twin prime conjecture
larry....@gmail.com
My name is Larry Freeman and I am a math amateur. I have a draft of a
proposed proof of the twin primes conjecture which can be found here:
http://proposedproofs.blogspot.com/2006/10/proposed-proof-twin-prime-conjecture.html
I would greatly appreciate it if someone who is an expert in number
theory could take a look at it and explain to me where I went wrong.
This solution came too easily to me for it to be correct.
I've spent close to 2 weeks time on the proof and the argument seems
very solid to me. It is based on very elementary number theory. The
proof is only 9 pages.
Thanks very much!
-Larry
larry....@gmail.com
To make it easier for people to follow my argument, I have put together
an informal version of the proof that reduces the main argument to 1
page:
http://proposedproofs.blogspot.com/2006/10/twin-prime-conjecture-informal-proof.html
You can still find the complete proof here:
http://proposedproofs.blogspot.com/2006/10/proposed-proof-twin-prime-conjecture.html
-Larry
Gerry Myerson
"larry....@gmail.com" <larry....@gmail.com> wrote:
Before I spent a second looking at your write-up, you have to
convince me that you can actually do some number theory. Here
are 5 problems for you, problems that would not be out of place
in an intro number theory course. If you can't do them, there
isn't the slightest chance that you can do something as hard
as the twin prime conjecture.
Good luck. And please don't top-post.
1. Show that if integers x and y can be expressed as a sum of
two squares of integers, then so can xy.
2. Prove that the cube root of 36 is irrational. Prove that
the base-10 logarithm of 2 is irrational.
3. Find the smallest positive integer n such that 2 n is the
square of an integer, 3 n is the 3rd power of an integer, and
5 n is the 5th power of an integer.
4. The Fibonacci numbers are given by F_0 = 0, F_1 = 1, and
F_n = F_(n - 1) + F_(n - 2) for n = 2, 3, ... (so they go
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, ...). Let phi be the
number (1 + sqrt 5) / 2. Prove that
phi^(n - 1) < F_(n + 1) < phi^n if n > 1.
Prove that F_1 + F_2 + F_3 + ... + F_n = F_(n + 2) - 1.
5. Find, with proof, an irreducible polynomial that has
sqrt 2 + sqrt 3 as a root.
--
Gerry Myerson (ge...@maths.mq.edi.ai) (i -> u for email)
lui...@yahoo.com
> larry....@gmail.com wrote:
> I have a draft of a proposed proof of the twin primes conjecture which can be found here.
If x and y are "twin pairs", x must be of the form 6n-1, and y of the
form 6n+1.
Its true that y = 1 mod(3). If y =x+2 then x = -1 mod(3) , but
never x = 2 mod(3)
Ludovicus
larry....@gmail.com
By the way, if x = -1 (mod 3), then x = 2 (mod 3) since -1 = 2 (mod 3).
-Larry
Eric Schmidt
In step (1) of Lemma 4 you use a variable x without explaining what it
is. Can you please change this to make it clearer?
--
Eric Schmidt
--
Posted via a free Usenet account from http://www.teranews.com
Gene Ward Smith
larry....@gmail.com wrote:
> Hi Everyone,
>
> To make it easier for people to follow my argument, I have put together
> an informal version of the proof that reduces the main argument to 1
> page:
>
> http://proposedproofs.blogspot.com/2006/10/twin-prime-conjecture-informal-proof.html
This claims there is a twin prime pair between 6n+3 and 10n+5.
Presuming this is true for small values of n, it is likely to be true,
as the Hardy-Littlewood conjecture would ential that, on average, the
number of prime pairs in such intervals goes to infinity O(n/ln(n)^2).
schoenf...@gmail.com
Nice work.
PS: There is a misprint in the the title of corollary 10.1.
John Roberts-Jones
"larry....@gmail.com" <larry....@gmail.com> wrote:
>Hi Math Experts,
>
>My name is Larry Freeman and I am a math amateur. I have a draft of a
>proposed proof of the twin primes conjecture which can be found here:
>http://proposedproofs.blogspot.com/2006/10/proposed-proof-twin-prime-conjecture.html
>
>I would greatly appreciate it if someone who is an expert in number
>theory could take a look at it and explain to me where I went wrong.
>This solution came too easily to me for it to be correct.
>
Lemma 8 considers two sets {6i+5, 6(i+1)+5, ..., 6(i+P-1)+5} and
{6i+7, 6(i+1)+7, ..., 6(i+P-1)+7}, where P = p1 * p2 * ... * pn is
a product of n distinct primes. Since each set is a complete
residue system modulo P, each contains phi(P) terms coprime to P,
where phi is Euler's totient function. You correctly derive this
as phi(P) = P * (1-1/p1) * (1-1/p2) * ... * (1-1/pn).
But how do you then justify the assertion that there are at least
P * (1-2/p1) * (1-2/p2) * ... * (1-2/pn) cases where corresponding
terms of the two sets are both coprime to P?
In the simplest case, if P = pq is the product of just two primes,
then there are phi(pq) = (p-1)*(q-1) terms coprime to pq in each
set, so we know that corresponding terms are both coprime to pq in
at least 2 * phi(pq) - pq = pq - 2p - 2q + 2 cases. Why do you say
that there are at least pq * (1-2/p) * (1-2/q) = pq - 2p - 2q + 4?
--
John Roberts-Jones
larry....@gmail.com
Thanks for noticing that. x should be c. I just updated it.
-Larry
larry....@gmail.com
Thanks very much for your comment. I need to review the
Hardy-Littlewood conjecture to better understand it.
-Larry
larry....@gmail.com
-Larry
larry....@gmail.com
Thanks very much for your question. I will do my best to explain why
your reasoning does not apply to my formula. If you see a flaw in my
reasoning, then please feel free to call it out.
Let's examine the situation where there are 7*5 twin pairs starting
with i=1 (that is, 6i+3 = 9 so we are looking at the numbers 6i+3+2 and
6i+3+4 = { 11, 13, 17, 19, ..., 215, 217 } so we let p=5,q=7 and
P=pq=35.
This is a total of 35 twin pairs (11,13), (17,19), ..., (215,217) or 70
different numbers.
