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Old 2018-02-25, 18:21   #1
Steve One
 
Feb 2018

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Default Twin Prime Conjecture Proof

TWIN PRIME CONJECTURE PROOF:

1. Unless 'twin prime' every twin prime possibilty, (11/13 41/43
71/73 etc) (17/19 47/49 77/79 etc)and(29/31 59/61 89/91)
has a lowest prime factor of 7 or above. Eg. 119/121 has
factors of 7 11 and 17, therefore it's lowest prime factor is 7.
Lowest prime factors are all that we use.

2. If twin primes end, then necessarilly, there exists a prime(n)
that closes the last possibility of a pair being twin prime:
By applying this prime in the series as 'lowest prime factor'
twin primes end, take it out, they go on. This would
necessarilly be so.

3. PRIME(1) = 7: PRIME(2) = 11 etc.

(Prime(1) - 2) / Prime(1)) × ((Prime(2) - 2)/ Prime(2)..........
... × (Prime(n) - 2) / Prime(n)...would have to equal zero if a
greatest prime number left zero gaps 'after' first position
placing as final lowest prime factor of a twin prime
possibilty.

5/7 × 9/11 × 11/13 × 15/17.....× (n - 2)/n) will never equal
zero. This is the number of 'gaps' (twin primes) left on
29/31 59/61 89/91 etc for example.

4. The equation has 'primes up to..' multiplied together,
dividing 'primes minus 2, up to..' multiplied together.
This cannot equal zero/leave zero gaps.
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Old 2018-02-25, 21:03   #2
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Quote:
Originally Posted by Steve One View Post
TWIN PRIME CONJECTURE PROOF:

1. Unless 'twin prime' every twin prime possibilty, (11/13 41/43
71/73 etc) (17/19 47/49 77/79 etc)and(29/31 59/61 89/91)
has a lowest prime factor of 7 or above. Eg. 119/121 has
factors of 7 11 and 17, therefore it's lowest prime factor is 7.
Lowest prime factors are all that we use.

2. If twin primes end, then necessarilly, there exists a prime(n)
that closes the last possibility of a pair being twin prime:
By applying this prime in the series as 'lowest prime factor'
twin primes end, take it out, they go on. This would
necessarilly be so.

3. PRIME(1) = 7: PRIME(2) = 11 etc.

(Prime(1) - 2) / Prime(1)) × ((Prime(2) - 2)/ Prime(2)..........
... × (Prime(n) - 2) / Prime(n)...would have to equal zero if a
greatest prime number left zero gaps 'after' first position
placing as final lowest prime factor of a twin prime
possibilty.

5/7 × 9/11 × 11/13 × 15/17.....× (n - 2)/n) will never equal
zero. This is the number of 'gaps' (twin primes) left on
29/31 59/61 89/91 etc for example.

4. The equation has 'primes up to..' multiplied together,
dividing 'primes minus 2, up to..' multiplied together.
This cannot equal zero/leave zero gaps.
Conveniently missed 23,25 as a "possibility". lowest prime factor ... 5.
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Old 2018-02-26, 04:30   #3
LaurV
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1. How about "9 and 11", "13 and 15", "25 and 27", etc...?
2. Oh, you don't like numbers ending in 5? How about "31 and 33", or "51 and 53", etc...?
3. When you say "lowest prime factors", keep in mind that this can be huuuuuuge... How about consecutive odd numbers ending in 1, 3, 7, 9, which are not divisible by 3, 5, 7, 11, 17, and not prime either? (such pair can be easily computed with modular calculus, even with a "lowest prime factor" of thousands of digits). That product you compute, is an infinite product, and it is always zero. As is the product for all primes p, of (p-2)/p which is a number under unit.
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Old 2018-02-26, 05:00   #4
axn
 
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Quote:
Originally Posted by Steve One View Post
1. Unless 'twin prime' every twin prime possibilty, (11/13 41/43
71/73 etc) (17/19 47/49 77/79 etc)and(29/31 59/61 89/91)
has a lowest prime factor of 7 or above. Eg. 119/121 has
factors of 7 11 and 17, therefore it's lowest prime factor is 7.
Lowest prime factors are all that we use.
Why do you start at 7? Hmmm... So you're considering only twin "possibilities" of the for 30n+12+/-1, 30n+18+/-1 and 30n+/-1, which, by construction would not have factors of 2,3 or 5. Cool. But you could've made it simpler and just stuck with 6n+/-1 form and used factors of 5,7,11,... But, whatever.

Quote:
Originally Posted by Steve One View Post
2. If twin primes end, then necessarilly, there exists a prime(n)
that closes the last possibility of a pair being twin prime:
By applying this prime in the series as 'lowest prime factor'
twin primes end, take it out, they go on. This would
necessarilly be so.
Whoa, there! Not so fast. Even if there are only finite number of twin primes, that doesn't mean that there are only a finite number of twin prime "possibilities". In fact, there is an infinite number of them in each series after the "last" twin prime. Therefore, there is no last prime(n) that "closes the last possibility". We have an infinite number of primes to choose from to close the infinite number of possibilities.
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Old 2018-02-26, 12:05   #5
Steve One
 
Feb 2018

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Default 2, 3 and 5 multiples

Quote:
Originally Posted by science_man_88 View Post
Conveniently missed 23,25 as a "possibility". lowest prime factor ... 5.
Multiples of 2, 3 and 5 are not included because they are instantly known as not possible. It serves no purpose to include them. I did not say anything was finite, only contained within the 3 paired series.

Last fiddled with by Steve One on 2018-02-26 at 12:40
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Old 2018-02-26, 12:30   #6
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Quote:
Originally Posted by Steve One View Post
Multiples of 2, 3 and 5 are not included because they are instantly known as not possible. It serves no purpose to include them.
Same could said for all primes under 1 million ...
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Old 2018-02-26, 12:41   #7
Steve One
 
Feb 2018

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Default Primes under 1 million

I don't actually know what you mean by that.
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Old 2018-02-26, 15:26   #8
CRGreathouse
 
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You're going to have to make your proof more formal before you discover that your mistake is that you've proved
Theorem: For any x, there are infinitely many pairs (p, p+2) such that neither p nor p+2 has a prime divisor <= x.
instead of
Conjecture: There are infinitely many pairs (p, p+2) such that neither p nor p+2 has a prime divisor <= sqrt(p+2).
This is the usual mistake to make when attempting a proof of the twin prime conjecture, so don't feel too bad.
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Old 2018-02-26, 16:46   #9
Steve One
 
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I don't see where l have even mentioned square roots. Nothing yet has been said to disprove what l wrote. Please show where specifically an error occurs.

Last fiddled with by Steve One on 2018-02-26 at 16:51 Reason: Missed out some writing
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Old 2018-02-26, 16:56   #10
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Quote:
Originally Posted by Steve One View Post
I don't see where l have even mentioned square roots. Nothing yet has been said to disprove what l wrote. Please show where specifically an error occurs.
The sqrt part is reference to the conjecture you say you've proved, not the one you have.
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Old 2018-02-26, 17:29   #11
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Quote:
Originally Posted by Steve One View Post
Nothing yet has been said to disprove what l wrote. Please show where specifically an error occurs.
Not worth my time, sorry. Maybe someone else will walk you through the mistake(s) in your proof.
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