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Thread: Divisibility (gcd) 2

  1. #1
    Sea
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    Divisibility (gcd) 2

    Show that: $\displaystyle b\neq 0$ and $\displaystyle a=bx+cy $ $\displaystyle \Rightarrow $ $\displaystyle (b,c)\mid (a,b)$
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  2. #2
    o_O
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    Let $\displaystyle d_1 = (a,b)$. This implies $\displaystyle d_1 = {\color{red}a}m + bn$ for some $\displaystyle m,n \in \mathbb{Z}$

    We're given that: $\displaystyle {\color{red}a} = bx+ cy$

    So: $\displaystyle d_1 = ({\color{red}bx + cy})m + bn \ \Leftrightarrow \ d_1 = {\color{magenta} b}(xm+n) + {\color{magenta}c}ym$

    Can you see why $\displaystyle (b,c) \mid d_1$ ?
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  3. #3
    Sea
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    Yes... ... I can see...


    $\displaystyle (b,c)=d \Rightarrow d|b $ and $\displaystyle d|c \Rightarrow d|b(xm+n)$ and $\displaystyle d|cym
    \Rightarrow d|b(xm+n)+cym \Rightarrow d| d_1$





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