************************ > > > > David C. Ullrich > > > >uhm.. i'm surprised to see something as ridiculous as that coming from > >you. > > Of course you are. None of the mathematicians finds it ridiculous. > >well that's not very logical. you basiclaly said this: > >a -> b is equivelant to ~a -> ~b > I did? Where? >x is a mathematician -> x doesn't find it ridiculous >from this you proceeded to >you see this is as ridiculous -> you are not a mathematematician >that's where Hint: you're making a fool of yourself. At least twice. First, I didn't say what you say I said. Second, the common error you accuse me of here is not an error at all, it's a perfectly valid logical step. a -> b _is_ equivalent to ~b -> ~a. I mean duh. Confusing an implication with its converse (which you're doing here, saying something about a -> b and then when asked about it saying something instead about ~b -> ~a) _is_ a common error. But not among people who expect to be taken seriously as mathematicians. >common error.. but no less an error for it's commonness > > >we're talking about infinitely repeating strings of decimal > >digits.. to the right of the decimal point.. which is a completely > >different thing from going the other way. > > Why is it completely different? > > (A: Because 0.111... is a real number and ...111.0 is not. > Q: Why is that? > A: Um, I don't really know, that's just what I was taught. > > The only reason one seems more ridiculous to you than the > other is that one looks familiar and the other doesn't. > There's no differennce in the _logic_ of the arguments; > so the fact that the conclusion of the second argument > is ridiculous is proof that there's something missing > from the first. What's missing is a _definition_ of > exactly what 0.111... _means_.) > > > i made no mention of this > >algorithm of proof applying to infinity. > > Huh? Neither did I. > > > ************************ > > David C. Ullrich > >my algorithm deals with digits going infinitely to the right. i never >claimed that it was a general algorithm for digits goin to the left >and digits going to the right. your counterexample is a >coutnerexample to a generalization that i never made. > Huh? We all know one example goes to the left and one to the > right, and we all know who said what. This doesn't say > anything about why one argument is valid and the other > is not. >you said that an example of digits going to the left was a >counterexample that showed my method to be wrong. i say that it was >not a counterexample because the statement i made had nothing to do >with instances where the digits proceed to the left. I know you said that. You're wrong - you need to show a difference in the logical structure of the two arguments. > i'm not talking >about infinitely large numbers. i'm talking about infinitely long >strings of digits that lead nevertheless to a finite value. brush up >on your arithmetic > Giggle. >*shudder* > How do you _know_ that one leads to a finite value? >ever heard of an epsilon-delta proof? Uh, yes. That's exactly what's missing from what you claim is a proof that 0.111...=1/9. That's why what you posted is not a proof. _That_ is the difference between 0.111...=1/9 and ...111 = -1/9; in the second case there is no epsilon-delta proof available. > What does that even mean? >it means it's not infinite. double negative. > ************************ > David C. Ullrich ************************ David C. Ullrich === Subject: Re: .9999... = 1 > > > > > > > >proof: > > > > > >x = .9999.... > > >10x = 9.9999.... > > >9x = 10x - x = 9.9999.... - .9999.... = 9 > > >x = 1 > > >.9999.... = 1 > > > > You weren't paying attention in that other thread. > > The same proof shows that > > > > ...111.0 = -1/9. > > > > > > > > ************************ > > > > David C. Ullrich > > > >uhm.. i'm surprised to see something as ridiculous as that coming from > >you. > > > > Of course you are. None of the mathematicians finds it ridiculous. > > > >well that's not very logical. you basiclaly said this: > > > >a -> b is equivelant to ~a -> ~b > > I did? Where? >x is a mathematician -> x doesn't find it ridiculous >from this you proceeded to >you see this is as ridiculous -> you are not a mathematematician >that's where > Hint: you're making a fool of yourself. At least twice. > First, I didn't say what you say I said. Second, the > common error you accuse me of here is not an error > at all, it's a perfectly valid logical step. a -> b > _is_ equivalent to ~b -> ~a. > I mean duh. Confusing an implication with its converse > (which you're doing here, saying something about a -> b > and then when asked about it saying something instead > about ~b -> ~a) _is_ a common error. But not among > people who expect to be taken seriously as mathematicians. > >common error.. but no less an error for it's commonness > > > >we're talking about infinitely repeating strings of decimal > >digits.. to the right of the decimal point.. which is a completely > >different thing from going the other way. > > > > Why is it completely different? > > > > (A: Because 0.111... is a real number and ...111.0 is not. > > Q: Why is that? > > A: Um, I don't really know, that's just what I was taught. > > > > The only reason one seems more ridiculous to you than the > > other is that one looks familiar and the other doesn't. > > There's no differennce in the _logic_ of the arguments; > > so the fact that the conclusion of the second argument > > is ridiculous is proof that there's something missing > > from the first. What's missing is a _definition_ of > > exactly what 0.111... _means_.) > > > > i made no mention of this > >algorithm of proof applying to infinity. > > > > Huh? Neither did I. > > > > > > ************************ > > > > David C. Ullrich > > > >my algorithm deals with digits going infinitely to the right. i never > >claimed that it was a general algorithm for digits goin to the left > >and digits going to the right. your counterexample is a > >coutnerexample to a generalization that i never made. > > Huh? We all know one example goes to the left and one to the > right, and we all know who said what. This doesn't say > anything about why one argument is valid and the other > is not. > >you said that an example of digits going to the left was a >counterexample that showed my method to be wrong. i say that it was >not a counterexample because the statement i made had nothing to do >with instances where the digits proceed to the left. > I know you said that. You're wrong - you need to show a difference > in the logical structure of the two arguments. > > i'm not talking > >about infinitely large numbers. i'm talking about infinitely long > >strings of digits that lead nevertheless to a finite value. brush up > >on your arithmetic > > Giggle. > *shudder* > How do you _know_ that one leads to a finite value? >ever heard of an epsilon-delta proof? > Uh, yes. That's exactly what's missing from what you claim > is a proof that 0.111...=1/9. That's why what you posted > is not a proof. > _That_ is the difference between 0.111...=1/9 and ...111 = -1/9; > in the second case there is no epsilon-delta proof available. > What does that even mean? >it means it's not infinite. double negative. > > ************************ > > David C. Ullrich > ************************ > David C. Ullrich read my formulation of YOUR logical error more carefully what i said was this: you implied this: a->b implies ~a -> ~b (i may have phrased the ~a -> ~b part as: b -> a which you know is equivelant and likewise NOT implied by a->b) this is a logical fallacy i am perfectly familiar with the fact that (a->b) -> (~b -> ~a) trust me. i'm at least as comfortable with logic as you are here is a difference in the logical structure of the two arguments: one has digits leading to the left of a decimal point. the other to the right. as you said, an epsilon-delta proof is unavailable for the former, which basically shatters your own argument that your suggestion of the case where digits lead to the left of the decimal point shatters my own argument. === Subject: Re: .9999... = 1 > > > > > > > >proof: > > > > > >x = .9999.... > > >10x = 9.9999.... > > >9x = 10x - x = 9.9999.... - .9999.... = 9 > > >x = 1 > > >.9999.... = 1 > > > > You weren't paying attention in that other thread. > > The same proof shows that > > > > ...111.0 = -1/9. > > > > > > > > ************************ > > > > David C. Ullrich > > > >uhm.. i'm surprised to see something as ridiculous as that coming from > >you. > > > > Of course you are. None of the mathematicians finds it ridiculous. > > > >well that's not very logical. you basiclaly said this: > > > >a -> b is equivelant to ~a -> ~b > > I did? Where? >x is a mathematician -> x doesn't find it ridiculous >from this you proceeded to >you see this is as ridiculous -> you are not a mathematematician >that's where > Hint: you're making a fool of yourself. At least twice. > First, I didn't say what you say I said. Second, the > common error you accuse me of here is not an error > at all, it's a perfectly valid logical step. a -> b > _is_ equivalent to ~b -> ~a. > I mean duh. Confusing an implication with its converse > (which you're doing here, saying something about a -> b > and then when asked about it saying something instead > about ~b -> ~a) _is_ a common error. But not among > people who expect to be taken seriously as mathematicians. > >common error.. but no less an error for it's commonness > > > >we're talking about infinitely repeating strings of decimal > >digits.. to the right of the decimal point.. which is a completely > >different thing from going the other way. > > > > Why is it completely different? > > > > (A: Because 0.111... is a real number and ...111.0 is not. > > Q: Why is that? > > A: Um, I don't really know, that's just what I was taught. > > > > The only reason one seems more ridiculous to you than the > > other is that one looks familiar and the other doesn't. > > There's no differennce in the _logic_ of the arguments; > > so the fact that the conclusion of the second argument > > is ridiculous is proof that there's something missing > > from the first. What's missing is a _definition_ of > > exactly what 0.111... _means_.) > > > > i made no mention of this > >algorithm of proof applying to infinity. > > > > Huh? Neither did I. > > > > > > ************************ > > > > David C. Ullrich > > > >my algorithm deals with digits going infinitely to the right. i never > >claimed that it was a general algorithm for digits goin to the left > >and digits going to the right. your counterexample is a > >coutnerexample to a generalization that i never made. > > Huh? We all know one example goes to the left and one to the > right, and we all know who said what. This doesn't say > anything about why one argument is valid and the other > is not. > >you said that an example of digits going to the left was a >counterexample that showed my method to be wrong. i say that it was >not a counterexample because the statement i made had nothing to do >with instances where the digits proceed to the left. > I know you said that. You're wrong - you need to show a difference > in the logical structure of the two arguments. > > i'm not talking > >about infinitely large numbers. i'm talking about infinitely long > >strings of digits that lead nevertheless to a finite value. brush up > >on your arithmetic > > Giggle. > *shudder* > How do you _know_ that one leads to a finite value? >ever heard of an epsilon-delta proof? > Uh, yes. That's exactly what's missing from what you claim > is a proof that 0.111...=1/9. That's why what you posted > is not a proof. > _That_ is the difference between 0.111...=1/9 and ...111 = -1/9; > in the second case there is no epsilon-delta proof available. > What does that even mean? >it means it's not infinite. double negative. > > > ************************ > > David C. Ullrich > ************************ > David C. Ullrich >read my formulation of YOUR logical error more carefully You should _really_ give this up. You're wrong about more or less all of it, and that's clear to everyone reading this. >what i said was this: >you implied this: No, you said that I _said_ that. I didn't. Let's suppose for the sake of argument that I did imply it: >a->b implies ~a -> ~b No, I didn't imply that. If I implied something like that it was a -> b implies ~b -> ~a. Claiming that I implied the other doesn't make it so, and is not going to convince anyone of anything (except possibly that you're either intentionally lying or unable to read English properly.) >(i may have phrased the ~a -> ~b part as: b -> a which you know is >equivelant and likewise NOT implied by a->b) >this is a logical fallacy >i am perfectly familiar with the fact that (a->b) -> (~b -> ~a) trust >me. i'm at least as comfortable with logic as you are >here is a difference in the logical structure of the two arguments: >one has digits leading to the left of a decimal point. the other to >the right. No, that's not a difference in the logical structure of the two. >as you said, an epsilon-delta proof is unavailable for the former, >which basically shatters your own argument that your suggestion of the >case where digits lead to the left of the decimal point shatters my >own argument. Giggle. You didn't _include_ that epsilon-delta proof. Yes, it's ************************ David C. Ullrich === Subject: Re: .9999... = 1 > > > > > > > > > > >proof: > > > > > >x = .9999.... > > >10x = 9.9999.... > > >9x = 10x - x = 9.9999.... - .9999.... = 9 > > >x = 1 > > >.9999.... = 1 > > > > > > You weren't paying attention in that other thread. > > > The same proof shows that > > > > > > ...111.0 = -1/9. > > > > > > > > > > > > ************************ > > > > > > David C. Ullrich > > > > > >uhm.. i'm surprised to see something as ridiculous as that coming from > > >you. > > > > Of course you are. None of the mathematicians finds it ridiculous. > > > >well that's not very logical. you basiclaly said this: > > > >a -> b is equivelant to ~a -> ~b > > > > I did? Where? > >x is a mathematician -> x doesn't find it ridiculous > >from this you proceeded to > >you see this is as ridiculous -> you are not a mathematematician > >that's where > > Hint: you're making a fool of yourself. At least twice. > First, I didn't say what you say I said. Second, the > common error you accuse me of here is not an error > at all, it's a perfectly valid logical step. a -> b > _is_ equivalent to ~b -> ~a. > > I mean duh. Confusing an implication with its converse > (which you're doing here, saying something about a -> b > and then when asked about it saying something instead > about ~b -> ~a) _is_ a common error. But not among > people who expect to be taken seriously as mathematicians. > > >common error.. but no less an error for it's commonness > > > > >we're talking about infinitely repeating strings of decimal > > >digits.. to the right of the decimal point.. which is a completely > > >different thing from going the other way. > > > > Why is it completely different? > > > > (A: Because 0.111... is a real number and ...111.0 is not. > > Q: Why is that? > > A: Um, I don't really know, that's just what I was taught. > > > > The only reason one seems more ridiculous to you than the > > other is that one looks familiar and the other doesn't. > > There's no differennce in the _logic_ of the arguments; > > so the fact that the conclusion of the second argument > > is ridiculous is proof that there's something missing > > from the first. What's missing is a _definition_ of > > exactly what 0.111... _means_.) > > > > > i made no mention of this > > >algorithm of proof applying to infinity. > > > > Huh? Neither did I. > > > > > > ************************ > > > > David C. Ullrich > > > >my algorithm deals with digits going infinitely to the right. i never > >claimed that it was a general algorithm for digits goin to the left > >and digits going to the right. your counterexample is a > >coutnerexample to a generalization that i never made. > > > > Huh? We all know one example goes to the left and one to the > > right, and we all know who said what. This doesn't say > > anything about why one argument is valid and the other > > is not. > > > >you said that an example of digits going to the left was a > >counterexample that showed my method to be wrong. i say that it was > >not a counterexample because the statement i made had nothing to do > >with instances where the digits proceed to the left. > > I know you said that. You're wrong - you need to show a difference > in the logical structure of the two arguments. > > > i'm not talking > >about infinitely large numbers. i'm talking about infinitely long > >strings of digits that lead nevertheless to a finite value. brush up > >on your arithmetic > > > > Giggle. > *shudder* > > How do you _know_ that one leads to a finite value? > >ever heard of an epsilon-delta proof? > > Uh, yes. That's exactly what's missing from what you claim > is a proof that 0.111...=1/9. That's why what you posted > is not a proof. > > _That_ is the difference between 0.111...=1/9 and ...111 = -1/9; > in the second case there is no epsilon-delta proof available. > > > What does that even mean? > >it means it's not infinite. double negative. > > > > > > ************************ > > > > David C. Ullrich > > > ************************ > > David C. Ullrich >read my formulation of YOUR logical error more carefully > You should _really_ give this up. You're wrong about > more or less all of it, and that's clear to everyone > reading this. >what i said was this: >you implied this: > No, you said that I _said_ that. I didn't. Let's > suppose for the sake of argument that I did imply > it: >a->b implies ~a -> ~b > No, I didn't imply that. If I implied something like > that it was a -> b implies ~b -> ~a. Claiming that I > implied the other doesn't make it so, and is not > going to convince anyone of anything (except possibly > that you're either intentionally lying or unable to > read English properly.) >(i may have phrased the ~a -> ~b part as: b -> a which you know is >equivelant and likewise NOT implied by a->b) >this is a logical fallacy >i am perfectly familiar with the fact that (a->b) -> (~b -> ~a) trust >me. i'm at least as comfortable with logic as you are >here is a difference in the logical structure of the two arguments: >one has digits leading to the left of a decimal point. the other to >the right. > No, that's not a difference in the logical structure of the two. >as you said, an epsilon-delta proof is unavailable for the former, >which basically shatters your own argument that your suggestion of the >case where digits lead to the left of the decimal point shatters my >own argument. > Giggle. You didn't _include_ that epsilon-delta proof. Yes, it's > ************************ > David C. Ullrich me: uhm.. i'm surprised to see something as ridiculous as that coming from you. you: Of course you are. None of the mathematicians finds it ridiculous. logic: i made a statement which we can just assume to be true: i, being what i am, am X (i am X) you made an implication: because, those who are mathematicians are ~X, you are X or more formally: (those who are mathematicians are ~X) -> (you are X) and obviously what you were implying was that I am not a mathematician. so look at the logic: you are saying that since i am not a mathematician. then the implication: (those who are mathematicians are ~X) reverses to: (those who are not mathematicians are X) which is a converse error. i'm a bit disgusted that i had to point that out in so much detail to you. you continue to lower your esteem. === Subject: Re: .9999... = 1 > >uhm.. i'm surprised to see something as ridiculous as that coming from > >you. > > Of course you are. None of the mathematicians finds it ridiculous. > >well that's not very logical. you basiclaly said this: > >a -> b is equivelant to ~a -> ~b > I did? Where? > x is a mathematician -> x doesn't find it ridiculous > from this you proceeded to > you see this is as ridiculous -> you are not a mathematematician > that's where Oh? Do you know the difference between an inverse and a contrapositive? Looks like you made the very type of mistake that you accused Ullrich of. