Mathematical-Reasoning · Unit 18 · Video 4 · Interactive Practice

Proving A × (B ∪ C) = (A × B) ∪ (A × C)

IKey Formulas

Statement	Name
$A \times B = \{(x, y) \mid x \in A \text{ and } y \in B\}$	Cartesian product
$(a, b) = (c, d) \iff a = c \text{ and } b = d$	Equality of ordered pairs
$A \times (B \cup C) = (A \times B) \cup (A \times C)$	Part (2): the product distributes over union
$(P \vee Q) \to R \equiv (P \to R) \wedge (Q \to R)$	Proof by cases

Key Insight: Write each chosen element as a pair $u = (x, y)$ . The first coordinate stays in $A$ throughout, and only the second coordinate needs the case split $y \in B$ or $y \in C$ .

IIBoth Sides, Pair by Pair

For every choice of $A$ , $B$ and $C$ , the pairs in $A \times (B \cup C)$ are exactly the pairs in $(A \times B) \cup (A \times C)$ .

A ⊆ {1, 2, 3, 4} B ⊆ {a, b, c, d} C ⊆ {a, b, c, d}

IIIThe Proof on a Picture

With $A = [0, 2]$ , $B = [2, 4]$ and $C = [3, 5]$ , each proof line is a fact about one coordinate of the chosen pair.

Inclusion Step 1

IVAn Ordered Pair Built from Sets

The set $\{\{a\}, \{a, b\}\}$ equals $\{\{c\}, \{c, d\}\}$ exactly when $a = c$ and $b = d$ .

a b c d

💡 Challenge: explain why the members of $\{\{a\}, \{a, b\}\}$ determine which of $a$ and $b$ is the first coordinate, including the case $a = b$ .

VQuiz Questions

Problem 1 · The Next Step

Given: the proof of $A \times (B \cup C) \subseteq (A \times B) \cup (A \times C)$ begins "Let $u \in A \times (B \cup C)$ ." Which statement follows next, directly from a definition?

✅ Correct! Unpacking the hypothesis with the definition of

A \times (B \cup C)

turns

u

into a pair whose coordinates have known homes:

x \in A

and

y \in B \cup C

❌ That is the goal. "

u \in A \times B

u \in A \times C

" is what must be shown (Step 3, worked backward from the conclusion), so writing it now assumes what the proof is supposed to prove.

❌ $u$ is an ordered pair. The elements of

A \times (B \cup C)

are pairs

(x, y)

; it is the coordinates that lie in

A

and in

B \cup C

, not

u

itself. For example

(1, c) \in \{1\} \times (\{a\} \cup \{c\})

but

(1, c) \notin \{1\}

❌ Too strong. The hypothesis only gives

y \in B \cup C

. Claiming

y \in B

silently drops the possibility

y \in C - B

, which is exactly why the proof needs two cases.

Show solution

By the definition of the Cartesian product, $A \times (B \cup C) = \{(x, y) \mid x \in A \text{ and } y \in B \cup C\}$ . So from $u \in A \times (B \cup C)$ we get:

\exists\, x \in A,\ \exists\, y \in B \cup C \text{ such that } u = (x, y).

This is Step 2 of the plan. Only afterwards does the proof split $y \in B \cup C$ into the cases $y \in B$ and $y \in C$ .

Problem 2 · Spot the Flaw

Given: a proposed proof of $A \times (B \cup C) \subseteq (A \times B) \cup (A \times C)$ : "Let $u \in A \times (B \cup C)$ . Then $u = (x, y)$ with $x \in A$ and $y \in B \cup C$ . (1) Since $y \in B \cup C$ , $y \in B$ and $y \in C$ . (2) So $u \in A \times B$ and $u \in A \times C$ . (3) Hence $u \in (A \times B) \cup (A \times C)$ ." Which part of the proof is invalid?

✅ Correct! By the definition of union,

y \in B \cup C

means

y \in B

y \in C

. With

A = \{1\}

B = \{a\}

C = \{c\}

and

u = (1, c)

, the hypothesis holds but

c \notin B

. The "or" calls for two cases,

y \in B

and

y \in C

❌ That proves the other inclusion. To prove

S \subseteq T

, choose an arbitrary element of

S

. Here

S = A \times (B \cup C)

, so starting with

u \in A \times (B \cup C)

is right; starting in

(A \times B) \cup (A \times C)

is the reverse inclusion.

❌ Step (2) is valid. If

x \in A

and

y \in B

, then

(x, y) \in A \times B

by the definition of the Cartesian product, and likewise

(x, y) \in A \times C

when

y \in C

. The trouble is the facts Step (2) is fed.

❌ Step (3) is valid. Every element of

A \times B

is in

(A \times B) \cup (A \times C)

by the definition of union, so a pair in both products is certainly in the union.