Your argument is that each of the two complete residue systems contain
phi(P) terms coprime to P so the total number of integers coprime to P
should be 2*phi(P).
That is correct but that is not what I am measuring. My goal is not to
count the *number of integers* coprime to P; my goal is to count the
*number of twin pairs* that are relatively prime to P.
It is my contention in Lemma 8 that the total number of twin pairs is
at least (1 - 2/p)(1 - 2/q).
The reason for this is the following:
[number of twin pairs relatively prime to P] = [total twin pairs] -
[twin pairs divided by p] - [twin pairs divided by q] + [twin pairs
divided by both]
[total twin pairs divided by p] = [total twin pairs]*2/p ; This is
established in Lemma 5
[total twin pairs divided by q] = [total twin pairs]*2/q ; This is
established in Lemma 5
[total twin pairs divided by both p and q] = [total twin
pairs]*(2/p)(2/q) = [total twin pairs]*(4/pq)
So that we have:
[number of twin pairs relatively prime to P] = [total twin pairs] -
[total twin pairs]*2/p - [total twin pairs]*2/q + [total twin
pairs]*4/pq = [total twin pairs]*[1 - 2/p][1 - 2/q]
Now, for each additional prime that we consider, we have:
[number of twin pairs relatively prime to p1,...,pn] = [total twin
pairs] - [total twin pairs]*[2/p1 + 2/p2 + ... + 2/pn] + [total twin
pairs]*[twin pairs divided simultaneous by 2] - [total twin
pairs]*[twin pairs divided simultaneously by 3] + ... [total twin
pairs]*(-1)^n*[total twin pairs divided simultaneously by all n primes]
This generalizes to [total twin
pairs]*[(p1-2)/p1][(p2-2)/p2]*...*[(pn-2)/pn]
I think the argument is best settled though by an empirical example.
Let's consider the first 35 twin pairs starting at 9. I will list them
all and put a * by any twin pair that is divisible by 5 or 7. I will
put a number by any pair that is not divisible by 5 or 7 (they may
still be divisible by some other prime, but this will show my formula
in the case of 2 primes).
By my equation, there should be at least 35*([(5-2)/5][(7-2)/7) =
35*(3/5)*(5/7) = 15 twin pairs that are not divisible by 5 or 7.
(11,13) [1]
(17,19) [2]
*(23,25)
(29,31) [3]
*(35,37)
(41,43) [4]
*(47,49)
*(53,55)
(59,61) [5]
*(65,67)
(71,73) [6]
*(77,79)
*(83,85)
*(89,91)
*(95,97)
(101,103) [7]
(107,109) [8]
*(113,115)
*(119,121)
*(125,127)
*(131,133)
(137,139) [9]
*(143,145)
(149,151) [10]
*(155,157)
*(161,163)
(167,169) [11]
*(173,175)
(179,181) [12]
*(185,187)
(191,193) [13]
(197,199) [14]
*(203,205)
(209,211) [15]
*(215,217)
Please let me know if I misunderstood your argument.
-Larry
larry....@gmail.com
I must admit that I am not certain that I have a proof so it may not be
worth your time to review it. If on the other hand, you would like to
help me find the flaw or help review it per chance that it is has any
validity, I will greatly appreciate it.
In the spirit of doing everything I can to encourage a number theorist
to help me review my proof, I will provide the answers to most of your
questions. I will try to give you the answers to the last questions
tomorrow or later this week. :-)
1. Show that if integers x and y can be expressed as a sum of two
squares of integers, then so can xy.
How's this:
(a^2 + b^2)(c^2 + d^2) =
= (a^2)(c^2) + (a^2)(d^2) + (b^2)(c^2) + (b^2)(d^2) =
= (a^2)(c^2) + 2abcd + (b^2)(d^2) + (a^2)(d^2) -2abcd +(b^2)(c^2) =
= (ac + bd)^2 + (ad - bc)^2
2. Prove that the cube root of 36 is irrational.
(a) Assume that that there exists x,y such that the cube root of 36 is
rational so that (x/y)^3 = 36
(b) We can assume that gcd(x,y)=1 since if they didn't we could divide
the common factor from (x/y). [All fractions can be reduced to a form
of gcd(x,y)=1 5/10 becomes 1/2 etc.]
(c) This gives us x^3 = 36(y^3)
(d) x is clearly even since 36 divides x^3.
(e) There exists x' such that x=2x' and we have (2x')^3 = 8x^3 =
36(y^3)
(f) We divide 4 from both sides to get: 2x^3 = 9(y^3) and we have a
contradiction since 2 must divide y which is impossible since
gcd(x,y)=1.
3. Prove that the base-10 logarithm of 2 is irrational.
Assume that log10(2) is rational for some positive integers m,n so that
log10(2)=m/n
It follows that 10^(m/n)=2
Raise each side ot the n power so that 2^n = 10^m
We now have 2^n = 2^m*5^m = 2^(n-m) = 5^m which is impossible.
4. Find the smallest positive integer n such that 2 n is the
square of an integer, 3 n is the 3rd power of an integer, and 5 n is
the 5th power of an integer.
This one was fun! Thanks very much since I had not seen it before. I
will skip some steps because the solution is more fun than explaining
how I came up with it.
The important insight is that there exists x,y,z such that:
(1/2)x^2 = (1/3)y^3 = (1/5)z^5 so that we have:
15x^2 = 10y^3 = 6z^5
The solution comes from realizing that we need to organize 2,5,3 so
that they meet the constraints. Lots of fun.
The solution that I came up with is:
z = 5^5*3^4*2^3
y = 5^8*3^7*2^5
x = 5^12*3^10*2^8
n = (1/2)x^2 = 5^24*3^20^2^15
The important point is that the number of powers of 5,4,2 must meet the
following constraints where (x',y',z' are the variables for x,y, and z)
Number of 5's = 2x'+1 = 3y'+1 = 5z'
Number of 3's = 2x'+1 = 3y' = 5z'+1
Number of 2's = 2x' = 3y'+1 = 5z'+1
5. The Fibonacci numbers are given by F_0 = 0, F_1 = 1, and
F_n = F_(n - 1) + F_(n - 2) for n = 2, 3, ... (so they go
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, ...). Let phi be the
number (1 + sqrt 5) / 2. Prove that
phi^(n - 1) < F_(n + 1) < phi^n if n > 1.
