\displaystyle \textbf{Theorem 1: }\text{For any three sets }A,\ B\text{ and }C,\text{ prove 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 \textbf{Proof:}
\displaystyle \text{(i) Let }(a,b)\text{ be an arbitrary element of }A\times(B\cup C).
\displaystyle (a,b)\in A\times(B\cup C)
\displaystyle \Rightarrow a\in A\text{ and }b\in B\cup C
\displaystyle \Rightarrow a\in A\text{ and }(b\in B\text{ or }b\in C)
\displaystyle \Rightarrow (a\in A\text{ and }b\in B)\text{ or }(a\in A\text{ and }b\in C)
\displaystyle \Rightarrow (a,b)\in A\times B\text{ or }(a,b)\in A\times C
\displaystyle \Rightarrow (a,b)\in(A\times B)\cup(A\times C)
\displaystyle \therefore A\times(B\cup C)\subseteq(A\times B)\cup(A\times C).\qquad\text{...(1)}
\displaystyle \text{Now, let }(x,y)\text{ be an arbitrary element of }(A\times B)\cup(A\times C).
\displaystyle (x,y)\in(A\times B)\cup(A\times C)
\displaystyle \Rightarrow (x,y)\in A\times B\text{ or }(x,y)\in A\times C
\displaystyle \Rightarrow (x\in A\text{ and }y\in B)\text{ or }(x\in A\text{ and }y\in C)
\displaystyle \Rightarrow x\in A\text{ and }(y\in B\text{ or }y\in C)
\displaystyle \Rightarrow x\in A\text{ and }y\in B\cup C
\displaystyle \Rightarrow (x,y)\in A\times(B\cup C)
\displaystyle \therefore (A\times B)\cup(A\times C)\subseteq A\times(B\cup C).\qquad\text{...(2)}
\displaystyle \text{Hence, from (1) and (2),}
\displaystyle A\times(B\cup C)=(A\times B)\cup(A\times C).
\displaystyle \\

\displaystyle \text{(ii) Let }(a,b)\text{ be an arbitrary element of }A\times(B\cap C).
\displaystyle (a,b)\in A\times(B\cap C)
\displaystyle \Rightarrow a\in A\text{ and }b\in B\cap C
\displaystyle \Rightarrow a\in A\text{ and }(b\in B\text{ and }b\in C)
\displaystyle \Rightarrow (a\in A\text{ and }b\in B)\text{ and }(a\in A\text{ and }b\in C)
\displaystyle \Rightarrow (a,b)\in A\times B\text{ and }(a,b)\in A\times C
\displaystyle \Rightarrow (a,b)\in(A\times B)\cap(A\times C)
\displaystyle \therefore A\times(B\cap C)\subseteq(A\times B)\cap(A\times C).\qquad\text{...(1)}
\displaystyle \text{Now, let }(x,y)\text{ be an arbitrary element of }(A\times B)\cap(A\times C).
\displaystyle (x,y)\in(A\times B)\cap(A\times C)
\displaystyle \Rightarrow (x,y)\in A\times B\text{ and }(x,y)\in A\times C
\displaystyle \Rightarrow (x\in A\text{ and }y\in B)\text{ and }(x\in A\text{ and }y\in C)
\displaystyle \Rightarrow x\in A\text{ and }(y\in B\text{ and }y\in C)
\displaystyle \Rightarrow x\in A\text{ and }y\in B\cap C
\displaystyle \Rightarrow (x,y)\in A\times(B\cap C)
\displaystyle \therefore (A\times B)\cap(A\times C)\subseteq A\times(B\cap C).\qquad\text{...(2)}
\displaystyle \text{Hence, from (1) and (2),}
\displaystyle A\times(B\cap C)=(A\times B)\cap(A\times C).
\displaystyle \\