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: When does this Cantor's (conversation) list end? X-RFC2646: Original > could you do us all a favor and actually study this before you start > thinking of answers. Learn what Cantor really Ôsaid' and what it > means, learn what the reals are, learn how to present a proof. If > after this you still think Cantor was wrong, present a proof that he > is. > It is *obvious* now that he is trolling. Look at all the tools he suddenly > master when some objection arise. Look at the careful way he answers only > the points where some of us make slight mistakes (not at all obvious), > while > pretending ...333 is an integer, or Peano (spelled Peanon, for god's > sake) > use the word infinite in the axioms, etc. etc. Also, he is progressively > shifting his arguments, and in the process using * all* the usual cranks > ideas (last one being the famous bijection 1->1, 12->0.21, 1234-> > 0.4321,... ), while most cranks are only using one or two... I agree with you. Michael === Subject: Re: Robyn Ellen Neufeldt - August 11th 1976 > N E U F E L D T > 14 5 21 6 5 12 4 20 = 87 > Of course, these findings (if you can call them that) only work in base 10 > and using the English translation of the bible. > Would the results be the same using the Hebrew version I wonder? Somehow I > doubt it! In any language, Daryl S. Kabatoff is still a ing kook. :O| === Subject: Re: God=G_uv proves 40k B.C. Creation > There you go! 3 eigenvectors in the beginning you said: > > [Hammond] > > Any way.... ALL 13 of these symmetry axes show up > > in Psychometric data at the 2nd order (Cattell 1975). > > And the REASON for it, is that the human brain is actually > > CUBIC. > > [Hammond] > [Hammond] > Hey..... you've lost the argument and now you're fishing around for > some semantic or typographical error to home in on. This is why we are having this discussion, and then you were adamant that they are eigenvectors, and insulted me for not understanding such a trivial matter. > In THURSTONE'S BOX PROBLEM there are 3-eigenvectors > which define the 3D Psychometry space Ahaa! A frog just jumped from your mouth. We were (at present at least) discussing the Thurstones box problem. Bringing in the psychometry space reveals why this stone wall is here: in your mind this is strongy associated with the real space causes psychometry space and stuff and I do not want to turn that rock. > and the other 17 measurements > are VECTORS when plotted in that space, and the other 17 vectors > ALL fall and ONLY FALL, on the 13 symmetry axes of the CUBE > defined in ENP space by the eigenvectors ENP which are the edges of the > BOX. A 20x20 matrix is not large enough to support 13 eigenvectors. Even if there were more vectors these would not be eigenvectors, as 2X+2Y is a linear combination of X and Y. > In the case of the human brain however, the Personality tests used have > hundreds of items and ALL 13 SYMMETRY AXES are extracted as > (orthogonal) eigenvectors. What!? (see the rest of the mail) > When they are rotated (by automated computer > program) to SIMPLE STRUCTURE it is found that they fall on the 13 > symmetry axes of a CUBE. > QED Rotated? You mean you first get some eigenvectors, and if they do not please you you change them. > > OK, this means that you can draw a 3-dimensional space and > > plot all 20 vectors in it. Of course you can rotate the coordinate > > system to any place you want.... but the first thing you notice > > is that there are 3-clusters of vectors in 3-orthogonal directions. > > So you rotate the coordinate system to there (rotation to simple > > structure). When you do this, you find that all the other vectors fall > > at 45 degree angles between two of the 3-eigenvectors (axes) > > and the rest of them fall midway between all 3 axes. > > This is physically interpreted to mean that the 3-eigenvectors > > are the edges of the Box, the cluster that falls 45-degrees > > between any two of them are the edge to edge symmetry axes > > of the Box, and the ones that fall midway between > > all 3 eigenvectors are the body diagonals of the Box. > > Hence you get 3+6+4 symmetry axes of the Box. > > Thurstone actually draws this 3D plot on pp 134-135. > Yes, but again. You claimed that these axes are eigenvectors. Do you > want withdraw that claim? > [Hammond] > No... and quit using pedantic academic debating terms with me, > this isn't a in academic debate.... and I don't talk to anyone who > uses such juvenile and asinine language. If you can't talk plain English > get the fcuk outta here. In this same mail you say both that thay are eigenvectors and are not. Which is it? Mark an X to one of the __'s below: I George Hammond believe that in the Thurstones Box Problem the vector corresponding to the measurement 2X+2Y is ___ a) an eigenvector of the correlation matrix. ___ b) is not an eigenvector of the correlation matrix. > [Hammond] > Eigenvectors are by definition ALWAYS ORTHOGONAL. Therefore Yes! Excellent! 2X+2Y is not orthogonal to X and Y, ergo cannot be an eigenvector, no matter now many measurements you add. > the 3 edges of the Box will always show up as the first largest eigenvalues > and will correspond to 2 edges of a box in Psychometry space. > Look, I've told you once before.... it doesn't MATTER what the actual value > of the other correlations with ENP are......... fact is 2x+2y has to > correlate > EQUALLY with either x or y, and xyz has to correlate EQUALLY with x > and y and z.... DOESN'T MATTER WHAT THE CORRELATION > ACTUALLY IS NUMERICALLY! So, no matter how things are algebraically, you are right? === Subject: Re: God=G_uv proves 40k B.C. Creation > > > > There you go! 3 eigenvectors in the beginning you said: > > > > [Hammond] > > Any way.... ALL 13 of these symmetry axes show up > > in Psychometric data at the 2nd order (Cattell 1975). > > And the REASON for it, is that the human brain is actually > > CUBIC. > > [Hammond] > > > [Hammond] > Hey..... you've lost the argument and now you're fishing around for > some semantic or typographical error to home in on. > This is why we are having this discussion, and then you were adamant > that they are eigenvectors, and insulted me for not understanding such > a trivial matter. > In THURSTONE'S BOX PROBLEM there are 3-eigenvectors > which define the 3D Psychometry space > Ahaa! A frog just jumped from your mouth. We were (at present at > least) discussing the Thurstones box problem. Bringing in the > psychometry space reveals why this stone wall is here: in your mind > this is strongy associated with the real space causes psychometry > space and stuff and I do not want to turn that rock. [Hammond] You're not making any sense.... the ARE 13-EIGENVECTORS in Psychometry, they are 3-eigenvectors plus 10 vectors in thurstone's Box Problem. For chrissakes Thurstone's Box Problem was made up before the invention of computers and they couldn't factor a 50x50 matrix which is needed to support 13 eigenvectors... the symmetry ases are composite factors in Thurstone's Box Problem, nevertheless they ARE THERE and appear in the graphical reconstruction of the BOX in PSYCHOMETRY SPACE. > and the other 17 measurements > are VECTORS when plotted in that space, and the other 17 vectors > ALL fall and ONLY FALL, on the 13 symmetry axes of the CUBE > defined in ENP space by the eigenvectors ENP which are the edges of the > BOX. A 20x20 matrix is not large enough to support 13 eigenvectors. > Even if there were more vectors these would not be eigenvectors, as > 2X+2Y is a linear combination of X and Y. [Hammond] It's a composite factor. lying symmetrically half way between X and Y. > In the case of the human brain however, the Personality tests used have > hundreds of items and ALL 13 SYMMETRY AXES are extracted as > (orthogonal) eigenvectors. > What!? (see the rest of the mail) > When they are rotated (by automated computer > program) to SIMPLE STRUCTURE it is found that they fall on the 13 > symmetry axes of a CUBE. > QED > Rotated? You mean you first get some eigenvectors, and if they do not > please you you change them. [Hammond] that's an incompetent statement by someone who DOES NOT KNOW what rotation to simple structure means.... and every Factor Analyst does. the rotation is NOT ARBITRARY and is doen AUTOMATICALLY by the computer using a lest squares hyperplane count maximization. The computer programs to do this (such as PROMAX) are famous in the industry and are built into modern statistical packages such as SPXX which is used daily by THOUSANDS of scientists world wide today. You don't know what you are talking about. > > > > OK, this means that you can draw a 3-dimensional space and > > plot all 20 vectors in it. Of course you can rotate the coordinate > > system to any place you want.... but the first thing you notice > > is that there are 3-clusters of vectors in 3-orthogonal directions. > > So you rotate the coordinate system to there (rotation to simple > > structure). When you do this, you find that all the other vectors fall > > at 45 degree angles between two of the 3-eigenvectors (axes) > > and the rest of them fall midway between all 3 axes. > > This is physically interpreted to mean that the 3-eigenvectors > > are the edges of the Box, the cluster that falls 45-degrees > > between any two of them are the edge to edge symmetry axes > > of the Box, and the ones that fall midway between > > all 3 eigenvectors are the body diagonals of the Box. > > Hence you get 3+6+4 symmetry axes of the Box. > > Thurstone actually draws this 3D plot on pp 134-135. > > > > Yes, but again. You claimed that these axes are eigenvectors. Do you > > want withdraw that claim? > [Hammond] > No... and quit using pedantic academic debating terms with me, > this isn't a in academic debate.... and I don't talk to anyone who > uses such juvenile and asinine language. If you can't talk plain English > get the fcuk outta here. SCIENTIFIC PROOF OF GOD WEBSITE 1st mirror site: http://geocities.com/scientific_proof_of_god 2nd mirror site: http://proof-of-god.freewebsitehosting.com new site (under construction): http://home.comcast.net/~proof-of-god === Subject: Re: God=G_uv proves 40k B.C. Creation > > > > There you go! 3 eigenvectors in the beginning you said: > > > > > [Hammond] > > > Any way.... ALL 13 of these symmetry axes show up > > > in Psychometric data at the 2nd order (Cattell 1975). > > > And the REASON for it, is that the human brain is actually > > > CUBIC. > > > [Hammond] > > > > > > [Hammond] > > Hey..... you've lost the argument and now you're fishing around for > > some semantic or typographical error to home in on. > This is why we are having this discussion, and then you were adamant > that they are eigenvectors, and insulted me for not understanding such > a trivial matter. > > In THURSTONE'S BOX PROBLEM there are 3-eigenvectors > > which define the 3D Psychometry space > Ahaa! A frog just jumped from your mouth. We were (at present at > least) discussing the Thurstones box problem. Bringing in the > psychometry space reveals why this stone wall is here: in your mind > this is strongy associated with the real space causes psychometry > space and stuff and I do not want to turn that rock. > [Hammond] > You're not making any sense.... the ARE 13-EIGENVECTORS > in Psychometry, they are 3-eigenvectors plus 10 vectors in > thurstone's Box Problem. For chrissakes Thurstone's Box Problem > was made up before the invention of computers and they couldn't > factor a 50x50 matrix which is needed to support 13 eigenvectors... > the symmetry ases are composite factors in Thurstone's Box Problem, > nevertheless they ARE THERE and appear in the graphical reconstruction > of the BOX in PSYCHOMETRY SPACE. So when the algebra does not match up, you move on to a more complex situation where inconvenient rules of algebra and logic are not in the way of your transcendential thinking. In the questionnaire: I George Hammond believe that in the Thurstones Box Problem the vector corresponding to the measurement 2X+2Y, ___ a) is an eigenvector of the correlation matrix. ___ b) is not an eigenvector of the correlation matrix. _X_ c) I get to choose on case to case basis. > > and the other 17 measurements > > are VECTORS when plotted in that space, and the other 17 vectors > > ALL fall and ONLY FALL, on the 13 symmetry axes of the CUBE > > defined in ENP space by the eigenvectors ENP which are the edges of the > > BOX. A 20x20 matrix is not large enough to support 13 eigenvectors. > Even if there were more vectors these would not be eigenvectors, as > 2X+2Y is a linear combination of X and Y. > [Hammond] > It's a composite factor. lying symmetrically half way between > X and Y. A composite eigenvector..... Sure George. Ingenious! We'll name it after you. > > In the case of the human brain however, the Personality tests used > have > > hundreds of items and ALL 13 SYMMETRY AXES are extracted as > > (orthogonal) eigenvectors. > What!? (see the rest of the mail) > > When they are rotated (by automated computer > > program) to SIMPLE STRUCTURE it is found that they fall on the 13 > > symmetry axes of a CUBE. > > QED > Rotated? You mean you first get some eigenvectors, and if they do not > please you you change them. > [Hammond] > that's an incompetent statement by someone who DOES NOT > KNOW what rotation to simple structure means.... and > every Factor Analyst does. the rotation is NOT ARBITRARY > and is doen AUTOMATICALLY by the computer using a > lest squares hyperplane count maximization. The computer > programs to do this (such as PROMAX) are famous in the > industry and are built into modern statistical packages such > as SPXX which is used daily by THOUSANDS of scientists > world wide today. > You don't know what you are talking about. Sure George. No, I do not know what rotation to a simple structure here means, you got me. === Subject: Re: God=G_uv proves 40k B.C. Creation > > > > > > There you go! 3 eigenvectors in the beginning you said: > > > > > > [Hammond] > > > Any way.... ALL 13 of these symmetry axes show up > > > in Psychometric data at the 2nd order (Cattell 1975). > > > And the REASON for it, is that the human brain is actually > > > CUBIC. > > > [Hammond] > > > > > > > [Hammond] > > Hey..... you've lost the argument and now you're fishing around for > > some semantic or typographical error to home in on. > > > > This is why we are having this discussion, and then you were adamant > > that they are eigenvectors, and insulted me for not understanding such > > a trivial matter. > > > > In THURSTONE'S BOX PROBLEM there are 3-eigenvectors > > which define the 3D Psychometry space > > > > Ahaa! A frog just jumped from your mouth. We were (at present at > > least) discussing the Thurstones box problem. Bringing in the > > psychometry space reveals why this stone wall is here: in your mind > > this is strongy associated with the real space causes psychometry > > space and stuff and I do not want to turn that rock. > [Hammond] > You're not making any sense.... the ARE 13-EIGENVECTORS > in Psychometry, they are 3-eigenvectors plus 10 vectors in > thurstone's Box Problem. For chrissakes Thurstone's Box Problem > was made up before the invention of computers and they couldn't > factor a 50x50 matrix which is needed to support 13 eigenvectors... > the symmetry ases are composite factors in Thurstone's Box Problem, > nevertheless they ARE THERE and appear in the graphical reconstruction > of the BOX in PSYCHOMETRY SPACE. > So when the algebra does not match up, you move on to a more complex > situation where inconvenient rules of algebra and logic are not in the > way of your transcendential thinking. [Hammond] No... that's just one of your forlorn wishes. You're like a drowning man grasping at straws when he knows the end is near. > > and the other 17 measurements > > are VECTORS when plotted in that space, and the other 17 vectors > > ALL fall and ONLY FALL, on the 13 symmetry axes of the CUBE > > defined in ENP space by the eigenvectors ENP which are the edges of the > > BOX. A 20x20 matrix is not large enough to support 13 eigenvectors. > > > > Even if there were more vectors these would not be eigenvectors, as > > 2X+2Y is a linear combination of X and Y. > [Hammond] > It's a composite factor. lying symmetrically half way between > X and Y. > A composite eigenvector..... Sure George. Ingenious! We'll name it > after you. [Hammond] No it's already named after Thrustone. See: Thurstone L.L. (1947) Multiple Factor Analysis, Chicago U. Press page 139 Correlated Factor. Yawnnnn................ > > > > In the case of the human brain however, the Personality tests used > have > > hundreds of items and ALL 13 SYMMETRY AXES are extracted as > > (orthogonal) eigenvectors. > > > > What!? (see the rest of the mail) > > > > When they are rotated (by automated computer > > program) to SIMPLE STRUCTURE it is found that they fall on the 13 > > symmetry axes of a CUBE. > > QED > > > > Rotated? You mean you first get some eigenvectors, and if they do not > > please you you change them. > [Hammond] > that's an incompetent statement by someone who DOES NOT > KNOW what rotation to simple structure means.... and > every Factor Analyst does. the rotation is NOT ARBITRARY > and is doen AUTOMATICALLY by the computer using a > lest squares hyperplane count maximization. The computer > programs to do this (such as PROMAX) are famous in the > industry and are built into modern statistical packages such > as SPXX which is used daily by THOUSANDS of scientists > world wide today. > You don't know what you are talking about. > Sure George. No, I do not know what rotation to a simple structure > here means, you got me. [Hammond] You really ought to know what Rotation to Simple Structure is.... everybody else in Factor Analysis knows what it is. It is universally used in all fields that use Factor Analysis, not only Psychometry. Look it up in any textbook on Factor Analysis, or even a statistics handbook. SCIENTIFIC PROOF OF GOD WEBSITE 1st mirror site: http://geocities.com/scientific_proof_of_god 2nd mirror site: http://proof-of-god.freewebsitehosting.com new site (under construction): http://home.comcast.net/~proof-of-god === Subject: Re: God=G_uv proves 40k B.C. Creation > > > > THURSTONE'S ORIGINAL BOX PROBLEM (which is > a Factor Analysisi) problem, DOES IDENTIFY ALL > 13 SYMMETRY AXES OF THE CUBE. > > > This discussion began with your statement that these 13 axes are > eigenvectors of the correlation matrix. Because I am so stupid and > incompetent, please explain how this happens. > > > [Hammond] > > BTW, while Thurstone was working with paper and pencil... > Cattell got into it not much later, and actually did the first > computerized Factor Analysis on the ENIAC computer > in the 1950's. The ENIAC was the world's first computer > No, it wasn't even the first electronic digital computer. Even when > you discount mechanical digital computers (remember Babbage's > Analytical Engine?), ENIAC wasn't the first. A common mistake. > http://www.scl.ameslab.gov/ABC/ > (503 - Connection failed) > What about the real pioneers in computer technology? > > > Zuse's computer in the museum (see link) still is working! > and had 20,000 12AU7 vacuum tubes in it (this was before > transistors) and the thing got so goddamned hot they had > to install 25 air conditioners in the room to keep it cool! > BTW, before he died, at the age of 92, Cattell used to write me > lengthy handwritten letters from Hawaii. > Put scans up on the web site. Remember the rules. > I think he knew I > was on something. > Any word and review from Waco? > Went to the shredder? Did you see any postings with the name > Dembski (this guy must be a German who was born in Poland, > BTW) in the thread? [Hammond] His picture is located at: http://www.iscid.org/william-dembski.php sort of looks like the Midwestern all American farmboy type.... better looking than any of us I'd wager. He's 44 btw. SCIENTIFIC PROOF OF GOD WEBSITE 1st mirror site: http://geocities.com/scientific_proof_of_god 2nd mirror site: http://proof-of-god.freewebsitehosting.com new site (under construction): http://home.comcast.net/~proof-of-god > Bernhard Schornak === Subject: Re: God=G_uv proves 40k B.C. Creation >Went to the shredder? Did you see any postings with the name >Dembski (this guy must be a German who was born in Poland, >BTW) in the thread? > [Hammond] > His picture is located at: > http://www.iscid.org/william-dembski.php > sort of looks like the Midwestern all American farmboy type.... > better looking than any of us I'd wager. He's 44 btw. Yet he looks nothing like Steven Segal - how odd. === Subject: Re: God=G_uv proves 40k B.C. Creation > >Went to the shredder? Did you see any postings with the name >Dembski (this guy must be a German who was born in Poland, >BTW) in the thread? > [Hammond] > His picture is located at: > http://www.iscid.org/william-dembski.php > sort of looks like the Midwestern all American farmboy type.... > better looking than any of us I'd wager. He's 44 btw. > Yet he looks nothing like Steven Segal - how odd. [Hammond] Looks like one of Billy Graham's kids.... another one of my heroes. SCIENTIFIC PROOF OF GOD WEBSITE 1st mirror site: http://geocities.com/scientific_proof_of_god 2nd mirror site: http://proof-of-god.freewebsitehosting.com new site (under construction): http://home.comcast.net/~proof-of-god === Subject: Re: God=G_uv proves 40k B.C. Creation > > >THURSTONE'S ORIGINAL BOX PROBLEM (which is >a Factor Analysisi) problem, DOES IDENTIFY ALL >13 SYMMETRY AXES OF THE CUBE. > > >This discussion began with your statement that these 13 axes are >eigenvectors of the correlation matrix. Because I am so stupid and >incompetent, please explain how this happens. > > >[Hammond] > > > BTW, while Thurstone was working with paper and pencil... >Cattell got into it not much later, and actually did the first >computerized Factor Analysis on the ENIAC computer >in the 1950's. The ENIAC was the world's first computer > >No, it wasn't even the first electronic digital computer. Even when >you discount mechanical digital computers (remember Babbage's >Analytical Engine?), ENIAC wasn't the first. A common mistake. >http://www.scl.ameslab.gov/ABC/ >(503 - Connection failed) >What about the real pioneers in computer technology? > > >Zuse's computer in the museum (see link) still is working! >and had 20,000 12AU7 vacuum tubes in it (this was before >transistors) and the thing got so goddamned hot they had >to install 25 air conditioners in the room to keep it cool! > BTW, before he died, at the age of 92, Cattell used to write me >lengthy handwritten letters from Hawaii. > >Put scans up on the web site. Remember the rules. >I think he knew I >was on something. > >Any word and review from Waco? >Went to the shredder? Did you see any postings with the name >Dembski (this guy must be a German who was born in Poland, >BTW) in the thread? >[Hammond] >His picture is located at: > http://www.iscid.org/william-dembski.php >sort of looks like the Midwestern all American farmboy type.... >better looking than any of us I'd wager. He's 44 btw. A matter of taste? I prefer women... ;) BTW - he looks like he has Polish origins (as his name suggests). Bernhard Schornak === Subject: Ostrowski's calculus book Ostrowski's calculus book was originally published in German, in three volumes. The English version only has one volume that I can find reference to. It does not contain all three volumes, as the last page of text contains a footnote refering to something that will be treated in volume 3. Does anyone have a reference to Ostrowski's volume 2 or 3 existing in English? David Ames === Subject: Re: EXACT calculation of integrals over 2D areas described by curvilinear polygons just like me :-) > The Mathematica program listed below evaluates exactly (numerically or > symbolically) the moment integral > int_A x^i y^j dA (i,j=nonnegative integers) > over an arbitrary straight-sided plane polygon, using decomposition > over triangles. This gives the area if i=j=0. The method for > integrating any polynomial exactly over a straight sided triangles is > explained in Exercise 15.8 of > http://caswww.colorado.edu/courses.d/IFEM.d/IFEM.Ch15.d/ IFEM.Ch15.pdf When the plane polygon has straight sides, things are easy; I suppose that the decomposition of a polygon into triangles is much - much easier (ear clipping?), and then the calculation of the integrals is straightforward. (The problem is that I cannot use Mathematica or any other program within my source code; I need C++ and/or Fortran source code. ). Decomposing into trapezoids is more difficult, but on the other hand, the trapezoids are less in number. The decomposition itself is not much of a problem, because the process will take place only once in my program. The decomposed domain will be stored into memory, therefore the overhead is minimal. > (I use this code for homework exercises in FEM courses.) > For curvilinear polygons it is tougher. One may decompose into > triangles one of which sides can be curved. If the curved side can > be described by polynomial interpolation via finite element iso-P > shape functions, Gauss quadrature formulas for triangles expressed in > natural coordinates can be used. But in general the integral is only > an approximation because the area metric is not described by > polynomial functions. For details on such formulas see Sec 24.2 of > http://caswww.colorado.edu/courses.d/IFEM.d/IFEM.Ch24.d/ IFEM.Ch24.pdf That is interesting... The sides of my triangles in fact will be arcs, however no-one can guarantee that only one side of the triangle will be curved. Moreover, the result will be an approximation, as you mention. Aristotelis === Subject: Re: EXACT calculation of integrals over 2D areas described by curvilinear polygons > just like me :-) > That is interesting... The sides of my triangles in fact will be arcs, > however no-one can guarantee that only one side of the triangle will be > curved. Moreover, the result will be an approximation, as you mention. > Aristotelis If the polygon is simply connected find the centroid of the corner points (those where arcs intersect). Joining the centroid to the corners gives triangles with only one curved side. Whether you can compute integrals exactly depends on how the arcs are defined. If they are just 2nd order parabolas an exact answer might be possible with a computer algebra system, and converted automatically to C. There was work from A. R. Mitchell at Dundee on such geometries (triangles with one curved side and quadrilaterals with one curved side) in the 70s. See his book The finite element method in partial differential equations' with R. Wait, Wiley, 1977. They used Lobatto numerical integration points to define the arcs since corners are then included. My degrees are in civil engineering (from Berkeley) but my work has been in aerospace since 1968. === Subject: Mitigating the Alabama Paradox The Senate must have exactly 100 members, but every state, regardless of population, gets 2 members, so apportionment is not a problem for it. At present, the House of Representatives always has 435 members. This could be varied, though; the requirement is that the average size of a district should be 30,000 voters, so it increased in size over time. Although a proof exists that all possible apportionment systems fail some reasonable condition, one can, by *changing the problem*, if a constraint can be removed, improve performance. What I would propose is this: Use Hamilton's method to apportion a House of every size from 430 members to 435 members. Choose the House size for which the sum of the squares of the differences from the ideal fractional apportionment and the actual integer apportionment is a minimum. This lets one retain all the desirable properties of the Hamilton system, while in addition ensuring that the Alabama paradox happens far less often. Of course, it has the consequence that the mix of population-based and constant weighting for each state in the Electoral College suddenly becomes somewhat randomized, and as that can be a significant variable in Presidential elections, that might be a grounds for objecting to the scheme. John Savard http://home.ecn.ab.ca/~jsavard/index.html === Subject: Re: Mitigating the Alabama Paradox >Of course, it has the consequence that the mix of population-based and >constant weighting for each state in the Electoral College suddenly >becomes somewhat randomized, and as that can be a significant variable >in Presidential elections, that might be a grounds for objecting to the >scheme. Since it can result in the House of Representatives shrinking, leading to seats being unavailable for many incumbents, this would be a far mor major barrier to the adoption of this scheme. John Savard http://home.ecn.ab.ca/~jsavard/index.html === Subject: Re: Mitigating the Alabama Paradox >The Senate must have exactly 100 members, but every state, regardless of >population, gets 2 members, so apportionment is not a problem for it. >Although a proof exists that all possible apportionment systems fail >some reasonable condition, one can, by *changing the problem*, if a >constraint can be removed, improve performance. >What I would propose is this: >Use Hamilton's method to apportion a House of every size from 430 >members to 435 members. >Choose the House size for which the sum of the squares of the >differences from the ideal fractional apportionment and the actual >integer apportionment is a minimum. stand some creative math too. A case in point is the reapportionment of NY State districts 28 and 29. I believe the effort was seriously ßawed. The gerrymandered shape of the district, linking Buffalo with Rochester by a thread is not the bulk of the problem--it was the math that allowed the report to state that EXACTLY 654,350 voters were in the new district 28 and 654,351 voters were in the new district 29. (reference: http://latfor.state.ny.us/maps/?sec=PC http://latfor.state.ny.us/maps/propcong/fc028.pdf http://www.pballew.net/arithm17.html#voteappr) The objective of reapportionment calculations, in addition to honesty, should be to achieve a measure of fair representation while minimizing the disruption of existing boundaries. Using a technique like yours over a number of districts would result in deficit or surplus representation. The deficit or surplus could be then reassigned to another region (presumably in the same state) and the process repeated. Hopefully the result would be the switching of whole towns or villages in or out of a congressional district rather than splitting them as the current apportionment has done. BTW-http://nycitizens.org/redistricting/scenarios.html offers an amusing opportunity to play the apportionment game. BTW, where did 35,000 come in? http://nycitizens.org/redistricting/usapportion.html It appears that 650,000 per seat is the current average with Wyoming's 495,000 as the most generous. John Bailey http://home.rochester.rr.com/jbxroads/mailto.html === Subject: Curve of Constant integral invariant Is it possible to solve for a 2D curve: Integral(ph)ds=constant, where ph is slope of curve and s its arc length? Arc length itself is a known integral invariant. Taking primes with respect to arc length s, ph'=constant (curvature) a circle, ph''= constant a Cornu spiral etc. are differential invariants. But is integration of slope/rotation along arc meaningful? Or is it possible as an isoperimetric problem: Integral (ph+lamda) ds =constant where lambda is a Lagrange multiplier? === Subject: Re: Field of Sets === Subject: Re: Field of Sets >Is a field of sets the same as an algebra of sets? >F field of sets over S when F subset P(S), >for all A,B in F, > A / B, A / B, and SA in F ? >Both Bartle and Billingsley say that a field and an algebra (of sets) >are the same thing. Both of these authors include the condition >that S be in the collection as part of their definition. Whoops, indeed S in F or even better to have definitions in parallel with ring of sets, nulset in F. >Halmos, in _Measure_Theory_, does not mention fields at all Good for him, Ôalgebra of sets' is preferred decor. insert a note defining an algebra of sets in Wikipedia's reference Ôsigma-algebra' and create a page for Ôfield of sets' as requested by some links in Wikipedia, pointing to the definition in Ôsigma-algebra'. ---- === Subject: Re: Field of Sets > insert a note defining an algebra of sets in Wikipedia's reference > Ôsigma-algebra' and create a page for Ôfield of sets' as requested by some > links in Wikipedia, pointing to the definition in Ôsigma-algebra'. Of course field is not sigma-algebra but only algebra. There is the term sigma-field for that. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: measure === Subject: Re: measure >Paul Halmos, Lecture 21 in _Lectures_on_Boolean_Algeras_ indicates >a Boolean algebra has a unique completion up to an isomorphism. >If what Halmos means by completion is an example of completion >in the sense of my post, then we have the following situation: >an embedding of B in a complete Boolean algebra B^ (the completion), >such that for any complete Boolean algebra X and any Boolean >algebra map f:B --> X, there exists a unique Boolean algebra >map f^: B^ --> X extending f. (The result of Solovay denies >that there could be an analogous situation where f^ preserves >all meets and joins.) I'm inclined to agree with him. When representing an order by lower sets we see the collection of lower sets are a complete lattice. However the embedding of an element a of the order, to it's lower set down a, tho preserving inf's doesn't preserve sup's. However one can always find for any so mapped sup, an embedded sup as an inf of upper bounds of the mapped components. Upon preforming this stunt for Q, do I get R? Naw, I get duplicates, the mapped down q = { x in Q | x <= q } and a covered or preceding lower set { x in Q | x < q } which, as Dedekind did, need weeding out. >Does Halmos give a universal property, and if so, is it the >same as given in the previous paragraph? Don't know, I've parroted some comments, made by Michael Hardy about what Halmos did, which had been synopsed out of a sci.math thread into my notes. If you the reference within reach, then perhaps the puzzle could be unpuzzled. Also within my notes is every Boolean algebra has a Dedekind completion. Roman Sikorski book on Boolean algebras gives an account of this. Sometimes I'm amazed how easily those guy's with books, algebras, spaces, or theorems sporting their names can convince me I'm not a mathematician. ---- === Subject: Re: How Is This Possible? Here is a Web page that Andy is crying about: At the bottom of that page is a link to a Google thread. > Anyone with a basic knowledge of economics could give a long list of > reasons > why your model is unrealistic. >Andy demonstrates a lack of basic knowledge of economics, I guess. >My model is a canonical model in capital theory. The references >in the previously-referred-to thread demonstrate that. > Anyone with a basic knowledge of economics would know that canonical > models > are often quite unrealistic. Here Andy has, through dishonesty or ignorance, deleted some of my remarks so he can argue against a position that I did not take. I do not claim that canonical models are realistic. Nor do I claim they are unrealistic. And Andy has still not demonstrated a basic knowledge of economics. >Andy insults the intelligence of the reader. >Notice that Andy has not suggested any mechanism that is likely >to lead to any of these conditions holding. Does Andy think that >equilibrium prices are likely to be proportional to labor values? > This would be unlikely in the real economy. However, in an unrealistic > abstract > model it would not be surprising if this turned out to be the general > case. As far as I am concerned, that unargued statement is just ignorant. The dualistic thinking does not seem helpful. Andy doesn't seem to mean anything by the general case; he is just ejaculating is not. And he shows no awareness of what special case assumptions are needed to make the labor theory of value hold - even though I specify them in the context of the model in the previously referenced thread. Maybe I'll add Andy's remark to my Usenet fumbles page. >He is too cowardly or ignorant to say. > Apparently, if I don't answer a question before it is asked, this is > Ôcowardly > or ignorant'. No. Andy still has not put forward anything substantial supporting any of the four special cases I listed. Whether he is too cowardly to take a position or just ignorant of what is being argued, I leave for our readers to decide. >Anyways, if Andy actually understood what he is saying - he is >just ejaculating is not without any understanding - he would >know that he has conceded he was wrong. Conditions (1) through >(4) are particular cases. In general - that is, unless one >such special case is true - the interest rate is unequal, in >equilibrium, to the marginal product of value capital. I have >proven that. > You merely assert that these conditions are Ôparticular' cases. Until you > demonstrate that they could not be general cases, your claim to have > proven > anything of substance will continue to be refuted. A particular case has an additional condition appended to the general case. For example, one might need to impose another condition on technology. (Technology is typically among the givens in neoclassical economics.) I have proven that these are special cases merely by listing extra conditions. Given what is at this link: Andy's denial that special case (3), in which the equilibrium rate of interest is equal to the equilibrium rate of growth, is a special case is particularly amusing in its ignorance. With that allusion to a closure of the model with neoclassical assumptions, I turn to this bit of ignorance from Andy: > A neoclassical economist would Ôclose' the model by making the > standard > assumptions of neoclassical theory, i.e. that investors are rational > profit > product > of capital is not equal to the interest rate, the economy will not be > in > equilibrium. >Andy is simply wrong, as was demonstrated in the first post on the >previously referenced thread. >My model had profit-maximizing firms. That does not close the model, >and Andy's claim does not follow. > Anyone with a basic knowledge of economics will know that a > profit-maximising > investor is not the same as a profit-maximising firm. Vienneau's > construction > of economic models is ßawed. If Andy were not ignorant of how to construct an argument, he might have specified how it matters what sort of agent is profit-maximizing. Andy simply made a mistake - his specification of a neoclassical closeure of the model does not close the model. > Vienneau keeps referring to Ôprice Wicksell effects' but there is no > indication > in his posts of what he imagines that to mean. >Andy insults the reader's intelligence. I explained repeatedly >what are in the post starting off the >thread previously referred to. Andy has edited the above quote to make it ungrammatical. > I don't think you explained it very clearly. Perhaps