How can tell if there is a prime number between any two numbers? Is there a conjecture, statement, formula or anything that determines whether or not there is a prime between two numbers? I'm not looking for a computer algorithm. Any help would be appreciated.
 A: I highly doubt this is possible. Otherwise you could simply check any number of being prime, just by considering the number itself and the number +1 or something along these lines. But checking numbers being prime is hard...
A: Bertrand's postulate in loose terms says:
There is alway a prime between $n$ and $2n$
Once we know this we can iterate it by shrinking or enlarging $n$ we get statements like:

*

*There is always a prime between $\lfloor {2n\over 3}\rfloor$ and $\lceil{4n\over 3}\rceil$

*There is always a prime between $\lfloor {3n\over 4}\rfloor$ and $\lceil{3n\over 2}\rceil$

*$\ldots$
rounding is simply to make sure we land on integers. However there's overlap between our intervals, therefore we get :

*

*Either there's a prime in the insection of these intervals, with the original, or there are two primes in the complement to their intersection.

We can break this down to 5 intervals that don't intersect( or minimally) with implications on each other:

*

*There is always either a prime between $n$ and $\lceil{4n\over 3}\rceil$  or one between $\lfloor {2n\over 3}\rfloor$ and $\lceil{3n\over 4}\rceil$  or $\lfloor {4n\over 3}\rfloor$ and $\lceil{3n\over 2}\rceil$  otherwise there's an interval where you can double a number without a prime.


*There is always either a prime between $n$ and $\lceil{3n\over 2}\rceil$  or one between $\lceil{3n\over 2}\rceil$ and $2n$.  otherwise there's an interval where you can double a number without a prime.


*There is always either a prime between $\lfloor {3n\over 4}\rfloor$ and $\lceil{4n\over 3}\rceil$  or one between $\lfloor{2n\over 3}\rfloor$ and $\lceil {3n\over 4}\rceil$ .  Otherwise there's an interval where you can double a number without a prime.
etc.
numeric example
if n=8
then
6 to 12 and 8 to 16 each have a prime, but it could be the same prime in each range because they overlap in between 8 and 12. If this weren't true  ( 11 is prime, but lets say it isn't for a moment), we'd get that both of the ranges 6 to 8 and 12 to 16 would require primes for Bertrands postulate to hold. I used a pseudoexample here, because
up to n=100 000 it's never true that center range doesn't have a prime.
A: The best we can do is that for any $x \ge 89693$ there is always at least one prime between $x$ and $x + \dfrac{1}{\log^3 x}$. This is much stronger than Bertrand's postulate and was proved by Pierre Dusart.