\displaystyle \textbf{Theorem 2: }\text{For any three sets }A,\ B\text{ and }C,\text{ prove that:}
\displaystyle A\times(B-C)=(A\times B)-(A\times C).
\displaystyle \textbf{Proof:}
\displaystyle \text{Let }(a,b)\text{ be an arbitrary element of }A\times(B-C).
\displaystyle (a,b)\in A\times(B-C)
\displaystyle \Rightarrow a\in A\text{ and }b\in B-C
\displaystyle \Rightarrow a\in A\text{ and }(b\in B\text{ and }b\notin C)
\displaystyle \Rightarrow (a\in A\text{ and }b\in B)\text{ and }b\notin C
\displaystyle \Rightarrow (a,b)\in A\times B\text{ and }(a,b)\notin A\times C
\displaystyle \Rightarrow (a,b)\in(A\times B)-(A\times C)
\displaystyle \therefore A\times(B-C)\subseteq(A\times B)-(A\times C).\qquad\text{...(1)}
\displaystyle \text{Again, let }(x,y)\text{ be an arbitrary element of }(A\times B)-(A\times C).
\displaystyle (x,y)\in(A\times B)-(A\times C)
\displaystyle \Rightarrow (x,y)\in A\times B\text{ and }(x,y)\notin A\times C
\displaystyle \Rightarrow x\in A,\ y\in B\text{ and }(x,y)\notin A\times C
\displaystyle \text{Since }x\in A\text{ and }(x,y)\notin A\times C,\text{ we must have }y\notin C.
\displaystyle \therefore x\in A\text{ and }(y\in B\text{ and }y\notin C)
\displaystyle \Rightarrow x\in A\text{ and }y\in B-C
\displaystyle \Rightarrow (x,y)\in A\times(B-C)
\displaystyle \therefore (A\times B)-(A\times C)\subseteq A\times(B-C).\qquad\text{...(2)}
\displaystyle \text{Hence, from (1) and (2),}
\displaystyle A\times(B-C)=(A\times B)-(A\times C).
\displaystyle \\

\displaystyle \textbf{Theorem 3: }\text{If }A\text{ and }B\text{ are any two non-empty sets, prove that:}
\displaystyle A\times B=B\times A\Longleftrightarrow A=B.
\displaystyle \textbf{Proof:}
\displaystyle \text{First, let }A=B.
\displaystyle \text{Then }A\times B=A\times A\text{ and }B\times A=A\times A.
\displaystyle \therefore A\times B=B\times A.
\displaystyle \text{Conversely, let }A\times B=B\times A.
\displaystyle \text{We shall prove that }A=B.
\displaystyle \text{Let }x\text{ be an arbitrary element of }A.
\displaystyle \text{Since }B\neq\varnothing,\text{ choose an element }b\in B.
\displaystyle x\in A\text{ and }b\in B
\displaystyle \Rightarrow (x,b)\in A\times B
\displaystyle \Rightarrow (x,b)\in B\times A\qquad[\because A\times B=B\times A]
\displaystyle \Rightarrow x\in B.
\displaystyle \therefore A\subseteq B.\qquad\text{...(1)}
\displaystyle \text{Again, let }y\text{ be an arbitrary element of }B.
\displaystyle \text{Since }A\neq\varnothing,\text{ choose an element }a\in A.
\displaystyle a\in A\text{ and }y\in B
\displaystyle \Rightarrow (a,y)\in A\times B
\displaystyle \Rightarrow (a,y)\in B\times A\qquad[\because A\times B=B\times A]
\displaystyle \Rightarrow y\in A.
\displaystyle \therefore B\subseteq A.\qquad\text{...(2)}
\displaystyle \text{Hence, from (1) and (2), }A=B.
\displaystyle \therefore A\times B=B\times A\Longleftrightarrow A=B.
\displaystyle \\

