mm-4739 === Subject: Re: Virgil proves (tests) Cantor (Was: Re: Yet another disproof of the > ... > All Cantor lists are square in the sense of being able to biject the > set > of members with the set of digit positions. So now, by Virgil's statement, we have a bijection from the members of > the list to the digits positions. Then the list is just as countable > as the sequences it lists. QDE Yes, by definition every list is countable. What is the problem with that? > What Cantor's proof *is* is that also every list of reals is incomplete. Indeed, by definition, how can a list ever be uncountable?? But lists can be incomplete in the sense of not listing all of the objects in the set from which they are taken. And that is the point. Which julio again missed. What Cantor's proof *is* is an apparently invalid proof It is apparent to those who are capable of undeestanding it that it is quite vcalid. To those like julio of insufficient understanding, it may forever remain beyond their comprehension. > from which one > rather *derives* that *there is such a thing called 'the reals' that > cannot be listed*. Julio's objection is clearly unral. OTOH, this is the proof that you, in the best case, don't even know > what you are talking about. Cantor did know what he was talking about, even though julio doesn't. When julio can produce a doctorate in mathematics and is a mathematics professor at a major university, as Cantor was, only then will julio's word alone, which is all we have so far from him, be worth anything. === Subject: Re: Virgil proves (tests) Cantor (Was: Re: Yet another disproof of the posting-account=cBXgtgkAAABzLyfMZLtdfwiGec9333UO CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) ... > All Cantor lists are square in the sense of being able to biject the > set > of members with the set of digit positions. So now, by Virgil's statement, we have a bijection from the members of > the list to the digits positions. Then the list is just as countable > as the sequences it lists. QDE Yes, by definition every list is countable. What is the problem with that? > What Cantor's proof *is* is that also every list of reals is incomplete. Indeed, by definition, how can a list ever be uncountable?? But lists can be incomplete in the sense of not listing all of the > objects in the set from which they are taken. And that is the point. Which julio again missed. What Cantor's proof *is* is an apparently invalid proof It is apparent to those who are capable of undeestanding it that it is > quite vcalid. To those like julio of insufficient understanding, it may > forever remain beyond their comprehension. from which one > rather *derives* that *there is such a thing called 'the reals' that > cannot be listed*. Julio's objection is clearly unral. OTOH, this is the proof that you, in the best case, don't even know > what you are talking about. Cantor did know what he was talking about, even though julio doesn't. > When julio can produce a doctorate in mathematics and is a mathematics > professor at a major university, as Cantor was, only then will julio's > word alone, which is all we have so far from him, be worth anything.- Hide quoted text - - Show quoted text - Virgil's version of logic puts George W Bush as one of the greatest leaders in the history of the world. === Subject: Re: Virgil disproves Cantor (Was: Re: Yet another disproof of the diagonal argument) It is a trivial property over induction that the list defined in the > OP is *not* square. Yours may not be square, but Cantor's certainly are square, at least > if that means that the length and width of all relevant lists are > equal. That's interesting actually. Come up with a square list and I'll buy you another beer. You have not yet bought me a first beer. Neither I'll ever have to buy you one, for the sake. More sensible than much of julio's postings. All Cantor lists are square in the sense of being able to biject the set > of members with the set of digit positions. So now, by Virgil's statement, we have a bijection from the members of > the list to the digits positions. Then the list is just as countable > as the sequences it lists. QDE If you mean QED, you are, for once, correct. Each string is of length omega as is the list. That julio tries to introduce the irrelevancies of nonsquare finite > lists shows how little of the Cantor theorem he understands. An *inductively defined list* is not finite lists. And whom do you allege said otherwise? === Subject: Re: Analysis with sequence and limit. > Hello teacher~ a_1 = 1 > a_2 = 8 > a_n = sqrt[{a_(n-1)}*{a_(n-2)}] Find the lim{n->oo} a_n Suppose b_0 = 0, b_1 = x, and b_n = (b_(n-1) + b_(n-2))/2 thereafter. (I.e., keep taking midpoints.) It's easy to check b_n = x[1 - 1/2 + 1/4 - ... + (-1/2)^n]. Therefore lim b_n = x*[sum(n=0,oo) (-1/2)^n] = 1/(1-(-1/2)) = x*(2/3). Set b_n = log(a_(n+1)). Then b_n is a midpoint sequence with x = log(8), so lim b_n = (2/3)*log(8) = log(8^(2/3)) = log(4). Therefore a_n -> 4. === Subject: Re: Analysis with sequence and limit. posting-account=Z3AipgkAAABkoMfyNwddSxsYhXHi5CDt CLR 1.1.4322; InfoPath.1; .NET CLR 2.0.50727; .NET CLR 3.0.04506.648; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) > Hello teacher~ a 1 = 1 > a 2 = 8 > a n = sqrt[{a (n-1)}*{a (n-2)}] Find the lim{n->oo} a n ----------------------------------------------------------------- > Let b n = log (2) a n. > Namely, a n = 2^(b n). > so, b 1 = 0 > and b 2 = 3 Since a n = sqrt[{a (n-1)}*{a (n-2)}], > 2^(b n) = [{2^(b (n-1))}*{2^(b (n-2))}]^(1/2) > so, b n = [{b (n-1)} + {b (n-2)}] / 2 so, b n - b (n-1) = ( [{b (n-1)} + {b (n-2)}] / 2 ) - b (n-1) > = (-1/2).[b (n-1) - b (n-2)] > = {(-1/2)^2}.[b (n-2) - b (n-2)] > = ... > = {(-1/2)^(n-2)}.[b 2 - b 1] > = {(-1/2)^(n-2)}.3 b 2 - b 1 = {(-1/2)^0}.3 > b 3 - b 2 = {(-1/2)^1}.3 > b 4 - b 3 = {(-1/2)^2}.3 > ... > b n - b (n-1) = {(-1/2)^(n-2)}.3 By sum, b n - b 1 = 3*[1 - (-1/2)^(n-1)] / [1 - (-1/2)] so, lim{n->oo} b n = 2. > so, lim{n->oo} a n = lim{n->oo} 2^(b n) = 2^2 = 4. OR... > Since a n = sqrt[{a (n-1)}*{a (n-2)}] > so, (a n)^2 = {a (n-1)}*{a (n-2)} [(a 1)^2].[(a 2)^2].[(a 3)^2].[(a 4)^2] ...... [(a n)^2] > = [(a 1)^2].[(a 2)^2].[(a 1)(a 2)].[(a 2)(a 3)].[(a 3)(a 4)] ...... > [{a (n-1)}{a (n-2)}] > = [(a 1)^3].[(a 2)^4].[(a 3)^2].[(a 4)^2].[(a 5)^2] ...... > [{a (n-2)}^2].[a (n-1)] so, [(a 1)^2].[(a 2)^2].[(a 3)^2].[(a 4)^2] ...... [(a n)^2] > = [(a 1)^3].[(a 2)^4].[(a 3)^2].[(a 4)^2].[(a 5)^2] ...... > [{a (n-2)}^2].[a (n-1)] so, [{a (n-2)}].[(a n)^2] = (a 1).[(a 2)^2] = 1.(8)^2 = 64 By lim{n->oo}, X^3 = 64. ( X = lim{n->oo} a n ) > so, X = 4.- Hide quoted text - - Show quoted text - Recast the problem by taking the log base 2 of everything. Then you have a 1 = 0 a 2 = 3 a n = (a n-1 + a n-2)/2 Working a few examples by hand gives a 3 = 3/2 a 4 = 9/4 a 5 = 15/8 a 6 = 33/16 ... The pattern is now clear. a n = 2 + ((-1)^(n-1) / 2^(n-2))) - in other words 2 plus or minus 1 over a growing power of 2. This pattern is quite easy to prove by induction. It is obvious that, in this form, a n -> 2, so in the original problem, a n -> 4 Achava === Subject: Motion equation Hi all, the motion equation for a point of mass m is given by the law mA = f (R, V, t) where A denotes the acceleration of the point, R is its position relatively to a fixed origin O, V is its velocity, t is a time parameter. In a special case, I found this notation f(R_0, 0, t) = 0 . What does that means? Maybe that force f (equal to 0) is independent from velocity V at the point determined by the vector R_0 ? Any help is appreciated. TIA === Subject: Re: Motion equation posting-account=zKSjoQoAAAC5wIsHKdIzIXzOHEFPgaXy Gecko/20080829 Firefox/2.0.0.17,gzip(gfe),gzip(gfe) > Hi all, the motion equation for a point of mass m is given by the law mA = f (R, V, t) where A denotes the acceleration of the point, R is its position > relatively to a fixed origin O, V is its velocity, t is a time parameter. > In a special case, I found this notation f(R_0, 0, t) = 0 . What does that means? Maybe that force f (equal to 0) is independent from > velocity V at the point determined by the vector R_0 ? It means that at any time t, the force is zero on a /stationary/ (non- points R. It quite possible to have a force with f(R_0,0,t) = 0 but f(R_0,V,t) =/= 0 for (some) V =/= 0; an example is that of a charged R.G. Vickson > Any help is appreciated. TIA === Subject: Re: integral and product but no sum ? <14254410.1222860089533.JavaMail.jakarta@nitrogen.mathforum.org>, > which smooth function f(z) can be expressed by an integral , an infinite > product , but not by an infinite sum ? > tommy1729 None: For any smooth f, f = integral f', f = f * 1 * 1 * ..., and f = f + 0 + 0 + ... === Subject: Re: integral and product but no sum ? <19973016.1222887490585.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=fwSgtAkAAACFnX70ssKwbvm9_oCZVHrx Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) <14254410.1222860089533.JavaMail.jaka...@nitrogen.math > forum.org>, which smooth function f(z) can be expressed by an > integral , an infinite > product , but not by an infinite sum ? > tommy1729 None: For any smooth f, f = integral f', f = f * 1 * > 1 * ..., and f = > f + 0 + 0 + ... here we go again ; 0 + 0 + 0 + ... is an infinite series. so when one talks about infinite sums and calculus , people reply with 0 + 0 + ... :s despite correct , you know darn well the OP is not intended that way. not really funny , rather childish ... anyways , to specify the question from the OP - not that i believe there is confusion , just you guys like to point out it could ' in principle ' be 0 + 0 + ... or other meaningless sh*t - when i talk about infinite products or infinite sums , i mean in terms of standard functions. ( of course ! ) > a + 0 is a standard function of a. === Subject: Re: integral and product but no sum ? <19973016.1222887490585.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=fwSgtAkAAACFnX70ssKwbvm9_oCZVHrx Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) <14254410.1222860089533.JavaMail.jaka...@nitrogen.math > forum.org>, which smooth function f(z) can be expressed by an > integral , an infinite > product , but not by an infinite sum ? > tommy1729 None: For any smooth f, f = integral f', f = f * 1 * > 1 * ..., and f = > f + 0 + 0 + ... here we go again ; 0 + 0 + 0 + ... is an infinite series. so when one talks about infinite sums and calculus , people reply with 0 + 0 + ... :s despite correct , you know darn well the OP is not intended that way. not really funny , rather childish ... anyways , to specify the question from the OP - not that i believe there is confusion , just you guys like to point out it could ' in principle ' be 0 + 0 + ... or other meaningless sh*t - > + 0 + 0 + ... is a meaningful mathematical expression. Sure, it's fairly trivial, but it's still meaningful nonetheless. Gotcha. Debunked. === Subject: Re: integral and product but no sum ? <19973016.1222887490585.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=fwSgtAkAAACFnX70ssKwbvm9_oCZVHrx Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) > <14254410.1222860089533.JavaMail.jaka...@nitrogen.math > forum.org>, which smooth function f(z) can be expressed by an > integral , an infinite > product , but not by an infinite sum ? > tommy1729 None: For any smooth f, f = integral f', f = f * 1 * > 1 * ..., and f = > f + 0 + 0 + ... here we go again ; 0 + 0 + 0 + ... is an infinite series. so when one talks about infinite sums and calculus , people reply with 0 + 0 + ... :s despite correct , you know darn well the OP is not intended that way. not really funny , rather childish ... anyways , to specify the question from the OP - not that i believe there is confusion , just you guys like to point out it could ' in principle ' be 0 + 0 + ... or other meaningless sh*t - + 0 + 0 + ... is a meaningful mathematical expression. Sure, it's > fairly trivial, but it's still meaningful > nonetheless. > Make that a = a + 0 + 0 + ..., actually. :) > Gotcha. Debunked. === Subject: Re: integral and product but no sum ? <19973016.1222887490585.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=lHNboAoAAACyasQ0uqX7OeM_tLuWGoQp SV1; .NET CLR 1.1.4322; .NET CLR 2.0.50727; .NET CLR 3.0.04506.648; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) spider-dtc-ta03.proxy.aol.com[CDBC7003] (Prism/1.2.1), HTTP/1.1 cache-dtc-ag12.proxy.aol.com[CDBC758C] (Traffic-Server/6.1.5 [uScM]) <14254410.1222860089533.JavaMail.jaka...@nitrogen.math > forum.org>, which smooth function f(z) can be expressed by an > integral , an infinite > product , but not by an infinite sum ? i mean in terms of standard functions. ( of course ! ) Actually, I rather doubt that even you know what you mean. And f(x) = 0, *IS* a standard function by any reasonable definition of 'standard function'. Now, I will give a hint: If a function has a representation as an infinite product, one can convert it to an infinite sum by taking logs. You do know what a logarithm is????? === Subject: Re: overlapping area of two ellipses After this thread, the question was asked again in comp.graphics.algorithms, which lead Dave Eberly to write a paper on a solution to the problem: http://www.geometrictools.com/Documentation/AreaIntersectingEllipses.pdf -- http://kaba.hilvi.org === Subject: Solution manual probability & statistics Myers Walpole 8th edition posting-account=6y26DwoAAADSChLzgbSSSS6Ul5-RI0IZ Web Browser from: http://bsalsa.com/; SLCC1; .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; .NET CLR 3.5.21022; InfoPath.2),gzip(gfe),gzip(gfe) I have the complete solution manual fo walpole myers for probability & statistics.. you can contact me at: jennicor@gmail.com Only 5$ === Subject: Re: A consideration concerning the diagonal argument of G. Cantor <87zlmrzxtn.fsf@phiwumbda.org> <1d79a$48bce687$82a1e228$604@news1.tudelft.nl> <6b855$48be3911$82a1e228$14221@news1.tudelft.nl> posting-account=6xUtvgkAAAD_jypmLa2oo2HnrV0e8X9q rv:1.7.13) Gecko/20060414,gzip(gfe),gzip(gfe) > Have you observed one lately? What did it look like? > If you are owner of two hands, put them together. What do you feel? > Maybe two hands? > Yes, but neither of them looks like a 1, and both of them > together don't look like a 2. They look like hands to me. > What do they look like to you? > And: Do tree looks for you like a tree? And Baum? Or arbre? It looks like a tree, but not a number. In my opinion, tree looks like a word. I've never seen a tree which looks like a word. You did? === Subject: Re: A consideration concerning the diagonal argument of G. Cantor <6k5cprF65986U1@mid.uni-berlin.de> posting-account=6xUtvgkAAAD_jypmLa2oo2HnrV0e8X9q rv:1.7.13) Gecko/20060414,gzip(gfe),gzip(gfe) > Tell me, Albrecht, if there was no creature in universe, which can think of > numbers, where are they then? Viele Gr??e > Klaus What a question. As good we can know they are there with and without creatures. Do you think there are not 10 planets around the sun if we are not? If there is no number, all is one. But since we live in a structured, membered universe, there are numbers of things. We distinguish between this, that, those and so on. So there are numbers as real as things. There is no ontological difference between numbers and things. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > There is no ontological difference between numbers and things. I guess, we are at the center of your stubborness here. If numbers were like things, there must be physical evidence of them. Energy, fields end the like are immaterial, however we detect their impact on the material objects. What are the direct physical impacts of numbers? You might continue to believe that there is a 10 around the sun, I don't. (I'm not even sure, if there is a planet, when that term doesn't exist. At least it's not a planet in our sense ... I mean, if there is sense.) > But since we live in a structured, > membered universe, there are numbers of things. > We distinguish between this, that, those and so on. Hey, again you're making the existence of numbers dependent of man. That's not a good argument for ontological equivalence with material objects ... If you could come up with something other, that numbers have in common with things, I mean, apart that we have images of both in mind, I mean, something provable physical? However, I confess, once I saw a physical number. It was in Sesame Street, when Lefty, the salesman (Schlemihl in German) tried to sell an 8 to Ernie ... Viele Gr.9f¤e Klaus === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=6xUtvgkAAAD_jypmLa2oo2HnrV0e8X9q rv:1.7.13) Gecko/20060414,gzip(gfe),gzip(gfe) Numbers are observable phenomena. If not, how would you observe when > you pay if your change is rigth? Let's say I buy an item worth 99 cents and I fork over a Two-euro > coin. I then observe two ones and conclude that the change is right? > Whatever are you smoking? Can I have some? Yours must be really bad, I see. > Has you ever realised that the amount of the coins in your pocket is > the sum of the numbers of a sort coins multiplied by the values of the > particular coins? Do you really argue that numbers aren't observable phenomena as > Virgil? Are there much more of you? Do you live together in a great > house? Numerals, when written, are observable and one can attach numerical > values to objects in ones mind, but the number themselves are all in the > mind, not in the numerals or the objects to which we attach values. > So, again, how are the coins not in your mind, but the numbers are? Having the coins in one's mind would be hazardous to one's health. > Having merely images of them in one's mind is the more usual way. And how do you distinguish between things which are only in your mind > and things which are in your mind and elsewhere? There are, to the best of my knowledge, no things both in my mind and > elsewhere. There are, on the other hand, many things elsewhere for which I have > IMAGES in my mind. If Albrecht's mind works otherwise, perhaps that is why he is having so > much trouble dealing with reality. Are you really as naive as you act or do you really have no clue about philsophical considerations in this concern? How could you know anything about the world? How could you know if there is a coin (or something else) or not? Do you really know if there is a coin, or do you only interpret some sensuous inputs as the existence of a coin? For sure, there are _just only_ IMAGES, nothing more. For sure, there is only something in your mind and nowhere else. Coins and numbers just as well. I ask you again: How would you distinguish between images which are only in your mind and images (trying to be exact) which have corresponding objects elsewhere. Okay, we know fictions. But number is no fiction. Ten bricks on your feed give another feeling to you than only one. A hand with just one finger is worse used than a hand with five fingers. Dream on and stick on your halfunderstanding. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor Numbers are observable phenomena. If not, how would you observe > when > you pay if your change is rigth? Let's say I buy an item worth 99 cents and I fork over a Two-euro > coin. I then observe two ones and conclude that the change is > right? > Whatever are you smoking? Can I have some? Yours must be really bad, I see. > Has you ever realised that the amount of the coins in your pocket is > the sum of the numbers of a sort coins multiplied by the values of > the > particular coins? Do you really argue that numbers aren't observable phenomena as > Virgil? Are there much more of you? Do you live together in a great > house? Numerals, when written, are observable and one can attach numerical > values to objects in ones mind, but the number themselves are all in > the > mind, not in the numerals or the objects to which we attach values. > So, again, how are the coins not in your mind, but the numbers are? Having the coins in one's mind would be hazardous to one's health. > Having merely images of them in one's mind is the more usual way. And how do you distinguish between things which are only in your mind > and things which are in your mind and elsewhere? There are, to the best of my knowledge, no things both in my mind and > elsewhere. There are, on the other hand, many things elsewhere for which I have > IMAGES in my mind. If Albrecht's mind works otherwise, perhaps that is why he is having so > much trouble dealing with reality. > Are you really as naive as you act or do you really have no clue about > philsophical considerations in this concern? How could you know anything about the world? How could you know if > there is a coin (or something else) or not? Do you really know if > there is a coin, or do you only interpret some sensuous inputs as the > existence of a coin? For sure, there are _just only_ IMAGES, nothing more. For sure, > there is only something in your mind and nowhere else. Coins and > numbers just as well. I ask you again: How would you distinguish between images which are > only in your mind and images (trying to be exact) which have > corresponding objects elsewhere. The sensory image of a coin I would treat as having a corresponding 'object' elsewhere, whereas the value of that coin need not have a corresponding 'object' anywhere, other than other minds. Okay, we know fictions. But number is no fiction. It is in some senses. It is an idea, but then, so is Santa Claus. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > And no one has any idea what it would mean to continue this process for > ever, to get some sort of result after infinitely many iterations. Not after infinitely many iterations. /At/ each of infinitely many > iterations. After any particular finite number of such iterations is completed, it > will be found that the anti-diagonal to the latest list is not a > member of the latest set of listed objects, and therefore not a member > of any of the prior ones either. Note that the meta-diagonals from the > latest list are irrelevant. The simpliest inductive argument (*) shows that the anti-diagonal, and > actually *any* constructed combination of the digits, *is* a member of > the list at every iteration. We disprove the diagonal argument. Intil you make that argument (which claiming its existence does not succeed in doing), you have no disproof. Besides which, there are several other proofs which have to be overturned before you can revoke the theorem(s) they establish. How about showing the alleged error in Cantor's first proof of uncountability of the reals? See http://en.wikipedia.org/wiki/Cantor%27s_first_uncountability_proof for details. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > Of course not. For that you need definitions for |R|, aleph_0 and > for > such numbers. But the common definitions together do show |R| > aleph_0. It is not at all a matter of definitions. It is a matter of aleph_0 > + aleph_0 = aleph_0. Fact is: Cantor's list-proof cannot exclude that all real numbers > belong to a sequence of sequences, i.e., belong to a countable set. And what sequence is that? In that sequence, which obviously includes 0, what real follows 0? It is the sequence of sequences, which is formed by the infinitely > repeated union of all regular list entries and all resulting meta- > diagonals. But while a countable sequence of countable sequences is countable, a countable sequence of incomplete sequences can be, and is in this case, still incomplete. If julio doubts it, let him try to disprove Cantor's first proof of the uncountability of the reals. http://en.wikipedia.org/wiki/Cantor%27s_first_uncountability_proof If he can even understand it. Or the theorem that there cannot be a surjection from any set to the set of all of its subsets. Julio is lost in a world which he does not even begin to understand. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > Actually, at EVERY stage one can be sure that the anti-diagonal > constructed from a list is not listed in that list. It is totally > irrelevant that there are other lists in which it may appear. And actually at every stage one cannot be sure that the antidiagonal > is not in a meta.-diagonale, and, therefore, in a countable set > tigether with the regular list entries. Since Cantor says nothing except that the anti-diagonal is not equal to > any of the listed elements of the list, WM's meta-diagonals are > irrelevant. It seems that Cantor's saying is incomplete. Only to incompetent kooks. > But should that be a > reason to stick to his wrong ideas? You have yet to show that any of Cantor's ideas are in anyway wrong. On the other hand, I am not the only one to point out some of the multitudinous errata in your ideas. And no one has any idea what it would mean to continue this process for > ever, to get some sort of result after infinitely many iterations. Not after infinitely many iterations. /At/ each of infinitely many > iterations. After any particular finite number of such iterations is completed, it > will be found that the anti-diagonal to the latest list is not a > member of the latest set of listed objects, and therefore not a member > of any of the prior ones either. Note that the meta-diagonals from the > latest list are irrelevant. I note that you insist so, but as I also see that the meta-diagonals > together with the ordinary entries form a countable set, I don't > believe you. That set is, in its turn, incomplete, and further iteration of that process can only produce forever incomplete lists. And this situation goes on So it is. All that exists is countable. All those list are countable but incomplete, at least as listing of ALL reals (or all binary sequences as in Cantors ACTUAL diagonal proof). === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > The set of ordinary entries and the set of meta-diagonals > together form a countable set. Those meta-diagonals, not being listed, are irrelevant. Why? Per ordre mufti? They are in a countable set together with the > regular entries of the list. Anything that need not be a member of a list, such as WM's irrelevant > meta-diagonals, does not have to be considered when analysing the > membership of that list. But they have to be considered when analysing the membership of the > diagonal to a countable set! If you mean the Cantor defined anti-diagonal, you are totally and > absolutely wrong. One only need consider things which have an index > position in the list, why? Because a listed element corresponds in the listing to a unique natural number position, but diagonals, anti-diagonals, and meta-diagonals do not. which your meta-diagonals do not. They have, for example a_m1, a_ (m+1)2, a_(m+2)3, ... with m =/= 1 is uniquely defined and can be incorporated into a list at position > 2m. What do you do with the string already in position 2m? Those meta-diagonals are spread over infinitely many positions > indexes, and Cantor's theorem says nothing about any such things at all That's why he failed. He did not consider these simple constructs. Cantor had a doctorate in mathematics and became professor a a major university in recognition of his understanding of such simple constructs. You could not even get a G.E.D. or teach kindergarten on the basis of the miniscule understanding you have shown here of such simple constructs. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > WM says... > There is no sequence containing > all real numbers. There is no pair of sequences containing all > real numbers. There is no triple of sequences containing all real > numbers. Unfounded assertion. It's provable. It cannot be excluded that a Cantor-list contains >all real numbers You just said that there is no sequence containg all the reals. You're contradicting yourself. No. It cannot be excluded that two sequences contain all real numbers. It can in mathematics even if it cannot in physics. > If the two sequences are unioned, you can construct another anti- > diagonal, but it cannot be excluded that this anti-diagoal is in the > sequencenof meta-diagonals simultaneously created by the union of the > sequences. No meta-diagonal as a meta-diagonal need occurs as a listed seqeunce Hence this (and every other real number) can be in a > countable set. If you create a listing which includes all the meta diagonals of the original list as well as the members of that original list, there will be an anti-diagonal not listed in that new list. In fact there will be as many anti-diagonals as listed elements. What is the reason of this paradoxical situation? The solution is > simple. All real numbers that exist can be written into one sequence. If you think so then do it. > Who should hinder us? But by forming such a sequence, we automatically > create many more real numbers by means of meta-diagonals etc. All > those numbers belong to a countable set, as can be proven by writing > them into one single infinite sequence. But since for EVERY list, there are as many anti-diagonals not in that list as things listed, julio's argument doesn't work. Yes, you certainly can know that, since Cantor proved that no countable > set contains all real numbers. He proved (and even his proof is in error) that no single sequence > contains all real numbers. Nothing more. That is quite sufficient, as for the reals to be countable requires that they be listable as a single sequence. A set whose members cannot be listed in a single sequence is, by definition, uncountable. What did you think 'uncountable' meant? Alas, he forgot to prove that all real numbers exist. Others have done that, e.g., via Dedekind cuts and equivalence classes of Cauchy seqeunces. === Subject: Re: Large Integer Math Calculation > can someone please calculate the value below? > Please note the negation of k. > b = 268435456 > k = 181584212 > x = 999298784357823475894317568943275692 > n = 19342813113834066796298817 > c = (b^(-k) * x ) (mod n) > What is the value of c? > I confirm once more: > b^(-k) * x `mod` n = b^(phi n - k) * x `mod` n > = 15290831880383268439929393 > It would be more elegant to use lambda(n) instead of phi(n). > What is the lambda function? Maybe I should have called the function > 'totient'. That would be a lot clearer. > But surely that's generally a rather slow way to do it? Finding b^k > mod n, then its modular inverse with respect to n, should be much > faster than factoring n to get phi(n) or lambda(n). > It's indeed much slower, but that wasn't the point. I just wanted to > confirm by going another correct way and getting the same result. Lambda(n) is the least common multiple of phi(p_i) >over all prime powers p_i which are divisors of n. >For example, when n = 825 = 3 * 5^2 * 11, phi(n) >is 2 * 20 * 10 = 400, but lambda(n) is only 20. It is unclear from your notation whether you intended to lambda(n) to be the LCM of phi(p_i) where p_i are the primes which are factors of n, or the LCM of phi(p_i^{k_i}) where the product of p_i^{k_i} is the prime factorization of n. I am sure that you meant the latter, but I thought that it might help others to make that clear. >The point is that a^lambda(n) mod n = 1 for all a >which are coprime to n. I just thought that anyone >who knew about modular exponentiation, and also was >able to factor such a large n, would already have >known about lambda(n)! That sounds a bit condescending. Either way works, and although your point about efficiency is correct, as you noted, neither way is very efficient. Better is to use the EEA to invert b or b^k mod n. >But yes, of course it gives the same result. Rob Johnson take out the trash before replying === Subject: Re: Large Integer Math Calculation > Hi All, can someone please calculate the value below? Please note the negation of k. b = 268435456 k = 181584212 x = 999298784357823475894317568943275692 n = 19342813113834066796298817 c = (b^(-k) * x ) (mod n) What is the value of c? > I think you can go to this page: http://gmplib.org/#TRY or probably one of many others on the Internet and do these calculations yourself. I don't think you should feel guilty about using the servers of others in this way. Anyone who puts stuff like this out as a CGI script is running a lightly-loaded server and they are grateful if you use the pages. The server is going to use about the same amount of electricity either way. There are probably a fair number of large integer calculation pages on the internet. === Subject: Re: Unilateral Laplace transform of 1/t <9o94e4d8rhh7eddjqm8q882nb2s8nn2iap@4ax.com> posting-account=j8_D9AoAAACuXBFn5S6zg07K9jikZU4_ CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > To the OP: Post the exact statement of the problem from the textbook > you referenced so we can consider what the author had in mind. I don't have the book handy, but it is along the following lines: >Knowing the transform of x(t) = 1 for t >= 0 to be 1/s, use the >integration property to obtain the transform of x(t)/t. This leads to >Int[s..oo] du/u, which I suspected didn't exist. I guess the textbook >author was not too careful when framing this problem. If someone has >the Solutions Manual to this book and post the solution given there, >maybe it will throw more light on what the author had in mind. --vv I believe the book you first made reference to was Linear Systems and > Signals by B. P. Lathi, and I'll guess the 2nd ed. I don't have the book, but after some digging I may have found the > exercise in question, which I've reproduced below. The problem begins by referring to the Laplace transform X(s) of > signal x(t) = 1/t u(t) That looks to be a mistake (typo?). I don't think the author meant to > write u(t), which most often refers to the unit step function, also > problem below, that line should probably read x(t) = 1/t y(t) It appears the point of the exercise is to derive the division by t > property. If the exercise below is the one you had in mind, look at > it again and see what you think. Page 480, exercise 4.2-9: > ------------------------ It is difficult to compute the Laplace transform X(s) of signal x(t) = 1/t u(t) by using direct integration. Instead, properties provide a simpler > method. (a) Use Laplace transform properties to express the Laplace transform > of t x(t) in terms of the unknown quantity X(s). (b) Use the definition to determine the Laplace transform of > y(t) = t x(t). (c) Solve for X(s) by using the two pieces from a and b. Simplify > your answer. Checking CRC tables, combining results for unit step function u(t) and 1/t*F(t) (F(t) some function) you get Laplace transform (1/t*u(t)) =integral (1/x dx) from s to inf, where s is the Laplace variable. Using definition of ln(), integral (1/x dx) from a to b = ln(b)- ln(a). Then integral (1/x dx) for x from s to inf = lim (as b=>inf) of [ln(b)- ln(s)], which is unbounded for any s. The unilateral Laplace transform of 1/t does not exist. Dirk === Subject: Re: Unilateral Laplace transform of 1/t (snip) > The problem begins by referring to the Laplace transform X(s) of > signal > x(t) = 1/t u(t) > That looks to be a mistake (typo?). I don't think the author meant to > write u(t), which most often refers to the unit step function, also > problem below, that line should probably read > x(t) = 1/t y(t) u is the variable most commonly used for change of variables problems in calculus. (v is second.) In any case, I agree it doesn't sound like the unit step function. (That was H in the books I remember, but others may use u.) -- glen === Subject: Re: Unilateral Laplace transform of 1/t >(snip) > The problem begins by referring to the Laplace transform X(s) of > signal > x(t) = 1/t u(t) > That looks to be a mistake (typo?). I don't think the author meant to > write u(t), which most often refers to the unit step function, also > problem below, that line should probably read > x(t) = 1/t y(t) u is the variable most commonly used for change of variables >problems in calculus. (v is second.) In any case, I agree it doesn't sound like the unit step function. >(That was H in the books I remember, but others may use u.) Indeed, H is still widely used. Some authors use the even longer name 'Heaviside unit step function'. MathWorld, Wikipedia, and PlanetMath all use the subject title 'Heaviside Step Function'. Abramowitz and Stegun use 'unit step function' and the notation u(t). Authors differ in their definition of the function. Simple. 