\displaystyle \textbf{Question 1: }\text{Let }A=\{a,b,c,d\},\ B=\{b,c,e\}\text{ and }C=\{a,b,e\}.\text{ Find:}
\displaystyle \text{(i) }A\cup B\qquad\text{(ii) }B\cup C\qquad\text{(iii) }A\cup C
\displaystyle \text{(iv) }A\cap B\qquad\text{(v) }B\cap C\qquad\text{(vi) }A\cap C

\displaystyle \text{Answer:}
\displaystyle \text{(i) }A\cup B=\{a,b,c,d,e\}
\displaystyle \text{(ii) }B\cup C=\{a,b,c,e\}
\displaystyle \text{(iii) }A\cup C=\{a,b,c,d,e\}
\displaystyle \text{(iv) }A\cap B=\{b,c\}
\displaystyle \text{(v) }B\cap C=\{b,e\}
\displaystyle \text{(vi) }A\cap C=\{a,b\}

\displaystyle \text{Note: The union of sets }A\text{ and }B,\text{ denoted by }A\cup B,\text{ is the set of all}
\displaystyle \text{elements which belong to }A\text{ or }B\text{ or both.}
\displaystyle \text{For example, if }A=\{2,3\}\text{ and }B=\{a,b\},\text{ then }A\cup B=\{2,3,a,b\}.
\displaystyle A\cup B=\{x:x\in A\text{ or }x\in B\}.
\displaystyle \text{Thus, }x\in A\cup B\text{ means }x\in A\text{ or }x\in B.
\displaystyle \text{Also, }x\notin A\cup B\text{ means }x\notin A\text{ and }x\notin B.

\displaystyle \text{The intersection of sets }A\text{ and }B,\text{ denoted by }A\cap B,\text{ is the set of all}
\displaystyle \text{elements which are common to both }A\text{ and }B.
\displaystyle \text{For example, if }A=\{1,2,3\}\text{ and }B=\{2,4,5\},\text{ then }A\cap B=\{2\},
\displaystyle \text{as }2\text{ is the only common element.}
\displaystyle A\cap B=\{x:x\in A\text{ and }x\in B\}.
\displaystyle \text{Thus, }x\in A\cap B\text{ means }x\in A\text{ and }x\in B.
\displaystyle \text{Also, }x\notin A\cap B\text{ means }x\notin A\text{ or }x\notin B.

\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Let }A=\{2,3,4,6\},\ B=\{4,6,8,9\}\text{ and }C=\{2,7,8,9\}.\text{ Find:}
\displaystyle \text{(i) }A\cup B\qquad\text{(ii) }B\cup C\qquad\text{(iii) }A\cup C
\displaystyle \text{(iv) }A\cap B\qquad\text{(v) }B\cap C\qquad\text{(vi) }A\cap C

\displaystyle \text{Answer:}
\displaystyle \text{(i) }A\cup B=\{2,3,4,6,8,9\}
\displaystyle \text{(ii) }B\cup C=\{2,4,6,7,8,9\}
\displaystyle \text{(iii) }A\cup C=\{2,3,4,6,7,8,9\}
\displaystyle \text{(iv) }A\cap B=\{4,6\}
\displaystyle \text{(v) }B\cap C=\{8,9\}
\displaystyle \text{(vi) }A\cap C=\{2\}

\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Let }A=\{1,4,7,8\}\text{ and }B=\{4,6,8,9\}.\text{ Find:}
\displaystyle \text{(i) }A-B\qquad\text{(ii) }B-A

\displaystyle \text{Answer:}
\displaystyle \text{(i) }A-B=\{1,4,7,8\}-\{4,6,8,9\}=\{1,7\}
\displaystyle \text{(ii) }B-A=\{4,6,8,9\}-\{1,4,7,8\}=\{6,9\}

\displaystyle \text{Note: For any two sets }A\text{ and }B,\text{ the difference }A-B\text{ is the set of all those}
\displaystyle \text{elements of }A\text{ which are not in }B.
\displaystyle \text{For example, if }A=\{1,2,3,4,5\}\text{ and }B=\{4,5,6\},\text{ then}
\displaystyle A-B=\{1,2,3\}\text{ and }B-A=\{6\}.
\displaystyle \text{Therefore, }A-B=\{x:x\in A\text{ and }x\notin B\}.
\displaystyle \text{Thus, }x\in A-B\text{ means }x\in A\text{ and }x\notin B.

