[PPT] - An Elementary Proof of Bertrands Postulate Daniel W. Cranston PowerPoint Presentation

SLIDE 1

An Elementary Proof of Bertrand’s Postulate

Daniel W. Cranston

Virginia Commonwealth University Davidson Math Coffee March 25, 2011

SLIDE 2

Overview

Bertrand’s Postulate For every positive n, there exists a prime p s.t. n < p ≤ 2n.

SLIDE 3

Overview

Bertrand’s Postulate For every positive n, there exists a prime p s.t. n < p ≤ 2n. Proof Outline Find upper and lower bounds for 2n

n

= (2n)!/(n!n!). If no such p

exists, then lower bound is larger than upper.

SLIDE 4

Overview

Bertrand’s Postulate For every positive n, there exists a prime p s.t. n < p ≤ 2n. Proof Outline Find upper and lower bounds for 2n

n

= (2n)!/(n!n!). If no such p

exists, then lower bound is larger than upper. Only works for big n.

SLIDE 5

Overview

Bertrand’s Postulate For every positive n, there exists a prime p s.t. n < p ≤ 2n. Proof Outline Find upper and lower bounds for 2n

n

= (2n)!/(n!n!). If no such p

exists, then lower bound is larger than upper. Only works for big n. Lemma 1: Bertrand’s Postulate holds for 1 ≤ n ≤ 4000.

SLIDE 6

Overview

Bertrand’s Postulate For every positive n, there exists a prime p s.t. n < p ≤ 2n. Proof Outline Find upper and lower bounds for 2n

n

= (2n)!/(n!n!). If no such p

exists, then lower bound is larger than upper. Only works for big n. Lemma 1: Bertrand’s Postulate holds for 1 ≤ n ≤ 4000. Pf: The following are primes: 2, 3, 5, 7, 13, 23, 43, 83, 163, 317, 631, 1259, 2503, 4001.

SLIDE 7

Overview

Bertrand’s Postulate For every positive n, there exists a prime p s.t. n < p ≤ 2n. Proof Outline Find upper and lower bounds for 2n

n

= (2n)!/(n!n!). If no such p

exists, then lower bound is larger than upper. Only works for big n. Lemma 1: Bertrand’s Postulate holds for 1 ≤ n ≤ 4000. Pf: The following are primes: 2, 3, 5, 7, 13, 23, 43, 83, 163, 317, 631, 1259, 2503, 4001. Lemma 2: For fixed n, let f (p) = pk s.t. pk| 2n

n

and pk+1

| 2n

n

.

SLIDE 8

Overview

Bertrand’s Postulate For every positive n, there exists a prime p s.t. n < p ≤ 2n. Proof Outline Find upper and lower bounds for 2n

n

= (2n)!/(n!n!). If no such p

exists, then lower bound is larger than upper. Only works for big n. Lemma 1: Bertrand’s Postulate holds for 1 ≤ n ≤ 4000. Pf: The following are primes: 2, 3, 5, 7, 13, 23, 43, 83, 163, 317, 631, 1259, 2503, 4001. Lemma 2: For fixed n, let f (p) = pk s.t. pk| 2n

n

and pk+1

| 2n

n

.

If Bertrand’s is false, then ∃ n > 4000 s.t. 4n/(2n) ≤ 2n n

SLIDE 9

Overview

Bertrand’s Postulate For every positive n, there exists a prime p s.t. n < p ≤ 2n. Proof Outline Find upper and lower bounds for 2n

n

= (2n)!/(n!n!). If no such p

exists, then lower bound is larger than upper. Only works for big n. Lemma 1: Bertrand’s Postulate holds for 1 ≤ n ≤ 4000. Pf: The following are primes: 2, 3, 5, 7, 13, 23, 43, 83, 163, 317, 631, 1259, 2503, 4001. Lemma 2: For fixed n, let f (p) = pk s.t. pk| 2n

n

and pk+1

| 2n

n

.

