Prove that if an integer n > 2, then there exist a prime number p such that n < p < n!
My proof.
Let p be a prime divisor of n!-1, then .
If p = n, then p|n!, but that is impossible since p|n!-1.
Now, how do I prove that p is not < n?
Prove that if an integer n > 2, then there exist a prime number p such that n < p < n!
My proof.
Let p be a prime divisor of n!-1, then .
If p = n, then p|n!, but that is impossible since p|n!-1.
Now, how do I prove that p is not < n?
Adapt Euclid's proof.
Suppose that for some there is no prime such that .
Now consider this is either a prime or composite the first of
these possibilities contradicts our assumption, so it must be composite.
Now let be the primes less than , then clearly .
But is composite, so it has a prime divisor, which therefore must be greater
than and less than , a contradiction.
So we conclude that it is not the case that that there are no primes such
that , that is for all there is a prime such that .
RonL