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Let A = \(\{ax : x \in \mathbb{N}\}\) and suppose \(B \cap C = \mathbb{N}\), where b and c are natural numbers that are relatively prime. Then which of the following must be true?

1. d = bc
2. c = bd
3. b = cd
4. None of these

Correct answer: b = cd

Solution

For the intersection to be all natural numbers, the generating parameters must align so that every natural number is representable in both sets. With b and c relatively prime, the only way this compatibility can hold is when one parameter equals the product of the other two in the given relation, namely b = cd. Hence that statement must be true.

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