If Bertrand’s is false, then ∃ n > 4000 s.t. 4n/(2n) ≤ 2n n

=
p≤2n

f (p)

SLIDE 10

Overview

Bertrand’s Postulate For every positive n, there exists a prime p s.t. n < p ≤ 2n. Proof Outline Find upper and lower bounds for 2n

n

= (2n)!/(n!n!). If no such p

exists, then lower bound is larger than upper. Only works for big n. Lemma 1: Bertrand’s Postulate holds for 1 ≤ n ≤ 4000. Pf: The following are primes: 2, 3, 5, 7, 13, 23, 43, 83, 163, 317, 631, 1259, 2503, 4001. Lemma 2: For fixed n, let f (p) = pk s.t. pk| 2n

n

and pk+1

| 2n

n

.

If Bertrand’s is false, then ∃ n > 4000 s.t. 4n/(2n) ≤ 2n n

=
p≤2n

f (p) =

p≤ 2n

3

f (p)

2n

3 <p≤n

f (p)

n<p≤2n

f (p).

SLIDE 11

Overview

Bertrand’s Postulate For every positive n, there exists a prime p s.t. n < p ≤ 2n. Proof Outline Find upper and lower bounds for 2n

n

= (2n)!/(n!n!). If no such p

exists, then lower bound is larger than upper. Only works for big n. Lemma 1: Bertrand’s Postulate holds for 1 ≤ n ≤ 4000. Pf: The following are primes: 2, 3, 5, 7, 13, 23, 43, 83, 163, 317, 631, 1259, 2503, 4001. Lemma 2: For fixed n, let f (p) = pk s.t. pk| 2n

n

and pk+1

| 2n

n

.

If Bertrand’s is false, then ∃ n > 4000 s.t. 4n/(2n) ≤ 2n n

=
p≤2n

f (p) =

p≤ 2n

3

f (p)

2n

3 <p≤n

f (p)

n<p≤2n

f (p). 1

SLIDE 12

Overview

Bertrand’s Postulate For every positive n, there exists a prime p s.t. n < p ≤ 2n. Proof Outline Find upper and lower bounds for 2n

n

= (2n)!/(n!n!). If no such p

exists, then lower bound is larger than upper. Only works for big n. Lemma 1: Bertrand’s Postulate holds for 1 ≤ n ≤ 4000. Pf: The following are primes: 2, 3, 5, 7, 13, 23, 43, 83, 163, 317, 631, 1259, 2503, 4001. Lemma 2: For fixed n, let f (p) = pk s.t. pk| 2n

n

and pk+1

| 2n

n

.

If Bertrand’s is false, then ∃ n > 4000 s.t. 4n/(2n) ≤ 2n n

=
p≤2n

f (p) =

p≤ 2n

3

f (p)

2n

3 <p≤n

f (p)

n<p≤2n

f (p). 1 1

SLIDE 13

Overview

Bertrand’s Postulate For every positive n, there exists a prime p s.t. n < p ≤ 2n. Proof Outline Find upper and lower bounds for 2n

n

= (2n)!/(n!n!). If no such p

exists, then lower bound is larger than upper. Only works for big n. Lemma 1: Bertrand’s Postulate holds for 1 ≤ n ≤ 4000. Pf: The following are primes: 2, 3, 5, 7, 13, 23, 43, 83, 163, 317, 631, 1259, 2503, 4001. Lemma 2: For fixed n, let f (p) = pk s.t. pk| 2n

n

and pk+1

| 2n

n

.

If Bertrand’s is false, then ∃ n > 4000 s.t. 4n/(2n) ≤ 2n n

=
p≤2n

f (p) =

p≤ 2n

3

f (p)

2n

3 <p≤n

f (p)

n<p≤2n

f (p). 1 1 So, we need bounds on f (p). . .

SLIDE 14

Upper Bounds on f (p)

Legendre’s Thm: n! contains factor p exactly

k≥1

n

pk

times.

SLIDE 15

Upper Bounds on f (p)

Legendre’s Thm: n! contains factor p exactly

k≥1

n

pk

times.

Ex: p = 3, n = 17.

SLIDE 16

Upper Bounds on f (p)

Legendre’s Thm: n! contains factor p exactly

k≥1

n

pk

times.

Ex: p = 3, n = 17. So: 17

3

+

17

32

+

17

33

+ . . . = 5 + 1 + 0 = 6.

SLIDE 17

Upper Bounds on f (p)

Legendre’s Thm: n! contains factor p exactly

k≥1

n

pk

times.