In other words, prove Binet's Formula. If you feel I am cheating here,
I am glad to rework it (but only if you really will review my proof and
send detailed feedback).
Here's a summary, I took from Dr. Math (I am always glad to promote
this wonderful web site!):
http://mathforum.org/library/drmath/view/52717.html
The Principle of Mathematical Induction states that if a certain
statement that depends on n is true for n = 0, and if its truth for
n = k implies its truth for n = k+1, then the statement is true for
all integers n >= 0.
There is an equivalent form, which appears superficially to be
different. It states that if a certain statement that depends on n is
true for n = 0, and if its truth for all n <= k implies its truth for
n = k+1, then the statement is true for all integers n >= 0.
To apply this Principle in either form is to prove the statement "by
induction."
Binet's formula is
F(n) = (a^n-b^n)/(a-b).
Here F(n) is the nth Fibonacci number, defined by
F(0) = 0,
F(1) = 1,
F(n) = F(n-1) + F(n-2), for all n >= 2.
The quantities a and b are defined by the formulas
a = (1+sqrt[5])/2,
b = (1-sqrt[5])/2 = -1/a.
They are the two roots of the quadratic equation x^2 - x - 1 = 0.
(Check this!)
To prove Binet's Formula, first show that it is true for n = 0 and
n = 1. Then for any k >= 1, assume it is true for all n <= k (in
particular for n = k and n = k-1), so that
F(k) = (a^k-b^k)/(a-b),
F(k-1) = (a^[k-1]-b^[k-1])/(a-b).
Now add these two equations together. The left-hand side becomes
F(k+1), according to the recursion defining the Fibonacci numbers.
Rearrange the right-hand side into the form
F(k+1) = (a^k+a^[k-1]-b^k-b^[k-1])/(a-b),
= (a^[k-1]*[a+1]-b^[k-1]*[b+1])/(a-b).
Now use the facts that a + 1 = a^2 and b + 1 = b^2, because a and b
are the roots of x^2 - x - 1 = 0. Then the above expression will
simplify into the form of Binet's Formula for n = k+1.
That establishes the hypotheses of the second form of the Principle of
Mathematical Induction. The conclusion of the Principle must therefore
hold, and Binet's Formula is true for all integers n >= 0.
-Larry
Phil Carmody
Equivalence to -1 modulo 3 is exactly the same as equivalent to
2 modulo 3. What is the distinction that you are trying to draw?
Phil
--
"Home taping is killing big business profits. We left this side blank
so you can help." -- Dead Kennedys, written upon the B-side of tapes of
/In God We Trust, Inc./.
Gene Ward Smith
larry....@gmail.com wrote:
> Hi Gene,
>
> Thanks very much for your comment. I need to review the
> Hardy-Littlewood conjecture to better understand it.
The conjecture is that the mean density of twin prime pairs (n-1, n+1)
is C2/ln(n)^2, where C2 is a numerical constant of approximately 2/3,
the Hardy-Littlewood constant.
jank...@hotmail.com
larry....@gmail.com wrote:
> Hi Math Experts,
>
> My name is Larry Freeman and I am a math amateur. I have a draft of a
> proposed proof of the twin primes conjecture which can be found here:
> http://proposedproofs.blogspot.com/2006/10/proposed-proof-twin-prime-conjecture.html
>
> I would greatly appreciate it if someone who is an expert in number
> theory could take a look at it and explain to me where I went wrong.
> This solution came too easily to me for it to be correct.
You are quite right about that. :-) The problem is the step from the
fact that the density of "pairs of odd integers relatively prime to 6
that are not divisible by any of the primes p1 through pn" is
(*) [1 - 2/p1][1 - 2/p2]*...*[1-2/pn]
(which appears to be correct) to the claim that the number of such
pairs among c consecutive pairs is floor(c*[1 - 2/p1][1 -
2/p2]*...*[1-2/pn]). It's just not that simple. I think you may have
been deceived by the use of a very crude lower bound for (*). It is
possible to give an asymptotically correct esimate of (*), and your
claim would then imply that the twin primes (or primes, with a modified
argument) are extremely evenly distributed. But that's not the case.
How about estimating the number of integers between 114 and 126 that
are coprime to 2, 3, 5, 7 and 11? Note that floor(13 *
(1-1/2)*(1-1/3)*(1-1/5)*(1-1/7)*(1-1/11)) = 2. But an integer less than
13^2 that is coprime to 2, 3, 5, 7 and 11 must be 1 or a prime, and
there are no primes in the interval.
---
J K Haugland
http://home.no.net/zamunda
Gerry Myerson
"larry....@gmail.com" <larry....@gmail.com> wrote:
Good work on 1, 2, 3, 4.
> 5. The Fibonacci numbers are given by F_0 = 0, F_1 = 1, and
> F_n = F_(n - 1) + F_(n - 2) for n = 2, 3, ... (so they go
> 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, ...). Let phi be the
> number (1 + sqrt 5) / 2. Prove that
> phi^(n - 1) < F_(n + 1) < phi^n if n > 1.
>
> In other words, prove Binet's Formula.
No. Well, I suppose you could first prove Binet, and then use it
to solve the problem, but I reckon you should be able to do the
problem without reference to Binet.
larry....@gmail.com
Thank you very much for your comment. I agree with you about the
mistake that is commonly made regarding the density of primes (just do
a Google search on the twin primes conjecture) but I don't believe that
your analysis applies to my argument.
If you take a look at Lemma 5 in my proof, for example, you will see
that the argument hinges on gcd(p,6)=1 and the argument is specific to
integers of the form 6(i+c) + 3 + h where h=2 or 4.
If you apply my argument to your example, you will see that my claim is
upheld.
I make the claim that you can apply my floor function only when you are
analyzing twin pairs (that is, numbers where x is congruent to 2 modulo
3 and y is congruent to 1 modulo 3 and y = x + 2).
