mm-4659 === Subject: Re: http://www.rasoulallah.net/index_english.asp http://www.rasoulallah.net/index_english.asp > What is this nonsense doing on sci.math? These fundamentalist idiots seem to think that by spreading their crap everywhere they will achieve world subjugation. === Subject: Re: http://www.rasoulallah.net/index_english.asp > http://www.rasoulallah.net/index_english.asp > What is this nonsense doing on sci.math? These fundamentalist idiots seem to think that by spreading their > crap everywhere they will achieve world subjugation. Or Christian Fundemontalists think that by posting pro-Islam crap everywhere, they will achieve everyone's disdain for all things Islam. How would you tell the difference? Phil -- -- Microsoft voice recognition live demonstration === Subject: Re: A conjecture related to Picard's theorem > In his PhD thesis, which was published here: > http://archive.numdam.org/ARCHIVE/AIF/AIF_1999__49_1/AIF_1999__49_1_303_0/AI F _1999__49_1_303_0.pdf > Bernhard Elsner (much better known as a film composer than as a > mathematician) stated a conjecture. It can be seen at the end of the > thesis, just before the acknowledgments, and it says: > Let D{0} be the punctured unit disk in the complex plane and let > U_1, U_2, ...,U_n be a finite open cover of D{0}. Suppose that on > each U_j there is an injective holomorphic function f_j, such that > df_j = df_k on each intersection U_j / U_k. Then the differentials > glue together to a meromorphic 1-form on the unit disk D. > So far so good. What I fail to understand is his next statement: > The conjecture is like a differential version of Picard's theorem. > *Which* Picard's theorem is he talking about and how is it related > with this conjecture? > > I'm really not sure about the answer to either question, but > I suspect that he's talking about the standard Big Picard > Theorem in complex analysis. > > Unless I'm missing something, if f is holomorphic then > df is just f' dz. So it's clear that the df_j above glue > together to form a holomorphic 1-form g dz in the > punctured disk. I'm not sure what it means to say > a 1-form is meromorphic, but it seems likely that > here it would mean just that g is meromorphic > in the disk, ie that g does not have an essential > singularity at the origin. > > Otoh the standard BPT can be phrased this way: > If g is holomorphic in the punctured disk and > g omits the two values 0 and 1 then g is > meromorphic in the disk. > > In the situation above we do know that g omits > the value 0, since each f_j is injective... Jose Carlos Santos === Subject: Re: A conjecture related to Picard's theorem <6f6vi0Fa7ec8U1@mid.individual.net> posting-account=d8k6zwoAAABcL3GgSll4g0K5gJnqUP3Z Gecko/2008070208 Firefox/3.0.1,gzip(gfe),gzip(gfe) concerning my conjecture David C. Ullrich got it right. A 1-form g dz is meromorphic if the function g is meromorphic. Now, with the hypothesis of the conjecture we know already that g is holomorphic on the punctured disk D0 and to prove the conjecture we only have to show that g doesn't have an essential singularity in O. There are two cases: 1) The residue of the 1-form g dz in 0 is zero. Then the 1-form has a primitive, ie, there exists f holomorphic on D0 such that df = g dz (that is f' = g ). Therefore, by hypothesis, the restrictions f|U_1 , ... , f|U_n are injective and with help of Big Picard we can easily deduce that f and g don't have an essential singularity. 2) The residue r of the 1-form g dz in 0 is not zero. Then g dz as a multi-valued primitive with the logarithmic component rln(z) . That is the case where I got stuck eleven years ago and therefore, with the help of Mikhail Zaidenberg, I used only a special setting of the conjecture which was sufficient for the needs of my PhD... I am positive that the conjecture is true, but I am also sure that the proof will not be easy at all! If someone found a counter-example I would be very, very surprised because it would be against my geometric intuition (I visualize f as a locally bi-holomorphic map p : V --> C where V is a simply connected infinite spiral turning around oo ...) Of course, I would be glad, if some one worked it out. Funnily this conjecture which I stated by pure coincidence seems today the only interesting part of my thesis written in 1997! I have left mathematics right after my PhD and I have no regrets, but sometimes when I want to get nostalgic I just open a beautiful book about algebraic geometry or topology or Riemann surfaces ;-) Have a nice day, B === Subject: Re: A conjecture related to Picard's theorem > concerning my conjecture David C. Ullrich got it right. A 1-form g > dz is meromorphic if the function g is meromorphic. Now, with the > hypothesis of the conjecture we know already that g is holomorphic > on the punctured disk D0 and to prove the conjecture we only have > to show that g doesn't have an essential singularity in O. There > are two cases: > > 1) The residue of the 1-form g dz in 0 is zero. Then the 1-form > has a primitive, ie, there exists f holomorphic on D0 such that > df = g dz (that is f' = g ). Therefore, by hypothesis, the > restrictions f|U_1 , ... , f|U_n are injective and with help of Big > Picard we can easily deduce that f and g don't have an essential > singularity. > > 2) The residue r of the 1-form g dz in 0 is not zero. Then g > dz as a multi-valued primitive with the logarithmic component > rln(z) . That is the case where I got stuck eleven years ago and > therefore, with the help of Mikhail Zaidenberg, I used only a special > setting of the conjecture which was sufficient for the needs of my > PhD... Jose Carlos Santos === Subject: Re: A conjecture related to Picard's theorem posting-account=d8k6zwoAAABcL3GgSll4g0K5gJnqUP3Z Gecko/2008070208 Firefox/3.0.1,gzip(gfe),gzip(gfe) > Otoh the standard BPT can be phrased this way: > If g is holomorphic in the punctured disk and > g omits the two values 0 and 1 then g is > meromorphic in the disk. What does Otoh stand for? > In the situation above we do know that g omits > the value 0, since each f_j is injective... There actually is no reason why g should omit the value 0 . Omit in the case of Big Picard's Theorem means omit as an image value not as a variable value. Big Picard says: If f is analytic and has an essential singularity in w then there exists a complex number v such that for every complex number u different from v and for every punctured neighborhood W of w there are infinitely many z in W with f(z) = u . In other words : If an analytic function has an essential singularity then it takes every value (with at most one exception) infinitely many times and that is true in any arbitrary small neighborhood of the singularity. Example : The function z -> exp(1/z) has an essential singularity at the origin and every value, except for 0 , appears infinitely many times as image; and this stays true in whatever small neighborhood of the origin. The value of Picard's Theorem can't be appreciated with this example because with z -> exp(1/z) it can also been shown by direct computation (using the well known geometric properties of the complex exponential). It is more interesting in general cases where you can't compute so easily. Just take away every third term from the series exp(1/z) = sum 1/(k!z^k) and you know what I mean... Bye, B === Subject: Re: A conjecture related to Picard's theorem posting-account=d8k6zwoAAABcL3GgSll4g0K5gJnqUP3Z Gecko/2008070208 Firefox/3.0.1,gzip(gfe),gzip(gfe) I agree! === Subject: card(P(N)) = card(N), rev 2: a preliminary result posting-account=F3H0JAgAAADcYVukktnHx7hFG5stjWse .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) card(P(N)) = card(N), rev 2: a preliminary result ------------------------------------------------- I shall only try to prove a preliminary property, from which it should then be trivial to prove the thesis, so I will leave the final steps out for now. The proof below is based on transfinite induction (see Wikipedia). All should be quite trivial apart from one of the last passages, which I have marked. To simplify the notation for the limit case, I shall take the set of the extended naturals as reference. To be explicit: N* := N U { 0, oo } 'A' is for-each 'U' is set-union 'e' is set-membership 'c=' is set-inclusion (improper) Let P(n) be the set: P(n) := { k e N* | k < 2^n }, n e N* By transfinite induction, we shall prove the following property: Prop.) A n e N* : P(n) c= N* -- Zero case: n = 0 P(0) = { 0 } c= c= N* -- Successor case: n+1, assuming P(n) c= N*, n e N P(n+1) = P(n) U { k e N* | 2^n <= k < 2^(n+1) } c= c= N* U { k e N* | 2^n <= k < 2^(n+1) } = = N* -- Limit case: n = oo, assuming A m e N : P(m) c= N* P(oo) = P(n) U { k e N* | 2^n <= k < 2^oo } c= c= N* U { k e N* | 2^n <= k < 2^oo } (The following passage should be the culprit:) Since k e N*, where N* := N U { 0, oo }, it must be: { k e N* | 2^n <= k < 2^oo } = { k e N* | 2^n <= k < oo } Then: P(oo) c= N* U { k e N* | 2^n <= k < 2^oo } = = N* U { k e N* | 2^n <= k < oo } = = N* Follows the thesis. QDE. A possible interpretation for the whole matter, in the form of a fast slogan: however fast, it's just going to infinity. Anyway, right or wrong, I hope we have narrowed it down... and that there just isn't some blatant mistake! Julio status: OPEN revision: 2 sender: Julio Di Egidio (aka LudovicoVan) sender-email: ju...@diegidio.aleph.name (del 'aleph.') copyright: 2008 (on behalf of) sci.math, sci.logic All rights reserved. > Second attempt. I'll try two equivalent formulations, in the hope > that, together, they will complement the eventual ambiguities. ----- 1st formulation (i) æ P(A) ~ { 0, 1, 2, ..., 2^n - 1}, AneN Let's simply call P(A) the set on the RHS of (i), keeping in mind that > the argument is going to hold due to this bijection and the > transitivity of the '~' relation. (ii) æP(A) c= N, AneN (iii) P(N) is infinite (trivial) We pass from P(A) to P(N) by transfinite induction, then: (iv) æP(N) ~ N, from (ii) and (iii) QDE. (1) ----- 2nd formulation Same comments as above. (i) æ card(P(A)) = 2^n, AneN (ii) æcard(P(A)) <= card(N), AneN > (iii) card(P(N)) >= oo (meaning: 'oo' as lower bound) (iv) æcard(P(N)) = card(N) QDE. (2) ----- Hmm, correct? The basic idea behind this argument is that P(A) is > always a subset of N in the finite case. Then, by transfinite > induction, while P(N) becomes infinite, its cardinality's upper bound > remains N's cardinality. (These informal expositions get me into more > troubles than they help, but I have to run the risk: no pain, no > gain!) Good luck. Julio status: OPEN > revision: 1 > sender: Julio Di Egidio (aka LudovicoVan) > sender-email: ju...@diegidio.aleph.name (del 'aleph.') > copyright: 2008 (on behalf of) sci.math, sci.logic > All rights reserved. card(P(N)) = card(N), an elementary proof (and some fuzzy notation > here and there). --- Preliminaries: 'P(A)' stands for the powerset of set 'A'. '~' stands for the existence of a bijection, which will be, in this > context, equivalent to stating that two sets are equinumerous. > Infinity of the set 'N' (the set of natural/ordinal numbers) is taken > to be defined as usual, by the existence of bijections between N and > its (infinite) subsets. No other kind of infinities are going to be > assumed: the only tool I have used is transfinite induction over the > naturals/ordinals, with omega as the limit ordinal. 'oo' stands for omega. --- Finite case: card(A) = n <=>def > æ æ A ~ { 0, 1, 2, .., n-1 } card(P(A)) = 2^n <= æ æ P(A) ~ { 1, 2, 4, ..., 2^n-1 } --- Infinite case (transf. induction, notation a bit fuzzy here): (1) n=oo => card(A) = card(N) = oo <=>def > æ æ A ~ N := { 0, 1, 2, ..., n }, n->oo (2) n=oo => card(P(A)) = card(P(N)) = 2^oo <= æ æ P(A) ~ P(N) ~ 2^N := { 1, 2, 4, ..., 2^n }, n->oo --- Theorem: card(P(N)) = card(N) --- Proof: (i) card(P(N)) = card(N) <=> P(N) ~ N, by def. of 'card' (ii) card(P(N)) = card(N) <=> 2^N ~ N, by (2) and trans. of '~' (iii) A bijection 2^N ~ N trivially exists: 2^N ~ N > -------- > æ 1 ~ 0 > æ 2 ~ 1 > æ 4 ~ 2 > ... Follows the thesis. QDE. It's THEN easy to get interesting lemmas. We get that omega is an > absolute upper bound and all infinities are equivalent. The diagonal > argument then assumes another shape, which will be the subject of > another post of mine, unless someone finds a flaw in the above quite > elementary argument. > Julio status: OPEN > sender: Julio Di Egidio (aka LudovicoVan) > sender-email: ju...@diegidio.aleph.name (del 'aleph.') > copyright: 2008 (on behalf of) sci.math, sci.logic > All rights reserved. === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result > > (The following passage should be the culprit:) > Since k e N*, where N* := N U { 0, oo }, it must be: > > { k e N* | 2^n <= k < 2^oo } = { k e N* | 2^n <= k < oo } Those two sets are only equal if the conditions 2^n <= k < 2^oo and 2^n <= k < oo are equivalent, also what is n here? -- He is not here; but far away The noise of life begins again And ghastly thro' the drizzling rain On the bald street breaks the blank day. === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 Gecko/2008070208 Firefox/3.0.1,gzip(gfe),gzip(gfe) > card(P(N)) = card(N), rev 2: a preliminary result > ------------------------------------------------- I shall only try to prove a preliminary property, from which it should > then be trivial to prove the thesis, so I will leave the final steps > out for now. The proof below is based on transfinite induction (see Wikipedia). All > should be quite trivial apart from one of the last passages, which I > have marked. To simplify the notation for the limit case, I shall take the set of > the extended naturals as reference. To be explicit: N* := N U { 0, oo } Usually we take N (w, for omega) to be {0 1 2 ...}. But I take it that you have N={1 2 3 .. }. Okay. Meanwhile, I take that your 'oo' stands for w = {0 1 2 ... }. So N* = wu{w}. > 'A' is for-each > 'U' is set-union > 'e' is set-membership > 'c=' is set-inclusion (improper) > Let P(n) be the set: P(n) := { k e N* | k < 2^n }, n e N* > By transfinite induction, we shall prove the following property: Prop.) A n e N* : P(n) c= N* Your proof is a waste of energy, since, BY DEFINITION, P(n) subset of N*. Sheesh. P(n) = { k e N* | k < 2^n }, so every member of P(n) is a member of N* since every member of P(n) is some k such that k e N* End of proof. By the way, just in case it comes up, keep in mind that for n in w or n=w, we have: 2^n = card({f | f is a function from n into {0 1}}). Unless you state otherwise, that is the definition of cardinal exponenitation (which, in the finite case, is equivalent with ordinary exponentiation on natural numbers). Anyway, it's not clear why you're thinking of transfinite induction. Transfinite induction proves something holds for all ordinals. The only things you want to prove for are the finite ordinals and w itself. > -- Zero case: n = 0 P(0) = { 0 } c= > æ æ c= N* Fine, but unneeded. > -- Successor case: n+1, assuming P(n) c= N*, n e N P(n+1) = P(n) U { k e N* | 2^n <= k < 2^(n+1) } c= > æ æ æ c= N* æ U { k e N* | 2^n <= k < 2^(n+1) } = > æ æ æ æ= N* Fine, but unneeded. > -- Limit case: n = oo, NO, that's wrong. The limit case is an arbitrary limit ordinal, not just w. But let's see what you do to try for w itself. However, it's unneeded. > assuming A m e N : P(m) c= N* We don't need to assume it. We HAVE it. But, anyway, that's not how transfinite induction works. Transfinite induction, in the form YOU referenced, allows assuming for an ARBITRARY limit ordinal that all its members have the property and then showing that that arbitrary limit ordinal has the property. It does NOT allow choosing a PARTICULAR limit ordinal. Here are some transfinite induction schemata: 1. F0 & Ak(k is a successor ordinal -> Fk) & Ak(k is a limit ordinal -> Fk) -> Ak(k is an ordinal -> Fk) 2. Ak((k is an ordinal & Am(m Fm)) -> Fk) -> Ak(k is an ordinal -> Fk) 3 F0 & Ak((k is an ordinal & Fk) -> Fk+1) & Ak((k is a limit ordinal & Am(m Fm)) -> Fk) -> Ak(k is an ordinal -> Fk) Please either put your transfinite induction in one of those forms, or PROVE another form. But anyway, your argument is not transfinite induction, and doesn't need to be. What you wanted to prove is IMMEDIATE just from you definitions. > P(oo) æ= P(n) U { k e N* | 2^n <= k < 2^oo } c= > æ æ æ c= N* æ U { k e N* | 2^n <= k < 2^oo } Fine, but unneeded. > (The following passage should be the culprit:) > Since k e N*, where N* := N U { 0, oo }, it must be: æ { k e N* | 2^n <= k < 2^oo } = { k e N* | 2^n <= k < oo } No. It is actually: { k e N* | 2^n <= k < 2^oo } = { k e N* | 2^n <= k <= oo } The very last inequality is '<=' not '<', since 'oo' is a member of N*. > Then: P(oo) c= N* æ U { k e N* | 2^n <= k < 2^oo } = > æ æ æ æ= N* æ U { k e N* | 2^n <= k < oo } = > æ æ æ æ= N* What a ridiculously convoluted proof just to show what is obvious. To prove P(w) subset of N*, it suffices just to observe that, BY DEFINITION, for all n in N* we have P(n) subset of N*. And so where is your proof that N is 1-1 with PN? MoeBlee === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result posting-account=F3H0JAgAAADcYVukktnHx7hFG5stjWse .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) What a ridiculously convoluted proof just to show what is obvious. Again, now I need some pondering. The next step should be as trivial, but we'll see. The basic idea is there is a straightforward bijection between the powerset and our set P(n) above. Anyway, we'll see. -LV === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result posting-account=F3H0JAgAAADcYVukktnHx7hFG5stjWse .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) > card(P(N)) = card(N), rev 2: a preliminary result > ------------------------------------------------- I shall only try to prove a preliminary property, from which it should > then be trivial to prove the thesis, so I will leave the final steps > out for now. The proof below is based on transfinite induction (see Wikipedia). All > should be quite trivial apart from one of the last passages, which I > have marked. To simplify the notation for the limit case, I shall take the set of > the extended naturals as reference. To be explicit: N* := N U { 0, oo } 'A' is for-each > 'U' is set-union > 'e' is set-membership > 'c=' is set-inclusion (improper) Let P(n) be the set: P(n) := { k e N* | k < 2^n }, n e N* By transfinite induction, we shall prove the following property: Prop.) A n e N* : P(n) c= N* -- Zero case: n = 0 P(0) = { 0 } c= > æ æ c= N* -- Successor case: n+1, assuming P(n) c= N*, n e N P(n+1) = P(n) U { k e N* | 2^n <= k < 2^(n+1) } c= > æ æ æ c= N* æ U { k e N* | 2^n <= k < 2^(n+1) } = > æ æ æ æ= N* -- Limit case: n = oo, assuming A m e N : P(m) c= N* P(oo) æ= P(n) U { k e N* | 2^n <= k < 2^oo } c= > æ æ æ c= N* æ U { k e N* | 2^n <= k < 2^oo } (The following passage should be the culprit:) > Since k e N*, where N* := N U { 0, oo }, it must be: æ { k e N* | 2^n <= k < 2^oo } = { k e N* | 2^n <= k < oo } Another one: Looking at the above equivalence, I'd maybe better have written: ... = { k e N* | 2^n <= k <= oo } I am not 100%, since here, as said, 'N*' helps in simplifying the notation for the limit case. The substance of the argument should remain unchanged. This said, I'll be waiting for feedback. -LV > Then: P(oo) c= N* æ U { k e N* | 2^n <= k < 2^oo } = > æ æ æ æ= N* æ U { k e N* | 2^n <= k < oo } = > æ æ æ æ= N* Follows the thesis. QDE. A possible interpretation for the whole matter, in the form of a fast > slogan: however fast, it's just going to infinity. Anyway, right or wrong, I hope we have narrowed it down... and that > there just isn't some blatant mistake! > Julio status: OPEN > revision: 2 > sender: Julio Di Egidio (aka LudovicoVan) > sender-email: ju...@diegidio.aleph.name (del 'aleph.') > copyright: 2008 (on behalf of) sci.math, sci.logic > All rights reserved. > Second attempt. I'll try two equivalent formulations, in the hope > that, together, they will complement the eventual ambiguities. ----- 1st formulation (i) æ P(A) ~ { 0, 1, 2, ..., 2^n - 1}, AneN Let's simply call P(A) the set on the RHS of (i), keeping in mind that > the argument is going to hold due to this bijection and the > transitivity of the '~' relation. (ii) æP(A) c= N, AneN (iii) P(N) is infinite (trivial) We pass from P(A) to P(N) by transfinite induction, then: (iv) æP(N) ~ N, from (ii) and (iii) QDE. (1) ----- 2nd formulation Same comments as above. (i) æ card(P(A)) = 2^n, AneN (ii) æcard(P(A)) <= card(N), AneN > (iii) card(P(N)) >= oo (meaning: 'oo' as lower bound) (iv) æcard(P(N)) = card(N) QDE. (2) ----- Hmm, correct? The basic idea behind this argument is that P(A) is > always a subset of N in the finite case. Then, by transfinite > induction, while P(N) becomes infinite, its cardinality's upper bound > remains N's cardinality. (These informal expositions get me into more > troubles than they help, but I have to run the risk: no pain, no > gain!) Good luck. Julio status: OPEN > revision: 1 > sender: Julio Di Egidio (aka LudovicoVan) > sender-email: ju...@diegidio.aleph.name (del 'aleph.') > copyright: 2008 (on behalf of) sci.math, sci.logic > All rights reserved. > card(P(N)) = card(N), an elementary proof (and some fuzzy notation > here and there). --- Preliminaries: 'P(A)' stands for the powerset of set 'A'. '~' stands for the existence of a bijection, which will be, in this > context, equivalent to stating that two sets are equinumerous. > Infinity of the set 'N' (the set of natural/ordinal numbers) is taken > to be defined as usual, by the existence of bijections between N and > its (infinite) subsets. No other kind of infinities are going to be > assumed: the only tool I have used is transfinite induction over the > naturals/ordinals, with omega as the limit ordinal. 'oo' stands for omega. --- Finite case: card(A) = n <=>def > æ æ A ~ { 0, 1, 2, .., n-1 } card(P(A)) = 2^n <= æ æ P(A) ~ { 1, 2, 4, ..., 2^n-1 } --- Infinite case (transf. induction, notation a bit fuzzy here): (1) n=oo => card(A) = card(N) = oo <=>def > æ æ A ~ N := { 0, 1, 2, ..., n }, n->oo (2) n=oo => card(P(A)) = card(P(N)) = 2^oo <= æ æ P(A) ~ P(N) ~ 2^N := { 1, 2, 4, ..., 2^n }, n->oo --- Theorem: card(P(N)) = card(N) --- Proof: (i) card(P(N)) = card(N) <=> P(N) ~ N, by def. of 'card' (ii) card(P(N)) = card(N) <=> 2^N ~ N, by (2) and trans. of '~' (iii) A bijection 2^N ~ N trivially exists: 2^N ~ N > -------- > æ 1 ~ 0 > æ 2 ~ 1 > æ 4 ~ 2 > ... Follows the thesis. QDE. It's THEN easy to get interesting lemmas. We get that omega is an > absolute upper bound and all infinities are equivalent. The diagonal > argument then assumes another shape, which will be the subject of > another post of mine, unless someone finds a flaw in the above quite > elementary argument. > Julio status: OPEN > sender: Julio Di Egidio (aka LudovicoVan) > sender-email: ju...@diegidio.aleph.name (del 'aleph.') > copyright: 2008 (on behalf of) sci.math, sci.logic > All rights reserved. === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result (The following passage should be the culprit:) > Since k e N*, where N* := N U { 0, oo }, it must be: { k e N* | 2^n <= k < 2^oo } = { k e N* | 2^n <= k < oo } > > Another one: > > Looking at the above equivalence, I'd maybe better have written: > > ... = { k e N* | 2^n <= k <= oo } > > I am not 100%, since here, as said, 'N*' helps in simplifying the > notation for the limit case. The substance of the argument should > remain unchanged. This { k e N* | 2^n <= k < 2^oo } only equals this { k e N* | 2^n <= k <= oo } if the conditions 2^n <= k < 2^oo and 2^n <= k <= oo are materially equivalent. Also, what is n? Have you defined 2^oo? -- He is not here; but far away The noise of life begins again And ghastly thro' the drizzling rain On the bald street breaks the blank day. === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result <48907C81.9CCB527F@tesco.net> posting-account=F3H0JAgAAADcYVukktnHx7hFG5stjWse .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) On 30 Jul, 15:36, Frederick Williams > On 30 Jul, 15:36, Frederick Williams This { k e N* | 2^n <= k < 2^oo } only equals this { k e N* | 2^n <= k <= oo } > > Yes, despite my abyss of ignorance, that one I had corrected myself. So what about the bits you snipped? > There is no point in discussing with you guys, you can't even read. -- He is not here; but far away The noise of life begins again And ghastly thro' the drizzling rain On the bald street breaks the blank day. === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result posting-account=yKimjgoAAACk5WwPVD4l9HmbpoR6Hmy4 en),gzip(gfe),gzip(gfe) > On 30 Jul, 15:36, Frederick Williams This ? { k e N* | 2^n <= k < 2^oo } only equals this ? { k e N* | 2^n <= k <= oo } Yes, despite my abyss of ignorance, that one I had corrected myself. There is no point in discussing with you guys, you can't even read. Communication is a two-way street, I believe. If everyone finds it hard to understand you, you may be doing something wrong, too. === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result posting-account=F3H0JAgAAADcYVukktnHx7hFG5stjWse .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) > Communication is a two-way street, I believe. If everyone finds it > hard to understand you, you may be doing something wrong, too. 1+1=2: that's again the logic of the billion flies. Just like peace should be a two-way street and not these self-blessing holy wars of the self-imposing global false-democracy of the true-idiots and the true-assassins and enslavers and exploiters, plain criminals against all intelligence and life. Anyway, it, this planet is just having what it's always deserved. You seem sincerely interested. Have a look a this then: > A n e N* : P(A_n) ~ P(n) > Is that correct? (I think that bijection *must* hold, > otherwise -say- we could not have our parings to > state Cantor's Theorem.) So how it is that Cantor's Theorem comes out to be just invalid? Is it possible to have those pairings in the general case, or is it not? (Apparently invalid, isn't it?) (That the answer to this question is not trivial, whichever the anser may be, I happen to know. For instance, we still haven't got a polynomial time algo for NP-complete problems, have we?) -LV === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result > > Communication is a two-way street, I believe. If everyone finds it > hard to understand you, you may be doing something wrong, too. > > 1+1=2: that's again the logic of the billion flies. Just like peace > should be a two-way street and not these self-blessing holy wars of > the self-imposing global false-democracy of the true-idiots and the > true-assassins and enslavers and exploiters, plain criminals against > all intelligence and life. Anyway, it, this planet is just having > what it's always deserved. > > You seem sincerely interested. Have a look a this then: > > A n e N* : P(A_n) ~ P(n) > Is that correct? That depends a great deal on what the symbols mean. IIRC, your N* means Union{{0},N,{oo}}, where oo is some sort of undefined entity not a member of Union{{0},N}, and N is the set of positive natural numbers. And IIRD< ~ is an infix operator meaning that there exists a bijection between the sets that it appears between. But I am uncertain about the meaning of A_n. And your P(.) predicate seems to be ambiguous, having at least two different meanings. === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result posting-account=Cbgh4AoAAAAr0dt1RqLOClWCyUWii2fU Gecko/2008071719 Firefox/3.0.1,gzip(gfe),gzip(gfe) Communication is a two-way street, I believe. If everyone finds it > hard to understand you, you may be doing something wrong, too. 1+1=2: that's again the logic of the billion flies. Just like peace > should be a two-way street and not these self-blessing holy wars of > the self-imposing global false-democracy of the true-idiots and the > true-assassins and enslavers and exploiters, plain criminals against > all intelligence and life. Anyway, it, this planet is just having > what it's always deserved. You seem sincerely interested. Have a look a this then: A n e N* : P(A_n) ~ P(n) > Is that correct? If A_n is an n-element set (i.e. A_n ~ n, i.e. we have a bijection f: A_n -> n), then of course We obtain a map P(f): P(A_n) -> P(n) x |-> { f(y) | y e x } and easily verify that it is bijective, hence P(A_n) ~ P(n) This hold without any restriction on n. >(I think that bijection *must* hold, > otherwise -say- we could not have our parings to > state Cantor's Theorem.) Which theorem of Cantor's are you referring to now? To state the theorem I have in mind(*) it is not required to have any pairings that could be inferred from a bijection P(A_n) ~ P(n). (*) If X is a set and f:X->P(X) is a map to its powerset, then f is not surjective. > So how it is that Cantor's Theorem comes out to be just invalid? Where does it come out just invalid (by a valid argument)? > Is it > possible to have those pairings in the general case, or is it not? > (Apparently invalid, isn't it?) (That the answer to this question is not trivial, whichever the > anser may be, I happen to know. For instance, we still haven't got a > polynomial time algo for NP-complete problems, have we?) P = NP ? has no relation to N ~ PN ? apart from typographical resemblance. The former question is as of yet unanswered (though it is widely believed that the answer is no); the latter has been answered with a definiete no a long time ago by a short, simple and convincing proof. A common hing, however, is that we find again and again people who either think they have found a fast algorithm to solve complicated algorithmical tasks (JSH factoring) or proofs of N~PN. -LV hagman === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result posting-account=F3H0JAgAAADcYVukktnHx7hFG5stjWse .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) You seem sincerely interested. Have a look a this then: A n e N* : P(A n) ~ P(n) > Is that correct? If A n is an n-element set (i.e. A n ~ n, i.e. we have a > bijection f: A n -> n), then of course > We obtain a map > P(f): P(A n) -> P(n) > æx |-> { f(y) | y e x } > and easily verify that it is bijective, hence > P(A n) ~ P(n) > This hold without any restriction on n. Perfect. >(I think that bijection *must* hold, > otherwise -say- we could not have our parings to > state Cantor's Theorem.) Which theorem of Cantor's are you referring to now? As I have said somewhere near to the beginning, I am refering to how Wikipedia puts it. I don't think that any alternative though equivalent formulation might change anything in that respect, but again: if there is any significant and well-known flaw in WP treatement, just let me know. > To state the theorem I have in mind(*) it is not required > to have any pairings that could be inferred from > a bijection P(A n) ~ P(n). (*) If X is a set and f:X->P(X) is a map to its powerset, > æ æ then f is not surjective. And how will you prove that? Cantor? > P = NP ? has no relation to N ~ PN ? apart from typographical > resemblance. That's of course not the relation I had in mind, but let's forget about it, it was just a side hint at the (nor necessarily direct) implications. -LV hagman === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result posting-account=AdyLXQoAAABgRay99CKv1O8Y_7jjivwq InfoPath.1),gzip(gfe),gzip(gfe) > As I have said somewhere near to the beginning, > I am refering to how Wikipedia puts it. If you truly think you have something nontrivial to say about something in mathematics this well known, you shouldn't be making references to something even first year college students are warned against using (as a primary reference). This is like someone trying to argue, in court of law, a legal precedent based on a previous court case, but using a newspaper story about the previous court case instead of the actual legal documents from the previous court case. Dave L. Renfro === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result You seem sincerely interested. Have a look a this then: A n e N* : P(A_n) ~ P(n) > Is that correct? If A_n is an n-element set (i.e. A_n ~ n, i.e. we have a > bijection f: A_n -> n), then of course > We obtain a map > P(f): P(A_n) -> P(n) > æx |-> { f(y) | y e x } > and easily verify that it is bijective, hence > P(A_n) ~ P(n) > This hold without any restriction on n. > > Perfect. > >(I think that bijection *must* hold, > otherwise -say- we could not have our parings to > state Cantor's Theorem.) Which theorem of Cantor's are you referring to now? > > As I have said somewhere near to the beginning, I am refering to how > Wikipedia puts it. I don't think that any alternative though > equivalent formulation might change anything in that respect, but > again: if there is any significant and well-known flaw in WP > treatement, just let me know. > > To state the theorem I have in mind(*) it is not required > to have any pairings that could be inferred from > a bijection P(A_n) ~ P(n). (*) If X is a set and f:X->P(X) is a map to its powerset, > æ æ then f is not surjective. > > And how will you prove that? Cantor? PROOF: Given any set S and and function f:S --> P(S), then T_f = {x e S: ~ x e f(x)} is a well defined subset of S and member of P(S), but A x e S ~(f(x) = T_f), so that f can never be surjective. Thus for every set S, Card(S) < Card(P(S)). === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 Gecko/2008070208 Firefox/3.0.1,gzip(gfe),gzip(gfe) > (*) If X is a set and f:X->P(X) is a map to its powerset, > æ æ then f is not surjective. And how will you prove that? Cantor? I told you already, with first order logic from the axioms of Z set theory. MoeBlee === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 Gecko/2008070208 Firefox/3.0.1,gzip(gfe),gzip(gfe) (*) If X is a set and f:X->P(X) is a map to its powerset, > æ æ then f is not surjective. And how will you prove that? Cantor? I told you already, with first order logic from the axioms of Z set > theory. Actually, it seems to me that, aside from trivial cases in which we don't have power sets, etc., we can do it with just intutitionistic logic and the axiom schema of separation: 1. Suppose f:x->Px. 2. EdAy(yed <-> (yex & ~yef(y))). ... axiom schema of separation 3. Ay(yed <-> (yex & ~yef(y))). ... 2, existential instantiation 4. dePx ... 3, definition of 'Px'. 5. Suppose Ezex d=f(z). 6. zex & d=f(z). ... 5, existential instantiation 7. zed <-> (zex & ~zef(z)). ... 3, universal instantiation 8. zed <-> ~zed. ... 6, 7 ... identity, sentential logic MoeBlee === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result posting-account=F3H0JAgAAADcYVukktnHx7hFG5stjWse .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) (*) If X is a set and f:X->P(X) is a map to its powerset, > æ æ then f is not surjective. And how will you prove that? Cantor? I told you already, with first order logic from the axioms of Z set > theory. I didn't miss it. Axioms are not proofs. You teach me a consistent system might still be completely useless. (Then, at an higher level, we might discuss what's the difference, if any, between correct although of no use and incorrect.) -LV > MoeBlee === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result posting-account=b59GCwoAAAAi5nnxSoswpnEj2IJlJetr Gecko/20080702 Firefox/2.0.0.16,gzip(gfe),gzip(gfe) > Axioms are not proofs. Axioms are proofs BY DEFINITION -- namely one-line proofs. (At least in the sense of mathematical logic. I know that a different definition is used for Whiskey.) Michel. === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 Gecko/2008070208 Firefox/3.0.1,gzip(gfe),gzip(gfe) Axioms are not proofs. Axioms are proofs BY DEFINITION -- namely one-line proofs. That's fine in an ordinary, informal sense. But strictly speaking (and admittedly pedantic), IF 'proof' is defined as a certain kind of SEQUENCE, then the axiom itself is not a proof, but rather the sequence in which the axiom is the first entry is the proof. I.e., if S is the axiom, then the proof is a SEQUENCE: {<0 S>} which is not S itself. MoeBlee === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result ... strictly speaking (and admittedly pedantic), > Well, I wouldn't call that pedantic. IF there's a place for rigor then in mathematical logic, I'd say. ... IF 'proof' is defined as a certain kind of SEQUENCE, then the axiom > itself is not a proof, but rather the sequence in which the axiom is the > first entry is the proof. I.e., if S is the axiom, then the proof is a > SEQUENCE: {<0 S>} which is not S itself. > A simple _fact_. (On the other hand, IF we refer to proofs as certain linguistic entities, THEN we might have a point in claiming that axioms *are* one-line proofs - in the sense that a proof in this case just consists of the very axiom.) B. -- For every line of Cantor's list it is true that this line does not contain the diagonal number. Nevertheless the diagonal number may be in the infinite list. (WM, sci.logic) === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result <7o9494pbmr1gisjupltrdikmvtsvaclhcg@4ax.com> posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 Gecko/2008070208 Firefox/3.0.1,gzip(gfe),gzip(gfe) > ... IF 'proof' is defined as a certain kind of SEQUENCE, then the axiom > itself is not a proof, but rather the sequence in which the axiom is the > first entry is the proof. I.e., if S is the axiom, then the proof is a > SEQUENCE: æ æ{<0 S>} which is not S itself. A simple fact . (On the other hand, IF we refer to proofs as certain linguistic > entities, THEN we might have a point in claiming that axioms *are* one-line proofs > - in the sense that a proof in this case just consists of the very axiom.) Wonderful. Now what do want me to do? Agree with it AGAIN for about the fourth time?! MoeBlee === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result > ... IF 'proof' is defined as a certain kind of SEQUENCE, then the axiom > itself is not a proof, but rather the sequence in which the axiom is the > first entry is the proof. I.e., if S is the axiom, then the proof is a > SEQUENCE: > æ æ{<0 S>} > which is not S itself. > A simple _fact_. (On the other hand, IF we refer to proofs as certain linguistic > entities, THEN we might have a point in claiming that axioms *are* one-line proofs > - in the sense that a proof in this case just consists of the very axiom.) Wonderful. Now what do want me to do? Agree with it AGAIN for about > the fourth time?! > Well, actually I am REALLY PLEASED with the result (outcome) of our little exchange. :-) B. -- For every line of Cantor's list it is true that this line does not contain the diagonal number. Nevertheless the diagonal number may be in the infinite list. (WM, sci.logic) === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result > Axioms are not proofs. > Axioms are proofs BY DEFINITION -- namely one-line proofs. > That's fine in an ordinary, informal sense. But strictly speaking (and admittedly pedantic), IF 'proof' is defined > as a certain kind of SEQUENCE, then the axiom itself is not a proof, > but rather the sequence in which the axiom is the first entry is the > proof. I.e., if S is the axiom, then the proof is a SEQUENCE: {<0, S>} which is not S itself. > Depends. If you encode your proof (i.e. the sequence of formulas) with n-tuples (S_1, S_2, ..., S_n) (where S_i is the formula at step i in the proof) then we would actually have (S_1) = S_1 , as required. :-) Since proofs (at least in FOPL) are required to be _finite_, n-tuples are appropriate mathematical objects to represent them (i.e. certain sequences of formulas). See: http://en.wikipedia.org/wiki/Tuple A mathematician would of course just identify {<0, S_1>} with (S_1) = S_1 in this case, I guess: Leading to the claim: Axioms are proofs BY DEFINITION -- namely one-line proofs. (Right, those guys are rather pragmatic, sometimes. :-) B. -- For every line of Cantor's list it is true that this line does not contain the diagonal number. Nevertheless the diagonal number may be in the infinite list. (WM, sci.logic) === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 Gecko/2008070208 Firefox/3.0.1,gzip(gfe),gzip(gfe) > Axioms are not proofs. > Axioms are proofs BY DEFINITION -- namely one-line proofs. That's fine in an ordinary, informal sense. But strictly speaking (and admittedly pedantic), IF 'proof' is defined > as a certain kind of SEQUENCE, then the axiom itself is not a proof, > but rather the sequence in which the axiom is the first entry is the > proof. I.e., if S is the axiom, then the proof is a SEQUENCE: {<0, S>} which is not S itself. Depends. If you encode your proof (i.e. the sequence of formulas) with n-tuples æ æ æ æ (S 1, S 2, ..., S n) (where S i is the formula at step i in the proof) then we would actually have æ æ æ æ (S 1) = S 1 , as required. :-) Yes, I ALREADY agreed to that. IF you take 'sequence' in the sense of 'iterated tuple'. > Since proofs (at least in FOPL) are required to be finite , n-tuples are > appropriate mathematical objects to represent them (i.e. certain sequences of > formulas). You may do that. But I see no indication in general that when 'proof' is defined as a certain kind of sequence, then INSTEAD of the literal sense of sequence, the author really means iterated tuples. > See:http://en.wikipedia.org/wiki/Tuple Great, more evidence of what a LOUSY rescource Wikipedia is. > A mathematician would of course just identify æ æ æ æ {<0, S 1>} > with > æ æ æ æ (S 1) = S 1 Yes, informally, that is quite commonly done. However, strictly speaking, the tuple and the sequence are different. > in this case, I guess: Leading to the claim: æ æ æ æ Axioms are proofs BY DEFINITION -- namely one-line proofs. (Right, those guys are rather pragmatic, sometimes. :-) Yes, I agreed from the outset that we say such things in an everyday, working sense. I don't know why you keep circling back to remind me of what I ALREADY agreed to from the beginning. (1) In an everyday informal sense, we take an axiom to be a proof of itself. (2) IF we take proofs as tuples (iterated ordered pairs) then, since a 1-tuple is just the object itself, an axiom is a proof of itself. (3) However, strictly speaking, if we take 'sequence' in the ordinary sense of a certain kind of function, then a single entry proof of the axiom S is {<0 S>}, which is not S itself. And this does not contradict (1) nor (2), since (1) is an informal sense and (3) is formal sense, and (2) is based on the IF that we take 'sequence' in the sense of iterated ordered pairs. By the way, in certain circumstances, there is an advantage to taking sequence in the literal sense of a certain kind of function and not as iterated ordered pairs. With sequences as functions, we can prove the existence of the set of all such finite sequences without the axiom schema of replacement, but to prove the set of all n-tuples (for n in omega) uses the axiom schema of replacement. MoeBlee === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result I don't know why you keep circling back to remind me of what I ALREADY > agreed to from the beginning. (1) In an everyday informal sense, we take an axiom to be a proof of > itself. (2) IF we take proofs as tuples (iterated ordered pairs) then, since a > 1-tuple is just the object itself, an axiom is a proof of itself. > Fine! This OF COURSE would relativize your claim: > Axioms are not proofs. > Of course they are not. [Moe] > Well, you see... Hence I dared to mention: In a _certain sense_ they are. One might consider an axiom to be a proof of length 1 (proving itself). It's just the of course-part of your claim, that triggered my reaction. But, of course: (3) However [...] if we take 'sequence' in the ordinary > sense of a certain kind of function, then a single entry > proof of the axiom S is {<0 S>}, which is not S itself. > Right. I guess from a type-theoretical point of view one might claim that a proof is of a higher type than a wff. Hence no wff can be a proof. (Of course, I know almost nothing about type theory...) By the way, in certain circumstances, there is an advantage to taking > sequence in the literal sense of a certain kind of function and not as > iterated ordered pairs. > I'm sure you are right. B. -- For every line of Cantor's list it is true that this line does not contain the diagonal number. Nevertheless the diagonal number may be in the infinite list. (WM, sci.logic) === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result <6p54945cd429hp3ucmri3qcsie8l6uvcn5@4ax.com> posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 Gecko/2008070208 Firefox/3.0.1,gzip(gfe),gzip(gfe) I don't know why you keep circling back to remind me of what I ALREADY > agreed to from the beginning. (1) In an everyday informal sense, we take an axiom to be a proof of > itself. (2) IF we take proofs as tuples (iterated ordered pairs) then, since a > 1-tuple is just the object itself, an axiom is a proof of itself. Fine! This OF COURSE would relativize your claim: > Axioms are not proofs. *I* relativized it MYSELF. Sheesh, you're telling me that my remarks stand only as qualified, when I had ALREADY made such a qualification. > Axioms are proofs BY DEFINITION -- namely one-line proofs. That's fine in an ordinary, informal sense. But strictly speaking (and admittedly pedantic), IF 'proof' is defined as a certain kind of SEQUENCE, then the axiom itself is not a proof, but rather the sequence in which the axiom is the first entry is the proof. I.e., if S is the axiom, then the proof is a SEQUENCE: {<0 S>} which is not S itself. And in my very first post I also included that as my context. > Of course they are not. [Moe] Well, you see... Hence I dared to mention: In a _certain sense_ they are. One might consider an axiom to be a proof of > length 1 (proving itself). It's just the of course-part of your claim, that triggered my reaction. But, of course: (3) However [...] if we take 'sequence' in the ordinary > sense of a certain kind of function, then a single entry > proof of the axiom S is {<0 S>}, which is not S itself. Right. This kind of thing is a mindless waste. I stated my context in my first post. You mentioned that we could adopt another context. I AGREED that in such other contexts that an axiom is a proof of itself. I don't see any point on which we are even in disagreement, yet you're going back over and over the matter. Please let's have a more interesting conversation. Oops, too late, I can that in post following you are STILL going on! MoeBlee === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result I stated my context in my first post. You mentioned that we could > adopt another context. I AGREED that in such other contexts that > an axiom is a proof of itself. > Great! So we have accomplished consent! :-) B. -- For every line of Cantor's list it is true that this line does not contain the diagonal number. Nevertheless the diagonal number may be in the infinite list. (WM, sci.logic) === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 Gecko/2008070208 Firefox/3.0.1,gzip(gfe),gzip(gfe) > I stated my context in my first post. You mentioned that we could > adopt another context. I AGREED that in such other contexts that > an axiom is a proof of itself. Great! So we have accomplished consent! :-) Since we agreed from the beginning, it wasn't hard to do. Though you sure maked it SEEM hard! MoeBlee === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result > I stated my context in my first post. You mentioned that we could > adopt another context. I AGREED that in such other contexts that > an axiom is a proof of itself. > Great! So we have accomplished consent! [Finally!] :-) > Since we agreed from the beginning, it wasn't hard to do. Though you > sure maked it SEEM hard! > You mean a proof of lenght 1? ;-) :-) B. -- For every line of Cantor's list it is true that this line does not contain the diagonal number. Nevertheless the diagonal number may be in the infinite list. (WM, sci.logic) === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 Gecko/2008070208 Firefox/3.0.1,gzip(gfe),gzip(gfe) > I stated my context in my first post. You mentioned that we could > adopt another context. I AGREED that in such other contexts that > an axiom is a proof of itself. > Great! So we have accomplished consent! [Finally!] :-) Since we agreed from the beginning, it wasn't hard to do. Though you > sure maked it SEEM hard! You mean a proof of lenght 1? ;-) :-) You want to send me to an early grave? That's what this is all about, right? %>) MoeBlee === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result (*) If X is a set and f:X->P(X) is a map to its powerset, > æ æ then f is not surjective. And how will you prove that? Cantor? I told you already, with first order logic from the axioms of Z set > theory. > > > I didn't miss it. > > Axioms are not proofs. You teach me a consistent system might still be > completely useless. In mathematics, if you cannot reduce what you are assuming (and you are always assuming something) to explicitly stated axioms, you have no idea of what you can count on. === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 Gecko/2008070208 Firefox/3.0.1,gzip(gfe),gzip(gfe) > (*) If X is a set and f:X->P(X) is a map to its powerset, > æ æ then f is not surjective. And how will you prove that? Cantor? I told you already, with first order logic from the axioms of Z set > theory. I didn't miss it. Axioms are not proofs. Of course they are not. A proof in a system is a SEQUENCE of sentences each of which is an axiom of the system or results from previous entries in the sequence by a rule of the system. And the sentence in question is provable by just intutionistic logic (even WEAKER than classical logic) with the axiom schema of separation. OF COURSE we don't claim to prove the sentence from NO axioms. > You teach me a consistent system might still be > completely useless. Whatever you mean by useless. Anyway, again, when we say we have a proof we mean that we have a proof from certain axioms and with certain rules. OF COURSE, for a sentence that is not a logical truth, we don't claim to prove it from NO axioms. We recognize that the sentence is not a logical truth (i.e., it is not true in every interpretation) but rather we claim that it is a THEOREM of certain axioms (i.e., it is true in every interpretation in which the axioms of true). And that is the case for ANY mathematical theorem that is not a logical truth, even the theorem 1+1=2. MoeBlee === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result > Axioms are not proofs. > Of course they are not. > In a _certain sense_ they are. One might consider an axiom to be a proof of length 1 (proving itself). A proof in a system is a SEQUENCE of sentences each of which is an axiom > of the system or results from previous entries in the sequence by a rule > of the system. > Right. (Ok, a sequence consisting just of one wff may be considered different from that wff, but ... well.) And the sentence in question is provable by just intutionistic logic > (even WEAKER than classical logic) with the axiom schema of separation. OF COURSE we don't claim to prove the sentence from NO axioms. > Right. :-) > You teach me a consistent system might still be completely useless. > I guess that's indeed the case. :-) B. -- For every line of Cantor's list it is true that this line does not contain the diagonal number. Nevertheless the diagonal number may be in the infinite list. (WM, sci.logic) === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 Gecko/2008070208 Firefox/3.0.1,gzip(gfe),gzip(gfe) > Axioms are not proofs. Of course they are not. In a certain sense they are. One might consider an axiom to be a proof of length > 1 (proving itself). Not in the sense I specify below: > A proof in a system is a SEQUENCE of sentences each of which is an axiom > of the system or results from previous entries in the sequence by a rule > of the system. Right. (Ok, a sequence consisting just of one wff may be considered different from > that wff, but ... well.) Yes, it is different. A sequence with just one entry is a set with one ordered pair, in this case, the second coordinate being the formula. That ordered pair is not the formula itself. MoeBlee === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result <7624941d21bcv69ah10cov3d40ohqv09rn@4ax.com> posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 Gecko/2008070208 Firefox/3.0.1,gzip(gfe),gzip(gfe) > Axioms are not proofs. > Of course they are not. > In a certain sense they are. One might consider an axiom to be a proof > of length 1 (proving itself). Not in the sense I specify below: > A proof in a system is a SEQUENCE of sentences each of which is an axiom > of the system or results from previous entries in the sequence by a rule > of the system. > Right. (Ok, a sequence consisting just of one wff may be considered different from > that wff, but ... well.) Yes, it is different. A sequence [consisting of just this one formula] > not the formula itself. difference between the axiom and the proof stated below? Axiom: > æ æ æ æ ExAy(y !e x). Proof [of the theorem ExAy(y !e x)]: æ æ æ æ ExAy(y !e x). Hardly, I guess. :-) Of course. But, as you recoginized, I'm not talking about such DISPLAYS INDICATING sequences, but rather the sequence ITSELF. If S is the axiom, then the proof of S is: {<0 S>} which is not S itself. I think you understood that from my remarks. > Ok, in this case one might require that a proof of length 1 (a proof-sequence > consisting of just one formula) is identified with the formula. Then my claim > would be correct even from a formal point of view. Sure, IF you add such a clause. But ordinarily, that clause is not added, and to add it unnecessarily complicates the definition. > Think n-tuples with n = 1, there we (usually) have (a 1) = a 1. That is correct. But THAT sense of n-tuple is different from the sense of, STRICTLY SPEAKING, a sequence, which is a certain kind of function. Ordinarily, we refer to a proof as a sequence and not as an n-tuple (where 'n-tuple' is in the sense of iterating the ordered pair operation). MoeBlee === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result > difference between the axiom and the proof stated below? > Axiom: > æ æ æ æ ExAy(y !e x). > Proof [of the theorem ExAy(y !e x)]: > æ æ æ æ ExAy(y !e x). > Hardly, I guess. :-) Of course. But, as you recoginized, I'm not talking about such > DISPLAYS INDICATING sequences, but rather the sequence ITSELF. > Right. The string a just consisting of the character 'a' is not identical with that character. :-) Or is it? :-o If S is the axiom, then the proof of S is: {<0, S>} which is not S itself. > Only if you encode proofs, i.e. sequences of wffs making up the proof, with /sequences defined as functions/ (in the usual set theoretic sense). I think you understood that from my remarks. > Sure. > Ok, in this case one might require that a proof of length 1 (a proof-sequence > consisting of just one formula) is _identified_ with the formula. Then my claim > would be correct even from a formal point of view. > Sure, IF you add such a clause. But ordinarily, that clause is not > added, and to add it unnecessarily complicates the definition. > Again, you right from a _formal_ point of view. Still many mathematicians would not hesitate to _assume_ (tacitly) exactly those clause, I guess. > Think of n-tuples with n = 1, there we (usually) have (a_1) = a_1. > That is correct. But THAT sense of n-tuple is different from the sense > of, STRICTLY SPEAKING, a sequence, which is a certain kind of > function. > At least, that's what you had in mind here (and of course, that's what we usually mean when talking about /sequences/, I guess). Ordinarily, we refer to a proof as a sequence and not as an n-tuple (where 'n- > tuple' is in the sense of iterating the ordered pair operation). > Right. But for finite many entries (i.e. n e IN) those two mathematical structures are isomorphic. Hence... Anyway, in any case we may claim that an axiom is a THEOREM (by definition). And that its proof consists of just one line containing exactly this axiom. (Note that I refer to a extra-set-theoretical reality here. If you like, I consider a proof to be a _linguistic entity_. And your /sequences/ are just mathematical means to analyze this entities. This would _justify_ the identification of a one-line proof with the formula making up this proof.) B. -- For every line of Cantor's list it is true that this line does not contain the diagonal number. Nevertheless the diagonal number may be in the infinite list. (WM, sci.logic) === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result <7b44941b5ep2kdr7v9hq6j1rmdm390t4uq@4ax.com> posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 Gecko/2008070208 Firefox/3.0.1,gzip(gfe),gzip(gfe) > The string a just consisting of the character 'a' is not identical with that > character. :-) Or is it? :-o It is not identical, IF we take 'string' to mean a 'finite sequence'. > If S is the axiom, then the proof of S is: {<0, S>} which is not S itself. Only if you encode proofs, i.e. sequences of wffs making up the proof, with > /sequences defined as functions/ (in the usual set theoretic sense). OF COURSE. I've said that about five times already. > I think you understood that from my remarks. Sure. Great. Then why are we going over it yet again? > Ok, in this case one might require that a proof of length 1 (a proof-sequence > consisting of just one formula) is identified with the formula. Then my claim > would be correct even from a formal point of view. Sure, IF you add such a clause. But ordinarily, that clause is not > added, and to add it unnecessarily complicates the definition. Again, you right from a formal point of view. Still many mathematicians would > not hesitate to assume (tacitly) exactly those clause, I guess. I've never seen evidence of that. Rather, we see all the time, as I said from the outset, that mathematicians ordinarily take these things in a more relaxed sense so that they don't fuss about the strict definition of 'sequence'. (And of course, as you know, there are technical definitions of 'proof' in which a proof is not a sequence but rather a certain kind of tree or other definitions.) > Think of n-tuples with n = 1, there we (usually) have (a 1) = a 1. That is correct. But THAT sense of n-tuple is different from the sense > of, STRICTLY SPEAKING, a sequence, which is a certain kind of > function. At least, that's what you had in mind here (and of course, that's what we usually > mean when talking about /sequences/, I guess). Yes, that's what I said I had in mind. And when the material does get real technical in mathematical logic, and an author says 'sequence', I bet that the author indeed means sequence in the technical sense of a certain kind of function. > Ordinarily, we refer to a proof as a sequence and not as an n-tuple (where 'n- > tuple' is in the sense of iterating the ordered pair operation). Right. But for finite many entries (i.e. n e IN) those two mathematical > structures are isomorphic. Hence... Yes, I already agreed to that (though for an isomporphism in an actual technical sense, you'd have to formulate the theorem). But see my comment in another post about the axiom schema of replacement. > Anyway, in any case we may claim that an axiom is a THEOREM (by definition). Indeed. And that was never at issue here. > And > that its proof consists of just one line containing exactly this axiom. Sure, in an informal sense where proofs are by line. > (Note > that I refer to a extra-set-theoretical reality here. If you like, I consider a > proof to be a linguistic entity . And your /sequences/ are just mathematical > means to analyze this entities. This would justify the identification of a > one-line proof with the formula making up this proof.) Of course, that is fine if you stipulate that informal sense. I merely cited the specific formal sense of a proof being a certain kind of sequence, as indeed in mathematical logic ordinarily we study proofs as mathamtical objects - a certain kind of sequence. (And, obviously, I'm using 'proof' here in such a 'technical sense' and not in its other non-technical sense as 'an argument or discourse that provides convincing grounds for believing a proposition or an argument or discourse that provides convincing grounds for believing that there exists a formal mathematical proof of a certain formula'.) MoeBlee === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result > Note that I refer to a extra-set-theoretical reality here. If you like, > I consider a proof to be a _linguistic entity_. And your /sequences/ are > just mathematical means to analyze this entities. This would _justify_ > the identification of a one-line proof with the formula making up > this proof.) > [...] I'm using 'proof' here in such a 'technical sense' and not in its > other non-technical sense as 'an argument or discourse that provides > convincing grounds for believing a proposition or an argument or > discourse that provides convincing grounds for believing that there > exists a formal mathematical proof of a certain formula'. > Same with me. The realistic view of a proof (as a certain linguistic entity) does not contradict the notion of proof in this formal sense (due to Frege). Anyway, I enjoyed our little exchange of ideas/thoughts. B. -- For every line of Cantor's list it is true that this line does not contain the diagonal number. Nevertheless the diagonal number may be in the infinite list. (WM, sci.logic) === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 Gecko/2008070208 Firefox/3.0.1,gzip(gfe),gzip(gfe) > Anyway, I enjoyed our little exchange of ideas/thoughts. I wish I could say the same. Let B = the set of all things you know. Let M = the set of all things I know. Now take the symmetric difference of B and M. Surely there is enough in that symmetric difference by which we could inform each other than what we've come up with today! MoeBlee === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result > Anyway, I enjoyed our little exchange of ideas/thoughts. > I wish I could say the same. > I'm sorry for you! :-/ Let B = the set of all things you know. Let M = the set of all things I know. Now take the symmetric difference of B and M. Surely there is enough in that symmetric difference by which we could > inform each other than what we've come up with today! > No doubt! :-) B. -- For every line of Cantor's list it is true that this line does not contain the diagonal number. Nevertheless the diagonal number may be in the infinite list. (WM, sci.logic) === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result posting-account=F3H0JAgAAADcYVukktnHx7hFG5stjWse .