# proof of inifinite primes

• Sep 19th 2007, 07:56 AM
DivideBy0
proof of inifinite primes
Could someone help me understand this proof?

It says, by assumption, $p=p_i$ for some $i=1,2,3,...,n$.

It also says $p_i | a$ and by assumption, $p_i | a+1$. It makes perfect sense if the $p_i$ in the first statement and the $p_i$ in the second are different values of $i$. But it declines this by saying that $p_i | (a+1)-a=1$ (the $p_i$ is the same in both statements). I don't see how this proves anything however, because you could simply choose a different prime for a+1, instead of sticking with the same prime and forcing a contradiction. Also I'm not confortable with all the assumptions being made.

N.B: Exercise 3.2 states that $n|a \wedge n|b \Rightarrow n|a-b$
and Axiom 3 states that $n|1 \Rightarrow n=\pm 1$.
• Sep 19th 2007, 08:13 AM
CaptainBlack
Quote:

Originally Posted by DivideBy0
Could someone help me understand this proof?

It says, by assumption, $p=p_i$ for some $i=1,2,3,...,n$.

It also says $p_i | a$ and by assumption, $p_i | a+1$. It makes perfect sense if the $p_i$ in the first statement and the $p_i$ in the second are different values of $i$. But it declines this by saying that $p_i | (a+1)-a=1$ (the $p_i$ is the same in both statements). I don't see how this proves anything however, because you could simply choose a different prime for a+1, instead of sticking with the same prime and forcing a contradiction. Also I'm not confortable with all the assumptions being made.

N.B: Exercise 3.2 states that $n|a \wedge n|b \Rightarrow n|a-b$
and Axiom 3 states that $n|1 \Rightarrow n=\pm 1$.

By construction every p_i divides a, as it is the product of all the primes.
Presumably Lemma 10.2 is that evey number >3 say is either a prime or
divisible by a prime. Hence N has a prime divisor which is amoung p_1..p_n,
and a+1=N. So if p_i is the prime divisor of N it divides both a and a+1.

RonL
• Sep 19th 2007, 08:36 AM
DivideBy0
Ah, I think it's coming together now...

So $N$ must be prime, right? Hmmm, then this is a bit like induction. Whenever you find an $n_k$, you have proof that $n_{k+1}$ exists.

Thanks.
• Sep 19th 2007, 10:30 AM
ThePerfectHacker
Quote:

Originally Posted by DivideBy0
Ah, I think it's coming together now...

So $N$ must be prime, right? Hmmm, then this is a bit like induction. Whenever you find an $n_k$, you have proof that $n_{k+1}$ exists.

Thanks.

No $N$ is not necessary prime. It has a prime factor which is not among the others.
• Dec 12th 2008, 10:19 PM
baboner
3 is a prime
7 is a prime
3*7 +1 =22 is not a prime

is'nt this conter the proof ?
• Dec 12th 2008, 10:53 PM
CaptainBlack
Quote:

Originally Posted by baboner
3 is a prime
7 is a prime
3*7 +1 =22 is not a prime

is'nt this conter the proof ?

No the assertion is not that $P=\left[1+\prod_{i=1}^n p_i\right]$ is prime but that it has a prime divisor not equal to any of the $p_i$'s. Now that may be $P$ itself but does not have to be.

You will note that $11|22$ and $11>3$ and $11>7$.

CB
• Dec 13th 2008, 02:10 AM
baboner
I see , thanks for explanation