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Page 1: Bertrand’s Postulate - Lola Thompson · Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33. Fun With Binomial Coe cients Another interesting observation that

Bertrand’s Postulate

Lola Thompson

Ross Program

July 3, 2009

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 1 / 33

Page 2: Bertrand’s Postulate - Lola Thompson · Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33. Fun With Binomial Coe cients Another interesting observation that

Bertrand’s Postulate

“I’ve said it once and I’ll say it again: There’s always a prime between nand 2n.”

-Joseph Bertrand, conjectured in 1845

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 2 / 33

Page 3: Bertrand’s Postulate - Lola Thompson · Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33. Fun With Binomial Coe cients Another interesting observation that

Bertrand’s Postulate

Bertrand did not prove his postulate. He verified the statement (by hand)for all positive integers n up to 6,000,000.

In 1850, Chebyshev proved Bertrand’s Postulate. For this reason, it is alsoreferred to as “Chebyshev’s Theorem.”

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 3 / 33

Page 4: Bertrand’s Postulate - Lola Thompson · Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33. Fun With Binomial Coe cients Another interesting observation that

Bertrand’s Postulate

Many other proofs have been found in the time since Chebyshev firstproved this theorem. We will follow a proof due to Ramanujan and Erdos.

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 4 / 33

Page 5: Bertrand’s Postulate - Lola Thompson · Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33. Fun With Binomial Coe cients Another interesting observation that

Outline

1 Fun With Binomial Coefficients

2 A Few New Arithmetic Functions

3 An Upper Bound for ψ(x)

4 Proving Bertrand’s PostulateThe Setup: ABC =

(2nn

)A Lower Bound for

(2nn

)An Upper Bound for CAn Upper Bound for BPutting Everything Together

5 Generalizations

6 Some Neat Applications

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 5 / 33

Page 6: Bertrand’s Postulate - Lola Thompson · Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33. Fun With Binomial Coe cients Another interesting observation that

Fun With Binomial Coefficients

Definition

n! = # of ways of ordering n elements

Definition(nk

)= # of ways of choosing k elements from a set containing n objects,

without worrying about order.

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 6 / 33

Page 7: Bertrand’s Postulate - Lola Thompson · Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33. Fun With Binomial Coe cients Another interesting observation that

Fun With Binomial Coefficients

There is a curious relationship between(nk

)and Pascal’s Triangle.

If we write down Pascal’s Triangle, the first few rows are:11 11 2 11 3 3 1

We can obtain the numbers in the next row by adding adjacent pairsof numbers from the previous row:1 3 3 11 4 6 4 1

The elements in the 4th row are precisely(40

)through

(44

).

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33

Page 8: Bertrand’s Postulate - Lola Thompson · Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33. Fun With Binomial Coe cients Another interesting observation that

Fun With Binomial Coefficients

There is a curious relationship between(nk

)and Pascal’s Triangle.

If we write down Pascal’s Triangle, the first few rows are:11 11 2 11 3 3 1

We can obtain the numbers in the next row by adding adjacent pairsof numbers from the previous row:1 3 3 11 4 6 4 1

The elements in the 4th row are precisely(40

)through

(44

).

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33

Page 9: Bertrand’s Postulate - Lola Thompson · Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33. Fun With Binomial Coe cients Another interesting observation that

Fun With Binomial Coefficients

There is a curious relationship between(nk

)and Pascal’s Triangle.

If we write down Pascal’s Triangle, the first few rows are:11 11 2 11 3 3 1

We can obtain the numbers in the next row by adding adjacent pairsof numbers from the previous row:1 3 3 11 4 6 4 1

The elements in the 4th row are precisely(40

)through

(44

).

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33

Page 10: Bertrand’s Postulate - Lola Thompson · Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33. Fun With Binomial Coe cients Another interesting observation that

Fun With Binomial Coefficients

There is a curious relationship between(nk

)and Pascal’s Triangle.

If we write down Pascal’s Triangle, the first few rows are:11 11 2 11 3 3 1

We can obtain the numbers in the next row by adding adjacent pairsof numbers from the previous row:1 3 3 11 4 6 4 1

The elements in the 4th row are precisely(40

)through

(44

).

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33

Page 11: Bertrand’s Postulate - Lola Thompson · Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33. Fun With Binomial Coe cients Another interesting observation that

Fun With Binomial Coefficients

Another interesting observation that you might make is that, at leastin the examples above, the sum of all of the elements in a row is apower of 2.

For example, 1 + 2 + 1 = 22 and 1 + 3 + 3 + 1 = 23.

Exercise:n∑

k=0

(nk

)= 2n.

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 8 / 33

Page 12: Bertrand’s Postulate - Lola Thompson · Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33. Fun With Binomial Coe cients Another interesting observation that

Fun With Binomial Coefficients

Another interesting observation that you might make is that, at leastin the examples above, the sum of all of the elements in a row is apower of 2.

For example, 1 + 2 + 1 = 22 and 1 + 3 + 3 + 1 = 23.

Exercise:n∑

k=0

(nk

)= 2n.

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 8 / 33

Page 13: Bertrand’s Postulate - Lola Thompson · Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33. Fun With Binomial Coe cients Another interesting observation that

Fun With Binomial Coefficients

Another interesting observation that you might make is that, at leastin the examples above, the sum of all of the elements in a row is apower of 2.

For example, 1 + 2 + 1 = 22 and 1 + 3 + 3 + 1 = 23.

Exercise:n∑

k=0

(nk

)= 2n.

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 8 / 33

Page 14: Bertrand’s Postulate - Lola Thompson · Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33. Fun With Binomial Coe cients Another interesting observation that

Fun With Binomial Coefficients

One particular type of binomial coefficient will be of interest to us aswe prove Bertrand’s Postulate. That coefficient is

(2nn

), the coefficient

in the center of the 2nth row of Pascal’s Triangle.

Which primes divide(2n

n

)? Let’s look at some examples.(4

2

)= 6. Which primes divide 6? What about

(63

)?(84

)?(10

5

)?

