/*! 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 If \(a \mid b\), show that \((-a... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 1

If \(a \mid b\), show that \((-a)|b, a|(-b)\), and \((-a) \mid(-b)\).

Short Answer

Expert verified
All combinations of signs result in divisors because integer multiplication allows for sign changes without solving divisibility.

Step by step solution

01

Understand what it means for a to divide b

When we say that \(a\) divides \(b\) (written as \(a \mid b\)), it means there is an integer \(k\) such that \(b = a \cdot k\). This will be the starting point to show how different signs of \(a\) and \(b\) can still satisfy similar divisibility properties.
02

Prove that (-a) divides b

Given \(a \mid b\), we have \(b = a \cdot k\) for some integer \(k\). To show that \(-a \mid b\), we must prove \(b = (-a) \cdot (-k)\). Since \((-a) \cdot (-k) = a \cdot k = b\), \(-a\) divides \(b\).
03

Prove that a divides (-b)

Given \(b = a \cdot k\), to show that \(a \mid -b\), we need to prove \(-b = a \cdot (-k)\). Since \(-b = - (a \cdot k) = a \cdot (-k)\), \(a\) divides \(-b\).
04

Prove that (-a) divides (-b)

From \(b = a \cdot k\), we need to show \(-b = (-a) \cdot k\). We have \(-b = - (a \cdot k) = (-a) \cdot k\). Thus, \(-a\) divides \(-b\).

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.

Integers
Integers are a family of numbers that include all whole numbers, both positive and negative, as well as zero. Unlike fractions or decimals, integers are whole and do not have parts. Understanding integers is essential to solving many math problems, especially those involving divisibility.

Here's how integers are categorized:
  • Positive Integers: These are numbers greater than zero, like 1, 2, and 3.
  • Negative Integers: These are numbers less than zero, like -1, -2, and -3.
  • Zero: This is the neutral integer that is neither positive nor negative.
When working with divisibility, it is crucial to recognize how integers, particularly negatives, behave under multiplication and division. This quality influences equations and proofs, as seen in exercises delving into dividing negative and positive integers.
Properties of divisibility
In mathematics, divisibility refers to one integer being able to be divided by another without leaving a remainder. Understanding these properties is critical for solving many types of problems, especially in number theory.

The properties of divisibility involve recognizing the relationship between a divisor and a dividend:
  • If an integer \(a\) divides another integer \(b\) (written as \(a \mid b\)), then there exists an integer \(k\) such that \(b = a \cdot k\).
  • This definition is consistent regardless of whether \(a\) and \(b\) are positive or negative.
The beauty of divisibility is that it remains robust under multiplication by negative numbers, which means if \(a \mid b\), then \(-a\) also divides \(b\), maintaining the integrity of integer multiplication relations.
Negative numbers
Negative numbers can be a bit tricky, especially when involving operations such as division or multiplication with other integers. They follow distinct rules that need to be carefully applied to solve mathematical problems.

When working with negative numbers in divisibility, there are a few crucial elements:
  • Multiplying two negative numbers results in a positive number, aligning with how negatives cancel each other out. This fact ensures that \((-a) \cdot (-k) = a \cdot k\) is still true.
  • Multiplying a negative number by a positive number results in a negative number, reinforcing that \((-a) \mid b\) translates to solving \(b = (-a) \cdot (-k)\).
  • The properties extend so that if \(a \mid b\), \(a\) divides \(-b\) and \(-a\) divides \(-b\), maintaining symmetry.
The rules are straightforward but must be consistently applied to ensure accurate results in calculations, especially when negatives are involved in proofs and equations.

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

Obtain the following version of the Division Algorthm: For integers \(a\) and \(b\), with \(b \neq 0\), there exist unique integers \(q\) and \(r\) that satisfy \(a=q b+r\), where \(-\frac{1}{2}|b|0\) or \(q=q^{\prime}-1\) if \(b<0 .]\)

Given integers \(a\) and \(b\), prove the following: (a) There exist integers \(x\) and \(y\) for which \(c=a x+b y\) if and only if \(\operatorname{gcd}(a, b) \mid c .\) (b) If there exist integers \(x\) and \(y\) for which \(a x+b y=\operatorname{gcd}(a, b)\), then \(\operatorname{gcd}(x, y)=1\).

For \(n \geq 1\), use mathematical induction to establish each of the following divisibility statements: (a) \(8 \mid 5^{2 n}+7\). \(\left[\right.\) Hint: \(5^{2(k+1)}+7=5^{2}\left(5^{2 k}+7\right)+\left(7-5^{2} \cdot 7\right)\).] (b) \(15 \mid 2^{4 n}-1\). (c) \(5 \mid 3^{3 n+1}+2^{n+1}\). (d) \(21 \mid 4^{n+1}+5^{2 n-1}\). (e) \(24 \mid 2 \cdot 7^{n}+3 \cdot 5^{n}-5\)

For any integer \(a\), show the following: (a) \(\operatorname{gcd}(2 a+1,9 a+4)=1\). (b) \(\operatorname{gcd}(5 a+2,7 a+3)=1\). (c) If \(a\) is odd, then \(\operatorname{gcd}(3 a, 3 a+2)=1\).

If \(a\) and \(b\) are integers, not both of which are zero, verify that $$ \operatorname{gcd}(a, b)=\operatorname{gcd}(-a, b)=\operatorname{gcd}(a,-b)=\operatorname{gcd}(-a,-b) $$
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Mechanics Maths

Read Explanation

Pure Maths

Read Explanation

Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Decision Maths

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.