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Math Help - Help with a proof - gcd

  1. #1
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    Help with a proof - gcd

    We define a\mathbb{Z}+b\mathbb{Z}=\left \{ au+bv:u,v\in\mathbb{Z} \right \}.

    I need to prove that a\mathbb{Z}+b\mathbb{Z}=(a,b)\cdot\mathbb{Z}.

    so, I need to prove a\mathbb{Z}+b\mathbb{Z}\subseteq (a,b)\cdot\mathbb{Z} and a\mathbb{Z}+b\mathbb{Z}\supseteq (a,b)\cdot\mathbb{Z}.

    One direction is easy:

    ( \supseteq) Let n\in (a,b)\cdot\mathbb{Z}, so n=(a,b)\cdot k, for some k\in\mathbb{Z}, and since there exist s,t\in\mathbb{Z} such that (a,b)=sa+tb, we get:
    n=(a,b)\cdot k
    n=(sa+tb)k
    n=ska+tkb
    n=(sk)a+(tk)b \Rightarrow n\in a\mathbb{Z}+b\mathbb{Z}

    ( \subseteq)
    I need help here...



    Thanks in advanced!
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  2. #2
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    Re: Help with a proof - gcd

    The ⊆ direction is easier: it does not even need Bézout's identity (a, b) = sa + tb.
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  3. #3
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    Re: Help with a proof - gcd

    Sorry, I couldn't think of anything.
    Can you give me another hint?
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  4. #4
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    Re: Help with a proof - gcd

    a=(a,b)a_1 and b=(a,b)b_1 with a_1,b_1\in\mathbb{Z}
    Thanks from Stormey
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