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) > Note that I refer to a extra-set-theoretical reality here. If you like, > I consider a proof to be a linguistic entity . And your /sequences/ are > just mathematical means to analyze this entities. This would justify > the identification of a one-line proof with the formula making up > this proof.) [...] I'm using 'proof' here in such a 'technical sense' and not in its > other non-technical sense as 'an argument or discourse that provides > convincing grounds for believing a proposition or an argument or > discourse that provides convincing grounds for believing that there > exists a formal mathematical proof of a certain formula'. I was in fact using proof vs axiom mostly in the non-technical sense. What I meant to say was exactly that stating a result as an axiom cannot provide convincing grounds in the face of an argument that states the opposite result from the basic properties of the natural numbers plus induction. -LV > Same with me. The realistic view of a proof (as a certain linguistic entity) > does not contradict the notion of proof in this formal sense (due to Frege). Anyway, I enjoyed our little exchange of ideas/thoughts. B. -- For every line of Cantor's list it is true that this line does not > æcontain the diagonal number. æNevertheless the diagonal number may > æbe in the infinite list. (WM, sci.logic) === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 Gecko/2008070208 Firefox/3.0.1,gzip(gfe),gzip(gfe) > I was in fact using proof vs axiom mostly in the non-technical > sense. What I meant to say was exactly that stating a result as an > axiom cannot provide convincing grounds in the face of an argument > that states the opposite result from the basic properties of the > natural numbers plus induction. But there IS no argument to the contrary from just the basic properties of natural number plus induction. It is obscure anyway what you would even mean by that. (1) There is no recursive axiomatization that proves all the properties of natural numbers. (2) So by 'basic', one might take you to mean first order PA. But proofs about power sets and things like that aren't even formalized in first order PA. (3) The proof we showed you used just ONE principle about sets (as far as I can tell upon a routine check of the argument): For any given set S and any formalized property P (that isn't formalized by mentioning D), there is a set D such that D is all aforementioned principle alone, we prove that for no set S is there a function from S onto PS. MoeBlee === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result > I was in fact using proof vs axiom mostly in the non-technical > sense. What I meant to say was exactly that stating a result as an > axiom cannot provide convincing grounds in the face of an argument > that states the opposite result from the basic properties of the > natural numbers plus induction. [Julio] > Isn't it funny, many cranks start out with rather reasonable requirements, but then, somehow, they get astray. Strange. @Julio: If we consider the Peano Axioms as a codification of the basic properties of the natural numbers plus induction, we clearly cannot prove the opposite result from them. Moreover no one ever claimed that Card(X) < Card(PX) is an _axiom_ of set theory. Actually it can be derived from the axioms (and definitions) of set theory, i.e. ZFC. B. -- For every line of Cantor's list it is true that this line does not contain the diagonal number. Nevertheless the diagonal number may be in the infinite list. (WM, sci.logic) === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 Gecko/2008070208 Firefox/3.0.1,gzip(gfe),gzip(gfe) > 1+1=2: that's again the logic of the billion flies. Just like peace > should be a two-way street and not these self-blessing holy wars of > the self-imposing global false-democracy of the true-idiots and the > true-assassins and enslavers and exploiters, plain criminals against > all intelligence and life. Anyway, it, this planet is just having > what it's always deserved. And to think I even doubted whether you'd ever give us your argument that N and PN are 1-1. But lo and behold you came through admirably. MoeBlee === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result posting-account=F3H0JAgAAADcYVukktnHx7hFG5stjWse .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) > card(P(N)) = card(N), rev 2: a preliminary result > ------------------------------------------------- I shall only try to prove a preliminary property, from which it should > then be trivial to prove the thesis, so I will leave the final steps > out for now. The proof below is based on transfinite induction (see Wikipedia). All > should be quite trivial apart from one of the last passages, which I > have marked. To simplify the notation for the limit case, I shall take the set of > the extended naturals as reference. To be explicit: N* := N U { 0, oo } 'A' is for-each > 'U' is set-union > 'e' is set-membership > 'c=' is set-inclusion (improper) Let P(n) be the set: P(n) := { k e N* | k < 2^n }, n e N* By transfinite induction, we shall prove the following property: Prop.) A n e N* : P(n) c= N* -- Zero case: n = 0 P(0) = { 0 } c= > æ æ c= N* -- Successor case: n+1, assuming P(n) c= N*, n e N P(n+1) = P(n) U { k e N* | 2^n <= k < 2^(n+1) } c= > æ æ æ c= N* æ U { k e N* | 2^n <= k < 2^(n+1) } = > æ æ æ æ= N* -- Limit case: n = oo, assuming A m e N : P(m) c= N* P(oo) æ= P(n) U { k e N* | 2^n <= k < 2^oo } c= Sorry, there is a typo there, better read: P(oo) = P(m) U { k e N* | 2^m <= k < 2^oo } c= and so on... -LV > æ æ æ c= N* æ U { k e N* | 2^n <= k < 2^oo } (The following passage should be the culprit:) > Since k e N*, where N* := N U { 0, oo }, it must be: æ { k e N* | 2^n <= k < 2^oo } = { k e N* | 2^n <= k < oo } Then: P(oo) c= N* æ U { k e N* | 2^n <= k < 2^oo } = > æ æ æ æ= N* æ U { k e N* | 2^n <= k < oo } = > æ æ æ æ= N* Follows the thesis. QDE. A possible interpretation for the whole matter, in the form of a fast > slogan: however fast, it's just going to infinity. Anyway, right or wrong, I hope we have narrowed it down... and that > there just isn't some blatant mistake! > Julio status: OPEN > revision: 2 > sender: Julio Di Egidio (aka LudovicoVan) > sender-email: ju...@diegidio.aleph.name (del 'aleph.') > copyright: 2008 (on behalf of) sci.math, sci.logic > All rights reserved. > Second attempt. I'll try two equivalent formulations, in the hope > that, together, they will complement the eventual ambiguities. ----- 1st formulation (i) æ P(A) ~ { 0, 1, 2, ..., 2^n - 1}, AneN Let's simply call P(A) the set on the RHS of (i), keeping in mind that > the argument is going to hold due to this bijection and the > transitivity of the '~' relation. (ii) æP(A) c= N, AneN (iii) P(N) is infinite (trivial) We pass from P(A) to P(N) by transfinite induction, then: (iv) æP(N) ~ N, from (ii) and (iii) QDE. (1) ----- 2nd formulation Same comments as above. (i) æ card(P(A)) = 2^n, AneN (ii) æcard(P(A)) <= card(N), AneN > (iii) card(P(N)) >= oo (meaning: 'oo' as lower bound) (iv) æcard(P(N)) = card(N) QDE. (2) ----- Hmm, correct? The basic idea behind this argument is that P(A) is > always a subset of N in the finite case. Then, by transfinite > induction, while P(N) becomes infinite, its cardinality's upper bound > remains N's cardinality. (These informal expositions get me into more > troubles than they help, but I have to run the risk: no pain, no > gain!) Good luck. Julio status: OPEN > revision: 1 > sender: Julio Di Egidio (aka LudovicoVan) > sender-email: ju...@diegidio.aleph.name (del 'aleph.') > copyright: 2008 (on behalf of) sci.math, sci.logic > All rights reserved. > card(P(N)) = card(N), an elementary proof (and some fuzzy notation > here and there). --- Preliminaries: 'P(A)' stands for the powerset of set 'A'. '~' stands for the existence of a bijection, which will be, in this > context, equivalent to stating that two sets are equinumerous. > Infinity of the set 'N' (the set of natural/ordinal numbers) is taken > to be defined as usual, by the existence of bijections between N and > its (infinite) subsets. No other kind of infinities are going to be > assumed: the only tool I have used is transfinite induction over the > naturals/ordinals, with omega as the limit ordinal. 'oo' stands for omega. --- Finite case: card(A) = n <=>def > æ æ A ~ { 0, 1, 2, .., n-1 } card(P(A)) = 2^n <= æ æ P(A) ~ { 1, 2, 4, ..., 2^n-1 } --- Infinite case (transf. induction, notation a bit fuzzy here): (1) n=oo => card(A) = card(N) = oo <=>def > æ æ A ~ N := { 0, 1, 2, ..., n }, n->oo (2) n=oo => card(P(A)) = card(P(N)) = 2^oo <= æ æ P(A) ~ P(N) ~ 2^N := { 1, 2, 4, ..., 2^n }, n->oo --- Theorem: card(P(N)) = card(N) --- Proof: (i) card(P(N)) = card(N) <=> P(N) ~ N, by def. of 'card' (ii) card(P(N)) = card(N) <=> 2^N ~ N, by (2) and trans. of '~' (iii) A bijection 2^N ~ N trivially exists: 2^N ~ N > -------- > æ 1 ~ 0 > æ 2 ~ 1 > æ 4 ~ 2 > ... Follows the thesis. QDE. It's THEN easy to get interesting lemmas. We get that omega is an > absolute upper bound and all infinities are equivalent. The diagonal > argument then assumes another shape, which will be the subject of > another post of mine, unless someone finds a flaw in the above quite > elementary argument. > Julio status: OPEN > sender: Julio Di Egidio (aka LudovicoVan) > sender-email: ju...@diegidio.aleph.name (del 'aleph.') > copyright: 2008 (on behalf of) sci.math, sci.logic > All rights reserved. === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result > card(P(N)) = card(N), rev 2: a preliminary result > ------------------------------------------------- > > I shall only try to prove a preliminary property, from which it should > then be trivial to prove the thesis, so I will leave the final steps > out for now. > > The proof below is based on transfinite induction (see Wikipedia). All > should be quite trivial apart from one of the last passages, which I > have marked. > > To simplify the notation for the limit case, I shall take the set of > the extended naturals as reference. To be explicit: > > N* := N U { 0, oo } One presumes that this N* is intended to be an ordered set with the usual order of N u {0} and with x < oo for all x =/= oo. In such things, it is better t be explicit about what one is assuming. > > 'A' is for-each > 'U' is set-union > 'e' is set-membership > 'c=' is set-inclusion (improper) > > Let P(n) be the set: > > P(n) := { k e N* | k < 2^n }, n e N* What has this to do with power sets? The point for power sets would be to have something like: for each n in N*, P(n) = {k : (k c= N*) and AxeN*(x e k => x <= n)} That would mean that each P(n), for any n on N*, is the power set for the set of x in N* no greater than that n. Since you do not have that, you are not showing anything about power sets at all. === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result posting-account=F3H0JAgAAADcYVukktnHx7hFG5stjWse .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) > To simplify the notation for the limit case, I shall take the set of > the extended naturals as reference. To be explicit: N* := N U { 0, oo } One presumes that this N* is intended to be an ordered set with the > usual order of N u {0} and with x < oo for all x =/= oo. In such things, it is better t be explicit about what one is assuming. That's why I have given it explicitly: N* := N U { 0, oo } which is different than what you are presuming above. > Let P(n) be the set: P(n) := { k e N* | k < 2^n }, n e N* What has this to do with power sets? So, does the property hold? If it holds, the rest should be trivial. -LV === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result > > > To simplify the notation for the limit case, I shall take the set of > the extended naturals as reference. To be explicit: N* := N U { 0, oo } One presumes that this N* is intended to be an ordered set with the > usual order of N u {0} and with x < oo for all x =/= oo. In such things, it is better t be explicit about what one is assuming. > > That's why I have given it explicitly: > > N* := N U { 0, oo } You also use the notation oo < n and n < oo so < needs defining on N*. -- He is not here; but far away The noise of life begins again And ghastly thro' the drizzling rain On the bald street breaks the blank day. === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result > > To simplify the notation for the limit case, I shall take the set of > the extended naturals as reference. To be explicit: N* := N U { 0, oo } One presumes that this N* is intended to be an ordered set with the > usual order of N u {0} and with x < oo for all x =/= oo. In such things, it is better t be explicit about what one is assuming. > > That's why I have given it explicitly: > > N* := N U { 0, oo } > > which is different than what you are presuming above. > > Let P(n) be the set: P(n) := { k e N* | k < 2^n }, n e N* What has this to do with power sets? > > So, does the property hold? > > If it holds, the rest should be trivial. Since as stated your symbols say nothing at all about power sets, what is it about power sets that is supposed to be trivial? === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result posting-account=F3H0JAgAAADcYVukktnHx7hFG5stjWse .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) > Since as stated your symbols say nothing at all about power sets, what > æis it about power sets that is supposed to be trivial? Are you kidding me? Does the property hold? The rest is trivial, but I won't get into the next step until this one has been cleared. -LV === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 Gecko/2008070208 Firefox/3.0.1,gzip(gfe),gzip(gfe) Since as stated your symbols say nothing at all about power sets, what > æis it about power sets that is supposed to be trivial? Are you kidding me? Does the property hold? The rest is trivial, but I won't get into the next step until this one > has been cleared. Unless I made a huge mistake (and therefore would need new glasses right away), it's trivial that it holds that P(n) subset of N*. You DEFINED P(n) explicitly to be a subset of N* when you put keN* in {keN* | ANYTHING HERE!}. Sheesh. MoeBlee === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result > Since as stated your symbols say nothing at all about power sets, what > is it about power sets that is supposed to be trivial? > Are you kidding me? > Does the property hold? > The rest is trivial, but I won't get into the next step until this one > has been cleared. > > Unless I made a huge mistake (and therefore would need new glasses > right away), it's trivial that it holds that P(n) subset of N*. You > DEFINED P(n) explicitly to be a subset of N* when you put keN* in > {keN* | ANYTHING_HERE!}. What is that 2^oo appearing in ANYTHING_HERE? His statement is not trivially true, because it doesn't even make any sense. It is going as usual with such cranky stuff: an endless repetition of attempts, each a greater mess than the previous one. The next step will be to smuggle that 2^oo which is floating around loosely into N*, and then to declare it countable because there is no place for anything uncountable in N*; and somehow this 2^oo is going to represent the power set of N. This then is the next version of this proof. While the initial problem in the previous version was the inappropriate usage of transfinite induction, now an additional and bigger problem arises: a basic misunderstanding about sound definitions. And this opens the door for another endless discussion about nonsense. R.B. === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 Gecko/2008070208 Firefox/3.0.1,gzip(gfe),gzip(gfe) > Since as stated your symbols say nothing at all about power sets, what > is it about power sets that is supposed to be trivial? > Are you kidding me? > Does the property hold? > The rest is trivial, but I won't get into the next step until this one > has been cleared. Unless I made a huge mistake (and therefore would need new glasses > right away), it's trivial that it holds that P(n) subset of N*. You > DEFINED P(n) explicitly to be a subset of N* when you put keN* in > {keN* | ANYTHING HERE!}. What is that 2^oo appearing in ANYTHING HERE? His statement is not trivially > true, because it doesn't even make any sense. I take it that 'oo' stands for omega (w). So '2^oo' stands for cardinal exponentiation: card({f | f is a function from w into {0 1}}). If he means otherwise, he can say so. But as I've described it, I don't see a problem. MoeBlee === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result > Since as stated your symbols say nothing at all about power sets, > what is it about power sets that is supposed to be trivial? > Are you kidding me? > Does the property hold? > The rest is trivial, but I won't get into the next step until this one > has been cleared. > Unless I made a huge mistake (and therefore would need new glasses > right away), it's trivial that it holds that P(n) subset of N*. You > DEFINED P(n) explicitly to be a subset of N* when you put keN* in > {keN* | ANYTHING_HERE!}. > What is that 2^oo appearing in ANYTHING_HERE? His statement is not > trivially true, because it doesn't even make any sense. > > I take it that 'oo' stands for omega (w). Yes, probably > So '2^oo' stands for > cardinal exponentiation: > card({f | f is a function from w into {0 1}}). Maybe. But I don't see that anywhere in ju...'s posting. Therefore his statements don't make sense. > If he means otherwise, he can say so. But as I've described it, I > don't see a problem. Did you ever do something like grading homework in a math course? Lets assume the students are asked to prove something. They hand in solutions which start with a couple of assumptions that look somehow weird, but are not outright wrong. Then things deteriorate. How would you handle this, as grader? Give credit up to but excluding the first statement which is plainly wrong, and accepting all statements for which /you/ can find some justification (isn't it the task of the student to formulate the proof in a way that it justifies itself)? Even if this involves to supplement definitions the student has not given? If you do so the student will never understand what has gone wrong in his proof. In the case at hand, do you have any doubt that the next step in the proof will be an attempt to smuggle 2^oo into N*? And if this has happened, at which point did the problem begin to surface? I'd say at that point where 2^oo appeared for the first time. Ralf === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result posting-account=F3H0JAgAAADcYVukktnHx7hFG5stjWse .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) > Did you ever do something like grading homework in a math course? Lets > assume the students are asked to prove something. This very assumption is flawed. You didn't give me no homework. This stuff is coming from my own pocket: the student got an idea! You just refuse to be listening at all because you are da boss and you say what can or cannot be proved, how, when, and by who. Fine. -LV === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result posting-account=9QOSvAoAAACEOWJVSDuswW7dB_0wApQO Gecko/2008061712 Fedora/3.0-1.fc9 Firefox/3.0,gzip(gfe),gzip(gfe) Did you ever do something like grading homework in a math course? Lets > assume the students are asked to prove something. This very assumption is flawed. You didn't give me no homework. This > stuff is coming from my own pocket: the student got an idea! You just > refuse to be listening at all because you are da boss and you say what > can or cannot be proved, how, when, and by who. Actually, I can very much relate to his comment. It very often happens, when dealing with some students, that what they do is not wrong, but meaningless. When they try to work their way away from failing, they usually demand that one point the precise point in their solutions where things go wrong. But it is quite often the case that there is no such point: it is the whole thing that is meaningless, so it is not wrong in any meaningful sense. -- m === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result Did you ever do something like grading homework in a math course? Lets > assume the students are asked to prove something. This very assumption is flawed. You didn't give me no homework. This > stuff is coming from my own pocket: the student got an idea! You just > refuse to be listening at all because you are da boss and you say what > can or cannot be proved, how, when, and by who. > > Actually, I can very much relate to his comment. > It very often happens, when dealing with some students, > that what they do is not wrong, but meaningless. > When they try to work their way away from failing, > they usually demand that one point the precise > point in their solutions where things go wrong. But > it is quite often the case that there is no such point: > it is the whole thing that is meaningless, so it is not > wrong in any meaningful sense. The commonly used phrase for that, dating back at least to Littlewood, is that he is not even wrong. === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result posting-account=F3H0JAgAAADcYVukktnHx7hFG5stjWse .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) > Since as stated your symbols say nothing at all about power sets, what > is it about power sets that is supposed to be trivial? > Are you kidding me? > Does the property hold? > The rest is trivial, but I won't get into the next step until this one > has been cleared. Unless I made a huge mistake (and therefore would need new glasses > right away), it's trivial that it holds that P(n) subset of N*. You > DEFINED P(n) explicitly to be a subset of N* when you put keN* in > {keN* | ANYTHING HERE!}. What is that 2^oo appearing in ANYTHING HERE? His statement is not trivially > true, because it doesn't even make any sense. I take it that 'oo' stands for omega (w). So '2^oo' stands for > cardinal exponentiation: > card({f | f is a function from w into {0 1}}). If he means otherwise, he can say so. But as I've described it, I > don't see a problem. I didn't even know what cardinal exponentiation is. What I believe is that the argument is over N, the naturals, and, that there is anything beyond and/or beside w, is rather a result we are supposed to get, not a starting point. So, to me, 2^oo just tells about a lower bound. The rest is what there is to prove. And my argument by induction seems to show there is an upper bound too. That is enough for it, and we have that there is only one 'w' or 'oo', and that it is -say- just: 2^oo = oo as it is: 2*oo = oo as it is: 1+oo = oo and so on. And then have a back-look at that 1+oo=oo because there is the diagonal argument, as well as the halting problem, as well as much of the rest. (BTW, those who won't get this one, please don't get mad.) At least, as I see it. No pretence. -LV > MoeBlee === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result > I didn't even know what cardinal exponentiation is. Then why try to prove something false about it? === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 Gecko/2008070208 Firefox/3.0.1,gzip(gfe),gzip(gfe) > I take it that 'oo' stands for omega (w). So '2^oo' stands for > cardinal exponentiation: > card({f | f is a function from w into {0 1}}). If he means otherwise, he can say so. But as I've described it, I > don't see a problem. I didn't even know what cardinal exponentiation is. Then you need to say what you mean by notation such as '2^oo'. > What I believe is that the argument is over N, the naturals, You mean 'the positive naturals', since you don't have 0 (which is a natural number (by most authors these days, I would bet)) in N. >and, that > there is anything beyond and/or beside w, is rather a result we are > supposed to get, not a starting point. I didn't assume anything about starting points except the axioms and primitives. > So, to me, 2^oo just tells about a lower bound. The rest is what there > is to prove. PLEASE STOP RIGHT THERE. There cannot be a coherent, meaningful discussion about your arguments until you DEFINE your terminology, in this case '2^oo'. And saying what a terminology tells about is not a definition. MoeBlee === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result posting-account=F3H0JAgAAADcYVukktnHx7hFG5stjWse .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) > It is going as usual with such cranky stuff Yeah, but this time the cranks is you. -LV === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result posting-account=F3H0JAgAAADcYVukktnHx7hFG5stjWse .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) Since as stated your symbols say nothing at all about power sets, what > æis it about power sets that is supposed to be trivial? Are you kidding me? Does the property hold? The rest is trivial, but I won't get into the next step until this one > has been cleared. Unless I made a huge mistake (and therefore would need new glasses > right away), it's trivial that it holds that P(n) subset of N*. You > DEFINED P(n) explicitly to be a subset of N* when you put keN* in > {keN* | ANYTHING HERE!}. Yes! :) I guess that's what was known by the back of my mind. The argument is indeed subtle: it seems to me it depends on how we define our domain. One only limit ordinal w is not in itself a problem (is it?), actually it seems we are sort of giving it substance with this argument. Anyway, never mind, thinking out loud. I'll try the next step. -LV Sheesh. MoeBlee === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 Gecko/2008070208 Firefox/3.0.1,gzip(gfe),gzip(gfe) > The argument is indeed subtle: it seems to me it depends on how we > define our domain. One only limit ordinal w is not in itself a problem > (is it?), actually it seems we are sort of giving it substance with > this argument. It's not even subtle. It's as obvious as the Eiffel Tower. You can take the following as a theorem schema: For any term T and any formula F, all closures of {keT | F} c= T are theorems. MoeBlee === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result posting-account=F3H0JAgAAADcYVukktnHx7hFG5stjWse .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) > It's not even subtle. It's as obvious as the Eiffel Tower. Still waiting to learn where you guys see the overall flaw. Oh yes, you wouldn't have tried it at all. That one I know. -LV === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result posting-account=F3H0JAgAAADcYVukktnHx7hFG5stjWse .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) > The _reason_ it doesn't follow is that N has many subsets > which are not subsets of any A_n. Theses are the > _infinite_ subsets. What you've actually proved is that > the _finite_ subsets of N are countable, which is true. You keep proving Cantor with Cantor. This is the only truth. -LV === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result > > >The _reason_ it doesn't follow is that N has many subsets >which are not subsets of any A_n. Theses are the >_infinite_ subsets. What you've actually proved is that >the _finite_ subsets of N are countable, which is true. > > > You keep proving Cantor with Cantor. No. The matter is proved with logic and set theory. Bob Kolker === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result > You keep proving Cantor with Cantor. [Julio] > Typical crank speak. No. The matter is proved with logic and set theory. > Or rather in the context of axiomatic set theory (i.e. using logic, from the set theoretic axioms and definitions). On the other hand, since _set theory_ originated from Cantor, and Cantor already proved that theorem (in his pre-axiomatic system of set theory), AND the theorem actually is called /Cantor's theorem/ it's hard NOT to refer to Cantor when dealing with this theorem. :-) And right, the typical proof for this theorem is still more or less identical with the proof given by Cantor. So what?! :-o B. Reference: http://en.wikipedia.org/wiki/Cantor's_theorem -- For every line of Cantor's list it is true that this line does not contain the diagonal number. Nevertheless the diagonal number may be in the infinite list. (WM, sci.logic) === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result > > The _reason_ it doesn't follow is that N has many subsets > which are not subsets of any A_n. Theses are the > _infinite_ subsets. What you've actually proved is that > the _finite_ subsets of N are countable, which is true. > > You keep proving Cantor with Cantor. > > This is the only truth. It is not even false. === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result posting-account=F3H0JAgAAADcYVukktnHx7hFG5stjWse .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) > You keep proving Cantor with Cantor. This is the only truth. It is not even false. Get lost. -LV === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result > > You keep proving Cantor with Cantor. This is the only truth. It is not even false. > > Get lost. > Just yesterday, you were asking for my opinion. Perhaps you should be more careful about what you ask for. === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result > It's not even subtle. It's as obvious as the Eiffel Tower. >Still waiting to learn where you guys see the overall flaw. > You haven't been reading the replies. A lot of people have > been correctly complaining about the fact that the notation > and terminology you use is all wrong. If we ignore that > and try to consider the underlying ideas, the overall > flaw is this: > Say A_n = {1,2...n}. It's true that card(P(A_n)) < aleph_0 > for every natural number n. And it's true that in some sense > N is the limit of A_n as n tends to infinity. But it simply > does not follow that card(P(N)) <= aleph_0. > The _reason_ it doesn't follow is that N has many subsets > which are not subsets of any A_n. Theses are the > _infinite_ subsets. What you've actually proved is that > the _finite_ subsets of N are countable, which is true. Or, putting it another way, the power set of the limit is not the limit of the power sets. -- Dave Seaman Third Circuit ignores precedent in Mumia Abu-Jamal ruling. === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result posting-account=yKimjgoAAACk5WwPVD4l9HmbpoR6Hmy4 en),gzip(gfe),gzip(gfe) It's not even subtle. It's as obvious as the Eiffel Tower. > Still waiting to learn where you guys see the overall flaw. I think you are trying to use transfinite induction incorrectly. If you explain what you think transfinite induction is then we can see. Try to be as clear as possible! For example if I was explaining induction on the naturals then I would state something like this: If we have a property, P and we have * P(0) is true and * For any natural number, k, if P is true of k then P is true of k+1 Then P is true of all natural numbers Can you explain your understanding of transfinite induction in a similar way? That is, the conditions and the conclusion.. === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result posting-account=F3H0JAgAAADcYVukktnHx7hFG5stjWse .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) > It's not even subtle. It's as obvious as the Eiffel Tower. > Still waiting to learn where you guys see the overall flaw. I think you are trying to use transfinite induction incorrectly. And I don't think so, but I might be mistaken. Everything has already been said, but here is again the substance of the argument, hopefully even more explicit (I just won't restate again the notation). I assume MoeBlee is correct in telling that the following property is trivial: A n e N* : P(n) c= N* Then there is an as trivial bijection between the set P(n) above and P(A n), the powerset of set A n, where n is the cardinality of some set A: A n e N* : P(A n) ~ P(n) Is that correct? (I think that bijection *must* hold, otherwise -say- we could not have our parings to state Cantor's Theorem.) I'd say the thesis just follows. (Was transfinite or even plain induction unneeded? You'll tell me.) > If > you explain what you think transfinite induction is then we can see. C'mon... -LV === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result > It's not even subtle. It's as obvious as the Eiffel Tower. > Still waiting to learn where you guys see the overall flaw. I think you are trying to use transfinite induction incorrectly. > > And I don't think so, but I might be mistaken. > > Everything has already been said, but here is again the substance of > the argument, hopefully even more explicit (I just won't restate again > the notation). > > I assume MoeBlee is correct in telling that the following property is > trivial: > > A n e N* : P(n) c= N* It is trivially false if P(S) is to mean the set of all subsets of S. Do you perhaps mean something like Card(P(n)) e N* ? Note that 'P(x)' as a predicate should have the same meaning for all arguments 'x', but you seem to be giving it two conflicting meanings depending on what sort of argument it is taking. That is certainly an abuse of notation sufficient to make it impossible for anyone else to be sure of what you are trying to say. === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result posting-account=yKimjgoAAACk5WwPVD4l9HmbpoR6Hmy4 en),gzip(gfe),gzip(gfe) It's not even subtle. It's as obvious as the Eiffel Tower. > Still waiting to learn where you guys see the overall flaw. I think you are trying to use transfinite induction incorrectly. And I don't think so, but I might be mistaken. Everything has already been said, but here is again the substance of > the argument, hopefully even more explicit (I just won't restate again > the notation). I assume MoeBlee is correct in telling that the following property is > trivial: æ A n e N* : P(n) c= N* Then there is an as trivial bijection between the set P(n) above and > P(A n), the powerset of set A n, where n is the cardinality of some > set A: æ A n e N* : P(A n) ~ P(n) Is that correct? (I think that bijection *must* hold, otherwise -say- > we could not have our parings to state Cantor's Theorem.) I'd say the thesis just follows. (Was transfinite or even plain induction unneeded? You'll tell me.) You seem to be using P() to denote two different things. One of them is the powerset, and one of them is something else to do with powers of natural numbers. Is this right? If so, could you rewrite the above so the two different things have different symbols denoting them? It's very confusing for me at the moment. Sorry. === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 Gecko/2008070208 Firefox/3.0.1,gzip(gfe),gzip(gfe) > I assume MoeBlee is correct in telling that the following property is > trivial: æ A n e N* : P(n) c= N* Why do you have to assume? Why don't you see for yourself that {keT | F} is a subset of T for any term T and any formula F. > Then there is an as trivial bijection between the set P(n) above and > P(A n), the powerset of set A n, STOP RIGHT THERE. You're using 'P' now for two DIFFERENT THINGS. That only bodes confusion. In your latest posts you had P serve as something quite different from power set. Please use 'P' for just one thing here. Since 'P' is common for power set, I suggest, you not use 'P' for your definition: P(n) := { k e N* | k < 2^n }, n e N* But rather use 'Q' (or whatever other than 'P'): For neN*, we put Q(n) = {keN* | k<2^n}. MoeBlee === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result posting-account=9QOSvAoAAACEOWJVSDuswW7dB_0wApQO Gecko/2008071615 Fedora/3.0.1-1.fc9 Firefox/3.0.1,gzip(gfe),gzip(gfe) It's not even subtle. It's as obvious as the Eiffel Tower. > Still waiting to learn where you guys see the overall flaw. Oh yes, you wouldn't have tried it at all. That one I know. -LV The overall flaw is that your argument is trying to prove something that is false (in `usual' set theory) -- m === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result posting-account=F3H0JAgAAADcYVukktnHx7hFG5stjWse .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) On 29 Jul, 18:00, Mariano Su.87rez-Alvarez > The overall flaw is that your argument is trying to > prove something that is false (in `usual' set > theory) It is just indirectly false, BTW. You have not even means to really disprove me. It's a funny thing. -LV -- m === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result posting-account=9QOSvAoAAACEOWJVSDuswW7dB_0wApQO Gecko/2008071615 Fedora/3.0.1-1.fc9 Firefox/3.0.1,gzip(gfe),gzip(gfe) > On 29 Jul, 18:00, Mariano Su.87rez-Alvarez The overall flaw is that your argument is trying to > prove something that is false (in `usual' set > theory) > It is just indirectly false, BTW. It is false in so far as one can actually prove the negation of your statement. I have no idea what `indirectly false' might mean. > You have not even means to really disprove me. I do not know what this means. > It's a funny thing. Not really. -- m === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 Gecko/2008070208 Firefox/3.0.1,gzip(gfe),gzip(gfe) > Still waiting to learn where you guys see the overall flaw. We're waiting to see your argument that N is 1-1 with PN. MoeBlee === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result posting-account=F3H0JAgAAADcYVukktnHx7hFG5stjWse .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) Still waiting to learn where you guys see the overall flaw. We're waiting to see your argument that N is 1-1 with PN. Now you are driving me crazy with this non-synchronized cross-posting. No, I didn't even attempt that!! (Although it might be a consequence.) How many times shall I still repeat this? -LV MoeBlee === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result Still waiting to learn where you guys see the overall flaw. We're waiting to see your argument that N is 1-1 with PN. > > Now you are driving me crazy with this non-synchronized cross-posting. > > No, I didn't even attempt that!! Then it is not at all clear what you are attempting. Except that it appears to be incoherent. === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 Gecko/2008070208 Firefox/3.0.1,gzip(gfe),gzip(gfe) > Still waiting to learn where you guys see the overall flaw. We're waiting to see your argument that N is 1-1 with PN. Now you are driving me crazy with this non-synchronized cross-posting. No, I didn't even attempt that!! (Although it might be a consequence.) How many times shall I still repeat this? The TITLE of this thread begins: card(PN) = card(N) and that is what you keep saying you're going to prove. That is equivalent to N is 1-1 with PN. Either way it is said: card(PN) = card(N) or N is 1-1 with PN we're waiting for you new argument. (It was made clear to you how your previous argument for card(PN) = card(N) was incorrect.) MoeBlee === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result posting-account=U44YcwkAAAAbGXB70Qr7gA3kornmKE4i Gecko/20080325 Ubuntu/7.10 (gutsy) Firefox/2.0.0.13,gzip(gfe),gzip(gfe) Since as stated your symbols say nothing at all about power sets, what > ?is it about power sets that is supposed to be trivial? Are you kidding me? Does the property hold? The rest is trivial, but I won't get into the next step until this one > has been cleared. Unless I made a huge mistake (and therefore would need new glasses > right away), it's trivial that it holds that P(n) subset of N*. You > DEFINED P(n) explicitly to be a subset of N* when you put keN* in > {keN* | ANYTHING HERE!}. Yes! :) I guess that's what was known by the back of my mind. The argument is indeed subtle: it seems to me it depends on how we > define our domain. One only limit ordinal w is not in itself a problem > (is it?), actually it seems we are sort of giving it substance with > this argument. Anyway, never mind, thinking out loud. I'll try the next step. I would recommend doing the first step first. You started all this with the claim in the title that suggests to mathematicians that you are saying something about powersets, yet powersets appear nowhere in your arguments. So, please start by saying -- in words only -- what you mean by the powerset of an arbitrary set X. You might also like to give a simple example by showing the powerset of the following simple set: A = {r, s, t} If we could see that you mean the same by powerset as everyone else, we could try to identify the discrepancy between your argument and, um, the normal mathematical one. Otherwise, we could perhaps try to thrash out mutually acceptable terminology for referring to what you mean. Brian Chandler === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result posting-account=F3H0JAgAAADcYVukktnHx7hFG5stjWse .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) Unless I made a huge mistake (and therefore would need new glasses > right away), it's trivial that it holds that P(n) subset of N*. You > DEFINED P(n) explicitly to be a subset of N* when you put keN* in > {keN* | ANYTHING_HERE!}. Yes! :) I guess that's what was known by the back of my mind. The argument is indeed subtle: it seems to me it depends on how we > define our domain. One only limit ordinal w is not in itself a problem > (is it?), actually it seems we are sort of giving it substance with > this argument. Anyway, never mind, thinking out loud. I'll try the next step. I would recommend doing the first step first. You started all this > with the claim in the title that suggests to mathematicians that you > are saying something about powersets, yet powersets appear nowhere in > your arguments. I am afraid you must have missed some bits. > So, please start by saying -- in words only -- what you mean by the > powerset of an arbitrary set X. You might also like to give a simple > example by showing the powerset of the following simple set: A = {r, s, t} You call: A = ( 1, 2, 3 } P(A) = { {}, { 1 }, { 2 }, { 3 }, { 1, 2 }, { 1, 3 }, { 2, 3 }, { 1, 2, 3 } } IMHO, to lower the level of noise, a good strategy might be to take people for serious and just tell what there is to tell: although one might straight discover there was nothing worth, the sooner the better in any case, isn't it? -LV > If we could see that _you_ mean the same by powerset as everyone > else, we could try to identify the discrepancy between your argument > and, um, the normal mathematical one. Otherwise, we could perhaps try > to thrash out mutually acceptable terminology for referring to what > _you_ mean. Brian Chandler === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result posting-account=U44YcwkAAAAbGXB70Qr7gA3kornmKE4i Gecko/20080325 Ubuntu/7.10 (gutsy) Firefox/2.0.0.13,gzip(gfe),gzip(gfe) with the claim in the title that suggests to mathematicians that you > are saying something about powersets, yet powersets appear nowhere in > your arguments. > I am afraid you must have missed some bits. > So, please start by saying -- in words only -- what you mean by the > powerset of an arbitrary set X. You haven't done this. Why not? It's quite simple, actually... > .., You might also like to give a simple > example by showing the powerset of the following simple set: A = {r, s, t} You call: A = ( 1, 2, 3 } P(A) = { > {}, > { 1 }, { 2 }, { 3 }, > { 1, 2 }, { 1, 3 }, { 2, 3 }, > { 1, 2, 3 } > } Why do you say You call?? Are you not quite sure if you agree? IMHO, to lower the level of noise, a good strategy might be to take > people for serious and just tell what there is to tell: although one > might straight discover there was nothing worth, the sooner the better > in any case, isn't it? That was my first reaction. I posted a proof of the opposite of what you claim to be trying to prove. Lots of people have given you straight comments, ranging from Wrong to Nonsense. What more do you want? Brian Chandler === Subject: Re: card(P(N)) = card(N), rev 2: a preliminary result posting-account=F3H0JAgAAADcYVukktnHx7hFG5stjWse .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) > Why do you say You call?? Are you not quite sure if you agree? We'll never agree, you are stupid. -LV === Subject: geometric mean calculation posting-account=7bdB9goAAACUdkVzwRjAJ-G7mqj5vDc- MathPlayer 2.10b; .NET CLR 1.0.3705; .NET CLR 1.1.4322; Media Center PC 4.0; .NET CLR 2.0.50727; IEMB3; IEMB3),gzip(gfe),gzip(gfe) how do you calculate geometric mean of this data: YEAR DOLLARS PERCENT CHG PERCENT CHNG (positive) 1979 15.6 1980 28 77.6 1.7756 1981 22 -20.6 0.7942 1982 4 -83.6 0.1636 1983 -23 -730.6 -6.3056 1984 -67 194.3 2.9427 1985 -90 34.0 1.3398 === Subject: Re: geometric mean calculation > how do you calculate geometric mean of this data: > > YEAR DOLLARS PERCENT CHG PERCENT CHNG > (positive) > 1979 15.6 > 1980 28 77.6 1.7756 > 1981 22 -20.6 0.7942 > 1982 4 -83.6 0.1636 > 1983 -23 -730.6 -6.3056 > 1984 -67 194.3 2.9427 > 1985 -90 34.0 1.3398 Well, leaving aside the fact that rising from 15.6 to 28 is not actually a 77.6% rise, and that -23 from 4 is not actually a change of -730.6%, you can't take the geometric mean of positive and negative numbers in any meaningful way. Perhaps you should tell us why you want to do this, and what sort of answer you are expecting. === Subject: Re: geometric mean calculation-- > how do you calculate geometric mean of this data: > YEAR DOLLARS PERCENT CHG PERCENT CHNG > (positive) > 1979 15.6 > 1980 28 77.6 1.7756 > 1981 22 -20.6 0.7942 > 1982 4 -83.6 0.1636 > 1983 -23 -730.6 -6.3056 > 1984 -67 194.3 2.9427 > 1985 -90 34.0 1.3398 > > Well, leaving aside the fact that rising from 15.6 > to 28 is not actually a 77.6% rise, and that -23 > from 4 is not actually a change of -730.6%, you > can't take the geometric mean of positive and > negative numbers in any meaningful way. Perhaps > you should tell us why you want to do this, and > what sort of answer you are expecting. Geometric mean calculation is common practice in the financial world. Performance indexes of stock markets, funds, portfolios etc. are (normally? often? of the relevant quantities and ratios. So interested persons can make profit unobserved, much like one can earn USD 0.005 per transaction on the average in normal money transfer by rounding to one direction only instead of rounding in the normal way. Whether this is legally allowed is a different issue. the components of the funds they buy. When selling they are paid according to the GM. The Happy profits: Johan E. Mebius === Subject: Re: geometric mean calculation-- <9128095.1217337231481.JavaMail.jakarta@nitrogen.mathforum.org> <488F6FFA.8090008@xs4all.nl> posting-account=CtWhuAoAAAAZZ9vwdovdqB3NNaiUa20_ SV1),gzip(gfe),gzip(gfe) > Geometric mean calculation is common practice in the financial world. > Performance indexes of stock markets, funds, portfolios etc. are (normally? often? > of the relevant quantities and ratios. > So interested persons can make profit unobserved, much like one can earn USD 0.005 per > transaction on the average in normal money transfer by rounding to one direction only > instead of rounding in the normal way. Whether this is legally allowed is a different issue. the components of the funds they buy. When selling they are paid according to the GM. The Happy profits: Johan E. Mebius More about Geometric Means on the Internet: http://en.wikipedia.org/wiki/Geometric_mean Anything free comes without profit: http://hdebruijn.soo.dto.tudelft.nl/jaar2008/ Han de Bruijn === Subject: Re: geometric mean calculation > how do you calculate geometric mean of this data: > > YEAR DOLLARS PERCENT CHG PERCENT CHNG > (positive) > 1979 15.6 > 1980 28 77.6 1.7756 > 1981 22 -20.6 0.7942 > 1982 4 -83.6 0.1636 > 1983 -23 -730.6 -6.3056 > 1984 -67 194.3 2.9427 > 1985 -90 34.0 1.3398 Well, leaving aside the fact that rising from 15.6 > to 28 is not actually a 77.6% rise, and that -23 > from 4 is not actually a change of -730.6%, you > can't take the geometric mean of positive and > negative numbers in any meaningful way. Perhaps > you should tell us why you want to do this, and > what sort of answer you are expecting. Well, the mean year is obviously 31st December 1981. Phil -- -- Microsoft voice recognition live demonstration === Subject: Re: geometric mean calculation Of what do you want the geometric mean? You have here more than one column of data. Some apepar to be dates. Do you want the geometric mean of the dates? Probably not. So what do you actually want? === Subject: Re: geometric mean calculation <13378941.1217319175151.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=7bdB9goAAACUdkVzwRjAJ-G7mqj5vDc- MathPlayer 2.10b; .NET CLR 1.0.3705; .NET CLR 1.1.4322; Media Center PC 4.0; .NET CLR 2.0.50727; IEMB3; IEMB3),gzip(gfe),gzip(gfe) On Jul 29, 4:12æam, riderofgiraffes more than one column of data. æSome apepar to be dates. > Do you want the geometric mean of the dates? æProbably > not. So what do you actually want? I'm trying to find the geometric mean of percentage change as well as the geometric mean of dollars. === Subject: Re: geometric mean calculation- > how do you calculate geometric mean of this data: > > YEAR DOLLARS PERCENT CHG PERCENT CHNG (positive) > 1979 15.6 > 1980 28 77.6 1.7756 > 1981 22 -20.6 0.7942 > 1982 4 -83.6 0.1636 > 1983 -23 -730.6 -6.3056 > 1984 -67 194.3 2.9427 > 1985 -90 34.0 1.3398 The definition formula of the geometric mean is GM (A1, A2, ..., An) = (A1 * A2 * ... * An)^(1/n) (A1 ... An all nonnegative) Computing the GM by computer directly by this formula is prone to overflow. To prevent overflow one can proceed in two ways: (1) Take first the n-th roots, then multiply. (2) If all numbers are positive then take the arithmetic mean M of log(A1), ..., log(An) and exponentiate M, i.e. take the antilogarithm of M. The specific logarithm base does not matter, at least not in theory. Good luck: Johan E. Mebius === Subject: Re: Directional derivative of a directional derivative of a Function posting-account=a93YEwoAAAClHm9Euy--V39SJpP16NI8 Gecko/2008070208 Firefox/3.0.1,gzip(gfe),gzip(gfe) > b'*H*a where H is the Hessian of C ... or am I missing something? > You're assuming C is twice differentiable, right? yesterday, and I've subsequently written out the solution for the 3D case I'm really interested in. === Subject: Re: the adjoint posting-account=mV9EXQoAAACmCMM9qg0N4eJlXyr2Z93U Gecko/2008070208 Firefox/3.0.1,gzip(gfe),gzip(gfe) > Also, what is an adjoint equation? Sometimes a certain problem, for example an evolution > equation on a Banach, admits an adjoint problem defined on the > dual Banach space (the Banach space of all bounded linear functionals > acting on the original space). It may be easier to study the adjoint > problem and then translate results back to the original space. The other way around is also possible. Sometimes one looks for a > pre-adjoint problem. An example of this kind of duality are the forward and backward Kolmogorov > equations: http://en.wikipedia.org/wiki/Kolmogorov backward equation Best wishes, > Sebastiaan. Not a particularly enlightening response - now the op will have to look at > the definition and meaning of Banach spaces. It will do him good. :-) === Subject: Area of an Off-Center Circle in Each Quadrant Say you have a circle who's diameter is the same as the width of one quadrant. How would you find the area of this circle that falls into each quadrant if the circle's center point was not at the origin? === Subject: Re: Area of an Off-Center Circle in Each Quadrant <15369562.1217289177446.JavaMail.jakarta@nitrogen.mathforum.org>, > Say you have a circle who's diameter is the same as the width of one > quadrant. How would you find .... You may not realize that you haven't explained the problem quite clearly enough. What do you mean by the width of one quadrant? Ken Pledger. === Subject: Re: Area of an Off-Center Circle in Each Quadrant <15369562.1217289177446.JavaMail.jakarta@nitrogen.mathforum.org>, > Say you have a circle who's diameter whose > is the same as the width of one > quadrant. How would you find the area of this circle that falls into each > quadrant if the circle's center point was not at the origin? Try again. Be coherent this time. === Subject: Re: How low can you jackass Beckwith get with your fraud. You supposedly have a physics PhD. You will never get over it! posting-account=OKTeIQkAAAAZk6JK1hK7-grwpoUDNy98 CLR 1.1.4322; .NET CLR 2.0.50727; InfoPath.1),gzip(gfe),gzip(gfe) Hey Becky, have you reviewed more of James Harris' masterpieces of > mathematics in your closed list (chortle) and found no errors? Becks, let me explain - you would lie for James Harris, so you lie > with James Harris, and now you have his fleas. Fraudy fraud Beckwith - ya can't live it down. Hahahahahahahahahahahahahahahahahaha! > closed list closed list closed list > personal space personal space personal space > Hey Quinn, I'm laughing at the superior intellect. > Hahahahahahahahahahahahahahahahahaha! Just think of all the spittle he has to clean off his computer screen after that foaming-at-the-mouth rant. Don't they make you take English as part of a PhD curricula, even for physics? After all, they expect you to be able to write papers, don't they? === Subject: Re: How low can you jackass Beckwith get with your fraud. You supposedly have a physics PhD. You will never get over it! posting-account=_VmC_goAAACU-xAuU22JkRqrgFylfYj- 1.1.4322),gzip(gfe),gzip(gfe) you to be able to write papers, don't they? Not if you put your name on someone else's paper, as Beckwith did for James Harris. ;-) === Subject: --- --- --- Peculiar Quadratic Equation Cc: deepkdeb@yahoo.com posting-account=iJfBPwgAAACaBAH7DreA6VfAOYvL5VDz 1.0.3705; .NET CLR 1.1.4322; Media Center PC 4.0; IEMB3; IEMB3),gzip(gfe),gzip(gfe) This is a modified version of the previous posting: Kindly consider the following equation (1) under the given conditions. mgx^2 - (gh^(x-1))x - n^x = 0 (1) Conditions: m, g, x, h, n are all integers such that m > 0; g, h, n > 1; prime x > 5 Assertion: (1) cannot be satisfied under the given conditions for some unique value of x x = [gh^(x-1) +/- sqrt((g^2h^2(x-1) + 4mgn^k)]/2mg (2) Argument: Since only unique value of x is sought so x will have only one value. So if such a solution exists the two values of x will be the same. In that case one gets (3) g^2h^2(x-1) + 4mgn^k) = 0 (3) But under the given conditions (3) is impossible. This justifies the assertion. Any helpful comments will be gratefully appreciated. === Subject: Re: --- --- --- Peculiar Quadratic Equation posting-account=a6woBRAAAADpNFZJBA7ZBx35zXaKmaP4 Gecko/2008070208 Firefox/3.0.1,gzip(gfe),gzip(gfe) > This is a modified version of the previous posting: Kindly consider the following equation (1) under the given conditions. æ æ æmgx^2 - (gh^(x-1))x - n^x æ= 0 æ æ æ æ æ æ(1) Conditions: m, g, x, h, n are all integers such that m > 0; g, h, n 1; prime x > 5 Assertion: (1) cannot be satisfied under the given conditions for some > unique value of x [snip approach & argument] I see ambiguity in your assertion regarding some unique value of x. Do you mean that given any integers m > 0 and g,h,n > 1, there either exists no prime x > 5 satisfying (1) or else there exists more than one prime x > 5 satisfying (1)? What effect on your assertion would it have if for choice of valid parameters m,g,h,n, there existed more than one real solution x, at least one of which happened to be an integer prime greater than 5? === Subject: Re: --- --- --- Peculiar Quadratic Equation Cc: deepkdeb@yahoo.com posting-account=iJfBPwgAAACaBAH7DreA6VfAOYvL5VDz 1.1.4322),gzip(gfe),gzip(gfe) This is a modified version of the previous posting: Kindly consider the following equation (1) under the given conditions. æ æ æmgx^2 - (gh^(x-1))x - n^x æ= 0 æ æ æ æ æ æ(1) Conditions: m, g, x, h, n are all integers such that m > 0; g, h, n 1; prime x > 5 Assertion: (1) cannot be satisfied under the given conditions for some > unique value of x [snip approach & argument] I see ambiguity in your assertion regarding some > unique value of x. æDo you mean that given any > integers m > 0 and g,h,n > 1, there either exists > no prime x > 5 satisfying (1) or else there exists > more than one prime x > 5 satisfying (1)? What effect on your assertion would it have if for > choice of valid parameters m,g,h,n, there existed > more than one real solution x, at least one of > which happened to be an integer prime greater than > 5? > ****** For better clarification the assertion may be restated as: There does not exist any prime x > 5 for which (1) as given above has any solution. This will help investigate a series of diophantine equations. ****** === Subject: Re: --- --- --- Peculiar Quadratic Equation > This is a modified version of the previous posting: Kindly consider the following equation (1) under the given conditions. æ æ æmgx^2 - (gh^(x-1))x - n^x æ= 0 æ æ æ æ æ æ(1) Conditions: m, g, x, h, n are all integers such that m > 0; g, h, n 1; prime x > 5 Assertion: (1) cannot be satisfied under the given conditions for some > unique value of x [snip approach & argument] I see ambiguity in your assertion regarding some > unique value of x. æDo you mean that given any > integers m > 0 and g,h,n > 1, there either exists > no prime x > 5 satisfying (1) or else there exists > more than one prime x > 5 satisfying (1)? What effect on your assertion would it have if for > choice of valid parameters m,g,h,n, there existed > more than one real solution x, at least one of > which happened to be an integer prime greater than > 5? > ****** > For better clarification the assertion may be restated as: There does > not exist any prime x > 5 for which (1) as given above has > any solution. This will help investigate a series of diophantine > equations. Except that you are not working with a diophantine equation. Theorems and techniques that apply to diophantine equations can not be applied (e.g. Baker's linear forms in logarithms) because you are restricting not only to integer solutions, but to PRIME solutions. I see no reason why anyone should care about this problem. === Subject: Re: --- --- --- Peculiar Quadratic Equation > This is a modified version of the previous posting: Kindly consider the following equation (1) under the given conditions. æ æ æmgx^2 - (gh^(x-1))x - n^x æ= 0 æ æ æ æ æ æ(1) Conditions: m, g, x, h, n are all integers such that m > 0; g, h, n 1; prime x > 5 Assertion: (1) cannot be satisfied under the given conditions for some > unique value of x [snip approach & argument] I see ambiguity in your assertion regarding some > unique value of x. æDo you mean that given any > integers m > 0 and g,h,n > 1, there either exists > no prime x > 5 satisfying (1) or else there exists > more than one prime x > 5 satisfying (1)? What effect on your assertion would it have if for > choice of valid parameters m,g,h,n, there existed > more than one real solution x, at least one of > which happened to be an integer prime greater than > 5? > ****** > For better clarification the assertion may be restated as: There does > not exist any prime x > 5 for which (1) as given above has > any solution. This will help investigate a series of diophantine > equations. > > Except that you are not working with a diophantine equation. > Theorems and techniques that apply to diophantine equations > can not be applied (e.g. Baker's linear forms in logarithms) > because you are restricting not only to integer solutions, but to > PRIME solutions. > > I see no reason why anyone should care about this problem. Well, there is a literature on problems where one or more variables are restricted to being prime. Anyway, it seems to me that n = g = h = x, m = x^(x-2) + x^(x-3) is a solution. Of course, OP will just come up with some other condition, rather than actually thinking about the problem & working it out on his/her own. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: --- --- --- Peculiar Quadratic Equation Cc: deepkdeb@yahoo.com posting-account=iJfBPwgAAACaBAH7DreA6VfAOYvL5VDz 1.1.4322),gzip(gfe),gzip(gfe) On Jul 30, 8:13æpm, Gerry Myerson This is a modified version of the previous posting: Kindly consider the following equation (1) under the given conditions. æ æ æmgx^2 - (gh^(x-1))x - n^x æ= 0 æ æ æ æ æ æ(1) Conditions: m, g, x, h, n are all integers such that m > 0; g, h, n 1; prime x > 5 Assertion: (1) cannot be satisfied under the given conditions for some > unique value of x [snip approach & argument] I see ambiguity in your assertion regarding some > unique value of x. æDo you mean that given any > integers m > 0 and g,h,n > 1, there either exists > no prime x > 5 satisfying (1) or else there exists > more than one prime x > 5 satisfying (1)? What effect on your assertion would it have if for > choice of valid parameters m,g,h,n, there existed > more than one real solution x, at least one of > which happened to be an integer prime greater than > 5? > ****** > For better clarification the assertion may be restated as: There does > not exist any prime x > 5 for which (1) as given above has > any solution. This will help investigate a series of diophantine > equations. Except that you are not working with a diophantine equation. > Theorems and techniques that apply to diophantine equations > can not be applied (e.g. Baker's linear forms in logarithms) > because you are restricting not only to integer solutions, but to > PRIME solutions. I see no reason why anyone should care about this problem. Well, there is a literature on problems where one or more variables > are restricted to being prime. Anyway, it seems to me that n = g = h = x, > m = x^(x-2) + x^(x-3) is a solution. (ABC) Of course, OP will just come up with some other condition, > rather than actually thinking about the problem & working > it out on his/her own. -- > Gerry Myerson (ge...