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Aug 15, 1993, 5:52:54 PM8/15/93

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TWO PROOFS OF THE RIEMANN HYPOTHESIS (both published in The Dartmouth

9/8/ŻŻ51 9Aug51---that is 1991 in the unscientific calendar, the

scientific calendar starts year 0000 with 1940 the first identification

of our Maker.

9/8/ŻŻ51 9Aug51---that is 1991 in the unscientific calendar, the

scientific calendar starts year 0000 with 1940 the first identification

of our Maker.

Discussion: Riemann conjectured that the real component for the

complex numbers at which the zeta function equals zero is 1/2. This

conjecture of Riemann is: the #1, major, most sought-after unsolved

problem in all of mathematics.

SORRY READERS but these two figures (macpaint did not copy translate,

but figure 1 is seen in Jacobs Mathematics a Human Endeavor. Figure 1:

A logarithmic spiral inside rectangles of whirling squares. The

squares and the rectangles go out to infinity and thus the spiral goes

out to infinity.

Figure 2: Collapsed wavefunction from a logarithmic spiral into

Riemannian space of an ellipse or sphere. And sorry that exponents and

symbol font does not copy translate.

PROOFS: Two proofs of the Riemann Hypothesis follows as (A) and (B).

Proof (A). Proof (A) 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/(2c))))x(1/(1-(1/(3c))))x(1/(1-(1/(5c))))x(1/(1-(1/(7c))))x(1/(

1-(1/(11c))))x . . . , where c is a complex variable, c=u+iv. The

Riemann zeta function is as follows: z(c) =

1+(1/(2c))+(1/(3c))+(1/(4c))+. . . , 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. Suppose

the Riemann Hypothesis is false then there is a Ż such that z(z)=Ż and

zą1/2 +iy, which implies there is another Ż 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 yą1, immediately transforms to a number (Ay)(1/x)), so

that exponents with a 1 in the numerator entail all of the Real

exponents). Then for exponent 1/3 there has to exist a number MąŻ where

(M+M+M) - (MXMXM) = Ż. Then for exponent 1/4 there has to exist a

number MąŻ where (M+M+M+M) -(MXMXMXM)=Ż, 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 2ąŻ, and (2+2) =(2X2),

and so (2+2) - (2X2)= Ż. In all of mathematics, 2 is the only number

where its sum equals its product and where the sum and product is a new

number 4. The property of zero (not a number) does not produce a new

number when added Ż+Ż or multiplied ŻxŻ. Therefore only real component

of 1/2 works for the Riemann zeta function to equal zero. Q.E.D.

As a check to see if there are any complex numbers which have the

property of (z+z)-(zxz)= Ż where z=x+iy

2(x+iy) - (x+iy)2 = Ż

2x +2iy - (x2 - y2 + 2ixy) = Ż gives a real component 2x- x2 + y2 = Ż

and an imaginary component 2iy - 2ixy = Ż

for imaginary component 2iy - 2ixy = Ż, implies 2iy(1-x) = Ż

thus y=Ż or x=1

for real component 2x-x2 + y2 =Ż when y=Ż implies 2x-x2 =Ż, (2-x)x = Ż

thus x=Ż or x=2

for x=1 implies 2x-x2+ y2 =Ż, (2-1)+y2 = Ż, y2= -1, thus y= + i,

substituting x=1 with y= + i, into z= x+iy gives z=1+ i2, thus z=Ż or

z=2. Therefore, only the numbers Ż and 2 satisfy (z+z)-(zxz)= Ż.

Q.E.D.

Proof (B). A geometrical proof follows. 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). That M(N) grows no faster than a constant multiple k

of N1/2Ne as N goes to infinity (e is arbitrary but greater than Ż).

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, . . . 2,2,4,6,1Ż,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/Logarithme 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. It is one of the most beautiful things in

all of the known world, that the distribution of prime numbers is

governed by a logarithmic function where these two mathematical

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 N1/2Ne 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

r=rŻeEj , and so depending on where the point of origin for the spiral

is taken rŻ determines k, and depending on the value of E, E determines

the e value for N, when E=Ż then the curve is a circle. The

logarithmic spiral inside rectangles of whirling squares implies that

for any number N then N1/2 is the limit of the factors for N, for

example, given the number 28, then 281/2=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 M1/2 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.

An electron has intrinsic spin of +1/2, only the positive value of

+1/2 works for the spin quantum number ms, no other number works. Spin

quantum number has positive values only, but spin states for an

electron or proton can correspond to s' = +1/2 and s'' = -1/2. I assert

our observable universe is the last one electron in the 5f6 of a

plutonium atom. Then the zeros of the zeta function are the charges

added-up and so protons cancel with electrons, no net charge remains,

because matter comes into existence from spontaneous neutron

materialization and thus there can not exist any net charge since

through radioactivities a neutron transforms into a proton plus

electron. Thus the zeta function is a quantum chart of every neutron,

proton, electron, and atom which came, or will come into existence. The

uncollapsed wave function (figure1) of quantum mechanics represents

numbers of mathematics such as irrational, transcendental numbers such

as e and ą. The collapsed wave function (figure2) is the

materialization of an atom or subatomic particle, where materialization

substitutes for existence of a rational number. The number 2 which is

1+1 represents the Plutonium Atom Totality itself, and the next term

represents perhaps the first hydrogen atom to exist, and so on for

every term in the zeta function.

The number 2 is the number for Bohr's complementary principle where

all matter has dual complements of particle-wave. Matter can not exist

without two things at once, thus 2 is the existor function. The

totality-- one plutonium atom exists in duality with atom parts and at

least one of those atom parts is itself. There are 4 quantum exclusion

numbers (n, L, mL, ms); 4 uncertainty conjugate variables of

position,momentum,energy,time; 4 interactions (forces) of physics; and

4 mathematical operators. All of these come as a byproduct of 2 where

2x2=2+2=4. Spin of an electron is a dynamical system, for without spin

then change would approach zero. Change in the observable electron

universe would approach zero without spin. Thus, the 1 Plutonium Atom

Totality divided by the existor function of 2, gives 1/2 for electron

spin.

