Mathematical-Reasoning · Unit 18 · Video 2 · Interactive Practice

The Cartesian Product A × B: Listing Pairs, Testing Laws

IKey Formulas

Statement	Name
$(x_1, y_1) = (x_2, y_2) \iff x_1 = x_2 \text{ and } y_1 = y_2$	Equality of ordered pairs
$A \times B = \{(x, y) \mid x \in A \text{ and } y \in B\}$ , $A^2 = A \times A$	Cartesian product
$(x, y) \notin C \times D \iff x \notin C \text{ or } y \notin D$	Negated membership (De Morgan)
$A \times (B \cap C) = (A \times B) \cap (A \times C)$ , and likewise with $\cup$ and $-$	Observed for $A = \{1, 2, 3\}$ , $B = \{a, b\}$ , $C = \{a, c\}$

Key Insight: Order matters: $(1, 2) \neq (2, 1)$ , and $B \times A \neq A \times B$ for $A = \{1, 2, 3\}$ , $B = \{a, b\}$ , even though both products have $3 \cdot 2 = 6$ elements.

IIMembership, One Coordinate at a Time

A pair lies in a product exactly when each coordinate belongs to its own factor.

Product

IIIProducts with $\cap$ , $\cup$ and $-$

Does $A \times (B \cap C)$ , $A \times (B \cup C)$ or $A \times (B - C)$ match the same operation on $A \times B$ and $A \times C$ ?

Operation

💡 Agreement for every choice of $A$ , $B$ and $C$ tried here is evidence; that each law holds for all sets is a claim that needs a proof.

IVQuiz Questions

Problem 1 · Build a Product

Given: $P = \{0, 1\}$ and $Q = \{x, y, z\}$ — find an element of $Q \times P$ and the number of elements of $Q \times P$ .

Which pair is an element of $Q \times P$ ?

How many elements does $Q \times P$ have?

✅ Correct! In

Q \times P

the first coordinate comes from

Q

and the second from

P

; there are

3 \cdot 2 = 6

such pairs.

❌ Order matters.

(0, x)

has first coordinate

0 \notin Q

, so it lies in

P \times Q

, not

Q \times P

❌ Not quite. The second coordinate

y

must be in

P = \{0, 1\}

, and

y \notin P

;

(x, y)

belongs to

Q \times Q

❌ Not quite. The first coordinate

1

must be in

Q = \{x, y, z\}

, and

1 \notin Q

;

(1, 1)

belongs to

P \times P

❌ That adds the sizes.

3 + 2 = 5

counts elements of

Q

and

P

together; each of the

3

first coordinates pairs with each of the

2

second coordinates, so multiply.

❌ That is $|Q \times Q|$ . The second coordinate has only

2

choices, from

P

, not

3

❌ That is $2^3$ . A product's size is

|Q| \cdot |P| = 3 \cdot 2

, not a power.

Show solution

By definition $Q \times P = \{(u, v) \mid u \in Q \text{ and } v \in P\}$ . Pair each element of $Q$ with each element of $P$ :

Q \times P = \{(x, 0), (x, 1), (y, 0), (y, 1), (z, 0), (z, 1)\}

So $(x, 1) \in Q \times P$ , since $x \in Q$ and $1 \in P$ . The roster has $3 \cdot 2 = 6$ pairs. The pair $(0, x)$ is its reversal and lies in $P \times Q$ instead.

Problem 2 · Negating Membership

Given: $C = \{1, 2, 3\}$ and $D = \{2, 4\}$ — find the meaning of $(x, y) \notin C \times D$ and a pair outside $C \times D$ .

$(x, y) \notin C \times D$ means

Which pair is not in $C \times D$ ?

✅ Correct! Membership is a conjunction, so its negation is a disjunction: one failing coordinate is enough, and in

(2, 3)

only the second coordinate fails.

❌ Too strong. By De Morgan,

\neg(P \wedge Q) \equiv \neg P \vee \neg Q

, not

\neg P \wedge \neg Q

. The pair

(5, 2)

is not in

C \times D

even though

2 \in D

❌ Not quite. This keeps

x \in C

un-negated. With

x \in C

and

y \in D

it is true, yet then

(x, y) \in C \times D

❌ Only one case. This describes pairs whose second coordinate alone fails; it misses pairs such as

(5, 2)

, whose first coordinate fails.

❌ It is in.

1 \in C

and

2 \in D

, so

(1, 2) \in C \times D

❌ It is in.

