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Relations and Their Properties

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Relations and Their Properties

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Relations and Their Properties

🎯 Learning Objectives

By the end of this topic, you will be able to:

Compute the Cartesian product of two or more sets
Define a binary relation as a subset of a Cartesian product
Classify relations as reflexive, symmetric, transitive, antisymmetric
Identify equivalence relations and partition sets into equivalence classes
Represent relations using ordered pairs, matrices, and graphs

📋 Prerequisites

Sets and Set Operations — sets, subsets, elements, set-builder notation
Arithmetic with integers — remainders (mod) and factors

📖 Core Content

2.1 Cartesian Product

2.1.1 Intuition: Combining Sets Systematically

Imagine you have two lists:

A = \{\text{Alice}, \text{Bob}\}

and

B = \{\text{Pizza}, \text{Pasta}\}

. How many possible ways can you pair one person with one dish? Every person can be paired with every dish:

(Alice, Pizza), (Alice, Pasta)
(Bob, Pizza), (Bob, Pasta) This all-possible-pairings is the Cartesian product $A \times B$ . The order matters: (Alice, Pizza) is different from (Pizza, Alice) — the first element always comes from $A$ , the second from $B$ .

💡 Why this matters: In databases, the CROSS JOIN of two tables is a Cartesian product. In probability, the sample space of rolling two dice is $\{1,2,3,4,5,6\} \times \{1,2,3,4,5,6\}$ .

2.1.2 Formal Definition

The Cartesian product of sets

A

and

B

is:

A \times B = \{(a, b) \mid a \in A,\ b \in B\}

Each element

(a, b)

is an ordered pair — the position matters. Key properties:

$|A \times B| = |A| \times |B|$
$A \times B \neq B \times A$ (unless $A = B$ or one is empty)
$A \times (B \times C) \neq (A \times B) \times C$ (triples vs. nested pairs)
$\emptyset \times A = A \times \emptyset = \emptyset$ For three sets:

A \times B \times C = \{(a, b, c) \mid a \in A,\ b \in B,\ c \in C\}

This generalizes to

n

sets (ordered

n

-tuples).

2.1.3 Worked Examples

Example 1.1 (Easy): Let

A = \{1, 2\}

B = \{3, 4\}

. Compute

A \times B

and

B \times A

. Step 1:

A \times B

pairs each element of

A

with each element of

B

A \times B = \{(1,3), (1,4), (2,3), (2,4)\}

Step 2:

B \times A

pairs each element of

B

with each element of

A

B \times A = \{(3,1), (3,2), (4,1), (4,2)\}

\boxed{A \times B \neq B \times A,\ \text{e.g.}\ (1,3) \in A \times B \text{ but } (1,3) \notin B \times A}

Example 1.2 (Medium): How many elements are in

A \times B \times C

|A| = 3

|B| = 4

|C| = 5

? Step 1:

|A \times B \times C| = |A| \times |B| \times |C| = 3 \times 4 \times 5 = 60

\boxed{60 \text{ ordered triples}}

Example 1.3 (Exam-style): Let

A = \{1\}

B = \{2, 3\}

. List all elements of

A \times B \times A

. Step 1: First coordinates from

A

, second from

B

, third from

A

A \times B \times A = \{(1,2,1), (1,3,1)\}

\boxed{\{(1,2,1), (1,3,1)\}}

2.2 Binary Relations

2.2.1 Intuition: Connections Between Elements

A relation is simply a rule that connects elements from one set to another. "Alice teaches Mathematics" defines a relation between people and subjects. "5 is less than 10" defines a relation between numbers. A relation doesn't have to connect every element — it's a subset of all possible pairings (the Cartesian product).

💡 Why this matters: Relations are the mathematical foundation for database tables (each row is a tuple), social networks (edges between users), and dependency graphs (prerequisite relationships).

2.2.2 Formal Definition

A binary relation

R

from set

A

to set

B

is a subset of

A \times B

R \subseteq A \times B

(a, b) \in R

, we write

a\,R\,b

(read: "

a

is related to

b

").