Any conjectures about when a prime has to divide(2n

n

)?

Exercise: p |(2n

n

)if there exists a positive integer j with n < pj ≤ 2n

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 9 / 33

Page 15: Bertrand’s Postulate - Lola Thompson · Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33. Fun With Binomial Coe cients Another interesting observation that

Fun With Binomial Coefficients

One particular type of binomial coefficient will be of interest to us aswe prove Bertrand’s Postulate. That coefficient is

(2nn

), the coefficient

in the center of the 2nth row of Pascal’s Triangle.

Which primes divide(2n

n

)? Let’s look at some examples.

(42

)= 6. Which primes divide 6? What about

(63

)?(84

)?(10

5

)?

Any conjectures about when a prime has to divide(2n

n

)?

Exercise: p |(2n

n

)if there exists a positive integer j with n < pj ≤ 2n

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 9 / 33

Page 16: Bertrand’s Postulate - Lola Thompson · Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33. Fun With Binomial Coe cients Another interesting observation that

Fun With Binomial Coefficients

One particular type of binomial coefficient will be of interest to us aswe prove Bertrand’s Postulate. That coefficient is

(2nn

), the coefficient

in the center of the 2nth row of Pascal’s Triangle.

Which primes divide(2n

n

)? Let’s look at some examples.(4

2

)= 6. Which primes divide 6? What about

(63

)?(84

)?(10

5

)?

Any conjectures about when a prime has to divide(2n

n

)?

Exercise: p |(2n

n

)if there exists a positive integer j with n < pj ≤ 2n

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 9 / 33

Page 17: Bertrand’s Postulate - Lola Thompson · Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33. Fun With Binomial Coe cients Another interesting observation that

Fun With Binomial Coefficients

One particular type of binomial coefficient will be of interest to us aswe prove Bertrand’s Postulate. That coefficient is

(2nn

), the coefficient

in the center of the 2nth row of Pascal’s Triangle.

Which primes divide(2n

n

)? Let’s look at some examples.(4

2

)= 6. Which primes divide 6? What about

(63

)?(84

)?(10

5

)?

Any conjectures about when a prime has to divide(2n

n

)?

Exercise: p |(2n

n

)if there exists a positive integer j with n < pj ≤ 2n

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 9 / 33

Page 18: Bertrand’s Postulate - Lola Thompson · Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33. Fun With Binomial Coe cients Another interesting observation that

Fun With Binomial CoefficientsAssuming that those exercises are true, we can prove two results:

Lemma (1)∏p:n<pj≤2n

p |(

2n

n

).

Proof We know that each prime p with n < pj ≤ 2n divides(2nn

).Since the primes are distinct, they are pairwise relatively

prime.Thus, their product must divide(2n

n

).

Lemma (2)(2nn

)≤ 4n for all n ≥ 0.

Proof Look at the 2nth row of Pascal’s Triangle:(2n

n

)is in the center,

22n = sum of all terms in the row.

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 10 / 33

Page 19: Bertrand’s Postulate - Lola Thompson · Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33. Fun With Binomial Coe cients Another interesting observation that

Fun With Binomial CoefficientsAssuming that those exercises are true, we can prove two results:

Lemma (1)∏p:n<pj≤2n

p |(

2n

n

).

Proof We know that each prime p with n < pj ≤ 2n divides(2nn

).Since the primes are distinct, they are pairwise relatively

prime.Thus, their product must divide(2n

n

).

Lemma (2)(2nn

)≤ 4n for all n ≥ 0.

Proof Look at the 2nth row of Pascal’s Triangle:(2n

n

)is in the center,

22n = sum of all terms in the row.

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 10 / 33

Page 20: Bertrand’s Postulate - Lola Thompson · Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33. Fun With Binomial Coe cients Another interesting observation that

Fun With Binomial CoefficientsAssuming that those exercises are true, we can prove two results:

Lemma (1)∏p:n<pj≤2n

p |(

2n

n

).

Proof We know that each prime p with n < pj ≤ 2n divides(2nn

).Since the primes are distinct, they are pairwise relatively

prime.Thus, their product must divide(2n

n

).

Lemma (2)(2nn

)≤ 4n for all n ≥ 0.

Proof Look at the 2nth row of Pascal’s Triangle:(2n

n

)is in the center,

22n = sum of all terms in the row.

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 10 / 33

Page 21: Bertrand’s Postulate - Lola Thompson · Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33. Fun With Binomial Coe cients Another interesting observation that

Fun With Binomial CoefficientsAssuming that those exercises are true, we can prove two results:

Lemma (1)∏p:n<pj≤2n

p |(

2n

n

).

Proof We know that each prime p with n < pj ≤ 2n divides(2nn

).Since the primes are distinct, they are pairwise relatively

prime.Thus, their product must divide(2n

n

).

Lemma (2)(2nn

)≤ 4n for all n ≥ 0.

Proof Look at the 2nth row of Pascal’s Triangle:(2n

n

)is in the center,

22n = sum of all terms in the row.Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 10 / 33

Page 22: Bertrand’s Postulate - Lola Thompson · Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33. Fun With Binomial Coe cients Another interesting observation that

A Few New Arithmetic FunctionsWe’re already familiar with σ, τ and ϕ. Let’s define some newarithmetic functions:

Definition

Λ(n) =

{log p, n = pk , k > 0

0, otherwise

Examples: Λ(2) = log 2, Λ(4) = log 2, Λ(6) = 0.

Definition

ψ(x) =∑n≤x

Λ(n)

Example:ψ(6) = 0 + log 2 + log 3 + log 2 + log 5 + 0 = log(22 · 3 · 5) = log(60).

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 11 / 33

Page 23: Bertrand’s Postulate - Lola Thompson · Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33. Fun With Binomial Coe cients Another interesting observation that

A Few New Arithmetic FunctionsWe’re already familiar with σ, τ and ϕ. Let’s define some newarithmetic functions:

Definition

Λ(n) =

{log p, n = pk , k > 0

0, otherwise

Examples: Λ(2) = log 2, Λ(4) = log 2, Λ(6) = 0.

