A79 == I have been thinking about prime numbers and using a Sin wave to work out if a number is divisable by another one. Working in Degs so I don't have to find the Pi button on my keyboard: Sin (180x/n) gives the answer 0 if you are testing number x to see if it divisable by number n. Hence not Prime. This can then be put into a product sequence with the set running from n=2 to INT(root x) sin(180x/n) Because you are multiplying then any 0 that comes out, hence a division occuring, gives a final answer as 0. Only a prime number gives a non-zero answer everytime and hence a non-zero product. What I want to know is is there anyway of creating a function to solve this? Zanziba ==== Subject: Re: Prime Number Checker? > This can then be put into a product sequence with the set running from > n=2 to INT(root x) sin(180x/n) What I want to know is is there anyway of creating a function to solve > this? Solve what? You have not stated a problem that needs solving. An integer x is composite iff G(x) = product n=2 to sqrt(x) sin(pi/2 x/n) is zero. This is true. So what? What do you want to solve? G(x), as defined above is a function such that G(x) = 0 if x is composite and G(x) != 0 if x is prime. Many such functions can be constructed. ==== Subject: Quadratic solution Is there an online site for solving quadratic equation using the 'factorising' method? Thnx, Big Show ==== Subject: Re: A question about Glivenko-Cantelli lemma >I know it is discrete. Great. >The definition of empirical distribtution is >http://planetmath.org/encyclopedia/EmpiricalDistributionFunction.html Fabulous. >It maps to the distribution function, I'm not sure what you mean by that. >so there should have a empirical density Huh? A discrete distribution simply does not _have_ a density function. It does have something that might be regarded as a density, but that density is a _singular_ measure, not a function. If you mean to be talking about the density in that sense, then the answer to your question is yes, _if_ you're asking about _vague_ convergence of the densities. But I have the impression that you're asking about pointwise convergence of the density _function_, and there simply is no such thing as the empirical density function! Let's back up a minute. Forget the lemma you were asking about. Define X to be a whatchacallit random variable: P(X = 0) = P(X = 1) = 1/2. That's a simple discrete random variable. Tell us what the density function for that X is. >(maybe should call it as >relative frequency) , so I wonder about the relationship between >relative frequency and the true pdf. that's what I am referring to. > ************************ C. Ullrich ==== Subject: Re: A question about Glivenko-Cantelli lemma <42C21026.1020408@netscape.net> asking about. Define X to be a whatchacallit > random variable: P(X = 0) = P(X = 1) = 1/2. > That's a simple discrete random variable. Tell > us what the density function for that X is. Is it 1/2 delta(x) + 1/2 delta(x-1)? where delta is defined as delta(x) = 1 when x = 0 =0 else ==== Subject: Re: A question about Glivenko-Cantelli lemma > Let's back up a minute. Forget the lemma you were >> asking about. Define X to be a whatchacallit >> random variable: P(X = 0) = P(X = 1) = 1/2. >> That's a simple discrete random variable. Tell >> us what the density function for that X is. >Is it >1/2 delta(x) + 1/2 delta(x-1)? where delta is defined as >delta(x) = 1 when x = 0 >=0 else No. If we'd specified that X was an integer-valued random variable this might make some sense, but the phrase random variable without qualification denotes a real-valued thing. That X does not _have_ a density function. ************************ C. Ullrich ==== Subject: Re: Cantor and the binary tree <61asb153qs3ek4trn37704ibuk99gee8d5@4ax.com> 0.111xxx... is a binary representation of a bunch, i.e. of a real >interval. And is therefore, no longer a binary representation of a number. > Yet, your conclusion was about binary representations of numbers. Before infinity is reached, the binary representations of numbers have paths in the tree (if these numbers do exist). >> Which would make it right. However, even I know that one can't prove >> the parallel postulate from the other axioms. >But in 1830 even the best mathematicians believed it could be proved, >except Gauss, Lobatchevski, and Bolyai, who did know the contrary >already. And this is relevant now because? Dwell on that story for another time. ==== Subject: Re: Cantor and the binary tree > >0.111xxx... is a binary representation of a bunch, i.e. of a real >interval. > > And is therefore, no longer a binary representation of a number. > Yet, your conclusion was about binary representations of numbers. > > Before infinity is reached, the binary representations of numbers have > paths in the tree (if these numbers do exist). > >> Which would make it right. However, even I know that one can't prove >> the parallel postulate from the other axioms. > >But in 1830 even the best mathematicians believed it could be proved, >except Gauss, Lobatchevski, and Bolyai, who did know the contrary >already. If you except Gauss, you are excluding the best off the top. And which good mathematicians actually claimed it could be proved? A lot of hacks provided what they called proofs, but a lot of hacks were (and still are) trisecting angles, squaring circles and duplicating cubes, too, at least according to their own evaluations. A lot of mathematicians, going all the way back to Euclid, have been dubious over the 'provability' of the so-called parallel postulate or it's equivalent forms, and Euclid himself specifically did not claim it to be provable but, with apparent reluctance, postulated it. ==== Subject: Re: Cantor and the binary tree <1119498995.344c6f1e27ade3c6c50e0ba879307012@teranews> <1119582534.f64abd486bab317ff81eb66addec82a9@teranews> Wrong. It covers all even numbers. Every even number is he largest one >of a finite set. There is no even number wich is not the largest one of >a finite set. So? In ZFC, there is nothing that is not the largest member of a > finite set. For every x, there is a set {x}. There is only one order > for {x}, and by that ordering, x is clearly the largest member of {x}. Read my text, then you will see that you agree to it. In order for your proof to work, every set of even number would have > to be a subset of {2m | m in N and m <= n} for some n in N. But, the > set of all even numbers {2m | m in N }, is not not such a set. > Therefore, your proof fails. Not, if every n is covered. And why shouldn't it? >I use the fact that any set of even numbers equal or less than 2n has >cardinality n. This fact is obviously correct. Why should it restrict >the validity of the proof? Because there are sets of even numbers that are not subsets of any the > sets in your proof. Your proof does not apply to them, yet you are > trying to conclude that they do. I do not at all need to consider sets which are not of the form {2,4,6,...,2n}. Because I do not conclude anything for other sets. ==== Subject: Re: Cantor and the binary tree > >Wrong. It covers all even numbers. Every even number is he largest one >of a finite set. There is no even number wich is not the largest one of >a finite set. > > So? In ZFC, there is nothing that is not the largest member of a > finite set. For every x, there is a set {x}. There is only one order > for {x}, and by that ordering, x is clearly the largest member of {x}. > > Read my text, then you will see that you agree to it. > > In order for your proof to work, every set of even number would have > to be a subset of {2m | m in N and m <= n} for some n in N. But, the > set of all even numbers {2m | m in N }, is not not such a set. > Therefore, your proof fails. > > Not, if every n is covered. And why shouldn't it? > >I use the fact that any set of even numbers equal or less than 2n has >cardinality n. This fact is obviously correct. Why should it restrict >the validity of the proof? > > Because there are sets of even numbers that are not subsets of any the > sets in your proof. Your proof does not apply to them, yet you are > trying to conclude that they do. > > > I do not at all need to consider sets which are not of the form > {2,4,6,...,2n}. Because I do not conclude anything for other sets. Then, as far as WM knows, there may be infinite sets of evens NOT of that form. Unless, of course, WM has a proof that EVERY set of evens must be a subset of one of that form, which requires that he prove a priori that EVERY set of evens has a maximal member. Presuming that, as WM so often does, begs the question, again. ==== Subject: Re: Cantor and the binary tree > I do not at all need to consider sets which are not of the form > {2,4,6,...,2n}. Because I do not conclude anything for other sets. Oh, thank goodness. I almost thought you had concluded something about the set of all even numbers, which is clearly not of that form. Glad to see I was wrong :-) Jan ==== Subject: Re: Cantor and the binary tree >> Nope, your proof also uses the fact that the set on the right hand side is >> finite, and has a maximal number that you conveniently put in the left >> hand side. > >I use the fact that any set of even numbers equal or less than 2n has >cardinality n. This fact is obviously correct. Why should it restrict >the validity of the proof? Because the set you are trying to apply it to doesn't have that > property. It has the only property which is relevant for the proof, namely, that each of its elements is finite and is the maximum of a finite sequence. ==== Subject: Re: Cantor and the binary tree > > > > >> Nope, your proof also uses the fact that the set on the right hand side >> is >> finite, and has a maximal number that you conveniently put in the left >> hand side. > >I use the fact that any set of even numbers equal or less than 2n has >cardinality n. This fact is obviously correct. Why should it restrict >the validity of the proof? > > Because the set you are trying to apply it to doesn't have that > property. > > It has the only property which is relevant for the proof, namely, that > each of its elements is finite and is the maximum of a finite sequence. Then your only legitimate conclusion is that every member (of the set of all evens) is finite and is the maximum of a finite sequence, which you already knew. ==== Subject: Re: Cantor and the binary tree That every set of finite even numbers is finite! But your proof didn't include any infinite sets. So how could you > have shown that every set of even numbers is finite. Every natural number defines a finite sequence. The set of natural numbers does not contain anything else but numbers which define finite sequences. Beyond finite numbers, and hence, beyond finite sequence, there is nothing in the set which could make it infinite. > (The finite as > presented is redundent, A set of finite numbers is finite. That is redundent. I do not understand how you can even spend a thought on the contrary. >> If he is trying to argue that the union of arbitrarily many finite sets >> must be finite, he is >complete induction turns out insufficient in cases of *uncountable* >sets. In fact, there is no reason why it shouldn't work for countable >sets. If I prove that for any set of even numbers the set constituting >them is finite, this proof is completely correct. Induction reaches >every natural number. There does not remain any one which could make a >set infinite. It is in fact the summit of nonsense if you insist that >complete induction were insufficient for all natural numbers. But your claim isn't about even natural numbers, it's about sets of > even natural numbers. And there are uncountably many such sets. So > induction does not work. Wrong. I consider only Abschnitte = sequences {2,4,6,...,2n} from 2 to 2n, and they are countable. ==== Subject: Re: Cantor and the binary tree > >That every set of finite even numbers is finite! > > But your proof didn't include any infinite sets. So how could you > have shown that every set of even numbers is finite. > > Every natural number defines a finite sequence. The set of natural > numbers does not contain anything else but numbers which define finite > sequences. Beyond finite numbers, and hence, beyond finite sequence, > there is nothing in the set which could make it infinite. The sets of values for each such sequence form an unending nested sequence of sets, whose union is a set containing more members that are in any single set of the union. If WM claims that the sequence of of naturals ends, he must be able to say something about when it ends, and give us some details on the natural with which it ends. For example, is that last natural even or odd? It has to be one or the other, and somone as knowing as WM should know the answer to that simple question. Is that largest natural a prime of not? If it is not a prime what is its largest prime factor? If it is prime, what does that do to Euclid's proof? > > A set of finite numbers is finite. That is redundent. I do not > understand how you can even spend a thought on the contrary. Then the set of real numbers between 0 and 1, being a set of numbers, must also be finite? ==== Subject: Re: Cantor and the binary tree The set of bunches includes all isolated paths. What is an isolated path? My guess would be one that has some node or > edge all to itself. Of course. If there are real numbers which can be distinguished from any other real number by a bit at a position which is enumerated by a natural number, then there are isolated paths in my tree. In particular, if Cantor's antidiagonal can be distinguished from any number of he list on such a place and if this antidiagonal is due to an isolated path, then such paths are in my tree. > In the tree in question, there are none, > >Either the paths remain forever in the bunch without isolating >themselves. Then the corresponding real numbers do not exist. Care to prove this? There is nothing to prove. The paths are only a picture of binary expansions of reals. If isolated binary expansions of reals do exist, then isolated paths do so. ==== Subject: Re: Cantor and the binary tree > >The set of bunches includes all isolated paths. > > What is an isolated path? My guess would be one that has some node or > edge all to itself. > > Of course. But in a maximal binary tree, there is no path that has any edge all to itself, since as many paths go through any edge as there are in the whole tree. > If there are real numbers which can be distinguished from > any other real number by a bit at a position which is enumerated by a > natural number, then there are isolated paths in my tree. There is no real number that can be distinguished from ALL other reals by any one bit position. For any bit position, there are uncountably may reals which agree up to and including that bit position. WM is again conflating the number of bunches, which is countable, with the number of paths in a bunch, which is not. In > particular, if Cantor's antidiagonal can be distinguished from any > number of he list on such a place and if this antidiagonal is due to an > isolated path, then such paths are in my tree. > > In the tree in question, there are none, > >Either the paths remain forever in the bunch without isolating >themselves. Then the corresponding real numbers do not exist. > > Care to prove this? > > There is nothing to prove. There is everything to prove. My claim, for which I have posted unrefuted proofs several times, is that the number of paths in any bunch is uncountable (uncountably infinite), just as the number of reals in any real interval, [a,b], with a < b, is uncountable. ==== Subject: Re: Cantor and the binary tree Complete induction will do so. Even Cantor was convinced that complete > > induction is sufficient for all countable sets, and with him every > > mathematician at his times. He may be convinced, in that case he was wrong. I rather think that you > misrepresent what Cantor has stated, because you do not understand the > distinction. Yes: I understand that erste Zahlenkasse means countable. Cantor understood that. > 2n > Card({2, 4, ...., 2n}) > This is true for *each* natural n. And that is what complete induction > is. It does not state anything about N as the right hand side of the > formula. You have a serious comprehension problem. Stating something > is valid for *all* n is the same as stating that something is valid for > *each* n. In Dutch (and I think in German) the same applies. It does > not state anything of the collection of *each* n, or of *all* n, it > only states something about the individual members. I know that each n does define a finite sequence. Fom that one can conclude that all n are in a finite sequence, in German and n Dutch. Your distinction is nothing than a helpless attempt to save set-theory as has those which nhave been aplied in the past several times. Think of L.9awenstein-Skolem: Every consistent first oder theory has a countable model. iIf set theoy is consistent, then it is countable, hence inconsistent. Therefore set theory canot be consistent. All your attempts are in vain. > So how you can get to your conclusion > > for the set of all even numbers using induction? > > because there is no even number at which the reasoning stops or gets > > incorrect. Yes, so what? This does not imply anything about the set of even numbers, > except that for each member of that set the set constructed by taking all > even numbers less than or equal to that number Why should I talk about any fixed number??? I speak of any number of the set, which is a finite numbers. And there are no further numbers. For finite numbers we can prove that they define a finite sequence. Why do you always consider larger numbers? Where do you get them from??? My proof covers each and every finite number, simultaneously, instantanneously, just like Cantor's antidiagonal. There is nothing beyond the numbers reached by complete induction. Of course, for every n there is an n+1, but that one is also covered (if Cantor can do so by his proof). ==== Subject: Re: Cantor and the binary tree > > > Complete induction will do so. Even Cantor was convinced that complete > > induction is sufficient for all countable sets, and with him every > > mathematician at his times. > > He may be convinced, in that case he was wrong. I rather think that you > misrepresent what Cantor has stated, because you do not understand the > distinction. Yes: > > I understand that erste Zahlenkasse means countable. Cantor understood > that. > > > 2n > Card({2, 4, ...., 2n}) > This is true for *each* natural n. And that is what complete induction > is. It does not state anything about N as the right hand side of the > formula. You have a serious comprehension problem. Stating something > is valid for *all* n is the same as stating that something is valid for > *each* n. In Dutch (and I think in German) the same applies. It does > not state anything of the collection of *each* n, or of *all* n, it > only states something about the individual members. > > I know that each n does define a finite sequence. Fom that one can > conclude that all n are in a finite sequence, in German and n Dutch. But it does not justify concluding that all are in the same finite sequence. For each n allows that there may be different finite sequences for different n's, as is actually the case. So that WM's deduction that one sequence will do for all is fallacious. As are so many of his deductions. > Why should I talk about any fixed number??? I speak of any number of > the set, which is a finite numbers. And there are no further numbers. What is this alleged largest natural number that is so large that one cannot add anything to it? I will have to be shown better evidence of its existence than WM has come up with yet to be cnvinced that it exists outside of WM's dreams. > For finite numbers we can prove that they define a finite sequence. Why > do you always consider larger numbers? Where do you get them from??? By adding one a number of times to the number WM says is the largest. ==== Subject: Re: Cantor and the binary tree Yup, and any set of even numbers equal or less than 2n is finite. *Every* natural number is the maximum of a *finite* sequence. By the sequence {1,2,3,...,n} for n --> oo we exhaust the set of all natural numbers. There is nothing to be expected beyond the sequence of every natural number. The same is true for the set of even numbers. You expect that there are more even numbers than any even number 2n, togh for every finite set you are wrong. You set is a chimera. The following limits are true (the only truth). For A n e N we have with LIM = lim {ntoinfty} = lim {n --> oo} LIM (Min({1,1/2,1/3,...,1/n}) * Card({1,1/2,1/3,...,1/n}) = 1 LIM (Min({1/2,1/4,...,1/2n}) * Card({1/2,1/4,1/6,...,1/2n}) = 1/2 LIM (Max({1,2,3,...,n}) / Card({1,2,3,...,n}) = 1 LIM (Max({2,4,6,...,2n}) * Card({2,4,6,...,2n}) = 2 Yes, so what? Your proof is (as you just admitted) about sets with > members less than some number 2n. In no way can you get at infinite sets > this way, by any form of induction, except transfinite induction, but you > are not using that. Wrong, the set includes only natural numbers and a countable set. Cantor explicitly required transfinite induction only for uncountable sets: ... es zeigt sich, da die vollst.8andige Induktion, wie sie bisher ausgebildet ist, bei gewissen Fragen nicht ausreicht, n.8amlich stets dann nicht ausreicht, wenn der Inbegriff aller F.8alle, die ein Satz darbieten kann, von einer h.9aheren als der ersten M.8achtigkeit ist. (16. Oct. 1883, 10 years after introducing uncountability). ==== Subject: Re: Cantor and the binary tree > > > Yup, and any set of even numbers equal or less than 2n is finite. > > *Every* natural number is the maximum of a *finite* sequence. By the > sequence {1,2,3,...,n} for n --> oo we exhaust the set of all natural > numbers. There is nothing to be expected beyond the sequence of every > natural number. If WM says n --> oo, then WM is conceding that naturals approach infinity, which cannot be the case if there is a largest or last one. So that WM is speaking with forked tougue. ==== Subject: Re: Cantor and the binary tree >Theorem. Any set of even numbers has a cadinality which is less than >>infinite. >>Proof 2n > Card({2,4,6,...,2n}). >> Your proof assumes that there is a largest even number in the set. You >> should really say Any set of even numbers with a largest member has >> a cardinality which is less than infinite. > >I assume that all numbers of the set are finite. So let's consider the case: 2 * ? > Card({2, 4, 6, .......}) All the members of the set on the right are finite. Promise. However, whatever > finite value I use in place of the question mark is going to be wrong. If you insert a fixed number on the left hand side, then the result is wrong. It s unfair, to do so, because on the right hand side, there is also no last number. All we know is, that every member of the set is finite. Hence we know that * can be replaced by variable which is not infinite. I do not assume a lagest member, but only that each member is finite. Your proof assumed that the set had a largest member, 2n. If the set does not > have a largest member then your proof fails. For A n e N we have with LIM = lim_{ntoinfty} = lim_{n --> oo} LIM (Min({1,1/2,1/3,...,1/n}) * Card({1,1/2,1/3,...,1/n}) = 1 LIM (Min({1/2,1/4,...,1/2n}) * Card({1/2,1/4,1/6,...,1/2n}) = 1/2 LIM (Max({1,2,3,...,n}) / Card({1,2,3,...,n}) = 1 LIM (Max({2,4,6,...,2n}) * Card({2,4,6,...,2n}) = 2 *Every* natural number is the maximum of a *finite* sequence. By the sequence {1,2,3,...,n} for n --> oo we exhaust the set of all natural numbers. There is nothing to be expected beyond the sequence of every natural number. ==== Subject: Re: Cantor and the binary tree > > >>Theorem. Any set of even numbers has a cadinality which is less than >>infinite. >>Proof 2n > Card({2,4,6,...,2n}). >> Your proof assumes that there is a largest even number in the set. You >> should really say Any set of even numbers with a largest member has >> a cardinality which is less than infinite. > >I assume that all numbers of the set are finite. > > So let's consider the case: > > 2 * ? > Card({2, 4, 6, .......}) > > All the members of the set on the right are finite. Promise. However, > whatever > finite value I use in place of the question mark is going to be wrong. > > If you insert a fixed number on the left hand side, then the result is > wrong. It s unfair, to do so, because on the right hand side, there is > also no last number. All we know is, that every member of the set is > finite. Hence we know that * can be replaced by variable which is not > infinite. I do not assume a lagest member, but only that each member is > finite. > > > Your proof assumed that the set had a largest member, 2n. If the set does > not > have a largest member then your proof fails. > > For A n e N we have with LIM = lim_{ntoinfty} = lim_{n --> oo} > > LIM (Min({1,1/2,1/3,...,1/n}) * Card({1,1/2,1/3,...,1/n}) = 1 > LIM (Min({1/2,1/4,...,1/2n}) * Card({1/2,1/4,1/6,...,1/2n}) = 1/2 > LIM (Max({1,2,3,...,n}) / Card({1,2,3,...,n}) = 1 > LIM (Max({2,4,6,...,2n}) * Card({2,4,6,...,2n}) = 2 > > *Every* natural number is the maximum of a *finite* sequence. *Every* natural number is the limit of *infinitely many* infinite sequences of rationals. One can even insist that they all be increasing sequences and still have infinitel many of them. > By the > sequence {1,2,3,...,n} for n --> oo we exhaust the set of all natural > numbers. There is nothing to be expected beyond the sequence of every > natural number. How can WM say this when he does denies that there is any oo to -->? He contradicts himself. Again! ==== Subject: Re: Cantor and the binary tree > > >>Theorem. Any set of even numbers has a cadinality which is less than >>infinite. >>Proof 2n > Card({2,4,6,...,2n}). >> Your proof assumes that there is a largest even number in the set. You >> should really say Any set of even numbers with a largest member has >> a cardinality which is less than infinite. > >I assume that all numbers of the set are finite. > > So let's consider the case: > > 2 * ? > Card({2, 4, 6, .......}) > > All the members of the set on the right are finite. Promise. However, > whatever > finite value I use in place of the question mark is going to be wrong. > > If you insert a fixed number on the left hand side, then the result is > wrong. It s unfair, to do so, because on the right hand side, there is > also no last number. All we know is, that every member of the set is > finite. Hence we know that * can be replaced by variable which is not > infinite. I do not assume a lagest member, but only that each member is > finite. Then WM is conceding that for EVERY natural number n, 2*n < Card({2, 4, 6, ...