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Math Help - Divisors

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

    Greetings

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

    x=dk

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

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

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

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

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