Question

Prove that:

(i)  (A ∪ B) × C = (A × C) ∪ (B × C)

(ii) (A ∩ B) × C = (A × C) ∩ (B×C)

 

Answer 1

(i) (A ∪ B) × C = (A × C) ∪ (B × C)
Let (a, b) be an arbitrary element of (A ∪ B) × C.
Thus, we have: \[(a, b) \in (A \cup B) \times C \]
\[ \Rightarrow a \in (A \cup B) \text{ and } b \in C \]
\[ \Rightarrow (a \in \text{ A or }a \in B) \text{ and } b \in C\]
\[ \Rightarrow (a \in A \text{ and }  b \in C ) \text{ or } (a \in B \text{ and } b \in C )\]
\[ \Rightarrow (a, b) \in (A \times C )\text{  or }  (a, b) \in (B \times C)\]
\[ \Rightarrow (a, b) \in (A \times C) \cup (B \times C )\]
\[ \therefore (A \cup B) \times C \subseteq (A \times C) \cup (B \times C)      . . . (i)\]

Again, let (x, y) be an arbitrary element of (A × C ) ∪ (B × C).
Thus, we have:

\[(x, y) \in (A \times C) \cup (B \times C)\]
\[ \Rightarrow (x, y) \text{ in }  (A \times C) \text{ or } (x, y) \in (B \times C) \]
\[ \Rightarrow (x \in A  y \in C) \text{ or } (x \in B  y \in C)\]
\[ \Rightarrow (x \in A \text{ or } x \in B) \text{ or } y \in C\]
\[ \Rightarrow (x \in A \cup B)   y \in C\]
\[ \Rightarrow (x, y) \in (A \cup B) \times C\]
\[ \therefore (A \times C) \cup (B \times C) \subseteq (A \cup B) \times C . . . (ii)\]

From (i) and (ii), we get:
(A ∪ B) × C = (A × C) ∪ (B × C)

(ii) (A ∩ B) × C = (A × C) ∩ (B×C)
Let (a, b) be an arbitrary element of (A ∩ B) × C.
Thus, we have:

\[ (a, b) \in (A \cap B) \times C\]
\[ \Rightarrow a \in (A \cap B)  b \in C\]
\[ \Rightarrow (a \in A  a \in B) b \in C\]
\[ \Rightarrow (a \in A  b \in C)  (a \in B  b \in C)\]
\[ \Rightarrow (a, b) \in (A \times C)  (a, b) \in (B \times C) \]
\[ \Rightarrow (a, b) \in (A \times C) \cap (B \times C) \]
\[ \therefore (A \cap B) \times C \subseteq (A \times C) \cap (B \times C) . . . \left( iii \right)\]

Again, let (x, y) be an arbitrary element of (A × C) ∩ (B × C).
Thus, we have:

\[(x, y) \in (A \times C) \cap (B \times C)\]
\[ \Rightarrow (x, y) \in (A \times C) (x, y) \in (B \times C)\]
\[ \Rightarrow (x \in A  y \in C)  (x \in B  y \in C)\]
\[ \Rightarrow (x \in A  x \in B)  y \in C\]
\[ \Rightarrow x \in (A \cap B)  y \in C\]
\[ \Rightarrow (x, y) \in (A \cap B) \times C\]
\[ \therefore (A \times C) \cap (B \times C) \subseteq (A \cap B) \times C . . . \left( iv \right)\]

From (iii) and (iv), we get:
 (A ∩ B) × C = (A × C) ∩ (B × C)