Aug 16, 1993, 10:59:32 AM8/16/93

to

So, Ludwig, if you look up a statement of the Riemann Hypothesis in

any decent book on complex analysis, you'll note that the zeta

function vanishes at the negative even integers. You claim to

have proven that all zeros lie on the line Re(s) = 1/2. Please

tell us all where you made your error.

Keith Conrad

Aug 16, 1993, 5:54:55 PM8/16/93

to

The zeta function in the interval (0,1). Sorry if I did not make

that clear. Thanks Keith for I must better anticipate the many

extraneous objections.

that clear. Thanks Keith for I must better anticipate the many

extraneous objections.

Aug 20, 1993, 10:06:55 PM8/20/93

to

No objections. (Disregarding Kieth's extraneous comment). Would the

Princeton Math Dept please do me the honor of calling Gina Kolata to

put me on the front cover of the New York Times. RH is exponentially

more important than old FLT. Just think, I could fit both of my RH

proofs on the front cover of NYT and put my FLT in the margin.

(Fermat's ghost is alive and kicking). Play the music of HALLELUJAH

CHORUS (hallelujah was just a nonsense term awaiting for PU to turn it

into its final form------------

ATOM PLUTONIUM CHORUS

lyrics by L. Plutonium

using the same music by G.F. Handel

______________________________________________________

ATOMPLUTONIUM, ATOMPLUTONIUM, PLUTONIUM, PLUTONIUM,

ATOM PLUTONIUM

ATOMPLUTONIUM, ATOMPLUTONIUM, PLUTONIUM, PLUTONIUM,

ATOM PLUTONIUM

FOR THE ATOM HAS INFINITE POTENTIAL

PLUTONIUM, PLUTONIUM, PLUTONIUM, PLUTONIUM, FOR THE ATOM HAS INFINITE

POSSIBILITIES

PLUTONIUM, PLUTONIUM, PLUTONIUM, PLUTONIUM, FOR THE ATOM HAS

INFINITE POTENTIAL

FOR THE ATOM HAS INFINITE POSSIBILITIES.

Princeton Math Dept please do me the honor of calling Gina Kolata to

put me on the front cover of the New York Times. RH is exponentially

more important than old FLT. Just think, I could fit both of my RH

proofs on the front cover of NYT and put my FLT in the margin.

(Fermat's ghost is alive and kicking). Play the music of HALLELUJAH

CHORUS (hallelujah was just a nonsense term awaiting for PU to turn it

into its final form------------

ATOM PLUTONIUM CHORUS

lyrics by L. Plutonium

using the same music by G.F. Handel

______________________________________________________

ATOMPLUTONIUM, ATOMPLUTONIUM, PLUTONIUM, PLUTONIUM,

ATOM PLUTONIUM

ATOMPLUTONIUM, ATOMPLUTONIUM, PLUTONIUM, PLUTONIUM,

ATOM PLUTONIUM

FOR THE ATOM HAS INFINITE POTENTIAL

PLUTONIUM, PLUTONIUM, PLUTONIUM, PLUTONIUM, FOR THE ATOM HAS INFINITE

POSSIBILITIES

PLUTONIUM, PLUTONIUM, PLUTONIUM, PLUTONIUM, FOR THE ATOM HAS

INFINITE POTENTIAL

FOR THE ATOM HAS INFINITE POSSIBILITIES.

A DOT OF THE ELECTRON PROBABILITY DENSITY DISTRIBUTION OF THE 5F6 FOR

THE LAST ELECTRON OF 231PU IS THE PLANET EARTH, ANOTHER DOT IS YOU,

ANOTHER DOT ME. AND ATOMS WILL NUCLEOSYNTHESIZE FOREVER AND EVER,

AND ATOMS WILL NUCLEOSYNTHESIZE FOREVER AND EVER, AND ATOMS WILL

NUCLEOSYNTHESIZE FOREVER AND EVER,

AND ATOMS WILL NUCLEOSYNTHESIZE FOREVER AND EVER.

ATOM OF ATOMS, FOREVER AND EVER, PLUTONIUM, PLUTONIUM, AND ATOM OF

ATOMS, FOREVER AND EVER, PLUTONIUM, PLUTONIUM, ATOM OF ATOMS, FOREVER

AND EVER,

PLUTONIUM, PLUTONIUM, AND ATOM OF ATOMS, FOREVER AND EVER,

PLUTONIUM, PLUTONIUM, ATOM OF ATOMS, FOREVER AND EVER,

PLUTONIUM, PLUTONIUM, AND ATOM OF ATOMS, ATOM OF ATOMS, FOREVER AND

EVER, AND ATOMS WILL NUCLEOSYNTHESIZE FOREVER AND EVER. ATOM OF ATOMS

AND ATOM OF ATOMS, PLUTONIUM, PLUTONIUM, AND ATOM OF ATOMS, FOREVER AND

EVER, AND ATOMS WILL NUCLEOSYNTHESIZE FOREVER AND EVER. ATOM OF ATOMS

AND ATOM OF ATOMS, ATOM OF ATOMS, AND ATOMS WILL NUCLEOSYNTHESIZE

FOREVER AND EVER. PLUTONIUM, PLUTONIUM, PLUTONIUM, PLUTONIUM.

ATOMPLUTONIUM!

Aug 20, 1993, 11:16:32 PM8/20/93

to

In article <CC377...@dartvax.dartmouth.edu>,

Ludwig Plutonium <Ludwig.P...@dartmouth.edu> wrote:

>No objections. (Disregarding Kieth's extraneous comment). Would the

Ludwig Plutonium <Ludwig.P...@dartmouth.edu> wrote:

>No objections. (Disregarding Kieth's extraneous comment). Would the

Probably, no one gives a shit.