Definition

ψ(x) =∑n≤x

Λ(n)

Example:ψ(6) = 0 + log 2 + log 3 + log 2 + log 5 + 0 = log(22 · 3 · 5) = log(60).

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 11 / 33

Page 24: Bertrand’s Postulate - Lola Thompson · Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33. Fun With Binomial Coe cients Another interesting observation that

A Few New Arithmetic FunctionsWe’re already familiar with σ, τ and ϕ. Let’s define some newarithmetic functions:

Definition

Λ(n) =

{log p, n = pk , k > 0

0, otherwise

Examples: Λ(2) = log 2, Λ(4) = log 2, Λ(6) = 0.

Definition

ψ(x) =∑n≤x

Λ(n)

Example:ψ(6) = 0 + log 2 + log 3 + log 2 + log 5 + 0 = log(22 · 3 · 5) = log(60).

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 11 / 33

Page 25: Bertrand’s Postulate - Lola Thompson · Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33. Fun With Binomial Coe cients Another interesting observation that

A Few New Arithmetic FunctionsWe’re already familiar with σ, τ and ϕ. Let’s define some newarithmetic functions:

Definition

Λ(n) =

{log p, n = pk , k > 0

0, otherwise

Examples: Λ(2) = log 2, Λ(4) = log 2, Λ(6) = 0.

Definition

ψ(x) =∑n≤x

Λ(n)

Example:ψ(6) = 0 + log 2 + log 3 + log 2 + log 5 + 0 = log(22 · 3 · 5) = log(60).

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 11 / 33

Page 26: Bertrand’s Postulate - Lola Thompson · Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33. Fun With Binomial Coe cients Another interesting observation that

An Upper Bound for ψ(x)

Lemma

ψ(x) ≤ x log 4

For now, let’s pretend that x ∈ Z. This will allow us to use theWell-Ordering Principle.

Base Case: If n = 1, then ψ(1) =∑n≤1

Λ(n) = Λ(1) = 0.

(This is certainly ≤ 1·log 4)

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 12 / 33

Page 27: Bertrand’s Postulate - Lola Thompson · Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33. Fun With Binomial Coe cients Another interesting observation that

Proof of Upper Bound for ψ(x)

Let S = {x ∈ Z+ | ψ(x) > x log 4}. Assume that S is non-empty. Then,by Well-Ordering Principle, S has a least element. Call it l .

Case 1: l = 2k

ψ(2k)− ψ(k)

=∑n≤2k

Λ(n)−∑n≤k

Λ(n) (from definition of ψ)

=∑

k<n≤2k

Λ(n) (canceling terms)

≤ log(2k

k

)(since Λ(n) = p if n = pj and 0 otherwise,so this line follows from Lemma (1)after taking log of both sides)

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 13 / 33

Page 28: Bertrand’s Postulate - Lola Thompson · Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33. Fun With Binomial Coe cients Another interesting observation that

Proof of Upper Bound for ψ(x)

Let S = {x ∈ Z+ | ψ(x) > x log 4}. Assume that S is non-empty. Then,by Well-Ordering Principle, S has a least element. Call it l .

Case 1: l = 2k

ψ(2k)− ψ(k)

=∑n≤2k

Λ(n)−∑n≤k

Λ(n) (from definition of ψ)

=∑

k<n≤2k

Λ(n) (canceling terms)

≤ log(2k

k

)(since Λ(n) = p if n = pj and 0 otherwise,so this line follows from Lemma (1)after taking log of both sides)

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 13 / 33

Page 29: Bertrand’s Postulate - Lola Thompson · Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33. Fun With Binomial Coe cients Another interesting observation that

Proof of Upper Bound for ψ(x)

Let S = {x ∈ Z+ | ψ(x) > x log 4}. Assume that S is non-empty. Then,by Well-Ordering Principle, S has a least element. Call it l .

Case 1: l = 2k

ψ(2k)− ψ(k)

=∑n≤2k

Λ(n)−∑n≤k

Λ(n) (from definition of ψ)

=∑

k<n≤2k

Λ(n) (canceling terms)

≤ log(2k

k

)(since Λ(n) = p if n = pj and 0 otherwise,so this line follows from Lemma (1)after taking log of both sides)

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 13 / 33

Page 30: Bertrand’s Postulate - Lola Thompson · Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33. Fun With Binomial Coe cients Another interesting observation that

Proof of Upper Bound for ψ(x)

Let S = {x ∈ Z+ | ψ(x) > x log 4}. Assume that S is non-empty. Then,by Well-Ordering Principle, S has a least element. Call it l .

Case 1: l = 2k

ψ(2k)− ψ(k)

=∑n≤2k

Λ(n)−∑n≤k

Λ(n) (from definition of ψ)

=∑

k<n≤2k

Λ(n) (canceling terms)

≤ log(2k

k

)(since Λ(n) = p if n = pj and 0 otherwise,so this line follows from Lemma (1)after taking log of both sides)

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 13 / 33

Page 31: Bertrand’s Postulate - Lola Thompson · Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33. Fun With Binomial Coe cients Another interesting observation that

Proof of Upper Bound for ψ(x)

From the previous slide: ψ(2k)− ψ(k) ≤ log(2k

k

).

From Lemma (2),(2k

k

)≤ 4k , i.e. ψ(2k)− ψ(k) ≤log(4k).

From high school math: log(4k) = k log 4.

So ψ(2k) ≤ ψ(k) + k log 4

≤ k log 4+k log 4 (since l = 2k was the smallest element in S ,so k /∈ S ⇒ ψ(k) ≤ k log 4)

∴ ψ(2k) ≤ 2k log 4, ie. l cannot be even.

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 14 / 33

Page 32: Bertrand’s Postulate - Lola Thompson · Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33. Fun With Binomial Coe cients Another interesting observation that

Proof of Upper Bound for ψ(x)

From the previous slide: ψ(2k)− ψ(k) ≤ log(2k

k

).

