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Chapter 1: Problem 12

Show that if \(a \mid b\) and \(b \mid a\) then \(a=\pm b\).

Short Answer

Expert verified
Question: Demonstrate that if \(a \mid b\) and \(b \mid a\), then \(a=\pm b\). Answer: If \(a \mid b\) and \(b \mid a\), there exist integers \(k_1\) and \(k_2\) such that \(b = a \cdot k_1\) and \(a = b \cdot k_2\). Upon substituting and simplifying, we find that \(1 = k_1 \cdot k_2\). Since both \(k_1\) and \(k_2\) are integers, there are two possibilities: \(k_1 = 1\) & \(k_2 = 1\) or \(k_1 = -1\) & \(k_2 = -1\). In either case, we can show that \(a=\pm b\).

Step by step solution

01

Understand the divisibility property definition

Divisibility property is defined as follows: for two integers \(a\) and \(b\), \(a\) is said to divide \(b\) (written as \(a \mid b\)) if there exists an integer \(k\) such that \(b = a \cdot k\).
02

Use the given conditions

According to the given information, \(a \mid b\) and \(b \mid a\). Let's use the definition of divisibility property for both cases. 1. \(a \mid b\) means there exists an integer \(k_1\) such that \(b = a \cdot k_1\). 2. \(b \mid a\) means there exists an integer \(k_2\) such that \(a = b \cdot k_2\).
03

Substitute \(b\) from the first equation into the second equation

We have \(b = a \cdot k_1\) from the first equation. Substitute this into the second equation \(a = b \cdot k_2\), we get: \(a = (a \cdot k_1) \cdot k_2\)
04

Simplify the equation and derive the result

We have \(a = (a \cdot k_1) \cdot k_2\). Now let's simplify the equation and get the result: \(a = a \cdot k_1 \cdot k_2\) Divide both sides by \(a\): \(1 = k_1 \cdot k_2\) As both \(k_1\) and \(k_2\) are integers, there are only two possibilities: \(k_1 = 1\) & \(k_2 = 1\) or \(k_1 = -1\) & \(k_2 = -1\). Now, let's use these values to find the relationship between \(a\) and \(b\). 1. If \(k_1 = 1\) and \(k_2 = 1\), using the first equation \(b = a \cdot k_1\), we have \(b = a \cdot 1\) or \(b = a\). 2. If \(k_1 = -1\) and \(k_2 = -1\), using the first equation \(b = a \cdot k_1\), we have \(b = a \cdot(-1)\) or \(b = -a\).
05

Final result

We have demonstrated that if \(a \mid b\) and \(b \mid a\), then \(a=\pm b\).

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Key Concepts

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

Integer Divisibility
Integer divisibility is a fundamental concept in number theory. It helps to determine how numbers relate to each other by one number being able to "fit perfectly" into another. To say that an integer \(a\) divides another integer \(b\) (written as \(a \mid b\)), means there exists some integer \(k\) such that \(b = a \cdot k\).
This concept tells us that \(b\) is a multiple of \(a\), with no remainder left over.
For students, understanding divisibility is key to solving many problems in math, especially those dealing with factors and multiples.

Key Points about Divisibility

  • It is essential in simplifying complex mathematical problems.
  • Helps in understanding how integers relate through factors and multiples.
  • Important for identifying number properties like prime numbers.

In the exercise, we use the divisibility property twice: if \(a \mid b\) and \(b \mid a\), then according to the definition, \(b = a \cdot k_1\) and \(a = b \cdot k_2\). By exploring these forms, it guides us to conclude that \(a\) and \(b\) are closely related, specifically as either exactly equal or opposites.
Mathematical Proof
A mathematical proof is a logical argument demonstrating the truth of a mathematical statement. It is structured step-by-step to lead from known truths to a new truth. In this exercise, we used proof by deduction.
The essence of this proof is to apply logical reasoning and known properties, like divisibility, to uncover a new relationship.
First, we assume the conditions of the premise鈥攏amely, that \(a\) divides \(b\) and vice versa. We then express these conditions in mathematical terms and perform algebraic manipulations to reach a conclusion.

Why Proof Matters

  • Helps to establish certainty and validity in mathematical statements.
  • Encourages logical thinking and problem-solving skills.
  • Necessary for advancing theories and applications in mathematics.

In this problem, proof steps include substituting and simplifying to show that the only possible conditions are when \(k_1 = k_2 = 1\), leading to \(a = b\), or \(k_1 = k_2 = -1\), resulting in \(a = -b\). Understanding proof strengthens your ability to tackle similar questions in mathematics.
Elementary Algebra
Elementary algebra is the branch of mathematics that deals with variables and the rules for manipulating these variables. It includes operations like addition, subtraction, multiplication, and division of both numbers and algebraic expressions.
In the step-by-step solution of the exercise, elementary algebra is used to manipulate the given conditions of divisibility. By substituting one expression into another, we simplify and deduce the relationships between the variables. This is a fundamental skill in algebra.

Understanding Algebraic Manipulations

  • Substitution: Replacing variables with other expressions or numbers to simplify equations.
  • Simplification: Reducing equations to their simplest form.
  • Identity Principles: Using basic algebraic principles, like division, to isolate variables and solve equations.
In our specific problem, by substituting \(b = a \cdot k_1\) into \(a = b \cdot k_2\), we developed \(a = a \cdot k_1 \cdot k_2\). With further algebraic identity manipulation, we reached \(1 = k_1 \cdot k_2\), showing that \(k_1\) and \(k_2\) must both be either 1 or -1, resulting in \(a = \pm b\). Mastery of such algebraic techniques provides an essential foundation that supports more complex mathematical concepts.

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