@maths.mq.edi.ai) (i -> u for email)- Hide quoted text - - Show quoted text - ***** equation (1) is for the unknown variable x which is restricted to be a prime > 5. It is not immediately clear to me how does your equation as given above ( I refer it by ABC) justifies that prime x > 5. A little additional explanation will be very helpful. ***** === Subject: Re: --- --- --- Peculiar Quadratic Equation > This is a modified version of the previous posting: Kindly consider the following equation (1) under the given conditions. mgx^2 - (gh^(x-1))x - n^x = 0 (1) Conditions: m, g, x, h, n are all integers such that m > 0; g, h, n 1; prime x > 5 mgx^2 = gxh^(x-1) + n^x; m,g,h,n,x in N; g,h,n > 1; x prime > 5 > Assertion: (1) cannot be satisfied under the given conditions for some > unique value of x What does that let's do the mind twist statement mean? Given m,g,b,n in N, g,h,n > 1, there's a unique prime x > 5 for which (1) does not hold? Is that what you mean? Let m = g = h = n = 2 4x^2 = 4x.2^(x-1) + 2^x x^2 = x.2^(x-1) + 2^(x-2) Clearly false for all odd integers x > 2. x = [gh^(x-1) +/- sqrt((g^2 h^2(x-1) + 4mgn^k)]/2mg (2) Argument: Since only unique value of x is sought so x will have only > one value. So if such a solution exists the two values of x will be > the same. In that case one gets (3) g^2 h^2(x-1) + 4mgn^k) = 0 (3) But under the given conditions (3) is impossible. This justifies the assertion. Any helpful comments will be gratefully appreciated. > === Subject: Re: --- --- --- Peculiar Quadratic Equation posting-account=O9zR9AkAAACmp918j6u5m5plppeILcze Filter 1.2.0.72; .NET CLR 1.0.3705; .NET CLR 1.1.4322; Media Center PC 4.0; .NET CLR 2.0.50727; .NET CLR 3.0.04506.648; .NET CLR 3.5.21022; WWTClient2),gzip(gfe),gzip(gfe) > This is a modified version of the previous posting: Kindly consider the following equation (1) under the given conditions. æ æ æmgx^2 - (gh^(x-1))x - n^x æ= 0 æ æ æ æ æ æ(1) Conditions: m, g, x, h, n are all integers such that m > 0; g, h, n 1; prime x > 5 Assertion: (1) cannot be satisfied under the given conditions for some > unique value of x > x = [gh^(x-1) +/- sqrt((g^2h^2(x-1) + 4mgn^k)]/2mg æ æ (2) Argument: Since only unique value of x is sought so x will have only > one ævalue. So if such a solution exists the two values of x will be > the same. In that case one gets (3) æ æ æg^2h^2(x-1) + 4mgn^k) = 0 æ æ æ æ æ æ æ (3) But under the given conditions (3) is impossible. This justifies the assertion. Any helpful comments will be gratefully appreciated. It is not a quadratic equation, so you can't justify using the quadratic formula to try to solve it. A quadratic equation is an equation of the form ax^2 + bx + c = 0, where a, b, and c are independent of x. In your case, b = -gh^(x-1), which is not independent of x. So your explanation is falacious. Dave === Subject: Re: --- --- --- Peculiar Quadratic Equation > This is a modified version of the previous posting: Kindly consider the following equation (1) under the given conditions. [All Google garbage has been removed.] mgx^2 - (gh^(x-1))x - n^x = 0 > Conditions: m, g, x, h, n are all integers such that m > 0; g, h, n 1; prime x > 5 Assertion: (1) cannot be satisfied under the given conditions for some > unique value of x > x = [gh^(x-1) +/- sqrt((g^2h^2(x-1) + 4mgn^k)]/2mg Argument: Since only unique value of x is sought so x will have only > one ævalue. So if such a solution exists the two values of x will be > the same. In that case one gets (3) g^2h^2(x-1) + 4mgn^k) = 0 No, that reasoning is incorrect. > But under the given conditions (3) is impossible. > It is not a quadratic equation, so you can't justify using the > quadratic formula to try to solve it. A quadratic equation is an > equation of the form ax^2 + bx + c = 0, where a, b, and c are > independent of x. In your case, b = -gh^(x-1), which is not > independent of x. So your explanation is falacious. > . . a(x).x^2 + b(x).x + c(x) = 0, you can complete the square and get for all x, a(x) /= 0 implies . . x = (-b(x) +- sqr(b(x)^2 - 4.a(x).c(x))/2a(x) which of course isn't a solution for x, but it is another equation involving x. === Subject: Re: --- --- --- Peculiar Quadratic Equation posting-account=O9zR9AkAAACmp918j6u5m5plppeILcze Filter 1.2.0.72; .NET CLR 1.0.3705; .NET CLR 1.1.4322; Media Center PC 4.0; .NET CLR 2.0.50727; .NET CLR 3.0.04506.648; .NET CLR 3.5.21022; WWTClient2),gzip(gfe),gzip(gfe) > It is not a quadratic equation, so you can't justify using the > quadratic formula to try to solve it. A quadratic equation is an > equation of the form ax^2 + bx + c = 0, where a, b, and c are > independent of x. In your case, b = -gh^(x-1), which is not > independent of x. So your explanation is falacious. . . a(x).x^2 + b(x).x + c(x) = 0, you can complete the square and get for all x, > a(x) /= 0 implies > . . x = (-b(x) +- sqr(b(x)^2 - 4.a(x).c(x))/2a(x) which of course isn't a solution for x, > but it is another equation involving x. Of course, that's what I said, as you can see from above... you can't justify using the quadratic formula to try to solve it. Dave === Subject: Re: --- --- --- Peculiar Quadratic Equation Cc: deepkdeb@yahoo.com posting-account=iJfBPwgAAACaBAH7DreA6VfAOYvL5VDz 1.0.3705; .NET CLR 1.1.4322; Media Center PC 4.0; IEMB3; IEMB3),gzip(gfe),gzip(gfe) It is not a quadratic equation, so you can't justify using the > quadratic formula to try to solve it. A quadratic equation is an > equation of the form ax^2 + bx + c = 0, where a, b, and c are > independent of x. In your case, b = -gh^(x-1), which is not > independent of x. So your explanation is falacious. . . a(x).x^2 + b(x).x + c(x) = 0, you can complete the square and get for all x, > a(x) /= 0 implies > . . x = (-b(x) +- sqr(b(x)^2 - 4.a(x).c(x))/2a(x) which of course isn't a solution for x, > but it is another equation involving x. Of course, that's what I said, as you can see from above... you can't > justify using the quadratic formula to try to solve it. Dave- Hide quoted text - - Show quoted text - **** So what is the conclusion? Does (1) have any real solution ? If not why not? I believe (1) has no unique solution. Kindly throw some light. **** === Subject: Re: --- --- --- Peculiar Quadratic Equation It is not a quadratic equation, so you can't justify using the > quadratic formula to try to solve it. A quadratic equation is an > equation of the form ax^2 + bx + c = 0, where a, b, and c are > independent of x. In your case, b = -gh^(x-1), which is not > independent of x. So your explanation is falacious. . . a(x).x^2 + b(x).x + c(x) = 0, you can complete the square and get for all x, > a(x) /= 0 implies > . . x = (-b(x) +- sqr(b(x)^2 - 4.a(x).c(x))/2a(x) which of course isn't a solution for x, > but it is another equation involving x. Of course, that's what I said, as you can see from above... you can't > justify using the quadratic formula to try to solve it. Dave- Hide quoted text - - Show quoted text - > > **** > So what is the conclusion? Does (1) have any real solution ? If not > why not? I believe (1) has no unique solution. Kindly throw some > light. > **** I don't see any (1). -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: --- --- --- Peculiar Quadratic Equation posting-account=iJfBPwgAAACaBAH7DreA6VfAOYvL5VDz 1.0.3705; .NET CLR 1.1.4322; Media Center PC 4.0; IEMB3; IEMB3),gzip(gfe),gzip(gfe) On Jul 29, 9:28æpm, Gerry Myerson quadratic formula to try to solve it. A quadratic equation is an > equation of the form ax^2 + bx + c = 0, where a, b, and c are > independent of x. In your case, b = -gh^(x-1), which is not > independent of x. So your explanation is falacious. . . a(x).x^2 + b(x).x + c(x) = 0, you can complete the square and get for all x, > a(x) /= 0 implies > . . x = (-b(x) +- sqr(b(x)^2 - 4.a(x).c(x))/2a(x) which of course isn't a solution for x, > but it is another equation involving x. Of course, that's what I said, as you can see from above... you can't > justify using the quadratic formula to try to solve it. Dave- Hide quoted text - - Show quoted text - **** > So what is the conclusion? Does (1) have any real solution ? If not > why not? I believe (1) has no unique solution. Kindly throw some > light. > **** I don't see any (1). -- > Gerry Myerson (ge...@maths.mq.edi.ai) (i -> u for email)- Hide quoted text - - Show quoted text - **** Kindly go to the top of the posting (initial). Just go up you will find (1), (2) maybe even (3) **** === Subject: Re: --- --- --- Peculiar Quadratic Equation > On Jul 29, 9:28æpm, Gerry Myerson quadratic formula to try to solve it. A quadratic equation is an > equation of the form ax^2 + bx + c = 0, where a, b, and c are > independent of x. In your case, b = -gh^(x-1), which is not > independent of x. So your explanation is falacious. . . a(x).x^2 + b(x).x + c(x) = 0, you can complete the square and get for all x, > a(x) /= 0 implies > . . x = (-b(x) +- sqr(b(x)^2 - 4.a(x).c(x))/2a(x) which of course isn't a solution for x, > but it is another equation involving x. Of course, that's what I said, as you can see from above... you can't > justify using the quadratic formula to try to solve it. Dave- Hide quoted text - - Show quoted text - **** > So what is the conclusion? Does (1) have any real solution ? If not > why not? I believe (1) has no unique solution. Kindly throw some > light. > **** I don't see any (1). -- > Gerry Myerson (ge...@maths.mq.edi.ai) (i -> u for email)- Hide quoted text - - Show quoted text - > > **** > Kindly go to the top of the posting (initial). Just go up you will > find (1), (2) maybe even (3) > **** That may work on your software. It doesn't work on mine, and it's not a good idea to assume that eveeryone reads news the same way you do. And just how hard is it to type in some equation once more, if you're serious about wanting to know something about it? -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: [] --- --- --- Peculiar Quadratic Equation u for email)- Hide > quoted text - - Show quoted text - **** > Kindly go to the top of the posting (initial). Just go up you will > find (1), (2) maybe even (3) > **** That may work on your software. It doesn't work on mine, > and it's not a good idea to assume that eveeryone reads news > the same way you do. And just how hard is it to type in some > equation once more, if you're serious about wanting to know > something about it? > The problem lies with using Google groups that do weird stuff with quotes, making them hard to read. Hence the reply get over clipped removing all that Google garbage. The portion of the garbage that you see is Hide quoted text - Show quoted text - There is also other garbage that shows up when the reader isn't using Google stuffits. I suggest OP get a read news source and reader. also a large source of off topic and/or idiotic posts. For that reason, a growing number of people and even some news servers, are now blocking all posts from Google Groups. http://www.improve-usenet.org/index.html First, use a news server other then Google. Second, download a news reader that allows filtering. all Google posts (most efficiently done filtering on Message-ID). Besides using the news server usually provided with your ISP internet account, you have a choice of a number of news servers and news readeres, some of which are listed below. In addition, for those needing a web base news source, a number math http://mathforum.org/kb -- NNTP news servers http://www.teranews.com (one time $3.95 set up cost, worth it) http://news.aioe.org text only, no cost http://news.motzarella.org no cost http://news.datemas.de no cost http://www.albasani.net no cost http://www.glorb.com/usenet.php $10/year http://individual.net 10 Euro/year http://www.usenet4all.se no cost http://www.databasix.com/groups $50/year News readers Dialog http://www.40tude.com/dialog/download.htm Xnews http://xnews.newsguy.com for file naming cf. http://blinkynet.net/comp/xnewsrels.html SeaMonkey (won't filter on Message-ID) http://www.seamonkey-project.org Thunderbird (won't filter on Message-ID) http://www.mozilla.com/en-US/thunderbird Forte Agent ($29, won't filter on Message-ID) Unix Pan http://pan.rebelbase.com XPN http://sourceforge.net/projects/xpn KDE http://kontact.kde.org/knode Mac: MT-Newswatcher, hogwash, MacSOUP, thoth Cross Platform http://www.claws-mail.org Proxy news servers NewsProxy http://bearware.info/NewsProxy/newsproxy.html Hamster Playground http://www.elbiah.de/hamster/pg Unix Leafnode http://leafnode.sourceforge.net/ Noffle http://noffle.sourceforge.net News sources, web and otherwise and references to news sites http://www.readfreenews.net http://en.wikipedia.org/wiki/List_of_news_clients http://dmoz.org/Computers/Usenet/Public_News_Servers http://www.alt.free.newsservers.net http://freenews.maxbaud.net http://news.motzarella.org http://www.newsreaders.com http://news.datemas.de commercial news servers Giganews http://www.giganews.com Supernews http://www.supernews.com (bought Giganews, yuck) Newsfeeds http://www.newsfeeds.com Usenetserve http://www.usenetserver.com Easynews http://www.easynews.com Altopia https://www.altopia.com Newsguy https://www.newsguy.com -- news.admin.net-abuse.usenet news.software.readers If your news server isn't carrying it, ask your system administrator to add it to the active file. Google should be given a Usenet Death Penalty (UDP) http://en.wikipedia.org/wiki/Usenet_Death_Penalty http://www.catb.org/jargon/html/U/Usenet-Death-Penalty.html http://www.catb.org/jargon/html/I/Internet-Death-Penalty.html Use another web search such as http://yahoo.com http://www.webcrawler.com http://www.ask.com and some other email service than Google's gmail. Then eventually Google may come to understand that it's doing something wrong. -- Essential features for a quality news reader. Correct quoting. Ascii-only posting. Correct referencing of prior messages in the thread. Filter capability to allow pruning of incoming messages. Spell checking option for outgoing messages. Offline news reading No ads, cookies, etc. Elimination of Google's hide quoted, show quoted nonsense. ---- === Subject: Re: #619 Applications as first chapter of AP-adics; new textbook: Mathematical Physics (AP-adic Primer) posting-account=yxbZkgkAAABQBvyYeebYQ-PAvi0uT3tG Gecko/20080311 Firefox/2.0.0.13,gzip(gfe),gzip(gfe) > David, do you have access to a Wimshurst generator, or can you buy > your own Wimshurst? > Have you ever set up a physics experiment? > Are you willing to give it a try? So that in August, we can report our > results to the Internet, whether > they agree or disagree. I can instruct you on the setup. > I have neither the time nor the interest. So you do not want to learn, and science does not mean to know for you. > Sorry I ever asked you, because you only show your deep ignorance of > physics. Well, I'd like to know if you made any measurements of the zero resistance in your superconductivity experiment. === Subject: #627 Sphere + Hyperbolic figure = Cube; new textbook: Mathematical Physics (AP-adic Primer) I seem to be bouncing around on this issue. Problem is that I did not actually take a physical sphere and see what the objects are. I took a round apple and cut it into half and set Northpole point contiguous to Southpole point and then put the two into a cube box. The empty space which the apple is not, is a hyperbolic space. I do not know if someone has given this hyperbolic space a name. It is not a pseudosphere or a cut up pseudosphere. So to make the physical toy would be three pieces. One half of the globe, and the other half of the globe and then this hyperbolic space wherein one can plug to two half globes. So we have this as the formula Elliptic geom union Hyperbolic geometry = Euclidean geom and we have this formula Sphere + Hyperbolic figure = Cube Complaint: some may complain that the edges of the hyperbolic figure are square edges where the casting of the toy took place in a cube box by filling in the space which was not the hemispheres. I would answer that complaint by saying the edges of the cube are not part of the hyperbolic figure. The *point of reasoning* is that if a elliptic figure joined with a hyperbolic figure consumes all of the space, then the union of the two is a Euclidean geometry. Like a picture frame, which is not part of the picture itself, and so the cube edges are not part of the hyperbolic figure. Does anyone know if this hyperbolic figure has a name? Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === posting-account=nPH_PQkAAACneDKT6RXopPWArC2We4Rq AppleWebKit/525.13 (KHTML, like Gecko) Version/3.1 Safari/525.13,gzip(gfe),gzip(gfe) On Jul 28, 4:06æpm, stevendaryl3...@yahoo.com (Daryl McCullough) > PD says... > Common sense is a liar and a cheat, and it clouds students' thinking > You have no right to blame your confusion on common sense. Rather, it > is your willful abandonment of common sense that is the cause of your > confusion. The irony here is that Shubee's paper deriving the Lorentz > transformations is much *farther* from common sense than > the *usual* derivation. That doesn't make it wrong, but > it makes it strange that Shubee would complain about > abandonment of common sense. EVERY author feels that their treatment is the clearest possible one and of course the one that makes the most sense to the author. It comes as a great shock to many an author that the lucidity of their presentation seems to be lost on their readers. Such books don't sell well, of course. Some of these same authors do not take this to be constructive criticism of their approach and curse the pool of their readers for being obstinately thick-headed, and sniff that the high quality of their presentation would be obviously wasted on such unwashed and ungrateful plebes. PD === posting-account=lBRURwoAAAB_-Q_b04pGziaymfr5yRFx Gecko/20080702 Firefox/2.0.0.16,gzip(gfe),gzip(gfe) > On Jul 28, 4:06 pm, stevendaryl3...@yahoo.com (Daryl McCullough) PD says... > Common sense is a liar and a cheat, and it clouds students' thinking > You have no right to blame your confusion on common sense. Rather, it > is your willful abandonment of common sense that is the cause of your > confusion. The irony here is that Shubee's paper deriving the Lorentz > transformations is much *farther* from common sense than > the *usual* derivation. That doesn't make it wrong, but > it makes it strange that Shubee would complain about > abandonment of common sense. EVERY author feels that their treatment is the clearest possible one > and of course the one that makes the most sense to the author. It comes as a great shock to many an author that the lucidity of their > presentation seems to be lost on their readers. Such books don't sell > well, of course. Some of these same authors do not take this to be constructive > criticism of their approach and curse the pool of their readers for > being obstinately thick-headed, and sniff that the high quality of > their presentation would be obviously wasted on such unwashed and > ungrateful plebes. PD I freely admit that my paper is greatly lacking in simplicity and that I need to improve the clarity of it but judging from your rage and contempt it seems that you have no problem understanding what I say in this newsgroup. Shubee === posting-account=nPH_PQkAAACneDKT6RXopPWArC2We4Rq AppleWebKit/525.13 (KHTML, like Gecko) Version/3.1 Safari/525.13,gzip(gfe),gzip(gfe) On Jul 28, 4:06 pm, stevendaryl3...@yahoo.com (Daryl McCullough) PD says... > Common sense is a liar and a cheat, and it clouds students' thinking > You have no right to blame your confusion on common sense. Rather, it > is your willful abandonment of common sense that is the cause of your > confusion. The irony here is that Shubee's paper deriving the Lorentz > transformations is much *farther* from common sense than > the *usual* derivation. That doesn't make it wrong, but > it makes it strange that Shubee would complain about > abandonment of common sense. EVERY author feels that their treatment is the clearest possible one > and of course the one that makes the most sense to the author. It comes as a great shock to many an author that the lucidity of their > presentation seems to be lost on their readers. Such books don't sell > well, of course. Some of these same authors do not take this to be constructive > criticism of their approach and curse the pool of their readers for > being obstinately thick-headed, and sniff that the high quality of > their presentation would be obviously wasted on such unwashed and > ungrateful plebes. PD I freely admit that my paper is greatly lacking in simplicity and that > I need to improve the clarity of it but judging from your rage and > contempt I don't recall ever feeling or expressing any rage or contempt whatsoever. The remark I just made is true for a lot of authors. You should not feel disparaged by that remark. It is a fact of life for authors and the publishers who back them, and this is why authors are paid by royalty rather than up front. The royalty arrangement is entirely a recognition that the proof of the pudding is in the eating. You are not immune to that adage any more than any other author. The test of your character will be what happens in the event things don't go as well as you had hoped. > it seems that you have no problem understanding what I say in > this newsgroup. Shubee === posting-account=lBRURwoAAAB_-Q_b04pGziaymfr5yRFx Gecko/20080702 Firefox/2.0.0.16,gzip(gfe),gzip(gfe) > I don't recall ever feeling or expressing any rage or contempt > whatsoever. Never? I recall one post of yours on another topic where you said that the longer you thought about my comments the angrier you got. > The remark I just made is true for a lot of authors. You should not > feel disparaged by that remark. It is a fact of life for authors and > the publishers who back them, and this is why authors are paid by > royalty rather than up front. Why must I insist on payment? Why can't I just finish writing up my novel approach to high school level special relativity and then find a physicist that believes in modus ponens, respects David Hilbert's philosophy of physics and can endorse me for a physics education submission to arXiv? Granted, there is a slim chance of satisfying all those requirements but it is possible. Shubee http://www.everythingimportant.org/relativity/special.pdf === posting-account=nPH_PQkAAACneDKT6RXopPWArC2We4Rq AppleWebKit/525.13 (KHTML, like Gecko) Version/3.1 Safari/525.13,gzip(gfe),gzip(gfe) I don't recall ever feeling or expressing any rage or contempt > whatsoever. Never? I recall one post of yours on another topic where you said that > the longer you thought about my comments the angrier you got. I just did a search on archived posts in s.p.r by me with the words angry or angrier and couldn't find anything about you at all. I certainly haven't expressed any anger or contempt for you in this conversation. The remark I just made is true for a lot of authors. You should not > feel disparaged by that remark. It is a fact of life for authors and > the publishers who back them, and this is why authors are paid by > royalty rather than up front. Why must I insist on payment? Why can't I just finish writing up my > novel approach to high school level special relativity and then find a > physicist that believes in modus ponens, respects David Hilbert's > philosophy of physics and can endorse me for a physics education > submission to arXiv? Granted, there is a slim chance of satisfying all > those requirements but it is possible. > You certainly can. But it would be useful for you to have some metric by which to judge the value of your contribution. Succeeding in getting it published is hardly a sufficient standard (though I know that for many, this is a significant accomplishment in itself) for that. Physicists judge the impact of each other's contributions not by how many papers a researcher publishes, but by how often those papers are referred to (that is, read and *used*) in subsequent papers. The impact of educational material is not judged by whether it is published, but by how often it is adopted and *used* in teaching. Even if you don't care about the compensation (donate it all or settle for negotiated royalty pittance), what the publisher is telling you via the royalty is how *valuable* your material is for teaching. Now, here is another opportunity for you to be honest with yourself, Shubee. If you say you are earnestly interested in the *educational* value of what you are doing, then you should hold yourself accountable to some metric by which to judge that value. If publishing it -- and that's all -- is sufficient for you as a metric for your success, then your *true* agenda (what's *really* important to you) is seeing your How do you hold yourself accountable, Shubee? PD === posting-account=lBRURwoAAAB_-Q_b04pGziaymfr5yRFx Gecko/20080702 Firefox/2.0.0.16,gzip(gfe),gzip(gfe) > I don't recall ever feeling or expressing any rage or contempt > whatsoever. Never? I recall one post of yours on another topic where you said that > the longer you thought about my comments the angrier you got. I just did a search on archived posts in s.p.r by me with the words > angry or angrier and couldn't find anything about you at all. I > certainly haven't expressed any anger or contempt for you in this > conversation. > The remark I just made is true for a lot of authors. You should not > feel disparaged by that remark. It is a fact of life for authors and > the publishers who back them, and this is why authors are paid by > royalty rather than up front. Why must I insist on payment? Why can't I just finish writing up my > novel approach to high school level special relativity and then find a > physicist that believes in modus ponens, respects David Hilbert's > philosophy of physics and can endorse me for a physics education > submission to arXiv? Granted, there is a slim chance of satisfying all > those requirements but it is possible. You certainly can. But it would be useful for you to have some metric > by which to judge the value of your contribution. Actually, I do have a metric. Can I get high school students to understand the subtle details of special relativity that professional physicists don't understand? Can I motive high school students to fear the insanity that says, Common sense is a liar and a cheat, and it clouds students' thinking? Finally, can I expose the spiritualistic philosophy of physicists with sufficient clarity to get the physics community to petition the pope for his blessing on my execution? > Now, here is another opportunity for you to be honest with yourself, > Shubee. If you say you are earnestly interested in the *educational* > value of what you are doing, then you should hold yourself accountable > to some metric by which to judge that value. Yes. I shall hold myself accountable to my metric. > How do you hold yourself accountable, Shubee? I shall keep pursuing my goal until it is accomplished. Shubee === posting-account=nPH_PQkAAACneDKT6RXopPWArC2We4Rq AppleWebKit/525.13 (KHTML, like Gecko) Version/3.1 Safari/525.13,gzip(gfe),gzip(gfe) I don't recall ever feeling or expressing any rage or contempt > whatsoever. Never? I recall one post of yours on another topic where you said that > the longer you thought about my comments the angrier you got. I just did a search on archived posts in s.p.r by me with the words > angry or angrier and couldn't find anything about you at all. I > certainly haven't expressed any anger or contempt for you in this > conversation. > Ah, so you can't find it either, though you're sure it's there. Never mind. I showed you a place where I did express some contempt. The remark I just made is true for a lot of authors. You should not > feel disparaged by that remark. It is a fact of life for authors and > the publishers who back them, and this is why authors are paid by > royalty rather than up front. Why must I insist on payment? Why can't I just finish writing up my > novel approach to high school level special relativity and then find a > physicist that believes in modus ponens, respects David Hilbert's > philosophy of physics and can endorse me for a physics education > submission to arXiv? Granted, there is a slim chance of satisfying all > those requirements but it is possible. You certainly can. But it would be useful for you to have some metric > by which to judge the value of your contribution. Actually, I do have a metric. Can I get high school students to > understand the subtle details of special relativity that professional > physicists don't understand? Can I motive high school students to fear > the insanity that says, Common sense is a liar and a cheat, and it > clouds students' thinking? Finally, can I expose the spiritualistic > philosophy of physicists with sufficient clarity to get the physics > community to petition the pope for his blessing on my execution? And how will you get them to do that if your material is not adopted and used in the classrooms where this is being taught? And how will you know if, in cases where it is being used, that students do see what you intend them to see? What are your CONCRETE metrics for ascertaining success? How do you hold yourself concretely accountable to your objectives? Now, here is another opportunity for you to be honest with yourself, > Shubee. If you say you are earnestly interested in the *educational* > value of what you are doing, then you should hold yourself accountable > to some metric by which to judge that value. Yes. I shall hold myself accountable to my metric. How do you hold yourself accountable, Shubee? I shall keep pursuing my goal until it is accomplished. Shubee === posting-account=lBRURwoAAAB_-Q_b04pGziaymfr5yRFx Gecko/20080702 Firefox/2.0.0.16,gzip(gfe),gzip(gfe) Actually, I do have a metric. Can I get high school students to > understand the subtle details of special relativity that professional > physicists don't understand? Can I motive high school students to fear > the insanity that says, Common sense is a liar and a cheat, and it > clouds students' thinking? Finally, can I expose the spiritualistic > philosophy of physicists with sufficient clarity to get the physics > community to petition the pope for his blessing on my execution? And how will you get them to do that if your material is not adopted > and used in the classrooms where this is being taught? And how will > you know if, in cases where it is being used, that students do see > what you intend them to see? What are your CONCRETE metrics for > ascertaining success? How do you hold yourself concretely accountable to your objectives? The Battle of Armageddon is a manifestly visible battle between light and darkness. It will appear eventually, just before the end of the world. Thousands will be fighting valiantly with great Light against darkness in the war. The book explaining the theological basis of this war is now being written. The whole earth will be illuminated by the power of the message (Rev 18:1). Hopefully you will know of it soon enough. Shubee === posting-account=nPH_PQkAAACneDKT6RXopPWArC2We4Rq AppleWebKit/525.13 (KHTML, like Gecko) Version/3.1 Safari/525.13,gzip(gfe),gzip(gfe) > Actually, I do have a metric. Can I get high school students to > understand the subtle details of special relativity that professional > physicists don't understand? Can I motive high school students to fear > the insanity that says, Common sense is a liar and a cheat, and it > clouds students' thinking? Finally, can I expose the spiritualistic > philosophy of physicists with sufficient clarity to get the physics > community to petition the pope for his blessing on my execution? And how will you get them to do that if your material is not adopted > and used in the classrooms where this is being taught? And how will > you know if, in cases where it is being used, that students do see > what you intend them to see? What are your CONCRETE metrics for > ascertaining success? How do you hold yourself concretely accountable to your objectives? The Battle of Armageddon is a manifestly visible battle between light > and darkness. It will appear eventually, just before the end of the > world. Thousands will be fighting valiantly with great Light against > darkness in the war. The book explaining the theological basis of this > war is now being written. The whole earth will be illuminated by the > power of the message (Rev 18:1). Hopefully you will know of it soon > enough. Shubee I suggest you read the book Don Quixote. PD === Actually, I do have a metric. Can I get high school students to > understand the subtle details of special relativity that professional > physicists don't understand? Can I motive high school students to fear > the insanity that says, Common sense is a liar and a cheat, and it > clouds students' thinking? Finally, can I expose the spiritualistic > philosophy of physicists with sufficient clarity to get the physics > community to petition the pope for his blessing on my execution? And how will you get them to do that if your material is not adopted > and used in the classrooms where this is being taught? And how will > you know if, in cases where it is being used, that students do see > what you intend them to see? What are your CONCRETE metrics for > ascertaining success? How do you hold yourself concretely accountable to your objectives? The Battle of Armageddon is a manifestly visible battle between light > and darkness. It will appear eventually, just before the end of the > world. Thousands will be fighting valiantly with great Light against > darkness in the war. The book explaining the theological basis of this > war is now being written. The whole earth will be illuminated by the > power of the message (Rev 18:1). Hopefully you will know of it soon > enough. Shubee I suggest you read the book Don Quixote. PD I suggest that you read the Bible. Shubee === posting-account=nPH_PQkAAACneDKT6RXopPWArC2We4Rq AppleWebKit/525.13 (KHTML, like Gecko) Version/3.1 Safari/525.13,gzip(gfe),gzip(gfe) > Actually, I do have a metric. Can I get high school students to > understand the subtle details of special relativity that professional > physicists don't understand? Can I motive high school students to fear > the insanity that says, Common sense is a liar and a cheat, and it > clouds students' thinking? Finally, can I expose the spiritualistic > philosophy of physicists with sufficient clarity to get the physics > community to petition the pope for his blessing on my execution? And how will you get them to do that if your material is not adopted > and used in the classrooms where this is being taught? And how will > you know if, in cases where it is being used, that students do see > what you intend them to see? What are your CONCRETE metrics for > ascertaining success? How do you hold yourself concretely accountable to your objectives? The Battle of Armageddon is a manifestly visible battle between light > and darkness. It will appear eventually, just before the end of the > world. Thousands will be fighting valiantly with great Light against > darkness in the war. The book explaining the theological basis of this > war is now being written. The whole earth will be illuminated by the > power of the message (Rev 18:1). Hopefully you will know of it soon > enough. Shubee I suggest you read the book Don Quixote. PD I suggest that you read the Bible. Oh, I have. Several times. But for *concrete* accountability, I do not hinge my plans on vindication at Armageddon. I try to be a little more practical about how to make a Christ-modeled impact in real terms in advance of that date. Granted, I have significant failings along those lines. But I do not hinge my plans on metrics solely in the hereafter. PD === posting-account=lBRURwoAAAB_-Q_b04pGziaymfr5yRFx Gecko/20080702 Firefox/2.0.0.16,gzip(gfe),gzip(gfe) Actually, I do have a metric. Can I get high school students to > understand the subtle details of special relativity that professional > physicists don't understand? Can I motive high school students to fear > the insanity that says, Common sense is a liar and a cheat, and it > clouds students' thinking? Finally, can I expose the spiritualistic > philosophy of physicists with sufficient clarity to get the physics > community to petition the pope for his blessing on my execution? And how will you get them to do that if your material is not adopted > and used in the classrooms where this is being taught? And how will > you know if, in cases where it is being used, that students do see > what you intend them to see? What are your CONCRETE metrics for > ascertaining success? How do you hold yourself concretely accountable to your objectives? The Battle of Armageddon is a manifestly visible battle between light > and darkness. It will appear eventually, just before the end of the > world. Thousands will be fighting valiantly with great Light against > darkness in the war. The book explaining the theological basis of this > war is now being written. The whole earth will be illuminated by the > power of the message (Rev 18:1). Hopefully you will know of it soon > enough. Shubee I suggest you read the book Don Quixote. PD I suggest that you read the Bible. Oh, I have. Several times. > But for *concrete* accountability, I do not > hinge my plans on vindication at Armageddon. It hasn't even occurred to you that you should think about it and investigate carefully to make sure that you are not on the losing side. > I try to be a little more practical about how to make a > Christ-modeled impact in real terms in advance of that > date. Yes, by constructing your own metric. > Granted, I have significant failings along those lines. Yes, I've noticed. > But I do not hinge my plans on metrics solely in the > hereafter. That is painfully obvious. You've devised your own metric for the here and now. Shubee === posting-account=nPH_PQkAAACneDKT6RXopPWArC2We4Rq AppleWebKit/525.13 (KHTML, like Gecko) Version/3.1 Safari/525.13,gzip(gfe),gzip(gfe) > I don't recall ever feeling or expressing any rage or contempt > whatsoever. Never? I recall one post of yours on another topic where you said that > the longer you thought about my comments the angrier you got. I just did a search on archived posts in s.p.r by me with the words > angry or angrier and couldn't find anything about you at all. I > certainly haven't expressed any anger or contempt for you in this > conversation. I'll correct myself on this. I certainly have expressed contempt for you in this conversation. In fact, here is a snippet: Shubee: > The reason Einstein's approach demonstrates incompetence is that > accelerating from one inertial frame to another can be thought of as > an instantaneous process or something that only takes a negligible > amount of time when compared to the entire trip. PD: where the acceleration of one of the clocks is *throughout* the entire trip. The GPS satellites are a fine example of the twin puzzle. Idiot. Correction: uninformed idiot. Correction: fastidiously perpetually uninformed idiot. I stand by this statement, and regret the previous one where I claimed that I've not shown you any contempt. The remark I just made is true for a lot of authors. You should not > feel disparaged by that remark. It is a fact of life for authors and > the publishers who back them, and this is why authors are paid by > royalty rather than up front. Why must I insist on payment? Why can't I just finish writing up my > novel approach to high school level special relativity and then find a > physicist that believes in modus ponens, respects David Hilbert's > philosophy of physics and can endorse me for a physics education > submission to arXiv? Granted, there is a slim chance of satisfying all > those requirements but it is possible. You certainly can. But it would be useful for you to have some metric > by which to judge the value of your contribution. Succeeding in > getting it published is hardly a sufficient standard (though I know > that for many, this is a significant accomplishment in itself) for > that. Physicists judge the impact of each other's contributions not by > how many papers a researcher publishes, but by how often those papers > are referred to (that is, read and *used*) in subsequent papers. The > impact of educational material is not judged by whether it is > published, but by how often it is adopted and *used* in teaching. Even > if you don't care about the compensation (donate it all or settle for > negotiated royalty pittance), what the publisher is telling you via > the royalty is how *valuable* your material is for teaching. Now, here is another opportunity for you to be honest with yourself, > Shubee. If you say you are earnestly interested in the *educational* > value of what you are doing, then you should hold yourself accountable > to some metric by which to judge that value. If publishing it -- and > that's all -- is sufficient for you as a metric for your success, then > your *true* agenda (what's *really* important to you) is seeing your How do you hold yourself accountable, Shubee? PD === posting-account=lBRURwoAAAB_-Q_b04pGziaymfr5yRFx Gecko/20080702 Firefox/2.0.0.16,gzip(gfe),gzip(gfe) It will also help if you can insert yourself as the lead instructor in > a college course on this subject, and use your approach as the > curriculum. Then departmental course evaluation surveys will be a > useful tool for validating your approach. It would be especially hilarious if we could both teach at the same college. I would teach special relativity to physics students as if the topic was an elementary mathematics course and you could teach lower level math to the math students to emphasize the typical physicists' philosophy. I, of course, would reduce the physics to an exercise in mathematics and you could go the other way and teach math to math students to validate the physicists' emphasis, which is to shake the students' confidence in common sense. As you have confessed, you believe that Common sense is a liar and a cheat, and it clouds students' thinking. Shubee === posting-account=rIfu6QoAAAD5nXG3h9QEE0J3dZn1U45R Gecko/2008070208 Firefox/3.0.1,gzip(gfe),gzip(gfe) It will also help if you can insert yourself as the lead instructor in > a college course on this subject, and use your approach as the > curriculum. Then departmental course evaluation surveys will be a > useful tool for validating your approach. It would be especially hilarious if we could both teach at the same > college. I would teach special relativity to physics students as if > the topic was an elementary mathematics course and you could teach > lower level math to the math students to emphasize the typical > physicists' philosophy. How could you teach SR to anyone when you are incapable of discussing SR as it relates to physics? I, of course, would reduce the physics to an exercise in mathematics > and you could go the other way and teach math to math students to > validate the physicists' emphasis, which is to shake the students' > confidence in common sense. As you have confessed, you believe that > Common sense is a liar and a cheat, and it clouds students' > thinking. Shubee Physics treats math as a tool, not a subject unto itself. You would get yourself lost in attempting to teach minutia. === posting-account=lBRURwoAAAB_-Q_b04pGziaymfr5yRFx Gecko/20080702 Firefox/2.0.0.16,gzip(gfe),gzip(gfe) On Jul 28, 4:06 pm, stevendaryl3...@yahoo.com (Daryl McCullough) > PD says... > Common sense is a liar and a cheat, and it clouds students' thinking > You have no right to blame your confusion on common sense. Rather, it > is your willful abandonment of common sense that is the cause of your > confusion. The irony here is that Shubee's paper deriving the Lorentz > transformations is much *farther* from common sense than > the *usual* derivation. It is true that I might have buried myself in a hole by presupposing the weakest axioms possible to derive the Lorentz transformation in the sense that it's not obvious that I can even calculate time dilation without adding more assumptions. The assumptions that I need to add are obvious. My dilemma is that I have derived the Lorentz transformation, it has been a while since I thought about the physics of my key postulate ( u_ij = -u_ji ) and it's not clear to me that I even have the assumptions that I need to compute time dilation for the traveling twin problem from what I have already assumed and derived. Actually, I think that's an amusing predicament to be in and it was the original purpose of this thread. Shubee === posting-account=rIfu6QoAAAD5nXG3h9QEE0J3dZn1U45R Gecko/2008070208 Firefox/3.0.1,gzip(gfe),gzip(gfe) > On Jul 28, 4:06 pm, stevendaryl3...@yahoo.com (Daryl McCullough) PD says... > Common sense is a liar and a cheat, and it clouds students' thinking > You have no right to blame your confusion on common sense. Rather, it > is your willful abandonment of common sense that is the cause of your > confusion. The irony here is that Shubee's paper deriving the Lorentz > transformations is much *farther* from common sense than > the *usual* derivation. It is true that I might have buried myself in a hole by presupposing > the weakest axioms possible to derive the Lorentz transformation in > the sense that it's not obvious that I can even calculate time > dilation without adding more assumptions. Yea sure - weakest axioms possible. If you say so. How many assumptions on top of your axiom set did you have to add in order to obtain the Lorentz transformations along one axis? I'm yet to see you actually connect what you are doing with special relativity. All you did was derive the Lorentz transformations - and that was done well over a century ago. What do you think you are adding to the discussion? === Subject: Re: Alexander's trick and homeomorphism extension posting-account=tCEoyAoAAAAkltU5zxOoI8uJ4lyz5-kv .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) > It is quite well known fact that if f:S^1 -> S^1 > is an orientation-preserving homeomorphism, > then it can be extended to an orientation > preserving homeomorphism of the closed unit > disk bar{D}, and the extension is unique up to > isotopy rel S^1. Could there be any analogous statement in case > æi have a topological disk U in C, but it's > boundary might not be Jordan ? any counter-examles? > Message was edited by: jane To answer your question.... There are many ways to create all kinds of paradoxes with the number zero. This seems directly analogous to the situation where we ask does nonexistence exist. If the physicist asks: I have zero apples. Do the apples exist ? We found that it is simply not logical to perform mathematical operations on the nonexistent. So, if there are no apples, then does it make any sense to assign a number to such a thing ? So we find ourselves at the question of whether zero is a number, or not. So, you would probably say that if you have zero apples then apples do exist, you simply have zero of them. But then you must agree that apples are oranges whenever you have zero of each. If having zero apples => apples do not exist, then [1] zero is not a number, [2] apples are never oranges If having zero apples => apples exist, then [1] zero is a number, [2] apples can be oranges You must either have that zero is not a number, or that apples can be oranges, both of which are very undesirable. And of course the way we resolve this paradox is by ignoring it. Well, it's no different than asking does noexistence exist. If nonexistence existed, then there would be no paradoxes. But there IS a paradox, because you cannot perform math on the nonexistent. And yes, I know that existence, nonexistence and existential indeterminacy are not the typical objects which even have the property of existing or not. However, it is possible to argue that you have 3 valid tautologies, and that all of them are well behaved. They make sense philosophically. And while math-like manipulations which employ existential indeterminacy are not quite mathematics, they are also not quite nonsense. They are somewhere inbetween. The tautology that existential indeterminacy exists indeterminately seems to validate that view. Ayn Rand was doing math. === Subject: Re: Alexander's trick and homeomorphism extension posting-account=tCEoyAoAAAAkltU5zxOoI8uJ4lyz5-kv .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) > It is quite well known fact that if f:S^1 -> S^1 > is an orientation-preserving homeomorphism, > then it can be extended to an orientation > preserving homeomorphism of the closed unit > disk bar{D}, and the extension is unique up to > isotopy rel S^1. Could there be any analogous statement in case > æi have a topological disk U in C, but it's > boundary might not be Jordan ? any counter-examles? > Message was edited by: jane To answer your question.... There are many ways to create all kinds of paradoxes with the number zero. This seems directly analogous to the situation where we ask does nonexistence exist. If the physicist asks: I have zero apples. Do the apples exist ? We found that it is simply not logical to perform mathematical operations on the nonexistent. So, if there are no apples, then does it make any sense to assign a number to such a thing ? So we find ourselves at the question of whether zero is a number, or not. So, you would probably say that if you have zero apples then apples do exist, you simply have zero of them. But then you must agree that apples are oranges whenever you have zero of each. If having zero apples => apples do not exist, then [1] zero is not a number, [2] apples are never oranges If having zero apples => apples exist, then [1] zero is a number, [2] apples can be oranges You must either have that zero is not a number, or that apples can be oranges, both of which are very undesirable. And of course the way we resolve this paradox is by ignoring it. Well, it's no different than asking does noexistence exist. If nonexistence existed, then there would be no paradoxes. But there IS a paradox, because you cannot perform math on the nonexistent. And yes, I know that existence, nonexistence and existential indeterminacy are not the typical objects which even have the property of existing or not. However, it is possible to argue that you have 3 valid tautologies, and that all of them are well behaved. They make sense philosophically. And while math-like manipulations which employ existential indeterminacy are not quite mathematics, they are also not quite nonsense. They are somewhere inbetween. The tautology that existential indeterminacy exists indeterminately seems to validate that view. Ayn Rand was doing math. === Subject: Re: Angle Problem posting-account=AjPAzwoAAACH0yVpXxeHvhLqHIy0VudH FunWebProducts),gzip(gfe),gzip(gfe) work? === Subject: Re: Angle Problem posting-account=zga2wgoAAAD_6fmi3XyA1bMyNINP0zBK WorldLynx),gzip(gfe),gzip(gfe) > work? Because the situation is that of the famous scissors problem, in which your shoreline is analogous to one blade of the scissors and the moving searchlight beam is analogous to the other blade. What is desired here is the velocity of the cutting point, which moves along both blades simultaneously. What you found is merely the projection onto the shoreline of the instantaneous point of coincidence of the two blade edges. Mr. Silvers' analysis is correct; the answer is 4 pi meters per second, as he said. Grover Hughes === Subject: Re: Angle Problem posting-account=AjPAzwoAAACH0yVpXxeHvhLqHIy0VudH Gecko/20080702 Firefox/2.0.0.16,gzip(gfe),gzip(gfe) Hi Grover, because not only is the angle changing, but the radius is, too. Nice problem. Anon. === Subject: Re: Manifold with finite fundamental group > On Jul 28, 11:59 am, jane Suppose a topological manifold has a finite > fundamental group. > Does this fact imply that the manifold is > compact ? Here's another reply to the same question: Let M > be > *any* manifold > with finite fundamental group. Either it is > compact > or not. If it is > not, then you got what you want. Otherwise, > consider > M x R, which > has the same fundamental group, but it is not > compact. > Suppose the following situation: M is a closed > 3-manifold with finite fundamental group. Does this > imply that the universal cover of M is compact ? > > Sure. Could you please explain why is that so. > -- m === Subject: Re: Manifold with finite fundamental group <26664908.1217309636096.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=9QOSvAoAAACEOWJVSDuswW7dB_0wApQO Gecko/2008071615 Fedora/3.0.1-1.fc9 Firefox/3.0.1,gzip(gfe),gzip(gfe) > On Jul 28, 11:59 am, jane Suppose a topological manifold has a finite > fundamental group. > Does this fact imply that the manifold is > compact ? Here's another reply to the same question: Let M > be > *any* manifold > with finite fundamental group. Either it is > compact > or not. If it is > not, then you got what you want. Otherwise, > consider > M x R, which > has the same fundamental group, but it is not > compact. > Suppose the following situation: M is a closed > 3-manifold with finite fundamental group. Does this > imply that the universal cover of M is compact ? Sure. Could you please explain why is that so. What have you tried doing in order to prove it? -- m === Subject: Re: Manifold with finite fundamental group > On Jul 28, 11:59 am, jane Suppose a topological manifold has a finite > fundamental group. > Does this fact imply that the manifold is > compact ? Here's another reply to the same question: > Let M > be > *any* manifold > with finite fundamental group. Either it is > compact > or not. If it is > not, then you got what you want. Otherwise, > consider > M x R, which > has the same fundamental group, but it is not > compact. > Suppose the following situation: M is a closed > 3-manifold with finite fundamental group. Does > this > imply that the universal cover of M is compact ? Sure. Could you please explain why is that so. > > What have you tried doing in order to prove it? I know that the fundamental group of M is isomorphic to the deck transfiormation group of the covering M' -> M. We know that it is finite. And i don't know how approach then, could this already imply that M' is compact or not ? > -- m === Subject: Re: Manifold with finite fundamental group <9108564.1217315490073.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=9QOSvAoAAACEOWJVSDuswW7dB_0wApQO Gecko/2008071615 Fedora/3.0.1-1.fc9 Firefox/3.0.1,gzip(gfe),gzip(gfe) > On Jul 28, 11:59 am, jane Suppose a topological manifold has a finite > fundamental group. > Does this fact imply that the manifold is > compact ? Here's another reply to the same question: > Let M > be > *any* manifold > with finite fundamental group. Either it is > compact > or not. If it is > not, then you got what you want. Otherwise, > consider > M x R, which > has the same fundamental group, but it is not > compact. > Suppose the following situation: M is a closed > 3-manifold with finite fundamental group. Does > this > imply that the universal cover of M is compact ? Sure. Could you please explain why is that so. What have you tried doing in order to prove it? I know that the fundamental group of M is isomorphic to the deck transfiormation group of the covering M' -> M. > We know that it is finite. And i don't know how approach then, could this already imply that M' is compact or not ? The usual approach of considering an open cover C of M' and using the hypothesis to find a finite subcover works in this case. -- m === Subject: Re: Manifold with finite fundamental group > On Jul 29, 2:33 am, jane Suppose a topological manifold has a > finite > fundamental group. > Does this fact imply that the manifold > is > compact ? Here's another reply to the same > question: > Let M > be > *any* manifold > with finite fundamental group. Either it > is > compact > or not. If it is > not, then you got what you want. > Otherwise, > consider > M x R, which > has the same fundamental group, but it is > not > compact. > Suppose the following situation: M is a > closed > 3-manifold with finite fundamental group. > Does > this > imply that the universal cover of M is > compact ? Sure. Could you please explain why is that so. What have you tried doing in order to prove it? I know that the fundamental group of M is > isomorphic to the deck transfiormation group of the > covering M' -> M. > We know that it is finite. And i don't know how approach then, could this > already imply that M' is compact or not ? > > The usual approach of considering an open cover C of > M' > and using the hypothesis to find a finite subcover > works in this case. Sorry for asking this again. I still don't see the proof. I thought about something like this: Let C be some cover of M' and denote by p the projection covering p: M' -> M. Then p(C) is also a cover of M and if it were possible to extract some finite subcover C' from p(C), then for any U from this C', p^{-1}= union V_i, and all the elements V_i are translates of some fixed V_0 by elements from the given finite fundamental group. hence there are finitely many of these V_i. And this way we would get that there is a finite subcover of C. But i am not sure for which reason we could extract finite subcover C' from the covering of M. That is why i think the proof should be different and more logical. I would appreciate if you could help with it. === Subject: Re: Three hats problem > Suppose that each of three men is given a black or a white hat. Each > man can see the two others' hats but not his own. Each man has the > option of guessing his own hat color or remaining silent, but this > choice must be made without knowledge of what the others do or say. > If at least one man guesses correctly and none guess incorrectly, they > win; otherwise they lose. > > To be mathematically precise, we suppose that each hat is chosen > independently and with probability p of being black. What, then, is > the best joint strategy for the three men to adopt? > > The men do not know the value of p, but, surprisingly, this doesn't > matter: there is a strategy they can adopt without knowing p which > cannot be improved upon by knowing it. Also surprising is just how > good this strategy is: by adopting it they would win 3/4 of the time > even for the worst-case value of p, and 5/6 of the time if p is chosen > uniformly at random from [0,1]. The problem may be discussed in this paper (note last paragraph of review): MR2226238 (2007e:68043) Guo, Wenge(1-CINC-EH); Kasala, Subramanyam(1-NCW); Rao, M. Bhaskara(1-CINC-GIF); Tucker, Brian The hat problem and some variations. (English summary) Advances in distribution theory, order statistics, and inference, 459--479, Stat. Ind. Technol., Birkh.8auser Boston, Boston, MA, 2006. 