3 \in C

and

4 \in D

, so

(3, 4) \in C \times D

❌ It is in. Both coordinates may be the same number:

2 \in C

and

2 \in D

, so

(2, 2) \in C \times D

Show solution

$(x, y) \in C \times D \iff x \in C \text{ and } y \in D$ . Negating both sides and applying De Morgan's law $\neg(P \wedge Q) \equiv \neg P \vee \neg Q$ :

(x, y) \notin C \times D \iff x \notin C \text{ or } y \notin D

Now test the pairs against $C = \{1, 2, 3\}$ and $D = \{2, 4\}$ :

$(1, 2)$ : $1 \in C$ , $2 \in D$ , so it is in $C \times D$ .
$(3, 4)$ : $3 \in C$ , $4 \in D$ , so it is in $C \times D$ .
$(2, 3)$ : $2 \in C$ but $3 \notin D$ , so it is not in $C \times D$ .
$(2, 2)$ : $2 \in C$ , $2 \in D$ , so it is in $C \times D$ .

Problem 3 · Difference and Union of Products

Given: $A = \{1, 2\}$ , $B = \{a, b, c\}$ , $C = \{b, c, d\}$ — find $(A \times B) - (A \times C)$ and the number of elements of $(A \times B) \cup (A \times C)$ .

$(A \times B) - (A \times C) =$

How many elements does $(A \times B) \cup (A \times C)$ have?

✅ Correct! The difference is

A \times (B - C) = A \times \{a\}

, and the union is

A \times (B \cup C) = A \times \{a, b, c, d\}

with

2 \cdot 4 = 8

pairs.

❌ Coordinates reversed. Every pair of

A \times B

has its number first, so

(a, 1)

cannot be in the difference.

❌ Not quite. The common factor

A

is not subtracted from itself; only pairs of

A \times B

that also lie in

A \times C

are removed, and

(1, a)

is not in

A \times C

❌ Too many pairs.

(1, d)

and

(2, d)

are not in

A \times B

at all, since

d \notin B

; a difference keeps only pairs from the first set.

❌ Overlap counted twice.

6 + 6 = 12

counts the

4

pairs with second coordinate

b

c

twice; subtract them once.

❌ That is the intersection.

(A \times B) \cap (A \times C)

has

4

pairs; the union also contains the pairs ending in

a

d

❌ Wrong overlap.

6 + 6 - 2

subtracts

|B \cap C| = 2

, but the shared pairs number

|A| \cdot |B \cap C| = 2 \cdot 2 = 4

Show solution

A \times B = \{(1, a), (1, b), (1, c), (2, a), (2, b), (2, c)\}

A \times C = \{(1, b), (1, c), (1, d), (2, b), (2, c), (2, d)\}

The shared pairs are the $4$ with second coordinate $b$ or $c$ . Removing them from $A \times B$ leaves

(A \times B) - (A \times C) = \{(1, a), (2, a)\} = A \times \{a\} = A \times (B - C).

For the union, count the shared pairs once:

|(A \times B) \cup (A \times C)| = 6 + 6 - 4 = 8,

which matches $A \times (B \cup C) = A \times \{a, b, c, d\}$ , with $2 \cdot 4 = 8$ pairs.

Problem 4 · Overlapping Factors in Both Orders

Given: $A = \{1, 2\}$ and $B = \{1, 2, 3\}$ — find the number of elements of $(A \times B) \cap (B \times A)$ .

✅ Correct! A pair in both products has each coordinate in both

A

and

B

, so the intersection is

\{1, 2\} \times \{1, 2\}

, with

4

pairs.

❌ Same size is not same set. Both products have

6

elements, but

(1, 3) \in A \times B

while

(1, 3) \notin B \times A

, since

3 \notin A

; they are not equal.

❌ Not disjoint here. For factors

\{1, 2, 3\}

and

\{a, b\}

the two orders share no pair, but here the factors overlap:

(1, 2)

lies in both products.

❌ That is $|B \times B|$ . Both coordinates must lie in

A

as well as in

B

, which rules out every pair containing

3

Show solution

$(x, y) \in A \times B$ means $x \in A$ and $y \in B$ ; $(x, y) \in B \times A$ means $x \in B$ and $y \in A$ . Both together say $x \in A \cap B$ and $y \in A \cap B$ , so

(A \times B) \cap (B \times A) = (A \cap B) \times (A \cap B) = \{1, 2\} \times \{1, 2\}.

That is $\{(1, 1), (1, 2), (2, 1), (2, 2)\}$ , with $2 \cdot 2 = 4$ elements. The pairs $(1, 3)$ and $(2, 3)$ of $A \times B$ , and $(3, 1)$ and $(3, 2)$ of $B \times A$ , are each in only one product.

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The Cartesian Product A × B: Listing Pairs, Testing Laws

IKey Formulas

IIMembership, One Coordinate at a Time

IIIProducts with ∩\cap∩, ∪\cup∪ and −-−

IVQuiz Questions

IIIProducts with $\cap$ , $\cup$ and $-$