Let's look at the twin pairs in the range 114 to 126 ("twin pair" is my
definition, I apologize for the nonstandard nature of it but I couldn't
think of a better term. Perhaps, "partial twin prime" since they are
relatively prime to 6 or "potential twin prime" would be better; I
rejected these other terms because I thought that they were too wordy)
The closest integer of the form is 111 so if we set i = 18, we get
6*18+3=111. I am also assume that the upper bound is of the form 6i+3
so I will use 123 which is 6*20+3.
So my analysis is regard to the number of twin pairs that are
relatively prime to 2,3,5,7,11 is equal to the total number of twin
pairs in the range multiplied by [(5-2)/5][(7-2)/7][(11-2)/11]
The total number of twin pairs (from Lemma 4) is: (123-111)/6 = 2.
So my formula for the minimum number of twin pairs relatively prime to
2,3,5,7,11 is:
floor(2*[3/5][5/7][9/11]) = 0
I apologize if I misunderstood your argument. Please let me know if
you have questions about any of my points. You might want to look at
my exchange with John Roberts-Jones for more details on the
justification for my formula:
http://groups.google.com/group/sci.math/browse_frm/thread/650388de35e91582/12cf790b87ab6022?lnk=raot#12cf790b87ab6022
Thank you very much for your feedback! If you have any suggestion for
how I can make it clearer about my assumptions regarding the prediction
of density of twin primes and I will be glad to update the proof. :-)
I suspect that many number theorists will assume that I am making a
probability argument when in fact I am not. There is no probability in
this argument. 1 argument is based on congruences modulo 6 (that is
where gcd(p,6)=1 comes in, see Lemma 5) and 1 argument is based on a
complete residue system (that is where the
floor(c*[p1-2/p1][p2-2/p2]*...*[pn-2/pn]) comes in (see Lemma 8).
Cheers,
-Larry
larry....@gmail.com
I will send answers to your problems 4 and 5 tomorrow.
Cheers,
-Larry
Proginoskes
larry....@gmail.com wrote:
> [...]
> If you take a look at Lemma 5 in my proof, for example, you will see
> that the argument hinges on gcd(p,6)=1 and the argument is specific to
In the corollary to Lemma 5 (the first nontrivial one, IMO), you state:
"Let x be an integer of the form 6i+3. Let p1,p2,...,pn be any primes
where gcd(p1,p2,...,pn,6)=1. Let c be the sequence of twin pairs
between 6i+3 and 6(i+c)+3. [...]"
I think "gcd(p1,p2,...,pn,6)=1" should be changed to "gcd(p_i, 6) = 1
for all i" (or
"gcd(p1 * p2 * ... * pn,6) = 1"), since
gcd(a(1),a(2),...,a(k)) is defined to be the largest number which
divides all of the a(i)'s. The condition
gcd(p1,p2,...,pn,6)=1 allows you to have p1 = 2 and p2 = 3. (In fact,
all you need are two non-equal primes for your gcd condition to be
true.)
This causes the proof of Corollary 5.1 to crash, at the point where you
have
6(a - b) = 0 (mod p1*p2*...*pn)
and want to cancel the 6. If you have either of the modified
conditions, the proof continues.
I don't know how this affects the rest of the proof; I haven't gotten
that far yet. (BTW, Lemma 6 is also "trivial".)
--- Christopher Heckman
Proginoskes
Proginoskes wrote:
> [...]
> I don't know how this affects the rest of the proof; I haven't gotten
> that far yet. (BTW, Lemma 6 is also "trivial".)
Typo: In Lemma 7, the list of numbers should be a a_1 + b, ..., a a_n
+ b. (The first
'+ b' is missing.)
In Lemma 8, do you need to change the gcd condition here as well?
Lemma 9 states: "Between an integer x of the form 18k+9 and 30k+15,
there is no twin pair that is divisible by a prime greater than 6k+3."
The first part does not parse correctly. Do you mean "Between integers
of the form 18k+9 and 30k+15, ..." instead?
--- Christopher Heckman
larry....@gmail.com
Thanks very much for your comment. I've updated the proof as you
suggested to make it more clear. I changed the corollary to Lemma 5 to
gcd(p_i,6)=1 for all i.
I agree with your judgment about the triviality of much of the lemmas
but I wanted to make sure that argument was as clear and as solid as
possible. I would prefer to err on the side of being too obvious
rather than being too cryptic.
Cheers,
-Larry
larry....@gmail.com
Thanks for all the comments! I've made the changes that you have
recommended! :-)
My answer is yes to both of your questions.
Cheers,
-Larry
larry....@gmail.com
I have revised Lemma 8 to make it more clear that I am not using
probabilities in my reasoning.
I hope that this resolves the issue that you raised. Please let me
know if it does not.
Regards,
-Larry
larry....@gmail.com
I have updated Lemma 8 to make my reasoning clearer. I hope that it
resolves the question that you raised. If it does not, please let me
know.
Regards,
-Larry
larry....@gmail.com
1. Prove that F_1 + F_2 + F_3 + ... + F_n = F_(n + 2) - 1
(a) F(4) - 1 = F(3) + F(2) - 1 = F(2) + F(1) + F(2) - 1 = F(2) + F(2)
= F(2) + F(1) + F(0) = F(2) + F(1)
(b) Assume that this is true up to n so that:
F(n) - 1 = F(n-2) + ... + F(1)
(c) Then F(n+1) - 1 = F(n) + F(n-1) - 1
(d) Applying (b) gives us:
F(n+1) - 1 = F(n-1) + F(n-2) + ... F(1)
(e) Using mathematical induction, we are done.
QED
-Larry
John Roberts-Jones
"larry....@gmail.com" <larry....@gmail.com> wrote:
>Hi John,
>
>Thanks very much for your question. I will do my best to explain why
>your reasoning does not apply to my formula. If you see a flaw in my
>reasoning, then please feel free to call it out.
>
>Let's examine the situation where there are 7*5 twin pairs starting
>with i=1 (that is, 6i+3 = 9 so we are looking at the numbers 6i+3+2 and
>6i+3+4 = { 11, 13, 17, 19, ..., 215, 217 } so we let p=5,q=7 and
>P=pq=35.
>
>This is a total of 35 twin pairs (11,13), (17,19), ..., (215,217) or 70
>different numbers.
>
>Your argument is that each of the two complete residue systems contain
>phi(P) terms coprime to P so the total number of integers coprime to P
>should be 2*phi(P).
