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

Suppose that gcd(a,b)=1, a|c and b|c. Prove that ab|c.

I've already got this part:
If gcd(a,b)=1 then there are integers r,s such that ra + sb = 1.
If a|n and b|n then there are integers k,l such that n = ka = lb.
rka + skb = k,
b(rl + sk) = k,
ab(rl + sk) = ka = c,
so ab | c

How can I show that the initial assumption of gcd(a,b)=1 is necessary for this to work?

Hello,

Let's say that gcd(a,b)=d then we have:



and



From this 2 realtions results that:





and



following the same procedure we will find that



and



this case is absurd if d is not equal to 1.

so gcd(a,b)=1;

Have a nice day,
Hush_Hush.