Ex: p = 3, n = 17. So: 17

3

+

17

32

+

17

33

+ . . . = 5 + 1 + 0 = 6.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

SLIDE 18

Upper Bounds on f (p)

Legendre’s Thm: n! contains factor p exactly

k≥1

n

pk

times.

Ex: p = 3, n = 17. So: 17

3

+

17

32

+

17

33

+ . . . = 5 + 1 + 0 = 6.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

SLIDE 19

Upper Bounds on f (p)

Legendre’s Thm: n! contains factor p exactly

k≥1

n

pk

times.

Ex: p = 3, n = 17. So: 17

3

+

17

32

+

17

33

+ . . . = 5 + 1 + 0 = 6.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

SLIDE 20

Upper Bounds on f (p)

Legendre’s Thm: n! contains factor p exactly

k≥1

n

pk

times.

Ex: p = 3, n = 17. So: 17

3

+

17

32

+

17

33

+ . . . = 5 + 1 + 0 = 6.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 Cor: f (p) = pr, where r =

k≥0

2n

pk

− 2
n

pk

.

SLIDE 21

Upper Bounds on f (p)

Legendre’s Thm: n! contains factor p exactly

k≥1

n

pk

times.

Ex: p = 3, n = 17. So: 17

3

+

17

32

+

17

33

+ . . . = 5 + 1 + 0 = 6.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 Cor: f (p) = pr, where r =

k≥0

2n

pk

− 2
n

pk

.

1.

2n

pk

− 2
n

pk

< 2n

pk − 2( n pk − 1) = 2.

SLIDE 22

Upper Bounds on f (p)

Legendre’s Thm: n! contains factor p exactly

k≥1

n

pk

times.

Ex: p = 3, n = 17. So: 17

3

+

17

32

+

17

33

+ . . . = 5 + 1 + 0 = 6.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 Cor: f (p) = pr, where r =

k≥0

2n

pk

− 2
n

pk

.

1.

2n

pk

− 2
n

pk

< 2n

pk − 2( n pk − 1) = 2.

2. If pk > 2n, then summand is 0, so f (p) ≤ 2n.

SLIDE 23

Upper Bounds on f (p)

Legendre’s Thm: n! contains factor p exactly

k≥1

n

pk

times.

Ex: p = 3, n = 17. So: 17

3

+

17

32

+

17

33

+ . . . = 5 + 1 + 0 = 6.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 Cor: f (p) = pr, where r =

k≥0

2n

pk

− 2
n

pk

.

1.

2n

pk

− 2
n

pk

< 2n

pk − 2( n pk − 1) = 2.

2. If pk > 2n, then summand is 0, so f (p) ≤ 2n.
3. If p >

√ 2n, then p2 > 2n, so f (p) ≤ p.

SLIDE 24

Upper Bounds on f (p)

Legendre’s Thm: n! contains factor p exactly

k≥1

n

pk

times.

Ex: p = 3, n = 17. So: 17

3

+

17

32

+

17

33

+ . . . = 5 + 1 + 0 = 6.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 Cor: f (p) = pr, where r =

k≥0

2n

pk

− 2
n

pk

.

1.

2n

pk

− 2
n

pk

< 2n

pk − 2( n pk − 1) = 2.

2. If pk > 2n, then summand is 0, so f (p) ≤ 2n.
3. If p >

√ 2n, then p2 > 2n, so f (p) ≤ p. Lemma 2’: If Bertrand’s is false, then ∃ n > 4000 s.t. 4n/(2n) ≤

p≤ 2n

3

f (p)

SLIDE 25

Upper Bounds on f (p)

Legendre’s Thm: n! contains factor p exactly

k≥1

n

pk

times.

Ex: p = 3, n = 17. So: 17

3

+

17

32

+

17

33

+ . . . = 5 + 1 + 0 = 6.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 Cor: f (p) = pr, where r =

k≥0

2n

pk

− 2
n

pk

.

1.

2n

pk

− 2
n

pk

< 2n

pk − 2( n pk − 1) = 2.

2. If pk > 2n, then summand is 0, so f (p) ≤ 2n.
3. If p >

√ 2n, then p2 > 2n, so f (p) ≤ p. Lemma 2’: If Bertrand’s is false, then ∃ n > 4000 s.t. 4n/(2n) ≤

p≤ 2n

3

f (p) ≤

p≤

√ 2n

2n

√

2n<p≤ 2n

3

p

SLIDE 26

Upper Bounds on f (p)

Legendre’s Thm: n! contains factor p exactly

k≥1

n

pk

times.

