I like this naive approach. I know that pi(x) =~ x/log(x), roughly.

and the denominators can't force 0 = 1, since (x^2)/ln(x^2) =~

((x+1)^2)/ln((x+1)^2). It appears adequate, since we know that

there's actually quite a gap between x^2 and (x+1)^2 as the pi's

are measured; not like just saying... between x and x+2.

From: Christ van Willegen <

cvwillegen@...>

To: Bernardo Boncompagni <

RedGolpe@...>

Cc: pgroup <

primenumbers@yahoogroups.com>

Sent: Thursday, September 15, 2011 8:19 AM

Subject: Re: [PrimeNumbers] Re: a prime between x^2 and (x+1)^2, proof?

On Thu, Sep 15, 2011 at 15:07, Bernardo Boncompagni

<

RedGolpe@...> wrote:

>> I think it's exactly Bill's point as well, that ln(x) can never be

>> equal to ln(y), if x <> y... that's what he's using for the argument!

>>

>> Basically, what Bill says is (am I right, Bill?): ln(x^2) <

>> ln((x+1)^2). Since Pi(w) ~ ln(w) (according to PNT), Pi(x^2) <

>> Pi((x+1)^2), so there must be a prime between x^2 and (x+1)^2.

>>

>

> So again, why there should be a prime between, say, x and x+2, since

> ln(x)<>ln(x+2)?

Good point, No idea.

Christ van Willegen

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