*p>n? set prime


by atsw
Tags: 2p>n, p>n, prime
atsw
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#1
Sep19-04, 09:16 PM
P: 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
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mathwonk
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#2
Sep19-04, 09:52 PM
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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.
NateTG
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#3
Sep20-04, 12:21 AM
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Quote Quote by atsw
is it true that for any set S:={1,....,n}

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

Thanks in advance
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.

robert Ihnot
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#4
Sep20-04, 01:36 AM
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*p>n? set prime


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/messag...tml?1095166598
Ethereal
#5
Sep20-04, 02:51 AM
P: n/a
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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