## RE reciprocal consecutive primes

*Expand Messages*

----- Original Message -----

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

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 )

----------------------------------------------------

Jose:

That's not true.

Consider b = a-1, then we have

m*n = (p+a)(p-b) = (p+a)(p-a+1) = p^2 +p-a(a+1)

If a(a+1) > p, then m*n < p^2, with b < a, m=p+a, n = p-b. You can get similar results

with different b < a.

----------------------------------------------

Jan:

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

------------------------------------------------------------

Jose:

Can you explain why? I can't see it at the moment.

Jose Brox

Your message has been successfully submitted and would be delivered to recipients shortly.