Prime Numbers in Set S: Is 2*p>n?

In summary, it appears that for any set S:={1,...,n} where p is the largest prime in S, n is smaller than the next larger prime.
  • #1
atsw
1
0
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:
 
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  • #2
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.
 
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  • #3
atsw said:
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:

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.
 
  • #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
 
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  • #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?
 

Related to Prime Numbers in Set S: Is 2*p>n?

1. What is the definition of a prime number?

A prime number is a positive integer that is only divisible by 1 and itself. In other words, it has no other factors besides 1 and itself.

2. How are prime numbers important in mathematics?

Prime numbers play a crucial role in number theory and cryptography. They are also used in various algorithms and calculations in fields such as computer science and physics.

3. What is the significance of the number 2 in the equation 2*p>n?

The number 2 in the equation represents the concept of a twin prime pair. These are two prime numbers that differ by 2, such as 3 and 5. This concept is important in understanding the distribution of prime numbers.

4. How does the value of p affect the equation 2*p>n?

The value of p determines the specific prime number that is being multiplied by 2. This can affect the overall value of the equation and determine whether or not the statement is true.

5. What is the significance of the comparison between 2*p and n in the equation 2*p>n?

The comparison between 2*p and n is important in determining whether or not the statement is true. If 2*p is greater than n, then the statement is true and there are no prime numbers in set S that satisfy the equation. If 2*p is less than n, then the statement is false and there is at least one prime number in set S that satisfies the equation.

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