8-) Refs: -- glen === Subject: Re: Unilateral Laplace transform of 1/t <9o94e4d8rhh7eddjqm8q882nb2s8nn2iap@4ax.com> posting-account=j8_D9AoAAACuXBFn5S6zg07K9jikZU4_ CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) To the OP: Post the exact statement of the problem from the textbook > you referenced so we can consider what the author had in mind. I don't have the book handy, but it is along the following lines: > Knowing the transform of x(t) = 1 for t >= 0 to be 1/s, use the > integration property to obtain the transform of x(t)/t. This leads to > Int[s..oo] du/u, which I suspected didn't exist. I guess the textbook > author was not too careful when framing this problem. If someone has > the Solutions Manual to this book and post the solution given there, > maybe it will throw more light on what the author had in mind. --vv w, (hopefully not THE 'w', on Pennsylvania Ave) Since your Laplace transform is unilateral, consider the possibility that the Laplace transform does not converge (i.e., does not exist). Unilateral LT(1/t)=integral(1/t*exp(-s*t)) integrated over t=[0, inf). Consider that there is an 'a' such that Re(exp(-s*t))>=a>0 for t in the interval [0,1], with s fixed, but unspecified. Also consider that integral(1/t) integrated over [0,1] = ( -ln(0) ), which is infinite; this follows from the definition of ln(t). Then it follows that Re( integral(1/t*exp(-s*t) ) integrated over [0,1] is also infinite. As the upper limit of integration is increased, the real part inside the integral stays strictly positive, so increasing the upper limit on integral(1/t*exp(-s*t)) strictly adds to Re( integral(1/t*exp(- s*t)) ). Therefore the real part of the unilateral Laplace transform of 1/t is unbounded for every s. So the unilateral Laplace transform of 1/t does not exit. Dirk P.S. The Fourier transform of this does not exist either when the function is considered to be 0 for t<0. The Fourier transform that you see computed is for ( 1/t, where t is from (-inf, inf), usually with the function redefined to be 0 for t=0). === Subject: Re: Unilateral Laplace transform of 1/t <9o94e4d8rhh7eddjqm8q882nb2s8nn2iap@4ax.com> posting-account=j8_D9AoAAACuXBFn5S6zg07K9jikZU4_ CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > To the OP: Post the exact statement of the problem from the textbook > you referenced so we can consider what the author had in mind. I don't have the book handy, but it is along the following lines: > Knowing the transform of x(t) = 1 for t >= 0 to be 1/s, use the > integration property to obtain the transform of x(t)/t. This leads to > Int[s..oo] du/u, which I suspected didn't exist. I guess the textbook > author was not too careful when framing this problem. If someone has > the Solutions Manual to this book and post the solution given there, > maybe it will throw more light on what the author had in mind. --vv w, (hopefully not THE 'w', on Pennsylvania Ave) Since your Laplace transform is unilateral, consider the possibility > that the Laplace transform does not converge (i.e., does not exist). Unilateral LT(1/t)=integral(1/t*exp(-s*t)) integrated over t=[0, > inf). Consider that there is an 'a' such that Re(exp(-s*t))>=a>0 for t in > the interval [0,1], with s fixed, but unspecified. Also consider that > integral(1/t) integrated over [0,1] = ( -ln(0) ), which is infinite; > this follows from the definition of ln(t). Then it follows that > Re( integral(1/t*exp(-s*t) ) integrated over [0,1] is also infinite. As the upper limit of integration is increased, the real part inside > the integral stays strictly positive, so increasing the upper limit on > integral(1/t*exp(-s*t)) strictly adds to Re( integral(1/t*exp(- > s*t)) ). Therefore the real part of the unilateral Laplace transform of 1/t is > unbounded for every s. So the unilateral Laplace transform of 1/t does > not exit. Dirk P.S. The Fourier transform of this does not exist either when the > function is considered to be 0 for t<0. The Fourier transform that > you see computed is for ( 1/t, where t is from (-inf, inf), usually > with the function redefined to be 0 for t=0). Does not exist, not Does not exit. Can't type. Dirk === Subject: Re: My academic credentials posting-account=aLpfCwoAAACh4BOs3HOlQBCoxUpEgyxc Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) >[...] a friend of mine while drunk after his finals briefly >believed that he was an intermediate field extension. Was he normal? Normal? To the floor he was standing on? > I doubt it. He was probably keeling over. Actually, if my memory serves me correctly, he was sitting in a chair > covered in his own vomit. Let's see if anyone can make a Galois theory > pun out of that. he certainly wasn't a class number i bet his local field was far from ideal hecke > if he needs representation > i know someone good at sweeping the ramifications > under the rug was his name norm? Damn, you're good. === Subject: Jump Diffusion Model Cc: bkuhlensch@aol.com posting-account=0GoAgwoAAAB6EiqKK1zXzJhwl0-sgfsD MathPlayer 2.10b; .NET CLR 1.1.4322; .NET CLR 2.0.50727; .NET CLR 3.0.04506.648; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) Hi! I am writing a thesis in quantitative finance in which I use the (Merton) Jump Diffusion Model. The model is described for example in www.econ.nthu.edu.tw/wp/0215E.pdf on page 5 and 6. My tutor told me that I could use a theorem by Khinchin to derive or motivate this formula. Unfortunately he could not exactly remember the theorem or its name. Does anybody know which theorem he could have meant and how I can use it in this context? === Subject: Re: Jump Diffusion Model posting-account=llxexwoAAADqntGDNkT4ytZG1pkEWVd9 SLCC1; .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; Zune 2.0; .NET CLR 1.1.4322),gzip(gfe),gzip(gfe) > Hi! I am writing a thesis in quantitative finance in which I use the > (Merton) Jump Diffusion Model. The model is described for example inwww.econ.nthu.edu.tw/wp/0215E.pdfon page 5 and 6. > My tutor told me that I could use a theorem by Khinchin to derive or > motivate this formula. Unfortunately he could not exactly remember the > theorem or its name. Does anybody know which theorem he could have > meant and how I can use it in this context? I do not know if this is it, I simply did a Google Search on Jump Diffusion and Khinchin. But I found a few references to L'evy processes and within that something called L'evy[CapitalEth]Khinchin Here are a few Financial Engineering papers that reference this in the context of Jump Diffusion models: http://www.fam.tuwien.ac.at/~papapan/papers/NTUA2008.pdf http://www.cmap.polytechnique.fr/~rama/papers/pide.pdf http://www.math.nyu.edu/research/carrp/papers/pdf/integtransform.pdf http://math.ucalgary.ca/~aswish/CollTalkJan24 08.pdf I am finishing 3 years of math (taking Linear Alg now) with the intent of studying Quant Fin next year via Columbia's CVN distance learning program. Where are you studying? Do you like the program? Alan === Subject: Re: Extension by Zero of a Sheaf > I am having some trouble to understand what B and BÁX > are. Are they the > .8espace .8etal.8e of ordinary sheaves as defined here: > http://en.wikipedia.org/wiki/Sheaf_(mathematics)#Sheav > es? I think so. I'll include the author's definition of sheaf, and some facts he deduces: X is a topological space. Definition: A sheaf (of abelian groups) on X is a pair (A,p) where: (i) A is a topological space (ii) p:A->X is a local homeomorphism onto X (iii) each A_x=p^(-1)(x) for x in X is an abelian group (and is called the stalk of A at x). (iv) the group operations are continuous .. If A is a sheaf on X and if Y C X, then a section of A over Y is a map s:Y->A such that ps=id. .. (a) p is an open map (b) Any section of A over an open set is an open map (c) Any element of A is in the range of some section over some open set. (d) The set of all images of sections over open sets is a base for the topology of A. (e) For any two sections s in A(U) and t in A(V), U and V open, the set W of points x in U_intersect_V such that s(x)=t(x) is open === Subject: Re: Photons -- Hidden Fundamentals > Tiny, in bunches of a few or in lots of many it comes > from across the universe, traveling for Billions of > years, (some say tiredlessly) rattling (something, > or itself), shaking, pulsating, oscillating or undulating > until it hits our eye and conveys to us the greatest > story ever told, Cosmology, the nature of the cosmos. > intrinsically related or the flip side of charge pairs > and magnetism; manifesting in gamma rays, light, > radio and IR etc, etc. & have held humankind in awe > forever. After all, these force carries make our vision > possible, so and such that we can behold nature. > We know a lot of the elementary properties of the > photon, which is capable of emission, absorption, > reflection, speed, diffraction and aberration. etc, etc.. > all well explained, which brings us the QUESTION: > Are there more still unknown & undiscovered properties > of the photon? Do we know ALL the properties that > can be attributed to the photon? ... Is there a theory > or conjecture that says, or even proves, that what we > currently know is all there is to know about the existence > of the photon? > NEWs-groupies tell me something NEW about the > photon, not regurgitating what is written in the WIKIs > or re-laments about c-v or c+v... or do not elborate > on what I have sketched out above. Tell me about your > suspicions about *** what else *** this cosmic vibration > could encode and still hide from us. > hanson > Dirk Bruere: > Quantum Archeology > Now that is state of the art fore-front stuff. If you have > followed the emergence-development of this - on your > radio program, could you post your impressions from > hanson > -- Unfortunately nobody knows how QA might eventually be achieved, just that it is possible in principle to recover information. http://www.geocities.com/john_f_ellis/QuantumArcheology.htm As for photons, I don't know of anything that rules out currently unknown properties. Maybe recasting physics in an information theoretic way might point to something. I also rather like the implications of the Wheeler Delayed Choice expt http://en.wikipedia.org/wiki/Wheeler%27s_delayed_choice_experiment and Quantum Non-demolition Imaging -- Dirk http://www.transcendence.me.uk/ - Transcendence UK http://www.theconsensus.org/ - A UK political party http://www.onetribe.me.uk/wordpress/?cat=5 - Our podcasts on weird stuff === Subject: Re: Photons -- Hidden Fundamentals > ------------- ahahahahahaha ----------- > ---------- ahahAHAHAahaha ---------- > -------- AHAHAHAHAHAHAHA ----------- ************************************************ * * * Androcles 5 cha-cha-hanson 0 * * * ************************************************ Maniacal laughter aside, what is YOUR hidden fundamental definition of a photon, sucker? === Subject: Re: Photons -- Hidden Fundamentals > ------------- ahahahahahaha ----------- > ---------- ahahAHAHAahaha ---------- > -------- AHAHAHAHAHAHAHA ----------- ************************************************ * * * Androcles 9 cha-cha-hanson 0 * * * ************************************************ Maniacal laughter aside, what is YOUR hidden fundamental definition of a photon, sucker? === Subject: Re: Photons -- Hidden Fundamentals > ahahaha... I have no **hidden fundamental** definition > of a photon. So you have no idea what you are talking about but can write reams of this corny crap: Tiny, in bunches of a few... Photons come in all sizes, cha-cha, they are quantities of energy. SOME of those that come from molecules are tiny because molecules are tiny, but that is really irrelevant. As physicists the hidden fundamental question we should ask is what gives the molecular photon its directionality?, the means of investigation being the phased array by which we already have the means of making photons of manageable quantities and sizes. Tell me about your suspicions about *** what else *** this cosmic vibration could encode and still hide from us. Ok, I will. Watson and Crick discovered the double helix of DNA from diffraction patterns. The photon contains the code for the simpler hydrogen atom. If you want to build a cathedral you should know what a brick is first. That's a hidden fundamental. === Subject: Re: Photons -- Hidden Fundamentals <18yEk.999$kI6.485@nwrddc01.gnilink.net> <3CEEk.95832$KV1.50410@newsfe25.ams2> <8KQEk.74$P5.16@nwrddc02.gnilink.net> posting-account=PTS84AoAAACr67p51zvy0Hlr3LkoIUcc SLCC1; .NET CLR 2.0.50727; .NET CLR 3.0.04506; Media Center PC 5.0; InfoPath.2),gzip(gfe),gzip(gfe) Discussing with Androcles will not bare any fruits. You are merely providing entertainment for the village idiots like moortel, Gisse, McCullough, PD, and other Einstein Dingleberries. === Subject: Re: Photons -- Hidden Fundamentals posting-account=UmSM6QoAAAAHQsbQSpqx55ht5J9R5UV_ CLR 1.1.4322; InfoPath.1),gzip(gfe),gzip(gfe) The MaxwellÍs equations predict the Aether. After all, magnetism does > not even satisfy the principle of relativity. My thought on the Aether and itÍs only a theory I pulled out of my twizzel stick With know evidence to substantiate the claim but it's interesting to me, and thatÍs that Aether or space time fabric is not curved but due to constraints of spherical bodies at the time the forefathers were trying to work this mystery out. the computations worked on or what they observed were all based on spherical bodies within the Aether, no one ever tried theoretically placing a big triangle or cone or square with comparable mass and density that displaces the same amount Aether and see the effect of gravitational lensing or any other interaction of how space-time is effected. is still curved, yes for spherical bodies within its confines but what about theoretical shapes not found normally in our universe. I think space time fabric or aether properties if it has any IÍm not convinced of any yet other that a universal resistance within it but thatÍs because space it's self is not a pure environment. I think the effects on the time continuum is more dependant on the way and shape of how much displacement occurs within the aether itself. === Subject: Re: Photons -- Hidden Fundamentals posting-account=7bF0GwoAAABMFHX6V4fON4-1F6LFJ834 1.1.4322; .NET CLR 2.0.50727; .NET CLR 3.0.04506.30; .NET CLR 3.0.04506.648),gzip(gfe),gzip(gfe) > Tiny, in bunches of a few or in lots of > many it comes from across the universe, > traveling for Billions of years, (some > say tiredlessly) rattling (something, > or itself), shaking, pulsating, > oscillating or undulating until it > hits our eye and conveys to us the > greatest story ever told, Cosmology, > the nature of the cosmos. Not sure about the rattling, shaking, and pulsing. Should we use our God-Vision (TM) and follow along with the photon, it might look very static... > packets are intrinsically related or > the flip side of charge pairs and > magnetism; manifesting in gamma rays, > light, radio and IR etc, etc. & have > held humankind in awe forever. After > all, these force carries make our > vision possible, ... and chemistry ... maybe even the illusion of spacetime (which is why c figures so prominently). > so and such that we can behold nature. We know a lot of the elementary > properties of the photon, which is > capable of emission, absorption, > reflection, speed, diffraction and > aberration. etc, etc.. all well > explained, ... well, described might be a better word choice, rather than explained ... > which brings us the QUESTION: Are the more still unknown & > undiscovered properties of the > photon? I hope so. > Do we know ALL the properties that > can be attributed to the photon? ... Probably not that evince in a 4D Universe. Inherent spin is an indication of a yet one dimension offset. > Is there a theory or conjecture that > says, or even proves, that what we > currently know is all there is to > know about the existence of the photon? Since all we know is based on subjective interpretation of meausrement, there is more to know. > NEWs-groupies tell me something NEW > about the photon, not regurgitating > what is written in the WIKIs or re- > laments about c-v or c+v... or do not > elborate on what I have sketched out > above. Tell me about your suspicions > about *** what else *** this cosmic > vibration could encode and still hide > from us. To do so would require that you eventually be killed, being as how you are not a member of the cabal of Men In Black. And then they'd have to come for me. Actually, there may be more to consider / cogitate from my particular pet theory, which edu-ma-cation in would probably take too long to arrive back at this thread topic. Namely that the CMBR is either what our container Universe looks like or is a normal shock of matter / energy infalling into a new partially orthogonal spacetime. Leaving light cones and light itself as features of the trip from past to reingested) at different external times. Or not. David A. Smith === Subject: Re: Photons -- Hidden Fundamentals <9KQEk.75$P5.4@nwrddc02.gnilink.net> posting-account=jPnQ2goAAAA461y3QD0lbyw0oKeThma1 AppleWebKit/525.18 (KHTML, like Gecko) Version/3.1.2 Safari/525.20.1,gzip(gfe),gzip(gfe) we all know that herr doktor-professor Albert got his prize for photo-electric dffect, but that is the only thing that requires this photon that you speak of; photons are otherwise totally useless, probably because they are zero-dimensional. where did you ever refute Uncle Al's experiments? === Subject: Re: Rubik's Cube solver program - correction >I can't answer your questions but I did one years back, >believe it or not, on a radio shack trs80 16k using >basic. >Using 91,92,93,94,95,96 covering all 6 sides of >the cubes using (6) 6*6 grids. The above should be (6) 9*9 grids. >I am going to have to try to fire it up to see how I >did it. I have no idea now how I did it back then but >I remember there was 2 options where you could try >to solve the cube or the other option was let it solve >a random mix on its own. I know it was a lot less than >51 moves for the program algorithm to solve it. >Dan === Subject: Re: Rubik's Cube solver program > I used to have a program that ran in Windows a few > years ago that > would solve the Rubik's Cube in the near optimal > number of moves, > maybe 22 or less. The solvers on the web use lots of > moves, over 50. > I've used Google to search for programs, but the ones > I did find I > could not get to run in Windows. Does anyone know > what program I am > applications these > days. Another question, what is the maximum-minimum number > of moves for a > perfectly scrambled 4x4x4 Rubik's Revenge Cube > (called God's > algorithm). --Tim923 I can't answer your questions but I did one years back, believe it or not, on a radio shack trs80 16k using basic. Using 91,92,93,94,95,96 covering all 6 sides of the cubes using (6) 6*6 grids. I am going to have to try to fire it up to see how I did it. I have no idea now how I did it back then but I remember there was 2 options where you could try to solve the cube or the other option was let it solve a random mix on its own. I know it was a lot less than 51 moves for the program algorithm to solve it. Dan === Subject: The Cure to the Genie Out of the Bottle (was: the situation is worse than you thought) <48E322E0.9600D416@hotmail.com> posting-account=G-TjQAkAAADYg6rno3bWQPnIwKFBrf1t MathPlayer 2.10b; .NET CLR 1.1.4322),gzip(gfe),gzip(gfe) > Good idea Graham. Lets also kill everyone who disagrees. That'd be difficult to do without first ensuring that (a) they're all together in one place and (b) nobody else is in that place. Otherwise you'd be facing a situation analogous to trying to nuke Israel and ending up killing off all the Palestianians in the process, while vitrifying the sands of grounds holy to billions of people. Hence, we find the ultimately solution to the modern-day analogue of Ring of Power -- the Genie out of the Bottle. To unmake it does not require finding the place of was made, and going back there to reverse its creation ... E = mc^2 will never be forgotten. Rather, unlike the Ring of Power, to neutralize this shadow of evil requires removing the key enabler of the Genie: geography. A nuclear bomb is impotent if the enemy to be targeted does not have a geography or is not associated with a location at all. Inasfar as there continue to be sovereigns, affiliations or people in this world which are identified with specific geographies, the Genie will remain potent. The world will never be cured of the curse of this scourge unleashed, until it removes the enabling situation of peoples'/nations/ affiliations centralized to locations. That is, by hook or crook, the Old World Order of geographically- defined nation states is, itself, the enabler of the Genie and is the most dangerous menace on this Earth. The very act of building walls around borders, and establishing purified geographic locales or enclaves buttresses the very thing that gives the Genie its power. Without condition (a) above, there will never be any such thing as killing them all. Even with conditionn (a), without condition (b), there will be no such thing as killing them all that doesn't also mean killing a lot of those who also agree with you, and alienating them too and making enemies of them too. It is for this reason that the Old World Order will soon be overturned -- either by dissolution of the geographic nation-states, or by the force of the nukes that everybody and their brother/sister will soon have in their possession. One way or the other, the Age of Geography is over. The only type of people, affiliation or organization that can exist in such a world remade is that based on decentralization at every turn. The central of the world order that will replace the nation-states is already right before your very eyes. It's Cyberspace. The Internet is, itself, the embryonic form of what will become the World Federation. It will not be some New World Order estbablished from top-down, by committee, by a remote convention in New York, Geneva or San Francisco by powerful elites. Rather, it will be the structure that is already in place, emerging spontaneously from bottom up by the collective results of your actions. You already are the key players in the World Federation. It is for this reason that the people who first conveyed the well- known equation that's seen its consequence in the nuclear bomb, ended his life as a World Federalist. Anyone who reads this is a de facto world federalist solely by virtue of their continued involvement in the entity that is destined to replace all the governments on Earth and subsume all geography and history. === Subject: Re: Wind Around Solitaire > Yes, the game as written is indeed too easy to play. Rule update: The even integers must go in squares immediately adjacent > to already filled in squares. I suspect that even with this rule update, the game is too easy to get > the maximum possible score. If we require that each even integer be placed immediately adjacent to the previous (odd) integer, that might make it difficult enough. Dan === Subject: cyclic group without sylow posting-account=gc2kDQoAAADMsLO9kJjQL9hCJkI0D8qJ .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506),gzip(gfe),gzip(gfe) How can you prove a group G of order 15 only using Cauchy's theorem or any more elementary way? i.e no Sylow's theorems. By Cauchy's theorem a group of order 15 contains an element of order x 3 and another of order y 5. Then if we can only show that xy has order 15 then we are done. So tried by cases, xy has order 1 , impossible otherwise y = x^(-1). But how to contradict the fact that it can't have order 3 neither 5? If we can show G is abelian we are done because (xy)^3 = x^3*y^3 = y^3, but this is also frustrating as I can't show that G is abelian..any suggestions, ideas? === Subject: Re: cyclic group without sylow posting-account=gc2kDQoAAADMsLO9kJjQL9hCJkI0D8qJ .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506),gzip(gfe),gzip(gfe) Sorry. I meant show G is cyclic. === Subject: Re: cyclic group without sylow posting-account=Yn5cwwoAAADntcMuRwk-EwLg-DMZ_hXN rv:1.9.0.2) Gecko/2008091618 Firefox/3.0.2,gzip(gfe),gzip(gfe) > Sorry. I meant show G is cyclic. enumeration? why avoid sylow here and in what sense? i like brute force methods so exhaustion of multiplication tables seems a pretty sure-cut way to solve that problem but sylow picks out your subgroups right away and very easy to prove do you want a proof avoiding finding the subgroups? or if allowed what would be considered substantialy different? -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- galathaea: prankster, fablist, magician, liar === Subject: Re: cyclic group without sylow posting-account=gc2kDQoAAADMsLO9kJjQL9hCJkI0D8qJ .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506),gzip(gfe),gzip(gfe) Sorry. I meant show G is cyclic. enumeration? why avoid sylow here > and in what sense? i like brute force methods > so exhaustion of multiplication tables > seems a pretty sure-cut way to solve that problem but sylow picks out your subgroups right away > and very easy to prove do you want a proof avoiding finding the subgroups? > or if allowed > what would be considered substantialy different? -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- > galathaea: prankster, fablist, magician, liar What I meant is that this problem is before Sylow's theorems so I think there might be another method. I saw the proof using Sylow's theorem and indeed is straightforward. === Subject: Re: Some of his math... for free! posting-account=n1ZfDgkAAABbCs44qOtz8dP-RkWuEBif Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) > Hi all, I want to to call your attention to a great event: has > decided to allow anyone to download the PDF version of his book Some > of My Math for free! It's right here: http://www.lulu.com/items/volume 63/2812000/2812588/4/print/Some of M... I'm so glad for this! For months that I had been struggling against this > temptation that I had of buying James' book (either as a book or as a > PDF file) for just $32.95. And now it is here for free! Now I can at > long last browse through James' discoveries just by reading a PDF file > which looks exactly like James' blog. And I am quite mystified that you are promoting the book in this way. You puzzle me... > Of course, all those readers would actually paid $32.95 for the book are > perhaps now feeling a bit frustrated. But just think how much better is > mankind now, with free access to all these revolutionary achievements! No one bought it. 0 sales. That proved to me that there was no shadow or hidden support around the world for my research which is what I was checking for. After the high Google search results for my research like with definition of mathematical proof I thought maybe... But I was wrong. The Google search results remain a mystery to me as to what they mean. But there seems to be no hidden support for my research out there. Looks like the negatives are completely dominant. I couldn't sell a single copy of the book, except to myself. === Subject: Re: Some of his math... for free! > I want to to call your attention to a great event: has > decided to allow anyone to download the PDF version of his book Some > of My Math for free! It's right here: > http://www.lulu.com/items/volume_63/2812000/2812588/4/print/Some_of_M... > I'm so glad for this! For months that I had been struggling against this > temptation that I had of buying James' book (either as a book or as a > PDF file) for just $32.95. And now it is here for free! Now I can at > long last browse through James' discoveries just by reading a PDF file > which looks exactly like James' blog. And I am quite mystified that you are promoting the book in this way. You puzzle me... Well, it's not the first time, as you know: http://mathforum.org/kb/thread.jspa?forumID=13&threadID=1778321&messageID=63 13640#6313640 Now, about the reason I am promoting the book in this way, I suggest the you focus your amazing analytical skills on the concept of irony. After spending as many time pondering about it as the time you spent so far on mathematics, come back and tell us which conclusions you reached. Of course, you will be as wrong as always, but meanwhile you will give us a break. > Of course, all those readers would actually paid $32.95 for the book are > perhaps now feeling a bit frustrated. But just think how much better is > mankind now, with free access to all these revolutionary achievements! No one bought it. 0 sales. That proved to me that there was no shadow or hidden support around > the world for my research which is what I was checking for. After the > high Google search results for my research like with definition of > mathematical proof I thought maybe... But I was wrong. The Google > search results remain a mystery to me as to what they mean. But there seems to be no hidden support for my research out there. Looks like the negatives are completely dominant. I couldn't sell a single copy of the book, except to myself. That is very strange indeed, James. I say this because when I told you, almost three years ago that you were alone in your research, your reply was What??!!! Are you deluded? You really think I am alone?: http://mathforum.org/kb/plaintext.jspa?messageID=4257429 Well, why is it then that your countless supporters don't go rushing to buy your book? Or, for that matter, why don't they rate each one of your posts at Usenet with five stars? According to Google Groups, your average rating, taken from 5384 ratings, is... one star! Jose Carlos Santos === Subject: Re: Some of his math... for free! posting-account=aLpfCwoAAACh4BOs3HOlQBCoxUpEgyxc Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) Hi all, I want to to call your attention to a great event: has > decided to allow anyone to download the PDF version of his book Some > of My Math for free! It's right here: http://www.lulu.com/items/volume 63/2812000/2812588/4/print/Some of M... I'm so glad for this! For months that I had been struggling against this > temptation that I had of buying James' book (either as a book or as a > PDF file) for just $32.95. And now it is here for free! Now I can at > long last browse through James' discoveries just by reading a PDF file > which looks exactly like James' blog. And I am quite mystified that you are promoting the book in this way. You puzzle me... If other people's actions are puzzling in light of your world-view, perhaps that is a sign that your world-view is wrong. === Subject: Re: Some of his math... for free! posting-account=n1ZfDgkAAABbCs44qOtz8dP-RkWuEBif Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) > Hi all, I want to to call your attention to a great event: has > decided to allow anyone to download the PDF version of his book Some > of My Math for free! It's right here: http://www.lulu.com/items/volume 63/2812000/2812588/4/print/Some of M... I'm so glad for this! For months that I had been struggling against this > temptation that I had of buying James' book (either as a book or as a > PDF file) for just $32.95. And now it is here for free! Now I can at > long last browse through James' discoveries just by reading a PDF file > which looks exactly like James' blog. And I am quite mystified that you are promoting the book in this way. You puzzle me... If other people's actions are puzzling in light of your world-view, > perhaps that is a sign that your world-view is wrong. Yes. It is. That is the reason for the post. I find his behavior puzzling. It seems obvious to simply ask him to explain himself. As I study the problem I am sure I will begin to solve it. It's just a matter of using various problem solving tools until I crack it. === Subject: Re: Some of his math... for free! posting-account=HaopWgoAAADs72-s8RQYwP_-ruRUuNzX .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; InfoPath.1),gzip(gfe),gzip(gfe) It's just a matter of using various problem solving tools until I > crack it. Perhaps you can analyze the causes of your delusional and narcissistic tendencies! === Subject: Re: Some of his math... for free! posting-account=OKTeIQkAAAAZk6JK1hK7-grwpoUDNy98 CLR 1.1.4322; .NET CLR 2.0.50727; InfoPath.1),gzip(gfe),gzip(gfe) Hi all, I want to to call your attention to a great event: has > decided to allow anyone to download the PDF version of his book Some > of My Math for free! It's right here: http://www.lulu.com/items/volume 63/2812000/2812588/4/print/Some of M... I'm so glad for this! For months that I had been struggling against this > temptation that I had of buying James' book (either as a book or as a > PDF file) for just $32.95. And now it is here for free! Now I can at > long last browse through James' discoveries just by reading a PDF file > which looks exactly like James' blog. And I am quite mystified that you are promoting the book in this way. You puzzle me... You thought the promotion would translate to sales? Of course, all those readers would actually paid $32.95 for the book are > perhaps now feeling a bit frustrated. But just think how much better is > mankind now, with free access to all these revolutionary achievements! No one bought it. Really? 0 sales. What a shame. That proved to me that there was no shadow or hidden support around > the world for my research which is what I was checking for. Looks like the world isn't going to beat a path to your door. > After the > high Google search results for my research like with definition of > mathematical proof I thought maybe... But I was wrong. The Google > search results remain a mystery to me as to what they mean. And how many times did people tell you they were meaningless? But there seems to be no hidden support for my research out there. And this surprises you? Looks like the negatives are completely dominant. I couldn't sell a single copy of the book, except to myself. There *IS* a way to sell books, but you probably won't go for it: tell the truth. === Subject: Re: Call to JSH posting-account=n1ZfDgkAAABbCs44qOtz8dP-RkWuEBif Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) > Hi JSH, Your book is very hard to read and understand. Oh. Sorry. > I call you to concentrate your efforts on clearing this text to make > it understandable even by non-geniuses. Maybe in a later draft... > Instead of posting modified research results to Usenet better > concentrate on the old in the book. Otherwise we cannot understand you > anyway. Well it's a necessary first step to finally pull everything into a book. I'll admit that I find it difficult to understand parts of it myself as I did the research years ago... === Subject: Re: question about prime partition Let triangle ABC , where the coordinats A( 0 , 0 ) , B( a , b ) , C( c , d ) then LARGE of triangle is LARGE = ( c - a )( b - d)/2 + d( c - a )/2 + a( b - d )/2 LARGE = ( bc - ad )/2 for A = ( p , q ) , please select to subtract B and C by A , redefine B and C since a b c d = matrix M ( 2x2 ) then LARGE = - det M /2 === Subject: Re: question about prime partition For M is matrix (nxn) , then det M = det M' , where M' is tranpose M . For matrix t b c t e f t h i = matrix M where e is not equal zero then we get tranpose M = M' Since det M = t< M*( n,2 ) , -M( n,3 ; sub ) > where M* was defined , and t is real. Let take t = 0 then det M = 0 , for all another choice in variable. Then we get det M' , also equal zero. Then determinant in here is not possible to calculate LARGE or even Volume , as useful in matrix 2x2. But we can conclutes there are M*( n,2 ) perpendicular subtract in set M( n,3;sub) , for n = 1 , 2 , 3 === Subject: Re: question about prime partition For ab = 0 , its true for a or b equal zero. Where a = b = 0 we cant redefinition as below equation 0/a = b , because 0/0 was registered for undefined , unacces or even expression was error for math. Some founder use this for condition to terminated. Where is sqrt3_(- 1) ? Since (-1)^3 = -1 its true for the answer. Then { (1+(-3)i)/2 , (1-(-3)i)/2 } two complexs where are in expression x^3 = - 1 , then x1 , x2 , x3 involves togather. === Subject: Re: question about prime partition Definition for Added in set A = { k1 , k2 , k3 } added in set is A ; add = { k2 + k3 , k1 + k3 , k1 + k2 } USAGE Expression x^3 = - 1 , for x complexs then for P = { x1 , x2 , x3 } roots of equation we get P = - P; add === Subject: Re: the situation is worse than you thought >I was at a cafe the other day, I asked the > girl at the counter for a cup of hot water. > It's a 12 oz. paper cup. I tell her, Fill it > about 3/4. > She hesitates, then Fractions aren't > my strong point, can you show me how > much you'd like? I had to point to the > level to dispense. Rich, that's a good story but what was the poor girl supposed to do? Fill it more than half, and less than max. It's not as if he was paying for it. John === Subject: Re: the situation is worse than you thought reply-type=response > Paul is roaring and this guy comes along and shoves a sliver of bamboo > under his toenail in a most unfriendly way ... and you come along and want > to cut off his foot. That is not a pleasant story at all. I'm at the point now where top posting is preferable... -- I believe that all government is evil, and that trying to improve it is largely a waste of time. - H.L. Mencken === Subject: Re: the situation is worse than you thought posting-account=UM3jRwkAAADTHFmJ20qgwageu031CeWA FunWebProducts; SLCC1; .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; PeoplePal 3.0; eMusic DLM/4; CNPVersion2 - Congoo NetPass; FDM; RRHSO_BLD1),gzip(gfe),gzip(gfe) > I was at a cafe the other day, I asked the > girl at the counter for a cup of hot water. > It's a 12 oz. paper cup. I tell her, Fill it > about 3/4. She hesitates, then Fractions aren't > my strong point, can you show me how > much you'd like? I had to point to the > level to dispense. That's right - 3/4 to the top literally > boggled her mind! I fall into a > melancholia, I fear for posterity... homo > sapiens had a good run, but nothing > lasts forever... I'll sell my bonds. - Ronald Reagan Do you mean all Homo Sapiens or just Americans? === Subject: Re: the situation is worse than you thought posting-account=fwSgtAkAAACFnX70ssKwbvm9_oCZVHrx Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) I was at a cafe the other day, I asked the > girl at the counter for a cup of hot water. > It's a 12 oz. paper cup. I tell her, Fill it > about 3/4. She hesitates, then Fractions aren't > my strong point, can you show me how > much you'd like? I had to point to the > level to dispense. That's right - 3/4 to the top literally > boggled her mind! I fall into a > melancholia, I fear for posterity... homo > sapiens had a good run, but nothing > lasts forever... I'll sell my bonds. - Ronald Reagan Do you mean all Homo Sapiens or just Americans? Probably anyone who just doesn't get enough education for whatever reason or another. Seems pretty common here in the Evil American Empire but it could also be in other places as well. === Subject: Re: the situation is worse than you thought posting-account=OKTeIQkAAAAZk6JK1hK7-grwpoUDNy98 CLR 1.1.4322; .NET CLR 2.0.50727; InfoPath.1),gzip(gfe),gzip(gfe) I was at a cafe the other day, I asked the > girl at the counter for a cup of hot water. > It's a 12 oz. paper cup. I tell her, Fill it > about 3/4. She hesitates, then Fractions aren't > my strong point, can you show me how > much you'd like? I had to point to the > level to dispense. That's right - 3/4 to the top literally > boggled her mind! I fall into a > melancholia, I fear for posterity... homo > sapiens had a good run, but nothing > lasts forever... I'll sell my bonds. - Ronald Reagan Not surprised. Most retail sales clerks today can't make change > without the help of a computer. Try it. Next time you make a cash > purchase for something like $4.15 cents, give them $5.15, and watch > them try to figure out how to make change for the difference. Most > will call the manager over, who might not have a clue either... Won't work, they'll just enter $5.15 as the amount tended and let the cash register figger the change. You have to find a place where the cash register isn't working if you want that kind of fun. I was at a record store once where the cash register stopped working. They closed the store. Tom Davidson > Richmond, VA- Hide quoted text - - Show quoted text - === Subject: Re: the situation is worse than you thought Really? Try again. Begin by reading instead of snipping, then you > won't have to think, all the information will be there.