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Thread: Divisors

  1. #1
    Senior Member I-Think's Avatar
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    Divisors

    Greetings

    Prove that any integer $\displaystyle x>1$ has either a divisor $\displaystyle >1$ and $\displaystyle \leq\sqrt{x}$ or no divisor $\displaystyle >1$ and$\displaystyle \leq\sqrt{x}$ (In the latter case it is called a prime).
    Prove this statement from the axioms governing the behavior of integers.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by I-Think View Post
    Greetings

    Prove that any integer $\displaystyle x>1$ has either a divisor $\displaystyle >1$ and $\displaystyle \leq\sqrt{x}$ or no divisor $\displaystyle >1$ and$\displaystyle \leq\sqrt{x}$ (In the latter case it is called a prime).
    Prove this statement from the axioms governing the behavior of integers.
    Suppose $\displaystyle x$ has a proper divisor $\displaystyle d>1$, then there exists an integer $\displaystyle k>1$ such that:

    $\displaystyle x=dk$

    Now if $\displaystyle d \le \sqrt{x}$ there is nothing left to do, so assume $\displaystyle d>\sqrt{x} $, then:

    $\displaystyle x=\sqrt{x}\sqrt{x}=dk>\sqrt{x}k $

    which implies that $\displaystyle k<\sqrt{x}$ which is a factor $\displaystyle \le \sqrt{x}.$

    Which leaves you tidy this up and sort out the details.

    CB
    Last edited by CaptainBlack; Jun 28th 2009 at 01:09 AM.
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