}) or does WM claim that he can name a number n for which it is false? > > > Your proof assumed that the set had a largest member, 2n. If the set does > not > have a largest member then your proof fails. > > For A n e N we have with LIM = lim_{ntoinfty} = lim_{n --> oo} > > LIM (Min({1,1/2,1/3,...,1/n}) * Card({1,1/2,1/3,...,1/n}) = 1 > LIM (Min({1/2,1/4,...,1/2n}) * Card({1/2,1/4,1/6,...,1/2n}) = 1/2 > LIM (Max({1,2,3,...,n}) / Card({1,2,3,...,n}) = 1 > LIM (Max({2,4,6,...,2n}) * Card({2,4,6,...,2n}) = 2 > > *Every* natural number is the maximum of a *finite* sequence. By the > sequence {1,2,3,...,n} for n --> oo we exhaust the set of all natural > numbers. There is nothing to be expected beyond the sequence of every > natural number. > ==== Subject: Re: Cantor and the binary tree >>Theorem. Any set of even numbers has a cadinality which is less than >infinite. >Proof 2n > Card({2,4,6,...,2n}). >> Your proof assumes that there is a largest even number in the set. You > should really say Any set of even numbers with a largest member has > a cardinality which is less than infinite. >>I assume that all numbers of the set are finite. >> So let's consider the case: >> 2 * ? > Card({2, 4, 6, .......}) >> All the members of the set on the right are finite. Promise. However, whatever >> finite value I use in place of the question mark is going to be wrong. If you insert a fixed number on the left hand side, then the result is >wrong. If there isn't a fixed number on the left hand side then what have you proven? ??? > Card({2, 4, 6, ....}) What is ??? You provided a nice, neat formula for finite cases. Doesn't the fact that you can't do this for infinite cases tell you something? >It s unfair, to do so, because on the right hand side, there is >also no last number. Which was my point. Your proof that the cardinality is finite, which implies that there is a fixed number on the left hand side, requires that there be a last number in the set. As for it being unfair; you were the one who claimed that your proof applied to infinite sets. >All we know is, that every member of the set is >finite. Hence we know that * can be replaced by variable which is not >infinite. No, we don't know that. Please prove it. >I do not assume a lagest member, but only that each member is >finite. If you don't have a largest number then your proof fails. You have already explained that things are different when there isn't a last number in the set, but you seem to believe that proving something for other cases also proves it for this case WHICH YOU HAVE ALREADY ADMITTED IS DIFFERENT. Alan -- Defendit numerus ==== Subject: Re: Cantor and the binary tree > Floor(Pi*10^10^100) has no reality at all. > > But Floor(Pi*10^10^100) mod 10 has reality. > > What kind of reality do you mean? The same kind of reality that 1, and 2 and so on, have. Ten you should be able at least to answer the question whether it is odd. ==== Subject: Re: Cantor and the binary tree > > > Floor(Pi*10^10^100) has no reality at all. > > But Floor(Pi*10^10^100) mod 10 has reality. > > > What kind of reality do you mean? > > The same kind of reality that 1, and 2 and so on, have. > > Ten you should be able at least to answer the question whether it is > odd. > I can say absolutely that WM is odd. That number is WM's number, so that in that sense at least, it is quite odd. ==== Subject: Re: Cantor and the binary tree No, it will never be possible, because of the restricted resources of > the universe (if pi is a normal irrational). Ah Ah. The universe is too restricted for a *one digit* number ????? A normal irrational is not a one-digit number, but is has a normal distribution of all ten digits. > What properties, other than that it has 10^100 decimal digits, do you > know of this number? For now, none. But it may be even or odd, etc. And if you just need (-1)^M > in some application, parity is enough. You will never know any property of that number, nor will anyone ever know. Therefore I say: It is not a number. If you consider chimeras numbers, then you may consider pi a number. ==== Subject: Re: Cantor and the binary tree > > > No, it will never be possible, because of the restricted resources of > the universe (if pi is a normal irrational). > > Ah Ah. The universe is too restricted for a *one digit* number ????? > > A normal irrational is not a one-digit number, but is has a normal > distribution of all ten digits. > > What properties, other than that it has 10^100 decimal digits, do you > know of this number? > > For now, none. But it may be even or odd, etc. And if you just need (-1)^M > in some application, parity is enough. > > You will never know any property of that number, nor will anyone ever > know. Therefore I say: It is not a number. You may say what you like, because we have now often seen that your saying a thing does not make it so. > If you consider chimeras > numbers, then you may consider pi a number. WE do not need your permission. ==== Subject: Re: Cantor and the binary tree Le 29/06/05 15:10, dans .82mueckenh@rz.fh-augsburg.de[Ca pitalEGrave] a .8ecrit: > > > > > No, it will never be possible, because of the restricted resources of > the universe (if pi is a normal irrational). >> >> Ah Ah. The universe is too restricted for a *one digit* number ????? > > A normal irrational is not a one-digit number, but is has a normal > distribution of all ten digits. Floor(pi*10^10^100) mod 10 is a one digit number. It is difficult to find its value, but how could you know it's impossible ? > What properties, other than that it has 10^100 decimal digits, do you > know of this number? >> >> For now, none. But it may be even or odd, etc. And if you just need (-1)^M >> in some application, parity is enough. > > You will never know any property of that number, nor will anyone ever > know. Therefore I say: It is not a number. If you consider chimeras > numbers, then you may consider pi a number. When do you believe a number deserves to be called a number ? What is the mathematical definition of a chimera number ? Do you know what a real number is (in the sense belongs to R), how they are constructed from rationnals ? ==== Subject: Re: Cantor and the binary tree The fact that each even number is divisible by 2 does not involve that > each element of the set of even numbers is divisible by 2. What? A number is in the set of even numbers iff it is even, right? > If it is even, it is divisible by 2, right? You are weird. No I sketch your arguing, distinguishing between properties of numbers and sets even in cases where there is no difference. If every natural number n is the last element of a finite sequence, then this holds also for n+1. It holds for every element of the set. And there is nothing beyond which could make the set of finite numbers infinite. ==== Subject: Re: Cantor and the binary tree > The fact that each even number is divisible by 2 does not involve that > each element of the set of even numbers is divisible by 2. > > What? A number is in the set of even numbers iff it is even, right? > If it is even, it is divisible by 2, right? > > You are weird. > > No I sketch your arguing, distinguishing between properties of numbers > and sets even in cases where there is no difference. There is always a difference between the members of a set and the set which cotains them as members, enven if WM it unaware of that difference. And what is true for every member of a set need not be true of the set itself > > If every natural number n is the last element of a finite sequence, > then this holds also for n+1. It holds for every element of the set. > And there is nothing beyond which could make the set of finite numbers > infinite. WM argues that if every member of a set is finite the set must be finite. Does he equally argue that if every member of a set is blue then the set must be blue? ==== Subject: Re: Cantor and the binary tree > So how you can get to your conclusion >> for the set of all even numbers using induction? > > because there is no even number at which the reasoning stops or gets > incorrect. Aargh. Pay attention now: The fact that something is true for each even number, does NOT mean it > is true for 'the set of even numbers'. Then the set of even numbers is worthless. Pay attention now: LIM = lim_{ntoinfty} = lim_{n --> oo} Minimum: Min({1,1/2,1/3,...,1/n}) = 1/n Maximum: Max({1,2,3,...,n}) = n Cardinality: Card({1,2,3,...,n}) = n Limits: For A n e N we have: LIM (Min({1,1/2,1/3,...,1/n}) * Card({1,1/2,1/3,...,1/n}) = 1 LIM (Min({1/2,1/4,...,1/2n}) * Card({1/2,1/4,1/6,...,1/2n}) = 1/2 LIM (Max({1,2,3,...,n}) / Card({1,2,3,...,n}) = 1 LIM (Max({2,4,6,...,2n}) * Card({2,4,6,...,2n}) = 2 *Every* natural number is the maximum of a *finite* sequence. By the sequence {1,2,3,...,n} for n --> oo we exhaust the set of all natural numbers. There is literally nothing to be expected beyond the sequence of every natural number. > 'All even numbers' refers to lots of even numbers, while 'the set of > even numbers' refers to a single mathematical object (which happens to > be a set). which happens to be a chimera like actual infinity. IF actual infinity in fact did exist, then we could take the set of well-ordered rational numbers (by Cantor or by Fraenkel or in some other form) and make the following transpositions: Le q_0 be the first element of the well-ordering. Let q_1 be the first rational, which is less than q_0. Exchange the positions of q_0 and q_1. Let now q_2 be the first element which is smaller than q_1. Exchange the positions of q_1 and q_2. Continue. According to Cantor one can perform infinitely many transpositions without changing the character of a well-ordered set. At the end we have a well-ordered set the first element of which is the smallest rational number. (And continuing, we can even obtain the well-ordered and simultaneously ordered set < of all rationals, because aleph_0 * aleph_0 = aleph_0 transpositions are sufficient.) IF Cantor can construct his antidiagonal number, then we can also construct the above set (in zero time, of course). But if we do never finish, having found the smallest rational, then Cantor will never finish his antidiagonal. Both attempts are nonsense, of course. But Cantor's approach is considered meaningful by some mathematicians, as yet. It is not necessary to comment that there is no smallest rational and therefore my approach is nonsense. Let me know what distinguishes Cantor's approach in principle from mine. ==== Subject: Re: Cantor and the binary tree > IF actual infinity in fact did exist, then we could take the set of > well-ordered rational numbers (by Cantor or by Fraenkel or in some > other form) and make the following transpositions: Le q_0 be the first element of the well-ordering. Let q_1 be the first > rational, which is less than q_0. Exchange the positions of q_0 and > q_1. Let now q_2 be the first element which is smaller than q_1. > Exchange the positions of q_1 and q_2. Continue. According to Cantor > one can perform infinitely many transpositions without changing the > character of a well-ordered set. Cantar never said you could actuall 'do' infinitely many things. > At the end we have a well-ordered set the first element of which is the > smallest rational number. (And continuing, we can even obtain the > well-ordered and simultaneously ordered set < of all rationals, because > aleph_0 * aleph_0 = aleph_0 transpositions are sufficient.) IF Cantor can construct his antidiagonal number, then we can also > construct the above set (in zero time, of course). But if we do never > finish, having found the smallest rational, then Cantor will never > finish his antidiagonal. Both attempts are nonsense, of course. But Cantor's approach is > considered meaningful by some mathematicians, as yet. > It is not necessary to comment that there is no smallest rational and > therefore my approach is nonsense. Let me know what distinguishes > Cantor's approach in principle from mine. What you are doing is defining some sort of sequence of infinite sets. You want to talk about the limit of that sequence, so firt, you would have to sort of specify what exactly you mean be 'limit' in this case, and then, you have to prove the limit exists, and has some properties you want it to have. More specific: I suppose you really want to talk about the positive rationals. If zero is in your set, it is also the minimal number, so your algorithm ends at some finite point. If it is not, look at the first element of the sets in the sequence. At any point in the sequence, that element has some value, so you can define the sequence that at each point has the value of the first element of the set at that point. The limit of thist sequence is zero, so intuitively, if your sequence of sets has a limit, you might want the first element of that set to be zero -- which is not in any of the sets in your sequence! The 'diagonal' in cantor's proof, on the other hand, can be seen as the limit of a simple sequence of numbers. This limit is a real number, precicely because real numbers are defined as limits of such sequences! This limit is finite, and each of its decimals is very well defined, so we know exactly what we are talking about. So the main difference between what you do and Cantor's proof, is that you actually don't have any definition of limit, and therefore really can't say anything about what it is, while in Cantor's proof that is all completely clear. In summary, nobody (with any mathematical schooling) ever claims that you can 'do' infinitely many things, but we can define limits. Once you have a good definition of what a 'limit' is, there is nothing wrong with reasoning about them, and sometimes, mathematicians get carried away and use sloppy wordings like 'at infinity' or 'after infinitely many iterations' when they really mean a limit. Which, as this thread indicates, can be a bit confusing for some people. Jan ==== Subject: Re: Cantor and the binary tree >> So how you can get to your conclusion >> for the set of all even numbers using induction? > > because there is no even number at which the reasoning stops or gets > incorrect. > > Aargh. Pay attention now: > > The fact that something is true for each even number, does NOT mean it > is true for 'the set of even numbers'. > > Then the set of even numbers is worthless. On the contrary, it is WM's monomania which forces him to produce such false claims that is worthless. > Pay attention now: > > LIM = lim_{ntoinfty} = lim_{n --> oo} > Minimum: Min({1,1/2,1/3,...,1/n}) = 1/n > Maximum: Max({1,2,3,...,n}) = n > Cardinality: Card({1,2,3,...,n}) = n > > Limits: For A n e N we have: > > LIM (Min({1,1/2,1/3,...,1/n}) * Card({1,1/2,1/3,...,1/n}) = 1 > LIM (Min({1/2,1/4,...,1/2n}) * Card({1/2,1/4,1/6,...,1/2n}) = 1/2 > LIM (Max({1,2,3,...,n}) / Card({1,2,3,...,n}) = 1 > LIM (Max({2,4,6,...,2n}) * Card({2,4,6,...,2n}) = 2 That last statement is quite wrong! > > *Every* natural number is the maximum of a *finite* sequence. By the > sequence {1,2,3,...,n} for n --> oo we exhaust the set of all natural > numbers. There is literally nothing to be expected beyond the sequence > of every natural number. > > 'All even numbers' refers to lots of even numbers, while 'the set of > even numbers' refers to a single mathematical object (which happens to > be a set). > > which happens to be a chimera like actual infinity. That set of all even numbers has as much reality as does any even number in it. They are all ideals in the world of the mind, and have no physical existence. > > IF actual infinity in fact did exist, then we could take the set of > well-ordered rational numbers First one would have to well-order them, as they are not well-ordered under the standard ordering. What is your well-ordering of the ratinals, WM? ==== Subject: Re: Cantor and the binary tree > LIM (Min({1,1/2,1/3,...,1/n}) * Card({1,1/2,1/3,...,1/n}) = 1 > LIM (Min({1/2,1/4,...,1/2n}) * Card({1/2,1/4,1/6,...,1/2n}) = 1/2 > LIM (Max({1,2,3,...,n}) / Card({1,2,3,...,n}) = 1 > LIM (Max({2,4,6,...,2n}) * Card({2,4,6,...,2n}) = 2 *Every* natural number is the maximum of a *finite* sequence. By the > sequence {1,2,3,...,n} for n --> oo we exhaust the set of all natural > numbers. There is literally nothing to be expected beyond the sequence > of every natural number. Right. So, you prove something for all natural numbers. Not for 'the set of natural numbers', seen as an individual object. Oh, one more thing you might not understand: If something is true for all elements of a sequence, that does not mean it has to be true for the limit of the sequence. For instance, all elements in the sequence {1/1, 1/2, 1/3, 1/4, ...} are larger than zero, while the limit is not. Jan ==== Subject: Re: Help with Fibonacci & Co No amateur of this silly formula ? x*F(1)+xO*F(2)+x^3*F(3)+x^4*(F4)+...+x^n*F(n) = x/(1-x-xO) for x>1/Phi Always unable to resolve or prove it even if I tried many times. ==== Subject: Re: Help with Fibonacci & Co Le 29/06/05 16:00, dans 14282253.1120053687383.JavaMail.jakarta@nitrogen.mathforum.org, .82bischare a .8ecrit: > No amateur of this silly formula ? > > x*F(1)+x?*F(2)+x^3*F(3)+x^4*(F4)+...+x^n*F(n) = x/(1-x-x?) for x>1/Phi > > Always unable to resolve or prove it even if I tried many times. Sure, it's wrong :-) It's F(1)*x+F(2)*x^2+... (infinite series) = x/(1-x-x^2) To prove that, you should try to expand x/(1-x-x^2) in Taylor series, sum(a(n)*x^n), and find a recurrence relation between a(), a(n+1) and a(n+2). And since F(n+1)/F(n) -> (1+sqrt(5))/2, you can easily compute the radius of convergence. ==== Subject: Re: Help with Fibonacci & Co Another method maybe without considering the result x/(1-x-xO) but just by developing the first part of the equation ?.. you cheater :) ==== Subject: Re: Help with Fibonacci & Co Le 29/06/05 16:46, dans 23563621.1120056439180.JavaMail.jakarta@nitrogen.mathforum.org, .82bischare a .8ecrit: > Another method maybe without considering the result x/(1-x-x?) but just by > developing the first part of the equation ?.. you cheater :) But your equation (stop at F(n)*x^n) is wrong... You could try (1-x-x^2)*(F(1)*x+F(2)*x^2+...+F(n)*x^n) The result is very simple, though not exactly x. Then you could let n -> infinity, with the assumption that |x| is small enough, but not zero. ==== Subject: Re: Help with Fibonacci & Co Why is that wrong ? I read about the result on wikipedia, I don't pose it as a truth. I don't understand better, sorry. I don't know, make as if you were the first man looking at the serie F(1)*x+F(2)*x^2+...+F(n)*x^n (fund maybe during the hunt or the gathering of berries behind a bush), without any other inscription in the surroundings: how would you develop, triturate, manipulate it with your neanderthalian tools to have a result respecting the standards of a common rupestrian form, excluding the main contemporary easier means consisting in proving that an egality already verified is true ? ==== Subject: Re: Help with Fibonacci & Co Le 29/06/05 17:53, dans 4769286.1120060439742.JavaMail.jakarta@nitrogen.mathforum.org, .82bischare a .8ecrit: > Why is that wrong ? I read about the result on wikipedia, I don't pose it as a > truth. So you don't understand Wikipedia http://en.wikipedia.org/wiki/Fibonacci_number On this page, the result is right... > I don't understand better, sorry. I already explained a manipulation that will give you a simpler formula. > I don't know, make as if you were the first man looking at the serie > F(1)*x+F(2)*x^2+...+F(n)*x^n (fund maybe during the hunt or the gathering of > berries behind a bush), without any other inscription in the surroundings: how > would you develop, triturate, manipulate it with your neanderthalian tools to > have a result respecting the standards of a common rupestrian form, > excluding > the main contemporary easier means consisting in proving that an egality > already verified is true ? Funny... If you have verified this equality, please explain how you did... ==== Subject: Re: Help with Fibonacci & Co Of course I didn't! That's why I asked this question :) ==== Subject: Re: Help with Fibonacci & Co Le 29/06/05 19:10, dans 16203883.1120065063652.JavaMail.jakarta@nitrogen.mathforum.org, .82bischare a .8ecrit: > Of course I didn't! That's why I asked this question :) Ok. Wikipedia does not give the same formula as yours, but an infinite series, as I already explained. Now, if you want to simplify F(1)*x+...+F(n)*x^n, you should expand (1-x-x^2)*(F(1)*x+...+F(n)*x^n). When it's done, you should be able to make n -> infinity. ==== Subject: Re: Numerical quadrature for uneven spacing? > > Another possibility is to treat the integration problem as solving an > ODE dy/dt = f(t), and then solve it using a Runge-Kutta method. > > >> The problem is that the Runge-Kutta methods require evaluation of the >> function at intermediate points. > > > Oops. OK, try one of those multistep methods, like Adams-Basforth. (But > I can see a problem that you need to 'seed' it.) No, this requires equal spacing. Ignore my messages with respect to this post. ==== Subject: Pytagorean Theorem X-RFC2646: Format=Flowed; Original I need a little input on a problem I have encountered regarding the Pytagorean Theoreme. We all know that Kati+Kati=Hypi (yes?) So therefore... If I know the hypotenuse, and the differnce ratio between the two sides, I should be able to calculate the lenght of each side... right? For instance... if the hypotenuse is 400(give or take 0.22-->) and the ratio between the sides are 1 (they are the same), the sides are both 283. Yes? So, if we keep (aprox) the same hypotenuse of 400, but make the ratio 2:1, or 2 for simplicity, one side is twice as long as the other, and the shortest being 179 and the longest being 358. Also, if the ratio was 1:2, or 0.5, the sides are still the same, only they have changed place. Are you still with me? What I like to do is to figure out is what sort of equation I need to be able to calculate the sides, when I know the hypotenuse and the ratio between the sides. Is this even possible to do whitout resorting to the trial and error methode? ==== Subject: Re: Pytagorean Theorem First, if cosine is the same as sine ,the hypotenuse will never be a rational number.The pythagorean formula to find rational lengths for the hypotenuse are,(A^2-B^2)^2 +(A*B*2)^2 = (A^2+B^2)^2 . Does this help you out? ==== Subject: Re: Pytagorean Theorem > First, if cosine is the same as sine ,the hypotenuse will never be a > rational number.The pythagorean formula to find rational lengths for > the hypotenuse are,(A^2-B^2)^2 +(A*B*2)^2 = (A^2+B^2)^2 . Does this > help you out? Has the Obfuscated C Code Contest been extended to mathematics? ==== Subject: Re: Pytagorean Theorem Le 29/06/05 16:37, dans 42c2b222$1@news.broadpark.no, .82Norleif SlettebAe a .8ecrit: > I need a little input on a problem I have encountered regarding the > Pytagorean Theoreme. > We all know that Kati+Kati=Hypi (yes?) > So therefore... If I know the hypotenuse, and the differnce ratio between > the two sides, I should be able to calculate the lenght of each side... > right? > For instance... if the hypotenuse is 400(give or take 0.22-->) and the ratio > between the sides are 1 (they are the same), the sides are both 283. Yes? > So, if we keep (aprox) the same hypotenuse of 400, but make the ratio 2:1, > or 2 for simplicity, one side is twice as long as the other, and the > shortest being 179 and the longest being 358. Also, if the ratio was 1:2, or > 0.5, the sides are still the same, only they have changed place. > Are you still with me? > What I like to do is to figure out is what sort of equation I need to be > able to calculate the sides, when I know the hypotenuse and the ratio > between the sides. > Is this even possible to do whitout resorting to the trial and error > methode? Not sure I understand your question... If you know c and r where a^2+b^2=c^2, b/a=r, then a^2*(1+r^2)=c^2 hence a=c/sqrt(1+r^2) and b=c*r/sqrt(1+r^2). With your examples: c=400, r=1: a = b = c/sqrt(2) = 282.842... c=400, r=2: a = c/sqrt(5) = 178.885... and b = c*2/sqrt(5) = 357.770... ==== Subject: Re: Pytagorean Theorem It might help us if we knew the extent of your mathematics education. This is a trivial exercise in 1st or 2nd year junior high school algebra. All you need to know is how to solve a quadratic equation. If the ratio between the sides is r and the hypotenuse is H, then x^2 + (rx)^2 = H^2. This equation is just a simple statement of the conditions you specified. One side is x. The other is r*x, the hypotenuse is H. Just solve for x. What's The Problem?? Perhaps instead you want to ask for help in how to formulate simple word problems? ==== Subject: Re: Pytagorean Theorem X-RFC2646: Format=Flowed; Original 0.o Well... I finished scool some time ago, and this is VERY far from what I usually do, so please forgive me if I have forgotten some of the basics of geometry/algebra. Still, I have to say when I got the first replie, I got that Ahh! Why didn't I think of that right away-feeling. > It might help us if we knew the extent of your mathematics education. This is a trivial exercise in 1st or 2nd year junior high school > algebra. > All you need to know is how to solve a quadratic equation. If the > ratio between the > sides is r and the hypotenuse is H, then x^2 + (rx)^2 = H^2. This equation is just a simple statement of the > conditions you > specified. One side is x. The other is r*x, the hypotenuse is H. Just solve for x. What's The Problem?? Perhaps instead you want to ask for help in how to formulate simple > word problems? > ==== Subject: Re: Pytagorean Theorem a=h/sqrt(1+k^2) where h is the hypotenuse and the sides are a and k*a. ==== Subject: Re: Pytagorean Theorem X-RFC2646: Format=Flowed; Original I'ts spot-on. Can't beleive I forgot about that. I had a bad case of fitting a square peg in a round hole syndrome. > a=h/sqrt(1+k^2) where h is the hypotenuse and the sides are a and k*a. > ==== Subject: Re: The factoring problem > It's not clear what would constitute a satisfactory answer > to such a why?. I suppose that a proof that factoring is > in the class NP-complete would be accepted by many as such > an answer, because that would mean that as the factors are > made arbitrarily large, the amount of work expected for > factoring the product would become larger at such a rate > that it would be impossible to accomplish on any classical > computing device (assuming P!