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al-mahed
al-mahed is offline
#40
Dec26-07, 06:04 PM
P: 258
Let all positive whole number be written as a sum of power of two:

[tex]1 = 2^0[/tex]
[tex]
2 = 2^1[/tex]
[tex]
3 = 2^0 + 2^1[/tex]
[tex]
4 = 2^2[/tex]
[tex]
5 = 2^0 + 2^2[/tex]
[tex]
6 = 2^1 + 2^2 [/tex]
[tex]
7 = 2^0 + 2^1 + 2^2[/tex]
[tex]
8 = 2^3 [/tex]
[tex]
9 = 2^0 + 2^3[/tex]
[tex]
10 = 2^1 + 2^3[/tex]
[tex]
11 = 2^0 + 2^1 + 2^3[/tex]
[tex]
12 = 2^2 + 2^3[/tex]
[tex]
13 = 2^0 + 2^2 + 2^3[/tex]
[tex]
14 = 2^1 + 2^2 + 2^3[/tex]
[tex]
15 = 2^0 + 2^1 + 2^2 + 2^3[/tex]
[tex]
16 = 2^4 [/tex]
[tex]
17 = 2^0 + 2^4[/tex]
[tex]
18 = 2^1 + 2^4[/tex]
[tex]
19 = 2^0 + 2^1 + 2^4[/tex]
[tex]
20 = 2^2 + 2^4

etc...[/tex]

We know by Tchebychef: for m > 1 there is at least one prime p such that m < p < 2m

if we put m = 2^n, n a natural > 0 ==> 2^n < p < 2^(n+1)

We know by Gauss: the number of primes from 1 to x is approximately =[tex]\frac{x}{ln(x)}[/tex]

The number of primes between [tex] 2^n[/tex] and [tex] 2^{n+1}[/tex] is approximately =

[tex]\frac{2^{n+1}}{ln(2^{n+1})} - \frac{2^n}{ln(2^n)}[/tex] = [tex]\frac{2^n}{ln(2)}*(\frac{2}{n+1} - \frac{1}{n})[/tex] = [tex]\frac{2^n}{ln(2)}*(\frac{n-1}{n^2+n})[/tex]

and we know that [tex]2^n > n^2[/tex] for [tex]\geq 3[/tex] ==> [tex]\frac{2^n}{ln(2)}*(\frac{n-1}{n^2+n})[/tex] diverges for n --> [tex]\infty[/tex]

for n increasing the Tchebychef's theorem underestimates the amount of primes