Exercise 3.1 — Ideal Textbook Solutions

Two ordered pairs are equal if and only if their corresponding coordinates are equal.

Equate the first components:
$$x - 5 = 4 \Rightarrow x = 4 + 5 \Rightarrow \mathbf{x = 9}$$

Equate the second components:
$$y + 1 = 6 \Rightarrow y = 6 - 1 \Rightarrow \mathbf{y = 5}$$

Therefore, the values are **$x = 9$** and **$y = 5$**.

1. Listing $A \times B$:
Map each element of $A = \{1, 2\}$ to each element of $B = \{2, 3, 5\}$: $$\mathbf{A \times B = \{(1, 2), (1, 3), (1, 5), (2, 2), (2, 3), (2, 5)\}}$$

2. Listing $B \times A$:
Map each element of $B$ to each element of $A$: $$\mathbf{B \times A = \{(2, 1), (2, 2), (3, 1), (3, 2), (5, 1), (5, 2)\}}$$

We know that the cardinality of a Cartesian product is given by the formula: $$n(A \times B) = n(A) \times n(B)$$

Substitute the given values into the equation:
$$20 = 4 \times n(B)$$

Solve for $n(B)$:
$$n(B) = \frac{20}{4} = \mathbf{5}$$

(a) $A \times B$:
$$\mathbf{A \times B = \{(1, x), (1, y), (2, x), (2, y), (3, x), (3, y)\}}$$

(b) $B \times A$:
$$\mathbf{B \times A = \{(x, 1), (x, 2), (x, 3), (y, 1), (y, 2), (y, 3)\}}$$

(c) $A \times A$:
$$\mathbf{A \times A = \{(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)\}}$$

(d) $B \times B$:
$$\mathbf{B \times B = \{(x, x), (x, y), (y, x), (y, y)\}}$$

Method 1: Listing elements
- $A \times B = \{(1,2), (1,3), (1,7), (2,2), (2,3), (2,7), (3,2), (3,3), (3,7)\}$
- $B \times A = \{(2,1), (2,2), (2,3), (3,1), (3,2), (3,3), (7,1), (7,2), (7,3)\}$
- Common pairs: $\mathbf{\{(2, 2), (2, 3), (3, 2), (3, 3)\}}$.

Method 2: Set Identity (Algebraic Proof)
We know that:
$$(A \times B) \cap (B \times A) = (A \cap B) \times (B \cap A)$$
First, find the intersection of the two sets $A$ and $B$:
$$A \cap B = \{1, 2, 3\} \cap \{2, 3, 7\} = \{2, 3\}$$
Now, find the Cartesian product of the intersection with itself:
$$(A \cap B) \times (A \cap B) = \{2, 3\} \times \{2, 3\} = \mathbf{\{(2, 2), (2, 3), (3, 2), (3, 3)\}}.$$

Step 1: Compute LHS $A \times (B \cup C)$
- Find $B \cup C$: $\{2, 3\} \cup \{4, 5\} = \{2, 3, 4, 5\}$.
- $A \times (B \cup C) = \{1, 2\} \times \{2, 3, 4, 5\}$: $$\text{LHS} = \mathbf{\{(1, 2), (1, 3), (1, 4), (1, 5), (2, 2), (2, 3), (2, 4), (2, 5)\}}$$

Step 2: Compute RHS $(A \times B) \cup (A \times C)$
- Find $A \times B$: $\{(1, 2), (1, 3), (2, 2), (2, 3)\}$.
- Find $A \times C$: $\{(1, 4), (1, 5), (2, 4), (2, 5)\}$.
- Union: $$\text{RHS} = \mathbf{\{(1, 2), (1, 3), (2, 2), (2, 3), (1, 4), (1, 5), (2, 4), (2, 5)\}}$$

Since LHS = RHS, the distributive property of Cartesian products over union is verified.