Lecture 3: Chebyshev's prime number theorem

Lecture 3:Chebyshev’s prime number theorem

Karl Dilcher

Dalhousie University, Halifax, Canada

December 15, 2018

Karl Dilcher Lecture 3:Chebyshev’s prime number theorem

1. Introduction

We begin with a basic definition.

Definition 1An integer p > 1 is called a prime number, or simply a prime, ifit has only 1 and itself as divisors.

An integer that is not a prime number is called composite.

Obvious question: How many primes are there?

Theorem 2 (Euclid)

There are infinitely many primes.

1. Introduction

Definition 1An integer p > 1 is called a prime number, or simply a prime, ifit has only 1 and itself as divisors.An integer that is not a prime number is called composite.

Theorem 2 (Euclid)

1. Introduction

Theorem 2 (Euclid)

1. Introduction

Theorem 2 (Euclid)

Proof: Suppose there were only finitely mny primes, namelyp1 = 2,p2 = 3,p3 = 5, . . . ,pr .

Now form the number

n = p1p2 · · · pr + 1.

n is not divisible byp1: remainder 1 upon division;p2: remainder 1 upon division;· · ·pr : remainder 1 upon division.

Hence any prime divisor of n must be distinct from thep1,p2, . . . ,pr , but this is a contradiction.

Now form the number

n = p1p2 · · · pr + 1.

Now form the number

n = p1p2 · · · pr + 1.

Now form the number

n = p1p2 · · · pr + 1.

Remark: Further to Euclid’s theorem, the following isnoteworthy.

(a) Variants of Euclid’s proof: There are infinitely many primesof the form 4k + 1, or 4k − 1, or 6k − 1.

(b) More generally, what about p = ak + b, where (a,b) = 1?

There are infinitely many primes in each class.First proved by P. G. Lejeune-Dirichlet (1805–1859) in 1837;this marked the beginning of analytic number theory.

There are infinitely many primes in each class.

First proved by P. G. Lejeune-Dirichlet (1805–1859) in 1837;this marked the beginning of analytic number theory.

(c) The prime numbers are both very regularly and veryirregularly distributed.

(i) “irregularity": two extreme situations.

First, there are “gaps" of arbitrary length between primes:Consider the integers

k ! + 2, k ! + 3, . . . , k ! + k − 1, k ! + k ;

all these k − 1 consecutive integers are composite becauseclearly

j | k ! + j for j = 2,3, . . . , k .

Second, there are the twin primes, namely pairs p,p + 2 ofprimes, such as

3,5 5,7 11,13 17,19 . . .

Are there infinitely many such pairs?Not known: The twin-prime conjecture, one of the mostfamous unsolved problems in number theory.

First, there are “gaps" of arbitrary length between primes:

Consider the integers

k ! + 2, k ! + 3, . . . , k ! + k − 1, k ! + k ;

j | k ! + j for j = 2,3, . . . , k .

3,5 5,7 11,13 17,19 . . .

k ! + 2, k ! + 3, . . . , k ! + k − 1, k ! + k ;

j | k ! + j for j = 2,3, . . . , k .

3,5 5,7 11,13 17,19 . . .

k ! + 2, k ! + 3, . . . , k ! + k − 1, k ! + k ;

j | k ! + j for j = 2,3, . . . , k .

3,5 5,7 11,13 17,19 . . .

k ! + 2, k ! + 3, . . . , k ! + k − 1, k ! + k ;

j | k ! + j for j = 2,3, . . . , k .

3,5 5,7 11,13 17,19 . . .

k ! + 2, k ! + 3, . . . , k ! + k − 1, k ! + k ;

j | k ! + j for j = 2,3, . . . , k .

3,5 5,7 11,13 17,19 . . .

Are there infinitely many such pairs?

Not known: The twin-prime conjecture, one of the mostfamous unsolved problems in number theory.

k ! + 2, k ! + 3, . . . , k ! + k − 1, k ! + k ;

j | k ! + j for j = 2,3, . . . , k .

3,5 5,7 11,13 17,19 . . .

(ii) “Regularity": Let’s look at the “Prime counting function" π(x):

1 2 3 4 5 6 7 8 9 10

(ii) “Regularity": Let’s look at the “Prime counting function" π(x):

1 2 3 4 5 6 7 8 9 10

0 10 20 30 40 50 60 70 80 90 100

0 100 200 300 400 500 600 700 800 900 1000

0 1000 2000 3000 4000 5000 6000 7000 8000 9000 10000

0 10000 20000 30000 40000 50000 60000 70000 80000 90000 100000

0 100000 200000 300000 400000 500000 600000 700000 800000 900000 1e+06

What is this function?

