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Old 2007-03-22, 10:59   #1
Patrick123
 
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Default Goldbach's Conjecture

Hi all,

At the risk of getting the , would this constitute as a proof of the Goldbach conjecture??

Code:
It can be shown that all primes greater than 3 have the form 6n+1 or 6n-1 where n is any positive integer > 1.
It can be shown that any even integer of 10 and greater has one of three forms, namely 6n-2, 6n and 6n+2

The sum of two primes > 3 will have one of 3 forms.

(6a-1) + (6b-1) = 6(a+b) -2

(6a-1) + (6b+1) = 6(a+b) or (6a+1) + (6b-1) = 6(a+b)

(6a+1) + (6b+1) = 6(a+b) +2

Letting a and b be any number > 0, it can be shown that all even numbers > 10 are the sum of two primes.

It can be shown manually for the even numbers 4,6 and 8 as the sum of their primes include the primes 2 and 3.
Regards
Patrick
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Old 2007-03-22, 11:40   #2
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Patrick,

Although your "proof" deserves the "crank" label, I will give you the benefit of doubt and simply point out that you have made a typical error in logic.

By your first axiom, IF p is prime, then p is of the form ….
You then claim IF n is of the form …, then n is prime.

http://en.wikipedia.org/wiki/Affirming_the_consequent

Last fiddled with by Wacky on 2007-03-22 at 11:42
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Old 2007-03-22, 11:59   #3
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Thanks Wacky,

I had a block but just could not see where, at this stage all I've shown is that all even numbers can be split in the form (6a +or- 1) + (6b +or- 1). I've not shown that they are prime.

Self-inflicted

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Patrick
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Old 2007-04-13, 05:35   #4
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Default A little more on Goldbach

Greetings to Johannesburg.
I come from a family (one branch)of american coal miners,and happen
to have attended nursery school in saxonwald, before retreating to east london as my father was in the nylon business: I am unfortunately a heavy coffee drinker, but have managed to keep the yellowing hands down to a minimum .
I want to point out something I am convinced is a must for a proof of Goldbach, that is not generally accepted:
It must be proven by parts,
a) for those evens divisible by 2 once
b) for those evens divisible by 4 or more
This divides all evens into exactly two seperate types, which behave
totally differently, so as to try for a proof without seperation, is totally ludicrous in my opinion. I can go into more detail if you wish.
Its been a long time since I witnessed the mine dances, by the way.
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Old 2007-04-13, 17:01   #5
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Quote:
Originally Posted by Patrick123 View Post
Code:
It can be shown that all primes greater than 3 have the form 6n+1 or 6n-1 where n is any positive integer > 1.
So the primes greater than 3 are congruent to either +1 or -1 mod 6. Okay.

Quote:
Code:
It can be shown that any even integer of 10 and greater has one of three forms, namely 6n-2, 6n and 6n+2
So the even integers, 10 or greater, are congruent to either -2, 0, or +2 mod 6. Okay.

Quote:
Code:
The sum of two primes > 3 will have one of 3 forms.
 
(6a-1) + (6b-1) = 6(a+b) -2
 
(6a-1) + (6b+1) = 6(a+b) or (6a+1) + (6b-1) = 6(a+b)
 
(6a+1) + (6b+1) = 6(a+b) +2
So the sum of two primes, each greater than 3, is congruent to either -2, 0, or +2 mod 6. Okay.

Quote:
Code:
Letting a and b be any number > 0, it can be shown that all even numbers > 10 are the sum of two primes.
Not okay.

Just because two sets of numbers (the set of even integers each 10 or greater, and the set of sums of two primes each greater than 3, in this case) map to the same set of equivalence classes doesn't mean that the sets of numbers are equal ... or that they have a single member in common!

Now, by examining specific cases, you can show that the sets share at least one member (e.g., 10 is in both sets). But showing that the sets are equal (i.e., that the sets share all of their members) is an (*ahem*) infinitely greater task.

Last fiddled with by cheesehead on 2007-04-13 at 17:04
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Old 2010-07-16, 15:19   #6
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I'm going to add my 2 cents (and yes I know it won't necessarily be counted in the dollars given). number freak talks of something similar and then says "it remains unsolved as of this writing" (about Goldbach's conjecture) could we use a similar method to the one on pg. 124 (with one major alteration) to help Goldbach's conjecture ?
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Old 2010-07-16, 17:20   #7
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.5x±n= prime for a certain n and a even x is basically what Goldbach's conjecture says if i understand it right.
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Old 2010-07-16, 17:29   #8
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Goldbach's conjecture states that any even number that is greater than 2 can be decomposed into the sum of two prime numbers.

Ex: 100 = 53 + 47

In most cases, there is more than 1 decomposition into prime numbers.

Last fiddled with by 3.14159 on 2010-07-16 at 17:33
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Old 2010-07-16, 18:24   #9
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Quote:
Originally Posted by 3.14159 View Post
Goldbach's conjecture states that any even number that is greater than 2 can be decomposed into the sum of two prime numbers.

Ex: 100 = 53 + 47

In most cases, there is more than 1 decomposition into prime numbers.
I realize that but .5x±n= prime just means that they have to be the same distance away from the mean of all the 2 number sums to a number which is obvious.
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Old 2010-07-16, 23:11   #10
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0.5x ± n = p? If you're referring to the Goldbach conjecture, it would be: p1 + p2 = 2n.

If I didn't understand your statement, apologies.

Quote:
.5x±n= prime for a certain n and a even x is basically what Goldbach's conjecture says if i understand it right.
How did you deduce that?

Last fiddled with by 3.14159 on 2010-07-16 at 23:13
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Old 2010-07-17, 00:37   #11
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Quote:
Originally Posted by 3.14159 View Post
0.5x ± n = p? If you're referring to the Goldbach conjecture, it would be: p1 + p2 = 2n.

If I didn't understand your statement, apologies.
don't worry i'm used to nobody knowing what i say lol. even my family doesn't understand me some days.



Quote:
Originally Posted by 3.14159 View Post
How did you deduce that?
my equation is logical as any 2 numbers that add to a third must have and average of half that third number. this in turn means to average correctly to that .5x that they are equally displaced (hence the +/-n) . so in my mind my equation is what would have to be solved for all even x and a given n (which seems to be 0 or an odd number)

Last fiddled with by science_man_88 on 2010-07-17 at 00:38
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