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

• February 7th 2013, 01:23 PM
dave52
proof if gcd(a,b)=1 and a|c b|c then ab|c
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• February 7th 2013, 01:35 PM
HallsofIvy
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.
• February 7th 2013, 08:10 PM
earthboy
Re: proof if gcd(a,b)=1 and a|c b|c then ab|c
Quote:

Originally Posted by dave52
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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 $ab|c$