\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Let }\xi=\{13,14,15,16,17,18,19,20,21\},
\displaystyle A=\{13,17,19\}\text{ and }B=\{14,16,18,20\}.\text{ Find:}
\displaystyle \text{(i) }A'\qquad\text{(ii) }B'

\displaystyle \text{Answer:}
\displaystyle \text{(i) }A'=\{14,15,16,18,20,21\}
\displaystyle \text{(ii) }B'=\{13,15,17,19,21\}

\displaystyle \text{Note: Let }\xi\text{ be the universal set and let }A\subseteq\xi.\text{ Then the complement of }A,
\displaystyle \text{denoted by }A',\text{ is the set of all those elements of }\xi\text{ which are not in }A.
\displaystyle \text{For example, let }\xi=\{1,2,3,4,5,6,7,8\}\text{ and }A=\{2,3,4\}.\text{ Then}
\displaystyle A'=\{1,5,6,7,8\}.
\displaystyle \text{Thus, }A'=\{x:x\in\xi\text{ and }x\notin A\}.
\displaystyle \text{Therefore, }x\in A'\text{ means }x\in\xi\text{ and }x\notin A.

\displaystyle \text{Please note:}
\displaystyle \phi'=\xi\text{ and }\xi'=\phi.
\displaystyle A\cup A'=\xi\text{ and }A\cap A'=\phi.

\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Let }\xi=\{x:x\in Z,\ -4\leq x\leq4\},
\displaystyle A=\{x:x\in W,\ x<4\}\text{ and }B=\{x:x\in N,\ 2<x\leq4\}.\text{ Find:}
\displaystyle \text{(i) }A'\qquad\text{(ii) }B'

\displaystyle \text{Answer:}
\displaystyle \text{Note: First find the elements of the three given sets.}
\displaystyle \xi=\{-4,-3,-2,-1,0,1,2,3,4\},\quad A=\{0,1,2,3\},\quad B=\{3,4\}.

\displaystyle \text{(i) }A'=\{-4,-3,-2,-1,4\}
\displaystyle \text{(ii) }B'=\{-4,-3,-2,-1,0,1,2\}

\displaystyle \\

\displaystyle \textbf{Question 6: }\text{Let }\xi=\{x:x\in N,\ x\text{ is a factor of }144\},
\displaystyle A=\{x:x\in N,\ x\text{ is a factor of }24\},
\displaystyle B=\{x:x\in N,\ x\text{ is a factor of }36\}\text{ and}
\displaystyle C=\{x:x\in N,\ x\text{ is a factor of }48\}.\text{ Find:}
\displaystyle \text{(i) }A'\qquad\text{(ii) }B'\qquad\text{(iii) }C'\qquad\text{(iv) }A\cup B
\displaystyle \text{(v) }B\cup C\qquad\text{(vi) }A\cup C\qquad\text{(vii) }A\cap B'
\displaystyle \text{(viii) }B\cap C'\qquad\text{(ix) }C-A\qquad\text{(x) }A-(B\cap C)

\displaystyle \text{Answer:}
\displaystyle \text{Note: First find the elements of each of the given sets.}
\displaystyle \xi=\{1,2,3,4,6,8,9,12,16,18,24,36,48,72,144\}.
\displaystyle \text{This is the universal set.}
\displaystyle A=\{1,2,3,4,6,8,12,24\}.
\displaystyle B=\{1,2,3,4,6,9,12,18,36\}.
\displaystyle C=\{1,2,3,4,6,8,12,16,24,48\}.