=NP). However, so far as I > know there has never been an accepted proof that factoring > has that asymptotic complexity class, and judging from > progress in factoring algorithms recently it doesn't feel > like an NP-complete problem. In fact, it is believed to be somewhere between 'NP' and 'P': the Number Field Sieve method has complexity O(e^{c(log n)^{1/3}(log log n)^{2/3}}). So as far as we know, factoring is easier than NP-complete problems. Of course, NP = P might still hold anyway :-) See http://mathworld.wolfram.com/NumberFieldSieve.html Jan ==== Subject: A property of the Fibonacci series? My son asked me about the Fibonacci series the other day, prompted by a generate the Fibonacci numbers. There were four columns of output from the program: the term number, its corresponding Fibonacci number and the sum of all the terms upto and including the current term. I noticed that the sum of the terms to (n-2) equalled the nth term, i.e F(n)= the sum to F(n-2). Is this a generally known property of the Fibonacci series? I've not come across it in the literature I've seen. The forth column of the computer program, by the way, gave the quotient F(n)/F(n-1) to show it approaches the Golden Ratio. I'm aware that this is a know property. ==== Subject: Re: A property of the Fibonacci series? Forget the above. I made a mistake. That should be F(n) = sum to F(n-2) + 1 which is what I observed and which a known property. ==== Subject: Re: A property of the Fibonacci series? > Forget the above. I made a mistake. That should be F(n) = sum to F(n-2) + 1 which is what I observed and > which a known property. Here are some more elementary properties that you and your son might enjoy exploring: As with a previous poster, for me the Fibonacci numbers are: F(1) = 1 F(2) = 1 F(3) = 2 and then 3, 5, 8, 13, 21, 34, 55, 89, 144, etc. 1) The square of any Fibonacci number differs by 1 from the product of the two surrounding Fibonacci numbers. In symbols, (F(n))^2 - F(n+1)*F(n-1) = (-1)^(n+1) 2) The product of any two consecutive Fibonacci numbers differs by 1 from the product of the 2 Fibonacci numbers that surround them. In symbols, F(n)*F(n+1) - F(n-1)*F(n+2) = (-1)^(n+1) 3) Fibonacci numbers of odd index are always the sum of the squares of two other Fibonacci numbers. In symbols F(2n+1) = (F(n))^2 + (F(n+1))^2 4) If a is a divisor of b, then F(a) is a divisor of F(b). It is almost true that if F(a) is a divisor of F(b) then a is a divisor of b, but there are some minor annoying details such as the fact that F(2) = 1 which is a divisor of everything but 2 is not a divisor of everything. One can easily get from these considerations that if F(n) is a prime number then either n is a prime number or n = 4. and so many more. There are many books that give all of these properties and the ways to derive them. Hope this is of interest, Achava ==== Subject: Re: A property of the Fibonacci series? And some I like: 2*F(n+1) = F(n) + sqrt(5*F(n)^2 + 4*(-1)^n) sum(atan(1/F(2*n+1)),n=1..infinity) = pi/4 | 1 1 |^n | 0 1 | Le 29/06/05 18:03, dans .82achava@hotmail.com[CapitalEGr ave] a .8ecrit: > > >> Forget the above. I made a mistake. >> >> That should be F(n) = sum to F(n-2) + 1 which is what I observed and >> which a known property. > > Here are some more elementary properties that you and your son might > enjoy exploring: > > As with a previous poster, for me the Fibonacci numbers are: > > F(1) = 1 > F(2) = 1 > F(3) = 2 > > and then 3, 5, 8, 13, 21, 34, 55, 89, 144, etc. > > 1) The square of any Fibonacci number differs by 1 from the product > of the two surrounding Fibonacci numbers. In symbols, > (F(n))^2 - F(n+1)*F(n-1) = (-1)^(n+1) > > 2) The product of any two consecutive Fibonacci numbers differs by 1 > from the product of the 2 Fibonacci numbers that surround them. > In symbols, > F(n)*F(n+1) - F(n-1)*F(n+2) = (-1)^(n+1) > > 3) Fibonacci numbers of odd index are always the sum of the squares of > two other Fibonacci numbers. In symbols > F(2n+1) = (F(n))^2 + (F(n+1))^2 > > 4) If a is a divisor of b, then F(a) is a divisor of F(b). It is > almost true that if F(a) is a divisor of F(b) then a is a divisor > of b, but there are some minor annoying details such as the fact > that F(2) = 1 which is a divisor of everything but 2 is not a > divisor of everything. One can easily get from these > considerations that if F(n) is a prime number then either n is a > prime number or n = 4. > > and so many more. > > There are many books that give all of these properties and the ways > to derive them. > > Hope this is of interest, > Achava > ==== Subject: Re: A property of the Fibonacci series? > My son asked me about the Fibonacci series the other day, prompted by a > generate the Fibonacci numbers. There were four columns of output from > the program: the term number, its corresponding Fibonacci number and > the sum of all the terms upto and including the current term. > > I noticed that the sum of the terms to (n-2) equalled the nth term, i.e > F(n)= the sum to F(n-2). For me, the first elements of the Fibonacci sequence are F(1) = 1 F(2) = 1 F(3) = 2 F(4) = 3 F(5) = 5 F(6) = 8 ... With this definition, what is true is that F(n) = F(1) + F(2) + ... + F(n - 2) +1 This result is due to Edouard Lucas, in 1876. Jose Carlos Santos ==== Subject: Re: A property of the Fibonacci series? Le 29/06/05 17:16, dans a .8ecrit: > My son asked me about the Fibonacci series the other day, prompted by a > generate the Fibonacci numbers. There were four columns of output from > the program: the term number, its corresponding Fibonacci number and > the sum of all the terms upto and including the current term. > > I noticed that the sum of the terms to (n-2) equalled the nth term, i.e > F(n)= the sum to F(n-2). Slight mistake: sum(F(k),k=0..n-2) = F(n) - 1 With F(0) = 0, F(1) = 1, F(n+1) = F(n) + F(n-1) > Is this a generally known property of the Fibonacci series? I've not > come across it in the literature I've seen. It's known and easily proven by induction: It's true for n=2: F(2) - 1 = 0 = F(0) If it's true for n, then F(n+1) - 1 = F(n) - 1 + F(n-1) = sum(F(k),k=0..n-2) + F(n-1) = sum(F(k),k=0..n-1) Hence it's true for n+1. > > The forth column of the computer program, by the way, gave the quotient > F(n)/F(n-1) to show it approaches the Golden Ratio. I'm aware that this > is a know property. > ==== Subject: Prgramming Contest: Geometry/Maps/Shapes Thought this might be of interest to the math geeks who program as well as the programming geeks who love math ... I'm afraid I'm a bit of both! GLOSSY: The Summer Programmer Of The Month Contest is underway! Deadline is September 30, 2005 http://dinsights.com/POTM ---------------------------------------------------------------------------- ---- I love taking digital pictures, but that nice glossy photo paper is expensive! So when my lovely wife asks for prints of a bunch of pictures, I always try and fit them onto a single piece of paper. The summer POTM challenges you to write a program that will place up to 26 pictures on a sheet of paper with the least amount of leftover space. ---------------------------------------------------------------------------- ---- The Programmer Of The Month (POTM) is a just-for-fun programming contest with an active forum and over 1100 members (we usually get 40-50 entries for each POTM challenge). Supported languages include C/C++/Perl/Ruby/PHP/Python/awk/shell. Emphasis is on the algorithms and the joy of programming. If it sounds like fun, please visit at http://dinsights.com/POTM ! =Fred (a.k.a. The POTM-MASTER) ==== Subject: Re: (N+1)^N/N^N > No, no, no, no Gheorge! You've messed things up! The > number e is BOTH > irrational and trascendental. There's NO dichotomy in > the real numbers > system: irrational OR (exclusive) trascendental. Some > numbers can be > both. The proof of the irrationality of e is a pretty > basic and easy > one, and any mild brained student of a first calculus > course could > understand it, so when you state that e is NOT > irrational it REALLY > makes you look bad and could make some people wonder > whether your > proofs of FLT are poorly expressed in english > because you can't > manage with the language or because of some lack of > depth in > matehmatics. > Tonio > According to classification of numbers Which I learned we have a set of two kinds of numbers: 1).numbers which can be solutions to a algebraic equation=Real Number 2)numbers which can not be a solution to a algebraic ecuation= =transcendental numbers Now nr 1).includes: a) number 0 a')natural numbers= positive integers b). [natural numbers+(negative integers]=Integers c) rational numbers=numbers which can be express as fraction of integers numbers d). irrational numbers=numbers which can not be express as a fraction of rational numbers e)complex numbers = a*i+b where a anb take values in a),a'),b),c),d) nr.2) transcendental numbers As you see ,it is only your definition of irrational of numbers that makes you confuse.But not a mathematician! Why ?Because a mathematicien knows this clasification of numbers. george ==== Subject: Re: (N+1)^N/N^N Le 29/06/05 17:49, dans 3551818.1120060213682.JavaMail.jakarta@nitrogen.mathforum.org, .82george ghiatae a .8ecrit: >> No, no, no, no Gheorge! You've messed things up! The >> number e is BOTH >> irrational and trascendental. There's NO dichotomy in >> the real numbers >> system: irrational OR (exclusive) trascendental. Some >> numbers can be >> both. The proof of the irrationality of e is a pretty >> basic and easy >> one, and any mild brained student of a first calculus >> course could >> understand it, so when you state that e is NOT >> irrational it REALLY >> makes you look bad and could make some people wonder >> whether your >> proofs of FLT are poorly expressed in english >> because you can't >> manage with the language or because of some lack of >> depth in >> matehmatics. >> Tonio >> > According to classification of numbers Which I learned > we have a set of two kinds of numbers: > 1).numbers which can be solutions to a algebraic equation=Real Number Wrong: they are called real algebraic numbers. > 2)numbers which can not be a solution to a algebraic ecuation= > =transcendental numbers Right. And the set of real algebraic and transcendent numbers is called R, the set of real numbers. > > Now nr 1).includes: > a) number 0 > a')natural numbers= positive integers > b). [natural numbers+(negative integers]=Integers > c) rational numbers=numbers which can be express as fraction > of integers numbers > d). irrational numbers=numbers which can not be express > as a fraction of rational numbers Wrong: irrational numbers are all numbers which are not rationnal, either taken in R or C according to context. Numbers in Z[i] are also called gaussian integers. > e)complex numbers = a*i+b where a anb take values in > a),a'),b),c),d) Wrong: complex algebraic numbers. > nr.2) transcendental numbers Right. Either real or complex according to context. > As you see ,it is only your definition of irrational of numbers > that makes you confuse. Who is confused ? You may have a look at http://mathworld.wolfram.com/IrrationalNumber.html http://mathworld.wolfram.com/TranscendentalNumber.html http://mathworld.wolfram.com/AlgebraicNumber.html http://mathworld.wolfram.com/RealNumber.html http://mathworld.wolfram.com/ComplexNumber.html http://mathworld.wolfram.com/RationalNumber.html http://mathworld.wolfram.com/GaussianInteger.html > But not a mathematician! > Why ?Because a mathematicien knows this clasification of numbers. So I can conclude you are not a mathematician. But it's well known on sci.math. ==== Subject: Re: (N+1)^N/N^N > > > > Le 29/06/05 17:49, dans > 3551818.1120060213682.JavaMail.jakarta@nitrogen.mathfo > rum.org, .82george > ghiatae a .8ecrit: > >> No, no, no, no Gheorge! You've messed things up! > The >> number e is BOTH >> irrational and trascendental. There's NO dichotomy > in >> the real numbers >> system: irrational OR (exclusive) trascendental. > Some >> numbers can be >> both. The proof of the irrationality of e is a > pretty >> basic and easy >> one, and any mild brained student of a first > calculus >> course could >> understand it, so when you state that e is NOT >> irrational it REALLY >> makes you look bad and could make some people > wonder >> whether your >> proofs of FLT are poorly expressed in english >> because you can't >> manage with the language or because of some lack > of >> depth in >> matehmatics. >> Tonio >> > According to classification of numbers Which I > learned > we have a set of two kinds of numbers: > > 1).numbers which can be solutions to a algebraic > equation=Real Number > > Wrong: they are called real algebraic numbers. > > 2)numbers which can not be a solution to a > algebraic ecuation= > =transcendental numbers > > Right. > > And the set of real algebraic and transcendent > numbers is called R, > the set of real numbers. > > > Now nr 1).includes: > a) number 0 > a')natural numbers= positive integers > b). [natural numbers+(negative integers]=Integers > c) rational numbers=numbers which can be express > as fraction > of integers numbers > d). irrational numbers=numbers which can not be > express > as a fraction of rational numbers > > Wrong: irrational numbers are all numbers which are > not rationnal, > either taken in R or C according to context. > > Numbers in Z[i] are also called gaussian integers. > > e)complex numbers = a*i+b where a anb take values > in > a),a'),b),c),d) > > Wrong: complex algebraic numbers. > > nr.2) transcendental numbers > > Right. Either real or complex according to context. > > As you see ,it is only your definition of > irrational of numbers > that makes you confuse. > > Who is confused ? > > You may have a look at > > http://mathworld.wolfram.com/IrrationalNumber.html > http://mathworld.wolfram.com/TranscendentalNumber.html > http://mathworld.wolfram.com/AlgebraicNumber.html > http://mathworld.wolfram.com/RealNumber.html > http://mathworld.wolfram.com/ComplexNumber.html > http://mathworld.wolfram.com/RationalNumber.html > http://mathworld.wolfram.com/GaussianInteger.html > > But not a mathematician! > Why ?Because a mathematicien knows this > clasification of numbers. > > So I can conclude you are not a mathematician. But > it's well known on > sci.math. > In short terms you have to say only that transcendental numbers are clasified as irrational numbers but not all irrational numbers are transcendental. In fact classification given by me makes that distinction. I am happy that you said that I proved myself that I am not a mathematician. george ==== Subject: Re: (N+1)^N/N^N > In short terms you have to say only that transcendental numbers > are clasified as irrational numbers but not all irrational numbers > are transcendental. Exactly. > In fact classification given by me makes that distinction. Great. But nobody cares, ya know. > I am happy that you said that I proved myself that I am not a mathematician. Oh, I can assure you that it wasn't difficult. ==== Subject: Re: (N+1)^N/N^N >According to classification of numbers Which I learned >we have a set of two kinds of numbers: >1).numbers which can be solutions to a algebraic equation=Real Number 2)numbers which can not be a solution to a algebraic ecuation= >=transcendental numbers According to classification of recent posters from Math Forum Which I learned we have a set of two kinds: 1) trolls 2) idiots. They are, of course, not mutually exclusive. Lee Rudolph ==== Subject: Re: (N+1)^N/N^N And the respones get larger and larger.You all out there in e land ,i just posted just one well known equaton and say that it is a vulgar fraction and i get some rather hostile reactions,MY MY.sOME OF YOU HERE IN MATHFORUM TAKE THIS MATH way too personal and serious.To me ,its a hobby and yes i too can obsess on some ideas,but to get hostile and carry out character assasinationa,come on ?So far ,i have not seen where i am wrong.Some of you here trear e as if it is RATIONAL,and its not.I suppose some of you here treat PI as if it is rational,especially when it has been DECIMALIZED.oOh ,i better stop. Im getting hostile myself. ==== Subject: Re: (N+1)^N/N^N Le 29/06/05 16:29, dans 21444724.1120055412321.JavaMail.jakarta@nitrogen.mathforum.org, .82Tonye a .8ecrit: > And the respones get larger and larger.You all out there in e land ,i just > posted just one well known equaton and say that it is a vulgar fraction and i > get some rather hostile reactions,MY MY.sOME OF YOU HERE IN MATHFORUM TAKE > THIS MATH way too personal and serious.To me ,its a hobby and yes i too can > obsess on some ideas,but to get hostile and carry out character > assasinationa,come on ?So far ,i have not seen where i am wrong.Some of you > here trear e as if it is RATIONAL,and its not.I suppose some of you here treat > PI as if it is rational,especially when it has been DECIMALIZED.oOh ,i better > stop. Im getting hostile myself. Manifestly you know nothing about rationnal numbers (hey, you asked if (n+1)^n/n^n is rationnal !). It's not hostility, just observation. Nobody in this thread ever said e is rationnal, the proof that it is not is almost trivial, BTW. ==== Subject: Re: (N+1)^N/N^N <21444724.1120055412321.JavaMail.jakarta@nitrogen.mathforum.org> Who's treating e as though it's rational? If you're going to make ridiculous accusation don't you think it's a bit absurd to take offense to others making the same accusations towards you? ==== Subject: Re: (N+1)^N/N^N <30476278.1119965359252.JavaMail.jakarta@nitrogen.mathforum.org> I see the principle of induction as one of the more non-intuitive > aspects of mathematics. I don't see how to show that such a finite > length algorithm exists without using induction. I mean I can do it for > numbers like 2, 10 or even 100. But I don't see how to do it for > seems to me that this is essentially using induction. So without the (non-intuitive at least for you) mathematical apparatus you can not reason about that? >> Is the concept of God logically consistent? This approach isn't likely to find an answer to the question. Why? Troy ==== Subject: Re: (N+1)^N/N^N Le 29/06/05 7:30, dans a .8ecrit: > No, no, no, no Gheorge! You've messed things up! The number e is BOTH > irrational and trascendental. There's NO dichotomy in the real numbers > system: irrational OR (exclusive) trascendental. Some numbers can be > both. The proof of the irrationality of e is a pretty basic and easy > one, and any mild brained student of a first calculus course could > understand it, so when you state that e is NOT irrational it REALLY > makes you look bad and could make some people wonder whether your > proofs of FLT are poorly expressed in english because you can't > manage with the language or because of some lack of depth in > matehmatics. Exactly ;-) ==== Subject: Re: (N+1)^N/N^N Discussion, linux) > mythical. isn't there? e is computable, if I'm not badly confused. Such algorithms cannot complete the task in a finite length of time or space, but the time and space needed to write out a googolplex (note spelling) is also unattainable in this particular universe. (Current a bit lower than a googol, so where do you plan to write all them zeroes?) I'm not sure I see a deep difference here. -- You are beneath contempt because you betray mathematics itself, and spit upon the truth, spit upon decency, and spit upon the intelligence of the world. You betrayed the world, and now it's time for the world to notice. -- James S. Harris awaits Justice for crimes against Math. ==== Subject: Re: (N+1)^N/N^N <30476278.1119965359252.JavaMail.jakarta@nitrogen.mathforum.org> <87ll4tshgh.fsf@phiwumbda.org > mythical. isn't there? e is computable, if I'm not badly confused. If you want the last digit of e then there is a problem. > Such > algorithms cannot complete the task in a finite length of time or > space, but the time and space needed to write out a googolplex (note > spelling) is also unattainable in this particular universe. (Current > a bit lower than a googol, so where do you plan to write all them > zeroes?) 0. Can you define the concept of not finite time? 1. the spelling is the same as that used in Stephen's message 3. It doesn't matter whether or not there is sufficient number of 4. Why not try more fresh cliche? Troy ==== Subject: Re: (N+1)^N/N^N >If you want the last digit of e then there is a problem. On the other hand, asking for the last digit of pi is a way to remove an evil creature from your computer (well, it worked for Spock). Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada ==== Subject: Re: (N+1)^N/N^N <87ll4tshgh.fsf@phiwumbda.org> Discussion, linux) >> mythical. >> isn't there? e is computable, if I'm not badly confused. If you want the last digit of e then there is a problem. If you want the last digit of e, then you are a moron. Why not ask for a round triangle? >> Such >> algorithms cannot complete the task in a finite length of time or >> space, but the time and space needed to write out a googolplex (note >> spelling) is also unattainable in this particular universe. (Current >> a bit lower than a googol, so where do you plan to write all them >> zeroes?) 0. Can you define the concept of not finite time? It's not necessary to define any such concept. The utterly standard way of defining computable reals doesn't define not finite time. > 1. the spelling is the same as that used in Stephen's message So what? It was wrong. See above. 3. It doesn't matter whether or not there is sufficient number of This seems utterly opaque. Yes, a googolplex is finite, and so is e. But e has an infinite number of digits and a googolplex has so many digits that it can't be written down in the actual universe (if written down means writing each decimal digit). So what? > 4. Why not try more fresh cliche? Go to hell. -- Jesse F. Hughes I'm ruler, said Yertle, of all that I see. But I don't see enough. That's the trouble with me. -- Yertle the Turtle, by Dr. Suess ==== Subject: Re: (N+1)^N/N^N > isn't there? e is computable, if I'm not badly confused. > > If you want the last digit of e then there is a problem. > Actually, I want the last digit of 1/7 ==== Subject: Re: (N+1)^N/N^N <30476278.1119965359252.JavaMail.jakarta@nitrogen.mathforum.org> <87ll4tshgh.fsf@phiwumbda.org> <290620051311538915%anniel@nym.alias.net.invalid > isn't there? e is computable, if I'm not badly confused. > > If you want the last digit of e then there is a problem. > > Actually, I want the last digit of 1/7 I'll bet it isn't divisible by 3 ==== Subject: Re: (N+1)^N/N^N > >mythical. >>isn't there? e is computable, if I'm not badly confused. > > > If you want the last digit of e then there is a problem. > Ah, now I see the mettle of the man with whom I am having an argument. :-) Here's a puzzle for you. What is the last digit of the decimal expansion of 1/2? ==== Subject: Re: (N+1)^N/N^N <30476278.1119965359252.JavaMail.jakarta@nitrogen.mathforum.org> <87ll4tshgh.fsf@phiwumbda.org> <9mywe.115037$nG6.88540@attbi_s22> What's your argument? ==== Subject: Re: (N+1)^N/N^N Discussion, linux) > No, no, no, no Gheorge! You've messed things up! The number e is BOTH > irrational and trascendental. There's NO dichotomy in the real numbers > system: irrational OR (exclusive) trascendental. Some numbers can be > both. Some? I reckon that *every* transcendental number is also irrational, no? -- I need to brief someone in government[...] Now if you acknowledge that my research MIGHT be important, then you can agree with me that it needs to be taken to areas off Usenet where some serious research can take place behind closed doors. I'd prefer the NSA. James S. Harris ==== Subject: Re: (N+1)^N/N^N <873br2tjo4.fsf@phiwumbda.org> <42c19bce$0$9305$626a14ce@news.free.fr> Discussion, linux) > 873br2tjo4.fsf@phiwumbda.org... >> Yeah, like we'd believe you. You're one of them EXPERTS, I'd wager. > You are EXPERTS! I'm not. But are you sure being an EXPERT carries such negative features? > Proove that $e^{sqrt{pi+1}}$ is irrational. Pass. -- Jesse F. Hughes This Trojan appears to utilize a function of the Windows Media DRM designed to enable license delivery scenarios as part of a social engineering attack. -- MS candidly explains the role of DRM licenses ==== Subject: Hurray for Kronecker The man was small in stature,but had the intellectual HONESTY to see what the irrationals were and what they stood for.I suppose this statement will generate hostililty. ==== Subject: Re: Hurray for Kronecker > > The man was small in stature,but had the intellectual HONESTY to see what the irrationals were and what they stood for.I suppose this statement will generate hostililty. Why hostility? Wouldn't one expect Kronecker to see what the irrationals were and what they stood for? Btw, what has his stature got to do anything? -- I don't know who you are Sir, or where you come from, but you've done me a power of good. ==== Subject: Re: Hurray for Kronecker Le 29/06/05 18:08, dans 4548443.1120061348674.JavaMail.jakarta@nitrogen.mathforum.org, .82Tonye a .8ecrit: > The man was small in stature,but had the intellectual HONESTY to see what the > irrationals were and what they stood for.I suppose this statement will > generate hostililty. Intellectual honesty has nothing to do with seeing something... ==== Subject: Re: Hurray for Kronecker > The man was small in stature,but had the intellectual HONESTY to > see what the irrationals were and what they stood for.I suppose > this statement will generate hostililty. Hostility towards *you*?! Why bother? Jose Carlos Santos ==== Subject: Re: Hurray for Kronecker >The man was small in stature,but had the intellectual HONESTY to see what the irrationals were and what they stood for.I suppose this statement will generate hostililty. No, just amusement. ************************ C. Ullrich ==== A first possible Hystorical reconstruction: Well,as I have shown on this disscusion group the origin of Pell's equation can be traced to the studies by trial and error of special Pytogora's triangles a^2+b^2=(b+1)^2 probably from 500.b.c to Archimedes time.Archimedes challenged his contemporaries to resolve a Pell's equation possible around 225 b.c.discovering very possible the Fermat's algorythm shown on this disscusion group. He knew how to calculate the summe of finite and inffinite of geometrical series.Before him this topic goes back to at least1200 b.c.as a papyrus shows ,and Euclid at least 100 years before Archimedes gives a clumsy formula to calculate this summe). Archimedes'formula is the same like the formula we used today. That shows that he knew that : X^n+Y^n=(X+Y)[X^(n-1)-Y*X^(n-2)+(Y^2)*X^(n-3).........+Y^(n-1)] For sure he knew how to expand (X+Y)^2 (it goes back at least to Euclid time). What we can detect from his work is that where in his time some problems which the mathematiciens before him tried to find long due answers to, and Archimede studied them and was succsefull about many of them. What kind the questions or problems? Well,to name 3: 1)Is there any solutions for any D(square free)to the Pell's equation? 2)Is there anyway an answer to what is the summ of infinite geometrical series? 