Infinite Primes Proof is complete ?

Eyala · #1 $Old$ November 2nd, 2009, 04:42 AM

Hello ,

The proof to infinite prime numbers is saying that if you create new
number N from factoring the known finite prime series(P1*P2..*Pn) and
add 1 ,you will have number that will be new prime ,or be factored
from new prime that is not included in the finite prime series
My question is ,how do I prove that the N cannot be factored from
P1*P1*P2*... which one or more of the primes from the finite series appears more than
once,the proof is talking only about the case that P1,P2,..Pn appears
once in the factoring series but what about the option that it appears
more than once ? .

Thanks

HallsofIvy · #2 $Old$ November 2nd, 2009, 05:44 AM

Quote:

Originally Posted by Eyala $View Post$

Hello ,

The proof to infinite prime numbers is saying that if you create new
number N from factoring the known finite prime series(P1*P2..*Pn) and
add 1 ,you will have number that will be new prime ,or be factored
from new prime that is not included in the finite prime series
My question is ,how do I prove that the N cannot be factored from
P1*P1*P2*... which one or more of the primes from the finite series appears more than
once,the proof is talking only about the case that P1,P2,..Pn appears
once in the factoring series but what about the option that it appears
more than once ? .

Thanks

What does a prime appearing more than once on the list have to do with anything? The proof does NOT say anything about a "factoring series" (unless you mean the purported "list of all primes" itself) or primes appearing once in a "factoring series". It simply use the fact that either a number is a prime or it is divisible by a prime (which is the definition of "prime" and "composite" numbers). If there were a finite number of primes, $P_1, P_2, \cdot\cdot\cdot, P_n$ , then none of those numbers divides $P1*P2*\cdot\cdot\cdot P_n+ 1$ . Either that number is itself prime or it is divisible by a prime that is NOT on that list. In either case that list does not include all primes.

If a number has $P1*P2$ as a factor, then it has P1 as a factor.

Eyala · #3 $Old$ November 2nd, 2009, 05:56 AM

Quote:

Originally Posted by HallsofIvy $View Post$

What does a prime appearing more than once on the list have to do with anything? The proof does NOT say anything about a "factoring series" (unless you mean the purported "list of all primes" itself) or primes appearing once in a "factoring series". It simply use the fact that either a number is a prime or it is divisible by a prime (which is the definition of "prime" and "composite" numbers). If there were a finite number of primes, $P_1, P_2, \cdot\cdot\cdot, P_n$ , then none of those numbers divides $P1*P2*\cdot\cdot\cdot P_n+ 1$ . Either that number is itself prime or it is divisible by a prime that is NOT on that list. In either case that list does not include all primes.

If a number has $P1*P2$ as a factor, then it has P1 as a factor.

The proof relies on the fact that if you divide N by one of the known Primes
you get 1 as remainder so the prime is not on our finite list,
but How could you prove that N is not divisable by P1*P1*P2 for example ?
sometimes numbers are factored by repeated occurance of primes like 4=2*2.

tonio · #4 $Old$ November 2nd, 2009, 06:49 AM

Quote:

Originally Posted by Eyala $View Post$

The proof relies on the fact that if you divide N by one of the known Primes
you get 1 as remainder so the prime is not on our finite list,
but How could you prove that N is not divisable by P1*P1*P2 for example ?
sometimes numbers are factored by repeated occurance of primes like 4=2*2.

If the number N is divisible by P1*P1*P2 then it is divisible by P1, something you already showed is impossible

Tonio