Thinking Mathematically (6th Edition)

Published by Pearson
ISBN 10: 0321867327
ISBN 13: 978-0-32186-732-2

Chapter 2 - Set Theory - 2.4 Set Operations and Venn Diagrams with Three Sets - Exercise Set 2.4 - Page 93: 69

Answer

given below

Work Step by Step

(a) To find the set \[A\cup \left( B'\cap C' \right)\], First perform the operation inside parenthesis: Now, compute \[B'\cap C'\] The complement of set Bcontaining all the elements of U which are not in B. So, set \[B'=\left\{ \text{c,d,e,f} \right\}\] Similarly, set \[C'=\left\{ \text{a,c,e,f} \right\}\] So, the intersection of both the sets containing their common elements. Therefore, \[\begin{align} & B'\cap C'=\left\{ \text{c,d,e,f} \right\}\cap \left\{ \text{a,c,e,f} \right\} \\ & =\left\{ \text{c,e,f} \right\} \end{align}\]. Now, the union of sets A and\[B'\cap C'\]containing elements either of the set A or \[B'\cap C'\]or both. So, \[\begin{align} & A\cup \left( B'\cap C' \right)=\left\{ \text{c} \right\}\cup \left\{ \text{c,e,f} \right\} \\ & =\left\{ \text{c,e,f} \right\} \end{align}\] Now, to find the set \[\left( A\cup B' \right)\cap \left( A\cup C' \right)\] First perform operation inside parenthesis: So, \[\begin{align} & A\cup B'=\left\{ \text{c} \right\}\cup \left\{ \text{c,d,e,f} \right\} \\ & =\left\{ \text{c,d,e,f} \right\} \\ & A\cup C'=\left\{ \text{c} \right\}\cup \left\{ \text{a,c,e,f} \right\} \\ & =\left\{ \text{a,c,e,f} \right\} \end{align}\] Their intersection is: \[\begin{align} & \left( A\cup B' \right)\cap \left( A\cup C' \right)=\left\{ \text{c,d,e,f} \right\}\cap \left\{ \text{a,c,e,f} \right\} \\ & =\left\{ \text{c,e,f} \right\} \end{align}\] The set \[A\cup \left( B'\cap C' \right)=\left\{ \text{c,e,f} \right\}\]and\[\left( A\cup B' \right)\cap \left( A\cup C' \right)=\left\{ \text{c,e,f} \right\}\]. (b) To find the set \[A\cup \left( B'\cap C' \right)\], First perform the operation inside parenthesis: Now, compute \[B'\cap C'\] The complement of set Bcontaining all the elements of U which are not in B. So, set \[B'=\left\{ \text{1,4,5,8} \right\}\] Similarly, set \[C'=\left\{ \text{1,2,3,5} \right\}\] So the intersection of both the sets containing their common elements. Therefore, \[\begin{align} & B'\cap C'=\left\{ 1,4,5,8 \right\}\cap \left\{ 1,2,3,5 \right\} \\ & =\left\{ 1,5 \right\} \end{align}\]. Now the union of sets A and\[B'\cap C'\]containing elements either of the set A or \[B'\cap C'\]or both. So, \[\begin{align} & A\cup \left( B'\cap C' \right)=\left\{ 1,3,7,8 \right\}\cup \left\{ \text{5} \right\} \\ & =\left\{ \text{1,3,5,7,8} \right\} \end{align}\] Now to find the set \[\left( A\cup B' \right)\cap \left( A\cup C' \right)\] First perform operation inside parenthesis: So, \[\begin{align} & A\cup B'=\left\{ 1,3,7,8 \right\}\cup \left\{ \text{1,4,5,8} \right\} \\ & =\left\{ \text{1,3,4,5,7,8} \right\} \\ & A\cup C'=\left\{ 1,3,7,8 \right\}\cup \left\{ 1,2,3,5 \right\} \\ & =\left\{ \text{1,2,3,5,7,8} \right\} \end{align}\] Their intersection is: \[\begin{align} & \left( A\cup B' \right)\cap \left( A\cup C' \right)=\left\{ \text{1,3,4,5,7,8} \right\}\cap \left\{ \text{1,2,3,5,7,8} \right\} \\ & =\left\{ \text{1,3,5,7,8} \right\} \end{align}\] The set \[A\cup \left( B'\cap C' \right)=\left\{ \text{1,3,5,7,8} \right\}\]and\[\left( A\cup B' \right)\cap \left( A\cup