3)Can the Pytagora's triangle formula be extended to powers greater than 2? I claim that Archimedes knew how to expand the bynomial (X+Y)^n and knew at least that the coeficient of X^(n-1)term and Y^(n-1) term is equal to n.This guy Was Smart(what a platitude!). He resolved much harder problems than this.Let's recognise! So he atacked The 3d problem Or question(Fermat last Theorem) and come out with Fermatlast theorem. What I Am saying is the in the time of Fermat where more of Archimedes's works manuscripts than today. And we know that Fermat had a collection of Archimedes works and in one of them (which did not survive )Fermat met the Archimedes*s challenges about Pell'Equation and Last theorem at least.And he went on and proved them. george ghiata Transform Y^2=D*X^2+1 into Pytogora triangle a^2+b^2=c^2 Here it is; Y^2=D*X^2+1 Case 1: D=2*C-1 a=Y-X b=X*(C*X-Y) c=X*(C*X-Y)+1 a^2+b^2=c^2 Case 2: D=C-1=even number X must be=2*K a=Y-X b=X*(C*K-Y) c=X*(C*K-Y)+1 a^2+b^2=c^2 Observation:Y and X are INTEGERS:We can invent a new problem,no? Create by george ghiata Here it is another one: Here is the theorem: If A,B,C are natural numbers then it takes less than 15 arithmetical operations( summ,difference,multiplication and division) tofind X,Y,Z natural numbers such that A*X+B*Y=C*Z; Let's take A , B , C then write : 2*A*B*C + Q - B^2 = C^2*(1 +2*A*B*C) Let's take x=A*B*C +Q and y =A*B*C - B^2 Now x+y is divisible by C^2 Now x - y = Q + B^2; and Q - B^2 =C*S Let's take :[ y*(Q +B^2) + 2*Q*B^2] and [ - x*(Q+B^2)+2*Q*B^2] If we add-up the two above expressions we get a result divisible by C^2 If we substract the two expressions we get a result divisible by C^2 Therefore [ y*(Q+B^2) + 2*Q*B^2] is divisible by C^2 If we substitute y with its value we get that [ A*B*C*(Q +B^2) - B^2*(Q +B^2) + 2*Q*B^2] is divisible by C^2 and B Therefore EQ1= A*B*C*(Q+B^2) + B^2*(Q - B^2) = C^2*B*Z Therefore A*(Q+B^2) + B*S=C*Z Therefore A*X + B*Y = C*Z If we count the number of arithmetical operations to get this result we find to be 11(eleven).QED-or may be 15 any way. Created by Gheorghe Ghiata HEllo Here is an experiment: symbol for ballbearing -> O symbol for solid board -> --------------- symbol for solenoid -> [] symbol for imaginary line = = = = -------------O----------- -------------O----------- Above we see two solid boards with the implented ballbearings and the solenoids between them Now we fix a rectangular magnet[tinny bar -like] in the ballbearing in the upper board (hangindown betwwen solenoids but above the imaginary double line shown inthe figure) Now we fix a round-flat magnet in the ballbearing in the lower board sitting upright between solenoids but below the imaginary line shown in the figure. Now we get closer and closer the two magnets by moving the boards verticaly toward each other. They will start to rotate inducing electric curent in the solenoids.I think that will be a selfreliant and cheap way to get electricity(thousands of small magnets can be used) Please describe how many things will be changed. Created by George ghiata My name is Gheorghe Ghiata- A second possible historical reconstruction Archimedes had a method to resolve Pell's equation Fermat Redescovere it and proved by infinite descent method that allways works. That method is shown first herein the text. Archimedes Proved FLT for n=3 by infinite Descend method and Fermat rediscovered it. and the formula: Cnk=n*(n-1)*(n-2)*(n-3)*.........*(n-k+1)/1*2*3*$.......*k which Was published by italian Math.Cardano in 1575 and Fermat's friend Merssene republish it in France in 1631.Archimedes probaly did not know the formula for large n's . That is why Fermat challenged his contemporaries FLT proof for n=3 The proof was the Same for any other odd prime n. The elementary Proof Of FLT by infinite Descend Method for any n=odd prime>1 is shown in this text First: About Pell's Equation I want to simplify my presentation of the method for which I have a proof by infinite descend methodthat It works allways to solve Pell's EQuation I presented it in writing to University Of victoria B.C CANADA and some of it before that to another professors.(In fact a sketch of it I sent 3-4 yaers ago to Mr.Prof Andrew Wiles!) You are going to see what Euler and many other mathematiciens miss to find. I never seen any where this :in no elemenetary theory of numbers books or advanced ones. Well ,it is only elementary high school algebra! Here is a simple exemple: Y^2=D*X^2+1 Let's choose D=29. Let's choose a<(29)^(1/2)>a+1 a=5. Now we write: 5^2-29=(1)*(-4) b^2-29=-4*k We choose b as being the largest number <(29)^(1/2)such that b+5 is divisible by factor 4(in this case) and we continuu the algorithm following this rule. So we have: 5^2-29=(1)*(-4) I 3^2-29=-4*5 II 2^2-29=-5*5 III 3^2-29=-5*4 5^2-29=-4*1 Now we know that (a^2+s*b^2)*(c^2+s*d^2)=(a*c-s*b*d)^2+s*(a*d+b*c)^2 So we multiply I*II=a^2-29*b^2=5*(4)^2 The proof shows that 4^2 simplifies and we get: I'=c^2-29*d^2=5 Now we multiply (I')*III= e^2-29*f^2=-5*5^2 After we divide by 5^2 we continuu this algorithm to the end and get x^2-29*y^2=-1 Now we multiply this identity by itself and get X^2-29*Y^2=1 The proof is elementary high school algebra technique. How in the world Euler could miss to find this algorythm and he never was able to proof Pell's equation ? More than that.It took for him 7 years to proof that a prime=4*k+1 can be represented as a^2+b^2 and to get a method how to find this representation. Whereas this algorithm which i called Fermat's-Archimedes Algorythm shows how to do it: As you see we got above the solution x^2+1=29*y^2 We can get from above a^2+1=29*b where a <29 and apply Fermat's-Archimedes algorythm c^2+1=b*e where c+a=b* ........ ........ x^2+1=y*1 Now we do the multiplication algorythm and get the representation X^2+Y^2=29 As you see we get y*1 as the last product. Well is needed only high school algebra to show that if we have x^2+s=D*b where s >0 then in the last product y*v we have 1= < v=< (s+1)/2 When s <0 there is a similar result but but different. In this way Fermat proved all his statements about the representations of diferrent primes .I proved them too. Try the one about, that all the primes 4*k+3 of X^2+5*Y^2 have as the last digit 3 or 7 Created by Gheorghe Ghiata Professor Harold Edwards in his book Fermat Last Theorem' Pell's eq. always have a infinity of solutios by proprely applyed infinit descend method.Mr Prof.says that at this time(1977) we do not see how Fermat could had aplyed that method for Pell's equation. Well, I rediscovered this Algorythm(may be somedoyelse too)and I named it Fermat Algorythm(Ishould had called it Archimedes algorythm) and I rediscovered the infinite descend method which proves that the algorythm always works(It is a high schoolAlgebra level). I would show now the begining of the proof by infinite descend method which Fermat talked about and which I send to Mr. Prof.Andrew Wiles in Dec.2002.The whole proof had be sent later to Mr.Prof.K Sound from University Of Michigan and Prof. Leeming from University Victoria B.C. Canada Here It Is : Let's take: X^2=A*Y^2+1 Let''take : (R0)^2-A=-(k0)*(k1) where (R0)^22 is impossible. OBSERVATION: X,Y ,Z relative Prime numbers Let's say that X^n+Y^n=Z^n Lets take: X+Y=W and R=X^(n-1)-(X^(n-2))*Y +(X^(n-3))*Y^2.........+Y^(n-1)= = W^(n-1)-(Cn1)*[W^(n-2)]*Y+(Cn2)*[W^(n-3]*Y^2- -(Cn3)*[W^(n-4)]*Y^ +(Cn4)*[W^(n-5)]*Y^4-............ ........-(Cn2)*[W^1)*Y^(n-2 ) +(Cn1)*Y^(n-1) where Cnk=[n*(n-1)*(n-2)*(n-3)*.........*(n-k+1)]/2*3*4*.... .*k When Z is not divisible by n we see that (X+Y)=W and R do not have any common divisor . Therefore they are relative prime and must be: R=z^n and W=X+Y=u^n and Z=u*z When Z is divisible by n,that is Z=u*n*z we see that W and R have as common divisor only n Therfore W=(u^n)*n^(n-1) and R=n*z^n The end of OBSERVATION CASE1 For simplicity of presentation lets take X*Y*Z not divisible by n When Z*X*Y is divisible by n the proof is almost the same. PROOF: Let's take : X^n+Y^n=Z^n X+Y- Z=B X=B+Q Y=B+P Therefore : EQ1: (B+Q)^n+(B+P)^n=(B+Q+P)^n X+Y=2*B+Q+P=u^n=W Z=B+Q+P=u*z B=b*u and Q+P=u*s and - z=-u^(n-1) +b and z-b=s 2*b+s=u^(n-1) b+s=z We see that we can write : X= - B+W-P and Q= -2*B +W-P Y= - B +W-Q and P= - 2*B+W-Q Z= - B+W Therefore we get : EQ2: (-B+W)^n +(-B-P)^n=(-B+W-P)^n Q=-2*B+W-P=q^n X=B+Q= -B+W-P=q*(z1) Therefore we get : B=c*q EQ3: (-B +W)^n +(-B-Q)^n=(-B+W-Q)^n P=-2*B+W-Q=p^n Y=B+P=-B+W-Q=p*(z2) B=d*p Since X,Y, Z are relative prime we have u,q,p relative prime. Therefore B=k*q*p*u=b*u Therefore b is divisible by q*p Lets take : X^n-X +Y^n-Y = Z^n-Z+Z-X-Y X+Y-Z=B is divisible By n Let's take: EQ1 : (B+Q)^n+(B+P)^n=(B+Q+P)^n EQ1: (2*B+Q+P)*T=(B+Q+P)^n where T=z^n can not be equal to 1 . Therefore z is not equal to 1 T= [B^(n-1)+(c1)*B^(n-2)+(c2)*B^(n-3)+....(ct)*B+(Q^n+P^n)/(Q+P)]=z^n and (c1),(c2).....(ct) are integers coeficients and (ct)= {n*Q^(n-1)+n*P^(n-1)-2*[Q^n+P^n]/(Q+P)}/(Q+P)=(Q+P)*V Since X^n-X+Y^n-Y is divisible by n we get that (T-1)=z^n-1 is divisible by n. Therefore (z-1) is divisible by n. Since B and Z have as common divisior only u We can write : T= Z*B*F(q,p)+[Q^n+P^n]/(Q+P)=z^n=[-b+u^(n-1)]^n where F(q,p) is a function=F of (q,p) When we develop the parantheses of z^n We see that z^n=G(q,p) where G(q,p) is a Function =G of (q,p) of (q,p) Let's multiply T by (Q+P): T*(Q+P)=(Q+P)*Z*B*F(q,p) +(Q^n+P^n)=(Q+P)*z^n=(u*s)*z^n T*(Q+P)= (Q+P)*Z*B*F(q,p) +(u*s)*(z*M)=(Q+P)*z^n Since z*u=Z we can divide by Z and get: T*(Q+P)/Z= (Q+P)*B*F(q,p) +s*M= s*z^(n-1) Since s=z-b we can write : T*(Q+P)/Z= (Q+P)*B*F(q,p)+s*M=z^n-b*z^(n-1) Since b is divisible by q*p we that s*M=A(q,p) where A(q,p) is a function =A of (q,p) Therefore: (Q^n+P^n)= (Q+P)*[Q^(n-1)-P*Q^(n-2) +........P^(n-1)] =Z*A(q,p)=(B+Q+P)*A(q,p) whereZ=B+Q+P=D(q,p) is a function D of (q,p) But we know that [Q^n+P^n] is a reducible polynom of (q,p) only one way to two factors. Therefore the only solution is [Q^n+P^n]/(Q+P) =z If we follow the same procedure with EQ2 we get the same results: 1) [(z1)-1] is divisible by n 2) (W^n-P^n)/(W-P)=(z1) 3) (z1) can not be equal to 1 If we follow the same procedure with EQ3 we get the same results: 1) [(z2)-1)] is divisible by n 2) (W^n-Q^n)/(W-Q)=(z2) 3)(z2) can not be equal 1 Now we take: EQ4 : X^n-Q^n+Y^n-P^n -Z^n+W^n=(-Q^n+Q)+(-P^n+P)+(W^n-W)+(W-Q-P) The EQ4 can be writen as : EQ4: n*m*B=(n^2)*J+2*B Therefore B is divisible by n^2 Therefore [(T/z)-1]=[z^(n-1)-1] is divisible by n^2 Since (z-1) is divisible by n we get from above that (z-1) is divisible by n^2 Since z+b=u^(n-1) and since b is divisible by n^2 we get that [ u^(n-1)-1] is divisible by n^2 The same way we get that 1) [(z1)-1] is divisible by n^2 2) [q^(n-1)-1] is divisible by n^2 The same way we get that 1) [(z2)-1] is divisible by n^2 2) {p^(n-1)-1} is divisible by n^2 With this new data we get that EQ4 can be written : n*B*m={n^3)*J'+2*B Therefore we started a Infinite descend. Case2: If X*Y*Z is divisible by n the proof is almost the same. Let's Take Z divisible by n. X+Y=2*B+Q+P=(u^n)*n^(n-1) Z=B+Q+P=n*u*z Therefore 2*B+Q+P=2*b*u*n+q^n +p^n=(u^n)*n^(n-1) Therefore from above and OBSERVATION we get that q^n+p^n is divisible by n^2. Therefore 2*b*n*u is divisible by n^2 . Therefore u is divisible by n. Using EQ2 and EQ3 we follow the same proof as in the case X*Y*Z not divisible by n and get to the stage of the proof in which we get that B is divisible by n^3. Since B=b*n*u we get that u is divisible by n^2. So as we continuu the infinite descent we will get u divisible by n^(m-1) when we get B divisible by n^m.That is how we get the infinite descent in the the case 2. Therefore Fermat Last Theorem is True. Created By GHEORGHE GHIATA- 06-06/05 Copy of this text has been sent to Mr.Andrew Wiles and Mr.Andrew Granville To MR. PROF.Andrew Wiles My goal has been greater than to proof Fermat Last theorem (like you.) This has been a EXPERIMENT which I described in one of the e-mail sent They might still have it in the Archive of Www.Mathforum.org Shortly it has to do whith proving that there is UNIVERSAL MEMORY. A budhist saying says: IF you try to find a solution to any kind of problem ,and you strugle, you better try to REMEMBER IT. AND all my proofs reflect that. And in the end I REMEMBEReD the PROOF of FLT. MY REGARDS, gheorghe ghiata 5/23/05 Hello. Here is another proof of FLT : The ARCHIMEDES PROOF OF FERMAT LAST THEOREM (No need for him to know formula of Cnk) Theorem: he EQUATION: X^n+Y^n=Z^n where X,Y,Z are relative prime Integers ,n=prime>2 is impossible. OBSERVATION: X,Y ,Z relative Prime numbers Let's say that X^n+Y^n=Z^n Lets take: X+Y=W and R=X^(n-1)-(X^(n-2))*Y +(X^(n-3))*Y^2.........+Y^(n-1)= = W^(n-1)-(Cn1)*[W^(n-2)]*Y+(Cn2)*[W^(n-3]*Y^2- -(Cn3)*[W^(n-4)]*Y^ +(Cn4)*[W^(n-5)]*Y^4-............ ........-(Cn2)*[W^1)*Y^(n-2 ) +(Cn1)*Y^(n-1) where Cnk=[n*(n-1)*(n-2)*(n-3)*.........*(n-k+1)]/2*3*4*.... .*k When Z is not divisible by n we see that (X+Y)=W and R do not have any common divisor . Therefore they are relative prime and must be: R=z^n and W=X+Y=u^n and Z=u*z When Z is divisible by n,that is Z=u*n*z we see that W and R have as common divisor only n Therfore W=(u^n)*n^(n-1) and R=n*z^n End of OBSERVATION Lets take Z not divisible by n When Z*X*Y is divisible by n the proof is almost the same. PROOF: Let's take : X^n+Y^n=Z^n X+Y- Z=B X=B+Q Y=B+P Therefore : EQ1: (B+Q)^n+(B+P)^n=(B+Q+P)^n X+Y=2*B+Q+P=u^n=W Z=B+Q+P=u*z B=b*u and Q+P=u*s and - z=-u^(n-1) +b and z-b=s 2*b+s=u^(n-1) b+s=z We see that we can write : X= - B+W-P and Q= -2*B +W-P Y= - B +W-Q and P= - 2*B+W-Q Z= - B+W Therefore we get : EQ2: (-B+W)^n +(-B-P)^n=(-B+W-P)^n Q=-2*B+W-P=q^n X=B+Q= -B+W-P=q*(z1) Therefore we get : B=c*q EQ3: (-B +W)^n +(-B-Q)^n=(-B+W-Q)^n P=-2*B+W-Q=m*p^n Y=B+P=-B+W-Q=r*p*(z2) WHen Y is not divisible by n then r=1 and m=1 When Y is divisible by n the r=n and m=n^(n-1) B=d*p*r Since X,Y, Z are relative prime we have u,q,p relative prime. Therefore B=k* r*q*p*u=b*u Therefore b is divisible by q*p*r Let's take: EQ1 : (B+Q)^n+(B+P)^n=(B+Q+P)^n EQ1: (2*B+Q+P)*T=(B+Q+P)^n where T=z^n can not be equal to 1 . Therefore z is not equal to 1 T= [B^(n-1)+(c1)*B^(n-2)+(c2)*B^(n-3)+....(ct)*B+(Q^n+P^n)/(Q+P)]=z^n and (c1),(c2).....(ct) are integers coeficients and (ct)= {n*Q^(n-1)+n*P^(n-1)-2*[Q^n+P^n]/(Q+P)}/(Q+P)=(Q+P)*V T can not be equal to 1 Therefore z must be not equal 1 Since X^n-X+Y^n-Y is divisible by n we get that (T-1)=z^n-1 is divisible by n. Therefore (z-1) is divisible by n. Since B and Z have as common divisior only u We can write : T= Z*B*F(q,p)+[Q^n+P^n]/(Q+P)=z^n=[-b+u^(n-1)]^n where F(q,p) is a function=F of (q,p) When we develop the parantheses of z^n We see that z^n=G(q,p) where G(q,p) is a Function =G of (q,p) of (q,p) Let's multiply T by (Q+P): T*(Q+P)=(Q+P)*Z*B*F(q,p) +(Q^n+P^n)=(Q+P)*z^n=(u*s)*z^n T*(Q+P)= (Q+P)*Z*B*F(q,p) +(u*s)*(z*M)=(Q+P)*z^n Since z*u=Z we can divide by Z and get: T*(Q+P)/Z= (Q+P)*B*F(q,p) +s*M= s*z^(n-1) Since s=z-b we can write : T*(Q+P)/Z= (Q+P)*B*F(q,p)+s*M=z^n-b*z^(n-1) Since b is divisible by q*p we that s*M=A(q,p) where A(q,p) is a function =A of (q,p) Therefore: (Q^n+P^n)= (Q+P)*[Q^(n-1)-P*Q^(n-2) +........P^(n-1)] =Z*A(q,p)=(B+Q+P)*A(q,p) whereZ=B+Q+P=D(q,p) is a function D of (q,p) But we know that [Q^n+P^n] is a reducible polynom of (q,p) only one way to two factors. Therefore the only solution is [Q^n+P^n]/(Q+P) =z Now if we substitute in T {Q^n+P^n]/(Q+P) with z we see that z=f(q,p) Now we have F= (Q^n+P^n)=Q+P)*z=s*Z=(z-b)*(B+Q+P) We know that z=f(q,p) and b is divisible by q*p. But Q^n+P^n is a polynomial of (q,p) which can be reducible only one way as the product of two polynomial factors of (q,p) ,but F shows that there is another way of decomposition in two polinomial factors of (q,p). That is impossible . Therefore Fermat Last Theorem is true. Created by GHEORGHE GHIATA-06/06/05 To AStrophysicists: When 5-D univers started burning from inside its center outwards, is that what I call Big Bang.And out of that continuu burning'of 5-D univers our 4-D univers, the inside circular uneven surface appeared and continuu to expand . Therefore our univers is a interface between two 5-D universes,sandwiched betwwen them. george ghiata THE END* *Author's note: This is the end of the real story which have taken place over a period of 4 years VIA electronic messages and postal service If you read all about it and you do not believe it ,then you can look I have to say TANKS to every PARTICIPANT. You might ask who the stage DIRECTOR has been. Here it is my answer: In a village, an unusual animal appeared and the people never seen before this kind of animal.So they decide to go and ask the sage of village.The sage told them to wait for the answer because he needs to meditate to find it.So,he fell in deep meditation and after awhile he start to cry and cry.Then at one moment he burst laughing and laugh and laugh.This made people very curios about these reactions and when the sage woke -up from the meditation they asked him why he cried so much His answer was :Well,when I see so many IGNORANTpeople who do not know what this strange animal is all about, I was suffering so much that I could not hold my tears. Why did you burst laughing then? the people asked. Well,I REALIZED that I did not know the Answer either. Created by GHeorghe ghiata Now Iam saying Good-by and I withdraw in my Ivory tower murmuring verses from the poem Lucifer by romanian poet Mihai Eminescu GHEORGHE GHIATA-06/06/05 _______________________________________________________________ About Mr. Prof.ANDREW WILES'PROOF OF FLT Plain Text Reply My understanding is that Mr. Prof. Andrew Wiles proved Fermat Last theorem for n>3. Does he had a original proof for n=3? If not, why he got Wolfsskehl Prize? If not ,that means I am the first person,the first time ,to give a more complet proof of FLT for n=prime >2.That is where the dispute is possible. We have to read Mr. Femat' statement very precise. GHEORGHE GHIATA GHEORGHE GHIATA (cod Nr.050322) _________________________________________________________________ ==== Subject: Re: About Mr. Prof.ANDREW WILES'PROOF OF FLT My understanding is that Mr. Prof. Andrew Wiles proved Fermat Last theorem for n>3. Does he had a original proof for n=3? If not, why he got Wolfsskehl Prize? If not ,that means I am the first person,the first time ,to give a more complet proof of FLT for n=prime >2.That is where the dispute is possible. We have to read Mr. Fermat' statement very precise. GHEORGHE GHIATA ==== Subject: Re: About Mr. Prof.ANDREW WILES'PROOF OF FLT Le 29/06/05 18:13, dans 20018479.1120061669412.JavaMail.jakarta@nitrogen.mathforum.org, .82george ghiatae a .8ecrit: > My understanding is that Mr. Prof. Andrew Wiles proved > Fermat Last theorem for n>3. > Does he had a original proof for n=3? > If not, why he got Wolfsskehl Prize? > If not ,that means I am the first person,the first time ,to give a more > complet proof of FLT for n=prime >2.That is where the dispute is possible. > We have to read Mr. Fermat' statement very precise. > GHEORGHE GHIATA bla bla bla :-) Not tired yet ? But *we* are ! ==== Subject: Question on a proof who want to type something in sci.math, who need to do routine problems to escape from a kind of madness of research, who love to teach etc, I have a trouble on understanding a proof of a proposition. The statement of the proposition is the following: Let X and Y be topological spaces. A mapping f of X into Y is continuous if and only if f is continuous at every point of X. The part of the proof which I am not sure of is this: If f is continuous at every point of X and if V is open in Y, every point x in (f(V))^{-1} has a neighborhood W_{x} such that f(W_{x}) subset V. Therefore W_{x} subset (f(V))^{-1}. [Up to this point, I am okay. However, the next statement is not obvious to me.] It follows that (f(V))^{-1} is the union of the open sets W_{x},... My question is, why does that follow? Would you, if you know why, consider that argument intuitive? Or would you know a rigorous argument? I might ask you more in the future (In fact, many). But I will state my guess too. Indeed, I think most deeply while I am typing my question. We have to show that each x is an interior point of (f(x))^{-1}, after all. I see. Indeed, if for every x in (f(V))^{-1}, there exists a neighborhood of x which is a subset of (f(V))^{-1}, then EVERY x is an interior point of (f(V))^{-1}; this in turn shows that (f(V))^{-1} is open. This is my answer. OnlyRH (This is my pseudo-name.) --------------------------------------------------- Someday, we will know how primes are distributed randomly. ==== Subject: Re: Question on a proof X-RFC2646: Format=Flowed; Original > who want to type something in sci.math, > who need to do routine problems to escape from a kind of madness of > research, > who love to teach > etc, I have a trouble on understanding a proof of a proposition. The statement of the proposition is the following: Let X and Y be topological spaces. A mapping f of X into Y is continuous > if and only if f is continuous at every point of X. The part of the proof which I am not sure of is this: If f is continuous > at every point of X and if V is open in Y, every point x in (f(V))^{-1} > has a neighborhood W_{x} such that > f(W_{x}) subset V. Therefore W_{x} subset (f(V))^{-1}. [Up to this > point, I am okay. However, the next statement is not obvious to me.] It > follows that > (f(V))^{-1} is the union of the open sets W_{x},... My question is, why does that follow? Would you, if you know why, consider > that argument intuitive? Or would you know a rigorous argument? It should be clear that the union of W_x is a subset of f^-1(V). , but each x in f^-1(V) is in W_x, thus the union equals f^-1(V). I might ask you more in the future (In fact, many). But I will state my > guess too. Indeed, I think most deeply while I am typing my question. We have to show that each x is an interior point of (f(x))^{-1}, after > all. I see. Indeed, if for every x in (f(V))^{-1}, there exists a > neighborhood of x which is a subset of (f(V))^{-1}, then EVERY x is an > interior point of (f(V))^{-1}; this in turn shows that > (f(V))^{-1} is open. This is my answer. OnlyRH (This is my pseudo-name.) --------------------------------------------------- > Someday, we will know how primes are distributed randomly. ==== Subject: Re: Question on a proof who should press the ENTER key once in a while. > I have a trouble on understanding a proof of a proposition. > > The statement of the proposition is the following: > > Let X and Y be topological spaces. A mapping f of X into Y is continuous > if and only if f is continuous at every point of X. OK. I'm assuming that you define continuous as if A is an open subset of Y, then f^{-1}(A) is an open subset of X. > The part of the proof which I am not sure of is this: If f is continuous > at every point of X and if V is open in Y, every point x in (f(V))^{-1} > has a neighborhood W_{x} such that f(W_{x}) subset V. Therefore > W_{x} subset (f(V))^{-1}. [Up to this point, I am okay. However, the next > statement is not obvious to me.] It follows that (f(V))^{-1} is the union > of the open sets W_{x},... > My question is, why does that follow? Would you, if you know why, consider > that argument intuitive? Or would you know a rigorous argument? Well, every W_x is a subset of f^{-1}(V) and therefore their union is also a subset of f^{-1}(V). On the other hand, f^{-1}(V) is the union of all sets of the form {x}, with x in f^{-1}(V) and therefore, since {x} is a subset of W_x, f^{-1}(V) is a subset of the union of all W_x. I've therefore proved that the two sets are equal. Jose Carlos Santos ==== Subject: Re: To Mr. Andrew Wiles:Do You Agree Fermat 's Proof Of FLT ever existed? HERE it is Fermat's Proof of FLT Theorem: he EQUATION: X^n+Y^n=Z^n where X,Y,Z are relative prime Integers ,n=prime>2 is impossible. OBSERVATION: X,Y ,Z relative Prime numbers Let's say that X^n+Y^n=Z^n Lets take: X+Y=W and R=X^(n-1)-(X^(n-2))*Y +(X^(n-3))*Y^2.........+Y^(n-1)= = W^(n-1)-(Cn1)*[W^(n-2)]*Y+(Cn2)*[W^(n-3]*Y^2- -(Cn3)*[W^(n-4)]*Y^ +(Cn4)*[W^(n-5)]*Y^4-............ ........-(Cn2)*[W^1)*Y^(n-2 ) +(Cn1)*Y^(n-1) where Cnk=[n*(n-1)*(n-2)*(n-3)*.........*(n-k+1)]/2*3*4*.... .*k When Z is not divisible by n we see that (X+Y)=W and R do not have any common divisor . Therefore they are relative prime and must be: R=z^n and W=X+Y=u^n and Z=u*z When Z is divisible by n,that is Z=u*n*z we see that W and R have as common divisor only n Therfore W=(u^n)*n^(n-1) and R=n*z^n The end of OBSERVATION CASE1 For simplicity of presentation lets take X*Y*Z not divisible by n When Z*X*Y is divisible by n the proof is almost the same. PROOF: Let's take : X^n+Y^n=Z^n X+Y- Z=B X=B+Q Y=B+P Therefore : EQ1: (B+Q)^n+(B+P)^n=(B+Q+P)^n X+Y=2*B+Q+P=u^n=W Z=B+Q+P=u*z B=b*u and Q+P=u*s and - z=-u^(n-1) +b and z-b=s 2*b+s=u^(n-1) b+s=z We see that we can write : X= - B+W-P and Q= -2*B +W-P Y= - B +W-Q and P= - 2*B+W-Q Z= - B+W Therefore we get : EQ2: (-B+W)^n +(-B-P)^n=(-B+W-P)^n Q=-2*B+W-P=q^n X=B+Q= -B+W-P=q*(z1) Therefore we get : B=c*q EQ3: (-B +W)^n +(-B-Q)^n=(-B+W-Q)^n P=-2*B+W-Q=p^n Y=B+P=-B+W-Q=p*(z2) B=d*p Since X,Y, Z are relative prime we have u,q,p relative prime. Therefore B=k*q*p*u=b*u Therefore b is divisible by q*p Lets take : X^n-X +Y^n-Y = Z^n-Z+Z-X-Y X+Y-Z=B is divisible By n Let's take: EQ1 : (B+Q)^n+(B+P)^n=(B+Q+P)^n EQ1: (2*B+Q+P)*T=(B+Q+P)^n where T=z^n can not be equal to 1 . Therefore z is not equal to 1 T= [B^(n-1)+(c1)*B^(n-2)+(c2)*B^(n-3)+....(ct)*B+(Q^n+P^n)/(Q+P)]=z^n and (c1),(c2).....(ct) are integers coeficients and (ct)= {n*Q^(n-1)+n*P^(n-1)-2*[Q^n+P^n]/(Q+P)}/(Q+P)=(Q+P)*V Since X^n-X+Y^n-Y is divisible by n we get that (T-1)=z^n-1 is divisible by n. Therefore (z-1) is divisible by n. Since B and Z have as common divisior only u We can write : T= Z*B*F(q,p)+[Q^n+P^n]/(Q+P)=z^n=[-b+u^(n-1)]^n where F(q,p) is a function=F of (q,p) When we develop the parantheses of z^n We see that z^n=G(q,p) where G(q,p) is a Function =G of (q,p) Let's multiply T by (Q+P): T*(Q+P)=(Q+P)*Z*B*F(q,p) +(Q^n+P^n)=(Q+P)*z^n=(u*s)*z^n T*(Q+P)= (Q+P)*Z*B*F(q,p) +(u*s)*(z*M)=(Q+P)*z^n Since z*u=Z we can divide by Z and get: T*(Q+P)/Z= (Q+P)*B*F(q,p) +s*M= s*z^(n-1) Since s=z-b we can write : T*(Q+P)/Z= (Q+P)*B*F(q,p)+s*M=z^n-b*z^(n-1) Since b is divisible by q*p we that s*M=A(q,p) where A(q,p) is a function =A of (q,p) Therefore: (Q^n+P^n)= (Q+P)*[Q^(n-1)-P*Q^(n-2) +........P^(n-1)] =Z*A(q,p)=(B+Q+P)*A(q,p) whereZ=B+Q+P=D(q,p) is a function D of (q,p) But we know that [Q^n+P^n] is a reducible polynom of (q,p) only one way to two factors. Therefore the only solution is [Q^n+P^n]/(Q+P) =z If we follow the same procedure with EQ2 we get the same results: 1) [(z1)-1] is divisible by n 2) (W^n-P^n)/(W-P)=(z1) 3) (z1) can not be equal to 1 If we follow the same procedure with EQ3 we get the same results: 1) [(z2)-1)] is divisible by n 2) (W^n-Q^n)/(W-Q)=(z2) 3)(z2) can not be equal 1 Now we take: EQ4 : X^n-Q^n+Y^n-P^n -Z^n+W^n=(-Q^n+Q)+(-P^n+P)+(W^n-W)+(W-Q-P) The EQ4 can be writen as : EQ4: n*m*B=(n^2)*J+2*B Therefore B is divisible by n^2 Therefore [(T/z)-1]=[z^(n-1)-1] is divisible by n^2 Since (z-1) is divisible by n we get from above that (z-1) is divisible by n^2 Since z+b=u^(n-1) and since b is divisible by n^2 we get that [ u^(n-1)-1] is divisible by n^2 The same way we get that 1) [(z1)-1] is divisible by n^2 2) [q^(n-1)-1] is divisible by n^2 The same way we get that 1) [(z2)-1] is divisible by n^2 2) {p^(n-1)-1} is divisible by n^2 With this new data we get that EQ4 can be written : n*B*m={n^3)*J'+2*B Therefore we started a Infinite descend. Case2: If X*Y*Z is divisible by n the proof is almost the same. Let's Take Z divisible by n. X+Y=2*B+Q+P=(u^n)*n^(n-1) Z=B+Q+P=n*u*z Therefore 2*B+Q+P=2*b*u*n+q^n +p^n=(u^n)*n^(n-1) Therefore from above and OBSERVATION we get that q^n+p^n is divisible by n^2. Therefore 2*b*n*u is divisible by n^2 . Therefore u is divisible by n. Using EQ2 and EQ3 we follow the same proof as in the case X*Y*Z not divisible by n and get to the stage of the proof in which we get that B is divisible by n^3. Since B=b*n*u we get that u is divisible by n^2. So as we continuu the infinite descent we will get u divisible by n^(m-1) when we get B divisible by n^m.That is how we get the infinite descent in the the case 2. Therefore Fermat Last Theorem is True. Created By GHEORGHE GHIATA- 06-06/05 Copy of this text has been sent to you via e-mail , Mr.PROF.Andrew Wiles What about this proof sent by e-mail to you: Here is another proof of FLT : The ARCHIMEDES PROOF OF FERMAT LAST THEOREM (No need for him to know formula of Cnk) Theorem: he EQUATION: X^n+Y^n=Z^n where X,Y,Z are relative prime Integers ,n=prime>2 is impossible. OBSERVATION: X,Y ,Z relative Prime numbers Let's say that X^n+Y^n=Z^n Lets take: X+Y=W and R=X^(n-1)-(X^(n-2))*Y +(X^(n-3))*Y^2.........