www.piclist.com/techref/index.htm?key=prime+numbers

BY : Sean Breheny email (remove spam text)

Peter has a good point - if you also show that all integers have prime

factorizations (easy to do) then your method DOES prove that there are

infinitely many primes because it proves that there exists an integer Q,

larger than Pn, which is not divisible by any of the primes below (or equal

to) Pn - so either this number Q is itself prime, in which case you've

found a prime which is not in your original set, or if Q is not prime, it

must be divisible by at least one prime factor which is not contained in

your original set. Since Q is not even, this prime factor is not 2. Since

your set contains all primes between 2 and Pn, inclusive of Pn, you have

shown that there is at least one prime which is not in your set but is

larger than Pn.

On Wed, Sep 6, 2017 at 3:56 PM, peter green <TakeThisOuTplugwashp10link.net> wrote:

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See also: www.piclist.com/techref/index.htm?key=prime+numbers