College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter P - Summary, Review, and Test - Review Exercises - Page 89: 8


$a.$ $\{\sqrt{81}\}$ $b.$ $\{0,\sqrt{81}\}$ $c.$ $\{-17,0,\sqrt{81}\}$ $d.$ $\Big\{-17,0,\sqrt{81},-\dfrac{9}{13},0.75\Big\}$ $e.$ $\{\pi,\sqrt{2}\}$ $f.$ $\Big\{-17,-\dfrac{9}{13},0,0.75,\sqrt{2},\pi,\sqrt{81}\Big\}$

Work Step by Step

$\Big\{-17,-\dfrac{9}{13},0,0.75,\sqrt{2},\pi,\sqrt{81}\Big\}$ $a.$ $\textbf{Natural numbers}$ The natural numbers are the positive integers and are commonly used for counting. In the given set, only $\sqrt{81}$ is a natural number because $\sqrt{81}=9$ and $9$ is a natural number. $b.$ $\textbf{Whole numbers}$ The whole numbers are all the positive integers including the number $0$. The whole numbers included in the given set are $0$ and $\sqrt{81}$ $c.$ $\textbf{Integers}$ An integer is a number that can be represented without a fractional component. In the given set, $-17$, $0$ and $\sqrt{81}$ are integers $d.$ $\textbf{Rational numbers}$ Any number that can be expressed as the quotient of two integers is a rational number. In the given set, $-17$, $-\dfrac{9}{13}$, $0$, $0.75$ and $\sqrt{81}$ are rational numbers. $e.$ $\textbf{Irrational numbers}$ If a real number is not rational, then it is irrational. In the given set, $\sqrt{2}$ and $\pi$ are irrational numbers. $f.$ $\textbf{Real numbers}$ Since the real numbers include all the numbers that represent a quantity along a line, the rational, irrational, whole and natural numbers, and integers are real numbers. All numbers in the set are real numbers.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.