Ex: p = 3, n = 17. So: 17

3

+

17

32

+

17

33

+ . . . = 5 + 1 + 0 = 6.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 Cor: f (p) = pr, where r =

k≥0

2n

pk

− 2
n

pk

.

1.

2n

pk

− 2
n

pk

< 2n

pk − 2( n pk − 1) = 2.

2. If pk > 2n, then summand is 0, so f (p) ≤ 2n.
3. If p >

√ 2n, then p2 > 2n, so f (p) ≤ p. Lemma 2’: If Bertrand’s is false, then ∃ n > 4000 s.t. 4n/(2n) ≤

p≤ 2n

3

f (p) ≤

p≤

√ 2n

2n

√

2n<p≤ 2n

3

p < (2n)

√ 2n p≤ 2n

3

p

SLIDE 27

Upper Bounds on f (p)

Legendre’s Thm: n! contains factor p exactly

k≥1

n

pk

times.

Ex: p = 3, n = 17. So: 17

3

+

17

32

+

17

33

+ . . . = 5 + 1 + 0 = 6.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 Cor: f (p) = pr, where r =

k≥0

2n

pk

− 2
n

pk

.

1.

2n

pk

− 2
n

pk

< 2n

pk − 2( n pk − 1) = 2.

2. If pk > 2n, then summand is 0, so f (p) ≤ 2n.
3. If p >

√ 2n, then p2 > 2n, so f (p) ≤ p. Lemma 2’: If Bertrand’s is false, then ∃ n > 4000 s.t. 4n/(2n) ≤

p≤ 2n

3

f (p) ≤

p≤

√ 2n

2n

√

2n<p≤ 2n

3

p < (2n)

√ 2n p≤ 2n

3

p So we need bounds on

p≤ 2n

3 p in terms of 4x. . .

SLIDE 28

Upper Bounds on

p≤x p

Lemma 3: For all real x ≥ 2, we have

p≤x p ≤ 4x−1.

SLIDE 29

Upper Bounds on

p≤x p

Lemma 3: For all real x ≥ 2, we have

p≤x p ≤ 4x−1.

Pf: If q is max prime ≤ x, then suffices to show:

p≤q p ≤ 4q−1,

SLIDE 30

Upper Bounds on

p≤x p

Lemma 3: For all real x ≥ 2, we have

p≤x p ≤ 4x−1.

Pf: If q is max prime ≤ x, then suffices to show:

p≤q p ≤ 4q−1,

since then

p≤x p = p≤q p ≤ 4q−1 ≤ 4x−1.

SLIDE 31

Upper Bounds on

p≤x p

Lemma 3: For all real x ≥ 2, we have

p≤x p ≤ 4x−1.

Pf: If q is max prime ≤ x, then suffices to show:

p≤q p ≤ 4q−1,

since then

p≤x p = p≤q p ≤ 4q−1 ≤ 4x−1.

We use strong induction on q and q = 2 is easy: 2 ≤ 4.

SLIDE 32

Upper Bounds on

p≤x p

Lemma 3: For all real x ≥ 2, we have

p≤x p ≤ 4x−1.

Pf: If q is max prime ≤ x, then suffices to show:

p≤q p ≤ 4q−1,

since then

p≤x p = p≤q p ≤ 4q−1 ≤ 4x−1.

We use strong induction on q and q = 2 is easy: 2 ≤ 4. Consider q = 2m + 1.

SLIDE 33

Upper Bounds on

p≤x p

Lemma 3: For all real x ≥ 2, we have

p≤x p ≤ 4x−1.

Pf: If q is max prime ≤ x, then suffices to show:

p≤q p ≤ 4q−1,

since then

p≤x p = p≤q p ≤ 4q−1 ≤ 4x−1.

We use strong induction on q and q = 2 is easy: 2 ≤ 4. Consider q = 2m + 1.

p≤2m+1

p =

SLIDE 34

Upper Bounds on

p≤x p

Lemma 3: For all real x ≥ 2, we have

p≤x p ≤ 4x−1.

Pf: If q is max prime ≤ x, then suffices to show:

p≤q p ≤ 4q−1,

since then

p≤x p = p≤q p ≤ 4q−1 ≤ 4x−1.

