/*! 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 Show that \(4+9 \sqrt{2}\) is an... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 0: Problem 5

Show that \(4+9 \sqrt{2}\) is an irrational number.

Short Answer

Expert verified
Assuming \(4 + 9\sqrt{2}\) is a rational number, we find a contradiction by showing that both \(a\) and \(b\) share a common factor other than 1 (which breaks the assumption that they are coprime). Thus, \(4 + 9 \sqrt{2}\) must be an irrational number.

Step by step solution

01

Assume it's a rational number

Let's assume that \(4 + 9\sqrt{2}\) is a rational number. Then it can be represented as the ratio of two integers, say \(a\) and \(b\), where \(b \neq 0\) and gcd\((a, b) = 1\) (i.e., \(a\) and \(b\) are coprime). So, \[ 4 + 9\sqrt{2} = \frac{a}{b} \]
02

Rearrange and square both sides

We want to isolate the radical (\(\sqrt{2}\)) on one side of the equation. So, we can rewrite the equation as \[ 9\sqrt{2} = \frac{a}{b} - 4 \] Next, square both sides of the equation to get rid of the radical: \[ (9\sqrt{2})^2 = \left(\frac{a}{b} - 4\right)^2 \]
03

Simplify and solve for a

Simplify the equation and solve for \(a\): \[ 81 \cdot 2 = \frac{a^2}{b^2} - 8 \cdot \frac{a}{b} + 16 \] \[ 162 = \frac{a^2 - 8ab + 16b^2}{b^2} \] Since \(b^2\) must divide the right side of the equation, we can see that \(b^2\) must divide into \(162\). Thus, \(a^2 - 8ab + 16b^2 = 162b^2\), and we can solve for \(a\): \[ a^2 = 8ab + (162 - 16) b^2 = 8ab + 146b^2 \]
04

Show a contradiction

Now we know that \(a^2\) can be written as a sum of terms with \(b\) as a factor. Since both \(8ab\) and \(146b^2\) have \(b\) as a factor, their sum (\(a^2\)) must also have \(b\) as a factor. This contradicts our initial assumption that \(a\) and \(b\) are coprime (gcd\((a, b) = 1\)), as now both \(a\) and \(b\) share a common factor other than 1, which is \(b\). Therefore, our assumption that \(4 + 9\sqrt{2}\) is a rational number must be incorrect, and so \(4 + 9\sqrt{2}\) is an irrational number.

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!

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

Give an example of an open interval and a closed interval whose intersection equals the interval (2,5) .

Suppose \(a\) and \(b\) are numbers. Explain why ei. ther \(ab\)

The first letters of the phrase "Please excuse my dear Aunt Sally" are used by some people to remember the order of operations: parentheses, exponentiation (which we will discuss in a later chapter), multiplication, division, addition, subtraction. Make up a catchy phrase that serves the same purpose but with exponentiation excluded.

Simplify the given expression as much as possible. $$ \frac{1}{x-y}\left(\frac{x}{y}-\frac{y}{x}\right) $$

The intersection of two sets of numbers consists of all numbers that are in both sets. If \(A\) and \(B\) are sets, then their intersection is denoted by \(A \cap B .\) In Exercises \(31-40,\) write each intersection as a single interval. $$ (-\infty,-3) \cap[-5, \infty) $$
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Pure Maths

Read Explanation

Logic and Functions

Read Explanation

Mechanics Maths

Read Explanation

Applied Mathematics

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.