Google Groups no longer supports new Usenet posts or subscriptions. Historical content remains viewable.
Dismiss

# TWO PROOFS OF THE RIEMANN HYPOTHESIS

1,597 views

### Ludwig Plutonium

Aug 15, 1993, 5:52:54â€¯PM8/15/93
to
TWO PROOFS OF THE RIEMANN HYPOTHESIS (both published in The Dartmouth
9/8/Å»Å»51 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 Å» such that z(z)=Å» and
zÄ…1/2 +iy, which implies there is another Å» 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 yÄ…1, 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 MÄ…Å» where
(M+M+M) - (MXMXM) = Å». Then for exponent 1/4 there has to exist a
number MÄ…Å» where (M+M+M+M) -(MXMXMXM)=Å», 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 2Ä…Å», and (2+2) =(2X2),
and so (2+2) - (2X2)= Å». 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 Å»+Å» or multiplied Å»xÅ». 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)= Å» where z=x+iy
2(x+iy) - (x+iy)2 = Å»
2x +2iy - (x2 - y2 + 2ixy) = Å» gives a real component 2x- x2 + y2 = Å»
and an imaginary component 2iy - 2ixy = Å»
for imaginary component 2iy - 2ixy = Å», implies 2iy(1-x) = Å»
thus y=Å» or x=1
for real component 2x-x2 + y2 =Å» when y=Å» implies 2x-x2 =Å», (2-x)x = Å»
thus x=Å» or x=2
for x=1 implies 2x-x2+ y2 =Å», (2-1)+y2 = Å», y2= -1, thus y= + i,
substituting x=1 with y= + i, into z= x+iy gives z=1+ i2, thus z=Å» or
z=2. Therefore, only the numbers Å» and 2 satisfy (z+z)-(zxz)= Å».
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 Å»).
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, . . . 2,2,4,6,1Å»,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=rÅ»eEj , and so depending on where the point of origin for the spiral
is taken rÅ» determines k, and depending on the value of E, E determines
the e value for N, when E=Å» 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 Ä…. 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, 10:59:32â€¯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

### Ludwig Plutonium

Aug 16, 1993, 5:54:55â€¯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.

### Ludwig Plutonium

Aug 20, 1993, 10:06:55â€¯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.

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!

### Charles Yeomans

Aug 20, 1993, 11:16:32â€¯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

Probably, no one gives a shit.

Charles Yeomans

### Bruce Ikenaga

Aug 21, 1993, 9:58:20â€¯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, 11:38:49â€¯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]

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
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,
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

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.

### Ludwig Plutonium

Aug 25, 1993, 11:34:09â€¯AM8/25/93
to
In article <1993Aug21.0...@Princeton.EDU>

> 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, . .

### Ludwig Plutonium

Aug 26, 1993, 11:04:06â€¯AM8/26/93
to
In article <CC3AF...@ms.uky.edu>
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.

### Charles Yeomans

Aug 26, 1993, 10:15:36â€¯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

### Ludwig Plutonium

Aug 26, 1993, 11:24:41â€¯PM8/26/93
to
In article <CCEBM...@ms.uky.edu>
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.

### Charles Yeomans

Aug 27, 1993, 10:08:31â€¯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.

Charles Yeomans

### Ludwig Plutonium

Sep 3, 1993, 12:29:54â€¯PM9/3/93
to
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.

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.

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.

### Benjamin J. Tilly

Sep 3, 1993, 2:48:54â€¯PM9/3/93
to
In article <CCsDt...@dartvax.dartmouth.edu>
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.

>
> 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

### Ludwig Plutonium

Sep 5, 1993, 5:44:53â€¯AM9/5/93
to
In article <26ance...@frodo.d.umn.edu>
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"
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.

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
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
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.

### john greene

Sep 4, 1993, 2:39:10â€¯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.

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

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

### Ludwig Plutonium

Sep 3, 1993, 4:07:09â€¯PM9/3/93
to
In article <CCsK9...@dartvax.dartmouth.edu>

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.

### Ludwig Plutonium

Sep 5, 1993, 10:55:03â€¯AM9/5/93
to
In article <CCvKE...@dartvax.dartmouth.edu>
Ludwig.P...@dartmouth.edu (Ludwig Plutonium) writes:

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

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

### Ludwig Plutonium

Sep 5, 1993, 7:26:48â€¯PM9/5/93
to
In article <CCvyr...@dartvax.dartmouth.edu>
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.

### Charles Yeomans

Sep 3, 1993, 11:28:01â€¯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.

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.

>
> 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

### Ludwig Plutonium

Sep 4, 1993, 7:20:14â€¯PM9/4/93
to
In article <CCt8A...@ms.uky.edu>
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

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.

### john greene

Sep 5, 1993, 1:14:40â€¯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:

(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

### Ludwig Plutonium

Sep 6, 1993, 11:12:35â€¯AM9/6/93
to
In article <CCwMG...@dartvax.dartmouth.edu>
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.

### Ludwig Plutonium

Sep 6, 1993, 6:48:43â€¯PM9/6/93
to
In article <CCxu9...@dartvax.dartmouth.edu>
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.

### Benjamin J. Tilly

Sep 6, 1993, 11:50:46â€¯AM9/6/93
to
In article <CCxu9...@dartvax.dartmouth.edu>
Ludwig.P...@dartmouth.edu (Ludwig Plutonium) writes:

>
> 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.

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.

### Charles Yeomans

Sep 6, 1993, 1:33:00â€¯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]

>
> 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

### Ludwig Plutonium

Sep 7, 1993, 6:13:58â€¯AM9/7/93
to
In article <CCy0r...@ms.uky.edu>
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??

0 new messages