\displaystyle \textbf{Question 1: }\text{Given }A=\{1,2,3\},\;B=\{3,4\},\;C=\{4,5,6\},\text{ find }
\displaystyle (A\times B)\cap(B\times C).
\displaystyle \text{Answer:}
\displaystyle \text{Given }A=\{1,2,3\},\;B=\{3,4\}\text{ and }C=\{4,5,6\}.
\displaystyle A\times B=\{(1,3),(1,4),(2,3),(2,4),(3,3),(3,4)\}.
\displaystyle B\times C=\{(3,4),(3,5),(3,6),(4,4),(4,5),(4,6)\}.
\displaystyle \therefore (A\times B)\cap(B\times C)=\{(3,4)\}.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{If }A=\{2,3\},\;B=\{4,5\},\;C=\{5,6\},\text{ find }A\times(B\cup C),
\displaystyle A\times(B\cap C)\text{ and }(A\times B)\cup(A\times C).
\displaystyle \text{Answer:}
\displaystyle \text{Given }A=\{2,3\},\;B=\{4,5\}\text{ and }C=\{5,6\}.
\displaystyle B\cup C=\{4,5,6\}.
\displaystyle \therefore A\times(B\cup C)=\{2,3\}\times\{4,5,6\}
\displaystyle =\{(2,4),(2,5),(2,6),(3,4),(3,5),(3,6)\}.
\displaystyle B\cap C=\{5\}.
\displaystyle \therefore A\times(B\cap C)=\{2,3\}\times\{5\}
\displaystyle =\{(2,5),(3,5)\}.
\displaystyle A\times B=\{(2,4),(2,5),(3,4),(3,5)\}.
\displaystyle A\times C=\{(2,5),(2,6),(3,5),(3,6)\}.
\displaystyle \therefore (A\times B)\cup(A\times C)
\displaystyle =\{(2,4),(2,5),(2,6),(3,4),(3,5),(3,6)\}.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{If }A=\{1,2,3\},\;B=\{4\}\text{ and }C=\{5\},\text{ verify that:}
\displaystyle \text{(i) }A\times(B\cup C)=(A\times B)\cup(A\times C)
\displaystyle \text{(ii) }A\times(B\cap C)=(A\times B)\cap(A\times C)
\displaystyle \text{(iii) }A\times(B-C)=(A\times B)-(A\times C).
\displaystyle \text{Answer:}
\displaystyle \text{Given }A=\{1,2,3\},\;B=\{4\}\text{ and }C=\{5\}.

\displaystyle \text{(i) }B\cup C=\{4,5\}.
\displaystyle A\times(B\cup C)=\{1,2,3\}\times\{4,5\}
\displaystyle =\{(1,4),(1,5),(2,4),(2,5),(3,4),(3,5)\}.\qquad\text{...(1)}
\displaystyle A\times B=\{(1,4),(2,4),(3,4)\}
\displaystyle A\times C=\{(1,5),(2,5),(3,5)\}
\displaystyle \therefore (A\times B)\cup(A\times C)
\displaystyle =\{(1,4),(1,5),(2,4),(2,5),(3,4),(3,5)\}.\qquad\text{...(2)}
\displaystyle \text{From (1) and (2),}
\displaystyle A\times(B\cup C)=(A\times B)\cup(A\times C).
\displaystyle \text{Hence verified.}

\displaystyle \text{(ii) }B\cap C=\{4\}\cap\{5\}=\phi.
\displaystyle \therefore A\times(B\cap C)=A\times\phi=\phi.\qquad\text{...(1)}
\displaystyle A\times B=\{(1,4),(2,4),(3,4)\}
\displaystyle A\times C=\{(1,5),(2,5),(3,5)\}
\displaystyle \therefore (A\times B)\cap(A\times C)=\phi.\qquad\text{...(2)}
\displaystyle \text{From (1) and (2),}
\displaystyle A\times(B\cap C)=(A\times B)\cap(A\times C).
\displaystyle \text{Hence verified.}

