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Math Help - Prime proof

  1. #1
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    Prime proof

    Euclid's Proof of Infinite Number of Primes: Suppose that  p_1=2 < p_2 = 3 < \dots < p_r are all of the primes. Let  P = p_1p_2...p_r+1 and let  p be a prime dividing  P ; then  p can not be any of  p_1 , p_2 , ..., p_r , otherwise p would divide the difference  P-p_1p_2\dots p_r=1 , which is impossible. So this prime  p is still another prime, and  p_1, p_2, ..., p_r would not be all of the primes.  \square

    (a) Why can't  p divide  1 , where  p is one of  \{p_{1}, p_{2}, p_{3}, \ldots, p_{r} \} ?

    Proof: We want to show that  P \equiv P-1 (\mod p) is false. To prove this, choose  a,b \in \mathbb{Z} with  a \neq 0 . Then  a divides  b if there is an element  c \in \mathbb{Z} such that  b = ac . Suppose for contradiction that  p divides  1 . Then  1 = pc where  c \in \mathbb{Z} . Since  p > 1 by definition,  c \not \in \mathbb{Z} . Contradiction. Thus  p cannot divide  1 .  \square

    Is this correct?
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  2. #2
    o_O
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    I think you're complicating things a bit. The only positive integer that can divide 1 is 1 which isn't a prime.
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  3. #3
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    But in a rigorous sense the above proof is correct right?
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  4. #4
    MHF Contributor kalagota's Avatar
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    you have to talk about P whether it is a prime or not.

    if P is prime, then you are done is P > p_i for all i.

    if P is not prime, then there is a p \in \{p_1, p_2,...,p_r\} that divides P.

    then p|P - p_1p_2...p_r implies p|1 \Longleftrightarrow p=1. this is your contradiction since p \in \{p_1, p_2,...,p_r\} and 1 is not in the set..
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