you could explain > again in > simple terms, for the benefit of your less intelligent readers, what > Ôprice > Wicksell effects' actually means. Does Andy state where he finds my explanation in the previous thread unclear? Does Andy look up any of the references I gave in the previous thread and comment on where he finds the reference unclear? Of course not. That would be to say something substantial, and we cannot have that. So why should I bother restating something I find quite trivial? Does Andy have any training in economics? Is Andy some sort of instructor on economics somewhere? Can his posts be used as data for the sociology of economics? These seem like trivial questions, but Andy remains too much a coward to answer them. -- r c v s a Whether strength of body or of mind, or wisdom, or i m p virtue, are found in proportion to the power or wealth e a e of a man is a question fit perhaps to be discussed by n e . slaves in the hearing of their masters, but highly @ r c m unbecoming to reasonable and free men in search of d o the truth. -- Rousseau === Subject: Re: How Is This Possible? >With that allusion to a closure of the model with neoclassical >assumptions, I turn to this bit of ignorance from Andy: > > A neoclassical economist would Ôclose' the model by making the > > standard > > assumptions of neoclassical theory, i.e. that investors are > > rational > > profit > > marginal > > product > > of capital is not equal to the interest rate, the economy will not > > be > > in > > equilibrium. > >Andy is simply wrong, as was demonstrated in the first post on the > >previously referenced thread. > >My model had profit-maximizing firms. That does not close the model, > >and Andy's claim does not follow. > Anyone with a basic knowledge of economics will know that a > profit-maximising > investor is not the same as a profit-maximising firm. Vienneau's > construction > of economic models is ßawed. >If Andy were not ignorant of how to construct an argument, he might >have specified how it matters what sort of agent is profit-maximizing. >Andy simply made a mistake - his specification of a neoclassical >closeure of the model does not close the model. Andy repeats a mistake: > Anyone with a basic knowledge of economics will understand how the > actions of > rational investors would be expected to lead to interest rates being > equal to > the marginal return to capital.It is quite remarkable that an expert such > as > Vienneau would not understand this. > This equality could be used to write another equation in the model in > Vienneau's post Andy remains simply wrong. Profit maximizing does not close the model. On the other hand, assuming an intertemporal utility-maximizing agent does close this sort of model. I demonstrate the latter point in: (An interactive version of the production side of these sort of models is at: The aggregate production function is on the last spreadsheet in that workbook. I know that function doesn't look like what many of you will have seen in economics classes. But mainstream economists often teach falsehoods.) > The implications of these equations can be worked out by anyone who > wishes to > do the maths. The above is just another manifestation of Andy's inability to construct an argument. Andy has not done the math: > If Vienneau's calculations are correct, Doesn't Andy know? Hasn't he worked through the model he keeps mentioning? > These implications would include > proof of the labour theory of value, given the assumptions in the model. I don't know how Andy manages to incorrectly reach that conclusion. I state in the post he references that the labor theory of value is generally not true. > Since this is a highly simplified model in which labour and capital are > the > only factors of production,this would not be a remarkable result. Ricardo > probably proved something similar in the 19th century. Andy probably doesn't understand Ricardo's demonstration in the first chapter of his Principles that the labor theory of value is generally not true. > > Vienneau keeps referring to Ôprice Wicksell effects' but there is > > no > > indication > > in his posts of what he imagines that to mean. > >Andy insults the reader's intelligence. I explained repeatedly > >what are in the post starting off the > >thread previously referred to. >Andy has edited the above quote to make it ungrammatical. > I don't think you explained it very clearly. Perhaps you could explain > again in > simple terms, for the benefit of your less intelligent readers, what > Ôprice > Wicksell effects' actually means. >Does Andy state where he finds my explanation in the previous thread >unclear? Does Andy look up any of the references I gave in the >previous thread and comment on where he finds the reference unclear? >Of course not. That would be to say something substantial, and we >cannot have that. So why should I bother restating something I >find quite trivial? > Observers might speculate that the bumbling amateur Vienneau is simply > parroting a phrase which he does not understand, if indeed it has any > meaning > at all. Does Andy state where he finds my explanation in the previous thread unclear? Of course not. He seems incapable of developing an argument. As for insulting the reader's intelligence - he's good at that. >Does Andy have any training in economics? > I have a bachelor's degree with some economics content. OK, so Andy's ignorance, confusion, and incapacities are not good for illustrating the sociology of academic economics. But, since he is being so stupid, I'll leave him in my Usenet fumbles page. I might even add more examples from this thread. -- r c v s a Whether strength of body or of mind, or wisdom, or i m p virtue, are found in proportion to the power or wealth e a e of a man is a question fit perhaps to be discussed by n e . slaves in the hearing of their masters, but highly @ r c m unbecoming to reasonable and free men in search of d o the truth. -- Rousseau === Subject: Philosophical meaning of possibility to separate variables In partial differential equations, a useful method of solving the equations arising is the method of separation of variables. In most situations for phycisists, this is possible. Is there a philosophical explanation for this? That is, what fundamental property of a phenomena is required for this to be possible? === Subject: Re: Philosophical meaning of possibility to separate variables > Remember the math is nothing more than modelling. The reality has does > not obey the maths, it is the simple math that is used to model the > reality, it will end up displaying some simple properties that are > inherent in the model used ( and not the reality itself). Allright but doesn't physicists choose to study phenomena which _do_ obey simple math? At least some subset of all theoretical physicists. And since they choose to model phenomena that are easily modeled, any property in the model is (idealy) a property of the phenomena? === Subject: Re: Philosophical meaning of possibility to separate variables > In partial differential equations, a useful method of solving the > equations arising is the method of separation of variables. In most > situations for phycisists, this is possible. Is there a philosophical > explanation for this? That is, what fundamental property of a phenomena is > required for this to be possible? Symmetry. An extensive discussion of separation of variables for equations of mathematical physics can be found in Section 5.1 of Morse & Feshbach's _Methods of Theoretical Physics, vol. I_. Igor === Subject: Optimization Problem I am working on a problem of maximizing a convex objective function (actually, a quadratic function) subject to non-smooth convex constriants. Does anyone know whether the solution is unique, i.e., any local maximum is the global maximum? Any suggestions, reference on Mike === Subject: Re: Optimization Problem >I am working on a problem of maximizing a convex objective function >(actually, a quadratic function) subject to non-smooth convex >constriants. Does anyone know whether the solution is unique, i.e., >any local maximum is the global maximum? Any suggestions, reference on Hint: try maximizing x^2 on subject to x >= -1 and x <= 1. Or did you mean the objective to be concave (or to minimize rather than maximize)? Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: The actual math, advanced polynomial factorization > The following factorization is useful: > For what? > f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) = > (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) > Note that x, m, f, and u are all independent variables. > No, they aren't independent. They are related by the equation > f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) = > (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) No. They are not. Notice that you can give any value you wish for x, m, f and u without checking each variable against the other because they are not constrained against each other. But notice that you can't just give a value for a_1, a_2 and a_3 because they ARE constrained by the values of x, m, f and u. James Harris === Subject: Re: The actual math, advanced polynomial factorization > > > The following factorization is useful: > > For what? > > > f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) = > > > > (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) > > > > Note that x, m, f, and u are all independent variables. > > No, they aren't independent. They are related by the equation > > f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) = > (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) > > No. They are not. Notice that you can give any value you wish for x, > m, f and u without checking each variable against the other because > they are not constrained against each other. > But notice that you can't just give a value for a_1, a_2 and a_3 > because they ARE constrained by the values of x, m, f and u. Mark the date on your calendar, folks: James Harris posted something that was correct. -- Christopher Heckman === Subject: Re: The actual math, advanced polynomial factorization > > > > The following factorization is useful: > > > > For what? > > > > f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) = > > > > (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) > > > > Note that x, m, f, and u are all independent variables. > > > > No, they aren't independent. They are related by the equation > > > > f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) = > > (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) > > > > No. They are not. Notice that you can give any value you wish for x, > m, f and u without checking each variable against the other because > they are not constrained against each other. > > But notice that you can't just give a value for a_1, a_2 and a_3 > because they ARE constrained by the values of x, m, f and u. > Mark the date on your calendar, folks: James Harris posted something that > was correct. > -- Christopher Heckman If you accept that x, m, f, and u are independent variables, while a_1, a_2, and a_3 are the ONLY dependent ones, then the rest of the argument is easy. Now then, as a second step, acknowledge that f^2 is a multiple of f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) and then maybe some progress can be made. Multiples don't divide off as a function, but posters trying to argue with me try to force a dependency on f. For instance, by claiming something like f^2 = w_1(m) w_2(m) w_3(m) where the w's vary as m varies. That's the sci.math objection. That group settled from what I gathered on the idea that there's this forced dependency on an independent variable. Interesting groupthink, eh? James Harris === Subject: Re: The actual math, advanced polynomial factorization ... > Multiples don't divide off as a function, but posters trying to argue > with me try to force a dependency on f. Why not? Do you have a proof that they do not divide off as a function? > For instance, by claiming something like > f^2 = w_1(m) w_2(m) w_3(m) > where the w's vary as m varies. Yes, so what? It has been shown to be true. But you never did acknowledge that, neither did you bother to attack the proofs, except by circumlocation. > That's the sci.math objection. That group settled from what I > gathered on the idea that there's this forced dependency on an > independent variable. There is *no* forced dependency. Whenever you have an irreducible primitive polynomial with integer coefficients, all of the roots are not coprime (within the algebraic integers) to the prime (in the integers) factors of the constant term. The forced dependency is on the roots; *they* depend on the independent variable, and as a result the factors depend on the independent variable. Or do you indeed think that while the roots depend on the independent variable, the factors should not do so? When you have the quadratic x^2 + a.x + 2, with Ôa' some integer, whenever the polynomial is irreducible (i.e., a^2 - 8 is not a square in the integers), both roots have a factor (in the algebraic integers) in common with Ôa', and the factors depend on Ôa'. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: The actual math, advanced polynomial factorization > > > The following factorization is useful: > > For what? > > > f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) = > > > > (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) > > > > Note that x, m, f, and u are all independent variables. > > No, they aren't independent. They are related by the equation > > f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) = > (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) > > No. They are not. Notice that you can give any value you wish for x, > m, f and u without checking each variable against the other because > they are not constrained against each other. > But notice that you can't just give a value for a_1, a_2 and a_3 > because they ARE constrained by the values of x, m, f and u. That's the mathematics, but notice what I face, which is what I've faced from sci.math posters for YEARS, in the reply by David Kastrup. > That's nothing to notice, that is what you may want to define. > Prima facie, the above equation just expresses a relation between all > of the variables and does not differentiate anything among them. In > the context of defining a particular task associated with the > equation, you can assign meaning to various variables. You always > muddy up things wildly. In particular, for the purpose you want, you > _choose_ the following characterizations: > a) x is a symbolic variable used for defining a polynomial. It does > not assume any value at all. > b) m, f and u are _declared_ to be free variables: you try making a > statement that will hold for all choices of them. > c) a_1, a_2 and a_3 are then (though not necessarily uniquely) > determined by that choice and [are in the coefficient space for the > polynomial defined over x] Correction in superseded post: this is > of course not the case when the polynomial can't be factored into > linear terms in the given coefficient space, like when factoring > the real polynomial x^2+1. > In order to make something like this make sense, you have to declare > _how_ m, f and u are supposed to be chosen: what possible values they > may assume. In addition, you have to declare the ring or other > algebraic structure in which a_1, a_2 and a_3 are supposed to be. > Only then can you start making meaningful statements. > And in fact, you have repeatedly changed your mind about those basic > premises you have not bothered to fix in APF, and are lying about > this when defending it. f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) and f, m, x and u are independent variables, while a_1, a_2, and a_3 are dependent variables. Such a simple thing, but notice all the verbiage from David Kastrup. It's not about what's mathematically correct with these people. I say they are YOU, and they are the modern math world. You've learned that math is just some way to b.s. people too intimidated to check you, while you've accepted things that are convenient but false, like the rather naive and wrong ideas that base what you call Galois Theory. Evariste Galois was right, but you people twisted his research. James Harris === Subject: Re: The actual math, advanced polynomial factorization > > > > > The following factorization is useful: > > > > For what? > > > > f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) = > > > > (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) > > > > Note that x, m, f, and u are all independent variables. > > > > No, they aren't independent. They are related by the equation > > > > f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) = > > (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) > > > > > > No. They are not. Notice that you can give any value you wish for x, > > m, f and u without checking each variable against the other because > > they are not constrained against each other. > > > > > > But notice that you can't just give a value for a_1, a_2 and a_3 > > because they ARE constrained by the values of x, m, f and u. > That's the mathematics, but notice what I face, which is what I've > faced from sci.math posters for YEARS, in the reply by David Kastrup. > > That's nothing to notice, that is what you may want to define. > Prima facie, the above equation just expresses a relation between all > of the variables and does not differentiate anything among them. In > the context of defining a particular task associated with the > equation, you can assign meaning to various variables. You always > muddy up things wildly. In particular, for the purpose you want, you > _choose_ the following characterizations: > > a) x is a symbolic variable used for defining a polynomial. It does > not assume any value at all. > b) m, f and u are _declared_ to be free variables: you try making a > statement that will hold for all choices of them. > c) a_1, a_2 and a_3 are then (though not necessarily uniquely) > determined by that choice and [are in the coefficient space for the > polynomial defined over x] Correction in superseded post: this is > of course not the case when the polynomial can't be factored into > linear terms in the given coefficient space, like when factoring > the real polynomial x^2+1. > > In order to make something like this make sense, you have to declare > _how_ m, f and u are supposed to be chosen: what possible values they > may assume. In addition, you have to declare the ring or other > algebraic structure in which a_1, a_2 and a_3 are supposed to be. > Only then can you start making meaningful statements. > > And in fact, you have repeatedly changed your mind about those basic > premises you have not bothered to fix in APF, and are lying about > this when defending it. > f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) = > (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) > and f, m, x and u are independent variables, while a_1, a_2, and a_3 > are dependent variables. > Such a simple thing, but notice all the verbiage from David Kastrup. Yes, f, m, x, and u are independent and a_1, a_2, and a_3 are not. But the statement you are discussing, i.e., divisibility of a_1, a_2 or a_3 by f in the algebraic integers is not really a statement about f. It is a statement about a_1, a_2, and a_3, conditional on f. That is, the statement itself is about *dependent* variables. So any vague principle you have in your head about independent variables does not apply. This is a bogus argument based on a nonmathematical principle that you think applies in this case. It does not. Andrzej > It's not about what's mathematically correct with these people. > I say they are YOU, and they are