Charles Yeomans

Aug 21, 1993, 9:58:20 AM8/21/93

to

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

I like it. Can I put it in my SIG? :-)

--

Bruce Ikenaga

US mail: Dept. of Math, CWRU, Cleveland, Ohio 44106

E-mail : b...@po.CWRU.edu

Aug 20, 1993, 11:38:49 PM8/20/93

to

In article <CC377...@dartvax.dartmouth.edu> Ludwig.P...@dartmouth.edu (Ludwig Plutonium) writes:

>No objections. (Disregarding Kieth's extraneous comment). Would the

>Princeton Math Dept please do me the honor of calling Gina Kolata to

>put me on the front cover of the New York Times. RH is exponentially

>more important than old FLT. Just think, I could fit both of my RH

>proofs on the front cover of NYT and put my FLT in the margin.

>(Fermat's ghost is alive and kicking). Play the music of HALLELUJAH

>CHORUS (hallelujah was just a nonsense term awaiting for PU to turn it

>into its final form------------

> ATOM PLUTONIUM CHORUS

> lyrics by L. Plutonium

> using the same music by G.F. Handel

>______________________________________________________

[silliness deleted]>No objections. (Disregarding Kieth's extraneous comment). Would the

>Princeton Math Dept please do me the honor of calling Gina Kolata to

>put me on the front cover of the New York Times. RH is exponentially

>more important than old FLT. Just think, I could fit both of my RH

>proofs on the front cover of NYT and put my FLT in the margin.

>(Fermat's ghost is alive and kicking). Play the music of HALLELUJAH

>CHORUS (hallelujah was just a nonsense term awaiting for PU to turn it

>into its final form------------

> ATOM PLUTONIUM CHORUS

> lyrics by L. Plutonium

> using the same music by G.F. Handel

>______________________________________________________

The Princeton Math Dept will do no such thing. They're still getting

letters from cranks claiming to have found the proof that fits

in a margin. Why honor you when all these others have found similarly

easy proofs?

Uh, let's see. For the umpteenth time, Wiles main mathematical achievement

was that he proved a partial version of the Taniyama-Shimura conjecture,

whose importance is pretty clear to number theorists. I have yet to

see you offer YOUR reason for why you think RH is so important. I agree

it's important, but do YOU know why (math reason, not metaphysical gibberish)?

As for dismissing Keith's objection so casually, you merely once again

demonstrate your total lack of understanding of rudimentary logic.

What he was so obviously pointing out was that your alleged proof,

were it even remotely reasonable, purported to DERIVE that

ALL the zeros of zeta lied on the line Re(z) = 1/2. You NEVER pointed

out a SINGLE step where you used the fact that you were considering zeta

in a region not containing the "trivial" zeros, and so

if your argument had any actual validity at all, it would

have to utilize such a hypothesis, lest when applied

to regions containing trivial zeros you would

be showing that the trivial zeros somehow did not exist,

a blatant contradiction. So SOMEWHERE is your argument, if

you think it's correct, you HAVE to be using somewhere the fact

that you're working in region, like maybe Re(z) > 0, where

there are no trivial zeros.

But alas, your "proof" is totally formal and you blindly manipulate

divergent expressions and you never make explicit any single step

at which point the alleged argument would collapse if

you tried to apply it to z with Re(z) < 0, say.

Until you can carefully explain why your logic is fine for

Re(z) > 0 but does not apply for various z with Re(z) < 0,

you must surely recognize there is something deeply wrong with

your technique.

Perhaps reading a detailed development of the basic

theorems of complex analysis (say, in Alfhors' book) is in order

for you. Then you'll gain some appreciation for why RH might

be as damn hard as everyone else thinks it is.

I suspect you might have been happier living 200 years ago when

mysticism and intuition were sometimes accepted in place of logical

proofs. But even then, the sorts of reasons you give for

things STILL would have been recognized for the shams that they are.

Brian Conrad

Aug 25, 1993, 11:34:09 AM8/25/93

to

In article <1993Aug21.0...@Princeton.EDU>

con...@fine.princeton.edu (Brian Conrad) writes:

con...@fine.princeton.edu (Brian Conrad) writes:

> and you never make explicit any single step

> at which point the alleged argument would collapse if

> you tried to apply it to z with Re(z) < 0, say.

You did not understand my proof. RH is true because one and only one

number in all of math has the property N+N=NxN=M, and that number is 2.

That is why only exp1/2 works. If exp1/3 works then there would have to

exist a number such that N+N+N=NxNxN=M.

By the way, why has Ronald Bruck taken down the proposed proof of RH

leaving only replies? Please inform as to why and when do these things

disappear?

Brian you did not understand either one of my 2 proposed proofs of RH.

Did you even read them? Brian it is obvious to most network readers

that you are on a scalping warpath. Tell me, were you the forced

Princeton volunteer to try to discredit LP or did you willingly

volunteer? If you chased math as much as your scalping warpath, you

might amount to something, someday, . .

Aug 26, 1993, 11:04:06 AM8/26/93

to

In article <CC3AF...@ms.uky.edu>

cyeo...@ms.uky.edu (Charles Yeomans) writes:

cyeo...@ms.uky.edu (Charles Yeomans) writes:

> Probably, no one gives a shit.

>

> Charles Yeomans

I respectfully request that you Charles Yeomans show us your

Euclid's proof of the Infinitude of Primes. Every working math major

worth his weight in salt can do that. I am awaiting.

Aug 26, 1993, 10:15:36 PM8/26/93

to

In article <CCDGI...@dartvax.dartmouth.edu>,

I can certainly parrot the proof of the infinitide of the primes attributed]

to Euclid, but here's another one for you.

Define a topology on the integers by taking the collection of all

arithmetic progressions as a basis. Each set whose elements form

an arithmetic progression is both open and closed. Thus the union

of any finite number of arithmetic progressions is again a closed set.

For a prime number p (and no, neither 1 nor -1 is prime), let A(p)

be the set of multiples of p. Let A be the union of all such A(p).

