Answer
See explanation
Work Step by Step
A concise way to see that the set of all positive squares \(S = \{1^2, 2^2, 3^2, \dots\}\) has the same cardinality as the set of all positive integers \(\mathbb{Z}^+\) is to exhibit an explicit bijection between them. One simple choice is the function
\[
f:\mathbb{Z}^+ \;\to\; S,
\quad f(n) \;=\; n^2.
\]
**Why this works**:
1. **Injective (one-to-one):**
If \(f(n_1) = f(n_2)\), then \(n_1^2 = n_2^2\). Because both \(n_1\) and \(n_2\) are positive integers, we must have \(n_1 = n_2\). Hence, \(f\) is injective.
2. **Surjective (onto):**
Any element of \(S\) is of the form \(k^2\) for some \(k \in \mathbb{Z}^+\). But \(f(k) = k^2\), so every square in \(S\) is hit by the function. Thus, \(f\) is surjective.
Since \(f\) is both injective and surjective, it is a **bijection**. Therefore, \(\mathbb{Z}^+\) and \(S\) have the same cardinality. Informally, “there are as many squares as there are positive integers,” even though squares are a proper subset of all positive integers. This is a classic illustration that infinite sets can behave counterintuitively with respect to size (cardinality).
