Question

If \(m, n \in \mathbb{Z},\) then \(\\{x \in \mathbb{Z}: m n \mid x\\} \subseteq\\{x \in \mathbb{Z}: m \mid x\\} \cap\\{x \in \mathbb{Z}: n \mid x\\}\).

Short answer

The short answer is that we have proven \(\{x \in \mathbb{Z}: m n \mid x\} \subseteq \{x \in \mathbb{Z}: m \mid x\} \cap \{x \in \mathbb{Z}: n \mid x\}\) and hence shown that if \(x\) can be divided evenly by the product of two integers \(m\) and \(n\), it can also be divided evenly by both \(m\) and \(n\).

Step-by-step solution

Understanding the symbols

Firstly, understand that \(m \mid x\) and \(n \mid x\) mean that \(x\) can be divided evenly by \(m\) and \(n\), respectively. The symbol \(\subseteq\) signifies 'is a subset of'. So for this problem, \(\{x \in \mathbb{Z}: m n \mid x\} \subseteq \{x \in \mathbb{Z}: m \mid x\} \cap \{x \in \mathbb{Z}: n \mid x\}\) means that if \(x\) can be divided evenly by \(mn\), it can be divided evenly by both \(m\) and \(n\).

Proof

Now, let's start the proof. Assume that \(x\) belongs to the set \(\{x \in \mathbb{Z}: m n \mid x\}\). That means, \(x\) is a multiple of \(mn\). Therefore, \(x = k(mn)\) for some \(k\) in integers. Here we note that \(k(mn)\) is divisible by both \(m\) and \(n\) as both \(m\) and \(n\) are factors of \(mn\). Therefore, \(x\) can belong to both the sets \(\{x \in \mathbb{Z}: m \mid x\}\) and \(\{x \in \mathbb{Z}: n \mid x\}\). Thus, if \(x\) belongs to the set \(\{x \in \mathbb{Z}: m n \mid x\}\), then \(x\) belongs to the set \(\{x \in \mathbb{Z}: m \mid x\} \cap \{x \in \mathbb{Z}: n \mid x\}\). This proves the subset relation.