From: Jan van Oort <glorifier@...>

Date: Nov 3, 2005 12:10 PM

Subject: Re: [PrimeNumbers] reciprocal consecutive primes

To: theo2357 <Theo.3.1415@...>

This is how far I got:

let

1/p < 1/q + 1/r ( theorem A )

then

p > qr/( r + q ) ( equivalence B )

with q = p + m

and

r = p + n

so we have

p > ( p + m ) ( p + n ) / ( 2p + m + n )

which yields

p^2 > mn

which is true iff

( m < p AND n < p ) OR ( m = p + a AND n = p - b AND b > a ) ( condition i )

now condition ( i ) only holds iff Bertrand's postulate holds

Bertrand's postulate has been proved

so condition ( i ) holds

so equivalence ( B ) holds

so theorem ( A ) holds

QFD

On 11/2/05, theo2357 <Theo.3.1415@...> wrote:

>

> Let p,q,r be three consecutive primes. How can be proven that

> 1/p < 1/q + 1/r? It's NOT trivial!

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