\displaystyle \text{(i) }A'=\{9,16,18,36,48,72,144\}

\displaystyle \text{(ii) }B'=\{8,16,24,48,72,144\}

\displaystyle \text{(iii) }C'=\{9,18,36,72,144\}

\displaystyle \text{(iv) }A\cup B=\{1,2,3,4,6,8,9,12,18,24,36\}

\displaystyle \text{(v) }B\cup C=\{1,2,3,4,6,8,9,12,16,18,24,36,48\}

\displaystyle \text{(vi) }A\cup C=\{1,2,3,4,6,8,12,16,24,48\}

\displaystyle \text{(vii) }A\cap B'=\{8,24\}

\displaystyle \text{(viii) }B\cap C'=\{9,18,36\}

\displaystyle \text{(ix) }C-A=\{16,48\}

\displaystyle \text{(x) }A-(B\cap C)
\displaystyle \text{First, }B\cap C=\{1,2,3,4,6,12\}.
\displaystyle \text{Therefore, }A-(B\cap C)=\{8,24\}.

\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Considering the sets given in Question 6, state whether the following}
\displaystyle \text{statements are true or false.}
\displaystyle \text{(i) }A\cap(B\cup C)=A\qquad\text{(ii) }A\subset C\qquad\text{(iii) }B\subseteq C
\displaystyle \text{(iv) }A\cap C'=\phi

\displaystyle \text{Answer:}
\displaystyle \text{Note: First find the elements of each of the given sets.}
\displaystyle \xi=\{1,2,3,4,6,8,9,12,16,18,24,36,48,72,144\}.
\displaystyle \text{This is the universal set.}
\displaystyle A=\{1,2,3,4,6,8,12,24\}.
\displaystyle B=\{1,2,3,4,6,9,12,18,36\}.
\displaystyle C=\{1,2,3,4,6,8,12,16,24,48\}.

\displaystyle \text{(i) }A\cap(B\cup C)=A
\displaystyle \text{Answer:}
\displaystyle \text{True}
\displaystyle B\cup C=\{1,2,3,4,6,8,9,12,16,18,24,36,48\}.
\displaystyle A\cap(B\cup C)=\{1,2,3,4,6,8,12,24\}=A.
\displaystyle \text{Hence, }A\cap(B\cup C)=A.

\displaystyle \text{(ii) }A\subset C
\displaystyle \text{Answer:}
\displaystyle \text{True}
\displaystyle A=\{1,2,3,4,6,8,12,24\}.
\displaystyle C=\{1,2,3,4,6,8,12,16,24,48\}.
\displaystyle \text{Every element of }A\text{ belongs to }C,\text{ while }16\text{ and }48\text{ belong to }C\text{ but not to }A.
\displaystyle \text{Therefore, }A\text{ is a proper subset of }C,\text{ i.e. }A\subset C.

\displaystyle \text{(iii) }B\subseteq C
\displaystyle \text{Answer:}
\displaystyle \text{False}
\displaystyle B=\{1,2,3,4,6,9,12,18,36\}.
\displaystyle C=\{1,2,3,4,6,8,12,16,24,48\}.
\displaystyle \text{The elements }9,18\text{ and }36\text{ belong to }B\text{ but do not belong to }C.
\displaystyle \text{Therefore, not every element of }B\text{ belongs to }C,\text{ and hence }B\not\subseteq C.

\displaystyle \text{(iv) }A\cap C'=\phi
\displaystyle \text{Answer:}
\displaystyle \text{True}
\displaystyle C'=\{9,18,36,72,144\}.
\displaystyle A=\{1,2,3,4,6,8,12,24\}.
\displaystyle \text{There are no elements common to }A\text{ and }C'.
\displaystyle \text{Therefore, }A\cap C'=\phi.

