Let \(A = \{x \in \mathbb{N}|\)\(2 \leq x \leq 3\}\), \(B = \{x \in \mathbb{W}|\)\(0 \leq x \leq 1\}\) and \(C = \{x \in \mathbb{N}|\)\(x \leq 2\}\). Then establish that \(A \times (B \cup C) = (A \times B) \cup (A \times C)\)
 
Answer:
 
To prove:
 
\(A \times (B \cup C) = (A \times B) \cup (A \times C)\)
 
Explanation:
 
\(B \cup C =\)
 
\(A \times (B \cup C) =\)
 
\(A \times B =\)
 
\(A \times C =\)
 
\((A \times B) \cup (A \times C) =\)
 
As a result, \(A \times (B \cup C) = (A \times B) \cup (A \times C)\)
 
Hence, we proved.
Answer variants:
(2,0),(2,1),(2,2),(3,0),(3,1),(3,2)
0,1,2
(2,1),(2,2),(3,1),(3,2)
(2,0),(2,1),(3,0),(3,1)