It is unfortunate that you have chosen to delete what I said,
replacing it with an inaccurate paraphrase. Restoring what I said:
>>Lemma 8 considers two sets {6i+5, 6(i+1)+5, ..., 6(i+P-1)+5} and
>>{6i+7, 6(i+1)+7, ..., 6(i+P-1)+7}, where P = p1 * p2 * ... * pn is
>>a product of n distinct primes. Since each set is a complete
>>residue system modulo P, each contains phi(P) terms coprime to P,
>>where phi is Euler's totient function. You correctly derive this
>>as phi(P) = P * (1-1/p1) * (1-1/p2) * ... * (1-1/pn).
>>
>>But how do you then justify the assertion that there are at least
>>P * (1-2/p1) * (1-2/p2) * ... * (1-2/pn) cases where corresponding
>>terms of the two sets are both coprime to P?
>>
>>In the simplest case, if P = pq is the product of just two primes,
>>then there are phi(pq) = (p-1)*(q-1) terms coprime to pq in each
>>set, so we know that corresponding terms are both coprime to pq in
>>at least 2 * phi(pq) - pq = pq - 2p - 2q + 2 cases. Why do you say
>>that there are at least pq * (1-2/p) * (1-2/q) = pq - 2p - 2q + 4?
>
>That is correct but that is not what I am measuring. My goal is not to
>count the *number of integers* coprime to P; my goal is to count the
>*number of twin pairs* that are relatively prime to P.
I was indeed addressing the question of counting the number of
cases where both members of a twin pair are coprime to P.
>
>It is my contention in Lemma 8 that the total number of twin pairs is
>at least (1 - 2/p)(1 - 2/q).
My principal difficulty was that you appeared to pluck this out of
thin air, without any justification.
>
>The reason for this is the following:
>
>[number of twin pairs relatively prime to P] = [total twin pairs] -
>[twin pairs divided by p] - [twin pairs divided by q] + [twin pairs
>divided by both]
Agreed.
>
>[total twin pairs divided by p] = [total twin pairs]*2/p ; This is
>established in Lemma 5
Agreed. Since each set is a complete residue system modulo pq, it
contains exactly q terms divisible by p. Since it is not possible
for corresponding terms both to be divisible by p, there are
exactly 2q twin pairs with one member divisible by p.
>
>[total twin pairs divided by q] = [total twin pairs]*2/q ; This is
>established in Lemma 5
Agreed.
>
>[total twin pairs divided by both p and q] = [total twin
>pairs]*(2/p)(2/q) = [total twin pairs]*(4/pq)
So you assert that there are exactly four twin pairs with one
member divisible by p and one member divisible by q. This can
happen in two ways, by having one member divisible by pq, or by
having one member divisible by p and the other divisible by q.
Each set contains exactly one term divisible by pq, so there are
exactly two twin pairs with one member divisible by pq.
I see no justification for assuming that there are also two twin
pairs with one member divisible by p and the other divisible by q.
Note that I claim not that this is wrong, but that, even if it is
right, you have not given a proof.
>
>So that we have:
>
>[number of twin pairs relatively prime to P] = [total twin pairs] -
>[total twin pairs]*2/p - [total twin pairs]*2/q + [total twin
>pairs]*4/pq = [total twin pairs]*[1 - 2/p][1 - 2/q]
>
>Now, for each additional prime that we consider, we have:
>
>[number of twin pairs relatively prime to p1,...,pn] = [total twin
>pairs] - [total twin pairs]*[2/p1 + 2/p2 + ... + 2/pn] + [total twin
>pairs]*[twin pairs divided simultaneous by 2] - [total twin
>pairs]*[twin pairs divided simultaneously by 3] + ... [total twin
>pairs]*(-1)^n*[total twin pairs divided simultaneously by all n primes]
>
>This generalizes to [total twin
>pairs]*[(p1-2)/p1][(p2-2)/p2]*...*[(pn-2)/pn]
>
>I think the argument is best settled though by an empirical example.
No! You need to prove that your assertion obtains for every
possible example, not in just one case.
--
John Roberts-Jones
fernando revilla
> Fair enough. I was getting tired.
>
> I will send answers to your problems 4 and 5
> tomorrow.
>
> Cheers,
You should not be forgiven for thinking honestly that
you have the proof. In a scientific and professional
discussion the principle of authority is irrelevant.
For that reason my congratulations, to you for being
honest and also to those who are helping you without
considering a previous cultural luggage.
Best for you.
Fernando.
jank...@hotmail.com
larry....@gmail.com wrote:
(...)
> Let's look at the twin pairs in the range 114 to 126 ("twin pair" is my
> definition, I apologize for the nonstandard nature of it but I couldn't
> think of a better term. Perhaps, "partial twin prime" since they are
> relatively prime to 6 or "potential twin prime" would be better; I
> rejected these other terms because I thought that they were too wordy)
>
> The closest integer of the form is 111 so if we set i = 18, we get
> 6*18+3=111. I am also assume that the upper bound is of the form 6i+3
> so I will use 123 which is 6*20+3.
>
> So my analysis is regard to the number of twin pairs that are
> relatively prime to 2,3,5,7,11 is equal to the total number of twin
> pairs in the range multiplied by [(5-2)/5][(7-2)/7][(11-2)/11]
>
> The total number of twin pairs (from Lemma 4) is: (123-111)/6 = 2.
>
> So my formula for the minimum number of twin pairs relatively prime to
> 2,3,5,7,11 is:
>
> floor(2*[3/5][5/7][9/11]) = 0
I picked this interval, and primes instead of twin primes, in order to
to keep things simple, but let's do it your way. The number of "twin
pairs" between 662 and 808 is 24. The number of such pairs that are
relatively prime to 5, 7, 11, 13, 17, 19 and 23 is then
floor(24 * 3/5 * 5/7 * 9/11 * 11/13 * 15/17 * 17/19 * 21/23)
= floor(24 * 3/7 * 9/13 * 15/19 * 21/23) = 5.
Right? But there are no twin primes at all in this interval.
larry....@gmail.com
I am very sorry if I misrepresented your argument. I want you to know
that I thought it was very insightful and really helped me to clarify
my argument in Lemma 8. I have already rewritten it. There was never
an intention on my part to delete anything. I apologize if I made a
mistake in my cut-and-paste.
