Is there always a prime between $n$ and $2n$?

Question

if we are interested to seek for the numbers of primes between $1-100$ and $100-1000$ or 1000..., why we don't asked if there is a always a prime between $n$ and $2n$ mayeb this interesting question help us to predict the numbers of primes between $1-100$,or $100-1000 $ or $1000-..$?

note: $n$ is natural number $ > 1$

I would be interest for any replies or comments .Thank you

Answer 1

$$\begin{array}{l}\text{Chebyshev said it,}\cr \text{And I say it again,}\cr \text{There is always a prime}\cr \text{Between n and 2n.}\end{array}$$

Erdos had made his first significant contribution to number theory when he was 20, and discovered an elegant proof for the theorem which states that for each number greater than 1, there is always at least one prime number between it and its double. The Russian mathematician Chebyshev had proved this in the 19th century, but Erdos's proof was far neater. News of his success was passed around Hungarian mathematicians, accompanied by a rhyme: "Chebyshev said it, and I say it again/There is always a prime between n and 2n."

http://en.wikipedia.org/wiki/Bertrand%27s_postulate

Answer 2

Yes. If $n > 1$, then there is always at least one odd prime number $p$ satisfying $n < p < 2n$. In other words, $\pi(2n - 1) > \pi(n)$ for all $n > 1$, where $\pi(x)$ is the prime counting function. Look at Sloane's A060715 and you'll see the only 0 is for $n = 1$.

This is called Bertrand's postulate, but Chebyshev is the one who proved it first. Ramanujan and Erdős also came up with proofs. (By the way, just in case you're wondering, this does not prove the Goldbach conjecture.)