Answer
See explanation
Work Step by Step
A succinct way to see that the nonnegative integers \(\{0,1,2,3,\dots\}\) have the same cardinality as the positive integers \(\{1,2,3,\dots\}\) is to give an explicit bijection. One of the simplest is:
\[
f: \mathbb{Z}^+ \;\to\; \mathbb{Z}_{\ge 0},
\quad f(n) \;=\; n - 1.
\]
- **Injectivity**: If \(f(n_1) = f(n_2)\), then \(n_1 - 1 = n_2 - 1\), so \(n_1 = n_2\).
- **Surjectivity**: Every nonnegative integer \(m \in \{0,1,2,\dots\}\) can be written as \(m = (m+1) - 1\), and \(m+1\) is a positive integer. Hence every nonnegative integer is hit by some \(n \in \mathbb{Z}^+\).
Because \(f\) is both one-to-one and onto, it is a **bijection**, so the set of nonnegative integers is countable.