I hope that I can convey clearly how much I appreciate the time you
have put in to evaluate my idea.
If I understand your point, you are unclear on the justification for
the argument that there at least 4 instances of integers divisible by
both pq. Here's the passage as I understand it:
>[total twin pairs divided by both p and q] = [total twin
> >pairs]*(2/p)(2/q) = [total twin pairs]*(4/pq)
>
> So you assert that there are exactly four twin pairs with one
> member divisible by p and one member divisible by q. This can
> happen in two ways, by having one member divisible by pq, or by
> having one member divisible by p and the other divisible by q.
>
> Each set contains exactly one term divisible by pq, so there are
> exactly two twin pairs with one member divisible by pq.
>
> I see no justification for assuming that there are also two twin
> pairs with one member divisible by p and the other divisible by q.
>
> Note that I claim not that this is wrong, but that, even if it is
> right, you have not given a proof.
The more I think about it, the more I think that you are right that I
have not proved this point as much as asserted it with confidence.
Here is my reasoning for how I came up with that formula:
We have 2 complete residue systems at 6(i+c) + 3 + 2 and 6(i+c) + 3 + 4
where c can be any value from {0,1,..., pq-1}
>From this perspective, there are exactly 2*pq numbers that make up the
sum of these two sets of numbers.
Of these, there are exactly (1/p)*(2*pq) = 2*q that are divisible by p
and (1/q)*(2pq)=2p that are divisible by q. [See my Lemma 5 for
details if needed]
Now, here's where its gets tricky. Even though we are talking about 2
complete residue systems, we are talking about 2 complete residue
systems that are not completely independent. That is, certain numbers
in 6(i+c)+3+2 are really the same as numbers in 6(i+c)+3+4. That is,
they make up the same twin pair.
If I was only counting numbers in the two residue systems and we make
the assumption that they are completely independent, then the number of
integers divisible by both is (1/q)*(1/p)(2pq) = 2. So that there are
2/pq as you correctly point out.
But we need to view prevent double counting for the case where
6(i+a)+3+2 is divisible by 1 prime and simultaneously 6(i+a)+3+4 is
divisible by another prime. In this case, there are 2 ways for this to
occur. It can occur where p divides 6(i+a)+3+2 and q divides
6(i+a)+3+4 or it can occur when q divides 6(i+a)+3+2 and p divides
6(i+a)+3+4. How do we prevent this double counting?
For considering only 1 prime, we don't need to worry about double
counting since neither p or q can divide both 6(i+a)+3+2 and
6(i+a)+3+4.
So, the problem comes down to figuring out the number of twin pairs
divisible by p or q is (2pq)[1/p + 1/q] = (pq)[2/p + 2/q] that
involve double counting.
For considering the number of times that both pq divide 6(i+a)+3+h, the
answer is as you point out: (pq)*(1/p)(1/q) so that we have:
2*(pq)*(1/p)(1/q) = 2.
Now, we also have to consider the number of times that p divides
6(i+a)+3+h and q divides 6(i+a)+3+(4-h). There are two ways that this
can occur (h=2 and h=4) and the odds of this occuring are 2*[pq][1/pq]
= 2
It seems to me that this is the detail that is missing from my
argument. I can prove this ratio based on the Chinese Remainder
Theorem (that's how I did it originally but I didn't include it in my
proof). If this detail is provided, then the equation becomes:
[number of twin pairs] = (pq)(1 - [2/p + 2/q] + [4/pq])
If I provide this detail, will it answer your concern?
Thanks very much for all your input! :-)
-Larry
larry....@gmail.com
are right.
The question you pose is how many twin pairs in the range 662 and 808
are relatively prime to 5,7,11,13,17,19,23.
Well, first we need to convert 662 and 808 to 6i+3 form. 663 is the
closest 6i+3 (i= 110) to 662 and 807 is the closest 6i+3 (i=134) form
to 808.
The number of twin pairs is (807-663)/6 = 24
floor(24*[3/5]*[5/7]*[9/11]*[11/13]*[15/17]*[17/19]*[21/23]) =
floor(24*[3/7]*[9/13]*[15/19]*[21/23])
The minimum number of twin pairs (by my formula) is
floor(24*[3/7]*[9/13]*[15/19]*[21/23]) = 5.
This gives us the following twin pairs (I will star those divisible by
5,7,11,13,17,19, or 23)
*(665,667)
*(671,673)
*(677,679)
*(683,685)
*(689,691)
*(695,697)
*(701,703)
*(707,709)
*(713,715)
*(719,721)
*(725.727)
*(731,733)
*(737,739)
*(743,745)
*(749,751)
*(755,757)
*(761,763)
*(767,769)
*(773,775)
*(779,781)
*(785,787)
*(791,793)
*(797,799)
*(803,805)
Looks like you are right! Thanks very much for pointing this out! I
need to analyze this data in more detail. If I cannot come up with a
good answer, I will withdraw my proof.
Thank you so much for your time! This is exactly the response that I
was looking for! :-)
-Larry
larry....@gmail.com
I found the problem with your reasoning (and I guess, with mine too).
This is really a great example! :-)
My formula hinges on complete residue systems so that means it only
applies when p1*p2*..*pn is less than c.
Since we are looking at a number 24, 5*7*11*13*17*19*23 is clearly
significantly larger than 24. In order for my formula to apply to this
situation, c would have to equal 5*7*11*13*17*19*23. My formula should
only work for analyzing the number of twin pairs that are relatively
prime to any of these primes independently (since only in this case do
we have a complete residue system)
Doing a quick test, I see that it does since it predicts that there
are:
floor(24*[3/5]) = 14 twin pairs not divisible by 5.
(671,673) [1]
(677,679) [2]
(689,691) [3]
(701,703) [4]
(707,709) [5]
(719,721) [6]
(731,733) [7]
(737,739) [8]
(749,751) [9]
(761,763) [10]
(767,769) [11]
(779,781) [12]
(791,793) [13]
(797,799) [14]
Your argument is fair since it applies directly to the formula as
presented in Draft 1 of my proof.
