Proof of Bertrand's postulate


Proof of Bertrand's postulate

In mathematics, Bertrand's postulate (actually a theorem) states that for each "n" ≥ 2 there is a prime "p" such that "n" < "p" < 2"n". It was first proven by Pafnuty Chebyshev, and a short but advanced proof was given by Srinivasa Ramanujan. The gist of the following elementary but involved proof by contradiction is due to Paul Erds; the basic idea of the proof is to show that a certain binomial coefficient needs to have a prime factor within the desired interval in order to be large enough.__TOC__The argument in the proof establishes a contradiction by comparing two facts:
* The binomial coefficient extstyleinom{2n}{n} is "large" in a precise sense (Lemma 1), and
* If Bertrand's postulate fails for "n", then all of the prime factors of this binomial coefficient are "small" (a combination of Lemma 3 and the negation of Bertrand's postulate).It is then shown by some extended computation (relying mostly on Lemma 2 and Lemma 4) that the second fact is inconsistent with the first one. Therefore Bertrand's postulate must hold. In order to present the main argument of the proof intelligibly, these lemmas and a few auxiliary claims are proved separately first.

Lemmas and computations

anchor|Lemma 1Lemma 1: A lower bound on the central binomial coefficients

Lemma: For any integer "n" &gt; 0, we have: frac{4^n}{2n+1} < inom{2n}{n}.

Proof: Applying the binomial theorem,:4^n = (1 + 1)^{2n} = sum_{k = 0}^{2n} inom{2n}{k}. Since extstyleinom{2n}{n} is the largest term in the sum, we have::4^n < (2n + 1) inom{2n}{n},as desired.

anchor|Lemma 2Lemma 2: An upper bound on prime powers dividing central binomial coefficients

For a fixed prime "p", define "R"("p","n") to be the highest natural number "x", such that "p""x" divides extstyleinom{2n}{n}.

Lemma: For any "p", we have "p""R"("p","n") ≤ 2"n".

Proof: The exponent of "p" in "n"! is (see Factorial#Number theory)::sum_{j = 1}^infty leftlfloor frac{n}{p^j} ight floor, so we get: R(p,n) =sum_{j = 1}^infty leftlfloor frac{2n}{p^j} ight floor - 2sum_{j = 1}^infty leftlfloor frac{n}{p^j} ight floor =sum_{j = 1}^infty left(leftlfloor frac{2n}{p^j} ight floor - 2leftlfloor frac{n}{p^j} ight floor ight). But each term of the last summation can either be 0 (if "n" / "p""j" mod 1 &lt; 1/2) or 1 (if "n" / "p""j" mod 1 ≥ 1/2) and all terms with "j" > log"p"(2"n") are 0. Therefore:R(p,n) leq log_p(2n), and we get::p^{R(p,n)} leq p^{log_p{2n = 2n. This completes the proof of the lemma.

anchor|Lemma 3Lemma 3: The exact power of a large prime in a central binomial coefficient

Lemma: For p > sqrt{2n}, we have:R(p,n) = leftlfloor frac{2n}{p} ight floor - 2leftlfloor frac{n}{p} ight floor.

Proof: Using the same sum for "R"("p","n") as in Lemma 2, we see that since "p"2 &gt; "2n", in fact only the first term is nonzero; this term is exactly the right-hand side of the above equation.

anchor|Lemma 4Lemma 4: An upper bound on the primorial

We estimate the primorial function,:n# = prod_{p leq n} p, where the product is taken over all "prime" numbers "p" less than or equal to "n".

Lemma: For all "n" ≥ 1, we have "n"# < 4"n".

Proof: The proof is by mathematical induction. First the base cases:
* "n" = 1: "n"# is the empty product:::n# = 1 < 4 = 4^1.
* "n" = 2: 2 is prime:::2# = 2 < 16 = 4^2.
* "n" > 2 and "n" is even: Since every even number greater than 2 is composite, we have by the induction hypothesis:::n# = (n-1)# < 4^{n-1} < 4^n.
* "n" &gt; 2 is odd. Let "n" = 2"m" + 1 with "m" &gt; 0; then since all the terms in the following binomial expansion are positive we have:::egin{align}4^m &= frac{1}{2}(1 + 1)^{2m + 1} = frac{1}{2}sum_{k = 0}^{2m+1} inom{2m + 1}{k} \ &> frac{1}{2} Biggl(inom{2m + 1}{m} + inom{2m + 1}{m + 1}Biggr) = inom{2m + 1}{m}.end{align}:Each prime "p" with "m" + 1 &lt; "p" &le; 2"m" + 1 divides extstyleinom{2m + 1}{m}, giving us:::frac{(2m + 1)#}{(m + 1)#} = prod_{p > m + 1}^{p leq 2m + 1} p leq inom{2m+1}{m} < 4^m. :By induction, ("m" + 1)# &lt; 4"m" + 1, so:::n# = (2m + 1)# < 4^m cdot (m + 1)# < 4^m cdot 4^{m + 1} = 4^{2m + 1} = 4^n. Thus the lemma is proved.

