The given statement is:

If $a$ divides $bc$, then $a$ divides $b$ or $a$ divides $c$.

To determine whether the statement is true or false, let's analyze it by considering a counterexample.

Analysis

The statement asserts that if $a \mid bc$, meaning that $a$ divides the product $bc$, then either $a \mid b$ or $a \mid c$ must be true. We need to check if this condition holds in all cases.

Counterexample:

Consider the integers $a = 6$, $b = 4$, and $c = 9$.

• First, check if $a \mid bc$. We calculate: $bc = 4 \times 9 = 36.$ Since $6 \mid 36$ (because $36 \div 6 = 6$), the condition $a \mid bc$ is satisfied.

• Now, check if $a \mid b$ or $a \mid c$:

  • For $a \mid b$, we check if $6 \mid 4$. This is false, because $4 \div 6$ is not an integer.
  • For $a \mid c$, we check if $6 \mid 9$. This is also false, because $9 \div 6$ is not an integer.

Conclusion

In this case, we have $a \mid bc$ (because $6 \mid 36$), but neither $a \mid b$ nor $a \mid c$. This provides a counterexample showing that the statement is false.

Thus, the conditional statement "If $a$ divides $bc$, then $a$ divides $b$ or $a$ divides $c$" is false.

Further Exploration

If you'd like, we can explore:

1. Why the distributive property of divisibility does not always apply.
2. Other counterexamples involving different integers.
3. Related properties of divisibility and how they can be used in proofs.
4. The logical structure of conditional statements and their negations.
5. The role of prime factorization in divisibility problems.

Tip: When dealing with divisibility, always test edge cases or small numbers to identify potential counterexamples!