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Thread: Little algebra proof

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

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

    Approach so far:

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

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

    I feel I can arrive at a contradiction, since $\displaystyle r \geq c$ from our initial statement, but not sure how to arrive at such.
    Last edited by rowe; Sep 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 $\displaystyle c > 0$ and not prime, then there is an integer $\displaystyle b > 0$ such that $\displaystyle b|c$ and $\displaystyle b \leq \sqrt{c}$.
    I am sure you mean that c is a positive integer.
    If $\displaystyle c$ is not prime then there are positive integers, $\displaystyle a~\&~b$ such that $\displaystyle c=a\cdot b$.
    What if $\displaystyle a>\sqrt{c}~\&~b>\sqrt{c}?$
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  3. #3
    Member rowe's Avatar
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    Then $\displaystyle ab \neq c$, but why do we assume $\displaystyle a > \sqrt{c}$?
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