most simple statement of Riemann Hypothesis and proofs #703 Correcting Math
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From: Archimedes Plutonium on 23 Jul 2010 19:15 Archimedes Plutonium wrote: > Archimedes Plutonium wrote: > > Archimedes Plutonium wrote: > > > Archimedes Plutonium wrote: > > > > I am looking for the best Riemann Hypothesis equivalent statement to > > > > tie in the Indirect Euclid Infinitude of Primes proof method. By > > > > correcting that flaw of logic that both P-1 > > > > and P+1 are necessarily prime, yielding the infinitude of Twin Primes, > > > > I suspect is a > > > > key to proving the Riemann Hypothesis RH. > > > > > > > > So I looked for equivalent RH statements: > > > > --- quoting Wikipedia in part --- > > > > Riemann's explicit formula for the number of primes less than a given > > > > number in terms of a sum over the zeros of the Riemann zeta function > > > > says that the magnitude of the oscillations of primes around their > > > > expected position is controlled by the real parts of the zeros of the > > > > zeta function. In particular the error term in the prime number > > > > theorem is closely related to the position of the zeros: for example, > > > > the supremum of real parts of the zeros is the infimum of numbers β > > > > such that the error is O(xβ) (Ingham 1932). > > > > > > > > > > > > Von Koch (1901) proved that the Riemann hypothesis is equivalent to > > > > the "best possible" bound for the error of the prime number theorem.. > > > > > > > > > > > > A precise version of Koch's result, due to Schoenfeld (1976), says > > > > that the Riemann hypothesis is equivalent to. . . > > > > > > > > --- end quoting --- > > > > > > Ingham, Von Koch, and Schoenfeld and others bespeak of the Riemann > > > Hypothesis > > > as the most efficient placing of primes in a prime distribution. As if > > > efficiency and accuracy > > > of placement of primes is what the Riemann Hypothesis is all about. > > > > > > > > > > > > > > Let me try to give an equivalent RH statement myself. > > > > > > > > It is already proven, I think it was Chebychev, that between n and 2n > > > > always exists another prime. > > > > > > > > So, let me focus on n+1 and 2n-1 > > > > > > > > We have: > > > > > > > > for 2, 2+1 = 3 and 4-1 = 3 > > > > > > > > for 3, 3+1=4 and 6-1=5 > > > > > > > > for 4, 4+1 =5 and 8-1=7 > > > > > > > > for 5, 5+1=6 and 10-1=9 > > > > > > > > etc etc > > > > > > > > Now, instead of Riemann getting involved with the Complex Number > > > > Plane, how about a > > > > Riemann Hypothesis more down to Earth. How about a Riemann Hypothesis > > > > with just the plain old Natural Numbers since we find billions and > > > > zillions of equivalent statements, but > > > > never the most simple statement. > > > > > > > > So let me proffer my own equivalent statement of the Riemann > > > > Hypothesis since the one > > > > thing that RH can never get away from is the distribution of prime > > > > numbers. > > > > > > > > Archimedes Plutonium's equivalent statement of the Riemann Hypothesis: > > > > The RH, if true says that as n becomes large, very large that both n+1 > > > > and 2n-1 > > > > are both prime numbers. If that is true, then a proof of that RH > > > > equivalent is easily > > > > begot from the Euclid Infinitude of Primes proof Indirect method for > > > > it makes > > > > n+1 and 2n-1 necessarily new primes as n goes to infinity. > > > > > > > > > Then this equivalent statement to the RH by myself is not efficient > > > and accurate enough. > > > > > > I should have said that the RH equivalent is such that n-1, n+1 and > > > 2n-1, 2n+1, all four > > > of those numbers are necessarily prime as n tends to infinity. > > > > > > An example of that is n=30 so that n-1 =29 and n+1=31, and > > > 2n-1=60--1=59 and 2n+1= > > > 60+1=61 are all four prime numbers. So that would be a Maximum density > > > of primes > > > given n goes to infinity. > > > > > > It is where the Infinitude of Primes proof conjoins with the Riemann > > > Hypothesis, and the > > > proof of this RH is simply a Indirect Method with Mathematical > > > Induction that yields four > > > Euclid Numbers, all four of which are necessarily prime numbers. > > > > > > > > > > > > > > > > > Now I am