68R05 (05A05 60C05 68P30 68Q25 90C60 94B05 94B65) In the hat problem, $n$ cooperating contestants are each given a hat with one of $m$ colors, chosen at random; each contestant can see only the hats on the heads of the others. Each contestant must either name the color of her own hat or pass. The group (collectively) wins if and only if at least one contestant does not pass and every contestant who does not pass correctly names the color of her hat. The contestants may not communicate once the hats are in place. What strategy should they use so as to maximize their chance of winning? It turns out that a best strategy gives a 75%chance of winning when $n=3$ and $m=2$ (with a uniform distribution for the color assignments). The hat problem is related to computational complexity, coding theory, and statistics. See [J. P. Buhler, Math. Intelligencer 24 (2002), no. 4, 44--49; MR1930867 The present paper presents an optimal strategy for $n=m=3$, say with colors red, white, and blue. Each player will guess blue if she sees two red hats, will guess red if she sees no red hats, and will pass if she sees one red hat. The resulting probability of winning is $15/27$, and it is proven that this strategy maximizes the probability of winning. The more general case of $m ge n = 3$ is also considered and a strategy presented, which is proved to be optimal among symmetric strategies (the probability of winning is $(3m^2-6m+6)/m^3$). The authors also prove an upper bound for the winning probability in the general case ($n/(n+m-1)$), which is, however, not always achievable. The paper also considers the case in which $n=3$ and $m=2$, but where the eight possible hat assignments are not equally likely (but their distribution is known to the contestants ahead of time). The paper concludes with other variations, open problems, and a potential application to cell biology. Reviewed by Jerrold W. Grossman -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Three hats problem > Suppose that each of three men is given a black or a white hat. Each > man can see the two others' hats but not his own. Each man has the > option of guessing his own hat color or remaining silent, but this > choice must be made without knowledge of what the others do or say. > If at least one man guesses correctly and none guess incorrectly, they > win; otherwise they lose. > > To be mathematically precise, we suppose that each hat is chosen > independently and with probability p of being black. What, then, is > the best joint strategy for the three men to adopt? > > The men do not know the value of p, but, surprisingly, this doesn't > matter: there is a strategy they can adopt without knowing p which > cannot be improved upon by knowing it. Also surprising is just how > good this strategy is: by adopting it they would win 3/4 of the time > even for the worst-case value of p, and 5/6 of the time if p is chosen > uniformly at random from [0,1]. The problem may be discussed in this paper (note last paragraph > of review): MR2226238 (2007e:68043) > Guo, Wenge(1-CINC-EH); Kasala, Subramanyam(1-NCW); Rao, M. > Bhaskara(1-CINC-GIF); Tucker, Brian > The hat problem and some variations. (English summary) Advances in > distribution theory, order statistics, and inference, 459--479, > Stat. Ind. Technol., Birkh.8auser Boston, Boston, MA, 2006. > 68R05 (05A05 60C05 68P30 68Q25 90C60 94B05 94B65) In the hat problem, $n$ cooperating contestants are each given a hat > with one of $m$ colors, chosen at random; each contestant can see only > the hats on the heads of the others. Each contestant must either name > the color of her own hat or pass. The group (collectively) wins if and > only if at least one contestant does not pass and every contestant who > does not pass correctly names the color of her hat. The contestants may > not communicate once the hats are in place. What strategy should they > use so as to maximize their chance of winning? It turns out that a best > strategy gives a 75%chance of winning when $n=3$ and $m=2$ (with a > uniform distribution for the color assignments). The hat problem is > related to computational complexity, coding theory, and statistics. See > [J. P. Buhler, Math. Intelligencer 24 (2002), no. 4, 44--49; MR1930867 How did that get past a reviewer? I imagine the average sci.mathers are OK with it, but expect rec.puzzlers to be able to spot the flaw instantly. Phil -- -- Microsoft voice recognition live demonstration === Subject: Re: Three hats problem > Suppose that each of three men is given a black or a white hat. Each > man can see the two others' hats but not his own. Each man has the > option of guessing his own hat color or remaining silent, but this > choice must be made without knowledge of what the others do or say. > If at least one man guesses correctly and none guess incorrectly, they > win; otherwise they lose. > > To be mathematically precise, we suppose that each hat is chosen > independently and with probability p of being black. What, then, is > the best joint strategy for the three men to adopt? > > The men do not know the value of p, but, surprisingly, this doesn't > matter: there is a strategy they can adopt without knowing p which > cannot be improved upon by knowing it. Also surprising is just how > good this strategy is: by adopting it they would win 3/4 of the time > even for the worst-case value of p, and 5/6 of the time if p is chosen > uniformly at random from [0,1]. The problem may be discussed in this paper (note last paragraph > of review): MR2226238 (2007e:68043) > Guo, Wenge(1-CINC-EH); Kasala, Subramanyam(1-NCW); Rao, M. > Bhaskara(1-CINC-GIF); Tucker, Brian > The hat problem and some variations. (English summary) Advances in > distribution theory, order statistics, and inference, 459--479, > Stat. Ind. Technol., Birkh.8auser Boston, Boston, MA, 2006. > 68R05 (05A05 60C05 68P30 68Q25 90C60 94B05 94B65) In the hat problem, $n$ cooperating contestants are each given a hat > with one of $m$ colors, chosen at random; each contestant can see only > the hats on the heads of the others. Each contestant must either name > the color of her own hat or pass. The group (collectively) wins if and > only if at least one contestant does not pass and every contestant who > does not pass correctly names the color of her hat. The contestants may > not communicate once the hats are in place. What strategy should they > use so as to maximize their chance of winning? It turns out that a best > strategy gives a 75%chance of winning when $n=3$ and $m=2$ (with a > uniform distribution for the color assignments). The hat problem is > related to computational complexity, coding theory, and statistics. See > [J. P. Buhler, Math. Intelligencer 24 (2002), no. 4, 44--49; MR1930867 > > > How did that get past a reviewer? I imagine the average > sci.mathers are OK with it, but expect rec.puzzlers to > be able to spot the flaw instantly. I have only seen the review, not the original paper (and I assume that's true for you, too). I wouldn't jump to any conclusions about what's in the paper from what's in the review. In particular, if what you are hinting is that the way the puzzle is stated in the review allows a better solution than the 75% solution, you may find that the authors of the paper were more careful in their statement of the puzzle than the author of the review. Or have I misunderstood your objection? -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Three hats problem > Suppose that each of three men is given a black or a white hat. Each > man can see the two others' hats but not his own. Each man has the > option of guessing his own hat color or remaining silent, but this > choice must be made without knowledge of what the others do or say. > If at least one man guesses correctly and none guess incorrectly, they > win; otherwise they lose. > > To be mathematically precise, we suppose that each hat is chosen > independently and with probability p of being black. What, then, is > the best joint strategy for the three men to adopt? > > The men do not know the value of p, but, surprisingly, this doesn't > matter: there is a strategy they can adopt without knowing p which > cannot be improved upon by knowing it. Also surprising is just how > good this strategy is: by adopting it they would win 3/4 of the time > even for the worst-case value of p, and 5/6 of the time if p is chosen > uniformly at random from [0,1]. > The problem may be discussed in this paper (note last paragraph > of review): > MR2226238 (2007e:68043) > Guo, Wenge(1-CINC-EH); Kasala, Subramanyam(1-NCW); Rao, M. > Bhaskara(1-CINC-GIF); Tucker, Brian > The hat problem and some variations. (English summary) Advances in > distribution theory, order statistics, and inference, 459--479, > Stat. Ind. Technol., Birkh.8auser Boston, Boston, MA, 2006. > 68R05 (05A05 60C05 68P30 68Q25 90C60 94B05 94B65) > In the hat problem, $n$ cooperating contestants are each given a hat > with one of $m$ colors, chosen at random; each contestant can see only > the hats on the heads of the others. Each contestant must either name > the color of her own hat or pass. The group (collectively) wins if and > only if at least one contestant does not pass and every contestant who > does not pass correctly names the color of her hat. The contestants may > not communicate once the hats are in place. What strategy should they > use so as to maximize their chance of winning? It turns out that a best > strategy gives a 75%chance of winning when $n=3$ and $m=2$ (with a > uniform distribution for the color assignments). The hat problem is > related to computational complexity, coding theory, and statistics. See > [J. P. Buhler, Math. Intelligencer 24 (2002), no. 4, 44--49; MR1930867 > > > How did that get past a reviewer? I imagine the average > sci.mathers are OK with it, but expect rec.puzzlers to > be able to spot the flaw instantly. I have only seen the review, not the original paper (and I assume > that's true for you, too). I wouldn't jump to any conclusions about > what's in the paper from what's in the review. In particular, if what > you are hinting is that the way the puzzle is stated in the review > allows a better solution than the 75% solution, you may find that > the authors of the paper were more careful in their statement of > the puzzle than the author of the review. Or have I misunderstood your objection? I presume not. Welcome to rec.puzzles :-) Phil -- -- Microsoft voice recognition live demonstration === Subject: Re: Three hats problem posting-account=G_G-iQoAAAB08LNQidt_LsMkopmIb4ZS Gecko/20060111 Firefox/1.5.0.1 Mnenhy/0.7.3.0,gzip(gfe),gzip(gfe) rules:- > A: if B same as C, guess same, else pass > B: if A not same as C, guess A, else pass > C: if A not same as B, guess A, else pass > so p(success) = 1-pq = 1-p(1-p) = 1-p+pp Yes! æVery nice, Ken! I originally heard this problem posed for the p=1/2 case, and I > believe the intended solution was for each player to pass when seeing > different hats, and to guess the opposite color when seeing same > hats. æResult: p(success)=3/4. æHowever, for general p, that strategy > has p(success)=3p(1-p), which is strictly worse than the solution you > give everywhere except at p=1/2. æDirect computation of p(success) for > all 3^12 possible strategies yields 45 distinct polynomials for > p(success), 43 of which are strictly less than 1-p+pp for all 0 (the other two being 3p(1-p), which is strictly less for all p except > 1/2, and 1-p+pp itself). æAlso, 1-p+pp is achieved only for the > strategy you give above (or its trivial re-labelings), so this > strategy does, indeed, dominate all others. A nice proof of this dominance would be welcome: æI have only a > computation -- one which surprised me by producing the result you give > above. æThis is a result which struck me as something I'd never have > thought of, so kudos again. -Jim Ferry > æMetron, Inc. > æf rr @m tsc .c m > æ e æy æe æ i æo æEvidently the conditions about not doing and not saying are not > that important. æPassing is doing and discussing strategy: > is saying. æBill J I took the reference to joint strategy to imply discussion before > the event. > The agreed strategy does not depend on anyone watching the others pass > or guess. You are correct. For the 8 possible arrangements, there are 6 correct guesses and 6 wrong guesses. Fortunately three wrong guesses occur on WBB and three occur on BWW. For each of the other six arrangements, there is one correct guess and. two passes. If my analysis is correct, the probability of winning is 0.75, Bill J === Subject: N-Step Mappings??? (was Re: Finding the Formula...) <6edll5F6hgqmU1@mid.dfncis.de> <6ee3euF6ju5qU1@mid.dfncis.de> posting-account=fwSgtAkAAACFnX70ssKwbvm9_oCZVHrx Gecko/2008070208 Firefox/3.0.1,gzip(gfe),gzip(gfe) > Mike - æI just skimmed through older notes and found one, > æwhich may interst you. Here I dealt with the > æderivatives, too. I've not edited this, it is > æjust raw material, where I usually fiddle with > æsome ideas. > æMay be it reflects some of your consideration or > æeven gives an idea. æYou may find it at > æhttp://go.helms-net.de/math/tetdocs/mike3.pdf > Nope, not really, sorry, as I still can't get what the pattern is in the coefficients a n... But I'm still wondering, what on Earth does that cryptic, cryptic phrase number of n-step mappings with so many inputs actually mean? As this might provide a short formula for at least the case of base-e U-tetration (which can be mapped to base-e^(1/e) T-tetration). It doesn't make any sense: mappings are usually functions, and the number of functions with so many inputs depends on the sets they are mapping to/from. Furthermore I also was unaware that functions had steps. Does anyone here have any idea as to what this admittedly strange thing means? === The First Law Of Economics (Issued 18 July 2008) (Version 3.1 on 29 July 2008) (view Summary by skipping indentation) (suits foreign language readers) ' The First Law Of Economics, is: People desire and enjoy to do a job of value they can be proud of. (Individual persons have the desire and enthusiasm or joy, to undertake something that is of value now or in the future, to their fellows of Mankind.) I have elaborated that in the HRI note 'Basic Economics - In Government and Enterprise, People Are Not Objects.' *(1) ' Also here, we can observe the whole scale *(2) from the top to the very bottom - where that natural desire and joy, gets more and more dominated, more and more prevented, more and more oppressed and destroyed, by Criminal Minds and by means of the false ideas that these create and force onto others, and thus establish as culture into the society, to evolve into customs, rules and laws that COUNTER the First Law of Economics, rules and systems of government and education that oppose your soul, and thus, that counter your and other people's joy of being alive, COUNTER the First Law of Economics, people's desire and joy to do a job of value they can and will be proud of. ' ' Criminal Minds - who do not exist - will counter and have countered the First Law of Economics, with You can NOT do what you yourself want and desire and know that is of true value to others, and that you thus will be proud of, but you must do, what the society demands of you, meaning, you must do, what Criminal Minds demand of you. Criminal Minds want you and everyone to see THEM as normal, and YOU as not normal if you oppose them: *(5)(6)(7) You must be regarded as someone who, through education or other means of coercion, has to be MADE normal, or by plain force of denial of your social and material (monetary) connections that you need in order to live, has to be convinced, that people only care for themselves, so you better let yourself be dominated 'in order to care for yourself': You have to be MADE to think and act as if Criminal Minds are normal people, and if you don't, you should agree, that you are considered crazy, or at least, that you are badly adjusted. Sane is in agreement with not seeing Criminal Minds, insane, crazy, is not in agreement with 'being blind,' with 'not seeing Criminal Minds', and sane is reliable, or 'maintaining stability,' as the Criminal Minds call it, if it suits them in politics or in an organization, and insane is destructive. Thus anyone who DOES see Criminal Minds, including law enforcement people who DO see Criminal Minds, must be dominated NOT to see Criminal Minds, or else lose their job. ' ' Nowadays, they indeed made it so - in universities, in education, in science, in medicine, in journalism, in politics, in most history text books even, and indeed in economics - that Criminal Minds DO NOT EXIST. That's THEIR, the Criminal Mind's definition of normal: You accept Criminal Minds above you, and you stop recognizing them. The Criminal Mind's definition of abnormal - you also know and practice obediently, as well, by avoiding to be abnormal - abnormal is: NOT agreeing, that Criminal Minds are above you; revolting against being dominated by a Criminal Mind, is showing you are abnormal; your feeling revulsion of a Criminal Mind, is, as defined above, not normal, but YOUR 'disease' or 'abnormality', and medical science will provide you with drugs in order NOT to feel, that someone is a Criminal Mind, and not to feel what he or she radiates at you and at - and into the body and soul of - anyone. So your or other's joy of being alive, is destroyed by them, and they leave you then to lead a life or the end of your life with an endless number of diseases for which the cause is not known, and that have no real cure. Which is NOT the purpose of life, but WHICH IS the purpose of Criminal Minds, that DO NOT EXIST. ' So the self-help gurus, and other Holinesses, are telling you, in their lectures and books, and in TV shows, how to adapt or adjust, how to live in the present or in harmony with Evil: They are coming up with ever-new methods of how you must adapt to the lies of Criminal Minds, WITHOUT THAT THEY EVER MENTION THE CRIMINAL MINDS ...isn't it. ' You don't know this, unless you read the Human Rights Issues, but actually you know it very well, because you experience this every day and night, you do see in front of you all that opposes the First Law of Economics, all who demand of people they are not allowed to and can not desire and enjoy to do a job of value they can be proud of, so you do actually know it very well, but, that is, you have not seen it formulated and voiced in a way that to understand (and thus to handle) what is actually going on, not the endless diversion into treating symptoms, symptoms and symptoms only, which is what Criminal Minds have largely diverted and perverted and degenerated the society into, but SEEING, understanding, and thus handling the CAUSE of what is going on, and that needs a very correct formulation and communication of it, as in the Definition of Peace, which is what I am doing, and what the Human Rights Issues are all about: To bare your soul right in front of you, with my poems, with my song 'reviving you softly,' strumming the cords of your soul, not now with heavenly music, but with words: To formulate and voice what is right in front of you, to make it visible and grasped, capable of being grasped and so understood, that you can actually handle it, as I did with all the things I defined for you, and that are defined for the first time correctly, defined by me straight from your own soul, and published in the Human Rights Issues. ' ' The First Law Of Economics, is: People desire and enjoy to do a job of value they can be proud of. (Individual persons have the desire and enthusiasm or joy, to undertake something that is of value now or in the future, to their fellows of Mankind.) ' You see then, that Criminal Minds have OPPOSED the simple and natural law, that comes straight from your soul - as does all that you read in the Human Rights Issues (see upcoming issue of HRI 20080726) does in actual fact come straight from your soul - OPPOSED by Criminal Minds, such as your economics guru Adam year 1775, and repeatedly thereafter - for which certain people paid him a lot of money - to write the lie, that: People can not feel others - which lie, he thought, would not get very wide acceptance, and therefore he added, so there must be some mysterious idea of charity inside people, and the next lie, which you are fiercely taught in schools and in the society, BY CRIMINAL MINDS and their associates, that: Everyone thinks only of himself, and works only for his own best interest - which very vicious lie also would not get full acceptance, so he, the economist Adam Smith added, that this automatically balances out into a free exchange between people, as if there is an 'invisible hand'. Interestingly, a would-be philosopher who wants to explain and sell his own Criminal Mind as normal, as how people are, like the Criminal Minds Freud, Einstein, and the like, they enforce THEIR OWN Criminal Mind, as that is what people are and what The Creation is, Adam Smith, a would-be philosopher and for the same reason as mentioned, telling people what economics is, explaining it, so as to make himself seem normal, like all people are, Adam Smith maintained a friendship with the Criminal Mind Benjamin Franklin, (*4) when this individual (Franklin) resided for some time in London, and the Criminal Mind Bill Gates presents his friend Warren Buffett with a copy of the infamous book by economist Adam Smith, 'The Wealth of Nations' - the struggle of Adam Smith for having Criminal Minds accepted as the economic motors of the society, including, of course, the Criminal Minds Bill Gates and Warren Buffett. ('It costs a penny to make. Sell it for a dollar. It's addictive. And there's fantastic brand loyalty.' - quoting Warren Buffett, on why he likes the cigarette business - 1987) ' ' The First Law Of Economics, is: People desire and enjoy to do a job of value they can be proud of. ' I repeat: Also here, we have the whole scale *(1) from the top to the very bottom, where that natural desire and joy gets more and more dominated, more and more prevented, more and more oppressed and destroyed, by Criminal Minds and by means of the false ideas, that these create and force onto others, and thus into the society, to become customs, rules and laws that COUNTER the First Law of Economics, that oppose your soul, and thus, that counter your and other people's joy of being alive. ' So it is good to keep knowing the First Law of Economics, as it resides in your soul - or in your heart, as some say. Koos Nolst Trenite 'Cause Trinity' human rights philosopher and poet 'Men of all nations came to listen to Solomon's wisdom, sent by all the kings of the world, who had heard of his wisdom.' 1 Kings 4:34 ________ Footnote: (1) 'Basic Economics - In Government and Enterprise, People Are Not Objects' {HRI note 20060201 version 3) (1 February 2006 - note version 3, on 2 Feb 2006) (2) 'Definition Of Insane - Relation To Humor' (DOI) (3) 'Famous European Life Energy Vampires - Unsuspected Of Destroying Your Emotional Integrity' (4) 'Famous American Life Energy Vampires - Unsuspected Of Destroying Your Emotional Integrity' (5) 'The Anti-Normal' {HRI 20071208-V1.0} (8 December 2007) (6) ' Insane Defined By Criminal Minds As 'Ability To Perceive Them' ' (7) ' Insane Defined By Criminal Minds - Part Twenty:(Conclusion) Others destroying people, is 'none of your business' ' (28 February 2008 - Version 1.0.1 on 28 Feb 2008) ' __________ 'The First Law of Human Rights' (FLOHR) {HRI 20060601-V3.1.2} (1 June 2006 - Version 3.1.2 on 31 Dec 2007) 'Obviously the Second Law Of Human Rights' (SLOHR) {HRI 20060924-V3.2.1} (24 September 2006 - Version 3.2.1 on 6 Aug 2007) (more to add) ' ____________ Verification: http://www.angelfire.com/space/platoworld Copyright 2008 by Koos Nolst Trenite - human rights philosopher and poet This is 'learnware' - it may not be altered, and it is free for anyone who learns from it and (even if he can not learn from it) who passes it on unaltered, and with this message included, to others who might be able to learn from it (but not to sociopaths, who vehemently oppose any true knowledge of life and of themselves). None of my writings may be used, ever, to support any political or religious or scientific 'agenda,' but only to educate, and to encourage people to judge un-dominated and for themselves, about any organizations or individuals. Send free-of-Envy and free-of-Hate, Beautiful e-mails to: PlatoWorld at Lycos.com === > The First Law Of Economics, is: > > People desire and enjoy to do a job of value they can be > proud of. To quote from Amadeus, There are two many notes in this post. The first two laws of economics are the same as the first an second law of thermodynamics. Any manipulation of political laws or sleight of hand that violates them is another fraud just like a a Ponzi scheme is. The current fuel crisis and all its accouterments are bringing out pseudoeconomists who are purporting to solve these problems by paper manipulation. Bill === Subject: Re: geometrical point > Hi all, > > when we put a dot with a lead pencil why is it not a > geometric point ? It consists of seeral carbon atoms and is, therefore, much too big By definition a point has position and no magnitude. A dot with a lead pencil has position AND magnitude You need three mutual touching faces to represent a point on a surface. Give the three faces a different color. Where two different colors touch each other you have a line (or edge). Where three different colors touch you have a point. === Subject: Re: geometrical point <10244539.1217261840288.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=jPnQ2goAAAA461y3QD0lbyw0oKeThma1 AppleWebKit/525.18 (KHTML, like Gecko) Version/3.1.2 Safari/525.20.1,gzip(gfe),gzip(gfe) this reminds of Bucky's definition of a point, a subvisible something, which is minimally a tetrahedron; sort of like a star, which is just a subvisible sun, til you apply spectral analysis. > You need three mutual touching faces to represent a point on a surface. Give the three faces a different color. Where two different colors touch each other you have a line (or edge). Where three different colors touch you have a point. thus: people should read the Bible to become literate, and to find out what Shakeepeare meant by that weird poem (KJVersion .-) why was the 900' Jesus depicted reading a new testament, or the 800# guerilla? or was it just a Book. well, Genesis is more of a bibiliography. I read that one SF author said that the ylived so long, but they were really dumb; like, there minds were pratically varves ... or some thing. now, what you were saying about salvation ... do you have a prefered application? === Subject: Wassfat posting-account=XHd4WgoAAADRnVQXs586cVoDxWlR8_O7 Gecko/20070508 Firefox/1.5.0.12,gzip(gfe),gzip(gfe) visiter ce blog si vous etes intirisser au recettes, http://wassfat.blogspot.com/ === Subject: Wassfat posting-account=XHd4WgoAAADRnVQXs586cVoDxWlR8_O7 Gecko/20070508 Firefox/1.5.0.12,gzip(gfe),gzip(gfe) visiter ce blog si vous etes intirisser au recettes, http://wassfat.blogspot.com/ === Subject: Re: A hole in space Let C cover S with open balls of R^2. > Proposition: some U in C with p in U or C is uncountable. Shouldn't be: C is countable? Covering S with balls with rational centers > and rational radiuses seems possible... ... or just the three open balls of radius 1 and centers on the > boundary of D^2, 120 degrees apart. > Whoops, I suppose so. Ok, ok, scratch the thread. === Subject: Re: A hole in space posting-account=jPnQ2goAAAA461y3QD0lbyw0oKeThma1 AppleWebKit/525.18 (KHTML, like Gecko) Version/3.1.2 Safari/525.20.1,gzip(gfe),gzip(gfe) this reminds of Bucky's definition of a point, a subvisible something, which is minimally a tetrahedron; sort of like a star, which is just a subvisible sun, til you apply spectral analysis. > You need three mutual touching faces to represent a point on a surface. Give the three > faces a different color. Where two different colors touch each other > you have a line (or edge). Where three different colors touch you have a point. thus: people should read the Bible to become literate, and to find out what Shakeepeare meant by that weird poem (KJVersion .-) why was the 900' Jesus depicted reading a new testament, or the 800# guerilla? or was it just a Book. well, Genesis is more of a bibiliography. I read that one SF author said that the ylived so long, but they were really dumb; like, there minds were pratically varves ... or some thing. now, what you were saying about salvation ... do you have a prefered application? > ... or just the three open balls of radius 1 and centers on the > boundary of D^2, 120 degrees apart. === Subject: Re: Question on algebraic curves posting-account=R7AgUAoAAADVFAtIe36IBmgohoHjZsKW Gecko/20061201 Firefox/2.0.0.16 (Ubuntu-feisty),gzip(gfe),gzip(gfe) > I have a problem with an exercise in a book. The exercise is the > following. Let k be an algebraically closed field with a subfield k'. Let C be an > algebraic curve in A^2(k) defined by a non-constant polynomial of > degree d and L a line that intersect C in exactly d points. Let the > minimal polynomials of L and C be in k'[x,y] aka. they are defined over k' -- C/k', L/k'. and let d-1 of the > intersection points be k' rational. Show that all d intersection > points are k' rational. C is given by f(x,y) = 0 and L by a*x+b*y=c. > Wlog. b!=0 (otherwise we have a!=0 and exchange x<->y in the > following). > Substituting y = (c-a*x)/b into f(x,y)=0 produces a polynomial > g(x) in K'[x] of degree <= d. > In fact, since g has exactly d roots in k, we see that deg(g)=d. > And if d-1 of the roots are in k', then all must be (since the > negative of the sum of all roots is a coefficient of g). Moreover, I can't imagine how one could handle such rationality > questions in fields different from R or Q or C where one have no > imagination. Observe that the argument above uses no special assumptions > about characteristic or other specialties of Q, R, C. > As a matter f fact, i still don't really have a good imagination > about C^2. For example if one considers the circle (which can not > imagined as a circle in an arbitrary alg. closed field) > C=V(x^2+y^2-1). Let k' be a subfield, then all the k' rational points > should be (0,1) and (2t/(t^2+1),(t^2-1)/(t^2+1)) with t in k'. How can > one see this without a picture? I am a little frustrated... S. === Subject: Question on quadrics posting-account=R7AgUAoAAADVFAtIe36IBmgohoHjZsKW Gecko/20061201 Firefox/2.0.0.16 (Ubuntu-feisty),gzip(gfe),gzip(gfe) let a projective quadric be a curve C in P^2(k) of degree 2 defined by a homogeneous polynomial f where k is an algebraically closed field of char different to 2. A mapping t: P^2(k) --> P^2(k) is called a coordinate transformation if there exists an invertible matrix A such that t(x:y:z)=((x:y:z)A) for all x,y,z and therefore let t(C) be the homogeneous zero set of the homogeneous polynomial f^A(x,y,z)=f((x,y,z)A^-1). Why does a coordinate transformation exists for C such that the image is defined by one of the homogeneous zero-sets x^2+y^2+z^2=0 (regular quadric) or x^2+y^2=0 (pair of lines) or x^2=0 (double line)? I am trying to learn these things from a book and unfortunately have no idea how to handle this. Therefore I am glad if someone could help me a little. S. === Subject: Re: Question on quadrics posting-account=R7AgUAoAAADVFAtIe36IBmgohoHjZsKW Gecko/20061201 Firefox/2.0.0.16 (Ubuntu-feisty),gzip(gfe),gzip(gfe) > Why does a coordinate transformation exists for C such that the image > is defined by one of the homogeneous zero-sets x^2+y^2+z^2=0 (regular > quadric) or x^2+y^2=0 (pair of lines) or x^2=0 (double line)? Sorry, I mean: x^2+y^2+z^2=0 (regular quadric) or z^2+x^2=0 (pair of lines) or z^2=0 (double line) === Subject: Analytic sheaf cohomology on the complex plane Mail-To-News-Contact: abuse@dizum.com I'm puzzled about how the first analytic sheaf cohomology group works out in practice. See http://camoo.freeshell.org/cohomquest.pdf please. Laura === Subject: Cancel Re: A consideration concerning the diagonal argument of G. Cantor -- For every line of Cantor's list it is true that this line does not contain the diagonal number. Nevertheless the diagonal number may be in the infinite list. (WM, sci.logic) === Subject: how to find appropriate and sufficiently simple approximation in 4 dimensions? posting-account=Lp215AoAAABaPqTOHMbNj9GpsECJOOv0 Gecko/20080702 Firefox/2.0.0.16,gzip(gfe),gzip(gfe) I have following problem: I have set of data dependent on three different variables, and I'd like to find some function which approximate this data, let's say f(x,y,z) ... I've tried to use FindFit function in Mathematica, but you have to choose f function on your own and this isn't easy task ... I want to minimalise maximum of relative difference between f(x,y,z) and data(x,y,z). My questions: 1) is there any suggestion how to find the form of f function? I've tried to finf it as a multiplicity of three one variable functions f(x,y,z)=f1(x)*f2(y)*f3(z) but it doesn't work ... another idea was to find 2D approximation g(x,y) with various coefficients dependent on z, but it also doesn't work in general ... 2) Are there any packages to find such approximations automaticly (like TableCurve and tableCurve3D for f(x) and f(x,y))? i hope somebody will help me ... === Subject: Re: how to find appropriate and sufficiently simple approximation in 4 dimensions? posting-account=CtWhuAoAAAAZZ9vwdovdqB3NNaiUa20_ SV1),gzip(gfe),gzip(gfe) > I have following problem: I have set of data dependent on three > different variables, and I'd like to find some function which > approximate this data, let's say f(x,y,z) ... I've tried to use > FindFit function in Mathematica, but you have to choose f function on > your own and this isn't easy task ... I want to minimalise maximum of > relative difference between f(x,y,z) and data(x,y,z). My questions: > 1) is there any suggestion how to find the form of f function? I've > tried to finf it as a multiplicity of three one variable functions > f(x,y,z)=f1(x)*f2(y)*f3(z) but it doesn't work ... another idea was to > find 2D approximation g(x,y) with various coefficients dependent on z, > but it also doesn't work in general ... > 2) Are there any packages to find such approximations automaticly > (like TableCurve and tableCurve3D for f(x) and f(x,y))? i hope somebody will help me ... Some work in 2-D has been done by this author. So maybe you can use ideas from the following reference: At: http://hdebruijn.soo.dto.tudelft.nl/www/programs/delphi.htm#SD Han de Bruijn === Subject: Re: how to find appropriate and sufficiently simple approximation in 4 dimensions? >I have following problem: I have set of data dependent on three >different variables, and I'd like to find some function which >approximate this data, let's say f(x,y,z) ... I've tried to use >FindFit function in Mathematica, but you have to choose f function on >your own and this isn't easy task ... I want to minimalise maximum of >relative difference between f(x,y,z) and data(x,y,z). My questions: >1) is there any suggestion how to find the form of f function? I've >tried to finf it as a multiplicity of three one variable functions >f(x,y,z)=f1(x)*f2(y)*f3(z) but it doesn't work ... another idea was to >find 2D approximation g(x,y) with various coefficients dependent on z, >but it also doesn't work in general ... >2) Are there any packages to find such approximations automaticly >(like TableCurve and tableCurve3D for f(x) and f(x,y))? i hope somebody will help me ... I haven't done anything like this myself, but, as no-one else has replied yet, I'll just mention that there is some Mathematica code for multivariate cubic spline interpolation here: and perhaps that (and/or its author) might be of some help? -- === Subject: Has James Harris gone mad? Hi all, Please don't answer right away to the question from the subject line. (Of course, if you would do that, the answer would be: Yes. Many years ago.). I am writing this post because James has been increasing the price of his book (both in the printed version as well as in the PDF one) and now the price he asks for any of them is $23.95: http://www.lulu.com/content/2812588 Does he really believe that someone will pay that amount for a PDF file with 50 pages of texts from publicly available posts from his blog? Jose Carlos Santos === Subject: Re: Has James Harris gone mad? posting-account=_k7REQoAAACDJL2M6OKCfSBj5_wTcvrO Gecko/20080702 Firefox/2.0.0.16,gzip(gfe),gzip(gfe) > Hi all, Please don't answer right away to the question from the subject line. > (Of course, if you would do that, the answer would be: Yes. Many years > ago.). I am writing this post because James has been increasing the > price of his book (both in the printed version as well as in the PDF > one) and now the price he asks for any of them is $23.95: http://www.lulu.com/content/2812588 Does he really believe that someone will pay that amount for a PDF file > with 50 pages of texts from publicly available posts from his blog? > Jose Carlos Santos It's the catchy title that works for me. Who wouldn't want this valuable first edition? In 200 years this'll be like a Shakespeare First Folio and ensure your descendants' fortune. === Subject: Re: Has James Harris gone mad? > Hi all, > > Please don't answer right away to the question from the subject line. > (Of course, if you would do that, the answer would be: Yes. Many years > ago.). I am writing this post because James has been increasing the > price of his book (both in the printed version as well as in the PDF > one) and now the price he asks for any of them is $23.95: > > http://www.lulu.com/content/2812588 > > Does he really believe that someone will pay that amount for a PDF file > with 50 pages of texts from publicly available posts from his blog? > > > Jose Carlos Santos Ha - certainly all 4 pieces which are sold will be personally signed! === Subject: Re: Has James Harris gone mad? posting-account=a6woBRAAAADpNFZJBA7ZBx35zXaKmaP4 Gecko/2008070208 Firefox/3.0.1,gzip(gfe),gzip(gfe) > Hi all, Please don't answer right away to the question from the subject line. > (Of course, if you would do that, the answer would be: Yes. Many years > ago.). I am writing this post because James has been increasing the > price of his book (both in the printed version as well as in the PDF > one) and now the price he asks for any of them is $23.95: http://www.lulu.com/content/2812588 Does he really believe that someone will pay that amount for a PDF file > with 50 pages of texts from publicly available posts from his blog? > Jose Carlos Santos Ha - certainly all 4 pieces which are sold will be personally signed! The other advantage an owner of this book would have is immunity from capricious deletions of JSH's blog posts and newsgroup contributions. book. === Subject: Re: Has James Harris gone mad? posting-account=3WPJYgoAAAA55VjhzK9i07RN8h8u8eEs Gecko/20080702 Firefox/2.0.0.16,gzip(gfe),gzip(gfe) book. > Now that's funny: I would thought it would read You CAN'T do my math, so buy my book. But if you have any thought of buying his book, I would suggest that you buy it at the current price. Day by day, James is realizing just how important his work is, and so there is no doubt that the price will only continue to go up. Also, once the world wakes up and realizes he is right, what textbook will be used to educate the youth? The Book of James, no doubt. Now with that type of demand, you can imagine what will happen to the price. You'll be fighting for a copy of this book for your kid at $1,000 a pop like Arnold and Sinbad fighting over a TurboMan toy. HTH, M === Subject: Re: Has James Harris gone mad? posting-account=AdyLXQoAAABgRay99CKv1O8Y_7jjivwq InfoPath.1),gzip(gfe),gzip(gfe) > Does he really believe that someone will pay that > amount [$23.95] for a PDF file with 50 pages of > texts from publicly available posts from his blog? This might make sense if you assume (and assume he also thinks, and I agree this last part is a little shaky) that the few people who would pay $3 or $4 would probably pay at least $23.95. That is, there is an extreme bimodal distribution in the price people are willing to pay for his book. Dave L. Renfro === Subject: Re: Has James Harris gone mad? > Does he really believe that someone will pay that > amount [$23.95] for a PDF file with 50 pages of > texts from publicly available posts from his blog? This might make sense if you assume (and assume he >also thinks, and I agree this last part is a little >shaky) that the few people who would pay $3 or $4 >would probably pay at least $23.95. That is, there >is an extreme bimodal distribution in the price >people are willing to pay for his book. I might read it if he paid me about $20.00. -- === Subject: Re: Has James Harris gone mad? posting-account=Cbgh4AoAAAAr0dt1RqLOClWCyUWii2fU Gecko/20080702 Firefox/2.0.0.16,gzip(gfe),gzip(gfe) > Hi all, Please don't answer right away to the question from the subject line. > (Of course, if you would do that, the answer would be: Yes. Many years > ago.). I am writing this post because James has been increasing the > price of his book (both in the printed version as well as in the PDF > one) and now the price he asks for any of them is $23.95: http://www.lulu.com/content/2812588 Does he really believe that someone will pay that amount for a PDF file > with 50 pages of texts from publicly available posts from his blog? > Jose Carlos Santos Where will you get a self-contained book about primes, logic, the object ring and a proof of FLT for thjat price? One odd thing for those used to read mathematical research or texts, there are few if any citations! Of course there are no citations because JSH is more important than any other potential source of knowledge (after all, when will miraculously referenc him and his ideas). hagman === Subject: Re: Has James Harris gone mad? posting-account=HaopWgoAAADs72-s8RQYwP_-ruRUuNzX Gecko/2008070208 Firefox/3.0.1,gzip(gfe),gzip(gfe) > Hi all, Please don't answer right away to the question from the subject line. > (Of course, if you would do that, the answer would be: Yes. Many years > ago.). I am writing this post because James has been increasing the > price of his book (both in the printed version as well as in the PDF > one) and now the price he asks for any of them is $23.95: http://www.lulu.com/content/2812588 Does he really believe that someone will pay that amount for a PDF file > with 50 pages of texts from publicly available posts from his blog? > Jose Carlos Santos I think this has been his scheme all along - to get rich from his math, have lots of women and finally get the world domination/ admiration that he so desperately desires. ;-) === Subject: Re: Has James Harris gone mad? > Please don't answer right away to the question from the subject line. > (Of course, if you would do that, the answer would be: Yes. Many years > ago.). I am writing this post because James has been increasing the > price of his book (both in the printed version as well as in the PDF > one) and now the price he asks for any of them is $23.95: > http://www.lulu.com/content/2812588 > Does he really believe that someone will pay that amount for a PDF file > with 50 pages of texts from publicly available posts from his blog? > > I think this has been his scheme all along - to get rich from his > math, have lots of women and finally get the world domination/ > admiration that he so desperately desires. > That's odd! I always thought that it would be the other way around, that is, the hammer first and the money, women and world domination after. :-) Jose Carlos Santos === Subject: Re: Has James Harris gone mad? <6f8qdqFadn0oU1@mid.individual.net> posting-account=gpERugkAAAB5_qKVhbO9UpGpOXFNrIYf 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > That's odd! I always thought that it would be the other way around, that > is, the hammer first and the money, women and world domination after. :-) First you get the money! Then you get the power! Then you get the women! Socks === Subject: Re: Has James Harris gone mad? posting-account=a6woBRAAAADpNFZJBA7ZBx35zXaKmaP4 Gecko/2008061017 Firefox/3.0,gzip(gfe),gzip(gfe) > Hi all, Please don't answer right away to the question from the subject line. > (Of course, if you would do that, the answer would be: Yes. Many years > ago.). I am writing this post because James has been increasing the > price of his book (both in the printed version as well as in the PDF > one) and now the price he asks for any of them is $23.95: http://www.lulu.com/content/2812588 Does he really believe that someone will pay that amount for a PDF file > with 50 pages of texts from publicly available posts from his blog? > Jose Carlos Santos This may actually be economically defensible. Although $23.95 is probably not the price a fully informed buyer and seller would reach through willing negotiation, there may be dozens of sales to be made with buyers who are impressed by the marketing hubris that sets a price of $23.95. What would an expert mathematician choose as the price that maximizes profit? === Subject: Re: Has James Harris gone mad? posting-account=n1ZfDgkAAABbCs44qOtz8dP-RkWuEBif Gecko/2008070208 Firefox/3.0.1,gzip(gfe),gzip(gfe) Hi all, Please don't answer right away to the question from the subject line. > (Of course, if you would do that, the answer would be: Yes. Many years > ago.). I am writing this post because James has been increasing the > price of his book (both in the printed version as well as in the PDF > one) and now the price he asks for any of them is $23.95: http://www.lulu.com/content/2812588 Does he really believe that someone will pay that amount for a PDF file > with 50 pages of texts from publicly available posts from his blog? > Jose Carlos Santos This may actually be economically defensible. Although $23.95 is probably not the price a > fully informed buyer and seller would reach > through willing negotiation, there may be > dozens of sales to be made with buyers who > are impressed by the marketing hubris that > sets a price of $23.95. What would an expert mathematician choose > as the price that maximizes profit? > What is the typical textbook price in number theory? I'm curious if anyone will answer honestly. JSH === Subject: Re: Has James Harris gone mad? posting-account=Cbgh4AoAAAAr0dt1RqLOClWCyUWii2fU Gecko/20080702 Firefox/2.0.0.16,gzip(gfe),gzip(gfe) > Hi all, Please don't answer right away to the question from the subject line. > (Of course, if you would do that, the answer would be: Yes. Many years > ago.). I am writing this post because James has been increasing the > price of his book (both in the printed version as well as in the PDF > one) and now the price he asks for any of them is $23.95: http://www.lulu.com/content/2812588 Does he really believe that someone will pay that amount for a PDF file > with 50 pages of texts from publicly available posts from his blog? > Jose Carlos Santos This may actually be economically defensible. Although $23.95 is probably not the price a > fully informed buyer and seller would reach > through willing negotiation, there may be > dozens of sales to be made with buyers who > are impressed by the marketing hubris that > sets a price of $23.95. What would an expert mathematician choose > as the price that maximizes profit? > What is the typical textbook price in number theory? I'm curious if anyone will answer honestly. JSH A 50 page collection of usenet posts? With a simple homemade ugly font (Arial?) title page? With mathematical symbols like the square root symbol simply being written sqrt? With unstructured blog-like texts? Without any valid proof? Without reference to other literature? The closest match with any type of printed text that comes to my mind is a free handout of a beginner accompanying his first talk in fornot of class. These are ususally handed out for free and are usually treated like waste paper by the other students (at least if the content quality is as in your case). hagaman === Subject: Re: Has James Harris gone mad? posting-account=a6woBRAAAADpNFZJBA7ZBx35zXaKmaP4 Gecko/2008070208 Firefox/3.0.1,gzip(gfe),gzip(gfe) > Hi all, Please don't answer right away to the question from the subject line. > (Of course, if you would do that, the answer would be: Yes. Many years > ago.). I am writing this post because James has been increasing the > price of his book (both in the printed version as well as in the PDF > one) and now the price he asks for any of them is $23.95: http://www.lulu.com/content/2812588 Does he really believe that someone will pay that amount for a PDF file > with 50 pages of texts from publicly available posts from his blog? > Jose Carlos Santos This may actually be economically defensible. Although $23.95 is probably not the price a > fully informed buyer and seller would reach > through willing negotiation, there may be > dozens of sales to be made with buyers who > are impressed by the marketing hubris that > sets a price of $23.95. What would an expert mathematician choose > as the price that maximizes profit? > What is the typical textbook price in number theory? Here's one that weighs in at almost $60 for almost 300 pages: http://www.amazon.com/Elementary-Introduction-Number-Theory-Calvin/dp/088133 8362 But is typical textbook really the best category for comparison? Isn't your listing at Lulu's as a reference book? Pricing for reference books, meant to merit institutional purchases, are at a premium to the gouging of students for required textbooks. I'm curious if anyone will answer honestly. JSH As always in matters of your curiousity, judge for yourself! === Subject: Re: Has James Harris gone mad? posting-account=n1ZfDgkAAABbCs44qOtz8dP-RkWuEBif Gecko/2008070208 Firefox/3.0.1,gzip(gfe),gzip(gfe) Hi all, Please don't answer right away to the question from the subject line. > (Of course, if you would do that, the answer would be: Yes. Many years > ago.). I am writing this post because James has been increasing the > price of his book (both in the printed version as well as in the PDF > one) and now the price he asks for any of them is $23.95: http://www.lulu.com/content/2812588 Does he really believe that someone will pay that amount for a PDF file > with 50 pages of texts from publicly available posts from his blog? > Jose Carlos Santos This may actually be economically defensible. Although $23.95 is probably not the price a > fully informed buyer and seller would reach > through willing negotiation, there may be > dozens of sales to be made with buyers who > are impressed by the marketing hubris that > sets a price of $23.95. What would an expert mathematician choose > as the price that maximizes profit? > What is the typical textbook price in number theory? Here's one that weighs in at almost $60 for almost > 300 pages: http://www.amazon.com/Elementary-Introduction-Number-Theory-Calvin/dp... But is typical textbook really the best category > for comparison? æIsn't your listing at Lulu's as a > reference book? æPricing for reference books, > meant to merit institutional purchases, are at a > premium to the gouging of students for required > textbooks. I'm curious if anyone will answer honestly. JSH As always in matters of your curiousity, judge > for yourself! > Yeah I notice you managed to pick a cheapie. That's an outlier. Here math students who I'm sure are getting ready to pay for textbooks as it gets closer to time for school to start again can finally get a chance to talk about price. How much are you paying for your math textbooks next semester? Seems the other sci.math people decided to get coy, like with the $60 U.S. example. ANYONE out there paying that amount for any of their math texts this year? James Harris === Subject: Re: Has James Harris gone mad? posting-account=OKTeIQkAAAAZk6JK1hK7-grwpoUDNy98 CLR 1.1.4322; .NET CLR 2.0.50727; InfoPath.1),gzip(gfe),gzip(gfe) > Hi all, Please don't answer right away to the question from the subject line. > (Of course, if you would do that, the answer would be: Yes. Many years > ago.). I am writing this post because James has been increasing the > price of his book (both in the printed version as well as in the PDF > one) and now the price he asks for any of them is $23.95: http://www.lulu.com/content/2812588 Does he really believe that someone will pay that amount for a PDF file > with 50 pages of texts from publicly available posts from his blog? > Jose Carlos Santos This may actually be economically defensible. Although $23.95 is probably not the price a > fully informed buyer and seller would reach > through willing negotiation, there may be > dozens of sales to be made with buyers who > are impressed by the marketing hubris that > sets a price of $23.95. What would an expert mathematician choose > as the price that maximizes profit? > What is the typical textbook price in number theory? And do they have more than 50 pages? I'm curious if anyone will answer honestly. How much is lulu charging YOU? I wonder if you'll answer honestly. JSH === Subject: Re: Congruence modulo p^2 posting-account=-_qNcQkAAACxv4j8qFSIU5L5V3znBi40 CLR 1.1.4322; .NET CLR 2.0.50727; .NET CLR 3.0.04506.30),gzip(gfe),gzip(gfe) Find x such that [(p+1)*...*(2p-1)]/[(p-1)!] is congruent to x modulo p^2 > where p is an odd prime. P.S. > I got that x is either 1 or p+1. I think it is 1 however I cannot prove it. ************************************************************** >snip Yes I thought of wislon's theorem, however this product succumbs more easily write the product as (p+1)/(p-1) * (p+2)/(p-2) *... multiply both numberator and denominator by (p-1)/(p-1) * (p-2)*(p-2) etc .. all non zero so, what is left atop, and what at the bottom? HTH JJ === Subject: Re: [OT] --- --- coding was: Peculiar quadratic equation > ax^2 + [mb^(x-1)]x + c = 0 ?(1) > > What appears on my screen several times after c = 0 is what > looks like and upside down U or intersection symbol, a small up > sidedown capital L subscript and a upside down pi exponent. What do you see ---- I see exactly what Pubkeybreaker has extracted from Google- : some spaces translated by Google into special spaces as usual, and misunderstood by Pubkeybreaker's browser as unicode FFFD : Replacement character - used to replace an incoming character whose value is unknown or unrepresentable in Unicode LOL, both use Google ! Google is unable to understand its own coding ! Only 7 spaces among the 14 spaces between the 0 and the (1) of Deep's post have been translated by Google. So that Pubkeybreaker's reply has 7 rep.chars intermixed with 7 ordinary spaces, which is exactly what *I* see : 7 space separated 'squares' (replacement chars). You see something else because your newsreader is unable to understand quoted-printable UTF8, as allready mentioned. I had to change the parameters of my own newsreader to be able to *send* this message (contains special chars not in the default font). -- Philippe Ch., mail : chephip+news@free.fr site : http://mathafou.free.fr/ (recreational mathematics) === Subject: Did I Find a Mistake in This Paper? A while back, somebody posted a link in sci.physics to the paper, Power-series solutions of the Lane-Emden equation, by Mohan & Bayaty [1], the Lane-Emden equation being f''(x)+2f'(x)/x+f^n=0. Mohan & Bayaty use Gredshteyn & Rhyzhik's [2] equations for raising a series to an integer power, to deal with the f^n term in this ODE. However, they use it to calculate solutions when n is a half integer, ie. n=m/2, even though Gredshteyn & Rhyzhik clearly state that n must be a natural number. Therefore, I think this is a mistake. This would be obvious, except for an unreferenced comment in Mohan & Bayaty's paper, that the series solution about the origin can be shown to assume the form f(x)=Sum{0 to inf}[a_k x^(2k)], which could possibly account for the necessary factor of two, in n=m/2. Since this comment is unreferenced, I have no way to see if it resolves the inconsistency. My intuition is that, even though it accounts for a missing power of powers of two by taking the square root of a series of even powers, because the algebra of square roots doesn't work like that. In other words, the equality, sqrt(Sum{0 to inf}[a_k x^(2k)]) = Sum{0 to inf}[a_k x^k], is wrong. I'd like to hear any intelligent input on whether I have found an actual mistake and therefore haven't been scooped on my original idea to solve Lane-Enden for n=m/2 in the correct way, by making a change of [1] http://adsabs.harvard.edu/abs/1980Ap%26SS..73..227M [2] http://snipurl.com/372d4 [www_amazon_com] === Subject: Math terminology for ignoratus apologies in advance if this is not an appropriate topic for this news group; I'd appreciate being referred to the correct one. My question is not a math problem per se. I am subtitling a movie with a scene from a math lecture, and I do not understand the terms used in the dialogue. I'd be glad if someone could pick some of the expressions apart in my search for more layman's terms (although I suspect that there really is no such thing). The dialogue is a professor stating: In Module 3 we noted that the collective Pythagorean angles embedded in our x to n ratio could be derived from the simple numinary A as the constant 10 and depicted thus: On screen, we see a circle divided into four equal segments, and one of these segments is in turn also divided into two. Module 3 I take to refer to a segment of the curriculum or class series Pythagorean angle .... is that merely one of the angles involved in the P theorem? Or did P do work on other kind of angles? collective Pythagorean angles ..... the sum of the 5 angles in the diagram...? embedded in our x to n ratio .... embed, yes, ratio yes; the whole thing = ?????? Again, I think that this cannot or at least should not be expressed in any other, less precise way, but still I'd appreciate some help with gettig this across to an audience. TiA, T === Subject: Re: Math terminology for ignoratus > > apologies in advance if this is not an appropriate topic for this news > group; > I'd appreciate being referred to the correct one. > > My question is not a math problem per se. I am subtitling a movie with a > scene from a math lecture, and I do not understand the terms used in the > dialogue. I'd be glad if someone could > pick some of the expressions apart in my search for more layman's terms > (although I suspect that there really is no such thing). A question: Why are you trying to improve on the dialog? When I read subtitles what I want to see is what's being _said_, not some improved or clarified version. In any case, here it seems to me that there's no point in trying to clarify the meaning of the dialogue, because it's actually just meaningless bull. That happens a lot with math-talk in movies. > The dialogue is a professor stating: > > In Module 3 we noted that the collective Pythagorean angles embedded in our > x to n ratio could be derived from the simple numinary A as the constant 10 > and depicted thus: > > On screen, we see a circle divided into four equal segments, and one of > these segments is in turn also divided into two. > > Module 3 I take to refer to a segment of the curriculum or class series > Pythagorean angle .... is that merely one of the angles involved in the P > theorem? > Or did P do work on other kind of angles? > > collective Pythagorean angles ..... the sum of the 5 angles in the > diagram...? > > embedded in our x to n ratio .... embed, yes, ratio yes; the whole thing > = ?????? > > > Again, I think that this cannot or at least should not be expressed in any > other, less precise way, but still I'd appreciate some help with gettig this > across to an audience. > > TiA, > > T -- David C. Ullrich === Subject: Re: Math terminology for ignoratus David C. Ullrich skrev i melding > A question: Why are you trying to improve on the dialog? When I read > subtitles what I want to see is what's being _said_, not some > improved or clarified version. Subtitles hardly ever improve or clarify ... The problem is that this has to be rendered in a different language, and it is hard to give a decent rendering in the target language if you don't understand what's being said in the source language. In any case, here it seems to me that there's no point in trying > to clarify the meaning of the dialogue, because it's actually just > meaningless bull. That happens a lot with math-talk in movies. Ah, that is indeed valuable information. That would render the passage above not so much dialogue as contentum, or almost contentum, where the point is not that the audience should understand, but precisely that they should not, in order to be properly awed. But still the problem arises of finding the equivalent hi-falutin' terms in the target language. There, I fear, I must turn to another group. T === Subject: Re: Math terminology for ignoratus > > > David C. Ullrich skrev i melding A question: Why are you trying to improve on the dialog? When I read > subtitles what I want to see is what's being _said_, not some > improved or clarified version. > > Subtitles hardly ever improve or clarify ... The problem is that this has to > be rendered in a different language, Oh - that's different. > and it is hard to give a decent > rendering in the target language if you don't understand what's being said > in the source language. > In any case, here it seems to me that there's no point in trying > to clarify the meaning of the dialogue, because it's actually just > meaningless bull. That happens a lot with math-talk in movies. > > Ah, that is indeed valuable information. That would render the passage above > not so much dialogue as contentum, or almost contentum, where the point is > not that the audience should understand, but precisely that they should not, > in order to be properly awed. > > But still the problem arises of finding the equivalent hi-falutin' terms in > the target language. There, I fear, I must turn to another group. > > > T -- David C. Ullrich === Subject: Re: Math terminology for ignoratus Perhaps you could tell us the source language, and the exact dialogue, and the target language, and then we can help you better. Mathematicians are often polyglots of necessity. === Subject: Re: Math terminology for ignoratus thx for the offer. riderofgiraffes skrev i melding > Perhaps you could tell us the source language, and the > exact dialogue, and the target language, and then we can > help you better. Mathematicians are often polyglots of > necessity. A long shot ...: Source: English Target: Norwegian The dialogie snippet is verbatim. TiA, T === Subject: Re: Can time dilation be computed with just the Lorentz transformation and no other assumptions? > Start with two coordinate systems in standard configuration. > In the one defined as at rest, put a clock at x=0. Each tick > of the clock marks a second in the rest frame, and the Lorentz > transform will map the tick events to a time (and place, but > that's not the question here) in the moving frame, which > tells us how fast the clock is running in the moving frame. > Obviously not. The first tick of the clock in the rest frame is > labeled as an event in a moving frame by the point x1' > To make the first step easy, might as will make the first tick > at t=t'=0. > and by a clock > at x1' that notes the time of that event as t1'. > I suggest keeping this as simple as possible to start. No need for a > second clock. That's the whole point. There is no way to avoid separate clocks. True, and it sure helps to understand how time dilation can be mutual. Now, as you seem to be aware, here you discuss the mapping of clock times and coordinate positions of one coordinate system to the times and positions on the other system. More precisely you discuss the mapping of (x1, t1) to (x1', t1') and the mapping of (x2, t2) to (x2', t2'). That is exactly what the LT is meant to be used for. Now you can write out what the relationship is for a certain case, as Dirk already did for you (perhaps you didn't like the D's?). You chose the case of x1=x2 (clock in rest in S). What you are looking for is t' = f(x, t) ; next you'll easily find (t2-t1) as function of (t2'-t1'). x=constant so that it falls out of the subtraction; it's just a few simple lines of equations! OK then, why not, I write it out for you: For x1=x2=x: t1' = gamma*(t1-v*x/c^2) t2' = gamma*(t2-v*x/c^2) => t2' - t1' = gamma*(t2 - t1) Harald === Subject: Re: Can time dilation be computed with just the Lorentz transformation and no other assumptions? <1217340962_821@sicinfo3.epfl.ch> posting-account=lBRURwoAAAB_-Q_b04pGziaymfr5yRFx Gecko/20080702 Firefox/2.0.0.16,gzip(gfe),gzip(gfe) On Jul 29, 9:16 am, harry Start with two coordinate systems in standard configuration. > In the one defined as at rest, put a clock at x=0. Each tick > of the clock marks a second in the rest frame, and the Lorentz > transform will map the tick events to a time (and place, but > that's not the question here) in the moving frame, which > tells us how fast the clock is running in the moving frame. > Obviously not. The first tick of the clock in the rest frame is > labeled as an event in a moving frame by the point x1' > To make the first step easy, might as will make the first tick > at t=t'=0. > and by a clock > at x1' that notes the time of that event as t1'. > I suggest keeping this as simple as possible to start. No need for a > second clock. That's the whole point. There is no way to avoid separate clocks. True, and it sure helps to understand how time dilation can be mutual. > Now, as you seem to be aware, here you discuss the mapping of clock times > and coordinate positions of one coordinate system to the times and positions > on the other system. More precisely you discuss the mapping of (x1, t1) to > (x1', t1') and the mapping of (x2, t2) to (x2', t2'). That is exactly what > the LT is meant to be used for. Now you can write out what the relationship > is for a certain case, as Dirk already did for you (perhaps you didn't like > the D's?). > You chose the case of x1=x2 (clock in rest in S). What you are looking for > is t' = f(x, t) ; next you'll easily find (t2-t1) as function of (t2'-t1'). > x=constant so that it falls out of the subtraction; it's just a few simple > lines of equations! OK then, why not, I write it out for you: For x1=x2=x: > t1' = gamma*(t1-v*x/c^2) > t2' = gamma*(t2-v*x/c^2) > => t2' - t1' = gamma*(t2 - t1) Harald, I'm not denying the mechanics of your computation. I'm only questioning its physical meaning in a general context. If you were to follow the same logic for