Q.The Cartesian product $A \times A$ has 9 elements among which are found $(-1, 0)$ and $(0, 1)$. Find the set $A$ and the remaining elements of $A \times A$.

Cartesian Product: From Intuition to Definition

Imagine you're ordering a pizza. You have two choices to make: the size (Small, Medium, Large) and the topping (Cheese, Pepperoni, Veggie). How many different pizzas can you order?

You can pair each size with each topping:

• Small + Cheese, Small + Pepperoni, Small + Veggie
• Medium + Cheese, Medium + Pepperoni, Medium + Veggie
• Large + Cheese, Large + Pepperoni, Large + Veggie

That's $3 \times 3 = 9$ possible pizzas. What you just did — systematically pairing every element of one set with every element of another — is the Cartesian product in action.

The Intuition

The Cartesian product is a way to combine two sets to create a new set of ordered pairs. The order matters: (Small, Cheese) is different from (Cheese, Small) — one is a pizza order, the other is nonsense.

Think of it like a multiplication table for sets. If set $A$ has $m$ items and set $B$ has $n$ items, their Cartesian product has $m \times n$ items.

The Precise Definition

Let $A$ and $B$ be two sets. The Cartesian product of $A$ and $B$, written $A \times B$, is the set of all ordered pairs $(a, b)$ where $a$ is from $A$ and $b$ is from $B$.

The vertical bar means "such that." So read it as: "The set of all ordered pairs (a, b) such that a belongs to A and b belongs to B."

Key Properties to Remember

1. Order matters: $A \times B$ is generally not the same as $B \times A$. For example, if $A = \{1, 2\}$ and $B = \{x, y\}$:

   • $A \times B = \{(1, x), (1, y), (2, x), (2, y)\}$
   • $B \times A = \{(x, 1), (x, 2), (y, 1), (y, 2)\}$

   These are different sets because the pairs are ordered differently.

2. Size formula: If $|A| = m$ and $|B| = n$, then $|A \times B| = m \times n$. This holds even if one set is empty — then the product is empty.

3. Empty set: $A \times \emptyset = \emptyset$ and $\emptyset \times B = \emptyset$. You can't form any pairs if one set has nothing to contribute.

Examples to Cement the Idea

Example 1: $A = \{1, 2\}$, $B = \{3, 4\}$

Four pairs, as expected ($2 \times 2 = 4$).

Example 2: $A = \{a\}$, $B = \{1, 2, 3\}$

Three pairs — every element of $B$ gets paired with the single element of $A$.

Example 3: $A = \{0, 1\}$, $B = \{0, 1\}$

This is the set of all possible 2-bit binary strings — a foundation for computer science.

Why This Matters

The Cartesian product is the mathematical backbone of: …