Answer
See explanation
Work Step by Step
To show \(\mathbb{Z}\) and \(3\mathbb{Z}\) have the same cardinality, it suffices to exhibit a bijection between them. A simple candidate is
\[
f: \mathbb{Z} \;\to\; 3\mathbb{Z}, \quad f(n) = 3n.
\]
1. **Injectivity**: If \(f(n_1) = f(n_2)\), then \(3n_1 = 3n_2\). Dividing both sides by 3 yields \(n_1 = n_2\), so \(f\) is injective.
2. **Surjectivity**: Every element of \(3\mathbb{Z}\) is of the form \(3k\) for some \(k \in \mathbb{Z}\). Given any \(3k \in 3\mathbb{Z}\), we see that \(f(k) = 3k\), so every element in \(3\mathbb{Z}\) is hit by \(f\). Hence, \(f\) is surjective.
Because \(f\) is both injective and surjective, it is a **bijection**, and thus \(\mathbb{Z}\) and \(3\mathbb{Z}\) have the same cardinality.