a nonzero-epsilon Lorentz-equivalent transformation, equations (43) and (44) of http://www.everythingimportant.org/relativity/special.pdf, you would get a different expression, yet the only difference between the Lorentz transformation and a nonzero-epsilon Lorentz-equivalent transformation is the way that clocks are synchronized. So your formula is a mere expression of a particular clock synchronization scheme. Shubee === Subject: Re: Can time dilation be computed with just the Lorentz transformation and no other assumptions? > On Jul 29, 9:16 am, harry Start with two coordinate systems in standard configuration. > In the one defined as at rest, put a clock at x=0. Each tick > of the clock marks a second in the rest frame, and the Lorentz > transform will map the tick events to a time (and place, but > that's not the question here) in the moving frame, which > tells us how fast the clock is running in the moving frame. > Obviously not. The first tick of the clock in the rest frame is > labeled as an event in a moving frame by the point x1' > To make the first step easy, might as will make the first tick > at t=t'=0. > and by a clock > at x1' that notes the time of that event as t1'. > I suggest keeping this as simple as possible to start. No need for a > second clock. > That's the whole point. There is no way to avoid separate clocks. > True, and it sure helps to understand how time dilation can be mutual. > Now, as you seem to be aware, here you discuss the mapping of clock times > and coordinate positions of one coordinate system to the times and > positions > on the other system. More precisely you discuss the mapping of (x1, t1) > to > (x1', t1') and the mapping of (x2, t2) to (x2', t2'). That is exactly > what > the LT is meant to be used for. Now you can write out what the > relationship > is for a certain case, as Dirk already did for you (perhaps you didn't > like > the D's?). > You chose the case of x1=x2 (clock in rest in S). What you are looking > for > is t' = f(x, t) ; next you'll easily find (t2-t1) as function of > (t2'-t1'). > x=constant so that it falls out of the subtraction; it's just a few > simple > lines of equations! OK then, why not, I write it out for you: > For x1=x2=x: > t1' = gamma*(t1-v*x/c^2) > t2' = gamma*(t2-v*x/c^2) > => t2' - t1' = gamma*(t2 - t1) Harald, I'm not denying the mechanics of your computation. I'm only > questioning its physical meaning in a general context. The above relationship between dt' and dt is what is called time dilation. There is no other physical meaning implied. > If you were to > follow the same logic for a nonzero-epsilon Lorentz-equivalent > transformation, equations (43) and (44) of > http://www.everythingimportant.org/relativity/special.pdf, you would > get a different expression, yet the only difference between the > Lorentz transformation and a nonzero-epsilon Lorentz-equivalent > transformation is the way that clocks are synchronized. So your > formula is a mere expression of a particular clock synchronization > scheme. No, it's more: it's a prediction of what results when we don't ascribe a preferred frame but simply define our frame to be in rest; until 1905 it was not clear that the factor should be gamma. Note that all this is (or was) rather well known. Harald === Subject: Re: Can time dilation be computed with just the Lorentz transformation and no other assumptions? posting-account=lBRURwoAAAB_-Q_b04pGziaymfr5yRFx Gecko/20080702 Firefox/2.0.0.16,gzip(gfe),gzip(gfe) > On Jul 29, 9:16 am, harry Start with two coordinate systems in standard configuration. > In the one defined as at rest, put a clock at x=0. Each tick > of the clock marks a second in the rest frame, and the Lorentz > transform will map the tick events to a time (and place, but > that's not the question here) in the moving frame, which > tells us how fast the clock is running in the moving frame. > Obviously not. The first tick of the clock in the rest frame is > labeled as an event in a moving frame by the point x1' > To make the first step easy, might as will make the first tick > at t=t'=0. > and by a clock > at x1' that notes the time of that event as t1'. > I suggest keeping this as simple as possible to start. No need for a > second clock. That's the whole point. There is no way to avoid separate clocks. True, and it sure helps to understand how time dilation can be mutual. > Now, as you seem to be aware, here you discuss the mapping of clock times > and coordinate positions of one coordinate system to the times and positions > on the other system. More precisely you discuss the mapping of (x1, t1) to > (x1', t1') and the mapping of (x2, t2) to (x2', t2'). That is exactly what > the LT is meant to be used for. Now you can write out what the relationship > is for a certain case, as Dirk already did for you (perhaps you didn't like > the D's?). > You chose the case of x1=x2 (clock in rest in S). What you are looking for > is t' = f(x, t) ; next you'll easily find (t2-t1) as function of (t2'-t1'). > x=constant so that it falls out of the subtraction; it's just a few simple > lines of equations! OK then, why not, I write it out for you: For x1=x2=x: > t1' = gamma*(t1-v*x/c^2) > t2' = gamma*(t2-v*x/c^2) > => t2' - t1' = gamma*(t2 - t1) Harald, I'm not denying the mechanics of your computation. I'm only > questioning its physical meaning in a general context. If you were to > follow the same logic for a nonzero-epsilon Lorentz-equivalent > transformation, equations (43) and (44) ofhttp://www.everythingimportant.org/relativity/special.pdf, you would > get a different expression, yet the only difference between the > Lorentz transformation and a nonzero-epsilon Lorentz-equivalent > transformation is the way that clocks are synchronized. So your > formula is a mere expression of a particular clock synchronization > scheme. Shubee If the Lorentz transform can be reduced to the Galilean transform by resetting clocks, rescaling distance measures and fiddling with clock rates, then don't we need to add another postulate? See page 11 of http://www.everythingimportant.org/relativity/special.pdf and equations (48) to (58). Shubee === Subject: Re: Can time dilation be computed with just the Lorentz transformation and no other assumptions? > On Jul 29, 9:16 am, harry [...] > If the Lorentz transform can be reduced to the Galilean transform by > resetting clocks, rescaling distance measures and fiddling with clock > rates, then don't we need to add another postulate? See page 11 of > http://www.everythingimportant.org/relativity/special.pdf and > equations (48) to (58). ?! I think you confuse inputs and outputs. For example, to solve an equation with two variables, if one is known as input then you can get the other out. Now, you appear to say that if we can do the inverse, by first determining the other we can find the one as output, and therefore don't we need to add one more input? No that doesn't make sense to me. Harald === Subject: Re: Can time dilation be computed with just the Lorentz transformation and no other assumptions? > Start with two coordinate systems in standard configuration. > In the one defined as at rest, put a clock at x=0. Each tick > of the clock marks a second in the rest frame, and the Lorentz > transform will map the tick events to a time (and place, but > that's not the question here) in the moving frame, which > tells us how fast the clock is running in the moving frame. > Obviously not. The first tick of the clock in the rest frame is > labeled as an event in a moving frame by the point x1' > To make the first step easy, might as will make the first tick > at t=t'=0. > and by a clock > at x1' that notes the time of that event as t1'. > I suggest keeping this as simple as possible to start. No need for a > second clock. > > That's the whole point. There is no way to avoid separate clocks. Since Dirk and I already showed you a way, that argument is a non-starter. > The second tick of > the clock in the rest frame is labeled by the event at a different > point x2' in the moving frame and the time of that event is recorded > by the clock at the point x2'. > You're just confusing yourself by adding complexity. > > You are obviously being confused by childish simplicity. > Your whole argument is based on the concept of the simultaneity of > spatially separated clocks, which is fallacious reasoning. > Instead of making up your own multi-clock example that doesn't help you > to understand, try looking at the simple example. A clock marks time in > the rest frame; the Lorentz transform maps its ticks to times (and > places) in the moving frame. > > There is no simpler example. If you don't want to know how to derive time dilation from the Lorentz transform and nothing more, you probably shouldn't have asked the question. -- --Bryan === Subject: Re: Can time dilation be computed with just the Lorentz transformation Bryan Olson says... >If you don't want to know how to derive time dilation from the Lorentz >transform and nothing more, you probably shouldn't have asked the question. I don't exactly get what Shubee is going on about, but whether you can derive time dilation from the Lorentz transform and nothing more depends on what you mean by time dilation, and what it means to assume the Lorentz transform. If you are asking the physical question about what happens when two clocks are separated and reunited again, then you can't answer that question without making assumptions about the nature of clocks. If you are asking the purely mathematical question of how the the time separation between two spacetime points transforms under a Lorentz transformation, then of course all you need to know is the Lorentz transform. -- Daryl McCullough Ithaca, NY === Subject: Re: Can time dilation be computed with just the Lorentz transformation > Bryan Olson says... > > If you don't want to know how to derive time dilation from the Lorentz > transform and nothing more, you probably shouldn't have asked the question. > > I don't exactly get what Shubee is going on about, but whether > you can derive time dilation from the Lorentz transform and > nothing more depends on what you mean by time dilation, and > what it means to assume the Lorentz transform. I don't think it makes sense to say you have the Lorentz transform, but do not know what the variables mean or what is being transformed. > If you are asking > the physical question about what happens when two clocks are > separated and reunited again, then you can't answer that > question without making assumptions about the nature of clocks. > > If you are asking the purely mathematical question of > how the the time separation between two spacetime points > transforms under a Lorentz transformation, then of course > all you need to know is the Lorentz transform. I think the most straightforward interpretation is basic SR factor-of-gamma time dilation. That doesn't seem to be the part to which Shubee objected. -- --Bryan === Subject: Re: Can time dilation be computed with just the Lorentz transformation Bryan Olson says... > Bryan Olson says... > > If you don't want to know how to derive time dilation from the Lorentz > transform and nothing more, you probably shouldn't have asked the question. > > I don't exactly get what Shubee is going on about, but whether > you can derive time dilation from the Lorentz transform and > nothing more depends on what you mean by time dilation, and > what it means to assume the Lorentz transform. I don't think it makes sense to say you have the Lorentz transform, but >do not know what the variables mean or what is being transformed. The Lorentz transformation is a transformation between two different coordinate systems. > If you are asking > the physical question about what happens when two clocks are > separated and reunited again, then you can't answer that > question without making assumptions about the nature of clocks. > > If you are asking the purely mathematical question of > how the the time separation between two spacetime points > transforms under a Lorentz transformation, then of course > all you need to know is the Lorentz transform. I think the most straightforward interpretation is basic SR >factor-of-gamma time dilation. That doesn't seem to be the part to which >Shubee objected. I'm not sure what part he is objecting to, but I thought it was the physical prediction (the twin paradox) that if you move a clock away at relativistic speed and then back again it will show less elapsed time than a clock that traveled inertially throughout. That prediction, as a prediction about what happens to an actual physical clock, doesn't follow from the Lorentz transformations alone. You also have to make assumptions about the behavior of clocks under inertial and accelerated motion. -- Daryl McCullough Ithaca, NY === Subject: Re: Can time dilation be computed with just the Lorentz transformation > Bryan Olson says... > Bryan Olson says... > > If you don't want to know how to derive time dilation from the Lorentz > transform and nothing more, you probably shouldn't have asked the question. > > I don't exactly get what Shubee is going on about, but whether > you can derive time dilation from the Lorentz transform and > nothing more depends on what you mean by time dilation, and > what it means to assume the Lorentz transform. I don't think it makes sense to say you have the Lorentz transform, but >do not know what the variables mean or what is being transformed. > > The Lorentz transformation is a transformation between two different > coordinate systems. > > If you are asking > the physical question about what happens when two clocks are > separated and reunited again, then you can't answer that > question without making assumptions about the nature of clocks. > > If you are asking the purely mathematical question of > how the the time separation between two spacetime points > transforms under a Lorentz transformation, then of course > all you need to know is the Lorentz transform. I think the most straightforward interpretation is basic SR >factor-of-gamma time dilation. That doesn't seem to be the part to which >Shubee objected. > > I'm not sure what part he is objecting to, but I thought it was > the physical prediction (the twin paradox) that if you move a > clock away at relativistic speed and then back again it will > show less elapsed time than a clock that traveled inertially > throughout. That prediction, as a prediction about what happens > to an actual physical clock, doesn't follow from the Lorentz > transformations alone. You also have to make assumptions about > the behavior of clocks under inertial and accelerated motion. No I do not. Uncle Al describe a chain of circumstance demonstrating this anomalous behavior of moving clocks in which the frames of all actors do not accelerate. _______________________BEGIN _______________________ The ratio by which the twins have aged at the end when they are back together again is the same in all reference frames: ratio = sqrt(t^2 - x^2 - y^2 - z^2)/t (with units of c=1) Inertial frames with relative *velocities* pursue different paths through spacetime. No clock anomaly is apparent in any of them until clocks are compared (by all being local when you do it, initial calibration then experiment). The situation is NOT symmetric. Past acceleration is irrelevant to the running of present clocks but not to the mixture of space and time in the reference frame that said clocks measure. You cannot synchronize clocks except by having them local. If they are local at the start, you can tell who was naughty thereafter without measuring acceleration. Given three identical clocks that are off (a state of not running, or of not even having been fabricated) and zeroed. Each clock has/will have a very short toggle jiggger switch sticking out. We load them (or their parts, or ore and a smelter and a machine shop) in individual spaceships and set up the experiment. CLOCK 1: That's our clock. It sits stationary in our inertial reference frame with a little jigger sticking out. Touch the jigger and the off state becomes on or the on state becomes off. Clock 1 is off. Or we can build it from parts just before we need it, and in the off state, zeroed. CLOCK 2: In a spaceship traveling at 0.999c relative to our inertial frame of reference. Clock 2 is off. It was built after all acceleration ceased, and set to zero. It skims past Clock 1 (our clock), the jiggers touch, both Clocks 1 and 2 are now on and locally synchronized by touching. Elapsed time accumulates in each one. The situation is NOT symmetric! CLOCK 3: In a spaceship traveling at 0.999c relative to our inertial frame of reference, but 180 degrees counter in direction to Clock 2. Clock 3 is zeroed and off. It was built after all acceleration ceased, and set to zero. Some arbitrary time after Clocks 1 and 2 synchronize and turn on by touching, Clocks 2 and 3 brush past each other, touching jiggers. Clock 2 is now off, Clock 3 is now on. Write down the elapsed time in now off Clock 2, then smash the clock with a sledgehammer. Or melt it down, or toss it over the side. The spaceship with Clock 3 is returning back over the path taken by the spaceship with Clock 2. CLOCK 1: That's our clock. It sits stationary in our inertial reference frame with a little jigger sticking out. Clock 3 rushes past, jiggers touch. Clocks 3 and 1 are now off. All clocks are off. No clock has accelerated while on or even while existing. Write down elapsed times, smash each clock with a sledgehammer. Or melt them down, or toss them. BOTTOM LINE: Get all three slips of paper together... Accelerate as you need. Or send all the results to all three folks by radio and never decelerate. All clocks have been smashed, melted, tossed. Their elapsed times were written down. The numbers on the papers won't change when you accelerate or broadcast the data. Finally.... compare elapsed times. Elapsed time #2+#3 does not equal #1, the local stationary reference frame summation. The sum of #2+#3 elasped time is only about 4.5% that than of #1's accumulated elapsed time. You have the Twin Paradox (Triplets) without any running clock having been accelerated - or having even existed during acceleration up or down. ________________________END_________________________ -- Michael Press === Subject: Re: Can time dilation be computed with just the Lorentz transformation Michael Press says... > I'm not sure what part he is objecting to, but I thought it was > the physical prediction (the twin paradox) that if you move a > clock away at relativistic speed and then back again it will > show less elapsed time than a clock that traveled inertially > throughout. That prediction, as a prediction about what happens > to an actual physical clock, doesn't follow from the Lorentz > transformations alone. You also have to make assumptions about > the behavior of clocks under inertial and accelerated motion. No I do not. Uncle Al describe a chain of circumstance >demonstrating this anomalous behavior of moving clocks >in which the frames of all actors do not accelerate. Sure. But then what prediction does it make about clocks that *do* accelerate? Under what circumstances is the formula elapsed time on clock = Integral along the path of the clock of square-root(1-(v/c)^2) dt valid? -- Daryl McCullough Ithaca, NY === Subject: Re: Can time dilation be computed with just the Lorentz transformation > Michael Press says... > I'm not sure what part he is objecting to, but I thought it was > the physical prediction (the twin paradox) that if you move a > clock away at relativistic speed and then back again it will > show less elapsed time than a clock that traveled inertially > throughout. That prediction, as a prediction about what happens > to an actual physical clock, doesn't follow from the Lorentz > transformations alone. You also have to make assumptions about > the behavior of clocks under inertial and accelerated motion. No I do not. Uncle Al describe a chain of circumstance >demonstrating this anomalous behavior of moving clocks >in which the frames of all actors do not accelerate. > > Sure. But then what prediction does it make about clocks > that *do* accelerate? Under what circumstances is the > formula > > elapsed time on clock = Integral along the path of the clock > of square-root(1-(v/c)^2) dt > > valid? That is as may be. I spoke to your claim: That prediction, as a prediction about what happens to an actual physical clock, doesn't follow from the Lorentz transformations alone. You also have to make assumptions about the behavior of clocks under inertial and accelerated motion I showed you a rendition of the twins paradox that does not, and need not, make assumptions about the behavior of clocks under accelerated motion. -- Michael Press === Subject: Re: Can time dilation be computed with just the Lorentz transformation posting-account=wigfZgkAAACDgITarXffzxJygX81YRSs elapsed time on clock = Integral along the path of the clock > of square-root(1-(v/c)^2) dt > valid? Of course that isn't valid. The person using that formula is correcting for a changing light path delay and mixing spatial and temporal components. Doppler Shift for Sound and Light http://www.mathpages.com/rr/s2-04/2-04.htm << It is to be found rather in the fact of his recognition that the four-dimensional space-time continuum of the theory of relativity, in its most essential formal properties, shows a pronounced relationship to the three-dimensional continuum of Euclidean geometrical space. 1 In order to give due prominence to this relationship, however, we must replace the usual time co-ordinate t by an imaginary magnitude sqrt(-1) ct proportional to it. Under these conditions, the natural laws satisfying the demands of the (special) theory of relativity assume mathematical forms, in which the time co-ordinate plays exactly the same r.99le as the three space co-ordinates. > http://www.bartleby.com/173/17.html A general Lorentz tranform passes that imaginary to tensor expressions which will preserve relevant symetries. A standard Lorentz transform drops the imaginary and yields nonsense. The Lorentz transformation http://farside.ph.utexas.edu/teaching/em/lectures/node109.html Sue... > -- > Daryl McCullough > Ithaca, NY === Subject: Re: Can time dilation be computed with just the Lorentz transformation posting-account=nPH_PQkAAACneDKT6RXopPWArC2We4Rq AppleWebKit/525.13 (KHTML, like Gecko) Version/3.1 Safari/525.13,gzip(gfe),gzip(gfe) On Jul 30, 3:39æpm, stevendaryl3...@yahoo.com (Daryl McCullough) > Bryan Olson says... > Bryan Olson says... > If you don't want to know how to derive time dilation from the Lorentz > transform and nothing more, you probably shouldn't have asked the question. > I don't exactly get what Shubee is going on about, æbut whether > you can derive time dilation from the Lorentz transform and > nothing more depends on what you mean by time dilation, and > what it means to assume the Lorentz transform. I don't think it makes sense to say you have the Lorentz transform, but >do not know what the variables mean or what is being transformed. The Lorentz transformation is a transformation between two different > coordinate systems. > If you are asking > the physical question about what happens when two clocks are > separated and reunited again, then you can't answer that > question without making assumptions about the nature of clocks. > If you are asking the purely mathematical question of > how the the time separation between two spacetime points > transforms under a Lorentz transformation, then of course > all you need to know is the Lorentz transform. I think the most straightforward interpretation is basic SR >factor-of-gamma time dilation. That doesn't seem to be the part to which >Shubee objected. I'm not sure what part he is objecting to, but I thought it was > the physical prediction (the twin paradox) that if you move a > clock away at relativistic speed and then back again it will > show less elapsed time than a clock that traveled inertially > throughout. That prediction, as a prediction about what happens > to an actual physical clock, doesn't follow from the Lorentz > transformations alone. You also have to make assumptions about > the behavior of clocks under inertial and accelerated motion. > Well, I believe you have to make some assumptions beyond the Lorentz transformation, but I'm not sure it's necessary to include the assumptions you cite. It actually follows directly from the invariant interval and what the length of a worldline is, I think. The invariant interval tells you what the infinitesimal for a timelike worldline segment looks like, and integrating that between two events A and B shows that the integral along that path is *maximized* for the straightest path worldlines which one has the shortest proper time. The only assumption you need to make is that the worldlines are drawn from an inertial reference frame, and that is determined by the fact that momentum conservation holds (or Newton's first law, if you like). This is what protects you from drawing the worldlines as the traveling twin sees it, because the traveling twin cannot draw the worldlines from an inertial reference frame without breaking momentum conservation. PD === Subject: Re: Can time dilation be computed with just the Lorentz PD says... > I'm not sure what part he is objecting to, but I thought it was > the physical prediction (the twin paradox) that if you move a > clock away at relativistic speed and then back again it will > show less elapsed time than a clock that traveled inertially > throughout. That prediction, as a prediction about what happens > to an actual physical clock, doesn't follow from the Lorentz > transformations alone. You also have to make assumptions about > the behavior of clocks under inertial and accelerated motion. Well, I believe you have to make some assumptions beyond the Lorentz >transformation, but I'm not sure it's necessary to include the >assumptions you cite. It actually follows directly from the invariant interval and what the >length of a worldline is, I think. The invariant interval tells you >what the infinitesimal for a timelike worldline segment looks like, >and integrating that between two events A and B shows that the >integral along that path is *maximized* for the straightest path >between A and B. >worldlines which one has the shortest proper time. >The only assumption you need to make is that the worldlines >are drawn from an inertial reference frame, and that is >determined by the fact that momentum conservation holds >(or Newton's first law, if you like). This is what >protects you from drawing the worldlines as the traveling twin sees >it, because the traveling twin cannot draw the worldlines from an >inertial reference frame without breaking momentum conservation. That's only half of the argument. Nothing that you've said implies anything about what happens when a clock goes on a high-velocity journey, turns around and returns. You haven't connected the mathematical quantity proper time with the time on a clock. That's where the assumptions that I gave come in. You have a mathematical quantity, Tau = Integral of square-root(1-(v/c)^2) dt. You have a physically measurable quantity, elapsed time = the time on the clock at the end of the journey minus the time on the clock at the start. What assumptions are needed in order to argue that the physical quantity is equal to the mathematical quantity? I believe that sufficient assumptions are: (1) If a clock is approximately at rest in a standard inertial coordinate system between times t1 and t2, then the elapsed time on the clock will be approximately equal to t2 - t1. (2) If as measured in a standard inertial coordinate system, a clock is accelerating between times t1 and t2, then the elapsed time on the clock will be at most t2 - t1. (3) The path of a clock is continuous, and its velocity in any finite interval of time has at most finitely many discontinuities. (4) Standard inertial coordinate systems are related by the Lorentz transformations (as well as translations and rotations). These 4 assumptions allow us to compute elapsed time on a clock by breaking the clock's path into segments that are approximately inertial, then using the Lorentz transforms to compute the elapsed time during each segment. In the limit as the number of segments goes to infinity, the elapsed time approaches the integral Integral of square-root(1-(v/c)^2) dt -- Daryl McCullough Ithaca, NY === Subject: Re: Can time dilation be computed with just the Lorentz posting-account=nPH_PQkAAACneDKT6RXopPWArC2We4Rq AppleWebKit/525.13 (KHTML, like Gecko) Version/3.1 Safari/525.13,gzip(gfe),gzip(gfe) On Jul 30, 6:49æpm, stevendaryl3...@yahoo.com (Daryl McCullough) > PD says... > I'm not sure what part he is objecting to, but I thought it was > the physical prediction (the twin paradox) that if you move a > clock away at relativistic speed and then back again it will > show less elapsed time than a clock that traveled inertially > throughout. That prediction, as a prediction about what happens > to an actual physical clock, doesn't follow from the Lorentz > transformations alone. You also have to make assumptions about > the behavior of clocks under inertial and accelerated motion. Well, I believe you have to make some assumptions beyond the Lorentz >transformation, but I'm not sure it's necessary to include the >assumptions you cite. It actually follows directly from the invariant interval and what the >length of a worldline is, I think. The invariant interval tells you >what the infinitesimal for a timelike worldline segment looks like, >and integrating that between two events A and B shows that the >integral along that path is *maximized* for the straightest path >between A and B. >worldlines which one has the shortest proper time. >The only assumption you need to make is that the worldlines >are drawn from an inertial reference frame, and that is >determined by the fact that momentum conservation holds >(or Newton's first law, if you like). This is what >protects you from drawing the worldlines as the traveling twin sees >it, because the traveling twin cannot draw the worldlines from an >inertial reference frame without breaking momentum conservation. That's only half of the argument. > Nothing that you've said implies anything about what happens > when a clock goes on a high-velocity journey, turns around and > returns. You haven't connected the mathematical quantity proper > time with the time on a clock. That's where the assumptions > that I gave come in. You have a mathematical quantity, > Tau = Integral of square-root(1-(v/c)^2) dt. > You have a physically measurable quantity, > elapsed time = the time on the clock at the end > of the journey minus the time on the clock at the > start. What assumptions are needed in order to argue > that the physical quantity is equal to the mathematical > quantity? Let me think about this. It seems like a very odd question. The proper time is the immediately local time where local means in the vicinity of the clock, and as many have said around here, time is what a clock measures. I believe that sufficient assumptions are: (1) If a clock is approximately at rest in a standard inertial > coordinate system between times t1 and t2, then the elapsed > time on the clock will be approximately equal to t2 - t1. Yes. And again, this is essentially the *definition* of time: Local time is what a local clock measures. > (2) If as measured in a standard inertial coordinate system, > a clock is accelerating between times t1 and t2, > then the elapsed time on the clock will be at most t2 - t1. Let me think about this one too. > (3) The path of a clock is continuous, and its velocity in > any finite interval of time has at most finitely many > discontinuities. Yes. > (4) Standard inertial coordinate systems are related by > the Lorentz transformations (as well as translations and > rotations). By definition of the Lorentz transform, yes. These 4 assumptions allow us to compute elapsed time on > a clock by breaking the clock's path into segments that > are approximately inertial, then using the Lorentz transforms > to compute the elapsed time during each segment. In the limit > as the number of segments goes to infinity, the elapsed time > approaches the integral Integral of square-root(1-(v/c)^2) dt Yes, that was certainly Einstein's pitch in the original paper. -- > Daryl McCullough > Ithaca, NY === Subject: Re: Can time dilation be computed with just the Lorentz PD says... >Let me think about this. It seems like a very odd question. The proper >time is the immediately local time where local means in the vicinity >of the clock, and as many have said around here, time is what a clock >measures. You have two different quantities here, which you are declaring to be equal, but you either have to *assume* that they are equal, or *prove* that they are equal using other assumptions. The fact that you use the word time to refer to both doesn't imply that they are equal. You set up a coordinate system. A coordinate system allows you to assign a 4-tuple (x,y,z,t) to each event (or point) in spacetime. So you can use the coordinate system in order to compute the time between two events e1 and e2: If e1 has coordinates (x1,y1,z1,t1) and e2 has coordinates (x2, y2, z2, t2), then the time between events e1 and e2 is just t2 - t1. That's one notion of time. A second notion of time is, as you say, what a clock measures. A clock is initially at some event e1, where the time on the clock is tau1. Then the clock travels to event e2, where the time on the clock is tau2. So you have a different notion of the time between events: tau2 - tau1. So the question is: what is the relationship between these two different notions of time? The first, and most obvious thing to say is that *if* the coordinate system is a standard inertial coordinate system, *and* the clock is at rest in that coordinate system (that is, if x1 = x2, y1 = y2, z1 = z2) *then* the two notions of time are equal: tau2 - tau1 = t2 - t1. But what about the case in which the clock is not at rest? What about the case in which the clock has a variable velocity? > I believe that sufficient assumptions are: > (1) If a clock is approximately at rest in a standard inertial > coordinate system between times t1 and t2, then the elapsed > time on the clock will be approximately equal to t2 - t1. Yes. And again, this is essentially the *definition* of time: Local >time is what a local clock measures. There are two *different* notions of time. What a clock measures is one notion, and coordinate time is a different notion. They aren't the same. The proper time formula relates two *different* notions of time. > (4) Standard inertial coordinate systems are related by > the Lorentz transformations (as well as translations and > rotations). By definition of the Lorentz transform, yes. No, it is *not* by definition of the Lorentz transform. It is an *assumption*. The definition of a standard inertial coordinate system (as I defined it, anyway) is a coordinate system such that (1) If a standard clock is at rest between times t1 and t2 then the elapsed time on the clock is given by t2 - t1, independent of the location of the clock. (2) If a standard measuring stick is at rest, and the endpoints of the stick are at (x1,y1,z1) and (x2,y2,z2), then (x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2 = 1. (3) If an object is undergoing force-free motion, then its path as a function of time satisfies dx/dt = constant, dy/dt = constant, dz/dt = constant. Just based on assumptions 1-3, we can't conclude that two standard inertial coordinate systems are related by the Lorentz transforms. It is an empirical fact that they are related by the Lorentz transforms. It's not true by definition. It doesn't follow from the definition that there *are* any standard inertial coordinate systems at all (as a matter of fact, when you take gravity into account, there aren't any standard inertial coordinate systems). The assumption made by Newton was that two standard inertial coordinate systems were related by Galilean transforms. The assumption made by Einstein was that, in addition to 1-3, we also have (4) light always has the same speed c in all directions. That assumption allows you to deduce that standard inertial coordinate systems are related by Lorentz transforms. -- Daryl McCullough Ithaca, NY === Subject: Re: Can time dilation be computed with just the Lorentz posting-account=nPH_PQkAAACneDKT6RXopPWArC2We4Rq AppleWebKit/525.13 (KHTML, like Gecko) Version/3.1 Safari/525.13,gzip(gfe),gzip(gfe) On Jul 31, 9:19 am, stevendaryl3...@yahoo.com (Daryl McCullough) > PD says... Let me think about this. It seems like a very odd question. The proper >time is the immediately local time where local means in the vicinity >of the clock, and as many have said around here, time is what a clock >measures. You have two different quantities here, which you are declaring to > be equal, but you either have to *assume* that they are equal, or > *prove* that they are equal using other assumptions. The fact that > you use the word time to refer to both doesn't imply that they > are equal. You set up a coordinate system. A coordinate system allows you > to assign a 4-tuple (x,y,z,t) to each event (or point) in > spacetime. So you can use the coordinate system in order to > compute the time between two events e1 and e2: If e1 has > coordinates (x1,y1,z1,t1) and e2 has coordinates (x2, y2, z2, t2), > then the time between events e1 and e2 is just t2 - t1. That's > one notion of time. A second notion of time is, as you say, what a clock measures. > A clock is initially at some event e1, where the time on the > clock is tau1. Then the clock travels to event e2, where the > time on the clock is tau2. So you have a different notion of > the time between events: tau2 - tau1. The previous two paragraphs are what bother me. To me, and maybe this is unjustified, when you say you set up a coordinate system, you are *saying* that those coordinates are what they are by virtue of an implicit measurement -- a ruler along a spatial axis or a clock along a time axis. Equivalently, as Einstein would say it, a coordinate system is an imaginary grid of rods and clocks BY WHICH the location and time of events are labeled. Frankly, I don't know how to set up a coordinate system that doesn't imply that. So the question is: what is the relationship between these > two different notions of time? The first, and most obvious thing to say is that *if* > the coordinate system is a standard inertial coordinate > system, *and* the clock is at rest in that coordinate > system (that is, if x1 = x2, y1 = y2, z1 = z2) *then* > the two notions of time are equal: tau2 - tau1 = t2 - t1. > But what about the case in which the clock is not at rest? > What about the case in which the clock has a variable > velocity? > I believe that sufficient assumptions are: > (1) If a clock is approximately at rest in a standard inertial > coordinate system between times t1 and t2, then the elapsed > time on the clock will be approximately equal to t2 - t1. Yes. And again, this is essentially the *definition* of time: Local >time is what a local clock measures. There are two *different* notions of time. What a clock measures > is one notion, and coordinate time is a different notion. They > aren't the same. The proper time formula relates two *different* > notions of time. > (4) Standard inertial coordinate systems are related by > the Lorentz transformations (as well as translations and > rotations). By definition of the Lorentz transform, yes. No, it is *not* by definition of the Lorentz transform. > It is an *assumption*. The definition of a standard > inertial coordinate system (as I defined it, anyway) > is a coordinate system such that (1) If a standard clock is at rest between times t1 and t2 > then the elapsed time on the clock is given by t2 - t1, > independent of the location of the clock. > (2) If a standard measuring stick is at rest, and the endpoints > of the stick are at (x1,y1,z1) and (x2,y2,z2), then > (x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2 = 1. > (3) If an object is undergoing force-free motion, then > its path as a function of time satisfies dx/dt = constant, > dy/dt = constant, dz/dt = constant. Yes, ok. Just based on assumptions 1-3, we can't conclude that > two standard inertial coordinate systems are related > by the Lorentz transforms. It is an empirical fact > that they are related by the Lorentz transforms. Yes, OK. I agree with that. > It's > not true by definition. It doesn't follow > from the definition that there *are* any standard > inertial coordinate systems at all (as a matter of > fact, when you take gravity into account, there aren't > any standard inertial coordinate systems). The assumption made by Newton was that two standard > inertial coordinate systems were related by Galilean > transforms. The assumption made by Einstein was that, > in addition to 1-3, we also have (4) light always has the > same speed c in all directions. That assumption allows > you to deduce that standard inertial coordinate systems > are related by Lorentz transforms. -- > Daryl McCullough > Ithaca, NY === Subject: Re: Can time dilation be computed with just the Lorentz PD says... On Jul 31, 9:19 am, stevendaryl3...@yahoo.com (Daryl McCullough) > You set up a coordinate system. A coordinate system allows you > to assign a 4-tuple (x,y,z,t) to each event (or point) in > spacetime. So you can use the coordinate system in order to > compute the time between two events e1 and e2: If e1 has > coordinates (x1,y1,z1,t1) and e2 has coordinates (x2, y2, z2, t2), > then the time between events e1 and e2 is just t2 - t1. That's > one notion of time. > A second notion of time is, as you say, what a clock measures. > A clock is initially at some event e1, where the time on the > clock is tau1. Then the clock travels to event e2, where the > time on the clock is tau2. So you have a different notion of > the time between events: tau2 - tau1. The previous two paragraphs are what bother me. To me, and maybe this >is unjustified, when you say you set up a coordinate system, you are >*saying* that those coordinates are what they are by virtue of an >implicit measurement -- a ruler along a spatial axis or a clock along >a time axis. As I said, a coordinate system is just a means of assigning 4-tuples to events in spacetime. There is no requirement that coordinates directly correspond to measurements, only that the mapping from spacetime to R^4 be continuous and invertible. For example, in spherical coordinates (r,theta,phi,t), the coordinates theta and phi don't correspond to distances as measured by a ruler (although they can be computed from measurements using rulers). >Equivalently, as Einstein would say it, a coordinate >system is an imaginary grid of rods and clocks BY WHICH the location >and time of events are labeled. Frankly, I don't know how to set up a >coordinate system that doesn't imply that. What you are talking about is what I was calling standard inertial coordinate systems. In such a coordinate system, the coordinates x,y,z correspond to distances from the origin as measured with a standard ruler, and t corresponds to time as measured with a standard clock. The problem is that t is not uniquely determined by clock measurements; you have to have some way to pick the zero for t at each spatial point. You can arbitrarily pick some event to be at time t=0, but that's only for one spot. To determine t=0 for other locations, you have to have some clock synchronization procedure. A standard inertial coordinate system is *not* the only possible way to set up a coordinate system. As a matter of fact, once you take gravity into account, there *is* no such thing as a standard inertial coordinate system. You can *locally* label events so that things look like a standard inertial coordinate system, but you can't get these local coordinate systems to fit together into a global inertial coordinate system. -- Daryl McCullough Ithaca, NY === Subject: Re: Can time dilation be computed with just the Lorentz transformation posting-account=lBRURwoAAAB_-Q_b04pGziaymfr5yRFx Gecko/20080702 Firefox/2.0.0.16,gzip(gfe),gzip(gfe) On Jul 30, 3:39 pm, stevendaryl3...@yahoo.com (Daryl McCullough) > It was > the physical prediction (the twin paradox) that if you move a > clock away at relativistic speed and then back again it will > show less elapsed time than a clock that traveled inertially > throughout. That prediction, as a prediction about what happens > to an actual physical clock, doesn't follow from the Lorentz > transformations alone. Exactly. > You also have to make assumptions about > the behavior of clocks under inertial and accelerated motion. Shubee === Subject: Re: Can time dilation be computed with just the Lorentz transformation > On Jul 30, 3:39 pm, stevendaryl3...@yahoo.com (Daryl McCullough) > > It was > the physical prediction (the twin paradox) that if you move a > clock away at relativistic speed and then back again it will > show less elapsed time than a clock that traveled inertially > throughout. That prediction, as a prediction about what happens > to an actual physical clock, doesn't follow from the Lorentz > transformations alone. > > Exactly. Bull. Time dilation (i.e. the subject of this thread) follows from the Lorentz Transformation alone anytime one of the LT variables is interpreted as time. > > You also have to make assumptions about > the behavior of clocks under inertial and accelerated motion. > > Shubee > === Subject: Re: Can time dilation be computed with just the Lorentz transformation jem says... >Bull. Time dilation (i.e. the subject of this thread) follows from >the Lorentz Transformation alone anytime one of the LT variables is >interpreted as time. As I said in another post, it depends on exactly what you mean by time dilation. There are two different notions of time: (1) elapsed time on a clock, and (2) coordinate time. You have to make additional assumptions in order to relate these two notions of time. The physically meaningful question is: If two standard clocks are initially together, and one is carried away at a relativistic speed and then brought back, while the other travels inertially throughout, then which will show the most elapsed time? The answer to that question does not follow from the Lorentz transforms without additional assumptions. -- Daryl McCullough Ithaca, NY === Subject: Re: Can time dilation be computed with just the Lorentz transformation and no other assumptions? > I have been arguing that my theory removes from special relativity > everything that is confused, unnecessary and not amenable to > experimental verification. Where by arguing you mean proclaiming your own greatness and calling others ugly names. -- --Bryan === Subject: Re: Can time dilation be computed with just the Lorentz transformation and no other assumptions? posting-account=lBRURwoAAAB_-Q_b04pGziaymfr5yRFx Gecko/20080702 Firefox/2.0.0.16,gzip(gfe),gzip(gfe) > I have been arguing that my theory removes from special relativity > everything that is confused, unnecessary and not amenable to > experimental verification. Where by arguing you mean proclaiming your own greatness and calling > others ugly names. -- > --Bryan You have every right to feel insulted. You made a point and I responded by saying that it was obviously false. And I supported my claim with an argument: Shubee === Subject: Re: Can time dilation be computed with just the Lorentz transformation and no other assumptions? > I have been arguing that my theory removes from special relativity > everything that is confused, unnecessary and not amenable to > experimental verification. > Where by arguing you mean proclaiming your own greatness and calling > others ugly names. > > You have every right to feel insulted. You made a point and I > responded by saying that it was obviously false. And I supported my > claim with an argument: And when I and others demonstrated that your arguments were bunk, you did exactly what I said you did. Proclaiming your own greatness: Within the next thousand years, it will be universally agreed that Shubertian physics is correct And calling others ugly names: -- --Bryan === Subject: Re: Can time dilation be computed with just the Lorentz transformation and no other assumptions? <3wxjk.15042$cW3.11886@nlpi064.nbdc.sbc.com> posting-account=lBRURwoAAAB_-Q_b04pGziaymfr5yRFx Gecko/20080702 Firefox/2.0.0.16,gzip(gfe),gzip(gfe) > You have every right to feel insulted. You made a point and I > responded by saying that it was obviously false. And I supported my > claim with an argument: And when I and others demonstrated that your arguments were bunk, I'm not aware of anyone even touching my argument, which requires understanding Xi_2, the simplest universe imaginable: but I now admit that I have no way to justify fiddling with clock rates and distance measures. > you did exactly what I said you did. Proclaiming your own greatness: Within the next thousand years, it will be universally agreed > that Shubertian physics is correct It seems like a reasonable response to a clueless critic when he insists on a prediction after demonstrating repeatedly that he has no understanding of what I'm saying. > And calling others ugly names: > I seriously doubt that you know my personal history with even one of those senseless detractors. Aren't you speaking out of total ignorance? Perhaps you presuppose that each of those chimpanzees must be human and therefore have some sort of reasonableness to them? Shubee === Subject: Re: Can time dilation be computed with just the Lorentz transformation and no other assumptions? > > You have every right to feel insulted. You made a point and I > responded by saying that it was obviously false. And I supported my > claim with an argument: > And when I and others demonstrated that your arguments were bunk, > > I'm not aware of anyone even touching my argument, which requires > understanding Xi_2, the simplest universe imaginable: You asked about the Lorentz transform, not about the universe in your head. > but I now admit that I have no way to justify fiddling with clock > rates and distance measures. And you're a little over a hundred years behind the times on that. > you did exactly what I said you did. > Proclaiming your own greatness: > Within the next thousand years, it will be universally agreed > that Shubertian physics is correct > > It seems like a reasonable response to a clueless critic when he > insists on a prediction after demonstrating repeatedly that he has no > understanding of what I'm saying. Are we raining on your parade? > And calling others ugly names: > > I seriously doubt that you know my personal history with even one of > those senseless detractors. Aren't you speaking out of total > ignorance? Perhaps you presuppose that each of those chimpanzees must > be human and therefore have some sort of reasonableness to them? You poor thing. -- --Bryan === Subject: Re: Can time dilation be computed with just the Lorentz transformation and no other assumptions? posting-account=lBRURwoAAAB_-Q_b04pGziaymfr5yRFx Gecko/20080702 Firefox/2.0.0.16,gzip(gfe),gzip(gfe) > You have every right to feel insulted. You made a point and I > responded by saying that it was obviously false. And I supported my > claim with an argument: > And when I and others demonstrated that your arguments were bunk, I'm not aware of anyone even touching my argument, which requires > understanding Xi_2, the simplest universe imaginable: You asked about the Lorentz transform, not about the universe in your head. The Lorentz transform is connected to a notion of spacetime. It can't be said that the Lorentz transform implies anything about time dilation unless it's true for all spacetimes for which the LT applies. If you don't understand the concept of counterexample, then you're in the wrong newsgroup. > but I now admit that I have no way to justify fiddling with clock > rates and distance measures. And you're a little over a hundred years behind the times on that. Why can't I adjust clock rates and distance measures? What if doing so leaves the fundamental spacetime structure constant unchanged? > you did exactly what I said you did. > Proclaiming your own greatness: > Within the next thousand years, it will be universally agreed > that Shubertian physics is correct It seems like a reasonable response to a clueless critic when he > insists on a prediction after demonstrating repeatedly that he has no > understanding of what I'm saying. Are we raining on your parade? You are not in my parade or your own. You have joined the chimpanzee charade. > And calling others ugly names: > I seriously doubt that you know my personal history with even one of > those senseless detractors. Aren't you speaking out of total > ignorance? Perhaps you presuppose that each of those chimpanzees must > be human and therefore have some sort of reasonableness to them? You poor thing. First you were crying about those poor chimpanzees. Now you're lamenting my unfortunate experiences with angry, smelly, -throwing chimpanzees. Make up your mind. Shubee === Subject: Re: Can time dilation be computed with just the Lorentz transformation and no other assumptions? > You have every right to feel insulted. You made a point and I > responded by saying that it was obviously false. And I supported my > claim with an argument: > And when I and others demonstrated that your arguments were bunk, > I'm not aware of anyone even touching my argument, which requires > understanding Xi_2, the simplest universe imaginable: > You asked about the Lorentz transform, not about the universe in your head. > > The Lorentz transform is connected to a notion of spacetime. It can't > be said that the Lorentz transform implies anything about time > dilation unless it's true for all spacetimes for which the LT applies. > If you don't understand the concept of counterexample, then you're in > the wrong newsgroup. Dirk showed you a general solution, using the derivative. If that doesn't work in the universe in your head, that's not shortcoming in the derivation. > but I now admit that I have no way to justify fiddling with clock > rates and distance measures. > And you're a little over a hundred years behind the times on that. > > Why can't I adjust clock rates and distance measures? What if doing so > leaves the fundamental spacetime structure constant unchanged? In in your own head you can do whatever you like. > you did exactly what I said you did. > Proclaiming your own greatness: > Within the next thousand years, it will be universally agreed > that Shubertian physics is correct > It seems like a reasonable response to a clueless critic when he > insists on a prediction after demonstrating repeatedly that he has no > understanding of what I'm saying. > Are we raining on your parade? > > You are not in my parade or your own. You have joined the chimpanzee > charade. > > And calling others ugly names: > I seriously doubt that you know my personal history with even one of > those senseless detractors. Aren't you speaking out of total > ignorance? Perhaps you presuppose that each of those chimpanzees must > be human and therefore have some sort of reasonableness to them? > You poor thing. > > First you were crying about those poor chimpanzees. No, that was about you, Shubee. When you call people chimpanzees, it is not they that look bad. That's a clue for you, whether you take it or not. > Now you're > lamenting my unfortunate experiences with angry, smelly, -throwing > chimpanzees. Make up your mind. No, no, Shubee. It's not your fights with your detractors that makes you pitiful; it's that you desperately crave attention and they're the only ones that even listen. -- --Bryan === Subject: Re: Can time dilation be computed with just the Lorentz transformation and no other assumptions? > You have every right to feel insulted. You made a point and I > responded by saying that it was obviously false. And I supported my > claim with an argument: > And when I and others demonstrated that your arguments were bunk, > I'm not aware of anyone even touching my argument, which requires > understanding Xi_2, the simplest universe imaginable: > You asked about the Lorentz transform, not about the universe in your head. > > The Lorentz transform is connected to a notion of spacetime. It can't > be said that the Lorentz transform implies anything about time > dilation unless it's true for all spacetimes for which the LT applies. > If you don't understand the concept of counterexample, then you're in > the wrong newsgroup. Nothing new there, Shoobo, just one more in a long string of demonstrations that you don't have a clue what you're taking about. The Lorentz Transformation, AS INCORPORATED IN SR (i.e. according to SR's definition of time and SR's associations between variables and measurements), necessarily implies time dilation. Period. And what ever gave you the silly idea that the LT (i.e. the mathematical transformation) is connected to the notion of spacetime. Do you seriously think the only things that can be LT related are space and time? If you understand the concept of a counterexample, just look at how LET uses the LT. Hint: not to relate space and time. > > but I now admit that I have no way to justify fiddling with clock > rates and distance measures. > And you're a little over a hundred years behind the times on that. > > Why can't I adjust clock rates and distance measures? You just finished saying I now admit that I have no way to justify fiddling with clock rates and distance measures. So suppose you tell us why you can't adjust them, Shooby. ? What if doing so > leaves the fundamental spacetime structure constant unchanged? Stop babbling, Shooby. > > you did exactly what I said you did. > Proclaiming your own greatness: > Within the next thousand years, it will be universally agreed > that Shubertian physics is correct > It seems like a reasonable response to a clueless critic when he > insists on a prediction after demonstrating repeatedly that he has no > understanding of what I'm saying. > Are we raining on your parade? > > You are not in my parade or your own. You have joined the chimpanzee > charade. > > And calling others ugly names: > I seriously doubt that you know my personal history with even one of > those senseless detractors. Aren't you speaking out of total > ignorance? Perhaps you presuppose that each of those chimpanzees must > be human and therefore have some sort of reasonableness to them? > You poor thing. > > First you were crying about those poor chimpanzees. Now you're > lamenting my unfortunate experiences with angry, smelly, -throwing > chimpanzees. Make up your mind. > > Shubee > === Subject: Re: Can time dilation be computed with just the Lorentz transformation and no other assumptions? posting-account=lBRURwoAAAB_-Q_b04pGziaymfr5yRFx Gecko/20080702 Firefox/2.0.0.16,gzip(gfe),gzip(gfe) The Lorentz transform is connected to a notion of spacetime. It can't > be said that the Lorentz transform implies anything about time > dilation unless it's true for all spacetimes for which the LT applies. > If you don't understand the concept of counterexample, then you're in > the wrong newsgroup. The Lorentz Transformation, AS INCORPORATED IN SR (i.e. according to > SR's definition of time and SR's associations between variables and > measurements), necessarily implies time dilation. Period. Why do you believe that my original question is about SR? > And what ever gave you the silly idea that the LT (i.e. the > mathematical transformation) is connected to the notion of > spacetime. All you have to know to grasp the context of my question is that general spacetime constructs that obey the Lorentz transformation exist. Shubee http://www.everythingimportant.org/relativity/special.pdf === Subject: Re: Can time dilation be computed with just the Lorentz transformation and no other assumptions? > The Lorentz transform is connected to a notion of spacetime. It can't > be said that the Lorentz transform implies anything about time > dilation unless it's true for all spacetimes for which the LT applies. > If you don't understand the concept of counterexample, then you're in > the wrong newsgroup. > The Lorentz Transformation, AS INCORPORATED IN SR (i.e. according to > SR's definition of time and SR's associations between variables and > measurements), necessarily implies time dilation. Period. > > Why do you believe that my original question is about SR? > > And what ever gave you the silly idea that the LT (i.e. the > mathematical transformation) is connected to the notion of > spacetime. > > All you have to know to grasp the context of my question is that > general spacetime constructs that obey the Lorentz transformation > exist. OK, Shoobo, so produce one that's not SR. > > Shubee > http://www.everythingimportant.org/relativity/special.pdf > > > === Subject: Re: Can time dilation be computed with just the Lorentz transformation and no other assumptions? <0DEjk.1317$wY7.1031@newsfe01.iad> posting-account=lBRURwoAAAB_-Q_b04pGziaymfr5yRFx Gecko/20080702 Firefox/2.0.0.16,gzip(gfe),gzip(gfe) > The Lorentz transform is connected to a notion of spacetime. It can't > be said that the Lorentz transform implies anything about time > dilation unless it's true for all spacetimes for which the LT applies. > If you don't understand the concept of counterexample, then you're in > the wrong newsgroup. > The Lorentz Transformation, AS INCORPORATED IN SR (i.e. according to > SR's definition of time and SR's associations between variables and > measurements), necessarily implies time dilation. Period. Why do you believe that my original question is about SR? > And what ever gave you the silly idea that the LT (i.e. the > mathematical transformation) is connected to the notion of > spacetime. All you have to know to grasp the context of my question is that > general spacetime constructs that obey the Lorentz transformation > exist. OK, Shoobo, so produce one that's not SR. That's easy: http://www.everythingimportant.org/relativity/special.pdf Note that a spacetime structure exists yet no clock-carrying travelers exist and I haven't created any physics to allow clock-carrying