Very nice problem, went against my initial guess anyway. I, however,
get the probability that the denominator is not divisible by prime p
to be p/(1+p), i.e. probability that it is divisible by prime p is
1/(1+p).
Argument goes something like this: the probability that p^k is the
highest power of p that divides an integer is P_k = (1/p^k)(1-1/p).
Probability that the denominator is not divisible by p is
Sum_{k>=0}Sum_{0<= j <=k}P_j P_k
which comes out to p/(1+p)
On Wed, May 7, 2008 at 2:37 PM, John Randall
<[EMAIL PROTECTED]> wrote:
> Raul Miller wrote:
> > I think that randomness might be something of a distraction
> > from the main point
>
> OK: here's my argument.
>
> Let p and q be random nonzero integers.
>
> Define events
>
> E : denominator of reduced fraction p/q is even.
>
> E(n): q is divisible by 2^n ; q is divisible by 2^n-1 but not 2^n.
>
> Then the E(n) constitute a partition of E.
>
> Since p(E(n))=1/2^(2*n),
>
> p(E)=1/4+1/16+1/64+....=1/3.
>
> The same argument gives the even more surprising result that the
> probability of the denominator being divisible by a prime p is
> 1/((p^2)-1.
>
> The problem of this argument is that it depends on p and q: the E(n)'s
> become empty after a while. Whether this is is important depends on
> how randomness is defined.
>
>
>
> Best wishes,
>
> John
>
>
>
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