To prove the statement:

If $ac \mid bc$, then $a \mid b$,

we will use the definition of divisibility. Here is the step-by-step proof:

Proof:

1. Assume that $ac \mid bc$:
   This means that there exists some integer $k$ such that:

   $bc = (ac)k.$

2. Simplify the equation:

   Rewrite the equation:

   $bc = ack.$

   Now, divide both sides of the equation by $c$ (assuming $c \neq 0$):

   $b = ak.$

3. Conclusion:

   The equation $b = ak$ shows that $b$ is a multiple of $a$.
   Therefore, by the definition of divisibility, $a \mid b$.

Summary:

We have shown that if $ac \mid bc$, then it must be true that $a \mid b$, assuming $c \neq 0$.

Would you like more details, or do you have any questions about this proof?

Here are 5 related questions that build on this topic:

1. What happens if $c = 0$ in the original statement?
2. Can you provide a counterexample to show that the statement does not hold if $c = 0$?
3. How does this proof change if we are considering divisibility in different number sets (like complex numbers)?
4. Is the converse of this statement true? If $a \mid b$, does $ac \mid bc$ for any $c$?
5. Can you extend this proof to show that if $ac \mid bc$ for any $c \neq 0$, then $\gcd(a, b) = a$?

Tip: Always verify divisibility by writing out the definitions and look for relationships between the divisors.