machines or travelers to jump from one inertial frame to another. Is there any relevant meaning to time dilation in this general context? Shubee === Subject: Re: Can time dilation be computed with just the Lorentz transformation and no other assumptions? > The Lorentz transform is connected to a notion of spacetime. It can't > be said that the Lorentz transform implies anything about time > dilation unless it's true for all spacetimes for which the LT applies. > If you don't understand the concept of counterexample, then you're in > the wrong newsgroup. > The Lorentz Transformation, AS INCORPORATED IN SR (i.e. according to > SR's definition of time and SR's associations between variables and > measurements), necessarily implies time dilation. Period. > Why do you believe that my original question is about SR? > And what ever gave you the silly idea that the LT (i.e. the > mathematical transformation) is connected to the notion of > spacetime. > All you have to know to grasp the context of my question is that > general spacetime constructs that obey the Lorentz transformation > exist. > OK, Shoobo, so produce one that's not SR. > > That's easy: http://www.everythingimportant.org/relativity/special.pdf What a surprise - Shooby's universal response to any question he can't answer. Have you incorporated the kitchen sink into that document yet, Shoobo? > > Note that a spacetime structure exists yet no clock-carrying travelers > exist and I haven't created any physics to allow clock-carrying > machines or travelers to jump from one inertial frame to another. Is > there any relevant meaning to time dilation in this general context? If time is one of the variables in the LT (regardless what LT spacetime that's not SR you think is in your silly pdf) then time dilation is necessarily implied. By definition. > Shubee > === Subject: Re: Can time dilation be computed with just the Lorentz transformation and no other assumptions? posting-account=lBRURwoAAAB_-Q_b04pGziaymfr5yRFx Gecko/20080702 Firefox/2.0.0.16,gzip(gfe),gzip(gfe) > All you have to know to grasp the context of my question is that > general spacetime constructs that obey the Lorentz transformation > exist. > OK, Shoobo, so produce one that's not SR. That's easy: http://www.everythingimportant.org/relativity/special.pdf Note that a spacetime structure exists yet no clock-carrying travelers > exist and I haven't created any physics to allow clock-carrying > machines or travelers to jump from one inertial frame to another. Is > there any relevant meaning to time dilation in this general context? If time is one of the variables in the LT (regardless what LT > spacetime that's not SR you think is in your silly pdf) then time > dilation is necessarily implied. By definition. It's not clear to me. Suppose I build a clock in one frame of reference and allow it to be carried into all inertial frames of reference. Note that in my derivation, each inertial frame has its own distance measure and definition of time. What if I suppose that instantaneousness exists and therefore a global simultaneity also exists. Why can't I declare that the transported clock never suffers time dilation and that the mathematically different time rates merely express a difference in clock calibration in a Newtonian universe? For the Lorentz-equivalent transformation equations that conveniently harmonize with instantaneousness, see my recently revised paper. http://www.everythingimportant.org/relativity/special.pdf Shubee === Subject: Re: Can time dilation be computed with just the Lorentz transformation and no other assumptions? > All you have to know to grasp the context of my question is that > general spacetime constructs that obey the Lorentz transformation > exist. > OK, Shoobo, so produce one that's not SR. > That's easy: http://www.everythingimportant.org/relativity/special.pdf > Note that a spacetime structure exists yet no clock-carrying travelers > exist and I haven't created any physics to allow clock-carrying > machines or travelers to jump from one inertial frame to another. Is > there any relevant meaning to time dilation in this general context? > If time is one of the variables in the LT (regardless what LT > spacetime that's not SR you think is in your silly pdf) then time > dilation is necessarily implied. By definition. > > It's not clear to me. If the amount of time between two events differs between reference frames then there's time dilation. Simple as that. > Suppose I build a clock in one frame of > reference and allow it to be carried into all inertial frames of > reference. Note that in my derivation, each inertial frame has its own > distance measure and definition of time. What if I suppose that > instantaneousness exists and therefore a global simultaneity also > exists. So each Inertial frame has its own time, but it's the same in every frame? What's that suppose to mean, Shooby? > Why can't I declare that the transported clock never suffers > time dilation and that the mathematically different time rates merely > express a difference in clock calibration in a Newtonian universe? That's what Lorentz did when he created LET, Shooby. There's no time dilation in LET, but neither does the LT relate time (or space) between reference frames in LET, i.e. this isn't an example of a spacetime construct that obeys the Lorentz transformation. So what's your point? > For the Lorentz-equivalent transformation equations that conveniently > harmonize with instantaneousness, see my recently revised paper. > > http://www.everythingimportant.org/relativity/special.pdf I've been through enough of your recently revised versions, Shooby, to know not to spend time hunting for what in all likelihood isn't even there. Copy and paste your equations here, if you have them. > > Shubee > > === Subject: Re: Can time dilation be computed with just the Lorentz transformation and no other assumptions? <1vZjk.5043$LF2.571@newsfe09.iad> posting-account=lBRURwoAAAB_-Q_b04pGziaymfr5yRFx Gecko/20080702 Firefox/2.0.0.16,gzip(gfe),gzip(gfe) There's no time dilation in LET, I'm sure Tom Roberts would disagree with you. Roberts is on record saying that LET is physically indistinguishable from special relativity. Shubee === Subject: Re: Can time dilation be computed with just the Lorentz transformation and no other assumptions? > There's no time dilation in LET, > > I'm sure Tom Roberts would disagree with you. Roberts is on record > saying that LET is physically indistinguishable from special > relativity. LET and SR are *experimentally* indistinguishable, Shooby, meaning that both SR and LET make identical predictions of the measurements of standard clocks and rulers. However, experimental indistinguishability doesn't conflict with no time dilation in LET, since time is defined differently in SR (i.e. by the measurement of standard clocks) than it is in LET (i.e. by the measurement of standard clocks at rest in the ether). Time dilation is the effect where the amount of time between events differs in different reference frames, which means that in LET, where the time between every pair of events is the same in every reference frame, there's no time dilation. === Subject: Re: Can time dilation be computed with just the Lorentz transformation jem says... >LET and SR are *experimentally* indistinguishable, Shooby, meaning >that both SR and LET make identical predictions of the measurements of >standard clocks and rulers. However, experimental >indistinguishability doesn't conflict with no time dilation in LET, >since time is defined differently in SR (i.e. by the measurement of >standard clocks) than it is in LET (i.e. by the measurement of >standard clocks at rest in the ether). I guess the meaning of time dilation is ambiguous. If you put it in the form of an empirical question: If two standard clocks are initially together, and one is taken far away at near light speed, and then brought back, while the other travels inertially throughout, then which clock will show the most elapsed time? then LET and SR will give the same answers. -- Daryl McCullough Ithaca, NY === Subject: Re: Can time dilation be computed with just the Lorentz transformation and no other assumptions? posting-account=lBRURwoAAAB_-Q_b04pGziaymfr5yRFx Gecko/20080702 Firefox/2.0.0.16,gzip(gfe),gzip(gfe) There's no time dilation in LET, I'm sure Tom Roberts would disagree with you. Roberts is on record > saying that LET is physically indistinguishable from special > relativity. Shubee Even the editors of Wikipedia agree with Roberts: Introducing the effects of length contraction and time dilation in a preferred frame of reference leads to the Lorentz transformation and therefore it is not possible to distinguish between LET and SR by experiment. - Wikipedia. http://en.wikipedia.org/wiki/Lorentz_ether_theory === Subject: Re: Can time dilation be computed with just the Lorentz transformation and no other assumptions? > There's no time dilation in LET, > I'm sure Tom Roberts would disagree with you. Roberts is on record > saying that LET is physically indistinguishable from special > relativity. > Shubee > > Even the editors of Wikipedia agree with Roberts: > > Introducing the effects of length contraction and time dilation in a > preferred frame of reference leads to the Lorentz transformation and > therefore it is not possible to distinguish between LET and SR by > experiment. - Wikipedia. > http://en.wikipedia.org/wiki/Lorentz_ether_theory > Like so much for popular consumption literature about this subject, that loosely worded sentence abuses the terminology. A correct statement would begin Introducing the effects of ruler contraction and clock slowing .... === Subject: Re: Can time dilation be computed with just the Lorentz transformation and no other assumptions? <1vZjk.5043$LF2.571@newsfe09.iad> posting-account=lBRURwoAAAB_-Q_b04pGziaymfr5yRFx Gecko/20080702 Firefox/2.0.0.16,gzip(gfe),gzip(gfe) > All you have to know to grasp the context of my question is that > general spacetime constructs that obey the Lorentz transformation > exist. > OK, Shoobo, so produce one that's not SR. > That's easy: http://www.everythingimportant.org/relativity/special.pdf > Note that a spacetime structure exists yet no clock-carrying travelers > exist and I haven't created any physics to allow clock-carrying > machines or travelers to jump from one inertial frame to another. Is > there any relevant meaning to time dilation in this general context? > If time is one of the variables in the LT (regardless what LT > spacetime that's not SR you think is in your silly pdf) then time > dilation is necessarily implied. By definition. It's not clear to me. If the amount of time between two events differs between reference > frames then there's time dilation. Simple as that. Look for the new thread that I started in your relativity reader. I believe that the riddle is clear enough and that you're not even close to answering it. Shubee === Subject: Re: Can time dilation be computed with just the Lorentz transformation and no other assumptions? posting-account=jPnQ2goAAAA461y3QD0lbyw0oKeThma1 AppleWebKit/525.18 (KHTML, like Gecko) Version/3.1.2 Safari/525.20.1,gzip(gfe),gzip(gfe) y'know, I didn't read the foregoing part, but this sounds like a joke. anyway. what if acceleration is simply a fastest time path, related to the brachistochrone ... which has the property, that if you start at any point on the path, it takes the same amount of time to be there; you can use the same exact sequence of are we There, yet? I think, not!... but we agree, that hte stay-at-home twin didn't experience any hard Gs, except fo the normal G of a type-F planet! > æ æ æThe twin so-called paradox is easily resolved this way. A force > is required to break the symmetry between the traveling twin and the > stay on earth twin. Basically, the stay at home twin has an extra > reset in his clock during the turn around. Thus, the time dilation > effect isn't real for him. The stay on earth twin had no such force > applied to him. So the time dilation effect is real to him. There is > no effect of force on the stay at home twin. The effect of the force > on the traveling twin is real, but different than one would expect > from Newtonian physics. thus: this reminds of Bucky's definition of a point, a subvisible something, which is minimally a tetrahedron; sort of like a star, which is just a subvisible sun, til you apply spectral analysis. > You need three mutual touching faces to represent a point on a surface. Give the three > faces a different color. Where two different colors touch each other > you have a line (or edge). Where three different colors touch you have a point. thus: people should read the Bible to become literate, and to find out what Shakeepeare meant by that weird poem (KJVersion .-) why was the 900' Jesus depicted reading a new testament, or the 800# guerilla? or was it just a Book. well, Genesis is more of a bibiliography. I read that one SF author said that the ylived so long, but they were really dumb; like, there minds were pratically varves ... or some thing. now, what you were saying about salvation ... do you have a prefered application? > ... or just the three open balls of radius 1 and centers on the > boundary of D^2, 120 degrees apart. === Subject: Re: Can time dilation be computed with just the Lorentz transformation and no other assumptions? posting-account=jPnQ2goAAAA461y3QD0lbyw0oKeThma1 AppleWebKit/525.18 (KHTML, like Gecko) Version/3.1.2 Safari/525.20.1,gzip(gfe),gzip(gfe) should have said, not the normal, but the formal gravity of a type-F planet, since that is nominally an unsolved problem ... other than Newton's fabulous algebraiztion of Kepler's orbital constraints: let there be dark! === Subject: Re: Can time dilation be computed with just the Lorentz transformation and no other assumptions? posting-account=jPnQ2goAAAA461y3QD0lbyw0oKeThma1 AppleWebKit/525.18 (KHTML, like Gecko) Version/3.1.2 Safari/525.20.1,gzip(gfe),gzip(gfe) to repeat the lesson: Schroedinger named the twins, Einstien killed the twins, Heisenberg ate one of the twins, but he asked for its name, first; the other twin is engaged in quantum teleportation, one way, no return, unless there's a REALLY BIG MIRROR OUT THERE AND IT'S FOCUSED ON ME, the Fast Twin, the Good, and my name is ... oops. --the Cheenysfear, 60/60/24/7/51. Week 52: Dick Hitchhikes to Darfur! === Subject: Re: Can time dilation be computed with just the Lorentz transformation and no other assumptions? posting-account=jPnQ2goAAAA461y3QD0lbyw0oKeThma1 AppleWebKit/525.18 (KHTML, like Gecko) Version/3.1.2 Safari/525.20.1,gzip(gfe),gzip(gfe) people should read the Bible to become literate, and to find out what Shakeepeare meant by that weird poem (KJVersion .-) why was the 900' Jesus depicted reading a new testament, or the 800# guerilla? or was it just a Book. well, Genesis is more of a bibiliography. I read that one SF author said that the ylived so long, but they were really dumb; like, there minds were pratically varves ... or some thing. now, what you were saying about salvation ... do you have a prefered application? > So what were you expecting? If you're looking for salvation, try === Subject: norm posting-account=1vQ5xwoAAADIDQUVBSMlqBb6NsFD508y SLCC1; .NET CLR 2.0.50727; .NET CLR 3.0.04506; InfoPath.2),gzip(gfe),gzip(gfe) let x,y are in R^n and ||.|| represent standard Eucledian norm. Then is it true that ||x+y||^2>=||x||^2-2||x||||y|| ? === Subject: Re: norm posting-account=Cbgh4AoAAAAr0dt1RqLOClWCyUWii2fU Gecko/20080702 Firefox/2.0.0.16,gzip(gfe),gzip(gfe) > let x,y are in R^n and ||.|| represent standard Eucledian norm. Then > is it true that ||x+y||^2>=||x||^2-2||x||||y|| ? This holds for *any* norm: ||x+y|| >= ||x|| - ||y|| by triangle inequality hence ||x+y||^2 >= (||x|| - ||y||)^2 = ||x||^2 - 2 ||x|| ||y|| + ||y||^2 >= ||x||^2 - 2 ||x|| ||y|| === Subject: Re: norm > let x,y are in R^n and ||.|| represent standard Eucledian norm. Then > is it true that ||x+y||^2>=||x||^2-2||x||||y|| ? Suffices to prove in R^2 === Subject: Re: Infinite 3d lattice > Of course, the forces F and G are STILL lower bounds (in terms of > magnitude), because the grid extends upwards and downwards to infinity and > there are contributions from there, too (although I suspect that because > the forces vary as 1/r^2, the corresponding series might converge). > I agree; but I haven't explored this yet. Do you agree that the problem can be completely reduced to 2D by considering a > point mass moving on the plane where the 2D infinite lattice L_{2D} = {(k,l): > k,l in Z} lies? I think that considering point masses on the plane is sufficient. What say ye? If you agree, we can start examining this case which is BY FAR simpler than > the 3D lattice. FWIW, I think the 2D case is doable. Consider the infinite lattice L = {(k,l): k,lin Z}, and x placed at (x,0): | | | ----F------D------E---- | | | ----C------A-x----B---- | |/ | ----I------G------H---- | | | It is clear that it suffices to consider only the upper half infinite lattice, for the bottom half part is completely symmetric, hence the total force on x will be 2F, where F is the force on x by the upper part. Ommitting the factors m*M*G_k in all subsequent calculations and denoting: f_k(x) = 1/(k^2 + x^2)/sqrt(1 + k^2/x^2), a bit of trigonometry shows that the k-th row above the row C/A/B (including row-0), exerts a total force of: f_k(x) + sum(f_k(n+x) - f_k(n-x), n=1..infinity) on x. Hence, the total force F exerted by the entire upper infinite lattice, will be: F(x) = sum(f_k(x) + sum(f_k(n+x) - f_k(n-x), n=1..infinity), k=0..infinity) (1) (The entire 2D infinite lattice then exerts a force 2F - f_0(x) - sum(f_0(n+x) - f_0(n-x)) on x). ***************** Let's proceed with the important step now. How to CRASH Maple (V rel 4 or 9). Enter: > f:=(k,x)->1/(k^2+x^2)/sqrt(1+k^2/x^2); > FR:=(k,x)->f(k,x)+sum(f(k,x+n)-f(k,x-n),n=1..infinity); > FT:=x->sum(F(k,x),k=0..infinity); > with(plots): > plot(FT(x),x=0..1/2) or, if you don't want to wait an eon, enter: > evalf(FT(0.25)) I have no idea if (1) converges for 0 < x < 1/2, but it looks like it does from some sparse numerical evidence. I could not force Maple to do anything useful with this expression. Perhaps some of the big guns here can find some tricks to simplify (1) so we can see what happens. -- I.N. Galidakis === Subject: Re: Infinite 3d lattice <1217239178.311206@athprx03> <1217342290.489954@athprx04> posting-account=oTDIagkAAACTxHurtPutBWvNQS8ZCNO9 Gecko/20080702 Firefox/2.0.0.16,gzip(gfe),gzip(gfe) > Of course, the forces F and G are STILL lower bounds (in terms of > magnitude), because the grid extends upwards and downwards to infinity and > there are contributions from there, too (although I suspect that because > the forces vary as 1/r^2, the corresponding series might converge). > I agree; but I haven't explored this yet. Do you agree that the problem can be completely reduced to 2D by considering a > point mass moving on the plane where the 2D infinite lattice L_{2D} = {(k,l): > k,l in Z} lies? I think that considering point masses on the plane is sufficient. What say ye? If you agree, we can start examining this case which is BY FAR simpler than > the 3D lattice. FWIW, I think the 2D case is doable. Consider the infinite lattice L = {(k,l): k,lin Z}, and x placed at (x,0): | | | > ----F------D------E---- > | | | > ----C------A-x----B---- > | |/ | > ----I------G------H---- > | | | It is clear that it suffices to consider only the upper half infinite lattice, > for the bottom half part is completely symmetric, hence the total force on x > will be 2F, where F is the force on x by the upper part. Ommitting the factors m*M*G_k in all subsequent calculations and denoting: f_k(x) = 1/(k^2 + x^2)/sqrt(1 + k^2/x^2), a bit of trigonometry shows that the > k-th row above the row C/A/B (including row-0), exerts a total force of: f_k(x) + sum(f_k(n+x) - f_k(n-x), n=1..infinity) on x. Hence, the total force F exerted by the entire upper infinite lattice, will be: F(x) = sum(f_k(x) + sum(f_k(n+x) - f_k(n-x), n=1..infinity), k=0..infinity) (1) (The entire 2D infinite lattice then exerts a force 2F - f_0(x) - sum(f_0(n+x) - > f_0(n-x)) on x). > ***************** Let's proceed with the important step now. How to CRASH Maple (V rel 4 or 9). > Enter: f:=(k,x)->1/(k^2+x^2)/sqrt(1+k^2/x^2); > FR:=(k,x)->f(k,x)+sum(f(k,x+n)-f(k,x-n),n=1..infinity); > FT:=x->sum(F(k,x),k=0..infinity); > with(plots): > plot(FT(x),x=0..1/2) > How about only going halfway to ininity? :) I'm out of town for a couple of days. I'll check back in on this on my return. === Subject: Re: Infinite 3d lattice posting-account=oTDIagkAAACTxHurtPutBWvNQS8ZCNO9 Gecko/20080702 Firefox/2.0.0.16,gzip(gfe),gzip(gfe) > would be /equal/ in all directions. Olber's paradox doesn't imply an / > infinitely/ bright sky; just a /uniformly/ bright sky. Right. Well, I think the strongest immediately correct statement > which can be made is fails to converge, which we note does not > necessarily mean blows up. Roughly speaking, concentric shells of > constant thickness and increasing radius include mass proportional to > r^2, so their 1/r^2 contributions to the force at a point fail to go > to zero. But... well, maybe not. Right. The naive spherical shell argument seems to forget that, contrary to what one might expect, at EVERY point WITHIN a spherical shell of arbitrary thickness and with constant density, the force of gravity exerted by that shell is 0. See, e.g.: http://hyperphysics.phy-astr.gsu.edu/Hbase/mechanics/sphshell2.html#wtls So the problem is rather: does a set of point sources sufficiently / approximate/ a spherical shell so as too have the property that within the (approximated) spherical shell something similar holds? My intuition says: yes. But intuition is not proof. === Subject: Re: Bounds on moment of density function posting-account=nwUM9QoAAACqBHn_2_rf8-VaiLMkDZmP SLCC1; .NET CLR 2.0.50727; .NET CLR 3.0.04506),gzip(gfe),gzip(gfe) Is there a bound for the moments a density function? >All I can assume is that I have only the first and secondmomentof >the distribution. Can I somehow bound the higher ordermoment? There are some lower bounds. æBut not upper bounds. æ > In fact, it may be that no higher moments exist, so > the even moments would be infinite. I found a paper A Lower Bound for a ProbabilityMomentof any >Absolutely Continuous Distribution with Finite Variance, however, I >am not quite sure what function B is. -- > This address is for information only. æI do not claim that these views > are those of the Statistics Department or of Purdue University. > Herman Rubin, Department of Statistics, Purdue University > hru...@stat.purdue.edu æ æ æ æ Phone: (765)494-6054 æ FAX: (765)494-0558 Apart from the bound given by the Markov Inequality, is there other bounds (requiring only the information of the first two moments)? === Subject: Re: card(P(N)) = card(N), final (finished) posting-account=F3H0JAgAAADcYVukktnHx7hFG5stjWse .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) The argument below is crystal clear already. Bye bye. -LV > Second attempt. I'll try two equivalent formulations, in the hope > that, together, they will complement the eventual ambiguities. ----- 1st formulation (i) æ P(A) ~ { 0, 1, 2, ..., 2^n - 1}, AneN Let's simply call P(A) the set on the RHS of (i), keeping in mind that > the argument is going to hold due to this bijection and the > transitivity of the '~' relation. (ii) æP(A) c= N, AneN (iii) P(N) is infinite (trivial) We pass from P(A) to P(N) by transfinite induction, then: (iv) æP(N) ~ N, from (ii) and (iii) QDE. (1) ----- 2nd formulation Same comments as above. (i) æ card(P(A)) = 2^n, AneN (ii) æcard(P(A)) <= card(N), AneN > (iii) card(P(N)) >= oo (meaning: 'oo' as lower bound) (iv) æcard(P(N)) = card(N) QDE. (2) ----- Hmm, correct? The basic idea behind this argument is that P(A) is > always a subset of N in the finite case. Then, by transfinite > induction, while P(N) becomes infinite, its cardinality's upper bound > remains N's cardinality. (These informal expositions get me into more > troubles than they help, but I have to run the risk: no pain, no > gain!) Good luck. Julio status: OPEN > revision: 1 > sender: Julio Di Egidio (aka LudovicoVan) > sender-email: ju...@diegidio.aleph.name (del 'aleph.') > copyright: 2008 (on behalf of) sci.math, sci.logic > All rights reserved. > card(P(N)) = card(N), an elementary proof (and some fuzzy notation > here and there). --- Preliminaries: 'P(A)' stands for the powerset of set 'A'. '~' stands for the existence of a bijection, which will be, in this > context, equivalent to stating that two sets are equinumerous. > Infinity of the set 'N' (the set of natural/ordinal numbers) is taken > to be defined as usual, by the existence of bijections between N and > its (infinite) subsets. No other kind of infinities are going to be > assumed: the only tool I have used is transfinite induction over the > naturals/ordinals, with omega as the limit ordinal. 'oo' stands for omega. --- Finite case: card(A) = n <=>def > æ æ A ~ { 0, 1, 2, .., n-1 } card(P(A)) = 2^n <= æ æ P(A) ~ { 1, 2, 4, ..., 2^n-1 } --- Infinite case (transf. induction, notation a bit fuzzy here): (1) n=oo => card(A) = card(N) = oo <=>def > æ æ A ~ N := { 0, 1, 2, ..., n }, n->oo (2) n=oo => card(P(A)) = card(P(N)) = 2^oo <= æ æ P(A) ~ P(N) ~ 2^N := { 1, 2, 4, ..., 2^n }, n->oo --- Theorem: card(P(N)) = card(N) --- Proof: (i) card(P(N)) = card(N) <=> P(N) ~ N, by def. of 'card' (ii) card(P(N)) = card(N) <=> 2^N ~ N, by (2) and trans. of '~' (iii) A bijection 2^N ~ N trivially exists: 2^N ~ N > -------- > æ 1 ~ 0 > æ 2 ~ 1 > æ 4 ~ 2 > ... Follows the thesis. QDE. It's THEN easy to get interesting lemmas. We get that omega is an > absolute upper bound and all infinities are equivalent. The diagonal > argument then assumes another shape, which will be the subject of > another post of mine, unless someone finds a flaw in the above quite > elementary argument. > Julio status: OPEN > sender: Julio Di Egidio (aka LudovicoVan) > sender-email: ju...@diegidio.aleph.name (del 'aleph.') > copyright: 2008 (on behalf of) sci.math, sci.logic > All rights reserved. === Subject: Re: card(P(N)) = card(N), final (finished) The argument below is crystal clear already. Bye bye. -LV > Second attempt. I'll try two equivalent formulations, in the hope > that, together, they will complement the eventual ambiguities. ----- 1st formulation (i) P(A) ~ { 0, 1, 2, ..., 2^n - 1}, AneN Let's simply call P(A) the set on the RHS of (i), keeping in mind that > the argument is going to hold due to this bijection and the > transitivity of the '~' relation. (ii) P(A) c= N, AneN (iii) P(N) is infinite (trivial) We pass from P(A) to P(N) by transfinite induction, then: (iv) P(N) ~ N, from (ii) and (iii) QDE. (1) ----- 2nd formulation Same comments as above. (i) card(P(A)) = 2^n, AneN (ii) card(P(A)) <= card(N), AneN > (iii) card(P(N)) >= oo (meaning: 'oo' as lower bound) (iv) card(P(N)) = card(N) QDE. (2) ----- Hmm, correct? The basic idea behind this argument is that P(A) is > always a subset of N in the finite case. Then, by transfinite > induction, while P(N) becomes infinite, its cardinality's upper bound > remains N's cardinality. (These informal expositions get me into more > troubles than they help, but I have to run the risk: no pain, no > gain!) Good luck. Julio > If you want to use transfinite induction, you have to show that the result is true for n=1, n+1, and the limit ordinal of n. The first two apply only to normal induction; the third is required for transfinite induction. You have not shown this third requirement is met in your proof, and nor can you, as its not. Consider the following proof which is similar in form to yours: All ordinals are finite. proof: 1 is finite. If n is finite, so is n+1. Therefore all numbers are finite. Of course, this is wrong, because w is not finite; the proof fails because w is not a successor to any other ordinal (its a limit ordinal), and the requirements of transfinite induction are not met. === Subject: EARN MONEY $1000-25000 PER MONTH posting-account=hZzRFwoAAADAQkeFVT0UNvUivsrwzt49 Gecko/20080404 Firefox/2.0.0.14,gzip(gfe),gzip(gfe) EARN MONEY $1000-25000 PER MONTH TAKE SIMPLE ONLINE SURVEYS CREATE FREE ACCOUNT OTHER DETAILS LOG ON TO **************************************************************************** **************************** http://www.AWSurveys.com/HomeMain.cfm?RefID=Venkatesh11 **************************************************************************** **************************** === Subject: This Week's Finds in Mathematical Physics (Week 267) Also available at http://math.ucr.edu/home/baez/week267.html July 23, 2008 This Week's Finds in Mathematical Physics (Week 267) John Baez After the workshop on categorical groups in Barcelona, I went to Granada - the world capital of categorical groups! Pilar Carrasco, an expert on this subject, had kindly invited me to spend a week there and give some talks. Even more kindly, she put me in a hotel right next to the Alhambra. So, this week I'll tell you about some categorical groups I saw in the Alhambra. I've long been fascinated by that melting-pot of cultures in southern invaded by Muslims in 711 AD, and became a center for mathematics and astronomy from around 930 AD to around the 1200s, when the city of Toledo, recaptured by Catholic Spaniards, became the center of a big translation industry - translating Arabic and Hebrew texts into Latin. This was important for the transmission of ancient Greek writings into Western Europe. The Alhambra was built after the true heyday of Andalusia, in the era when Muslims had almost been pushed out by the Catholics. Its construction was begun by Muhammed ibn Nasr, founder of the Nasrid Dynasty - the last Muslim dynasty in Spain. In 1236, Ferdinand III of Castile captured the marvelous city of Cordoba, ornament of the world. Ibn Nasr saw which way the wind was blowing, and arranged to pay tribute to Ferdinand and even help him take the city of Seville in return for leaving his city - Granada - alone. He started building the Alhambra in 1238. It was completed in the late 1300s. For the mathematician, one striking thing about the Alhambra is the marvelous tile patterns. On my visit, I took photos of all the tiles I could see: 1) John Baez, Alhambra tiles, http://math.ucr.edu/home/baez/alhambra Some people claim that tilings with all 17 possible wallpaper groups claim with all the vehemence such an academic issue deserves, saying that only 13 wallpaper groups are visible: 2) Branko Gr.9fnbaum, The emperor's new clothes: full regalia, G-string, or nothing?, with comments by Peter Hilton and Jean Pedersen, Math. Intelligencer 6 (1984), 47-56. As mentioned in week221, this page shows 13 of the 17: 3) Steve Edwards, Tilings from the Alhambra, http://www2.spsu.edu/math/tile/grammar/moor.htm Of the remaining four, two seem completely absent in Islamic art - the groups called pgg and pg. Both are fairly low on symmetries, so they might have been avoided for lack of visual interest. Let me describe them, just for fun. You can learn the definition of wallpaper groups, and learn a lot more about them, from this rather 4) Wikipedia, Wallpaper group, http://en.wikipedia.org/wiki/Wallpaper_group The group pgg is the symmetry group of a popular zig-zag method of group are 180-degree rotations: you can rotate any brick 180 degrees around its center, and the pattern comes back looking the same. There are no reflections in this group. But, there are glide reflections in two diagonal directions: a glide reflection is a combination of a translation along some line and a reflection across that line. The group pg is a subgroup of pgg. If we take our zig-zag pattern of bricks and break the 180-degree rotation symmetry somehow, the remaining symmetry group is pg. This group contains no reflections and no rotations. It contains translations along one axis and glide reflections along another. For more on tilings, try this book. Among other things, it points out that there's a lot more beauty and mathematical structure in tilings than is captured by their symmetry groups! 5) Branko Gr.9fnbaum and G. C. Shephard, Tilings and Patterns, New York, Freeman, 1987. The mathemagician John Horton Conway has come up with a very nice proof that there are only 17 wallpaper groups. This is nicely 6) John H. Conway, The orbifold notation for surface groups, in Groups, Combinatorics and Geometry, London Math. Soc. Lecture Notes Series 165, Cambridge U. Press, Cambridge, 1990, pp. 438-447 Here's the basic idea. Take a wallpaper pattern and count two points as the same if they're related by a symmetry. In other words - in math jargon - take the plane and mod out by the wallpaper group. The result is a 2-dimensional orbifold. In a 2d manifold, every point has a little neighborhood that looks like the plane. In a 2d orbifold, every point has a little neighborhood that looks either like the plane, or the plane mod a finite group of rotations and/or reflections. Let's see how this works for a few simple wallpaper groups. I'll start with the most boring wallpaper group in the world, p1. If you thought pg was dull, wait until you see p1. It's the symmetry group of this wallpaper pattern: R R R R R R RRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR R R R R R R R R R R R R R R R R R R RRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR R R R R R R R R R R R R R R R R R R RRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR R R R R R R This group doesn't contain any rotations, reflections or glide reflections - I used the letter R to rule those out. It only contains translations in two directions, the bare minimum allowed by the definition of a wallpaper group. If we take the plane and mod out by this group, all the points labelled x get counted as the same: R R R R R R RRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR R R R R R R R x R x R x R x R x R R R R R R R RRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR R R R R R R R x R x R x R x R x R R R R R R R RRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR R R R R R R Similarly, all these points labelled y get counted as the same: R R R R R R RRRRRyRRRRRyRRRRRyRRRRRyRRRRRyRRR R R R R R R R R R R R R R R R R R R RRRRRyRRRRRyRRRRRyRRRRRyRRRRRyRRR R R R R R R R R R R R R R R R R R R RRRRRyRRRRRyRRRRRyRRRRRyRRRRRyRRR R R R R R R So, when we take the plane and mod out by the group p1, we get a rectangle with its right and left edges glued together, and with its top and bottom edges glued together. This is just a torus. A torus is a 2d manifold, which is a specially dull case of a 2d orbifold. Now let's do a slightly more interesting example: T T T T T T TTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTT T T T T T T T T T T T T T T T T T T TTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTT T T T T T T T T T T T T T T T T T T TTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTT T T T T T T The letter T is more symmetrical than the letter R: you can reflect it, and it still looks the same. (If you're viewing this using some font where the letter T *doesn't* have this symmetry, switch fonts!) So, the symmetry group of this wallpaper pattern, called pm, is bigger than p1: it also contains reflections and glide reflections along a bunch of parallel lines. So now, all these points labelled x get counted as the same when we mod out: T T T T T T TTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTT T T T T T T T x x T x x T x x T x x T x x T T T T T T T TTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTT T T T T T T T x x T x x T x x T x x T x x T T T T T T T TTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTT T T T T T T and similarly for all these points labelled y: T T T T T T TTTyTyTTTyTyTTTyTyTTTyTyTTTyTyTTT T T T T T T T T T T T T T T T T T T TTTyTyTTTyTyTTTyTyTTTyTyTTTyTyTTT T T T T T T T T T T T T T T T T T T TTTyTyTTTyTyTTTyTyTTTyTyTTTyTyTTT T T T T T T but look how these points labelled z work: T T T T T T TTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTT T z T z T z T z T z T T T T T T T T T T T T T TTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTT T z T z T z T z T z T T T T T T T T T T T T T TTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTT T z T z T z T z T z T There are only half as many z's per rectangle, since they lie on reflection lines. Because of this subtlety, this time when we mod out we get an orbifold that's not a manifold! It's the torus of the previous example, but now folded in half. We can draw it as *half* of one of the rectangles above, with the top and bottom glued together, but not the sides: TTTT T . T . T . TTTT So, it's a cylinder... but in a certain technical sense the points at the ends of this cylinder count as half points: they lie on reflection lines, so they've been folded in half. This particular orbifold looks a lot like a 2d manifold with boundary. That's a generalization of a 2d manifold where some points - the boundary points - have a neighborhood that looks like a half-plane. But 2d orbifolds can also have cone points and mirror reflector points. What's a cone point? It's like the tip of a cone. Take a piece of paper, cut it like a pie into n equal wedges, take one wedge, and glue its edges together! This gives a cone - and the tip of this cone is a cone point. We say it has order n, because the angle around it is not 2pi but 2pi/n. Here's a more sophisticated way to say the same thing: take a regular n-gon and mod out by its rotational symmetries, which form a group with n elements. When we're done, the point in the center is a cone point of order n. We could also mod out by the rotation *and* reflection symmetries of the n-gon, which form a group with 2n elements. This is harder to visualize, but when we're done, the point in the center is a corner reflector of order 2n. To see one of these fancier possibilities, let's look at the orbifold coming from a wallpaper pattern with even more symmetries: . . . . . . ................................. . . . . . . . . . . . . . . . . . . ................................. . . . . . . . . . . . . . . . . . . ................................. . . . . . . These are supposed to be rectangles, not squares. So, 90-degree rotations are not symmetries of this pattern. But, in addition to all the symmetries we had last time, now we have reflections about a bunch of horizontal lines. We get a wallpaper group called pmm. What orbifold do we get now? It's a torus folded in half *twice*! That sounds scary, but it's not really. We can draw it as a *quarter* of one of the rectangles above: .... . . .... Now no points on the edges are glued together. So, it's just a rectangle. The points on the edges are boundary points, and the corners are corner reflection points of order 4. In a certain technical sense - soon to be explained - points on the edges of this rectangle count as half points, since they lie on a reflection line and have been folded in half. But the corners count as 1/4 points, since they lie on *two* reflection lines, so they've been folded in half *twice*! This is where it gets really cool. There's a way to define an Euler characteristic for orbifolds that generalizes the usual formula for 2d manifolds. And, it can be a fraction! The usual formula says to chop our 2d manifold into polygons and compute V - E + F That is: the number of vertices, minus the number of edges, plus the number of faces. In a 2d orbifold, we use the same formula, but with some modifications. First, we require that every cone point or corner reflector be one of our vertices. Then: We count edges and vertices on the boundary for 1/2 the usual amount. We count a cone point of order n as 1/nth of a point. We count a corner reflector of order 2n as 1/(2n)th of a point. The idea is that these features have been folded over by a certain amount, so they count for a fraction of what they otherwise would. It turns out that if we calculate the Euler characteristic of a 2d orbifold coming from a wallpaper group, we always get zero. And, there are just 17 possibilities! In fact, wallpaper groups are secretly *the same* as 2d orbifolds with vanishing Euler characteristic! So, they're not just mathematical curiosities: they're almost as fundamental as 2d manifolds. The torus and the cylinder, which we've already seen, are two examples. These are well-known to have Euler characteristic zero. Of course, we should be careful: now we're dealing with the cylinder as an orbifold, so the points on the boundary count as half points - but its Euler characteristic still vanishes. A more interesting example is the square we get from the group pmm. Let's chop it into vertices, edges, and one face, and figure out how much each of these count: 1/2 1/4-------1/4 | | | | 1/2| 1 |1/2 | | | | 1/4-------1/4 1/2 So, the Euler characteristic of this orbifold is (1/4 + 1/4 + 1/4 + 1/4) - (1/2 + 1/2 + 1/2 + 1/2) + 1 = 0 This is different than the usual Euler characteristic of a rectangle! As usual, Conway has figured out a charming way to explain all this: 7) John Conway, Peter Doyle, Jane Gilman and Bill Thurston, Geometry and the Imagination in Minneapolis, available at http://www.geom.uiuc.edu/docs/doyle/mpls/handouts/handouts.html Especially see the sections near the end entitled Symmetry and orbifolds, Names for features of symmetrical pattern, Names for symmetry groups and orbifolds, The orbifold shop and The Euler characteristic of an orbifold. He has a way of naming 2d orbifolds that lets you easily see how they cost. Any orbifold that costs 2 dollars corresponds to a wallpaper group, and if you list them, you see there are exactly 17! (I only know two places where the number 17 played an important role in mathematics - the other is much more famous.) Why exactly 2 dollars? This is related to this formula for the Euler characteristic of a g-handled torus: V - E + F = 2 - 2g where g is the number of handles. At the orbifold shop, each handle costs 2 dollars. So, if you buy one handle, you're done: you get an ordinary torus, which has Euler characteristic zero. This is the result of taking the plane and modding out by a spectacularly dull wallpaper group. If you don't waste all your money on a handle, you can buy more interesting orbifolds. Devoted readers of This Week's Finds can guess why I'm talking about this. It's not just that I like the Alhambra. The usual Euler characteristic is a generalization of the cardinality of a finite set that allows *negative* values - but not fractional ones. There's also something called the homotopy cardinality of a space, which allows *fractional* values - but not negative ones! If we combine these ideas, we get the orbifold Euler characteristic, which allow both negative and fractional values. This has various further generalizations, like the Euler characteristic of a differentiable stack - and Leinster's Euler characteristic of a category, explained in week244. We should be able to use these to categorify a lot of math involving rational numbers. But, it's especially cool how this game of listing all 2d orbifolds with Euler characteristic 0 fits together with things like the Egyptian fractions approach to ADE Dynkin diagrams - as explained in week182. Here I used the Euler characteristic to list all ways ways to regularly tile a sphere with regular polygons. This gave the McKay correspondence linking Platonic solids to the simply-laced Lie algebras A_n, D_n, E_6, E_7 and E_8. Taking different values of the Euler characteristic, the same idea let us classify regular tilings of the plane or hyperbolic plane by regular polygons, and see how these correspond to affine or hyperbolic simply-laced Kac-Moody algebras. Even better, compact quotients of these tilings give some very nice modular curves, like Klein's quartic curve: 8) John Baez, Klein's quartic curve, http://math.ucr.edu/home/baez/klein.html So, there a lot of connections to be made here... and I can tell I haven't made them all yet! Why don't you give it a try? To add to the fun, my friend Eugene Lerman has just written a very nice survey paper on orbifolds: 9) Eugene Lerman, Orbifolds as stacks?, available as arXiv:0806.4160. This describes some deeper ways to think about orbifolds. For example, when we form an orbifold by taking the plane and modding out by a wallpaper group, we shouldn't really say two points on the plane become *equal* if there's a symmetry carrying one to the other. Instead, we should say they are *isomorphic* - with the symmetry being the isomorphism. This gives us a groupoid, whose objects are points on the plane and whose morphisms are symmetries taking one point to another. It's a Lie groupoid, since there's a manifold of objects and a manifold of morphisms, and everything in sight is smooth. So, orbifolds can be thought of as Lie groupoids. This leads to the real point of Lerman's paper: orbifolds form a 2-category! This should be easy to believe, since there's a 2-category with groupoids as objects, functors as morphisms, and natural transformations as 2-morphisms. In week75 and week80 I explained the closely related 2-category with *categories* as objects; the same idea works for groupoids. So, to get a 2-category of *Lie* groupoids, we just need to take this idea and make everything smooth in a suitable sense. This turns out to be trickier than you might at first think - and that's where differentiable stacks come in. I should explain these someday, but not today. For now, try these nice introductions: 10) J. Heinloth, Some notes on differentiable stacks, Mathematisches 1-32. Available as http://www.math.nyu.edu/~tschinke/WS04/pdf/book.pdf or separately as http://www.uni-essen.de/~hm0002/stacks.pdf 12) Kai Behrend and Ping Xu, Differentiable stacks and gerbes, available as arXiv:math/0605694. Now, besides the Alhambra, Granada also has a wonderful Department of Algebra. Yes - a whole department, just for algebra! And this department has many experts on categorical groups, also known as 2-groups. So it's worth noting that there are 2-groups lurking in the Alhambra. Any object in any category has a group of symmetries. Similarly, any object in any 2-category has a 2-group of symmetries. So, any orbifold has a 2-group of symmetries. We should be able to get some interesting 2-groups this way. The group of *all* symmetries of a manifold - its diffeomorphism group - is quite huge. That's because you can warp it and bend it any way you like, as long as that way is smooth. Similarly, the 2-group of *all* symmetries of an orbifold will often be quite huge. To cut down the symmetry group of a manifold, we can equip it with a Riemannian metric - a nice distance function - and then consider only symmetries that preserve distances. We can get lots of nice groups this way, called isometry groups. For example, the group E8, which has been in the news disturbingly often of late, is the isometry group of a 128-dimensional Riemannian manifold called the octooctonionic projective plane. So, maybe we can get some nice 2-groups as isometry 2-groups of Riemannian orbifolds. Of course, for this to make sense, we need to know what we mean by a Riemannian metric on on orbifold! I'm no expert on this, but I'm pretty sure the idea makes sense. And I'm pretty sure that the 2d orbifold we get from a specific wallpaper pattern has a Riemannian metric coming from the usual distance function on the plane. (Warning: the same wallpaper group can arise as symmetries of wallpaper patterns that are different enough to give different Riemannian orbifolds!) So, here's a potentially fun question: what 2-groups show up as isometry 2-groups of Riemannian orbifolds coming from wallpaper patterns? Try to work out some examples. I don't expect the answer to be staggeringly profound - but it sets up a link between the Alhambra and 2-groups, and that's cool enough for me! By the way, I obtained copies of some very interesting theses in Granada: 13) Antonio Martinez Ceggara, Cohomologia Varietal, Ph.D. thesis, Departamento de Algebra y Fundamentos, Universidad de Santiago de Compostela. 14) Pilar Carrasco, Complejos Hipercruzados: Cohomologia y Extensiones, Ph. D. theis, Cuadernos de Algebra 6, Departamento de Algebra y Fundamentos, Universidad de Granada, 1987. Antonio Cegarra was the one who brought 2-group theory to Granada, and Pilar Carrasco was his student. It's unfortunate that these theses come from the day before electronic typesetting. Luckily, Carrasco's was later turned into a paper: 15) Pilar Carrasco and Antonio Martinez Ceggara, Group-theoretic algebraic models for homotopy types, Jour. Pure Appl. Algebra 75 (1991), 195-235. Also available as http://www.ugr.es/~acegarra/Group theoretic algebraic models for homotopy types.pdf This tackles the ever-fascinating, ever-elusive problem of taking the information in the homotopy type of a topological space and packaging it in some manageable way. If our space is connected, with a chosen basepoint, and it has vanishing homotopy groups above the 2nd, a 2-group will do the job quite nicely! The same idea should work for numbers larger than 2, but n-groups get more and more elaborate as n increases. Carrasco and Cegarra package all the information into a hypercrossed complex, and I would really like to understand this better. Carrasco and Cegarra's paper is quite dense. So, I'm very happy to hear that Carrasco plans to translate her thesis into English! Before I finish, let me mention one more paper about 2-groups: 16) Jo.8bo Faria Martins, The fundamental crossed module of the complement of a knotted surface, available as arXiv:0801.3921. Martins was unable to attend the Barcelona workshop on 2-groups, but I met him later in Lisbon, and he explained some of the ideas here to me. A crossed module is just another way of thinking about a 2-group. So, translating the language a bit, the basic concept behind this paper is the fundamental 2-group of (X,A,*). Here X is a topological space that contains a subspace A that contains a point *. Here's how it goes. A 2-group is a 2-category with one object: * a bunch of morphisms: f *-------->* (which must all be invertible), and a bunch of 2-morphisms: f ---->--- / * T * / ---->---- g (which must also be invertible). So, we get the fundamental 2-group of (X,A,*) as follows: Let the only object be the point *. Let the morphisms be paths in A starting and ending at *. Let the 2-morphisms be homotopy classes of paths of paths in X. If you let A be all of X, you get the fundamental 2-group of (X,*), and this is what people mean when they say connected pointed homotopy 2-types are classified by 2-groups. But the generalization is also quite nice. In the above paper, Martins uses this generalization, and a bunch of other ideas, to give an explicit presentation of the fundamental 2-group of the complement of a 2-knot (a sphere embedded in 4d Euclidean space). In a certain sense this generalizes the usual Wirtinger presentation of the fundamental group of the complement of a knot. But, it's a bit different. ----------------------------------------------------------------------- Addenda: I should mention the definition of a wallpaper group: it's a discrete subgroup of the isometry group of the plane that includes translations in two linearly independent directions. We need the right equivalence relation on these groups to get just 17 of them: they're equivalent if you can conjugate one by an affine transformation of the plane and get the other. Alas, over at the n-Category Cafe Richard Hepworth has shown that all the isometry 2-groups of orbifolds coming from wallpaper groups are equivalent to mere groups. It's a pity! But, at least his remarks shed a lot of light on the general theory of isometry 2-groups. Here is a recipe for computing the isometry 2-groups of a Riemannian orbifold X//G, with G a discrete group acting on a connected simply connected manifold X. I think these are precisely the sorts of orbifolds you are interested in. Apologies for the nasty presentation! There is one object. The arrows are pairs (f,phi) where f is an isometry of X, phi an automorphism of G, and f is equivariant with respect to phi, i.e. f(gx) = phi(g)f(x). The 2-arrows from (f,phi) to (k,kappa) are the elements g of G for which k = gf and kappa = g phi g^{-1}. I'll leave you to work out the various composition maps. Of course, I haven't told you what a Riemannian metric on an orbifold (or groupoid or stack) is. You can find definitions in various places, but in the case in point these things are all equivalent to putting a G-invariant metric on X. (Does one always exist? Yes, but the fact that G acts with finite stabilizers is essential here.) I also haven't said how I get the above answer. If you already knew the answer for X = point then the above would be your first guess. The connected and simply connected: all G-valued functions on X are constant, and all G-bundles are trivial. Maybe if you're interested I can go into more detail. Hmm, John and David's discussion has reminded me of a result I once proved: Suppose we are given a morphism f:X -> Y of orbifolds and a 2-automorphism phi:f => f. Then phi is trivial if and only if its restriction to any point of X is trivial. One consequence of this is the following: Let X be an effective orbifold. Then any self-equivalence of X has no nontrivial automorphisms. And in particular: The isometry 2-group of an effective orbifold is equivalent to a group. What does effective mean? It means that the automorphism group of any point in X acts effectively on its tangent space, and consequently that almost all points of the orbifold have no inertia at all. This includes all of the orbifolds you are discussing. The fact that the isometry 2-groups are all equivalent to groups is apparent in the description I gave earlier: if we have a 2-arrow g:(f,phi) => (f,phi) then gf = f, so that g is the identity. Of course, there are many interesting noneffective orbifolds. Some of them are called gerbes. (Gerbes with band a finite group.) Maybe you want to compute their isometry 2-groups? Here's a fact for you: The isometry 2-group of the nontrivial Z/2-gerbe over S^2 is itself a Z/2-gerbe over O(3), the isometry group of S^2. In particular, it is not equivalent to any group. The gerbe over O(3), however, is trivial. But is there a trivialization that respects the 2-group structure? Actually not *all* the orbifolds coming from wallpaper groups are effective in the above sense. For the orbifold to be effective, I believe the corresponding group must include eitherreflections or glide reflections across two different axes, or a rotation by less than 180 degrees. For example, the torus and cylinder described above are noneffective orbifolds. But, most of the interesting examples are covered by Hepworth's results, and the others seem also to have isometry 2-groups that are equivalent to mere groups. For details try: 17) Richard Hepworth, The age grading and the Chen-Ruan cup product, available as arXiv:0706.4326. Richard Hepworth, Morse inequalities for orbifold cohomology, available as arXiv:0712.2432. According to Hepworth, the first paper contains the little fact that implies that the automorphism 2-group of an effective orbifold is equivalent to a group. Of course it contains lots of other stuff, too! The second discusses Morse theory on differentiable Deligne- Mumford stacks (these are the proper etale ones). It defines Morse functions, vector fields and Riemannian metrics on differentiable DM stacks. It also proves that Morse functions are generic and that vector fields can be integrated. You can see more discussion of this Week's Finds at the n-Category Cafe: http://golem.ph.utexas.edu/category/2008/07/this_weeks_finds_in_mathematic_2 8.html ----------------------------------------------------------------------- mathematics and physics, as well as some of my research papers, can be obtained at http://math.ucr.edu/home/baez/ For a table of contents of all the issues of This Week's Finds, try http://math.ucr.edu/home/baez/twfcontents.html A simple jumping-off point to the old issues is available at http://math.ucr.edu/home/baez/twfshort.html If you just want the latest issue, go to http://math.ucr.edu/home/baez/this.week.html === Subject: Re: This Week's Finds in Mathematical Physics (Week 267) 2008 16:22:53, John Baez posted: Also available at http://math.ucr.edu/home/baez/week267.html July 23, 2008 >This Week's Finds in Mathematical Physics (Week 267) >John Baez After the workshop on categorical groups in Barcelona, I went to >Granada - the world capital of categorical groups! Pilar Carrasco, >an expert on this subject, had kindly invited me to spend a week >there and give some talks. Even more kindly, she put me in a hotel >right next to the Alhambra. So, this week I'll tell you about some >categorical groups I saw in the Alhambra. > ... Du Sautoy, Marcus Finding moonshine : a mathematician's journey through symmetry / Marcus du S . - London : Fourth Estate, 2007 includes his observations on Alhambra groups, IIRC. -- (c) John Stockton, near London. *@merlyn.demon.co.uk/?.?.Stockton@physics.org Web - FAQish topics, acronyms, & links. Correct <= 4-line sig. separator as above, a line precisely -- (SoRFC1036) Do not Mail News to me. Before a reply, quote with > or > (SoRFC1036) === Subject: Is our math only good in our space-time realm? Here's something to think about: posting-account=A451vgoAAABycJv5ugpkfX8QG96-xuRZ .NET CLR 2.0.50727; .NET CLR 3.0.04506),gzip(gfe),gzip(gfe) (Click link below:) http://www.rbduncan.com/schrod.htm This book was written by a mathematical physcist who was one of that group who got us on the moon. === Subject: Checking for alternate solutions in exact differential equations posting-account=mtv4agoAAAA7l_0rNFOQ8uq0qtHbz76I Gecko/20080528 Epiphany/2.22,gzip(gfe),gzip(gfe) I am learning first order exact differential equations. While solving some equations, say: y' - x^2 = x^2 * y, where y is a function of x we get one solution, i.e. 3ln(1+y) = x^3 + c, where c is a constant But this also has another solution y = -1. But then there are other types of equations, e.g. y dx + x dy = 0, which has only one solution xy = c. So, in which type of equations, do I have to be careful about the extra solutions? Is there something like a thumb-rule which I can use to identify if there is any extra solution or not? === Subject: Re: Checking for alternate solutions in exact differential equations- > I am learning first order exact differential equations. While solving > some equations, say: > > y' - x^2 = x^2 * y, where y is a function of x > > we get one solution, i.e. 3ln(1+y) = x^3 + c, where c is a constant > > But this also has another solution y = -1. > > But then there are other types of equations, e.g. y dx + x dy = 0, > which has only one solution xy = c. > > So, in which type of equations, do I have to be careful about the > extra solutions? Is there something like a thumb-rule which I can use > to identify if there is any extra solution or not? Indeed, first-order ODVs normally have a continuous one-parameter family of solutions. If the curves represented by this family have an enveloping curve then the equation of the enveloping curve also represents a solution of the given ODV. Prototypical example is (dy/dx)^2 = 1 - y^2. The one-parameter family of solutions is x -> y = sin (x + C), the singular solutions are y = 1 and y = -1. Literature: one of the best is E.L.Ince: Ordinary Differential Equations. Dover 1956, unabridged and unaltered republication of the original edition by Longmans, Green and Co., 1926 Good luck: Johan E. Mebius === Subject: incoherent wave posting-account=H-IscAoAAABkDNrURGSxo9jPN3MJ3a8A 1.0.3705; .NET CLR 1.1.4322),gzip(gfe),gzip(gfe) Hi: Given wave equation div( grad(u) )+k^2*u = 0. This is a linear eqn and thus if we have two wave sources u_1 and u_2, then both of them satisfy the wave equation and we have u_total = u_1+u_2. But the energy cannot be superimposed. By definition, if there are 2 wave sources and they are incoherent, then their energy can be superimposed. But what is the math descirption for incoherence? Also, do they satisfy the wave eqn above? === Subject: Re: incoherent wave > Hi: > > Given wave equation div( grad(u) )+k^2*u = 0. This is a linear eqn > and thus > > if we have two wave sources u_1 and u_2, then both of them satisfy the > > wave equation and we have u_total = u_1+u_2. But the energy cannot be > > superimposed. > Right, because the energy is proportional to the -square- of the wave amplitude. > By definition, if there are 2 wave sources and they are incoherent, > then > > their energy can be superimposed. I'm not sure about this. If two waves are incoherent, it does not mean that they cannot interfere at some point in space, does it? It does mean that their phase difference is not a constant in time. Maybe here you can find something of interest: http://en.wikipedia.org/wiki/Coherence_%28physics%29 But what is the math descirption for > > incoherence? Also, do they satisfy the wave eqn above? > Best wishes, Sebastiaan. -- Replace .dz by .nl === Subject: Integral domain result from Ax's The elementary theory of finite fields posting-account=vlHY6AoAAACylqY38Om9mNYqDPbekcM1 Gecko/2008072820 Firefox/3.0.1,gzip(gfe),gzip(gfe) We say that an extension L of a field K is a regular extension of K if Lotimes_Kbar{K}, with bar{K} the algebraic closure of K, is an integral domain. The following have been proven in Masayoshi Nagata: Theory of Commutative Fields on pages 94-96: For a subfield K of a field L, the following three conditions are equivalent. (i) L is a regular extension of K. (ii) K is algebraically closed in L, and L is separable over K. (iii) Lotimes_K L' is an integral domain for every extension L' of K. Can this theorem be used to prove the following theorem which can be found on Ax's paper stated as well-known: Let K be a field, bar{K} the algebraic closure of K, and R a commutative K-algebra. Then bar{K} otimes_K R is an integral domain if and only if Aotimes_K R is an integral domain for every extension field A of K. I'm not sure because Nagata uses field extensions but I have to prove the theorem for K-algebras. === Subject: Re: Integral domain result from Ax's The elementary theory of finite fields