Wednesday, November 18, 2009

Bertrand's Postulate (Theorem)

In today's blog, I will present a proof for Bertrand's Postulate which is really a theorem.

The content is taken primarily from an excellent Wikipedia article.

Lemma 1:

If n is greater than 5, then 2n/3 is greater than 2n


(1) For n ≥ 5, 2n(2n-9) is greater than 0 since 2n ≥ 10 and 2n-9 ≥ 1.

(2) 2n(2n - 9) = 4n2 - 18n.

(3) Since 4n2 - 18n is greater than 0, 4n2 is greater than 18n.

(4) Dividing both sides by 9 gives us 4n2/9 is greater than 2n.

(5) Taking the square root of both sides gives us 2n/3 is greater than 2n


Lemma 2:

Let p be a prime number that is greater than 2n

Let x be the highest nonnegative integer such that px divides 2n!/[(n!)(n!)]



(1) We know that in general, the maximum power of p that divides n! is (see Theorem, here):

(2) So the highest power of p that divides (2n)!/[(n!)(n!)] is:

(3) Since p is greater than 2n, it follows that p2 is greater than 2n. So we only need to consider the case where i=1.

(4) This leaves us with:


Lemma 3:

Bertrand's Posulate is true for n less than 2048


Between any n less than 2048 and its double, one will find one of the following primes:

3, 5, 7, 13, 23, 43, 83, 163, 317, 631, 1259 and 2503


n = 2 --> 3

n between 3 and 4 --> 5

n between 5 and 6 --> 7

n between 7 and 12 --> 13

n between 13 and 22 --> 23

n between 23 and 42 --> 43

n between 43 and 82 --> 83

n between 83 and 162 --> 163

n between 163 and 316 --> 317

n between 317 and 630 --> 631

n between 631 and 1258 --> 1259

n between 1259 and 2047 --> 2503


Definition 1: R(p,n): maximum divisor of 2n!/(n!)(n!)

Let R(p,n) be the function that returns the maximum power of a given prime that divides 2n!/(n!)(n!)

For any prime p, pR(p,n) divides 2n!/(n!)(n!) but pR(p,n)+1 does not.

Example 1: R(3,5)

(2*5)!/(5!)(5!) = 10*9*8*7*6/5*4*3*2 = 2*9*2*7*2*2 = 24*32*7.

R(3,5) = 2 since 32 divides 24*32*7 but 33 = 27 does not.

Lemma 4:

For all p, pR(p,n) ≤ 2n


(1) The multiplicity m of p in a factorial n! (see Theorem 1, here) is:

so that pm divides n! but pm+1 does not.

(2) Based on this, the formula for R(p,n) is:

which is equivalent to:

(3) We can see that if pi is greater than 2n, then:

floor(2n/pi) = 0 and floor(n/pi) = 0 so that we have:

floor(2n/pi) - 2*(n/pi) = 0

so that we have:

(4) Using the Division Algorithm (see Theorem 1, here), there exists q,r such that:

n = qpi + r

where r is less than pi

(5) From this, we can see that:


Case 1: r = 0

In this case, floor(2n/pi) = 2q and 2*floor(n/pi) = 2*q so that:

floor(2n/pi) - 2*floor(2n/pi) = 0

Case 2: r is less than (1/2)pi

In this case, floor(2n/pi) = 2q and 2*floor(n/pi) = 2*q so that:

floor(2n/pi) - 2*floor(2n/pi) = 0

Case 3: r ≥ (1/2)pi

In this case, floor(2n/pi) = 2q+1 and 2*floor(n/pi) = 2*q so that:

floor(2n/pi) - 2*floor(2n/pi) = 1

(5) So that we have:

pi ≤ plogp(2n) = 2n.


Lemma 5:

For all i:

0 ≤ i ≤ 2n → (2n!)/(n!)(n!) ≥ (2n!)/(i!)(2n-i)!


(1) We know that (see Definition 3, here for review of C(n,r) if needed):

C(2n,i-1) = (2n!)/(i-1)!(2n-i+1)!

C(2n,i) = (2n!)/(i!)(2n-i)!

(2) So, we see that:

C(2n,i) = (2n-i+1)/(i)*C(2n,i-1)


C(2n,i-1) = (i)/(2n-i+1)*C(2n,i)

(3) If i ≤ n, it follows that:

2n - i + 1 is greater than i since 2n - i + 1 ≥ n+1

So up to i=n, C(2n,i) is greater than C(2n,i-1)

(5) If i ≥ n+1, it follows that:

i is greater than 2n - i + 1 since 2n - i + 1 ≤ n

So down to i=n+1, C(2n,i-1) is greater than C(2n,i)

(6) So, we have shown that C(2n,n) is greater than C(2n,i) for i less than n and C(2n,n) is greater than C(2n,i) for i greater than n.