The only integers not belonging to A are 1 and -1. It is not

hard for me to show that the set {1, -1} is not open; thus A

cannot be a closed set. Thus A can not be a finite union of

closed sets and we see that there is an infinity of primes.

This proof certainly isn't mine, though I wish it were.

You should see if you can apply these ideas to Goldbach's conjecture

and get a nice, clean proof.

Charles Yeomans

Aug 26, 1993, 11:24:41 PM8/26/93

to

In article <CCEBM...@ms.uky.edu>

cyeo...@ms.uky.edu (Charles Yeomans) writes:

cyeo...@ms.uky.edu (Charles Yeomans) writes:

> I can certainly parrot the proof of the infinitide of the primes attributed]

> to Euclid, but here's another one for you.

Please do Sir. Only your version of it. Thank You.

Aug 27, 1993, 10:08:31 PM8/27/93

to

In article <CCEEt...@dartvax.dartmouth.edu>,

In my own words, here it is:

Suppose you are given a finite list of primes, p_1, ..., p_n. THen

the product p_1*...*p_n + 1 is not divisible by any of p_1, ..., p_n;

this follows from the division algorithm. Since this number is greater

than 1 (I should have observed that a) I assume the existence of

a prime number, from which the existence of a positive prime number

follws, and b) p_1, ..., p_n are all positive.), it is either prime

or is divisible by a prime. In the latter case, unique factorization

implies that any prime factor of p_1*...*p_n + 1 is not in the

aforementioned list. Thus we obtain a new prime number. Now the

tinest bit of induction allows one to deduce the infinitude of the

primes.

WHat's your point?

Charles Yeomans

Sep 3, 1993, 12:29:54 PM9/3/93

to

In article <CCG5y...@ms.uky.edu>

cyeo...@ms.uky.edu (Charles Yeomans) writes:

cyeo...@ms.uky.edu (Charles Yeomans) writes:

> latter case, unique factorization

> implies that any prime factor of p_1*...*p_n + 1 is not in the

> aforementioned list.

My whole contention with proofs of IP is summarized into this:

1) There is a direct proof of IP which increases the cardinality of

any finite set of primes such as you have done here Charles. Only I

would guess that you would call your version the indirect method.

Usually all of the flawed proofs of IP mix both the direct and indirect

within the body of the proof.

2) There is an indirect proof of IP and in its STRONGEST FORM it

yields twin primes. Most every indirect proof version of IP is flawed

by holding TWO DEPENDENT CONDITIONALS simultaneously. So when the step

asserting a contradiction is reached one cannot safely say which of the

two conditionals (two suppositions) have you discharged. Have you in

fact disposed of the first supposition that the set of all primes is

finite or the second supposition--suppose this new number add 1 is

composite.

> WHat's your point?

The point of all of this is that I am asking as many people to have

a relook at the standard gospel indirect proof of IP, in hopes that

more people, not just Ben Tilly, can see the wrinkle in the gospel. It

is this wrinkle which has left the proof of the Infinitude of Twin

primes stranded on some forlorn and deserted math island.

Sep 3, 1993, 2:48:54 PM9/3/93

to

In article <CCsDt...@dartvax.dartmouth.edu>

Ludwig.P...@dartmouth.edu (Ludwig Plutonium) writes:

Ludwig.P...@dartmouth.edu (Ludwig Plutonium) writes:

> In article <CCG5y...@ms.uky.edu>

> cyeo...@ms.uky.edu (Charles Yeomans) writes:

>

> > latter case, unique factorization

> > implies that any prime factor of p_1*...*p_n + 1 is not in the

> > aforementioned list.

>

> My whole contention with proofs of IP is summarized into this:

> 1) There is a direct proof of IP which increases the cardinality of

> any finite set of primes such as you have done here Charles. Only I

> would guess that you would call your version the indirect method.

> Usually all of the flawed proofs of IP mix both the direct and indirect

> within the body of the proof.

>

I would say that very few people would agree with you about your

characterization of direct proofs vs indirect proofs.

> 2) There is an indirect proof of IP and in its STRONGEST FORM it

> yields twin primes. Most every indirect proof version of IP is flawed

> by holding TWO DEPENDENT CONDITIONALS simultaneously. So when the step

> asserting a contradiction is reached one cannot safely say which of the

> two conditionals (two suppositions) have you discharged. Have you in

> fact disposed of the first supposition that the set of all primes is

> finite or the second supposition--suppose this new number add 1 is

> composite.

>

Actually he had two assumptions. One was that you believed unique

factorization, the other was that there were a finite number of primes

which you can therefore put on a complete list. The point is that this

new number add 1 is either a prime not on your list, or it is a

composite number with a factor not on your list. Either way there is a

prime that is not on your list, so your list is not complete. Therefore

a contradiction holds. Since the first assumption of unique

factorization can actually be proved, there is only one assumption that

could be false.

> > WHat's your point?

>

> The point of all of this is that I am asking as many people to have

> a relook at the standard gospel indirect proof of IP, in hopes that

> more people, not just Ben Tilly, can see the wrinkle in the gospel. It

> is this wrinkle which has left the proof of the Infinitude of Twin

> primes stranded on some forlorn and deserted math island.

Please do not think that I think that there is a problem with the

standard proof. What I did is wrote the standard proof out carefully,

and then you said that it was right. Furthermore I do _not_ think that

your attempt at a proof that there are an infinite numbr of paris of

twin primes is correct for reasons that I have statd elsewhere.

BTW a piece of nettiquite. It is not considered proper to say that

someone agrees with you on something, unless you have their permission

to do so. In this case I most certainly did not agree with you, and I

do not like others misrepresenting my views.

Ben Tilly

Sep 5, 1993, 5:44:53 AM9/5/93

to

In article <26ance...@frodo.d.umn.edu>

jgr...@frodo.d.umn.edu (john greene) writes:

jgr...@frodo.d.umn.edu (john greene) writes:

> Theorem: There are infinitely many primes.

>

> Proof: Suppose the list of primes is finite, call them

> p1, p2, ..., Pl. Let n = p1*p2*...*Pl + 1. What kind of

> number is n? Is it composite? NO, since it has no prime

> factors (not being divisible by any prime on the list.)