\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Let }A=\{a,b,c,d,e\},\ B=\{a,c,e,g\}\text{ and }C=\{b,e,f,g\}.
\displaystyle \text{Then verify the following identities:}
\displaystyle \text{(i) }B\cup C=C\cup B\qquad\text{(ii) }B\cap C=C\cap B
\displaystyle \text{(iii) }A\cup(B\cup C)=(A\cup B)\cup C
\displaystyle \text{(iv) }A\cap(B\cap C)=(A\cap B)\cap C
\displaystyle \text{(v) }A\cup(B\cap C)=(A\cup B)\cap(A\cup C)
\displaystyle \text{(vi) }A\cap(B\cup C)=(A\cap B)\cup(A\cap C)

\displaystyle \text{Answer:}

\displaystyle \text{(i) }B\cup C=C\cup B
\displaystyle B=\{a,c,e,g\}\text{ and }C=\{b,e,f,g\}.
\displaystyle B\cup C=\{a,b,c,e,f,g\}.
\displaystyle C\cup B=\{a,b,c,e,f,g\}.
\displaystyle \therefore B\cup C=C\cup B.\text{ Hence proved.}

\displaystyle \text{(ii) }B\cap C=C\cap B
\displaystyle B=\{a,c,e,g\}\text{ and }C=\{b,e,f,g\}.
\displaystyle B\cap C=\{e,g\}.
\displaystyle C\cap B=\{e,g\}.
\displaystyle \therefore B\cap C=C\cap B.\text{ Hence proved.}

\displaystyle \text{(iii) }A\cup(B\cup C)=(A\cup B)\cup C
\displaystyle A=\{a,b,c,d,e\},\quad B=\{a,c,e,g\},\quad C=\{b,e,f,g\}.
\displaystyle \text{LHS: }B\cup C=\{a,b,c,e,f,g\}.
\displaystyle A\cup(B\cup C)=\{a,b,c,d,e\}\cup\{a,b,c,e,f,g\}
\displaystyle \phantom{A\cup(B\cup C)}=\{a,b,c,d,e,f,g\}.
\displaystyle \text{RHS: }A\cup B=\{a,b,c,d,e,g\}.
\displaystyle (A\cup B)\cup C=\{a,b,c,d,e,g\}\cup\{b,e,f,g\}
\displaystyle \phantom{(A\cup B)\cup C}=\{a,b,c,d,e,f,g\}.
\displaystyle \therefore \text{LHS}=\text{RHS}.\text{ Hence proved.}

\displaystyle \text{(iv) }A\cap(B\cap C)=(A\cap B)\cap C
\displaystyle A=\{a,b,c,d,e\},\quad B=\{a,c,e,g\},\quad C=\{b,e,f,g\}.
\displaystyle \text{LHS: }B\cap C=\{e,g\}.
\displaystyle A\cap(B\cap C)=\{a,b,c,d,e\}\cap\{e,g\}=\{e\}.
\displaystyle \text{RHS: }A\cap B=\{a,c,e\}.
\displaystyle (A\cap B)\cap C=\{a,c,e\}\cap\{b,e,f,g\}=\{e\}.
\displaystyle \therefore \text{LHS}=\text{RHS}.\text{ Hence proved.}

\displaystyle \text{(v) }A\cup(B\cap C)=(A\cup B)\cap(A\cup C)
\displaystyle A=\{a,b,c,d,e\},\quad B=\{a,c,e,g\},\quad C=\{b,e,f,g\}.
\displaystyle \text{LHS: }B\cap C=\{e,g\}.
\displaystyle A\cup(B\cap C)=\{a,b,c,d,e\}\cup\{e,g\}=\{a,b,c,d,e,g\}.
\displaystyle \text{RHS: }A\cup B=\{a,b,c,d,e,g\}.
\displaystyle A\cup C=\{a,b,c,d,e,f,g\}.
\displaystyle (A\cup B)\cap(A\cup C)
\displaystyle =\{a,b,c,d,e,g\}\cap\{a,b,c,d,e,f,g\}=\{a,b,c,d,e,g\}.
\displaystyle \therefore \text{LHS}=\text{RHS}.\text{ Hence proved.}

