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Math Help - Little algebra proof

  1. #1
    Member rowe's Avatar
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    Little algebra proof

    If c > 0 is an integer and not prime, then there is an integer b > 0 such that b|c and b \leq \sqrt{c}.

    Approach so far:

    Contradiction, let us assume that b > \sqrt{c}.

    Since br = c for some integer r, then r < c.

    I feel I can arrive at a contradiction, since r \geq c from our initial statement, but not sure how to arrive at such.
    Last edited by rowe; September 7th 2010 at 10:09 AM. Reason: correction
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  2. #2
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    Quote Originally Posted by rowe View Post
    If c > 0 and not prime, then there is an integer b > 0 such that b|c and b \leq \sqrt{c}.
    I am sure you mean that c is a positive integer.
    If c is not prime then there are positive integers, a~\&~b such that c=a\cdot b.
    What if a>\sqrt{c}~\&~b>\sqrt{c}?
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  3. #3
    Member rowe's Avatar
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    Then ab \neq c, but why do we assume a > \sqrt{c}?
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