Proof of Bertrand's postulate

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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.[1] 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.

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 Bertrand's postulate.

Lemmas and computation[edit]

Lemma 1: A lower bound on the central binomial coefficients[edit]

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 outside the summation).

Lemma 2: An upper bound on prime powers dividing central binomial coefficients[edit]

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[edit]

Lemma: If is odd and , then

Proof: There are exactly two factors of in the numerator of the expression , coming from the two terms and in , and also two factors of in the denominator from two copies of the term in . These factors all cancel, leaving no factors of in . (The bound on in the preconditions of the lemma ensures that is too large to be a term of the numerator, and the assumption that is odd is needed to ensure that contributes only one factor of to the numerator.)

Lemma 4: An upper bound on the primorial[edit]

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 , .[2]

Proof: Since and , it suffices to prove the result under the assumption that is an integer, . Since is an integer and all the primes appear in its numerator but not in its denumerator, we have


The proof proceeds by mathematical induction.

  • If , then
  • If , then
  • If is odd and , then by the above estimate and the induction assumtion for it's
  • If is even and , then by the above estimate and the induction assumtion for it's

Thus the lemma is proven.

Proof of Bertrand's Postulate[edit]

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 < pn: 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.

Proof by Shigenori Tochiori[edit]

Using Lemma 4, Tochiori refined Erdos's method and proved if there exists a positive integer such that there is no prime number then . [3]

First, refine lemma 1 to:

Lemma 1': For any integer , we have

Proof: By induction: and assuming the truth of the lemma for ,

Then, refine the estimate of the product of all small primes via a better estimate on (the number of primes at most ):

Lemma 5: For any natural number , we have

Proof: Except for , every prime number has or . Thus is upper bounded by the number of natural numbers with or , plus one (since this counts and misses ). Thus

Now, calculating the binomial coefficient as in the previous section, we can use the improved bounds to get (for , which implies so that ):

Taking logarithms to get

and dividing both sides by :

Now the function is decreasing for , so is decreasing when . But

so . The remaining cases are proven by an explicit list of primes, as above.


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