
Re: the probability theory has holes!
Posted:
Dec 1, 2009 2:39 PM


eestath wrote: > What i had in mind is Eratosthenes Sieve: > > for 2: we exlude the first 1/2 > > for 3:then we exclude 1/3 but half of them > have factor 2 so we actually exclude 1/2*3 > > for 5:then we exclude 1/5 but half of them > have factor 2 and 1/3 have factor 3 so we > actually exclude 1/2*3*5
Right here is the problem: "1/3 have factor 3". What you want is "2/3 DO NOT have factor 3". The 1/3 with factor 3 have already been excluded. Another 1/5 of the fraction ( 1  1/2 )*( 1  1/3 ) that do not have either 2 or 3 as a factor is what you want to exclude for the next step.
And ( 1  1/2 )*( 1  1/3 )*1/5 = 1/15.
> So the Series is 1=1/2 + 1/(2*3) + 1/(2*3*5)+....
Another way to look at this:
(fraction not div by 2)*(fraction not div by 3)* (fraction not div by 5)* ... = (fraction not div by any prime)
which means
( 1  1/2 )*( 1  1/3 )*( 1  1/5 )* ... = 0
If you are not already familiar with the Riemann zeta function, then I think you would find this very interesting: http://en.wikipedia.org/wiki/Riemann_zeta_function#Euler_product_formula
> You answer but you are actually wrong!!!!!!!!!!
One of the more pleasant surprises on the Internet is finding someone who will admit that they were wrong (when they finally understand that they were wrong).
I wonder, are you going to surprise us?
Jim Burns