C' \right)=\left\{ \text{1,3,5,7,8} \right\}\] (c) Inductive reasoning is the process of arriving at a general conclusion based on observations of specific examples. So, we consider examples given in part (a)and part (b). In part (a), observe that set\[A\cup \left( B'\cap C' \right)=\left\{ \text{c,e,f} \right\}\]and set\[\left( A\cup B' \right)\cap \left( A\cup C' \right)=\left\{ \text{c,e,f} \right\}\] So, they are equal. Also, in part (b), set \[A\cup \left( B'\cap C' \right)=\left\{ \text{1,3,5,7,8} \right\}\]and\[\left( A\cup B' \right)\cap \left( A\cup C' \right)=\left\{ \text{1,3,5,7,8} \right\}\]. So they are equal. Generalize the concept from the above two observations that for any set A, B and CIntersection is Distributive over Union of sets. So, \[A\cup \left( B'\cap C' \right)=\left( A\cup B' \right)\cap \left( A\cup C' \right)\]. For any set A, B and C: \[A\cup \left( B'\cap C' \right)=\left( A\cup B' \right)\cap \left( A\cup C' \right)\] (d) Deductive reasoning is the process of proving a specific conclusion from one or more general statements. A conclusion that is proved to be true by deductive reasoning is called a theorem. Now to prove \[A\cup \left( B'\cap C' \right)=\left( A\cup B' \right)\cap \left( A\cup C' \right)\] Consider a Venn diagram of set A, B, C and universal set U. To find the region which represent the set \[A\cup \left( B'\cap C' \right)\]in the above Venn diagram, First perform the operation inside parenthesis: Now, find the region of set \[B'\cap C'\]. The complement of set Bcontaining all the elements of U which are not in B. Regions II, III, V and VI together represents set B. Then removing these regions from universal set U (containing all the regions), the left part is complement of set B. So, regions I, IV, VII and VIII represents the set\[B'\]. Similar way, Regions IV, V, VI, and VII together represent set C. Then removing these regions from universal set U (containing all the regions), the left part is complement of set C. So, regions I, II, III and VIII represents the set\[C'\]. Then, intersection\[B'\cap C'\] represented by the common regions of both the sets, which are I and VIII. Now, set A is represented by the regions I, II, IV and V. Then,\[A\cup \left( B'\cap C' \right)\]is represented by union of the regions of the set A and set\[B'\cap C'\]. So, regions I, II, IV, V and VIII represent the set\[A\cup \left( B'\cap C' \right)\]. Now, to find the set\[\left( A\cup B' \right)\cap \left( A\cup C' \right)\]. First perform operation inside parenthesis: So, set\[A\cup B'\]is represented by I, II, IV, V, VII and VIII and set\[A\cup C'\]is represented by I, II, III, IV, V and VIII. Then, their intersection \[\left( A\cup B' \right)\cap \left( A\cup C' \right)\] is represented by I, II, IV, V and VIII. Since, set\[A\cup \left( B'\cap C' \right)\]is represented by regions I, II, IV, V and VIII and also the set\[\left( A\cup B' \right)\cap \left( A\cup C' \right)\]represented by regions I, II, IV, V and VIII. So, they are equal for any sets A, B and C. i.e.\[A\cup \left( B'\cap C' \right)=\left( A\cup B' \right)\cap \left( A\cup C' \right)\] \[A\cup \left( B'\cap C' \right)=\left( A\cup B' \right)\cap \left( A\cup C' \right)\]for any set A, B and C. Hence, it is a theorem.
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