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For example $8$ is in the middle of the interval between $5$ and $11$, $9$ is at equal distance between $7$ and $11$; $10$ between $7$ and $13$.

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3 Answers 3

If so, then every even number is a sum of two primes. But this is a notorious open problem, known as the Goldbach conjecture.

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For those to whom it's not obvious why: $n$ is exactly between $p$ and $q$ is equivalent to $2n = p+q$. And to be precise, the conjecture in the question is only equivalent to the conjecture that every even number that is not twice a prime is a sum of two primes, which is, strictly speaking, weaker than the Goldbach conjecture (though of course there's no reason to expect it to be easier or the answer to be different). –  ShreevatsaR Nov 16 '11 at 12:36
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Also, it may be worth adding that the Goldbach conjecture is believed to be true and has been verified up to very large numbers, which means (for the OP) that (it is believed that) every number is at equal distance between two prime numbers. –  ShreevatsaR Nov 16 '11 at 12:57
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@ShreevatsaR: It's not weaker than the Goldbach conjecture -- numbers that are twice a prime are trivially the sum of two primes... –  Charles Nov 16 '11 at 14:22
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@kojiro, 3 is equidistant from 3 and 3. –  Gerry Myerson Nov 17 '11 at 22:46
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@GerryMyerson that happens to me all the time. But to be sure, I think I'm right to infer that the numbers should be [low, high] relative to the original nonprime, lest the proof become trivial and meaningless: 4 is equidistant from 3 and 3; 6 from 5 and 5; 8 from 7 and 7, etc… "Every nonprime int is equidistant from a prime and that same prime." –  kojiro Nov 18 '11 at 13:09
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1 is a positive nonprime number not between any prime numbers at all. If you consider that cheating (I wouldn't know why), then see Gerry Myerson's anwer.

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Surely 1 is a prime number? It's only divisible by 1 and itself (1) –  LordScree Nov 16 '11 at 14:25
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1 isn't considered prime because, otherwise, various theorems would be inelagently phrased ("all primes except 1"). –  David Mitra Nov 16 '11 at 14:52
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@LordScree: No, $1$ is not prime, it is a unit. You could see math.stackexchange.com/questions/120/is-1-a-prime-number –  Ross Millikan Nov 16 '11 at 14:54
    
OK, thanks for clarifying for me :) –  LordScree Nov 16 '11 at 17:12
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@AAA Consider posing questions or providing suitable answers to posted questions in order to gain the 50 reputation points required to begin commenting. (And note that to be perfect a positive integer must equal the sum of all its proper positive factors, not just the prime factors; e.g., $28 = 1 + 2 + 4 + 7 + 14$ is perfect.) –  Arthur Fischer Jul 28 '12 at 13:43
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Check out a related theory: 'Green-Tao Theorem' which is a special case of Erdős conjecture and 'Primes in arithmetic progression' - in short, the primes contain arbitrarily long arithmetic progressions.

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Copy + paste error, thanks for noticing my mistake. Re-edit time. –  Alvin K. Nov 17 '11 at 4:28
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protected by t.b. Jul 28 '12 at 13:30

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