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Transcript of Bertrand's Postulate
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Bertrand’s Postulate
Lola Thompson
Ross Program
July 3, 2009
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Bertrand’s Postulate
“I’ve said it once and I’ll say it again: There’s always a prime between n
and 2n.”-Joseph Bertrand, conjectured in 1845
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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.”
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Bertrand’s Postulate
Many other proofs have been found in the time since Chebyshev first
proved this theorem. We will follow a proof due to Ramanujan and Erdos.
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Outline
1 Fun With Binomial Coefficients
2 A Few New Arithmetic Functions
3 An Upper Bound for ψ(x )
4 Proving Bertrand’s Postulate
The Setup: ABC =
2nn
A Lower Bound for2n
n
An Upper Bound for CAn Upper Bound for BPutting Everything Together
5 Generalizations
6 Some Neat Applications
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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.
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Fun With Binomial Coefficients
There is a curious relationship between nk and Pascal’s Triangle.
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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
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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
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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 1
1 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 precisely4
0
through
44
.
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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.
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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.
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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:
nk =0
nk
= 2n.
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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.
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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 divide2n
n
? Let’s look at some examples.
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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 divide2n
n
? Let’s look at some examples.42
= 6. Which primes divide 6? What about
63
?8
4
?10
5
?
Any conjectures about when a prime has to divide2n
n
?
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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 divide2n
n
? Let’s look at some examples.42
= 6. Which primes divide 6? What about
63
?8
4
?10
5
?
Any conjectures about when a prime has to divide2n
n
?
Exercise: p |2nn
if there exists a positive integer j with n < p j
≤ 2n
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C ffi
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Fun With Binomial CoefficientsAssuming that those exercises are true, we can prove two results:
Lemma (1)p :n<p j ≤2n
p |
2n
n
.
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F Wi h Bi i l C ffi i
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Fun With Binomial CoefficientsAssuming that those exercises are true, we can prove two results:
Lemma (1)p :n<p j ≤2n
p |
2n
n
.
Proof We know that each prime p with n < p
j
≤ 2n divides2nn
.Since the primes are distinct, they are pairwise relatively
prime.Thus, their product must divide2n
n
.
Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 10 / 33
F Wi h Bi i l C ffi i
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Fun With Binomial CoefficientsAssuming that those exercises are true, we can prove two results:
Lemma (1)p :n<p j ≤2n
p |
2n
n
.
Proof We know that each prime p with n < p
j
≤ 2n divides2nn
.Since the primes are distinct, they are pairwise relatively
prime.Thus, their product must divide2n
n
.
Lemma (2)2nn
≤ 4n for all n ≥ 0.
Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 10 / 33
F With Bi i l C ffi i t
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Fun With Binomial CoefficientsAssuming that those exercises are true, we can prove two results:
Lemma (1)p :n<p j ≤2n
p |
2n
n
.
Proof We know that each prime p with n < p
j
≤ 2n divides2nn
.Since the primes are distinct, they are pairwise relatively
prime.Thus, their product must divide2n
n
.
Lemma (2)2nn
≤ 4n for all n ≥ 0.
Proof Look at the 2nth row of Pascal’s Triangle:2nn is in the center,
22n
= sum of all terms in the row.Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 10 / 33
A F N A ith ti F ti
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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 = p k , k > 0
0, otherwise
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A Fe Ne A ith etic F ctio s
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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 = p k , k > 0
0, otherwise
Examples: Λ(2) = log 2, Λ(4) = log 2, Λ(6) = 0.
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A Few New Arithmetic Functions
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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 = p k , k > 0
0, otherwise
Examples: Λ(2) = log 2, Λ(4) = log 2, Λ(6) = 0.
Definition
ψ(x ) =n≤x
Λ(n)
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A Few New Arithmetic Functions
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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 = p k , 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).
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An Upper Bound for ψ(x)
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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)
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Proof of Upper Bound for ψ(x)
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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 )
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Proof of Upper Bound for ψ(x)
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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)
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Proof of Upper Bound for ψ(x)
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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 itl .
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)
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Proof of Upper Bound for ψ(x)
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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 itl .