> Snipping is good, What was it you said again? The snipper has to have enough common sense to _leave in_ the relevant comments to which he or she or it is replying. Otherwise, it's just trolling. Rich === Subject: Re: the situation is worse than you thought > Really? Try again. Begin by reading instead of snipping, then you > won't have to think, all the information will be there. > Snipping is good, What was it you said again? The snipper has to have enough common sense to _leave in_ the relevant >comments to which he or she or it is replying. Otherwise, it's just >trolling. Rich All I have to say is that I have gained a whole new appreciation for the job done by Paul Popinjay, RGP's Ass't Coordinator. The patience he has shown to people insisting on arguing irrelevancies is above and beyond the call of duty. And he doesn't even get paid for taking all that abuse from people who just like to argue. I wouldn't take that job for all the money in the world. === Subject: Re: the situation is worse than you thought > Really? Try again. Begin by reading instead of snipping, then you > won't have to think, all the information will be there.> Snipping is good, What was it you said again? >The snipper has to have enough common sense to _leave in_ the relevant >comments to which he or she or it is replying. Otherwise, it's just >trolling. >Rich All I have to say is that I have gained a whole new appreciation for > the job done by Paul Popinjay, RGP's Ass't Coordinator. The patience he has shown to people insisting on arguing irrelevancies > is above and beyond the call of duty. > And he doesn't even get paid for taking all that abuse from people who > just like to argue. I wouldn't take that job for all the money in the world. Ok, so you don't want to be paid either... What's poker about? === Subject: Re: the situation is worse than you thought > Really? Try again. Begin by reading instead of snipping, then you > won't have to think, all the information will be there. > Snipping is good, > What was it you said again? The snipper has to have enough common sense to _leave in_ the relevant > comments to which he or she or it is replying. Otherwise, it's just > trolling. Rich And what was the relevant comments to which he or she or it is replying? === Subject: Re: the situation is worse than you thought > I was at a cafe the other day, I asked the > girl at the counter for a cup of hot water. > It's a 12 oz. paper cup. I tell her, Fill it > about 3/4. She hesitates, then Fractions aren't > my strong point, can you show me how > much you'd like? I had to point to the > level to dispense. That's right - 3/4 to the top literally > boggled her mind! I fall into a > melancholia, I fear for posterity... homo > sapiens had a good run, but nothing > lasts forever... Not to worry. In few month, you will be able to see people with advanced degrees working as the waiters. VLV === Subject: Re: the situation is worse than you thought posting-account=o6eA3woAAAAV18f5Qf__iJBROGTMZWh8 AppleWebKit/525.18 (KHTML, like Gecko) Version/3.1.2 Safari/525.22,gzip(gfe),gzip(gfe) > That's right - 3/4 to the top literally > boggled her mind! I fall into a > melancholia, I fear for posterity... homo > sapiens had a good run, but nothing > lasts forever... Not to worry. In few month, you will be able to see people with advanced > degrees working as the waiters. VLV Who's to say the waitress in question doesn't hold one now. And isn't reading the petty self-indulgent sentimentality here with a grin? Like a lot of us are.... === Subject: Re: the situation is worse than you thought > Keyword: about. > It means approximately. The information It's a 12 oz. paper cup was > redundant. I thought he only mentioned 12 oz paper cup once, but even so. > Obviously he didn't expect the girl to measure it. But what if the cup was > actually more than 12 ounces? What if it held 14 ounces, even though it was > assumed to hold 12? Would you end up with the cup being under-filled? Why > not just order a medium and hope that it was close to the 7 ounce mark > that Rich was hoping to get? The point is, she couldn't even figure out how to _guesstimate_ 3/4 full. Geez, the kids on Cyberchase do this level of math routinely, and they're, like, seven! The problem, of course, has been the collusion of government and the unions, who have been dumbing-down the schools for at least two generations now; whether it's intentional or unintentonial remains to be determined. Rich === Subject: Re: the situation is worse than you thought Distribution: world > Keyword: about. > It means approximately. The information It's a 12 oz. paper cup was > redundant. I thought he only mentioned 12 oz paper cup once, but even so. He mentioned it once, and it's redundant to filling it 3/4 full. > But what if the cup was >actually more than 12 ounces? It doesn't matter, since he didn't ask her to put a particular amount in. He asked her to fill it 3/4 full. If he asked her to fill it halfway full, would it matter how big it was? No, you look at the side and eyeball a halfway point, kind of like folding a piece of paper in half. Similarly, to fill a cup 3/4 full, you look at the side of the cup, split the difference between top and bottom, and then split the difference between that point and the top. > What if it held 14 ounces, even though it was >assumed to hold 12? Would you end up with the cup being under-filled? Not if you filled it 3/4 full, as requested. > Why >not just order a medium and hope that it was close to the 7 ounce mark >that Rich was hoping to get? He didn't say anything abuout seven ounces; you pulled that out of some orifice. -- Michael F. Stemper #include Why doesn't anybody care about apathy? === Subject: Re: the situation is worse than you thought It doesn't matter, since he didn't ask her to put a particular amount > in. He asked her to fill it 3/4 full. > I didn't say that he said to put a certain amount in it. But if it was a 12 ounce cup and she was supposed to fill it 3/4th full, it would be under-filling it if unbeknownst to them it was actually 14 ounces. It would be short by an amount of 3/4 times the difference of 2 ounces when adjusting. If 3/4 of 12 is 7 then I don't know what 3/4 of 14 is but it would have to be different because two different numbers cannot be the same unless they are the same number. -Paul P. === Subject: Re: the situation is worse than you thought Distribution: world >I was at a cafe the other day, I asked the > girl at the counter for a cup of hot water. > It's a 12 oz. paper cup. I tell her, Fill it > about 3/4. > She hesitates, then Fractions aren't > my strong point, can you show me how > much you'd like? I had to point to the > level to dispense. Rich, that's a good story but what was the poor girl supposed to do? Make a >rough guess as to when it was 7 ounces of hot water? Seven ounces has nothing to do with it. About three-quarters was the request. In other words, about one-fourth of the way down from the top. >A lot of people can't just eyeball a certain amount of ounces in a paper >cup. You don't need to eyeball a certain number of ounces. You need to look at the side, go half-way down, and then half-way from there back up. > How do you know how wide the cup is? How wide the cup is doesn't matter! -- Michael F. Stemper #include Why doesn't anybody care about apathy? === Subject: Re: the situation is worse than you thought Seven ounces has nothing to do with it. About three-quarters was the > request. In other words, about one-fourth of the way down from the top. > In the case that the cup was really 12 ounces it would have something to do with it. If the cup was larger than 12 ounces then obviously seven ounces would not be 3/4 of the cup. It would be less because the cup is bigger. -Paul P. === Subject: Re: the situation is worse than you thought posting-account=G-TjQAkAAADYg6rno3bWQPnIwKFBrf1t MathPlayer 2.10b; .NET CLR 1.1.4322),gzip(gfe),gzip(gfe) On Oct 1, 2:09am, Pepe Papon six? He confirmed that this was correct. I ordered six You can't fault someone for a relative lack of familiarity with a largely outmoded term whose very use dates the person uttering it. It'd be like asking a used car dealer for how many pence or shilling can one procure a crank-free horseless carriage. === Subject: Re: the situation is worse than you thought >On Oct 1, 2:09am, Pepe Papon The guy at the register asked me what size McNuggets I would like, and I replied A half-dozen. > Just to be sure, I asked him, I can't get a half-dozen, but I can get > six? He confirmed that this was correct. > I ordered six You can't fault someone for a relative lack of familiarity with a >largely outmoded term whose very use dates the person uttering it. It'd be like asking a used car dealer for how many pence or shilling >can one procure a crank-free horseless carriage. Dozen is outmoded? You're comparing dozen to horseless carriage? Wow. === Subject: Re: the situation is worse than you thought posting-account=G-TjQAkAAADYg6rno3bWQPnIwKFBrf1t MathPlayer 2.10b; .NET CLR 1.1.4322),gzip(gfe),gzip(gfe) > I was at a cafe the other day, I asked the girl at the counter for a cup of hot water... > Fill it about 3/4. She hesitates, then Fractions aren't my strong point... I fall into a melancholia, Who's the more numerically illiterate? The one who only feigns to stumble over numbers by directive of a Pop Culture that, nonetheless betrays its true nature by being awash in numeracy at every turn, or the one who doesn't even have the presence of mind to seek out the 7 digits that count -- her number. Everybody knows what 3/4 means. It's as much a part of pop culture as everything else (e.g. the 3rd turn on the 400 meter run). But it's also equally part of Pop Culture to go out of your way to manifest the exact antithesis of nerdy bookishness. It's proof by case-in-point that the last thing you ever want to do to find out what a human says, believes or means is to directly put it to them. You can't trust a human to give you a straight answer or account on anything they hold to any further than you can throw one. Instead, to elicit the true mathematical ability requirest getting udner the consciously-guarding- against-nerdiness-appearance hood and catching them in the act. I.e., setting up a Busted! on them. And yet, here we are. Not a single one of the 7 digits (or 10 in crowded US metro areas; assuming this is in the US). === Subject: Re: the situation is worse than you thought to dispense. That's right - 3/4 to the top literally > boggled her mind! I fall into a > melancholia, I fear for posterity... homo > sapiens had a good run, but nothing > lasts forever... I'll sell my bonds. - Ronald Reagan I will weep for the Republic -- Bob Kolker === Subject: Re: the situation is worse than you thought On 2008-09-30 19:36:57 -0700, RichD said: > She hesitates, then Fractions aren't > my strong point, can you show me how > much you'd like? I had to point to the > level to dispense. You're always better off just letting 'em fill it up. Pour the rest off yourself. There's definitely a brain trust at most fast food places. I recently had the misfortune of going through the McDonald's drive through. I ordered a Big 'n Tasty, add cheese and bacon, no sauce of any kind (I hate catsup & mayo on burgers). What did I get? 1. Bacon Cheeseburger, with mayo and catsup. 2. Big and Tasty, no cheese or bacon, with mayo and catsup. -- thepixelfreak === Subject: Re: the situation is worse than you thought >I was at a cafe the other day, I asked the >girl at the counter for a cup of hot water. >It's a 12 oz. paper cup. I tell her, Fill it >about 3/4. She hesitates, then Fractions aren't >my strong point, can you show me how >much you'd like? I had to point to the >level to dispense. I actually came across someone who couldn't tell how many hours are between 9:00 and 17:00... insisted on 9 hours. -- Programmeren in Almere? E-mail naar nico@nctdevpuntnl (punt=.) === Subject: Re: the situation is worse than you thought >I was at a cafe the other day, I asked the >girl at the counter for a cup of hot water. >It's a 12 oz. paper cup. I tell her, Fill it >about 3/4. >She hesitates, then Fractions aren't >my strong point, can you show me how >much you'd like? I had to point to the >level to dispense. I actually came across someone who couldn't tell how many hours are > between 9:00 and 17:00... insisted on 9 hours. > 17:00 .. hmm... that's 5 o'clock Counting down, 1) 5 o'clock 2) 4 o'clock 3) 3 o'clock 4) 2 o'clock 5) 1 o'clock 6) 12 o'clock and three hours between 9:00 and midday makes 9 hours. :-) === Subject: Re: the situation is worse than you thought It was pretty clear to the rest of us. > The girl is a waitress and doesn't know what 3/4 of a cupful is, right? > An airhead who loses at poker needs a job like that. The insinuation was > her trait is hereditary. Of course you could be really intelligent and > are only bluffing that you are as stupid as she is. > Androcles, I think I have been quite patient with you so far. Need I remind you that I can suspend your posting permissions on rec.gambling.poker? But back to the original topic, the girl did NOT have trouble figuring out what 3/4 of the cupful was, as you imply. The orignal post assumes that the cup was only 12 ounces. Is the girl supposed to be able to judge where 7 ounces would stop at? That would be unreasonably demanding and unreasonably unreasonable. Please tone it down a little, Androcles, if you wish to continue posting to our rec.gambling.poker. I welcome your point of view, but just don't get snotty with me. -Paul Popinjay Asst. Coordinator, Rec.Gambling.Poker === Subject: Re: the situation is worse than you thought > I'm getting this ing crap blasted at me on sci.electronics.design > because you arogant stupid bastards insist on cross-posting your > introverted personal battles all over the place! OFF POPINJAY, YOU CUNT, AND LEAVE THE REST OF US IN PEACE! > Ok, this was really too much! I can't believe how nasty people are today. Did you all wake up on the wrong side of bed this morning? Look, Terry, I don't think you realize the spot you have put me in. First off, I did not start this cross-posting stuff. I merely responded to the original post, which apparently was cross-posted. If you have an issue, why not take it up with him? And further, people on rec.gambling.poker already know that I hold an important position there as the coordinator. If it were not for me, the newsgroup threads would be completely uncoordinated. I was tasked with this responsibility because of the respect I have earned on our newsgroup and people know that I am up for the responsibility. So you cannot just talk to me in that tone of voice. You cannot just call me the c-word and expect to continue with your posting as if nothing had happened. I am seriously considering reporting you. Don't be surprised if your keyboard suddenly starts typing all those little stars like ********. -Paul P. === Subject: Re: the situation is worse than you thought It was pretty clear to the rest of us. > The girl is a waitress and doesn't know what 3/4 of a cupful is, right? > An airhead who loses at poker needs a job like that. The insinuation was > her trait is hereditary. Of course you could be really intelligent and > are only bluffing that you are as stupid as she is. Androcles, I think I have been quite patient with you so far. Need I > remind you that I can suspend your posting permissions on > rec.gambling.poker? Please do, I've never heard of it before and have no interest in popinjays. > But back to the original topic, the girl did NOT have trouble figuring out > what 3/4 of the cupful was, as you imply. The orignal post assumes that > the cup was only 12 ounces. Is the girl supposed to be able to judge > where 7 ounces would stop at? That would be unreasonably demanding and > unreasonably unreasonable. Please tone it down a little, Androcles, if > you wish to continue posting to our rec.gambling.poker. I welcome your > point of view, but just don't get snotty with me. Hmm... I'm really worried now. How about... umm... off, you prat? === Subject: Re: the situation is worse than you thought <_yBEk.2144$YU2.1893@nlpi066.nbdc.sbc.com> posting-account=qHNgPQoAAAA40SLlrmjQxSJhnyKS76HY Gecko/2008070208 Firefox/3.0.1,gzip(gfe),gzip(gfe) > Please tone it down a little, Androcles, if > you wish to continue posting to our rec.gambling.poker. I welcome your > point of view, but just don't get snotty with me. Hmm... I'm really worried now. How about... umm... off, you prat? http://www.youtube.com/v/1M17RvxsZHc&autoplay=1&hl=en&fs=1 === Subject: Re: the situation is worse than you thought > Hmm... I'm really worried now. How about... umm... off > Ok, I don't like this ONE bit! I've been quite patient with you, Androcles. I don't think I want you reading or replying to any of my posts from now on. Especially if you are going to spew such venom. Like I said, I don't get paid to do this, and the last thing I want to do for free is take some from you. Please don't make me ban you from reading and/or replying to my posts, Androcles. Just take it upon yourself, out of courtesy, do not read or reply to me again. -Paul Popinjay Asst. Coordiantor, Rec.Gambling.Poker === Subject: Re: the situation is worse than you thought Hmm... I'm really worried now. How about... umm... off Ok, I don't like this ONE bit! I've been quite patient with you, > Androcles. > I don't think I want you reading or replying to any of my posts from now > on. > Especially if you are going to spew such venom. Like I said, I don't get > paid to do this, and the last thing I want to do for free is take some from you. Please don't make me ban you from reading and/or replying to my > posts, Androcles. Just take it upon yourself, out of courtesy, do not > read > or reply to me again. -Paul Popinjay > Asst. Coordiantor, Rec.Gambling.Poker > Ok, your boon is granted. *plonk* === Subject: Re: the situation is worse than you thought *plonk* > No, Androcles, you cannot killfile me because I have already banned YOU! I don't believe in killfiling people but if I ban you then you will not be able to read or reply to my posts. Don't try to change my mind, either. You have no one to blame but yourself. -Paul P. === Subject: Re: the situation is worse than you thought <_yBEk.2144$YU2.1893@nlpi066.nbdc.sbc.com> posting-account=qHNgPQoAAAA40SLlrmjQxSJhnyKS76HY Gecko/2008070208 Firefox/3.0.1,gzip(gfe),gzip(gfe) headwall.stanford.edu!newshub.sdsu.edu!flpi089.ffdc.sbc.com! prodigy.net!flpi088.ffdc.sbc.com!prodigy.com!flpi107.ffdc.sbc.com! flpi145.ffdc.sbc.com.POSTED!7b401650!not-for-mail === >Subject: Re: the situation is worse than you thought >X-RFC2646: Original > Hmm... I'm really worried now. How about... umm... off Ok, I don't like this ONE bit! I've been quite patient with you, Androcles. >I don't think I want you reading or replying to any of my posts from now on. >Especially if you are going to spew such venom. Like I said, I don't get >paid to do this, and the last thing I want to do for free is take some >from you. Please don't make me ban you from reading and/or replying to my >posts, Androcles. Just take it upon yourself, out of courtesy, do not read >or reply to me again. -Paul Popinjay >Asst. Coordiantor, Rec.Gambling.Poker > Who is the new guy? Is he seriously? o well.... ok.... here is the stories Paul: Usenet has online bulletin board system that began at Duke University in 1979. Usenet users post messages to teh newsgroups that can be read (and responded to) by anyone who has access to teh internets. rly! Newsgroups grown to the thousands; they're hosted all over the world specially to troll every conceivable topic. You to want to post your own comments to any existing Usenet newsgroups. Preferably about half a dozen at a time. Welcome you very much. -gaby de wilde === Subject: Re: the situation is worse than you thought Who is the new guy? Is he seriously? o well.... ok.... here is the stories Paul: Usenet has online bulletin board system that began at Duke University > in 1979. Usenet users post messages to teh newsgroups that can be read > (and responded to) by anyone who has access to teh internets. fact that people SHOULD be civil and polite. Sure, they don't HAVE to, but they SHOULD. Which brings us to Androcles. I never said anything to warrant being cussed at. So now, I just don't want him to read my posts any more. That's all. He doesn't realize how many people LOVE to read my posts. So it is his loss. He's out! He will NOT read or reply to my posts anymore, or I will report him. He made his bed, now he can lie in it. -Paul P. === Subject: Re: the situation is worse than you thought > AHAHAHAHAHAHA... AHAHAHA... Paul, if/since you > post from rec.gambling.poker, why don't you simply put > on your pokerface and ignore Andro.. or even simpler, > scratch rec.gambling.poker in the follow-up and you will > lose him. He only posts to sci.physics, AFAIK... & to boot > Paul, have some heart and compassion: Androcles, the > retired Truck Driver had a very bad decompression > mode when he ended and unwound from his career. > Within one year (2007/8), he lost his daughter to cancer, > he had a heart attack/aneurysm that got him a stent > implant. The poor sucker is in constant pain, must take > Hi-BP meds that aggravate him, a situation which he > inflames with his constant intake of booze. --- Then the > Brit's DMV revoked his driver's license... & last month > he crushed/damaged his feet and ankles... Read it in his > posts... Give him a break and play with him. You are all > that he's got left Or, Paul, are you worse off then Andro is? > ... ahahahaha... ahahahanson > I'm sorry. I didn't know any of that. He is not a regular on rec.gambling.poker and that is where I am posting from. I don't think I was rude to him but I'll try to be more careful. Most of the posters on rec.gambling.poker know me and they can tell you that I am one of the nicest and most considerate people on the newsgroup. I don't want to hurt Androcle's feelings so maybe it is better that I just ignore him. I'd rather he just did not read or reply to my posts. I guess I could report him if he keeps replying to me after I already told him not to. I have that right to demand that he not reply in a thread if that thread is started by me. -Paul P. === Subject: Re: the situation is worse than you thought PS: > AFA your reporting thing/intention... ahahahahaha.... > Do you really think that any ISP is going to drop > any of their own customers who pay them a monthly > fee, just because some angry schmuck complains > ...but who does NOT pay them anything?... ahahaha... > No Hanson, it is not the ISP I will report Androcles to. For one thing, I can report him to the newsgroup certification agency. By the way, you can tell that he is still reading my posts by his reply to your post, even though I already banned him from reading or replying to my posts about three hours ago. He's going to get himself in hot water and he'll have only himself to blame. Right now I don't even want to talk about it. I am so frustrated right now by the abuse that I have had to endure today from some of these people on other newsgroups who are cross-posting to rec.gambling.poker. I am not used to such hostility because rec.gambling.poker is usually MUCH more civil. -Paul P. === Subject: Re: the situation is worse than you thought <_yBEk.2144$YU2.1893@nlpi066.nbdc.sbc.com> <1MREk.2768$yr3.1045@nlpi068.nbdc.sbc.com> <4kTEk.122$P5.10@nwrddc02.gnilink.net> posting-account=qHNgPQoAAAA40SLlrmjQxSJhnyKS76HY Gecko/2008070208 Firefox/3.0.1,gzip(gfe),gzip(gfe) On Oct 2, 1:39am, Paul Popinjay > PS: > AFA your reporting thing/intention... ahahahahaha.... > Do you really think that any ISP is going to drop > any of their own customers who pay them a monthly > fee, just because some angry schmuck complains > ...but who does NOT pay them anything?... ahahaha... No Hanson, it is not the ISP I will report Androcles to. For one thing, I > can report him to the newsgroup certification agency. By the way, you can > tell that he is still reading my posts by his reply to your post, even > though I already banned him from reading or replying to my posts about three > hours ago. He's going to get himself in hot water and he'll have only > himself to blame. Right now I don't even want to talk about it. Did your illegal gambling outfit rob you from the rent monies again Paul? And now you are recruiting some half dozen depressed physicists here? Isn't it a tad to much like applied physics? (defeats the prurpose) Does that really work? === Subject: Re: the situation is worse than you thought AHAHAHAHAHAHA... AHAHAHA... Paul, if/since you > post from rec.gambling.poker, why don't you simply put > on your pokerface and ignore Andro.. or even simpler, > scratch rec.gambling.poker in the follow-up and you will > lose him. He only posts to sci.physics, AFAIK... & to boot > Paul, have some heart and compassion: Androcles, the > retired Truck Driver had a very bad decompression > mode when he ended and unwound from his career. > Within one year (2007/8), he lost his daughter to cancer, > he had a heart attack/aneurysm that got him a stent > implant. The poor sucker is in constant pain, must take > Hi-BP meds that aggravate him, a situation which he > inflames with his constant intake of booze. --- Then the > Brit's DMV revoked his driver's license... & last month > he crushed/damaged his feet and ankles... Read it in his > posts... Give him a break and play with him. You are all > that he's got left Or, Paul, are you worse off then Andro is? > ... ahahahaha... ahahahanson > I'm sorry. I didn't know any of that. He is not a regular on > rec.gambling.poker and that is where I am posting from. I don't think I > was rude to him but I'll try to be more careful. Most of the posters on > rec.gambling.poker know me and they can tell you that I am one of the > nicest and most considerate people on the newsgroup. I don't want to > hurt > Androcle's feelings so maybe it is better that I just ignore him. I'd > rather he just did not read or reply to my posts. I guess I could report > him if he keeps replying to me after I already told him not to. I have > that right to demand that he not reply in a thread if that thread is Hawhawhawhaw! let's have a pity party for Androcles! You bring the vodka, I'll supply the peach yoghurt! Better yet, make it a wake! Your compassion (emotional drivel) is underwhelming, cha-cha, === Subject: Re: the situation is worse than you thought Sweeping the dirt under the carpet is not the solution, Kyle. I don't often have opinion (preferring proof) but in this case I will state mine. Cross posting is limited to four groups by my own ISP. It should cease altogether, or at the very least limited to two newsgroups, otherwise there is little point in having categories at all. If it is desirable to post to more than one group the user should be compelled to post twice, two separate posts. It won't happen, of course. I don't approve of top posting either, but since you did, so shall I. We have standards here that are not adhered to! > Androcles, my dear, dear friend, there was this time, oh, 10/1/2008 10:20 > Keyword: about. > It means approximately. The information It's a 12 oz. paper cup was > redundant. > I thought > Really? Try again. Begin by reading instead of snipping, then you > won't have to think, all the information will be there. > threads on rec.gambling.poker and I try my best. I'm not even being > paid for this. > Egads... well, that explains it. I thought I had a royal flush but when I > looked > it was only two pair. > Our group started out many years ago as mostly discussions about poker. > But as you can see it has blossomed into a place where we have many > off-topic threads and especially some very interesting and insightful > political discussions. When I first started coordinating the threads it > sci.physics is a usenet newgroups and it hasn't been upgraded, although > you crossposted to it. You can upgrade it if you want to, it certainly > needs > it. > The group was much more manageable. Until that time the senior > coordinator had been doing much of the work by himself. Please check > your Profile, Androcles, check to see if the confirmation number is in > the box in the upper righthand corner, or if it is grayed out. > Profile? What profile? The only grey I've got is my beard. > I have to go to my real job soon and I will try to straighten some > things out here when I get home. Obviously we have a sudden surge in > the traffic on rec.gambling.poker. That is a good thing! I just want > to see that it gets done right. I take pride in how I have coordinated > this group even though I don't get paid for it and sometimes it seems > thankless. later. -Paul Popinjay > Asst. Coordinator, Rec.Gambling.Poker > God, this should be interesting. This is the worst crosspost ever... RGP > with sci.math and sci.logic? We have standards here! Pluseth, WORLDS ARE COLLIDING. To RichD: please to not be crossposting RGP with those cesspools, sir! > === Subject: Re: the situation is worse than you thought Androcles, my dear, dear friend, there was this time, oh, 10/1/2008 12:53 PM, or thereabouts, when you let the following craziness loose on Usenet: > Sweeping the dirt under the carpet is not the solution, Kyle. > I don't often have opinion (preferring proof) but in this case > I will state mine. Cross posting is limited to four groups by > my own ISP. It should cease altogether, or at the very least > limited to two newsgroups, otherwise there is little point in > having categories at all. If it is desirable to post to more than > one group the user should be compelled to post twice, two > separate posts. It won't happen, of course. I don't approve > of top posting either, but since you did, so shall I. We have > standards here that are not adhered to! > This would help with the Usenet/Spam problems, at a minimum. Of course, it'd be way too easy to write a script to post the same message repeatedly to a list... which would have the same result. When did I top-post? I do, occasionally, but I didn't in the post to which you directed this response. === Subject: Re: the situation is worse than you thought correspondent. > Androcles, my dear, dear friend, there was this time, oh, 10/1/2008 12:53 > PM, or thereabouts, when you let the following craziness loose on Usenet: > Sweeping the dirt under the carpet is not the solution, Kyle. > I don't often have opinion (preferring proof) but in this case > I will state mine. Cross posting is limited to four groups by > my own ISP. It should cease altogether, or at the very least > limited to two newsgroups, otherwise there is little point in > having categories at all. If it is desirable to post to more than > one group the user should be compelled to post twice, two > separate posts. It won't happen, of course. I don't approve > of top posting either, but since you did, so shall I. We have > standards here that are not adhered to! This would help with the Usenet/Spam problems, at a minimum. Of course, > it'd be way too easy to write a script to post the same message repeatedly > to a list... which would have the same result. When did I top-post? I do, occasionally, but I didn't in the post to which > you directed this response. > === Subject: Re: the situation is worse than you thought Androcles, my dear, dear friend, there was this time, oh, 10/1/2008 1:20 PM, or thereabouts, when you let the following craziness loose on Usenet: > correspondent. > Androcles, my dear, dear friend, there was this time, oh, 10/1/2008 12:53 > PM, or thereabouts, when you let the following craziness loose on Usenet: > Sweeping the dirt under the carpet is not the solution, Kyle. > I don't often have opinion (preferring proof) but in this case > I will state mine. Cross posting is limited to four groups by > my own ISP. It should cease altogether, or at the very least > limited to two newsgroups, otherwise there is little point in > having categories at all. If it is desirable to post to more than > one group the user should be compelled to post twice, two > separate posts. It won't happen, of course. I don't approve > of top posting either, but since you did, so shall I. We have > standards here that are not adhered to! > This would help with the Usenet/Spam problems, at a minimum. Of course, > it'd be way too easy to write a script to post the same message repeatedly > to a list... which would have the same result. > When did I top-post? I do, occasionally, but I didn't in the post to which > you directed this response. Oh, that's just a script. Normally the header is Androcles them!!!1!@! === Subject: Re: the situation is worse than you thought > Androcles, my dear, dear friend, there was this time, oh, 10/1/2008 1:20 > PM, or thereabouts, when you let the following craziness loose on Usenet: > correspondent. > Androcles, my dear, dear friend, there was this time, oh, 10/1/2008 > 12:53 PM, or thereabouts, when you let the following craziness loose on > Usenet: > Sweeping the dirt under the carpet is not the solution, Kyle. > I don't often have opinion (preferring proof) but in this case > I will state mine. Cross posting is limited to four groups by > my own ISP. It should cease altogether, or at the very least > limited to two newsgroups, otherwise there is little point in > having categories at all. If it is desirable to post to more than > one group the user should be compelled to post twice, two > separate posts. It won't happen, of course. I don't approve > of top posting either, but since you did, so shall I. We have > standards here that are not adhered to! > This would help with the Usenet/Spam problems, at a minimum. Of course, > it'd be way too easy to write a script to post the same message > repeatedly to a list... which would have the same result. When did I top-post? I do, occasionally, but I didn't in the post to > which you directed this response. > in message etc.; I just like people to know how much I appreciate > them!!!1!@! > Just a script? Oh... but that was MY point. :-) Of course, it'd be way too easy to write a script to post the same message repeatedly to a list... which would have the same result... and it has. === Subject: Re: the situation is worse than you thought Distribution: world >I was at a cafe the other day, I asked the >girl at the counter for a cup of hot water. >It's a 12 oz. paper cup. I tell her, Fill it >about 3/4. She hesitates, then Fractions aren't >my strong point, can you show me how >much you'd like? > Not everybody has math instincts. Maybe she has other talents, like > speaking three languages or something. Or maybe she's just dumb... > there's nothing inherently wrong with that. John, could YOU tell when it would have been 3/4 of the paper cup full just >by sight and without measuring it? Even though a paper cup in a cafe is >supposed to be the 12-oz size, a lot of them aren't really 12 ounces. What >if it had been 14 ounces if filled all the way to the top? What's it matter? 3/4 of the way to the top is still halfway up the side and then halfway from there to the top. > I agree that the girl probably just wasn't very bright. But >a lot of women are like that. I just wanted to set this statement off and look at it, all by itself. -- Michael F. Stemper #include 91.2% of all statistics are made up by the person quoting them. === Subject: Re: the situation is worse than you thought Distribution: world I was at a cafe the other day, I asked the >girl at the counter for a cup of hot water. >It's a 12 oz. paper cup. I tell her, Fill it >about 3/4. She hesitates, then Fractions aren't >my strong point, can you show me how >much you'd like? Not everybody has math instincts. Maybe she has other talents, like > speaking three languages or something. Or maybe she's just dumb... > there's nothing inherently wrong with that. John, could YOU tell when it would have been 3/4 of the paper cup full just >by sight and without measuring it? Even though a paper cup in a cafe is >supposed to be the 12-oz size, a lot of them aren't really 12 ounces. What >if it had been 14 ounces if filled all the way to the top? What's it matter? 3/4 of the way to the top is still halfway up the > side and then halfway from there to the top. I agree that the girl probably just wasn't very bright. But >a lot of women are like that. I just wanted to set this statement off and look at it, all by itself. Have you looked up popinjay in a dictionary? -- Michael Press === Subject: Re: the situation is worse than you thought What's it matter? 