> Is it prime? NO (this is where LP gets inconsistant) since

> it is not in the list of primes. Clearly it is not a unit

> either, so n is neither a unit, no prime, nor composite.

> Since this contradicts unique factorization, we have a

> contradiction, which establishes the result.

John, yours is the newest twist. Not only do you hold two

suppositions at once but you invoke UPFAT twice and you have the record

to date of holding three contradictions at once.

To pick apart your argument stepwise is this:

> Is it composite?

You are really saying Suppose n is composite, and then your answer "No"

relies on UPFAT for justification yielding your first contradiction

which you did not want to admit to, instead you glided right on to your

next step.

> Is it prime? NO (this is where LP gets inconsistant)

Your justification for this step John, which you forgot to state is

that your finite list p1, p2, ..., Pl is all the primes that exist,

and so n cannot be prime.

And John, by the way where you state "(this is where LP gets

inconsistant)", stands corrected to (this is where John gets into

spelling difficulty).

And now for your fireworks finale, you state.

> Since this contradicts unique factorization, we have a

> contradiction, which establishes the result.

You already had a contradiction much earlier when you supposed n was

composite. And now that you finally state another one here "this

contradicts unique factorization". What are you logically connecting

"this" with? Are you saying that it is a contradiction for n to be

a-number-at-all since it is neither prime,composite, or unit. You see

John, stating your contradiction of UPFAT at this point of your

argument is misplaced, it should have been stated earlier when you were

looking at n as composite. And then your third contradiction "we have a

contradiction" is quite needed in order to say the set of primes is

infinite. But where did you pull that contradiction from? Out of the

thin blue air, since your most embarrassing problem at this juncture is

that you do not have any new prime to augment to your original list,

since you contended that n was not prime. And so your argument fails at

the very end, for you are stuck with the same primes that you started

with, and you should restate your last line "which cannot establish the

desired result."

John, math proofs are not like some dime fiction novel where you

have a main plot and for those hapless readers who were not paying

close attention then a summary or recap will provide a happy

denouement. You must learn not to treat a math proof like a fiction

novel.

John, see Ronald Bruck's indirect proof of IP which is a valid

proof. The genuine indirect proof of IP has only one supposition and

two contradictions.

In article <25ock4$7...@mathj.usc.edu>

br...@mathj.usc.edu (Ronald Bruck) writes:

> If W-1 or (mutatis mutandi) W+1 is not prime then it has a prime factor.

> But none of p1, p2, ..., pL is a divisor, and these are the only primes;

> contradiction; so W-1 and W+1 are prime.

Sep 4, 1993, 2:39:10 PM9/4/93

to

I tried this once and expected a response from LP. I'll make

this one a little stronger. LP's proof of what he calls IP

is flawed by his own reasoning. He says that the usual proof

is wrong because one has not dispenced with the finiteness

of the list of primes before assuming a number is composite.

Here is the proof that I claim LP should come up with if he

wants to be consistant. Note that he can in no way obtain

twin primes from this approach.

this one a little stronger. LP's proof of what he calls IP

is flawed by his own reasoning. He says that the usual proof

is wrong because one has not dispenced with the finiteness

of the list of primes before assuming a number is composite.

Here is the proof that I claim LP should come up with if he

wants to be consistant. Note that he can in no way obtain

twin primes from this approach.

Theorem: There are infinitely many primes.

Proof: Suppose the list of primes is finite, call them

p1, p2, ..., Pl. Let n = p1*p2*...*Pl + 1. What kind of

number is n? Is it composite? NO, since it has no prime

factors (not being divisible by any prime on the list.)

Is it prime? NO (this is where LP gets inconsistant) since

it is not in the list of primes. Clearly it is not a unit

either, so n is neither a unit, no prime, nor composite.

Since this contradicts unique factorization, we have a

contradiction, which establishes the result.

To sumerize, if LP insists that it is wrong to assume n

is composite in order to obtain a contradiction, it is

certainly just as wrong to assume that n is prime.

John

Sep 3, 1993, 4:07:09 PM9/3/93

to

In article <CCsK9...@dartvax.dartmouth.edu>

Benjamin...@dartmouth.edu (Benjamin J. Tilly) writes:

Benjamin...@dartmouth.edu (Benjamin J. Tilly) writes:

> Furthermore I do _not_ think that

> your attempt at a proof that there are an infinite numbr of paris

How many Eiffel Towers then?

In article <CCsK9...@dartvax.dartmouth.edu>

Benjamin...@dartmouth.edu (Benjamin J. Tilly) writes:

> Actually he had two assumptions. One was that you believed unique

> factorization, the other was that there were a finite number of primes

> which you can therefore put on a complete list. The point is that this

> new number add 1 is either a prime not on your list, or it is a

> composite number with a factor not on your list. Either way there is a

> prime that is not on your list, so your list is not complete. Therefore

> a contradiction holds. Since the first assumption of unique

> factorization can actually be proved, there is only one assumption that

> could be false.

No, unique prime factorization was not another assumption. Ben have

you forgotten your pretty indirect proof of IP? Math proofs often are

slippery and slide away without notice. I can relay it if you like.

Sep 5, 1993, 10:55:03 AM9/5/93

to

In article <CCvKE...@dartvax.dartmouth.edu>

Ludwig.P...@dartmouth.edu (Ludwig Plutonium) writes:

Ludwig.P...@dartmouth.edu (Ludwig Plutonium) writes:

> The genuine indirect proof of IP has only one supposition and

> two contradictions.

Sorry, in my haste I made an error. The valid indirect proof of IP has

only one supposition and one contradiction, most invalid proofs have

two suppositions with two contradictions.

