/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 6 Let \(d: \mathbb{N} \rightarrow ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 6

Let \(d: \mathbb{N} \rightarrow \mathbb{N},\) where \(d(n)\) is the number of natural number divisors of \(n\). This is the number of divisors function introduced in Exercise (6) from Section \(6.1 .\) Is the function \(d\) an injection? Is the function \(d\) a surjection? Justify your conclusions.

Short Answer

Expert verified
The function \(d: \mathbb{N} \rightarrow \mathbb{N}\), which gives the number of natural number divisors of \(n\), is not an injection, as we found a counterexample with \(d(2) = d(3)\) but \(2 \neq 3\). However, it is a surjection, as we showed that for any natural number \(k\), we can find an \(n \in \mathbb{N}\) such that \(d(n) = k\).

Step by step solution

01

Check for Injection

: To check if the function \(d\) is an injection, we must prove either that for every \(x, y \in \mathbb{N}\), \(d(x) = d(y)\) implies \(x = y\), or we need to find a counterexample showing that there exist \(x, y \in \mathbb{N}\) such that \(d(x) = d(y)\) but \(x \neq y\). To find a counterexample, we can check the values of function \(d\) for some small numbers, like \(d(2) =2\) since \(2\) has two divisors: \(1\) and \(2\). And when we look at \(d(3)\), it also equals \(2\), because \(3\) has two divisors: \(1\) and \(3\). So, we get \(d(2) = d(3)\), but \(2 \neq 3\). Hence, the function \(d\) is not an injection.
02

Check for Surjection

: A function is surjective if every element in the codomain (here, \(\mathbb{N}\)) has at least one corresponding element in the domain (also \(\mathbb{N}\)). In other words, we need to show that for each natural number \(k\), we can find an \(n \in \mathbb{N}\) such that \(d(n) = k\). To prove this, let's consider the prime factorization of any natural number \(n\). Prime factorization is unique for each \(n\), and we can define it as: \(n = p_1^{e_1} \cdot p_2^{e_2} \cdot ... \cdot p_k^{e_k}\), where \(p_i\) are prime numbers and \(e_i\) are the corresponding exponents. Now, the total number of divisors for \(n\) can be represented as: \(d(n) = (e_1 + 1)(e_2 + 1)...(e_k + 1)\), as we can choose any combination of the exponents of the prime factors. We know that the multiplication of two or more natural numbers gives another natural number. Therefore, no matter which exponents we choose for the prime factors, the result of the multiplication, i.e., the total number of divisors, will be a natural number. So, for any natural number \(k\), we can find an \(n \in \mathbb{N}\) such that \(d(n) = k\). This proves that the function \(d\) is a surjection. In conclusion, the function \(d: \mathbb{N} \rightarrow \mathbb{N}\) is not an injection, but it is a surjection.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Injection Function
An injection function, also known as a one-to-one function, maps elements from the domain to the codomain in such a way that no two distinct elements in the domain have the same image in the codomain. In simpler terms, every item in the domain is unique and gets its unique output in the codomain.

For the number of divisors function, we consider if this uniqueness holds. By examining small natural numbers, we find that both 2 and 3 as inputs give the same output, 2 divisors each. This reveals that the function is not one-to-one because it fails the basic requirement of an injection: distinct inputs should not lead to a common output. Hence, multiple natural numbers can have the same number of divisors, indicating that the number of divisors function is not an injection.
Surjection Function
A surjection function, also known as an onto function, has a domain where every element in the codomain is the image of at least one element from the domain. Put simply, there are no leftovers in the codomain; everything gets covered by the function from the domain.

In the case of the number of divisors function, for every natural number, we can find at least one natural number whose divisors count up to that number. This is possible due to the unique prime factorization of natural numbers, which allows arranging the primes and their powers in a way to produce any desired count of divisors. Calculating the divisors involves multiplying the incremented exponents of the prime factors of a number. Since the product of these increments is also a natural number, every natural number is attainable as the number of divisors of some natural number. Therefore, the number of divisors function is a surjection.
Prime Factorization
Prime factorization is a key concept in understanding natural number divisors. It states that every natural number greater than 1 can be written as the product of prime numbers in a unique way鈥攖his is the Fundamental Theorem of Arithmetic. For instance, the number 18 can be expressed as the product of primes as 18 = 2 x 3 x 3 or 18 = 2 x 3^2.

Understanding prime factorization is crucial because it provides insight into the structure of numbers and is essential for calculating the number of divisors. The exponents in the prime factorization are especially important, as they help establish the divisor count formula \(d(n) = (e_1 + 1)(e_2 + 1)...(e_k + 1)\), where \(e_i\) are the powers of prime factors \(p_i\). Prime factorization not only unravels the composition of a number but also lays the groundwork for advanced mathematical concepts such as least common multiples (LCM), greatest common divisors (GCD), and the classification of numbers based on their divisors.
Natural Number Divisors
Counting the natural number divisors of a given number is a fascinating exploration of number theory. By default, every number n has at least two divisors, 1 and itself. As we look into larger numbers, their divisors can include a mixture of prime and composite numbers, depending on their structure.

The factors of a number n can be counted using its prime factorization, where the formula for the number of divisors encompasses choosing different combinations of exponents for the prime factors. For a number \(n = p_1^{e_1} \times p_2^{e_2} \times ... \times p_k^{e_k}\), the formula \(d(n) = (e_1 + 1)(e_2 + 1)...(e_k + 1)\) is utilized to find the total possible combinations, thus representing the count of divisors of n. The beauty lies in how these combinations encompass all potential divisors, showing a deep interconnectedness between prime factorization and divisor counts, revealing the intricate tapestry of number theory.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

For each of the following functions, determine if the function is an injection and determine if the function is a surjection. Justify all conclusions. (a) \(f: \mathbb{Z} \rightarrow \mathbb{Z}\) defined by \(f(x)=3 x+1,\) for all \(x \in \mathbb{Z}\). (b) \(F: \mathbb{Q} \rightarrow \mathbb{Q}\) defined by \(F(x)=3 x+1,\) for all \(x \in \mathbb{Q}\) (c) \(g: \mathbb{R} \rightarrow \mathbb{R}\) defined by \(g(x)=x^{3},\) for all \(x \in \mathbb{R}\) (d) \(G: \mathbb{Q} \rightarrow \mathbb{Q}\) defined by \(G(x)=x^{3},\) for all \(x \in \mathbb{Q}\). (e) \(k: \mathbb{R} \rightarrow \mathbb{R}\) defined by \(k(x)=e^{-x^{2}},\) for all \(x \in \mathbb{R}\) (f) \(K: \mathbb{R}^{*} \rightarrow \mathbb{R}\) defined by \(K(x)=e^{-x^{2}},\) for all \(x \in \mathbb{R}^{*}\). Note: \(\mathbb{R}^{*}=\\{x \in \mathbb{R} \mid x \geq 0\\}\). (g) \(K_{1}: \mathbb{R}^{*} \rightarrow T\) defined by \(K_{1}(x)=e^{-x^{2}},\) for all \(x \in \mathbb{R}^{*},\) where \(T=\) \(\\{y \in \mathbb{R} \mid 0

Let \(C\) be the set of all real functions that are continuous on the closed interval \([0,1] .\) Define the function \(A: C \rightarrow \mathbb{R}\) as follows: For each \(f \in C,\) $$A(f)=\int_{0}^{1} f(x) d x$$ Is the function \(A\) an injection? Is it a surjection? Justify your conclusions.

The number of divisors function. Let \(d\) be the function that associates with each natural number the number of its natural number divisors. That is, \(d: \mathbb{N} \rightarrow \mathbb{N}\) where \(d(n)\) is the number of natural number divisors of \(n\). For example, \(d(6)=4\) since \(1,2,3,\) and 6 are the natural number divisors of 6 (a) Calculate \(d(k)\) for each natural number \(k\) from 1 through 12 . (b) Does there exist a natural number \(n\) such that \(d(n)=1 ?\) What is the set of preimages of the natural number \(1 ?\) (c) Does there exist a natural number \(n\) such that \(d(n)=2 ?\) If so, determine the set of all preimages of the natural number \(2 .\) (d) Is the following statement true or false? Justify your conclusion. For all \(m, n \in \mathbb{N},\) if \(m \neq n,\) then \(d(m) \neq d(n) .\) (e) Calculate \(d\left(2^{k}\right)\) for \(k=0\) and for each natural number \(k\) from 1 through 6 (f) Based on your work in Exercise (6e), make a conjecture for a formula for \(d\left(2^{n}\right)\) where \(n\) is a nonnegative integer. Then explain why your conjecture is correct. (g) Is the following statement is true or false? For each \(n \in \mathbb{N},\) there exists a natural number \(m\) such that \(d(m)=n\)

Creating Functions with Finite Domains. Let \(A=\\{a, b, c, d\\}, B=\) \(\\{a, b, c\\},\) and \(C=\\{s, t, u, v\\} .\) In each of the following exercises, draw an arrow diagram to represent your function when it is appropriate. (a) Create a function \(f: A \rightarrow C\) whose range is the set \(C\) or explain why it is not possible to construct such a function. (b) Create a function \(f: A \rightarrow C\) whose range is the set \(\\{u, v\\}\) or explain why it is not possible to construct such a function. (c) Create a function \(f: B \rightarrow C\) whose range is the set \(C\) or explain why it is not possible to construct such a function. (d) Create a function \(f: A \rightarrow C\) whose range is the set \(\\{u\\}\) or explain why it is not possible to construct such a function. (e) If possible, create a function \(f: A \rightarrow C\) that satisfies the following condition: For all \(x, y \in A,\) if \(x \neq y,\) then \(f(x) \neq f(y)\) If it is not possible to create such a function, explain why. (f) If possible, create a function \(f: A \rightarrow\\{s, l, u\\}\) that satisfies the following condition: For all \(x, y \in A,\) if \(x \neq y,\) then \(f(x) \neq f(y) .\) If it is not possible to create such a function, explain why.

Let \(f: A \rightarrow B\) and \(g: B \rightarrow A\). Let \(I_{A}\) and \(I_{B}\) be the identity functions on the sets \(A\) and \(B\), respectively. Prove each of the following: (a) If \(g \circ f=I_{A},\) then \(f\) is an injection. (b) If \(f \circ g=I_{B},\) then \(f\) is a surjection. (c) If \(g \circ f=I_{A}\) and \(f \circ g=I_{B},\) then \(f\) and \(g\) are bijections and \(g=f^{-1}\)
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Pure Maths

Read Explanation

Logic and Functions

Read Explanation

Discrete Mathematics

Read Explanation

Statistics

Read Explanation

Probability and Statistics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.