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

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

• See the first result: Google Commented Nov 16, 2014 at 19:27
• If you want $n$ to be a natural number and the prime to be strictly between, I can give you a very simple counterexample. Maybe you mean, $n \gt 1$.
– user137731
Commented Nov 16, 2014 at 19:33
• let us see ur counterexample Commented Nov 16, 2014 at 19:37
• user51189 - The counter example is n=1. There is no prime number strictly between 1 and 2. Your restriction to the naturals fails. You can easily fix this by restricting your conjecture to naturals greater 1.
– 123
Commented Nov 16, 2014 at 19:45
• @JpMcCarthy no need to use the double factorial there. Commented Nov 16, 2014 at 20:05

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

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

• thank you for this, i think that the same answer for this conjecture found in this book:ams.org\bookspages.acalc Commented Nov 16, 2014 at 20:35
• Is there always a prime between n and 3n/2, for n >= 4 ? Commented Nov 25, 2014 at 17:16

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.)

• thank you for your answer but i don't know why you wrote :one odd prime number which you wrote in the same time $p$ satisfying $n < p < 2n$ . I think all primes greater then 2 are odd . Commented Nov 16, 2014 at 21:18
• You're correct, all primes except 2 are odd. I wanted to emphasize the inequality is strict: $n < p < 2n$ rather than $n \leq p \leq 2n$. There are no odd primes between 1 and 2, and there is only one odd prime between 2 and 4. Commented Nov 17, 2014 at 0:40
• Oops, should've said "all primes in $\mathbb{Z}^+$ except 2 are odd." Some people around here go looking for the slightest inaccuracy to beat you up for. Commented Nov 17, 2014 at 0:52