
Divides proof
I'm struggling with the following question:
let a, b, c, k, l be integers
Prove that if: ab and ac then a(kb + lc)
I'm not looking for the answer outright just a little help.
I realise that:
b = ka
c = la
(kb + lc) = m x a
The approach I've been taking so far is to go: b + c = ka + kl and then try to rearrange (ka + kl) to (kb + lc) since I know that
if ab and ac then a(b + c)
Am I on the right track? If so then I know it's my algebra skills which are the problem.

Re: Divides proof
$\displaystyle k$ and $\displaystyle l$ are arbitrary integers. It doesn't make sense to let $\displaystyle b = ka$ and $\displaystyle c = la$. Plus it confuses readers like me....
You can use different integers, e.g. $\displaystyle b = qa$ and $\displaystyle c = ra$, in which $\displaystyle kb + lc = k(qa) + l(ra) = a(kq + lr)$, clearly, $\displaystyle a$ divides this.
An alternate solution is to use mods, which works just as well (and it doesn't need random variables!).

Re: Divides proof
Thanks for that Richard, makes much more sense.
We have been studying mods lately so out of interest, how would you approach this using mods?

Re: Divides proof
Basically, $\displaystyle b \equiv 0 (\mod a)$ and $\displaystyle c \equiv 0 (\mod a)$, so for any integers $\displaystyle k$ and $\displaystyle l$,
$\displaystyle bk \equiv 0 (\mod a)$
$\displaystyle cl \equiv 0 (\mod a)$
Therefore the sum $\displaystyle bk + cl$ is also congruent to 0 (mod a).