curious since I define with precision the finite-number > > > > versus the infinite-number > > > > as the boundary at 10^500. So I am curious as to whether 10^500 (+1) > > > > is a prime number > > > > and its associate of 2x(10^500) -1. If not, then let us chose as the > > > > boundary where n+1 > > > > and 2n-1 in the region of 10^500 are both prime numbers. So that > > > > mathematics does share > > > > a input into the selection of the boundary between finite and infinite- > > > > number. > > > > > > > > Perhaps a major reason the RH was never proven or steered into a > > > > correct path to prove it, was that it was too much cloaked in the > > > > Complex Number Plane and if someone had retrieved it out of that > > > > cloaking, would have seen it in its more basic form that n+1 and 2n-1 > > > > are both > > > > primes when n tends to infinity. They may not have realized that a > > > > simple tinker to fix the logic flaw of Euclid IP indirect, but at > > > > least they would have made RH more understandable. > > > > Mathematicians are like artists, once they paint legs on a snake, they > > > > refuse to remove the legs and rather increase the complexity. > > > > > > > > > > Now Physics is the king of sciences and mathematics is only a room, a > > > tiny > > > room in the house of physics. And Physics would define the boundary > > > between > > > finite number versus infinite-number and it would be the largest > > > Planck unit > > > which is the Coulomb Interactions in element 100 of about 10^500. But > > > here is where > > > mathematics has a "say at the table". Since the RH of above would have > > > four primes > > > at n, 2n, the question is does 10^500 plus and minus 1 yield twin > > > primes and does > > > 2x10^500 plus and minus 1 yield twin primes? If so, then we assuredly > > > take 10^500 as > > > the boundary between finite-number versus infinite-number. Or if there > > > is another large > > > number in the vicinity of 10^500 that yields those four primes. > > > > > > Carbon in me, carbon of plutonium, fill me with life anew, that I may > > > love what thou dost > > > love. Oxygen in me, oxygen of plutonium. . > > > > > > > I suppose there is even more symmetry to add to the above. In my > > example of > > 30 with twin primes astride 30 and then 2n as 60 with twin primes > > astride, which > > is very symmetrical because the interval has two primes but also the > > interval > > below 30 and the interval above 60 already have their primes set to > > go. > > > > But there is additional symmetry. Keep in mind I am striving for a > > maximum Riemann Hypothesis as to the maximum efficiency of primes. So > > in the interval 30 to 60 there is > > a possibility of having the midpoint be a prime, in this case 45 is > > not prime, but in another > > case of n to 2n have (n + 2n)/2 be a prime. > > > > Off hand I cannot provide an example of that. And should the boundary > > marker of finite-number > > with infinite-numbers such as 10^500 be an example of that situation? > > > > I think not, because the Infinitude of Primes Proof, indirect does not > > venture into whether the > > midpoint is prime. > > > > The whole idea of this most simple and elegant statement for the > > equivalent of the Riemann Hypothesis is to measure RH against the full > > potential of the Indirect IP proof. The full potential is that we end > > up with four new necessarily primes as Euclid Numbers of the twin > > primes on each end node of n and 2n. Euclid IP Indirect is silent > > about a midpoint prime. > > So I think I have captured the full extent of RH with Euclid IP. > > > > If I am correct, then the Euclid Infinitude of Primes Proof, Indirect > > proves the Riemann Hypothesis as true. > > > > Now why would RH escape that attention? Probably because noone saw the > > mistake in > > Euclid IP Indirect and how much it influences the rest of mathematics. > > > > But that leaves me in a funny position, for it leaves me only with FLT > > to complain about that it has no proof until a precision definition of > > finite versus infinite-number is given as 10^500 or > > thereabouts. When I had RH to include with FLT, I felt a bit better of > > a stronger case. But now it looks as though FLT is the only one > > standing