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1. Give an example of a function f N N which is injective but not surjective.

1. Give an example of a function f N N which is injective but not surjective.

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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**Exercise 1: Functions** **Task:** Give an example of a function \( f : \mathbb{N} \to \mathbb{N} \) which is injective but not surjective. ### Explanation - **Injective Function (One-to-One):** A function \( f \) is injective if different elements in the domain map to different elements in the codomain. Formally, \( f(a) = f(b) \) implies \( a = b \). - **Surjective Function (Onto):** A function \( f \) is surjective if every element in the codomain has a pre-image in the domain. There are no elements in the codomain left unmapped. **Example Solution:** Consider the function \( f(n) = n + 1 \). - **Injectivity:** If \( f(a) = f(b) \), then \( a + 1 = b + 1 \), which implies \( a = b \). Therefore, \( f \) is injective. - **Not Surjective:** The element 0 in the codomain \(\mathbb{N}\) (if we include 0 in natural numbers) does not have a pre-image, as no natural number \( n \) satisfies \( n + 1 = 0 \). Thus, \( f \) is not surjective.
Transcribed Image Text:**Exercise 1: Functions** **Task:** Give an example of a function \( f : \mathbb{N} \to \mathbb{N} \) which is injective but not surjective. ### Explanation - **Injective Function (One-to-One):** A function \( f \) is injective if different elements in the domain map to different elements in the codomain. Formally, \( f(a) = f(b) \) implies \( a = b \). - **Surjective Function (Onto):** A function \( f \) is surjective if every element in the codomain has a pre-image in the domain. There are no elements in the codomain left unmapped. **Example Solution:** Consider the function \( f(n) = n + 1 \). - **Injectivity:** If \( f(a) = f(b) \), then \( a + 1 = b + 1 \), which implies \( a = b \). Therefore, \( f \) is injective. - **Not Surjective:** The element 0 in the codomain \(\mathbb{N}\) (if we include 0 in natural numbers) does not have a pre-image, as no natural number \( n \) satisfies \( n + 1 = 0 \). Thus, \( f \) is not surjective.
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