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January 1st, 2017, 06:09 PM   #11
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Quote:
Originally Posted by MrAwojobi View Post
So why do you have a problem with the fact that I have made my proof accessible to college kids by stating 5 lines that you have compressed into 1 line. At least you didn't have a problem with the mathematics. Make comments about the mathematics instead of criticizing the presentation.
This is entirely wrong. Conciseness is the result of careful expression. That which is unnecessarily prolix invites confusion.

Perhaps I am incorrect that all you intended by your first five lines is

"All primes > 3 can be expressed in the form $6n \pm 1$" and

"Consider the sequence of all primes > 3 in ascending order."

If I am incorrect, what did you mean?

If I am correct, however, then there is no question that 2117 and 2119 are not in that sequence because they are not prime. So skipjack's example is not on point. But who is at fault for making obscure what could be clear? You are. And as skipjack says, your saying that something is clear does not make it so. It is the writer's obligation to present things as clearly as possible. You have failed to do so as demonstrated by the responses you have received.

Last edited by skipjack; October 31st, 2018 at 03:37 AM.
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January 1st, 2017, 10:33 PM   #12
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Quote:
Originally Posted by MrAwojobi View Post
In the sieve described here, it should be clear that the number of multiples of 5 eliminated in the number sequence > the number of non-multiples of 5 eliminated for any N.
Things that are "clear" in math often have a bad habit of being wrong. In this case, what you are claiming is indeed wrong. The prime number theorem gives asymptotic estimates for the number of composite numbers less than $N$ explicitly. This is provably $N - \pi(N) \approx N - \frac{N}{\log N}$. You may notice that as a ratio of $N$, this approaches 1.

On the other hand, the number of multiples of 5 less than $N$ is approximately $\frac{N}{5}$ which you may notice as a ratio of $N$ is bounded away from 1.

In fact, the density of multiples of 5 to non-multiples of 5 is 0 which you might recognize contradicts your assertion that it is "clear" that the multiples of 5 is greater than non-multiples of 5.

Quote:
Originally Posted by MrAwojobi View Post
Also the minimum difference between any 2 multiples of 5 is 10 and there will be a pair of numbers, 6n-1 and 6n+1, in between these 2 multiples of 5.
I hope it is clear why this statement is false given the flaw in the previous statement.
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January 2nd, 2017, 05:35 AM   #13
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Quote:
Originally Posted by MrAwojobi View Post
. . . the minimum difference between any 2 multiples of 5 is 10 and there will be a pair of numbers, 6n-1 and 6n+1, in between these 2 multiples of 5.
Yes, but such a pair needn't be prime. If the odd multiples of 5 are 2115 and 2125, the only such pair lying between them is (2117, 2119), and 2117 and 2119 are both composite.

Twin primes greater than (3, 5) must have the form (6n - 1, 6n + 1). From all such number pairs, MrAwojobi considers sieving out those which aren't prime pairs by removing (composite) multiples of successive odd primes. As there are infinitely many primes, there are infinitely many pairs of the form (6n - 1, 6n + 1) for which at least one of these numbers is prime. It seems highly plausible that the other number will sometimes be prime as well, even if one uses only very large values of n, but this is currently unproved. The sieving method is a very simple way of finding primes, and can in principle be used to find all primes. Using this method to find twin primes will in principle find all such pairs, but examining this process doesn't seem to provide any proof that there are infinitely many of them. The attempted proof by MrAwojobi has flaws, as has been pointed out.
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January 2nd, 2017, 06:56 AM   #14
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I have noticed an error concerning a claim I made in the proof. I will reword that part of the proof and post an update.
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January 4th, 2017, 03:31 AM   #15
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I think 2117 and 2119 are not in the final sequence because they have already been eliminated by 29 and 13 sieving.

The way I understand his post is ... he list's the sequence of positive integers of the form $6n \pm 1$ ... then he applies the sieve (like Eratosthenes) on this list ... then he examines the numbers that remain and draws his conclusions on those remaining numbers.

For example 35 gets eliminated by 5 so is NOT there to be eliminated by 7 and since 35 is gone by the time we get around to eliminating multiples of 7 , it does not add to the count of numbers eliminated by 7. So his claim ...

"it should be clear that the number of multiples of 5 eliminated in the number sequence > the number of non-multiples of 5 eliminated for any N."

seems true to me

However , I don't think the final claim of his post follows logically IMHO

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January 4th, 2017, 07:01 AM   #16
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Originally Posted by agentredlum View Post

For example 35 gets eliminated by 5 so is NOT there to be eliminated by 7 and since 35 is gone by the time we get around to eliminating multiples of 7 , it does not add to the count of numbers eliminated by 7. So his claim ...

"it should be clear that the number of multiples of 5 eliminated in the number sequence > the number of non-multiples of 5 eliminated for any N."

seems true to me
"Seems true to me" has a tad less force than QED.

