The End of an Era - That of The Natural Numbers
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From: A on 2 May 2010 17:44 On May 2, 4:21 pm, "Tim Golden BandTech.com" <tttppp...(a)yahoo.com> wrote: > On May 1, 11:49 am, A <anonymous.rubbert...(a)yahoo.com> wrote: > > > You know, if m divides n, you have a quotient map from Z/nZ to Z/mZ, > > which is a ring morphism. You can take Z/nZ for all n, together with > > all those quotient maps, and "let n go to infinity" by taking the > > inverse limit in the category of commutative rings. But you don't get > > the integers when you do this; instead, you get the profinite > > integers, the Cartesian product of one copies of the p-adic integers > > for every prime number p. (In some ways the profinite integers are > > extremely natural and important themselves; e.g. the absolute Galois > > group of any finite field k is isomorphic to the profinite integers. > > But the profinite integers are still not the integers themselves.) > > > You can try to go the other way, taking a direct limit, instead of an > > inverse limit. Here's how that works out: if m divides n, you have an > > inclusion from Z/mZ to Z/nZ (the "multiplication by n/m" map), which > > is a morphism of abelian groups, but not a ring morphism, since (for > > example) it doesn't send the unit element 1 to the unit element 1. But > > okay, you can still take Z/nZ for every n, equipped with these > > inclusion maps, and "let n go to infinity" by taking the direct limit > > in the category of abelian groups (you can't do it in rings, since > > these maps aren't ring morphisms). But again, you don't get the > > integers. When you take the direct limit of Z/(p^n)Z for all n, where > > p is a prime number, you get the abelian group Z with p inverted; if > > you're not familiar with that construction, you can read about it in > > any book that talks about localization of rings and modules, e.g. > > Atiyah and Macdonald's excellent, short, and very readable > > "Introduction to Commutative Algebra." So when you try to take a > > direct limit of Z/nZ for all n, "letting n go to infinity" in that > > way, you get...the integers, but with every prime number p > > inverted...in other words, the rational numbers. > > > So I think the notion of "modulo infinity" is not as straightforward > > as it seems! > > There is no need to use all of this quotient language. It is not used > in real analysis and I do not see it as a clean construction. My own Quotients are certainly used in real analysis--for example, quotient vector spaces. The real numbers themselves are a quotient ring: the ring of Cauchy sequences in the rationals, modulo equivalence. > impedance at this level prevents me from following your argument, but > certainly I accept your conclusion. The idea of a wrapping count which > never wraps would seem inherently conflicted, but because its lower > forms are simple and clean and have no need of pushing for higher > numbers these low forms do not deserve to be treated as constructions > of an infinite bearing number system. I am saying simply that it makes > no sense to define modulo numbers in terms of the natural numbers. It > is instead cleaner to do the reverse. Structured information will > build more complicated things from simpler things, both formally and > informally. Informally, every natural number that you use beyond nine > is already making use of modulo principles. > > > > > > > This is a strange thing to say. The ring of polynomials in one > > variable with coefficients in any ring R is itself a ring, and this > > isn't hard to prove; most undergraduate mathematics majors prove it as > > an exercise in an algebra course. It's elementary and it's used every > > day, in mathematical practice. > > Consider the care with which the product was only just treated in the > ring definition. > This new product of marrying values of different types is not > congruent and so to claim that this branch of math is coherent one > would be forced to discuss this second product form to the same level > of formality as the first, rather than just slide it under the table. > Is a polynomial coefficient a product? If so, then you tell me where > it is handled formally. It certainly is not in the ring definition. > The X in these polynomial products is not declared to belong to any > set and so cannot fit the ring definition. My own conclusion on this > detail is that when this coefficient business is handled formally that > the product of a real value with a complex value will remain as such > and not resolve to a complex value, so that a value > a in the Reals > in product with > z in the Complex > will yield an unresolvable > a z > and this then is some tidbit of new mathematics, likely more useful to > physicists than to mathematicians. In the higher dimension associative > algebras one sees larger spaces like > C x C > and it would be foolhardy to claim that