Corollary 5.1:


(1) Using the Binomial Theorem (see Theorem, here):

(2) Using Lemma 5 above, we know that:

(3) So that we have:

(4) And dividing both sides by (2n+1) finishes the proof.


Lemma 6:

If 2t is less than 8t, then t is less than 6


(1) If t=6, then 2t is greater than 8t since:

26 = 64

8(6) = 48

(2) Assume it is true up to some t ≥ 6 so that: 2t is greater than 8t.

(3) Then 2*(2t)=2t+1 is greater than 2*8t = 16t

(4) 16t is greater than 8t+8 since 16=8t + 8t and for t ≥ 6, 8t is greater than 8.

(5) So that we have 2t+1 is greater than 8t+8 = 8(t+1).


Lemma 7:



n is less than 2048.


(1) Let n = 22t-1

(2) So 2n = 22t

(3) 22t = 2t

(4) So 2nln(2) = 2tln(2)

(5) (4)ln(2n) = 4ln(22t) = 2t*4*ln(2)

(6) Dividing both sides by ln(2) gives us:

2t is less than 8t

(7) Using Lemma 6 above, we conclude that t is less than 6.

(8) So, it follows that n is less than 22(6)-1 = 211 = 2048.


Theorem: Bertrand's Postulate

For any n, there exists a prime p such that p is greater than n and less than 2n.


(1) We know that Bertrand's Postulate is true for n less than 2048 [see Lemma 3 above]

(2) Assume for some n ≥ 2048, there is no prime between n and 2n.

(3) Let p be the largest prime factor of (2n!)/[(n!)(n!)] that is less than or equal to n.

(4) Using Lemma 1 above, we know that p ≤ 2n/3 since:

(a) Assume that that p is greater than 2n/3.

(b) Since n ≥ 2048, using Lemma 1 above, p is also greater than 2n

(c) This means that the highest power of x is given by (see Lemma 2 above):

(d) Since 2n/3 is less than p, it follows that 2n is less than 3p, and further that 2n/p is less than 3 so that 2n/p ≤ 2.

(e) But from step #3 above, we know that n ≥ p so that we have n/p ≥ 1.

(f) This gives us that 2n/p ≤ 2 and n/p ≥ 1 so that we have:

x = floor(2n/p) - 2*floor(n/p) ≤ 2 - 2*1 = 0.

(g) So, under these circumstances, the highest power of x is 0 which means that there is no such p and we can conclude that p ≤ 2n/3.

(5) From step #4, we have:

(2n!)/(n!n!) = ∏ (p ≤ 2n/3) pR(p,n)

where R(p,n) is a function on p,n such that pR(p,n) divides (2n!)/(n!n!) but pR(p,n)+1 does not. [see Definition 1 above]

(6) From Lemma 2 above, we know that if p is greater than 2n, then R(p,n)=1 so that we have:

(7) Since there are less than 2n primes less than 2n, using Lemma 4 above, we have:

(8) If we use a limit for primorials (see Theorem, here), we have:

(9) Using Corollary 5.1 above, we also have:

(10) So, putting it all together, we have:

(11) If we multiply 4(-2n/3) to both sides we get:

(12) Now, since n ≥ 1, we note that 4n2 is greater than 2n+1

(a) For n=1, 4(1)2 is greater than 2(1)+1

(b) Assume that 4n2 is greater than 2n+1

(c) Since n ≥ 1, 4n2 + 8n is greater than 2n+1+2

(d) 4n2 + 8n + 1 is greater than 2(n+1) + 1

(e) And:

4(n+1)2 is greater than 2(n+1)+1

(13) So it follows that:

(14) Since n is greater than 18, it follows that (4/3)√2n is greater than 2 + √2n since:

(a) 2(n) is greater than 6

(b) (4/3)√2n = (1/3)√2n + √2n

(c) And (1/3)√2n is greater than (1/3)*6 = 2.

(15) This gives us:

(16) Putting it all together gives us:

(17) Taking the logarithm of both sides, gives us:

(18) We can simplify this further by multiplying 3 to both sides and noting that 4=22 so that we get:

(19) We can further simplify it by dividing both sides by 2n to get:

(20) But now we use Lemma 7 above to reach a contradiction.




The primorial is the product of primes less than or equal to a number n.

