1. ## proving ab|c

I dont know why im having trouble with this problem i sware i did it last semester in number theory now for some reason i cant figure it out in modern algebra.

Given a|b and b|c and gcd(a,b)=1 then ab|c. I know we get b = aj, c = bk and 1 = am + bn i cant figure out how to put them together to get ab|c

2. Originally Posted by ChrisBickle
I dont know why im having trouble with this problem i sware i did it last semester in number theory now for some reason i cant figure it out in modern algebra.

Given a|b and b|c and gcd(a,b)=1 then ab|c. I know we get b = aj, c = bk and 1 = am + bn i cant figure out how to put them together to get ab|c
Since $(a,b)=1$ there exists $x,y\in\mathbb{Z}$ such that $ax+by=1\implies axc+byc=c$. So then, since $ab\mid acx$ and $ab\mid byc$ we can conclude that $ab\mid acx+byc=c$

3. dont mean to sound stupid but im not seeing how we get ab|acx and ab|byc

edit: are we using the axc + byc =c then saying byc = ajybk?

4. Originally Posted by ChrisBickle
dont mean to sound stupid but im not seeing how we get ab|acx and ab|byc
$ab\mid acx$ since $a\mid a$ and $b\mid c$. $ab\mid byc$ since $a\mid b$ and $b\mid c$.

5. thanks alot

6. Well ... if you think about it, if a|b and gcd(a,b)=1 => a=1, so if b|c, then ab|c ....

Why does a=1?

a|b => ma = b

So, gcd(a,b)=gcd(a,ma)=a

Since gcd(a,b)=1 => a=1

7. I think it is a typo. it should be "a|c,b|c,and gcd(a,b)=1...".

8. ah, haha ... probably ... I was kind of skeptical about the question