one can assimilate a real or > complex coefficient into this 4D space, for which of these two sets > will you interperet them as? In general dimension the procedure does > not hold up. > > > > > I seem to recall there being a thread where many people weren't (after > > much time and effort was spent) able to convince you that polynomial > > rings are a valid construction; and I doubt I could succeed where they > > failed, so I'd better not try. But really, polynomial rings do make > > sense. > > No, they do not. The value X, or rather, the X, is completely > undefined. It is familiarity with real valued X that allows acceptance > of the undefined X. This usage is an important instance that goes > unadressed in the mathematics texts that I have studied, which I admit > is limited. The actual ability to take the product of a real value and > this ill-defined X is nonexistent. What then is this notation? Shall > you accept from me the construction of > 2.34 Y > where Y has no set that it is declared to belong to? No, this math has > stretched into a terrain that deserves more scrutiny. You would not > accept the line above from me, and the only reason that you accept it > from them is sheer mimicry. > You have a lot of misunderstandings about what a polynomial ring is, what algebras are, how the real numbers can be embedded in the complex numbers, and so on. I could spend the next 30 minutes typing out an explanation of each of these ideas and addressing each of your objections; but, as I recall, others have tried to explain such things to you before, and you didn't respond well to this. I am going to spend the next 30 minutes doing other things instead. I suppose you'll take this as my weak-minded subservience to some conspiracy of mathematicians, or something; but I imagine that even if I explained these ideas to you, you would still make the same accusation at me, and neither of us would be any better off. > This devolution is far away from the original intent, and you have > provided no falsification of my argument. Instead you dodge; and the > long way around. You've not even addressed multiplication as addition > here. The level of dodge is absurd. Your own credibility is exposed. I > take that vacuous space as loose proof. I'm not sure what you think I dodged, or what the "original intent" was (or what "loose proof" is). What you said a few posts ago about multiplication as addition is fine, as I said last time I posted here.
From: Nam Nguyen on 2 May 2010 17:46 Alan Smaill wrote: > Nam Nguyen <namducnguyen(a)shaw.ca> writes: > >> Nam Nguyen wrote: >>> Alan Smaill wrote: >>>> People differe on their intuitions; >>>> surely you admit that? And diverging intuitions can both be reasonable. >>> Of course that's reasonable - _in general_ . But appropriate contexts >>> will cap the limit how much divergent opinions could be. For instance, >>> if you have the intuition that ~(x=x) then that's fine and you'd have >>> to make logical/mathematical deductions in a FOL different from FOL=. >>> Or if you'd would prefer the intuitions of Genzen over Hilbert's on >>> rules of inferences that's fine too. >>> >>> But there should be _no blank-check on intuitions_, if we're talking >>> about mathematics and the reasoning about it. >> For example, given the context of FOL= and of the kind of intuition >> of the naturals that PA's non-induction-schema axioms be true, > > well, of course you can find many accounts of the intuitive > acceptability of the induction principle. See Poincar�, > for example. > >> then >> one has to _appropriately in those contexts_ explain (or plan to >> explain) why, e.g., cGC is true/false, and not just give any-what-so >> -ever kind of explanation, as WH did in his recent post. > > There is absolutely no reason to require any such explanation. > I mean, no-one is claiming that Peano arithmetic, or > heyting arithemtic, provides the answer to all the > questions about arithmetic, are they? > > In the absence of such an explanation, do you want to rule this > intuition inadmissable, then? It might have been my fault in not making it clear to others but ruling in or out admissibility of such purported truth or falsehood is not my genuine intention. I did express my intention before in some other threads. What I'm really ... really .. after is _formalizing_ what is impossible to know or infer. Would that be too much to ask for, when we're already receptive to the idea we can't answer all the "questions about arithmetic". I mean, even though we can't answer all the mathematical questions about arithmetic, wouldn't it be in our own reasoning interest to _formally_ categorize them in to which one we'd know and which one we'd NOT know? My effort here is an attempt to formalize the categorization, classification. And I think there would be benefits to such effort, such as being able to describe phenomena of SR, AI, etc... within some revised FOL framework.
