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*p>n? set prime

  1. Sep 19, 2004 #1
    is it true that for any set S:={1,....,n}

    2*p>n , where p is the largest prime in S?

    Thanks in advance :biggrin:
  2. jcsd
  3. Sep 19, 2004 #2


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    lets try it. {1} does not seem to work. {1,2} ok

    {1,2,3,4} ok,

    it looks likely from now on, but why don't we use the theorem behind "mills constant"? (I just heard of this today, on this site.)

    i.e. there is a constant K such that for any prime p, the next prime is closer than K p^(5/8), in particular it is closer than Kp.

    so once p gets larger than K, we have that the next prime is closer than p^2.

    hence the next prime is smaller than p+p^2 = p(1+p), which is a lot smaller than 2^p. so this says that in any sequence of integers, {1,2,....,p,....,n} where p is the alrgest prime, then n is smaller than the next larger prime hence n is smaller than 2^p. so this is certainly true eventually, i.e. for large n.

    but it is probably easy to prove it is always true. i just do not see it right now.
    Last edited: Sep 19, 2004
  4. Sep 20, 2004 #3


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    If you mean something like {1,2,....n} then it's true for n>1. What comes to mind immediately is IIRC Bertrand's Postulate which indicates that for any [tex]m \in \mathbb{N}[/tex], there is a prime [tex]p[/tex] with [tex]m \leq p \leq 2m[/tex] , and since [tex]2^{\frac{n}{2}}>n[/tex] for n sufficiently large.
  5. Sep 20, 2004 #4
    i think this problem might relate to the matter of whether there is always a prime between any positive integer X in the interval X^2 and (X+1)^2. Call such a prime, p. If so then 2p, which at the smallest would be 2*(X^2+1), and assumed the only prime in that interval, would be bounded by one less than the smallest prime q, being also assumed the largest prime, in the interval (X+1)^2, (X+2)^2, which would have size at the most of (X+1)^2 + 2X+2.

    This is going to be true when 2X^2+2 > (X+1)^2 +2X+2, or X^2>4X+1 or X>4.

    This may be a good start, but while this proposition seems obvious, it is not know that any such prime exists! This indicates just how difficult this problem might be.

    See http://nrich.maths.org/discus/messages/7601/19862.html?1095166598 [Broken]
    Last edited by a moderator: May 1, 2017
  6. Sep 20, 2004 #5
    While you're at it, how about generalising the conjecture, that between x^n and (x+1)^n, there would always be a prime number?
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