+Y^(n-1)= = W^(n-1)-(Cn1)*[W^(n-2)]*Y+(Cn2)*[W^(n-3]*Y^2- -(Cn3)*[W^(n-4)]*Y^ +(Cn4)*[W^(n-5)]*Y^4-............ ........-(Cn2)*[W^1)*Y^(n-2 ) +(Cn1)*Y^(n-1) where Cnk=[n*(n-1)*(n-2)*(n-3)*.........*(n-k+1)]/2*3*4*.... .*k When Z is not divisible by n we see that (X+Y)=W and R do not have any common divisor . Therefore they are relative prime and must be: R=z^n and W=X+Y=u^n and Z=u*z When Z is divisible by n,that is Z=u*n*z we see that W and R have as common divisor only n Therfore W=(u^n)*n^(n-1) and R=n*z^n End of OBSERVATION Lets take Z not divisible by n When Z*X*Y is divisible by n the proof is almost the same. PROOF: Let's take : X^n+Y^n=Z^n X+Y- Z=B X=B+Q Y=B+P Therefore : EQ1: (B+Q)^n+(B+P)^n=(B+Q+P)^n X+Y=2*B+Q+P=u^n=W Z=B+Q+P=u*z B=b*u and Q+P=u*s and - z=-u^(n-1) +b and z-b=s 2*b+s=u^(n-1) b+s=z We see that we can write : X= - B+W-P and Q= -2*B +W-P Y= - B +W-Q and P= - 2*B+W-Q Z= - B+W Therefore we get : EQ2: (-B+W)^n +(-B-P)^n=(-B+W-P)^n Q=-2*B+W-P=q^n X=B+Q= -B+W-P=q*(z1) Therefore we get : B=c*q EQ3: (-B +W)^n +(-B-Q)^n=(-B+W-Q)^n P=-2*B+W-Q=m*p^n Y=B+P=-B+W-Q=r*p*(z2) WHen Y is not divisible by n then r=1 and m=1 When Y is divisible by n the r=n and m=n^(n-1) B=d*p*r Since X,Y, Z are relative prime we have u,q,p relative prime. Therefore B=k* r*q*p*u=b*u Therefore b is divisible by q*p*r Let's take: EQ1 : (B+Q)^n+(B+P)^n=(B+Q+P)^n EQ1: (2*B+Q+P)*T=(B+Q+P)^n where T=z^n can not be equal to 1 . Therefore z is not equal to 1 T= [B^(n-1)+(c1)*B^(n-2)+(c2)*B^(n-3)+....(ct)*B+(Q^n+P^n)/(Q+P)]=z^n and (c1),(c2).....(ct) are integers coeficients and (ct)= {n*Q^(n-1)+n*P^(n-1)-2*[Q^n+P^n]/(Q+P)}/(Q+P)=(Q+P)*V T can not be equal to 1 Therefore z must be not equal 1 Since X^n-X+Y^n-Y is divisible by n we get that (T-1)=z^n-1 is divisible by n. Therefore (z-1) is divisible by n. Since B and Z have as common divisior only u We can write : T= Z*B*F(q,p)+[Q^n+P^n]/(Q+P)=z^n=[-b+u^(n-1)]^n where F(q,p) is a function=F of (q,p) When we develop the parantheses of z^n We see that z^n=G(q,p) where G(q,p) is a Function =G of (q,p) Let's multiply T by (Q+P): T*(Q+P)=(Q+P)*Z*B*F(q,p) +(Q^n+P^n)=(Q+P)*z^n=(u*s)*z^n T*(Q+P)= (Q+P)*Z*B*F(q,p) +(u*s)*(z*M)=(Q+P)*z^n Since z*u=Z we can divide by Z and get: T*(Q+P)/Z= (Q+P)*B*F(q,p) +s*M= s*z^(n-1) Since s=z-b we can write : T*(Q+P)/Z= (Q+P)*B*F(q,p)+s*M=z^n-b*z^(n-1) Since b is divisible by q*p we that s*M=A(q,p) where A(q,p) is a function =A of (q,p) Therefore: (Q^n+P^n)= (Q+P)*[Q^(n-1)-P*Q^(n-2) +........P^(n-1)] =Z*A(q,p)=(B+Q+P)*A(q,p) whereZ=B+Q+P=D(q,p) is a function D of (q,p) But we know that [Q^n+P^n] is a reducible polynom of (q,p) only one way to two factors. Therefore the only solution is [Q^n+P^n]/(Q+P) =z Now if we substitute in T {Q^n+P^n]/(Q+P) with z we see that z=f(q,p) Now we have F= (Q^n+P^n)=Q+P)*z=s*Z=(z-b)*(B+Q+P) We know that z=f(q,p) and b is divisible by q*p. But Q^n+P^n is a polynomial of (q,p) which can be reducible only one way as the product of two polynomial factors of (q,p) ,but F shows that there is another way of decomposition in two polinomial factors of (q,p). That is impossible . Therefore Fermat Last Theorem is true. Created by GHEORGHE GHIATA-06/06 (cod Nr.050322 ==== Subject: Halmos proof question In Halmos explanation of Russell's Paradox he calls the statement: B belongs to B ----> not (B belongs to B) a contradiction. I was under the impression that a statement of the form p and (not p) is a contradiction and a statement of the form p ----> not p is always false but not a contradiction. ==== Subject: Re: Halmos proof question In Halmos explanation of Russell's Paradox he calls the statement: > > B belongs to B ----> not (B belongs to B) > > a contradiction. > Does he? Please give a quote. A. ==== Subject: Re: Halmos proof question Le 29/06/05 18:33, dans .82agapito6314@aol.com[CapitalEG rave] a .8ecrit: > In Halmos explanation of Russell's Paradox he calls the statement: > > B belongs to B ----> not (B belongs to B) > > a contradiction. I was under the impression that a statement of the > form > > p and (not p) is a contradiction is always false > and a statement of the form > p ----> not p is always false but not a contradiction. is equivalent to not p, since A => B is equivalent to B or not A > > ==== Subject: Re: Halmos proof question X-RFC2646: Format=Flowed; Original > In Halmos explanation of Russell's Paradox he calls the statement: B belongs to B ----> not (B belongs to B) a contradiction. I was under the impression that a statement of the > form p and (not p) is a contradiction and a statement of the form p ----> not p is always false but not a contradiction. i believe he also has not (B in B) implies B in B. so we have p implies not p, and not p implies p. if you unpack this as: (not p or not p) and (p or p), we have p and not p. ==== Subject: Re: Halmos proof question >In Halmos explanation of Russell's Paradox he calls the statement: B belongs to B ----> not (B belongs to B) a contradiction. I was under the impression that a statement of the >form p and (not p) is a contradiction and a statement of the form p ----> not p is always false but not a contradiction. > If -> is as in propositional logic then p -> not p is not a contradiction, and it's also not always false! If p is false then p -> not p is false. Anyway, I don't have the book in front of me, but I suspect that you're leaving a little bit out. What I suspect he says is that B belongs to B ----> not (B belongs to B) and _also_ not(B belongs to B) ----> B belongs to B. The conjunction of both those statements is a contradiction. (Maybe he just says and the converse is similar or something...) ************************ C. Ullrich ==== Subject: Re: An Approximate Circle ... >> A circle to draw. ... >> A circle of finite points is the outcome. ... > Fairly simple; given a computer-approximatible radius ... > simply taking y=sqrt(r^2-x^2) to the allowed precision ... Two responders have suggested how to draw highly accurate approximate circles, in response to Eagleson's somewhat elliptical proposal. Here's a quick way to draw an inaccurate, but not bad-looking for being handdrawn, circle on graph paper. Given radius r, choose n so r ~ n(n+1)/2. Let L=n(n+1)/2. Draw the first quadrant by placing points at (0,L), (n,L-1), (n+n-1,L-1-2), ... (L,0). Translate to chosen center, connect the dots, mirror and flip for other quadrants. -jiw ==== Subject: Re: The product of all primes less than some x >> No. The product p# of the primes less than p, can be >> greater than e^x. Such a prime was found or it was proved that it should exist? > Rosser and Schoenfeld proved (1975) that: >> 0.998684p < log(p#) < 1.001102p >> The lower inequality for p>1319007 >> The upper inequality for p>2. >> See The New Book of Prime Number Records >> page 245. >> Ludovicus Where can I find such references? >I mean The New Book of Prime Number Records. >There is something posted on-line? >Thx, >Dan See also the entry and references here: http://mathworld.wolfram.com/RossersTheorem.html The matter of crossings of pi(x) and Li(x), and the value of the smallest such x, is a different matter. For more references, see: http://mathworld.wolfram.com/SkewesNumber.html ==== Subject: Re: The product of all primes less than some x Le 29/06/05 4:59, dans 32804181.1120014022801.JavaMail.jakarta@nitrogen.mathforum.org, .82Dan Dimae a .8ecrit: >> No. The product p# of the primes less than p, can be >> greater than e^x. > > Such a prime was found or it was proved that it should exist? > >> Rosser and Schoenfeld proved (1975) that: >> 0.998684p < log(p#) < 1.001102p >> The lower inequality for p>1319007 >> The upper inequality for p>2. >> See The New Book of Prime Number Records >> page 245. >> Ludovicus > > Where can I find such references? > I mean The New Book of Prime Number Records. > There is something posted on-line? http://www.amazon.com/exec/obidos/tg/detail/-/0387944575/ref=pd_sxp_f/002-01 97399-5645627?v=glance&s=books ==== Subject: Re: maximizing the largest singular value of a product power constraint on F is abs(trace(F*F)) so it concerns the set of singular values and not only the largest one (as you thought), Jonathan Stephen Montgomery-Smith a .8ecrit : > hello, > > can anyone tell me how to choose F to maximize largest singular value > of > > (H x F) for a given H and a power constaint on F : norm(F*F')<=P So I am guessing that H and F are matrices. The largest singular value > of A is the same as norm(A), and your condition is equivalent to > norm(F)<=sqrt(P) (because its a C*-algebra?). The inequality norm(HF) <= norm(H)norm(F) is going to tell you that the largest attainable is sqrt(P) norm(H). The way to do this is as follows. Write H = UDV where U and V are > orthogonal, and D is diagonal with the diagonal entries positive in > non-increasing order (i.e the signular values in order). Let E be the > projection onto the first co-ordinate, i.e. the diagonal matrix with > entries (1,0,...,0). Then F = sqrt(P) V^{-1} E U{^-1} is going to do > the job. > > I hope I am understanding you correctly. > > Stephen ==== Subject: Re: maximizing the largest singular value of a product OK, your constraint is the so called Hilbert-Schmidt norm (that is, presuming that you have a bit of a misprint and meant trace(F*F'), the abs is unnecessary). There is something called, I think, Horner's inequality which tells you that you basically only need to consider diagonal matrixes with positive entries arranged in non-increasing order. Thus, more or less, you are asking to maximise h_1 f_1 where the sequence (h_i) is given, and the constraint is that sum f_i^2 <= P, and h_i and f_i are positive non-increasing. You can see from this consideration that the answer will be no different > power constraint on F is > > abs(trace(F*F)) > > so it concerns the set of singular values and not only the largest one > (as you thought), > > > Jonathan > > Stephen Montgomery-Smith a .8ecrit : > >hello, >>can anyone tell me how to choose F to maximize largest singular value >of >>(H x F) for a given H and a power constaint on F : norm(F*F')<=P >>So I am guessing that H and F are matrices. The largest singular value >>of A is the same as norm(A), and your condition is equivalent to >>norm(F)<=sqrt(P) (because its a C*-algebra?). The inequality >>norm(HF) <= norm(H)norm(F) >>is going to tell you that the largest attainable is sqrt(P) norm(H). >>The way to do this is as follows. Write H = UDV where U and V are >>orthogonal, and D is diagonal with the diagonal entries positive in >>non-increasing order (i.e the signular values in order). Let E be the >>projection onto the first co-ordinate, i.e. the diagonal matrix with >>entries (1,0,...,0). Then F = sqrt(P) V^{-1} E U{^-1} is going to do >>the job. >>I hope I am understanding you correctly. >>Stephen > > > ==== Subject: Parallel conjungate gradient method for linear system from FEM method for elastic problems Hallo, I am working on my PhD, being stuck with following problem (and I desperedly need a solution [as many people]): I am writing finite element code for the elastic problems (for later use for the aeroelasticity computations). I use only elastic continuum formulation with Lagrange elements (P1, Q1, Q2. No shell or beam elements). Therefore the stiffness matrix is sometimes not very well conditioned. (Don't worry, boundary conditions are OK, the matrix is not singular.) The stiffness and mass matrix can be either non-symmetric, or (in the case of better treatment of the boundary conditions) symmetric, positive definite. I need to compute eigenvalues and also to do time-dependent analysis (using Newmark method). As the eigensolver, I use Arpack software (Arnoldi method, package has a SLEPc interface) and to solve linear system I use PETSc (with additional libraries, such as SuperLU). For small problems I use either LU decomposition, or, for symmetric problems conjungate gradients (CG). However, CG does not really converge, even with incomplete Cholesky preconditioner, unless some filling of the matrix is allowed (e.g. ICC(50)). For single-processor computations LU decomposition is fast enough (at least for the moment). (I have worked out some better renumbering algorithm than reverse Cuthill-McKee, so the band is quite low.) Now I am stacked with parallel version. I have tried Swartz method with overlapping for the preconditioning and each sub-domain solve with the incomplete Cholesky factorization with filling + CG. Not convergent at all. The _safe_ version - LU decomposition in parallel (SuperLU package) is _very_ slow. I have a couple of questions: 1. Why my CG method does not converge??? The matrix is symmetric, problem not very tough (e.g. 2D beam clamped on one edge, with force on the other side, with length ratio 1/20)??? If you check comertial software packages, they claim to use CG without any preconditioning (e.g. Ansys). 2. Which numerical method do you recomend to use for single-processor computations? 3. Which numerical method to solve sparse linear system do you recommend for parallel computations? (This is actually the most important question, because I really don't know how to continue with my PhD for the moment). 4. Do you have problems with the speed of the SuperLU package for parallel computations? 5. Do you have an idea, why GMRES with ILU preconditioning does not work? I have believed that it also should work. Jiri Dobes ==== Subject: Re: Is this expression viewable on sci.math? >> Or just set up a web forum (blecch). Or just use ascii like we have been doing successfully for years. Newton, Cauchy, Taylor and a bunch of others have been successful too - without any computers... Being successful one way should not be a reason to avoid testing new things. There might be an even better way (which way it would be, i have no idea, hehe). By the way - what is so bad with HTML? I'm not arguing that it's good - i simply would like to be enlighted on that matter. I hear there's a site that will create small gif's provided you send them LaTeX-code. That would be good... -- V.8anligen Konrad --------------------------------------------------- Sleep - thing used by ineffective people as a substitute for coffee Ambition - a poor excuse for not having enough sense to be lazy --------------------------------------------------- ==== Subject: Re: Is this expression viewable on sci.math? In message , >> The tedious thing for me would be typing non-keyboard characters. >> Windows does provide a graphical interface called Character Map but >> I have never worked out how to use it efficiently. Perhaps an >> interested programmer could produce something easier? If you know the hexadecimal code for the character you want, you can >hold Alt, type + on the keypad, followed by the code, and release >Alt. I'm not sure what to do if your keyboard doesn't have a numeric >keypad with + (e.g. most laptops). That seems to me a moderately efficient way to access more than a >million potential characters. > That is too easy. The obvious way is to type each character as a 16 bit binary number, giving access to more than one million actual (not potential) characters. -- Jeremy Boden ==== Subject: Re: Is this expression viewable on sci.math? >> If you know the hexadecimal code for the character you want, you >> can hold Alt, type + on the keypad, followed by the code, and >> release Alt. I'm not sure what to do if your keyboard doesn't >> have a numeric keypad with + (e.g. most laptops). >> That seems to me a moderately efficient way to access more than a >> million potential characters. > That is too easy. > The obvious way is to type each character as a 16 bit binary number, > giving access to more than one million actual (not potential) > characters. How? My MatLab claims that 2^16 is merely 65536... That's just enough for Japense characters but that's an other story... :) -- V.8anligen Konrad --------------------------------------------------- Sleep - thing used by ineffective people as a substitute for coffee Ambition - a poor excuse for not having enough sense to be lazy --------------------------------------------------- ==== Subject: Re: Is this expression viewable on sci.math? Le 29/06/05 19:01, dans 3ig1d2Fla3l4U1@individual.net, .82Konrad Vilterstene a .8ecrit: > If you know the hexadecimal code for the character you want, you > can hold Alt, type + on the keypad, followed by the code, and > release Alt. I'm not sure what to do if your keyboard doesn't > have a numeric keypad with + (e.g. most laptops). > > That seems to me a moderately efficient way to access more than a > million potential characters. > >> That is too easy. >> The obvious way is to type each character as a 16 bit binary number, >> giving access to more than one million actual (not potential) >> characters. > > > How? My MatLab claims that 2^16 is merely 65536... > That's just enough for Japense characters but that's > an other story... :) Because of the ways characters are encoded in UTF16. It isn't a true 2 bytes encoding... The exact encoding scheme may be found at http://www.unicode.org/ There is also UTF8 (1 to 6 bytes I think), and UTF32. ==== Subject: Re: Three collinear points >Given two parallel lines p and q, and three collinear points >A, B, C, all laying in the same plane. Wait -- are they points or chickens? >Let P be an arbitrary point on line p and let A1, B1, C1 be >intersections of line q with lines PA, PB, PC, respectively. There's a problem if A, B, or C already lies on p. I'll assume they don't. >It seems that ratio A1B1 : B1C1 doesn't depend on the choice of >point P. Is that true and if yes, how can it be proven? Well, it's ugly but you can use coordinate geometry. Impose coordinates on the plane so that p is the y-axis and q is the line x = 1 . Then P is a point of the form (0,t). The fact that A B C are collinear means precisely that the vectors B-A and C-A are parallel (I'm assuming A, B, and C are distinct), so we can write the coordinates of these points in the form A = (x0,y0) B = (x0+u,y0+v) C = (x0 + k u, y0 + k v) for some x0 y0 u v k . So now we can compute A1, the intersection of the lines x=1 and (y-t)(x0-0) = (x-0)(y0-t); A1 is the point ( 1, t + (y0-t)/x0 ) and similarly B1 = ( 1, t + (y0+v - t)/(x0+u) ) and C1 = ( 1, t + (y0 + k v - t)/(x0 + k u) ) . So the lengths A1B1 and B1C1 are respectively (y0-t)/x0 - (y0+v - t)/(x0+u) = (y0 u - x0 v -t u)/( x0 (x0 + u) ) and (y0+v - t)/(x0+u) - (y0 + k v - t)/(x0 + k u) = (k-1)(y0 u - x0 v - t u)/( (x0+ku)(x0+u) ). The ratio of these two lengths is then (k-1) x0 / (x0+ku) , which is independent of t (i.e. of P). So there's a proof, though I'll be the first to admit it's pretty unenlightening. dave ==== Subject: Re: Three collinear points Excuse me, I wasn't precise enough. Suppose that neither A, B nor C are on p or q. Suppose also that A, B and C are on one side of q while p is on another side. ==== Subject: Re: Three collinear points > Excuse me, I wasn't precise enough. Suppose that neither A, B nor C are > on p or q. Suppose also that A, B and C are on one side of q while p is > on another side. > this is thales theorem. -- clement ==== Subject: Re: Three collinear points on p or q. Suppose also that A, B and C are on one side of q while p is > on another side. > Suppose you include the context so I can follow the line of thought? ==== Subject: Re: Orlow cardinality question Dik T. Winter said: > > Dik T. Winter said: > ... > > Nope. Start with the (true) statement that 1 is finite. Next we can proof > > (really simple) that the successor of a finite number is a finite number. > So by induction all (natural) numbers are finite. An easy proof by > > induction this appears to be. The next question is, is the number of > > finite numbers finite? Suppose it is, in that case there is a finite > > number m, such that the number of finite numbers is m. But the successor > > of m is also finite, but not in the set. So we have a contradiction > > and so the number of finite numbers can not be finite. > > > > You are going stepwise, and also ignoring what being finite is. It is not an > > equality, but an inequality. A number is finite if it is less than any > > infinite number. > > Wrong, in my proof above there are no infinite numbers, and they are not > needed. Right, they are excluded by the term finite, which means not infinite. > > > So, you are > > basically proving n > the fact n > proof is declared to prove a property for ALL n in N, the entire infinite > > set, which means it holds for an infinite number of iterations, not some > > finite number. > > No, for each number there is only a finite number of iterations. You can > reach every element of N by counting, and that is actually precisely what > the induction axiom does assert. So, there are only a finite number of them, since you can count to any one in a finite number of steps. > > > However, over this infinite set of interations which each include an > > increment on value of the maximal element of the set, that value has > > been incremented by 1 an infinite number of times, > > Nope, at no time will it be incremented an infinite number of times. And > for each finite number of increments the maximal value is still finite. So, you are really asserting that the set is NOT infinite? Maybe you claim that some iterations produce more than one natural number, maybe an infinite number? I mean, is the set the result of an infinite number of iterations, or a finite number? > > > As far as the argument that there are an infinite number of such finite > > numbers, it goes right back to your maximal element, largest finite > > mantra. Just because you cannot specify the largest finite natural is > > not just cause to claim the set is infinite. Not all finite sets have > > a largest element. You need to apply some known analysis and arithmetic > > to your problem, and stop using purely axiomatic methods. > > On the other hand, it is your mantra that states that the number of finite > naturals is finite. But every finite set has a maximal member, and you can > not show that maximal member. That is the only thing I state. Yes, Virgil has been harping on the maximal member, which is your largest finite mantra. He still hasn't commented on the infinite ordered set from 000...000 through 999...999, with maximal and minimal elements. Would you care to take a shot at that? > > > No it doesn't. Finiteness is determined by axioms of finiteness in > > arithmetic contexts. S^L is finite if S and L are finite, for instance. > > The finiteness of the set of finite naturals is demonstrable through > > various methods I have offered. > > Various methods, none of them proof. But now you state it again, the set > of finite naturals is finite, by that logic there should be a maximal > finite element. Which is it? Not by MY logic. That is a generalization that is unwarranted. It rests on unfounded assumptions. > > > You are talking about three different kinds of infinity at the same time, > > confusing one with the other. (And actually there are more.) > > I am not confusing anything, Dik. > > > > 1. oo is only conceptual, it is not a number. You may see it as > > encompassing all the other kinds of infinities. When you look at the > > real line, you can bend it to a circle and name the missing point oo > > (the one-point compactification of the reals). You can do something > > similar with the plane, and n-dimensional spaces. Because it is not > > a number, arithmetic with it is not really defined, because any way > > you want to define arithmetic on it violates one of the arithmetic > > axioms. > > Infinity is a numerical concept, and can be decalred as the inverse of zero, > > it's opposite complement. There really is no non-numerical concept of > > infinity. > > There is no numerical concept of infinity. If you declare it as the inverse > of zero in your mathematical system, and as a number, there are a few > axioms that do no longer hold. What is 0 * oo? When you say I do not > know, that means that in your system you will never know whether there is > an answer to the multiplication of two variables. The product of unit 0 and oo is the finite unit 1. > > > Just because the arithmetic hasn't been well defined yet doesn't mean it > > can't. Most of the groundwork is there, and it is only this kind of > > resistance that makes it so hard to achieve. > > Go ahead and put in the arithmetic rules such that the basic arithmetical > axioms remain valid. > > > 2. omega is an ordinal number. Sets have the same ordinal number when > > there is an order-preserving bijection between them. Only part of > > arithmetic on them is defined: addition, multiplication, but not > > subtraction or division. And arithmetic on ordinals have strange > > rules: 1 + omega = omega != omega + 1. > > Yes, what arithmetic you have for infinite numbers so far seems pretty weak. > > Too bad your method doesn't preserve more arithmetic. > > Note that this is a different kind of infinities, and also that it is not > my method. Duly noted. So, do you have any methods of your own? > > > 3. aleph_0 is a cardinal number. Sets have the same cardinal number when > > there is an arbitrary bijection between them. Here also only part of > > arithmetic is defined, and again, not subtraction. On the other hand, > > arithmetic is much simpler. Addition is commutative, and aleph0 + k, > > where k is any cardinal less than or equal to aleph0 is aleph0. So > > also 2 * aleph0 = aleph0 + aleph0 = aleph0. > > Yeah, still pretty weak. > > > > 4. In the surreals omega is also defined, but with a different meaning. > > Full arithmetic is defined with them, and they in fact form a field. > > So omega - 1 != omega != omega + 1. > > Only in (2) and (3) can we talk about a smallest infinity. > > Yes, 2 and 3 allow such nonsense due to the inherent inconsistencies in the > > approach. Surreals seem like a very good step in the right direction. > > You will find that in (4) you still have that the natural numbers are all > finite... And that most numbers do not have a decimal expansion. But indeed > it can be shown that every archimedean field can be embedded in the surreals. > And the largest of those fields are the reals themselves. I will have to look into that. Maybe John Conway needs an email. > > > > of pofnats. oh, and if you don't like that, here's a rule for you: > > > alpha+1=alpha. Does that make you happy? Why not? So hard to please > > > > No I do not like it. alpha + 1 = alpha violates the axioms for natural > > numbers, so alpha can not be a natural number. > > Oh I already responded to Virgil. Just start the naturals at 0, say alpha is > > the size of the set, so the largest element if alpha-1, which obeys Peano, > > but alpha+1=alpha, becase alpha's outside of the set, and doesn't have to > > obey Peano. > > And what is the size of the set if we add alpha to the set? I dunno. the whole idea is stupid anyway. You're supposed to reject it so I can point out how it'sentirely symmetrical with your omega, but now you're considering it. Alas! > > > Oh yeah, you don't HAVE infinite > > numbers, really. > > No, it is more that there are so many different kinds of infinite numbers. > In my summary I missed still at least two. > -- Smiles, Tony ==== Subject: Re: Orlow cardinality question Virgil said: > > Virgil said: > > finish ... never. > Okay, so each natural number is the last one, incremented, but you > never increment an infinite number of times? > Right! > > Then you must not have an infinite > number of natural numbers, since one element is added to the set each > of these times. > > How does it follow? The set of values S = {1, 1/2, 1/3, ...