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matt grime

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6=2+3 does it?

What about the smallest natural greater than 2 that isn't a prime? isn't that a counter example too?

What about the smallest natural greater than 2 that isn't a prime? isn't that a counter example too?

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I forgot to mention that 1 and 2 can be used in the sum. I made a mistake. Not 6 = 2 + 3, 5 = 2 + 3. Once you define 1 and 2 you get 3. From 3+1 you get 4 and so on. 5=2+3....matt grime said:6=2+3 does it?

What about the smallest natural greater than 2 that isn't a prime? isn't that a counter example too?

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HallsofIvy

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So you are NOT talking about a "sum of primes" because 1 is not prime.

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matt grime

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For every n>1 there is a prime number p such that n<p<2n.

So, given m an integer, if it is even pick an prime in the range m/2 to m, and if it is odd pick a prime in the range (m+1)/2 to m+1, call this prime p(1).

Then m-p(1) < m/2, and we proceed by indcuction - the next prime we pick must be distinct from p(1) since it must be less than m/2, and p(1) is greater than m/2.

We just need to show how to write the sum for "small m" to get the full proof, ie m=1,2,3 which yo'uve done.

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Yeah?

So you are saying the Goldbach Conjecture has been proven?

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Icebreaker

What if we added up the first 300 002 primes?Schnirelman (1939) proved that every even number can be written as the sum of not more than 300 000 primes (Dunham 1990)

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Zurtex

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What is that supposed to mean :uhh: ?Icebreaker said:What if we added up the first 300 002 primes?

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Icebreaker

Nevermind.

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matt grime

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robert Ihnot said:Matt grime: the result follows quite easily from Russell's postulate.

Yeah?As re-expressed by Euler, an equivalent form of this conjecture (called the "strong" or "binary" Goldbach conjecture) asserts that all positive even integers can be expressed as the sum of two primes.http://mathworld.wolfram.com/GoldbachConjecture.html

So you are saying the Goldbach Conjecture has been proven?

Of course not, but the OP didn't ask for the sum of TWO primes, he asked for the sum to be expressed as *some* sum of N distinct primes, allowing for 1 to be included.

The possibly dodgy assumption I made was that it wasn't supposed to be a fixed N, which seems reasonable, given the other hidden assumptions in the post, after all it would be required that N=1 or 2 otherwise, which we know to be false and bloody hard respectively.

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Oh, that is correct! I just followed his examples, you know: 6=2+3 and 18 =13+5. But do you mean,

http://mathforum.org/library/drmath/view/51505.html

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matt grime

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Zurtex

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It has been proved there exists a prime for any natural number n > 2 there exists a prime (and I'm really dragging this one from memory) p such that:matt grime said:

[tex]n - n^{\frac{23}{42}} < p < n[/tex]

Which is quite a bit stronger . Although I'm not sure how useful.

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CRGreathouse

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http://functions.wolfram.com/NumberTheoryFunctions/Prime/29/0003/Zurtex said:It has been proved there exists a prime for any natural number n > 2 there exists a prime (and I'm really dragging this one from memory) p such that:

[tex]n - n^{\frac{23}{42}} < p < n[/tex]

Which is quite a bit stronger . Although I'm not sure how useful.

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