Identifying Injective Functions

An injective function, also known as a one-to-one function, is a function where distinct elements in the domain map to distinct elements in the codomain. In this problem, the domain is the set of ordered pairs of natural numbers, \( \mathbb{N} \times \mathbb{N} = \{ (m, n) \mid m \in \mathbb{N}, n \in \mathbb{N} \} \), and the codomain is the set of natural numbers, \( \mathbb{N} = \{1, 2, 3, \dots\} \).

For a function \( f: \mathbb{N} \times \mathbb{N} \to \mathbb{N} \) to be injective, for any two distinct pairs \( (m_1, n_1) \) and \( (m_2, n_2) \) in \( \mathbb{N} \times \mathbb{N} \), their function values must be distinct. Equivalently, if \( f(m_1, n_1) = f(m_2, n_2) \), then it must imply that \( (m_1, n_1) = (m_2, n_2) \), which means \( m_1 = m_2 \) and \( n_1 = n_2 \).

Analyzing Each Function for Injectivity

Let's examine each given function:

Function \( f_1(m, n) = 2^m 3^n \)

Consider two pairs \( (m_1, n_1) \) and \( (m_2, n_2) \) from \( \mathbb{N} \times \mathbb{N} \) such that \( f_1(m_1, n_1) = f_1(m_2, n_2) \). This means \( 2^{m_1} 3^{n_1} = 2^{m_2} 3^{n_2} \). By the Fundamental Theorem of Arithmetic (the unique prime factorization theorem), every integer greater than 1 has a unique prime factorization. Since 2 and 3 are prime numbers, the equality \( 2^{m_1} 3^{n_1} = 2^{m_2} 3^{n_2} \) can only hold if the exponents of the corresponding prime factors are equal. Therefore, we must have \( m_1 = m_2 \) and \( n_1 = n_2 \). Since \( f_1(m_1, n_1) = f_1(m_2, n_2) \) implies \( m_1 = m_2 \) and \( n_1 = n_2 \), the pairs must be identical, i.e., \( (m_1, n_1) = (m_2, n_2) \). Thus, \( f_1 \) is an injective function.

Function \( f_2(m, n) = mn + m + n \)

Let's check if there exist two distinct pairs that map to the same value. Consider the pairs \( (1, 2) \) and \( (2, 1) \). Both are in \( \mathbb{N} \times \mathbb{N} \) and are distinct. Let's evaluate \( f_2 \) for these pairs: \( f_2(1, 2) = (1)(2) + 1 + 2 = 2 + 1 + 2 = 5 \) \( f_2(2, 1) = (2)(1) + 2 + 1 = 2 + 2 + 1 = 5 \) Since \( f_2(1, 2) = f_2(2, 1) \) but \( (1, 2) \neq (2, 1) \), the function \( f_2 \) is not injective.

Function \( f_3(m, n) = m^2 + n^3 \)

Let's check if there exist two distinct pairs that map to the same value. Consider the pairs \( (3, 2) \) and \( (4, 1) \). Both are in \( \mathbb{N} \times \mathbb{N} \) and are distinct. Let's evaluate \( f_3 \) for these pairs: \( f_3(3, 2) = 3^2 + 2^3 = 9 + 8 = 17 \) \( f_3(4, 1) = 4^2 + 1^3 = 16 + 1 = 17 \) Since \( f_3(3, 2) = f_3(4, 1) \) but \( (3, 2) \neq (4, 1) \), the function \( f_3 \) is not injective.

Function \( f_4(m, n) = m^2 n^3 \)

Let's check if there exist two distinct pairs that map to the same value. Consider the pairs \( (8, 1) \) and \( (1, 4) \). Both are in \( \mathbb{N} \times \mathbb{N} \) and are distinct. Let's evaluate \( f_4 \) for these pairs: \( f_4(8, 1) = 8^2 \cdot 1^3 = 64 \cdot 1 = 64 \) \( f_4(1, 4) = 1^2 \cdot 4^3 = 1 \cdot 64 = 64 \) Since \( f_4(8, 1) = f_4(1, 4) \) but \( (8, 1) \neq (1, 4) \), the function \( f_4 \) is not injective.

Conclusion

Based on the analysis, only function \( f_1(m, n) = 2^m 3^n \) is injective among the given options.