} is an > infinite set of finite values. The correspondence f(x) = 1/x, is a > bijection between N and S, and is, in a sense, its own inverse. If > there is any infinite natural in N, there must be an infinitesimal > rational number in S that is greater than zero but less that every > positive real number. But the properties of the reals do not allow any > such numbers. > You numbskull!!! Are those natural numbers? Why do you even continue this > rope- > a-dope nonsense? > > In the apparently vain hope that some truths will eventually penetrate > TO's carefully defended ignorance. Truths that consist of changing the subject? Nice approach, or retreat, rather. > > Do you have an infinite set or not? > Yes! > Of naturals? > > How do you get an infinite set without infinite increments? > > > > By finite increments without end. > Infinitesimal you mean. > No! By having each increment the same finite size. And an infinite number of them, summing to an overall finite incrementation? This is what is not possible. > > What is 1/n-1/n+1 as n goes to infinity? You have one end of your > finite range densely populated in the reals, which is the reason you > have an infinite set. The only way for a totally sparsely populated > set like the naturals to have an infinite number of elements is to > have an infinite range. > > By having an unbounded range, meaning that there is no maximum distance > bertween members but there is also no infinite distance between members. Then you do not have an infinite set, but an unbounded set. > > TO has this delusion that when something is endless it still somehow > must have an end. Assuming that self-contradiction is the root of all > TO's errors. Virgil lives under some kind of delusion that all other mathematics is somehow subjugated to the rule of cardinality, but that is an unjustified pompous position to take. Stop violating basic rules with your proofs. > > Finite-range sets with infinite elements are dense in the reals at at > least one point. > > As subsets of the reals they need not be dense anywhere, but they must > have, topologically speaking, at least one condensation point in the > reals. Uh, that's what I mean by being dense at at least one point, I believe. > But the topology if the reals does not contain points at > infinity, so no such condensation to an infinite point is either > necessary or possible, at least in the reals, and every natural is also > a real! I was talking about a finite range set of reals, which may all be finite, such as [0,1], which is dense throughout, or the set of 1/n for n in the naturals, which is dense at n=oo, or 1/oo, or the point at 0. Without at least one such point of density in the reals, you cannot have an infinite number of elements within a finite range of values. Since the naturals have no such condensation point, or location with density in the reals, it cannot contain an infinite number of elements within a finite range of values, which means some values must be infinitely greater than others in the set, and must therefore be infinite, if the set is infinite. > > TO wants the set of naturals, N, to be bounded in N, but that requires > to a maximum member in N. > No it doesn't. > > What is TO's definition of boundedness of a set in itself that allows it > to have members larger that its upper bounds? Okay, I think i misspoke in those three words. The standard set N has maximum member N, but that doesn't mean that one can't count higher. Think of N as 999...999, and N+1 as 000...001:000...000 > > If there is no need for a natural which is a bound on the set of all > finite naturals, then there is no need for an infinite natural. > > > > You have a bad attitude. No cookie for you. > > My bad attitude consists in pointing out the myriad flaws and > self-contradictions in TO's arguments. It is only bad from TO's point > is TO who has the bad attitude. You still haven't told me what I did wrong in my S^L proof. How do you get infinite numbers of strings without having infinitely long strings? > > To oppose kookdom is only viewed as a bad attitude by kooks. > > > I have defended my ideas while developing them further right here > before you. It is not my fault if you can't follow simple math from > any real area of math that contradicts your hocus pocus. I mean, the > above proof is so straightforward, my 11-year-old and my > history-major boss can understand it, but you can't. It's incredible > the damage this theory does to one's mind. > > And you are responsible for the damage you have done to your 11 year > old's mind (is he by any chance named Nathan?) and your boss's mind. > > I suspect that in the long run both will castigate you for it. > > They are happy to hear of all sorts of things from me. > > We can wait til they learn how wrong they, and you, are. But you will > have managed to do considerable damage to them before tha happens. > That's not the way those around me see it. I suppose my mother's cardiologist would think I was doing her damage by telling her to dump the statins and stick to the vitamin C and lysine, but then again, he didn't expect her to live when she was in the hospital, and is amazed at her strength a year later. No, he didn't appreciate being called to the carpet, but he seems to understand it now. One day, if you're lucky, you'll feel the same way. -- Smiles, Tony ==== Subject: Re: Orlow cardinality question Virgil said: > > Virgil said: > > > We have already discussed the problems with declaring the whole > numbers to necessarily be finite, and I have rejected that notion. > > I do not know how whole numbers differ from natural numbers in TO's > mind, but the set of finite natural numbers, where a natural is taken > to be finite if the set of naturals up to and including it is a finite > set by the Cantor definition, is , by the Cantor definition, an infinite > set. The size, meaning cardinality, of that set of naturals is not a > member of the set. > > > Without that restriction, the size of the set is N. > > Then the size of the set is not a member of the set because TO agrees > that the set cannot have any maximum member, but that size would be a > maximum member if a member at all. > > > You are confused. > > Not about naturals, but possibly almost as confused about what goes on > in TO's head as TO is. > > Your pofnats have no maximal member. The standard unit set does, but numbers continue beyond that, even though it's infinite. > > My set of naturals is the same as Peano's and, except possibly for > including 0, the same as that of everybody but TO. > And my N is infinite in cardinality though the set of its members > prededing any one of its members is of finite cardinality. Oh, that makes sense, or not. > > What? You comment in the middle, but when I derive the expected > result in alternative ways you don't respond? I guess no one has > any problem with this part. It's interesting people complain, but > never agree or concede. > > Why should the vast majority, being in the right, concede anything to a > minority of one who is in the wrong? If you thought there was something wrong with it, you might have commented on that, but there was no response. Still, no one has raised any real objections to my S^L argument, except to give the same old Cantorian counter-arguments. Where in my proof did I err? > > > Every specific number is either finite or infinite, depending on the > finiteness of the position of its most significant digit. > > Every basal representation of every real number, including all naturals, > has a first digit. > Every finite number has a most significant digit in a finite position. -- Smiles, Tony ==== Subject: Re: Orlow cardinality question stephen@nomail.com said: > stephen@nomail.com said: >> Given a set of symbols with size S, we can produce a set of all strings using >> those symbols that have length L, and the size of this set will be S^L. Digital >> number systems fall into this category, with S being the number base, which is >> always finite (2 for binary, 10 for decimal). If we want to have an infinite >> set of digital strings, therefore, S^L needs to be infinite, but S is finite, >> so L, the length of the strings, needs to be infinite to have an infinite set >> of such strings. >> >> You are assuming in your proof that there is a fixed L that is the >> maximum sized string. There is no maximum sized string. There >> is no L for which the number of strings is S^L. This has nothing >> to do with information theory. For any finite alphabet, there >> is a infinite number of finite strings. > What???? You better chack that statement with your buddies. That's way out in > left field. I mean, that's just patently false. You claim S^L can be infinite, > for finite S and L??? You're beyond hope, if that's not just a typo. > > No, I claim that there is no L such that S^L is the number of > strings in the language. There is no maximum length to > a string. There is no single L such that S^L is the number > of strings. So, then, there is no size to the set, as I have been saying. Still, if you claim L is finite, and S is finite, the S^L is finite. > >> >> Do you know anything about the theory of computation? A >> language is a set of finite strings. > Yes, I looked that up, and saw that, alas, it states strings are finite and > made of a finite set of symbols, but the set of strings is infinite. This needs > to be fixed, now. It's absolutely incorrect, and violates such basic math that > You are incurably stupid. Bye. Yeah, whatever. > > Stephen > -- Smiles, Tony ==== Subject: Re: Orlow cardinality question stephen@nomail.com said: > stephen@nomail.com said: > imaginatorium@despammed.com said: > What is a finite dividing line? Consider the subset of the surreals > consisting of > {0, 1, 2, 3, ...} U {w, w-1, w-2, w-3, ...} > > This is totally ordered (not in the order above, obviously) and > includes an unending sequence of integers increasing in steps of 1 from > 0, and an unending decreasing set stepping down from w (omega). Very > obviously no element of this set is a dividing line, but there is a > very clear division between numbers of the form n a natural number and > numbers of the form w-m for m a natural number. This might look vaguely > like your twilight zone, but one difference is that if you start at 0 > and successively add 1, you never get out of the finite part (and vice > versa from w downwards). Adding 1 infinitely many times isn't > defined, and adding infinity requires a clarification of what you > mean by infinity. > Read up on infinite series. >> >> Where, please? Since nothing I have ever read so far about infinite >> series agrees with any of your stuff about adding things infinitely >> many times, approaching infinity, and so on, it's plainly a waste >> of time reading my own books. Have you written one on Orlovian infinite >> series yet? Or where do you suggest I should look? >> >> Tony seems to think that because some series diverge, that is >> proof that there are natural numbers that are infinite in size. >> Apparently a series cannot diverge unless you are able to reach >> infinity by repeatedly adding 1. >> >> Stephen >> > In an infinite series where each term is 1, the series diverges. The series can > only converge if the limit of the terms is zero. > http://mathworld.wolfram.com/DivergenceTests.html > -- > Smiles, > > That has nothing to do with the question at hand however. > The fact that series diverge does not mean that any > natural number is infinite. > > Stephen > It means the sum of an infinite number of 1's is an infinite sum, as any schoolchild can tell you. -- Smiles, Tony ==== Subject: Re: Orlow cardinality question stephen@nomail.com said: > stephen@nomail.com said: >> Every specific number is either finite or infinite, depending on the finiteness >> of the position of its most significant digit. >> >> And how do you determine if the position of its most significant digit >> is finite or not? It sounds like you are defining finite in terms >> of finite, which is circular. Presumably the position of a digit >> is a number. So the number that equals the position of the the most >> significant digit is finite if the position of its most significant digit >> is finite. >> >> With that definition you could not even prove that 1 is finite. >> The position of the most significant digit in 1 is 1. Therefore >> 1 is finite if 1 is finite. Brilliant! >> >> Stephen >> > Finite natural numbers start inductively with 1 and follow the rules of finity, > such that a+b is finite for finite a and b. Inductive proof doesn't allow this > to continue for infinity, however, since finiteness is not an equality but a > vague inequality. > > Finiteness is a vague inequality? Where do you get this > bullshit from? You are sounding more and more like Lester > everyday. > > Finiteness is not an inequality. It is a property. > Something is finite, or it is not. There are no > shades of finiteness. Of course there maybe according > to whatever strange definition of finite you are using, > but as you have never defined any of your terms, such > as finite, natural number, etc. all you say is nonsense. > > Stephen > A finite number is less than any infinite number. It is less than an infinite distance from zero. But there is no smallest infinity, so this definition does not pinpoint a place where finite numbers end, since there is no place where infinite numbers begin. Finite doesn't mean equal to anything, but less thn anything infinite, as infinite means greater than anything finite. -- Smiles, Tony ==== Subject: Re: Orlow cardinality question stephen@nomail.com said: >> Every specific number is either finite or infinite, depending on the finiteness >> of the position of its most significant digit. >> And how do you determine if the position of its most significant digit >> is finite or not? It sounds like you are defining finite in terms >> of finite, which is circular. Presumably the position of a digit >> is a number. So the number that equals the position of the the most >> significant digit is finite if the position of its most significant digit >> is finite. >> With that definition you could not even prove that 1 is finite. >> The position of the most significant digit in 1 is 1. Therefore >> 1 is finite if 1 is finite. Brilliant! >> Stephen > Finite natural numbers start inductively with 1 and follow the rules of finity, > such that a+b is finite for finite a and b. Inductive proof doesn't allow this > to continue for infinity, however, since finiteness is not an equality but a > vague inequality. > > Finiteness is a vague inequality? Where do you get this > bullshit from? You are sounding more and more like Lester > everyday. > > Finiteness is not an inequality. It is a property. > Something is finite, or it is not. There are no > shades of finiteness. Of course there maybe according > to whatever strange definition of finite you are using, > but as you have never defined any of your terms, such > as finite, natural number, etc. all you say is nonsense. > > Stephen > A finite number is less than any infinite number. It is less than an infinite > distance from zero. But there is no smallest infinity, so this definition does > not pinpoint a place where finite numbers end, since there is no place where > infinite numbers begin. Finite doesn't mean equal to anything, but less thn > anything infinite, as infinite means greater than anything finite. Right, brilliant... um, but how would anyone working from just your definitions go about understanding *anything* concrete about which numbers were finite and which infinite? I mean, suppose I declared that infinity equals 57. Then all numbers up to and including 56 are finite, and are of course less than all infinite numbers, which start at 57. These in turn are more than all finite numbers. Well, I expect you have some feeling that this isn't quite right. OK, let me consider your worms: you say somewhere that the Tonats range as follows: 000...000 to 999...999 Do I understand that these worms consist of so many digits that if I started at the right, and moved leftwards reciting a ditty (one-ping-two-ping-three...) I would never stop? (Remember that a fairly simple program can be written to generate the next bit of the ditty from the preceding segment, without going back more than a finite distance to the last ping, so there's no way the ditty will run out on the last verse, because the last verse doesn't exist.) Can I assume that the Tonat 000...000 is the same as normal mathematical zero, and for example 000...057 is 57? Do I understand correctly that 999...999 is an infinite Tonat, and is the greatest Tonat in the set of Tonats, except that it has a successor, which is not in the set of Tonats? (And you do, is this right, claim that the Tonats satisfy Peano's axioms? Which say that for any (To)nat x in the set, the set also includes the successor of x, don't they? But this isn't a contradiction, for reasons still to hard for me to understand?) OK, so if 000...000 is finite and 999...999 is infinite, even though we understand that the twilight zone is vague and indefinite, can't you say anything about at least some of the numbers in between? What about 111...111: is this finite? In normal maths, the pofnats, if they could be written in this strange way would all start 000..., because only a finite (that's _our_ finite!) number of digits from the right can be nonzero. But I'd be interested to hear your teachings about this. There's no chance ever, I presume, that you would actually write a definition that anyone else could use to answer the question? Oh yes, group theory: You believe that in a sense the Tints form an infinite cyclic group. You also claim that for any two finite integers, their sum is finite; this seems to imply that the pofnats form a subgroup of this infinite cyclic group. I believe this implies that the generator of the infinite cyclic group cannot be an element of the subgroup, and in fact this means that the generator of your infinite cyclic group is an infinite integer. Any clue which one? (No, I know, you haven't a clue.) Brian Chandler http://imaginatorium.org ==== Subject: Re: Orlow cardinality question stephen@nomail.com said: > stephen@nomail.com said: >> imaginatorium@despammed.com said: > So it *appears* to me that the set of Tonats has a maximal element (a T > for which T+1 is not a Tonat) and also for any Tonat called Q, Q+1 is a > Tonat. Well, this *appears* to me to be a contradiction (independent of > whether the Tonats are otherwise well-defined). Can you clear up this > apparent contradiction for me? >> That would be a contradiction if I had ever said that. I never claimed there is >> any last integer, finite or infinite, but that the size of the set of integers >> is simply to be taken as a standard discrete infinity with which other discrete >> infinities may be compared. >> >> To quote you: >> Whatever size the set is, that value is a value of an element in the set. >> You really can't deny this constant equality between the range of element >> values and the size of the set. >> So according to you the size of the set of integers is an integer. >> But now you are saying that the size is not the last integer, >> which seems to contradict your inductive proof that the size >> of { 1, ... n } is n even when n is infinite. > No, it contradicts your notion that the successor to an integer is an integer > in the standard set. > > By definition the successor of an integer is an integer. > If you disagree, then you are not talking about integers, > but something else of your own devising. Are you incapable > of talking about the integers as they are defined by > the rest of the world? > >> >> Here are three simple questions: >> 1. what is the size of the set of all integers > N >> 2. is the size of the set of all integers an integer? > yes >> 3. are there any integers larger than the size of the >> set of all integers? > yes. This is what is confusing people. There are whole numbers larger than the > standard set. There are discrete infinite set size larger than N, but less than > R. There is a whole spectrum of infinities. > > So the size of the set of all integers is not equal to > the value of the largest integer. But I thought you had > inductively proven that for any set of integers, the > size equalled the largest element? > > Your inconsistencies are rapidly multiplying. > > Stephen > In the standard set of integers, the largest is N, the size, but one can still talk about larger whole numbers than what is in that standard set. 999...999 can have a successor in 000...001:000...000, but that adds another dimension of infinity in a sense. It's a different approach. -- Smiles, Tony ==== Subject: Re: Orlow cardinality question talk about larger whole numbers than what is in that standard set. 999...999 > can have a successor in 000...001:000...000, but that adds another dimension of > infinity in a sense. It's a different approach. So what is the size of ALL your whole numbers. Surely you could just take the set of all of these, are there infinitely many infinite dimensions? What if I have 9's in all the positions and have infinitely many dimensions, what happens when I add 1 to that number, where does it carry over? Especially since I have now ALL of your whole numbers in one place (set), then I assume that if I add 1 I should get another whole number right? Jiri ==== Subject: Re: Orlow cardinality question stephen@nomail.com said: > > >> N is in N, but N+1 is not. The successor of N is zero, with a carry over into >> the next infinity digit. > > Only in computer arithmentic with finite registers. > > You mean only in Tony's fantasy land. I have never seen > a computer with an 'infinity digit'. > > Stephen > Every adding circuit has a carry bit. This way, numbers can be chained together infinitely, if desired, for arbitrarily long binary numbers, through software. -- Smiles, Tony ==== Subject: Re: Orlow cardinality question stephen@nomail.com said: > stephen@nomail.com said: >> Kastrup said: > > Kastrup said: >> >> Actually, I have no problem with the naturals going on forever, >> as long as they achieve infinite values, which I have no problem >> with either. >> >> Unfortunately, the naturals have a problem achieving infinite >> values. They can't. While every possible limit gets exceeded by >> some naturals, there is no single natural that could exceed all >> limits. > > Then one can say the same about set size, which is incremented in > exactly the same steps as the maximal value. > > Rubbish. The set of natural numbers is defined by five axioms, not by > somebody putting numbers into a bag. The size of the sets of naturals > is not incremented, it just is. >> it's defined recursively with a base case and each element defined with respect >> to its predecessor. It's incremented. > > The set of all finite naturals is finite, indeterminate as its size > may be. > > Stomping your feet and sulking is not going to make it so. >> No, the restriction of finiteness does that all on its own. > >> I really don't see what problems that introduces. If you want to >> say the series of finite naturals has no end because the end is >> not identifiable, then that is okay, but it is of the boundless >> variety, and not infinite. >> >> Infinite as a set measure is defined as the ability to place a >> set into bijection with a proper subset. The successor relation >> does just that with the naturals. So the set is infinite. > > That's one way to look at it. So? What does having infinite whole > numbers break in your world? > > The fifth Peano axiom. Any set containing 0 and including for each of > its numbers its successors, contains _all_ elements of the naturals. > Since the successor operation will give a finite number from a finite > number, the induction axiom strictly rules out infinite numbers in the > set of naturals. >> Not after the infinite number of increments required to produce the infinite >> set. >> >> The set is not produced. There is no time involved. >> > > Now if you want to, you can call the set size of the naturals a > number, or you can call it nonexistent. But what you _can't_ call > it is a natural number. That name is already taken for the set of > numbers defined by the five Peano axioms, and the size of this set > can't be a member of this set itself, because it does not obey the > axioms. >> Yes it does. I have already shown how the size of the set IS the maximal >> element. Same thing, a natural number, which is used to count things, like >> natural numbers. Natural numbers are things used to count things, like natural >> numbers, which are things used to count things, like natural numbers......... >> >> So you admit that your theory is fundamentally inconsistent. >> N is a natural number according to you, but N+1 is not, in >> direct contradiction with the definition of the natural numbers. >> >> I wonder if you think N-1 is a natural number or not? >> Given your totally arbitrary system, who knows? >> >> > > > But whatever such numbers are, however useful or > enlightening they might be: one thing they aren't by definition: > members of the set of naturals. >> The size of the set of naturals is in the set. >> >> And therefore N+1 is in the set. >> > > The axioms for that set leave no place for such a number designating > the size of the set _as_ _a_ _member_ of the set. >> Poppycock! >> >> Explain how N can be in the set of natural numbers but N+1 >> is not? The axioms clearly state that if x is a natural number, >> then x+1 is a natural number. It does not get much simpler >> than that. >> >> Stephen >> > Because N is declared to be the size of the set, and therefore its maximal > member. > > You are the one declaring N is the size of the set, and therefore > its maximal member, despite the fact that this contradicts > the definition of the natural numbers. Why declare something > that contradicts the definition? That means that you > do not even have a definition of natural number other > than some vague hand-wavy thing you have never specified. > > Stephen > I am not contradicting the definition, but using it in a slightly different way by choosing a unit infinite set N of the whole numbers, with a unit infinite size N. Say we have all the whole numbers, and add 3/4 to the set. Then, don't we have