Let's address each of the statements one by one:

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(a) If $a \mid b$ and $c \mid d$, then $ab \mid cd$.

Analysis:

• $a \mid b$ means $b = ka$ for some integer $k$.
• $c \mid d$ means $d = lc$ for some integer $l$.

Thus, we know:

• $b = ka$
• $d = lc$

Now consider the product $cd$: $cd = (ka)(lc) = kl(ac)$ This shows that $cd$ is a multiple of $ab$, because $cd = (kl)(ab)$.

Hence, $ab \mid cd$.

Conclusion:

The statement is true.

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(b) If $a \mid bc$, then $a \mid b$ or $a \mid c$.

Analysis:

This statement suggests that if $a$ divides the product $bc$, then $a$ must divide either $b$ or $c$.

This is false, and a counterexample will demonstrate this.

Counterexample:

Let $a = 6$, $b = 2$, and $c = 3$.

• $bc = 2 \times 3 = 6$.
• Clearly, $a \mid bc$, because $6 \mid 6$.

However:

• $a = 6$ does not divide $b = 2$ (since $6$ does not divide $2$).
• $a = 6$ does not divide $c = 3$ (since $6$ does not divide $3$).

Thus, $a \mid bc$, but neither $a \mid b$ nor $a \mid c$.

Conclusion:

The statement is false, as shown by the counterexample.

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(c) If $a$ and $b$ are positive and $a^2 \mid b^2$, then $a \mid b$.

Analysis:

The statement suggests that if $a^2$ divides $b^2$, then $a$ must divide $b$.

Let’s analyze this. If $a^2 \mid b^2$, then there exists an integer $k$ such that: $b^2 = k \cdot a^2$ This implies: $b = a \cdot \sqrt{k}$ For $b$ to be an integer, $k$ must be a perfect square. Let $k = m^2$, so: $b = a \cdot m$ Thus, $a \mid b$.

Conclusion:

The statement is true.

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Final Summary:

• (a) True. Proven.
• (b) False. Counterexample provided.
• (c) True. Proven.

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Do you want more details or have any questions? Here are some related questions:

1. Can you provide other counterexamples for statement (b)?
2. What happens if one of $a$, $b$, or $c$ is negative?
3. How would the statements change if $a$, $b$, or $c$ were not integers?
4. Can you provide examples where $a^2 \mid b^2$ but the quotient $b/a$ is not 1?
5. What would happen if $a = 0$ in statement (a)?

Tip: When working with divisibility, always keep in mind how multiplication and factorization behave in relation to prime factorizations.