There are natural numbers $a,b,c$ such that $ac\mid bc$, while $a\nmid b$

Question

There are natural numbers $a,b,c$ such that $ac\mid bc$, while $a\nmid b$.

My answer was False. I substituted numbers for $a,b,c$. But I think I have confused my self. so $a = 2 , b = 4 , c = 6$.

Answer 1

I assumed that $\Bbb N=\{1,2,3,\dots\}$.

Given Statement: $\exists a,b,c\in\Bbb N$ ; $ac|bc\implies a\not|\ \ b$

Negation Statement: $\forall a,b,c\in\Bbb N$, $ac|bc\implies a|b$.

Which of the two statements is true?

We claim that the negation statement is true.

Let $a,b,c\in\Bbb N$ such that $ab|bc$. Then $$\frac{b}{a}=\frac{bc}{ac}\in\Bbb N.$$ This means that $a|b$.

Because the negation statement is true, what can be said to the given statement? The answer is FALSE.

Answer 2

The statement $ac|bc$ means $ack = bc$ for some $k\in\mathbb{Z}$. Therefore $ak = b$, or equivalently $a|b$. So the original statement is not true.

Edit: it has been pointed out that if either $a$ or $b$ is 0 then the statement is trivially true.