3/4 of the way to the top is still halfway up the > side and then halfway from there to the top. > This would be true, but you are making the same mistake in assuming that the cup was actually 12 ounces. If it was in fact 14 ounces, wouldn't you end up under-filling the cup if you did it as you just suggested? Think about it. I have plenty of patience and I like helping people. -Paul Popinjay Asst. Coordinator, Rec.Gambling.Poker === Subject: Re: the situation is worse than you thought Hmmmmmm. Can you confirm one of the following? A. You are a single parent raising two children, and 2/3 of the household > is male and 1/3 is female. Or B. You are married and you and your wife are raising two children, and > 50% of the household is male and 50% is female. > Don, I don't think either of these two completely different scenarios is any of anyone's business. I have already told you I have two children. My daughter Penelope Popinjay is 16, and my son Augusto Pinochet Popinjay is 12. They're both very bright, and I'll tellya the Popinjays is one good looking family! You should see Penelope in those tight little jeans she likes to wear! However, Don, I'd rather keep the discussion on topic with the original post. I am the Assistant Coordinator on rec.gambling.poker and it makes my job much easier if the threads are coordinated. Try to keep -Paul Popinjay Asst. Coordinator, Rec.Gambling.Poker === Subject: Re: the situation is worse than you thought Distribution: world > Maybe it would be easier if she filled three cups to the normal level, > and then divided the total amount into four equal portions using a > fourth empty cup, then discarded the left-over quarter and then > combined the contents of the remaining three into one of the cups. ;-) I'm not trying to be argumentative and I don't want to drag this thread out >unnecessarily. But originally he wanted the 12-oz cup to be about 3/4 full. >How could he expect a normal everyday waitress to judge and stop filling it >at 7 ounces? I could see if he had asked her for half full. Nobody said anything about seven ounces. Go check the original post. All that he said was about 3/4 full. So, she eyeballs halfway up the side and then about halfway from there to the top. Bingo, about 3/4. No number of ounces, seven or otherwise, was in the original post. Somebody who's qualified to be the waitress's co-worker made that up in a followup. -- Michael F. Stemper #include 91.2% of all statistics are made up by the person quoting them. === Subject: Re: the situation is worse than you thought Nobody said anything about seven ounces. Go check the original post. All > that he said was about 3/4 full. So, she eyeballs halfway up the side > and then about halfway from there to the top. Bingo, about 3/4. No number of ounces, seven or otherwise, was in the original post. > Somebody who's qualified to be the waitress's co-worker made that > up in a followup. > No, Michael, I think someone DID mention seven ounces. Maybe it is YOU who needs to go back and check the original post. But before you do that, Michael, let me refresh everyone's memory that I am the Assistant Coordinator on Rec.Gambling.Poker. And I am not used to being told to go back and check or re-read something. If I comment on something first and you do not understand it then I suggest that YOU need to do the re-checking. Back to the original post, in case you were too lazy to pay attention, it SAID that the paper cup was 12-ounces! Ok? Then he asked her to fill it about 3/4 full. NOW do you see where I got the seven ounces from? It was IMPLIED! And yes, I realize he said about. I'm not so anal that I can't give or take a dram here or there. So for all you nit pickers, let's just call it ABOUT 7 ounces! Ok? Sheesh! -Paul Popinjay Asst. Coordinator, Rec.Gambling.Poker === Subject: Re: the situation is worse than you thought > Nobody said anything about seven ounces. Go check the original post. All > that he said was about 3/4 full. So, she eyeballs halfway up the side > and then about halfway from there to the top. Bingo, about 3/4. > No number of ounces, seven or otherwise, was in the original post. > Somebody who's qualified to be the waitress's co-worker made that > up in a followup. > No, Michael, I think someone DID mention seven ounces. Maybe it is YOU who >needs to go back and check the original post. But before you do that, >Michael, let me refresh everyone's memory that I am the Assistant >Coordinator on Rec.Gambling.Poker. And I am not used to being told to go >back and check or re-read something. If I comment on something first and >you do not understand it then I suggest that YOU need to do the re-checking. >Back to the original post, in case you were too lazy to pay attention, it >SAID that the paper cup was 12-ounces! Ok? Then he asked her to fill it >about 3/4 full. NOW do you see where I got the seven ounces from? It was >IMPLIED! And yes, I realize he said about. I'm not so anal that I can't >give or take a dram here or there. So for all you nit pickers, let's just >call it ABOUT 7 ounces! Ok? Sheesh! -Paul Popinjay >Asst. Coordinator, Rec.Gambling.Poker Mr. Esteemed RPG Asst. Coordinator - I think you need to go back to elementary school, and take some basic fractions classes - 3/4 of 12 is (by all my calculators) 9, not 7. So where did the 7 ounces come from? -- Peter Bennett, VE7CEI peterbb4 (at) interchange.ubc.ca GPS and NMEA info: http://vancouver-webpages.com/peter Vancouver Power Squadron: http://vancouver.powersquadron.ca === Subject: Re: the situation is worse than you thought Distribution: world <_yBEk.2144$YU2.1893@nlpi066.nbdc.sbc.com> posting-account=OKTeIQkAAAAZk6JK1hK7-grwpoUDNy98 CLR 1.1.4322; .NET CLR 2.0.50727; InfoPath.1),gzip(gfe),gzip(gfe) On Oct 1, 12:45 pm, mstem...@walkabout.empros.com (Michael Stemper) > Maybe it would be easier if she filled three cups to the normal level, > and then divided the total amount into four equal portions using a > fourth empty cup, then discarded the left-over quarter and then > combined the contents of the remaining three into one of the cups. ;-) I'm not trying to be argumentative and I don't want to drag this thread out >unnecessarily. But originally he wanted the 12-oz cup to be about 3/4 full. >How could he expect a normal everyday waitress to judge and stop filling it >at 7 ounces? I could see if he had asked her for half full. Nobody said anything about seven ounces. Go check the original post. All > that he said was about 3/4 full. So, she eyeballs halfway up the side > and then about halfway from there to the top. Bingo, about 3/4. No number of ounces, seven or otherwise, was in the original post. What part of 12 oz. paper cup don't you understand? I was at a cafe the other day, I asked the girl at the counter for a cup of hot water. It's a 12 oz. paper cup. I tell her, Fill it about 3/4. up in a followup. -- > Michael F. Stemper > #include Maybe it would be easier if she filled three cups to the normal level, > and then divided the total amount into four equal portions using a > fourth empty cup, then discarded the left-over quarter and then > combined the contents of the remaining three into one of the cups. ;-) >I'm not trying to be argumentative and I don't want to drag this thread out >unnecessarily. But originally he wanted the 12-oz cup to be about 3/4 full. >How could he expect a normal everyday waitress to judge and stop filling it >at 7 ounces? I could see if he had asked her for half full. > Nobody said anything about seven ounces. Go check the original post. All > that he said was about 3/4 full. So, she eyeballs halfway up the side > and then about halfway from there to the top. Bingo, about 3/4. > No number of ounces, seven or otherwise, was in the original post. What part of 12 oz. paper cup don't you understand? girl at the counter for a cup of hot water. >It's a 12 oz. paper cup. I tell her, Fill it >about 3/4. > Which says nothing about seven ounces. As Androcles mentioned, the fact that the cup had a twelve ounce capacity was unneeded information. The request was to fill it 3/4 full. -- Michael F. Stemper #include You can lead a horse to water, but you can't make him talk like Mr. Ed by rubbing peanut butter on his gums. === Subject: Re: the situation is worse than you thought Distribution: world <_yBEk.2144$YU2.1893@nlpi066.nbdc.sbc.com> posting-account=OKTeIQkAAAAZk6JK1hK7-grwpoUDNy98 CLR 1.1.4322; .NET CLR 2.0.50727; InfoPath.1),gzip(gfe),gzip(gfe) On Oct 1, 5:16pm, mstem...@walkabout.empros.com (Michael Stemper) Maybe it would be easier if she filled three cups to the normal level, > and then divided the total amount into four equal portions using a > fourth empty cup, then discarded the left-over quarter and then > combined the contents of the remaining three into one of the cups. ;-) >I'm not trying to be argumentative and I don't want to drag this thread out >unnecessarily. But originally he wanted the 12-oz cup to be about 3/4 full. >How could he expect a normal everyday waitress to judge and stop filling it >at 7 ounces? I could see if he had asked her for half full. > Nobody said anything about seven ounces. Go check the original post. All > that he said was about 3/4 full. So, she eyeballs halfway up the side > and then about halfway from there to the top. Bingo, about 3/4. > No number of ounces, seven or otherwise, was in the original post. What part of 12 oz. paper cup don't you understand? girl at the counter for a cup of hot water. >It's a 12 oz. paper cup. I tell her, Fill it >about 3/4. > Which says nothing about seven ounces. As Androcles mentioned, the fact > that the cup had a twelve ounce capacity was unneeded information. The > request was to fill it 3/4 full. What part of No number of ounces, seven or otherwise, was in the original post. don't you understand? -- > Michael F. Stemper > #include by rubbing peanut butter on his gums. === Subject: Re: the situation is worse than you thought God, this should be interesting. This is the worst crosspost ever... RGP > with sci.math and sci.logic? We have standards here! Pluseth, WORLDS ARE COLLIDING. To RichD: please to not be crossposting RGP with those cesspools, sir! > What is THIS, Kyle? Are you suddenly the new coordinator here? What makes you think I don't have this situation already under control? Wtf is your problem? -Paul Popinjay Asst. Coordinator, Rec.Gambling.Poker === Subject: Re: the situation is worse than you thought Paul Popinjay, my dear, dear friend, there was this time, oh, 10/1/2008 12:32 PM, or thereabouts, when you let the following craziness loose on Usenet: God, this should be interesting. This is the worst crosspost ever... RGP > with sci.math and sci.logic? > We have standards here! > Pluseth, WORLDS ARE COLLIDING. > To RichD: please to not be crossposting RGP with those cesspools, sir! > What is THIS, Kyle? Are you suddenly the new coordinator here? What makes > you think I don't have this situation already under control? Wtf is your > problem? -Paul Popinjay > Asst. Coordinator, Rec.Gambling.Poker I'm doing meta-work here, Popinjay. This is very confusing for me... don't know which sock to use! === Subject: 3rd dimension posting-account=f7H7PgoAAADzb4HPxF4K6ZCvraN06mLv 98),gzip(gfe),gzip(gfe) In Theory x = y = z is the third dimension. Therefore 8 sectors of the xyz axis exist. For a cubical shape to exist in equation is; x = y = a (first sector) or 1 = 1 = 1 x = y = b(second sector) or 1 = 1 = 1 x = y = c(third sector) or 1 = 1 = 1 x = y = d(fourth sector) or 1 = 1 = 1 x = y = e(fifth sector) or 1 = 1 = 1 x = y = f(sixth sector) or 1 = 1 = 1 x = y = g(seventh sector) or 1 = 1 = 1 x = y = h(eighth sector) or 1 = 1 = 1 The eight sectors or octorants have each a number making a cubical shape!!! Jon Riley Toronto === Subject: Re: 3rd dimension posting-account=51499wkAAAAr0khaExEjlNiEzK2ubANQ (KHTML, like Gecko) Safari/419.3,gzip(gfe),gzip(gfe) > In Theory x = y = z is the third dimension. No. Try learning before making pronouncements. === Subject: Re: What if: the Church had NOT condemned Galileo posting-account=jPnQ2goAAAA461y3QD0lbyw0oKeThma1 AppleWebKit/525.18 (KHTML, like Gecko) Version/3.1.2 Safari/525.20.1,gzip(gfe),gzip(gfe) > There is a scientific inquisition, albeit a non-violent one. In 1597, Johannes Kepler had sent a copy of his new book, Mysterium Cosmographicum, to Galileo. This was the work in which Kepler proposed the Platonic solids as the basis for understanding the harmonic ordering of the planetary orbits around the Sun. Galileo thereupon sent a letter to Kepler, explaining that he, too, was a follower of the Copernican or heliocentric view, but that he ``had not dared'' to come forward with this view because of fear, and preferred to sit on the whole business because of the climate of opinion. Kepler had written back urging Galileo to be confident and to go forward with the struggle for truth, offering to find publishers in Germany if the Italian climate were too oppressive. Galileo did not do this, and refused to comment in detail on Kepler's book. According to Kepler's biographer Max Caspar, in the following years Galileo used material from Kepler in his lectures, but without giving Kepler credit. Kepler and Galileo were in frequent contact for over 30 years. Kepler commented with benevolent interest--and with subtle polemics--about Galileo's published works. But Galileo never commented systematically on Kepler's laws. In 1609, Kepler published his Astronomia Nova, expounding his first and second laws of planetary motion--that the planets move in ellipses of which the Sun is one focus, and that the planets sweep out equal areas in equal times between themselves and the Sun as they revolve. In Galileo's Dialogues on the Two Great World Systems, published in 1533, Kepler is hardly mentioned, while the discussion centers on Copernicus, with his perfect circle orbits of observed positions of the planets. At the end, one of the characters says that he is surprised at Kepler for being so ``puerile'' as to attribute the tides to the attraction of the Moon. During the first years of the pontificate of Pope Urban VIII Barberini, Galileo was the semi-official scientist for the pope. But in 1631, when the Swedish Protestant army of Gustavus Adolphus fought its way through Germany, reached the Alps, and seemed ready to sweep down on Rome, Urban VIII turned abruptly from a pro-French to a pro- Spanish policy. The Spanish ascendancy is the backdrop for the trial of Galileo carried out by the Dominicans with Jesuit support. Some years earlier, Sarpi had forecast that if Galileo went to Rome, the Jesuits and others were likely to ``turn ... the question of physics and astronomy into a theological question,'' so as to condemn Galileo as ``an excommunicated heretic'' and force him to ``recant all his views on this subject.'' Sarpi in 1616 seemed to know very well what would happen more than 15 years later, well after his own death. It is evident that the scenario sketched here corresponded to Sarpi's own long-term plan. For Galileo, the trial was one of the greatest public relations successes of all time. The gesture of repression against Galileo carried out by the Dominicans of Santa Maria Sopra Minerva in Rome established the equation Galileo=modern experimental science struggling against benighted obscurantism. That equation has stood ever since, and this tragic misunderstanding has had terrible consequences for human thought. Lost in the brouhaha about Galileo is the more relevant fact that Kepler had been condemned by the Inquisition more than a decade before. Sarpi's philosophical and scientific writings were not published until after World War II. These are the Pensieri, or Thoughts, and the Arte di Ben Pensare, the Art of Thinking Well. Sarpi's achievement for Venetian intelligence was to abstract the method of Aristotle from the mass of opinions expressed by Aristotle on this or that particular issue. In this way, sense certainty could be kept as the basis of scientific experiments, and Aristotle's embarrassingly outdated views on certain natural phenomena could be jettisoned. This allowed the Venetians to preserve the essential Aristotle, while attacking exponents of the Aristotelian or Peripatetic school, such as the Jesuits of the Collegio Romano. These writings by Sarpi have not been translated, but they are the basis of everything written by Sir Francis Bacon. The Bacon-Hobbes menage was in close contact with Sarpi and Micanzio. Sarpi can also be found in Locke, who took almost 1,000 pages to write what Sarpi had put down in 30. In the Art of Thinking Well, Sarpi starts from sense perception and sense certainty. He suggests that an impression made on our sensory apparatus by outside objects has to be distinguished from those objects. Especially he points to tastes, odors, and sounds, which he thinks are a matter of our nervous system, not of outside reality. In a different category are ideas of quantity, size, and time, which are objective. In the same manuscript, Sarpi lists the immortality of the soul as one on a list of wrong ideas. Sarpi repeats the argument of Pomponazzi that since there is no knowledge without sensation, the soul dies with the body. Again, the trademark of the Venetian dead souls faction. Galileo's epistemology comes straight from Sarpi. We can see this in Galileo's 1623 essay Il Saggiatore, The Assayer. For Galileo, colors, tastes, sounds, smells, are mere words. They exist only for our bodies. Galileo makes the famous comparison of these to tickling. If you brush a feather over the soles of the feet or the armpits of a marble statue, you will not produce a tickle. But if you do this to a human being, you will cause that tickling sensation. So, Galileo says, it is time to get rid of ears, tongues, and noses, and go for shapes, he proceeds quickly to a reductionist theory of atoms, in which heat is explained as the effect a ``fiery minims'' of igneous atoms. Galileo's epistemology is identical with that of Sarpi. This is what Galileo means when he denies Aristotle to say that the truth is written in the book of nature, and written in mathematical characters. Galileo was a reductionist. Sarpi died in 1623, and Galileo's case officer became the Servite monk Fulgenzio Micanzio. After Galileo had been condemned, Micanzio reminded Galileo of the assignment he had received from Sarpi 20 years earlier: to write a treatise on motion. And by the way, added Micanzio, I have 258 pounds here for you. Later, Micanzio would procure Galileo a pension of 60 scudi per year from the coffers of the Venetian state. Galileo responded to Micanzio's orders with the 1638 Discourses on Two New Sciences, Mechanics and Local Motion. Because Galileo had been condemned by the Inquisition, he could not be published anywhere that papal authority was strong. Micanzio therefore arranged for Galileo's book to be printed by the Dutch Elsevir press in Leyden. expert in science and philosophy--called a virtuoso in the parlance of the day--who had commented that although he did not deny Galileo's scientific ability, ``the things that you bring are not new, but are answer to this virtuoso is that although Galileo and Kepler may sometimes seem to agree about certain astronomical phenomena, ``my way of philosophizing is so different from his.'' (Nov. 19, 1634). In letters written in 1640, Galileo threw further light on his own scientific method. Galileo complained that he had been misunderstood: ``Against all the reason in the world, I am accused of impugning Peripatetic doctrine, whereas I profess and am certain of observing more religiously the Peripatetic--or, to put it better, Aristotelian-- teachings than many others....'' (Aug. 24, 1640). Galileo asserted that he had tried to study phenomena: ``that in all natural effects assure me of their existence, their an sit [if it be], whereas I gain nothing from their how, their quomodo.'' (June 23, 1640). Some might try to dismiss these admissions as a distortion of Galileo's outlook caused by the crackdown of which he was still a victim, but I would submit that this is the real Galileo talking. What Galileo is trying to express here is the same thing Isaac Newton meant with his infamous ``hypotheses non fingo,'' ``I do not fabricate hypotheses.'' Which brings us to Newton. Newton: A Cultist Kook http://members.tripod.com/~american_almanac/venscien.htm === Subject: Re: JSH: One of the great relations? posting-account=fzmnpgoAAAAmyKTlxv-T7BPlBuylv6TZ Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) thank you for your honest kindness === Subject: Re: JSH: One of the great relations? posting-account=fzmnpgoAAAAmyKTlxv-T7BPlBuylv6TZ Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) just don't do drugs. and if you haven't then you are probably manic. === Subject: Re: JSH: One of the great relations? posting-account=n1ZfDgkAAABbCs44qOtz8dP-RkWuEBif Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) My point is that solutions in general to x^2 + Dy^2 = F, can be > explained now with the relation x^2 + Dy^2 = F, requiring that (x-Dy)^2 + D(x+y)^2 = F*(D+1), as look > again at (x-Dy + D(x+y))^2 + D(x-Dy+x+y)^2 = F*(D+1)^2 The first + in the first term should be -. as simplifying a bit that's ((D+1)x)^2 + D(2x+(1-D)y)^2 = F*(D+1)^2 Hey, did I get the math right? That should be two iterations starting > from No. First term should be (x(1-D)-2Dy)^2. Ok, that makes more sense. > You should consider using a computer to check this kind of thing: sage: F=x^2+D*y^2 > sage: expand((x*(1-D)-2*D*y)^2 + D*(2*x+y*(1-D))^2-F*(D+1)^2) > 0 It was early in the morning. I still hadn't had my first cup of coffee! Regardless, this result is so incredible I'm becoming reticent about discussing it all on newsgroups. Is there more to getting sage to do that expansion. Like where did you tell it that (x-Dy)^2 + D(x+y)^2 = F*(D+1)? Can you show the first 10? James === Subject: Re: JSH: One of the great relations? posting-account=n1ZfDgkAAABbCs44qOtz8dP-RkWuEBif Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) > Regardless, this result is so incredible I'm becoming reticent about > discussing it all on newsgroups. And this is why you never succeed in convincing anyone of the worth of > your results. When people don't understand your results, or attack > them, you need to defend them. (I mean defend them mathematically, not > defend them by accusing mathematicians of conspiring against you). But as usual I changed my mind. Amazingly to me there is finally also a case where I can cite someone else's research as being along the same line: http://arxiv.org/abs/0806.2490v1 For what it's worth that paper is moving up in Google search results when you do a search on Pell's equation. Gist of it all for those wondering is: you don't need irrational numbers or non-rational numbers AT ALL to understand Pell's equation. My current research takes you a step forward into the full explanation of all the behavior associated with solutions to x^2 + Dy^2 = F. So there could be a historic shift in an area that has 2000 years of mathematical history with the dropping entire of non-rationals as just a historical legacy, when it comes to understanding Pell's equation. === Subject: Re: Infinite Binary Strings: A Question > Sure it does. Maybe I just need a little time to come round to this > way of thinking. I've been looking up probability, randomness and such, to > see if I can find a better way to express what still niggles. I can't say > I've found anything. Except there seems to be something very odd about a > notion of probability in whch, where infinite sampling spaces are > concerned, something can have zero probability and not be impossible (and > conversely probability 1 and not be certain). But maybe understanding that > is just understanding the subject. > Leon While there is indeed something very odd about it, there is nothing >self-contradictory given the usual mathematical definitions. There is, after all, nothing in those definitions that requires >mathematical probability to conform with our notions of reality, at >least for infinite sample spaces. It's comforting to know that we needn't require our mathematical definitions to conform to reality. Joking apart, I accept that the relevance of the observation about probablility to the foregoing discussion has yet to be made out, indeed, may not exist. leon === Subject: Re: Infinite Binary Strings: A Question ... > On the contrary, the implementing of a bijection between finite > collections is a form of counting. >Where is the contradiction? >Huh? Don't understand who is claiming that what is a contradiction, >but never mind... Again, not my words/ >and to which you responded with on the contrary. I've lost track of who's attributing what to who, here. Is it > important? > Well Leon, many of us have lost track of your lengthy and convoluted >posts :-) Sincere apologies. >Can I ask you a question? Do you believe in uncountable sets? Yes > Do you >believe the set of infinite binary strings -- that is, the set of >functions from N to {0,1} -- is uncountable? Yes > (That's two separate >questions.) I'm trying to understand where you're coming from. Why? (Is it yes in those two questions.) What does it mean? I guess that's where I'm coming from. (Plus a double-dose of: What the hell am I doing here?) leon === Subject: Re: Infinite Binary Strings: A Question > ... > On the contrary, the implementing of a bijection between > finite > collections is a form of counting. >Where is the contradiction? >Huh? Don't understand who is claiming that what is a contradiction, >but never mind... Again, not my words/ >write, >and to which you responded with on the contrary. I've lost track of who's attributing what to who, here. Is it > important? > Well Leon, many of us have lost track of your lengthy and convoluted >posts :-) Sincere apologies. Can I ask you a question? Do you believe in uncountable sets? Yes Do you >believe the set of infinite binary strings -- that is, the set of >functions from N to {0,1} -- is uncountable? Yes (That's two separate >questions.) I'm trying to understand where you're coming from. Why? (Is it yes in those two questions.) What does it mean? I > guess that's where I'm coming from. (Plus a double-dose of: What the hell > am I doing here?) > leon Does that here refer to being in sci.math or to being alive on earth? The second is considerably off topic. === Subject: Re: Infinite Binary Strings: A Question > On the contrary, the implementing of a bijection between > finite > collections is a form of counting. >Where is the contradiction? Huh? Don't understand who is claiming that what is a contradiction, >but never mind... Again, not my words/ write, >and to which you responded with on the contrary. I've lost track of who's attributing what to who, here. Is it > important? Well Leon, many of us have lost track of your lengthy and convoluted >posts :-) Sincere apologies. Can I ask you a question? Do you believe in uncountable sets? Yes Do you >believe the set of infinite binary strings -- that is, the set of >functions from N to {0,1} -- is uncountable? Yes (That's two separate >questions.) I'm trying to understand where you're coming from. Why? (Is it yes in those two questions.) What does it mean? I > guess that's where I'm coming from. (Plus a double-dose of: What the hell > am I doing here?) > leon Does that here refer to being in sci.math or to being alive on earth? >The second is considerably off topic. Now Virgil, don't pretend you don't know the answer to this. Lead me on to Beatrice, And in the contemplation of her countenance, the vision of mathematical divinity. leon === Subject: Re: Infinite Binary Strings: A Question On the contrary, the implementing of a bijection between > finite > collections is a form of counting. >Where is the contradiction? Huh? Don't understand who is claiming that what is a contradiction, >but never mind... Again, not my words/ write, >and to which you responded with on the contrary. I've lost track of who's attributing what to who, here. Is it > important? Well Leon, many of us have lost track of your lengthy and convoluted >posts :-) Sincere apologies. Can I ask you a question? Do you believe in uncountable sets? Yes Do you >believe the set of infinite binary strings -- that is, the set of >functions from N to {0,1} -- is uncountable? Yes (That's two separate >questions.) I'm trying to understand where you're coming from. Why? (Is it yes in those two questions.) What does it mean? I > guess that's where I'm coming from. (Plus a double-dose of: What the hell > am I doing here?) > leon Does that here refer to being in sci.math or to being alive on earth? >The second is considerably off topic. Now Virgil, don't pretend you don't know the answer to this. Lead > me on to Beatrice, And in the contemplation of her countenance, the vision > of mathematical divinity. leon These threads have been more like the earlier conversations between Beatrice and Benedict. And I very much misdoubt that it will resolve itself is as amicably as they did. === Subject: Re: Infinite Binary Strings: A Question Ok. If what you want is to gain a better understanding of infinite >strings, I have an example for you to contemplate. >An infinite sequence of zeros and ones is just a function from N, the >set of natural numbers, to the 2-element set {0,1}. Do you understand >what I mean by that? For each N, you have either a 0 or a 1. So you have >a 0 or a 1 in place 1, in place 2, in place 3, ... >Now, suppose we have an infinite string -- that is, a function f from N >to {0,1}. By picking out those n for which f(n) = 1, we define a >particular subset of N. >For example, the string 10101010101010... with 1's and 0's alternating, >defines the set {1, , 5, 7, ...} because we have a 1 in positions 1, 3, >5, ... >With me so far? Absolutely. In fact I think I mentioned somewhere along the line > the equivalence between selecting from an (denumerably) infinite set and > choosing an infinite binary string. Leon, you couldn't be more wrong on that point. There are uncountably >many infinite binary strings. If you don't agree with that, there's no >point in the conversation. Possibly a misunderstanding here. I only meant to agree with you that an infinite binary string is the same thing as an (infinite) selection from an infinite set. leon === Subject: Re: Infinite Binary Strings: A Question >Ok. If what you want is to gain a better understanding of infinite >strings, I have an example for you to contemplate. >An infinite sequence of zeros and ones is just a function from N, the >set of natural numbers, to the 2-element set {0,1}. Do you understand >what I mean by that? For each N, you have either a 0 or a 1. So you have >a 0 or a 1 in place 1, in place 2, in place 3, ... >Now, suppose we have an infinite string -- that is, a function f from N >to {0,1}. By picking out those n for which f(n) = 1, we define a >particular subset of N. >For example, the string 10101010101010... with 1's and 0's alternating, >defines the set {1, , 5, 7, ...} because we have a 1 in positions 1, 3, >5, ... >With me so far? Absolutely. In fact I think I mentioned somewhere along the line > the equivalence between selecting from an (denumerably) infinite set and > choosing an infinite binary string. Leon, you couldn't be more wrong on that point. There are uncountably >many infinite binary strings. If you don't agree with that, there's no >point in the conversation. Possibly a misunderstanding here. I only meant to agree with you > that an infinite binary string is the same thing as an (infinite) selection > from an infinite set. One can select objects by some property they have that other members of that same set do not have, and an infinite selection by quality is not necessarily sequential. For example, select the set of irrationals from the set of reals. If there really are uncountably many of them then this sort of selection cannot possibly be sequential. === Subject: Re: Infinite Binary Strings: A Question <0ec8e4thqvoghia5v3aonk9i6d0lq1ml5n@4ax.com> 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) >Ok. If what you want is to gain a better understanding of infinite >strings, I have an example for you to contemplate. >An infinite sequence of zeros and ones is just a function from N, the >set of natural numbers, to the 2-element set {0,1}. Do you understand >what I mean by that? For each N, you have either a 0 or a 1. So you have >a 0 or a 1 in place 1, in place 2, in place 3, ... >Now, suppose we have an infinite string -- that is, a function f from N >to {0,1}. By picking out those n for which f(n) = 1, we define a >particular subset of N. >For example, the string 10101010101010... with 1's and 0's alternating, >defines the set {1, , 5, 7, ...} because we have a 1 in positions 1, 3, >5, ... >With me so far? > Absolutely. In fact I think I mentioned somewhere along the line > the equivalence between selecting from an (denumerably) infinite set and > choosing an infinite binary string. Leon, you couldn't be more wrong on that point. There are uncountably >many infinite binary strings. If you don't agree with that, there's no >point in the conversation. Possibly a misunderstanding here. I only meant to agree with you > that an infinite binary string is the same thing as an (infinite) selection > from an infinite set. One can select objects by some property they have that other members of > that same set do not have, and an infinite selection by quality is not > necessarily sequential. Still, *you* deny that an infinite string or list has an end at all, and your selection by quality is just another trick: it holds when you like, it does not hold when you don't. > For example, select the set of irrationals from the set of reals. > If there really are uncountably many of them then this sort of selection > cannot possibly be sequential. What fun your soups of not necessarily this, one can do that, you can select, but not sequentially!. How interesting. === Subject: Re: Infinite Binary Strings: A Question Let's skip a bit, or we'll be here all day... (And again...) Anyway, according to my defintion of equality in > number, two finite collections have the same number if they have the same > count. There is no reference to bijection. >No, sure there is no reference to bijection. But you cannot give an >example of two finite collections which have the same count (I'm >assuming you're not using this phrase with any surprising meaning) for >which there is not a bijection between the two sets. And neither can >you give an example of two finite collections which do not have the >same count for which there _is_ a bijection between them. So your >same count definition and the bijection definition define the same >relation. > But I am not defining the natural numbers in terms of a relation. > And the absence of a reference to bijection is crucial. Well, I'm sorry, but I don't think the absence of you saying something >can cause it not to exist. You talk about counting -- that's fine. But >you can't stop someone else coming along and pointing out that what >you have said can also be expressed (probably more tersely) using >standard terminology. That is, of course, insofar as we can understand >what you mean. But _if_ you say something with a mathematical meaning, >it can be transformed into all sorts of different expressions (some of >which, like the category theory version, may well be beyond both you >and me!) -- and the mathematical meaning will be the same. > I'll come back to this if I may. It occurred to me, whilst thinking about this off and on today, I mean Wednesday, (Usenet seems such a rush! I'm probably just a slow thinker.), that what I was trying to do with the folding exercise really depended on the understanding of the counting numbers I'm trying to present, and was not an independent intuition. In other words, perhaps I can try and clarify the whole discussion by clarifying a part of it, the folding business, whilst leaving an aspect of it unclarified. The further-to-be clarified business is the nature of natural numbers. My claim (further to be defended) is that the natural numbers require a definition in terms of process, where process here is irreducible to and inequivalent to standard forms of definition of number. In other words, whilst we can by all means have the ordinary notion of a (unordered) set as primitive, the idea of an /ordered/ set is to be understood as equally primitive, and not reducible to the former. But as I say, later. If that does not tax your admittedly capacious patience. I am defining the > natural numbers in terms of a process, a human process (unpalatable as this > may be), a particular kind of sorting, where all that is required is the > grasp of an individual thing, the understanding that a thing is a thing of > a particular kind, that here we have such a thing, and now we have another. > Does not the understanding of a bijection presuppose that? We cannot even > talk without presupposing such basic notions. The concept of a counting > number 'falls out' of such a process, without appealing to bijection. Of > course there will be a bijection between two collections which have the > same count, because testing for a bijection is already a (specialized) form > of counting. I don't know if this conception is novel or not, but I don't > see what's difficult to understand. Sorry, I'm slightly lost: which is the conception whose novelty I'm >supposed to check? I don't immediately see anything new in what you >have said. Anyway, now we seem to move on to constructing the reals, or >similar... >Hmm, this is actually the order in which the numbers are born (his >terminology) in Conways surreals (see p. 11 of ONAG). But he does give >a construction to account for this, rather than just assuming the >completion of inking in, which I can't claim to be able to make much >sense of. > Another poster made reference to something like this. I'll look it > up. Be warned that On Numbers and Games (ONAG), the book in which Conway >introduced the surreals (later thus named by [Donald Knuth?? I >think]), is pretty hard. He absolutely assumes you have a grasp of >elementary set theory, and can understand (for example) the set {0, 1, >2, ...