/*! 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 32 Let \(n \in \mathbb{Z}\) with \(... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 32

Let \(n \in \mathbb{Z}\) with \(n>1\). Show that \(\sum_{i=1}^{n} 1 / i\) is not an integer.

Short Answer

Expert verified
Question: Prove that the sum of the reciprocals of integers from 1 to n is not an integer for any positive integer n > 1. Answer: We used mathematical induction to prove that the sum of the reciprocals of integers from 1 to n is not an integer. We proved the statement for the base case n=2. Then, we assumed that the statement holds for n=k and showed that it also holds for n=k+1.

Step by step solution

01

Base Case: Proving for n = 2

Given \(n=2\), we have the following sum of reciprocals of the integers from \(1\) to \(2\): $$\sum_{i=1}^{2} \frac{1}{i} = \frac{1}{1} + \frac{1}{2} = 1 + \frac{1}{2} = \frac{3}{2}$$ The sum \(\frac{3}{2}\) is not an integer since the denominator is greater than \(1\). Therefore, the statement holds true for the base case, \(n=2\).
02

Assumption: The Statement Holds for n = k

Now, let's assume the statement holds true for \(n=k\) where \(k \in \mathbb{Z}\) and \(k>1\). It means, for \(n=k\), the sum of the reciprocals of integers from \(1\) to \(k\) is not an integer: $$\sum_{i=1}^{k} \frac{1}{i} \neq \text{integer}$$
03

Proof: The Statement Holds for n = k+1

Let's consider the case \(n=k+1\). We will include the reciprocal of \((k+1)\) in the sum and try to show that this sum is still not an integer: $$\sum_{i=1}^{k+1} \frac{1}{i} = \frac{1}{1} + \frac{1}{2} + \cdots + \frac{1}{k} + \frac{1}{k+1}$$ Now, since we assumed that the sum of the first \(k\) terms is not an integer, adding a non-integer value to the sum, \(\frac{1}{k+1}\), will still leave the sum as a non-integer. Formally, we can say $$\sum_{i=1}^{k+1} \frac{1}{i} = \sum_{i=1}^{k} \frac{1}{i} + \frac{1}{k+1} \neq \text{integer}$$ By induction, we have proved that the sum of the reciprocals of integers from \(1\) to \(n\) is not an integer for all \(n \in \mathbb{Z}\) with \(n>1\).

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.

Proof by Induction
Proof by induction is a fundamental method of mathematical proof typically used to establish a given statement for all natural numbers. It consists of two critical steps: the base case and the inductive step. The base case verifies that the statement holds true for an initial value, which is usually the smallest value for which the statement is claimed to be true. Regarding the sum of reciprocals, we proved the base case for the statement when the integer is 2.

After the base case, the inductive step involves assuming the statement is true for an arbitrary natural number, k, which we referred to as the induction hypothesis. Then, we must show that if the statement holds for k, it must also hold for k+1. Applying this method compels us to a conclusion: if the statement holds for the base case and the inductive step is true, then the statement is true for all natural numbers greater than or equal to the base case. In our exercise, this means that the sum of the reciprocals of integers from 1 to any number, n, does not produce an integer result鈥攁 sophisticated but beautiful manifestation of number theory.
Non-Integer Sum
Understanding non-integer sums can be challenging, but they are crucial when dealing with continuous quantities in both pure and applied mathematics. A non-integer sum results from adding together numbers that do not perfectly fit into the integer set - whole numbers including negatives, zero, and positives. In the instance of the sum of reciprocals of integers, adding a fraction like \(\frac{1}{k+1}\) to a sum that is already non-integer guarantees that the result remains non-integer. This is because integers do not contain fractions, and adding or subtracting fractions from an integer, or another non-integer, will not magically produce an integer result.

The subtle beauty in formulating such sums often illuminates deeper properties of numbers, as we can observe in our proof by induction for the sum of reciprocals, where even without calculating the exact value, we establish the nature of the sum as non-integer.
Reciprocals of Integers
The reciprocal of an integer \(n\) is expressed as \(\frac{1}{n}\). When dealing with reciprocals of integers, particularly in series, intriguing patterns and properties can emerge. For instance, the harmonic series, which is the sum of reciprocals of all natural numbers, grows without bound and yet does so very slowly compared to other infinite series.

The step by step solution presented questions the integrality of the sum of these reciprocals for integers greater than 1. Notably, each term after the first increases the number of factors in the denominator, inherently ensuring the sum does not simplify to an integer. This ties directly to a deeper understanding of how fractions and division fit into the broader context of number theory, enriching our appreciation for seemingly simple operations like taking reciprocals.
Number Theory
Number theory is a branch of pure mathematics devoted primarily to the study of the integers and integer-valued functions. It is deeply rooted in mathematical proof, including techniques like proof by induction. The sum of reciprocals exercise is a classic example of a number theoretic problem that explores the nature of integers and their relationships, particularly with regards to division and the properties of fractions.

In number theory, understanding how sums of reciprocals behave is not just an isolated exercise. It relates to prime numbers, divisibility, and even to the depths of modern cryptography and computational mathematics. By examining simple concepts such as the sum of reciprocals of integers, we unleash a whole world of sophisticated relationships that yield insights spanning the basic fundamentals to the most advanced theorems in mathematics.

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

Show that \(\operatorname{gcd}(a+b, \operatorname{lcm}(a, b))=\operatorname{gcd}(a, b)\) for all \(a, b \in \mathbb{Z}\).

Let \(n\) be an integer. Show that if \(a, b\) are relatively prime integers, each of which divides \(n,\) then \(a b\) divides \(n\).

Let \(x \in \mathbb{R}\) and \(n \in \mathbb{Z}\) with \(n>0 .\) Show that \(\lfloor\lfloor x\rfloor / n\rfloor=\lfloor x / n\rfloor ;\) in particular, \(\lfloor\lfloor a / b\rfloor / c\rfloor=\lfloor a / b c\rfloor\) for all positive integers \(a, b, c .\)

Let \(a_{1}, \ldots, a_{k} \in \mathbb{Z}\) with \(k>1\). Show that: $$ \operatorname{gcd}\left(a_{1}, \ldots, a_{k}\right)=\operatorname{gcd}\left(a_{1}, \operatorname{gcd}\left(a_{2}, \ldots, a_{k}\right)\right)=\operatorname{gcd}\left(\operatorname{gcd}\left(a_{1}, \ldots, a_{k-1}\right), a_{k}\right) $$ \(\operatorname{lcm}\left(a_{1}, \ldots, a_{k}\right)=\operatorname{lcm}\left(a_{1}, \operatorname{lcm}\left(a_{2}, \ldots, a_{k}\right)\right)=\operatorname{lcm}\left(\operatorname{lcm}\left(a_{1}, \ldots, a_{k-1}\right), a_{k}\right)\)

Show that for all integers \(a, b, c,\) we have: (a) \(\operatorname{gcd}(a, b)=\operatorname{gcd}(b, a) ;\) (b) \(\operatorname{gcd}(a, b)=|a| \Longleftrightarrow a \mid b ;\) (c) \(\operatorname{gcd}(a, 0)=\operatorname{gcd}(a, a)=|a|\) and \(\operatorname{gcd}(a, 1)=1 ;\) (d) \(\operatorname{gcd}(c a, c b)=|c| \operatorname{gcd}(a, b)\).
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Discrete Mathematics

Read Explanation

Mechanics Maths

Read Explanation

Pure Maths

Read Explanation

Applied Mathematics

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.