N+1 elements? If we have the multiples of 1/2, two for every natural, then don't we have 2N 1/2's? That's my thinking on the matter. -- Smiles, Tony ==== Subject: Re: Orlow cardinality question stephen@nomail.com said: > stephen@nomail.com said: >> stephen@nomail.com said: > imaginatorium@despammed.com said: >> Stopping after first (oops! can't promise it's the first, can I?) >> error... >> >> Dik T. Winter said: >> > stephen@nomail.com said: >> ... >> > What I proved is that for all n in N, the set defined as the whole >> > numbers from 1 to n has the property that its maximal element and >> > its size are equal to n. >> > This is a property of n, since the set is defined by n, according to >> > a simple rule. There is absolutely nothing wrong with the way I have >> > applied induction in this case. >> > >> > But you have not proven anything about all sets. There are >> > sets that are not defined as the whole numbers from 1 to n. >> > In particular, the set of natural numbers is not defined as the >> > whole numbers from 1 to n. There is no n such that 1 to n is >> > all the natural numbers. Therefore your proof says nothing >> > about the set of all natural numbers. >> >> > It does as n approaches N (aleph_0). If aleph_0 is the size of the set, >> > then it is also the maximal element. >> >> Sorry, but you are guilty of circular reasoning. Aleph_0 is the size of the >> set, but not a number according to Peano's axioms. The induction axiom is >> about numbers only, so the axiom does not apply. >> >> I don't care what you think is in the set or out of it, but if induction proves >> anything then the size of the set of naturals is also the value of its maximal >> element. >> >> It doesn't have a maximal element, so this nonexistent element is not >> the size of the naturals. Because suppose P were the maximal element, >> then by the axioms for the pofnats, P+1 is also a member of the set, so >> P is not maximal. Contradiction. >> >> Brian Chandler >> http://imaginatorium.org >> >> > largest finite. Ommmmmm largest finite..... (shakes rattle and drinks snake > blood) > > Is that supposed to mean something? You are the one who > just keeps repeating things because you know they are right > without proof. > > Whatever you can say about my maximal element, I can say about your set size. > So, you might want to watch your tongue. > > Remember, I don't have a problem with N+1 anyway. > > So then N+1 must also be a natural number according to you, and > therefore N is not the maximal value. N+1 is the maximal value. > No wait, N+2 is the maximal value. No, it must be N+3. It > has to be there somewhere, according to your logic. Maybe > it is 2N, or 3N, or N*N. Gee, do your natural numbers ever stop? > > Stephen > >> Uh, no, they're an infinite set, with infinite values and everything. N is a >> standard infinity, declared as the size of the set of whole numbers. There are >> bigger and smaller infinities. >> -- >> >> But you claim that N is a natural number. You claim to have >> proven that every set of natural numbers must contain >> a maximal element and that maximal element is the size of the >> set. However N+1 > N, and if N is a natural number, then N+1 >> is a natural number, and so N is not the maximal element and is >> not the size of the natural numbers. That same argument >> shows that N+1 is not the largest either, and so on >> >> Is N a natural number? Is N the size of the natural numbers? >> Are there natural numbers larger than N? You seem to have a fundamental >> inconsistency here. >> >> Stephen >> > Yes, you detect a paradox. We can always add more digits onto whatever we are > talking about. N is the unit discrete 1-D infinity. We declare it to be equal > to size of the set of integers, but we can change it using any arithmetic > operation. Being the unit infinity doesn't make it the largest number, or the > smallest infinity. It makes it an infinity unit, declared somewhat arbitrarily, > as any unit is, for the purposes of relative measure. You may think of oo+1=oo > as one would think of 1+0=1, but does adding one point to the end of a line > really add nothing at all? > > So now oo+1=oo? There really is no consistency to your > claims. > > Stephen > > No, that is not what I am saying. I said YOU may think of it like that, like most people also think of 1+0=1, but in my perspective, that addition of 1 is significant, and if the 0 represents a point, without length, perhaps the addition of that 0 actually adds something to the line, even if it's imperceptible on the scale of the line as a whole. So, no, that was not what I was claiming. -- Smiles, Tony ==== Subject: Re: Orlow cardinality question > stephen@nomail.com said: >> Yes, you detect a paradox. We can always add more digits onto whatever we are >> talking about. N is the unit discrete 1-D infinity. We declare it to be equal >> to size of the set of integers, but we can change it using any arithmetic >> operation. Being the unit infinity doesn't make it the largest number, or the >> smallest infinity. It makes it an infinity unit, declared somewhat arbitrarily, >> as any unit is, for the purposes of relative measure. You may think of oo+1=oo >> as one would think of 1+0=1, but does adding one point to the end of a line >> really add nothing at all? >> >> So now oo+1=oo? There really is no consistency to your >> claims. >> >> Stephen >> >> > No, that is not what I am saying. I said YOU may think of it like that, like > most people also think of 1+0=1, but in my perspective, that addition of 1 is > significant, and if the 0 represents a point, without length, perhaps the > addition of that 0 actually adds something to the line, even if it's > imperceptible on the scale of the line as a whole. So, no, that was not what I > was claiming. > -- > Smiles, Oh, I see. You are claiming that 1+0 <> 1. That makes a lot more sense. There really is no consistency to your claims. Stephen ==== Subject: Re: Orlow cardinality question Virgil said: > > Jesse F. Hughes said: > > Jesse F. Hughes said: >> Instead, I expect one of the following to occur: >> >> (1) Tony denies that the induction axiom proves that every natural >> number is finite. > Correct! >> >> (2) Tony agrees entirely, but adds that of course that axiom is the >> reason that Peano is self-contradictory. > > Peano is not self-contradictory, nor implies that all naturals are > finite, except by excluding the qualifier that the property proven > by inductive proof be constant, such as an equality, or at least not > decreasing in truth at each iteration. This should be a condition of > inductive proof. > > Oh, that's cute. So it's *less true* that 14 is finite than that 13 > is. And it's even *less true* that 15 is finite. And so on. And > eventually, we reach numbers for which it is just barely true that the > are finite. > As a number grows, the differenc ebetween it and a larger number grows less. > > Not if the larger number is not fninte. > > distance. That is the point of infinite numbers if they are to have > any point at all. > That is precisely the point with which I disagree. That point is pointless. What if we did that with finite numbers? We'd have no math at all. People disagreed with negative numbers, with imaginary numbers, with surreal numbers.... Maybe there is a place for infinite wholes, and the idea that removing a finite amount from an infinite amount makes it a smaller infinity. You might want to consider the possibility. -- Smiles, Tony ==== Subject: Re: Orlow cardinality question Virgil said: > > stephen@nomail.com said: > > > You clearly believe they do have a maximal element, and it is a > direct consequence of your proofs. To quote you: > Whatever size the set is, that value is a value of an element in the > set. You > really can't deny this constant equality between the range of element > values > and the size of the set. > So the set of all naturals contains an element that equals the size > of the set of all naturals. You call this N. You claim that N+1 > is not a natural (despite the obvious contradiction). Presumably > N is greater than all the other naturals, because that seems to > be your definition of size. So following your logic there > is a maximal natural number which you call N. > > Of course you could change your mind and decide that N+1 is a natural > number, but now the naturals contain an element larger than the size, > which contradicts your claim that the size equals the maximum element. > > Stephen > > Okay the contradiction you see is not within my system, but between mine and > yours. There is no maximal element to the set of finite naturals. My N is not > finite, but has a maximal element, but that element has a successor, since > every whole number does. It's a very different system from yours. > > If the maximal element in N has a successor, that successor is in N too, > and is larger than the maximal element. > > I certainly do not want to work withon any such a self-contradictory > number system. > No, there is always a larger number, a successor, even to N. N is chosen as a unit infinity because it is simple, not the largest or smallest. N+1 is considered natural, but additional to the unit set of counting numbers. My definition of the basic set is different from yours, since you consider any bijections to idicate equivalence, while I do not. You consider oo+1=oo, but I do not. I say N is the set of whole numbers, but consider numbers beyond that, since that is the only way we can talk about larger sets, like the multiples of 1/2 have a set size of 2N. It's a different paradigm, but not internally contradictory. -- Smiles, Tony ==== Subject: Re: Orlow cardinality question Virgil said: > > > The second > refers to the grid as some formless quarter-plane, but that is > incorrect, > > Since each digit in that quarter plane grid has a grid location > indicated by a pair of naturals , one for row and one for column, how is > it not a quarter plane? > The row is indicated by a real, not a natural. Remember, you are listing the reals, each with a number of digits equal to the naturals. The rectangle is infinite, but is still governed by the same geometry as the finite lists of digital numbers or complete sets of strings. -- Smiles, Tony ==== Subject: Re: Orlow cardinality question > Virgil said: > > > The second > refers to the grid as some formless quarter-plane, but that is > incorrect, > > Since each digit in that quarter plane grid has a grid location > indicated by a pair of naturals , one for row and one for column, how is > it not a quarter plane? > > The row is indicated by a real, not a natural. Remember, you are listing the > reals, each with a number of digits equal to the naturals. The rectangle is > infinite, but is still governed by the same geometry as the finite lists of > digital numbers or complete sets of strings. If that row is on of the listed numbers, then the row is indicated by its position in that list, a natural number, just as the column is the position of some digit in some listed number, also a natural number. ==== Subject: Re: Orlow cardinality question Virgil said: > > Jan de Vos said: > > > This sequence can converge to a finite value, or it can diverge -- in > which case we say that the series (or sometimes the sum) diverges. > It can diverge in two ways: by adding to an infinite sum, or by oscillating > around some number or function endlessly. > > There is /nothing/ that says the sum should be a natural number. > The sum is either oscillating, or some number, finite or infinite. > > Begs the question of whether infinite applies to numbers. > It does not apply to naturals or integers or rationals or reals or > complexes, at least in any standard model of them. Infinity is a quantitative concept. Of course it applies to numbers. I have already shown how integers can be infinite. It's not a very hard trick. Comparison tests are common in infinite series, like determining a series diverges because each of its terms is larger than the corresponding term of another series known to diverge to infinity. This is pretty standard stuff, even though mathematicians are scared to bring it back home to number systems for fear Kronecker will haunt them. > > If TO wants to have some non-standard model, he must define that model > before he is allowed to use it. That's kind of what I'm doing here. > > Saying 'but after doing X for an infinite number of times...' make no > sence, because it already assumes that 'infinity' is some number, so > that you can repeat something that many times. That assumption is > what makes your proof not a proof, but a statement of the form 'A > implies A'. > That's a pile of bunk. Infinite series deal with what happens to sums as the > number of terms goes to infinity. > > Goes to infinity should be gets large without limit, since however > far one goes towards infinity, one is still infinitely far away from > infinity. > > AT infinity, sum(1/2^n) is 1. > > At infinity is where one never is. You need to change your name to Zeno. > > AT infinity, sum(1/n) is infinite, and sum(1) is infinitely larger. > All this pansy-ass talk about things never getting to infinity is > Kronecker disease. Get over it. > > If TO thinks he can get there, let him go there and report back what it > was like, but with some concrete evidence that he was really there. > Here, I sent you a postcard. Wish you were here!! -- Smiles, Tony ==== Subject: Re: Orlow cardinality question > Virgil said: > > Jan de Vos said: > > > This sequence can converge to a finite value, or it can diverge -- in > which case we say that the series (or sometimes the sum) diverges. > It can diverge in two ways: by adding to an infinite sum, or by > oscillating > around some number or function endlessly. > > There is /nothing/ that says the sum should be a natural number. > The sum is either oscillating, or some number, finite or infinite. > > Begs the question of whether infinite applies to numbers. > It does not apply to naturals or integers or rationals or reals or > complexes, at least in any standard model of them. > Infinity is a quantitative concept. Of course it applies to numbers. I have > already shown how integers can be infinite. It's not a very hard trick. TO has claimed it, but his claims require such numbers to violate the Peano properties, so whatever they may be, they cannot be natural numbers. > Comparison tests are common in infinite series, like determining a series > diverges because each of its terms is larger than the corresponding term of > another series known to diverge to infinity. This only says that one series dominates another, it does not compare non-existent limits. > If TO wants to have some non-standard model, he must define that model > before he is allowed to use it. > That's kind of what I'm doing here. That is what TO is trying, but failing, to do here. > At infinity is where one never is. > You need to change your name to Zeno. perhaps TO should study Zeno a bit more carefully. > > AT infinity, sum(1/n) is infinite, and sum(1) is infinitely larger. > All this pansy-ass talk about things never getting to infinity is > Kronecker disease. Get over it. > > If TO thinks he can get there, let him go there and report back what it > was like, but with some concrete evidence that he was really there. > > Here, I sent you a postcard. Wish you were here!! Wish TO would stay there. ==== Subject: Re: Orlow cardinality question Virgil said: > > Virgil said: > > > Bull. I used established math in two other areas to show that an > infinite set of naturals requires infinite values, as well as an > inductive proof that shows the same. You are in defiance of what > math says about finite vs. infinite values. In defiance and > denial. I am only in defiance of your self- contradictory > nonsense. > > Speaking of BULL! > > TO has not used established math in any area to prove that a set of > finite objects must contain an infinite object. > > I proved that an infinite set of distinct whole numbers MUST contain > elements with infinite vales, and that a set of whole numbers which > are all finite MUST be finite, to be specific. > > Non-empty well-ordered sets, such as N, have the property that every > non-empty subset has a minimal member. > What is the minimal member of the set of infinite naturals? > And consider the set N* of all naturals less than that minimal infinite > natural. (sigh) There IS no smallest infinity, just as there IS no largest finite number. Defining your set along that non-boundary gives you an ill-defined set. Now, if I consider ALL the whole numbers, from (in binary) 000...000 to 111...111, there is an infinite set where each specified number has a successor (especially if the successor to 111...111 is 000...000, which is it in one way), and there is a maximal and minimal element. What do you say about such a set definition? Doesn't it contradict the idea that a max and min means the set is finite? > > I can prove that any ordered set which is infinite, in the sense that it > allows injection to a proper subset, if and only if it has a non-empty > subset having no maximum member or a non-empty subset containing no > minimum member. Since non-empty subsets of the set of naturals must > always have minimal members, a set of naturals is infinite if and only > if it has a subset with no maximal member. > > So that either N*, with no 'TOnaturals' in it, has a maximal member or > N* is infinite. This all hinges on your method of proof. You make statements that I am supposed to accept, but I don't. What about the set described above, which you never commented on? > > TO has only managed to support his own belief in his own delusions, > and has not managed to impose those delusions on anyone else. > > blah blah blah..... > > TO has repeatedly rejected sound mathematical disproofs of his > counter-mathematical claims. > > Incorrect. > > Only in the not too reliable opinion of TO himself. > > > For example, TO claims that the set of naturals, N, must contain > what he calls an infinite natural, with properties that seem to > differ from post to post. > > Not true. I have proven it is the case, three different ways, and > your only proof to the contrary contradicts itself. > > Only in the not too reliable opinion of TO himself. > > N is easily represented by a union of nicely finite initial sets, > each containing only finitely many finite elements none greater > than its last and largest element. Then every member of that union > must be a member of at least one of those finite sets capped by a > finite value. > > Yeah and your maximal value is in that largest finite set. (sigh) > > Show me that largest finite set which is so large that adding one to > its maximal value and appending this new natural to the set does not > create a larger finite set. (sigh) > > > Or does TO claim that a union of sets can contain an object not in > any one of the sets of that union? > > Of course not. Duh. > > Since O has claimed things equally idiotic, one cannot be sure without > asking what TO will claim next. > Your lack of understanding is yours to own. -- Smiles, Tony ==== Subject: Re: Orlow cardinality question > Virgil said: > Non-empty well-ordered sets, such as N, have the property that every > non-empty subset has a minimal member. > What is the minimal member of the set of infinite naturals? > And consider the set N* of all naturals less than that minimal infinite > natural. > (sigh) There IS no smallest infinity, If the set of infinite naturals does not have a smallest member, then, as a set of naturals, it must be empty, since EVERY non-empty set of naturals has a smallest natural as a member. As can be easily proved from the Peano postulates. > just as there IS no largest finite number. If number' here means natural number then TO is correect, and from this, one can easily show that the set of finite natural numbers is an infinite set according to the Cantor definition of infinite for sets. >Defining your set along that non-boundary gives you an ill-defined > set. Ill-defined sets are not sets at all, but if one cannot distinguish finite naturals from infinite naturals, how does TO distinguish between them? And if TO can distinguish between them, they can be separated into well-defined sets. > Now, if I consider ALL the whole numbers, from (in binary) 000...000 to > 111...111, there is an infinite set where each specified number has a > successor > (especially if the successor to 111...111 is 000...000, which is it in one > way), and there is a maximal and minimal element. If the successor to 111...111 is 000...000 then the set of such objects is not an ordered set according to the definition of an ordered set, since we have both 111...111 > 000...000 and 000...000 > 111...111 > What do you say about such a set definition? Doesn't it contradict > the idea that a max and min means the set is finite? How is the set ordered? Given two 'members', give the algorithm for comparing their sizes. > > I can prove that any ordered set which is infinite, in the sense that it > allows injection to a proper subset, if and only if it has a non-empty > subset having no maximum member or a non-empty subset containing no > minimum member. Since non-empty subsets of the set of naturals must > always have minimal members, a set of naturals is infinite if and only > if it has a subset with no maximal member. > > So that either N*, with no 'TOnaturals' in it, has a maximal member or > N* is infinite. > This all hinges on your method of proof. You make statements that I > am supposed to accept, but I don't. What about the set described > above, which you never commented on? I have, but TO has ignored my comments. In the first place, I do not believe that To can show that his set can be made to satisfy any description of mathematical linear ordering. Various partial orders are easily possible, but TO must show that his ordering is also complete, so that ANY two members may be compared for size. For example, given two members each with infinitely many zeros preceding and trailing a single 1, how does one compare them for size without a count of the number of preceding zeros or a count of the number of trailing zeros, or both, for both? > > Your lack of understanding is yours to own. Actually, my understanding seems to be widely shared, whereas TO's is widely rejected. ==== Subject: Re: Orlow cardinality question Virgil said: > > Jiri Lebl said: > We > define a set N as the collection of all things which > meet certain conditions. That set IS, it isn't > incremented. > You define the set N using Peano's successor operator and a starting > point of 1 > or 0, after which each is defined as the successor to the last. So, > you're full > of it. > > One of these days Tony, you should really read Peano's definition of N. > It's good to know about something before arguing about it. Yes Peano > defines successor operator, but the definition says nothing about > constructing N by using the successor. Peano constructs N from a > LARGER set, not from SMALLER sets. > > Jiri > > > No, Peano defines the set as containing 1 (or 0), and for every element it > contains, it also contains the successor. Now, what I am suggesting is in > violation of that rule, unless we consider the naturals to be a circular set. > What is that, a ring? That is the way integers work in computers. Then every > element has a successor and a predecessor, within that ring. Now, every time > we > go around the ring, we send a carry bit to the next ring, as an increment, so > you have a base-N system. This turns out to be the true representation of the > real numbers, an infinite-digit infinite-base number system. Remember S^L? > What > if S AND L are infinite? Now you have a true digital representation of the > reals, finite, infinite, and infintiesimal. Tadaa! > > When computers have registers that can record infinitely many bits, only > then will they be reasonable models for dealing with the set of all > naturals. > > Until then, TO is using the wrong sort of model, and the quality of his > conclusions demonstrates that wrongness. > Actually, someone else made the same comment a while back in email, and while computers are finite, like everything we create, there is a way of looking at computer registers that says a lot about number systems and infinity. I think I have alluded to 2's complement here, when I had my little epiphany about why it works the way it does, which I had just sort of accepted, since it works so well. It's nice to see clearly WHY it works after decades. I haven't found the explanation yet anywhere on the web, either. Discovering is fun! In 2's complement, the ubiquitous signed integer system in computers, the negative of a number is the bitwise inverse (change all 0's to 1's, and 1's to 0's), plus 1. The negative numbers all begin with a 1 and the positives (and 0) with a 0. -1 is all 1's, and when you add 1, you end up carrying a 1 all the way to the left, and dropping it, but it works perfectly, producing all 0's, which is 0. Likewise, subtracting 1 from 0 produces all 1's, borrowing all the way to the left. These overflows don't seem to break anything, but produce proper results around zero. Meanwhile, at the other end of the circle, opposite from 000...000, we seem to have an extra negative number, 100...000, giving us (for n digits altogether), 2^(n-1) negatives and 2^(n-1)-1 positives, and zero. This number, 100...000, is 011...111 + 1, which doesn't involve an overflow carry out of the available bits, but does represent the greatest positive number, plus 1, equalling the greatest negative number. Obviously, the greatest negative number minus 1 is the greatest positive number. So, while there is no overflow, there is a quantity discontinuity at that point. To me, this number, 100...000, really represents both positive and negative infinity, as 000...000 represents both positive and negative zero. Now, of course, binary numbers can be used the way you are probably more used to, as unsigned integers, from 000...000 up to 111...111. In that case, with a finite number of digits, 100...000 doesn't represent any infinity, but just the middle of the finite number line. However, we still have a discontinuity in our number circle, since 111...111 + 1 still equals 000...000, so the successor of the largest number is zero. In this case, the overflow or undeflow (carry or borrow out the left digit) indicates the discontinuity, and infinity, or the closest thing to it, is considered to be 111...111. Now, if we have infinite digits, 111...111 is our largest number and obviously infinite, and 100...000 is about half that (plus 1/2). Any number with 1's in positions infinitely far from the right are infinite, and those comprise most of the infinitely long numbers. So, there are two ways of looking at the binary number circle, which give us two ways of seeing infinity, as the number of bits approaches infinity. The computer is finite, but so are we, and we have some notion of infinity. It's not surprising that a math machine would too. Besides, sloppy programmers experience the Turing-machine variety of infinity every day, when they FORGET to increment their values, and watch their programs go into infinite loops, which are really only potentially infinite, of course, since the programmers soon reboot their machines or kill the process, to find their mistake. :D -- Smiles, Tony ==== Subject: Re: salaries in Hong Kong <42BCE95A.961263B9@netvigator.com> <42bdfdf6$0$2019$ed2e19e4@ptn-nntp-reader04.plus.net How long is a piece of string? I'm earning well over US2 million per year > and my senior colleagues earn over 3 million which is the norm. >> How much do investment bankers and financial analysts make in Hong Kong >> nowadays? > > At least $50,000 per month at a guess. I know one that had his Lexus > stolen and it was not insured against theft. Not too bothered, he said > it was just a month's salary > > Bob > humanist Brit. > Hong Kong >> Mark Demers >> EquityValue Investments Wow, $2 million a year. Do you have to be a senior banker, or just one with 3 years experience to earn at that level? I presume you must have an MBA or law degree to get the job in the first place. MD > ==== Subject: Re: mathematician salaries <05vte.2$45.931@news.uchicago.edu> Aldar Hong Kong Chan, if you can make $2 mil as an investment banker in Hong Kong, why are you wasting your time in a second tier math phd program with no job prospects in my country? Mark EquityValue Investments ==== Subject: proving an infinite series is divergent Hello I'd like some tips to prove the following: Let x_n be a sequence of positive numbers such that Sum(n=1, oo)x_n diverges. Let s_n = x_1...