the modern math world. > You've learned that math is just some way to b.s. people too > intimidated to check you, while you've accepted things that are > convenient but false, like the rather naive and wrong ideas that base > what you call Galois Theory. > Evariste Galois was right, but you people twisted his research. > James Harris === Subject: Re: The actual math, advanced polynomial factorization > The following factorization is useful: > f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) = > (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) > Note that x, m, f, and u are all independent variables. It's a lot of > variables, yes, but they're independent of each other. > Also note that a_1, a_2 and a_3 are dependent variables. > Now divide both sides by f^2 to get > (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f = > (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)/f^2 > and all the arguing for over TWO YEARS NOW has been over what that > means. > Let m=0, and you get > (0*f^4 - 3* 0*f^2 + 3*0) x^3 - 3(-1 + 0*f^2 )x u^2 + u^3 f = > (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)/f^2 > which is > -3(-1)x u^2 + u^3 f = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)/f^2 > which is > u^2(3x + uf) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)/f^2 > and it's clear that two of the a's have to equal 0, while one equals > 3. > But that means that the f^2 divides through two of the factors > (a_1 x + uf), (a_2 x + uf), or (a_3 x + uf) > and, remember, x, m, f and u are all independent variables. > Therefore, two of the factors have f as a factor in general without > regard to the value of m, but it turns out that IN THE RING OF > ALGEBRAIC INTEGERS they do not always have f as a factor. Notice that > not always, it's important. > Some posters rather than accept that x, m, f and u are independent > variables so their value doesn't matter to each other, claim my result > only holds if m=0, as a special case, but if variables are independent > of each other, how can a particular value be a special case? > So why fight the result? Well Evariste Galois came up with the theory > for how you can *represent* roots of irrational polynomials. > What I've shown is an analysis tool for actually figuring out factors > with irrationals, but no one had tools of this power before, and with > the results of Galois they made a leap: They decided that the work of > Galois was not just about representation, but also about actual > factors. What you say is not true. Galois theory does indeed talk about factors, but you have to understand what Galois theory actually says; it helps to know a little algebra, but your grasp of algebra is insufficient. For instance, Galois theory says that, given any two roots r and s of an irreducible polynomial P(x) over a field F, for any arithmetic statement in the elements of F(r) (The field F with r attached) that is true, must also be true with r replaced by s. This forms a general symmetry principle that applies for roots of polymomials. Note that it is *NOT* just a statement about representing roots. For instance, as an example of an arithmetic statement in F(r), one could imagine finding a factorization (Q,R,S: polynomials in F) Q(r) = R(r)S(r) then one would be able to deduce that this factorization is also valid: Q(s) = R(s)S(s). Aside: Actually, these facts don't really require the full power of Galois theory. They are a result of the fact that the field F(r) is canonically isomorphic to the quotient of the polynomial ring F[x] by the ideal enerated by P(x). However you might note that this quotient, written: F[x]/ makes no mention of the root r at all. In this model of F(r), it is the class of x modulo the ideal P(x) that serves as our abstracted quantity r. In considering the other root s, we also have this: F(s) is canonically isomorphic to F[x]/, with the class of x modulo being identified with the root s. As a corollary, one finds that F(r) is canonically isomorphic to F(s), with r corresponding to s under the canonical isomorphism. Now, suppose the polynomial in question has integer coefficients, and is monic. Then if w is any factor of r in the ring of algebraic integers, that fact can be expressed via a true arithmetic statement involving only the value r and the ring of integers. By the general symmetry principle I mentioned in the preceding paragraph, this same statement with r replaced by s will also be true. The result is that the number s will *necessarily* be divisible by an algebraic integer w'. The two numbers w and w' are not totally independent of each other, for they have the same irreducible polynomial (that is, for any polynomial P(x) having P(w) = 0, we will also have P(w') = 0). In other words, if you find *any* algebraic integer that divides one root r of an irreducible monic polynomial P(x), you'll find an algebraic integer w' *with the same irreducible polynomial* that divides any other root s of P(x). It goes even farther than that, in fact. If this factor w is also a factor of an ordinary integer n in the ring of algebraic integers, w' will *also* be a factor of n in the ring of algebraic integers. > Now the algebra supporting my case is simple enough, but I've been > arguing for over two years on the same points and posters have > routinely rejected basic algebra, like that independent variables are > independent, while making up their own kind of weird mathematics I've > at times called voodoo mathematics. There is no doubt in my mind that you deny what I've written above. I'll make you a deal: pose a factorization that you wish to claim causes the above general principle (w divides r <==> w' divides s) or the minimal extension (w divides r and n <==> w' divides s and n) to fail. If it fails, then I'm wrong and will retract the above statements. If it passes, you may wish to try again. And again. And again. My claim is that it won't fail, because I claim the symmetry principle stated above is a valid method. > Let me show you how it works. Like they'll say there exists algebraic > integer functions w_1(m), w_2(m), and w_3(m), where > w_1(m) w_2(m) w_3(m) = f, and > a_1 = b_1 w_1(m), a_2 = b_2 w_2(m), and a_3 = b_3 w_3(m) > where the b's are algebraic integers. > Now with > (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f = > (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)/f^2 > you then have > (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f = > (b_1 x + u w_2(m)w_3(m))(b_2 + uw_1(m)w_3(m))(b_3 x + uw_1(m)w_2(m)) > where all might seem well and good, like why can't you just have these > functions that somehow relate INDEPENDENT VARIABLES but I've said > that before and the sci.math'ers make fun of me. > There's just one small little problem with that setup, f^2 was > multiplied times both sides before, but what if I play a trick on the > voodoo math? > Let's say that rather than multiply both sides by f^2, I multiply > times 13^2? > You see, the point of have some constant multiple like f^2 is that if > you divide it off it's gone, like if I have > 4(x^2 + 3x + 2) = 4(x+1)(x+2) > and divide 4 off, it's gone. But not in the voodoo math of sci.math > posters...if the factorization is irrational. They always come back > to say, yeah, if you're talking about polynomial factors like in my > simple example, then yeah, that's how constant multiples behave!!! > It gets weirder. My result holds for a family of polynomials. And just what is that result? Please state the result that you're claiming. So far, I see many references to an alleged result, but no statement of what it is. > To see that let's make some adjustments: > f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) > and multiplying through by f^2 gives > (m^3 f^4 - 3m^2 f^2 + 3m)f^2 x^3 - 3(-1+mf^2 )x u^2 f^2 + u^3 f^3 > where notice in the last two terms you have u and f paired, so let > y=uf, so you get > (m^3 f^4 - 3m^2 f^2 + 3m)f^2 x^3 - 3(-1+mf^2 )x y^2 + y^3 > and the factorization now is > (m^3 f^4 - 3m^2 f^2 + 3m)f^2 x^3 - 3(-1+mf^2 )x y^2 + y^3 = > (a_1 x + y)(a_2 x + y)(a_3 x + y) > and my result from before still applies because the a's are not > dependent on the value of y, while still having the same values as > before, so now I can change y, as yet another independent variable, so > let's put some numbers in. > Let f=sqrt(2), m=1, y=1, which gives > (4 - 3(2) + 3)(2)x^3 - 3(-1 + 2)x + 1 = (a_1 x + 1)(a_2 x + 1)(a_3 x > + 1) > which is > 2x^3 - 3x + 1 = (a_1 x + 1)(a_2 x + 1)(a_3 x + 1) > and you can just look at it and see that x=1, works as a root of the > polynomial 2x^3 - 3x + 1. > That leaves room for only two of the a's to have sqrt(2) as a factor, > and in this case you can see it work IN the ring of algebraic > integers. > Notice that if you wish you can try your own numbers. As in the polynomial factorization: 65 x^3 - 12 x + 1 = (a_1 x + 1)(a_2 x + 1)(a_3 x + 1) for which you maintain that one of the a_i is coprime to 5 in the ring of algebraic integers? You know that each of the a_i's has a common non-unit factor with 5, in the ring of algebraic integers. You have seen the proof, which involves only ordinary arithmetic. How can any one of them be coprime to 5 in that ring? Answer that one. No, you won't, will you? You continue to sell the same pile of irrelevant junk, over and over again, but instead of producing the following sort of nonsense: > Mathematically the results are always the same without regard to how > reducible over rationals the primary cubic may be. > That's how math works. Problem is, your so-called math doesn't even work. All you can put together are toy problems in which the numbers that are supposed to be coprime to something are the roots of an irreducible polynomial, and the other root is not related to that polynomial. Your claims that it doesn't matter, well, they don't matter. All roots of an irreducible polynomial are indistinguishable using only arithmetic in the field of coefficients together with the root in question. Any relation held in that domain by one root will hold for any other, but in the analogous domain (i.e., the field of coefficients together with the new root). > If Galois Theory wasn't such a big deal there wouldn't have been all > the arguing, which has gone on for over two years now. This sentence doesn't mean anything, does it? > So, mathematically, my case is solid and absolute, but math society is > protecting its wrong beliefs about roots of polynomials irreducible > over Q, and to do that the people engaging in that activity cannot be > real mathematicians. Not. > They're social animals. A real mathematician would just accept the > math as correct and disdain the earlier false assumption as it's just > wrong. A real mathematician would stop this dancing around and answer the simple question I've asked for months: why, when you know that each a_i shares a non-unit algebraic integer factor with 5, in the ring of algebraic integers, do you continue to claim that one of the a_i is coprime to 5? Do you maintain that the ring of algebraic integers is so wacky that now the ordinary integers no longer work? After all, my proof only involves ordinary arithmetic. > Mathematicians don't defend wrong beliefs about mathematics, as they > actually love mathematics itself. I am not defending any belief. I'm asking a simple question: why do you claim to be correct, when ordinary arithmetic shows you that you are just wrong? > So, if you fight a correct result, to hold on to a false belief, you > prove you're not really a mathematician, no matter what people call > you or you call yourself. Do you claim that my result, which involves only the multiplication of ordinary integer polynomials, and the application of the simple fact that the a_i's satisfy a specific polynomial equation of degree 3, is not correct? My result says that each a_i has a non-unit factor in the ring of algebraic integers, which is also a factor of 5 in the same ring. You claim that one of the a_i's is coprime to 5 in that same ring. The two assertions are patently incompatible, being in direct contradiction of each other, and my result is backed up by the five steps that only depend on one using ordinary arithmetic, and logically entail the result. Your assertion is backed up by loosely-worded argument, in which nothing is asserted to belong to the ring of algebraic integers, and nothing actually proven, and by your continuing libelous* remarks concerning the honesty of an entire profession (*they would indeed be libelous if there were the slightest possibility that anyone could believe a word you write). If my result or my logic are not correct, what step is incorrect: In this outline, a refers to one of the coefficients that you claim must be coprime to 5. P(x) refers to the polynomial x^3 - 12 x^2 + 65, and I note that P(-a) = 0, for each of your values a. 1. The following formulas are true: q(x)r(x) = (64 x + 128)P(x) + 5 r(x)s(x) = (32 x + 72)P(x) + x, where q,r,s are defined as follows: q(x) = 8 x^2 - 76 x - 185 r(x) = 8 x^2 - 4 x - 45 s(x) = 4 x^2 - 37 x - 104 2. Since -a is a root of P(x), we have the following factorizations: q(-a)r(-a) = 5 r(-a)s(-a) = -a 3. The minimal polynomial of r(-a) is given as: MP_r = x^3 - 969 x^2 + 315 x + 5 which is irreducible over Q. This shows that r(-a) is an algebraic integer, but not a unit in that ring. shared by a and 5 is not a unit in the ring of algebraic integers. 5. a and 5 are not coprime in the ring of algebraic integers, since they share a non-unit factor in common. > That's the problem I think with the modern math world as somehow the > belief has taken hold that you have a social position, not a calling, > and that as long as society calls you a mathematician, then you are, > no matter how false mathematically what you believe is. The problem that I think is that you are so filled with the love of your own failed attempts at making something of your life that you cannot change. If you claim I am stating any mathematical falsehood, you'll stand up and find the error. The statements (1-5) of my argument are there for everyone to see, and if you cannot find error in them, then by your own standards, you must accede to their truth. > James Harris Dale. === Subject: Re: The actual math, advanced polynomial factorization > and all the arguing for over TWO YEARS NOW has been over what that means. What arguing? It goes like this: 1. You post something that is wrong. 2. Several people post counterexamples. 3. You ignore them and go back to step #1, occasionally pausing to tell everyone how great you are. The only thing of interest here is that after step #2, sometimes there is interesting discussion about the counterexamples, and how they are found, and how general they are. -- --Tim Smith === Subject: Re: The actual math, advanced polynomial factorization Discussion, linux) > and all the arguing for over TWO YEARS NOW has been over what that means. > What arguing? It goes like this: > 1. You post something that is wrong. > 2. Several people post counterexamples. > 3. You ignore them and go back to step #1, occasionally pausing to tell > everyone how great you are. > The only thing of interest here is that after step #2, sometimes there is > interesting discussion about the counterexamples, and how they are found, > and how general they are. *That's* the only thing of interest? Hey, I come for the occasional monologues on James's greatness. -- If you see math knowledge as a tool--as a hammer--with which you can attack other people then ... you defeat rational discourse. I get to call my proof the Hammer. It's more powerful than *any* physical object. It is overwhelming force. -- Two JSH quotes === Subject: The actual math, advanced polynomial factorization > 2x^3 - 3x + 1 = (a_1 x + 1)(a_2 x + 1)(a_3 x + 1) > and you can just look at it and see that x=1, works as a root of the > polynomial 2x^3 - 3x + 1. > That leaves room for only two of the a's to have sqrt(2) as a factor, > and in this case you can see it work IN the ring of algebraic > integers. The polynomial factors into 2 x^3 - 3 x + 1 = (x - 1)(2 x^2 + 2 x - 1) and the roots of 2 x^2 + 2 x - 1 are a = -1/2 - sqrt(3)/2 b = -1/2 + sqrt(3)/2 The product a b = -1/2, not an algebraic integer, so the roots are not algebraic integers. > So, mathematically, my case is solid and absolute You are trying to find a counterexample to theorems that were proved a century ago. A case based on such an attempt is doomed to failure. Stop digging your heels in like Balaam's ass and burying your head in the sand like an ostrich. Move on. -- === Subject: Units in algebraic integers I've given the example of x^2 + (2s_1 + 3s_2)x + s_1 s_2 6 = (x + 2s_1))(x + 3s_2) where s_1 s_2 = 1, and 2s_1 + 3s_2 is an integer to highlight the oddity that if s_1 and s_2 are rational, then trivially you can have both be units as both can equal 1, but if s_1 and s_2 are irrational, then they cannot be units in the ring of algebraic integers. It gets weirder. I picked 2 and 3 because they're the first two primes--and therefore coprime to each other--and are small, but let's replace them with Ôa' and b, non-unit coprime algebraic integers: x^2 + (as_1 + bs_2)x + s_1 s_2 ab = (x + as_1))(x + bs_2) where s_1 s_2 = 1, as_1 + bs_2 is an integer, and ab is an integer. Question: Can s_1 and s_2 be irrational and algebraic integers? James Harris === Subject: Re: Units in algebraic integers > I've given the example of > x^2 + (2s_1 + 3s_2)x + s_1 s_2 6 = (x + 2s_1))(x + 3s_2) > where s_1 s_2 = 1, and 2s_1 + 3s_2 is an integer > to highlight the oddity that if s_1 and s_2 are rational, then > trivially you can have both be units as both can equal 1, but if s_1 > and s_2 are irrational, then they cannot be units in the ring of > algebraic integers. It's silly to think of this as being in any way a fundamental, or even desirable, property of the ring of integers that should be preserved in the ring of algebraic integers. > It gets weirder. ^ ^ Typos > I picked 2 and 3 because they're the first two primes--and therefore > coprime to each other--and are small, but let's replace them with Ôa' > and b, non-unit coprime algebraic integers: > x^2 + (as_1 + bs_2)x + s_1 s_2 ab = (x + as_1))(x + bs_2) > where s_1 s_2 = 1, as_1 + bs_2 is an integer, and ab is an integer. > Question: Can s_1 and s_2 be irrational and algebraic integers? > James Harris Question: Why should it matter? What is it that makes you believe that the ring of algebraic integers should have whatever numbers *you* decree it should have? Bill Dubuque has pointed out an amusing fact. If you adjoin numbers to the ring of algebraic integers to form a larger ring, you are faced with a dilemma: Either the ring will favor *some* solutions of a polynomial equation over others (i.e., there will be an irreducible integer polynomial for which only some of the roots are in the bigger ring, and the rest won't be), or it will consist of the full field of algebraic numbers. I would regard this to be a fatal blow to a useful generalization of the notion of integrality. Imagine having a newinteger r, for which its irreducible polynomial had a non-newinteger root s. By Galois symmetry, the ring Q(r) is canonically isomorphic to Q(s). Every statement you can make in arithmetic that is true in Q(r) has a corresponding true statement in Q(s) that is obtained by simply replacing all r's by s's, yet s is not a newinteger. How on earth could this newinteger concept even be defined algebraically? It couldn't be a definition that involved the equality of any arithmetic expressions involving solely the rational numbers and the newinteger in question. Talk about rings with screwy properties, that would take the cake. All determination of membership would have to be done by fiat; I suppose that might bolster ßagging car sales, but