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Math Help - 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, 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.
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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, 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
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  3. #3
    Senior Member DivideBy0's Avatar
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    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.
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    Quote Originally Posted by DivideBy0 View Post
    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.
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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 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
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  7. #7
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    I see , thanks for explanation
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