The End of an Era - That of The Natural Numbers
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From: William Hughes on 1 May 2010 03:11 On May 1, 3:01 am, Nam Nguyen <namducngu...(a)shaw.ca> wrote: > William Hughes wrote: <snip> > > > Whether or not (1) is decidable depends crucially on the > > status of GC. > > So, if GC is false, would that "crucially" tell us if (1) decidable? No. (1) is decidable, iff, cGC is decidable (pGC is easily seen to be true). Since this is independent of the decidability of GC, the use of "crucially" is inappropriate (cGC might be termed the "Generalized Goldbach Conjecture"). However, I cannot see why you should think the fact that you have "any intuition about the naturals" means that cGC is undecidable. - William Hughes
From: Nam Nguyen on 1 May 2010 11:11 William Hughes wrote: > On May 1, 3:01 am, Nam Nguyen <namducngu...(a)shaw.ca> wrote: >> William Hughes wrote: > > <snip> > >>> Whether or not (1) is decidable depends crucially on the >>> status of GC. >> So, if GC is false, would that "crucially" tell us if (1) decidable? > > No. (1) is decidable, iff, cGC is decidable > (pGC is easily seen to be true). Since this is > independent of the decidability of GC, the use of > "crucially" is inappropriate (cGC might be > termed the "Generalized Goldbach Conjecture"). You don't seem to know what you're talking about. First, _you_ are the one who introduced "crucially" (in the sentence above), why then do you now say "the use of 'crucially' is inappropriate"? Secondly, you've not defended what you yourself stated (again above which I just repeat here): William Hughes wrote: >>> Whether or not (1) is decidable depends crucially on the >>> status of GC. My question to you is suppose "status of GC" is false, why would (1) "crucially" depend on GC's being false? > However, I cannot see why you should think the > fact that you have "any intuition about the > naturals" means that cGC is undecidable. My memory might be bad, but where did I say that?
From: William Hughes on 1 May 2010 11:27 On May 1, 12:11 pm, Nam Nguyen <namducngu...(a)shaw.ca> wrote: <snip> > You don't seem to know what you're talking about. First, _you_ > are the one who introduced "crucially" (in the sentence above), > why then do you now say "the use of 'crucially' is > inappropriate"? I am pointing out an error I made. <snip> > > > However, I cannot see why you should think the > > fact that you have "any intuition about the > > naturals" means that cGC is undecidable. > > My memory might be bad, but where did I say that? You said "if we have any intuition about the naturals then we'd also have the intuition that we can't know the arithmetic truth or falsehood of (1)" Now apply the fact that (1) is decidable iff cGC is decidable
From: A on 1 May 2010 11:33 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. > > > 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: A on 1 May 2010 11:49
On May 1, 7:04 am, "Tim Golden BandTech.com" <tttppp...(a)yahoo.com> wrote: > The original point was that we can regard the natural numbers as the > modulo-infinity numbers, and this yields primitives prior to the > naturals. I say again, we do not need the number four in order to 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! > count to three. It so happens that these modulo forms are meaningful. > Though you are used to relying upon the polynomial to gain their > behavior, these discrete constructions can provide similar results > from a primitive level. > Beyond that abstract algebra is falsifiable. The very phrase > 'polynomials with real coefficients' > is incompatible with the ring definition. That meaningless X that only > carries a sequential positional value... that is toxic. You already > had what you needed back here in the modulo number, and it is sign > where it makes its presence felt. The standard +/- sign of the real 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. 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. > number is modulo-two behaved. Will you require another image of the > natural numbers be deployed in order to provide sign mechanics? I > don't think so. > > Really, this is so simple I find it difficult to believe that you > would insist on inflating a compact (informal) set of values out to an > infinite length set. You are going to insist on infinity now when it > is not at all necessary. Shame on you. You are fine instances of what > mathematics has become. Furthermore, there is no direct falsification > of my statement. There is no conflict in looking at discrete > multiplication as addition. I used to think you were fair TP, but now > when I wipe I'll be remembering your acronym. > > - Tim |