On Sun, 20 Jul 2008, Brian M. Scott wrote:
> Apparently what you were really trying to say was something
> like this, which is another matter altogether:
>
> For any y in N let P(y) be the set of primes not
> exceeding sqrt(y), and let n(y) be the number of
> integers in [1, y] not divisible by any member of
> P(y). Then for sufficiently large values of y,
> n(y) is approximately p(y), the number of primes
> in [1, y].
>
Typo? n(y) is approximately P(y), the number of primes
> For any k in [1, y], either k is divisible by some prime in
> P(y), or k is a prime greater than sqrt(y). Thus, n(y) is
> simply the number of primes in [1, y] that are strictly
> greater than sqrt(y), and hence n(y) = p(y) - p(sqrt(y)).
>
> Now n(y)/p(y) = 1 - p(sqrt(y))/p(y). For large y we have
> p(y) approximately equal to y/ln(y) and p(sqrt(y))
> approximately equal to sqrt(y)/ln(sqrt(y)) =
> 2 * sqrt(y)/ln(y), so p(sqrt(y))/p(y) is approximately
> 2 * sqrt(y)/y = 2/sqrt(y), which does indeed tend to 0 as y
> increases.
>
|
