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AnalyticalNumberTheory

Chapter 4

Elementary Results on the Distribution of Primes

4.1

Introduction

Definition 4.1. For real number x > 0, let π(x) denote the number of primes not exceeding x. The behavior of π(x) as the function of x has been studied by many mathematicians ever since the eighteenth century. Inspection of tables of primes led C.F. Gauss (1792) and A.M. Legendre (1798) to conjecture that x . (4.1) π(x) ∼ ln x This conjecture was first proved independently by J. Hadamard and de la Vall´ee Poussin in 1896 and is now known as the Prime Number Theorem. We record the theorem as follows: Theorem 4.1 (Prime Number Theorem). Let x be a real positive number and π(x) be the number of primes less than x. Then x π(x) ∼ . ln x Proofs of the Prime Number Theorem are often classified as elementary or analytic. The proofs of J. Hadamard and de la Vallee Poussin are analytic, using complex function theory and properties of the Riemann zeta function ζ(s) (see Definition 3.3 for the definition of ζ(s) when s ∈ R and s > 1). Elementary proofs were discovered around 1949 by A. Selberg and P. Erd¨os. Their proofs do not involve ζ(s) and complex function theory, hence the name “elementary”. There are other elementary proofs of the prime number theorem since the appearance of the work of Selberg and Erd¨os, one of which is [4]. The 41

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AnalyticalNumberTheory

Analytic Number Theory for Undergraduates

proof given in [4] relies on proving an equivalent statement of the Prime Number Theorem and the mean value of µ(n). This equivalent statement of the Prime Number Theorem will be established in Section 4.5. In this chapter, we derive some basic properties of π(x) and establish several statements equivalent to the Prime Number Theorem. We will also use the results discussed in this chapter to study Bertrand’s Postulate, which states that for n ≥ 2, there exists a prime between n and 2n. 4.2

The function ψ(x)

We recall the definition of Mangoldt’s function (see Exercise 2.6, Problem 1). Definition 4.2. Let n be a positive integer and let ( ln p, if n is a prime power Λ(n) = 0, otherwise. Definition 4.3. For real number x ≥ 1, X X ψ(x) = Λ(n) = ln p. pm ≤x

n≤x

Theorem 4.2. There exist positive constants c1 and c2 such that c1 x ≤ ψ(x) ≤ c2 x. Proof.

For x ≥ 4, let S=

X

n≤x

ln n − 2

X

ln n.

n≤ x 2

By Theorem 3.2 with f (n) = ln n, we find that Z x Z x X 1 ln n = ln tdt + {t} dt − {x} ln x + {y} ln y t 1 1 n≤x

= x ln x − x + O(ln x).

(4.2)

This implies that S = x ln 2 + O(ln x). Therefore, there exists an x0 ≥ 4 such that x ≤S≤x 2

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(4.3)

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Elementary Results on the Distribution of Primes

43

whenever x ≥ x0 ≥ 4. Next, since (see Section 2.6, Problem 1) X ln n = Λ(d), d|n

we find that S=

XX

n≤x d|n

Λ(d) − 2

XX

Λ(d)

n≤ x 2 d|n

hxi d 2d d≤x d≤ x 2 nh i h io hxi X X x x = Λ(d) −2 + Λ(d) . d 2d d x x =

X

Λ(d)

hxi

−2

X

Λ(d)

d≤ 2

2

Hence, S=

X

Λ(d)θd +

X

750,

(4.24)

if x > 0,

(4.25)

if x > 750,

(4.26)

if x > 0.

(4.27)

and

ψ(x) − ψ

x 2


3, we conclude that   1 ln[x]! − 2 ln[x/2]! ≥ ln Γ(x) − 2 ln Γ x+1 . 2 To prove (4.24), it suffices to show that for x > 750, ln Γ(x) − 2 ln Γ

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1 x+1 2



>

2x . 3

(4.29)

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By (4.28), we deduce that   1 x+1 ln Γ(x) − 2 ln Γ 2   √ √ 1 ϑ1 = ln 2π + x − ln x − x + − 2 ln 2π 2 12x     x  x 1 x ϑ2 −2 + ln +1 +2 +1 − , 2 2 2 2 3x + 6

where both ϑ1 , ϑ2 belong to the interval (0, 1). Simplifying the above inequality, we deduce that     1 2x ln Γ(x) − 2 ln Γ x + 1 > x ln − 2 ln x. 2 x+2 For x > 750, x ln



2x x+2



− 2 ln x >

2x . 3

This completes the proof of (4.29).



Proof of (4.25). The proof is similar to that for (4.24). We use the inequality   1 1 ln[x]! − 2 ln[x/2]! ≤ ln Γ(x + 1) − 2 ln Γ x+ 2 2 and Stirling’s formula to conclude that (see Problem 8 of Exercise 4.7) ln[x]! − 2 ln[x/2]! ≤

3 x 4

for all x > 0.