We use strong induction on q and q = 2 is easy: 2 ≤ 4. Consider q = 2m + 1.

p≤2m+1

p =

p≤m+1

p

m+1<p

≤2m+1

p

SLIDE 35

Upper Bounds on

p≤x p

Lemma 3: For all real x ≥ 2, we have

p≤x p ≤ 4x−1.

Pf: If q is max prime ≤ x, then suffices to show:

p≤q p ≤ 4q−1,

since then

p≤x p = p≤q p ≤ 4q−1 ≤ 4x−1.

We use strong induction on q and q = 2 is easy: 2 ≤ 4. Consider q = 2m + 1.

p≤2m+1

p =

p≤m+1

p

m+1<p

≤2m+1

p ≤ 4(m+1)−1 2m + 1 m

SLIDE 36

Upper Bounds on

p≤x p

Lemma 3: For all real x ≥ 2, we have

p≤x p ≤ 4x−1.

Pf: If q is max prime ≤ x, then suffices to show:

p≤q p ≤ 4q−1,

since then

p≤x p = p≤q p ≤ 4q−1 ≤ 4x−1.

We use strong induction on q and q = 2 is easy: 2 ≤ 4. Consider q = 2m + 1.

p≤2m+1

p =

p≤m+1

p

m+1<p

≤2m+1

p ≤ 4(m+1)−1 2m + 1 m

≤ 4m(1

222m+1)

SLIDE 37

Upper Bounds on

p≤x p

Lemma 3: For all real x ≥ 2, we have

p≤x p ≤ 4x−1.

Pf: If q is max prime ≤ x, then suffices to show:

p≤q p ≤ 4q−1,

since then

p≤x p = p≤q p ≤ 4q−1 ≤ 4x−1.

We use strong induction on q and q = 2 is easy: 2 ≤ 4. Consider q = 2m + 1.

p≤2m+1

p =

p≤m+1

p

m+1<p

≤2m+1

p ≤ 4(m+1)−1 2m + 1 m

≤ 4m(1

222m+1) = 42m = 4q−1

SLIDE 38

Upper Bounds on

p≤x p

Lemma 3: For all real x ≥ 2, we have

p≤x p ≤ 4x−1.

Pf: If q is max prime ≤ x, then suffices to show:

p≤q p ≤ 4q−1,

since then

p≤x p = p≤q p ≤ 4q−1 ≤ 4x−1.

We use strong induction on q and q = 2 is easy: 2 ≤ 4. Consider q = 2m + 1.

p≤2m+1

p =

p≤m+1

p

m+1<p

≤2m+1

p ≤ 4(m+1)−1 2m + 1 m

≤ 4m(1

222m+1) = 42m = 4q−1 Lemma 2”: If Bertrand’s is false, then ∃ n > 4000 s.t. 4n/(2n) < (2n)

√ 2n4

2n 3

SLIDE 39

Upper Bounds on

p≤x p

Lemma 3: For all real x ≥ 2, we have

p≤x p ≤ 4x−1.

Pf: If q is max prime ≤ x, then suffices to show:

p≤q p ≤ 4q−1,

since then

p≤x p = p≤q p ≤ 4q−1 ≤ 4x−1.

We use strong induction on q and q = 2 is easy: 2 ≤ 4. Consider q = 2m + 1.

p≤2m+1

p =

p≤m+1

p

m+1<p

≤2m+1

p ≤ 4(m+1)−1 2m + 1 m

≤ 4m(1

222m+1) = 42m = 4q−1 Lemma 2”: If Bertrand’s is false, then ∃ n > 4000 s.t. 4n/(2n) < (2n)

√ 2n4

2n 3

4

n 3 < (2n)1+

√ 2n

SLIDE 40

Upper Bounds on

p≤x p

Lemma 3: For all real x ≥ 2, we have

p≤x p ≤ 4x−1.

Pf: If q is max prime ≤ x, then suffices to show:

p≤q p ≤ 4q−1,

since then

p≤x p = p≤q p ≤ 4q−1 ≤ 4x−1.

We use strong induction on q and q = 2 is easy: 2 ≤ 4. Consider q = 2m + 1.

p≤2m+1

p =

p≤m+1

p

m+1<p

≤2m+1

p ≤ 4(m+1)−1 2m + 1 m

≤ 4m(1

222m+1) = 42m = 4q−1 Lemma 2”: If Bertrand’s is false, then ∃ n > 4000 s.t. 4n/(2n) < (2n)

√ 2n4

2n 3

4

n 3 < (2n)1+

√ 2n

n 3 lg 4 < (1 + √ 2n) lg(2n)

SLIDE 41

Ugly inequalities

Obs. a + 1 < 2a for all a ≥ 2 (by induction)

SLIDE 42

Ugly inequalities

Obs. a + 1 < 2a for all a ≥ 2 (by induction)

So 2n

SLIDE 43

Ugly inequalities

Obs. a + 1 < 2a for all a ≥ 2 (by induction)

So 2n = (

6

√ 2n)6

SLIDE 44

Ugly inequalities

Obs. a + 1 < 2a for all a ≥ 2 (by induction)

So 2n = (

6

√ 2n)6 < ( 6 √ 2n

+ 1)6

SLIDE 45

Ugly inequalities

Obs. a + 1 < 2a for all a ≥ 2 (by induction)

So 2n = (

6

√ 2n)6 < ( 6 √ 2n

+ 1)6 < 26⌊ 6

√ 2n⌋

SLIDE 46

Ugly inequalities

Obs. a + 1 < 2a for all a ≥ 2 (by induction)

So 2n = (

6

√ 2n)6 < ( 6 √ 2n

+ 1)6 < 26⌊ 6

√ 2n⌋ ≤ 26 6 √ 2n.

SLIDE 47

Ugly inequalities

Obs. a + 1 < 2a for all a ≥ 2 (by induction)

So 2n = (

6

√ 2n)6 < ( 6 √ 2n

+ 1)6 < 26⌊ 6

√ 2n⌋ ≤ 26 6 √ 2n.

22n

SLIDE 48

Ugly inequalities

Obs. a + 1 < 2a for all a ≥ 2 (by induction)

So 2n = (

6

√ 2n)6 < ( 6 √ 2n

+ 1)6 < 26⌊ 6

√ 2n⌋ ≤ 26 6 √ 2n.

22n =

4n/33

SLIDE 49

Ugly inequalities

Obs. a + 1 < 2a for all a ≥ 2 (by induction)

So 2n = (

6

√ 2n)6 < ( 6 √ 2n

+ 1)6 < 26⌊ 6

√ 2n⌋ ≤ 26 6 √ 2n.

22n =

4n/33 ≤ (2n)3(1+

√ 2n)

SLIDE 50

Ugly inequalities

Obs. a + 1 < 2a for all a ≥ 2 (by induction)

So 2n = (

6

√ 2n)6 < ( 6 √ 2n

+ 1)6 < 26⌊ 6

√ 2n⌋ ≤ 26 6 √ 2n.

22n =

4n/33 ≤ (2n)3(1+

√ 2n) ≤

26 6

√ 2n3(1+ √ 2n)

SLIDE 51

Ugly inequalities

Obs. a + 1 < 2a for all a ≥ 2 (by induction)

So 2n = (

6

√ 2n)6 < ( 6 √ 2n

+ 1)6 < 26⌊ 6

√ 2n⌋ ≤ 26 6 √ 2n.

22n =

4n/33 ≤ (2n)3(1+

√ 2n) ≤

26 6

√ 2n3(1+ √ 2n)

Since n ≥ 50, we have 18 < 2 √ 2n, so

SLIDE 52

Ugly inequalities

Obs. a + 1 < 2a for all a ≥ 2 (by induction)

So 2n = (

6

√ 2n)6 < ( 6 √ 2n

+ 1)6 < 26⌊ 6

√ 2n⌋ ≤ 26 6 √ 2n.

22n =

4n/33 ≤ (2n)3(1+

√ 2n) ≤

26 6

√ 2n3(1+ √ 2n)

Since n ≥ 50, we have 18 < 2 √ 2n, so 22n = 2

6

√ 2n(18+18 √ 2n)

SLIDE 53

Ugly inequalities

Obs. a + 1 < 2a for all a ≥ 2 (by induction)

So 2n = (

6

√ 2n)6 < ( 6 √ 2n

+ 1)6 < 26⌊ 6

√ 2n⌋ ≤ 26 6 √ 2n.

