## Thinking Mathematically (6th Edition)

(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. 