From Lemma (2),(2k

k

)≤ 4k , i.e. ψ(2k)− ψ(k) ≤log(4k).

From high school math: log(4k) = k log 4.

So ψ(2k) ≤ ψ(k) + k log 4

≤ k log 4+k log 4 (since l = 2k was the smallest element in S ,so k /∈ S ⇒ ψ(k) ≤ k log 4)

∴ ψ(2k) ≤ 2k log 4, ie. l cannot be even.

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 14 / 33

Page 33: Bertrand’s Postulate - Lola Thompson · Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33. Fun With Binomial Coe cients Another interesting observation that

Proof of Upper Bound for ψ(x)

From the previous slide: ψ(2k)− ψ(k) ≤ log(2k

k

).

From Lemma (2),(2k

k

)≤ 4k , i.e. ψ(2k)− ψ(k) ≤log(4k).

From high school math: log(4k) = k log 4.

So ψ(2k) ≤ ψ(k) + k log 4

≤ k log 4+k log 4 (since l = 2k was the smallest element in S ,so k /∈ S ⇒ ψ(k) ≤ k log 4)

∴ ψ(2k) ≤ 2k log 4, ie. l cannot be even.

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 14 / 33

Page 34: Bertrand’s Postulate - Lola Thompson · Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33. Fun With Binomial Coe cients Another interesting observation that

Proof of Upper Bound for ψ(x)

From the previous slide: ψ(2k)− ψ(k) ≤ log(2k

k

).

From Lemma (2),(2k

k

)≤ 4k , i.e. ψ(2k)− ψ(k) ≤log(4k).

From high school math: log(4k) = k log 4.

So ψ(2k) ≤ ψ(k) + k log 4

≤ k log 4+k log 4 (since l = 2k was the smallest element in S ,so k /∈ S ⇒ ψ(k) ≤ k log 4)

∴ ψ(2k) ≤ 2k log 4, ie. l cannot be even.

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 14 / 33

Page 35: Bertrand’s Postulate - Lola Thompson · Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33. Fun With Binomial Coe cients Another interesting observation that

Proof of Upper Bound for ψ(x)

From the previous slide: ψ(2k)− ψ(k) ≤ log(2k

k

).

From Lemma (2),(2k

k

)≤ 4k , i.e. ψ(2k)− ψ(k) ≤log(4k).

From high school math: log(4k) = k log 4.

So ψ(2k) ≤ ψ(k) + k log 4

≤ k log 4+k log 4 (since l = 2k was the smallest element in S ,so k /∈ S ⇒ ψ(k) ≤ k log 4)

∴ ψ(2k) ≤ 2k log 4, ie. l cannot be even.

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 14 / 33

Page 36: Bertrand’s Postulate - Lola Thompson · Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33. Fun With Binomial Coe cients Another interesting observation that

Proof of Upper Bound for ψ(x)

From the previous slide: ψ(2k)− ψ(k) ≤ log(2k

k

).

From Lemma (2),(2k

k

)≤ 4k , i.e. ψ(2k)− ψ(k) ≤log(4k).

From high school math: log(4k) = k log 4.

So ψ(2k) ≤ ψ(k) + k log 4

≤ k log 4+k log 4 (since l = 2k was the smallest element in S ,so k /∈ S ⇒ ψ(k) ≤ k log 4)

∴ ψ(2k) ≤ 2k log 4, ie. l cannot be even.

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 14 / 33

Page 37: Bertrand’s Postulate - Lola Thompson · Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33. Fun With Binomial Coe cients Another interesting observation that

Proof of Upper Bound for ψ(x)

Case 2: l = 2k + 1.

The argument for the case where l is odd is similar to the case where l iseven. The result is the same, i.e. it shows that l cannot be odd.Since the least element of S is neither even nor odd then S must empty.

Remark: The inequality ψ(x) ≤ x log 4 holds for ALL x ≥ 1 (not justintegers). Why?

ψ(x) = ψ(bxc) ≤ bxclog4 ≤ x log 4.

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 15 / 33

Page 38: Bertrand’s Postulate - Lola Thompson · Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33. Fun With Binomial Coe cients Another interesting observation that

Proof of Upper Bound for ψ(x)

Case 2: l = 2k + 1.

The argument for the case where l is odd is similar to the case where l iseven. The result is the same, i.e. it shows that l cannot be odd.Since the least element of S is neither even nor odd then S must empty.

Remark: The inequality ψ(x) ≤ x log 4 holds for ALL x ≥ 1 (not justintegers). Why?

ψ(x) = ψ(bxc) ≤ bxclog4 ≤ x log 4.

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 15 / 33

Page 39: Bertrand’s Postulate - Lola Thompson · Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33. Fun With Binomial Coe cients Another interesting observation that

Proving Bertrand’s Postulate

We will use what we have learned about(2n

n

)and ψ(x) in order to prove

our main result:

Theorem (Bertrand’s Postulate)

For every n ∈ Z+, there exists a prime p such that n < p ≤ 2n.

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 16 / 33

Page 40: Bertrand’s Postulate - Lola Thompson · Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33. Fun With Binomial Coe cients Another interesting observation that

The Setup : ABC =(

2nn

)Let A =

∏n<p≤2n p.

From Lemma (1), we know that A |(2n

n

). So, there exists m ∈ Z st.

A ·m =(2n

n

).

By the Unique Factorization Theorem, we can factor m into aproduct of primes (uniquely).

Let B = contribution to(2n

n

)from primes p ∈ (

√2n, n].

Let C = contribution to(2n

n

)from primes p ≤

√2n.

Then ABC =(2n

n

).

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 17 / 33

Page 41: Bertrand’s Postulate - Lola Thompson · Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33. Fun With Binomial Coe cients Another interesting observation that

The Setup : ABC =(

2nn

)Let A =

∏n<p≤2n p.

From Lemma (1), we know that A |(2n

n

). So, there exists m ∈ Z st.

A ·m =(2n

n

).

