(One statement of) Bertrand's postulate is that for each n ≥ 2 there is a prime p such that n < p < 2n. It was first proven by Pafnuty Chebyshev, but the gist of the following elementary but involved proof by contradiction is due to Paul Erdös. We denote the set of prime numbers with and define:
- n = 1:
- n = 2:
- n > 2 and n is even:
- and n is odd. Let n = 2m+1 with m > 0:
Now for the proof of Bertrand's postulate. Assume there is a counterexample: an integer n ≥ 2 such that there is no prime p with n < p < 2n.
If 2 ≤ n;; < 2048, then one of the prime numbers 3, 5, 7, 13, 23, 43, 83, 163, 317, 631, 1259 and 2503 (each being less than twice its predecessor), call it p, will satisfy n < p < 2n''. Therefore n ≥ 2048.
has no prime factors p such that:
- 2n < p, because 2n is the largest factor.
- , because of a trivial expansion of the original assumption.
- , because (since ) which gives us .
has at most one factor of every prime . As , the product of over all other primes is at most . Since is the product of over all primes p, we get:
Q.E.D.