# Thread: proof of inifinite primes

1. ## proof of inifinite primes

Could someone help me understand this proof?

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

It also says $\displaystyle p_i | a$ and by assumption, $\displaystyle p_i | a+1$. It makes perfect sense if the $\displaystyle p_i$ in the first statement and the $\displaystyle p_i$ in the second are different values of $\displaystyle i$. But it declines this by saying that $\displaystyle p_i | (a+1)-a=1$ (the $\displaystyle p_i$ 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 $\displaystyle n|a \wedge n|b \Rightarrow n|a-b$
and Axiom 3 states that $\displaystyle n|1 \Rightarrow n=\pm 1$.

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

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

It also says $\displaystyle p_i | a$ and by assumption, $\displaystyle p_i | a+1$. It makes perfect sense if the $\displaystyle p_i$ in the first statement and the $\displaystyle p_i$ in the second are different values of $\displaystyle i$. But it declines this by saying that $\displaystyle p_i | (a+1)-a=1$ (the $\displaystyle p_i$ 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 $\displaystyle n|a \wedge n|b \Rightarrow n|a-b$
and Axiom 3 states that $\displaystyle n|1 \Rightarrow n=\pm 1$.
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

3. Ah, I think it's coming together now...

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

Thanks.

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

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

Thanks.
No $\displaystyle N$ is not necessary prime. It has a prime factor which is not among the others.

5. 3 is a prime
7 is a prime
3*7 +1 =22 is not a prime

is'nt this conter the proof ?

6. 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 $\displaystyle P=\left[1+\prod_{i=1}^n p_i\right]$ is prime but that it has a prime divisor not equal to any of the $\displaystyle p_i$'s. Now that may be $\displaystyle P$ itself but does not have to be.

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

CB

7. I see , thanks for explanation