JSH: Depressing reality, prime reality
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From: Transfer Principle on 13 Mar 2010 03:02 On Mar 10, 4:48 am, "Ostap S. B. M. Bender Jr." <ostap_bender_1...(a)hotmail.com> wrote: > On Mar 9, 9:08 pm, Transfer Principle <lwal...(a)lausd.net> wrote: > > which is a point in favor of the standard theorists. > I am not sure what you mean by "standard theorists" in this context. > Are "standard theorists" - people who are familiar with Dirichlet > Theorem and Chebotarev's Theorem? > Then who are "non-standard theorists" in this context? People, who > are NOT familiar with Dirichlet Theorem and Chebotarev's Theorem? An interesting interpretion of my term "standard theorists." > Then isn't your term "non-standard theorists" just a synonym to > "ignoramuses"? I prefer to think of "non-standard theorists" as those who don't automatically accept ZFC or PA as the dominant theory. I prefer to think that one can accept alternate theories and new ideas without being "ignoramuses." > Then why do you say that it is one point " in favor of the standard > theorists"? Wouldn't competent people have MORE THAN ONE point over > ignoramuses? > In fact, why are we still discussing this open-and-shut case and > keeping score? Clearly, as always, the score is: > Knowledge: infinity > Ignorance: 0 An interesting interpretation. Then the only way to "score" would be to learn more knowledge. And I definitely did. I learned more about Dirichlet and Chebotarev and what they say about primes' distribution. > > On the other hand, suppose Wiles had proved that FLT is actually > > _undecidable_ in ZFC. > Sorry, but that's an impossibility. As long as ZFC is not self- > contradictory, hypotheses like FLT can NOT be proven undecidable. > Don't you see that? The two posters with the last name Hughes (Jesse and William) can explain this much better than I can. Refer to their posts earlier in this thread (at least concerning Goldbach's conjecture -- FLT is of course a moot point since Wiles did prove FLT).
From: master1729 on 21 Mar 2010 10:49 ostap bender 1900 wrote : > > Most so-called "mathematicians" have not discovered a > single new axiom > in their entire miserable lives. So, you are ahead of > everybody. > > But don't stop at one. Go for at least 157 new > axioms. > are you advising james to become a set theorist ?
From: master1729 on 21 Mar 2010 11:19 lwalke wrote : > On Mar 8, 5:12 am, "Ostap S. B. M. Bender Jr." > <ostap_bender_1...(a)hotmail.com> wrote: > > On Mar 7, 7:50 am, JSH <jst...(a)gmail.com> wrote: > > > Reality of course is that I first wondered if > > > primes might have no preference with each other > > Wait. Didn't you earlier discover that primes > prefer composites? > > I usually don't bother to post in JSH threads, but > there's > something I'd like to point out here: > > > you derive that some primes "might have preference > with each other". > > In another JSH thread last month, a standard theorist > found a way > to express (not _prove_, of course, but just express) > JSH's claim > using more rigorous mathematical terminology. Instead > of referring > to "preferences," the standard theorist mentioned the > notion of > asymptotic density, as follows. > > Let P(k) denote the kth prime. That is to say, P(k) > is defined > recursively as: > > P(1) = 2 > P(k+1) = min({peN | p>P(k) & p prime}) > > Then JSH's claim can be stated as follows: if m and n > are coprime, > then the set > > {keN | (P(k) == n) mod m} > > has asymptotic density 1/phi(m) (phi is Euler's > totient). > > One standard theorist claimed that the result is true > and follows > from Dirichlet's theorem, but Dirichlet's theorem > only shows that > these sets are _infinite_, not what their density is. > > Another standard theorist claimed that the result is > false, and > used quadratic residues in his argument. Indeed, he > pointed out > that if r is a quadratic residue mod m, and n is a > quadratic > nonresidue mod m, then for many natural numbers M, > > card({keN | k<M & (P(k) == n) mod m}) > card({keN | > k<M & (P(k) == r) > mod m}) > > One can see why this may be the case -- since r is a > quadratic > residue mod m, by definition, the set on the right > contains > squares, and by definition squares are not primes. > However, the > poster failed to mention whether this affects the > asymptotic > density at all, so that: > > density({keN | (P(k) == n) mod m}) > 1/phi(m), n a > nonresidue mod m > density({keN | (P(k) == r) mod m}) < 1/phi(m), r a > residue mod m > > > > The best never stop at one. > > Andrew Wiles has zero. Unlike you, he can prove > theorems, so he has no > > need to "discover" axioms. > > Until a standard theorist gives a proof, or at least > a reference > to a proof, of either JSH's claim or its negation, > the possibility > remains that his claim is undecidable in ZFC. In that > case, JSH > would be justified in calling his claim an "axiom." > Indeed, since > Euclid's Fifth Postulate and AC have been shown to be > undecidable > in neutral geometry and ZF, respectively, one can add > either these > or their negations as _axioms_ and produce a new > theory. > > > > There are no one-hit wonders in major > mathematical discovery. > > Most so-called "mathematicians" have not discovered > a single new axiom > > in their entire miserable lives. > > "Discovering" axioms isn't meaningful, but > discovering nontrivial > statements that are undecidable in ZFC is -- > especially if one can > derive interesting results from them. > > Consider Riemann's Hypothesis. So far neither RH nor > its negation > have been proved in ZFC, and the possibility remains > that RH might > be undecidable in ZFC. But there are some results in > number theory > that require RH. Therefore, if RH is shown to be > undecidable in ZFC, > some number theorists might prefer to work in ZFC+RH > and not ZFC. > > It's conceivable that JSH's claim might also lead to > interesting > results in number theory. Although I don't really > believe that JSH > has a proof of the Twin Primes Conjecture, if someday > in the future > someone proves the Twin Primes Conjecture, but the > proof requires > JSH's claim, and that claim has been proved > undecidable in ZFC, > then number theorists might choose to work in ZFC+JSH > rather than > ZFC, thus vindicating JSH's axiom once and for all. we know that prime factorization is not random , it can produce primes with preferences. a good example is given by my 911 polynomial. in the 911 polynomial take a positive random x , and notice all prime factors are 1 mod 10. quadratic residues are also a good example of non-random factorization. so an axiom ? rather a conjecture with hope for proof or disproof. however all such ideas are polynomial ideas and thus we might conjecture , not an axiom , the following : p mod q is random for p a non-quadratic residue mod q and q a non-quadratic residue mod p. notice that a sqrt deviation does not imply non-random , see e.g. random walk. so james idea is not ' rescued ' because it was not in ' danger '. btw i doubt the correctness of the proof you mention. the paper does not seem convincing , quadratic residues seem ignored and the result proven seems too strong ( close to GRH ). i know i did not define random formally , but then again did anyone ever ? ;) have you looked at my website btw ? regards tommy1729
From: master1729 on 21 Mar 2010 11:37
> On Mar 9, 12:43 am, "Ostap S. B. M. Bender Jr." > <ostap_bender_1...(a)hotmail.com> wrote: > > On Mar 9, 12:30 am, "Ostap S. B. M. Bender Jr." > > > > if someday in the future > > > > someone proves the Twin Primes Conjecture, but > the proof requires > > > > JSH's claim, and that claim has been proved > undecidable in ZFC, > > > > then number theorists might choose to work in > ZFC+JSH rather than > > > > ZFC, thus vindicating JSH's axiom once and for > all. > > I am too lazy to check, but wouldn't the proof > (using only ZFC) that > > JSH doesn't contradict ZFC, be actually a proof > that JSH is true in > > ZFC? > > It could be the case that _neither_ JSH _nor_ its > negation contradicts > ZFC, so that JSH would be _undecidable_ in ZFC. > > Actually, since Magadin established that JSH isn't > really an "axiom" > at all, let's use Riemann's hypothesis intead. In > other words, it > could be the case that _neither_ RH _nor_ its > negation contradicts > ZFC, so that RH would be _undecidable_ in ZFC. > Obviously, ZFC doesn't > prove both RH _and_ its negation (unless ZFC is > inconsistent). > > There have been a few threads here at sci.math which > discuss the > possibility that RH is undecidable in ZFC, or that > Goldbach's > Conjecture is undecidable in ZFC or PA. Some standard > theorists > wonder, if RH or GC is undecidable in ZFC, does this > mean that RH or > GC are "true"? no. |