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.

Explore the proof that if ac divides bc, then a divides b. This number theory problem demonstrates key concepts of divisibility using algebraic manipulation. Suitable for high school students in Grades 10-12.

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