Definition: Primorial: n#

The primorial n# is defined as follows:


1# = 1

3# = 3*(2#) = 3*2*(1#)=3*2

6# = 5# = 5*3*2

Theorem: if n≥ 1, then n# is less than 4n


(1) It is true for (n=1 and n=2)

For n=1, 1#=1 is less than 4.

For n=2, 2#=2 is less than 42 = 16

(2) If n is even and n is greater than 2, then n is composite and we have:

n# = (n-1)# which is proven if we show that (n-1)# is less than 4n-1 since 4n-1 is less than 4n.

(3) To complete the proof, we show that n# is less than 4n if n is odd:

(a) By the inductive hypothesis, we can assume that it is true for all integers less than n and we assume that n is an odd integer where n=2x+1 and x ≥ 1 so that n ≥ 3.

(b) Now, we note that 4m = (1+1)2m = (2-1)*(1+1)2m+1 = (1/2)(1+1)2m+1

(c) Using the Binomial Theorem (see here), we note that:

(1+1)2m+1 = ∑ (i=0, 2m+1) (2m+1!)/[(i!)(2m+1-i)!]

(d) So,

(1/2)(1+1)2m+1 = (1/2) ∑ (i=0, 2m+1) (2m+1)!/[(i!)(2m+1-i)!]

is greater than:

(1/2) [(2m+1)!/(m!)(m+1)! + (2m+1)!/(m+1)!(m)!] = (2m+1)!/(m!)(m+1)!

(e) So, from (b) and (d), we can conclude that:

4m is greater than (2m+1)!/(m!)(m+1)!

since 4m = (1/2)(1+1)2m+1 which is greater than (2m+1)!/(m!)(m+1)!

(f) Each prime p that is greater than m+1 and less than 2m+1 divides (2m+1)!/(m+1)!(m!) since:

(2m+1)!/(m+1)!(m!) is an integer. [see Lemma 5, here]

Let u = (2m+1)!/(m+1)!(m!)

(2m+1)! = u(m+1)!(m!)

Since p does not divide (m+1)! and does not divide (m!), by Euclid's Generalized Lemma (see Corollary 2.1, here), p must divide u.

(g) This gives us:

From the definition of primorial [see Definition above], step #3f above, and step #3e above.

(h) From the inductive hypothesis in step #3a, we can assume that:

(m+1)# is less than 4m+1

(i) So, we have (using step #3g above):

n# = (2m+1)# which is less (m+1)#*4m

(j) Using step #3h above, we have:

(2m+1)# is less than 4m+1*4m = 42m+1



Pascal's Rule

Theorem: Pascal's Rule

C(n,k) + C(n,k-1) = C(n+1,k)



(1) We can find a common denominator for C(n,k) and C(n,k-1) so that we have:

(2) So that:

(3) To complete the proof, we note that: n-k+1 = n+1-k so that we have:

= C(n,k+1)



Monday, November 16, 2009

Multiplicity of a Prime Factor

Definition 1: Multiplicity of a prime factor

The multiplicity of a prime factor of a given integer is the highest power of that factor that divides the integer.

Example 1: 60

The prime factorization of is 2*2*3*5

The multiplicity of 3 is 1. The multiplicity of 5 is 1. The multiplicity of 2 is 2.

Adrien-Marie Legendre came up with the following theorem regarding the multiplicity of a prime for n!. (For review of factorials (n!), see here)

Theorem 1: The multiplicity of a given prime p in n! is:

That is, let x = the multiplicity of a prime p.

Then px divides n! but px+1 does not.


(1) If p is greater than n, then p doesn't divide n! and the answer is 0.

(2) So, we can assume p ≤ n.

(3) Let k be the largest power of p such that pk ≤ n.

(4) For each 1 ≤ i ≤ k, the number of integers from 1 to n that are divisible by pi but not divisible by pi+1 is:

If no numbers are divisible by pi, then floor(n/pi) = 0 [as does floor(n/pi+1)].

floor(n/pi+1) equals the number of integers that are greater than pi+1 and divisible by pi+1.

So the number of integers that are uniquely divisible by pi is the difference of these two floor functions.

(5) The value of px which divides n! is equal to the product of all the powers of p that divide any of the integers in the sequence from 1 .. n.

(6) So, the we can compute the value of x with:

(7) Putting this together, we see the following pattern:

(8) We can see for i=1, we have floor(n/p1), for i=2, we have 2*floor(n/p2) - 1*floor(n/p2), and for i=3, we have 3*floor(n/p3) - 2*floor(n/p3) and so on.

(9) The net result of this is that we rewrite the equation in step #7 to the following: