Bertrand's Theorem
For all n, there exists a prime p, s.t. n<=p<2n
Proof:
Either: For all n, there exists m s.t. prime, p = n+m, with: 0<=m<n
Or: For all m, there exists n s.t. prime, p = m+n, with: 0<=n<m
Therefore
Either: n<=p<2n
Or: m<=p<2m
From which result follows
Copyright 1998 James Wanless