Question

If \(k \in \mathbb{Z},\) then \(\\{n \in \mathbb{Z}: n \mid k\\} \subseteq\left\\{n \in \mathbb{Z}: n \mid k^{2}\right\\}\).

Short answer

Yes, it is true that for all \( k \) that are integers, all integers \( n \) that divide \( k \) are a subset of the integers that divide \( k^2 \). This is proven through the transitive property of divisibility.

Step-by-step solution

Understand the notion of sets and subsets

A set \( A \) is a subset of a set \( B \) if every element of \( A \) is also an element of \( B \). So to prove $\{n \in \mathbb{Z}: n \mid k\} \subseteq \left\{n \in \mathbb{Z}: n \mid k^2 \right\}$, it means that every integer, \( n \), that divides \( k \) must also divide \( k^2 \).

Proof of the subset

Let \( n \) be any integer that divides \( k \), i.e. \( n \mid k \). Then there exists some integer \( m \) such that \( k = n \cdot m \). Thus \( k^2 = (n \cdot m)^2 = n^2 \cdot m^2 \). It is clear that \( n \) divides \( k^2 \), i.e. \( n \mid k^2 \). Hence, every element of the set $\{n \in \mathbb{Z}: n \mid k \}$ is also an element of the set $\{n \in \mathbb{Z}: n \mid k^2 \}$. Therefore, we can conclude that $\{n \in \mathbb{Z}: n \mid k\}$ is a subset of $\{n \in \mathbb{Z}: n \mid k^2\}$.