So, to fix my proof, I need to provide conditions under which the
formula applies and then show that the formula applies in the case
where the range is 6i+3 and 10i+5 which, in all fairness, I didn't do.
It looks like to fix my proof, I need to quickly introduce draft 2.
:-)
Regards,
-Larry
jank...@hotmail.com
larry....@gmail.com wrote:
> Looks like you are right! Thanks very much for pointing this out! I
> need to analyze this data in more detail. If I cannot come up with a
> good answer, I will withdraw my proof.
>
> Thank you so much for your time! This is exactly the response that I
> was looking for! :-)
No problem. Glad to be able to help. ;-)
Aluminium Holocene Holodeck Zoroaster
"All the arguments can be made quantitative, and we know that
the first progression of primes of length k has entries less than
2^2^2^2^2^2^2^100k.... Two of the exponentials come
from number-theoretic pseudorandomness bounds on the almostprimes.
Four come from the best known bounds on Szemeridi's theorem
(from the work of Gowers). The last one comes from the structure
theorm.
"If the Riemann hypothesis is true, we can remve one
of the exponentials."
> Of these, there are exactly (1/p)*(2*pq) = 2*q that are divisible by p
> and (1/q)*(2pq)=2p that are divisible by q. [See my Lemma 5 for
> details if needed]
thus:
you get five golden sevenpointed stars for that;
which kinds of heptagram do you want?
7 is a Pierpont prime, anyway, so that
it is constructable indeed, without compasses,
straightedge or trisector (form of 2^r 3^s p_1 p_2 ... .-)
> the tomahawk.
> Pretty clearly that is what the Pawnee used.
thus:
did anyone see that article, quoted in the references,
below, in *American Scientist* from '98?
thus:
References
Fuller, R. B. (1961) "Tensegrity" Portfolio Artnews Annual 4: 112-127
Connelly, R., Back, A. (1998) "Mathematics and tensegrity" Am.
Scientist 86: 142-151
thus:
funny you should mention The Wheel; unfortunately,
for those of us interested in preinventing it,
_The Bicycle Wheel_ was taken offline & put back in print,
maybe because of our dyscussion of it,
on a Bucky Fuller listserver. anyway, you can see, why,
the Wright bros. were so savvy, knowing this little system,
which we have preliminarily classified in terms
of "tensegrity."...
as for top-posting,
ettiquette is less important than repetitive strain injury
from unneeded "scrolling" --
work it out on your life's palimpsest!...
(that is to say, if you can't follow implications
in a "thread" in written English, there's a class for you
at the local community college: EFL.)
thus:
oops, I'd put the wrong URL in my sig, because
of the "W05" in two of them (well,
the other article is good, two). I want to quote
from the PDF -- I will!...
I discovered this construction while contemplating
a fragment from Diophantus' lost work "On Polygonal
Numbers," which describes a proposition
by the classical Greek geometer Hypsicles:
"There has also been proved what was stated
by Hypsicles in a definition, namely, that 'if
there be as many numbers as we please beginning
from 1 and increasing by the same common difference ...'"
--it takes some to jitterbug!
http://members.tripod.com/~american_almanac
http://www.21stcenturysciencetech.com/2006_articles/Keplerian.W05.pdf
http://www.rwgrayprojects.com/synergetics/plates/figs/plate01.html
http://larouchepub.com/other/2006/3322_ethanol_no_science.html
http://www.wlym.com/pdf/iclc/howthenation.pdf
Gerry Myerson
"larry....@gmail.com" <larry....@gmail.com> wrote:
> OK, here's the solution to one more of your problems (2 more to go):
>
> 1. Prove that F_1 + F_2 + F_3 + ... + F_n = F_(n + 2) - 1
>
> (a) F(4) - 1 = F(3) + F(2) - 1 = F(2) + F(1) + F(2) - 1 = F(2) + F(2)
> = F(2) + F(1) + F(0) = F(2) + F(1)
>
> (b) Assume that this is true up to n so that:
>
> F(n) - 1 = F(n-2) + ... + F(1)
>
> (c) Then F(n+1) - 1 = F(n) + F(n-1) - 1
>
> (d) Applying (b) gives us:
>
> F(n+1) - 1 = F(n-1) + F(n-2) + ... F(1)
>
> (e) Using mathematical induction, we are done.
>
> QED
OK. But you can stop now - you've made your point
(unless of course you're enjoying this, in which case,
be my guest).
Phil Carmody
> In article <1160567009.3...@k70g2000cwa.googlegroups.com>,
> "larry....@gmail.com" <larry....@gmail.com> wrote:
>
> > OK, here's the solution to one more of your problems (2 more to go):
>
> (unless of course you're enjoying this, in which case,
> be my guest).
I suspect Larry was unfamiliar with the number of incompetant
doofi we get here in sci.math who make bold claims. Almost
all of which are unable to demonstrate _any_ ability in the
field in which they claim to have made a breakthrough.
Hence your challenge, which I'm sure you didn't expect any
sensible replies to, and you'd be right 99% of the time.
So Gerry - are you actually going to waste your time looking
at the paper, now that you've wasted Larry's time with those
questions? :-D
I've not looked at it, but I would be willing to bet that
it's "sieve method that doesn't consider error bounds". It
almost always is. There is a standard sieve methods text,
it's almost certainly way above my level:
H. Halberstam and H. E. Richert. (1974). Sieve Methods.
Academic Press, London, 1974. ISBN 0-12-318250-6
If anyone could recommend a text which aims a little lower,
introducing the subject to someone like myself, I'd be very
grateful.
Proginoskes
Phil Carmody wrote:
> Gerry Myerson <ge...@maths.mq.edi.ai.i2u4email> writes:
> > In article <1160567009.3...@k70g2000cwa.googlegroups.com>,
> > "larry....@gmail.com" <larry....@gmail.com> wrote:
> >
> > > OK, here's the solution to one more of your problems (2 more to go):
> >
> > OK. But you can stop now - you've made your point
> > (unless of course you're enjoying this, in which case,
> > be my guest).
>
> I suspect Larry was unfamiliar with the number of incompetant
> doofi we get here in sci.math who make bold claims. Almost
> all of which are unable to demonstrate _any_ ability in the
> field in which they claim to have made a breakthrough.
