Question

Prove that \(\left\\{9^{n}: n \in \mathbb{Z}\right\\} \subseteq\left\\{3^{n}: n \in \mathbb{Z}\right\\},\) but \(\left\\{9^{n}: n \in \mathbb{Z}\right\\} \neq\left\\{3^{n}: n \in \mathbb{Z}\right\\}\).

Short answer

The subset relationship \( \{9^n: n \in \mathbb{Z}\} \subseteq \{3^n: n \in \mathbb{Z}\} \) is proven by demonstrating that every element in the first set is also in the second set. To show that the sets are not identical, it suffices to identify at least one element in the second set that does not belong to the first set, such as \( 3^1 = 3 \).

Step-by-step solution

Prove Subset

Given \( n \in \mathbb{Z} \), observe that \( 9^n = (3^2)^n = 3^{2n} \). Since \( 2n \) is also an integer when \( n \) is an integer, it is clear that every element in the set \( \{9^n: n \in \mathbb{Z}\} \) is also in the set \( \{3^n: n \in \mathbb{Z}\} \), so the first set is a subset of the second.

Prove Sets Are Not Equal

To prove that the two sets are not equal, just need to find at least one element that belongs to the second set \( \{3^n: n \in \mathbb{Z}\} \) but not the first set \( \{9^n: n \in \mathbb{Z}\} \). For instance, when \( n = 1 \), then \( 3^1 = 3 \) which belongs to the second set but not the first as there is no integer \( n \) for which \( 9^n = 3 \). Thus, the two sets are not identical.