posting-account=9QOSvAoAAACEOWJVSDuswW7dB_0wApQO Gecko/2008061712 Fedora/3.0-1.fc9 Firefox/3.0,gzip(gfe),gzip(gfe) > We say that an extension L of a field K is a regular extension of K if > Lotimes Kbar{K}, with bar{K} the algebraic closure of K, is an > integral domain. The following have been proven in Masayoshi Nagata: Theory of > Commutative Fields on pages 94-96: For a subfield K of a field L, the following three conditions are > equivalent. (i) L is a regular extension of K. > (ii) K is algebraically closed in L, and L is separable over K. > (iii) Lotimes K L' is an integral domain for every extension L' of K. Can this theorem be used to prove the following theorem which can be > found on Ax's paper stated as well-known: Let K be a field, bar{K} the algebraic closure of K, and R a > commutative K-algebra. Then bar{K} otimes K R is an integral domain > if and only if Aotimes K R is an integral domain for every extension > field A of K. I'm not sure because Nagata uses field extensions but I have to prove > the theorem for K-algebras. It is a rather simpler thing. Suppose A otimes K R is a domain for all extensions A of the field K, and suppose x, y are non-zero elements of bar{K} such that x y = 0 in bar{K} otimes K R. Since x and y can be written as finite sums of elementary tensors in bar{K} otimes K R, you can find a finite extension of K, contained in bar{K}, such that x and y are also in A otimes K R. This is absurd since we are supposing that A otimes K R is a domain. The converse is similar. -- m === Subject: Re: Integral domain result from Ax's The elementary theory of finite fields posting-account=vlHY6AoAAACylqY38Om9mNYqDPbekcM1 Gecko/2008072820 Firefox/3.0.1,gzip(gfe),gzip(gfe) On 29 hein.8a, 22:17, Mariano Su.87rez-Alvarez We say that an extension L of a field K is a regular extension of K if > Lotimes Kbar{K}, with bar{K} the algebraic closure of K, is an > integral domain. The following have been proven in Masayoshi Nagata: Theory of > Commutative Fields on pages 94-96: For a subfield K of a field L, the following three conditions are > equivalent. (i) L is a regular extension of K. > (ii) K is algebraically closed in L, and L is separable over K. > (iii) Lotimes K L' is an integral domain for every extension L' of K. Can this theorem be used to prove the following theorem which can be > found on Ax's paper stated as well-known: Let K be a field, bar{K} the algebraic closure of K, and R a > commutative K-algebra. Then bar{K} otimes K R is an integral domain > if and only if Aotimes K R is an integral domain for every extension > field A of K. I'm not sure because Nagata uses field extensions but I have to prove > the theorem for K-algebras. It is a rather simpler thing. Suppose A otimes K R is a domain for all extensions A > of the field K, and suppose x, y are non-zero elements > of æbar{K} such that x y = 0 in bar{K} otimes K R. > Since x and y can be written as finite sums of elementary > tensors in bar{K} otimes K R, you can find a finite extension > of K, contained in bar{K}, such that x and y are also in > A otimes K R. This is absurd since we are supposing that > A otimes K R is a domain. The converse is similar. -- m What about the transcendental part over an algebraic extension? :) I don't see why it is similar. === Subject: Re: Integral domain result from Ax's The elementary theory of finite fields posting-account=vlHY6AoAAACylqY38Om9mNYqDPbekcM1 Gecko/2008072820 Firefox/3.0.1,gzip(gfe),gzip(gfe) > On 29 hein.8a, 22:17, Mariano Su.87rez-Alvarez > We say that an extension L of a field K is a regular extension of K if > Lotimes Kbar{K}, with bar{K} the algebraic closure of K, is an > integral domain. The following have been proven in Masayoshi Nagata: Theory of > Commutative Fields on pages 94-96: For a subfield K of a field L, the following three conditions are > equivalent. (i) L is a regular extension of K. > (ii) K is algebraically closed in L, and L is separable over K. > (iii) Lotimes K L' is an integral domain for every extension L' of K. Can this theorem be used to prove the following theorem which can be > found on Ax's paper stated as well-known: Let K be a field, bar{K} the algebraic closure of K, and R a > commutative K-algebra. Then bar{K} otimes K R is an integral domain > if and only if Aotimes K R is an integral domain for every extension > field A of K. I'm not sure because Nagata uses field extensions but I have to prove > the theorem for K-algebras. It is a rather simpler thing. Suppose A otimes K R is a domain for all extensions A > of the field K, and suppose x, y are non-zero elements > of æbar{K} such that x y = 0 in bar{K} otimes K R. > Since x and y can be written as finite sums of elementary > tensors in bar{K} otimes K R, you can find a finite extension > of K, contained in bar{K}, such that x and y are also in > A otimes K R. This is absurd since we are supposing that > A otimes K R is a domain. The converse is similar. -- m What about the transcendental part over an algebraic extension? :) I > don't see why it is similar. In my opinion, if A otimes R is a domain for every extension it is clear that bar{K} otimes R is a domain. The other inclusion is more interesting. I don't see why the theorem is true for a field, say A, which is an algebraic extension of purely transcendental extension. Could it be solved by Nulstellensats. === Subject: Re: Integral domain result from Ax's The elementary theory of finite fields posting-account=vlHY6AoAAACylqY38Om9mNYqDPbekcM1 Gecko/2008072820 Firefox/3.0.1,gzip(gfe),gzip(gfe) > On 29 hein.8a, 22:17, Mariano Su.87rez-Alvarez > We say that an extension L of a field K is a regular extension of K if > Lotimes Kbar{K}, with bar{K} the algebraic closure of K, is an > integral domain. The following have been proven in Masayoshi Nagata: Theory of > Commutative Fields on pages 94-96: For a subfield K of a field L, the following three conditions are > equivalent. (i) L is a regular extension of K. > (ii) K is algebraically closed in L, and L is separable over K. > (iii) Lotimes K L' is an integral domain for every extension L' of K. Can this theorem be used to prove the following theorem which can be > found on Ax's paper stated as well-known: Let K be a field, bar{K} the algebraic closure of K, and R a > commutative K-algebra. Then bar{K} otimes K R is an integral domain > if and only if Aotimes K R is an integral domain for every extension > field A of K. I'm not sure because Nagata uses field extensions but I have to prove > the theorem for K-algebras. It is a rather simpler thing. Suppose A otimes K R is a domain for all extensions A > of the field K, and suppose x, y are non-zero elements > of æbar{K} such that x y = 0 in bar{K} otimes K R. > Since x and y can be written as finite sums of elementary > tensors in bar{K} otimes K R, you can find a finite extension > of K, contained in bar{K}, such that x and y are also in > A otimes K R. This is absurd since we are supposing that > A otimes K R is a domain. The converse is similar. -- m What about the transcendental part over an algebraic extension? :) I > don't see why it is similar. In my opinion, if A otimes R is a domain for every extension it is clear that bar{K} otimes R is a domain. The other inclusion is more interesting. I don't see why the theorem is true for a field, say A, which is an algebraic extension of purely transcendental extension. Could it be solved by Nulstellensats. === Subject: Re: Roundess as a shape coefficent posting-account=CtWhuAoAAAAZZ9vwdovdqB3NNaiUa20_ SV1),gzip(gfe),gzip(gfe) Roundness of the object bounded with closed curve is defined as: > r = perimeter^2/(4*pi*area) > and is said to be minimal and equals 1 for a perfect circle. Do you know (or at least where to find) the proof that this is indeed > minimal? This would be the solution for task of finding maximal area that we > can bound closing the string of given length and for some integral > inequities as well (to which it leads when we substitute formulas for > perimeter and area). > http://en.wikipedia.org/wiki/Isoperimetric_inequality > http://www-groups.dcs.st-and.ac.uk/~history/Extras/Calculus_of_Variat... Quote: for today we know: Of all surfaces bounded by curves of a given > length, > the circle is the one of largest area. The branch of mathematics which > establishes a rigorous proof of this statement is the calculus of > variations. http://en.wikipedia.org/wiki/Calculus_of_variations Could'nt find back an old reproduction of the relevant proof. Sorry. Found back my old notes. Use the magic word if you want more. Han de Bruijn === Subject: Re: Roundess as a shape coefficent posting-account=tRBaOAoAAAB7qCLaIYH-yq0oSzOukouW After this it made me easy to find a brilliant proof that I am satisfied with and which doesn't use calculus of variation in: Differential Geometry of Curves and Surfaces, Victor Andreevich Toponogov, (page 18, Problem 1.5.5 (Isoperimetric problem)). Cauchy inequality, formula for arc length and Greens theorem and it was done by Hurwitz. This was how I tried to solve it by myself. If somebody has this proof I would appreciate any opportunity to see it. Ryszard === Subject: ortogonal matrices with dim=n containing only n variables posting-account=Lp215AoAAABaPqTOHMbNj9GpsECJOOv0 Gecko/2008072820 Firefox/3.0.1,gzip(gfe),gzip(gfe) We have matrix {{x,y},{-y,x}} which is 2-dimensional matrix and is ortogonal and contains only variables x and y. Are there such matrices for all n, which means if there are matrices with dim=n containing only x1, x2, ..., xn which are ortogonal? If not for all n, then what is criteria and reason that not for all? and if for some n such matrces exists is there only one or there exist more, how many (of course I don't take into account matrices which are created from givem by row or column permutations)? === Subject: Re: ortogonal matrices with dim=n containing only n variables posting-account=JbgxqAoAAABtpHDW0Y3Rj06f1KAxfL77 Gecko/20080721 Firefox/2.0.0.16,gzip(gfe),gzip(gfe) > We have matrix {{x,y},{-y,x}} which is 2-dimensional matrix and is > ortogonal and contains only variables x and y. I am not sure what you mean. This matrix is orthogonal for some values of (x,y) but not others. (A matrix A is orthogonal if A A^T = A^T A = I.) It does have orthogonal columns (and rows) however. > Are there such matrices for all n, which means if there are matrices > with dim=n containing only x1, x2, ..., xn which are ortogonal? If not > for all n, then what is criteria and reason that not for all? and if > for some n such matrces exists is there only one or there exist more, > how many (of course I don't take into account matrices which are > created from givem by row or column permutations)? Given the appearance of -y in your example, I suppose you will be satisfied if the entries of the matrix are functions of x1, ..., xn. In that case the answer is yes. Take the vector (x1, ..., xn) and extend it to an orthogonal basis by something like Gram-Schmidt (omit the normalization if you like), and use the basis vectors as the columns (or rows) of your matrix. === Subject: Re: ortogonal matrices with dim=n containing only n variables posting-account=Lp215AoAAABaPqTOHMbNj9GpsECJOOv0 1.1.4322; .NET CLR 2.0.50727; .NET CLR 3.0.04506.30),gzip(gfe),gzip(gfe) > Given the appearance of -y in your example, I suppose you will be > satisfied if the entries of the matrix are functions of x1, ..., xn. > In that case the answer is yes. æTake the vector (x1, ..., xn) and > extend it to an orthogonal basis by something like Gram-Schmidt (omit > the normalization if you like), and use the basis vectors as the > columns (or rows) of your matrix. sorry again, I didn't read your post till the end. I have to precise my requirements: I'm not looking matrices which contain as their elements something like 2*x1-3*x3+x2-10*x4, but only one variable with minus or plus (see examples from previous post) === Subject: Re: ortogonal matrices with dim=n containing only n variables posting-account=Lp215AoAAABaPqTOHMbNj9GpsECJOOv0 1.1.4322; .NET CLR 2.0.50727; .NET CLR 3.0.04506.30),gzip(gfe),gzip(gfe) We have matrix {{x,y},{-y,x}} which is 2-dimensional matrix and is > ortogonal and contains only variables x and y. I am not sure what you mean. æThis matrix is orthogonal for some > values of (x,y) but not others. æ(A matrix A is orthogonal if A A^T = > A^T A = I.) æIt does have orthogonal columns (and rows) however. Are there such matrices for all n, which means if there are matrices > with dim=n containing only x1, x2, ..., xn which are ortogonal? If not > for all n, then what is criteria and reason that not for all? and if > for some n such matrces exists is there only one or there exist more, > how many (of course I don't take into account matrices which are > created from givem by row or æcolumn permutations)? Given the appearance of -y in your example, I suppose you will be > satisfied if the entries of the matrix are functions of x1, ..., xn. > In that case the answer is yes. æTake the vector (x1, ..., xn) and > extend it to an orthogonal basis by something like Gram-Schmidt (omit > the normalization if you like), and use the basis vectors as the > columns (or rows) of your matrix. sorry ... I forgot definition of the ortogonality of the matrix. I don't really looking ortogonal matrices but matrices in which each row is ortogonal to other row, so if r 1, r 2, ..., r n are the rows of this matrix then r i*r j=0 for i different than j, where * denotes scalar product. For example for n=2 matrix {{x,y},{-y,x}} is such matrix, becouse: (x,y)*(-y,x)=-xy+xy=0 for n=4 {{x,y,z,t},{z,t,-x,-y},{-y,x,t,-z},{-t,z,-y,x}} is one of such matrices, becouse: r 1*r 2=(x,y,z,t)*(z,t,-x,-y)=xz+yt-xz-yt=0, r 1 r 3=(x,y,z,t)*(-y,x,t,-z)=-xy+xy+tz-tz=0, r 1*r 4=(x,y,z,t)*(-t,z,-y,x)=-xt+yz-yz+xt=0, r 2*r 3=(z,t,-x,-y)*(-y,x,t,-z)=-zy+xt-xt+yz=0, r 2*r 4=(z,t,-x,-y)*(-t,z,-y,x)=-zt+tz+xy-xy=0, r 3*r 4=(-y,x,t,-z)*(-t,z,-y,x)=yt+xz-yt-yz=0 for n=2k-1, k=1,2,3,... such matrix doesn't exist, becouse there are odd number of terms in scalar product. I suppose that such matrices exist only for n=2^k, k is an positive integer, but I couldn't know why and is it really true ... and if there exist such matrices for all n=2^k or only for 2, 4 and some other powers of 2? === Subject: Re: ortogonal matrices with dim=n containing only n variables posting-account=JeJlQwoAAADyvmg0o61ss8D4PmHJbp0h rv:1.8.1.16) Gecko/20080702 Firefox/2.0.0.16,gzip(gfe),gzip(gfe) > For example for n=2 matrix {{x,y},{-y,x}} is such matrix, becouse: > (x,y)*(-y,x)=-xy+xy=0 for n=4 {{x,y,z,t},{z,t,-x,-y},{-y,x,t,-z},{-t,z,-y,x}} is one of suchmatrices, becouse: > r_1*r_2=(x,y,z,t)*(z,t,-x,-y)=xz+yt-xz-yt=0, > r_1_r_3=(x,y,z,t)*(-y,x,t,-z)=-xy+xy+tz-tz=0, > r_1*r_4=(x,y,z,t)*(-t,z,-y,x)=-xt+yz-yz+xt=0, > r_2*r_3=(z,t,-x,-y)*(-y,x,t,-z)=-zy+xt-xt+yz=0, > r_2*r_4=(z,t,-x,-y)*(-t,z,-y,x)=-zt+tz+xy-xy=0, > r_3*r_4=(-y,x,t,-z)*(-t,z,-y,x)=yt+xz-yt-yz=0 Your matrix is - up to negations of variable and row/column permutations/negations - known in the Hadamard matrix literature as the Williamson array. Williamson's original reason for considering such a matrix was to construct Hadamard matrices. If one can find symmetric, commuting matrices x, y, z, t of size n with entries 1 and -1 satisfying x^2+y+2+z^2+t^2 = 4nI, then by plugging these matrices into your array, one gets a 4n by 4n Hadamard matrix. > for n=2k-1, k=1,2,3,... such matrix doesn't exist, becouse there are > odd number of terms in scalar product. I suppose that suchmatrices > exist only for n=2^k, k is an positive integer, but I couldn't know > why and is it really true ... and if there exist suchmatricesfor all > n=2^k or only for 2, 4 and some other powers of 2? If one insists that each variable appear exactly once in each row and column, then such arrays exist only in dimensions 1, 2, 4, and 8. This was proved by a number of different people. References are given in A. Hedayat and W. D. Wallis, Hadamard matrices and their applications, Annals of Statistics 6 (1978) 1184-1238. (See page 1196.) One of the proofs was given by Ted Spence, Edward Spence, Hadamard designs, Proceedings of the American Mathematical Society, Vol. 32, No. 1 (Mar., 1972), pp. 29-31. Hedayat and Wallis quote Storer as stating that the result is essentially contained in work of Hurwitz from 1898. If you allow repeated variables in each row, you can find other solutions. For example, Williamson's construction was generalized by Baumert and Hall, who found the following 12x12 array: {{a, a, a, b, -b, c, -c, -d, b, c, -d, -d}, {a, -a, b, -a, -b, -d, d, -c, -b, -d, -c, -c}, {a, -b, -a, a, -d, d, -b, b, -c, -d, c, -c}, {b, a, -a, -a, d, d, d, c, c, -b, -b, -c}, {b, -d, d, d, a, a, a, c, -c, b, -c, b}, {b, c, -d, d, a, -a, c, -a, -d, c, b, -b}, {d, -c, b, -b, a, -c, -a, a, b, c, d, -d}, {-c, -d, -c, -d, c, a, -a, -a, -d, b, -b, -b}, {d, -c, -b, -b, -b, c, c, -d, a, a, a, d}, {-d, -b, c, c, c, b, b, -d, a, -a, d, -a}, {c, -b, -c, c, d, -b, -d, -b, a, -d, -a, a}, {-c, -d, -d, c, -c, -b, b, b, d, a, -a, -a}} This uses only four variables, rather than 12. If a,b,c,d satisfy the same condition as was imposed on x,y,z,t in Williamson's construction, one gets a 12n by 12n Hadamard matrix. This is how the first Hadamard matrix of size 156 was found. -Will Orrick === Subject: Re: ortogonal matrices with dim=n containing only n variables posting-account=JeJlQwoAAADyvmg0o61ss8D4PmHJbp0h rv:1.8.1.16) Gecko/20080702 Firefox/2.0.0.16,gzip(gfe),gzip(gfe) The Hurwitz result mentioned in the previous post is the famous result that the only division algebras over the reals for which distance is preserved under multiplication by a unit vector are the real numbers, complex numbers, quaternions, and octonions. The corresponding matrices can be thought of as representations of these algebras. For example, the quaternion q = a+bi+cj+dk can be represented by phi(q) = {{ a,-b,-c,-d}, { b, a,-d, c}, { c, d, a,-b}, { d,-c, b, a}}. This is a representation in the sense that phi(qr) = phi(q)phi(r). An alternative representation is tau(q) = {{ a,-b,-c,-d}, { b, a, d,-c}, { c,-d, a, b}, { d, c,-b, a}}. The 8x8 self-reciprocal arrays represent the octonions, although one must be careful to interpret this properly since octonion multiplication is not associative. See http://arxiv.org/abs/math.RA/0003166 . -Will === Subject: Re: ortogonal matrices with dim=n containing only n variables > sorry ... I forgot definition of the ortogonality of the matrix. I > don't really looking ortogonal matrices but matrices in which each row > is ortogonal to other row, so if r 1, r 2, ..., r n are the rows of > this matrix then r i*r j=0 for i different than j, where * denotes > scalar product. > > For example for n=2 matrix {{x,y},{-y,x}} is such matrix, becouse: > (x,y)*(-y,x)=-xy+xy=0 > > for n=4 {{x,y,z,t},{z,t,-x,-y},{-y,x,t,-z},{-t,z,-y,x}} is one of such > matrices, becouse: > r 1*r 2=(x,y,z,t)*(z,t,-x,-y)=xz+yt-xz-yt=0, > r 1 r 3=(x,y,z,t)*(-y,x,t,-z)=-xy+xy+tz-tz=0, > r 1*r 4=(x,y,z,t)*(-t,z,-y,x)=-xt+yz-yz+xt=0, > r 2*r 3=(z,t,-x,-y)*(-y,x,t,-z)=-zy+xt-xt+yz=0, > r 2*r 4=(z,t,-x,-y)*(-t,z,-y,x)=-zt+tz+xy-xy=0, > r 3*r 4=(-y,x,t,-z)*(-t,z,-y,x)=yt+xz-yt-yz=0 > > for n=2k-1, k=1,2,3,... such matrix doesn't exist, becouse there are > odd number of terms in scalar product. I suppose that such matrices > exist only for n=2^k, k is an positive integer, but I couldn't know > why and is it really true ... and if there exist such matrices for all > n=2^k or only for 2, 4 and some other powers of 2? It may be you'll find something useful if you look up Hadamard matrices. Sometimes these are expressed as 0-1 matrices, but if you replace each 0 with -1 you get a matrix with orthogonal rows (and columns). Now you have the extra problem of trying to fit the variables in, but it gives you a starting place. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: ortogonal matrices with dim=n containing only n variables posting-account=9QOSvAoAAACEOWJVSDuswW7dB_0wApQO Gecko/2008071615 Fedora/3.0.1-1.fc9 Firefox/3.0.1,gzip(gfe),gzip(gfe) On Jul 30, 3:36æam, Gerry Myerson sorry ... I forgot definition of the ortogonality of the matrix. I > don't really looking ortogonal matrices but matrices in which each row > is ortogonal to other row, so if r 1, r 2, ..., r n are the rows of > this matrix then r i*r j=0 for i different than j, where * denotes > scalar product. For example for n=2 matrix {{x,y},{-y,x}} is such matrix, becouse: > (x,y)*(-y,x)=-xy+xy=0 for n=4 {{x,y,z,t},{z,t,-x,-y},{-y,x,t,-z},{-t,z,-y,x}} is one of such > matrices, becouse: > r 1*r 2=(x,y,z,t)*(z,t,-x,-y)=xz+yt-xz-yt=0, > r 1 r 3=(x,y,z,t)*(-y,x,t,-z)=-xy+xy+tz-tz=0, > r 1*r 4=(x,y,z,t)*(-t,z,-y,x)=-xt+yz-yz+xt=0, > r 2*r 3=(z,t,-x,-y)*(-y,x,t,-z)=-zy+xt-xt+yz=0, > r 2*r 4=(z,t,-x,-y)*(-t,z,-y,x)=-zt+tz+xy-xy=0, > r 3*r 4=(-y,x,t,-z)*(-t,z,-y,x)=yt+xz-yt-yz=0 for n=2k-1, k=1,2,3,... such matrix doesn't exist, becouse there are > odd number of terms in scalar product. I suppose that such matrices > exist only for n=2^k, k is an positive integer, but I couldn't know > why and is it really true ... and if there exist such matrices for all > n=2^k or only for 2, 4 and some other powers of 2? It may be you'll find something useful if you look up Hadamard > matrices. Sometimes these are expressed as 0-1 matrices, > but if you replace each 0 with -1 you get a matrix with orthogonal > rows (and columns). Now you have the extra problem of trying to > fit the variables in, but it gives you a starting place. Indeed, for each matrix like the one Pawel wants one gets a few Hadamard matrices (by evaluating the variables at 1 or -1) This puts a rather strong condition on the sizes of Pawel' matrices. It'd be interesting to know which Hadamard matrices can be lifted to Pawel matrices. -- m === Subject: Re: ortogonal matrices with dim=n containing only n variables posting-account=zKSjoQoAAAC5wIsHKdIzIXzOHEFPgaXy Gecko/20080702 Firefox/2.0.0.16,gzip(gfe),gzip(gfe) > We have matrix {{x,y},{-y,x}} which is 2-dimensional matrix and is > ortogonal and contains only variables x and y. Are there such matrices for all n, which means if there are matrices > with dim=n containing only x1, x2, ..., xn which are ortogonal? If not > for all n, then what is criteria and reason that not for all? and if > for some n such matrces exists is there only one or there exist more, > how many (of course I don't take into account matrices which are > created from givem by row or column permutations)? You also need A^T A = I, so det(A)^2 = 1. In your case this says x^2 + y^2 = 1, so you do not really have two independent variables x and y: you must have x in [-1,1] and y = +-sqrt(1-x^2). R.G. Vickson === Subject: first eigenvalue i am trying to find the numerical value of the first eigenfunction for the Bi-laplacian with zero Dirichlet boundary conditions, ie. u=normal derivative of u =0 on boundary of B in the unit ball B in R^N. I can use the variational principle and then use Maple and test on test functions to get a one sided bound on this... Is there any way to use Maple to get the answer? craig === Subject: Re: first eigenvalue > i am trying to find the numerical value of the first > eigenfunction this should read eigenvalue > for the Bi-laplacian with zero > Dirichlet boundary conditions, ie. > > u=normal derivative of u =0 on boundary of B > > > in the unit ball B in R^N. > > > I can use the variational principle and then use > Maple and test on test functions to get a one sided > bound on this... > > > Is there any way to use Maple to get the answer? > > > > > craig === Subject: Re: Solutions Manuals & Test banks AVAILABLE - Accounting, Finance, Taxation hi there i would be interested in 1. Intermediate Accounting, 12th edition by Kieso, Weygandt, > Warfield please email me at hrksams[at]gmail[dot]com > For this one the TEST BANK > COMPLETE SOLUTIONS MANUALS in PDF and WORD for the > following textbooks: 1. Intermediate Accounting, 12th edition by Kieso, Weygandt, > Warfield; > For this one the TEST BANK is available too. 2. Auditing and Assurance Services, 12th edition by Arens, Elder, > Beasley > For this one the TEST BANK is available too. 3. Advanced Accounting, 9th edition by Fischer, Taylor, Cheng; > For this one the TEST BANK is available too. 4. Contemporary Financial Management, 10th edition by Moyer McGuigan, > Kretlow > 5. West Federal Taxation 2007, 30th edition by Willis, Hoffman, > Maloney, Raabe Email 7sirius7(at)gmail(dot)com to order. > === Subject: q p posting-account=KlY0KgoAAADW-V8Odf9VBWNcNSYrL4RR AntivirXP08; Tablet PC 1.7; .NET CLR 1.0.3705; .NET CLR 1.1.4322; .NET CLR 2.0.50727; InfoPath.2),gzip(gfe),gzip(gfe) http://www.rasoulallah.net/index_english.asp www.nubdalislam.net/msg/1.htm http://www.nubdalislam.net/msg/2.htm http://www.nubdalislam.net/msg/3.htm http://www.nubdalislam.net/msg/4.htm http://www.nubdalislam.net/msg/5.htm http://www.nubdalislam.net/msg/6.htm http://www.nubdalislam.net/msg/7.htm === Subject: the problem with Cantor When you consider computable reals 0<=cr<1 what variety do you get in the decimal expansions? Every possible sequence of digits to infinite length is represented as a computable number. You might argue Omega is not represented in it entirety, but Omega to 10 decimal places is computable, Omega to 100 decimal places is computable, Omega to a googol decimal places is computable, ... What does that ... mean? It means Omega is computable to infinite decimal places. OK you argue, show me the actual computable real where it is apparent to oo decimal places and in theory there is none. But the set of CR does contain this sequence of digits. ALL sequences of digits to oo are computable. And I think that's good enough to refute the existence of sets that are uncountably larger than oo. FACT: All sequences of digits are computable to infinite length. Herc -- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~ === Subject: Re: the problem with Cantor posting-account=euF15goAAACbw3KIqEWxZHCIPUc2KPmU .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506),gzip(gfe),gzip(gfe) > When you consider computable reals 0<=cr<1 > what variety do you get in the decimal expansions? And so another nonstandard mathematician |-|erc, has appeared out of lurkdom to refute Cantor. I've only glanced at this thread, but it appears that |-|erc's argument is another one based on the assumption (that doesn't hold in ZFC) that if for every natural number n, phi(n) holds, then we must have phi(N) holding as well. === Subject: Re: the problem with Cantor posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 Gecko/2008070208 Firefox/3.0.1,gzip(gfe),gzip(gfe) > And so another nonstandard mathematician |-|erc, has > appeared out of lurkdom to refute Cantor. Funny that you think Herc is a MATHEMATICIAN. MoeBlee === Subject: Re: the problem with Cantor posting-account=suWj4AkAAADE1IvGmj55Nmq3f98qb17e SIMBAR Enabled; SIMBAR={70306B22-CB8C-4d52-BFF4-18424E217075}; MathPlayer 2.10b; FunWebProducts; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) When you consider computable reals 0<=cr<1 > what variety do you get in the decimal expansions? And so another nonstandard mathematician |-|erc, has > appeared out of lurkdom to refute Cantor. I've only glanced at this thread, but it appears that > |-|erc's argument is another one based on the > assumption (that doesn't hold in ZFC) that if for every > natural number n, phi(n) holds, then we must have > phi(N) holding as well. *************************************************************** I can't be sure (pretty confussing with all those non-defined terms and stuff), but at the bottom line it seems |-|erc tries somehow to turn over the argument used in Cantor's Diagonal proof and he says: OK, I can write down all the computable numbers and that way I get ALL the possible real numbers, [quote: Computable reals displays EVERY type of decimal expansion imaginable, there is nothing it misses.]. Proof? Very simple: tell me which number would I miss doing this! Oh, but if you can point such a number then it is computable, and thus I would have written it...taraaaaan! Well, there seems to exist a rather huge logical flaw up there: how can you know a priori whether you've written down all the conmputable real numbers? Or even better, and turning over the tortilla once again: what if you present me the list (because it will be a list...right??) of ALL computable real numbers, and then I use Cantor's Diagonal argument and pinpoint a real number which is NOT in that list? Because believe me: in any list of real numbers there will be a number I can pinpoint and prove it is NOT in that list... Of course, it could be |-|erc didn't actually mean the above... Tonio Well === Subject: Re: the problem with Cantor posting-account=F3H0JAgAAADcYVukktnHx7hFG5stjWse .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) > When you consider computable reals 0<=cr<1 > what variety do you get in the decimal expansions? And so another nonstandard mathematician |-|erc, has > appeared out of lurkdom to refute Cantor. I've only glanced at this thread, but it appears that > |-|erc's argument is another one based on the > assumption (that doesn't hold in ZFC) that if for every > natural number n, phi(n) holds, then we must have > phi(N) holding as well. *************************************************************** I can't be sure (pretty confussing with all those non-defined terms > and stuff), but at the bottom line it seems |-|erc tries somehow to > turn over the argument used in Cantor's Diagonal proof and he says: OK, I can write down all the computable numbers and that way I get > ALL the possible real numbers, [quote: Computable reals displays EVERY > type of decimal expansion imaginable, there is nothing it misses.]. > Proof? Very simple: tell me which number would I miss doing this! Oh, > but if you can point such a number then it is computable, and thus I > would have written it...taraaaaan! Well, there seems to exist a rather huge logical flaw up there: how > can you know a priori whether you've written down all the conmputable > real numbers? > Or even better, and turning over the tortilla once again: what if > you present me the list (because it will be a list...right??) of ALL > computable real numbers, and then I use Cantor's Diagonal argument and > pinpoint a real number which is NOT in that list? Because believe me: > in any list of real numbers there will be a number I can pinpoint and > prove it is NOT in that list... I guess the argument goes like: IF there is such a list, THEN no sequence can ever escape it. Let's assume we have such a list. Then we can apply a suitable diagonalisation function and get a sequence that differs from the first sequence in the first place, from the second sequence in the second place, and so on. The original list is infinite. By extending the argument: Any such diagonal sequence differs from the last sequence in the last place. That is to say that, at the limit, any diagonal sequence is still a sequence in the list, namely the limit sequence. -LV > Of course, it could be |-|erc didn't actually mean the above... Tonio Well === Subject: Re: the problem with Cantor posting-account=suWj4AkAAADE1IvGmj55Nmq3f98qb17e SIMBAR Enabled; SIMBAR={70306B22-CB8C-4d52-BFF4-18424E217075}; MathPlayer 2.10b; FunWebProducts; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) When you consider computable reals 0<=cr<1 > what variety do you get in the decimal expansions? And so another nonstandard mathematician |-|erc, has > appeared out of lurkdom to refute Cantor. I've only glanced at this thread, but it appears that > |-|erc's argument is another one based on the > assumption (that doesn't hold in ZFC) that if for every > natural number n, phi(n) holds, then we must have > phi(N) holding as well. *************************************************************** I can't be sure (pretty confussing with all those non-defined terms > and stuff), but at the bottom line it seems |-|erc tries somehow to > turn over the argument used in Cantor's Diagonal proof and he says: OK, I can write down all the computable numbers and that way I get > ALL the possible real numbers, [quote: Computable reals displays EVERY > type of decimal expansion imaginable, there is nothing it misses.]. > Proof? Very simple: tell me which number would I miss doing this! Oh, > but if you can point such a number then it is computable, and thus I > would have written it...taraaaaan! Well, there seems to exist a rather huge logical flaw up there: how > can you know a priori whether you've written down all the conmputable > real numbers? > Or even better, and turning over the tortilla once again: what if > you present me the list (because it will be a list...right??) of ALL > computable real numbers, and then I use Cantor's Diagonal argument and > pinpoint a real number which is NOT in that list? Because believe me: > in any list of real numbers there will be a number I can pinpoint and > prove it is NOT in that list... I guess the argument goes like: IF there is such a list, THEN no > sequence can ever escape it. Let's assume we have such a list. Then we can apply a suitable > diagonalisation function and get a sequence that differs from the > first sequence in the first place, from the second sequence in the > second place, and so on. The original list is infinite. By extending the argument: Any such > diagonal sequence differs from the last sequence in the last > place. That is to say that, at the limit, any diagonal sequence is > still a sequence in the list, namely the limit sequence. -LV ************************************************************** Last sequence? Where is that? At the limit? At the limit of what and when?? Limit sequence? Limit of what? I really don't understand what you meant. Tonio === Subject: Re: the problem with Cantor > I guess the argument goes like: IF there is such a list, THEN no > sequence can ever escape it. > > Let's assume we have such a list. If one assumes a falsehood, one can prove anything, including that that assumption is false. So that if assuming that every sequence is in the list still allows us to prove that there is a sequence not in the list, as it does, then it is not in the list. > Then we can apply a suitable > diagonalisation function and get a sequence that differs from the > first sequence in the first place, from the second sequence in the > second place, and so on. > > The original list is infinite. By extending the argument: Any such > diagonal sequence differs from the last sequence in the last > place. That is to say that, at the limit, any diagonal sequence is > still a sequence in the list, namely the limit sequence. Just what in hell is that supposed to mean? To be in the list means that it has a finite position in the list, so there is no such thing as a limit member of such a list. === Subject: Re: the problem with Cantor posting-account=AdyLXQoAAABgRay99CKv1O8Y_7jjivwq InfoPath.1),gzip(gfe),gzip(gfe) > Let's assume we have such a list. Then we can apply > a suitable diagonalisation function and get a sequence > that differs from the first sequence in the first place, > from the second sequence in the second place, and so on. The original list is infinite. By extending the argument: > Any such diagonal sequence differs from the last sequence > in the last place. That is to say that, at the limit, > any diagonal sequence is still a sequence in the list, > namely the limit sequence. Just what do you mean by a list? If you mean a correspondence with 1, 2, 3, ... (what one would expect if you're using a diagonal argument with the positive integers), then there is no last place since there is no last positive integer, something most 12-year olds could have told you. Dave L. Renfro === Subject: Re: the problem with Cantor > Let's assume we have such a list. Then we can apply > a suitable diagonalisation function and get a sequence > that differs from the first sequence in the first place, > from the second sequence in the second place, and so on. > And so on in this case means: it differs from ALL entries (i.e sequences) in the list. And that in turn means, it is not itself a sequence in the list. Period. Though of course in the strange worlds of cranks there are always additional possibilities: For every line of Cantor's list it is true that this line does not contain the diagonal number. Nevertheless the diagonal number may be in the infinite list. (WM, sci.logic) Of course, this contradicts ANY form of logic, but who cares?! > The original list is infinite. By extending the argument: > Any such diagonal sequence differs from the last sequence > in the last place. > Errr... what last sequence? Since the considered list does not have a last sequence, it's hard to see what this idiot might mean here. > That is to say ... > Yes? > that, at the limit, ... > Limit? What limit? Hell!!! > any diagonal sequence is still a sequence in the list, > namely the limit sequence. > What EXACTLY _is_ /the limit sequence/ (->definition). And where's the _proof_ that this /limit sequence/ is 1. identical with the diagonal sequence and 2. a sequence in the list? (->proof) Unproven statements carry little weight in the world of mathematics. - Amir D. Aczel Just what do you mean by a list? If you mean a correspondence > with 1, 2, 3, ... (what one would expect if you're using a > diagonal argument with the positive integers), then there > is no last place since there is no last positive integer, > something most 12-year olds could have told you. > Trying to ARGUE with a crank? :-o B. -- For every line of Cantor's list it is true that this line does not contain the diagonal number. Nevertheless the diagonal number may be in the infinite list. (WM, sci.logic) === Subject: Re: the problem with Cantor > The original list is infinite. By extending the argument: > Any such diagonal sequence differs from the last sequence > in the last place. > Errr... what last sequence? Since the considered list does not have a > last sequence, it's hard to see what this idiot might mean here. And to compound the idiot's idiocy of a last sequence in a list without any last sequence, he refers to a last place in a set of sequences none of which have last places. At least doubly idiotic. === Subject: Re: the problem with Cantor posting-account=F3H0JAgAAADcYVukktnHx7hFG5stjWse .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) > Let's assume we have such a list. Then we can apply > a suitable diagonalisation function and get a sequence > that differs from the first sequence in the first place, > from the second sequence in the second place, and so on. The original list is infinite. By extending the argument: > Any such diagonal sequence differs from the last sequence > in the last place. That is to say that, at the limit, > any diagonal sequence is still a sequence in the list, > namely the limit sequence. Just what do you mean by a list? If you mean a correspondence > with 1, 2, 3, ... (what one would expect if you're using a > diagonal argument with the positive integers), then there > is no last place since there is no last positive integer, > something most 12-year olds could have told you. Dave L. Renfro Just and nise as always. IF there is such a list, THEN no sequence can ever escape it. We can then call all possible diagonal sequences for any enumeration rule of such putative list: the class of omega-sequences for that rule. (And this is now a very well defined notion.) That such complete infinite list itself exists, might finally be a matter for an axiom. -LV === Subject: Re: the problem with Cantor posting-account=AdyLXQoAAABgRay99CKv1O8Y_7jjivwq InfoPath.1),gzip(gfe),gzip(gfe) > We can then call all possible diagonal sequences for any > enumeration rule of such putative list: the class of > omega-sequences for that rule. (And this is now a very > well defined notion.) You can do this, but that's like saying we can consider a basket of apples when someone points out that the rocketship you were describing doesn't exist. Anyway, for what it's worth, this has been studied: Robert Gray, Georg Cantor and transcendental numbers, American Mathematical Monthly 101 #1 (November 1994), 819-832. > That such complete infinite list itself exists, might > finally be a matter for an axiom. Why should this require an axiom? It's a subset of the collection of all sequences of elements from some fixed set, so it's existence follows from the subset selection axiom. Dave L. Renfro === Subject: Re: the problem with Cantor posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 Gecko/2008070208 Firefox/3.0.1,gzip(gfe),gzip(gfe) > That such complete infinite list itself exists, might > finally be a matter for an axiom. Why should this require an axiom? It's a subset of the > collection of all sequences of elements from some fixed > set, so it's existence follows from the subset selection > axiom. If I understand what he's saying, yes, it's a theorem that the set exists. But, contrary, to his description, it is not listable (enumerable, i.e. countable). MoeBlee === Subject: Re: the problem with Cantor > Let's assume we have such a list. Then we can apply > a suitable diagonalisation function and get a sequence > that differs from the first sequence in the first place, > from the second sequence in the second place, and so on. The original list is infinite. By extending the argument: > Any such diagonal sequence differs from the last sequence > in the last place. That is to say that, at the limit, > any diagonal sequence is still a sequence in the list, > namely the limit sequence. Just what do you mean by a list? If you mean a correspondence > with 1, 2, 3, ... (what one would expect if you're using a > diagonal argument with the positive integers), then there > is no last place since there is no last positive integer, > something most 12-year olds could have told you. Dave L. Renfro > > Just and nise as always. > > IF there is such a list, THEN no sequence can ever escape it. > > We can then call all possible diagonal sequences for any enumeration > rule of such putative list: the class of omega-sequences for that > rule. (And this is now a very well defined notion.) > > That such complete infinite list itself exists, might finally be a > matter for an axiom. > But requiring the addition of such an axiom to most set theory axiom systems allows proof of P / ~P, which characteristic is quite generally regarded as being deleterious to an axiom system's worth. === Subject: Re: the problem with Cantor posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 Gecko/2008070208 Firefox/3.0.1,gzip(gfe),gzip(gfe) Let's assume we have such a list. Then we can apply > a suitable diagonalisation function and get a sequence > that differs from the first sequence in the first place, > from the second sequence in the second place, and so on. The original list is infinite. By extending the argument: > Any such diagonal sequence differs from the last sequence > in the last place. That is to say that, at the limit, > any diagonal sequence is still a sequence in the list, > namely the limit sequence. Just what do you mean by a list? If you mean a correspondence > with 1, 2, 3, ... (what one would expect if you're using a > diagonal argument with the positive integers), then there > is no last place since there is no last positive integer, > something most 12-year olds could have told you. Dave L. Renfro Just and nise as always. IF there is such a list, THEN no sequence can ever escape it. We can then call all possible diagonal sequences for any enumeration > rule of such putative list: the class of omega-sequences for that > rule. (And this is now a very well defined notion.) That such complete infinite list itself exists, might finally be a > matter for an axiom. You mean the existence of a list (an ENUMERATION) of the set of anti- diagonals of all lists of computable reals? Then make that an axiom along with WHAT OTHER axioms for mathematics? It's inconsistent with the axioms of Z set theory. So if you make it an axiom, you'll need to tell us what the rest of your axioms are. MoeBlee === Subject: Re: the problem with Cantor So if you make it an axiom, you'll need to > tell us what the rest of your axioms are. > WHY?! Since he's a crank, he needn't. :-) B. -- For every line of Cantor's list it is true that this line does not contain the diagonal number. Nevertheless the diagonal number may be in the infinite list. (WM, sci.logic) === Subject: Re: the problem with Cantor posting-account=F3H0JAgAAADcYVukktnHx7hFG5stjWse .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) When you consider computable reals 0<=cr<1 > what variety do you get in the decimal expansions? And so another nonstandard mathematician |-|erc, has > appeared out of lurkdom to refute Cantor. I've only glanced at this thread, but it appears that > |-|erc's argument is another one based on the > assumption (that doesn't hold in ZFC) that if for every > natural number n, phi(n) holds, then we must have > phi(N) holding as well. *************************************************************** I can't be sure (pretty confussing with all those non-defined terms > and stuff), but at the bottom line it seems |-|erc tries somehow to > turn over the argument used in Cantor's Diagonal proof and he says: OK, I can write down all the computable numbers and that way I get > ALL the possible real numbers, [quote: Computable reals displays EVERY > type of decimal expansion imaginable, there is nothing it misses.]. > Proof? Very simple: tell me which number would I miss doing this! Oh, > but if you can point such a number then it is computable, and thus I > would have written it...taraaaaan! Well, there seems to exist a rather huge logical flaw up there: how > can you know a priori whether you've written down all the conmputable > real numbers? > Or even better, and turning over the tortilla once again: what if > you present me the list (because it will be a list...right??) of ALL > computable real numbers, and then I use Cantor's Diagonal argument and > pinpoint a real number which is NOT in that list? Because believe me: > in any list of real numbers there will be a number I can pinpoint and > prove it is NOT in that list... I guess the argument goes like: IF there is such a list, THEN no > sequence can ever escape it. Let's assume we have such a list. Then we can apply a suitable > diagonalisation function and get a sequence that differs from the > first sequence in the first place, from the second sequence in the > second place, and so on. The original list is infinite. By extending the argument: Any such > diagonal sequence differs from the last sequence in the last > place. That is to say that, at the limit, any diagonal sequence is > still a sequence in the list, namely the limit sequence. In this sense, I guess we might call all possible diagonal sequences for any enumeration rule of the list: the class of omega sequences for that rule. -LV -LV Of course, it could be |-|erc didn't actually mean the above... Tonio Well === Subject: Re: the problem with Cantor posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 Gecko/2008070208 Firefox/3.0.1,gzip(gfe),gzip(gfe) > I guess the argument goes like: IF there is such a list, THEN no > sequence can ever escape it. Let's assume we have such a list. Then we can apply a suitable > diagonalisation function and get a sequence that differs from the > first sequence in the first place, from the second sequence in the > second place, and so on. The original list is infinite. By extending the argument: Any such > diagonal sequence differs from the last sequence in the last > place. That is to say that, at the limit, any diagonal sequence is > still a sequence in the list, namely the limit sequence. the last sequence, the last place, and the limit sequence. All undefined nonsense. We give you our EXPLICIT axioms, rules, and primitives, as well as our EXPLICIT definitions. What are yours, including your explicit RIGOROUS definition of the last sequence, the last place, and the limit sequence? MoeBlee === Subject: Re: the problem with Cantor posting-account=F3H0JAgAAADcYVukktnHx7hFG5stjWse .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) I guess the argument goes like: IF there is such a list, THEN no > sequence can ever escape it. Let's assume we have such a list. Then we can apply a suitable > diagonalisation function and get a sequence that differs from the > first sequence in the first place, from the second sequence in the > second place, and so on. The original list is infinite. By extending the argument: Any such > diagonal sequence differs from the last sequence in the last > place. That is to say that, at the limit, any diagonal sequence is > still a sequence in the list, namely the limit sequence. the last sequence, the last place, and the limit sequence. All > undefined nonsense. It is not undefined nonsense: That is to say that, at the limit, any diagonal sequence is still a sequence in the list, namely the limit sequence. I am not trying to be ambiguous: what is unclear? -LV > We give you our EXPLICIT axioms, rules, and primitives, as well as our > EXPLICIT definitions. What are yours, including your explicit RIGOROUS > definition of the last sequence, the last place, and the limit > sequence? MoeBlee === Subject: Re: the problem with Cantor posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 Gecko/2008070208 Firefox/3.0.1,gzip(gfe),gzip(gfe) I guess the argument goes like: IF there is such a list, THEN no > sequence can ever escape it. Let's assume we have such a list. Then we can apply a suitable > diagonalisation function and get a sequence that differs from the > first sequence in the first place, from the second sequence in the > second place, and so on. The original list is infinite. By extending the argument: Any such > diagonal sequence differs from the last sequence in the last > place. That is to say that, at the limit, any diagonal sequence is > still a sequence in the list, namely the limit sequence. the last sequence, the last place, and the limit sequence. All > undefined nonsense. It is not undefined nonsense: That is to say that, at the limit, any diagonal sequence is still a > sequence in the list, namely the limit sequence. I am not trying to be ambiguous: what is unclear? > We give you our EXPLICIT axioms, rules, and primitives, as well as our > EXPLICIT definitions. What are yours, including your explicit RIGOROUS > definition of the last sequence, the last place, and the limit > sequence? There is NO last sequence. There is NO last place. And 'limit' is meaningless unless you define the specific sense of taking such limits (by some topology, metric, ordering, or other explicit definition). It is meaningless (and typically CRANK) to just wave one's hand and to speak of the limit case when one has not given an actual mathematical definition of such a limit. MoeBlee === Subject: Re: the problem with Cantor posting-account=F3H0JAgAAADcYVukktnHx7hFG5stjWse .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) > I guess the argument goes like: IF there is such a list, THEN no > sequence can ever escape it. Let's assume we have such a list. Then we can apply a suitable > diagonalisation function and get a sequence that differs from the > first sequence in the first place, from the second sequence in the > second place, and so on. The original list is infinite. By extending the argument: Any such > diagonal sequence differs from the last sequence in the last > place. That is to say that, at the limit, any diagonal sequence is > still a sequence in the list, namely the limit sequence. the last sequence, the last place, and the limit sequence. All > undefined nonsense. It is not undefined nonsense: That is to say that, at the limit, any diagonal sequence is still a > sequence in the list, namely the limit sequence. I am not trying to be ambiguous: what is unclear? > We give you our EXPLICIT axioms, rules, and primitives, as well as our > EXPLICIT definitions. What are yours, including your explicit RIGOROUS > definition of the last sequence, the last place, and the limit > sequence? There is NO last sequence. There is NO last place. And 'limit' is > meaningless unless you define the specific sense of taking such limits > (by some topology, metric, ordering, or other explicit definition). It > is meaningless (and typically CRANK) to just wave one's hand and to > speak of the limit case when one has not given an actual > mathematical definition of such a limit. MoeBlee I have given an argument based on the basic properties of the natural numbers, and on induction to give a very specific and again basic sense to the limit case. Here I have furtherly given a very specific and unambiguous sense to the diagonal procedure up to a definition for the class of omega- sequences for any given *rule* to enumerate the original, complete, infinite, and putative enumerable collection (was, a bit ambiguously, list: just inherited from the previous discussion). No such *rule*, no diagonal function to build! No such *collection* (the list) to begin with... not even the computables!? So that your objections, in general, are a contradiction to your own language and tools, and not even near to be real objections. (Why do you keep with this reminding me of *your* axioms? You are just pointing out the *problems* with the accepted approaches, while the objections to the above are really immaterial: you deny your own language and tools when you deny that the above makes any sense as it stands. In the name of your axioms. And that is all your objections keep amounting to.) -LV === Subject: Re: the problem with Cantor posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 Gecko/2008070208 Firefox/3.0.1,gzip(gfe),gzip(gfe) > I guess the argument goes like: IF there is such a list, THEN no > sequence can ever escape it. Let's assume we have such a list. Then we can apply a suitable > diagonalisation function and get a sequence that differs from the > first sequence in the first place, from the second sequence in the > second place, and so on. The original list is infinite. By extending the argument: Any such > diagonal sequence differs from the last sequence in the last > place. That is to say that, at the limit, any diagonal sequence is > still a sequence in the list, namely the limit sequence. the last sequence, the last place, and the limit sequence. All > undefined nonsense. It is not undefined nonsense: That is to say that, at the limit, any diagonal sequence is still a > sequence in the list, namely the limit sequence. I am not trying to be ambiguous: what is unclear? > We give you our EXPLICIT axioms, rules, and primitives, as well as our > EXPLICIT definitions. What are yours, including your explicit RIGOROUS > definition of the last sequence, the last place, and the limit > sequence? There is NO last sequence. There is NO last place. And 'limit' is > meaningless unless you define the specific sense of taking such limits > (by some topology, metric, ordering, or other explicit definition). It > is meaningless (and typically CRANK) to just wave one's hand and to > speak of the limit case when one has not given an actual > mathematical definition of such a limit. > I have given an argument based on the basic properties of the natural > numbers, and on induction to give a very specific and again basic > sense to the limit case. Your argument is NONSENSE. It's been explained to you. > Here I have furtherly given a very specific and unambiguous sense to > the diagonal procedure up to a definition for the class of omega- > sequences for any given *rule* to enumerate the original, complete, > infinite, and putative enumerable collection (was, a bit ambiguously, > list: just inherited from the previous discussion). No such *rule*, no diagonal function to build! > No such *collection* (the list) to begin with... not even the > computables!? So that your objections, in general, are a contradiction to your own > language and tools, and not even near to be real objections. > (Why do you keep with this reminding me of *your* axioms? You are just > pointing out the *problems* with the accepted approaches, while the > objections to the above are really immaterial: you deny your own > language and tools when you deny that the above makes any sense as it > stands. In the name of your axioms. And that is all your objections > keep amounting to.) No, you contradict two principles combined: (1) Intuitionisitc logic (weaker even than classical logic) combined with (2) The principle that for any formalizable property P (not mentioning D) and for any set S there is a set D that is the set of members of S that have property P. THEN, my argument is not that you are not allowed to contradict those, but rather that if you DO contradict those, then what principles do you offer in their place? But you have no appreciation really of that question, since you have no understanding of what is involved in formulating a sufficient body of logical and mathematical principles to prove the theorems that are used in calculus, which mathematics for the sciences and technology. And you contradict yourself: You said you wanted to learn. But instead, your posting reflects only someone doggedly committed to repeating his most basic misconceptions. MoeBlee === Subject: Re: the problem with Cantor posting-account=F3H0JAgAAADcYVukktnHx7hFG5stjWse .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) > I have given an argument based on the basic properties of the natural > numbers, and on induction to give a very specific and again basic > sense to the limit case. Your argument is NONSENSE. It's been explained to you. Here I have furtherly given a very specific and unambiguous sense to > the diagonal procedure up to a definition for the class of omega- > sequences for any given *rule* to enumerate the original, complete, > infinite, and putative enumerable collection (was, a bit ambiguously, > list: just inherited from the previous discussion). No such *rule*, no diagonal function to build! > No such *collection* (the list) to begin with... not even the > computables!? So that your objections, in general, are a contradiction to your own > language and tools, and not even near to be real objections. > (Why do you keep with this reminding me of *your* axioms? You are just > pointing out the *problems* with the accepted approaches, while the > objections to the above are really immaterial: you deny your own > language and tools when you deny that the above makes any sense as it > stands. In the name of your axioms. And that is all your objections > keep amounting to.) No, you contradict two principles combined: (1) Intuitionisitc logic (weaker even than classical logic) I am afraid I cannot see where I am contradicting intuitionistic logic. Could you be explicit? (Can *I* really be the one to perform such comparisons?) > combined > with (2) The principle that for any formalizable property P (not > mentioning D) and for any set S there is a set D that is the set of > members of S that have property P. How can I ever be contradicting that one? In a fully-computable realm, any formalizable property does express a set, and viceversa. It is rather the accepted approach that is problematic in this respect. -LV > THEN, my argument is not that you are not allowed to contradict those, > but rather that if you DO contradict those, then what principles do > you offer in their place? === Subject: Re: the problem with Cantor posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 Gecko/2008070208 Firefox/3.0.1,gzip(gfe),gzip(gfe) > No, you contradict two principles combined: (1) Intuitionisitc logic (weaker even than classical logic) I am afraid I cannot see where I am contradicting intuitionistic > logic. Could you be explicit? (Can *I* really be the one to perform > such comparisons?) Not intuitionistic logic ALONE. I said COMBINED. You are contradicting intuitionistic logic COMBINED with (2): > combined > with (2) The principle that for any formalizable property P (not > mentioning D) and for any set S there is a set D that is the set of > members of S that have property P. How can I ever be contradicting that one? Because (2) combined with intuitionistic logic, entails that there is no set that can map onto its power set. So when you claim that there is a set that maps onto its power set then you are contradicting intutionistic logic combined with (2). > In a fully-computable realm, > any formalizable property does express a set, and viceversa. It is > rather the accepted approach that is problematic in this respect. Whatever you mean by accepted approach, you are not coming to grips with this fact: Intutionistic logic and the axiom schema of separation prove that no set maps onto its power set. So if you think there is a set that maps onto its power set then you must discard either intutionistic logic or the axiom schema of separation. > THEN, my argument is not that you are not allowed to contradict those, > but rather that if you DO contradict those, then what principles do > you offer in their place? No answer from you. MoeBlee === Subject: Re: the problem with Cantor > In a fully-computable realm, ... > Do we have a formal definition of /fully-computable realm/, Moe? Moreover does this idiot ASSUME that the context of our considerations is a fully-computable realm? - Whatever that may be. > ... any formalizable property does express a set [...]. > Fascinating. One might think that (at least in a standard realm) /being not an element of itself/ is a formalizable property: x !e x. Still it's hard to see how in a fully-computable realm the Russell-set can exist, i.e. a set y such that Ax(x e y <-> x !e x). Note that this property (in a class theory) does express (determine) a proper class. (Though in NF, where there are only sets, the predicate x !e x is not admissible, since it's not stratified.) > It is rather the accepted approach that is problematic in this respect. > Crank speak. B. -- For every line of Cantor's list it is true that this line does not contain the diagonal number. Nevertheless the diagonal number may be in the infinite list. (WM, sci.logic) === Subject: Re: the problem with Cantor posting-account=F3H0JAgAAADcYVukktnHx7hFG5stjWse .