that demands and crys out for a precision boundary definition > > of finite versus infinite. But maybe, just maybe the nonmath > > conjecture of engineering called the NP conjecture is mincemeat once > > 10^500 is made the boundary. > > > Now when I gave my two proof offerings of the Riemann Hypothesis of > the early > 1990s, I had it published in a newspaper and do not recall the exact > year whether > it was 1991 or 1992 or 1993, but that is neither here or there. In > those two proof > offerings I was guided by the Moebius equivalent to the Riemann > Hypothesis and > constructed a logarithmic spiral inside rectangles of whirling > squares. Many of us > have seen this geometrical rectangles of whirling squares. I argued > that the Riemann Hypothesis was true for the log curve was contained > inside the rectangles and would > never get out and that was the bounds of the Moebius equivalent. > > Well here, let me just copy paste my two old proof offerings of RH: > TWO PROOFS OF THE RIEMANN HYPOTHESIS > > PROOFS: Two proofs of the Riemann Hypothesis follows as A > and B. > > Proof (A) is a geometrical proof. It was proved that the Riemann > Hypothesis is equivalent to the following-- the Moebius function mu of > x, m(x), and adding-up the values of m(x) for all n less than or equal > to N giving M(N). Then M(N) grows no faster than a constant multiple k > of (N^1/2)(N^E) as N goes to infinity (E is arbitrary but greater than > 0). Figure1, by setting-up a logarithmic spiral in a rectangle of > whirling squares where the squares are the sequences: > 1,1,2,3,5,8,13,21,34,55,89, . . . 2,2,4,6,10,16,26, . . . > 3,3,6,9,15,24,39, . . . then every number appears in at least one of > these sequences because every number will start a sequence. Since all > numbers are represented uniquely by prime factors (the unique prime > factorization theorem or called the fundamental theorem of arithmetic) > and The Prime Number Theorem: the distribution of prime numbers is > governed by a logarithmic function, where (An/n)/(1/Ln of n) tends to > 1 > as n increases, where An denotes the number of primes below the > positive integer n, and where An/n is called the density of the primes > in the first n positive integers. The density of the primes, An/n, is > approximated by 1/(Ln of n), and as n increases, the approximation > gets > better. The distribution of prime numbers is governed by a > logarithmic function where these two math concepts-- one of prime > numbers, and the other, logarithms seem unconnected at first > appearance, but in reality they are totally connected. Geometrically, > the logarithmic spiral exhausts every positive integer, see figure 1. > The area of the rectangles containing the logarithmic spiral is always > greater, since the spiral is always inside the rectangles. Thus the > Moebius function k (N^1/2)(N^E) is satisfied since the area of the > logarithmic spiral is less than the rectangle whose area represents > the > number N, and whose sides represent its factors. The area of a > logarithmic spiral is represented by A=(r)(e^(Hj)) , and so depending > on where the point of origin for the spiral is taken rsubO determines > k, and depending on the value of H, H determines the E value for N, > when H=0 then the curve is a circle. The logarithmic spiral inside > rectangles of whirling squares implies that for any number N then > N^1/2 > is the limit of the factors for N, for example, given the number 28, > then 28^1/2 = 5.2915. . and so looking for the factors of 28, it is > useless to try beyond 5 because the factors repeat, 4x7 then repeats > as > 7x4. But if the Moebius function was false then there must exist a > number M such that M^1/2 is not the limit of the factors for M and the > spiral is outside of the square, which is impossible, hence the > Moebius > function is true. Therefore the Riemann Hypothesis is proved. Q.E.D. > > My second proof (B) of the Riemann Hypothesis uses a reductio > ad absurdum argument. Euler proved that a formula encoding the > multiplication of primes was equal to the zeta function. Euler's > formula in complex variable form is as follows: > (1/(1-(1/(2^c))))x(1/(1-(1/(3^c))))x(1/(1-(1/(5^c))))x(1/(1-(1/ > (7^c))))x > > (1/(1-(1/(11^c))))x . . . , where