Sep 5, 1993, 7:26:48 PM9/5/93

to

In article <CCvyr...@dartvax.dartmouth.edu>

Ludwig.P...@dartmouth.edu (Ludwig Plutonium) writes:

Ludwig.P...@dartmouth.edu (Ludwig Plutonium) writes:

> Sorry, in my haste I made an error. The valid indirect proof of IP has

> only one supposition and one contradiction, most invalid proofs have

> two suppositions with two contradictions.

Sorry, I am now of the opinion that there two valid indirect proofs

of IP. Correct me if wrong. One is stronger than the other. This is not

uncommon in math to have to proofs, one a strong form and another a

weak proof form. I can cite that there is a strong mathematical

induction and a weak mathematical induction.

The strong form of indirect proof of IP proves that a pair of twin

primes is generated. The weak form of indirect proof of IP proves that

either a prime or a prime factor exists which was not in the original

list. The weak form of the indirect proof of IP uses the algorithm of

the direct proof method which increases the cardinality of any finite

set of primes.

I came to this opinion for I am tired of arguing over logical

conditionals. Forgive me if I sounded rude or insulting to anyone, my

intent was only to nail down IP and ITP.

My feelings are that through Usenet sci.math I am able to sharpen

my proposed proofs via this democratic assault on each one of my

proofs/assertions. Thanks due to readers and objectors.

Sep 3, 1993, 11:28:01 PM9/3/93

to

In article <CCsDt...@dartvax.dartmouth.edu>,

Ludwig Plutonium <Ludwig.P...@dartmouth.edu> wrote:

>In article <CCG5y...@ms.uky.edu>

>cyeo...@ms.uky.edu (Charles Yeomans) writes:

>

>> latter case, unique factorization

>> implies that any prime factor of p_1*...*p_n + 1 is not in the

>> aforementioned list.

>

> My whole contention with proofs of IP is summarized into this:

> 1) There is a direct proof of IP which increases the cardinality of

>any finite set of primes such as you have done here Charles. Only I

>would guess that you would call your version the indirect method.

>Usually all of the flawed proofs of IP mix both the direct and indirect

>within the body of the proof.

Ludwig Plutonium <Ludwig.P...@dartmouth.edu> wrote:

>In article <CCG5y...@ms.uky.edu>

>cyeo...@ms.uky.edu (Charles Yeomans) writes:

>

>> latter case, unique factorization

>> implies that any prime factor of p_1*...*p_n + 1 is not in the

>> aforementioned list.

>

> My whole contention with proofs of IP is summarized into this:

> 1) There is a direct proof of IP which increases the cardinality of

>any finite set of primes such as you have done here Charles. Only I

>would guess that you would call your version the indirect method.

>Usually all of the flawed proofs of IP mix both the direct and indirect

>within the body of the proof.

What is indirect about my argument? Given a list of primes, I show

how to generate a prime not on the list. It's positively

algorithmic.

>

> 2) There is an indirect proof of IP and in its STRONGEST FORM it

>yields twin primes. Most every indirect proof version of IP is flawed

>by holding TWO DEPENDENT CONDITIONALS simultaneously. So when the step

>asserting a contradiction is reached one cannot safely say which of the

>two conditionals (two suppositions) have you discharged. Have you in

>fact disposed of the first supposition that the set of all primes is

>finite or the second supposition--suppose this new number add 1 is

>composite.

>

What if it is composite? THen factor it into primes and use

unique factorization to conclude that all of its factors aren;t on

the given list of primes.

>> WHat's your point?

>

> The point of all of this is that I am asking as many people to have

>a relook at the standard gospel indirect proof of IP, in hopes that

>more people, not just Ben Tilly, can see the wrinkle in the gospel. It

>is this wrinkle which has left the proof of the Infinitude of Twin

>primes stranded on some forlorn and deserted math island.

I reckon I know as much number theory as the next guy, and in my

opinion your argument for the twin prime conjecture is is no

argument at all. No amount of agonzing over Euclid's argument

will get you a proof of that conjecture, I think.

Charles Yeomans

Sep 4, 1993, 7:20:14 PM9/4/93

to

In article <CCt8A...@ms.uky.edu>

cyeo...@ms.uky.edu (Charles Yeomans) writes:

cyeo...@ms.uky.edu (Charles Yeomans) writes:

> I reckon I know as much number theory as the next guy, and in my

> opinion your argument for the twin prime conjecture is is no

> argument at all. No amount of agonzing over Euclid's argument

> will get you a proof of that conjecture, I think.

>

> Charles Yeomans

Charles I ask you to please consider this:

In article <25ock4$7...@mathj.usc.edu>

br...@mathj.usc.edu (Ronald Bruck) writes:

> If W-1 or (mutatis mutandi) W+1 is not prime then it has a prime factor.

> But none of p1, p2, ..., pL is a divisor, and these are the only primes;

> contradiction; so W-1 and W+1 are prime.

Beautifully stated Ron.

Yours is even shorter than Ben Tilly's pretty proof and much shorter

than mine. This is the indirect proof method of IP. The direct proof

method for IP is this: given any finite set of primes, then multiply

them and by either adding 1, or subtracting 1 then either a new prime

or a new prime factor (this is where the prime factor comes into the

proof of the infinitude of primes, in the direct method, never the

indirect) is produced to augment the original finite set, increasing

the cardinality of the finite set. The direct proof method of IP is

analogous to the direct proof that the positive integers are infinite.

Simply add 1 to the largest member of any finite set of positive

integers producing a new number not in the original set, hence the

cardinality can always be increased.

Sep 5, 1993, 1:14:40 PM9/5/93

to

I was wondering what kind of responce I would get from LP. What I

find interesting is that he can take a logically correct proof

and dismiss it for a logically twisted proof. I quote:

find interesting is that he can take a logically correct proof

and dismiss it for a logically twisted proof. I quote:

(from Ronald Bruck) If W-1 or (mutatis mutandi) W+1 is

not prime then it has a prime factor. But none of p1

p2, ..., pL is a divisor, and these are THE ONLY PRIMES;

contradiction; so W-1 and W+1 are prime.

My problem with this proof is its absurd conclusion. Let's see if

I can reword my proof to meet some of LP's objections.