Computations

We collect here four numerical claims which come up in the proof. Here "n" is always an integer, and "t" is a real number.
#anchor|Claim 1 If "n" &gt; 5, then 2n/3 > sqrt{2n};
#anchor|Claim 2 If "n" &gt; 0, then 2"n" + 1 &lt; (2"n")2;
#anchor|Claim 3 If "n" &gt; 18, then 2 < sqrt{2n}/3;
#anchor|Claim 4 If 2"t" &lt; 8"t", then "t" &lt; 6.

To prove 1, square both sides and solve::frac{2n}{3} > sqrt{2n} quadLeftrightarrowquad frac{4n^2}{9} > 2n quadLeftrightarrowquad 4n^2 - 18n = 2n(2n - 9) > 0and if "n" &gt; 5, both factors are positive, so this is in fact true.

To prove 2, collect the sides together and complete the square::2n + 1 < (2n)^2 quadLeftrightarrowquad (2n)^2 - (2n) - 1 > 0 quadLeftrightarrowquad left(2n - frac{1}{2} ight)^2 - frac{5}{4} > 0.If "n" &gt; 0 is an integer, then 2"n" ≥ 2, so the left-hand side is indeed always positive.

To prove 3, square both sides::2 < frac{sqrt{2n{3} quadLeftrightarrowquad 36 < 2n quadLeftrightarrowquad n > 18.To prove 4, note that 26 = 64 > 48 = 8 &times; 6, so the inequality is true for "t" = 6. Now comparing derivatives::frac{d}{dt} 2^t = 2^t ln 2 > 8 = frac{d}{dt} 8tfor "t" &gt; 6, since then the left side is 64 ln 2, or about 44. Thus 2"t" increases more quickly than 8"t" as well as beginning greater, so remains greater.

Proof of Bertrand's Postulate

Assume there is a counterexample: an integer "n" ≥ 2 such that there is no prime "p" with "n" ≤ "p" &lt; 2"n".

If 2 ≤ "n" &lt; 2048, then "p" can be chosen from among the prime numbers 3, 5, 7, 13, 23, 43, 83, 163, 317, 631, 1259 and 2503 (each being less than twice its predecessor) such that "n" &lt; "p" &lt; 2"n". Therefore "n" ≥ 2048.

There are no prime factors "p" of extstyleinom{2n}{n} such that:
* 2"n" &lt; "p", because 2"n" is the largest factor;
* "n" ≤ "p" ≤ 2"n", because we assumed there is no such prime number;
* 2"n" / 3 &lt; "p" ≤ "n": by computation 1 we have p > sqrt{2n} , so Lemma 3 applies, and by rearranging the inequalities 2"n" / 3 &lt; "p" and "n" ≥ "p" we deduce that "n" / "p" ≥ 1 and 2"n" / "p" &lt; 3. Combining these results, we get::R(p,n) = leftlfloor frac{2n}{p} ight floor - 2leftlfloor frac{n}{p} ight floor leq 2 - 2 = 0. Therefore, every prime factor "p" satisfies "p" ≤ 2"n" / 3.

When p > sqrt{2n}, the number extstyle {2n choose n} has at most one factor of "p". By Lemma 2, for any prime "p" we have "p""R"("p","n") ≤ 2"n", so the product of the "p""R"("p","n") over all other primes is at most (2n)^{sqrt{2n. Then, starting with Lemma 1 and decomposing the right-hand side into its prime factorization, these bounds give::frac{4^n}{2n + 1} < inom{2n}{n} = left(prod_{p le sqrt{2n p^{R(p,n)} ight) left(prod_{p > sqrt{2n^{p le 2n/3} p^{R(p,n)} ight) < (2n)^{sqrt{2n prod_{p > 1}^{p leq 2n/3} p = (2n)^{sqrt{2n cdot lfloor 2n/3 floor#. Using Lemma 4, this simplifies::frac{4^n}{2n + 1} < (2n)^{sqrt{2n 4^{lfloor 2n/3 floor} leq (2n)^{sqrt{2n 4^{2n/3}. Clearing the denominator and applying computations 2 and 3::4^{n/3} < (2n + 1)(2n)^{sqrt{2n < (2n)^{2 + sqrt{2n < (2n)^{(4/3)sqrt{2n. Taking the logarithm of both sides::frac{n}{3}ln 4 < frac{4}{3}sqrt{2n} ln(2n) quadLeftrightarrowquad sqrt{2n} ln{2} < 4 cdot ln(2n). Substituting 22"t" for 2"n"::frac{2^t}{t} < 8. This gives us "t" &lt; 6 by computation 4, and the contradiction::n = frac{2^{2t{2} < frac{2^{2 cdot 6{2} = 2048. Thus no counterexample to the postulate is possible. Q.E.D.

References

This proof of Bertrand's postulate is based on the one in Proofs from THE BOOK.


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