/*! 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 57 Explain why $$ \left|\frac{a... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 0: Problem 57

Explain why $$ \left|\frac{a}{b}\right|=\frac{|a|}{|b|} $$ for all real numbers \(a\) and \(b\) (with \(b \neq 0\) ).

Short Answer

Expert verified
In order to show that \(\left|\frac{a}{b}\right|=\frac{|a|}{|b|}\) for all real numbers a and b, with the condition that b is not equal to zero, we can discuss the possible cases based on the signs of a and b: (1) a and b are both positive, (2) a is positive and b is negative, (3) a is negative and b is positive and, (4) a and b are both negative. In each case, we can manipulate the fractions with absolute values using their properties, such as \(|-a|=|a|\) and \(\frac{x}{y} = \frac{-x}{-y}\). By showing that the equation holds true in all cases, we conclude that \(\left|\frac{a}{b}\right|=\frac{|a|}{|b|}\) for all real numbers a and b, with the condition that \(b \neq 0\).

Step by step solution

01

Mention the properties of absolute values

Absolute value has some important properties that we should be aware of before we dive into the proof. Recall that |x| denotes the absolute value of x. The main properties include: 1. \(|x| \geq 0\) and \(|x|=0\) if and only if \(x=0\) 2. \(|xy| = |x||y|\) for all real numbers x and y 3. \(|\frac{x}{y}| = \frac{|x|}{|y|}\) for all real numbers x and y, provided y 鈮� 0.
02

Consider the cases based on the signs of a and b

Since we want to show that \(\left|\frac{a}{b}\right|=\frac{|a|}{|b|}\), it's best to discuss the possible cases based on the signs of a and b. 1. If a and b are both positive numbers. 2. If a is positive and b is negative. 3. If a is negative and b is positive. 4. If a and b are both negative numbers. Now, let's break down these cases and show that the given equation holds true for all of them.
03

Case 1 - a and b are both positive numbers

If a and b are both positive numbers, then their absolute values are the same as their original values. In this case: \[\left|\frac{a}{b}\right| = \frac{a}{b}\] Since |a|=a and |b|=b, then: \[\frac{|a|}{|b|} = \frac{a}{b}\] So, in this case, the given equation holds true.
04

Case 2 - a is positive and b is negative

If a is positive and b is negative, then |a|=a and |b|=-b. The equation becomes: \[\left|\frac{a}{b}\right| = \left|\frac{a}{-b}\right|\] Since multiplying the numerator and denominator of a fraction by the same number doesn't change the value of the fraction, and |-1|=1, this is equivalent to: \[\left|\frac{-a}{b}\right| = \frac{|-a|}{|b|}\] Now, since |-a|=|a|=a, |b|=-b, and |-a(-1)|=(-a)(-1)=a, we can rewrite the equation as: \[\left|\frac{a}{b}\right| = \frac{a}{-b}\] This shows that in this case, too, the given equation holds true.
05

Case 3 - a is negative and b is positive

If a is negative and b is positive, then |a|=-a and |b|=b. In this case, the equation becomes: \[\left|\frac{-a}{b}\right| = \frac{|-a|}{|b|}\] Since |-a|=|a|=-a and |b|=b, and the product |-a(-1)|=(-a)(-1)=a, we can rewrite the equation as: \[\left|\frac{a}{b}\right| = \frac{-a}{b}\] This shows that in this case, too, the given equation holds true.
06

Case 4 - a and b are both negative numbers

If a and b are both negative numbers, then their absolute values are -a and -b, respectively. In this case, the equation becomes: \[\left|\frac{-a}{-b}\right| = \left|\frac{a}{b}\right|\] Since both the numerator and denominator are multiplied by the same negative number, we can apply the third property of absolute values mentioned in Step 1: \[\frac{|-a|}{|-b|} = \frac{|a|}{|b|}\] Thus, in this case, the given equation holds true as well.
07

Conclusion

In summary, we have shown that the given equation \(\left|\frac{a}{b}\right|=\frac{|a|}{|b|}\) holds true for all possible combinations of a and b's signs, as long as \(b \neq 0\). This means the given equation holds true for all real numbers a and b (with \(b \neq 0\)).

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

Simplify the given expression as much as possible. $$ 3(2 m+4(n+5 p))+6 n $$

(a) True or false: If \(0

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. $$ (-3, \infty) \cap[-5, \infty) $$

Simplify the given expression as much as possible. $$ \frac{\frac{6}{5}}{\frac{7}{4}} $$

Give an example of an open interval and a closed interval whose union equals the interval (2,5) .
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Applied Mathematics

Read Explanation

Decision Maths

Read Explanation

Pure Maths

Read Explanation

Discrete Mathematics

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.