Theorem There are infinitely many primes.

Idea of proof: Show that if the list of primes is finite, Unique

factorization is contradicted in some manor. First, let me state

my understanding of unique factorization. UFT: every positive

integer is either a unit (1) or up to the order of its factors,

is uniquely expressible as a product of primes. I will require

only that every integer > 1 is a product of primes.

Proof: Suppose that there are finitely many primes, and that

p1, p2, ..., pL is a complete list. Let n = p1*p2...pL + 1.

n is an integer > 1. n is not one of p1, ..., pL nor can it

be expressed as a product of numbers p1, .., pL. Thus, n

is not expressible as a product of primes. By UFT, n IS

expressible as a product of primes, contradiction.

John

Sep 6, 1993, 11:12:35 AM9/6/93

to

In article <CCwMG...@dartvax.dartmouth.edu>

Ludwig.P...@dartmouth.edu (Ludwig Plutonium) writes:

Ludwig.P...@dartmouth.edu (Ludwig Plutonium) writes:

> My feelings are that through Usenet sci.math I am able to sharpen

> my proposed proofs via this democratic assault on each one of my

> proofs/assertions. Thanks due to readers and objectors.

Math proofs are beautiful for that in just a tiny strip of fact,

opens up an entire proof. My Strong form of the proof of IP, in which

two primes are generated W-1 and W+1 not on the original list, is

enough to yield the proof of ITP. (By the way, the tiny strip of fact

that 2+2=2x2=4, only two numbers in math with this feature is the

eventual proof of FLT, I just have to build the logical muscle around

this heart.)

My first posted proof of ITP is unacceptable to every reader. I

myself have not yet rejected it completely overboard. However, at the

moment I offer this line of attack.

In the proof of IP, as Ronald Bruck states it: In article

<25ock4$7...@mathj.usc.edu>

br...@mathj.usc.edu (Ronald Bruck) writes:

> If W-1 or (mutatis mutandi) W+1 is not prime then it has a prime factor.

> But none of p1, p2, ..., pL is a divisor, and these are the only primes;

> contradiction; so W-1 and W+1 are prime.

PROOF OF ITP: Take the Strong form of the proof of IP and with W-1

and W+1. Start a new augmented list over with p1, p2, ..., pL,W-1,W+1

as all the primes that exist. Go through the same Strong argument

producing another new twin primes, call them Z-1,Z+1. Augment a new

list and go through the Strong argument generating more twin primes, ad

infinitum. WhooooooLa!!!

Come on, argue me out of this one.

Sep 6, 1993, 6:48:43 PM9/6/93

to

In article <CCxu9...@dartvax.dartmouth.edu>

Ludwig.P...@dartmouth.edu (Ludwig Plutonium) writes:

Ludwig.P...@dartmouth.edu (Ludwig Plutonium) writes:

> PROOF OF ITP: Take the Strong form of the proof of IP and with W-1

> and W+1. Start a new augmented list over with p1, p2, ..., pL,W-1,W+1

> as all the primes that exist. Go through the same Strong argument

> producing another new twin primes, call them Z-1,Z+1. Augment a new

> list and go through the Strong argument generating more twin primes, ad

> infinitum. WhooooooLa!!!

>

> Come on, argue me out of this one.

I see this line of attack is evaporating or has already evaporated.

I have another line of attack which looks promising. I do not have a

coherent write-up for it but the lines of attack are similar to my

proof of the infinitude of perfect numbers/infinitude of regular

n-sided polygons.

If you had read those proofs, recall that I used the factorial

prime-free sequence. In the factorial prime-free sequence the only

prime candidates were N!-1 and N!+1. The Strong form of the indirect

proof of IP generated a twin primes W-1 and W+1. Perhaps there can be a

linkage here. That at the property of infinity the only prime number

candidates at all are N!-1 and N!+1.

Here is what I am thinking is a highly lucrative assault on ITP.

Consider the prime-free factorial sequence and find the first twin

primes N!-1 and N!+1 in it. Find the next twin primes to get a feel for

how far apart twin primes are in this prime-free factorial sequence.

Now comes the ugly part.

This is a take-home exercise. (Hints: perhaps IP will help or perhaps

some restrictions such as there has to exist at least one prime between

n and 2n using both positive and negative primes. I do not know at this

point.

Sep 6, 1993, 11:50:46 AM9/6/93

to

>

> Math proofs are beautiful for that in just a tiny strip of fact,

> opens up an entire proof. My Strong form of the proof of IP, in which

> two primes are generated W-1 and W+1 not on the original list, is

> enough to yield the proof of ITP. (By the way, the tiny strip of fact

> that 2+2=2x2=4, only two numbers in math with this feature is the

> eventual proof of FLT, I just have to build the logical muscle around

> this heart.)

That logical muscle is *very/ unlikely to come through. Let us let c be

a primitive 9'th root of 1. Consider the ring Z[c]. In it the only

solution to N+N+N=N*N*N is 0. But you have a nontrivial solution to FLT

because

c^3 + (c^2)^3 = (-1)^3. Because of that example I think that you are

going to have a *very* hard time showing that FLT follows for Z from

the fact that the only solution in Z to N+N+N=N*N*N is 0.

> My first posted proof of ITP is unacceptable to every reader. I

> myself have not yet rejected it completely overboard. However, at the

> moment I offer this line of attack.

>

> In the proof of IP, as Ronald Bruck states it: In article

> <25ock4$7...@mathj.usc.edu>

> br...@mathj.usc.edu (Ronald Bruck) writes:

>

> > If W-1 or (mutatis mutandi) W+1 is not prime then it has a prime factor.

> > But none of p1, p2, ..., pL is a divisor, and these are the only primes;

> > contradiction; so W-1 and W+1 are prime.

>

> PROOF OF ITP: Take the Strong form of the proof of IP and with W-1

> and W+1. Start a new augmented list over with p1, p2, ..., pL,W-1,W+1

> as all the primes that exist. Go through the same Strong argument

> producing another new twin primes, call them Z-1,Z+1. Augment a new

> list and go through the Strong argument generating more twin primes, ad

> infinitum. WhooooooLa!!!

>

> Come on, argue me out of this one.

Your strong argument explicitly needed the fact that your finite list

of primes contained *all* of the primes. I think that you should think

about what that means because the rest of us have, and it means that

you made a mistake. You ar just more obstinate about it than most.

Sep 6, 1993, 1:33:00 PM9/6/93

to

In article <CCxu9...@dartvax.dartmouth.edu>,

Ludwig Plutonium <Ludwig.P...@dartmouth.edu> wrote:

>In article <CCwMG...@dartvax.dartmouth.edu>

>Ludwig.P...@dartmouth.edu (Ludwig Plutonium) writes:

[some stuff which I have deleted]

Ludwig Plutonium <Ludwig.P...@dartmouth.edu> wrote:

>In article <CCwMG...@dartvax.dartmouth.edu>

>Ludwig.P...@dartmouth.edu (Ludwig Plutonium) writes:

>

> In the proof of IP, as Ronald Bruck states it: In article

><25ock4$7...@mathj.usc.edu>

>br...@mathj.usc.edu (Ronald Bruck) writes:

>

>> If W-1 or (mutatis mutandi) W+1 is not prime then it has a prime factor.

>> But none of p1, p2, ..., pL is a divisor, and these are the only primes;

>> contradiction; so W-1 and W+1 are prime.

>

> PROOF OF ITP: Take the Strong form of the proof of IP and with W-1

>and W+1. Start a new augmented list over with p1, p2, ..., pL,W-1,W+1

>as all the primes that exist. Go through the same Strong argument

>producing another new twin primes, call them Z-1,Z+1. Augment a new

>list and go through the Strong argument generating more twin primes, ad

>infinitum. WhooooooLa!!!

>

> Come on, argue me out of this one.

u appear to be quoting Ron Bruck out of context. As readers have

observed ad nauseum, the algorithm you claim does not

produce sets of twin primes. Indeed, it is not at all clear to

me that there are infinitely many primes of the form 2*p_1*...*p_n + 1.

Why don't you see if you can show that this is true before

claiming that you have a proof of the twin prime conjecture?

I do appear to have been mistaken in my belief that you understood Euclid's

proof of the infinitude of the primes. SO let's go over it again, shall we?

Let us start from the assumption that every nonempty set of natural

numbers has a least member. We then have the following

THEOREM. Let p and d be natural numbers. Then there exist natural

numbers q and r such that

a) p = q*d + r

b) 0 <= r < d

c) q and r are the unique numbers satisfying a) and b).

The proof of this follows immediately from the afore-mentioned assumption

about the natural numbers and I leave it as an exercise.

DEFINITION. Let a and b be natural numbers. One says that"a divides b"

if there exists a natural number q such that a = b*q + 0.

DEFINITION. A natural number U is called a *unit* if the equation

U*x = 1 has a solution in natural numbers..

One knows that 1 is the only unit inthe set of natural numbers.

DEFINITION. A natural number p is said to be prime if whenever

p divides a product a*b of natural numbers, then p divides a or

p divides b (or possibly both). In addition, p must not be a unit.

THEOREM. A natural number p is prime if it has no factors

other than itself and units.

PROOF. Suppose a divides p and suppose a is not a unit. We show that in

fact a must be p.

Since a divides p there is a number b such that p = a*b. Clearly, p

divides a*b and must therefore divide a or b. Suppose that p

divides a. Then there exists a number c such that a = pc, from

which we obtain a = abc. Divide both sides by a to get bc = 1, from

which one concludes that b is a unit and thus a must be p.

Let N be a natural number and suppose N is not a unit. Then either

N is prime or not. If N is not prime, then one can argue that

N may be written as a product of primes.

THEOREM. Every natural number other than 1 can be written as the product

of primes (from the set of natural numbers) is exactly

one way, up to rearrangement of the factors.

This is generally known as the unique factorization theorem.

I will also omit the proof of this theorem. It follows directly from

the definition of prime.

THEOREM. There are infinitely many primes.

PROOF. Let p_1, ..., p_n be a list of prime natural numbers. We show

that this is not a complete list.

Consider the number N = p_1*...*p_n + 1. It follows from the first

theorem that none of the p_i divide N. Clearly N is not a unit

and thus must be divisible by a prime. By the theorem of unique

factorization, it cannot be one of the primes p_1, ..., p_n. Thus we

conclude that there is a prime not on the list.

NOTE that one cannot conclude that p_1*...*p_n+1 is prime, and indeed

one knows that in many cases such numbers are not prime. Thus

I do not see how this can be used to attack the twin prime conjecture.

Charles Yeomans

Sep 7, 1993, 6:13:58 AM9/7/93

to

In article <CCy0r...@ms.uky.edu>

cyeo...@ms.uky.edu (Charles Yeomans) writes:

cyeo...@ms.uky.edu (Charles Yeomans) writes:

> THEOREM. There are infinitely many primes.

>

> PROOF. Let p_1, ..., p_n be a list of prime natural numbers. We show

> that this is not a complete list.

>

> Consider the number N = p_1*...*p_n + 1. It follows from the first

> theorem that none of the p_i divide N. Clearly N is not a unit

> and thus must be divisible by a prime. By the theorem of unique

> factorization, it cannot be one of the primes p_1, ..., p_n. Thus we

> conclude that there is a prime not on the list.

Thank you Charles, I buy this as another version of the Weak form of

the indirect proof method for IP. I myself like the Strong form better,

since it extracts all the possible fruits out of the argument. What I

mean is that we can arrive at the fact that W-1 and W+1 are prime and

not on the original list. Whereas you at most arrive at W+1 is prime or

a prime factor of W+1 (which you did not state but implied) exists

which is not in the original list.

Please Charles, I would like your opinion of perhaps a new math

technique under the topic. NEW FORM OF MATHEMATICAL INDUCTION??

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