By the Unique Factorization Theorem, we can factor m into aproduct of primes (uniquely).

Let B = contribution to(2n

n

)from primes p ∈ (

√2n, n].

Let C = contribution to(2n

n

)from primes p ≤

√2n.

Then ABC =(2n

n

).

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 17 / 33

Page 42: Bertrand’s Postulate - Lola Thompson · Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33. Fun With Binomial Coe cients Another interesting observation that

The Setup : ABC =(

2nn

)Let A =

∏n<p≤2n p.

From Lemma (1), we know that A |(2n

n

). So, there exists m ∈ Z st.

A ·m =(2n

n

).

By the Unique Factorization Theorem, we can factor m into aproduct of primes (uniquely).

Let B = contribution to(2n

n

)from primes p ∈ (

√2n, n].

Let C = contribution to(2n

n

)from primes p ≤

√2n.

Then ABC =(2n

n

).

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 17 / 33

Page 43: Bertrand’s Postulate - Lola Thompson · Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33. Fun With Binomial Coe cients Another interesting observation that

The Setup : ABC =(

2nn

)Let A =

∏n<p≤2n p.

From Lemma (1), we know that A |(2n

n

). So, there exists m ∈ Z st.

A ·m =(2n

n

).

By the Unique Factorization Theorem, we can factor m into aproduct of primes (uniquely).

Let B = contribution to(2n

n

)from primes p ∈ (

√2n, n].

Let C = contribution to(2n

n

)from primes p ≤

√2n.

Then ABC =(2n

n

).

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 17 / 33

Page 44: Bertrand’s Postulate - Lola Thompson · Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33. Fun With Binomial Coe cients Another interesting observation that

The Setup : ABC =(

2nn

)Let A =

∏n<p≤2n p.

From Lemma (1), we know that A |(2n

n

). So, there exists m ∈ Z st.

A ·m =(2n

n

).

By the Unique Factorization Theorem, we can factor m into aproduct of primes (uniquely).

Let B = contribution to(2n

n

)from primes p ∈ (

√2n, n].

Let C = contribution to(2n

n

)from primes p ≤

√2n.

Then ABC =(2n

n

).

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 17 / 33

Page 45: Bertrand’s Postulate - Lola Thompson · Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33. Fun With Binomial Coe cients Another interesting observation that

The Setup : ABC =(

2nn

)Goal: We want to show that BC <

(2nn

).

Why does this imply that Bertrand’s Postulate holds?

If BC <(2n

n

)then A 6= 1.

But A =∏

n≤p≤2n p, so there must be a prime dividing A.

In other words, there must be a prime between n and 2n dividing(2n

n

).

This proves Bertrand’s Postulate because it proves the existence of aprime between n and 2n.

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 18 / 33

Page 46: Bertrand’s Postulate - Lola Thompson · Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33. Fun With Binomial Coe cients Another interesting observation that

The Setup : ABC =(

2nn

)Goal: We want to show that BC <

(2nn

).

Why does this imply that Bertrand’s Postulate holds?

If BC <(2n

n

)then A 6= 1.

But A =∏

n≤p≤2n p, so there must be a prime dividing A.

In other words, there must be a prime between n and 2n dividing(2n

n

).

This proves Bertrand’s Postulate because it proves the existence of aprime between n and 2n.

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 18 / 33

Page 47: Bertrand’s Postulate - Lola Thompson · Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33. Fun With Binomial Coe cients Another interesting observation that

The Setup : ABC =(

2nn

)Goal: We want to show that BC <

(2nn

).

Why does this imply that Bertrand’s Postulate holds?

If BC <(2n

n

)then A 6= 1.

But A =∏

n≤p≤2n p, so there must be a prime dividing A.

In other words, there must be a prime between n and 2n dividing(2n

n

).

This proves Bertrand’s Postulate because it proves the existence of aprime between n and 2n.

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 18 / 33

Page 48: Bertrand’s Postulate - Lola Thompson · Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33. Fun With Binomial Coe cients Another interesting observation that

The Setup : ABC =(

2nn

)Goal: We want to show that BC <

(2nn

).

Why does this imply that Bertrand’s Postulate holds?

If BC <(2n

n

)then A 6= 1.

But A =∏

n≤p≤2n p, so there must be a prime dividing A.

In other words, there must be a prime between n and 2n dividing(2n

n

).

This proves Bertrand’s Postulate because it proves the existence of aprime between n and 2n.

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 18 / 33

Page 49: Bertrand’s Postulate - Lola Thompson · Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33. Fun With Binomial Coe cients Another interesting observation that

The Setup : ABC =(

2nn

)

In order to show that BC <(2n

n

), we will find upper bounds for B and C

and we will find a lower bound for(2n

n

).

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 19 / 33

Page 50: Bertrand’s Postulate - Lola Thompson · Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33. Fun With Binomial Coe cients Another interesting observation that

A Lower Bound for(

2nn

)Lemma(2n

n

)≥ 4n

2n

Proof

In an earlier exercise, we showed that2n∑j=0

(2nj

)= 4n.

Since the two end terms in a row of Pascal’s triangle are both 1, then2n∑j=0

(2nj

)= 2 +

2n−1∑j=1

(2nj

).

We know that the middle term,(2n

n

), is the largest term in the sum.

Thus, 2 +2n−1∑j=1

(2nj

)≤ (2n)

(2nn

)since we are summing 2n terms.

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 20 / 33

Page 51: Bertrand’s Postulate - Lola Thompson · Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33. Fun With Binomial Coe cients Another interesting observation that

A Lower Bound for(

2nn

)Lemma(2n

n

)≥ 4n

2n

Proof

In an earlier exercise, we showed that2n∑j=0

(2nj

)= 4n.

Since the two end terms in a row of Pascal’s triangle are both 1, then2n∑j=0

(2nj

)= 2 +

2n−1∑j=1

(2nj

).

We know that the middle term,(2n

n

), is the largest term in the sum.

Thus, 2 +2n−1∑j=1

(2nj

)≤ (2n)

(2nn

)since we are summing 2n terms.

