Question

Let $A = \{5n - 4n - 1 : n \in \mathbb{N}\}$ and $B = \{16(n - 1): n \in \mathbb{N}\}$ be sets. Then:
 

• A. Neither $A \subseteq B$ nor $B \subseteq A$
• B. $A \subseteq B$
• C. $B \subseteq A$
• D. $A \cap B$ is a finite set

Hint:

To compare sets, express the elements explicitly and check if one set's elements can be represented as multiples of the other.

Answer 1

The Correct Option is B

Given the sets: \[ A = \{5n - 4n - 1 : n \in \mathbb{N}\} = \{n - 1 : n \in \mathbb{N}\} \] and \[ B = \{16(n - 1) : n \in \mathbb{N}\} \] The elements of set $A$ are all natural numbers minus one, and the elements of set $B$ are multiples of 16 minus 16. We can see that for every value of $n$ in set $A$, we can find a corresponding value in set $B$, thus making $A$ a subset of $B$. Thus, $A \subseteq B$.