this is the original question prove: $\forall c \in Z, a\neq 0 $and b both $ \in Z$ $a|b \iff c\cdot a|c\cdot b$
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 can someone help me get started on this
Edit: I understand how to get to a|b now but how do I write this using formal proof logic?
Update: 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
If you have that ca | cb thats like saying that $bc=caq$ for some $q \in \mathbb{Z}$ so, if $c\neq 0$ 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 $b=aq$ wich means that $a|b$