>
> Hence your challenge, which I'm sure you didn't expect any
> sensible replies to, and you'd be right 99% of the time.
>
> So Gerry - are you actually going to waste your time looking
> at the paper, now that you've wasted Larry's time with those
Hopefully not, since an error has already been found:
--- Christopher Heckman
larry....@gmail.com
I was enjoying it. :-)
Someone already found a flaw in my reasoning so there's no need to look
at the proof. I knew that it existed. I had been unable to find it so
I was hoping that someone would be able to help me find it.
The flaw does not appear to be fatal but it may be beyond my
mathematical ability to work past it.
The argument on twin primes depends on viewing twin primes as two
parallel but interacting complete residue systems and then using
assumptions about complete residue systems to establish that there must
exist at least 1 twin prime in the range 6i+3 and 10i+5 (I wrote a
computer program that seemed to confirm that this relation works for
i=1 .. 40M+ using data on primes from
http://primes.utm.edu/lists/small/millions/; testing on the program was
minimal but I tested both the positive condition and one negative
condition by removing some primes from my list).
I have also been corresponding with one of the top experts in twin
prime theory, Professor Dan Goldston at San Jose State University. Of
course, I didn't want to waste his time until I got some additional
eyeballs on a proof that none of my immediate friends or family could
understand.
Cheers,
-Larry
larry....@gmail.com
Thanks very much for the book suggestion. I will take a look.
I am also reviewing Dan Goldston et al's approach which looks so
promising that I suspect that he will resolve the twin prime conjecture
once and for all:
http://arxiv.org/PS_cache/math/pdf/0605/0605696.pdf
In my spare time, I am still working on the sieve objections that were
raised (see Professor Haugland's thread if you are interested). There
is one interesting note. The example that disproves my argument was
well chosen. If you add +6 to the counter example given to me, you
find a twin prime. :-)
But it follows my reasoning exactly so it was a great counterexample!
-Larry
jank...@hotmail.com
larry....@gmail.com skrev:
> In my spare time, I am still working on the sieve objections that were
> raised (see Professor Haugland's thread if you are interested). There
> is one interesting note. The example that disproves my argument was
> well chosen. If you add +6 to the counter example given to me, you
> find a twin prime. :-)
Not a coincidence; I wrote a program that looked up large gaps between
twin primes before posting!
Gerry Myerson
Phil Carmody <thefatphi...@yahoo.co.uk> wrote:
> Gerry Myerson <ge...@maths.mq.edi.ai.i2u4email> writes:
> > In article <1160567009.3...@k70g2000cwa.googlegroups.com>,
> > "larry....@gmail.com" <larry....@gmail.com> wrote:
> >
> > > OK, here's the solution to one more of your problems (2 more to go):
> >
> > OK. But you can stop now - you've made your point
> > (unless of course you're enjoying this, in which case,
> > be my guest).
>
> I suspect Larry was unfamiliar with the number of incompetant
> doofi we get here in sci.math who make bold claims. Almost
> all of which are unable to demonstrate _any_ ability in the
> field in which they claim to have made a breakthrough.
>
> Hence your challenge, which I'm sure you didn't expect any
> sensible replies to, and you'd be right 99% of the time.
>
> So Gerry - are you actually going to waste your time looking
> at the paper, now that you've wasted Larry's time with those
> questions? :-D
I would feel obliged to look at the proof, but someone else
spared me the effort.
> I've not looked at it, but I would be willing to bet that
> it's "sieve method that doesn't consider error bounds". It
> almost always is. There is a standard sieve methods text,
> it's almost certainly way above my level:
>
> H. Halberstam and H. E. Richert. (1974). Sieve Methods.
> Academic Press, London, 1974. ISBN 0-12-318250-6
>
> If anyone could recommend a text which aims a little lower,
> introducing the subject to someone like myself, I'd be very
> grateful.
There are a couple of chapters about sieves in Bateman & Diamond,
Analytic Number Theory, An Introductory Course, World Scientific,
2004. Might be worth a look.
Phil Carmody
I'll see if I can peek inside at some point. "Readership: Graduate
students" implies that it might be beyond my rusty brain. I think
it'll probably sit just behind Bach and Shallit in my queue.
Gerry Myerson
Phil Carmody <thefatphi...@yahoo.co.uk> wrote:
> Gerry Myerson <ge...@maths.mq.edi.ai.i2u4email> writes:
> > In article <87fydu6...@nonospaz.fatphil.org>,
> > Phil Carmody <thefatphi...@yahoo.co.uk> wrote:
> > > H. Halberstam and H. E. Richert. (1974). Sieve Methods.
> > > Academic Press, London, 1974. ISBN 0-12-318250-6
> > >
> > > If anyone could recommend a text which aims a little lower,
> > > introducing the subject to someone like myself, I'd be very
> > > grateful.
> >
> > There are a couple of chapters about sieves in Bateman & Diamond,
> > Analytic Number Theory, An Introductory Course, World Scientific,
> > 2004. Might be worth a look.
>
> I'll see if I can peek inside at some point. "Readership: Graduate
> students" implies that it might be beyond my rusty brain. I think
> it'll probably sit just behind Bach and Shallit in my queue.
I don't think you'll find anything about sieves (other than
Eratosthenes) for undergrads - but if you do, please let me know!
larry....@gmail.com
One of my favorite authors, M. Ram Murty, has written a book on sieves,
An Introduction to Sieve Methods and their Application. I've really
enjoyed his book on number theory (which is partially responsible for
my original posting). The description says that it's appropriate for
undergraduates.
Here's the link:
http://www.cambridge.org/catalogue/catalogue.asp?isbn=0521612756
It also has an e-book form.
Cheers,
-Larry
Phil Carmody
> Hi Gerry, Phil,
>
> One of my favorite authors, M. Ram Murty, has written a book on sieves,
> An Introduction to Sieve Methods and their Application. I've really
> enjoyed his book on number theory (which is partially responsible for
> my original posting). The description says that it's appropriate for
> undergraduates.
>
> Here's the link:
> http://www.cambridge.org/catalogue/catalogue.asp?isbn=0521612756
>
> It also has an e-book form.
That looks like it will fill a hole on my bookshelf very nicely.
Thanks for the recommendation.