\displaystyle \text{(vi) }A\cap(B\cup C)=(A\cap B)\cup(A\cap C)
\displaystyle A=\{a,b,c,d,e\},\quad B=\{a,c,e,g\},\quad C=\{b,e,f,g\}.
\displaystyle \text{LHS: }B\cup C=\{a,b,c,e,f,g\}.
\displaystyle A\cap(B\cup C)=\{a,b,c,d,e\}\cap\{a,b,c,e,f,g\}=\{a,b,c,e\}.
\displaystyle \text{RHS: }A\cap B=\{a,c,e\}\text{ and }A\cap C=\{b,e\}.
\displaystyle (A\cap B)\cup(A\cap C)=\{a,c,e\}\cup\{b,e\}=\{a,b,c,e\}.
\displaystyle \therefore \text{LHS}=\text{RHS}.\text{ Hence proved.}

\displaystyle \\

\displaystyle \textbf{Question 9: }\text{Let }A=\{b,c,d,e\}\text{ and }B=\{d,e,f,g\}\text{ be two subsets of}
\displaystyle \text{the universal set }\xi=\{b,c,d,e,f,g\}.\text{ Then verify the following:}
\displaystyle \text{(i) }(A\cup B)'=A'\cap B'\qquad\text{(ii) }(A\cap B)'=A'\cup B'

\displaystyle \text{Answer:}

\displaystyle \text{(i) }(A\cup B)'=A'\cap B'
\displaystyle \text{LHS: }A\cup B=\{b,c,d,e\}\cup\{d,e,f,g\}=\{b,c,d,e,f,g\}=\xi.
\displaystyle \therefore (A\cup B)'=\xi'=\phi.
\displaystyle \text{RHS: }A'=\{f,g\}\text{ and }B'=\{b,c\}.
\displaystyle \therefore A'\cap B'=\{f,g\}\cap\{b,c\}=\phi.
\displaystyle \therefore \text{LHS}=\text{RHS}.\text{ Hence proved.}

\displaystyle \text{(ii) }(A\cap B)'=A'\cup B'
\displaystyle \xi=\{b,c,d,e,f,g\}.
\displaystyle \text{LHS: }A\cap B=\{b,c,d,e\}\cap\{d,e,f,g\}=\{d,e\}.
\displaystyle \therefore (A\cap B)'=\{b,c,f,g\}.
\displaystyle \text{RHS: }A'=\{f,g\}\text{ and }B'=\{b,c\}.
\displaystyle \therefore A'\cup B'=\{f,g\}\cup\{b,c\}=\{b,c,f,g\}.
\displaystyle \therefore \text{LHS}=\text{RHS}.\text{ Hence proved.}

\displaystyle \\

\displaystyle \textbf{Question 10: }\text{Fill in the blanks:}

\displaystyle \text{(i) }A\cup A=\ldots\qquad\text{(ii) }A\cap A=\ldots\qquad\text{(iii) }A\cup\phi=\ldots
\displaystyle \text{(iv) }A\cap\phi=\ldots\qquad\text{(v) }(A\cup B)'=\ldots\qquad\text{(vi) }(A\cap B)'=\ldots

\displaystyle \text{Answer:}
\displaystyle \text{(i) }A\cup A=A
\displaystyle \text{(ii) }A\cap A=A
\displaystyle \text{(iii) }A\cup\phi=A
\displaystyle \text{(iv) }A\cap\phi=\phi
\displaystyle \text{(v) }(A\cup B)'=A'\cap B'
\displaystyle \text{(vi) }(A\cap B)'=A'\cup B'

