Answer
See explanation
Work Step by Step
**Solution Outline**
1. **Definition of \(g\)**
The function in question is
\[
g\colon \mathbb{Z}^+ \times \mathbb{Z}^+ \;\to\; \mathbb{Z}^+,
\quad
g(m,n) \;=\; 2^m\,3^n.
\]
We want to show \(g\) is **one‐to‐one** (injective) and then use this to conclude \(\mathbb{Z}^+ \times \mathbb{Z}^+\) is countable.
---
## 1. Proving \(g\) is One‐to‐One
Suppose
\[
g(m_1,n_1) \;=\; g(m_2,n_2),
\]
i.e.,
\[
2^{m_1}\,3^{n_1}
\;=\;
2^{m_2}\,3^{n_2}.
\]
Since \(2\) and \(3\) are **distinct prime bases**, the only way these two products can be equal is if the exponents of \(2\) match and the exponents of \(3\) match. Formally, prime factorization is unique, so we must have
\[
m_1 \;=\; m_2
\quad\text{and}\quad
n_1 \;=\; n_2.
\]
Thus \((m_1,n_1)=(m_2,n_2)\). This shows \(g\) is injective.
---
## 2. Using \(g\) to Prove \(\mathbb{Z}^+ \times \mathbb{Z}^+\) Is Countable
Since \(g\) is an **injection** from \(\mathbb{Z}^+ \times \mathbb{Z}^+\) **into** \(\mathbb{Z}^+\), we see that
\[
\{\, (m,n)\mid m,n\in \mathbb{Z}^+\}\;\;
\text{injects into}
\;\;\{\,2^m 3^n \mid m,n\in \mathbb{Z}^+\}\;\subseteq\;\mathbb{Z}^+.
\]
1. \(\mathbb{Z}^+\) (the positive integers) is **countable**.
2. Any **subset** of a countable set is at most countable.
3. Therefore, \(\mathbb{Z}^+ \times \mathbb{Z}^+\) is **at most** countable.
But \(\mathbb{Z}^+ \times \mathbb{Z}^+\) is clearly **infinite** (since there are infinitely many pairs \((m,n)\)), so it is **countably infinite**.
Hence we conclude:
\[
\boxed{\mathbb{Z}^+ \times \mathbb{Z}^+\text{ is countable.}}
\]