The prime number theorem shows that the primes are veryregularly behaved “in the large".

More exactly, we define the function

π(x) =∑p≤x

1 = #{p ≤ x | p prime}.

It was proved by Jacques Hadamard (1865–1963) and(independently) Charles de la Vallée Poussin (1866–1962) in1896 that

π(x) ∼ xlog x

as x →∞,

which is a short form for writing

limx→∞

π(x)log x

What is this function?The prime number theorem shows that the primes are veryregularly behaved “in the large".

π(x) =∑p≤x

1 = #{p ≤ x | p prime}.

π(x) ∼ xlog x

as x →∞,

limx→∞

π(x)log x

π(x) =∑p≤x

1 = #{p ≤ x | p prime}.

π(x) ∼ xlog x

as x →∞,

limx→∞

π(x)log x

π(x) =∑p≤x

1 = #{p ≤ x | p prime}.

π(x) ∼ xlog x

as x →∞,

limx→∞

π(x)log x

π(x) =∑p≤x

1 = #{p ≤ x | p prime}.

π(x) ∼ xlog x

as x →∞,

limx→∞

π(x)log x

Proof relies heavily on complex analysis and is beyond thescope of these lectures.

However, a weaker version, which still shows how the primesbehave “in the large", is easier to obtain.

This will be the main objective of this lecture.

Before the eventual proof in 1896, the most important advancewas made by the Russian mathematician Pavnutii LvovichChebyshev (1821–1894) in 1850.

Theorem 3

For all n ≥ 2,18≤ π(n)

n/ log n≤ 12.

To prove this theorem, we first state another one:

Theorem 4

H(n) =n∑

In other words: π(n) is of the same order of magnitude as thereciprocal of the average of (1

2 ,13 ,

14 , . . . ,

Theorem 3

n/ log n≤ 12.

Theorem 4

H(n) =n∑

2 ,13 ,

14 , . . . ,

Theorem 3

n/ log n≤ 12.

Theorem 4

H(n) =n∑

2 ,13 ,

14 , . . . ,

Proof of Theorem 3 For n ≥ 2 we have

+ · · ·+ 1n<

= log n.

For n ≥ 4 we have

logn2≥ 1

2log n,

and also12

log 3 ≤ 12

log 2 ≤ 12.

Hence for all n ≥ 2,

log n ≤ H(n) ≤ log n,

and therefore Theorem 3 follows from Theorem 4.

+ · · ·+ 1n<

= log n.

For n ≥ 4 we have

logn2≥ 1

2log n,

and also12

log 3 ≤ 12

log 2 ≤ 12.

+ · · ·+ 1n<

= log n.

For n ≥ 4 we have

logn2≥ 1

2log n,

and also12

log 3 ≤ 12

log 2 ≤ 12.

+ · · ·+ 1n<

= log n.

For n ≥ 4 we have

logn2≥ 1

2log n,

and also12

log 3 ≤ 12

log 2 ≤ 12.

Remarks: (1) The numbers H(n) are closely related to theharmonic numbers

+ · · ·+ 1n.

(2) In view of the first line of the above proof it is of interest tonote that the limit

γ := limn→∞

+ · · ·+ 1n− log n

)actually exists. γ is called Euler’s constant, also known as theEuler-Mascheroni constant.

After the numbers π and e it is probably the most importantspecial constant in mathematics, and it is also closely related tothe Gamma function Γ(x).

+ · · ·+ 1n.

γ := limn→∞

+ · · ·+ 1n− log n

)actually exists.

γ is called Euler’s constant, also known as theEuler-Mascheroni constant.

+ · · ·+ 1n.

γ := limn→∞

+ · · ·+ 1n− log n

+ · · ·+ 1n.

γ := limn→∞

+ · · ·+ 1n− log n

In preparation for the proof of Theorem 4, we first prove severallemmas.

Lemma 5

2k+1)≤ 2k .

Proof. Even numbers cannot be primes, except 2.For n > 9 we clearly have π(n) ≤ n/2.For the smaller cases we count the primes:

π(2) = 1 = 20, π(4) = 2 = 21, π(8) = 4 = 22.

Lemma 5

2k+1)≤ 2k .

π(2) = 1 = 20, π(4) = 2 = 21, π(8) = 4 = 22.

Lemma 5

2k+1)≤ 2k .

Proof. Even numbers cannot be primes, except 2.