Proof of (4.26) and (4.27). These two inequalities follow immediately from (4.23)–(4.25).



Lemma 4.14. For each positive real number x, we have 3 ψ(x) < x if x > 0 (4.30) x x 2     √ x x ψ(x) − ψ +ψ ≤ θ(x) + 2ψ( x) − θ +ψ 2 3 3  x  x 2√ < θ(x) − θ + + 3 x. (4.31) 2 2

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Analytic Number Theory for Undergraduates

Proof of (4.30). To prove (4.30), we use (4.27) repeatedly, with x replaced by x/2, x/4, · · · and add up the results. We find that   3 1 3 ψ(x) ≤ x 1 + + · · · < x. 4 2 2 Proof of (4.31). From (4.22), we find that √ ψ(x) − 2ψ( x) ≤ θ(x). Hence √ ψ(x) ≤ θ(x) + 2ψ( x). Next, from (4.22), θ(x/2) ≤ ψ(x/2). Using the above inequalities, we deduce that √ ψ(x) − ψ(x/2) + ψ(x/3) ≤ θ(x) + 2ψ( x) − θ(x/2) + ψ(x/3). For the second inequality, we use (4.30) to deduce that √ √ 2ψ( x) + ψ(x/3) ≤ 3 x + x/2. We are now ready to prove Bertrand’s Postulate. By (4.26), ψ(x) − ψ(x/2) + ψ(x/3) ≥

2 x 3

for x > 750. Hence we deduce from (4.31) that √ θ(x) − θ(x/2) ≥ 2x/3 − x/2 − 3 x, and there is a prime between x and 2x for x > 750. We are now left with verifying that Bertrand’s Postulate is true for x ≤ 750. This is straightforward and we leave it as an exercise. Example 4.1. Suppose S is a set of consecutive integers which contains a prime. Show that there exists an integer in S that is relatively prime to all the other integers in S. Conversely, show that if for every finite set of consecutive integers with at least a prime contains a number that is relatively prime to the others, then Bertrand’s Postulate is true.

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Elementary Results on the Distribution of Primes

AnalyticalNumberTheory

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Proof. Let S := {n, n + 1, · · · , n + k} be any set of consecutive integers with at least a prime. Suppose p is the largest prime in the set S. If 2p ≤ n + k, then by Bertrand’s Postulate, there exists another prime q such that p < q < 2p and this contradicts our assumption that p is the largest prime in the set. Hence 2p > n + k and p is relatively prime to all the other integers in S. Conversely, let n > 1 be given and consider the set T = {2, 3, · · · , 2n}. Obviously there is at least a prime in T and hence by hypothesis, there exists an integer q such that q is relatively prime to the other numbers in T . This implies that 2q > 2n (for, if 2q < 2n then q will not be prime to 2q.) Hence n < q < 2n. To prove that q is prime, we suppose a|q and a 6= 1. Then a ∈ T and so, a is not relatively prime to q. This contradicts the choice of q. Hence q must be prime. This means that there is a prime between n and 2n if n > 1, and this is Bertrand’s Postulate.  4.7

Exercises

1. Prove that the following two relations are equivalent.   x x +O (a) π(x) = ln x  ln2 x x  (b) θ(x) = x + O . ln x

2. Show that if p, q are primes, then X 1 = (ln ln x)2 + O(ln ln x). pq pq≤x

3. Let ω(n) be defined as in Definition 3.6. Show that X ω(n) = x ln ln x + O (x) . n≤x

4. Show that for x ≥ 2,

X

n≤x

ψ

x n

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= x ln x + O(x).

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Analytic Number Theory for Undergraduates

5. Show that for x ≥ 1,

X θ(n) = ln x + O(1). n2

n≤x

6. Show that

X ln p = ln x + O(1). p−1

p≤x

7. Let x be a positive real number. Verify using Stirling’s formula that 3 ln[x]! − 2 ln[x/2]! < x if x > 0. 4

8. Follow Erd¨os’ proof of Bertrand’s Postulate by completing the following exercises. (a) Let r(p) be such that Show that

pr(p) ≤ 2n < pr(p)+1 .   Y 2n | pr(p) . n p≤2n

2n (b) Show that if p > 2 and < p ≤ n then 3   2n p6| . n (c) Show that

Y

p < 4n .

p≤n

(d) Use the above results to deduce Bertrand’s Postulate. 9. Show that if n is any positive integer greater than 1, then n! is never a perfect square. 10. Let n be a positive integer. Assume that when m > 20, there exists a prime between m/2 and m − 6. Show that n can be expressed as a sum of distinct primes when n > 6. (Challenge: Try to prove the above result without making the first assumption.)

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11. Use Bertrand’s Postulate to show that for every positive integer n, the number 1 1 1 1 + + + ··· + 2 3 n can never be an integer. 12. Use induction and Bertrand’s Postulate to show that if pn denote the n-th prime, then for n > 3, pn < p1 + p2 + · · · + pn−1 .

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