Theorem 1: For any three sets $A, B, C$, prove that:

i)  $A \times (B \cup C) = (A \times B) \cup (A \times C)$

(ii) $A \times (B \cap C) = (A \times B) \cap (A \times C)$.

Proof:

(i)  Let $(a, b)$ be an arbitrary element of $A \times (B \cup C)$. Then

$(a,b) \cup A \times (B \cup C)$

$\Rightarrow a \cup A \ and \ b \cup B \cup C$     [By Definition]

$\Rightarrow a \in A \ and \ (b \in B \ or \ b \in C)$     [By Definition of Union]

$\Rightarrow (a \in A \ and \ b \in B) \ or \ ( a \in A \ and \ b \in c)$

$\Rightarrow (a,b) \in A \times B \ or \ (a,b) \in A \times C$

$\Rightarrow (a,b) \in (A \times B) \cup (A \times C)$

$\therefore A \times (B \cup C) \subseteq (A \times B) \cup (A \times C)$   … … … … … i)

Now, let $(x, y)$ be an arbitrary element of $(A \times B) \cup (A \times C)$.

Then, $(x,y) \in (A \times B) \cup ( A \times C)$

$\Rightarrow (x,y) \in (A \times B) \ or \ (x,y) \in (A \times C)$

$\Rightarrow (x \in A \ and \ y \in B) \ or \ (x \in A \ and \ y \in C)$

$\Rightarrow x \in A \ and \ ( y \in B \ or \ y \in C)$

$\Rightarrow x \in A \ and \ y \in (B \cup C)$

$\Rightarrow (x, y) \in A \times (B \cup C)$

$\therefore ( A \times B) \cup ( A \times C) \subseteq A \times ( B \cup C)$  … … … … … ii)

Hence from i) and ii) we get $A \times (B \cup C) = (A \times B) \cup (A \times C)$

ii)   Let $(a, b)$ be an arbitrary element of $A \times (B \cap C)$. Then,

$(a,b) \in A \times (B \cap C)$

$\Rightarrow a \in A \ and \ b \in (B \cap C)$

$\Rightarrow a \in A \ and \ (b \in B \ and \ b \in C)$

$\Rightarrow (a \in A \ and \ b \in B) \ and \ (a \in A \ and \ b \in C)$

$\Rightarrow (a,b) \in A \times B \ and \ (a,b) \in A \times C$

$\Rightarrow (a,b) \in (A \times B) \cap (A \times C)$

$\therefore A \times (B \cap C) \subseteq (A \times B) \cap (A \times C)$  … … … … … i)

Let $(x, y)$ be an arbitrary element of $(A \times B) \cap (A \times C)$. Then,

$(x, y) \in (A \times B) \cap (A \times C)$

$\Rightarrow (x,y) \in (A \times B) \ and \ (x, y) \in A \times C$

$\Rightarrow (x \in A \ and \ y \in B) \ and \ (x \in A \ and \ y \in C)$

$\Rightarrow x \in A \ and \ (y \in B \ and \ y \in C)$

$\Rightarrow x \in A \ and \ y \in (B \cap C)$

$\Rightarrow (x,y) \in A \times ( B \cap C)$

$\therefore (A \times B) \cap (A \times C) \subseteq A \times (B \cap C)$  … … … … … ii)

Hence, from (i) and (ii), we get $A \times (B \cap C) = (A \times B) \cap (A \times C)$.

Theorem 2: For any three sets $A, B, C$, prove that: $A \times (B-C) = (A \times B)-(A \times C)$.

Proof:

Let $(a,b)$ be an arbitrary element of $A \times (B -C)$. Then

$(a,b) \in A \times (B-C)$

$\Rightarrow a \in A \ and \ b \in (B-C)$

$\Rightarrow a \in A \ and \ (b \in B \ and \ b \notin C)$

$\Rightarrow (a \in A \ and \ b \in B) \ and \ (a \in A \ and \ b \notin C)$

$(a,b) \in (A \times B) \ and \ (a,b) \notin (A \times C)$

$\Rightarrow (a,b) \in (A \times B) - (A \times C)$

$\therefore A \times (B-C) \subseteq (A \times B)- (A \times C)$  … … … … … i)

Again, let $(x,y)$ be an arbitrary element of $(A \times B) -(A \times C)$. Then,

$(x, y) \in (A \times B) - (A \times C)$

$(x,y) \in A \times B \ and \ (x,y) \notin A \times C$

$\Rightarrow (x \in A \ and \ y \in B) \ and \ (x \in A \ and \ y \notin C)$

$\Rightarrow x \in A \ and \ (y \in B \ and \ y \notin C)$

$\Rightarrow x \in A \ and \ y \in (B -C)$

$\Rightarrow (x,y) \in A \times (B -C)$

$\therefore (A \times B)-(A \times C) \subseteq A \times (B-C)$  … … … … … ii)

Hence, from (i) and (ii), we get

$A \times (B -C) = (A \times B) -(A \times C)$

Theorem 3: If $A$ and $B$ are any two non-empty sets, then prove that: $A \times B=B \times A \Leftrightarrow A=B$.

Proof:

First, let $A = B$. Then we have to prove that $A \times B=B \times A$.

$A=B \Rightarrow A \times B=A \times A$ and, $B \times A=A \times A \Rightarrow A \times B = B \times A$

Conversely, let $A \times B = B \times A$. Then we have to Prove that $A = B$.

Let $x$ be an arbitrary element of $A$. Then,

$x \in A + (x,b) \in A \times B$ for all $b \in B \Rightarrow (x,b) \in B \times A$     $[ \because A \times B = B \times A ]$

$\Rightarrow x \in B$

$\therefore A \subseteq B$

Again, let $y$ be an arbitrary element of $B$. Then,

$y \in B \Rightarrow (a,y) \in A \times B$ for all $a \in A \Rightarrow (a,y) \in B \times A$    $[ \because A \times B = B \times A ]$

$\Rightarrow y \in A$

$\therefore B \subseteq A$

Hence, $A = B$.