For n > 9 we clearly have π(n) ≤ n/2.For the smaller cases we count the primes:

π(2) = 1 = 20, π(4) = 2 = 21, π(8) = 4 = 22.

Lemma 5

2k+1)≤ 2k .

Proof. Even numbers cannot be primes, except 2.For n > 9 we clearly have π(n) ≤ n/2.

For the smaller cases we count the primes:

π(2) = 1 = 20, π(4) = 2 = 21, π(8) = 4 = 22.

Lemma 5

2k+1)≤ 2k .

π(2) = 1 = 20, π(4) = 2 = 21, π(8) = 4 = 22.

Lemma 6

l ≤ H(2l) ≤ l .

Proof: Group the terms of the sum H(2l) in 2 different ways:

H(2l) =12

)+ · · ·

2l−1 + 1+ · · ·+ 1

)≥ 1

)+ · · ·

(12l + · · ·+ 1

Lemma 6

l ≤ H(2l) ≤ l .

Proof: Group the terms of the sum H(2l) in 2 different ways:

H(2l) =12

)+ · · ·

2l−1 + 1+ · · ·+ 1

)≥ 1

)+ · · ·

(12l + · · ·+ 1

On the other hand,

H(2l) =

)+ · · ·+ 1

)+ · · ·

2l−1 + · · ·+ 12l−1 +

)≤ l .

Lemma 7

The canonical factorization of n! is

n! =∏p≤n

pbn/pc+bn/p2c+···.

Proof: Only primes p ≤ n can occur in the factorization of n!.Fix such a prime p. In the sequence of factors

1, 2, 3, . . . , n − 1, n of n!,

– every pth number (namely p,2p, . . .) is divisible by p;there are bn/pc of them;– every p2th number (namely p,2p, . . .) is divisible by another p;there are bn/p2c of them;– and so on, so that the power of p in n! is

bnpc+ b n

p2 c+ b np3 c+ · · ·

Lemma 7

n! =∏p≤n

Proof: Only primes p ≤ n can occur in the factorization of n!.

Fix such a prime p. In the sequence of factors

1, 2, 3, . . . , n − 1, n of n!,

bnpc+ b n

p2 c+ b np3 c+ · · ·

Lemma 7

n! =∏p≤n

1, 2, 3, . . . , n − 1, n of n!,

– every pth number (namely p,2p, . . .) is divisible by p;there are bn/pc of them;

– every p2th number (namely p,2p, . . .) is divisible by another p;there are bn/p2c of them;– and so on, so that the power of p in n! is

bnpc+ b n

p2 c+ b np3 c+ · · ·

Lemma 7

n! =∏p≤n

1, 2, 3, . . . , n − 1, n of n!,

– every pth number (namely p,2p, . . .) is divisible by p;there are bn/pc of them;– every p2th number (namely p,2p, . . .) is divisible by another p;there are bn/p2c of them;

– and so on, so that the power of p in n! is

bnpc+ b n

p2 c+ b np3 c+ · · ·

Lemma 7

n! =∏p≤n

1, 2, 3, . . . , n − 1, n of n!,

bnpc+ b n

p2 c+ b np3 c+ · · ·

Lemma 8

The power of a prime p in(N

∑m≥1

(⌊Npm

⌋−⌊

N − npm

Proof: We use the fact that(Nn

n!(N − n)!

and apply Lemma 7.

Lemma 8

The power of a prime p in(N

∑m≥1

(⌊Npm

⌋−⌊

N − npm

Proof: We use the fact that(Nn

n!(N − n)!

and apply Lemma 7.

Proof of Theorem 4

(i) We begin with the following assertion:∏n<p≤2n

p∣∣∣∣(2n

)∣∣∣∣ ∏pr≤2n<pr+1

pr . (1)

Proof: We use the fact that(2nn

If n < p ≤ 2n then p | (2n)!, while p - n!.This proves the left part of (1).

Proof of Theorem 4(i) We begin with the following assertion:∏

n<p≤2n

p∣∣∣∣(2n

)∣∣∣∣ ∏pr≤2n<pr+1

pr . (1)

n<p≤2n

p∣∣∣∣(2n

)∣∣∣∣ ∏pr≤2n<pr+1

pr . (1)

n<p≤2n

p∣∣∣∣(2n

)∣∣∣∣ ∏pr≤2n<pr+1

pr . (1)

If n < p ≤ 2n then p | (2n)!, while p - n!.

