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Thread: Divisibility proof

  1. #1
    rf0
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    Divisibility proof

    gcd(a,b) = 1 and a|m and b|m. How to prove ab|m without using unique-prime-factorization theorem?
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by rf0 View Post
    gcd(a,b) = 1 and a|m and b|m. How to prove ab|m without using unique-prime-factorization theorem?
    $\displaystyle a \mid m \implies m=am' ~,~ \text{where } m' \in \mathbb{Z}$

    $\displaystyle b \mid m \implies m=bm'' ~,~ \text{where } m'' \in \mathbb{Z}$

    Therefore $\displaystyle am'=bm''$ and hence a divides $\displaystyle bm''$

    By Gauss theorem, if $\displaystyle \text{gcd}(a,b)=1$ and $\displaystyle a \mid bc$, then $\displaystyle a \mid c$
    So here, we can conclude that $\displaystyle a \mid m''$.
    So we can write $\displaystyle m''=am''' ~,~ \text{where } m''' \in \mathbb{Z}$

    So $\displaystyle m=bm''=b(am''')=(ab)m'''$

    Therefore $\displaystyle ab \mid m \quad \quad \square$
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  3. #3
    rf0
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    Thanks a bunch, I understand it now
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