c is a complex variable, c=u+iv. The > Riemann zeta function is as follows. Re(c) = > 1+(1/(2^c))+(1/(3^c))+(1/(4^c))+. . . , where c is a complex variable, > c=u+iv. Euler's formula involves multiplication of terms and the > Riemann zeta function involves addition of terms of a sequence. Taking > Re(c) > 0, suppose the Riemann Hypothesis is false then there is a 0 > such that Re(c)=0 and c not equal 1/2 +iy, which implies there is > another 0 which is not on the 1/2 real line. Which means another real > number other than 1/2 works as an exponent resulting in a zero for the > Riemann zeta function, and a zero in the Euler formula. Thus, Riemann > zeta function subtract Euler formula must equal zero. This implies > for > any other real number exponent, either rational or irrational numbers, > such as for example the rational exponents: 1/3,1/4,1/5, . . . (Note: > any other exponent y/x , where y and x are Real numbers and where the > Real number of A^(y/x) such that y not equal 1, immediately transforms > to a number A^y(1/x), so that exponents with a 1 in the numerator > entail all of the Real exponents). To make clear of the above, for > example, 2^2/3 is 4^1/3. So then back to the proof. Then for exponent > 1/3 there has to exist a number M not equal 0 where (M+M+M)^1/M = > (MXMXM)^1/M = M. Then for exponent 1/4 there has to exist a number M > not equal 0 where (M+M+M+M)^1/M = (MXMXMXM)^1/M = M, and so on. > Including the infinite number of cases where the x denominator is > irrational are impossible. Only the real number 1/2 works since 2 does > not equal 0, and (2+2)^1/2 = (2X2)^1/2 = 2, and so (2+2)^1/2 > - (2X2)^1/2 = 0. In all of Reals and the Complex numbers, 2 is the > only number N which has the encoding ((N+N)^1/N) = ((NxN)^1/N) = N. > Unlike 0, the number 2, its sum equals its product and where the sum > and product is a new number 4. If RH were false, then another number > other than 2 would satisfy a generalized encoding formula ((N+N)^1/N) > = ((NxN)^1/N) = N. False, hence the proof. QED > -------- > > Now let me just get to the point. I am saying that the correct Euclid > Infinitude of Primes > Proof Indirect gives us four points or four prime numbers at maximum > gives us four new > prime numbers as Euclid Numbers in the Indirect Proof. Those four > primes are two sets of > Twin Primes. > > Now to make a rectangle you need four points as the four corners. So > you can see > where I am going with this. > > That the proof of the Riemann Hypothesis all hinges on a complete, > correct and valid > Euclid Infinitude of Primes proof Indirect Method which fetches four > new primes. > > Is Infintude of Twin Primes true? Of course it is because IP Indirect > fetches W-1 > and W+1 as IP primes indirect method. > > Is Riemann Hypothesis true? Of course it is true because IP indirect > method fetches > at maximum and symmetrical two pairs of twin primes located at n and > 2n. > > So that when Ingham, Von Koch, and Schoenfeld were praising the > Riemann Hypothesis > for its efficiency of prime placement along the number-line, that > efficiency can be > made into a most simple restatement and equivalency of the Riemann > Hypothesis. > > Riemann Hypothesis: given the Natural Numbers which are well defined > with a boundary > of finite-number and infinite-number such as 10^500 as the boundary of > where infinite-number > begins. That as the Natural Numbers and primes tend to infinity and > where there always exists at least one prime between the intervals of > n and 2n, that at infinity we have the case > where we have four primes as that of n-1, n+1, 2n-1 and 2n+1. > > P.S. I need to verify that either 10^500 is this boundary that has > these four primes or some > other number within the vicinity of 10^500. > I do not know whether the Twin Primes can construct rectangles in whirling rectangles as an independent proof of the Riemann Hypothesis, or, whether the Twin Primes proof which yields very long thin rectangles acts to bolster the above proof of RH. In other words, I am not sure whether all of this goes to making one proof of RH or whether I have two independent proofs of RH. Archimedes Plutonium http://www.iw.net/~a_plutonium/ whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies |