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Prove that the function f from Nx N to N defined by f(m, n) = 2m3 is one-to-one but not onto.

Prove that the function f from Nx N to N defined by f(m, n) = 2m3 is one-to-one but not onto.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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**Problem Statement**: Prove that the function \( f \) from \(\mathbb{N} \times \mathbb{N}\) to \( \mathbb{N} \) defined by \( f(m, n) = 2^m 3^n \) is one-to-one but not onto. --- **Explanation**: 1. **Function Definition**: The function \( f \) takes two natural numbers \( m \) and \( n \), and maps them to another natural number by calculating \( 2^m 3^n \). 2. **One-to-One (Injective)**: - A function is one-to-one if different inputs map to different outputs. - To show \( f \) is one-to-one, assume \( f(m_1, n_1) = f(m_2, n_2) \), which means \( 2^{m_1} 3^{n_1} = 2^{m_2} 3^{n_2} \). - This implies that \( 2^{m_1 - m_2} = 3^{n_2 - n_1} \). - For this equality to hold, both exponents need to be zero, i.e., \( m_1 = m_2 \) and \( n_1 = n_2 \). - Therefore, the function is injective. 3. **Not Onto (Surjective)**: - A function is onto if every element in the codomain has a pre-image in the domain. - Consider a number like 5, which cannot be expressed as \( 2^m 3^n \); thus, 5 is not in the image of \( f \). - Consequently, not all natural numbers can be obtained by \( f(m, n) \), proving it is not onto. This problem demonstrates how certain mapping functions work and provides insight into the concepts of injective and surjective functions in mathematics.
Transcribed Image Text:**Problem Statement**: Prove that the function \( f \) from \(\mathbb{N} \times \mathbb{N}\) to \( \mathbb{N} \) defined by \( f(m, n) = 2^m 3^n \) is one-to-one but not onto. --- **Explanation**: 1. **Function Definition**: The function \( f \) takes two natural numbers \( m \) and \( n \), and maps them to another natural number by calculating \( 2^m 3^n \). 2. **One-to-One (Injective)**: - A function is one-to-one if different inputs map to different outputs. - To show \( f \) is one-to-one, assume \( f(m_1, n_1) = f(m_2, n_2) \), which means \( 2^{m_1} 3^{n_1} = 2^{m_2} 3^{n_2} \). - This implies that \( 2^{m_1 - m_2} = 3^{n_2 - n_1} \). - For this equality to hold, both exponents need to be zero, i.e., \( m_1 = m_2 \) and \( n_1 = n_2 \). - Therefore, the function is injective. 3. **Not Onto (Surjective)**: - A function is onto if every element in the codomain has a pre-image in the domain. - Consider a number like 5, which cannot be expressed as \( 2^m 3^n \); thus, 5 is not in the image of \( f \). - Consequently, not all natural numbers can be obtained by \( f(m, n) \), proving it is not onto. This problem demonstrates how certain mapping functions work and provides insight into the concepts of injective and surjective functions in mathematics.
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