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Thread: Divisibility (gcd) 4

  1. #1
    Sea
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    Divisibility (gcd) 4

    Show that: $\displaystyle (a,b)=(a,c)=1 \Rightarrow (a,b,c)=1$
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    Let $\displaystyle d = (a,b,c)$. This implies $\displaystyle d \mid a$, $\displaystyle d \mid b$, and $\displaystyle d \mid c$.

    But this means $\displaystyle d$ is a common divisor of $\displaystyle a$ and $\displaystyle b$. So $\displaystyle d \mid (a,b)$.

    Can you finish?
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  3. #3
    Sea
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    I think...


    $\displaystyle (a,b)=1$ and $\displaystyle (a,c)=1$

    $\displaystyle (a,b,c)=d \Rightarrow d|a ,d|b$ and $\displaystyle d|c \Rightarrow d|a, d|b$ and $\displaystyle d|a,d|c \Rightarrow d|(a,b) $and $\displaystyle d|(a,c)\Rightarrow d|1 \Rightarrow
    d=1$


    $\displaystyle \therefore$ $\displaystyle (a,b)=1$ and $\displaystyle (a,c)=1 \Rightarrow (a,b,c)=1$
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