\displaystyle \\

\displaystyle \textbf{Question 11: }\text{Let }\xi=\{x:x\in N,\ 4\leq x<18\}\text{ and }A,B\text{ and }C\text{ be}
\displaystyle \text{subsets of }\xi.\text{ Given }A=\{x:x\text{ is a multiple of }2\},
\displaystyle B=\{x:x\text{ is a multiple of }3\}\text{ and }C=\{x:x\in N,\ x<11\}.\text{ Then verify:}
\displaystyle \text{(i) }(A\cup B)'=A'\cap B'\qquad\text{(ii) }(A\cap B)'=A'\cup B'
\displaystyle \text{(iii) }A-B=A\cap B'\qquad\text{(iv) }A\cup(B\cap C)=(A\cup B)\cap(A\cup C)

\displaystyle \text{Answer:}
\displaystyle \text{Note: First identify the elements of all four sets. This makes the solution easier.}
\displaystyle \xi=\{4,5,6,7,8,9,10,11,12,13,14,15,16,17\}.
\displaystyle A=\{4,6,8,10,12,14,16\}.
\displaystyle B=\{6,9,12,15\}.
\displaystyle C=\{4,5,6,7,8,9,10\}.

\displaystyle \text{(i) }(A\cup B)'=A'\cap B'
\displaystyle \text{LHS: }A\cup B=\{4,6,8,9,10,12,14,15,16\}.
\displaystyle \therefore (A\cup B)'=\{5,7,11,13,17\}.
\displaystyle \text{RHS: }A'=\{5,7,9,11,13,15,17\}.
\displaystyle B'=\{4,5,7,8,10,11,13,14,16,17\}.
\displaystyle A'\cap B'=\{5,7,11,13,17\}.
\displaystyle \therefore \text{LHS}=\text{RHS}.\text{ Hence proved.}

\displaystyle \text{(ii) }(A\cap B)'=A'\cup B'
\displaystyle \text{LHS: }A\cap B=\{6,12\}.
\displaystyle \therefore (A\cap B)'=\{4,5,7,8,9,10,11,13,14,15,16,17\}.
\displaystyle \text{RHS: }A'=\{5,7,9,11,13,15,17\}.
\displaystyle B'=\{4,5,7,8,10,11,13,14,16,17\}.
\displaystyle A'\cup B'=\{4,5,7,8,9,10,11,13,14,15,16,17\}.
\displaystyle \therefore \text{LHS}=\text{RHS}.\text{ Hence proved.}

\displaystyle \text{(iii) }A-B=A\cap B'
\displaystyle \text{LHS: }A-B=\{4,6,8,10,12,14,16\}-\{6,9,12,15\}
\displaystyle \phantom{\text{LHS: }A-B}=\{4,8,10,14,16\}.
\displaystyle \text{RHS: }A\cap B'
\displaystyle =\{4,6,8,10,12,14,16\}\cap\{4,5,7,8,10,11,13,14,16,17\}
\displaystyle =\{4,8,10,14,16\}.
\displaystyle \therefore \text{LHS}=\text{RHS}.\text{ Hence proved.}

\displaystyle \text{(iv) }A\cup(B\cap C)=(A\cup B)\cap(A\cup C)
\displaystyle \text{LHS: }B\cap C=\{6,9\}.
\displaystyle A\cup(B\cap C)=\{4,6,8,10,12,14,16\}\cup\{6,9\}
\displaystyle \phantom{A\cup(B\cap C)}=\{4,6,8,9,10,12,14,16\}.
\displaystyle \text{RHS: }A\cup B=\{4,6,8,9,10,12,14,15,16\}.
\displaystyle A\cup C=\{4,5,6,7,8,9,10,12,14,16\}.
\displaystyle (A\cup B)\cap(A\cup C)
\displaystyle =\{4,6,8,9,10,12,14,15,16\}\cap\{4,5,6,7,8,9,10,12,14,16\}
\displaystyle =\{4,6,8,9,10,12,14,16\}.
\displaystyle \therefore \text{LHS}=\text{RHS}.\text{ Hence proved.}

\displaystyle \\

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