From: Tim Golden BandTech.com on 2 May 2010 22:16 On May 1, 11:33 am, A <anonymous.rubbert...(a)yahoo.com> wrote: > On Apr 30, 12:56 pm, "Tim Golden BandTech.com" <tttppp...(a)yahoo.com> > wrote: > > > > > On Apr 30, 11:37 am, A <anonymous.rubbert...(a)yahoo.com> wrote: > > > This looks to me like you weren't able to write down the > > > multiplication in the integers modulo 3 without referring to the > > > multiplication on the natural numbers, i.e., the Cartesian product > > > (what you are calling "superposition") of finite sets. When you write > > > 11 for 2, for example, you are thinking of 2 as the cardinality of any > > > set in bijection with the set of symbols 1 in the expression 11. Then > > > 2 * 3 is the cardinality of any set in bijection with the set of > > > symbols in the expression 111 111, i.e., the cardinality of any set in > > > bijection with the set of symbols 1 in the Cartesian product of the > > > sets 11 and 111. > > > No number larger than > > 111 > > in the superposition format was used. > > You give the example of > > 2 * 3 > > which in the superpositional form is the product > > (11)(111) > > which is the sum > > 111 111 > > or in the ordinary representation > > 3 + 3 > > We could also have written > > 2 + 2 + 2 > > or > > 11 11 11 > > and so we are free to define multiplication in terms of summation. > > There is no need to introduce any cartesian product within the modulo > > numbers. I think you should be able to find a direct contradiction > > within my logic if I am wrong. But you yourself opened thinking it is > > possible. > > No, you're right about this--I think this does work, as a way of > defining the multiplication on the integers modulo n. So I do believe > you now, that you can define the integers modulo n, as a ring, without > invoking the natural numbers, and without a really undue amount of > effort. OK. Thanks. My apologies. I did not catch this post. - Tim > > > > > > The multiplication on the natural numbers comes directly from the > > > Cartesian product of finite sets; likewise, the addition on the > > > natural numbers comes directly from the disjoint union of finite sets. > > > This is because, in a fundamental way, the natural numbers arise from > > > COUNTING finite collections of things. > > > > I think you will have a difficult time writing down the multiplication > > > on the integers modulo n without reference to the multiplication on > > > the natural numbers. The integers modulo n are certainly important > > > but, keeping in mind their addition and multiplication, they do not > > > arise in such a fundamental way as the natural numbers. > > > No, this is not true at all. We can do out an entire table if you > > like. Mod-3 there are only a few possibilities > > (1)(1), (1)(2), (1)(0), > > (2)(2), (2)(0), > > (0)(0) . > > Again the triangle. Each of these expressions can be regarded as a > > sum. If you find no conflict with the sum mod-3 that I described then > > I see no way for you to dispute these products as sums. I have no need > > to rely upon the cartesian product as you present. Neither does ring > > theory make this requirement. It states simply that elements in a set > > yield results in that set under the operations of product and sum, or > > whatever operators you like. It is only one set that is required. > > > It is instead true that you need the modulo number in order to express > > your products and sums of large natural numbers. For instance, if I > > ask you for the product of > > (21)(11) > > in modulo-ten notation then we have already made use of the modulo > > concept and it cannot follow that the modulo concept be built out of > > these natural numbers. > > > - Tim
From: Tim Golden BandTech.com on 2 May 2010 22:37 On May 2, 5:44 pm, A <anonymous.rubbert...(a)yahoo.com> wrote: > On May 2, 4:21 pm, "Tim Golden BandTech.com" <tttppp...(a)yahoo.com> > wrote: > > > > > On May 1, 11:49 am, A <anonymous.rubbert...(a)yahoo.com> wrote: > > > > You know, if m divides n, you have a quotient map from Z/nZ to Z/mZ, > > > which is a ring morphism. You can take Z/nZ for all n, together with > > > all those quotient maps, and "let n go to infinity" by taking the > > > inverse limit in the category of commutative rings. But you don't get > > > the integers when you do this; instead, you get the profinite > > > integers, the Cartesian product of one copies of the p-adic integers > > > for every prime number p. (In some ways the profinite integers are > > > extremely natural and important themselves; e.g. the absolute Galois > > > group of any finite field k is isomorphic to the profinite integers. > > > But the profinite integers are still not the integers themselves.) > > > > You can try to go the other way, taking a direct limit, instead