/*! 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 5 There are distinct integers \(m\... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 5

There are distinct integers \(m\) and \(n\) such that \(\frac{1}{m}+\frac{1}{n}\) is an integer.

Short Answer

Expert verified
There are two possible cases: 1. If \(m = 1\) and \(n = 2\), then \(\frac{1}{m} + \frac{1}{n} = 1 + \frac{1}{2} = \frac{3}{2}\), which is not an integer. 2. If \(n = 1\) and \(m = 2\), then \(\frac{1}{m} + \frac{1}{n} = \frac{1}{2} + 1 = \frac{3}{2}\), which is not an integer. Thus, there are no distinct integers \(m\) and \(n\) such that \(\frac{1}{m} + \frac{1}{n}\) is an integer.

Step by step solution

01

Find a common denominator

The common denominator for the given fractions would be \(mn\). So, let's rewrite both fractions with the common denominator. Step 2: Rewrite the fractions with a common denominator
02

Rewrite the fractions with a common denominator

Rewrite the fractions \(\frac{1}{m}\) and \(\frac{1}{n}\) with their common denominator \(mn\), so the equation looks like this: \[\frac{n}{mn} + \frac{m}{mn}\]. Step 3: Combine the fractions
03

Combine the fractions

Let's combine the fractions under the same denominator: \[\frac{n + m}{mn}\]. Step 4: Set the equation equal to an integer
04

Set the equation equal to an integer

Now, we need \(\frac{n + m}{mn}\) to be an integer. Let's set the equation equal to an integer and call it \(k\): \[\frac{n + m}{mn} = k\]. Step 5: Rearrange equation to isolate k
05

Rearrange equation to isolate k

As we need the sum \((n + m)\) to be divisible by the product of \(mn\), let's rearrange the equation to isolate \(k\): \[n+m = k(mn)\]. Step 6: Analyze the equation
06

Analyze the equation

Since we are dealing with integers and \(m\) and \(n\) are distinct, observe that either one of \(m\) or \(n\) must be equal to 1 (or else \(n + m\) could never equal to the product \((mn)\)). Let's analyze the two possibilities. Step 7: Set either m or n equal to 1
07

Set either m or n equal to 1

Possibility 1: If \(m = 1\), then \(n + 1 = k(n)\). Solving for \(n\), we get \(n = \frac{k}{k - 1}\). Since \(n\) is an integer, \(k - 1\) must divide \(k\), which means that \(k = 2\). So, \(n = 2\). Possibility 2: If \(n = 1\), then \(m + 1 = k(m)\). Solving for \(m\), we get \(m = \frac{k}{k - 1}\). Since \(m\) is an integer, \(k - 1\) must divide \(k\), which means that \(k = 2\). So, \(m = 2\). These possibilities show that there are distinct integers \(m\) and \(n\) such that \(\frac{1}{m} + \frac{1}{n}\) is an integer. For example, if \(m = 1\) and \(n = 2\), then \(\frac{1}{m} + \frac{1}{n} = 1 + \frac{1}{2} = \frac{3}{2}\).

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.

Integer Solutions
When we speak about integer solutions, we mean finding values of variables that are whole numbers. In this context, our task is to find distinct integer values for the variables \( m \) and \( n \) such that \( \frac{1}{m} + \frac{1}{n} \) results in an integer. This involves some clever math analysis because we must satisfy the condition for it to turn out as an integer value.
In the problem at hand, we found that either \( m \) or \( n \) must equal 1. By setting one of these to 1, the equation simplifies significantly, making it easier to find other integer solutions. In our solution, when \( m = 1 \), \( n = 2 \) was one possible integer solution, and if \( n = 1 \), then \( m = 2 \) is another.
Fractions with Common Denominator
To solve the given problem, we first need to convert the fractions \( \frac{1}{m} \) and \( \frac{1}{n} \) to have a common denominator. The common denominator for these fractions is the product \( mn \).
By rewriting the fractions as \( \frac{n}{mn} + \frac{m}{mn} \), we can easily combine them into a single fraction: \( \frac{n+m}{mn} \). Using a common denominator allows us to work with fractions in a simplified way, facilitating their combination and making them easier to analyze further.
Equation Analysis
Equation analysis involves dissecting the equation to understand its components and behavior. In our exercise, the key was to make the resulting fraction \( \frac{n+m}{mn} \) equal to an integer \( k \).
This requires that the numerator, \( n+m \), must be divisible by the denominator, \( mn \). By equating \( \frac{n+m}{mn} = k \), and rearranging, we came to \( n+m = k(mn) \). This rearrangement is crucial as it sets the groundwork to find integer solutions and analyze conditions under which the equation holds true.
Distinct Integers
Distinct integers are different whole numbers. The term is used here to focus on finding different values for \( m \) and \( n \) such that they are not the same, thus excluding solutions like \( m = n \).
This adds an additional constraint to our problem. It forces us to carefully choose values so that \( m \) and \( n \) remain different yet ensure \( \frac{1}{m} + \frac{1}{n} \) still results in an integer. Our solution confirms that \( m = 1 \) and \( n = 2 \) or vice versa satisfies this condition, thereby achieving distinct integers.

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

When expressions of the form \((x-r)(x-s)\) are multiplied out, a quadratic polynomial is obtained. For instance, \((x-2)(x-(-7))=(x-2)(x+7)=x^{2}+5 x-14 .\) \(H\) a. What can be said about the coefficients of the polynomial obtained by multiplying out \((x-r)(x-s)\) when both \(r\) and \(s\) are odd integers? when both \(r\) and \(s\) are even integers? when one of \(r\) and \(s\) is even and the other is odd? b. It follows from part (a) that \(x^{3}-1253 x+255\) cannot be written as a product of two polynomials with integer coefficients. Explain why this is so.

For all real numbers \(x\), if \(0

Theorem: The sum of any two even integers equals \(4 k\) for some integer \(k\). "Proof: Suppose \(m\) and \(n\) are any two even integers. By definition of even, \(m=2 k\) for some integer \(k\) and \(n=2 k\) for some integer \(k\). By substitution, \(m+n=2 k+2 k=4 k\). This is what was to be shown."

The sum of any two odd integers is even.

Compute \(\lfloor x\rfloor\) and \(\lceil x\rceil\) for each of the values of \(x\) in \(1-4\). $$ 17 / 4 $$
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Applied Mathematics

Read Explanation

Discrete Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Decision Maths

Read Explanation

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.