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:

$\displaystyle a=a_1*d \Rightarrow (a_1d|c)$

and

$\displaystyle b= b_1*d \Rightarrow (b_1d|c)$

From this 2 realtions results that:

$\displaystyle ab|c \Rightarrow d^2a_1b_1|c \Rightarrow $

$\displaystyle d^2a_1|c $

and

$\displaystyle d^2b_1|c$

following the same procedure we will find that

$\displaystyle d^na_1|c $

and

$\displaystyle d^nb_1|c$

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

so gcd(a,b)=1;

Have a nice day,
Hush_Hush.