\displaystyle \textbf{Theorem 4: }\text{If }A\subseteq B,\text{ show that}
\displaystyle A\times A\subseteq(A\times B)\cap(B\times A).
\displaystyle \textbf{Proof:}
\displaystyle \text{Let }(a,b)\text{ be an arbitrary element of }A\times A.
\displaystyle (a,b)\in A\times A
\displaystyle \Rightarrow a\in A\text{ and }b\in A.
\displaystyle \text{Since }A\subseteq B,\text{ we also have }a\in B\text{ and }b\in B.
\displaystyle \therefore a\in A,\ b\in B\text{ and }a\in B,\ b\in A.
\displaystyle \Rightarrow (a,b)\in A\times B\text{ and }(a,b)\in B\times A
\displaystyle \Rightarrow (a,b)\in(A\times B)\cap(B\times A).
\displaystyle \therefore A\times A\subseteq(A\times B)\cap(B\times A).
\displaystyle \\

\displaystyle \textbf{Theorem 5: }\text{If }A\subseteq B,\text{ prove that }A\times C\subseteq B\times C
\displaystyle \text{for any set }C.
\displaystyle \textbf{Proof:}
\displaystyle \text{Let }(a,c)\text{ be an arbitrary element of }A\times C.
\displaystyle (a,c)\in A\times C
\displaystyle \Rightarrow a\in A\text{ and }c\in C.
\displaystyle \text{Since }A\subseteq B,\ a\in A\Rightarrow a\in B.
\displaystyle \therefore a\in B\text{ and }c\in C
\displaystyle \Rightarrow (a,c)\in B\times C.
\displaystyle \therefore A\times C\subseteq B\times C.
\displaystyle \\

\displaystyle \textbf{Theorem 6: }\text{If }A\subseteq B\text{ and }C\subseteq D,\text{ prove that}
\displaystyle A\times C\subseteq B\times D.
\displaystyle \textbf{Proof:}
\displaystyle \text{Let }(a,c)\text{ be an arbitrary element of }A\times C.
\displaystyle (a,c)\in A\times C
\displaystyle \Rightarrow a\in A\text{ and }c\in C.
\displaystyle \text{Since }A\subseteq B\text{ and }C\subseteq D,
\displaystyle a\in A\Rightarrow a\in B\text{ and }c\in C\Rightarrow c\in D.
\displaystyle \therefore a\in B\text{ and }c\in D
\displaystyle \Rightarrow (a,c)\in B\times D.
\displaystyle \therefore A\times C\subseteq B\times D.
\displaystyle \\

\displaystyle \textbf{Theorem 7: }\text{For any sets }A,\ B,\ C\text{ and }D,\text{ prove that}
\displaystyle (A\times B)\cap(C\times D)=(A\cap C)\times(B\cap D).
\displaystyle \textbf{Proof:}
\displaystyle \text{Let }(a,b)\text{ be an arbitrary element of }(A\times B)\cap(C\times D).
\displaystyle (a,b)\in(A\times B)\cap(C\times D)
\displaystyle \Rightarrow (a,b)\in A\times B\text{ and }(a,b)\in C\times D
\displaystyle \Rightarrow (a\in A\text{ and }b\in B)\text{ and }(a\in C\text{ and }b\in D)
\displaystyle \Rightarrow (a\in A\cap C)\text{ and }(b\in B\cap D)
\displaystyle \Rightarrow (a,b)\in(A\cap C)\times(B\cap D)
\displaystyle \therefore (A\times B)\cap(C\times D)\subseteq(A\cap C)\times(B\cap D).\qquad\text{...(1)}
\displaystyle \text{Now, let }(x,y)\text{ be an arbitrary element of }(A\cap C)\times(B\cap D).
\displaystyle (x,y)\in(A\cap C)\times(B\cap D)
\displaystyle \Rightarrow x\in A\cap C\text{ and }y\in B\cap D
\displaystyle \Rightarrow (x\in A\text{ and }x\in C)\text{ and }(y\in B\text{ and }y\in D)
\displaystyle \Rightarrow (x,y)\in A\times B\text{ and }(x,y)\in C\times D
\displaystyle \Rightarrow (x,y)\in(A\times B)\cap(C\times D)
\displaystyle \therefore (A\cap C)\times(B\cap D)\subseteq(A\times B)\cap(C\times D).\qquad\text{...(2)}
\displaystyle \text{Hence, from (1) and (2),}
\displaystyle (A\times B)\cap(C\times D)=(A\cap C)\times(B\cap D).
\displaystyle \\  