This proves the left part of (1).

n<p≤2n

p∣∣∣∣(2n

)∣∣∣∣ ∏pr≤2n<pr+1

pr . (1)

On the other hand, by Lemma 8 the power of p in(2n

r∑m=1

(⌊2npn

⌋− 2

⌊npn

⌋)︸︷︷︸

≤ r .

This proves the right part of (1).

(ii) Changing divisibility in (1) into inequalities, we get for n ≥ 1,

nπ(2n)−π(n) <∏

n<p≤2n

p ≤(

∏pr≤2n<pr+1

pr ≤ (2n)π(2n). (2)

Left inequality follows from: π(2n)− π(n) primes between n and2n, and the smallest one is at least n.

Right inequality follows from: Number of p in the precedingproduct is π(2n), while each factor pr is at most 2n.

r∑m=1

(⌊2npn

⌋− 2

⌊npn

⌋)︸︷︷︸

≤ r .

nπ(2n)−π(n) <∏

n<p≤2n

p ≤(

∏pr≤2n<pr+1

pr ≤ (2n)π(2n). (2)

r∑m=1

(⌊2npn

⌋− 2

⌊npn

⌋)︸︷︷︸

≤ r .

nπ(2n)−π(n) <∏

n<p≤2n

p ≤(

∏pr≤2n<pr+1

pr ≤ (2n)π(2n). (2)

r∑m=1

(⌊2npn

⌋− 2

⌊npn

⌋)︸︷︷︸

≤ r .

nπ(2n)−π(n) <∏

n<p≤2n

p ≤(

∏pr≤2n<pr+1

pr ≤ (2n)π(2n). (2)

r∑m=1

(⌊2npn

⌋− 2

⌊npn

⌋)︸︷︷︸

≤ r .

nπ(2n)−π(n) <∏

n<p≤2n

p ≤(

∏pr≤2n<pr+1

pr ≤ (2n)π(2n). (2)

Now(2nn

2n(2n − 1) · · · (n + 1)

n(n − 1) · · · 1

n − 1

)· · ·(

n − j

)· · ·(

2 +n − 1

)≥ 2n,

and (2nn

)≤ (1 + 1)2n = 22n.

Hence, with (2) we get for n ≥ 1,

nπ(2n)−π(n) < 22n, 22n ≤ (2n)π(2n).

Now(2nn

2n(2n − 1) · · · (n + 1)

n(n − 1) · · · 1

n − 1

)· · ·(

n − j

)· · ·(

2 +n − 1

)≥ 2n,

and (2nn

)≤ (1 + 1)2n = 22n.

nπ(2n)−π(n) < 22n, 22n ≤ (2n)π(2n).

Now(2nn

2n(2n − 1) · · · (n + 1)

n(n − 1) · · · 1

n − 1

)· · ·(

n − j

)· · ·(

2 +n − 1

)≥ 2n,

and (2nn

)≤ (1 + 1)2n = 22n.

nπ(2n)−π(n) < 22n, 22n ≤ (2n)π(2n).

Now let n = 2k , k = 0,1,2, . . ..

Then the last inequalities become

2k(π(2k+1)−π(2k )) < 22k+1, 22k ≤ 2(k+1)π(2k+1),

k(π(2k+1)− π(2k )) < 2k+1, 2k ≤ (k + 1)π(2k+1). (3)

(k + 1)π(2k+1)− kπ(2k ) < 2k+1 + π(2k+1) ≤ 3 · 2k ,

where the first inequality follows from (3), and the second onefrom Lemma 5.

Now let n = 2k , k = 0,1,2, . . ..Then the last inequalities become

2k(π(2k+1)−π(2k )) < 22k+1, 22k ≤ 2(k+1)π(2k+1),

k(π(2k+1)− π(2k )) < 2k+1, 2k ≤ (k + 1)π(2k+1). (3)

(k + 1)π(2k+1)− kπ(2k ) < 2k+1 + π(2k+1) ≤ 3 · 2k ,

Now let n = 2k , k = 0,1,2, . . ..Then the last inequalities become

2k(π(2k+1)−π(2k )) < 22k+1, 22k ≤ 2(k+1)π(2k+1),

k(π(2k+1)− π(2k )) < 2k+1, 2k ≤ (k + 1)π(2k+1). (3)

(k + 1)π(2k+1)− kπ(2k ) < 2k+1 + π(2k+1) ≤ 3 · 2k ,

(k + 1)π(2k+1)− kπ(2k ) < 2k+1 + π(2k+1) ≤ 3 · 2k

Replace k successively by 0,1, . . . , k and add.On the left we have a “telescoping sum" so that

(k + 1)π(2k+1) < 3(

20 + 21 + · · ·+ 2k)< 3 · 2k+1.