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 = p j and 0 otherwise,so this line follows from Lemma (1)after taking log of both sides)
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Proof of Upper Bound for ψ(x)
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Proof of Upper Bound for ψ(x )
From the previous slide: ψ(2k ) − ψ(k ) ≤ log
2k k
.
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Proof of Upper Bound for ψ(x)
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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 ).
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Proof of Upper Bound for ψ(x )
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pp ψ( )
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.
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Proof of Upper Bound for ψ(x )
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pp ψ( )
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
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Proof of Upper Bound for ψ(x )
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pp ψ( )
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)
Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 14 / 33
Proof of Upper Bound for ψ(x )
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pp ψ( )
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.
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Proof of Upper Bound for ψ(x )
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( )
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?
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Proof of Upper Bound for ψ(x )
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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(
x
)≤
x
log4≤
x log 4.
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Proving Bertrand’s Postulate
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We will use what we have learned about2n
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.
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The Setup : ABC =
2nn
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Let A = n<p
≤2n p .
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The Setup : ABC =
2nn
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Let A = n<p
≤2n p .
From Lemma (1), we know that A |2nn
. So, there exists m ∈ Z st.
A · m =
2nn
.
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The Setup : ABC =
2nn
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Let A = n<p
≤2n p .
From Lemma (1), we know that A |2nn
. So, there exists m ∈ Z st.
A · m =
2nn
.
By the Unique Factorization Theorem, we can factor m into aproduct of primes (uniquely).
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The Setup : ABC =
2nn
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Let A = n<p
≤2n p .
From Lemma (1), we know that A |2nn
. So, there exists m ∈ Z st.
A · m =
2nn
.
By the Unique Factorization Theorem, we can factor m into aproduct of primes (uniquely).
Let B = contribution to
2nn
from primes p ∈ (√
2n, n].
Let C = contribution to 2n
n from primes p
≤√
2n.
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The Setup : ABC =
2nn
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Let A = n<p
≤2n p .
From Lemma (1), we know that A |2nn
. So, there exists m ∈ Z st.
A · m =
2nn
.
By the Unique Factorization Theorem, we can factor m into aproduct of primes (uniquely).
Let B = contribution to
2nn
from primes p ∈ (√
2n, n].
Let C = contribution to 2n
n from primes p
≤√
2n.
Then ABC =2n
n
.
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The Setup : ABC =
2nn
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Goal: We want to show that BC <2n
n
.Why does this imply that Bertrand’s Postulate holds?
If BC <
2nn
then A = 1.
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The Setup : ABC =
2nn
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Goal: We want to show thatBC
<2n
n
.Why does this imply that Bertrand’s Postulate holds?
If BC <
2nn
then A = 1.
But A =
n≤p ≤2n p , so there must be a prime dividing A.
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The Setup : ABC =
2nn
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Goal: We want to show thatBC
<2n
n
.Why does this imply that Bertrand’s Postulate holds?
If BC <
2nn
then A = 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 dividing2n
n
.
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The Setup : ABC =
2nn
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Goal: We want to show thatBC
<2n
n
.Why does this imply that Bertrand’s Postulate holds?
If BC <
2nn
then A = 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 dividing2n
n
.
This proves Bertrand’s Postulate because it proves the existence of a
prime between n and 2n.
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The Setup : ABC =
2nn
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In order to show that BC <
2nn
, we will find upper bounds for B and C
and we will find a lower bound for2n
n
.
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A Lower Bound for
2nn
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Lemma
2nn ≥
4n
2n
Proof
In an earlier exercise, we showed that2n
j =0
2n j = 4n.
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A Lower Bound for
2nn
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Lemma
2nn ≥
4n
2n
Proof
In an earlier exercise, we showed that2n
j =0
2n j = 4n.
Since the two end terms in a row of Pascal’s triangle are both 1, then2n
j =0
2n j
= 2 +
2n−1 j =1
2n j
.
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A Lower Bound for
2nn
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Lemma
2nn ≥
4n
2n
Proof
In an earlier exercise, we showed that2n
j =0
2n j = 4n.
Since the two end terms in a row of Pascal’s triangle are both 1, then2n
j =0
2n j
= 2 +
2n−1 j =1
2n j
.
We know that the middle term,
2nn
, is the largest term in the sum.