+ x_n. Then, Sum(n=1, oo)(x_n/s_n) diverges too. I tried applying the ratio test, but it failed. Amanda ==== Subject: Re: proving an infinite series is divergent > Hello > I'd like some tips to prove the following: > > Let x_n be a sequence of positive numbers such that Sum(n=1, oo)x_n > diverges. Let s_n = x_1...+ x_n. Then, Sum(n=1, oo)(x_n/s_n) diverges > too. > I tried applying the ratio test, but it failed. > > Amanda > Is there a number c > 0 such that x_n/s_n > c x_n for all n ? ==== Subject: proving an infinite series is divergent Hello I'd like some tips to prove the following: Let x_n be a sequence of positive numbers such that Sum(i=1, oo) x_n is divergent. Let s_n = x_1....+ x_n. Then Sum(i=1, oo) (x_n)/(s_n) is divergent too. I tried applying the ratio test, but it failed. Amanda ==== Subject: Call for Undergraduate research papers The Morehead Electronic Journal of Applicable Mathematics http://www.moreheadstate.edu/colleges/science/math/mejam/ interdisciplinary journal, sponsored by Morehead State University, which provide a refereed outlet for undergraduates in any discipline to publish quality papers and see the results quickly. undergraduate curriculum and which emphasize the applicability of mathematics while maintaining significant mathematical interest. Papers may be historical, expository, or completely original in nature but must adhere to strict academic standards and should explore some aspect of the applications of mathematics. Papers will typically be the result of summer research programs, internships, or other school projects. Papers from all disciplines will be considered for publication. disciplines, and a network of referees. Each student submitting a paper must have the support and guidance of a faculty sponsor whose name will appear in the Journal. Faculty sponsors are full-time faculty members who enforce rigorous academic standards and attest that the work was done before the student received a bachelor's degree. Faculty sponsors may be asked to join the pool of referees in their areas of expertise. The Department of Mathematical Sciences and Morehead State University are devoted to quality undergraduate programs and provide support to The editorial board of the journal reserves the right to reject any paper it deems inappropriate. ==== Subject: Re: Tic-tac-toe probability >A random game of tic-tac-toe (on a 3 x 3 grid) is played with each player making a random choice from the legal moves available at his turn. A player wins if he gets 3 in a line (row or diagonal). 1) What is the probability that the first player wins? >2) That the second player wins? There are 9! possible complete games, but only choose(9,4) ending positions. Each of those ending positions represents 4!5! different games. Ending positions can be separated into 4 categories: 1. 3-in-a-row for first player only -- all 4!5! games go to 1 2. 3-in-a-row for second player only -- all 4!5! games go to 2 3. No 3-in-a-row 4. Each player has exactly one 3-in-a-row Category 2 is the set of cases where player 2 wins on a diagonal -- 12*4!5! Category 1 contains 29*4!5! diagonal wins (one contains both diagonals), plus (6*9-9-6*2)4!5! non-diagonal wins. Category 3 contains 44*4!5! games. Category 4 contains 36*4!5! games, and of each 4!5! player 2 wins 2268, leaving 612 for player 1. Hmmmm, my total end-state count is 121, and it should be 126. There's a mistake in here somewhere. --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. ==== Subject: Re: Tic-tac-toe probability > A random game of tic-tac-toe (on a 3 x 3 grid) is played with each player making a random choice from the legal moves available at his turn. A player wins if he gets 3 in a line (row or diagonal). 1) What is the probability that the first player wins? > 2) That the second player wins? I've written two Mathematica programs to solve this problem and some of its generalizations. They may be found at http://rec-puzzles.org/solutions/contrib/tictactoe.html The first program breaks down the probabilities into single and double wins (i.e., winning with one or two n-in-a-rows), and rational and irrational games (rational games being those which neither player could win from any position occuring in the game) for the 3x3 and 4x4 cases. *** 3x3 case *** P(x wins) = 737/1260 (singly: 11/20, doubly: 11/315) P(o wins) = 121/420 P(draw) = 8/63 (rational game: 4/405) *** 4x4 case *** P(x wins) = 23768/75075 (singly: 23528/75075, doubly: 16/5005) P(o wins) = 60191/225225 (singly: 59471/225225, doubly: 16/5005) P(draw) = 206/495 (rational game: 79548797/432432000) The second program computes only the (exact) probabilities of the three basic outcomes (x wins, o wins, draw), but very efficiently. Computations for an nxn board are presented for n <= 30. The exact results quickly become unwieldly. E.g., for n = 8 we have P(x wins) = 668180850847705345608012278006/ 16092333328365244761241249093275 P(o wins) = 11244354325023279758612949641/ 272751412345173640021038120225 P(draw) = 17152850719229794/ 18700246336148883 As n grows large, the probability of x winning remains only very slightly higher than o winning for even n, with x having more of an advantage for odd n. However, the probability of x or o winning decreases by a factor of 2 (roughly) for every increment of n. E.g., n x wins o wins draw 26 0.00000048817765798 0.00000048817764572 0.99999902364469630 27 0.00000026303297352 0.00000024357574111 0.99999949339128538 28 0.00000013108006192 0.00000013108006110 0.99999973783987698 29 0.00000070260588751 0.00000065420785809 0.99999986431862544 30 0.00000035029006183 0.00000035029006128 0.99999992994198769 -Jim Ferry Metron, Inc. f rr @m tsc .c m e y e i o ==== Subject: Re: Tic-tac-toe probability > A random game of tic-tac-toe (on a 3 x 3 grid) is played with each player making a random choice from the legal moves available at > his turn. A player wins if he gets 3 in a line (row or diagonal). > 1) What is the probability that the first player wins? > 2) That the second player wins? This question was posed by F. E. Clark of Rutgers University as Problem > E1324 in the American Mathematical Monthly, June-July 1958, Vol. 65, > No. 6, p. 447. A solution by T. M. Little of the University of > California was published on pp. 144-145 of the February 1959 issue. > According to Little's calculation the first player wins with > probability 737/1260, the second player wins with probability 363/1260, > the probability of a draw is 160/1260; the probability of the game > ending on the 5th, 6th 7th, 8th, 9th turn is 120/1260, 111/1260, > 333/1260, 252/1260, 284/1260. Little's method is to classify the 126 > positions with 5 X's and 4 O's into five groups depending on how many > winning combinations of X's and O's they contain. Does that method account for the fact that the winner in some 5x,4o combinations will depend on which order the squares are filled? > ==== Subject: Re: Tic-tac-toe probability Le 29/06/05 3:21, dans 3254356.1120008109864.JavaMail.jakarta@nitrogen.mathforum.org, .82Dan Dimae a .8ecrit: > No ideas? Well... I see no trick, so I would try a direct computation of all cases: 9 squares, so 2^9 possibilities (actually, that's not exact since a player can win before the end). If you take symetries into account, you may be able to write all solutions and compute your proba of win. Otherwise, a computer can do that very quickly. I don't know if there are clever tricks for this problem. ==== Subject: Re: cranky challenge - beating a game >Le 29/06/05 1:58, dans >22134911.1120003163981.JavaMail.jakarta@nitrogen.mathforum.org, .82Gary >Z.e a .8ecrit: > two players : >> >> >> number A is shown to player 2, knowing nothing about how A & B were generated. >> now player 2 has to guess which one is larger, A or B? >> >> Question: can player 2 guess it correctly in the probability > 50%? how can >> he? If player 1 takes A & B at random, the probability of correct guess (and >optimal strategy) greatly depends on the distribution. If A & B are not >random, and are taken from a computed series of number, it's up to player 2 >to find a good strategy: with an optimal strategy, he wins each time (he has >found the series used by player 1). I think we have not enough information >to conclude. It could even happen that player 1 decides to always let A=1... If player 1 chooses 2 numbers and then assigns them randomly to A and B with equal probability, there is no strategy for player 2 for which p <> 50%. --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. ==== Subject: Re: cranky challenge - beating a game > two players : > > > number A is shown to player 2, knowing nothing about how A & B were generated. > now player 2 has to guess which one is larger, A or B? > > Question: can player 2 guess it correctly in the probability > 50%? how can he? > > have fun, > > Gary In the version I was told, player 2 was randomly shown one of A and B. In that case, the way to proceed is as follows: Player 2, before seeing a number, chooses some increasing function from [0,infty) to (0,1). After seeing a number, of value x, player 2 says this number is larger with probability f(x), and the other number is larger otherwise. If player 1 gets to choose which number player 2 is shown, I don't think this works. ==== Subject: Re: cranky challenge - beating a game > two players : > number A is shown to player 2, knowing nothing about how A & B were generated. now player 2 has to guess which one is larger, A or B? Question: can player 2 guess it correctly in the probability > 50%? how can he? have fun, Gary I don't see any way to guess if it were really true that nothing is known about how A and B are generated. However, suppose that A was a very very small positive number - so small that Player 1 could not have written it down in less than, say, 60 years. If B were an even smaller number, Player 1 would have to be over 120 years old to have written down both A and B. Since no one is that old, Player 2 concludes that A is smaller than B. A ridiculously extreme case to be sure. Less extreme cases give less information, but *never* zero information. The point being, you never know *nothing* about how the numbers were generated. An optimal use of even the slightest, most trivial fact should increase the chances of guessing right to a minuscule amount over 50%. Nora B. ==== Subject: Re: cranky challenge - beating a game Le 29/06/05 20:05, dans a .8ecrit: >> two players : >> >> >> number A is shown to player 2, knowing nothing about how A & B were >> generated. now player 2 has to guess which one is larger, A or B? >> >> Question: can player 2 guess it correctly in the probability > 50%? how can >> he? >> >> have fun, >> >> Gary > > I don't see any way to guess if it were really true that nothing is > known about how A and B are generated. > > However, suppose that A was a very very small positive number - so > small that Player 1 could not have written it down in less than, say, > 60 years. If B were an even smaller number, Player 1 would have to > be over 120 years old to have written down both A and B. Since no > one is that old, Player 2 concludes that A is smaller than B. > > A ridiculously extreme case to be sure. Less extreme cases give > less information, but *never* zero information. The point being, > you never know *nothing* about how the numbers were generated. An > optimal use of even the slightest, most trivial fact should increase > the chances of guessing right to a minuscule amount over 50%. Wrong. See example given by James Walby. It's impossible to beat (that kind of) chance. ==== Subject: Re: cranky challenge - beating a game > two players : > > > number A is shown to player 2, knowing nothing about how A & B were > generated. now player 2 has to guess which one is larger, A or B? > > Question: can player 2 guess it correctly in the probability > 50%? As others have noted, if P2 knows how P1 generated A and B, then for some methods of generation P2 can guess correctly with high probability. Here is a method where such knowledge doesn't aid P2: P1 chooses A=8, then chooses B=7 or B=9, each with p=1/2. (This requires one random bit, which can be based on a hardware random number generator such as a noisy diode, knowledge of which would not aid P2.) -jiw ==== Subject: Re: Help me master mathematics! > Mark is a FORMER mathematician. Now, he's a multimillionaire financier > and philanthropist. Ah, so you've gone down in the world. -- Wayne Brown (HPCC #1104) | When your tail's in a crack, you improvise fwbrown@bellsouth.net | if you're good enough. Otherwise you give | your pelt to the trapper. e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock ==== Subject: Re: Help me master mathematics! <4EFve.1032264$w62.31182@bgtnsc05-news.ops.worldnet.att.net> Huygens discovered wave packets. Some drone programmed wavelets into computers after computers were >invented, but are you mathematicians really so proud of this as your >major invention in the past 20 years? > Apart from the fact that an idea isn't patentable, somehow RSA got patented. -- Jeremy Boden ==== Subject: Re: Help me master mathematics! <4EFve.1032264$w62.31182@bgtnsc05-news.ops.worldnet.att.net> <9hblyJCCOpwCFwnJ@jboden.demon.co.uk> Patenting is irrelevant and doesn't mean an idea is truly useful. A pizza company patented a way to insert cheese inside the crust of pizza for chrissakes. And there are millions of other worthless patents. Mark knows, since he's invested in patents. EquityValue Investments ==== Subject: Re: Help me master mathematics! <4EFve.1032264$w62.31182@bgtnsc05-news.ops.worldnet.att.net> Oh yeah? Name one useful pure math discovery in the last 20 years that is of use to taxpayers? Mark EquityValue Investments ==== Subject: Re: Help me master mathematics! > Oh yeah? Name one useful pure math discovery in the last 20 years that > is of use to taxpayers? Who cares? Mathematics -- like other intellectual pursuits -- needs no justification. -- Wayne Brown (HPCC #1104) | When your tail's in a crack, you improvise fwbrown@bellsouth.net | if you're good enough. Otherwise you give | your pelt to the trapper. e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock ==== Subject: Re: Help me master mathematics! <4EFve.1032264$w62.31182@bgtnsc05-news.ops.worldnet.att.net> <87r7emlubh.fsf@phiwumbda.org> Discussion, linux) > Mark is a FORMER mathematician. Now, he's a multimillionaire financier > and philanthropist[...] who doesn't exist. -- Jesse F. Hughes How come there's still apes running around loose and there are humans? Why did some of them decide to evolve and some did not? Did they choose to stay as a monkey or what? -Kans. Board of Ed member ==== Subject: Re: intermediate value theorem for the lattices? > I wonder if I'm using the correct term here. (Google indicates pretty > much that I'm not). Given a <= c > b <= c > a <= d > b <= d Prove that there exists x a <= x > b <= x > x <= c > x <= d The part inter is there: Riesz Interpolation Property. Lattices have it, but a partially ordered set need not be a lattice to have this property. Take real polynomial functions on the real line, with the ordering p <= q means either p=q, or p(x) < q(x) for all real x. (Caution: it's not the most often used ordering.) Harder exercise: With polynomial functions as above, but with customary ordering, will Riesz Interpolation Property still hold? ==== Subject: Re: Is the set N of natural numbers well defined? Re. Virgil's comments.On the notation issue, I had, at least in my mind, correctly differentiated between the infinite union of intervals and the finite union;perhaps the notation was clumsy. On the substantive issue, the point that N belongs to S was not adequately demonstrated is well taken.But it is not, perhaps, fatal to the main direction of the argument, as the following would appear to show. We have, as before,S=(I1,I2,I3, . . .} Let U1=I1, U2=I1UI2,U3=I1UI2UI3, and so on, and US=union of all the Intervals. Let S*=(U1,U2,U3, . . .} Then, there is a 1-1 map F between the two sets such that F(U1)=I1,F(U2)=I2,F(U3)=I3 and so on. Now, the members of S* are formed by taking unions of successive members of S, and the process can continue till All the members are exhausted (shades of the diagonal argument!).Then US is a member of the set S*. Now, what is the image of US under F? It cannot be anything other than N; F(US)=N. Then, N belongs to S. As I stated earlier, all diagonal arguments assume that N belongs to S. They would appear to be vulnerable on this count. ==== Subject: Re: Is the set N of natural numbers well defined? Do you think - is there any formal concept how to say, which number is FINITE? Yes, Natural numbers are finite, but, as I said before, 0,f(0),f(f(0)),... could be much more longer then any kardinal...! Yes, we all now, that 12 is finite, but what about 11111.....1? Where the finite number ends? All formal mathematic begins with - you get FINITE formulas - but which WERE long formula is still finite and which is infinite? And that is the point - the CIRCLE - if you wanna to know what Finete number are, you must know it BEFORE - you have to know what finite formalas are! And if you know it in another way then others do, if your finite numbers are infinite in other systems - what can you do? Nothing, because there is no way how to express what finite is - in todays math! So naturals are not well defined - only small - REALLY small numbers - like 10 or 10^10 are. ==== Subject: Re: Is the set N of natural numbers well defined? > Do you think - is there any formal concept how to > say, which number is FINITE? Yes, an equivalence class defined by bijectivity among all finite sets ( these one defined as impossibility of finding bijectivity with all proper subsets ). Here, defining N in this way we obtain the formal concept of finite cardinal. >Yes, Natural numbers are > finite, but, as I said before, 0,f(0),f(f(0)),... > could be much more longer then any kardinal...! It could not it you associate for example f(f(0)) with the equivalence class consisting of all sets with three elements. > Yes, we all now, that 12 is finite, but what about > 11111.....1? Where the finite number ends? Never. Is there any problem with this ? > All formal mathematic begins with - you get FINITE > formulas - but which WERE long formula is still > finite and which is infinite? And that is the point - > the CIRCLE - if you wanna to know what Finete number > are, you must know it BEFORE - you have to know what > finite formalas are! I think this is only as passing from mathematical physiology to mathematical anatomy. I mean all depens on the starting point. By the way, the circle provides a good interpretation of infinite in the sense of real numbers but not in the naturals one ( or intuitively speaking, in terms of continuum and not in the discrete ) >And if you know it in another > way then others do, if your finite numbers are > infinite in other systems - what can you do? Nothing, > because there is no way how to express what finite is > - in todays math! So naturals are not well defined - > only small - REALLY small numbers - like 10 or 10^10 > are. Well, I think nobody believe such a thing. Fernando. ==== Subject: Re: Is the set N of natural numbers well defined? > It could not it you associate for example f(f(0)) > ) with the equivalence class consisting of all sets > with three elements. > > Yes, we all now, that 12 is finite, but what about > 11111.....1? Where the finite number ends? > > Never. Is there any problem with this ? Yes, there is. Take any nonstandard model of N*. In this model all sequences of natural numbers (from N) are natural - and this means also finite - numbers from (N*) - for example sequence n0 = 1,2,3,4,5,6,7,...,n,n+1,.... is finite number from N* - this number is finite in the model of N* - and there is no way of finding bijectivity with all proper subsets (in this model) of this number - but this number is greater than all numbers from N! Or the same example in an other point of view - take countable sequence 1,1,1,1,1,1,1,.... this means the function F(n) = 1 for all n e N and make the sequence F(0),F(1),F(2),.... for all n e N. Then there is natural number in N* which is F(0)+F(1)+F(2)+.... := F(0),F(0)+F(1),F(0)+F(1)+F(2),.... = n0. Is n0 infinite? Why? Or some says there is no simple mapping from reals of <0,1> to natural numbers. Yes, there is: Take any real number (normal R) - for example (PI-3) and make his 10-adic sequence: (PI-3) = 0,1415926.... Then Natural image of (PI-3) is (PI-3):= 1+40+100+5000+9000+20000+.... and this is natural number in (every!!) N*. Why this number is not finite? So infinite for one could be finite for an other and vice versa. MS ==== Subject: Re: Is the set N of natural numbers well defined? > It could not it you associate for example f(f(0)) > ) with the equivalence class consisting of all > sets > with three elements. > > Yes, we all now, that 12 is finite, but what > about > 11111.....1? Where the finite number ends? > > Never. Is there any problem with this ? > > Yes, there is. Take any nonstandard model of N*. In > this model all sequences of natural numbers (from N) > are natural - and this means also finite - numbers > from (N*) - for example > > sequence n0 = 1,2,3,4,5,6,7,...,n,n+1,.... is finite > number from N* - this number is finite in the model > of N* - and there is no way of finding bijectivity > with all proper subsets (in this model) of this > number - but this number is greater than all numbers > from N! Or the same example in an other point of view > - take countable sequence 1,1,1,1,1,1,1,.... this > means the function F(n) = 1 for all n e N and make > the sequence F(0),F(1),F(2),.... for all n e N. Then > there is natural number in N* which is > F(0)+F(1)+F(2)+.... := > F(0),F(0)+F(1),F(0)+F(1)+F(2),.... = n0. Is n0 > infinite? Why? > > Or some says there is no simple mapping from reals of > <0,1> to natural numbers. Yes, there is: > > Take any real number (normal R) - for example (PI-3) > and make his 10-adic sequence: > > (PI-3) = 0,1415926.... > > Then Natural image of (PI-3) is (PI-3):= > 1+40+100+5000+9000+20000+.... and this is natural > number in (every!!) N*. Why this number is not > finite? > > So infinite for one could be finite for an other and > vice versa. > > MS At the present stage of the discussion I think we may settle what we are talking about, otherwise it seems to be as two monologues. A good starting point would be to define what N is, in the sense that being a primitive concept would be determined by a set of axioms. Of course I have all the rigths to think that those hipothetycal axioms do not contain all the substance but at least I'd know what your naturals are. Les us see if we are talking about the same things. In any case I should repeat a sentence in the book of Douglas R. Hofstadter ( Godel, Escher, Bach ) : when we are concerned about foundations of Logic, we usually change brain by heart. Fernando. ==== Subject: Re: Is the set N of natural numbers well defined? When ICm talking about Natural numbers - I used following definition: Finite natural number is every GROUP of ONEs, which doesnCt change in time. This could be definition. Axiom N1 is: For every Set, which doesnCt change!, there exist Natural number with cardinality bigger to this set. There is no Finite Set of all natural number - because there than by Axiom N1 there is bigger number than cardinality of this set. The Class of Finite natural number is still change and is strictly unFinite! Yes, there must be more specific definition of GROUP of ONEs, but in general it is clear what it says - Everything, what doesnCt change - is countable! There is no thing (not even ordinal numbers, and not any Set from standard theory) - which is uncountable! Yes, there could be set, which is called N - and that is the smallest set of all inductive sets (standard axiom of infinity from ZF teory), but then there exist also Finite natural number - called it N too - which one can take as Sum of Ones over the set N. So when I was talking about sum (PI-3) = 1+40+100+.... this is for me finite number - because PI is not changing. There is no time in math of today - and that means, than everything is finite. ==== Subject: Re: Solving Simultaneous Equations X-RFC2646: Format=Flowed; Original Le 29/06/05 0:47, dans vskwe.75567$Vj3.50239@fe2.news.blueyonder.co.uk, > .82 Michael Hims e a .8ecrit : > I have the equations: >> 8a +4b +2c+d=1 >> 27a +9b +3c+d=4 >> 64a +16b+4c+d=10 >> 125a+25b+5c+d=20 >> How do I solve them all? > > 1st solution: Gauss method 2nd solution: 1 4 10 20... is binomial(n+1,3) for n=2,3,4,5. > The solutions (a,b,c,d) are the coefficients of an interpolation > polynomial > of degree 4 at abscissas 2,3,4,5. Here the polynomial is binomial(X+1,3), > or > (X+1)*X*(X-1)/6 = X^3/6 - X/6, hence > d=b=0 > a=-c=1/6 > However I'm only 15 years old and that explanation is confusing. Could you please explain that a little clearer please? ==== Subject: Re: Solving Simultaneous Equations >> >> >> I have the equations: >> 8a +4b +2c+d=1 >> 27a +9b +3c+d=4 >> 64a +16b+4c+d=10 >> 125a+25b+5c+d=20 >> >> How do I solve them all? One method is Gaussian Elimination: >http://mathworld.wolfram.com/GaussianElimination.html. That's a good method in general, but for these equations there's a short cut. Take the difference between each consecutive pair of equations; that'll give you three equations not involving d and with constant coefficient of c. Take the differences of these three equations, to get two equations involving a and b only, with constant coefficient of b. Take the diffenence of these two to get an equation for a only. Now substitute back to determine a, b, c, d. Mike Guy