On Sun, 20 Jul 2008 19:26:12 +0100, Jack <jj@[EMAIL PROTECTED]
>
wrote in <news:6rLgk.13933$9S6.934@[EMAIL PROTECTED]
> in
alt.algebra.help:
>>>>>>> Never mind. I'll guess what you're trying to say.
>>>>>>> Let P be a set of primes.
>>>>>>> Let n(y,P) be the number of integers n, in [1,y]
>>>>>>> with for all p in P and p <= sqr y, p not divide n.
>>>>>>> Then n(y,P) converges on being negligibly different from
>>>>>>> the number of primes found in that interval. Hm. Mostly
>>>>>>> gibberish. Let's see.
>>>>>>> Let p(y) be the number of primes in [1,y]
>>>>>>> Then lim(y->oo) n(y,P)/p(y) = 1.
[...]
> I don't know why you are saying <<Take P = {2}. For all
> y >= 4, n(y, P) is simply the number of odd integers in
> the interval [1, y], which is ceil(y/2) ~ y/2>> and
> subsequently <<More generally, let P be any finite set of
> primes, and let m = prod(P)>> when I made it clear that
> I want P to be the set of primes of value less than the
> square root of y, [...]
You did not make it clear to William. Look at his attempt
to make sense of your claim, which is still present at the
top of this post (and which you later said was correct): it
does NOT say any such thing. It STARTS with a set P of
primes, and lets y be arbitrary. And as I pointed out
several times, I was refuting THAT claim (e.g., 'This shows
that Jack's claim is false, assuming that your translation
of it is correct').
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].
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.
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