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 20 / 33

Page 52: Bertrand’s Postulate - Lola Thompson · Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33. Fun With Binomial Coe cients Another interesting observation that

A Lower Bound for(

2nn

)Lemma(2n

n

)≥ 4n

2n

Proof

In an earlier exercise, we showed that2n∑j=0

(2nj

)= 4n.

Since the two end terms in a row of Pascal’s triangle are both 1, then2n∑j=0

(2nj

)= 2 +

2n−1∑j=1

(2nj

).

We know that the middle term,(2n

n

), is the largest term in the sum.

Thus, 2 +2n−1∑j=1

(2nj

)≤ (2n)

(2nn

)since we are summing 2n terms.

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 20 / 33

Page 53: Bertrand’s Postulate - Lola Thompson · Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33. Fun With Binomial Coe cients Another interesting observation that

A Lower Bound for(

2nn

)Lemma(2n

n

)≥ 4n

2n

Proof

In an earlier exercise, we showed that2n∑j=0

(2nj

)= 4n.

Since the two end terms in a row of Pascal’s triangle are both 1, then2n∑j=0

(2nj

)= 2 +

2n−1∑j=1

(2nj

).

We know that the middle term,(2n

n

), is the largest term in the sum.

Thus, 2 +2n−1∑j=1

(2nj

)≤ (2n)

(2nn

)since we are summing 2n terms.

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 20 / 33

Page 54: Bertrand’s Postulate - Lola Thompson · Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33. Fun With Binomial Coe cients Another interesting observation that

Another Pair of Useful Exercises

The following exercises will also be useful in bounding BC :

Exercise Prove or disprove and salvage if possible: If x ∈ R, thenb2xc − 2bxc = 0 or 1.

Exercise (a) What power of the prime p appears in the primefactorization of n!?(b) What power of p appears in the factorization of

(nk

)?

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 21 / 33

Page 55: Bertrand’s Postulate - Lola Thompson · Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33. Fun With Binomial Coe cients Another interesting observation that

An Upper Bound for C

Lemma

C ≤ (2n)√

2n−1

Proof

Let k be the highest power of p dividing(2n

n

). Have you solved both

of the exercises on the previous slide? Once you have, you will see

that k ≤∑

j :pj≤2n

1.

The sum on the right counts the number of j that satisfy pj ≤ 2n.

In order to determine that number, we need to solve the equationpx ≤ 2n. But this is the same as solving x log p ≤ log 2n.

Thus, x ≤ log 2nlog p .

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 22 / 33

Page 56: Bertrand’s Postulate - Lola Thompson · Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33. Fun With Binomial Coe cients Another interesting observation that

An Upper Bound for C

Lemma

C ≤ (2n)√

2n−1

Proof

Let k be the highest power of p dividing(2n

n

). Have you solved both

of the exercises on the previous slide? Once you have, you will see

that k ≤∑

j :pj≤2n

1.

The sum on the right counts the number of j that satisfy pj ≤ 2n.

In order to determine that number, we need to solve the equationpx ≤ 2n. But this is the same as solving x log p ≤ log 2n.

Thus, x ≤ log 2nlog p .

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 22 / 33

Page 57: Bertrand’s Postulate - Lola Thompson · Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33. Fun With Binomial Coe cients Another interesting observation that

An Upper Bound for C

Lemma

C ≤ (2n)√

2n−1

Proof

Let k be the highest power of p dividing(2n

n

). Have you solved both

of the exercises on the previous slide? Once you have, you will see

that k ≤∑

j :pj≤2n

1.

The sum on the right counts the number of j that satisfy pj ≤ 2n.

In order to determine that number, we need to solve the equationpx ≤ 2n. But this is the same as solving x log p ≤ log 2n.

Thus, x ≤ log 2nlog p .

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 22 / 33

Page 58: Bertrand’s Postulate - Lola Thompson · Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33. Fun With Binomial Coe cients Another interesting observation that

An Upper Bound for C

Lemma

C ≤ (2n)√

2n−1

Proof

Let k be the highest power of p dividing(2n

n

). Have you solved both

of the exercises on the previous slide? Once you have, you will see

that k ≤∑

j :pj≤2n

1.

The sum on the right counts the number of j that satisfy pj ≤ 2n.

In order to determine that number, we need to solve the equationpx ≤ 2n. But this is the same as solving x log p ≤ log 2n.

Thus, x ≤ log 2nlog p .

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 22 / 33

Page 59: Bertrand’s Postulate - Lola Thompson · Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33. Fun With Binomial Coe cients Another interesting observation that

An Upper Bound for C

Since the j that satisfy pj ≤ 2n must be integers,

then∑

j :pj≤2n

1 = b log 2n

log pc.

Recall that C =∏

p≤√

2n

p|(2n

n

) p.

Suppose that a prime p is not included in C , i.e. p >√

2n and p ≤ 2n.

Then b log 2nlog p c = 1.

So all of the primes p such that pk |(2n

n

)for k > 1 must occur in C .

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 23 / 33

Page 60: Bertrand’s Postulate - Lola Thompson · Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33. Fun With Binomial Coe cients Another interesting observation that

An Upper Bound for C

Recalling that C =∏

p≤√

2n

p|(2n

n

) p, we have

C ≤∏

p≤√

2n

pblog 2nlog pc

≤∏

p≤√

2n

plog 2nlog p

=∏

p≤√

2n

e log 2n (log rule: pα = eα log p)

=∏

p≤√

2n

2n

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 24 / 33

Page 61: Bertrand’s Postulate - Lola Thompson · Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33. Fun With Binomial Coe cients Another interesting observation that

An Upper Bound for C

Recalling that C =∏

p≤√

2n

p|(2n

n

) p, we have

C ≤∏

p≤√

2n

pblog 2nlog pc

≤∏

p≤√

2n

plog 2nlog p

=∏

p≤√

2n

e log 2n (log rule: pα = eα log p)

=∏

p≤√

2n

2n

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 24 / 33

Page 62: Bertrand’s Postulate - Lola Thompson · Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33. Fun With Binomial Coe cients Another interesting observation that

An Upper Bound for C

Recalling that C =∏

p≤√

2n

p|(2n

n

) p, we have

C ≤∏

p≤√

2n

pblog 2nlog pc

≤∏

p≤√

2n

plog 2nlog p

=∏

p≤√

2n

e log 2n (log rule: pα = eα log p)

=∏

p≤√

2n

2n

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 24 / 33

Page 63: Bertrand’s Postulate - Lola Thompson · Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33. Fun With Binomial Coe cients Another interesting observation that

An Upper Bound for C

But∏

p≤√

2n

2n = (2n)π(√

2n), where π(√

2n) = # of primes ≤√

2n.

