Discrete Mathematics and Its Applications, Seventh Edition

Published by McGraw-Hill Education
ISBN 10: 0073383090
ISBN 13: 978-0-07338-309-5

Chapter 2 - Section 2.5 - Cardinality of Sets - Exercises - Page 176: 10

Answer

(a) $A =\cal{R}$ (real numbers), $B =\cal{R}-${$0$} (nonzero real numbers) (b) $A= (0,1) \cup N$ (all real numbers between 0 and 1 and all nonnegative integers) , $B= (0, 1)$ (all real numbers between 0 and 1) c) $A=\cal{ R}$ (real numbers) ,$B= R^- \cup \text{{0}}$

Work Step by Step

(a) For example, the set $A =\cal{R}$ (real numbers) is uncountable and the set $B =\cal{R}-${$0$} (nonzero real numbers) is also uncountable. Using the definition of the difference and the fact that 0 is the only element of A that is not in B: $A- B= \cal{R} -(\cal{R}- ${$0$}) = {0} Since the set {$0$} contains only 1 element, the set {$0$} is finite and thus $A- B$ is finite as well. (b) For example, the set $A= (0,1) \cup N$ (all real numbers between 0 and 1 and all nonnegative integers) is uncountable and the set $B= (0, 1)$ (all real numbers between 0 and 1) is also uncountable. Using the definition of the difference and the fact that the nonnegative integers are the only element of $A$ that is not in $B$ $A-B=((0, 1) \cup N)- (0, 1) =N$ The set $N$ of non negative integers is countably infinite. c) For example, the set $A=\cal{ R}$ (real numbers) is uncountable and the set $B= R^- \cup \text{{0}}$ (negative real numbers and zero) is also uncountable. Using the definition of the difference and the fact that the positive real numbers are the only element of $A$ that is not in $B$ $A- B= \cal{R}- (\cal{R}^- \cup $ { $0$ } $= \cal{R}^+$ The set $\cal{R}^-$ of positive real numbers is uncountable.
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