Question

Use the methods introduced in this chapter to prove the following statements. Prove that \(\\{12 n: n \in \mathbb{Z}\\} \subseteq\\{2 n: n \in \mathbb{Z}\\} \cap\\{3 n: n \in \mathbb{Z}\\}\).

Short answer

The statement is true. Any multiple of 12 is also a multiple of both 2 and 3.

Step-by-step solution

Understanding the set notations

The notation \(\{12 n: n \in \mathbb{Z}\}\) means the set of all integers multiplied by 12. Likewise \(\{2 n: n \in \mathbb{Z}\}\) and \(\{3 n: n \in \mathbb{Z}\}\) denote the sets of all integers multiplied by 2 and 3, respectively.

Demonstrating that an element of the first set is in both of the other sets

Firstly, pick an arbitrary element from the set \(\{12 n: n \in \mathbb{Z}\}\). This would be some integer multiplied by 12, or 12n. 12n can be rewritten as 2(6n), meaning 12n is in the set \(\{2 n: n \in \mathbb{Z}\}\), because 6n is an integer. 12n can also be rewritten as 3(4n), meaning 12n is in the set \(\{3 n: n \in \mathbb{Z}\}\), because 4n is also an integer.

Finalizing the proof

As we have shown that 12n, an arbitrary element from the set \(\{12 n: n \in \mathbb{Z}\}\), is in both\(\{2 n: n \in \mathbb{Z}\}\) and \(\{3 n: n \in \mathbb{Z}\}\), it means that every element in \(\{12 n: n \in \mathbb{Z}\}\) lies in the intersection of the latter two sets. As this is true for any arbitrary element of \(\{12 n: n \in \mathbb{Z}\}\), we have shown that \(\{12 n: n \in \mathbb{Z}\} \subseteq\{2 n: n \in \mathbb{Z}\} \cap \{3 n: n \in \mathbb{Z}\}\).