Since 1 is not prime, then π(√

2n) ≤√

2n − 1.

Thus, we see that C ≤ (2n)√

2n−1.

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 25 / 33

Page 64: Bertrand’s Postulate - Lola Thompson · Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33. Fun With Binomial Coe cients Another interesting observation that

An Upper Bound for C

But∏

p≤√

2n

2n = (2n)π(√

2n), where π(√

2n) = # of primes ≤√

2n.

Since 1 is not prime, then π(√

2n) ≤√

2n − 1.

Thus, we see that C ≤ (2n)√

2n−1.

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 25 / 33

Page 65: Bertrand’s Postulate - Lola Thompson · Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33. Fun With Binomial Coe cients Another interesting observation that

An Upper Bound for C

But∏

p≤√

2n

2n = (2n)π(√

2n), where π(√

2n) = # of primes ≤√

2n.

Since 1 is not prime, then π(√

2n) ≤√

2n − 1.

Thus, we see that C ≤ (2n)√

2n−1.

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 25 / 33

Page 66: Bertrand’s Postulate - Lola Thompson · Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33. Fun With Binomial Coe cients Another interesting observation that

An Upper Bound for B

Lemma

B ≤ 423n

Proof

Recall that B =∏

√2n<p≤n

p|(2n

n

) p.

For all n > 4.5,√

2n < 23n

(square both sides, then divide both sides by n)

We will separate B into two products:∏

√2n<p≤ 2

3n

p and∏

23n<p≤n

p.

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 26 / 33

Page 67: Bertrand’s Postulate - Lola Thompson · Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33. Fun With Binomial Coe cients Another interesting observation that

An Upper Bound for B

Lemma

B ≤ 423n

Proof

Recall that B =∏

√2n<p≤n

p|(2n

n

) p.

For all n > 4.5,√

2n < 23n

(square both sides, then divide both sides by n)

We will separate B into two products:∏

√2n<p≤ 2

3n

p and∏

23n<p≤n

p.

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 26 / 33

Page 68: Bertrand’s Postulate - Lola Thompson · Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33. Fun With Binomial Coe cients Another interesting observation that

An Upper Bound for B

Lemma

B ≤ 423n

Proof

Recall that B =∏

√2n<p≤n

p|(2n

n

) p.

For all n > 4.5,√

2n < 23n

(square both sides, then divide both sides by n)

We will separate B into two products:∏

√2n<p≤ 2

3n

p and∏

23n<p≤n

p.

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 26 / 33

Page 69: Bertrand’s Postulate - Lola Thompson · Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33. Fun With Binomial Coe cients Another interesting observation that

An Upper Bound for B

Let p ∈ (23n, n].

We will show that k = 0 when p is in this range, where k is thehighest power of p dividing

(2nn

).

Since p ∈ (23n, n], we have 1 ≤ n

p <32 .

Thus np = 1 + r , with 0 ≤ r < 1

2 .

Using this fact with the formula for k that you found in the problem

set yields k = 0. So, the product∏

23n<p≤n

p doesn’t contribute any

primes to B.

∴ B =∏

√2n<p≤n

p|(2n

n

) p ≤∏

p≤ 23n

p ≤ 423n (since ψ(x) ≤ x log 4).

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 27 / 33

Page 70: Bertrand’s Postulate - Lola Thompson · Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33. Fun With Binomial Coe cients Another interesting observation that

An Upper Bound for B

Let p ∈ (23n, n].

We will show that k = 0 when p is in this range, where k is thehighest power of p dividing

(2nn

).

Since p ∈ (23n, n], we have 1 ≤ n

p <32 .

Thus np = 1 + r , with 0 ≤ r < 1

2 .

Using this fact with the formula for k that you found in the problem

set yields k = 0. So, the product∏

23n<p≤n

p doesn’t contribute any

primes to B.

∴ B =∏

√2n<p≤n

p|(2n

n

) p ≤∏

p≤ 23n

p ≤ 423n (since ψ(x) ≤ x log 4).

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 27 / 33

Page 71: Bertrand’s Postulate - Lola Thompson · Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33. Fun With Binomial Coe cients Another interesting observation that

An Upper Bound for B

Let p ∈ (23n, n].

We will show that k = 0 when p is in this range, where k is thehighest power of p dividing

(2nn

).

Since p ∈ (23n, n], we have 1 ≤ n

p <32 .

Thus np = 1 + r , with 0 ≤ r < 1

2 .

Using this fact with the formula for k that you found in the problem

set yields k = 0. So, the product∏

23n<p≤n

p doesn’t contribute any

primes to B.

∴ B =∏

√2n<p≤n

p|(2n

n

) p ≤∏

p≤ 23n

p ≤ 423n (since ψ(x) ≤ x log 4).

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 27 / 33

Page 72: Bertrand’s Postulate - Lola Thompson · Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33. Fun With Binomial Coe cients Another interesting observation that

An Upper Bound for B

Let p ∈ (23n, n].

We will show that k = 0 when p is in this range, where k is thehighest power of p dividing

(2nn

).

Since p ∈ (23n, n], we have 1 ≤ n

p <32 .

Thus np = 1 + r , with 0 ≤ r < 1

2 .

