College Algebra (6th Edition)

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

Chapter P - Prerequisites: Fundamental Concepts of Algebra - Exercise Set P.1: 35

Answer

a. Natural Numbers : $\sqrt 100 $ b. Whole Numbers : 0, $\sqrt 100 $ c. Integers : -9, 0, $\sqrt 100 $ d. Rational Numbers : -4/5, 0.25, 9.2, 0, $\sqrt 100 $,-9 e. Irrational Numbers : $\sqrt 3 $ f. Real Numbers : -4/5, 0.25, 9.2, 0, $\sqrt 100 $, $\sqrt 3 $,-9

Work Step by Step

a. Natural Numbers are the numbers used for counting. The only number in the set { -9, -4/5, 0, 0.25, $\sqrt 3 $, 9.2, $\sqrt 100 $ } is $\sqrt 100 $ because $\sqrt 100 $ is 10. b. Whole numbers consists of 0 and Natural numbers. So, the whole numbers in the set { -9, -4/5, 0, 0.25, $\sqrt 3 $, 9.2, $\sqrt 100 $ } are 0, $\sqrt 100 $ c. The set of integers includes the negative of the natural numbers and the whole numbers. So the set { -9, -4/5, 0, 0.25, $\sqrt 3 $, 9.2, $\sqrt 100 $ } contains the integers 0, $\sqrt 100 $ and -9. d. The set of rational numbers is the set of all numbers that can be expressed as a quotient of two integers, with the denominator not 0. Rational numbers can be expressed as terminating or repeating decimals. So, -4/5, 0.25, 9.2, 0 (0=0/1), $\sqrt 100 $ ($\sqrt 100 $ =10= 10 /1), -9 (-9/1 = -9) are rational numbers. e. The set of irrational numbers is the set of all numbers whose decimal representations are neither terminating nor repeating and also cannot be expressed as a quotient of integers. In the given set { -9, -4/5, 0, 0.25, $\sqrt 3 $, 9.2, $\sqrt 100 $ }, $\sqrt 3 $ $\approx$ 1.732. (1.7320508075688..). In decimal form, $\sqrt 3 $ neither terminate nor have blocks of repeating digits. So the only irrational number in the given set is $\sqrt 3 $. f. Real Numbers are either rational or irrational. So all the numbers in the given set { -9, -4/5, 0, 0.25, $\sqrt 3 $, 9.2, $\sqrt 100 $ } are real numbers.
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