22n =

4n/33 ≤ (2n)3(1+

√ 2n) ≤

26 6

√ 2n3(1+ √ 2n)

Since n ≥ 50, we have 18 < 2 √ 2n, so 22n = 2

6

√ 2n(18+18 √ 2n) ≤ 2

6

√ 2n(20 √ 2n)

SLIDE 54

Ugly inequalities

Obs. a + 1 < 2a for all a ≥ 2 (by induction)

So 2n = (

6

√ 2n)6 < ( 6 √ 2n

+ 1)6 < 26⌊ 6

√ 2n⌋ ≤ 26 6 √ 2n.

22n =

4n/33 ≤ (2n)3(1+

√ 2n) ≤

26 6

√ 2n3(1+ √ 2n)

Since n ≥ 50, we have 18 < 2 √ 2n, so 22n = 2

6

√ 2n(18+18 √ 2n) ≤ 2

6

√ 2n(20 √ 2n) = 220(2n)2/3

SLIDE 55

Ugly inequalities

Obs. a + 1 < 2a for all a ≥ 2 (by induction)

So 2n = (

6

√ 2n)6 < ( 6 √ 2n

+ 1)6 < 26⌊ 6

√ 2n⌋ ≤ 26 6 √ 2n.

22n =

4n/33 ≤ (2n)3(1+

√ 2n) ≤

26 6

√ 2n3(1+ √ 2n)

Since n ≥ 50, we have 18 < 2 √ 2n, so 22n = 2

6

√ 2n(18+18 √ 2n) ≤ 2

6

√ 2n(20 √ 2n) = 220(2n)2/3

Taking logs gives: 2n ≤ 20(2n)2/3

SLIDE 56

Ugly inequalities

Obs. a + 1 < 2a for all a ≥ 2 (by induction)

So 2n = (

6

√ 2n)6 < ( 6 √ 2n

+ 1)6 < 26⌊ 6

√ 2n⌋ ≤ 26 6 √ 2n.

22n =

4n/33 ≤ (2n)3(1+

√ 2n) ≤

26 6

√ 2n3(1+ √ 2n)

Since n ≥ 50, we have 18 < 2 √ 2n, so 22n = 2

6

√ 2n(18+18 √ 2n) ≤ 2

6

√ 2n(20 √ 2n) = 220(2n)2/3

Taking logs gives: 2n ≤ 20(2n)2/3 2n ≤ 8000

SLIDE 57

Ugly inequalities

Obs. a + 1 < 2a for all a ≥ 2 (by induction)

So 2n = (

6

√ 2n)6 < ( 6 √ 2n

+ 1)6 < 26⌊ 6

√ 2n⌋ ≤ 26 6 √ 2n.

22n =

4n/33 ≤ (2n)3(1+

√ 2n) ≤

26 6

√ 2n3(1+ √ 2n)

Since n ≥ 50, we have 18 < 2 √ 2n, so 22n = 2

6

√ 2n(18+18 √ 2n) ≤ 2

6

√ 2n(20 √ 2n) = 220(2n)2/3

Taking logs gives: 2n ≤ 20(2n)2/3 2n ≤ 8000 n ≤ 4000.

SLIDE 58

Overview Redux

Bertrand’s Postulate For every positive n, there exists a prime p s.t. n < p ≤ 2n.

SLIDE 59

Overview Redux

Bertrand’s Postulate For every positive n, there exists a prime p s.t. n < p ≤ 2n. Lemma 1: Bertrand’s Postulate holds for 1 ≤ n ≤ 4000.

SLIDE 60

Overview Redux

Bertrand’s Postulate For every positive n, there exists a prime p s.t. n < p ≤ 2n. Lemma 1: Bertrand’s Postulate holds for 1 ≤ n ≤ 4000. Legendre’s Theorem implies

SLIDE 61

Overview Redux

Bertrand’s Postulate For every positive n, there exists a prime p s.t. n < p ≤ 2n. Lemma 1: Bertrand’s Postulate holds for 1 ≤ n ≤ 4000. Legendre’s Theorem implies For all p, f (p) ≤ 2n;

SLIDE 62

Overview Redux

Bertrand’s Postulate For every positive n, there exists a prime p s.t. n √ 2n, then f (p) ≤ p.

