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Math Help - 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
    .
    little rigorous proof:

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