Proof of Bertrand's postulate
In mathematics, Bertrand's postulate (actually a theorem) states that for each there is a prime such that . It was first proven by Pafnuty Chebyshev, and a short but advanced proof was given by Srinivasa Ramanujan. The gist of the following elementary proof is due to Paul Erdős. The basic idea of the proof is to show that a certain central binomial coefficient needs to have a prime factor within the desired interval in order to be large enough. This is made possible by a careful analysis of the prime factorization of central binomial coefficients.
- 1 Lemmas and computation
- 2 Proof of Bertrand's Postulate
- 3 References
The main steps of the proof are as follows. First, one shows that every prime power factor that enters into the prime decomposition of the central binomial coefficient is at most . In particular, every prime larger than can enter at most once into this decomposition; that is, its exponent is at most one. The next step is to prove that has no prime factors at all in the gap interval . As a consequence of these two bounds, the contribution to the size of coming from all the prime factors that are at most grows asymptotically as for some . Since the asymptotic growth of the central binomial coefficient is at least , one concludes that for large enough the binomial coefficient must have another prime factor, which can only lie between and . Indeed, making these estimates quantitative, one obtains that this argument is valid for all . The remaining smaller values of are easily settled by direct inspection, completing the proof of the Bertrand's postulate.
Lemmas and computation
Lemma 1: A lower bound on the central binomial coefficients
Lemma: For any integer , we have
Proof: Applying the binomial theorem,
since is the largest term in the sum in the right-hand side, and the sum has terms (including the initial two outside the summation).
Lemma 2: An upper bound on prime powers dividing central binomial coefficients
For a fixed prime , define to be the largest natural number such that divides .
Lemma: For any prime , .
Proof: The exponent of in is (see Factorial#Number theory):
But each term of the last summation can either be zero (if ) or 1 (if ) and all terms with are zero. Therefore
This completes the proof of the lemma.
Lemma 3: The exact power of a large prime in a central binomial coefficient
Lemma: If is odd and , then
Proof: The factors of in the numerator come from the terms and , and in the denominator from two factors of . These cancel since is odd.
Lemma 4: An upper bound on the primorial
We estimate the primorial function,
where the product is taken over all prime numbers less than or equal to the real number .
Lemma: For all real numbers ,
Proof: Considering it is sufficient to prove the lemma for natural numbers . The proof is by mathematical induction.
- is even:
- is odd. Let , then by binomial theorem:
- Each prime p with divides , giving us:
- By induction for :
Thus the lemma is proven.
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 < 468, then p can be chosen from among the prime numbers 3, 5, 7, 13, 23, 43, 83, 163, 317, 631 (each being less than twice its predecessor) such that n < p < 2n. Therefore n ≥ 468.
There are no prime factors p of such that:
- 2n < p, because every factor must divide (2n)!;
- p = 2n, because 2n is not prime;
- n < p < 2n, because we assumed there is no such prime number;
- 2n / 3 < p ≤ n: by Lemma 3.
Therefore, every prime factor p satisfies p ≤ 2n/3.
When the number has at most one factor of p. By Lemma 2, for any prime p we have pR(p,n) ≤ 2n, so the product of the pR(p,n) over the primes less than or equal to is at most . Then, starting with Lemma 1 and decomposing the right-hand side into its prime factorization, and finally using Lemma 4, these bounds give:
Taking logarithms yields to
By concavity of the right-hand side as a function of n, the last inequality is necessarily verified on an interval. Since it holds true for n=467 and it does not for n=468, we obtain
But these cases have already been settled, and we conclude that no counterexample to the postulate is possible.
- Ramanujan, S. (1919), "A proof of Bertrand's postulate", Journal of the Indian Mathematical Society 11: 181–182