SLIDE 63

Overview Redux

Bertrand’s Postulate For every positive n, there exists a prime p s.t. n √ 2n, then f (p) ≤ p. Lemma 3: For all real x ≥ 2, we have

p≤x p ≤ 4x−1.

SLIDE 64

Overview Redux

Bertrand’s Postulate For every positive n, there exists a prime p s.t. n √ 2n, then f (p) ≤ p. Lemma 3: For all real x ≥ 2, we have

p≤x p ≤ 4x−1.

Lemma 2∗: If Bertrand’s is false, then ∃ n > 4000 s.t. 4n/(2n) ≤ 2n n

SLIDE 65

Overview Redux

Bertrand’s Postulate For every positive n, there exists a prime p s.t. n √ 2n, then f (p) ≤ p. Lemma 3: For all real x ≥ 2, we have

p≤x p ≤ 4x−1.

Lemma 2∗: If Bertrand’s is false, then ∃ n > 4000 s.t. 4n/(2n) ≤ 2n n

=
p≤

√ 2n

f (p)

√

2n<p≤ 2n

3

f (p)

2n

3 <p≤n

f (p)

n<p≤2n

f (p).

SLIDE 66

Overview Redux

Bertrand’s Postulate For every positive n, there exists a prime p s.t. n √ 2n, then f (p) ≤ p. Lemma 3: For all real x ≥ 2, we have

p≤x p ≤ 4x−1.

Lemma 2∗: If Bertrand’s is false, then ∃ n > 4000 s.t. 4n/(2n) ≤ 2n n

=
p≤

√ 2n

f (p)

√

2n<p≤ 2n

3

f (p)

2n

3 <p≤n

f (p)

n<p≤2n

f (p). 1

SLIDE 67

Overview Redux

Bertrand’s Postulate For every positive n, there exists a prime p s.t. n √ 2n, then f (p) ≤ p. Lemma 3: For all real x ≥ 2, we have

p≤x p ≤ 4x−1.

Lemma 2∗: If Bertrand’s is false, then ∃ n > 4000 s.t. 4n/(2n) ≤ 2n n

=
p≤

√ 2n

f (p)

√

2n<p≤ 2n

3

f (p)

2n

3 <p≤n

f (p)

n<p≤2n

f (p). 1 1

SLIDE 68

Overview Redux

Bertrand’s Postulate For every positive n, there exists a prime p s.t. n √ 2n, then f (p) ≤ p. Lemma 3: For all real x ≥ 2, we have

p≤x p ≤ 4x−1.

Lemma 2∗: If Bertrand’s is false, then ∃ n > 4000 s.t. 4n/(2n) ≤ 2n n

≤
p≤

√ 2n

f (p)

√

2n<p≤ 2n

3

f (p)

2n

3 <p≤n

f (p)

n<p≤2n

f (p). 1 1 2n

SLIDE 69

Overview Redux

Bertrand’s Postulate For every positive n, there exists a prime p s.t. n √ 2n, then f (p) ≤ p. Lemma 3: For all real x ≥ 2, we have

p≤x p ≤ 4x−1.

Lemma 2∗: If Bertrand’s is false, then ∃ n > 4000 s.t. 4n/(2n) ≤ 2n n

≤
p≤

√ 2n

f (p)

√

2n<p≤ 2n

3

f (p)

2n

3 <p≤n

f (p)

n<p≤2n

f (p). 1 1 2n p

SLIDE 70

Overview Redux

Bertrand’s Postulate For every positive n, there exists a prime p s.t. n √ 2n, then f (p) ≤ p. Lemma 3: For all real x ≥ 2, we have

p≤x p ≤ 4x−1.

Lemma 2∗: If Bertrand’s is false, then ∃ n > 4000 s.t. 4n/(2n) ≤ 2n n

≤
p≤

√ 2n

f (p)

√

2n<p≤ 2n

3

f (p)

2n

3 <p≤n

f (p)

n<p≤2n

f (p). 1 1 2n p But 4n/(2n) ≤ (2n)

√ 2n42n/3 implies n ≤ 4000.

SLIDE 71

Overview Redux

Bertrand’s Postulate For every positive n, there exists a prime p s.t. n √ 2n, then f (p) ≤ p. Lemma 3: For all real x ≥ 2, we have