Domain of $R$ : $\{a \in A \mid \exists b \in B,\ (a,b) \in R\}$
Range of $R$ : $\{b \in B \mid \exists a \in A,\ (a,b) \in R\}$
If $A = B$ , we say $R$ is a relation on $A$ . (Diagram)

2.2.3 Examples of Relations

Set $A$	Set $B$	Relation $R$	$R \subseteq A \times B$
Teachers	Courses	$t$ teaches $c$	$\{(t,c) \mid t \text{ teaches } c\}$
$\mathbb{N}$	$\mathbb{N}$	$a \mid b$ ( $a$ divides $b$ )	$\{(a,b) \mid a \mid b\}$
$\mathbb{R}$	$\mathbb{R}$	$a < b$	$\{(a,b) \mid a < b\}$
Students	GPAs	$\leq 8.5$ GPA	$\{(s,g) \mid \text{GPA}(s) \leq 8.5\}$

Pythagorean triples as a relation:

R = \{(a,b,c) \mid a,b,c \in \mathbb{N},\ a,b,c > 0,\ a^2 + b^2 = c^2\}

Example triples:

(3,4,5)

(5,12,13)

(6,8,10)

2.3 Properties of Relations (on a Set)

When

R

is a relation on a set

A

(i.e.,

R \subseteq A \times A

), we classify it by these properties:

2.3.1 Reflexive

Intuition: Every element is related to itself. "Every number is equal to itself." Definition:

R

is reflexive if for every

a \in A

(a, a) \in R

. Examples:

$\leq$ on $\mathbb{R}$ : $a \leq a$ for all $a$ → reflexive ✓
divides on $\mathbb{N}$ : $a \mid a$ for all $a > 0$ → reflexive ✓
$<$ on $\mathbb{R}$ : $a < a$ is false → NOT reflexive ✗

2.3.2 Symmetric

Intuition: If

a

relates to

b

, then

b

relates back to

a

. Like a two-way street. Definition:

R

is symmetric if whenever

(a,b) \in R

, then

(b,a) \in R

. Examples:

= on $\mathbb{R}$ : $a = b \implies b = a$ → symmetric ✓
gcd(a,b) = 1 on $\mathbb{N}$ : $\gcd(a,b) = 1 \iff \gcd(b,a) = 1$ → symmetric ✓
$\leq$ on $\mathbb{R}$ : $2 \leq 3$ does NOT imply $3 \leq 2$ → NOT symmetric ✗

2.3.3 Transitive

Intuition: If

a

relates to

b

and

b

relates to

c

, then

a

relates to

c

. Like a chain. Definition:

R

is transitive if whenever

(a,b) \in R

and

(b,c) \in R

, then

(a,c) \in R

. Examples:

$<$ on $\mathbb{R}$ : $a < b$ and $b < c$ implies $a < c$ → transitive ✓
divides on $\mathbb{N}$ : $a \mid b$ and $b \mid c$ implies $a \mid c$ → transitive ✓
"is a parent of": NOT transitive (my parent's parent is my grandparent, not my parent) ✗

2.3.4 Antisymmetric

Intuition: If

a

relates to

b

and

b

relates to

a

, then

a

and

b

must be the same. No distinct two-way relationships. Definition:

R

is antisymmetric if whenever

(a,b) \in R

and

(b,a) \in R

, then

a = b

. Examples:

$\leq$ on $\mathbb{R}$ : if $a \leq b$ and $b \leq a$ , then $a = b$ → antisymmetric ✓
divides on $\mathbb{N}$ : if $a \mid b$ and $b \mid a$ , then $a = b$ → antisymmetric ✓
a - b = 2 on $\mathbb{N}$ : neither symmetric nor antisymmetric ✗ (Diagram)

2.3.5 Worked Example: Classifying Relations

Example 2.1:

R = \{(a,b) \mid a,b \in \mathbb{N},\ a \leq b\}

. Classify

R

Property	Check	Result
Reflexive	$a \leq a$ for all $a$ ? ✓ Yes	✓
Symmetric	If $2 \leq 3$ , does $3 \leq 2$ ? No	✗
Transitive	If $a \leq b$ and $b \leq c$ , then $a \leq c$ ? ✓ Yes	✓
Antisymmetric	If $a \leq b$ and $b \leq a$ , then $a = b$ ? ✓ Yes	✓