Show solution

Step (1) replaces "or" with "and". From $y \in B \cup C$ the definition of union gives only $y \in B$ or $y \in C$ , and the pair $(1, c)$ with $A = \{1\}$ , $B = \{a\}$ , $C = \{c\}$ shows that $y$ need not be in both. The theorem is still true; this argument just does not prove it. The repair is a proof by cases:

Case 1: $y \in B$ . Then $x \in A$ and $y \in B$ , so $u \in A \times B \subseteq (A \times B) \cup (A \times C)$ .

Case 2: $y \in C$ . Then $x \in A$ and $y \in C$ , so $u \in A \times C \subseteq (A \times B) \cup (A \times C)$ .

In both cases $u \in (A \times B) \cup (A \times C)$ .

Problem 3 · The Reverse Inclusion

Given: to prove $(A \times B) \cup (A \times C) \subseteq A \times (B \cup C)$ , the proof begins "Let $v \in (A \times B) \cup (A \times C)$ ." Answer both parts.

Where do the two cases come from?

Case 2: $v \in A \times C$ , so $v = (s, t)$ with $s \in A$ , $t \in C$ . Which step finishes this case?

✅ Correct! The cases come from the union of the two products, and in each case the second coordinate lands in

B \cup C

while

s

stays in

A

❌ Not quite. Start from the hypothesis

v \in (A \times B) \cup (A \times C)

and apply the definition of union.

❌ That is circular.

t \in B \cup C

is what this inclusion must establish; at the start the only fact available is

v \in (A \times B) \cup (A \times C)

❌ The first coordinate needs no split. In both

A \times B

and

A \times C

the first coordinate is in

A

, so "

s \in A

s \notin A

" adds nothing; the cases come from which product contains

v

❌ $v$ is an ordered pair. Elements of

(A \times B) \cup (A \times C)

are pairs; statements about membership in

A

B \cup C

belong to its coordinates.

❌ Not quite. The goal is a statement about the second coordinate: show

t \in B \cup C

❌ Wrong coordinate. Nothing says

s \in C

, and the product

A \times (B \cup C)

needs the second coordinate

t

B \cup C

❌ An extra assumption.

C \subseteq B

is not given; with

B = \{a\}

C = \{c\}

it fails. The case only needs

C \subseteq B \cup C

, which always holds.

❌ Too strong. Case 2 gives

t \in C

only;

t

need not be in

B

. Membership in one of

B

C

is enough for

t \in B \cup C

Show solution

By the definition of union, $v \in (A \times B) \cup (A \times C)$ means $v \in A \times B$ or $v \in A \times C$ : these are the two cases.

Case 1: $v \in A \times B$ . Then $v = (s, t)$ with $s \in A$ and $t \in B$ . Since $B \subseteq B \cup C$ , $t \in B \cup C$ , so $v \in A \times (B \cup C)$ .

Case 2: $v \in A \times C$ . Then $v = (s, t)$ with $s \in A$ and $t \in C$ . Since $C \subseteq B \cup C$ , $t \in B \cup C$ , so $v \in A \times (B \cup C)$ .

In both cases $v \in A \times (B \cup C)$ , which proves the reverse inclusion.

Problem 4 · Transfer to Part (4)

Given: to prove $(A \cup B) \times C \subseteq (A \times C) \cup (B \times C)$ , let $u \in (A \cup B) \times C$ , so $u = (x, y)$ with $x \in A \cup B$ and $y \in C$ . Which case split finishes this inclusion?

✅ Correct! In part (4) the union sits in the first factor, so the roles swap: the first coordinate carries the case split and the second stays in

C

❌ Coordinates swapped. Here

x

is the coordinate in

A \cup B

and

y

is in

C

. Since

(A \cup B) \times C

and

C \times (A \cup B)

are in general different sets, the coordinates cannot trade places.

❌ Not every $x$ is in $A$ . With

A = \{1\}

B = \{2\}

C = \{a\}

, the pair

(2, a)

is in

(A \cup B) \times C

but not in

A \times C

❌ Those cases come from the goal. "

u \in A \times C

u \in B \times C

" is what must be shown; splitting on it assumes the conclusion. The split must come from the hypothesis

x \in A \cup B

Show solution

Let $u \in (A \cup B) \times C$ . By the definition of the Cartesian product, $u = (x, y)$ with $x \in A \cup B$ and $y \in C$ . So $x \in A$ or $x \in B$ .

Case 1: $x \in A$ . Then $x \in A$ and $y \in C$ , so $u \in A \times C \subseteq (A \times C) \cup (B \times C)$ .

Case 2: $x \in B$ . Then $x \in B$ and $y \in C$ , so $u \in B \times C \subseteq (A \times C) \cup (B \times C)$ .

This is the proof of part (2) with the factors swapped: now $y$ stays in $C$ throughout and only $x$ needs the cases.

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