Let's try to figure out what is being asserted. First note that N and n are different variables: n is any natural number whereas N is a natural number such that $N = 6k \pm 1,\ k \in \mathbb N^+.$

Anyway, the assertion is (assuming that I am reading this prolix stuff correctly) that in the finite sequence of numbers of the form $6n\pm 1$ from 5 to N, the number that are evenly divisible by 5 exceeds the number that are not evenly divisible by 5 but are evenly divisible by EVERY other odd prime < N.

I suspect it is fairly easy to prove formally that the number of such numbers divisible by 5 exceeds the number of such numbers evenly divisible by 7 but not 5 (though the OP has not bothered to do so). It does not, however, obviously follow from that that the number of numbers in the form of $6n \pm 1$ evenly divisible by 5 exceeds the number of such numbers evenly divisible by 7 but by no lower prime plus the number of such numbers evenly divisible by 11 but by no lower prime plus the number of such numbers evenly divisible by 13 but by no lower prime, etc. However intuitive it may or may not be (and it is not intuitive to me), an appeal to intuition is not a proof.

Last edited by JeffM1; January 4th, 2017 at 07:03 AM.
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October 30th, 2018, 01:56 PM   #17
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Quote:
Originally Posted by MrAwojobi View Post
TWIN PRIME CONJECTURE PROOF
Primes > 3 and their odd multiples that are not multiples of 3 are of the form 6n ± 1, where n is a positive integer. Consider a sieving process operating on numbers of the form 6n ± 1 that are arranged in ascending order i.e. 5,7,11,13,17,19,23,25,29,31,35,37, … In this sieving process, a number will be eliminated only once. So 5 eliminates its multiples ≠ 5 in ascending order i.e. 5x5, 5x7, 5x11, 5x13, 5x17, 5x19, 5x23, 5x5x5, 5x29, 5x31, 5x5x7, 5x37, 5x41, 5x43, 5x47, 5x7x7, 5x53, 5x5x11, … and then 7 eliminates its multiples ≠ 7 that are not multiples of 5 i.e. 7x7, 7x11, 7x13, ... and then 11 eliminates its multiples ≠ 11 that are not multiples of 5 and 7 starting from 11x11 and so on. So the numbers that will not be eliminated in this process are the prime numbers > 3. Consider a finite part of the infinite sequence described above. Let the last number in this sequence be N, where N is some multiple of 5 of the form 6n ± 1. In the sieve described here, it should be clear that the number of multiples of 5 eliminated in the number sequence > the number of non-multiples of 5 eliminated for any N. Also the minimum difference between any 2 multiples of 5 is 10 and there will be a pair of numbers, 6n-1 and 6n+1, in between these 2 multiples of 5. For instance, 5x23 =115 and 5x5x5 =125. 119 and 121 are of the form 6n-1 and 6n+1 respectively and in between 115 and 125. Larger differences mean more than 1 pair of 6n ± 1 e.g. consecutive multiples 5x7=35 and 5x11=55 in the sequence above have a difference of 20 between them and will have 41, 43 (of the form 6n-1, 6n+1 respectively) and 47, 49 (of the form 6n-1, 6n+1 respectively) in between them. This means that, for any N > 5x13=65 (see sequence above), the number of multiples of 5 of the form 6n ± 1 < the number of pairs of 6n ± 1. This means that the non-multiples of 5 of the form 6n ± 1 eliminated < the number of pairs of 6n ± 1. This therefore means that there will always be pairs of 6n ± 1, the twin primes, that will not be eliminated. The twin prime conjecture is true i.e. there are infinitely many twin primes.
Up to 2003 there seem to be more prime pairs of form 6n+/-5, so are there other larger primes such as 6n+/- 50?
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October 30th, 2018, 05:43 PM   #18
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I'm quite hesitant to make any comment here since most people around here are cranks and the first (and only) move a crank will make when presented with objections to their proof is to drag the entire conversation into a never ending tragedy of semantics and bad math.

In case you are the rare student who just happened to not notice your error, I'll point it out.

Quote:
Originally Posted by MrAwojobi View Post
In the sieve described here, it should be clear that the number of multiples of 5 eliminated in the number sequence > the number of non-multiples of 5 eliminated for any N.
While it is true that more composites are sieved out in the "5 step" than any larger prime, the remainder of your proof relies on more composites being sieved out in this step than every later step COMBINED which is certainly not true.

For example, there are 74 composite numbers less than or equal to 100 but only 13 of these are sieved in the 5 step (6 were already removed in the 3 step).
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October 30th, 2018, 06:37 PM   #19
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Originally Posted by Joppy View Post
Just a pet peeve of mine, regardless of the content's validity. In this case, it would aid in the reader's ability to easily correct any mistakes made by OP. Unless that was intentional.
Not if the OP didn't make any misteaks!

Last edited by skipjack; October 31st, 2018 at 03:59 AM.
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October 30th, 2018, 06:40 PM   #20
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Originally Posted by Maschke View Post
Not if the OP didn't make any misteaks!
Mmm, steak.
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