/*! 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 1 Use a direct proof to show that ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Use a direct proof to show that the sum of two odd integers is even.

Short Answer

Expert verified
The sum of two odd integers is even: \((2k + 1) + (2m + 1) = 2(k + m + 1)\).

Step by step solution

01

- Define Odd Integers

Define two odd integers. Let the first odd integer be represented as \(2k + 1\) and the second odd integer be represented as \(2m + 1\), where \(k\) and \(m\) are integers.
02

- Add the Two Odd Integers

Add the expressions for the two odd integers: \((2k + 1) + (2m + 1)\).
03

- Simplify the Expression

Simplify the resulting expression: \((2k + 1) + (2m + 1) = 2k + 2m + 2\).
04

- Factor Out Common Terms

Factor out the common factor of 2: \(2k + 2m + 2 = 2(k + m + 1)\).
05

- Conclude the Result

Recognize that since \(k + m + 1\) is an integer, \(2(k + m + 1)\) is an even number. Therefore, the sum of two odd integers is even.

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 91Ó°ÊÓ!

Key Concepts

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

Odd Integers
Odd integers are numbers that are not divisible by 2. In other words, when you divide an odd number by 2, you get a remainder of 1. Examples of odd integers include 1, 3, 5, 7, and 9. Mathematically, we can express any odd integer as \(2k + 1\), where \(k\) is an integer.

This expression means that if you start from an even number \(2k\) and add 1 to it, you will get an odd number. For instance, if \(k = 2\), then \(2k + 1 = 2 \cdot 2 + 1 = 5\), which is an odd number.

When dealing with problems involving odd integers, such as adding them together or multiplying them, recognizing this basic expression \(2k + 1\) is crucial. It simplifies our calculations and helps prove various properties of odd numbers.
Even Numbers
Even numbers are integers that are divisible by 2 without any remainder. Examples of even numbers include 2, 4, 6, 8, and 10.

Mathematically, any even number can be represented as \(2n\), where \(n\) is also an integer. This representation makes it easy to identify and work with even numbers. For instance, if \(n = 3\), then \(2n = 2 \cdot 3 = 6\), which is an even number.

One important property of even numbers is that when you add or subtract even numbers, the result is always even. For example, adding two even numbers, like \(6 + 4\), gives \(10\). Similarly, multiplying an even number by any integer will always result in an even number.
Integer Factorization
Integer factorization involves expressing an integer as a product of other integers, which are called its factors. For instance, the factors of 6 are 1, 2, 3, and 6 because \(6 = 2 \cdot 3\).

In the context of our exercise, factorization played a critical role in simplifying the sum of two odd integers. After adding the two expressions for odd integers \((2k + 1) + (2m + 1)\), the result was simplified to \(2(k + m + 1)\).

This factorization helped us conclude that the sum is even, as \(2(k + m + 1)\) is divisible by 2. Whenever we highlight common factors within algebraic expressions, factorization becomes a valuable tool. It simplifies our work and helps to understand the underlying properties of numbers and their operations.

It's good practice to familiarize yourself with recognizing and extracting common factors, as it proves helpful in a variety of mathematical problems.

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

Express the negation of these propositions using quantifiers, and then express the negation in English. a) Some drivers do not obey the speed limit. b) All Swedish movies are serious. c) No one can keep a secret. d) There is someone in this class who does not have a good attitude.

Use a proof by exhaustion to show that a tiling using dominoes of a \(4 \times 4\) checkerboard with opposite corners removed does not exist. [Hint: First show that you can assume that the squares in the upper left and lower right corners are removed. Number the squares of the original checkerboard from 1 to \(16,\) starting in the first row, moving right in this row, then starting in the leftmost square in the second row and moving right, and so on. Remove squares 1 and \(16 .\) To begin the proof, note that square 2 is covered either by a domino laid horizontally, which covers squares 2 and \(3,\) or vertically, which covers squares 2 and \(6 .\) Consider each of these cases separately, and work through all the subcases that arise. \(]\)

Prove that given a real number \(x\) there exist unique numbers \(n\) and \(\epsilon\) such that \(x=n+\epsilon, n\) is an integer, and \(0 \leq \epsilon<1 .\)

Determine the truth value of each of these statements if the domain for all variables consists of all integers. $$ \begin{array}{ll}{\text { a) } \forall n\left(n^{2} \geq 0\right)} & {\text { b) } \exists n\left(n^{2}=2\right)} \\ {\text { c) } \forall n\left(n^{2} \geq n\right)} & {\text { d) } \exists n\left(n^{2}<0\right)}\end{array} $$

Exercises \(48-51\) establish rules for null quantification that we can use when a quantified variable does not appear in part of a statement. Establish these logical equivalences, where \(x\) does not occur as a free variable in \(A\) . Assume that the domain is nonempty. $$ \begin{array}{l}{\text { a) } \forall x(A \rightarrow P(x)) \equiv A \rightarrow \forall x P(x)} \\ {\text { b) } \exists x(A \rightarrow P(x)) \equiv A \rightarrow \exists x P(x)}\end{array} $$

See all solutions

Recommended explanations on Math Textbooks

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.