can't imagine any further advantage. One wonders whether JSH himself should decide this issue for each and every algebraic number. I think so. Does anyone care to second the nomination? Dale === Subject: Re: Units in algebraic integers > Bill Dubuque has pointed out an amusing fact. If you adjoin numbers to > the ring of algebraic integers to form a larger ring, you are faced with > a dilemma: Either the ring will favor some solutions of a polynomial > equation over others (i.e., there will be an irreducible integer > polynomial for which only some of the roots are in the bigger ring, > and the rest won't be), or it will consist of the full field of > algebraic numbers. That's an incorrect characterization of the dilemma. I suggest you review my prior post which discusses this topic at length, see For a somewhat deeper viewpoint on integral extensions see also --Bill Dubuque === Subject: Re: Units in algebraic integers >Bill Dubuque has pointed out an amusing fact. If you adjoin numbers to >the ring of algebraic integers to form a larger ring, you are faced with >a dilemma: Either the ring will favor some solutions of a polynomial >equation over others (i.e., there will be an irreducible integer >polynomial for which only some of the roots are in the bigger ring, >and the rest won't be), or it will consist of the full field of >algebraic numbers. > That's an incorrect characterization of the dilemma. I suggest you > review my prior post which discusses this topic at length, see > For a somewhat deeper viewpoint on integral extensions see also > --Bill Dubuque who also was kind enough to set the record straight, this was clearly my error in mis-remembering something that had seemed clear at the time of posting (but, during the several hours in the meantime, had become progressively less clear). Sorry to have required both you guys to step in to clean up my mess. I'll try to be a bit more thorough in my use of the Google record in checking details. Dale. === Subject: Re: Units in algebraic integers days. My association with the Department is that of an alumnus. [.snip.] >Bill Dubuque has pointed out an amusing fact. If you adjoin numbers to >the ring of algebraic integers to form a larger ring, you are faced with >a dilemma: Either the ring will favor *some* solutions of a polynomial >equation over others (i.e., there will be an irreducible integer >polynomial for which only some of the roots are in the bigger ring, >and the rest won't be), or it will consist of the full field of >algebraic numbers. This last statement is incorrect. The alternative is not that the ring will be the full field of algebraic numbers; the alternative is that the ring will necessarily contain some rationals which are not integers. That is: some rational primes will become units. Because of that, this new ring will fail to reßect some of the arithmetical properties of the integers; and so it can only give partial information when dealing with a diophantine equation. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Units in algebraic integers > [.snip.] >Bill Dubuque has pointed out an amusing fact. If you adjoin numbers to >the ring of algebraic integers to form a larger ring, you are faced with >a dilemma: Either the ring will favor *some* solutions of a polynomial >equation over others (i.e., there will be an irreducible integer >polynomial for which only some of the roots are in the bigger ring, >and the rest won't be), or it will consist of the full field of >algebraic numbers. > This last statement is incorrect. The alternative is not that the > ring will be the full field of algebraic numbers; the alternative is > that the ring will necessarily contain some rationals which are not > integers. That is: some rational primes will become units. > Because of that, this new ring will fail to reßect some of the > arithmetical properties of the integers; and so it can only give > partial information when dealing with a diophantine equation. Oh, right. In the interim (since posting), I had been thinking I was missing something (else how could one obtain intermediate rings between the algebraic integers and the full set of algebraic numbers, of which there are plenty, including those which behave properly wrt the appropriate Galois groups). I made the mistake of relying on my own faulty memory of Dubuque's Dale. === Subject: Polynomial functions and equations discovered Polynomial functions and equations fully discovered and explained! Free to learn, study and use: linear, quadratic, cubic, quartic, quintic,...and the nth-degree polynomial. See http://www.nabla.hr === Subject: Re: Infinte strings of digits > No it isn't. It does *imply* the existence > of such a bijection, using the Axiom of Choice, > but it doesn't let you *construct* one, in > the sense I suspect Peter was looking for. > In fact no reasonably definable such bijection > exists. That's the interesting mathematical > point to be seen here, and it can be quite well-formulated > at the level of formality from Peter's original > post. Interestingly, although the existence of these bijections can't be proved without using something like the Axiom of Choice, it's not at all hard to find surjections in each direction. It's the injections that are difficult. Alan Stern === Subject: Zenkin's paper on Cantor Here is a much more refined attack on Cantor's proof than what has lately occured in some threads that I have avoided: http://www.com2com.ru/alexzen/papers/Cantor/Fatal_Mistake_of_ Cantor.html Does Alexander Zenkin have a good point? -- Eray === Subject: Re: Zenkin's paper on Cantor I think that it is wothwhile to consider Cantor's arguments with Seen from different perspectives, the real numbers actually vary in their composition. It's like the blind men and the elephant, except different: the elephant is the snake, hippo, and zebra. In this case the elephant is an elephant, but if you need a snake it'll do. My arguments that I can biject the naturals and reals are somewhat different than Zenkin's, I don't think I've ever argued that the expansion was finite, but many or most other lines of argument, some bad, some good, some dependent on other conditions, have been addressed over the years here on sci.math math among us. I don't validate Zenkin's argument at issue here. When I was first asked to provide a bijection between the naturals and reals, I replied with the Equivalency Function, so named because its purpose is to exhibit a bijective mapping between the naturals and unit interval of reals, among other things. That's working well not only in the face of the antidiagonal argument, but also Cantor's first proof, about which I say between any two definite rational endpoints at some finite iteration there exists a pair of irrationals, between which is, you guessed it: a rational. Anyways, EF introduces the notion of iota, which is basically defined as the smallest positive real number, that is basically Leibniz' dx from the integral calculus. Neither the antidiagonal argument nor the nested intervals arguments apply to EF, except perhaps in showing that EF is the only way to biject N and R[0,1]. As well, the reals are discussed vis-a-vis Dedekind cuts, Cauchy sequences, and the hyperreals of non-standard analysis, and about how Dedekind cuts and Cauchy sequences do not represent all of the hyperreals, each of which is a real number. There's still consideration of the powerset mapping result and Cantor-Bernstein transitivity, it is a broad consideration that in my view best fits within a foundationless set theory. A variety of analytical results pertaining to EF are discussed, for Most considerations of transfinite cardinals have little or no application in empirical or observable results. The ones that do have alternatives. Basically one way to consider the real numbers is that essentially regardless of their field attributes, that there is a contiguous sequence of them. There is a continuous sequence of them. Ross F. -- Also, consider this: the unit impulse function times one less twice the unit step function times plus/minus one is the mother of all wavelets. === Subject: Re: Zenkin's paper on Cantor > I think that it is wothwhile to consider Cantor's arguments with > Seen from different perspectives, the real numbers actually vary in > their composition. It's like the blind men and the elephant, except > different: the elephant is the snake, hippo, and zebra. In this case > the elephant is an elephant, but if you need a snake it'll do. This is a work of genius. Please don't ever explain this, you would only spoil it. > Ross F. === Subject: Re: Zenkin's paper on Cantor > Here is a much more refined attack on Cantor's proof than what has > lately occured in some threads that I have avoided: > http://www.com2com.ru/alexzen/papers/Cantor/Fatal_Mistake_of_ Cantor.html > Does Alexander Zenkin have a good point? I really didn't try to understand Zenkin's paper, but from what I have gathered at looking at all of the Cantor-bashers, they simply don't like the idea of infinite sets existing, period. This tradition goes back to the ancient Greeks. This thinking even permeated religious circles, Jewish and Christian (and probably Muslim too) during the Middle Ages. If you read texts written during that era which give arguments for the existence of God (proof that since there must be a first cause, there must be God), you'll see that none of them even consider the possibility of an infinite chain of causes; they roughly say that everything has a cause and since the number of causes must be finite, it must end somewhere and that somewhere is God, the First Cause. These types of arguments were given by Maimonides and Aquinas. See http://www.princeton.edu/~grosen/puc/phi203/cosmological.html The thinking is this: Because no one has ever observed an infinite set, there is no evidence for such a set existing in reality. Only potential infinity (the idea of infinity) is real, but actual infinity has never been demonstrated. If you believe this, then it is only natural that you will find faults in Cantor's argument. Craig === Subject: Re: Zenkin's paper on Cantor > Here is a much more refined attack on Cantor's proof than what has > lately occured in some threads that I have avoided: > > http://www.com2com.ru/alexzen/papers/Cantor/Fatal_Mistake_of_ Cantor.html > > Does Alexander Zenkin have a good point? > > I really didn't try to understand Zenkin's paper, but from what I have > gathered at looking at all of the Cantor-bashers, they simply don't > like the idea of infinite sets existing, period. This tradition goes > back to the ancient Greeks. > This thinking even permeated religious circles, Jewish and Christian > (and probably Muslim too) during the Middle Ages. If you read texts > written during that era which give arguments for the existence of God > (proof that since there must be a first cause, there must be God), > you'll see that none of them even consider the possibility of an > infinite chain of causes; they roughly say that everything has a cause > and since the number of causes must be finite, it must end somewhere > and that somewhere is God, the First Cause. These types of arguments > were given by Maimonides and Aquinas. > See http://www.princeton.edu/~grosen/puc/phi203/cosmological.html > The thinking is this: Because no one has ever observed an infinite > set, there is no evidence for such a set existing in reality. Only > potential infinity (the idea of infinity) is real, but actual infinity > has never been demonstrated. > If you believe this, then it is only natural that you will find faults > in Cantor's argument. Hello Craig, cosmological argument, Aristotle's view of infinity and Cantor's theory. The interesting thing is how these strange philosophical views can find their way into mathematics. If mathematics were objective, then this would have been impossible, e.g. no metaphysical theory should inßuence the findings in molecular biology. At any rate, Zenkin does not seem to be rigorous at all in this paper (perhaps because his english wasn't that good?), but it seems that he has taken the effort to present his arguments on the FOM list. That's why I paid attention to his paper. a journal, which mentions the same incommensurability between potential and actual infinity. More philosophical than Zenkin's, but the same lack of rigor, I think... (I didn't find his refutation convincing) Ouch, in fact, this is downright bad, in the section named mixes and muddles he does sound pretty muddled :) http://www.cantorsdonutparadox.co.uk/index.asp -- Eray === Subject: Re: Zenkin's paper on Cantor > Here is a much more refined attack on Cantor's proof than what has > lately occured in some threads that I have avoided: > http://www.com2com.ru/alexzen/papers/Cantor/Fatal_Mistake_of_ Cantor.html > Does Alexander Zenkin have a good point? > I really didn't try to understand Zenkin's paper, but from what I have > gathered at looking at all of the Cantor-bashers, they simply don't > like the idea of infinite sets existing, period. This tradition goes > back to the ancient Greeks. No, that is not what most anti-Cantor cranks believe. They believe that R is equinumerous with N. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: Zenkin's paper on Cantor > No, that is not what most anti-Cantor cranks believe. They believe > that R is equinumerous with N. For every method you can come up with of _not_ listing all members of R I can come up with a method of _not_ listing all members of N. The conclusion is obvious. Phil ;-) -- They no longer do my traditional winks tournament lunch - liver and bacon. It's just what you need during a winks tournament lunchtime to replace lost === Subject: Re: Zenkin's paper on Cantor X-URL: http://mygate.mailgate.org/mynews/comp/comp.theory/ af7c172412d96b0049e84a507e b8eef1.48257%40mygate.mailgate.org [inane request for assistance understanding self-evident nonsense] > http://www.com2com.ru/alexzen/papers/Cantor/Fatal_Mistake_of_ Cantor.html What an excellent measure of your lack of credentials as a comuputer theorist that you had to ask for help in evaluating the merits of such drivel. Do play on pretending to belong here, but the evidence of your (lack of any meaningful) skills has the stench of Olcottism rising from it in ever more suffocating waves. Despite having your kindergarden-level errors of comprehension of the simplest concepts pointed out to you by many competent commentators, you continue to misuse the most clearly defined terminology, you continue to let intuition replace reason, science, and logic, in choosing what to believe, you continue to defend your profound ignorance as if it were your most valued possession, with insults, sneering, handwaving, muddle commentary and deliberate obfuscation. Where have we seen that behavior before? Newsgroup comp.ai.philosophy, where no computer theory skills are required or expected, feels all your attention really belongs right there. Hear it calling you home? xanthian. By the way, there are half a dozen kinds of limit trees, in fields from formal language theory to constructivist set theory to transfinite number theory to graph theory, and none of them make your ideas make sense. With six semesters of advanced calculus, I at least understand what a limit is, something you have so far failed to grasp, as Owen Jacobson highlighted for you in his recent response. I'm working on limit trees, but I have to go clear back to relearn set theory to get there. -- === Subject: Re: Zenkin's paper on Cantor X-RFC2646: Original [Eray Ozkural] > Here is a much more refined attack on Cantor's proof than what has > lately occured in some threads that I have avoided: > http://www.com2com.ru/alexzen/papers/Cantor/Fatal_Mistake_of_ Cantor.html > Does Alexander Zenkin have a good point? I expect he stumbled into a truth when he remarked that Zigmund Freud should be consulted for a real understanding of papers like his . === Subject: Re: Zenkin's paper on Cantor >Here is a much more refined attack on Cantor's proof than what has >lately occured in some threads that I have avoided: >http://www.com2com.ru/alexzen/papers/Cantor/Fatal_Mistake_of _Cantor.html Refined. Right. He says he's _quoting_ the proof, but the version of the proof that he gives actually _is_ wrong (although of course he doesn't mention the actual error.) >Does Alexander Zenkin have a good point? What a stupid question. ************************ David C. Ullrich === Subject: Re: Zenkin's paper on Cantor >Here is a much more refined attack on Cantor's proof than what has >lately occured in some threads that I have avoided: > >http://www.com2com.ru/alexzen/papers/Cantor/Fatal_Mistake_of _Cantor.html > Refined. Right. He says he's _quoting_ the proof, but the version > of the proof that he gives actually _is_ wrong (although of course > he doesn't mention the actual error.) >Does Alexander Zenkin have a good point? > What a stupid question. Well, his paper seems to have been published somewhere, so I did not want to categorize his work either way. Do you mean that the journal is full of crackpots? I've been told there are such journals, but its name looked kind of serious, I can't tell :) -- Eray === Subject: Re: Zenkin's paper on Cantor >Here is a much more refined attack on Cantor's proof than what has >lately occured in some threads that I have avoided: > > >http://www.com2com.ru/alexzen/papers/Cantor/Fatal_Mistake_of _Cantor.html > Refined. Right. He says he's _quoting_ the proof, but the version > of the proof that he gives actually _is_ wrong (although of course > he doesn't mention the actual error.) >Does Alexander Zenkin have a good point? > What a stupid question. >Well, his paper seems to have been published somewhere, so I did not >want to categorize his work either way. Do you mean that the journal >is full of crackpots? Uh, no, I meant that asking whether someone pointing out the fatal ßaw in a simple theorem has a good point is stupid. You should really, um, ... I didn't pay attention to much of those threads on Chaitin's Omega. My impression was you were probably making some sort of sense and the disputes were due to the fact that things like completely random and true for no reason are not well defined, so that A could say this is true for no reason, B could say that that's false, and what A meant and what B meant could both be true. But lately, talking about infinite decimals and related things, you've been sounding totally incompetent and/or totally ignorant. (I point this out just to suggest that you might want to avoid looking that way, by avoiding discussion of questions that you really have absolutely no understanding of.) >I've been told there are such journals, but its >name looked kind of serious, I can't tell :) Most of us don't decide whether a mathematical argument is valid based on the name of the journal. (Of course nobody knows all of mathematics, so there are plenty of papers that any given person is unable to evaluate for himself. But the topic of the present paper is utterly elementary...) ************************ David C. Ullrich === Subject: Re: Zenkin's paper on Cantor >Well, his paper seems to have been published somewhere, so I did not >want to categorize his work either way. Do you mean that the journal >is full of crackpots? > Uh, no, I meant that asking whether someone pointing out the > fatal ßaw in a simple theorem has a good point is stupid. You view the theorem as simple. Ok. Then, perhaps the truth of the continuum hypothesis is a simple question, too? I suppose you view it as false, since modern set theorists view it as false. -- Eray Ozkural === Subject: Re: Zenkin's paper on Cantor > Then, perhaps the truth of the continuum hypothesis is a simple > question, too? What is your definition of truth in this context? I