We are given that:

• $c \mid b$ (i.e., $c$ divides $b$),
• $b \mid a$ (i.e., $b$ divides $a$).

We need to prove that $c \mid a$ (i.e., $c$ divides $a$).

Proof:

1. From $c \mid b$: By the definition of divisibility, $c \mid b$ means there exists some integer $k_1$ such that: $b = c \cdot k_1$ where $k_1 \in \mathbb{Z}$ (the set of integers).

2. From $b \mid a$: Similarly, $b \mid a$ means there exists some integer $k_2$ such that: $a = b \cdot k_2$ where $k_2 \in \mathbb{Z}$.

3. Substitute $b$ from Step 1 into the equation for $a$ in Step 2: Substituting $b = c \cdot k_1$ into $a = b \cdot k_2$, we get: $a = (c \cdot k_1) \cdot k_2$ Simplifying this expression: $a = c \cdot (k_1 \cdot k_2)$ Let $k = k_1 \cdot k_2$, which is also an integer because the product of two integers is an integer.

4. Conclusion: Since $a = c \cdot k$, where $k$ is an integer, it follows by the definition of divisibility that $c \mid a$.

Thus, we have proven that if $c \mid b$ and $b \mid a$, then $c \mid a$.

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Would you like more details or have any questions?

Here are 5 related questions to expand on this concept:

1. What are some examples where this theorem can be applied?
2. How does this proof change if $a$, $b$, and $c$ are negative integers?
3. Can this result be generalized to more than three numbers?
4. How does the greatest common divisor relate to this proof?
5. What happens if one of the numbers is zero?

Tip: Always remember, the definition of divisibility can be extended to any integral domain, not just integers!