Re: Calculus XOR Probability

Han de Bruijn wrote:

Good ideas. But instead, I'm tempted to proceed towards an answer to the
_original_ question. Given infinitesimal probabilities e , can we define
now in *R the probability that any natural chosen at random is divisible
by a fixed natural number a ? And prove that the outcome is a real 1/a ?

If D(a,n) = '(number of naturals <= n divisible by fixed natural a)
divided by n'. Then D(a,n) = floor(n/a)/n

This gives rise to the following hyperreal number:

d = (0, 0, 0, 0, 0, 0, 0, ... : (a-1) zeroes for n < a
,1/a, 1/(a+1) ,1/(a+2), ... , 1/(2a-1) : for a <= n < 2a
,2/(2a), 2/(2a+1) ,2/(2a+2), ... , 2/(3a-1) : for 2a <= n < 3a
, ... , and for k.a <= n < k.(a+1) :
,k/(k.a), k/(k.a+1), k/(k.a+2), ... , k/(k.a+a-1)
, ... )

Where it is observed that 0 <= b < a in:

1/a - k/(k.a+b) = b/(k.a^2 + a.b) < 1/(k.a)

Looking upon the real value p = 1/a as a hyperreal:

p = (1/a, 1/a, 1/a, ... )

and e = (1, 1/2, 1/3, ... , 1/n) being an infinitesimal

it's easy to see that, in the hyperreals *R : 0 < p - d < e/a

Thus the probability in *R that any random natural is divisible by a
fixed natural (a) is infinitely close to 1/a (is the halo of 1/a).

Provided that probability theory can be and has been extended to the
hyperreal numbers.

Hope I've made no mistakes.

Han de Bruijn

.

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