
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
and are arbitrary integers. It doesn't make sense to let and . Plus it confuses readers like me....
You can use different integers, e.g. and , in which , clearly, 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, and , so for any integers and ,
Therefore the sum is also congruent to 0 (mod a).