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Thread: proof if gcd(a,b)=1 and a|c b|c then ab|c

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    proof if gcd(a,b)=1 and a|c b|c then ab|c

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    Re: proof if gcd(a,b)=1 and a|c b|c then ab|c

    That follows pretty quickly from the definition of "greatest common dominator", doesn't it? If gcd(a, b)= 1, then a and b have no common factors. You might try, for one thing, consider the prime factorization of c.
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    Re: proof if gcd(a,b)=1 and a|c b|c then ab|c

    Quote Originally Posted by dave52 View Post
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    little rigorous proof:

    if $\displaystyle a|c$ and $\displaystyle b|c$ there exists some integers $\displaystyle r$ and $\displaystyle s$ such that $\displaystyle c=ar=bs$
    now as $\displaystyle gcd(a,b)=1$, we can write: $\displaystyle 1=ax+by \implies$ (after multiplying both sides by $\displaystyle c$) $\displaystyle c=acx+bcy$
    after substituting the right expressions for $\displaystyle c $ we get $\displaystyle c=a(bs)x+b(ar)y=(ab)(sx)+(ab)(ry)=ab(sx+ry)$, which in other words mean $\displaystyle ab|c$
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