.

Printable View

- Feb 7th 2013, 01:23 PMdave52proof if gcd(a,b)=1 and a|c b|c then ab|c
.

- Feb 7th 2013, 01:35 PMHallsofIvyRe: 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. - Feb 7th 2013, 08:10 PMearthboyRe: proof if gcd(a,b)=1 and a|c b|c then ab|c
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 $\displaystyle ab|c$