\displaystyle \text{(iii) }B-C=\{4\}-\{5\}=\{4\}.
\displaystyle A\times(B-C)=\{1,2,3\}\times\{4\}
\displaystyle =\{(1,4),(2,4),(3,4)\}.\qquad\text{...(1)}
\displaystyle A\times B=\{(1,4),(2,4),(3,4)\}
\displaystyle A\times C=\{(1,5),(2,5),(3,5)\}
\displaystyle \therefore (A\times B)-(A\times C)
\displaystyle =\{(1,4),(2,4),(3,4)\}.\qquad\text{...(2)}
\displaystyle \text{From (1) and (2),}
\displaystyle A\times(B-C)=(A\times B)-(A\times C).
\displaystyle \text{Hence verified.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Let }A=\{1,2\},\;B=\{1,2,3,4\},\;C=\{5,6\}\text{ and }D=\{5,6,7,8\}.
\displaystyle \text{Verify that: (i) }A\times C\subset B\times D\qquad\text{(ii) }A\times(B\cap C)=(A\times B)\cap(A\times C).
\displaystyle \text{Answer:}

\displaystyle \text{(i) Given }A=\{1,2\},\;B=\{1,2,3,4\},\;C=\{5,6\}\text{ and }D=\{5,6,7,8\}.
\displaystyle A\times C=\{(1,5),(1,6),(2,5),(2,6)\}.\qquad\text{...(1)}
\displaystyle B\times D=\{(1,5),(1,6),(1,7),(1,8),(2,5),(2,6),(2,7),(2,8),
\displaystyle (3,5),(3,6),(3,7),(3,8),(4,5),(4,6),(4,7),(4,8)\}.\qquad\text{...(2)}
\displaystyle \text{From (1) and (2), }A\times C\subset B\times D.
\displaystyle \text{Hence verified.}

\displaystyle \text{(ii) }B\cap C=\{1,2,3,4\}\cap\{5,6\}=\phi.
\displaystyle \therefore A\times(B\cap C)=A\times\phi=\phi.\qquad\text{...(1)}
\displaystyle A\times B=\{(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4)\}.
\displaystyle A\times C=\{(1,5),(1,6),(2,5),(2,6)\}.
\displaystyle \therefore (A\times B)\cap(A\times C)=\phi.\qquad\text{...(2)}
\displaystyle \text{From (1) and (2),}
\displaystyle A\times(B\cap C)=(A\times B)\cap(A\times C).
\displaystyle \text{Hence verified.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{If }A=\{1,2,3\},\;B=\{3,4\}\text{ and }C=\{4,5,6\},\text{ find}
\displaystyle \text{(i) }A\times(B\cap C)\qquad\text{(ii) }(A\times B)\cap(A\times C)
\displaystyle \text{(iii) }A\times(B\cup C)\qquad\text{(iv) }(A\times B)\cup(A\times C).
\displaystyle \text{Answer:}

\displaystyle \text{(i) Given }A=\{1,2,3\},\;B=\{3,4\}\text{ and }C=\{4,5,6\}.
\displaystyle B\cap C=\{4\}.
\displaystyle \therefore A\times(B\cap C)=\{(1,4),(2,4),(3,4)\}.

\displaystyle \text{(ii) }A\times B=\{(1,3),(1,4),(2,3),(2,4),(3,3),(3,4)\}.
\displaystyle A\times C=\{(1,4),(1,5),(1,6),(2,4),(2,5),(2,6),(3,4),(3,5),(3,6)\}.
\displaystyle \therefore (A\times B)\cap(A\times C)=\{(1,4),(2,4),(3,4)\}.

\displaystyle \text{(iii) }B\cup C=\{3,4,5,6\}.
\displaystyle \therefore A\times(B\cup C)
\displaystyle =\{(1,3),(1,4),(1,5),(1,6),(2,3),(2,4),(2,5),(2,6),
\displaystyle (3,3),(3,4),(3,5),(3,6)\}.
\displaystyle \text{(iv) }A\times B=\{(1,3),(1,4),(2,3),(2,4),(3,3),(3,4)\}.
\displaystyle A\times C=\{(1,4),(1,5),(1,6),(2,4),(2,5),(2,6),(3,4),(3,5),(3,6)\}.
\displaystyle \therefore (A\times B)\cup(A\times C)
\displaystyle =\{(1,3),(1,4),(1,5),(1,6),(2,3),(2,4),(2,5),(2,6),
\displaystyle (3,3),(3,4),(3,5),(3,6)\}.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{Prove that:}
\displaystyle \text{(i) }(A\cup B)\times C=(A\times C)\cup(B\times C)
\displaystyle \text{(ii) }(A\cap B)\times C=(A\times C)\cap(B\times C).
\displaystyle \text{Answer:}

