Proving Divisibility Property: ab|ac implies b|c

[SeanThatOneGuy]

I'm really having trouble with this proof. at first I thought, oh easy the a's cancel then I realized I am proving that property so that was no help at all. Here is my work so far:

--snip--
Let a, b, and c be integers with a≠0.
If ab|ac, we know from the definition of "divides" that there is an integer k, such that ac=ab⋅k

Then (a - c⋅k) = 0 so k(a - b⋅k) = 0

Since we know that a≠0, then b|c.
--snip--

I'm pretty sure that I'm off the rails after "such that ac=ab⋅k" anyone care to help me? please?

 

[mathman]

Then (a - c⋅k) = 0 so k(a - b⋅k) = 0

Where did that come from?

ac=ab.k => c=b.k

 

[SeanThatOneGuy]

I got a hint from the professor and it really threw me for a loop

--snip--
Then (�� - ��⋅�) = 0 so �(� - �⋅�)
--snip--

so, I was trying to make that form work.

 

[mathman]

ac-kab=0 therefore a(c-kb)=0. Since a≠0, c-kb=0 or b|c.