1. **Problem Statement:** Prove that for the sets $$A = \{a \mid a^2 - 1 < 10, a \in \mathbb{Z}\},\quad B = \{b \mid b - 1 < 2, b \in \mathbb{N}\},\quad C = \{c \mid |c| \leq 1, c \in \mathbb{Z}\},$$ the equality $$A \times (B \cap C) = (A \times B) \cap (A \times C)$$ holds. 2. **Recall the Cartesian product and intersection definitions:** - Cartesian product: $$X \times Y = \{(x,y) \mid x \in X, y \in Y\}$$ - Intersection: $$B \cap C = \{x \mid x \in B \text{ and } x \in C\}$$ 3. **Step 1: Find the sets explicitly.** - For $A$: Solve $a^2 - 1 < 10 \Rightarrow a^2 < 11$. Since $a \in \mathbb{Z}$, possible $a$ values satisfy $-3 \leq a \leq 3$. So, $$A = \{-3, -2, -1, 0, 1, 2, 3\}.$$ - For $B$: Solve $b - 1 < 2 \Rightarrow b < 3$ with $b \in \mathbb{N}$. Since $\mathbb{N} = \{1, 2, 3, \ldots\}$, possible $b$ values are $1$ and $2$. So, $$B = \{1, 2\}.$$ - For $C$: $|c| \leq 1$ with $c \in \mathbb{Z}$ means $c \in \{-1, 0, 1\}$. So, $$C = \{-1, 0, 1\}.$$ 4. **Step 2: Find $B \cap C$.** $$B \cap C = \{x \mid x \in B \text{ and } x \in C\} = \{1, 2\} \cap \{-1, 0, 1\} = \{1\}.$$ 5. **Step 3: Compute $A \times (B \cap C)$.** $$A \times (B \cap C) = A \times \{1\} = \{(a,1) \mid a \in A\} = \{(-3,1), (-2,1), (-1,1), (0,1), (1,1), (2,1), (3,1)\}.$$ 6. **Step 4: Compute $(A \times B) \cap (A \times C)$.** - First, $$A \times B = \{(a,b) \mid a \in A, b \in B\} = \{(a,1), (a,2) \mid a \in A\}.$$ - Second, $$A \times C = \{(a,c) \mid a \in A, c \in C\} = \{(a,-1), (a,0), (a,1) \mid a \in A\}.$$ - Their intersection is $$ (A \times B) \cap (A \times C) = \{(a,x) \mid a \in A, x \in B \cap C\} = A \times (B \cap C).$$ 7. **Conclusion:** We have shown that $$A \times (B \cap C) = (A \times B) \cap (A \times C),$$ which completes the proof. **Final answer:** The equality holds true for the given sets.