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Thread: Prime Theorem-Proof?

  1. #1
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    Prime Theorem-Proof?

    P\a
    Last edited by thegarden; Dec 2nd 2008 at 08:26 AM.
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  2. #2
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    Quote Originally Posted by thegarden View Post
    Proof required for this theorem:
    Let p be prime, p|ab implies p|a or p|b.
    Thanks.
    Surely p isn't a prime then?
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  3. #3
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    I think we can assume p is prime
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  4. #4
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    If p is prime, then p does not divide a. And p does not divide b. By definition of being prime. (Unless a and b are 1 and p).
    So proving that p divides a or b would be rather difficult, no?

    Are you sure you've got the question right?

    a and b are integers, I'm assuming...
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  5. #5
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    Quote Originally Posted by thegarden View Post
    Proof required for this theorem:
    Let p be prime, p|ab implies p|a or p|b.
    Thanks.
    This theorem is correct. ("$\displaystyle p|a$" means that $\displaystyle p$ divides $\displaystyle a$, or that $\displaystyle a$ is a multiple of $\displaystyle p$, not the contrary)

    It results from Gauss Theorem: We assume that $\displaystyle p|ab$. If $\displaystyle p|a$, we are done. Suppose $\displaystyle p\not|a$. Then $\displaystyle a$ and $\displaystyle p$ are relatively prime.
    (If $\displaystyle d$ divides $\displaystyle a$ and $\displaystyle p$, then $\displaystyle d=1$ or $\displaystyle d=p$ because $\displaystyle p$ is prime, hence necessarily $\displaystyle d=1$ since $\displaystyle p\not|a$; hence $\displaystyle 1$ is the only positive common divisor of $\displaystyle a$ and $\displaystyle p$)
    Because $\displaystyle p|ab$, you deduce from Gauss Theorem that $\displaystyle p|b$. qed
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