of an > > > inverse limit. Here's how that works out: if m divides n, you have an > > > inclusion from Z/mZ to Z/nZ (the "multiplication by n/m" map), which > > > is a morphism of abelian groups, but not a ring morphism, since (for > > > example) it doesn't send the unit element 1 to the unit element 1. But > > > okay, you can still take Z/nZ for every n, equipped with these > > > inclusion maps, and "let n go to infinity" by taking the direct limit > > > in the category of abelian groups (you can't do it in rings, since > > > these maps aren't ring morphisms). But again, you don't get the > > > integers. When you take the direct limit of Z/(p^n)Z for all n, where > > > p is a prime number, you get the abelian group Z with p inverted; if > > > you're not familiar with that construction, you can read about it in > > > any book that talks about localization of rings and modules, e.g. > > > Atiyah and Macdonald's excellent, short, and very readable > > > "Introduction to Commutative Algebra." So when you try to take a > > > direct limit of Z/nZ for all n, "letting n go to infinity" in that > > > way, you get...the integers, but with every prime number p > > > inverted...in other words, the rational numbers. > > > > So I think the notion of "modulo infinity" is not as straightforward > > > as it seems! > > > There is no need to use all of this quotient language. It is not used > > in real analysis and I do not see it as a clean construction. My own > > Quotients are certainly used in real analysis--for example, quotient > vector spaces. The real numbers themselves are a quotient ring: the > ring of Cauchy sequences in the rationals, modulo > equivalence. > > > > > impedance at this level prevents me from following your argument, but > > certainly I accept your conclusion. The idea of a wrapping count which > > never wraps would seem inherently conflicted, but because its lower > > forms are simple and clean and have no need of pushing for higher > > numbers these low forms do not deserve to be treated as constructions > > of an infinite bearing number system. I am saying simply that it makes > > no sense to define modulo numbers in terms of the natural numbers. It > > is instead cleaner to do the reverse. Structured information will > > build more complicated things from simpler things, both formally and > > informally. Informally, every natural number that you use beyond nine > > is already making use of modulo principles. > > > > This is a strange thing to say. The ring of polynomials in one > > > variable with coefficients in any ring R is itself a ring, and this > > > isn't hard to prove; most undergraduate mathematics majors prove it as > > > an exercise in an algebra course. It's elementary and it's used every > > > day, in mathematical practice. > > > Consider the care with which the product was only just treated in the > > ring definition. > > This new product of marrying values of different types is not > > congruent and so to claim that this branch of math is coherent one > > would be forced to discuss this second product form to the same level > > of formality as the first, rather than just slide it under the table. > > Is a polynomial coefficient a product? If so, then you tell me where > > it is handled formally. It certainly is not in the ring definition. > > The X in these polynomial products is not declared to belong to any > > set and so cannot fit the ring definition. My own conclusion on this > > detail is that when this coefficient business is handled formally that > > the product of a real value with a complex value will remain as such > > and not resolve to a complex value, so that a value > > a in the Reals > > in product with > > z in the Complex > > will yield an unresolvable > > a z > > and this then is some tidbit of new mathematics, likely more useful to > > physicists than to mathematicians. In the higher dimension associative > > algebras one sees larger spaces like > > C x C > > and it would be foolhardy to claim that one can assimilate a real or > > complex coefficient into this 4D space, for which of these two sets > > will you interperet them as? In general dimension the procedure does > > not hold up. > > > > I seem to recall there being a thread where many people weren't (after > > > much time and effort was spent) able to convince you that polynomial > > > rings are a valid construction; and I doubt I could succeed where they > > > failed, so I'd better not try. But really, polynomial rings do make > > > sense. > > > No, they do not. The value X, or rather, the X, is completely > > undefined. It is familiarity with real valued X that allows acceptance > > of the undefined X. This usage is an important instance that goes > > unadressed in the mathematics texts that I