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: 8

Answer

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

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.
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