/*! 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 4 Determine whether the indicated ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 4

Determine whether the indicated pairs of elements are associates in the indicated domains. \(\begin{array}{lll}3+2 \sqrt{2} & \text { and } 1-\sqrt{2} & \text { in } \mathbb{Z}[\sqrt{2}]\end{array}\)

Short Answer

Expert verified
Yes, these numbers are associates because their ratio is a unit.

Step by step solution

01

Understand the Context

The goal is to determine if two numbers, \(3+2\sqrt{2}\) and \(1-\sqrt{2}\), are associates in the ring \(\mathbb{Z}[\sqrt{2}]\). Two elements are associates in this domain if one is a multiple of the other by a unit in \(\mathbb{Z}[\sqrt{2}]\). A unit is a number whose inverse also lies in \(\mathbb{Z}[\sqrt{2}]\).
02

Define the Relationship

If \(3 + 2\sqrt{2}\) and \(1 - \sqrt{2}\) are associates, then there exists a unit \(u\) in \(\mathbb{Z}[\sqrt{2}]\) such that \((3 + 2\sqrt{2}) = u(1 - \sqrt{2})\). Therefore, \(u = \frac{3 + 2\sqrt{2}}{1 - \sqrt{2}}\) must be a unit.
03

Calculate the Unit

First, find \(u = \frac{3 + 2\sqrt{2}}{1 - \sqrt{2}}\). To simplify, multiply the numerator and the denominator by the conjugate of \((1 - \sqrt{2})\), which is \((1 + \sqrt{2})\), to rationalize the denominator.
04

Rationalize the Denominator

Compute the new expression: \[\frac{(3 + 2\sqrt{2})(1 + \sqrt{2})}{(1 - \sqrt{2})(1 + \sqrt{2})} = \frac{(3 + 2\sqrt{2})(1 + \sqrt{2})}{1 - 2} = \frac{3 + 3\sqrt{2} + 2\sqrt{2} + 4}{-1}\]The denominator simplifies to \(-1\), so the expression becomes:\[-(3 + 3\sqrt{2} + 2\sqrt{2} + 4) = -(7 + 5\sqrt{2})\].
05

Determine If the Result Is a Unit

The expression simplifies to \(-(7 + 5\sqrt{2})\). For this to be a unit, its inverse must also be in \(\mathbb{Z}[\sqrt{2}]\). Calculating the norm can help determine this. The norm \(N(a + b\sqrt{2}) = a^2 - 2b^2\). For \(-7 - 5\sqrt{2}\), the norm is \((-7)^2 - 2(-5)^2 = 49 - 50 = -1\). The norm being 1 or -1 indicates a unit.

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.

Units in Algebraic Domains
In algebraic domains, a unit is an incredibly important concept. Imagine it as a magical number that can reverse itself!
Units have an inverse that, when multiplied with the original number, results in 1. This means, for a number to be a unit in an algebraic domain, its inverse must also belong to the same domain.
For instance, when dealing with the domain \(\mathbb{Z}[\sqrt{2}]\), a number like \(a + b\sqrt{2}\) is a unit if its inverse is also of the form \(c + d\sqrt{2}\), where \(c\) and \(d\) are integers.
  • The first step is usually to find the norm, \(N(a + b\sqrt{2}) = a^2 - 2b^2\).
  • If this norm equals \(1\) or \(-1\), then you likely have a unit!
Understanding units is critical for determining relationships like being associates.
Ring Theory
Ring theory is like the playground for algebraists. It involves studying sets equipped with two operations: addition and multiplication.
Rings include familiar number sets like integers, but they expand to more complex sets too.
When you encounter a problem asking whether numbers are associates in a specific domain, this is often rooted in ring theory.
  • A ring must satisfy certain criteria: closure under addition and subtraction, an additive identity (like zero), and multiplicative associativity.
  • Some rings also have a multiplicative identity, like 1 for integers, which adds an interesting twist!
In our specific case, being in the ring \(\mathbb{Z}[\sqrt{2}]\) means dealing with numbers of the form \(a + b\sqrt{2}\). Making sure these fit within the rules of the ring helps ensure they can properly be manipulated as associate elements.
Rationalizing Denominators
Rationalizing denominators might sound complicated, but it's a simple process to simplify expressions.
When you have a fraction like \(\frac{3 + 2\sqrt{2}}{1 - \sqrt{2}}\), you want the denominator to be free from surds or square roots.
The trick is to multiply both the numerator and denominator by the conjugate of the denominator. For \(1 - \sqrt{2}\), the conjugate is \(1 + \sqrt{2}\).
  • Multiply the numerator and denominator by \(1 + \sqrt{2}\).
  • This is like using magic to make the square roots vanish!
  • The denominator then transforms into a nice, simple integer, making the expression easier to handle.
This step is crucial in determining whether a fraction equals a unit in certain algebraic domains. Simplifying it aids in comparing the elements in the ring effectively.

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

Let \(D\) be a Euclidean domain, \(a, b,\) and \(c\) nonzero elements of \(D,\) and \(d\) a greatest common divisor of \(a\) and \(b\). Show that if \(a \mid b c\), then \((a / d) \mid c\).

Let \(I\) be a nontrivial ideal in \(\mathbb{Z}[i]\). Show that \(\mathbb{Z}[i] / I\) is a finite ring.

In Exercises 5 through 8 factor the indicated Gaussian integers into a product of irreducibles in \(\mathbb{Z}[i]\) $$ -1+5 i $$

In \(\mathbb{Z}[\sqrt{2}]\) show that (a) \(1+\sqrt{2}\) is a unit. (b) \(\pm(1+\sqrt{2})^{n}\) for \(n \in \mathbb{Z}\) is a unit. (c) \(\pm(1+\sqrt{2})^{n}\) for \(n \in \mathbb{Z}\) are all the units.

Prove or disprove that \(\mathbb{Z}[x]\) is a Euclidean domain.
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Calculus

Read Explanation

Mechanics Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Geometry

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.