have studied, which I admit > > is limited. The actual ability to take the product of a real value and > > this ill-defined X is nonexistent. What then is this notation? Shall > > you accept from me the construction of > > 2.34 Y > > where Y has no set that it is declared to belong to? No, this math has > > stretched into a terrain that deserves more scrutiny. You would not > > accept the line above from me, and the only reason that you accept it > > from them is sheer mimicry. > > You have a lot of misunderstandings about what a polynomial ring is, > what algebras are, how the real numbers can be embedded in the complex > numbers, and so on. I could spend the next 30 minutes typing out an > explanation of each of these ideas and addressing each of your > objections; but, as I recall, others have tried to explain such things > to you before, and you didn't respond well to this. I am going to > spend the next 30 minutes doing other things instead. I suppose you'll > take this as my weak-minded subservience to some conspiracy of > mathematicians, or something; but I imagine that even if I explained > these ideas to you, you would still make the same accusation at me, > and neither of us would be any better off. > > > This devolution is far away from the original intent, and you have > > provided no falsification of my argument. Instead you dodge; and the > > long way around. You've not even addressed multiplication as addition > > here. The level of dodge is absurd. Your own credibility is exposed. I > > take that vacuous space as loose proof. > > I'm not sure what you think I dodged, or what the "original intent" > was (or what "loose proof" is). What you said a few posts ago about > multiplication as addition is fine, as I said last time I posted here. I see. I missed that post. My apologies. Still, my disgust with the lack of sense of abstract algebra(AA) beyond the ring definition (entering the polynomial and quotient ring) stands. That the subject also concerns itself with these modulo behaviors is an especially strong signal when coupled with the fact that sign has been overlooked as a modulo-two behavior. I appreciate your lack of desire to engage in a discussion on it. Still, anytime I see usage of the quotient ring construction to make an argument I will state my position. The arithmetic product, only just given formal treatment in the ring definition, is immediately abused in AA. I took a real analysis course that was before abstract algebra. The traditional constructions of the various number systems leading up to the reals was there. Are you claiming that this rendition was incorrect? I suppose the explanation is that so many things are expressible in terms of the quotient ring that it becomes a standard. Still, in terms of fundamental understanding, this is the wrong course, even if it is not falsifiable. Anyway A, I am sorry to have missed your other post and would take back some of my harsh language. I am so used to the usenet dodge that I made a bad assumption. - Tim
From: William Hughes on 3 May 2010 08:12
On May 2, 6:46 pm, Nam Nguyen <namducngu...(a)shaw.ca> wrote: > What I'm > really ... really .. after is _formalizing_ what is impossible to > know or infer. Would that be too much to ask for, My guess is yes, but what do I know? Certainly a good start would be to find a class of unknowable statements. A good start is not proclaiming the end of natural numbers and posting an "observation" that twists the phrase "any intuition" past the breaking point. It is interesting to note that while you have been presented with many putative "intuitions" given which the truth or falsehood of (1) is knowable (you have accepted none and explicitly rejected one), you have not presented a single "intuition" under which the truth or falsehood of (1) is not knowable. Given that the truth or falsehood of (1) is knowable, iff the truth or falsehood of cGC is knowable, I think that finding an interesting example will be hard. <snip> > I mean, even though we can't answer all the mathematical questions > about arithmetic, wouldn't it be in our own reasoning interest to > _formally_ categorize them in to which one we'd know and which one we'd > NOT know? It would also be in our own reasoning interest to divide the questions into those which have a proof that can be written down in a century, and those which do not. I don't see any hope here either. Given an unsolved mathematical question Q, there are three possibilities -there is a solution to Q that is reasonably short (reasonably short may be many volumes) -there is a solution to Q but the shortest solution is too long to write down -there is no solution to Q Except for contrived examples, I cannot see how to distinguish among the three. - William Hughes |