\displaystyle \text{(i) Let }(a,b)\in(A\cup B)\times C.
\displaystyle \Rightarrow a\in A\cup B\text{ and }b\in C
\displaystyle \Rightarrow (a\in A\text{ or }a\in B)\text{ and }b\in C
\displaystyle \Rightarrow (a\in A\text{ and }b\in C)\text{ or }(a\in B\text{ and }b\in C)
\displaystyle \Rightarrow (a,b)\in A\times C\text{ or }(a,b)\in B\times C
\displaystyle \Rightarrow (a,b)\in(A\times C)\cup(B\times C).
\displaystyle \therefore (A\cup B)\times C\subseteq(A\times C)\cup(B\times C).\qquad\text{...(1)}
\displaystyle \text{Conversely, let }(x,y)\in(A\times C)\cup(B\times C).
\displaystyle \Rightarrow (x,y)\in A\times C\text{ or }(x,y)\in B\times C
\displaystyle \Rightarrow (x\in A\text{ and }y\in C)\text{ or }(x\in B\text{ and }y\in C)
\displaystyle \Rightarrow (x\in A\text{ or }x\in B)\text{ and }y\in C
\displaystyle \Rightarrow x\in A\cup B\text{ and }y\in C
\displaystyle \Rightarrow (x,y)\in(A\cup B)\times C.
\displaystyle \therefore (A\times C)\cup(B\times C)\subseteq(A\cup B)\times C.\qquad\text{...(2)}
\displaystyle \text{From (1) and (2),}
\displaystyle (A\cup B)\times C=(A\times C)\cup(B\times C).
\displaystyle \text{Hence proved.}

\displaystyle \text{(ii) Let }(a,b)\in(A\cap B)\times C.
\displaystyle \Rightarrow a\in A\cap B\text{ and }b\in C
\displaystyle \Rightarrow (a\in A\text{ and }a\in B)\text{ and }b\in C
\displaystyle \Rightarrow (a\in A\text{ and }b\in C)\text{ and }(a\in B\text{ and }b\in C)
\displaystyle \Rightarrow (a,b)\in A\times C\text{ and }(a,b)\in B\times C
\displaystyle \Rightarrow (a,b)\in(A\times C)\cap(B\times C).
\displaystyle \therefore (A\cap B)\times C\subseteq(A\times C)\cap(B\times C).\qquad\text{...(1)}
\displaystyle \text{Conversely, let }(x,y)\in(A\times C)\cap(B\times C).
\displaystyle \Rightarrow (x,y)\in A\times C\text{ and }(x,y)\in B\times C
\displaystyle \Rightarrow (x\in A\text{ and }y\in C)\text{ and }(x\in B\text{ and }y\in C)
\displaystyle \Rightarrow (x\in A\text{ and }x\in B)\text{ and }y\in C
\displaystyle \Rightarrow x\in A\cap B\text{ and }y\in C
\displaystyle \Rightarrow (x,y)\in(A\cap B)\times C.
\displaystyle \therefore (A\times C)\cap(B\times C)\subseteq(A\cap B)\times C.\qquad\text{...(2)}
\displaystyle \text{From (1) and (2),}
\displaystyle (A\cap B)\times C=(A\times C)\cap(B\times C).
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{If }A\times B\subseteq C\times D\text{ and }A\times B\neq\phi,\text{ prove that}
\displaystyle A\subseteq C\text{ and }B\subseteq D.
\displaystyle \text{Answer:}
\displaystyle \text{Since }A\times B\neq\phi,\text{ both }A\text{ and }B\text{ are non-empty.}
\displaystyle \text{Let }a\in A.
\displaystyle \text{Since }B\neq\phi,\text{ choose }b\in B.
\displaystyle \therefore (a,b)\in A\times B.
\displaystyle \text{Since }A\times B\subseteq C\times D,
\displaystyle (a,b)\in C\times D.
\displaystyle \Rightarrow a\in C\text{ and }b\in D.
\displaystyle \therefore a\in A\Rightarrow a\in C.
\displaystyle \therefore A\subseteq C.
\displaystyle \text{Similarly, let }b\in B.
\displaystyle \text{Since }A\neq\phi,\text{ choose }a\in A.
\displaystyle \therefore (a,b)\in A\times B\subseteq C\times D.
\displaystyle \Rightarrow a\in C\text{ and }b\in D.
\displaystyle \therefore b\in B\Rightarrow b\in D.
\displaystyle \therefore B\subseteq D.
\displaystyle \text{Hence, }A\subseteq C\text{ and }B\subseteq D.
\displaystyle \\

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