seems to remember that Goedel and Cohen have prooven it is undecidable in the standard set theory (and if my memory is good, with or without the axiom of choice). A+ -- Jean-Marc === Subject: Re: Zenkin's paper on Cantor >Well, his paper seems to have been published somewhere, so I did not >want to categorize his work either way. Do you mean that the journal >is full of crackpots? > Uh, no, I meant that asking whether someone pointing out the > fatal ßaw in a simple theorem has a good point is stupid. > You view the theorem as simple. I think almost all mathematicians do. > Ok. Then, perhaps the truth of the > continuum hypothesis is a simple question, too? Now David didn't say that, did he? You are setting up another straw man :-( > I suppose you view it > as false, since modern set theorists view it as false. How they view it is irrelevant since they know it is independent of ZFC (assuming con(ZFC)). -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: Zenkin's paper on Cantor >Here is a much more refined attack on Cantor's proof than what has >lately occured in some threads that I have avoided: > > >http://www.com2com.ru/alexzen/papers/Cantor/Fatal_Mistake_of _Cantor.html > Refined. Right. He says he's _quoting_ the proof, but the version > of the proof that he gives actually _is_ wrong (although of course > he doesn't mention the actual error.) >Does Alexander Zenkin have a good point? > What a stupid question. > Well, his paper seems to have been published somewhere, so I did not > want to categorize his work either way. How pathetic. You suspend your judgment because a paper has been published somewhere. This is craven deference to authority indeed, > Do you mean that the journal > is full of crackpots? I've been told there are such journals, but its > name looked kind of serious, I can't tell :) No judgment at all then :-( -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: Zenkin's paper on Cantor Originator: mtx014@linux.services.coventry.ac.uk (Robert Low) >Well, his paper seems to have been published somewhere, so I did not >want to categorize his work either way. Do you mean that the journal Have confidence in your understanding, man. If it's nonsense, it's nonsense, even if it's published somewhere. I doubt there's any journal that has never published nonsense. >is full of crackpots? Can't say. But don't you think it's kind of telling that a claim that Cantor's diagonalisation argument is technically ßawed should be published in what looks like a general philosophy journal, rather than a mathematics or logic one? (I'm sure there's a technical term for Ôrefutation by innuendo'; if not, I need one now.) -- Rob. http://www.mis.coventry.ac.uk/~mtx014/ === Subject: Re: Zenkin's paper on Cantor >Well, his paper seems to have been published somewhere, so I did not >want to categorize his work either way. Do you mean that the journal > Have confidence in your understanding, man. If it's nonsense, > it's nonsense, even if it's published somewhere. I doubt > there's any journal that has never published nonsense. >is full of crackpots? > Can't say. But don't you think it's kind of telling > that a claim that Cantor's diagonalisation argument > is technically ßawed should be published in what > looks like a general philosophy journal, rather than > a mathematics or logic one? (I'm sure there's a technical > term for Ôrefutation by innuendo'; if not, I need > one now.) Yes, I fully agree. It should have been published in a professional math journal. And due to two or more ßaws demonstrated on this thread, this would have been impossible. -- Eray Ozkural === Subject: Re: Zenkin's paper on Cantor >Well, his paper seems to have been published somewhere, so I did not >want to categorize his work either way. Do you mean that the journal > Have confidence in your understanding, man. If it's nonsense, > it's nonsense, even if it's published somewhere. I doubt > there's any journal that has never published nonsense. >is full of crackpots? > Can't say. But don't you think it's kind of telling > that a claim that Cantor's diagonalisation argument > is technically ßawed should be published in what > looks like a general philosophy journal, rather than > a mathematics or logic one? (I'm sure there's a technical > term for Ôrefutation by innuendo'; if not, I need > one now.) > Yes, I fully agree. It should have been published in a professional > math journal. Do you mean mathematics/mathematical when you say math? Anyway, it should not have been published in a mathematics journal --- it would have been a waste of paper. > And due to two or more ßaws demonstrated on this > thread, ßaws? > this would have been impossible. Impossible surely since no mathematician would want to publish such drivel. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: Zenkin's paper on Cantor >Here is a much more refined attack on Cantor's proof than what has >lately occured in some threads that I have avoided: >http://www.com2com.ru/alexzen/papers/Cantor/Fatal_Mistake_ of_Cantor.html > Refined. Right. He says he's _quoting_ the proof, but the version > of the proof that he gives actually _is_ wrong (although of course > he doesn't mention the actual error.) >Does Alexander Zenkin have a good point? > What a stupid question. David, Didn't anyone tell you There are no stupid questions....there are only stupid people ? -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: Zenkin's paper on Cantor > Here is a much more refined attack Hardly. > on Cantor's proof than what has > lately occured in some threads that I have avoided: Then how can you compare? > http://www.com2com.ru/alexzen/papers/Cantor/Fatal_Mistake_of_ Cantor.html > Does Alexander Zenkin have a good point? No. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: Zenkin's paper on Cantor > Here is a much more refined attack on Cantor's proof than what has > lately occured in some threads that I have avoided: It's exactly as refined as all such arguments. === Subject: Re: Zenkin's paper on Cantor > Here is a much more refined attack on Cantor's proof than what has > lately occured in some threads that I have avoided: > http://www.com2com.ru/alexzen/papers/Cantor/Fatal_Mistake_of_ Cantor.html > Does Alexander Zenkin have a good point? I quit reading the paper about halfway through when he started claiming that Cantor's method was invalid because it doesn't distinguish finite sets from infintie ones. It was obvious anyway from the beginning that he's a crank because of the method of his arguments: he discusses Cantor's proof, and indeed set theory and metamathematics in general, with a lofty air of slight disdain for the reader, using terms like famous diagonal method to make the reader feel either impressed or ignorant (or both) and making very general claims like The one and only basis for the differentiation of such infinities is [Cantor's theorem] without ever providing evidence. The idea in such a tactic is to con the reader into thinking that the author is an authority who need not be bothered with trivialities, so that the reader doesn't notice that all the author ever discusses are trivialities. He never says anything deep about the theorem or its proof, but merely latches onto a single irrelevant feature which he perceives to be of critical importance (the finite/infinite disparity) because he totally misunderstands set theory. After that (okay, I chickened out and read the rest) he goes on to trot out the same argument that all the cranks do, namely that adding in the diagonal number must make the list complete after all. His justification demonstrates a lack of understanding of the nature of infinity and also of logic itself. In fact, he says that the list is complete at this moment and then goes ahead and uses the diagonal argument again, which indicates that he's heard the usual objections to this method but, having failed to understand them, he just sits on the fence and both accepts and denies them at once, hoping no doubt that when someone tries to tear him down he can answer them no matter what their problem is. Once he goes through some faulty logic based mostly on the emphatic restatement of certain semantically ambiguous phrases like actuality he considers himself to be finished, and concludes his paper with the traditional appeal to authority, exhorting the reader to his point of view on the grounds that such luminaries as Aristotle, Leibniz, Cauchy, etc. all had personal dislikes for infinite objects, as if this made much difference to the proof itself. Then, bizarrely, he closes with a reference to Freud. I'd have to say that the paper is mostly a shell game involving sophisticated-sounding philosophical terms, hair-fine distinctions that have no bearing on the question at hand, and self-contradictions. They should frame this and use it in an exhibit on detecting frauds. -- Ryan Reich ryanr@uchicago.edu === Subject: Re: Zenkin's paper on Cantor > Here is a much more refined attack on Cantor's proof than what has > lately occured in some threads that I have avoided: > > http://www.com2com.ru/alexzen/papers/Cantor/Fatal_Mistake_of_ Cantor.html > > Does Alexander Zenkin have a good point? > I quit reading the paper about halfway through when he started claiming that > Cantor's method was invalid because it doesn't distinguish finite sets from > infintie ones. It was obvious anyway from the beginning that he's a crank > because of the method of his arguments: he discusses Cantor's proof, and > indeed set theory and metamathematics in general, with a lofty air of > slight disdain for the reader, using terms like famous diagonal method to > make the reader feel either impressed or ignorant (or both) and making very > general claims like The one and only basis for the differentiation of such > infinities is [Cantor's theorem] without ever providing evidence. The > idea in such a tactic is to con the reader into thinking that the author is > an authority who need not be bothered with trivialities, so that the reader > doesn't notice that all the author ever discusses are trivialities. He > never says anything deep about the theorem or its proof, but merely latches > onto a single irrelevant feature which he perceives to be of critical > importance (the finite/infinite disparity) because he totally > misunderstands set theory. Yes, yes, Cantor does not have only a single proof, right? But he distorts this fact. Interesting. He must be a crank, then. -- Eray === Subject: Re: Zenkin's paper on Cantor Originator: mtx014@linux.services.coventry.ac.uk (Robert Low) >Here is a much more refined attack on Cantor's proof than what has >lately occured in some threads that I have avoided: >http://www.com2com.ru/alexzen/papers/Cantor/Fatal_Mistake_of _Cantor.html >Does Alexander Zenkin have a good point? Maybe the one on the top of his head. For example, he seems to be under the impression that if you list all the finite expansions of the form 0.b1b2b3...bn where each bi is either 0 or 1, then there are only n of them. -- Rob. http://www.mis.coventry.ac.uk/~mtx014/ === Subject: Re: Zenkin's paper on Cantor >Here is a much more refined attack on Cantor's proof than what has >lately occured in some threads that I have avoided: > >http://www.com2com.ru/alexzen/papers/Cantor/Fatal_Mistake_of _Cantor.html >Does Alexander Zenkin have a good point? > Maybe the one on the top of his head. For example, he > seems to be under the impression that if you list all > the finite expansions of the form > 0.b1b2b3...bn > where each bi is either 0 or 1, then there are only > n of them. I just skimmed over the paper, and I hadn't noticed this. That would be a folly... -- Eray === Subject: Re: Zenkin's paper on Cantor >Here is a much more refined attack on Cantor's proof than what has >lately occured in some threads that I have avoided: > >http://www.com2com.ru/alexzen/papers/Cantor/Fatal_Mistake_of _Cantor.html >Does Alexander Zenkin have a good point? > Maybe the one on the top of his head. For example, he > seems to be under the impression that if you list all > the finite expansions of the form > 0.b1b2b3...bn > where each bi is either 0 or 1, then there are only > n of them. > I just skimmed over the paper, and I hadn't noticed this. That would be a > folly... Don't you ever pay attention? :-( -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: um.....this... hello.....doctor~ let x > 0 show that 0< (1/2)*ln[{(e^x)-1} / x] <1 --------------------------------- 0 < .......i can show it. let f(x) = (e^x) - 1 by mean value theorem, (e^x)-1 / x = f'(c) , c in (0,00) so, (e^x)-1 / x = f'(c) = e^c > 1 thus, ln[{(e^x)-1} / x] > ln 1 = 0 but, i can't show < 1. i need your advice, please. thank you very much. === Subject: Re: um.....this... X-RFC2646: Original mina_world escribi.97: > hello.....doctor~ > let x > 0 > show that 0< (1/2)*ln[{(e^x)-1} / x] <1 > --------------------------------- > 0 < .......i can show it. > let f(x) = (e^x) - 1 > by mean value theorem, > (e^x)-1 / x = f'(c) , c in (0,00) > so, > (e^x)-1 / x = f'(c) = e^c > 1 > thus, ln[{(e^x)-1} / x] > ln 1 = 0 > but, i can't show < 1. It would be very curious if you can do .... Because Lim(ln((e^x - 1)/x), x, +inf) = +inf .... ! -- Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: Re: um.....this... > mina_world escribi? > hello.....doctor~ > let x > 0 > show that 0< (1/2)*ln[{(e^x)-1} / x] <1 > --------------------------------- > 0 < .......i can show it. > let f(x) = (e^x) - 1 > by mean value theorem, > (e^x)-1 / x = f'(c) , c in (0,00) > so, > (e^x)-1 / x = f'(c) = e^c > 1 > thus, ln[{(e^x)-1} / x] > ln 1 = 0 > but, i can't show < 1. > It would be very curious if you can do .... > Because Lim(ln((e^x - 1)/x), x, +inf) = +inf .... ! oh....i am sorry. it looks like that this problem has the error itself. thank you very much. === Subject: Multi dimensional geometry Hi! In 2 dimensions a squares diagonal is root 2 if the sides are 1. In 3d it is root 3. I can not visualize 4d but is the diagonal root 4 then? Is it just the same all the way so a 100 d cube is root 100? Is there a way to visualize 4d and 5d cubes for example? Softwares? === Subject: Re: Multi dimensional geometry >Hi! In 2 dimensions a squares diagonal is root 2 if the sides are 1. In 3d >it is root 3. I can not visualize 4d but is the diagonal root 4 then? >Is it just the same all the way so a 100 d cube is root 100? Yes. >Is there a way to visualize 4d and 5d cubes for example? Softwares? Sometimes it's useful to think of time as a 4th dimension. You could imagine a 4-d hypercube by imagining a 3-d cube which appears, stays for 1 unit of time (which you somehow must equate with 1 unit of space), and then disappears. This is a hypercube which has one set of edges parallel to the time axis. If you oriented the hypercube differently you'd get some shape (I'm not sure what) which expanded from a point to something roughly cube-sized and then contracted to a point again, after something between 1 and sqrt(4) units of time. You can also project a hyperdimensional object onto a plane or hyperplane, as we commonly do to view 3-d objects in a 2-d space such as a computer screen. --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. === Subject: Re: Multi dimensional geometry > Hi! In 2 dimensions a squares diagonal is root 2 if the sides are 1. In 3d > it is root 3. I guess that what you mean is that the length of an inner diagonal of a 3-dimensinal cube whose side measures 1 is the square root of 3. Yes, you're right. > I can not visualize 4d but is the diagonal root 4 then? The analogous diagonal has length square root of 4 (=2), yes. > Is it just the same all the way so a 100 d cube is root 100? Yes, if you make the same kind of adjustment as I made above. Jose Carlos Santos === Subject: easy integral.... hello...doctor~ integral 1 / [sqrt{(x^2) + (a^2)}] dx ---------------------------------- x= a tan u (-pi/2 < u < pi/2) [sqrt{(x^2) + (a^2)}] = a sin u dx = a (sec u)^2 du so, integral 1 / [sqrt{(x^2) + (a^2)}] dx = integral (1 / [a sin u])*{a (sec u)^2} du = integral sec u du = ln |sec u + tan u| + C (this process is omission) = ln |[sqrt{(x^2) + (a^2)}/a] + [x/a]| + C --------------------------------------- um.....but answer is ln |sqrt{(x^2) + (a^2)} + x| + C i can't find my error utterly. help me, please. thank you very much for your advice. === Subject: Re: easy integral.... ETAtAhUApkLz7j+uFVd4N9MEHtlC/dgc55YCFD+ K6Dvbdb72LPszm1dqiyXS9HHr Consider the indefinite integral of (x-1) dx. Term by term it's (1/2) x^2 - x + C. Or let u = x-1, so (x-1) dx = u du and the indefinite integral then becomes (1/2)(u^2) + C = (1/2)(x-1)^2 + C. But the two C's are arbitraty and don't really need to be the same in the two formulas. Perhaps the second formula should be written with C' instead of C. If, for example, you select C = 1 in the first antiderivative I give above, and decide upon C' = 1/2 in the second, how then do the funcitons compare? Why? What is the difference, for all x, between (1/2)x^2 - x and (1/2)(x-1)^2? Now look up the properties of logarthms and observe that the same type of situation arises in your case, if a is regarded as a constant. --OL === Subject: Re: easy integral.... > hello...doctor~ > integral 1 / [sqrt{(x^2) + (a^2)}] dx > = ln |[sqrt{(x^2) + (a^2)}/a] + [x/a]| + C > --------------------------------------- > um.....but answer is ln |sqrt{(x^2) + (a^2)} + x| + C > i can't find my error utterly. They're both correct, differing only but a quantity (depending on the parameter a) which is constant with respect to x. The latter form is nicer, however, since it allows us to use a=0. BTW, let me suggest that you avoid needless grouping symbols. For example, ln | sqrt(x^2 + a^2)/a + x/a | + C is easier to read than your ln |[sqrt{(x^2) + (a^2)}/a] + [x/a]| + C. David === Subject: Re: easy integral.... > hello...doctor~ > integral 1 / [sqrt{(x^2) + (a^2)}] dx > = ln |[sqrt{(x^2) + (a^2)}/a] + [x/a]| + C > --------------------------------------- > um.....but answer is ln |sqrt{(x^2) + (a^2)} + x| + C > i can't find my error utterly. > They're both correct, differing only but a quantity (depending on the > parameter a) which is constant with respect to x. The latter form is > nicer, however, since it allows us to use a=0. On second thought, that's only half true, so to speak. If we say that an antiderivative is log|Sqrt(x^2 + a^2) + x| and then try to use a = 0, everything's fine if x is positive, but it's not valid if x is negative. (This is starting to remind me of my recent thread A Ôbetter' antiderivative of x^p?) So we might now ask if we can write an antiderivative of 1/Sqrt(x^2 + a^2) with respect to x which, if a = 0, will work correctly regardless of whether x is positive or negative. I suggest that sign(x) log(Sqrt(x^2 + a^2) + |x|) is such an antiderivative. Alas, having fixed that problem has created another one, which is arguably worse: This new antiderivative is not continuous at x = 0 (except in the special case when a^2 = 1). David Cantrell === Subject: Re: easy integral.... X-RFC2646: Original mina_world escribi.97: > hello...doctor~ > integral 1 / [sqrt{(x^2) + (a^2)}] dx > ---------------------------------- > x= a tan u (-pi/2 < u < pi/2) > [sqrt{(x^2) + (a^2)}] = a sin u > dx = a (sec u)^2 du > so, > integral 1 / [sqrt{(x^2) + (a^2)}] dx > = integral (1 / [a sin u])*{a (sec u)^2} du > = integral sec u du > = ln |sec u + tan u| + C (this process is omission) > = ln |[sqrt{(x^2) + (a^2)}/a] + [x/a]| + C It must be = ln |[sqrt{(x^2) + (a^2)}/a + x/a]| + C An that is = ln |[sqrt{(x^2) + (a^2)} + x]/a| + C = ln |sqrt{(x^2) + (a^2)} + x| - ln(a) + C = ln |sqrt{(x^2) + (a^2)} + x| + C' with C' = -ln(a) + C -- Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com > --------------------------------------- > um.....but answer is ln |sqrt{(x^2) + (a^2)} + x| + C > i can't find my error utterly. > help me, please. > thank you very much for your advice. === Subject: Re: Philosophical meaning of possibility to separate variables by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i8QGZBC31147; >In partial differential equations, a useful method of solving the >equations arising is the method of separation of variables. In most >situations for phycisists, this is possible. Is there a >philosophical explanation for this? That is, what fundamental >property of a phenomena is required for this to be possible? Philosophical?? Whether an equation is separable or not depends entirely on the coordinate system which is determined by the person, not the physical situation. The philosophical point is that physicists like doing things the easy way so they set up their coordinate systems to make the problem easy. === Subject: Re: Philosophical meaning of possibility to separate variables >Philosophical?? Whether an equation is separable or not depends entirely on >the coordinate system which is determined by the person, not the physical >situation. With philosophical I meant an explanation involving the phenomena rather than the definition a DE in m variables is separable if it can be written as a product of etc... which only involves the DE modelling the phenomena. >The philosophical point is that physicists like doing things the easy way so >they set up their coordinate systems to make the problem easy. And the essence is that there _is_ such a coordinate system. I believe the other post by Igor answered my question but I don't see the connection between symmetry and separability. === Subject: Re: Fake (imaginary) numbers by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i8QGZBP31142; >Are their any guides to Fake (imaginary) numbers for a Neophite and >10th grader like me ???? === Subject: Re: Complex, singularities, exponentials, roots. by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i8QGZB931138; > 2)Show that there is a holomorphic root of z^2-1 > in the region C[-1,1]. > > This seems confusing, since there is no holomorphic > log of z^2-1 in this region and the sqr. is defined > in function of the log: > > If there was a hol. log g(z), then > > e^g(z)=z^2-1 =>e^g(z)*g'(z)=(z^2-1)*g'(z)=2z => > > g'(z)=2z/(z^2-1) . With a change of variables, > > this last becomes g'(z)=du/u , but there is no > > log here, since C[-1,1] winds around 0. > > How can there be a root if the root is defined > in function of the log, but the log does not exist? > > Yes, this one is a little tricky. But no, the square root is > not _defined_ in terms of the log. i)Re: definition of z^a , not in functionof exp: Don't we define z^a as e^a*logz ? ii)Re the necessary condition on the region R that R not wind around the origin: Is the reason that winding around the origin implies a change of 2*pi in the argument forcing a discontinuity in the value of the argument?. > > Anyway, this fact is a consequence of the following: > > Theorem. Suppose V is an open set, f is holomorphic in V, > and f has no zero in V. There exists g holomorphic in V > with g^2 = f if and only if > > (*) (1/2pi i) int_C f'(z)/f(z) dz = an _even_ integer > > for every closed curve C in V. > > Pf: One direction is easy: if f = g^2 then f'/f = 2g'/g. > > The other direction involves analytic continuation. I'm > going to describe the argument a little informally: > > Suppose (*) holds. Fix a point p in V, and _attemmpt_ > to define F = log(f) in V by setting > > (**) F(z) = int_C f'(w)/f(w) dw + log(f(p)), > > where log(f(p)) is any logarithm of f(p) and C is a > curve from p to z. > > Now if that actually worked you'd have F -= log(f), > because F(p) = log(f(p)) and F' = f'/f. But of course > it doesn't work: Since the integral of f'/f need not > be zero, (**) can give many different definitions > for F(z) depending on _which_ curve C from p to z > you use. > > So (**) defines a multi-valued function F, which > does have the property that exp(F) = f. Now note > that (*) shows that any two values of F(z) (for > a fixed z) differ by n*2*pi*i, where n is an > _even_ integer. This shows that g = exp(F/2) > is a single-valued function, an actual function > holomorphic in V. And g^2 = exp(2*F/2) = exp(F) = f, > qed. > > (Hint for using this in your problem: If C is a > closed curve in V then the function Ind(z,C), > the winding number of C about z, is continuous > in the complement of V (and integer-valued, > hence constant in connected subsets of the > complement of V.) >There's a more elementary approach. Take the usual square roots of z-1 and >z+1 and multiply them to get > f(z) = |z^2-1|^(1/2)*exp(i[arg(z-1) + arg(z+1)]/2) >for z in C (-oo,1]. f is analytic on this domain and f(z)^2 = z^2-1 >there. But now check that f(z) = -f(-z) for all z in C R (it's enough to >check the positive imaginary axis). Because f is analytic across (1,oo), f >extends analytically across (-oo,-1), yielding an analytic square root of >z^2-1 on C [-1,1]. >Well of course that's right, and probably better in the context of >qualifying exams. Seems interesting to know exactly what functions >have square roots, though. (Actually it was about a year ago that >exactly the current question came up on the complex-exam committee >here - what you write is the first thing we came up with, but I >didn't feel I really understood the story until I came up with >************************ >David C. Ullrich === Subject: Re: Complex, singularities, exponentials, roots. > 2)Show that there is a holomorphic root of z^2-1 > in the region C[-1,1]. > > This seems confusing, since there is no holomorphic > log of z^2-1 in this region and the sqr. is defined > in function of the log: > > If there was a hol. log g(z), then > > e^g(z)=z^2-1 =>e^g(z)*g'(z)=(z^2-1)*g'(z)=2z => > > g'(z)=2z/(z^2-1) . With a change of variables, > > this last becomes g'(z)=du/u , but there is no > > log here, since C[-1,1] winds around 0. > > How can there be a root if the root is defined > in function of the log, but the log does not exist? > > Yes, this one is a little tricky. But no, the square root is > not _defined_ in terms of the log. > i)Re: definition of z^a , not in functionof exp: > Don't we define z^a as e^a*logz ? g is a holomorphic square root of f if g^2 = f. > ii)Re the necessary condition on the region R that R > not wind around the origin: ??? Necessary for what? > Is the reason that winding > around the origin implies a change of 2*pi in the argument > forcing a discontinuity in the value of the argument?. > > > Anyway, this fact is a consequence of the following: > > Theorem. Suppose V is an open set, f is holomorphic in V, > and f has no zero in V. There exists g holomorphic in V > with g^2 = f if and only if > > (*) (1/2pi i) int_C f'(z)/f(z) dz = an _even_ integer > > for every closed curve C in V. > > Pf: One direction is easy: if f = g^2 then f'/f = 2g'/g. > > The other direction involves analytic continuation. I'm > going to describe the argument a little informally: > > Suppose (*) holds. Fix a point p in V, and _attemmpt_ > to define F = log(f) in V by setting > > (**) F(z) = int_C f'(w)/f(w) dw + log(f(p)), > > where log(f(p)) is any logarithm of f(p) and C is a > curve from p to z. > > Now if that actually worked you'd have F -= log(f), > because F(p) = log(f(p)) and F' = f'/f. But of course > it doesn't work: Since the integral of f'/f need not > be zero, (**) can give many different definitions > for F(z) depending on _which_ curve C from p to z > you use. > > So (**) defines a multi-valued function F, which > does have the property that exp(F) = f. Now note > that (*) shows that any two values of F(z) (for > a fixed z) differ by n*2*pi*i, where n is an > _even_ integer. This shows that g = exp(F/2) > is a single-valued function, an actual function > holomorphic in V. And g^2 = exp(2*F/2) = exp(F) = f, > qed. > > (Hint for using this in your problem: If C is a > closed curve in V then the function Ind(z,C), > the winding number of C about z, is continuous > in the complement of V (and integer-valued, > hence constant in connected subsets of the > complement of V.) >There's a more elementary approach. Take the usual square roots of z-1 and >z+1 and multiply them to get > f(z) = |z^2-1|^(1/2)*exp(i[arg(z-1) + arg(z+1)]/2) >for z in C (-oo,1]. f is analytic on this domain and f(z)^2 = z^2-1 >there. But now check that f(z) = -f(-z) for all z in C R (it's enough to >check the positive imaginary axis). Because f is analytic across (1,oo), f >extends analytically across (-oo,-1), yielding an analytic square root of >z^2-1 on C [-1,1]. >Well of course that's right, and probably better in the context of >qualifying exams. Seems interesting to know exactly what functions >have square roots, though. (Actually it was about a year ago that >exactly the current question came up on the complex-exam committee >here - what you write is the first thing we came up with, but I >didn't feel I really understood the story until I came up with >************************ >David C. Ullrich ************************ David C. Ullrich === Subject: solving x^ ( x^ (x +1) ) ) = y , y known. by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i8QGZCp31206; I propose a Ôsimple' exercise to you. Here y >1. A generalization can be found... Alain. === Subject: Re: solving x^ ( x^ (x +1) ) ) = y , y known. > I propose a Ôsimple' exercise to you. > Here y >1. > A generalization can be found... > Alain. x^x^(x+1)=(x^x)^(x^x)=w^w=y The last one can be solved for w via Lambert's W function as: w=ln(y)/W(ln(y))=A so now solving for x: x^x=A => x=ln(A)/W(ln(A)) for a final answer of: ln(ln(y)/W(ln(y)))/W(ln(ln(y)/W(ln(y)))) -- I. N. Galidakis --- http://users.forthnet.gr/ath/jgal/ === Subject: Re: solving x^ ( x^ (x +1) ) ) = y , y known. > I propose a Ôsimple' exercise to you. > Here y >1. > A generalization can be found... > Alain. ln(x) = W(ln ( W ( ln(y) ) ) ), where W is the lambert- W function. I am pretty crappy at maths, so don't be surprised if im wrong. === Subject: Re: solving x^ ( x^ (x +1) ) ) = y , y known. > > I propose a Ôsimple' exercise to you. > Here y >1. > A generalization can be found... > > Alain. > ln(x) = W(ln ( W ( ln(y) ) ) ), where W is the lambert- W function. Sorry, had a typo: ln(x) = W(W(ln(y))) === Subject: Re: solving x^ ( x^ (x +1) ) ) = y , y known. > x^ ( x^ (x +1) ) ) = y Unbalanced parentheses = bad -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: Fake (imaginary) numbers by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i8QGZCm31215; >Are their any guides to Fake (imaginary) numbers for a Neophite and >10th grader like me ???? You might get a kick out of the book The Book of Numbers by Conway and Guy. It's about numbers generally, in their many different guises; it's fun, informative, and can be read and appreciated on many levels, from the high school level to the level of professional mathematicians. There's a section on complex numbers where they are described as ways of moving the Euclidean plane. For example, you can think of -1 as representing a 180 degree rotation, and -i and i as 90 degree rotations (one clockwise, one counterclockwise, although which is which is purely a matter of convention). Generally, the complex number a + bi represents a kind of twirling motion -- read the book. There's a beautiful intuitive explanation of the equation e^{pi*i} = -1 in terms of these motions. Complex numbers are an amazing and fascinating subject -- I hope you will learn to love them (and not think of them as particularly fake!). Todd Trimble === Subject: Re: Fake (imaginary) numbers >Are their any guides to Fake (imaginary) numbers for a Neophite and >10th grader like me ???? > You might get a kick out of the book The Book of Numbers by > Conway and Guy. It's about numbers generally, in their many > different guises; it's fun, informative, and can be read and > appreciated on many levels, from the high school level to the > level of professional mathematicians. > There's a section on complex numbers where they are described > as ways of moving the Euclidean plane. For example, you can > think of -1 as representing a 180 degree rotation, and -i and > i as 90 degree rotations (one clockwise, one counterclockwise, > although which is which is purely a matter of convention). > Generally, the complex number a + bi represents a kind of > twirling motion -- read the book. > There's a beautiful intuitive explanation of the equation > e^{pi*i} = -1 > in terms of these motions. Complex numbers are an amazing and > fascinating subject -- I hope you will learn to love them (and > not think of them as particularly fake!). > Todd Trimble I am sorry for calling complex numbers fake it's just that is what my teacher called it ... so I thought that's what they were called but I seem to be wrong. Also, I'll listen to your advice and read The Book of Numbers by Conway and Guy (when I have the time) ^_^ === Subject: Re: Fake (imaginary) numbers numbers for a Neophite and 10th grader like me ???? Take a look at my website: http://1iz.de or- the same - http://i-is-no-longer-imaginary.gmxhome.de (If it's useful, send me an e-mail) Have fun Hero === Subject: Re: Internal language of a category by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i8QGZCJ31235; >There are different notions of internal languages such as the one for >Cartesian closed categories or the one for toposes. But, each time, the >definition of the internal language is given ad hoc. Is there an abstract >definition of what is an (or the?) internal language of a category? >David. Not quite sure what you mean ad hoc, or even what you are looking for really, since the language of category theory stands on its own, but if I had to guess, you might be looking for general ways of arguing as if objects had elements. A general idea (as you probably know) is to replace elements, in the usual sense of arrows 1 --> A from a terminal object, by generalized elements X --> A, i.e. X-parametrized points of A. But this is pretty superficial: it really amounts to considering the Yoneda embedding A |--> hom(-, A), and AFAIK there's not a great deal more you can say which applies to general categories. Taking this a step further, the general idea would then be embedding an abstract category X into something more Set-based, e.g., Set^{X^op} or some subcategory thereof (just as Cayley embeds an abstract group into a permutation group), but to make this idea really effective, it helps to consider the exactness properties of the embedding (to what extent does the embedding preserve colimits, etc.) The Freyd-Mitchell embedding theorems are an early example of such a program for abelian categories. Freyd has been one of the main champions of embedding theorems, and his book with Scedrov (Categories, Allegories) gives a number of embedding or representation theorems which, depending on how much structure or exactness properties one is willing to assume of the category, permit one to argue as though with elements (to put it very roughly). They take this idea quite some distance, without getting bogged down in details of internal formal languages. (Since some of your other posts indicate that you are interested in the more linguistic aspects of category theory, you might be interested in Freyd's formalization of diagrammatic arguments using Q-sequences, if you don't know this already. See the book, or the 1976 collection of papers dedicated to Eilenberg, titled something like Algebra, Topology, and Category Theory, (ed. Heller & Tierney). Categories for the Working Mathematician has a little bit about elementary rules for diagram chasing in abelian categories. Don't know if this has helped. You might ask your question at the categories mailing list. Todd Trimble === Subject: Re: Who thinks Goldbach's Conjecture is unprovable? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i8QGZCC31249; >|>False. It is possible that Goldbach's Conjecture is independent of >|>ZFC, i.e., that Con(ZFC) implies Con(ZFC+GC) and Con(ZFC) implies >|>Con(ZFC+not GC). (Just as the Continuum Hypothesis is independent of >|>ZFC.) I suspect most mathematicians believe that this independence is >|>not the case. >There is already a result of Chen that every large enough >even number > 2 is either a sum of two primes, or the sum >of a prime and a product of two primes. It seems implausible >to me that we have been unable to strengthen this to a proof >of Goldbach due to some essential barrier to doing so, as >opposed to the ordinary difficulties in pushing forward a >subject like this. One of the posters opined that GC is not amenable to analytic methods (without offering any argument), but didn't Vinogradov prove, using analytic arguments, that every suffiently large odd number is a sum of three odds? Does Chen also use analytic methods? >|Maybe I'm confused, but suppose that (not GC) is unprovable >|(say in PA). Then it would seem that each even constant > 2 is >|in fact the sum of two primes, for if there were a counterexample, >|then there would exist a finite calculation which showed that >|(e.g. use Eratosthenes sieve to list all primes less than the >|constant, and check that no two summed to the constant). >|I.e. that a proof of unprovability of (not GC) would perforce >|rule out the possibility of such a finite calculation, no matter >|how lengthy. >That's right. If GC is false, then it is possible to disprove in ZF, >PA, or even very weak systems of arithmetic. >It doesn't eliminate the possibility that GC is true but can't >be proven in your favorite axiom system, whether PA or ZF, >however. Of course. I was really addressing undecidability (GC and [not GC] both unprovable?), not the possibility of just GC unprovable. Todd Trimble === Subject: Re: Who thinks Goldbach's Conjecture is unprovable? |One of the posters opined that GC is not amenable to analytic |methods (without offering any argument), but didn't Vinogradov |prove, using analytic arguments, that every suffiently large |odd number is a sum of three odds? Does Chen also use analytic |methods? I have never actually seen either proof, but I had always had the impression that they were essentially analytic number theory proofs. Keith Ramsay === Subject: Re: Who thinks Goldbach's Conjecture is unprovable? > |One of the posters opined that GC is not amenable to analytic > |methods (without offering any argument), but didn't Vinogradov > |prove, using analytic arguments, that every suffiently large > |odd number is a sum of three odds? Does Chen also use analytic > |methods? > I have never actually seen either proof, but I had always had > the impression that they were essentially analytic number > theory proofs. > Keith Ramsay The first of these isn't too hard. For odd n >= 3, write n = (n-2) + 1 + 1. Harry === Subject: Re: Who thinks Goldbach's Conjecture is unprovable? >That's right. If GC is false, then it is possible to disprove in ZF, >PA, or even very weak systems of arithmetic. >It doesn't eliminate the possibility that GC is true but can't >be proven in your favorite axiom system, whether PA or ZF, >however. >Of course. I was really addressing undecidability (GC and [not GC] >both unprovable?), not the possibility of just GC unprovable. I think what you're missing is that from this statement of yours: >|I.e. that a proof of unprovability of (not GC) would perforce >|rule out the possibility of such a finite calculation, no matter >|how lengthy. one may further conclude that (not GC), if true, cannot be unprovable. So a proof that GC is undecidable is also a proof (in a more powerful system, presumably) that GC is true. The same applies to any conjecture for which a counterexample, if it exists, would necessarily be reachable by some algorithm in a finite number of steps. /dan