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) I guess the argument goes like: IF there is such a list, THEN no > sequence can ever escape it. Let's assume we have such a list. Then we can apply a suitable > diagonalisation function and get a sequence that differs from the > first sequence in the first place, from the second sequence in the > second place, and so on. The original list is infinite. By extending the argument: Any such > diagonal sequence differs from the last sequence in the last > place. That is to say that, at the limit, any diagonal sequence is > still a sequence in the list, namely the limit sequence. the last sequence, the last place, and the limit sequence. All > undefined nonsense. It is not undefined nonsense: That is to say that, at the limit, any diagonal sequence is still a > sequence in the list, namely the limit sequence. I am not trying to be ambiguous: what is unclear? > We give you our EXPLICIT axioms, rules, and primitives, as well as our > EXPLICIT definitions. What are yours, including your explicit RIGOROUS > definition of the last sequence, the last place, and the limit > sequence? There is NO last sequence. There is NO last place. And 'limit' is > meaningless unless you define the specific sense of taking such limits > (by some topology, metric, ordering, or other explicit definition). It > is meaningless (and typically CRANK) to just wave one's hand and to > speak of the limit case when one has not given an actual > mathematical definition of such a limit. > I have given an argument based on the basic properties of the natural > numbers, and on induction to give a very specific and again basic > sense to the limit case. Your argument is NONSENSE. It's been explained to you. Here I have furtherly given a very specific and unambiguous sense to > the diagonal procedure up to a definition for the class of omega- > sequences for any given *rule* to enumerate the original, complete, > infinite, and putative enumerable collection (was, a bit ambiguously, > list: just inherited from the previous discussion). No such *rule*, no diagonal function to build! > No such *collection* (the list) to begin with... not even the > computables!? So that your objections, in general, are a contradiction to your own > language and tools, and not even near to be real objections. > (Why do you keep with this reminding me of *your* axioms? You are just > pointing out the *problems* with the accepted approaches, while the > objections to the above are really immaterial: you deny your own > language and tools when you deny that the above makes any sense as it > stands. In the name of your axioms. And that is all your objections > keep amounting to.) No, you contradict two principles combined: (1) Intuitionisitc logic (weaker even than classical logic) combined > with (2) The principle that for any formalizable property P (not > mentioning D) and for any set S there is a set D that is the set of > members of S that have property P. THEN, my argument is not that you are not allowed to contradict those, > but rather that if you DO contradict those, then what principles do > you offer in their place? But you have no appreciation really of that question, since you have > no understanding of what is involved in formulating a sufficient body > of logical and mathematical principles to prove the theorems that are > used in calculus, which mathematics for the sciences and technology. And you contradict yourself: You said you wanted to learn. But > instead, your posting reflects only someone doggedly committed to > repeating his most basic misconceptions. MoeBlee Now you are getting just and nise yourself. I am glad you have given something explicit I can work upon. I am simply not (yet?) able to explicitly state the list of my axioms. I keep telling you: what you get is what you see, natural numbers plus induction. If anything else is needed, then let's name it. Anyway, I'll tell you what I can get from your argument above. -LV === Subject: Re: the problem with Cantor > The original list is infinite. By extending the argument: Any such > diagonal sequence differs from the last sequence in the last > place. That is to say that, at the limit, any diagonal sequence is > still a sequence in the list, namely the limit sequence. the last sequence, the last place, and the limit sequence. All > undefined nonsense. > > > It is not undefined nonsense: > > That is to say that, at the limit, any diagonal sequence is still a > sequence in the list, namely the limit sequence. > > I am not trying to be ambiguous: what is unclear? Among other things, how a sequence which differs from every member of a list can still be a member of that list. === Subject: Re: the problem with Cantor > When you consider computable reals 0<=cr<1 > what variety do you get in the decimal expansions? > > And so another nonstandard mathematician |-|erc, has > appeared out of lurkdom to refute Cantor. > > I've only glanced at this thread, but it appears that > |-|erc's argument is another one based on the > assumption (that doesn't hold in ZFC) that if for every > natural number n, phi(n) holds, then we must have > phi(N) holding as well. Where ZF's universe is the set of all sets that don't contain themselves, which it is in some other theory or ZF would be complete, which via Goedel's incompleness would have it inconsistent, then ZF's universe is the Russell set. Skolemize, it's countable: bijecting to the natural numbers while containing itself, so they contain themselves. It's the contrapositive (else it's not), it does hold in ZFC because if it didn't then ZFC would yield a contradiction, so it does: contradiction: ZFC is inconsistent. programs is quite remarkable. http://plus.maths.org/issue37/features/omega/index.html For example, he suggests that a program of length n can be sampled by sampling n-many bits at random, but, then he has sampled n-many different programs, of length 1, ..., n. Consider sampling a real number, uniformly from the unit interval, where like any number it's sampled by successive bits of its expansion. In sampling a rational number a particular rational is sampled infinitely many times, in an irrational, each distinctly once. In the consideration of infinite programs, eg loops, there are many ways to consider statically why a program, even if it never halts (which basically has that it loops in place, the instruction at address x is: goto x) instead loopingly halts or it can be compressed. Borel vs. Combinatorics, anyone? There's only one theory with no axioms, consistent and complete, it's a theory of nothing. Coincidentally, that's everything. Ross F. === Subject: Re: the problem with Cantor posting-account=Cbgh4AoAAAAr0dt1RqLOClWCyUWii2fU Gecko/2008071719 Firefox/3.0.1,gzip(gfe),gzip(gfe) > When you consider computable reals 0<=cr<1 > what variety do you get in the decimal expansions? Every possible sequence of digits to infinite length > is represented as a computable number. æYou might > argue Omega is not represented in it entirety, but > Omega to 10 decimal places is computable, > Omega to 100 decimal places is computable, > Omega to a googol decimal places is computable, > ... What does that ... mean? æIt means Omega is computable > to infinite decimal places. æOK you argue, show me the > actual computable real where it is apparent to oo decimal > places and in theory there is none. æBut the set of CR does > contain this sequence of digits. æALL sequences of digits > to oo are computable. æAnd I think that's good enough to > refute the existence of sets that are uncountably larger than oo. FACT: All sequences of digits are computable to infinite length. Herc > -- > ~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Whoa! This looks a bit like the WM tree in disguise! hagman === Subject: Re: the problem with Cantor > When you consider computable reals 0<=cr<1 > what variety do you get in the decimal expansions? > Every possible sequence of digits to infinite length > is represented as a computable number. æYou might > argue Omega is not represented in it entirety, but > Omega to 10 decimal places is computable, > Omega to 100 decimal places is computable, > Omega to a googol decimal places is computable, > ... > What does that ... mean? æIt means Omega is computable > to infinite decimal places. æOK you argue, show me the > actual computable real where it is apparent to oo decimal > places and in theory there is none. æBut the set of CR does > contain this sequence of digits. æALL sequences of digits > to oo are computable. æAnd I think that's good enough to > refute the existence of sets that are uncountably larger than oo. > FACT: All sequences of digits are computable to infinite length. > Herc > -- > ~~~~~~~~~~~~~~~~~~~~~~~~~~~~ > > Whoa! This looks a bit like the WM tree in disguise! He has been asked what he means by omega, but didn't give a sensible answer. Maybe it is something like what is called omega here: http://plus.maths.org/issue37/features/omega/index.html But who knows. Ralf === Subject: Re: the problem with Cantor posting-account=AdyLXQoAAABgRay99CKv1O8Y_7jjivwq InfoPath.1),gzip(gfe),gzip(gfe) > When you consider computable reals 0<=cr<1 > what variety do you get in the decimal expansions? Every possible sequence of digits to infinite length > is represented as a computable number. You might > argue Omega is not represented in it entirety, but > Omega to 10 decimal places is computable, > Omega to 100 decimal places is computable, > Omega to a googol decimal places is computable, > ... What does that ... mean? It means Omega is computable > to infinite decimal places. OK you argue, show me the > actual computable real where it is apparent to oo decimal > places and in theory there is none. But the set of CR does > contain this sequence of digits. ALL sequences of digits > to oo are computable. And I think that's good enough to > refute the existence of sets that are uncountably larger > than oo. FACT: All sequences of digits are computable to infinite > length. I believe the issue is the existence of a uniform computability procedure for each n-digit truncation of the number. Sure, there is such a procedure for the first 100 digits, such a procedure for the first 1000 digits, such a procedure for the first 10000 digits, and so on, but these procedures have to be essentially the same for all such n-digit truncations if we're going to use them to verify the computability of the number. For computable numbers this can be arranged, but for non-computable numbers this can't be arranged. Kind of like uniform continuity vs. continuity, or any situation in which a statement of the form (for all x)(there exists y) is true, but not the corresponding statement where the quantifiers are switched, (there exists y)(for all x). Anyway, I don't know much about formal computability, so maybe someone who does can comment on whether this general idea has any bearing on the actual issue at hand. Dave L. Renfro === Subject: Re: the problem with Cantor posting-account=McZ3aQkAAADz6LV-boDe1LcriRhf3lj3 Gecko/2008070208 Firefox/2.0.0.9;MEGAUPLOAD 1.0,gzip(gfe),gzip(gfe) > When you consider computable reals 0<=cr<1 > what variety do you get in the decimal expansions? Every possible sequence of digits to infinite length > is represented as a computable number. æYou might > argue Omega is not represented in it entirety, but > Omega to 10 decimal places is computable, > Omega to 100 decimal places is computable, > Omega to a googol decimal places is computable, > ... What does that ... mean? æIt means Omega is computable > to infinite decimal places. æOK you argue, show me the > actual computable real where it is apparent to oo decimal > places and in theory there is none. æBut the set of CR does > contain this sequence of digits. æALL sequences of digits > to oo are computable. æAnd I think that's good enough to > refute the existence of sets that are uncountably larger than oo. FACT: All sequences of digits are computable to infinite length. Herc > -- > ~~~~~~~~~~~~~~~~~~~~~~~~~~~~ I tried to prove something similar. But, when we define the rational matrix R, r(i,j) = j/i, we see that there are as many rational numbers lesser than 1, as rational numbers greater than 1. Since in the first row of R, all natural numbers n are enumerated, rational numbers greater than 1 overflow capacity of natural numbers to count, regardless to the infinite length of the string of natural numbers. This infinity increases the size of the rational matrix R, too. kunzmilan === Subject: Re: the problem with Cantor > When you consider computable reals 0<=cr<1 > what variety do you get in the decimal expansions? > > Every possible sequence of digits to infinite length > is represented as a computable number. No, that's just not true. There are uncountablyl many such sequences, but only countably many computable numbers. You might > argue Omega is not represented in it entirety, but > Omega to 10 decimal places is computable, > Omega to 100 decimal places is computable, > Omega to a googol decimal places is computable, > ... What do you mean here by Omega? Anyway, your reasoning is flawed. 10 is a finite natural number; and 10^100 is a finite natural number; and 10^googol is a finite natural number; but the set of all finite natural numbers, N, is not a fiinite natural number. You have to be careful when passing to infinity. > > What does that ... mean? It means Omega is computable > to infinite decimal places. You haven't used any properties of computable numbers here. Whatever you mean by Omega (do you mean lower-case omega, 'w' in ASCII, to represent the order type of N?), you haven't shown anything about its computability. >OK you argue, show me the > actual computable real where it is apparent to oo decimal > places and in theory there is none. But the set of CR does > contain this sequence of digits. ALL sequences of digits > to oo are computable. And I think that's good enough to > refute the existence of sets that are uncountably larger than oo. > > FACT: All sequences of digits are computable to infinite length. > Clearly not. === Subject: Re: the problem with Cantor When you consider computable reals 0<=cr<1 > what variety do you get in the decimal expansions? Every possible sequence of digits to infinite length > is represented as a computable number. No, that's just not true. There are uncountablyl many such sequences, > but only countably many computable numbers. > You might > argue Omega is not represented in it entirety, but > Omega to 10 decimal places is computable, > Omega to 100 decimal places is computable, > Omega to a googol decimal places is computable, > ... What do you mean here by Omega? Isn't the 1st digit of Omega 1 if the 1st Turing machine halts, 0 otherwise something like that Anyway, your reasoning is flawed. 10 is a finite natural number; and > 10^100 is a finite natural number; and 10^googol is a finite natural > number; but the set of all finite natural numbers, N, is not a fiinite > natural number. You have to be careful when passing to infinity. How many digits is Omega computable to? > What does that ... mean? It means Omega is computable > to infinite decimal places. You haven't used any properties of computable numbers here. Whatever you > mean by Omega (do you mean lower-case omega, 'w' in ASCII, to represent > the order type of N?), you haven't shown anything about its > computability. >OK you argue, show me the > actual computable real where it is apparent to oo decimal > places and in theory there is none. But the set of CR does > contain this sequence of digits. ALL sequences of digits > to oo are computable. And I think that's good enough to > refute the existence of sets that are uncountably larger than oo. FACT: All sequences of digits are computable to infinite length. > Clearly not. OK, the set of possible infinite sequences are computable to how many digits? Herc === Subject: Re: the problem with Cantor posting-account=9QOSvAoAAACEOWJVSDuswW7dB_0wApQO Gecko/2008071615 Fedora/3.0.1-1.fc9 Firefox/3.0.1,gzip(gfe),gzip(gfe) > When you consider computable reals 0<=cr<1 > what variety do you get in the decimal expansions? Every possible sequence of digits to infinite length > is represented as a computable number. No, that's just not true. There are uncountablyl many such sequences, > but only countably many computable numbers. æYou might > argue Omega is not represented in it entirety, but > Omega to 10 decimal places is computable, > Omega to 100 decimal places is computable, > Omega to a googol decimal places is computable, > ... What do you mean here by Omega? Isn't the 1st digit of Omega 1 if the 1st Turing machine halts, 0 otherwise > something like that Are you supposed to *know* what Omega is? That's a very mild requirement for your original post to have any sense at all... Now you are asking what the first bit is? > [ ... yet another instance of the apparently endemic > sci.math disease causing people to `pass to the limit' > manners, snipped. ] -- m === Subject: Re: the problem with Cantor > When you consider computable reals 0<=cr<1 > what variety do you get in the decimal expansions? Every possible sequence of digits to infinite length > is represented as a computable number. No, that's just not true. There are uncountablyl many such sequences, > but only countably many computable numbers. You might > argue Omega is not represented in it entirety, but > Omega to 10 decimal places is computable, > Omega to 100 decimal places is computable, > Omega to a googol decimal places is computable, > ... What do you mean here by Omega? Isn't the 1st digit of Omega 1 if the 1st Turing machine halts, 0 otherwise > something like that -Are you supposed to *know* what Omega is? That's a -very mild requirement for your original post to have -any sense at all... -Now you are asking what the first bit is? No we don't need to know any values of Omega. We know by redundancy that we have computed Omega to n digits because every possible sequence to n digits has been computed. Herc === Subject: Re: the problem with Cantor posting-account=9QOSvAoAAACEOWJVSDuswW7dB_0wApQO Gecko/2008071615 Fedora/3.0.1-1.fc9 Firefox/3.0.1,gzip(gfe),gzip(gfe) When you consider computable reals 0<=cr<1 > what variety do you get in the decimal expansions? Every possible sequence of digits to infinite length > is represented as a computable number. No, that's just not true. There are uncountablyl many such sequences, > but only countably many computable numbers. You might > argue Omega is not represented in it entirety, but > Omega to 10 decimal places is computable, > Omega to 100 decimal places is computable, > Omega to a googol decimal places is computable, > ... What do you mean here by Omega? Isn't the 1st digit of Omega 1 if the 1st Turing machine halts, 0 otherwise > something like that -Are you supposed to *know* what Omega is? That's a > -very mild requirement for your original post to have > -any sense at all... -Now you are asking what the first bit is? No we don't need to know any values of Omega. æWe know by redundancy > that we have computed Omega to n digits because every possible sequence > to n digits has been computed. So... Say that I pick a number between 1 and 10, but I do not tell you which. Then you make a list of all thenumbers between 1 and 10. Do you think that after making that list you know which number I picked? -- m === Subject: Re: the problem with Cantor So... Say that I pick a number between 1 and 10, but I do not tell you which. Then you make a list of all thenumbers between 1 and 10. Do you think that after making that list you know which number I picked? -- m ******************** Nicely argued. === Subject: Re: the problem with Cantor <48903c01$0$2275$afc38c87@news.optusnet.com.au> posting-account=9QOSvAoAAACEOWJVSDuswW7dB_0wApQO Gecko/2008061712 Fedora/3.0-1.fc9 Firefox/3.0,gzip(gfe),gzip(gfe) On Jul 30, 7:02æam, Peter Webb > So... Say that I pick a number between 1 and 10, but > I do not tell you which. Then you make a list of all > thenumbers between 1 and 10. Do you think that after > making that list you know which number I picked? -- m ******************** > Nicely argued. And he agreed! -- m === Subject: Re: the problem with Cantor On Jul 30, 7:02 am, Peter Webb > So... Say that I pick a number between 1 and 10, but > I do not tell you which. Then you make a list of all > thenumbers between 1 and 10. Do you think that after > making that list you know which number I picked? -- m ******************** > Nicely argued. -And he agreed! No, I said I could *write down* your number (by writing down all 10) Just like I can write down all the digits of Omega in sequence, by computing the set of all computable reals. If you think I can't, then which digit am I going to miss? Computable reals displays EVERY type of decimal expansion imaginable, there is nothing it misses. Herc === Subject: Re: the problem with Cantor > On Jul 30, 7:02 am, Peter Webb > So... > Say that I pick a number between 1 and 10, but > I do not tell you which. Then you make a list of all > thenumbers between 1 and 10. Do you think that after > making that list you know which number I picked? > -- m > ******************** > Nicely argued. -And he agreed! No, I said I could *write down* your number (by writing down all 10) Just like I can write down all the digits of Omega in sequence, > by computing the set of all computable reals. You can't compute the set of computable reals. If you think you can, give me an algorithm for doing so. > If you think I can't, > then which digit am I going to miss? > It depends upon how you define Omega; ie, the rules for the TM that you use. > Computable reals displays EVERY type of decimal expansion imaginable, > there is nothing it misses. > That statement may or may not be true, depending upon how you define the word imaginable. Give me a definition, and I will tell you if your statement is correct. === Subject: Re: the problem with Cantor When you consider computable reals 0<=cr<1 > what variety do you get in the decimal expansions? Every possible sequence of digits to infinite length > is represented as a computable number. No, that's just not true. There are uncountablyl many such sequences, > but only countably many computable numbers. You might > argue Omega is not represented in it entirety, but > Omega to 10 decimal places is computable, > Omega to 100 decimal places is computable, > Omega to a googol decimal places is computable, > ... What do you mean here by Omega? Isn't the 1st digit of Omega 1 if the 1st Turing machine halts, 0 otherwise > something like that -Are you supposed to *know* what Omega is? That's a > -very mild requirement for your original post to have > -any sense at all... -Now you are asking what the first bit is? No we don't need to know any values of Omega. We know by redundancy > that we have computed Omega to n digits because every possible sequence > to n digits has been computed. =So... = =Say that I pick a number between 1 and 10, but =I do not tell you which. Then you make a list of all =thenumbers between 1 and 10. Do you think that after =making that list you know which number I picked? Now you get it! What I said was, in this context, is that I can write down your number. Do you agree that the 1st 10 digits of Omega appear in sequence in the right position, somewhere in the list of computable reals? Herc === Subject: Re: the problem with Cantor > Do you agree that the 1st 10 digits of Omega appear in sequence in > the right position, somewhere in the list of computable reals? I do not even agree that Omega has any digits appearing any place. === Subject: Re: the problem with Cantor posting-account=AdyLXQoAAABgRay99CKv1O8Y_7jjivwq InfoPath.1),gzip(gfe),gzip(gfe) > Do you agree that the 1st 10 digits of Omega appear > in sequence in the right position, somewhere in the > list of computable reals? > I do not even agree that Omega has any digits appearing any place. I believe he's talking about Chaitin's omega. Dave L. Renfro === Subject: Re: the problem with Cantor > > Do you agree that the 1st 10 digits of Omega appear > in sequence in the right position, somewhere in the > list of computable reals? > > > I do not even agree that Omega has any digits appearing any place. > > I believe he's talking about Chaitin's omega. > > > Dave L. Renfro If he refuses to say which one, the he is hardly in a position to object to my counter based on an omega rather more relevant to Cantor. === Subject: Re: the problem with Cantor > Do you agree that the 1st 10 digits of Omega appear > in sequence in the right position, somewhere in the > list of computable reals? > I do not even agree that Omega has any digits appearing any place. I believe he's talking about Chaitin's omega. > Dave L. Renfro If he refuses to say which one, the he is hardly in a position to object > to my counter based on an omega rather more relevant to Cantor. I'm using Herc's Omega, I defined it already. its a real number. if the 1st Turing machine halts, Omega begins 0.1, if the 1st Turing Machine runs forever, Omega begins 0.0. Similarly for the second digit and so on, giving it a binary representation. Herc === Subject: Re: the problem with Cantor > When you consider computable reals 0<=cr<1 > what variety do you get in the decimal expansions? > Every possible sequence of digits to infinite length > is represented as a computable number. > No, that's just not true. There are uncountablyl many such sequences, > but only countably many computable numbers. > You might > argue Omega is not represented in it entirety, but > Omega to 10 decimal places is computable, > Omega to 100 decimal places is computable, > Omega to a googol decimal places is computable, > ... > What do you mean here by Omega? Isn't the 1st digit of Omega 1 if the 1st Turing machine halts, 0 > otherwise > something like that > Anyway, your reasoning is flawed. 10 is a finite natural number; and > 10^100 is a finite natural number; and 10^googol is a finite natural > number; but the set of all finite natural numbers, N, is not a fiinite > natural number. > You have to be careful when passing to infinity. How many digits is Omega computable to? > That depends upon exactly how Omega is defined, ie the TM rules. You might be able to work out the first n digits (with n finite) if you can prove that for all n digits you can use some argument to prove that all of these definitely halt or don't halt. However, no such proof is possible in general; at some point in the expansion you will just have to say that you don't know whether the TM terminates, and at that point you don't know what the rest of Omega looks like. > What does that ... mean? It means Omega is computable > to infinite decimal places. > You haven't used any properties of computable numbers here. Whatever you > mean by Omega (do you mean lower-case omega, 'w' in ASCII, to represent > the order type of N?), you haven't shown anything about its > computability. >OK you argue, show me the > actual computable real where it is apparent to oo decimal > places and in theory there is none. But the set of CR does > contain this sequence of digits. ALL sequences of digits > to oo are computable. And I think that's good enough to > refute the existence of sets that are uncountably larger than oo. > FACT: All sequences of digits are computable to infinite length. > Clearly not. OK, the set of possible infinite sequences are computable to how many > digits? Herc As I said before, any finite number. === Subject: Re: the problem with Cantor > When you consider computable reals 0<=cr<1 > what variety do you get in the decimal expansions? > Every possible sequence of digits to infinite length > is represented as a computable number. > No, that's just not true. There are uncountablyl many such sequences, > but only countably many computable numbers. > You might > argue Omega is not represented in it entirety, but > Omega to 10 decimal places is computable, > Omega to 100 decimal places is computable, > Omega to a googol decimal places is computable, > ... > What do you mean here by Omega? Isn't the 1st digit of Omega 1 if the 1st Turing machine halts, 0 > otherwise > something like that > Anyway, your reasoning is flawed. 10 is a finite natural number; and > 10^100 is a finite natural number; and 10^googol is a finite natural > number; but the set of all finite natural numbers, N, is not a fiinite > natural number. > You have to be careful when passing to infinity. How many digits is Omega computable to? > That depends upon exactly how Omega is defined, ie the TM rules. You might be able to work out the first n digits (with n finite) if you can > prove that for all n digits you can use some argument to prove that all of > these definitely halt or don't halt. However, no such proof is possible in > general; at some point in the expansion you will just have to say that you > don't know whether the TM terminates, and at that point you don't know what > the rest of Omega looks like. > In a binary system of numbers, roughly 1/2 of all computable reals (0<=cr<1) begin with 1, and the other 1/2 begin with 0. Say Omega starts with 0.1 1/2 of CR also begin with 0.1. Say Omega starts with 0.10 1/4 of CR also begin with 0.10 Omega then becomes 0.101 1/8 of CR also begin with 0.101 No matter how many digits of Omega you consider, a percentage of computable reals will compute that string of digits. As the number of reals in the list of computable reals approaches infinity, the number of digits of Omega that are computed approaches infinity. Considering the entire infinite set of computable reals, it contains the sequence of digits contained in Omega (to infinite length) Herc === Subject: Re: the problem with Cantor > When you consider computable reals 0<=cr<1 > what variety do you get in the decimal expansions? > Every possible sequence of digits to infinite length > is represented as a computable number. > No, that's just not true. There are uncountablyl many such sequences, > but only countably many computable numbers. > You might > argue Omega is not represented in it entirety, but > Omega to 10 decimal places is computable, > Omega to 100 decimal places is computable, > Omega to a googol decimal places is computable, > ... > What do you mean here by Omega? Isn't the 1st digit of Omega 1 if the 1st Turing machine halts, 0 > otherwise > something like that > Anyway, your reasoning is flawed. 10 is a finite natural number; and > 10^100 is a finite natural number; and 10^googol is a finite natural > number; but the set of all finite natural numbers, N, is not a fiinite > natural number. > You have to be careful when passing to infinity. How many digits is Omega computable to? > That depends upon exactly how Omega is defined, ie the TM rules. You might be able to work out the first n digits (with n finite) if you can > prove that for all n digits you can use some argument to prove that all of > these definitely halt or don't halt. However, no such proof is possible in > general; at some point in the expansion you will just have to say that you > don't know whether the TM terminates, and at that point you don't know what > the rest of Omega looks like. > In a binary system of numbers, roughly 1/2 of all computable reals (0<=cr<1) > begin with 1, and the other 1/2 begin with 0 > > Say Omega starts with 0.1 What requires Omega to have any binary representation at all? Is |-| uck claiming that omega is a computable real that falls between 0 and 1? Unless he is, the rest of his post is garbage. === Subject: Re: the problem with Cantor > When you consider computable reals 0<=cr<1 > what variety do you get in the decimal expansions? > Every possible sequence of digits to infinite length > is represented as a computable number. > No, that's just not true. There are uncountablyl many such sequences, > but only countably many computable numbers. > You might > argue Omega is not represented in it entirety, but > Omega to 10 decimal places is computable, > Omega to 100 decimal places is computable, > Omega to a googol decimal places is computable, > ... > What do you mean here by Omega? Isn't the 1st digit of Omega 1 if the 1st Turing machine halts, 0 > otherwise > something like that > Anyway, your reasoning is flawed. 10 is a finite natural number; and > 10^100 is a finite natural number; and 10^googol is a finite natural > number; but the set of all finite natural numbers, N, is not a fiinite > natural number. > You have to be careful when passing to infinity. How many digits is Omega computable to? > That depends upon exactly how Omega is defined, ie the TM rules. You might be able to work out the first n digits (with n finite) if you can > prove that for all n digits you can use some argument to prove that all of > these definitely halt or don't halt. However, no such proof is possible in > general; at some point in the expansion you will just have to say that you > don't know whether the TM terminates, and at that point you don't know what > the rest of Omega looks like. > In a binary system of numbers, roughly 1/2 of all computable reals (0<=cr<1) > begin with 1, and the other 1/2 begin with 0 > Say Omega starts with 0.1 What requires Omega to have any binary representation at all? its a real number. if the 1st Turing machine halts, Omega begins 0.1, if the 1st Turing Machine runs forever, Omega begins 0.0. Similarly for the second digit and so on, giving it a binary representation. > Is |-| uck claiming that omega is a computable real that falls between 0 > and 1? no. I'm claiming all the digits of Omega appear in sequence in the set of computable reals. Unless he is, the rest of his post is garbage. True or False? As the number of reals in the list of computable reals approaches infinity, the number of digits of Omega that are computed approaches infinity. Herc === Subject: Re: the problem with Cantor > Say Omega starts with 0.1 What requires Omega to have any binary representation at all? > > its a real number. By what definition of real number? The most commonly accepted definitions of real number involve either Cauchy sequences or Dedekind cuts of rationals, neither of which allow omega as a real number, so what is your definitions of real number that includes omega? > > > Is |-| uck claiming that omega is a computable real that falls between 0 > and 1? > > no. I'm claiming all the digits of Omega appear in sequence in the set of > computable reals. You first have to establish that omega has digits. > Unless he is, the rest of his post is garbage. > > True or False? No, garbage > As the number of reals in the list of computable reals approaches > infinity, > the number of digits of Omega that are computed approaches infinity. Not even false. === Subject: Re: the problem with Cantor <488ff344$0$1027$afc38c87@news.optusnet.com.au> <5JTjk.24056$IK1.18640@news-server.bigpond.net.au> posting-account=9QOSvAoAAACEOWJVSDuswW7dB_0wApQO Gecko/2008071615 Fedora/3.0.1-1.fc9 Firefox/3.0.1,gzip(gfe),gzip(gfe) > When you consider computable reals 0<=cr<1 > what variety do you get in the decimal expansions? > Every possible sequence of digits to infinite length > is represented as a computable number. > No, that's just not true. There are uncountablyl many such sequences, > but only countably many computable numbers. > æYou might > argue Omega is not represented in it entirety, but > Omega to 10 decimal places is computable, > Omega to 100 decimal places is computable, > Omega to a googol decimal places is computable, > ... > What do you mean here by Omega? Isn't the 1st digit of Omega 1 if the 1st Turing machine halts, 0 > otherwise > something like that > Anyway, your reasoning is flawed. 10 is a finite natural number; and > 10^100 is a finite natural number; and 10^googol is a finite natural > number; but the set of all finite natural numbers, N, is not a fiinite > natural number. > You have to be careful when passing to infinity. How many digits is Omega computable to? That depends upon exactly how Omega is defined, ie the TM rules. You might be able to work out the first n digits (with n finite) if you can > prove that for all n digits you can use some argument to prove that all of > these definitely halt or don't halt. However, no such proof is possible in > general; at some point in the expansion you will just have to say that you > don't know whether the TM terminates, and at that point you don't know what > the rest of Omega looks like. In a binary system of numbers, roughly 1/2 of all computable reals (0<=cr<1) > begin with 1, and the other 1/2 begin with 0 Say Omega starts with 0.1 What requires Omega to have any binary representation at all? its a real number. æif the 1st Turing machine halts, Omega begins 0.1, > if the 1st Turing Machine runs forever, Omega begins 0.0. æSimilarly > for the second digit and so on, giving it a binary representation. Is |-| uck claiming that omega is a computable real that falls between 0 > and 1? no. æI'm claiming all the digits of Omega appear in sequence in the set of > computable reals. Unless he is, the rest of his post is garbage. True or False? > æ æ As the number of reals in the list of computable reals approaches infinity, > æ æ the number of digits of Omega that are computed approaches infinity. There is another option: meaningless. -- m === Subject: Re: the problem with Cantor [Herc] > True or False? > As the number of reals in the list of computable reals approaches infinity, > the number of digits of Omega that are computed approaches infinity. [m] -There is another option: meaningless. So you're saying no digits of Omega are ever computed, by any computer ever? Can anyone else evaluate whether my proposition above is true or false? Herc === Subject: Re: the problem with Cantor > [Herc] > True or False? > As the number of reals in the list of computable reals approaches infinity, > the number of digits of Omega that are computed approaches infinity. > > [m] > -There is another option: meaningless. > > So you're saying no digits of Omega are ever computed, by any computer ever? One would be well advised to avoid use of any computer on which any such digits have been computed. > > Can anyone else evaluate whether my proposition above is true or false? I see questions, but no proposition in this posting. === Subject: Re: the problem with Cantor > When you consider computable reals 0<=cr<1 > what variety do you get in the decimal expansions? > > Every possible sequence of digits to infinite length > is represented as a computable number. You might > argue Omega is not represented in it entirety, but > Omega to 10 decimal places is computable, > Omega to 100 decimal places is computable, > Omega to a googol decimal places is computable, > ... > > What does that ... mean? It means Omega is computable > to infinite decimal places. OK you argue, show me the > actual computable real where it is apparent to oo decimal > places and in theory there is none. But the set of CR does > contain this sequence of digits. ALL sequences of digits > to oo are computable. And I think that's good enough to > refute the existence of sets that are uncountably larger than oo. > > FACT: All sequences of digits are computable to infinite length. Not true. The set of possible infinite sequences is a non-countable set. The set of computable (i.e. Turing computable) infinite sequences is a countable set. Bob Kolker === Subject: Re: the problem with Cantor > When you consider computable reals 0<=cr<1 > what variety do you get in the decimal expansions? Every possible sequence of digits to infinite length > is represented as a computable number. You might > argue Omega is not represented in it entirety, but > Omega to 10 decimal places is computable, > Omega to 100 decimal places is computable, > Omega to a googol decimal places is computable, > ... What does that ... mean? It means Omega is computable > to infinite decimal places. OK you argue, show me the > actual computable real where it is apparent to oo decimal > places and in theory there is none. But the set of CR does > contain this sequence of digits. ALL sequences of digits > to oo are computable. And I think that's good enough to > refute the existence of sets that are uncountably larger than oo. FACT: All sequences of digits are computable to infinite length. Not true. The set of possible infinite sequences is a non-countable set. > The set of computable (i.e. Turing computable) infinite sequences is a > countable set. Bob Kolker OK, the set of possible infinite sequences are computable to how many digits? Herc === Subject: Re: the problem with Cantor > When you consider computable reals 0<=cr<1 > what variety do you get in the decimal expansions? > Every possible sequence of digits to infinite length > is represented as a computable number. You might > argue Omega is not represented in it entirety, but > Omega to 10 decimal places is computable, > Omega to 100 decimal places is computable, > Omega to a googol decimal places is computable, > ... > What does that ... mean? It means Omega is computable > to infinite decimal places. OK you argue, show me the > actual computable real where it is apparent to oo decimal > places and in theory there is none. But the set of CR does > contain this sequence of digits. ALL sequences of digits > to oo are computable. And I think that's good enough to > refute the existence of sets that are uncountably larger than oo. > FACT: All sequences of digits are computable to infinite length. > Not true. The set of possible infinite sequences is a non-countable set. > The set of computable (i.e. Turing computable) infinite sequences is a > countable set. > Bob Kolker OK, the set of possible infinite sequences are computable to how many > digits? Herc Any finite number. === Subject: Re: the problem with Cantor > OK, the set of possible infinite sequences are computable to how many > digits? Herc Any finite number. Bill Goates has so much money in the bank. If you asked whether he had any finite number of money in the bank the answer was always yes, how much money total does he have? Infinite. Herc === Subject: Re: the problem with Cantor > When you consider computable reals 0<=cr<1 > what variety do you get in the decimal expansions? > Every possible sequence of digits to infinite length > is represented as a computable number. You might > argue Omega is not represented in it entirety, but > Omega to 10 decimal places is computable, > Omega to 100 decimal places is computable, > Omega to a googol decimal places is computable, > ... > What does that ... mean? It means Omega is computable > to infinite decimal places. OK you argue, show me the > actual computable real where it is apparent to oo decimal > places and in theory there is none. But the set of CR does > contain this sequence of digits. ALL sequences of digits > to oo are computable. And I think that's good enough to > refute the existence of sets that are uncountably larger than oo. > FACT: All sequences of digits are computable to infinite length. > Not true. The set of possible infinite sequences is a non-countable set. > The set of computable (i.e. Turing computable) infinite sequences is a > countable set. > Bob Kolker OK, the set of possible infinite sequences are computable to how many > digits? Herc Any finite number. None, some, or all? Herc === Subject: Re: the problem with Cantor > When you consider computable reals 0<=cr<1 > what variety do you get in the decimal expansions? > Every possible sequence of digits to infinite length > is represented as a computable number. You might > argue Omega is not represented in it entirety, but > Omega to 10 decimal places is computable, > Omega to 100 decimal places is computable, > Omega to a googol decimal places is computable, > ... > What does that ... mean? It means Omega is computable > to infinite decimal places. OK you argue, show me the > actual computable real where it is apparent to oo decimal > places and in theory there is none. But the set of CR does > contain this sequence of digits. ALL sequences of digits > to oo are computable. And I think that's good enough to > refute the existence of sets that are uncountably larger than oo. > FACT: All sequences of digits are computable to infinite length. > Not true. The set of possible infinite sequences is a non-countable > set. > The set of computable (i.e. Turing computable) infinite sequences is a > countable set. > Bob Kolker > OK, the set of possible infinite sequences are computable to how many > digits? > Herc > Any finite number. None, some, or all? Herc Any = for all n e N. Which somewhat equates to all, except in common English all is slightly ambiguous in that it can mean either any or the set of. But the deal is, you give me any finite number and I can work it out to that precision. === Subject: Re: the problem with Cantor > When you consider computable reals 0<=cr<1 > what variety do you get in the decimal expansions? > > Every possible sequence of digits to infinite length > is represented as a computable number. You might > argue Omega is not represented in it entirety, but > Omega to 10 decimal places is computable, > Omega to 100 decimal places is computable, > Omega to a googol decimal places is computable, > ... > > What does that ... mean? It means Omega is > computable > to infinite decimal places. OK you argue, show me > the > actual computable real where it is apparent to oo > decimal > places and in theory there is none. But the set of > CR does > contain this sequence of digits. ALL sequences of > digits > to oo are computable. And I think that's good enough > to > refute the existence of sets that are uncountably > larger than oo. > > FACT: All sequences of digits are computable to > infinite length. Say we have a list of all rationals. We take #1 and #2 on the list and put the smaller one on the left, and the larger one on the right. Then, we pick off #3, #4, and so on in succession. If the number is greater than both numbers at the top of the two piles (in construction), or if it's less than both, it's discarded. Otherwise, it's between the number at the top of the left pile and the number at the top of the right pile. Rule: if the last number placed was placed atop the right pile, the current one (assumed to be in between) is placed above the left pile, and vice versa. After #1 and #2 have been placed, first turn goes to the left pile; from then on, alternation on placing numbers that are in between. So does this result in a new number right between the two piles? Can it work for a list of the computables? David Bernier === Subject: Cancel Re: A consideration concerning the diagonal argument of G. Cantor -- For every line of Cantor's list it is true that this line does not contain the diagonal number. Nevertheless the diagonal number may be in the infinite list. (WM, sci.logic) === Subject: Re: A consideration concerning the diagonal argument of G. Cantor Again you confuse physical and mathematical -- > you confuse 7 chairs with the number 7. > Not very surprising. In de.sci.mathematik he once claimed that three trees in a certain garden ARE the number three. (Right, that's indeed what he claimed - not that the three trees are an instance of the number three, ore anything like that, but...) B. -- For every line of Cantor's list it is true that this line does not contain the diagonal number. Nevertheless the diagonal number may be in the infinite list. (WM, sci.logic) === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > ... you confuse 7 chairs with the number 7. > Not very suprising. In de.sci.mathematik WM once claimed that three > trees in a certain garden ARE the number three. (Right, that's indeed > what he claimed - not that the three trees are an instance of the > number three, ore anything like that, but...) > Die drei B.8aume *sind* die Zahl drei. [WM, 12 Apr. 2007] (The three trees *are* the number three.) Well, sure, in a world where identical things can differ *everything* is possible. :-) B. -- For every line of Cantor's list it is true that this line does not contain the diagonal number. Nevertheless the diagonal number may be in the infinite list. (WM, sci.logic) === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=yKimjgoAAACk5WwPVD4l9HmbpoR6Hmy4 en),gzip(gfe),gzip(gfe) æ> æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ It is not > æ> uncommon that technical terms in mathematics have very different > æ> meanings. Right. But that is not so frequent in a single field of mathematics. æBut > the term heap had only a single meaning in mathematics, and that was > *not* the meaning you intende. The term heap was used already in the Papyrus Rhind æto denote a > collection of similar objects. And in the present discussion there is > no danger to mistake such a heap with an algebraic structure. What's this? WM is not using the definition of heap that can be found on wikipedia? What's this exchange about then?: It is also valid for heaps. Define heap. Explain what it is valid means. Discussion with you is too tedious. Come back when you will have > finished basic courses. (An explanation of heap you can even find in > Wikipedia.) Why would I assume that you are using the standard meaning for a term? > You so rarely do! You are incapable of giving definitions or sticking to them and I claim my five pounds. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor It's a pity that Robinson did not adopt Frege's terminology which would have helped to avoid misunderstandings. denotation. Clearly the term the set of all natural numbers DOES have a /meaning/ (/Sinn/ due to Frege), but it might lack a denotation (Bedeutung). It might be the case -as Robinson advocates- that actually there IS NO set of all natural numbers in reality. (3) (i) Infinite totalities do not exist in any proper sense of the > word (i.e., either really or ideally). More precisely, any mention, or > purported mention, of infinite totalities is, literally MEANINGLESS. > ... i.e. lacks a denotation. (ii) Nevertheless, we should continue the business of Mathematics as > usual, i.e., we should act as IF infinite totalities really existed. > Right. We see: those terms ARE meaningful (at least in mathematics). (4) [...] I must regard a theory which refers to an infinite totality > as MEANINGLESS in the sense that its terms and sentences cannot posses > the direct interpretation in an actual structure that we should expect > them to have by analogy with concrete (e.g., empirical) situations. > This is not to say that such a theory is pointless of devoid of > significance. > Right. Let's assume that Pegasus does not exist, with other words, that the word/name Pegasus does not have a denotation (since there is no winged horse). Still the claim (*) Pegasus is a winged horse does have a MEANING (sense, Sinn), expresses a THOUGHT (Gedanke). One might even assign a truth value (a denotation/Bedeutung) to it, namely /false/, since Pegasus does not exist. To claim that the statement (*) is meaningless seems slightly odd to me. (Well, ok, I'm used to Fregean terminology, I have to admit.) So he says more about the exact sense in which he thinks the notion of > infinity is meaningless but goes on to say that this does NOT entail > that a theory of infinity is devoid of SIGNIFICANCE. And that is > exactly UNlike [WM's] polemics. > Right. (5) [...] the second basic principle of my formalist philosophy. > Which I personally would rather call a /fictionalistic/ position. This the prescription, or suggestion, or advice, to continue to do > Mathematics in the classical way, i.e., in particular, to use terms > which purport to refer to infinite totalities as if they really > existed. > Right. Actually, he might have replaced the word infinite totalities with mathematical objects. There is no substantial difference between /infinite totalities/ and other /mathematical objects/ - none of them is REAL (imho). So let me paraphrase Robinson's statements: (i') Mathematical objects do not exist in any proper sense of the word (i.e., either really or ideally). More precisely, any mention, or purported mention, of mathematical objects lacks a denotation. (ii) Nevertheless, we should continue the business of Mathematics as usual, i.e., we should act as IF mathematical objects really existed. (6) The formalist holds that direct interpretability is not a > necessary condition for the acceptability of a mathematical theory. > Same with the fictionalistic view. (8) To sum up [...]. To UNDERSTAND a theory means to be able to > follow its logical development and not necessarily, to interpret, > or give denotation for, its [...] terms. > Oh, FINALLY he got the right word!!! :-) (10) [...] it is undeniable that in certain important instances they > [certain formal uninterpreted theories] include fragments which are > indeed capable of interpretetation, more particularly empirical > interpretation. For example, [...] the axiomatic Set Theory of Zermelo > and Fraenkel. According to our formalist point of view this theory is > uninterpreted. But it is nevertheless the case that the rules for > addition and multiplication of finite cardinals, which form part of > that theory, are applicable to concrete, more particularly, empirical > situations. > Right, of course. Sets are fictitious, but not necessarily (all of) their applications. B. -- For every line of Cantor's list it is true that this line does not contain the diagonal number. Nevertheless the diagonal number may be in the infinite list. (WM, sci.logic) === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322),gzip(gfe),gzip(gfe) > But WM is so satisfied by those incomplete trees So let us complete the tree. What would be your proposal how to > proceed? Include each of Cantor's uncountably many infinite binary sequences as > corresponding to a separate path, Really, I would very much like to do so in order to make you happier > than you seem to be, alas, I cannot see any missing node. Where did I ever say any nodes need to be missing? It is quite possible > to include every node in at least one path but simultaneously to exclude > most paths. Ah, that's a new aspect. S there are paths which have all nodes in common? WM's incomplete tree in which the only paths are those limited to > finitely many right branchings does that very nicely (as each node can > be reached after at most finitely many right branchings) Nice to hear. So your further arguing is not based upon nodes or digits, i.e., it is not based upon mathematics, but on mathemagics with numbers that exist without definition and without a path of their own in the tree. We should stop at this point, because it is idle to argue about religious belief. > but that leaves > out uncountably many paths having infinitely many right branchings. If you were consequent, you would say that that leaves out infinitely many nodes. Or you could say that Canor's diagonal proof leaves out infinitely many lines. And any other WM construction which limits the allowable patterns of > left and right branchings may include every node and every edge but > simultaneously exclude most paths. Magic, magic! But Cantor's list-proof stands undefeated? We should stop at this point, because it is idle to argue about religious belief. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > But WM is so satisfied by those incomplete trees So let us complete the tree. What would be your proposal how to > proceed? Include each of Cantor's uncountably many infinite binary sequences as > corresponding to a separate path, Really, I would very much like to do so in order to make you happier > than you seem to be, alas, I cannot see any missing node. Where did I ever say any nodes need to be missing? It is quite possible > to include every node in at least one path but simultaneously to exclude > most paths. > > Ah, that's a new aspect. Hardly! That has been an obvious aspect, at least to everyone who understands what is really going on, from the beginning. > S there are paths which have all nodes in > common? There might be in WM's trees, I do not speak for them except to note their necessarily missing paths. WM's incomplete tree in which the only paths are those limited to > finitely many right branchings does that very nicely (as each node can > be reached after at most finitely many right branchings) > > Nice to hear. So your further arguing is not based upon nodes or > digits On the contrary, it is precisely based on binary digits in the sense that WM's tree does not allow most digit sequences to occur in his tree, whereas an honestly complete tree must require every sequence of digits to occur in it as a path. , > > We should stop at this point, because it is idle to argue about > religious belief. And WM won't give his up, however unreal it is shown to be. > > but that leaves > out uncountably many paths having infinitely many right branchings. > > If you were consequent, you would say that that leaves out infinitely > many nodes. Every node is the end of a finite path, and in an infinite tree, it is only infinite paths we need look at. > Or you could say that Canor's diagonal proof leaves out > infinitely many lines. If I were as much a fool as WM there are all sorts of silly things like that that I could say, but I am not so illogical. And any other WM construction which limits the allowable patterns of > left and right branchings may include every node and every edge but > simultaneously exclude most paths. > > Magic, magic! But Cantor's list-proof stands undefeated? Quite so. The magic of Cantor's (and other's) logic quite overpowers the slight slight of hand of WM's mytheology. > > We should stop at this point, because it is idle to argue about > religious belief. And WM's proselytizing is too sloppy to capture new believers anyway. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > but that leaves > out uncountably many paths having infinitely many right branchings. If you were consequent, you would say that that leaves out infinitely > many nodes. Every node is the end of a finite path, and in an infinite tree, it is > only infinite paths we need look at. Or you could say that Canor's diagonal proof leaves out > infinitely many lines. Every line is the end of a finite number, and in an infinite list, it is only infinite numbers we need look at. > Magic, magic! But Cantor's list-proof stands undefeated? Quite so. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > > but that leaves > out uncountably many paths having infinitely many right branchings. If you were consequent, you would say that that leaves out infinitely > many nodes. Every node is the end of a finite path, and in an infinite tree, it is > only infinite paths we need look at. Or you could say that Cantor's diagonal proof leaves out > infinitely many lines. > > Every line is the end of a finite number, and in an infinite list, it > is > only infinite numbers we need look at. > > Magic, magic! But Cantor's list-proof stands undefeated? Quite so. > measure of his ineptness. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322),gzip(gfe),gzip(gfe) > Given any countable set of paths in one of WM's phony trees, I can show > at least as many missing paths. What the tree gives or contains is every infinite combination of the > binary digits zero and one. Which one do you have in mind as missing? If all of the paths in WM's tree are those which contain only finitely > many 1's, as in one of WM's constructions, then that tree is missing > all of the uncountably many paths which contain infinitely many 1's. I cannot determine a node that would prove your assertion. As far as I can see the path 0.111... stretches further and further. Before I can tell which paths are missing, WM must tell me which > countable set of paths his pseudotree contains. It contains all nodes that can be generated by the countable set of paths each of which has a tail of infinitely many zeros. > One of those pseudotrees > was the one in which each path had at nost finitly many right > branchings (or equivalently finitely many 1's for seqeunces of 0's and > 1's) If none of the uncountably many paths many are missing, then it isn't > one of WM's trees after all, but is the actually complete tree. My tree seems to be complete. But, of course, I will refrain from that assertion in case you come up with the name of a missing node. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > Given any countable set of paths in one of WM's phony trees, I can show > at least as many missing paths. What the tree gives or contains is every infinite combination of the > binary digits zero and one. Which one do you have in mind as missing? If all of the paths in WM's tree are those which contain only finitely > many 1's, as in one of WM's constructions, then that tree is missing > all of the uncountably many paths which contain infinitely many 1's. > > I cannot determine a node that would prove your assertion. As far as I > can see the path 0.111... stretches further and further. My assertion does not depend on any single node but on an infinite set of right-child nodes in a tree which does not contain or allow any such set, so that WM is being, as expected terminally myopic. Before I can tell which paths are missing, WM must tell me which > countable set of paths his pseudotree contains. > > It contains all nodes that can be generated by the countable set of > paths each of which has a tail of infinitely many zeros. Then it excludes all uncountbly many paths containing both infinitely many right child nodes and infinitely many left child nodes. > > One of those pseudotrees > was the one in which each path had at nost finitely many right > branchings (or equivalently finitely many 1's for seqeunces of 0's and > 1's) If none of the uncountably many paths are missing, then it isn't > one of WM's trees after all, but is the actually complete tree. > > My tree seems to be complete. But, of course, I will refrain from that > assertion in case you come up with the name of a missing node. Why does WM keep harping on missing nodes? Each node corresponds to a binary fraction, and no one is saying that any of those are missing, but each missing path corresponds to one of the uncountably many real numbers between 0 and 1 which is not a binary fraction, and all such paths, along with those numbers, are missing. And any rearrangement of WM's limited set of paths, even if it still covers all nodes, excludes most paths