Divisibility of a Product

Problem Statement

Prove that for all positive integers $a, b,$ and $c$ , if $a$ divides the product $bc$ , then $a$ divides $b$ or $a$ divides $c$ .

Hints

Attempt to set up a direct proof using the definition of divisibility: $bc = ak$ for some integer $k$ . Can you algebraically force $a$ to divide $b$ or $c$ ?

Wait—is this statement actually true? Try testing it with some small integer values. What happens if $a$ is a composite number (like $4$ or $6$ ) rather than a prime number?

Exam Strategy: If you ever encounter an exam question that seems fundamentally incorrect or impossible, do not leave the page blank! A blank page guarantees 0 marks. Instead, prove that the statement is false. The absolute best way to completely destroy a ``for all'' statement is to provide a single, concrete counterexample.

Solutions

The problem statement is strictly false. We will disprove it by providing a counterexample.

To disprove the statement ``if $a$ divides $bc$ , then $a|b$ or $a|c$ '', we must find specific positive integers $a, b,$ and $c$ such that $a$ divides $bc$ , but $a \nmid b$ and $a \nmid c$ .

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

Check the condition (the premise):

bc = 2 \times 3 = 6

Since $6$ divides $6$ , the premise that $a$ divides $bc$ is satisfied.

Check the conclusion:

Does $a$ divide $b$ ? No, $6$ does not divide $2$ .
Does $a$ divide $c$ ? No, $6$ does not divide $3$ .

Since both parts of the conclusion are false, the entire statement is invalid.

Therefore, it is not true for all positive integers $a, b, c$ that $a|bc \implies a|b \text{ or } a|c$ . $\blacksquare$

Takeaways

Trust Your Math, Not the Paper: Examiners are human and occasionally make mistakes. If you leave a flawed question blank, you receive 0 marks. If you mathematically prove that the question is flawed, you demonstrate mastery and will be awarded marks.
The Power of Counterexamples: To prove a universal statement (``for all $x$ \ldots'') you must prove it algebraically for every number in existence. But to disprove it, you only need to find one single case where it fails.
Euclid's Lemma: The statement in this problem becomes completely true if we add one crucial condition: $a$ must be a prime number. Composite numbers can have their factors split across $b$ and $c$ , which is why $a=6$ failed above.

Further Readings

If you found this proof interesting, be sure to check out these relevant HSC booklets to sharpen your reasoning skills:

HSC Proofs: https://vumaths.com/booklets/hsc-proofs/
HSC Mechanics: https://vumaths.com/booklets/hsc-mechanics/
HSC Inequalities: https://vumaths.com/booklets/hsc-inequalities/

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