}, or {0, 1, 2, ..., w} and why the latter is not the same as >(the nonexistent) {0, 1, 2, ...w}. >You ask: > How do we get /all/ numbers of the form m/2^n? >Then say: > Only it the width is actually zero, do we get, on unfolding, the > real numbers. >If this was the binary folding, then on unfolding you get >(surprise!): /all/ numbers of the form m/2^n. >Not the reals. > I'm sorry, but I don't follow how you're not following me here. > Folding the wire to any finite width, however small, will leave some > rational numbers out. After folding the wire to zero width, it is kinked > everywhere, and not merely every rational number of the form m/2^n, but > every rational number and every irrational number ( all infinite binary > strings, periodic and aperiodic alike) is positioned. Perhaps I will > understand what you mean when I follow up the references above. The most fundamental problem is that the metaphor of folding a wire >simply does not allow for all the mathematical distinctions we need to >make. More concretely, I have to ask, what does folding the wire to zero >width mean? Suppose we count the foldings with natural numbers, so >after 5 folds we have reached 1/32 (we're looking at a unit length >piece of wire, right?). After some vast number, we have reached a >minuscule fraction. But there is no last natural number called (for >example!) infinity, so we never reach a width of zero. We fold the >wire indefinitely finely, and that's it. (FWIW, the surreals form a >field, so there are no zero-divisors, so although 1/w is a perfectly >fine surreal, it is not equal to zero. So if you go hoping that the >surreals let you get to infinity, where numbers have zero >reciprocals, you will be disappointed. Incidentally, this is actually >in Chapter 1 of ONAG, the second, since the first is Chapter 0.) You see, you are using the wire as a metaphor, and I'm trying to point >out that it simply isn't a fine enough tool. Surely, if a wire has >been bent enough that there are no unbent pieces of more than zero >length, you would say it is completely bent? Your binary folding >described above (if carried out as a supertask or whatever) puts a >kink at every binary fraction position (i.e. all m/2^2 in the >appropriate range). So there are no nonzero length unbent bits. But >you have not put a kink at 1/3, nor 1/root(2), nor 1/pi. In some >sense you have filled the interval, yet, there are indefinitely many >more fillings (ok... inkings in if you like!) that can be done, >each adding (infinitely many) more points, each transforming the >interval from one with no nonzero gaps to a denser one with no >nonzero gaps. I simply do not think that mental pictures of wires are >any use at all for trying to grok this. At the end of your last paragraph you said: Perhaps I will understand >what you mean when I follow up the references above. Respectfully, I >don't think you will if you try reading ONAG, because it's too hard. >(But don't be put off looking) I hope this is some sort of answer to your previous post. I understand, I think, that there are an infinite (un-ending) sequence of half-foldings of the wire, thus giving, in some sense, all of the numbers of the form m/2^n in the unit interval. And since we can also have 1/3 foldings, and 1/5 foldings, and so on, and since it should be possbile to combine these foldings (or we could just drop the wire metaphor and talk about all the rationals in the unit interval), then I understand that in this way we can talk about all the rationals, without having resorted to zero-width folding, which exists as a kind of external bound on such foldings. The point about foldings is that the quarter foldings include the half foldings, the eighth foldings include the quarter foldings, and so on, so that each 'degree' of folding includes all former degrees of folding. The foldings themselves, apart from the numbers created by them, form a sequence. But there is no folding which includes all previous foldings. Think of it this way: given a cake and an infinite number of people, there isn't actually any way in which I could give each of them a piece. There is no method of sharing out the cake short of a division of the cake into infinitely small pieces, equivalent to a zero-width folding of the wire (real or surreal, you decide.) But I think I am relying on this idea of a process, here. (Which is perhaps the point about supertasks.) But I accept that reliance. As counting is a process, so is listing. There is no list of all the rationals because there is no end to the process of listing them, just as there is no end to the process of counting. (I suspect there is a curious skew effect here. Can we quantify over the rationals? Of course we can. In the (meta?) context of understanding number, can we refer to all the rationals? No. Pragmatism, even when raised to level of philosophical creed, is not the answer.) >Here's another book, which I do recommend (very strongly) as a good >introduction to how to jump out of school algebra (which is, in the >end, the problem here). your analysis of the problem is correct. >W W Sawyer, A concrete approach to abstract algebra (written 50 >years ago, and you can probably find a cheap copy at http://abe.com or >similar) In the end, these different sorts of inking in are algebraic: the >first at least slightly unexpected jump being from rationals to >irrationals (goes back to Pythagoras), but if you work through the 220 >(small) pages of this book, in principle you will understand why it is >not possible to trisect an angle by ruler and compass construction. >After this somewhat out-of-fashion bit of crank-busting you should >find it really easy to sort out the errors in crank threads on >Cantor (or at least those that are coherent enough to be wrong). Brian Chandler I've probably muddied the waters. That was useful for me, anyhow. If I can substantiate what I'm saying about the natural numbers, maybe the conversation won't have been completely one-sided. leon === Subject: Re: Infinite Binary Strings: A Question posting-account=U44YcwkAAAAbGXB70Qr7gA3kornmKE4i Gecko/20080325 Ubuntu/7.10 (gutsy) Firefox/2.0.0.13,gzip(gfe),gzip(gfe) > Let's skip a bit, or we'll be here all day... [interlude] > But I am not defining the natural numbers in terms of a relation. > And the absence of a reference to bijection is crucial. Well, I'm sorry, but I don't think the absence of you saying something >can cause it not to exist. [/interlude] > My claim (further to be defended) is that the natural > numbers require a definition in terms of process, where process here is > irreducible to and inequivalent to standard forms of definition of number. Require a particular form of definition? I think that is fundamentally opposed to the spirit of mathematics. You can define something called the [PREFIX]-naturals, where you get to choose PREFIX, using any method you like - using set theory or not, by appeal to something in geometry, whatever. People do, in fact define the naturals different ways all the time, but usually the object, after completing the definition, is to show that these are in fact isomorphic to the usual naturals, after which you drop PREFIX. So in Conway's book, he defines a number as a structure {L | R} where L and R are sets of numbers, and for no members l and r of L and R respectively is l >= r. OK, here I do suggest if you can find a copy you read the few relevant pages of ONAG, to see if you can see how he gets the Conway-naturals, and then shows them to be isomorphic to the usual naturals. Maths is only interested in things up to isomorphism, which is why mathematical content is culture-free: ichi-ni-san, un-deux-trois, whatever, refer to _exactly_ the same thing. So you are free to define your Leon-naturals, but if they are to be much use you will need to show them isomorphic to the usual ones, and in the end you will have contributed a new definition but not a new definiens (I like pseudo-latin, but I wonder is that's right?) Incidentally, you might ask - if Conway just defines the same naturals, what's the point? But of course a different definition technique enables different extensions, which is where it gets interesting. In fact the definition of a _game_ (Part 1, the second in the book) turns out to be the same, { L | R } without the condition that no l >= r. understand, I think, that there are an infinite (un-ending) sequence of > half-foldings of the wire, thus giving, in some sense, all of the numbers > of the form m/2^n in the unit interval. And since we can also have 1/3 > foldings, and 1/5 foldings, and so on, and since it should be possbile to > combine these foldings (or we could just drop the wire metaphor and talk > about all the rationals in the unit interval), then I understand that in > this way we can talk about all the rationals, without having resorted to > zero-width folding, which exists as a kind of external bound on such > foldings. Right. > The point about foldings is that the quarter foldings include the > half foldings, the eighth foldings include the quarter foldings, and so on, > so that each 'degree' of folding includes all former degrees of folding. > The foldings themselves, apart from the numbers created by them, form a > sequence. But there is no folding which includes all previous foldings. If you write the sequences as follows: 1/2, 1/4, 3/4, 1/8, 3/8, 5/8, 7/8, 1/16, 3/16, 5/16, 7/16, ... 1/3, 2/3, 1/9, 2/9, 4/9, 5/9, 7/9, 8/9, 1/27, 2/27, ... 1/5, 2/5, ... 1/7, ... ... then indeed once you go below any set of dots there is no folding that includes all the preceding ones. But note that you could always interleave the sequences (in any number of different ways) so that there are only dots at the bottom, in which case for any folding A there is another folding B that includes all the values obtained up to folding A. > Think of it this way: given a cake and an infinite number of people, there > isn't actually any way in which I could give each of them a piece. Given an infinite number of people there isn't any way you could actually afford to hire a room to hold them all. This is the problem with physical metaphors. But given an unending sequence of requests, there is a way you could assign a fraction of a cake to each request, in such a way that no request would ever go unfulfilled. > no method of sharing out the cake short of a division of the cake into > infinitely small pieces, equivalent to a zero-width folding of the wire > (real or surreal, you decide.) There's no clear meaning to infinitely small pieces... and please note that the surreal name (not in Conway's book, of course) is a technical one. It is meant to indicate that Conway's number include a large collection than the reals; it doesn't just mean a Magritte-like something funny here. In particular, the surreals form a Field (a field that is a proper class), and there is no reciprocal of zero, so no number equivalent to zero-width folding. > But I think I am relying on this idea of a process, here. (Which is > perhaps the point about supertasks.) But I accept that reliance. As > counting is a process, so is listing. There is no list of all the rationals > because there is no end to the process of listing them, just as there is no > end to the process of counting. Hmm, this meaning of list is too confusingly different from the normal mathematical meaning. Can you find another word? Counting (the naturals) is an unending process, but there is no natural we cannot count to, so usually we say we can make a list of them. Well, it's just like this: 0, 1, 2, 3, 4, 5, ... That list will eventually reach _any_ natural. In the usual sense, there _is_ a list of rationals, because we can make one, just like the list of the naturals above, so that there is no rational which will never be reached. Here's an example way of listing the rationals between 0 and 1, using the lists above: 1/2, 1/4, 3/4, * 1/8, 3/8, 5/8, 7/8, 1/16, 3/16, 5/16, 7/16, ... 15/16, * 1/32, ... 1/3, 2/3, 1/9, 2/9, 4/9, 5/9, 7/9, 8/9, 1/27, 2/27, ... 1/5, 2/5, ... 1/7, ... ... In the first list I've put an asterisk before 1/2^p, for all p a prime. At each asterisk start folding in the list that starts with 1/ p, the corresponding prime. Once a list is folded in, after each element of the top list we include the next element of each of the folded in lists in turn. Can you see that every one of the rationals in (0,1) is now in this created list. This isn't the usual way of combining them, but hey, maths is a liberal discipline! > (I suspect there is a curious skew effect > here. Can we quantify over the rationals? Of course we can. In the (meta?) > context of understanding number, can we refer to all the rationals? No. > Pragmatism, even when raised to level of philosophical creed, is not the > answer.) I don't understand what you mean by quantify over. And I don't understand how one can ever say: Can we refer to P? No!, because for any P you have already referred to it in asking the question. Brian Chandler === Subject: Re: Approaching the infinite binary tree <290920081547112928%edgar@math.ohio-state.edu.invalid> posting-account=yxbZkgkAAABQBvyYeebYQ-PAvi0uT3tG Gecko/20071127 Firefox/2.0.0.11,gzip(gfe),gzip(gfe) > Anyway I am puzzled: The number of nodes is Aleph 0?? > Yes, aleph 0 is the cardinality of the set of natural numbers, which > means that the nodes can be counted. > Are you sure? One then could conclude that the set of paths must be > countable too, since it can be put in a bijection with a subset of the > nodes, namely the set of the leafs. > That might be what you'd expect intuitively, but in fact that is > not the case. The leaves (nodes) are not commensurate with > the paths. > For every node there is a finite path leading to that node. > But there are infinite paths in the complete binary tree that do > not have terminal nodes, i.e., there are endless paths having no > leaves. And who says that? Well, that's what the complete binary tree is. Unless you're considering the binary tree composed only of finite-length paths. Are you? If so, then there is a bijection between the nodes and the (finite) paths, and they are countable, and there are Aleph 0 of them. So which binary tree are you talking about? > Another thing to consider is that if all the paths (including the > infinite paths) were countable, then you could arrange them in > their natural order (from left to right), and then you could show > what the successor was for any given path in the tree. But you > can't do this. For example, what is the successor of (the path next > to) the path defined as alternating left/right turns (i.e., the path > given as L,R,L,R,L,R,..., or equivalently, the binary sequence > 01010101...)? Of course I can do that, and I have already done. Below is few members > before and after the one you mention: > ... > (01)0010 > (01)0011 > (01)0100 > (01) > (01)0110 > (01)0111 > (01)1000 > ... Ah, now I see the problem. First, I'm assuming that your notation (01)1000 means the digit sequence 0101010101...1000. (If not, please say so.) If that's the case, then your sequences have not been well-defined. For one thing, given (01)00 and (01)0001, and assuming that the (01) represent an infinite-length sequence of repeating digit pairs, how do we know when we've reached the end of the repeating part and have reached the rightmost digits? Assuming that we can number the repeating digits with natural indices, then what indices do you give the digits that follow the repeating pairs? There are several ways to do this, but I don't know which indexing scheme you're using. Also, since you're using infinite sequences of digits corresponding to the paths in the binary tree, you seem to be talking about the binary tree that includes infinite-length paths. Are you? > There is no problem in answering *your* question, the problems are > elsewhere, but we can't get there until we put aside the dogma. There > is NO problem in answering YOUR question as given above, but in YOUR > dogmatic framework of reference, which is anyway not mine. Well, you're going to have to be a little more specific about things. Which binary tree are you referring to? Which indexing scheme are you using for your digit sequences? > This question has no answer, as a consequence of the fact > that the infinite paths in the tree are not countable. You are just under a spell, so to say. No, I'm just making the assumption that you were talking about the complete binary tree. Are you? > Because just as you can number the levels (as 1, 2, 3, 4, etc.), > you can number the nodes in all the levels (as 1, then 2, 3, then > 4, 5, 6, 7, and so on). ?Each level can be given a unique natural > index, and likewise each node can be given a unique natural > index. > Since each index is a natural, and there is no last index in > either case, there are Aleph 0 levels and Aleph 0 nodes. And so what? The problem is not that the nodes are countable, it's > that your dogma imposes the paths not to be: which is simply > ridiculous. No, again I'm making an assumption because you didn't say anything to lead me to assume otherwise. === Subject: Re: Approaching the infinite binary tree > [quote] > An infinite complete binary tree is a tree with Aleph_0 levels, where > for each level d the number of existing nodes at level d is equal to > 2^d. The cardinal number of the set of all nodes is Aleph_0. The > cardinal number of the set of all paths is 2^Aleph_0. > [/quote] > Anyway I am puzzled: The number of nodes is Aleph_0?? > Yes, aleph_0 is the cardinality of the set of natural numbers, which > means that the nodes can be counted. Are you sure? One then could conclude that the set of paths must be > countable too, since it can be put in a bijection with a subset of the > nodes, namely the set of the leafs. That might be what you'd expect intuitively, but in fact that is > not the case. The leaves (nodes) are not commensurate with > the paths. For every node there is a finite path leading to that node. > But there are infinite paths in the complete binary tree that do > not have terminal nodes, i.e., there are endless paths having no > leaves. [*] And who says that? Further argument shows that there can be no bijection > between the nodes and the infinite paths. In fact, there are > far more infinite paths than finite paths in the tree. And who says that? Lots of us agree that the definitions say that. Another thing to consider is that if all the paths (including the > infinite paths) were countable, then you could arrange them in > their natural order (from left to right), and then you could show > what the successor was for any given path in the tree. But you > can't do this. For example, what is the successor of (the path next > to) the path defined as alternating left/right turns (i.e., the path > given as L,R,L,R,L,R,..., or equivalently, the binary sequence > 01010101...)? Of course I can do that, and I have already done. Below is few members > before and after the one you mention: ... > (01)0010 > (01)0011 > (01)0100 > (01) > (01)0110 > (01)0111 > (01)1000 > ... If (01)0010 represents a sequence endless on the left and ending in 00010, then none of julio's seqeinces are of the correct form. In their corrected reverse form, between 0100(01) and 1100(01), ther are lots of strings, including 0110(xx) for any x and y om {0,1}. There is no problem in answering *your* question, the problems are > elsewhere, but we can't get there until we put aside the dogma. It is your dogma, not ours, that you are being bit by. > There > is NO problem in answering YOUR question as given above, but in YOUR > dogmatic framework of reference, which is anyway not mine. This question has no answer, as a consequence of the fact > that the infinite paths in the tree are not countable. You are just under a spell, so to say. I'd indeed had expected the cardinality of the set of nodes to look > something like [2^(Aleph_0+1)]-1, maybe equal to 2^Aleph_0, but how > can it just be Aleph_0, where Aleph_0 is the cardinality of the > levels? Because just as you can number the levels (as 1, 2, 3, 4, etc.), > you can number the nodes in all the levels (as 1, then 2, 3, then > 4, 5, 6, 7, and so on). Each level can be given a unique natural > index, and likewise each node can be given a unique natural > index. Since each index is a natural, and there is no last index in > either case, there are Aleph_0 levels and Aleph_0 nodes. And so what? The problem is not that the nodes are countable, it's > that your dogma imposes the paths not to be: which is simply > ridiculous. It may seem ridiculous to julio, it it is nevertheless true. There are many sets besides N that have cardinality Aleph_0, > and all of them can be bijected with N. The set of even naturals, > for example, where each natural k in N is mapped to a unique > even natural 2k. Another example is Galileo's set of squares, > where each natural k in N is mapped to k^2. All of these sets have > the same cardinality of Aleph_0. Ditto. It may seem ridiculous to julio, it it is nevertheless true. === Subject: Re: Approaching the infinite binary tree <290920081547112928%edgar@math.ohio-state.edu.invalid> 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) > [quote] > An infinite complete binary tree is a tree with Aleph 0 levels, where > for each level d the number of existing nodes at level d is equal to > 2^d. The cardinal number of the set of all nodes is Aleph 0. The > cardinal number of the set of all paths is 2^Aleph 0. > [/quote] > Anyway I am puzzled: The number of nodes is Aleph 0?? > Yes, aleph 0 is the cardinality of the set of natural numbers, which > means that the nodes can be counted. Are you sure? One then could conclude that the set of paths must be > countable too, since it can be put in a bijection with a subset of the > nodes, namely the set of the leafs. That might be what you'd expect intuitively, but in fact that is > not the case. The leaves (nodes) are not commensurate with > the paths. For every node there is a finite path leading to that node. > But there are infinite paths in the complete binary tree that do > not have terminal nodes, i.e., there are endless paths having no > leaves. [*] And who says that? Further argument shows that there can be no bijection > between the nodes and the infinite paths. In fact, there are > far more infinite paths than finite paths in the tree. And who says that? Lots of us agree that the definitions say that. Liar! It's CANTOR that says that!! No bloody definitions of anything involved here. > Another thing to consider is that if all the paths (including the > infinite paths) were countable, then you could arrange them in > their natural order (from left to right), and then you could show > what the successor was for any given path in the tree. But you > can't do this. For example, what is the successor of (the path next > to) the path defined as alternating left/right turns (i.e., the path > given as L,R,L,R,L,R,..., or equivalently, the binary sequence > 01010101...)? Of course I can do that, and I have already done. Below is few members > before and after the one you mention: ... > (01)0010 > (01)0011 > (01)0100 > (01) > (01)0110 > (01)0111 > (01)1000 > ... If (01)0010 represents a sequence endless on the left and ending in > 00010 No, you just and still confuse a numeral for a number. You again confirm that you don't even know what you are talking about. > , then none of julio's seqeinces are of the correct form. In their corrected reverse form, between 0100(01) and 1100(01), ther > are lots of strings, including 0110(xx) for any x and y om {0,1}. There is no problem in answering *your* question, the problems are > elsewhere, but we can't get there until we put aside the dogma. It is your dogma, not ours, that you are being bit by. My dogma?? My dogma is that you guys are a waste of time, with an intellectal level of a 4 years old kid. > There > is NO problem in answering YOUR question as given above, but in YOUR > dogmatic framework of reference, which is anyway not mine. This question has no answer, as a consequence of the fact > that the infinite paths in the tree are not countable. You are just under a spell, so to say. I'd indeed had expected the cardinality of the set of nodes to look > something like [2^(Aleph 0+1)]-1, maybe equal to 2^Aleph 0, but how > can it just be Aleph 0, where Aleph 0 is the cardinality of the > levels? Because just as you can number the levels (as 1, 2, 3, 4, etc.), > you can number the nodes in all the levels (as 1, then 2, 3, then > 4, 5, 6, 7, and so on). .95ËüEach level can be given a unique natural > index, and likewise each node can be given a unique natural > index. Since each index is a natural, and there is no last index in > either case, there are Aleph 0 levels and Aleph 0 nodes. And so what? The problem is not that the nodes are countable, it's > that your dogma imposes the paths not to be: which is simply > ridiculous. It may seem ridiculous to julio, it it is nevertheless true. True this balls! YOUR DOGMA IMPOSES THE PATHS NOT TO BE!!! Moron and liar. > There are many sets besides N that have cardinality Aleph 0, > and all of them can be bijected with N. The set of even naturals, > for example, where each natural k in N is mapped to a unique > even natural 2k. Another example is Galileo's set of squares, > where each natural k in N is mapped to k^2. All of these sets have > the same cardinality of Aleph 0. Ditto. It may seem ridiculous to julio, it it is nevertheless true. Yeah, it's true that on this argument too there is nothing more to add: YOUR MATH IS PROVABLY A JOKE. === Subject: Re: Approaching the infinite binary tree > [quote] > An infinite complete binary tree is a tree with Aleph_0 levels, where > for each level d the number of existing nodes at level d is equal to > 2^d. The cardinal number of the set of all nodes is Aleph_0. The > cardinal number of the set of all paths is 2^Aleph_0. > [/quote] > Anyway I am puzzled: The number of nodes is Aleph_0?? > Yes, aleph_0 is the cardinality of the set of natural numbers, which > means that the nodes can be counted. Are you sure? One then could conclude that the set of paths must be > countable too, since it can be put in a bijection with a subset of the > nodes, namely the set of the leafs. That might be what you'd expect intuitively, but in fact that is > not the case. The leaves (nodes) are not commensurate with > the paths. For every node there is a finite path leading to that node. > But there are infinite paths in the complete binary tree that do > not have terminal nodes, i.e., there are endless paths having no > leaves. [*] And who says that? Further argument shows that there can be no bijection > between the nodes and the infinite paths. In fact, there are > far more infinite paths than finite paths in the tree. And who says that? Lots of us agree that the definitions say that. Liar! It's CANTOR that says that!! No bloody definitions of anything > involved here. Actually, Cantor didn't say much of anything about infinite binary tree, though he might well have if he had been questioned about them. The sort of complete infinite binary trees under discussions as in http://en.wikipedia.org/wiki/Binary_tree Quote: An infinite complete binary tree is a tree with aleph_0 levels, where for each level d the number of existing nodes at level d is equal to 2^d. The cardinal number of the set of all nodes is aleph_0. The cardinal number of the set of all paths is 2^ aleph_0. > Another thing to consider is that if all the paths (including the > infinite paths) were countable, then you could arrange them in > their natural order (from left to right), and then you could show > what the successor was for any given path in the tree. But you > can't do this. For example, what is the successor of (the path next > to) the path defined as alternating left/right turns (i.e., the path > given as L,R,L,R,L,R,..., or equivalently, the binary sequence > 01010101...)? Of course I can do that, and I have already done. Below is few members > before and after the one you mention: ... > (01)0010 > (01)0011 > (01)0100 > (01) > (01)0110 > (01)0111 > (01)1000 > ... If (01)0010 represents a sequence endless on the left and ending in > 00010 No, you just and still confuse a numeral for a number. You again > confirm that you don't even know what you are talking about. I certainly do not know what you are talking about. , then none of julio's seqeinces are of the correct form. In their corrected reverse form, between 0100(01) and 1100(01), ther > are lots of strings, including 0110(xx) for any x and y om {0,1}. There is no problem in answering *your* question, the problems are > elsewhere, but we can't get there until we put aside the dogma. It is your dogma, not ours, that you are being bit by. My dogma?? My dogma is that you guys are a waste of time, with an > intellectal level of a 4 years old kid. Based on that dogma, you should stop posting, as you can hardly expect learn anything from us not teach us anything according to that dogma. There > is NO problem in answering YOUR question as given above, but in YOUR > dogmatic framework of reference, which is anyway not mine. This question has no answer, as a consequence of the fact > that the infinite paths in the tree are not countable. You are just under a spell, so to say. I'd indeed had expected the cardinality of the set of nodes to look > something like [2^(Aleph_0+1)]-1, maybe equal to 2^Aleph_0, but how > can it just be Aleph_0, where Aleph_0 is the cardinality of the > levels? Because just as you can number the levels (as 1, 2, 3, 4, etc.), > you can number the nodes in all the levels (as 1, then 2, 3, then > 4, 5, 6, 7, and so on). .95ËüEach level can be given a unique natural > index, and likewise each node can be given a unique natural > index. Since each index is a natural, and there is no last index in > either case, there are Aleph_0 levels and Aleph_0 nodes. And so what? The problem is not that the nodes are countable, it's > that your dogma imposes the paths not to be: which is simply > ridiculous. It may seem ridiculous to julio, it is nevertheless true. True this balls! YOUR DOGMA IMPOSES THE PATHS NOT TO BE!!! Moron and > liar. Such incoherence! There are many sets besides N that have cardinality Aleph_0, > and all of them can be bijected with N. The set of even naturals, > for example, where each natural k in N is mapped to a unique > even natural 2k. Another example is Galileo's set of squares, > where each natural k in N is mapped to k^2. All of these sets have > the same cardinality of Aleph_0. Ditto. It may seem ridiculous to julio, it it is nevertheless true. Yeah, it's true that on this argument too there is nothing more to > add: YOUR MATH IS PROVABLY A JOKE. You have not provided any such proofs, and have carefully avoided presenting anything of the nature of mathematics based counter-argument. === Subject: Re: Approaching the infinite binary tree <290920081547112928%edgar@math.ohio-state.edu.invalid> 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 sort of complete infinite binary trees under discussions as in http://en.wikipedia.org/wiki/Binary tree Liar as ever. > Yeah, it's true that on this argument too there is nothing more to > add: YOUR MATH IS PROVABLY A JOKE. You have not provided any such proofs, and have carefully avoided > presenting anything of the nature of mathematics based counter-argument. You prove yourself. Liar and moron. === Subject: Re: Approaching the infinite binary tree The sort of complete infinite binary trees under discussions as in http://en.wikipedia.org/wiki/Binary_tree Liar as ever. Yeah, it's true that on this argument too there is nothing more to > add: YOUR MATH IS PROVABLY A JOKE. You have not provided any such proofs, and have carefully avoided > presenting anything of the nature of mathematics based counter-argument. You prove yourself. Liar and moron. Those who have no relevant arguments to make in their favor are forced to personal attacks. === Subject: Re: Approaching the infinite binary tree <290920081547112928%edgar@math.ohio-state.edu.invalid> 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 sort of complete infinite binary trees under discussions as in http://en.wikipedia.org/wiki/Binary tree Liar as ever. Yeah, it's true that on this argument too there is nothing more to > add: YOUR MATH IS PROVABLY A JOKE. You have not provided any such proofs, and have carefully avoided > presenting anything of the nature of mathematics based counter-argument. You prove yourself. Liar and moron. Those who have no relevant arguments to make in their favor are forced > to personal attacks. That's what you are: a liar, and overall just another moron. === Subject: Re: Approaching the infinite binary tree >, > Mariano Suarez-Alvarez > You seem to mean by this that you have a last level, the one > indexed by w. I can drop the informally: A tree with w levels! What is > the > problem > with that? Nothing at all. However, you have a huge problem if you are saying there > is a level that has index w. It may make perfectly good sense, depending on your definition of > tree. > http://en.wikipedia.org/wiki/Tree_(set_theory) Just one of the many problems this creates for you: > what is the index of the level immediately before level w? Depending on your definition of tree, it can happen that level > immediately before need not be defined. Probably means something like an infinite complete binary tree as > defined inhttp://en.wikipedia.org/wiki/Binary_tree, in which case > each > node other than the root must have a parent node in the level > immediately before it. what you furtherly remark is of course the usual nonsense). > And my nonsense is considerably more sensible than you inept attempts > at sense. The only mention to infinite trees is here: [quote] > An infinite complete binary tree is a tree with Aleph_0 levels, where > for each level d the number of existing nodes at level d is equal to > 2^d. The cardinal number of the set of all nodes is Aleph_0. The > cardinal number of the set of all paths is 2^Aleph_0. > [/quote] Anyway I am puzzled: The number of nodes is Aleph_0?? Yes, aleph_0 is the cardinality of the set of natural numbers, which > means that the nodes can be counted. Are you sure? One then could conclude that the set of paths must be > countable too, since it can be put in a bijection with a subset of the > nodes, namely the set of the leafs. That leaf property holds nicely only for finite trees in which every > path ends in a leaf node, but in the complete infinite binary tree none > of them have leaf nodes, just as endless strings of characters do not > have last characters. Since the only meaning in your infinite tree is just the same as my > mentioning leaves, namely an inductive-based meaning, your objection > is the usual nonsense. It is easy to biject the set of all paths of the infinite complete binary tree with the set of all functions from the 0 origined N to {left,right}. Then each node at level n corresponds to a function from N_n = {x in N: x <= n} to {left,right}, but there is no last node to any path. Over induction we have a bijection between paths and the nodes at each > step. Over induction we thus show that, as far as the nodes are > countable, so it must be the number of paths, being it in bijection > with a subset of the nodes. According to your type of inductive argument the fact that all finite sets are finite implies by your induction that infinite sets must also finite. Your claim that such bijection does not hold must be spelled out: On the contrary, it is your claim that must be spelled out unless we are also to conclude that infinite sets must be finite. > the > ONLY reason why such an almost trivial result must be contradicted is > simply and again Cantor: that is, your dogma. I'd indeed had expected the cardinality of the set of nodes to look > something like [2^(Aleph_0+1)]-1, maybe equal to 2^Aleph_0, but how > can it just be Aleph_0, where Aleph_0 is the cardinality of the > levels? In much the same way as the set of even naturals can have the same > cardinality as the set of all naturals. > In fact, one can have an endless chain of equally countable sets with > each being a proper subset of its predecessor, simply by removing one > more element from each successive link in the chain. It is clear that if you cannot sort out issues like this on your own, > you will never be able to deal with infinite cardinalities without error. These three paragraphs of yours are complete and irrelevent nonsense. That they may well seem so to you does not mean that they are so to anyone else. You seem often unable to grasp what is trivial to most pothers here, especially if it does not fit into your personal worldview. === Subject: Solution Manual to Fundamentals of Fluids Mechanics 5th (Munson & Young) posting-account=faN68woAAAB1Ld4vL1HYXoLoGwiqmKsO 1.1.4322; InfoPath.2),gzip(gfe),gzip(gfe) Looking for the solution manual to Fundamentals of Fluids Mechanics 5th (Munson & Young)? It has Chapter 1~12 and is in PDF format. If you need it, email collegemanual [at] gmail [dot] com for more info. === Subject: Re: Algebra with order a, b >Hello teacher~ >Group G = b =/= e. >Find the order of b. >Answer : 7 >------------------------------------------------------------- >Since answer, >I must show that b^7 = e. >so, the order of b is 1 or 7. of course, 7. >But I can't deduce well that b^7 = e. >so, I need your advice. What happens if you to a^2 b a^{-2}? Hm... obscure hint. This is not like you. a^3 = e b^2 = a.b.a^{-1} ==> b^4 = a.(b^2).a^{-1} ==> b^8 = a.(b^4).a^{-1} b = (a^3).b.(a^{-3}) = (a^2).(a.b.a^{-1}).(a^{-2}) = (a^2).b^2.(a^{-2}) = a.(a.b^2.a^{-1}).a^{-1} = a.(b^4).a^{-1} = b^8 so, b^7 = e. === Subject: Re: Algebra with order a, b days. My association with the Department is that of an alumnus. >Hello teacher~ Group G = Since answer, >I must show that b^7 = e. >so, the order of b is 1 or 7. of course, 7. But I can't deduce well that b^7 = e. >so, I need your advice. > What happens if you to a^2 b a^{-2}? Hm... obscure hint. Not at all. It should be enough to get you through to the end. > This is not like you. What can I say? I have the sneaking suspicion that mina is really a shared address that is being used by someone other than the person who began posting several years ago. That person asked and seemed to understand questions of equal or higher difficulty back then; it is hard to believe that person could have regressed so far since then. Perhaps that person simply built some good will and now is trying to inherit that good will to a new generation of students. >a^3 = e >b^2 = a.b.a^{-1} ==> b^4 = a.(b^2).a^{-1} >==> b^8 = a.(b^4).a^{-1} >b = (a^3).b.(a^{-3}) >= (a^2).(a.b.a^{-1}).(a^{-2}) >= (a^2).b^2.(a^{-2}) >= a.(a.b^2.a^{-1}).a^{-1} >= a.(b^4).a^{-1} >= b^8 so, b^7 = e. Not so obscure after all, then. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes by Bill Watterson) Arturo Magidin magidin-at-member-ams-org === Subject: Re: Algebra with order a, b >Hello teacher~ >Group G = b =/= e. >Find the order of b. >Answer : 7 >------------------------------------------------------------- >Since answer, >I must show that b^7 = e. >so, the order of b is 1 or 7. of course, 7. >But I can't deduce well that b^7 = e. >so, I need your advice. What happens if you to a^2 b a^{-2}? >Hm... obscure hint. Not at all. It should be enough to get you through to the end. In fact, the criteria for enough is very subjective. > This is not like you. What can I say? I have the sneaking suspicion that mina is really a > shared address that is being used by someone other than the person who > began posting several years ago. That person asked and seemed to > understand questions of equal or higher difficulty back then; it is > hard to believe that person could have regressed so far since > then. Perhaps that person simply built some good will and now is > trying to inherit that good will to a new generation of students. Hm... Sherlock Holmes ? But I am me. >a^3 = e >b^2 = a.b.a^{-1} ==> b^4 = a.(b^2).a^{-1} >==> b^8 = a.(b^4).a^{-1} >b = (a^3).b.(a^{-3}) >= (a^2).(a.b.a^{-1}).(a^{-2}) >= (a^2).b^2.(a^{-2}) >= a.(a.b^2.a^{-1}).a^{-1} >= a.(b^4).a^{-1} >= b^8 >so, b^7 = e. Not so obscure after all, then. More thinking... G = < a, b | a^3 = e, b^7 = e, b^2 = a.b.a^{-1} > G is really order 21. Because, H = {e, a, a^2} N = {e, b, b^2, b^3, b^4, b^5, b^6} H / N is a subgroup of G. Of cousre, H / N is a subgroup of H and N. Since |H / N| | 3 and 7, so, H / N = {e} It means that a^i =/= b^j except e. Consequently, the elements of G is form (a^i).