Prime numbers from (n) to (2n)

[limitkiller]

is there any proofs for:
"for any natural (n) there are prime numbers from n to 2n,including" ??

 

[Gib Z]

This is known as "Bertrand's Postulate", and a long but elementary proof is contained here: http://en.wikipedia.org/wiki/Proof_of_Bertrand's_postulate

 

[limitkiller]

thanks
but i had trouble understanding last part :"This gives us the contradiction:n < 468."
could anyone help me with that

 

[limitkiller]

help...no one answered for a long time...

 

[CellCoree]

I can confirm that there is indeed a proof for the statement "for any natural (n) there are prime numbers from n to 2n, including." This proof is known as the Bertrand's postulate, also known as the Bertrand-Chebyshev theorem. It states that for any integer n≥2, there exists at least one prime number between n and 2n.

The proof of this theorem was first given by Joseph Bertrand in 1845, and later improved by Pafnuty Chebyshev in 1852. The proof uses techniques from number theory and combinatorics, and relies on the properties of the prime numbers and their distribution.

One key aspect of the proof is the use of the prime number theorem, which states that the number of prime numbers less than or equal to a given number n is approximately equal to n/ln(n), where ln(n) is the natural logarithm of n. This theorem helps in estimating the number of primes between n and 2n, and shows that there must be at least one prime in this range.

Furthermore, the proof also uses a concept known as the Sieve of Eratosthenes, which is a method for finding prime numbers. This technique involves systematically eliminating composite numbers from a list of consecutive integers, leaving behind only the prime numbers.

In conclusion, the existence of prime numbers from n to 2n, including, is mathematically proven by the Bertrand's postulate. This result has numerous applications in mathematics and other fields, and is an important concept in number theory.