Using this and (3), we find

12· 2k+1

k + 1≤ π(2k+1) < 3

k + 1.

Now let n ∈ N, n > 1, be such that

2k+1 ≤ n < 2k+2.

π(n) ≤ π(2k+2) < 32k+2

k + 2≤ 6

H(2k+2)≤ 6

where we have used Lemma 6 in the third inequality.

(k + 1)π(2k+1)− kπ(2k ) < 2k+1 + π(2k+1) ≤ 3 · 2k

Replace k successively by 0,1, . . . , k and add.

On the left we have a “telescoping sum" so that

(k + 1)π(2k+1) < 3(

20 + 21 + · · ·+ 2k)< 3 · 2k+1.

12· 2k+1

k + 1≤ π(2k+1) < 3

k + 1.

2k+1 ≤ n < 2k+2.

π(n) ≤ π(2k+2) < 32k+2

k + 2≤ 6

H(2k+2)≤ 6

(k + 1)π(2k+1)− kπ(2k ) < 2k+1 + π(2k+1) ≤ 3 · 2k

(k + 1)π(2k+1) < 3(

20 + 21 + · · ·+ 2k)< 3 · 2k+1.

12· 2k+1

k + 1≤ π(2k+1) < 3

k + 1.

2k+1 ≤ n < 2k+2.

π(n) ≤ π(2k+2) < 32k+2

k + 2≤ 6

H(2k+2)≤ 6

(k + 1)π(2k+1)− kπ(2k ) < 2k+1 + π(2k+1) ≤ 3 · 2k

(k + 1)π(2k+1) < 3(

20 + 21 + · · ·+ 2k)< 3 · 2k+1.

12· 2k+1

k + 1≤ π(2k+1) < 3

k + 1.

2k+1 ≤ n < 2k+2.

π(n) ≤ π(2k+2) < 32k+2

k + 2≤ 6

H(2k+2)≤ 6

(k + 1)π(2k+1)− kπ(2k ) < 2k+1 + π(2k+1) ≤ 3 · 2k

(k + 1)π(2k+1) < 3(

20 + 21 + · · ·+ 2k)< 3 · 2k+1.

12· 2k+1

k + 1≤ π(2k+1) < 3

k + 1.

2k+1 ≤ n < 2k+2.

π(n) ≤ π(2k+2) < 32k+2

k + 2≤ 6

H(2k+2)≤ 6

(k + 1)π(2k+1)− kπ(2k ) < 2k+1 + π(2k+1) ≤ 3 · 2k

(k + 1)π(2k+1) < 3(

20 + 21 + · · ·+ 2k)< 3 · 2k+1.

12· 2k+1

k + 1≤ π(2k+1) < 3

k + 1.

2k+1 ≤ n < 2k+2.

π(n) ≤ π(2k+2) < 32k+2

k + 2≤ 6

H(2k+2)≤ 6

where we have used Lemma 6 in the third inequality.Karl Dilcher Lecture 3:Chebyshev’s prime number theorem

On the other hand,

π(n) ≥ π(2k+1) ≥ 12· 2k+1

k + 1=

18· 2k+2

12(k + 1)

≥ 18· 2k+2

H(2k+1)≥ 1

where we have used Lemma 6 for the second-last inequality.

Putting the last two strings of inequalities together, we finallyobtain the statement of Theorem 4.

On the other hand,

π(n) ≥ π(2k+1) ≥ 12· 2k+1

k + 1=

18· 2k+2

12(k + 1)

≥ 18· 2k+2

H(2k+1)≥ 1

where we have used Lemma 6 for the second-last inequality.

Putting the last two strings of inequalities together, we finallyobtain the statement of Theorem 4.

Consequence of Chebyshev’s theorem:

Proof of a famous and at that time unsolved problem, namely

Bertrand’s Postulate: (J. L. F. Bertrand, 1845)

If x ∈ R, x > 1, then there is at least one prime in the openinterval (x ,2x).

Chebyshev said,And I’ll say it again,There’s always a primeBetween n and 2n.

Consequence of Chebyshev’s theorem:

Proof of a famous and at that time unsolved problem, namely

Bertrand’s Postulate: (J. L. F. Bertrand, 1845)

If x ∈ R, x > 1, then there is at least one prime in the openinterval (x ,2x).

Chebyshev said,And I’ll say it again,There’s always a primeBetween n and 2n.

Lecture 3: Chebyshev's prime number theorem

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