(b^j) , 0<=i<=2 , 0<=j<=6. If (a^i).(b^j) = (a^k).(b^l), then a^(i-k) = b^(l-j). so, it must be a^(i-k) = b^(l-j) = e. so, a^i = a^k , b^l = b^j. It means that 21 elements of G are distinct. so, G is a group of order 21. Of cousre, G is non-abelian. If G is abelian, b^2 = a.b.a^{-1} ==> b^2 = b ==> b = e. ----------------------------------------------------------------- Theorem) If p and q are distict primes with p < q, then every group G of order pq has a single subgroup of order q and this subgroup is normal in G. Hence G is not simple. If q is not congruent to 1 modulo p, then G is abelian and cycle. |G| = 21 = 3*7, (3 < 7) G has a Sylow 7-subgroup and that the number of such subgroups is congruent to 1 modulo 7 and divides 3*7. so, 1 or 8 or 15. so, the only possibility is the number 1. It means that this Sylow 7-subgroup is normal in G. (Of course, this subgroup is precisely subgroup of order 7.) and 7 is congruent to 1 modulo 3. === Subject: Re: Algebra with order a, b > What can I say? I have the sneaking suspicion that mina is really a > shared address that is being used by someone other than the person who > began posting several years ago. That person asked and seemed to > understand questions of equal or higher difficulty back then; it is > hard to believe that person could have regressed so far since > then. Perhaps that person simply built some good will and now is > trying to inherit that good will to a new generation of students. Students indeed. Her students: http://tinyurl.com/49a7de -- I.N. Galidakis === Subject: Re: Algebra with order a, b posting-account=-PngCgkAAAD2yUjosqWv1Nf1lkqWP4lp rv:1.8.1.16) Gecko/20080702 Firefox/2.0.0.16,gzip(gfe),gzip(gfe) > Hello teacher~ Group G = Since answer, > I must show that b^7 = e. > so, the order of b is 1 or 7. of course, 7. You cannot just say of course 7. Assuming that you can prove that b^7 = e (and Arturo gave you a hint about that), you still have to prove that the order really is 7, and not 1. Derek Holt. > But I can't deduce well that b^7 = e. > so, I need your advice. === Subject: Re: Algebra with order a, b > Hello teacher~ > Group G = b =/= e. > Find the order of b. > Answer : 7 > ------------------------------------------------------------- > Since answer, > I must show that b^7 = e. > so, the order of b is 1 or 7. of course, 7. You cannot just say of course 7. Assuming that you can prove that > b^7 = e (and Arturo gave you a hint about that), you still have to > prove that the order really is 7, and not 1. But we are told that b =/= e, so surely the order of b is not 1. Well, maybe it depends on how you defined order, but if b has order 1, then we should certainly have b^1 = e, i.e. b=e. === Subject: Re: Algebra with order a, b <86k5cs3pcp.fsf@vulcan.lan> posting-account=-PngCgkAAAD2yUjosqWv1Nf1lkqWP4lp Gecko/20080710 SUSE/2.0.0.16-4.1 Firefox/2.0.0.16,gzip(gfe),gzip(gfe) > Hello teacher~ > Group G = b =/= e. > Find the order of b. > Answer : 7 > ------------------------------------------------------------- > Since answer, > I must show that b^7 = e. > so, the order of b is 1 or 7. of course, 7. You cannot just say of course 7. Assuming that you can prove that > b^7 = e (and Arturo gave you a hint about that), you still have to > prove that the order really is 7, and not 1. But we are told that b =/= e, so surely the order of b is not 1. > Well, maybe it depends on how you defined order, but if b has order > 1, then we should certainly have b^1 = e, i.e. b=e. Sorry! You are right - I replied too late at night. However, since it is possible to prove that b =/e without being told, I am surprised that the information is given. Consider the subgroup H of S_7 generated by the permutations A = (2,3,5)(4,7,6), B = (1,2,3,4,5,6,7) Then A^3=1, and (composing from right to left), A B A^-1 = B^2. So there is a homomorphism f:G->H with f(a)=A, f(b)=B, and b =/= 1 follows from B =/= 1. In fact f is an isomorphism, and |G|=|H|=21. Derek Holt. === Subject: Re: Algebra with order a, b > Hello teacher~ > Group G = Find the order of b. > Answer : 7 > ------------------------------------------------------------- > Since answer, > I must show that b^7 = e. > so, the order of b is 1 or 7. of course, 7. > You cannot just say of course 7. Assuming that you can prove that > b^7 = e (and Arturo gave you a hint about that), you still have to > prove that the order really is 7, and not 1. > But we are told that b =/= e, so surely the order of b is not 1. > Well, maybe it depends on how you defined order, but if b has order > 1, then we should certainly have b^1 = e, i.e. b=e. Sorry! You are right - I replied too late at night. However, since it is possible to prove that b =/e without being told, >I am surprised that the information is given. But b = e satisfies the 2nd identity. quasi === Subject: free groups posting-account=4l8zHQoAAADdlkoqMdUVhRbP5dh_2A-D Gecko/2008092417 Firefox/2.0;MEGAUPLOAD 1.0,gzip(gfe),gzip(gfe) Hallo! The following facts about a free group F are well known. (1) F is torsion-free; (2) F is non-commutative (if F has rank >1); (3) an abelian subgroup of F is cyclic; (4) the center of F is trivial (if F is not the integers); (5) every subgroup of F is free. My question is: under what conditions is F necessarily free? Fred. === Subject: Re: free groups days. My association with the Department is that of an alumnus. >Hallo! The following facts about a free group F are well known. >(1) F is torsion-free; >(2) F is non-commutative (if F has rank >1); >(3) an abelian subgroup of F is cyclic; >(4) the center of F is trivial (if F is not the integers); >(5) every subgroup of F is free. My question is: under what conditions is F necessarily free? Huh? What do you mean? -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes by Bill Watterson) Arturo Magidin magidin-at-member-ams-org === Subject: Re: free groups posting-account=suWj4AkAAADE1IvGmj55Nmq3f98qb17e InfoPath.1; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) Hallo! The following facts about a free group F are well known. >(1) F is torsion-free; >(2) F is non-commutative (if F has rank >1); >(3) an abelian subgroup of F is cyclic; >(4) the center of F is trivial (if F is not the integers); >(5) every subgroup of F is free. My question is: under what conditions is F necessarily free? Huh? What do you mean? > **************************************************************** I'm guessing the OP asks under what conditions can we assure a group F is free? I'd say that F is free iff we can find a generating set X of F such that F = < X ; Empty Set > , meaning: F is generated by X and there are no non-trivial relations on F. Tonio === Subject: Re: free groups posting-account=5i_ghQoAAABbiBbHHRb14_4kV5Y2uwJY Gecko/2008092417 Firefox/2.0;MEGAUPLOAD 1.0,gzip(gfe),gzip(gfe) Note that (1)+(2) does not imply the freeness of the group. Take the free product of the free abelian group of rank 2 by itself. Besides, (3) implies (2). I guess (5) does not imply freeness also: the converse of Nielsen-Schreier is not true. Best, Louis Andrew. === Subject: -- rational distances Does there exist a closed, nonempty subset S of R^2 such that d(p,S) is in Q for all points p in Q^2? quasi === Subject: Re: -- rational distances posting-account=8wyvFgoAAAAJYWyfLHRzREe3lxFCHRTd MathPlayer 2.10b; .NET CLR 1.1.4322; .NET CLR 2.0.50727; .NET CLR 3.0.04506.30),gzip(gfe),gzip(gfe) > Does there exist a closed, nonempty subset S of R^2 such that d(p,S) > is in Q for all points p in Q^2? quasi S=R^2 smn === Subject: Re: -- rational distances > Does there exist a closed, nonempty subset S of R^2 such that d(p,S) > is in Q for all points p in Q^2? > quasi S=R^2 smn Yep, I forgot to bar that. Ok, consided it barred ... Does there exist a closed, nonempty proper subset S of R^2 such that d(p,S) is in Q for all points p in Q^2? quasi === Subject: Re: -- rational distances Does there exist a closed, nonempty subset S of R^2 such that d(p,S) > is in Q for all points p in Q^2? quasi >S=R^2 smn Yep, I forgot to bar that. Ok, consided it barred ... Does there exist a closed, nonempty proper subset S of R^2 such that >d(p,S) is in Q for all points p in Q^2? No, that can also be easily achieved. For example, let S = {(x,y) in R^2 | y <=0}. Ok, here's a fixed version ... Does there exist a nonempty compact subset S of R^2 such that d(p,S) is in Q for all points p in Q^2? quasi === Subject: Re: -- rational distances 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) > Does there exist a closed, nonempty subset S of R^2 such that d(p,S) > is in Q for all points p in Q^2? quasi >S=R^2 smn Yep, I forgot to bar that. Ok, consided it barred ... Does there exist a closed, nonempty proper subset S of R^2 such that >d(p,S) is in Q for all points p in Q^2? No, that can also be easily achieved. For example, let S = {(x,y) in R^2 | y <=0}. This seems to work because you don't have to take square roots, and anyway require that the frontier of S is only made of rational points, which I suppose means the frontier stays always parallel to one or the other axis. > Ok, here's a fixed version ... Does there exist a nonempty compact subset S of R^2 such that d(p,S) > is in Q for all points p in Q^2? As soon as you make S bounded, I wouldn't see how you can fulfil the required property at all, because you'll have anyway to take square roots for the distance from almost all points... === Subject: Re: -- rational distances is in Q for all points p in Q^2? > D = { (x,y) | x^2 + y^2 <= 1 } Let r = inf{ r | S subset rD } Some point p in S / rD. Let L be the tangent to rD at p. L divides the plane into a region that cnntains S and one that doesn't. In the latter region you can find some points s in QxQ with d(s,S) = d(p,s) not in Q. In addition notice d(p,nulset) not in Q. === Subject: Re: -- rational distances > Does there exist a nonempty compact subset S of R^2 such that d(p,S) > is in Q for all points p in Q^2? D = { (x,y) | x^2 + y^2 <= 1 } >Let r = inf{ r | S subset rD } Some point p in S / rD. You're mixing up letters -- in the stated problem, p is in Q^2. Then later you take s in Q^2 (rather than s in S). OK, so aside from that mixup, when you say some point p in S / rD, I assume you meant some point p in S intersect (the circular boundary of rD). Right? >Let L be the tangent to rD at p. L divides the plane into a region that cnntains S and one >that doesn't. In the latter region you can find some points > s in QxQ with d(s,S) = d(p,s) not in Q. Seem plausible. But that's the key -- can you always find some such points in Q^2? What if S is convex? Sure, but for distances to sets, it suffices to consider closed sets. >In addition notice d(p,nulset) not in Q. The problem explicitly requires that S be nonempty. quasi === Subject: Re: -- rational distances <5gv8e498v477akii9pdi68kn5vef2jj2k5@4ax.com > Does there exist a nonempty compact subset S of R^2 such that d(p,S) > is in Q for all points p in Q^2? D = { (x,y) | x^2 + y^2 <= 1 } >Let r = inf{ r | S subset rD } Some point p in S / rD. You're mixing up letters -- in the stated problem, p is in Q^2. Then later you take s in Q^2 (rather than s in S). OK, so aside from that mixup, when you say some point p in S / rD, I assume you meant some point p in S intersect (the circular boundary of rD). Right? > Yes, it works out to that. >Let L be the tangent to rD at p. L divides the plane into a region that cnntains S and one >that doesn't. In the latter region you can find some points > s in QxQ with d(s,S) = d(p,s) not in Q. Seem plausible. But that's the key -- can you always find some such points in Q^2? What if S is convex? > What's the problem? It's still in rD. Oh, the point p may not be unique. What the heck? S could be rD itself and all of bd rD would be 'p' points. Sure, but for distances to sets, it suffices to consider closed sets. In addition notice d(p,nulset) not in Q. The problem explicitly requires that S be nonempty. > It's slight generalization. ;-) === Subject: Re: -- rational distances Does there exist a nonempty compact subset S of R^2 such that d(p,S) > is in Q for all points p in Q^2? D = { (x,y) | x^2 + y^2 <= 1 } > Let r = inf{ r | S subset rD } Too many r's... > Some point p in S / rD. I presume you want p with ||p|| = r = sup {||s||: s in S}. > Let L be the tangent to rD at p. L divides the plane into a region that cnntains S and one > that doesn't. In the latter region you can find some points > s in QxQ with d(s,S) = d(p,s) not in Q. How so? The only points s where d(s,S) = d(p,s) might be the ray {tp: 1 <= t < infty}, and this might not intersect Q^2. > In addition notice d(p,nulset) not in Q. -- Robert Israel israel@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: -- rational distances is in Q for all points p in Q^2? D = { (x,y) | x^2 + y^2 <= 1 } > Let r = inf{ r | S subset rD } Too many r's... Some point p in S / rD. I presume you want p with ||p|| = r = sup {||s||: s in S}. > It works out to that. > Let L be the tangent to rD at p. L divides the plane into a region that cnntains S and one > that doesn't. In the latter region you can find some points > s in QxQ with d(s,S) = d(p,s) not in Q. How so? The only points s where d(s,S) = d(p,s) might be the > ray {tp: 1 <= t < infty}, and this might not intersect Q^2. > Take the normal to L at p, and vary the slope by +-epsilon. Some of the outside points in QxQ and the wedge between the two lines will have irrational distance to p. ---- === Subject: solution manual posting-account=8tdwzAoAAACrDhR8o5Q_XoOKVlRjTXi6 5.0),gzip(gfe),gzip(gfe) hi iam need for a solution manual of the book computer networks by larry peterson and bruce davie could you provide me with links? === Subject: solution manual posting-account=8tdwzAoAAACrDhR8o5Q_XoOKVlRjTXi6 5.0),gzip(gfe),gzip(gfe) hi iam need for a solution manual of the book computer networks by larry peterson and bruce davie could you provide me with links? === Subject: Approximating the Branches of the Complex Lambert's W function Using Analytic Continuation With the rise in popularity of Lambert's W function, people often ask me for a fast and efficient algorithm for W. I usually refer them to Wiki's page for W, which lists several good ones: http://en.wikipedia.org/wiki/Lambert_W_function Maple contains a complete algorithm for W(k,z) based on Halley's method, but as far as I know, its exact implementation details have never been made public. Here's an algorithm for W(1,1+i) based on the analytic continuation of W and the notion of germ, which can be easily modified to calculate W(k,z), for any complex branch k in Z and any z in C, provided z is fairly far away from the corresponding branch's branch points: http://ioannis.virtualcomposer2000.com/math/LWCalculating2.html If anyone is interested in porting the above to C/FORTRAN, feel free to email me to discuss the details. --- I.N. Galidakis === Subject: Re: tomic polynomial: CONJECTURE 3 > Every algebraic number is a ( finite ) sum of zero's > of tommy polynomials. I suspect your conjecture is true. Moreover, I think it would still be true if the coefficient set {1,0,-1} was replaced by {0,1}, or for that matter {0,a} for any nonzero a in Q. As far as proving it (or disproving it), that may be quite hard. While thinking about your conjecture, I came up with a few related questions, but since they're not directly about tomic polynomials, I'll pose those questions in a separate thread. quasi === Subject: Re: tomic polynomial: CONJECTURE 3 Every algebraic number is a ( finite ) sum of zero's > of tommy polynomials. I suspect your conjecture is true. Moreover, I think it would still be true if the coefficient set > {1,0,-1} was replaced by {0,1}, or for that matter {0,a} for any > nonzero a in Q. So a tommy polynomial is a polynomial (of degree >= 1) whose coefficients are in {1,0,-1}? Then any root of such a thing is an algebraic integer. A sum of algebraic integers is an algebraic integer. But not every algebraic number is an algebraic integer, e.g. the only rationals that are algebraic integers are integers. -- Robert Israel israel@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: tomic polynomial: CONJECTURE 3 Every algebraic number is a ( finite ) sum of zero's > of tommy polynomials. I suspect your conjecture is true. Moreover, I think it would still be true if the coefficient set >{1,0,-1} was replaced by {0,1}, or for that matter {0,a} for any >nonzero a in Q. As far as proving it (or disproving it), that may be quite hard. While thinking about your conjecture, I came up with a few related >questions, but since they're not directly about tomic polynomials, >I'll pose those questions in a separate thread. The counterexample of 1/3, noted by galathaea in my thread about the field generated by the set of roots of unity works just as well as a counterexample here. The roots of {-1,0,1}-polynomials are algebraic integers, hence so is any finite sum. I guess you could revise the conjecture to claim that you can get all algebraic integers. But even with that restriction, I've changed my opinion as to the truth of your conjecture -- I now strongly doubt it. quasi === Subject: Re: tomic polynomial: CONJECTURE 3 quasi a .8ecrit : Every algebraic number is a ( finite ) sum of zero's > of tommy polynomials. I suspect your conjecture is true. For which reasons? Moreover, I think it would still be true if the coefficient set > {1,0,-1} was replaced by {0,1}, or for that matter {0,a} for any > nonzero a in Q. Again, for which reasons As far as proving it (or disproving it), that may be quite hard. > Which is why I ask again : why makes you believe it is true? > While thinking about your conjecture, I came up with a few related > questions, but since they're not directly about tomic polynomials, > I'll pose those questions in a separate thread. quasi === Subject: Re: tomic polynomial: CONJECTURE 3 >quasi a .8ecrit : Every algebraic number is a ( finite ) sum of zero's > of tommy polynomials. I suspect your conjecture is true. For which reasons? Intuition. > Moreover, I think it would still be true if the coefficient set > {1,0,-1} was replaced by {0,1}, or for that matter {0,a} for any > nonzero a in Q. Again, for which reasons. The same intuition. > As far as proving it (or disproving it), that may be quite hard. Which is why I ask again : why makes you believe it is true? Well, that was then. As of now, I think tommy's conjecture is false (even with the restriction to algebraic integers). But it may not be so easy to disprove. As a simple test case, try this ... Is sqrt(2) a finite sum of roots of {-1,0,1}-polynomials? quasi === Subject: Re: tomic polynomial: CONJECTURE 3 quasi a .8ecrit : quasi a .8ecrit : Every algebraic number is a ( finite ) sum of zero's > of tommy polynomials. > I suspect your conjecture is true. > For which reasons? Intuition. Moreover, I think it would still be true if the coefficient set > {1,0,-1} was replaced by {0,1}, or for that matter {0,a} for any > nonzero a in Q. > Again, for which reasons. The same intuition. As far as proving it (or disproving it), that may be quite hard. > Which is why I ask again : why makes you believe it is true? Well, that was then. As of now, I think tommy's conjecture is false (even with the > restriction to algebraic integers). But it may not be so easy to disprove. As a simple test case, try this ... Is sqrt(2) a finite sum of roots of {-1,0,1}-polynomials? quasi Sure, but all this goes to show 1) that your intuition is not a very reliable tool 2) that tommy's conjectures are similar to JSH ones : not very interesting, often very easy to refutate, not thought upon at all, and worst of all, formulated as ego-booster and not out of ny genious interest... === Subject: Re: tomic polynomial: CONJECTURE 3 >quasi a .8ecrit : quasi a .8ecrit : > Every algebraic number is a ( finite ) sum of zero's > of tommy polynomials. > I suspect your conjecture is true. > For which reasons? Intuition. Moreover, I think it would still be true if the coefficient set > {1,0,-1} was replaced by {0,1}, or for that matter {0,a} for any > nonzero a in Q. > Again, for which reasons. The same intuition. As far as proving it (or disproving it), that may be quite hard. > Which is why I ask again : why makes you believe it is true? Well, that was then. As of now, I think tommy's conjecture is false (even with the > restriction to algebraic integers). But it may not be so easy to disprove. As a simple test case, try this ... Is sqrt(2) a finite sum of roots of {-1,0,1}-polynomials? quasi Sure, but all this goes to show 1) that your intuition is not a very >reliable tool Sure, my intuition is sometimes wrong, and it's subject to change as new information is obtained -- that's just the nature of it. But it's right enough of the time to have my trust, at least as a guidance as to what to expect. Besides, right or wrong, it typically leads to more questions, further insights, and hence it evolves, but always leading. >2) that tommy's conjectures are similar to JSH ones Not by a long shot. > : not very interesting, I find some of them very interesting, >often very easy to refutate, Sure, so what. As long as the underlying concept has some validity, the conjectures can usually be revised to reach the boundary where trivially true or trivially false suddenly becomes non-trivial. >not thought upon at all, value, that's enough. >and worst of all, formulated as ego-booster and not out of ny genious >interest... In my opinion, egoless posts are often very boring. Sure, tommy's ego is sometimes way out of control, but then again, we all have flaws. In tommy's case, as I see it, there are many redeeming virtues. tommy1729 loves math -- there's no question about that, and in my opinion, his questions are driven not just by ego, but also by a powerful curiosity, the lure of possible discovery, and the challenge and joy of the hunt. I can easily relate to that. quasi === Subject: Re: tomic polynomial: CONJECTURE 3 <4411947.1222871746307.JavaMail.jakarta@nitrogen.mathforum.org> <48e466ab$0$8011$7a628cd7@news.club-internet.fr> <48e46f27$0$8009$7a628cd7@news.club-internet.fr> posting-account=06BQLAoAAADoC7Y4z9FWcUwGvMa7xMG9 7.4),gzip(gfe),gzip(gfe) quasi a .8ecrit : quasi a .8ecrit : > Every algebraic number is a ( finite ) sum of zero's > of tommy polynomials. > I suspect your conjecture is true. > For which reasons? > Intuition. > Moreover, I think it would still be true if the coefficient set > {1,0,-1} was replaced by {0,1}, or for that matter {0,a} for any > nonzero a in Q. > Again, for which reasons. > The same intuition. > As far as proving it (or disproving it), that may be quite hard. > Which is why I ask again : why makes you believe it is true? > Well, that was then. > As of now, I think tommy's conjecture is false (even with the > restriction to algebraic integers). > But it may not be so easy to disprove. > As a simple test case, try this ... > Is sqrt(2) a finite sum of roots of {-1,0,1}-polynomials? > quasi Sure, but all this goes to show 1) that your intuition is not a very >reliable tool Sure, my intuition is sometimes wrong, and it's subject to change as > new information is obtained -- that's just the nature of it. But it's > right enough of the time to have my trust, at least as a guidance as > to what to expect. Besides, right or wrong, it typically leads to more > questions, further insights, and hence it evolves, but always leading. 2) that tommy's conjectures are similar to JSH ones Not by a long shot. : not very interesting, I find some of them very interesting, often very easy to refutate, Sure, so what. As long as the underlying concept has some validity, the conjectures > can usually be revised to reach the boundary where trivially true or > trivially false suddenly becomes non-trivial. not thought upon at all, value, that's enough. and worst of all, formulated as ego-booster and not out of ny genious >interest... In my opinion, egoless posts are often very boring. Sure, tommy's ego is sometimes way out of control, but then again, we > all have flaws. In tommy's case, as I see it, there are many redeeming > virtues. tommy1729 loves math -- there's no question about that, and in my > opinion, his questions are driven not just by ego, but also by a > powerful curiosity, the lure of possible discovery, and the challenge > and joy of the hunt. I can easily relate to that. quasi- Masquer le texte des messages pr.8ec.8edents - - Afficher le texte des messages pr.8ec.8edents - Bonjour, Like most of us on this site , but : Does math love tommy1729 ...and any of us? Alain === Subject: Re: tomic polynomial: CONJECTURE 3 > tommy1729 loves math -- there's no question about that, and in my > opinion, his questions are driven not just by ego, but also by a > powerful curiosity, the lure of possible discovery, and the challenge > and joy of the hunt. I can easily relate to that. Like most of us on this site , but : >Does math love tommy1729 ...and any of us? Haha. quasi === Subject: Re: tomic polynomial: CONJECTURE 3 <4411947.1222871746307.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=Yn5cwwoAAADntcMuRwk-EwLg-DMZ_hXN rv:1.9.0.2) Gecko/2008091618 Firefox/3.0.2,gzip(gfe),gzip(gfe) > Every algebraic number is a ( finite ) sum of zero's > of tommy polynomials. I suspect your conjecture is true. Moreover, I think it would still be true if the coefficient set > {1,0,-1} was replaced by {0,1}, or for that matter {0,a} for any > nonzero a in Q. As far as proving it (or disproving it), that may be quite hard. While thinking about your conjecture, I came up with a few related > questions, but since they're not directly about tomic polynomials, > I'll pose those questions in a separate thread. well since all abelian extensions are cyclotomic the only algebraic numbers one needs to study are those that are members of nonabelian extensions do the {0,1}- or {-1,0,1}-polynomials exhaust them? i don't know i'm just stating some obvious things... -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- galathaea: prankster, fablist, magician, liar === Subject: Re: tomic polynomial: CONJECTURE 3 <4411947.1222871746307.JavaMail.jakarta@nitrogen.mathforum.org>, > Every algebraic number is a ( finite ) sum of zero's > of tommy polynomials. > tommy1729 as you can see , i am a man of my word , coming back to this concept of tomic > polynomials as mentioned in the OP. despite its been a while , i havent forgotten ^_^ I have. What is a tomic polynomial? What is a tommy polynomial? -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: tomic polynomial: CONJECTURE 3 ><4411947.1222871746307.JavaMail.jakarta@nitrogen.mathforum.org>, > Every algebraic number is a ( finite ) sum of zero's > of tommy polynomials. > tommy1729 as you can see , i am a man of my word , coming back to this concept of tomic > polynomials as mentioned in the OP. despite its been a while , i havent forgotten ^_^ I have. What is a tomic polynomial? What is a tommy polynomial? Presumably this: quasi === Subject: field generated by the set of roots of unity Here are some questions, related to and inspired by tommy1729's conjecture about tomic polynomials ... Let K be the field of algebraic numbers, and let F = Q(S) where S = {x in K | x^n = 1 for some n in N}. In other words, F is the subfield of K generated by the set of roots of unity. Questions: (1) Does F = K? (2) Is every element of F a finite linear combination, over Q, of elements of S? (3) Assuming the answer to (2) is yes, is every element of F a finite sum of elements of S? quasi === Subject: Re: field generated by the set of roots of unity posting-account=Yn5cwwoAAADntcMuRwk-EwLg-DMZ_hXN rv:1.9.0.2) Gecko/2008091618 Firefox/3.0.2,gzip(gfe),gzip(gfe) > Here are some questions, related to and inspired by tommy1729's > conjecture about tomic polynomials ... Let K be the field of algebraic numbers, and let F = Q(S) where S = {x in K | x^n = 1 for some n in N}. In other words, F is the subfield of K generated by the set of roots > of unity. Questions: (1) Does F = K? no examples can be generated from methods that solve the embedding problem in galois theory examples of low degree are difficult i don't think an extension exists for instance for the quaternion group of order 8 but i may be wrong... > (2) Is every element of F a finite linear combination, over Q, of > elements of S? by definition of generated by? maybe i'm not understanding... > (3) Assuming the answer to (2) is yes, is every element of F a finite > sum of elements of S? integers? or rationals? if i understand then the answer is no no sum would give 1/3 (w 11) (i think) -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- galathaea: prankster, fablist, magician, liar === Subject: Re: field generated by the set of roots of unity > Here are some questions, related to and inspired by > tommy1729's > conjecture about tomic polynomials ... Let K be the field of algebraic numbers, and let F > = Q(S) where S = {x in K | x^n = 1 for some n in N}. In other words, F is the subfield of K generated by > the set of roots > of unity. Questions: (1) Does F = K? no examples can be generated from methods > that solve the embedding problem > in galois theory examples of low degree are difficult i don't think an extension exists > for instance > for the quaternion group of order 8 > but i may be wrong... > The Galois extension F|Q is abelian - it is in fact the maximal abelian extension of Q. Hence F cannot contain a Galois extension having a non-abelian Galois group. > (2) Is every element of F a finite linear > combination, over Q, of > elements of S? by definition of generated by? maybe i'm not understanding... (3) Assuming the answer to (2) is yes, is every > element of F a finite > sum of elements of S? integers? > or rationals? if i understand > then the answer is no no sum would give 1/3 (w_11) > (i think) -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- > galathaea: prankster, fablist, magician, liar === Subject: Re: field generated by the set of roots of unity > Here are some questions, related to and inspired by tommy1729's > conjecture about tomic polynomials ... > Let K be the field of algebraic numbers, and let F = Q(S) where > S = {x in K | x^n = 1 for some n in N}. > In other words, F is the subfield of K generated by the set of roots > of unity. > Questions: > (1) Does F = K? no examples can be generated from methods > that solve the embedding problem > in galois theory examples of low degree are difficult i don't think an extension exists > for instance >for the quaternion group of order 8 >but i may be wrong... > (2) Is every element of F a finite linear combination, over Q, of > elements of S? by definition of generated by? Well, S generates F as a field, but question (2) asks whether S generates S as a vector space. But as Arturo Magidin outlines, the answer to question (2) is yes. >maybe i'm not understanding... > (3) Assuming the answer to (2) is yes, is every element of F a finite > sum of elements of S? integers? >or rationals? if i understand > then the answer is no no sum would give 1/3 (w_11) > (i think) Yes, of course -- you're absolutely right -- 1/3 can't be a finite sum of elements of S. Every element of S is an algebraic integer, hence so is any finite sum of elements of S. Then I guess the revised question should be ... (3) [revised]: Is every algebraic integer in F a finite sum of elements of S? quasi === Subject: Re: field generated by the set of roots of unity Here are some questions, related to and inspired > by tommy1729's > conjecture about tomic polynomials ... > Let K be the field of algebraic numbers, and let F > = Q(S) where > S = {x in K | x^n = 1 for some n in N}. > In other words, F is the subfield of K generated > by the set of roots > of unity. > Questions: > (1) Does F = K? no examples can be generated from methods > that solve the embedding problem > in galois theory examples of low degree are difficult i don't think an extension exists > for instance >for the quaternion group of order 8 >but i may be wrong... > (2) Is every element of F a finite linear > combination, over Q, of > elements of S? by definition of generated by? Well, S generates F as a field, but question (2) asks > whether S > generates S as a vector space. But as Arturo Magidin > outlines, the > answer to question (2) is yes. maybe i'm not understanding... > (3) Assuming the answer to (2) is yes, is every > element of F a finite > sum of elements of S? integers? >or rationals? if i understand > then the answer is no no sum would give 1/3 (w_11) > (i think) Yes, of course -- you're absolutely right -- 1/3 > can't be a finite sum > of elements of S. Every element of S is an algebraic integer, hence so > is any finite sum > of elements of S. Then I guess the revised question should be ... (3) [revised]: Is every algebraic integer in F a finite sum of > elements of S? quasi Yes. Every algebraic integer x is contained in a cyclotomic extension Q(z), where z denotes a primitive nth root of unity for some n. The ring of algebraic integers in Q(z) is known to be Z[z], so that x can be written as a linear combination of powers of z with integer coefficients. H === Subject: Re: field generated by the set of roots of unity days. My association with the Department is that of an alumnus. >Here are some questions, related to and inspired by tommy1729's >conjecture about tomic polynomials ... Let K be the field of algebraic numbers, and let F = Q(S) where S = {x in K | x^n = 1 for some n in N}. In other words, F is the subfield of K generated by the set of roots >of unity. Questions: (1) Does F = K? No. F is the inverse limit of the cyclotomic fields, under the usual Galois correspondence in the infinite case. >(2) Is every element of F a finite linear combination, over Q, of >elements of S? I believe so; look at Q(a), which will be contained in a finite extension, given by a finite number of elements of S, which can always be enriched to include powers of those elements since S is closed under powers to get a spanning set. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes by Bill Watterson) Arturo Magidin magidin-at-member-ams-org === Subject: Re: field generated by the set of roots of unity >Here are some questions, related to and inspired by tommy1729's >conjecture about tomic polynomials ... >Let K be the field of algebraic numbers, and let F = Q(S) where > S = {x in K | x^n = 1 for some n in N}. >In other words, F is the subfield of K generated by the set of roots >of unity. >Questions: >(1) Does F = K? No. F is the inverse limit of the cyclotomic fields, under the usual >Galois correspondence in the infinite case. Yes, I follow. But I should have seen that F can't equal K, for the simple reason that every element of F can be expressed as an arithmetic combination of radicals. >(2) Is every element of F a finite linear combination, over Q, of >elements of S? I believe so; look at Q(a), which will be contained in a finite >extension, given by a finite number of elements of S, which can always >be enriched to include powers of those elements since S is closed >under powers to get a spanning set. Yes, I see. So that takes care of questions (1) and (2). So only question (3) remains. quasi === Subject: Re: floor function ?? > > mathforum.org... > does somewhere in mathematical formulas or > functions, the expression floor(x/a)*floor(x/b) - floor(x^2/ab) occur ? tommy1729 I believe that in Knuth's book Concrete > Mathematics > some of the properties > are covered. If you are looking to bound the expression above > and > if x,a,and b are > integers, let me make the observation that: x/a >= floor(x/a) >= x/a - (a-1)/a x/b >= floor(x/b) >= x/b - (b-1)/b x**2/ab >= floor(x**2/ab) >= x**2/ab - (ab-1)/ab A little bit of number theory comes in because not > all combinations may be > possible. For example, if x is one less than a > multiple of a and also one > less than a multiple of b, this may restrict x^2 > mod ab. thank you , but i dont have access to that book :( the properties you have given are nice but trivial. anything more known ? What are you trying to do? calculus , group theory and number theory ^_^ A function that might have an interesting graph could be: w(x) = e*x - floor(e*x) - pi*x + floor(pi*x) . David Bernier > and less practical , just looking for a pleasing - > nontrivial - formula with that particular expression > , in e.g. an infinite series. > Dave A. tommy1729 === Subject: Re: floor function ?? > does somewhere in mathematical formulas or functions, the expression floor(x/a)*floor(x/b) - floor(x^2/ab) occur ? tommy1729 I believe that in Knuth's book Concrete Mathematics some of the properties are covered. If you are looking to bound the expression above and if x,a,and b are integers, let me make the observation that: x/a >= floor(x/a) >= x/a - (a-1)/a x/b >= floor(x/b) >= x/b - (b-1)/b x**2/ab >= floor(x**2/ab) >= x**2/ab - (ab-1)/ab A little bit of number theory comes in because not all combinations may be possible. For example, if x is one less than a multiple of a and also one less than a multiple of b, this may restrict x^2 mod ab. What are you trying to do? Dave A. === Subject: Re: floor function ?? > mathforum.org... > does somewhere in mathematical formulas or > functions, the expression > floor(x/a)*floor(x/b) - floor(x^2/ab) > occur ? > tommy1729 > I believe that in Knuth's book Concrete Mathematics > some of the properties > are covered. > If you are looking to bound the expression above and > if x,a,and b are > integers, let me make the observation that: > x/a >= floor(x/a) >= x/a - (a-1)/a > x/b >= floor(x/b) >= x/b - (b-1)/b > x**2/ab >= floor(x**2/ab) >= x**2/ab - (ab-1)/ab > A little bit of number theory comes in because not > all combinations may be > possible. For example, if x is one less than a > multiple of a and also one > less than a multiple of b, this may restrict x^2 > mod ab. thank you , but i dont have access to that book :( the properties you have given are nice but trivial. anything more known ? > What are you trying to do? calculus , group theory and number theory ^_^ and less practical , just looking for a pleasing - nontrivial - formula > with that particular expression , in e.g. an infinite series. Ah, OK, and I assume that a and b are constants but you'll vary x, i.e. the series would be: f(x, a, b), f(x+1, a, b), f(x+2, a, b), etc. I think the series may be fairly easy to derive. Note that whenever x is a multiple of a*b, the result will be 0 (I think). The question is what it does in between. Please feel free to write me directly at d a s h l e y _at_ g m a i l dot c o m. I'm interested in any results you derive. Dave. === Subject: Re: floor function ?? reply-type=response > mathforum.org... > does somewhere in mathematical formulas or > functions, the expression floor(x/a)*floor(x/b) - floor(x^2/ab) occur ? tommy1729 I believe that in Knuth's book Concrete Mathematics > some of the properties > are covered. If you are looking to bound the expression above and > if x,a,and b are > integers, let me make the observation that: x/a >= floor(x/a) >= x/a - (a-1)/a x/b >= floor(x/b) >= x/b - (b-1)/b x**2/ab >= floor(x**2/ab) >= x**2/ab - (ab-1)/ab A little bit of number theory comes in because not > all combinations may be > possible. For example, if x is one less than a > multiple of a and also one > less than a multiple of b, this may restrict x^2 > mod ab. > thank you , but i dont have access to that book :( > the properties you have given are nice but trivial. > anything more known ? What are you trying to do? > calculus , group theory and number theory ^_^ > and less practical , just looking for a pleasing - nontrivial - formula > with that particular expression , in e.g. an infinite series. Ah, OK, and I assume that a and b are constants but you'll vary x, i.e. > the series would be: f(x, a, b), f(x+1, a, b), f(x+2, a, b), etc. I think the series may be fairly easy to derive. Note that whenever x is > a multiple of a*b, the result will be 0 (I think). The question is what > it does in between. Please feel free to write me directly at d a s h l e y _at_ g m a i > l dot c o m. I'm interested in any results you derive. Dave. Well, here we go: #include int main(void) { int a, b, x; int term1, term2, term3; a = 2; b = 5; for (x=0; x<250; x++) { term1 = x/a; term2 = x/b; term3 = (x * x) / (a * b); printf(a:%5d b:%5d x:%5d t1:%5d t2:%5d t3:%5d r:%5dn, a, b, x, term1, term2, term3, (term1*term2)-term3); } return(0); } a: 2 b: 5 x: 0 t1: 0 t2: 0 t3: 0 r: 0 a: 2 b: 5 x: 1 t1: 0 t2: 0 t3: 0 r: 0 a: 2 b: 5 x: 2 t1: 1 t2: 0 t3: 0 r: 0 a: 2 b: 5 x: 3 t1: 1 t2: 0 t3: 0 r: 0 a: 2 b: 5 x: 4 t1: 2 t2: 0 t3: 1 r: -1 a: 2 b: 5 x: 5 t1: 2 t2: 1 t3: 2 r: 0 a: 2 b: 5 x: 6 t1: 3 t2: 1 t3: 3 r: 0 a: 2 b: 5 x: 7 t1: 3 t2: 1 t3: 4 r: -1 a: 2 b: 5 x: 8 t1: 4 t2: 1 t3: 6 r: -2 a: 2 b: 5 x: 9 t1: 4 t2: 1 t3: 8 r: -4 a: 2 b: 5 x: 10 t1: 5 t2: 2 t3: 10 r: 0 a: 2 b: 5 x: 11 t1: 5 t2: 2 t3: 12 r: -2 a: 2 b: 5 x: 12 t1: 6 t2: 2 t3: 14 r: -2 a: 2 b: 5 x: 13 t1: 6 t2: 2 t3: 16 r: -4 a: 2 b: 5 x: 14 t1: 7 t2: 2 t3: 19 r: -5 a: 2 b: 5 x: 15 t1: 7 t2: 3 t3: 22 r: -1 a: 2 b: 5 x: 16 t1: 8 t2: 3 t3: 25 r: -1 a: 2 b: 5 x: 17 t1: 8 t2: 3 t3: 28 r: -4 a: 2 b: 5 x: 18 t1: 9 t2: 3 t3: 32 r: -5 a: 2 b: 5 x: 19 t1: 9 t2: 3 t3: 36 r: -9 a: 2 b: 5 x: 20 t1: 10 t2: 4 t3: 40 r: 0 a: 2 b: 5 x: 21 t1: 10 t2: 4 t3: 44 r: -4 a: 2 b: 5 x: 22 t1: 11 t2: 4 t3: 48 r: -4 a: 2 b: 5 x: 23 t1: 11 t2: 4 t3: 52 r: -8 a: 2 b: 5 x: 24 t1: 12 t2: 4 t3: 57 r: -9 a: 2 b: 5 x: 25 t1: 12 t2: 5 t3: 62 r: -2 a: 2 b: 5 x: 26 t1: 13 t2: 5 t3: 67 r: -2 a: 2 b: 5 x: 27 t1: 13 t2: 5 t3: 72 r: -7 a: 2 b: 5 x: 28 t1: 14 t2: 5 t3: 78 r: -8 a: 2 b: 5 x: 29 t1: 14 t2: 5 t3: 84 r: -14 a: 2 b: 5 x: 30 t1: 15 t2: 6 t3: 90 r: 0 a: 2 b: 5 x: 31 t1: 15 t2: 6 t3: 96 r: -6 a: 2 b: 5 x: 32 t1: 16 t2: 6 t3: 102 r: -6 a: 2 b: 5 x: 33 t1: 16 t2: 6 t3: 108 r: -12 a: 2 b: 5 x: 34 t1: 17 t2: 6 t3: 115 r: -13 a: 2 b: 5 x: 35 t1: 17 t2: 7 t3: 122 r: -3 a: 2 b: 5 x: 36 t1: 18 t2: 7 t3: 129 r: -3 a: 2 b: 5 x: 37 t1: 18 t2: 7 t3: 136 r: -10 a: 2 b: 5 x: 38 t1: 19 t2: 7 t3: 144 r: -11 a: 2 b: 5 x: 39 t1: 19 t2: 7 t3: 152 r: -19 a: 2 b: 5 x: 40 t1: 20 t2: 8 t3: 160 r: 0 a: 2 b: 5 x: 41 t1: 20 t2: 8 t3: 168 r: -8 a: 2 b: 5 x: 42 t1: 21 t2: 8 t3: 176 r: -8 a: 2 b: 5 x: 43 t1: 21 t2: 8 t3: 184 r: -16 a: 2 b: 5 x: 44 t1: 22 t2: 8 t3: 193 r: -17 a: 2 b: 5 x: 45 t1: 22 t2: 9 t3: 202 r: -4 a: 2 b: 5 x: 46 t1: 23 t2: 9 t3: 211 r: -4 a: 2 b: 5 x: 47 t1: 23 t2: 9 t3: 220 r: -13 a: 2 b: 5 x: 48 t1: 24 t2: 9 t3: 230 r: -14 a: 2 b: 5 x: 49 t1: 24 t2: 9 t3: 240 r: -24 a: 2 b: 5 x: 50 t1: 25 t2: 10 t3: 250 r: 0 a: 2 b: 5 x: 51 t1: 25 t2: 10 t3: 260 r: -10 a: 2 b: 5 x: 52 t1: 26 t2: 10 t3: 270 r: -10 a: 2 b: 5 x: 53 t1: 26 t2: 10 t3: 280 r: -20 a: 2 b: 5 x: 54 t1: 27 t2: 10 t3: 291 r: -21 a: 2 b: 5 x: 55 t1: 27 t2: 11 t3: 302 r: -5 a: 2 b: 5 x: 56 t1: 28 t2: 11 t3: 313 r: -5 a: 2 b: 5 x: 57 t1: 28 t2: 11 t3: 324 r: -16 a: 2 b: 5 x: 58 t1: 29 t2: 11 t3: 336 r: -17 a: 2 b: 5 x: 59 t1: 29 t2: 11 t3: 348 r: -29 a: 2 b: 5 x: 60 t1: 30 t2: 12 t3: 360 r: 0 a: 2 b: 5 x: 61 t1: 30 t2: 12 t3: 372 r: -12 a: 2 b: 5 x: 62 t1: 31 t2: 12 t3: 384 r: -12 a: 2 b: 5 x: 63 t1: 31 t2: 12 t3: 396 r: -24 a: 2 b: 5 x: 64 t1: 32 t2: 12 t3: 409 r: -25 a: 2 b: 5 x: 65 t1: 32 t2: 13 t3: 422 r: -6 a: 2 b: 5 x: 66 t1: 33 t2: 13 t3: 435 r: -6 a: 2 b: 5 x: 67 t1: 33 t2: 13 t3: 448 r: -19 a: 2 b: 5 x: 68 t1: 34 t2: 13 t3: 462 r: -20 a: 2 b: 5 x: 69 t1: 34 t2: 13 t3: 476 r: -34 a: 2 b: 5 x: 70 t1: 35 t2: 14 t3: 490 r: 0 a: 2 b: 5 x: 71 t1: 35 t2: 14 t3: 504 r: -14 a: 2 b: 5 x: 72 t1: 36 t2: 14 t3: 518 r: -14 a: 2 b: 5 x: 73 t1: 36 t2: 14 t3: 532 r: -28 a: 2 b: 5 x: 74 t1: 37 t2: 14 t3: 547 r: -29 a: 2 b: 5 x: 75 t1: 37 t2: 15 t3: 562 r: -7 a: 2 b: 5 x: 76 t1: 38 t2: 15 t3: 577 r: -7 a: 2 b: 5 x: 77 t1: 38 t2: 15 t3: 592 r: -22 a: 2 b: 5 x: 78 t1: 39 t2: 15 t3: 608 r: -23 a: 2 b: 5 x: 79 t1: 39 t2: 15 t3: 624 r: -39 a: 2 b: 5 x: 80 t1: 40 t2: 16 t3: 640 r: 0 a: 2 b: 5 x: 81 t1: 40 t2: 16 t3: 656 r: -16 a: 2 b: 5 x: 82 t1: 41 t2: 16 t3: 672 r: -16 a: 2 b: 5 x: 83 t1: 41 t2: 16 t3: 688 r: -32 a: 2 b: 5 x: 84 t1: 42 t2: 16 t3: 705 r: -33 a: 2 b: 5 x: 85 t1: 42 t2: 17 t3: 722 r: -8 a: 2 b: 5 x: 86 t1: 43 t2: 17 t3: 739 r: -8 a: 2 b: 5 x: 87 t1: 43 t2: 17 t3: 756 r: -25 a: 2 b: 5 x: 88 t1: 44 t2: 17 t3: 774 r: -26 a: 2 b: 5 x: 89 t1: 44 t2: 17 t3: 792 r: -44 a: 2 b: 5 x: 90 t1: 45 t2: 18 t3: 810 r: 0 a: 2 b: 5 x: 91 t1: 45 t2: 18 t3: 828 r: -18 a: 2 b: 5 x: 92 t1: 46 t2: 18 t3: 846 r: -18 a: 2 b: 5 x: 93 t1: 46 t2: 18 t3: 864 r: -36 a: 2 b: 5 x: 94 t1: 47 t2: 18 t3: 883 r: -37 a: 2 b: 5 x: 95 t1: 47 t2: 19 t3: 902 r: -9 a: 2 b: 5 x: 96 t1: 48 t2: 19 t3: 921 r: -9 a: 2 b: 5 x: 97 t1: 48 t2: 19 t3: 940 r: -28 a: 2 b: 5 x: 98 t1: 49 t2: 19 t3: 960 r: -29 a: 2 b: 5 x: 99 t1: 49 t2: 19 t3: 980 r: -49 a: 2 b: 5 x: 100 t1: 50 t2: 20 t3: 1000 r: 0 a: 2 b: 5 x: 101 t1: 50 t2: 20 t3: 1020 r: -20 a: 2 b: 5 x: 102 t1: 51 t2: 20 t3: 1040 r: -20 a: 2 b: 5 x: 103 t1: 51 t2: 20 t3: 1060 r: -40 a: 2 b: 5 x: 104 t1: 52 t2: 20 t3: 1081 r: -41 a: 2 b: 5 x: 105 t1: 52 t2: 21 t3: 1102 r: -10 a: 2 b: 5 x: 106 t1: 53 t2: 21 t3: 1123 r: -10 a: 2 b: 5 x: 107 t1: 53 t2: 21 t3: 1144 r: -31 a: 2 b: 5 x: 108 t1: 54 t2: 21 t3: 1166 r: -32 a: 2 b: 5 x: 109 t1: 54 t2: 21 t3: 1188 r: -54 a: 2 b: 5 x: 110 t1: 55 t2: 22 t3: 1210 r: 0 a: 2 b: 5 x: 111 t1: 55 t2: 22 t3: 1232 r: -22 a: 2 b: 5 x: 112 t1: 56 t2: 22 t3: 1254 r: -22 a: 2 b: 5 x: 113 t1: 56 t2: 22 t3: 1276 r: -44 a: 2 b: 5 x: 114 t1: 57 t2: 22 t3: 1299 r: -45 a: 2 b: 5 x: 115 t1: 57 t2: 23 t3: 1322 r: -11 a: 2 b: 5 x: 116 t1: 58 t2: 23 t3: 1345 r: -11 a: 2 b: 5 x: 117 t1: 58 t2: 23 t3: 1368 r: -34 a: 2 b: 5 x: 118 t1: 59 t2: 23 t3: 1392 r: -35 a: 2 b: 5 x: 119 t1: 59 t2: 23 t3: 1416 r: -59 a: 2 b: 5 x: 120 t1: 60 t2: 24 t3: 1440 r: 0 a: 2 b: 5 x: 121 t1: 60 t2: 24 t3: 1464 r: -24 a: 2 b: 5 x: 122 t1: 61 t2: 24 t3: 1488 r: -24 a: 2 b: 5 x: 123 t1: 61 t2: 24 t3: 1512 r: -48 a: 2 b: 5 x: 124 t1: 62 t2: 24 t3: 1537 r: -49 a: 2 b: 5 x: 125 t1: 62 t2: 25 t3: 1562 r: -12 a: 2 b: 5 x: 126 t1: 63 t2: 25 t3: 1587 r: -12 a: 2 b: 5 x: 127 t1: 63 t2: 25 t3: 1612 r: -37 a: 2 b: 5 x: 128 t1: 64 t2: 25 t3: 1638 r: -38 a: 2 b: 5 x: 129 t1: 64 t2: 25 t3: 1664 r: -64 a: 2 b: 5 x: 130 t1: 65 t2: 26 t3: 1690 r: 0 a: 2 b: 5 x: 131 t1: 65 t2: 26 t3: 1716 r: -26 a: 2 b: 5 x: 132 t1: 66 t2: 26 t3: 1742 r: -26 a: 2 b: 5 x: 133 t1: 66 t2: 26 t3: 1768 r: -52 a: 2 b: 5 x: 134 t1: 67 t2: 26 t3: 1795 r: -53 a: 2 b: 5 x: 135 t1: 67 t2: 27 t3: 1822 r: -13 a: 2 b: 5 x: 136 t1: 68 t2: 27 t3: 1849 r: -13 a: 2 b: 5 x: 137 t1: 68 t2: 27 t3: 1876 r: -40 a: 2 b: 5 x: 138 t1: 69 t2: 27 t3: 1904 r: -41 a: 2 b: 5 x: 139 t1: 69 t2: 27 t3: 1932 r: -69 a: 2 b: 5 x: 140 t1: 70 t2: 28 t3: 1960 r: 0 a: 2 b: 5 x: 141 t1: 70 t2: 28 t3: 1988 r: -28 a: 2 b: 5 x: 142 t1: 71 t2: 28 t3: 2016 r: -28 a: 2 b: 5 x: 143 t1: 71 t2: 28 t3: 2044 r: -56 a: 2 b: 5 x: 144 t1: 72 t2: 28 t3: 2073 r: -57 a: 2 b: 5 x: 145 t1: 72 t2: 29 t3: 2102 r: -14 a: 2 b: 5 x: 146 t1: 73 t2: 29 t3: 2131 r: -14 a: 2 b: 5 x: 147 t1: 73 t2: 29 t3: 2160 r: -43 a: 2 b: 5 x: 148 t1: 74 t2: 29 t3: 2190 r: -44 a: 2 b: 5 x: 149 t1: 74 t2: 29 t3: 2220 r: -74 a: 2 b: 5 x: 150 t1: 75 t2: 30 t3: 2250 r: 0 a: 2 b: 5 x: 151 t1: 75 t2: 30 t3: 2280 r: -30 a: 2 b: 5 x: 152 t1: 76 t2: 30 t3: 2310 r: -30 a: 2 b: 5 x: 153 t1: 76 t2: 30 t3: 2340 r: -60 a: 2 b: 5 x: 154 t1: 77 t2: 30 t3: 2371 r: -61 a: 2 b: 5 x: 155 t1: 77 t2: 31 t3: 2402 r: -15 a: 2 b: 5 x: 156 t1: 78 t2: 31 t3: 2433 r: -15 a: 2 b: 5 x: 157 t1: 78 t2: 31 t3: 2464 r: -46 a: 2 b: 5 x: 158 t1: 79 t2: 31 t3: 2496 r: -47 a: 2 b: 5 x: 159 t1: 79 t2: 31 t3: 2528 r: -79 a: 2 b: 5 x: 160 t1: 80 t2: 32 t3: 2560 r: 0 a: 2 b: 5 x: 161 t1: 80 t2: 32 t3: 2592 r: -32 a: 2 b: 5 x: 162 t1: 81 t2: 32 t3: 2624 r: -32 a: 2 b: 5 x: 163 t1: 81 t2: 32 t3: 2656 r: -64 a: 2 b: 5 x: 164 t1: 82 t2: 32 t3: 2689 r: -65 a: 2 b: 5 x: 165 t1: 82 t2: 33 t3: 2722 r: -16 a: 2 b: 5 x: 166 t1: 83 t2: 33 t3: 2755 r: -16 a: 2 b: 5 x: 167 t1: 83 t2: 33 t3: 2788 r: -49 a: 2 b: 5 x: 168 t1: 84 t2: 33 t3: 2822 r: -50 a: 2 b: 5 x: 169 t1: 84 t2: 33 t3: 2856 r: -84 a: 2 b: 5 x: 170 t1: 85 t2: 34 t3: 2890 r: 0 a: 2 b: 5 x: 171 t1: 85 t2: 34 t3: 2924 r: -34 a: 2 b: 5 x: 172 t1: 86 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r: -76 a: 2 b: 5 x: 194 t1: 97 t2: 38 t3: 3763 r: -77 a: 2 b: 5 x: 195 t1: 97 t2: 39 t3: 3802 r: -19 a: 2 b: 5 x: 196 t1: 98 t2: 39 t3: 3841 r: -19 a: 2 b: 5 x: 197 t1: 98 t2: 39 t3: 3880 r: -58 a: 2 b: 5 x: 198 t1: 99 t2: 39 t3: 3920 r: -59 a: 2 b: 5 x: 199 t1: 99 t2: 39 t3: 3960 r: -99 a: 2 b: 5 x: 200 t1: 100 t2: 40 t3: 4000 r: 0 a: 2 b: 5 x: 201 t1: 100 t2: 40 t3: 4040 r: -40 a: 2 b: 5 x: 202 t1: 101 t2: 40 t3: 4080 r: -40 a: 2 b: 5 x: 203 t1: 101 t2: 40 t3: 4120 r: -80 a: 2 b: 5 x: 204 t1: 102 t2: 40 t3: 4161 r: -81 a: 2 b: 5 x: 205 t1: 102 t2: 41 t3: 4202 r: -20 a: 2 b: 5 x: 206 t1: 103 t2: 41 t3: 4243 r: -20 a: 2 b: 5 x: 207 t1: 103 t2: 41 t3: 4284 r: -61 a: 2 b: 5 x: 208 t1: 104 t2: 41 t3: 4326 r: -62 a: 2 b: 5 x: 209 t1: 104 t2: 41 t3: 4368 r: -104 a: 2 b: 5 x: 210 t1: 105 t2: 42 t3: 4410 r: 0 a: 2 b: 5 x: 211 t1: 105 t2: 42 t3: 4452 r: -42 a: 2 b: 5 x: 212 t1: 106 t2: 42 t3: 4494 r: -42 a: 2 b: 5 x: 213 t1: 106 t2: 42 t3: 4536 r: -84 a: 2 b: 5 x: 214 t1: 107 t2: 42 t3: 4579 r: -85 a: 2 b: 5 x: 215 t1: 107 t2: 43 t3: 4622 r: -21 a: 2 b: 5 x: 216 t1: 108 t2: 43 t3: 4665 r: -21 a: 2 b: 5 x: 217 t1: 108 t2: 43 t3: 4708 r: -64 a: 2 b: 5 x: 218 t1: 109 t2: 43 t3: 4752 r: -65 a: 2 b: 5 x: 219 t1: 109 t2: 43 t3: 4796 r: -109 a: 2 b: 5 x: 220 t1: 110 t2: 44 t3: 4840 r: 0 a: 2 b: 5 x: 221 t1: 110 t2: 44 t3: 4884 r: -44 a: 2 b: 5 x: 222 t1: 111 t2: 44 t3: 4928 r: -44 a: 2 b: 5 x: 223 t1: 111 t2: 44 t3: 4972 r: -88 a: 2 b: 5 x: 224 t1: 112 t2: 44 t3: 5017 r: -89 a: 2 b: 5 x: 225 t1: 112 t2: 45 t3: 5062 r: -22 a: 2 b: 5 x: 226 t1: 113 t2: 45 t3: 5107 r: -22 a: 2 b: 5 x: 227 t1: 113 t2: 45 t3: 5152 r: -67 a: 2 b: 5 x: 228 t1: 114 t2: 45 t3: 5198 r: -68 a: 2 b: 5 x: 229 t1: 114 t2: 45 t3: 5244 r: -114 a: 2 b: 5 x: 230 t1: 115 t2: 46 t3: 5290 r: 0 a: 2 b: 5 x: 231 t1: 115 t2: 46 t3: 5336 r: -46 a: 2 b: 5 x: 232 t1: 116 t2: 46 t3: 5382 r: -46 a: 2 b: 5 x: 233 t1: 116 t2: 46 t3: 5428 r: -92 a: 2 b: 5 x: 234 t1: 117 t2: 46 t3: 5475 r: -93 a: 2 b: 5 x: 235 t1: 117 t2: 47 t3: 5522 r: -23 a: 2 b: 5 x: 236 t1: 118 t2: 47 t3: 5569 r: -23 a: 2 b: 5 x: 237 t1: 118 t2: 47 t3: 5616 r: -70 a: 2 b: 5 x: 238 t1: 119 t2: 47 t3: 5664 r: -71 a: 2 b: 5 x: 239 t1: 119 t2: 47 t3: 5712 r: -119 a: 2 b: 5 x: 240 t1: 120 t2: 48 t3: 5760 r: 0 a: 2 b: 5 x: 241 t1: 120 t2: 48 t3: 5808 r: -48 a: 2 b: 5 x: 242 t1: 121 t2: 48 t3: 5856 r: -48 a: 2 b: 5 x: 243 t1: 121 t2: 48 t3: 5904 r: -96 a: 2 b: 5 x: 244 t1: 122 t2: 48 t3: 5953 r: -97 a: 2 b: 5 x: 245 t1: 122 t2: 49 t3: 6002 r: -24 a: 2 b: 5 x: 246 t1: 123 t2: 49 t3: 6051 r: -24 a: 2 b: 5 x: 247 t1: 123 t2: 49 t3: 6100 r: -73 a: 2 b: 5 x: 248 t1: 124 t2: 49 t3: 6150 r: -74 a: 2 b: 5 x: 249 t1: 124 t2: 49 t3: 6200 r: -124 Dave. === Subject: Re: dark geometry >consider two solid objects on a plane with a smooth edge. Solid objects? Do you mean 3-dimensional regions, or 2-dimensional? >how should one move the two objects such that they touch eachother with maximal shared lenght of their edges ? Edges? Then I guess by solid objects you mean regions in the plane. But in general, 2 such regions can't share edges. Perhaps you mean polygonal regions? But that would be trivial. >the two solid objects are given by r1 = f1(q1) and r2 = f2(q2) where q1 and q2 are the angles , f1 and f2 are taylor series ( mapping [0,2pi] -> real >= 0 ) and r1 and r2 are the distances too their relative centers. Those curves r1 = f1(q1) r2 = f2(q2) are the boundary curves for the respective regions, expressed in polar coordinates -- right? >any kind of movement consistant in 2D is allowed. of course multiple solutions might exist , even uncountable. but a single method for a working solution is all that is desired. Can you give a concrete example, using specific regions, to illustrate the issues? >conjectures : A) no efficient algoritm exist too solve this. B) the problem can be undecidable. >tommy1729 ps : analogues exist for unclosed curves , higher dimensions etc quasi === Subject: URGENT HELP NEEDED TO SOLVE STATISTIC PROBLEMS Show all steps and give details interpretation 1. A beer distributor claims that a new display, featuring a life-size picture of a well-known rock singer will increase product sales in supermarket by an average of 50 cases in a week. For a random sample of 20 high-volume liquor outlets, the average sales increase was 41.3 cases, and the sample standard deviation was 12.2 cases. Test at the 5% level the null hypothesis that the population mean sales increase is at least 50 cases, stating any assumptions you make. 2. Plastic sheets produced by a machine are periodically monitored for possible fluctuations in thickness. If the true variance in thickness exceeds 2.25 square millimeters, there is cause for concern about product quality. Thickness measurements for a random sample of 10 sheets produced in a particular shift were taken, giving the following results (in millimeters): 226 226 232 227 225 228 225 228 229 230 a) Find the sample variance. b) Test at the 5% significance level the null hypothesis that the population variance is at most 2.25. 3. Some nurses in County Public Health conducted a survey of women who had received inadequate prenatal care. They used information from birth certificate to select mothers for the survey. The mothers that were selected were divided into two groups: 14 mothers who said they had 5 or fewer prenatal visits and 14 mothers who said they had 6 or more prenatal visits. Let X and Y equal the respective birthweights of the babies from these two sets of mothers and assume that the distribution of X is N ([Micro]x, mu square) and the distribution of Y is N([Micro]y, & mu square). (a) Define the test statistic and critical region for testing Ho : [Micro]x - [Micro]y = 0 Against H1 : [Micro]x - [Micro]y < 0. Let alpha = 0.05 (b) Given that the observations of X were 49 108 110 82 93 114 134 114 96 52 101 114 120 116 And the observations of Y were 133 108 93 119 119 98 106 131 87 153 116 129 97 110 Calculate the value of the test statistic and state your conclusion. (c) Approximate the p-value. (d) Construct a histogram on the same figure for these two sets of data. Do the histogram support your conclusion? (e) Test whether the assumption of equal variances is valid. Let alpha = 0.05 === Subject: Re: URGENT HELP NEEDED TO SOLVE STATISTIC PROBLEMS posting-account=K5WE3woAAAAXArsybjkbN6LjMxWdHtbX Gecko/20080829 Firefox/2.0.0.17,gzip(gfe),gzip(gfe) > Show all steps and give details interpretation 1. A beer distributor claims that a new display, featuring a life-size picture of a well-known rock singer will increase product sales in supermarket by an average of 50 cases in a week. For a random sample of 20 high-volume liquor outlets, the average sales increase was 41.3 cases, and the sample standard deviation was 12.2 cases. Test at the 5% level the null hypothesis that the population mean sales increase is at least 50 cases, stating any assumptions you make. Step 1. Go to your course notes and/or your textbook and look up similar problems. Step 2. Translate the conditions of this new problem into the language and notation of the previous analysis. Step 3. Carry out the numerical computations, and consult the relevant statistical tables or use the relevant menu items in a spreadsheet to complete the analysis. 2. Plastic sheets produced by a machine are periodically monitored for possible fluctuations in thickness. If the true variance in thickness exceeds 2.25 square millimeters, there is cause for concern about product quality. Thickness measurements for a random sample of 10 sheets produced in a particular shift were taken, giving the following results (in millimeters): 226 226 232 227 225 > 228 225 228 229 230 a) Find the sample variance. Step 1. Look up the formula in the textbook or course notes. Step 2. Apply the formula. b) Test at the 5% significance level the null hypothesis that the population variance is at most 2.25. 3. Some nurses in County Public Health conducted a survey of women who had received inadequate prenatal care. They used information from birth certificate to select mothers for the survey. The mothers that were selected were divided into two groups: 14 mothers who said they had 5 or fewer prenatal visits and 14 mothers who said they had 6 or more prenatal visits. Let X and Y equal the respective birthweights of the babies from these two sets of mothers and assume that the distribution of X is N ([Micro]x, mu square) and the distribution of Y is N([Micro]y, & mu square). (a) Define the test statistic and critical region for testing Ho : [Micro]x - [Micro]y = 0 > Against H1 : [Micro]x - [Micro]y < 0. Let alpha = 0.05 (b) Given that the observations of X were 49 108 110 82 93 114 134 > 114 96 52 101 114 120 116 And the observations of Y were 133 108 93 119 119 98 106 > 131 87 153 116 129 97 110 Calculate the value of the test statistic and state your conclusion. (c) Approximate the p-value. > (d) Construct a histogram on the same figure for these two sets of data. Do the histogram support your conclusion? > (e) Test whether the assumption of equal variances is valid. Let alpha = 0.05 Similar advice applies to all the other cases above. I have already passed all my courses, so don't need to practice any more by doing stats homework. However, IF I were taking a course, I would certainly want to do the work myself (perhaps with some help) in order to cement my understanding of the material, and to get ready for writing the exams. But that's just me. R.G. Vickson === Subject: > A group of order 15 is cyclic To prove: a group G of order 15 is cyclic. The proof I am studying runs as follows: Since G contains an element x of order 3 and an element y of order 5, the map Z/3 x Z/5 --> G defined by (i, j) --> x^i y^j is an isomorphism, provided xy = yx. It's enough to show that Z(G) is nontrivial, for then G/Z(G) is cyclic, hence G is abelian. If Z(G) = 1, then G must contain 1 conjugacy class of size 5, and 3 of size 4. Set z = xyx^{-1} and w = x^{-1} y x. Assume first z = y^j; then y = y^{j^3} so that j^3 = 1 mod 5 ==> j = 1 (mod 5) so that z = y ==> xy = yx and we're done. Otherwise, assume z =/= y^j for any j; then generates a subgroup distinct from , as does . Furthermore, and are also distinct. Do we really have 12 elements of order 5, namely z^j, w^j and y^j, as j runs from 1 to 4? If so, how does this contradict the class equation (whenever Z(G) is assumed to be trivial)? === Subject: Re: > A group of order 15 is cyclic If Z(G) = 1, then G must contain 1 conjugacy class of > size 5, and ... 3 of size 3, of course. === Subject: Re: Yet another disproof of the diagonal argument 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 1 Oct, 18:07, Achava Nakhash, the Loving Snake WHERE IS THE BLOODY FLAW IN THE PROOF I HAVE GIVEN! You haven't given a proof of anything. Quite the opposite: you just confirm that noone can show nor has shown a single flaw in my almost trivial argument; all objections amount to zero math and straight denial on the lines of: how you dare criticize such a wanderful amazing magic sacred eternal true thing as Cantor's, and where is your argument at all (plus all the usual ad hominem, etc. etc.). So for me the question is surely settled, and for the rest I simply won't find here and in these conditions any better math than *I* have already provided: actually any math at all. So, for what I am concerned, the discussion is over, although honest comments and questions are still welcome: I say I got *no* objections, and I say all Cantor's related mythology is gonna stay *out* of any framework of mine, though I don't say my argument is perfect in itself, that's *not even the point*. === Subject: Re: Yet another disproof of the diagonal argument > On 1 Oct, 18:07, Achava Nakhash, the Loving Snake WHERE IS THE BLOODY FLAW IN THE PROOF I HAVE GIVEN! You haven't given a proof of anything. Quite the opposite: you just confirm that noone can show nor has shown > a single flaw in my almost trivial argument; all objections amount to > zero math and straight denial on the lines of: how you dare criticize > such a wanderful amazing magic sacred eternal true thing as Cantor's, > and where is your argument at all (plus all the usual ad hominem, etc. > etc.). So for me the question is surely settled, and for the rest I > simply won't find here and in these conditions any better math than > *I* have already provided: actually any math at all. So, for what I am concerned, the discussion is over, although honest > comments and questions are still welcome: I say I got *no* objections, > and I say all Cantor's related mythology is gonna stay *out* of any > framework of mine, though I don't say my argument is perfect in > itself, that's *not even the point*. Julio's absence from further discussions will raise the level of those discussions measurably. === Subject: Re: Yet another disproof of the diagonal argument 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) > Julio's absence from further discussions will raise the level of those > discussions measurably. No absence from anything, Moron. Now that I've grasped the little math language needed to understand what an amount of bull YOU produce, the fun has just begun. === Subject: Re: Yet another disproof of the diagonal argument Julio's absence from further discussions will raise the level of those > discussions measurably. No absence from anything, Moron. Now that I've grasped the little math language needed to understand > what an amount of bull YOU produce, the fun has just begun. Unless your posts begin to show some mathematics, they will soon become too boring to attract any further attention. === Subject: Re: Yet another disproof of the diagonal argument 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) > Julio's absence from further discussions will raise the level of those > discussions measurably. No absence from anything, Moron. Now that I've grasped the little math language needed to understand > what an amount of bull YOU produce, the fun has just begun. Unless your posts begin to show some mathematics, they will soon become > too boring to attract any further attention. Moron, and liar. You and math have nothing to do together. === Subject: Re: Yet another disproof of the diagonal argument Cantor starts with a supposed complete list of countably many > infinite sequences that includes all infinite sequences of 0's and > 1's. Actually, that is not quite how Cantor's original diagonal proof was > stated, though quite similar. Cantor's Original Diagonal Proof dealt with infinite strings of only the > two characters 'm' and 'w'. Cantor's Original Diagonal Proof made no assumptions about the > completeness of the list, nor about anything except that it was a > counting of the strings listed in it. It as a direct proof, not an > indirect proof. Then, when Cantor's construction produced a string not in the list (and > could easily have been extended to produce a countable infinity of such > non-member strings) he was able to conclude that every such list omitted > strings. > I stand corrected. I was taking Ludovico Van's word for how Cantor's > actual diagonal proof went. I have never actually looked at Cantor's > work or seen it quoted. The diagonal technique is nice, and it is > useful in other contexts (this is for the general audience). I > generally prefer other proofs of the uncountability of the reals, such > as the one I posted earlier, since the way I was taught the diagonal > proof you had to mess with decimal representations and worry about > sequences that are all nines from some point on. This can be dealt > with easily enough, but it is not pretty. Achava There is a lot of mythology abounding about Cantor and his work, and it is easy enough to get details wrong. === Subject: Re: Yet another disproof of the diagonal argument > On 1 Oct, 14:12, Mariano Su.87rez-Alvarez WHERE IS THE BLOODY FLAW IN THE PROOF I HAVE GIVEN! As pretty much every one has tried to explain to you, > leading to a proof. It falls very much in the > not-even-wrong category, so no one is going to be > able to point to a specific flaw in it, to a specific > step at which things go bad: as given, it does not > is very much as asking where the arithmetic error > occurs in the following: hfdjshjuwevjvjvjkhjshjahfadjfnvnckjsfd Yeah, sure, hope the spell is gonna work. Its spelling is on a par with your mathematics. === Subject: Re: Yet another disproof of the diagonal argument > ... > All that has zero relevance to the OP always magically becomes the > main objection in every discussion. But of course the bubbling is mine, eh?? It is due to the aggressive and insulting way you respond to objections. No, it's due to the LIES, BOYCOTTING and AD HOMINEM one consistently > gets from the morons like YOU. You are certainly the one who started the ad hominem arguments, which a backtracing of these threads will easily establish. And it is a truism, that the one that starts ad hominem arguments is the one who has no better arguments. === Subject: Re: Yet another disproof of the diagonal argument > WHERE IS THE BLOODY FLAW IN THE PROOF I HAVE GIVEN! Among other things, it requires something true for finite sets to be true for an infinite set without any valid justification. === Subject: Re: Yet another disproof of the diagonal argument >On 1 Oct, 06:07, Achava Nakhash, the Loving Snake >You claim to have an undergraduate math degree >No, I've never claimed anything like that. My apologies for getting that wrong. > The conclusion I drew from that is therefore invalid. Rather than invalid, shouldn't we say that the argument you made was unsound? Yours, Jim Burns === Subject: Re: Yet another disproof of the diagonal argument <48E3BC20.5020902@osu.edu> posting-account=Z3AipgkAAABkoMfyNwddSxsYhXHi5CDt CLR 1.1.4322; InfoPath.1; .NET CLR 2.0.50727; .NET CLR 3.0.04506.648; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) >On 1 Oct, 06:07, Achava Nakhash, the Loving Snake You claim to have an undergraduate math degree >No, I've never claimed anything like that. My apologies for getting that wrong. > The conclusion I drew from that is therefore invalid. Rather than invalid, shouldn't we say that > the argument you made was unsound? Yours, > Jim Burns Drawing an incorrect conclusion from a correct premise is unsound as it is illogical. Drawing a logically correct conclusion from an invalid premise renders the conclusion invalid. At least that is my understanding of the correct use of English. Achava === Subject: Re: Yet another disproof of the diagonal argument >On 1 Oct, 06:07, Achava Nakhash, the Loving Snake >You claim to have an undergraduate math degree >No, I've never claimed anything like that. >My apologies for getting that wrong. >The conclusion I drew from that is therefore invalid. >Rather than invalid, shouldn't we say that >the argument you made was unsound? > Drawing an incorrect conclusion from a correct premise is > unsound as it is illogical. Drawing a logically correct > conclusion from an invalid premise renders the conclusion > invalid. At least that is my understanding of the correct > use of English. I've only acquired my own understanding of the meaning of valid and sound informally, basically by reading in places such as sci.math and sci.logic. Therefore, I feel the need to refer to a better authority than my own self: http://en.wikipedia.org/wiki/Soundness#Sound_arguments : An argument is sound if and only if : 1. The argument is valid. : 2. All of its premises are true. http://en.wikipedia.org/wiki/Validity_(logic)#Validity_of_arguments ; An argument is deductively valid if, ; whenever all premises are true, ; the conclusion is also necessarily true. It seems to me that, since you conceded that one of your premises may not be true (that LudovicoVan had an undergraduate math degree), your argument would properly be called unsound. It could conceivably be invalid as well (for example, if your argument were applied to someone who truly did have an undergraduate degree in math but who, nonetheless, had no experience writing proofs). (The truth is that I was trying to be a little bit funny -- in natural discourse, the distinction does not seem to be drawn as sharply -- at least, in the conversations I've seen.) Jim Burns === Subject: Re: Yet another disproof of the diagonal argument days. My association with the Department is that of an alumnus. >On 1 Oct, 06:07, Achava Nakhash, the Loving Snake >You claim to have an undergraduate math degree >No, I've never claimed anything like that. > My apologies for getting that wrong. > The conclusion I drew from that is therefore invalid. > Rather than invalid, shouldn't we say that > the argument you made was unsound? Drawing an incorrect conclusion from a correct premise is unsound as >it is illogical. Drawing a logically correct conclusion from an >invalid premise renders the conclusion invalid. At least that is my >understanding of the correct use of English. Sound and unsound refer to the truth or lack thereof of the premises. Valid and invalid refers to whether the premises logically imply the conclusion. An unsound, valid argument is one that begins with false premises, but in which the premises do logically imply the conclusion. That conclusion may be a true or a false statement, of course. The distinction, however, is somewhat technical and colloquially many people use valid and invalid to refer both to the truth and falsity of a given proposition, and to whether the premises do or do not logically imply the conclusions in an argument. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes by Bill Watterson) Arturo Magidin magidin-at-member-ams-org === Subject: Re: Yet another disproof of the diagonal argument Dik T. Winter a .8ecrit : > ... > All that has zero relevance to the OP always magically becomes the > main objection in every discussion. But of course the bubbling is mine, eh?? It is due to the aggressive and insulting way you respond to objections. Stop feeding the troll === Subject: Re: Yet another disproof of the diagonal argument > where the arithmetic error > occurs in the following: hfdjshjuwevjvjvjkhjshjahfadjfnvnckjsfd ^^^ I thought it was quite obvious... ;-) === Subject: Re: sup C= sup A + sup B Let A and B be subsets of R. Let C={a + b, a in A, b in B}. Show > that sup C= sup A + sup B. > sup A+B = sup A + sup B (R,+) is an ordered group with the axiom a <= b ==> a + x <= b + y. > (R,*) is also an ordered group with the same axiom > . . a <= b ==> ax + bx. No, but true for the multiplicative group of _positive_ real numbers. > (R+)^* as I used below. > Thus the proof of one is the proof of the other. Show sup A+B <= sup A + sup B > Show for all b in B, b + sup A <= sup A+B > b <= sup A+B - sup A > Thus sup B <= sup A+B - sup A > . . sup A + sup B <= sup A+B And the reverse inequality? > It needs detailing? for all a in A, b in B, a + b <= sup A + sup B If sup A, sup B exist then (A <= c for c uppper bound A) A + B <= sup A + sup B. if A + B <= c, then for all a in A, a + B <= c B <= c - a; sup B <= c - a; a <= c - sup B; sup A <= c - sup B sup A + sup B <= c Hence sup A+B exists and equals sup A + sup B. Notice proof works in all (partially) ordered groups. If sup A+B exists, then does sup A and sup B exist? Of course, in a bounded complete group they do. Yet the question remains in an ordered group such as Q or QxQ, does sup A+B exist imply sup A, sup B exist? No, for Z[1/3], let A = B = { x in Z[1/3] | 2x < 1 } or for (Q+)^* let A = B = { q | q^2 <= 2 }. Presuming OP is with the reals, his question lacks in clarifying that A and B are bounded. So the correct and complete theorem for all ordered groups G, is sup A, sup B in G ==> sup A+B in G, sup A+B = sup A + sup B and of course the order dual inf A, inf B in G ==> inf A+B in G, inf A+B = inf A + inf B > Indeed, the results are the same be the group R,Q, Z, (R+)^* or (Q+)^*. No. sup is not defined within Q - here you need kind of completion in > the general case (see other thread on MacNeille completion. > sup{ x in Q | x < 2 } = 2 in Q > Notice that Z doesn't have half elements for many of the elements. > (Q+)^* is the multiplicative group of positive rationals where > 'half' elements are square roots and theorem is stated the same, ie > sup A*B = sup A * sup B Sigh, you are leaving the structure here as well. E.g. sqrt(2) is not > rational. > I haven't at all. > Basically you need to be very careful whether your structures are closed > under the operations involved. Otherwise the technique by analysing the > logic of the proof will not work in general. > All that's required is sup A, sup B exist in an ordered group. No more. Riddle of the day. What is the bound of a bounded group? === Subject: Re: Subgroups of direct products <24648942.1222739471553.JavaMail.jakarta@nitrogen.mathforum.org> the subgoups mZ x nZ for all n,m in Z0? If still no, have you counter example. The one generated by (1,1) and (-1,1), for example. > (n,m) -> n(1,1) + m(-1,1) = (n - m, n + m). Ok. === Subject: Re: -- Wrong limits do not commute On Sep 30, 10:37 am, Han de Bruijn No dream, after you take the limit the x is no longer there. >I wouldn't put it like this. It sounds as if taking a limit is an >action, and I think that's part of Han's confusion. >Hey ! We're getting somewhere ! Substitute convinction for confusion >and we're almost there ! Does constructivism ring a bell ? >It doesn't appear relevant to anything I've said. I don't know of any >constructivist who would doubt that lim_x f(x,y) is a unary function >of y (if it exists), or that if > lim_x f(x,y) = g(y), >then lim_y lim_x f(x,y) = lim_y g(y). >In fact, you're the only constructivist I know who thinks that >constructivism is a magic wand that makes whatever he says correct. >Please read, I mean: _read_, my last posting in response to Virgil. Right. In that posting, you express surprise that lim_y lim_x f(x,y) != lim_y f(X,y), where X is big. So, you're mighty confused about iterated limits, but > I don't see any reason to blame this confusion on the poor > constructivists. The fact is that lim_y lim_x f(x,y) != lim_y f(X,y) because lim_x f(x,y) != f(X,y), not even when X is big. Yes. I've confessed that, very much explicitly, in the posting. So again, Jesse: > Please read, I mean: _read_, my last posting in response to Virgil. Suppose that you are not interested in numerical approximations, right ? As a philosopher, you cannot be blamed. But I can assure you that the rest of the world _is_ interested in numerical approximations. Han de Bruijn === Subject: Re: -- Wrong limits do not commute > ... > And (2x - 1)/(x - 1) =?= oo/oo: unstable. With these heuritics there are not many stable limits. OOPS ! But ah, there is only _one_ variable up there .. Yes, so what? But let me introduce another variable: > (2x - y)/(x - y) =?= oo/oo: unstable? Okay ! I should never have done this ! Instead, let x->oo and y->oo . Thinking that (x/y) means anything then gives, heuristically: x^2 - y^2 (x/y)^2 - 1 --------- -> ----------- : unstable x^2 + y^2 (x/y)^2 + 1 x.y (x/y) --------- -> ----------- : unstable x^2 + y^2 (x/y)^2 + 1 x.y (x/y) --------- -> ----------- : unstable x^2 - y^2 (x/y)^2 - 1 (1 - 1/y)^x -> [ (1 - 1/y)^y ]^(x/y) -> e^(-x/y) : unstable x + y ----- -> 1/y + 1/x = 0 : stable x.y Han de Bruijn === Subject: Re: -- Wrong limits do not commute You would be completely wrong. An example of limits >that do not commute was given. They are neither wrong, >nor numerically unstable. You are confusing the >related concepts of continuity and stability. >Nope. You gave a wrong example. Why is it wrong? You admit that the limits are meaningful. You admit that the limits do not commute. Your example with the chip and the transistors resulted in a limit approaching infinity. Blame it on HdB that he cannot handle such limits. If you nevertheless keep insisting on this, then this debate between the two of us is not going to reveal anything interesting anymore. So I'm going to finish this part of the thread, here and *NOW*. Han de Bruijn === Subject: Re: -- Wrong limits do not commute ... > Time to come to a Conclusion. And _this_ one IMO cannot be denied. There is a HUGE difference between ALMOST reaching the inner limit > and then proceeding with the outer limit, in comparison with first > completing the inner limit, and then proceeding with the outer one. Yeah, right. There is indeed a HUGE difference between not taking the > limit and taking the limit. No surprise here. Yes, I _know_ that the above procedure should be dismissed as illegal, > but it reveals an _important_ point, which I'm trying to communicate > with you, all the time, all the thread. What if I say the following: Limits that do not commute are NUMERICALLY UNSTABLE. See ? No offending terminology anymore. Many limits are numerically unstable. And that is not only iterated > limits. Sure. But the latter are it in some special way. Because it's not so > much the outcome but rather the _process_ that leads to instability. > It's difficult to find anything on the internet that is resemblant to > my notion of an unstable limit. Could'nt find much: http://arxiv.org/PS_cache/math/pdf/0311/0311448v1.pdf : page 4 bottom ? Han de Bruijn It is physics, so is irrelevant. === Subject: Re: -- Wrong limits do not commute > Tip, hint: lim f(x) does not mean f(large number) > x->oo nor does it mean f(oo), however it does not depend on the valuse of x. It may be illegal, but the notation f(oo) gives us a very easy way to > establish whether such limits are numerically unstable or not. Thinking > that (oo/oo) means anything gives, heuristically: And (2x - 1)/(x - 1) =?