Theorem 4: If $A \in B$, show that $A \times A \subseteq (A \times B ) \cap (B \times A)$.

Proof: Let $(a, b)$ be an arbitrary element of $A \times A$

Then, $(a, b) \in A \times A$

$\Rightarrow a \in A \ and \ b \in A$

$\Rightarrow (a \in A, b \in A) \ and \ (a \in A, b \in A)$

$\Rightarrow a \in A, b \in B) \ and \ (a \in B, b \in A)$

$\Rightarrow (a,b) \in (A \times B) \ and \ (a,b) \in (B \times A)$     $[ \because A \subseteq B \therefore a, b \in A \Rightarrow a, b \in B ]$

$\Rightarrow (a,b) \in (A \times B) \cap (B \times A)$

$\therefore A \times A \subseteq (A \times B) \cap (B \times A)$

Hence, $A \subset B \Rightarrow A \times A \subseteq (A \times B) \cap (B \times A)$

Theorem 5: If $A \subseteq B$, prove that: $A \times C \subseteq B \times C$ for any set $C$.
Proof:  Let $(a, b)$ be an arbitrary element of $A \times C$.

Then, $(a,b) \in A \times C$

$\Rightarrow a \in A \ and \ b \in C$

$\Rightarrow a \in B \ and \ b \in C$     $[ \because A \subseteq B \therefore a \in A \Rightarrow a \in B ]$

$\Rightarrow (a,b) \in B \times C$

$\therefore A \times C \subseteq B \times C$

Theorem 6:  If $A \subseteq B$ and $C \subseteq D$, prove that: $A \times C \subseteq B \times D$

Proof:  Let $(a,b)$ be an arbitrary element of $A \times C$,

Then $(a,b) \in A \times C$

$a \in A \ and \ b \in C$

$a \in B \ and \ b \in D$     $[\because A \subseteq B \ and \ C \subseteq D ]$

$(a,b) \in B \times D$

$\therefore A \times C \subseteq B \times D$

Theorem 7: For any sets $A, B,C, D$ prove that: $(A \times B) \cap (C \times D) =(A \cap C) \times (B \cap D)$.

Proof:

Let $(a,b)$ be an arbitrary element of $(A \times B) \cap (C \times D)$. Then,

$(a,b) \in (A \times B) \cap (C \times D)$

$\Rightarrow (a,b) \in A \times B \ and \ (a,b) \in C \times D$

$\Rightarrow (a \in A \ and \ b \in B) \ and \ (a \in C \ and \ b \in D)$

$\Rightarrow (a \in A \ and \ a \in C) \ and \ (b \in B \ and \ b \in D)$

$\Rightarrow a \in (A \cap C) \ and \ b \in B \cap D$

$\Rightarrow (a,b) \in (A \cap C) \times (B \cap D)$

$\therefore (A \times B) \cap (C \times D) \subseteq (A \cap C) \times (B \cap D )$

Similarly, $(A \cap C) \times (B \cap D) \in (A \times B) \cap (C \times D)$

Hence, $(A \times B) \cap (C \times D) = (A \cap C) \times (B \cap D)$

Theorem 8: For any three sets $A, B, C$ prove that:

(i) $A \times (B' \cup C')'= (A \times B) \cap (A \times C)$

ii) $A \times (B' \cap C')' = (A \times B) \cup (A \times C)$

Proof:

i)  $A \times (B' \cup C')' = A \times ((B')' \cap (C')') = A \times (B \cap C) = (A \times B) \cap (A \times C)$         [ By De-Morgan’s law ]

ii) $A \times (B' \cap C')' = A \times (B')' \cup (C')' = A \times (B \cup C) = ( A \times B) \cup (A \times C)$         [ By De-Morgan’s law ]

Theorem 9:  Let $A$ and $B$ be two non-empty sets having $n$ elements in common, then prove that $A \times B$ and $B \times A$ have $n^2$ elements in common.

Proof:

We have, $(A \times B ) \cap (C x D) = (A \cap C) \times (B \cap D)$

$\Rightarrow (A \times B) \cap (B \times A) = (A \cap B) \times (B \cap A)$

$\Rightarrow (A \times B) \cap (B \times A) = ( A \cap B) \times (A \cap B)$    [On replacing $C$ by $B$ and $D$ by $A$ ]

It is given that $A \cap B has$latex n elements, so $(A \cap B) \times (B \cap A)$ has $n^2$ elements.

But $(A \times B) \cap (B \times A) = (A \cap B) \times (A \cap B)$

$(A \times B) \cap (B \times A)$ has $n^2$ elements

Hence $A \times B \ and \ B \times A$ have $n^2$ elements in common

Theorem 10: Let $A$ be a non-empty set such that $A \times B=A \times C$. Show that $B=C$.

Proof:

Let $b$ be an arbitrary element of $B$.

Then, $(a,b) \in A \times B$ for all $a \in A$

$\Rightarrow (a,b) \in A \times C$ for all $a \in A$

$\Rightarrow b \in C$

Thus, $b \in B \Rightarrow b \in C$

$\therefore B \subset C$

Let $c$ be an arbitrary element of $C$.

Then, $(a,c) \in A \times C$ for all $a \in A$

$\Rightarrow (a,c) \in A \times B$ for all $a \in A$

$\Rightarrow c \in B$

$c \in C \Rightarrow c \in B$

$\therefore C \subset B$

From i) and ii) we get $B = C$.