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Thread: proof of inifinite primes

  1. #1
    Senior Member DivideBy0's Avatar
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    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 $\displaystyle n|1 \Rightarrow n=\pm 1$.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by DivideBy0 View Post
    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
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  3. #3
    Senior Member DivideBy0's Avatar
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    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.
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  4. #4
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    Quote Originally Posted by DivideBy0 View Post
    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.
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  5. #5
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    3 is a prime
    7 is a prime
    3*7 +1 =22 is not a prime

    is'nt this conter the proof ?
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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by baboner View Post
    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
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  7. #7
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    I see , thanks for explanation
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