\displaystyle \textbf{Theorem 8: }\text{For any three sets }A,\ B\text{ and }C,\text{ prove that:}
\displaystyle \text{(i) }A\times(B'\cup C')'=(A\times B)\cap(A\times C)
\displaystyle \text{(ii) }A\times(B'\cap C')'=(A\times B)\cup(A\times C).
\displaystyle \textbf{Proof:}
\displaystyle \text{(i) }A\times(B'\cup C')'
\displaystyle =A\times\big((B')'\cap(C')'\big)\qquad[\text{By De Morgan's law}]
\displaystyle =A\times(B\cap C)
\displaystyle =(A\times B)\cap(A\times C).
\displaystyle \text{(ii) }A\times(B'\cap C')'
\displaystyle =A\times\big((B')'\cup(C')'\big)\qquad[\text{By De Morgan's law}]
\displaystyle =A\times(B\cup C)
\displaystyle =(A\times B)\cup(A\times C).
\displaystyle \\

\displaystyle \textbf{Theorem 9: }\text{Let }A\text{ and }B\text{ be two sets having }n\text{ elements in}
\displaystyle \text{common. Prove that }A\times B\text{ and }B\times A\text{ have }n^2\text{ elements in common.}
\displaystyle \textbf{Proof:}
\displaystyle \text{We know that}
\displaystyle (A\times B)\cap(C\times D)=(A\cap C)\times(B\cap D).
\displaystyle \text{Replacing }C\text{ by }B\text{ and }D\text{ by }A,\text{ we get}
\displaystyle (A\times B)\cap(B\times A)=(A\cap B)\times(B\cap A)
\displaystyle =(A\cap B)\times(A\cap B).
\displaystyle \text{Since }A\text{ and }B\text{ have }n\text{ elements in common,}
\displaystyle n(A\cap B)=n.
\displaystyle \therefore n\big((A\times B)\cap(B\times A)\big)
\displaystyle =n\big((A\cap B)\times(A\cap B)\big)
\displaystyle =n(A\cap B)\cdot n(A\cap B)
\displaystyle =n\cdot n=n^2.
\displaystyle \therefore A\times B\text{ and }B\times A\text{ have }n^2\text{ elements in common.}
\displaystyle \\

\displaystyle \textbf{Theorem 10: }\text{Let }A\text{ be a non-empty set such that }A\times B=A\times C.
\displaystyle \text{Show that }B=C.
\displaystyle \textbf{Proof:}
\displaystyle \text{Let }b\text{ be an arbitrary element of }B.
\displaystyle \text{Since }A\neq\varnothing,\text{ choose an element }a\in A.
\displaystyle b\in B\Rightarrow(a,b)\in A\times B
\displaystyle \Rightarrow(a,b)\in A\times C\qquad[\because A\times B=A\times C]
\displaystyle \Rightarrow b\in C.
\displaystyle \therefore B\subseteq C.\qquad\text{...(1)}
\displaystyle \text{Now, let }c\text{ be an arbitrary element of }C.
\displaystyle \text{Choose the same }a\in A.
\displaystyle c\in C\Rightarrow(a,c)\in A\times C
\displaystyle \Rightarrow(a,c)\in A\times B\qquad[\because A\times B=A\times C]
\displaystyle \Rightarrow c\in B.
\displaystyle \therefore C\subseteq B.\qquad\text{...(2)}
\displaystyle \text{Hence, from (1) and (2), }B=C.
\displaystyle \\

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