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Math Help - Coprime Proof

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    Coprime Proof

    Hi all,

    a, b are coprime integers such that a|m and b|m, for some integer m. Prove, using Euclid’s lemma, that ab|m.
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  2. #2
    Member courteous's Avatar
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    Red face

    a|m \Rightarrow m=xa
    b|m \Rightarrow m=yb
    ab|m \Rightarrow m=z(ab)=z(\frac{m}{x}\frac{m}{y})=mz(\frac{1}{x}\f  rac{1}{y})

    This probably doesn't suffice, you'd might want to wait for a reply from one of the forum's masters. (Waiting for their admonishment.)

    Meanwhile, is there a [tex] tag keyboard shortcut?
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  3. #3
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    Consider: a|m, so m=xa for some x. Since b|m, b|xa. Since a,b are coprime, b\not|a, nor do they share any factors whatsoever, therefore b|x, and x=yb for some y. Now m=xa=yab, so ab|m. QED

    Courteous: Consider yourself trivially admonished.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by sirellwood View Post
    Hi all,

    a, b are coprime integers such that a|m and b|m, for some integer m. Prove, using Euclid’s lemma, that ab|m.
    Alternatively

    Problem: Suppose \text{gcd}\left(a,b\right)=1. Furthermore suppose that a\mid m and b\mid m. Prove that ab\mid m.

    Proof: Since \text{gcd}(a,b)=1 we know \exists x,y\in\mathbb{Z} such that ax+by=1. Therefore for the same x,y we'd have that axm+bmy=m. Since b|m it is clear that ab|axm similarly since a|m is it clear that ab|bym. Therefore ab|axm+bym=m\quad\blacksquare
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