Using this fact with the formula for k that you found in the problem

set yields k = 0. So, the product∏

23n<p≤n

p doesn’t contribute any

primes to B.

∴ B =∏

√2n<p≤n

p|(2n

n

) p ≤∏

p≤ 23n

p ≤ 423n (since ψ(x) ≤ x log 4).

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 27 / 33

Page 73: Bertrand’s Postulate - Lola Thompson · Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33. Fun With Binomial Coe cients Another interesting observation that

An Upper Bound for B

Let p ∈ (23n, n].

We will show that k = 0 when p is in this range, where k is thehighest power of p dividing

(2nn

).

Since p ∈ (23n, n], we have 1 ≤ n

p <32 .

Thus np = 1 + r , with 0 ≤ r < 1

2 .

Using this fact with the formula for k that you found in the problem

set yields k = 0. So, the product∏

23n<p≤n

p doesn’t contribute any

primes to B.

∴ B =∏

√2n<p≤n

p|(2n

n

) p ≤∏

p≤ 23n

p ≤ 423n (since ψ(x) ≤ x log 4).

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 27 / 33

Page 74: Bertrand’s Postulate - Lola Thompson · Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33. Fun With Binomial Coe cients Another interesting observation that

Putting Everything Together

What we have shown:(2nn

)= ABC(2n

n

)≥ 4n

2n

C ≤ (2n)√

2n−1

B ≤ 423n

So, A =(2n

n

)/(BC )

≥4n

2n

(2n)√

2n−1·423 n

= 413 n

(2n)√

2n

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 28 / 33

Page 75: Bertrand’s Postulate - Lola Thompson · Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 7 / 33. Fun With Binomial Coe cients Another interesting observation that

Putting Everything Together

What we have shown:(2nn

)= ABC(2n

n

)≥ 4n

2n

C ≤ (2n)√

2n−1

B ≤ 423n

So, A =(2n

n

)/(BC )

≥4n

2n

(2n)√

2n−1·423 n

= 413 n

(2n)√

2n

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Putting Everything Together

What we have shown:(2nn

)= ABC(2n

n

)≥ 4n

2n

C ≤ (2n)√

2n−1

B ≤ 423n

So, A =(2n

n

)/(BC )

≥4n

2n

(2n)√

2n−1·423 n

= 413 n

(2n)√

2n

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Putting Everything Together

Remember that in order for Bertrand’s Postulate to hold, we need toshow that A > 1. Thus, we need to determine when 4

13n > (2n)

√2n.

Using high school math, we can show that 413n > (2n)

√2n holds when

n > 450.

∴ Bertrand’s Postulate holds for all n > 450.

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 29 / 33

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Putting Everything Together

Remember that in order for Bertrand’s Postulate to hold, we need toshow that A > 1. Thus, we need to determine when 4

13n > (2n)

√2n.

Using high school math, we can show that 413n > (2n)

√2n holds when

n > 450.

∴ Bertrand’s Postulate holds for all n > 450.

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 29 / 33

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Putting Everything Together

Remember that in order for Bertrand’s Postulate to hold, we need toshow that A > 1. Thus, we need to determine when 4

13n > (2n)

√2n.

Using high school math, we can show that 413n > (2n)

√2n holds when

n > 450.

∴ Bertrand’s Postulate holds for all n > 450.

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 29 / 33

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Finishing Up

In order to conclude that Bertrand’s Postulate is true for all n ∈ Z+, wejust need to check values of n ≤ 450.

Remember that Bertrand checked all n up to 6,000,000, so if you believehim then we’re done!

If not, consider the list of primes 2, 3, 5, 7, 13, 23, 43, 83, 163, 317, 631,where each is less than twice the preceding. This proves Bertrand’sPostulate for all n < 631, since any such n can be squeezed between twonumbers on the list.

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Generalizations

One way that a mathematician finds new problems to solve is by lookingat a result that has already been proven and asking “Does this hold in amore general setting?”

Of course, “more general” can mean many different things. For example,we showed that

(2nn

)≤ 4n for all n ≥ 0. Perhaps we could have shown a

similar result for any integer n. Another generalization would be to try tobound

(knn

), k ∈ Z+.

Can you think of a “more general” statement of Bertrand’s Postulate?

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Generalizations

Here are a few well-known generalizations of Bertrand’s Postulate:

Theorem (Sylvester)

The product of k consecutive integers greater than k is divisible by aprime greater than k.

Theorem (Erdos)

For any positive integer k, there is a natural number N such that for alln > N, there are at least k primes between n and 2n.

Conjecture (Legendre) For every n > 1, there is a prime p suchthat n2 < p < (n + 1)2.

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Generalizations

Here are a few well-known generalizations of Bertrand’s Postulate:

Theorem (Sylvester)

The product of k consecutive integers greater than k is divisible by aprime greater than k.

Theorem (Erdos)

For any positive integer k, there is a natural number N such that for alln > N, there are at least k primes between n and 2n.

Conjecture (Legendre) For every n > 1, there is a prime p suchthat n2 < p < (n + 1)2.

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 32 / 33

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Generalizations

Here are a few well-known generalizations of Bertrand’s Postulate:

Theorem (Sylvester)

The product of k consecutive integers greater than k is divisible by aprime greater than k.

Theorem (Erdos)

For any positive integer k, there is a natural number N such that for alln > N, there are at least k primes between n and 2n.

Conjecture (Legendre) For every n > 1, there is a prime p suchthat n2 < p < (n + 1)2.

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Some Neat Applications

Using Bertrand’s Postulate, we can also prove many other interestingresults, including:

Every integer n > 6 can be written as a sum of distinct primes.

∀N ∈ N, there exists an even integer k > 0 for which there are atleast N prime pairs p, p + k .

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 33 / 33

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Some Neat Applications

Using Bertrand’s Postulate, we can also prove many other interestingresults, including:

Every integer n > 6 can be written as a sum of distinct primes.

∀N ∈ N, there exists an even integer k > 0 for which there are atleast N prime pairs p, p + k .

Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 33 / 33