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A Lower Bound for
2nn
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Lemma
2nn ≥
4n
2n
Proof
In an earlier exercise, we showed that2n
j =0
2n j = 4n.
Since the two end terms in a row of Pascal’s triangle are both 1, then2n
j =0
2n j
= 2 +
2n−1 j =1
2n j
.
We know that the middle term,
2nn
, is the largest term in the sum.
Thus, 2 +2n−1 j =1
2n j
≤ (2n)2n
n
since we are summing 2n terms.
Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 20 / 33
Another Pair of Useful Exercises
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The following exercises will also be useful in bounding BC :
Exercise Prove or disprove and salvage if possible: If x ∈ R, then
2x
−2
x
= 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
An Upper Bound for C
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Lemma
C ≤ (2n)√2n−1
Proof
Let k be the highest power of p dividing
2nn
. Have you solved both
of the exercises on the previous slide? Once you have, you will seethat k ≤
j :p j ≤2n
1.
Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 22 / 33
An Upper Bound for C
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Lemma
C ≤ (2n)√2n−1
Proof
Let k be the highest power of p dividing
2nn
. Have you solved both
of the exercises on the previous slide? Once you have, you will seethat k ≤
j :p j ≤2n
1.
The sum on the right counts the number of j that satisfy p j ≤ 2n.
Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 22 / 33
An Upper Bound for C
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Lemma
C ≤ (2n)√2n−1
Proof
Let k be the highest power of p dividing
2nn
. Have you solved bothof the exercises on the previous slide? Once you have, you will see
that k ≤
j :p j ≤2n
1.
The sum on the right counts the number of j that satisfy p j ≤ 2n.
In order to determine that number, we need to solve the equationp x ≤ 2n. But this is the same as solving x log p ≤ log2n.
Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 22 / 33
An Upper Bound for C
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Lemma
C ≤ (2n)√2n−1
Proof
Let k be the highest power of p dividing
2nn
. Have you solved bothof the exercises on the previous slide? Once you have, you will see
that k ≤
j :p j ≤2n
1.
The sum on the right counts the number of j that satisfy p j ≤ 2n.
In order to determine that number, we need to solve the equationp x ≤ 2n. But this is the same as solving x log p ≤ log2n.
Thus, x ≤ log2nlog p
.
Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 22 / 33
An Upper Bound for C
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Since the j that satisfy p j ≤ 2n must be integers,
then
j :p j ≤2n
1 = log2n
log p .
Recall that C = p ≤√2np |2n
n
p .
Suppose that a prime p is not included in C , i.e. p >√
2n and p ≤ 2n.
Then
log2n
logp
= 1.
So all of the primes p such that p k |2nn
for k > 1 must occur in C .
Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 23 / 33
An Upper Bound for C
Recalling that C =
p we have
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Recalling that C
p ≤√2n
p
|2n
n
p , we have
C ≤
p ≤√2n
p log 2n
log p
≤
p ≤√2n
p log 2nlog p
Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 24 / 33
An Upper Bound for C
Recalling that C =
p, we have
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Recalling that C
p ≤√2n
p
|2n
n
p , we have
C ≤
p ≤√2n
p log 2n
log p
≤
p ≤√2n
p log 2nlog p
= p ≤√2n
e log2n (log rule: p α = e α log p )
Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 24 / 33
An Upper Bound for C
Recalling that C =
p , we have
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g
p ≤√2n
p
|2n
n
p,
C ≤
p ≤√2n
p log 2n
log p
≤
p ≤√2n
p log 2nlog p
= p ≤√2n
e log2n (log rule: p α = e α log p )
=
p ≤√2n
2n
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An Upper Bound for C
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But
p ≤√2n
2n = (2n)π(√
2n), where π(√
2n) = # of primes ≤ √2n.
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An Upper Bound for C
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But
p ≤√2n
2n = (2n)π(√
2n), where π(√
2n) = # of primes ≤ √2n.
Since 1 is not prime, then π(√2n) ≤ √2n − 1.
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An Upper Bound for C
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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
An Upper Bound for B
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Lemma
B ≤ 423 n
Proof
Recall that B =
√2n<p
≤n
p |
2nn
p .
Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 26 / 33
An Upper Bound for B
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Lemma
B ≤ 423 n
Proof
Recall that B =
√2n<p
≤n
p |
2nn
p .
For all n > 4.5,√
2n < 23 n
(square both sides, then divide both sides by n)
Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 26 / 33
An Upper Bound for B
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Lemma
B ≤ 423 n
Proof
Recall that B =
√2n<p
≤n
p |
2nn
p .
For all n > 4.5,√
2n < 23 n
(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
An Upper Bound for B
Let p ∈ ( 23 n, n].
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Let p ∈ ( 3 n, n].
Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 27 / 33
An Upper Bound for B
Let p ∈ ( 23 n, n].
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p ∈ ( 3 , ]
We will show that k = 0 when p is in this range, where k is thehighest power of p dividing
2nn
.
Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 27 / 33
An Upper Bound for B
Let p ∈ ( 23 n, n].
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p ( 3 , ]
We will show that k = 0 when p is in this range, where k is thehighest power of p dividing
2nn
.
Since p ∈ ( 23 n, n], we have 1 ≤ n
p < 3
2 .
Thus np
= 1 + r , with 0 ≤ r < 12 .
Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 27 / 33
An Upper Bound for B
Let p ∈ ( 23 n, n].
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p ( 3 , ]
We will show that k = 0 when p is in this range, where k is thehighest power of p dividing
2nn
.
Since p ∈ ( 23 n, n], we have 1 ≤ n
p < 3
2 .
Thus np
= 1 + r , with 0 ≤ r < 12 .
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 .
Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 27 / 33
An Upper Bound for B
Let p ∈ ( 23 n, n].
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( 3 ]
We will show that k = 0 when p is in this range, where k is thehighest power of p dividing
2nn
.
Since p ∈ ( 23 n, n], we have 1 ≤ n
p < 3
2 .
Thus np
= 1 + r , with 0 ≤ r < 12 .
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 ≤ 2
3n
p ≤ 423n (since ψ(x ) ≤ x log 4).
Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 27 / 33
Putting Everything Together
What we have shown:
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2nn
= ABC 2nn
≥ 4n
2n
C ≤ (2n)√
2n−1
B ≤ 42
3 n
So, A =2n
n
/(BC )
Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 28 / 33
Putting Everything Together
What we have shown:
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2n
n
= ABC 2nn
≥ 4n
2n
C ≤ (2n)√
2n−1
B ≤ 42
3 n
So, A =2n
n
/(BC )
≥4n
2n
(2n)√ 2n−1·4 23 n
Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 28 / 33
Putting Everything Together
What we have shown:
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2n
n
= ABC 2nn
≥ 4n
2n
C ≤ (2n)√
2n−1
B ≤ 42
3 n
So, A =2n
n
/(BC )
≥4n
2n
(2n)√ 2n−1·4 23 n
= 413 n
(2n)√
2n
Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 28 / 33
Putting Everything Together
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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.
Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 29 / 33
Putting Everything Together
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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.
Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 29 / 33
Putting Everything Together
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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
Finishing Up
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In order to conclude that Bertrand’s Postulate is true for all n ∈ Z+
, we just 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.
Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 30 / 33
Generalizations
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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?
Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 31 / 33
Generalizations
Here are a few well-known generalizations of Bertrand’s Postulate:
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Theorem (Sylvester)The product of k consecutive integers greater than k is divisible by a
prime greater than k.
Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 32 / 33
Generalizations
Here are a few well-known generalizations of Bertrand’s Postulate:
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Theorem (Sylvester)The product of k consecutive integers greater than k is divisible by a
prime greater than k.
Theorem (Erdos)
For any positive integer k, there is a natural number N such that for all
n > N, there are at least k primes between n and 2n.
Lola Thompson (Ross Program) Bertrand’s Postulate July 3, 2009 32 / 33
Generalizations
Here are a few well-known generalizations of Bertrand’s Postulate:
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Theorem (Sylvester)The product of k consecutive integers greater than k is divisible by a
prime greater than k.
Theorem (Erdos)
For any positive integer k, there is a natural number N such that for all
n > 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
Some Neat Applications
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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.
Lola Thompson (Ross Program) Bertrand’s Postulate July 3 2009 33 / 33
Some Neat Applications
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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