How Does the Cancellation Property Prove Divisibility in Z?

[mikky05v]

Homework Statement


this is the original question
prove: $\forall$ c $\in$ Z, a≠ 0 and b both $\in$ Z$
a|b ⇔ c*a|c*b

Then he corrected himself by saying for problem 1: to show that ca | cb implies a | b ... you must assume c NOT = 0 and invoke "Cancellation Property" of Z.

This kind of confused me but i think I get what he means

The attempt at a solution

so I understand that If you have that ca | cb that's like saying that ac=cbq for some q∈ℤ so, if c≠0 you can just take out those c in the both sides of the expression(because of "Cancellation Property" as he said) and you got left a=bq which means that a|b

my problem is how do I translate this into a formal proof if and only if proof.

 

[mikky05v]

ok so this is what I've got
Prove: ∀c∈Z, c≠0 and b both∈Z a|b⇔ca|cb
a|b if and only if b=ak for some k∈Z
if and only if cb=cak for some c∈Z
if and only if ac|cb

Is this a valid proof? It seems kind of short and it's lacking the "cancelation property" but I'm not sure I understand how to write it any other way