= oo/oo: unstable. With these heuritics there are not many stable limits. OOPS ! But ah, there is only _one_ variable up there .. Then look at [(2x - 1)(2y - 1)]/[(x - 1)(y - 1)] =?= oo/oo === Subject: Re: -- Wrong limits do not commute You would be completely wrong. An example of limits > that do not commute was given. They are neither wrong, > nor numerically unstable. You are confusing the > related concepts of continuity and stability. Nope. You gave a wrong example. It seems that everything that HdB does not entirely agree with i WRONG to him. Then by symmetry, everything of his weird theories that we do not agree with is equally WRONG to all of us. And since nothing as sane as logic is allowed to be a guide in HdB's judgments, our majority rules! === Subject: Re: -- Wrong limits do not commute >On Sep 30, 10:37 am, Han de Bruijn No dream, after you take the limit the x is no longer there. >I wouldn't put it like this. It sounds as if taking a limit is an >action, and I think that's part of Han's confusion. >Hey ! We're getting somewhere ! Substitute convinction for confusion >and we're almost there ! Does constructivism ring a bell ? It doesn't appear relevant to anything I've said. I don't know of any > constructivist who would doubt that lim_x f(x,y) is a unary function > of y (if it exists), or that if lim_x f(x,y) = g(y), then lim_y lim_x f(x,y) = lim_y g(y). In fact, you're the only constructivist I know who thinks that > constructivism is a magic wand that makes whatever he says correct. Please read, I mean: _read_, my last posting in response to Virgil. Don't bother. If is worth saying at all, HdB will say it again. And again. And again. And may do so anyway. === Subject: Re: -- Wrong limits do not commute > On Sep 30, 10:37 am, Han de Bruijn No dream, after you take the limit the x is no longer there. >I wouldn't put it like this. It sounds as if taking a limit is an >action, and I think that's part of Han's confusion. >Hey ! We're getting somewhere ! Substitute convinction for confusion >and we're almost there ! Does constructivism ring a bell ? It doesn't appear relevant to anything I've said. I don't know of any >constructivist who would doubt that lim_x f(x,y) is a unary function >of y (if it exists), or that if lim_x f(x,y) = g(y), then lim_y lim_x f(x,y) = lim_y g(y). In fact, you're the only constructivist I know who thinks that >constructivism is a magic wand that makes whatever he says correct. >Please read, I mean: _read_, my last posting in response to Virgil. Don't bother. If is worth saying at all, HdB will say it again. And again. And again. And may do so anyway. No. As far as HdB is concerned, this thread will come to an end, soon. Han de Bruijn === Subject: Re: -- Wrong limits do not commute Try |x| > 2*10^(10^1000)/sqrt(10^(-10^1000)) > Yes, it is. /I/ give /you/ y = 10^(10^1000) and eps = 10^(-10^1000). > Deal with it. These values as proposed by Horand Gassmann are fine, because they are > _abstract_. Meaning that they won't fit into any computer, not even if > the computing power of the whole universe could be claimed. By fitting > into I mean: as a sort of floating point representation. I'm going to > reveal another secret behind this thread now, at hand of the Gassmann > numbers. It will give rise to a new terminology as well, which is much > less offending than, say, blind limits. Let's first repeat our iterated non commuting limits, with the outcome > according to the mainstreamers: x^2 - y^2 > lim lim --------- = + 1 > y->oo x->oo x^2 + y^2 x^2 - y^2 > lim lim --------- = - 1 > x->oo y->oo x^2 + y^2 The inner limit of the first iterated limit is defined as follows: | x^2 - y^2 | __________ > | --------- - 1 | < eps if x > y.M where M = V 2/eps - 1 > | x^2 + y^2 | There is nothing in the definition of limits that requires anyone to use M = sqrt(2/eps - 1). One can just as easily use 2/sqrt(eps) or 100/sqrt(eps) > The inner limit of the second iterated limit is defined as follows: | x^2 - y^2 | __________ > | --------- + 1 | < eps if y > x.M where M = V 2/eps - 1 > | x^2 + y^2 | Then we have the stupid limit game, where Horand insisted on the above > values for eps and y , so we're talking about the _first_ iterated > limit (where y is fixed so to speak). So far so good. But we are going to do something quite unexpected now. > We are going to insert a HALT here. Yes ! Enough is enough ! We have > travelled to the other end of the universe in the x-direction and we > have obtained an error eps which is much, much smaller than even the > Planck length. So we have decided to STOP. A very reasonable decision. > The value of x where we stop is called X . We can safely substitute > here, for example, the value suggested by Virgil: > ___ > |x| > 2*|y|/sqrt(eps) -> X = 2.y / Veps Okay. Now that we have arrived at this incredible and immensely large > value of x , the outcome of the inner limit has become so astonishing > accurate that no digital machine in the universe can possibly compute > the difference between the approximation and the exact value. Thus at > this point, we decide to process the outer limit. Which seems to be a > reasonable decision. Your decisions may seem reasonable to you, but need not be binding on anyone else. Mathematicians do not need to allow their imaginations to be bounded by the grubby physical considerations of physicists. Therefore we start to increase the values for y > and the stupid limit game .. Oh no ! Don't let start it all again ! That HdB cannot out think Time to come to a Conclusion. And _this_ one IMO cannot be denied. There is a HUGE difference between ALMOST reaching the inner limit > and then proceeding with the outer limit, in comparison with first > completing the inner limit, and then proceeding with the outer one. However huge that difference may be, it is irrelevant in evaluating mathemtical iterated limits, in which each limit must be evaluated exactly, if possible, before the next is initiated. Mathematicians are not responsible for the peculiarities of physicists. Yes, I _know_ that the above procedure should be dismissed as illegal, > but it reveals an _important_ point, which I'm trying to communicate > with you, all the time, all the thread. What if I say the following: Limits that do not commute are NUMERICALLY UNSTABLE. See ? No offending terminology anymore. Though it is a well known fact > that numerically unstable results _are_ virtually worthless in Applied > Mathematics. We're back home. So that doing physics with valid pure mathematics must be done with some care if it is to do what the physicists want it to do. But that is only a problem for physicists, not for mathematicians nor mathematics. HdB wants to limit mathematics to being an indentured servant to physics, but that is not how mathematics works. If mathematics is not free from such indenture, it is not mathematics at all but just physics. === Subject: Re: -- Wrong limits do not commute [ .. stuff where HdB more or less agrees with snipped .. ] [ .. stuff that's typical for Virgil snipped as well .. ] > So that doing physics with valid pure mathematics must be done with some > care if it is to do what the physicists want it to do. It's Applied Mathematics I've been talking about, not just physics. > But that is only a problem for physicists, not for mathematicians nor > mathematics. HdB wants to limit mathematics to being an indentured servant to > physics, but that is not how mathematics works. If mathematics is not free from such indenture, it is not mathematics at > all but just physics. As a collegue of mine said yesterday: mathematics is _far more general_ than will ever be needed in practice. I'm not denying pure mathematics it's existence. But perhaps it would be useful to know what mathematics is relevant for the Applied and what not. A conclusion from this thread could be that there is not yet a well defined answer to this question. And perhaps there will never be. Good for you. Han de Bruijn === Subject: Re: -- Wrong limits do not commute And _you_ have never even run the risk of trying to produce one flawless >mathematical formula in this group. You're just good in faire blah blah. An astute description of my shortcomings no doubt. What you have in >mind with producing flawless mathematical formulas is obscure to me, >though. >Something like this ? The limit of the second derivative of the function >f(x) = x^4.sin(1/x) for x -> 0 ? >Even better, the limiting process as a whole can be written as just ONE, >non-iterated limit, namely: f(h) - 2.f(0) + f(-h) > lim --------------------- where we use a very well known numerical > h->0 h^2 approximation for the second derivative. Resulting in: lim h^2.[ sin(1/h) + sin(-1/h) ] = 0 . > h->0 >I run the risk. You don't. You regularly fail, Aatu doesn't. Of course Aatu doesn't. How can you fail if you don't take risks ? Han de Bruijn The difference is a matter of whether one is pontificating about what one is ignorant of. === Subject: Re: -- Wrong limits do not commute >On Sep 30, 10:37 am, Han de Bruijn No dream, after you take the limit the x is no longer there. I wouldn't put it like this. It sounds as if taking a limit is an > action, and I think that's part of Han's confusion. Hey ! We're getting somewhere ! Substitute convinction for confusion > and we're almost there ! Does constructivism ring a bell ? The point is much simpler: if > lim f(x,y) = g(y), then the two terms denote the same function. Since > the right hand side is clearly a function of y (and not x), the same > is true for the left hand side. It's known that constructivist mathematics maybe a bit more complicated > than formalist mathematics. Please hang on. As WE are not constructionists, we are not constrained by the limitations they choose to impose on themselves. === Subject: Re: -- Wrong limits do not commute On Sep 30, 10:37 am, Han de Bruijn ... > Therefore, due to the existence of the inner limit, there is > an > upper bound on the values of y , meaning that it cannot > assume > all values. > Eh? You lost me here. The outer limit is over a function that > does > not contain x. (The inner limit removes x.) > Yes, but it does not remove the _condition_ on x and y . > How can there be a condition on x and y if there is no x? >step there is no longer an x. >Keep dreaming about the x that has disappeared suddenly, mysteriously. >Tip, hint: have completed infinities even infected limits ? I hope not. You really haven't a clue what the definition of limit means. [ .. irrelevant wisdom snipped .. ] Definition, definition .. That's not the point. _Meaning_ is the point. Han de Bruijn In math, meanings are largely delimited by definitions, and are almost never coherently explained without reference to some definitions. === Subject: Re: -- Wrong limits do not commute On Sep 30, 10:37 am, Han de Bruijn Keep dreaming about the x that has disappeared suddenly, mysteriously. No dream, after you take the limit the x is no longer there. Tip, hint: have completed infinities even infected limits ? I hope not. Tip, hint: The process of taking a limit does not involve > completed infinities (whatever they are). You're wrong. And evidence proving this shall be published very soon. Please hang on. Han de Bruijn Pie in the Sky again? === Subject: Re: -- Wrong limits do not commute On Sep 30, 10:37 am, Han de Bruijn Keep dreaming about the x that has disappeared suddenly, mysteriously. No dream, after you take the limit the x is no longer there. >Tip, hint: have completed infinities even infected limits ? I hope not. Tip, hint: The process of taking a limit does not involve >completed infinities (whatever they are). >You're wrong. And evidence proving this shall be published very soon. >Please hang on. Pie in the Sky again? The _inner_ limit in my published very soon has been truncated to an incompleted infinity, meaning that x is a finite, very large, but doesn't matter how large, number. With other words: potential infinite. What is actually demonstrated is that resoning with _potential_ infinite gives a HUGE difference when compared with common reasoning; where it is assumed that the infinity of the inner limit is _completed_. Of course, you would say .. Han de Bruijn === Subject: Re: -- Wrong limits do not commute On Sep 30, 10:37 am, Han de Bruijn Keep dreaming about the x that has disappeared suddenly, mysteriously. No dream, after you take the limit the x is no longer there. Tip, hint: have completed infinities even infected limits ? I hope not. Tip, hint: The process of taking a limit does not involve > completed infinities (whatever they are). You're wrong. And evidence proving this shall be published very soon. Please hang on. Han de Bruijn In Robinson's non-standard analysis, or equivalent, one properly may base limiting processes on completed infinities infinites and infinitesimals, but not in any form of standard analysis. And the only wrong limits, either simple, multiple or iterated are the ones which according to the standard mathematical definitions do not exist. What may seem wrong about mathematics to a physicist is mathematically irrelevant, unless he/she can express that wrongness in mathematically valid terms. === Subject: solution manual of basic principle and calculation in chemical engineering by David Himmelblau posting-account=I1XvfQoAAAC1w4134GSshrRouS2lTkDb SV1),gzip(gfe),gzip(gfe) plz help me where i can get the solution manual of basic principle and calculation in chemical engineering by David Himmelblau === Subject: Re: CERN concern But if CERN can > produce non-decaying micro black holes, then > cosmic ray events should produce some too. > === Subject: Re: Embedding orders into complete lattices Let S be a (partially) ordered set. S can be ordered embeded into > a complete subset of P(S) in such a way that meets and finite > joins are preserved. The subset of P(S) is the collection of up directed, down (ie > lower) sets of S. The mapping is d(p) = { x | x <= p }. Is there a way to order embed S into a complete lattice that > preserves all meets and joins? > Yes, it is called the Dedekind-MacNeille completion of (S,<). > Can you give any details? I don't know that kind of research you usually do, but if it is an > internet research then your search machine needs a lot to learn. See here: > Oh my, I under estimated Wikipedia. > Is the completion still a subset of P(S) Yes. Moreover it is a universal object since it fulfils the universal > property of a completion. > Indeed, it's the collection of the upper bound sets of a set. Huh? That's order reversing. Perhaps it's sets of lower bounds. What's a universal thingy? > Then again for dense orders and for linear dense orders. I speculate that the universal property of the MacNeille completion will > help you here. > What property is that? Riddle of the day. Is the universal property in foreclosure? ---- === Subject: exponential e posting-account=JDHTogoAAACdrvMn2VgFOzlkHioeb9OS AppleWebKit/525.18 (KHTML, like Gecko) Version/3.1.2 Safari/525.20.1,gzip(gfe),gzip(gfe) Hello all, I found a problem that asks what SUM( k^2/k!) from 1 -> infinite is equal to. The correct answer is 2e. I am completely stumped!!! All I'm working with is that e = Sum(1/k!), what am i missing? R === Subject: Re: exponential e > Hello all, I found a problem that asks what SUM( k^2/k!) from 1 -> infinite is > equal to. The correct answer is 2e. I am completely stumped!!! All I'm working with is that e = Sum(1/k!), what am i missing? R sum(k^2/k !; k = 1 .. oo) = sum(k/(k - 1)! ; k = 1 .. oo) = [Note index change] sum((k + 1)/k ! ; k = 0 .. oo) = sum(k/k! ; k = 0 .. oo) + sum(1/k! ; k = 0 .. oo) = [Note 0/0! = 0] sum(k/k! ; k = 1 .. oo) + e = sum(1/(k - 1)! ; k = 1 .. oo) + e = [Index change again] sum(1/k! ; k = 0 .. oo) + e = e + e. (Ordinarily, I would have just given a hint but I couldn't think of one that was useful.) -- Paul Sperry Columbia, SC (USA) === Subject: Re: exponential e <011020081656234340%plsperry@sc.rr.com> posting-account=JDHTogoAAACdrvMn2VgFOzlkHioeb9OS AppleWebKit/525.18 (KHTML, like Gecko) Version/3.1.2 Safari/525.20.1,gzip(gfe),gzip(gfe) Hello all, I found a problem that asks what SUM( k^2/k!) from 1 -> infinite is > equal to. The correct answer is 2e. I am completely stumped!!! All I'm working with is that e = Sum(1/k!), what am i missing? R sum(k^2/k !; k = 1 .. oo) = sum(k/(k - 1)! ; k = 1 .. oo) = [Note index change] sum((k + 1)/k ! ; k = 0 .. oo) = sum(k/k! ; k = 0 .. oo) + sum(1/k! ; k = 0 .. oo) = [Note 0/0! = 0] sum(k/k! ; k = 1 .. oo) + e = sum(1/(k - 1)! ; k = 1 .. oo) + e = [Index change again] sum(1/k! ; k = 0 .. oo) + e = e + e. (Ordinarily, I would have just given a hint but I couldn't think of one > that was useful.) -- > Paul Sperry > Columbia, SC (USA) Merci Beaucoup! Was my initial attempt, but messed up with shifting my index - subtracted one instead adding one - quick sloppy rookie mistake. === Subject: nees solution posting-account=FPDN7goAAAB-Qg8BG5RArtx_ycOa38ZO .NET CLR 2.0.50727; .NET CLR 3.0.04506; 3P_UPCPC 1.0.20.1),gzip(gfe),gzip(gfe) I need solution manual for fundameantals of statistical and thermal physics by reif === Subject: nees solution posting-account=FPDN7goAAAB-Qg8BG5RArtx_ycOa38ZO .NET CLR 2.0.50727; .NET CLR 3.0.04506; 3P_UPCPC 1.0.20.1),gzip(gfe),gzip(gfe) I need solution manual for fundameantals of statistical and thermal physics by reif === Subject: Re: One of the great relations? > Or am I just in love with one more of my discoveries. More of my > math? > I don't even want to read much of any more math texts any more, which > I guess is wrong, but it's weird. you never read or owned a math text. > you stated such previously. > I won. I killed the other guy inside of me. So now there is only me. Finally!!, a clear admission of Guilt. > We of sci.math are taking JSH to court for murdering himself. Will you ask for the death penalty? === Subject: Linear Algebra Problem posting-account=JDHTogoAAACdrvMn2VgFOzlkHioeb9OS AppleWebKit/525.18 (KHTML, like Gecko) Version/3.1.2 Safari/525.20.1,gzip(gfe),gzip(gfe) Hello all, I am entertaining myself lately reviewing the basics of mathematics and keep finding myself stumped. Any direction with this problem would be appreciated Let A be 3x3 matrix where A[0,1,2]^T = [1,0,0]^T A[3,4,5]^T = [0,1,0]^T then A[6,7,8]^T must be equal to... All I keep thinking is that [6,7,8] is a linear combo of the two given vectors -> -1[0,1,2]^T + 2[3,4,5]^T = [6,7,8]^T So, obviously this is important but what am I missing? Does it have to do with the fact that A took the first 2 vectors to e1 and e2 (euclid base)? R === Subject: Re: Linear Algebra Problem > Hello all, I am entertaining myself lately reviewing the basics of mathematics > and keep finding myself stumped. Any direction with this problem > would be appreciated Let A be 3x3 matrix where A[0,1,2]^T = [1,0,0]^T > A[3,4,5]^T = [0,1,0]^T then A[6,7,8]^T must be equal to... > All I keep thinking is that [6,7,8] is a linear combo of the two given > vectors > -> -1[0,1,2]^T + 2[3,4,5]^T = [6,7,8]^T So, obviously this is important but what am I missing? Does it have to > do with the fact that A took the first 2 vectors to e1 and e2 (euclid > base)? R The linearity properties of matrix multiplication means that if A*p = u and A*q = v then A*(q - p) = v - u. === Subject: Re: Linear Algebra Problem posting-account=JDHTogoAAACdrvMn2VgFOzlkHioeb9OS AppleWebKit/525.18 (KHTML, like Gecko) Version/3.1.2 Safari/525.20.1,gzip(gfe),gzip(gfe) > Hello all, I am entertaining myself lately reviewing the basics of mathematics > and keep finding myself stumped. Any direction with this problem > would be appreciated Let A be 3x3 matrix where A[0,1,2]^T = [1,0,0]^T > A[3,4,5]^T = [0,1,0]^T then A[6,7,8]^T must be equal to... All I keep thinking is that [6,7,8] is a linear combo of the two given > vectors > -> -1[0,1,2]^T + 2[3,4,5]^T = [6,7,8]^T So, obviously this is important but what am I missing? Does it have to > do with the fact that A took the first 2 vectors to e1 and e2 (euclid > base)? R Never-mind. Wow, I've lost my edge - mathematics is not very forgiving of deserters! A[ -x1 + 2x2] = -Ax1 + 2Ax2 = [-1,2,0]^T === Subject: proof that if u on (0,1) is uniformly distributed, 1-u is also uniform? Looking for a standard proof of this. Ref or outline appreciated. Continuous or discrete (or both). === Subject: Re: proof that if u on (0,1) is uniformly distributed, 1-u is also uniform? > Looking for a standard proof of this. Ref or outline > appreciated. Continuous or discrete (or both). Hint: P(1-u <= x) = P(u >= 1-x) Now consider the cases x < 0 , 0 <= x <= 1 and x > 1 and use the fact that u is uniformly distributed on (0;1). Best wishes Torsten. === Subject: Huge number theoretic structure posting-account=n1ZfDgkAAABbCs44qOtz8dP-RkWuEBif Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) I've found that given x^2 + Dy^2 = F it is forced that z^2 + D(x+y)^2 = F(D+1) and you can verify trivially that z=x-Dy, by substituting out F, from the first equation into the second: z^2 + D(x+y)^2 = (x^2 + Dy^2)(D+1) so, expanding and simplifying: z2 = (D+1)x^2 + D^2 y^2 + Dy^2 - Dx^2 - 2Dxy - Dy^2 gives z^2 = x^2 - 2Dxy + D^2 y^2 = (x-Dy)^2. And it may seem like a trivial result, but now you have a chain where the first 3 equations are: x^2 + Dy^2 = F (x-Dy)^2 + D(x+y)^2 = F(D+1) and (x-Dy - D(x+y))^2 + D(x-Dy + x+y)^2 = F(D+1)^2 which is ((1-D)x-2Dy)^2 + D(2x + (1-D)y)^2 = F(D+1)^2. And that may not seem remarkable unless you realize that for every solution to the first equation, you have a solution to that last with (D+1)x, and (D+1)y, so what if you go backwards and ask, can (1-D)x-2Dy = (D+1)x or (D+1)y and 2x + (1-D)y = (D+1)y or (D+1)x? But if so, then you have two solutions for x in terms of y, so BOTH must give the same answer or there is a contradiction. For instance if (1-D)x-2Dy = (D+1)x, then y = x, is a solution, and then 2x + (1-D)y = (D+1)y, gives that x = Dy, and only D=1 can work. So there is a way to solve for D. Wow. But then I have x^2 + y^2 = 2x^2 = F, and a set of solutions forced on F for which all the relation are valid, but there are solutions, for instance, F=2, will work. So, it's now possible to make the conjecture that all integer solutions to equations of the form x^2 + Dy^2 = F, are part of this immensely huge number theoretic structure, which is infinitely sized, where depending on the value of D, and the value of F, at various points in the chain you will have situations when you have multiples of x and y. What's remarkable about that conjecture is it allows an explanation for all previously noted behavior and indicates that solutions have to do with linear equations and nothing else. Intriguingly there is other research noting a lack of a need for non- rational numbers with these equations: Pell's equation without irrational numbers Authors: N. J. Wildberger (Submitted on 16 Jun 2008) Abstract: We solve Pell's equation in a simple way without continued fractions or irrational numbers, and relate the algorithm to the Stern Brocot tree. Comments: 9 pages, 3 figures Subjects: Number Theory (math.NT) Cite as: arXiv:0806.2490v1 [math.NT] http://arxiv.org/abs/0806.2490v1 If my conjecture is correct then there is an infinitely sized and static number theoretic structure which is just an infinite expansion where the first three equations are: 1. x^2 + Dy^2 = F 2. (x-Dy)^2 + D(x+y)^2 = F(D+1) 3. ((1-D)x-2Dy)^2 + D(2x + (1-D)y)^2 = F(D+1)^2 (can you add more?) and where whenever the exponent of (D+1) is even, you can have a case where you just have a multiple of x and y, which will define a set of possible values for F, so the previously seen behavior with solutions to these equations can be explained as finding points in the expansion where you have a multiple with a given D, which will allow your F. So continued fraction solutions can be explained as just places along the chain where for a given D, your F will work where you get a multiple of x and y. If correct, then the structure can be mapped out to whatever level you want, and at each point with an even exponent you can solve for possible D and F as I demonstrated above, and completely explain Pell's Equation. That number theoretic structure would be just this HUGE static object, unimaginably complex in a way, but also remarkably simple as it just starts with the well-known equation: x^2 + Dy^2 = F. I don't know about you but I like simple explanations. Nice to know the why of Pell's Equation solutions like why some of them are so hard to find while others are not. I like knowing the why. === Subject: Re: Huge number theoretic structure > I've found that given x^2 + Dy^2 = F Let's work an example. I choose D=-3, F=1. it is forced that z^2 + D(x+y)^2 = F(D+1) and you can verify trivially that z=x-Dy, by substituting out F, from > the first equation into the second: z^2 + D(x+y)^2 = (x^2 + Dy^2)(D+1) so, expanding and simplifying: z2 = (D+1)x^2 + D^2 y^2 + Dy^2 - Dx^2 - 2Dxy - Dy^2 gives z^2 = x^2 - 2Dxy + D^2 y^2 = (x-Dy)^2. And it may seem like a trivial result, but now you have a chain where > the first 3 equations are: x^2 + Dy^2 = F So, x^2 - 3y^2 = 1 (x-Dy)^2 + D(x+y)^2 = F(D+1) (x+3y)^2 - 3(x+y)^2 = -2 and (x-Dy - D(x+y))^2 + D(x-Dy + x+y)^2 = F(D+1)^2 which is ((1-D)x-2Dy)^2 + D(2x + (1-D)y)^2 = F(D+1)^2. (4x+6y)^2 - 3(2x+4y)^2 = 4 And that may not seem remarkable unless you realize that for every > solution to the first equation, you have a solution to that last with > (D+1)x, and (D+1)y, so what if you go backwards and ask, can (1-D)x-2Dy = (D+1)x or (D+1)y 4x+6y = -2x or -2y > and 2x + (1-D)y = (D+1)y or (D+1)x? 2x+4y = -2x or -2y But if so, then you have two solutions for x in terms of y, so BOTH > must give the same answer or there is a contradiction. For instance if (1-D)x-2Dy = (D+1)x, then y = x, is a solution, and then If 4x+6y = -2x, then y = -x. (You got the sign wrong). It's easy to see that x^2-3y^2 = 1 has no solutions where |x| = |y|, so this does not give us a solution. > 2x + (1-D)y = (D+1)y, gives that x = Dy, and only D=1 can work. 2x+4y = -2y gives that x = -3y. No sign of a solution to the original equation lurking in there that I can see. So there is a way to solve for D. Wow. Why do I want to solve for D? I know what D is: -3. > But then I have x^2 + y^2 = 2x^2 = F, and a set of solutions forced on > F for which all the relation are valid, but there are solutions, for > instance, F=2, will work. Too bad that's not the F from the equation we are trying to solve. The rest of your post was vague and unclear, and I am unable to continue applying it to x^2-3y^2=1. -- --Tim Smith === Subject: Re: Any ideas what to do? So computer checking of claims does not meaningfully exist in the math > field. Just curious, but who would you trust better than mathematicians (who you seem to abhor) to write those computer programs checking mathematical proofs? > So what can be done? I'm really wondering. Open your eyes in between dreams. Be less arrogant. Back on topic, and very narrowly answering the particular question in the title... Maybe the kind of mechanical substitutions and algebraic manipulations you appear to be so fond of, could lead to a mid-to-high grade level 101 Offbeat Math Problems book. Provided of course that you filtered-out the junk claims, filled-in some resemblance of formal proofs/answers to what's left, and followed the rest of the rules to get it published someplace reputable. Liviu === Subject: Re: Any ideas what to do? <0LVEk.1448$fD.944@flpi145.ffdc.sbc.com> posting-account=n1ZfDgkAAABbCs44qOtz8dP-RkWuEBif Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) So computer checking of claims does not meaningfully exist in the math > field. Just curious, but who would you trust better than mathematicians (who > you seem to abhor) to write those computer programs checking > mathematical proofs? Expert systems are not written entirely by the expert. Your question is like wondering how medical diagnosis programs can get written if medical doctors only would write them. Or do you naively believe that only medical doctors write such programs? I'm curious as this issue repeatedly comes up, where math people seem bizarrely to believe that mathematicians would actually write the programs!!! Do you all think that? Are you all so naive about modern computer science? Who do you think write these programs? I'll tell you: computer scientists. Like mathematicians do mathematical research, computer scientists write programs for expert systems and can do so in areas where they themselves are NOT expert. Sometimes the math community seems remarkably obtuse about the simplest things!!! Or are you playing stupid? Like mathematicians are experts with mathematics. Computer scientists you, you, I find it hard to describe you people, are the experts with computer systems!!! Duh. === Subject: Re: Any ideas what to do? JSH the math field. > Just curious, but who would you trust better than mathematicians > (who you seem to abhor) to write those computer programs checking > mathematical proofs? I'll tell you: computer scientists. [...] Computer scientists you, you, I find it hard to describe you people, > are the experts with computer systems!!! I'll take that as a generic you since I've never claimed being an expert, in either computers or math for that matter. I was, and still am, just curious who you'd hypothetically put your trust on, if not the people familiar with the subject being checked. If you say computer scientists with no further qualification, then that's simply plain laughable. If you qualify it with guided/directed by ... then you have still not filled in the blanks, and only evaded the question. > Your question is like wondering how medical diagnosis programs > can get written if medical doctors only would write them. Or do you naively believe that only medical doctors write such > programs? Naive or not, I would myself keep a safe distance from any medical program written by a c.s. alone ;-) Liviu P.S. You seem to have brushed over my problems book suggestion. Just for the record, that _was_ serious. === Subject: Re: JSH: Any ideas what to do? > I don't know much about either contribution, but I do understand that > theorems about constructing geometric objects were a central object of > study in the past. I don't know that whether her theorem contributed > to an active area of research. In any case, I still can't see what you expect. If she never > submitted her work to an mathematical journal, how can you suggest > that the math community wronged her by not publishing it? As usual, JSH is greatly exaggerating the worth of some minor work. At least it is someone else's minor work, rather than his own. The difference between Britney Gallivan's minor result and JSH's minor results is that her's was simultaneously correct and new. JSH usually fails on one or both of those. Briefly, she derived the following formula for folding paper: L = pi d (2^n + 4) (2^n - 1) / 6 where d is the thickness of the paper, and n is how many times you fold it in half over itself, and L is how long your piece of paper has to be in order to be folded n times. A proof of this is available here: In addition to deriving the above formula, she also set the world record for paper folding, by successfully folding a piece of paper 12 times (it was a very long piece of paper!). It had been widely believed that you could not fold a piece of paper in half over itself more than 8 times: Gallivan made a worthwhile contribution to mathematical knowledge, and has been appropriately acknowledged. If JSH would try doing the same, he would be similarly acknowledged. -- --Tim Smith === Subject: Re: JSH: One of the great relations? posting-account=fzmnpgoAAAAmyKTlxv-T7BPlBuylv6TZ Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) thank you for your honest kindness === Subject: Re: JSH: One of the great relations? posting-account=fzmnpgoAAAAmyKTlxv-T7BPlBuylv6TZ Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) just don't do drugs. and if you haven't then you are probably manic. === Subject: Re: JSH: One of the great relations? > My point is that solutions in general to x^2 + Dy^2 = F, can be > explained now with the relation x^2 + Dy^2 = F, requiring that (x-Dy)^2 + D(x+y)^2 = F*(D+1), as look > again at (x-Dy + D(x+y))^2 + D(x-Dy+x+y)^2 = F*(D+1)^2 The first + in the first term should be -. as simplifying a bit that's ((D+1)x)^2 + D(2x+(1-D)y)^2 = F*(D+1)^2 Hey, did I get the math right? That should be two iterations starting > from No. First term should be (x(1-D)-2Dy)^2. You should consider using a computer to check this kind of thing: sage: F=x^2+D*y^2 sage: expand((x*(1-D)-2*D*y)^2 + D*(2*x+y*(1-D))^2-F*(D+1)^2) 0 -- --Tim Smith === Subject: Re: JSH: One of the great relations? posting-account=n1ZfDgkAAABbCs44qOtz8dP-RkWuEBif Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) My point is that solutions in general to x^2 + Dy^2 = F, can be > explained now with the relation x^2 + Dy^2 = F, requiring that (x-Dy)^2 + D(x+y)^2 = F*(D+1), as look > again at (x-Dy + D(x+y))^2 + D(x-Dy+x+y)^2 = F*(D+1)^2 The first + in the first term should be -. as simplifying a bit that's ((D+1)x)^2 + D(2x+(1-D)y)^2 = F*(D+1)^2 Hey, did I get the math right? That should be two iterations starting > from No. First term should be (x(1-D)-2Dy)^2. Ok, that makes more sense. > You should consider using a computer to check this kind of thing: sage: F=x^2+D*y^2 > sage: expand((x*(1-D)-2*D*y)^2 + D*(2*x+y*(1-D))^2-F*(D+1)^2) > 0 It was early in the morning. I still hadn't had my first cup of coffee! Regardless, this result is so incredible I'm becoming reticent about discussing it all on newsgroups. Is there more to getting sage to do that expansion. Like where did you tell it that (x-Dy)^2 + D(x+y)^2 = F*(D+1)? Can you show the first 10? James === Subject: Re: JSH: One of the great relations? Regardless, this result is so incredible I'm becoming reticent about > discussing it all on newsgroups. And this is why you never succeed in convincing anyone of the worth of your results. When people don't understand your results, or attack them, you need to defend them. (I mean defend them mathematically, not defend them by accusing mathematicians of conspiring against you). -- --Tim Smith === Subject: Re: JSH: One of the great relations? posting-account=n1ZfDgkAAABbCs44qOtz8dP-RkWuEBif Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) > Regardless, this result is so incredible I'm becoming reticent about > discussing it all on newsgroups. And this is why you never succeed in convincing anyone of the worth of > your results. When people don't understand your results, or attack > them, you need to defend them. (I mean defend them mathematically, not > defend them by accusing mathematicians of conspiring against you). But as usual I changed my mind. Amazingly to me there is finally also a case where I can cite someone else's research as being along the same line: http://arxiv.org/abs/0806.2490v1 For what it's worth that paper is moving up in Google search results when you do a search on Pell's equation. Gist of it all for those wondering is: you don't need irrational numbers or non-rational numbers AT ALL to understand Pell's equation. My current research takes you a step forward into the full explanation of all the behavior associated with solutions to x^2 + Dy^2 = F. So there could be a historic shift in an area that has 2000 years of mathematical history with the dropping entire of non-rationals as just a historical legacy, when it comes to understanding Pell's equation. === Subject: Re: JSH: One of the great relations? > You should consider using a computer to check this kind of thing: sage: F=x^2+D*y^2 > sage: expand((x*(1-D)-2*D*y)^2 + D*(2*x+y*(1-D))^2-F*(D+1)^2) > 0 coffee! Regardless, this result is so incredible I'm becoming reticent about > discussing it all on newsgroups. Is there more to getting sage to do that expansion. Like where did > you tell it that (x-Dy)^2 + D(x+y)^2 = F*(D+1)? Can you show the first 10? Here's my complete Sage session: ---------------------------------------------------------------------- | Type notebook() for the GUI, and license() for information. | ---------------------------------------------------------------------- sage: x,y,D=var('x y D') sage: F=x^2+D*y^2 sage: expand((x*(1-D)-2*D*y)^2 + D*(2*x+y*(1-D))^2 - F*(D+1)^2) 0 I didn't tell it that (x-Dy)^2 + D(x+y)^2 = F*(D+1) That's not needed to verify the equation for F*(D+1)^2. -- --Tim Smith === Subject: Re: archimedean property question > would you mind giving me an example (and solution) of an archimedean > property in a simple problem. > im having trouble finding it online. thank you The Archimedean Property is simply that for any real number r there is a bigger natural number n. I think that you could probably think of a lot of numerical examples. Like, say, for real r=.9999998 and n=1 then n>r You'll see this property applied when you start to examine the supremum and infimum of a set. Hope that helps! === Subject: Re: archimedean property question > would you mind giving me an example (and solution) of an archimedean > property in a simple problem. > im having trouble finding it online. Try . > thank you You're welcome. -- --------------------------- | BBB b Barbara at LivingHistory stop co stop uk | B B aa rrr b | | BBB a a r bbb | Quidquid latine dictum sit, | B B a a r b b | altum viditur. | BBB aa a r bbb | -----------------------------