\boxed{\text{Reflexive, Transitive, Antisymmetric}}

Example 2.2:

R = \{(a,b) \mid a,b \in \mathbb{Z},\ |a - b| = 2\}

. Classify

R

Property	Check	Result
Reflexive	$	a - a
Symmetric	$	a - b
Transitive	$(0,2) \in R$ and $(2,4) \in R$ , but $(0,4) \notin R$ since $	0-4

\boxed{\text{Only Symmetric}}

2.4 Equivalence Relations

2.4.1 Intuition: Grouping Similar Things

An equivalence relation captures the idea of "sameness" under some criterion. If we group integers by "remainder when divided by 5", all numbers with remainder 2 are equivalent (like 2, 7, 12, 17, ...). They behave the same way modulo 5.

💡 Why this matters: Equivalence relations partition data into buckets. In clustering, "same cluster" is an equivalence relation. In cryptography, "same residue class mod n" is fundamental.

2.4.2 Formal Definition

A relation

R

on a set

A

is an equivalence relation if it is:

Reflexive: $(a,a) \in R$ for all $a \in A$
Symmetric: $(a,b) \in R \iff (b,a) \in R$
Transitive: $(a,b) \in R$ and $(b,c) \in R \implies (a,c) \in R$

2.4.3 Equivalence Classes

An equivalence relation partitions

A

into disjoint subsets called equivalence classes. The equivalence class of

a \in A

is:

[a] = \{b \in A \mid (a,b) \in R\}

Properties:

Every element belongs to exactly one equivalence class
$[a] = [b]$ iff $(a,b) \in R$
Different equivalence classes are disjoint
The union of all equivalence classes equals $A$

2.4.4 Worked Examples

Example 3.1 (Easy): Define

R

\mathbb{Z}

by:

a\,R\,b

a \bmod 5 = b \bmod 5

. Check reflexive:

a \bmod 5 = a \bmod 5

→ ✓ Check symmetric: If

a \bmod 5 = b \bmod 5

, then

b \bmod 5 = a \bmod 5

→ ✓ Check transitive: If

a \bmod 5 = b \bmod 5

and

b \bmod 5 = c \bmod 5

, then

a \bmod 5 = c \bmod 5

→ ✓ Equivalence classes (remainders 0, 1, 2, 3, 4):

$[0] = \{\ldots, -10, -5, 0, 5, 10, \ldots\}$
$[1] = \{\ldots, -9, -4, 1, 6, 11, \ldots\}$
$[2] = \{\ldots, -8, -3, 2, 7, 12, \ldots\}$
$[3] = \{\ldots, -7, -2, 3, 8, 13, \ldots\}$
$[4] = \{\ldots, -6, -1, 4, 9, 14, \ldots\}$

\boxed{\text{Equivalence relation with 5 classes}}

Example 3.2 (Medium): Let

A = \{1,2,3,4,5,6\}

. Define

R = \{(a,b) \mid a,b \in A,\ a \equiv b \pmod{3}\}

. Find the equivalence classes. Step 1: Numbers with same remainder modulo 3:

Remainder 0: $\{3,6\}$
Remainder 1: $\{1,4\}$
Remainder 2: $\{2,5\}$ Step 2: Verify: every element of $A$ appears in exactly one class. ✓

\boxed{\{[3,6], [1,4], [2,5]\}}

Example 3.3 (Hard): Is "same parity" (both even or both odd) an equivalence relation on

\mathbb{Z}

? Step 1 — Reflexive: Every integer has the same parity as itself (even×even or odd×odd). ✓ Step 2 — Symmetric: If

a

and

b

have same parity, then

b

and

a

do too. ✓ Step 3 — Transitive: If

a,b

have same parity and

b,c

have same parity, then

a,c

have same parity. ✓ Equivalence classes:

Even numbers: $\{\ldots, -4, -2, 0, 2, 4, \ldots\}$
Odd numbers: $\{\ldots, -3, -1, 1, 3, 5, \ldots\}$

\boxed{\text{Yes, it is an equivalence relation with 2 classes (evens and odds)}}

2.5 Representing Relations

2.5.1 Matrix Representation

For a relation

R

on a finite set

A = \{a_1, a_2, \ldots, a_n\}

, build an

n \times n

matrix

M_R

where:

M_R[i][j] = \begin{cases} 1 & \text{if } (a_i, a_j) \in R \\ 0 & \text{otherwise} \end{cases}

Example:

A = \{1,2,3\}

R = \{(1,1), (1,2), (2,3), (3,1)\}

M_R = \begin{bmatrix} 1 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{bmatrix}

Properties visible from matrix:

Reflexive: All diagonal entries are 1
Symmetric: Matrix is symmetric ( $M_R = M_R^T$ )
Antisymmetric: If $M_R[i][j] = 1$ and $i \neq j$ , then $M_R[j][i] = 0$

2.5.2 Graph Representation (Directed Graphs)

A relation

R

A

can be drawn as a directed graph where:

Elements of $A$ are vertices
Each $(a,b) \in R$ is a directed edge $a \rightarrow b$ (Diagram) This graph corresponds to the matrix above.

📐 Key Formulas — Summary Table

Concept	Formula/Definition	When to Use
Cartesian product	$A \times B = \{(a,b) \mid a \in A, b \in B\}$	Generating all ordered pairs
Size of product	$\	A \times B\
Binary relation	$R \subseteq A \times B$	Modeling connections
Reflexive	$\forall a \in A: (a,a) \in R$	Check diagonal in matrix
Symmetric	$(a,b) \in R \implies (b,a) \in R$	Two-way relationships
Transitive	$(a,b),(b,c) \in R \implies (a,c) \in R$	Chain relationships
Antisymmetric	$(a,b),(b,a) \in R \implies a = b$	Partial orders
Equivalence relation	Reflexive + Symmetric + Transitive	Grouping/classification
Equivalence class	$[a] = \{b \mid (a,b) \in R\}$	Partition of a set
Matrix entry	$M_R[i][j] = 1$ if $(a_i,a_j) \in R$	Computing properties

⚠️ Common Pitfalls

Pitfall 1: Confusing Symmetric with Antisymmetric

Mistake: Thinking a relation cannot be both symmetric and antisymmetric. Why: A relation CAN be both — but only if it's a subset of the identity relation. Example:

R = \{(1,1), (2,2)\}

\{1,2\}

is both symmetric (each pair has its reverse, which is itself) and antisymmetric (if

(a,b)

and

(b,a)

are in

R

, then

a=b

). Most relations are neither symmetric nor antisymmetric. Eg:

R = \{(1,2), (1,3)\}

is neither.

Pitfall 2: Forgetting That Transitive Requires ALL Chains

Mistake: Checking only one chain and concluding transitivity. Correct approach: You must check every pair

(a,b),(b,c)

. If ANY chain exists without its corresponding

(a,c)

, the relation is NOT transitive. Example:

R = \{(1,2), (2,3), (1,3), (3,4)\}

. Is

(1,2),(2,3) \to (1,3)

✓. But

(1,2),(2,4)

(2,4) \notin R

, so no check needed. However

(1,3),(3,4) \to (1,4)

(1,4) \notin R

→ NOT transitive!

Pitfall 3: Thinking $\{(1,1), (2,2)\}$ is Reflexive on $\{1,2,3\}$

Mistake: Reflexive requires

(a,a)

for every

a \in A

. If

A = \{1,2,3\}

, then

(3,3)

must be present. Without it, the relation is not reflexive. Correct: Check each element in the set

A

, not just those that appear in

R

Pitfall 4: Confusing "divides" Relation Domain

Mistake: Thinking

0 \mid 5

(0 divides 5) is valid. Correct: Division by zero is undefined. So the divides relation is defined only on positive integers or non-zero integers.

a \mid b

requires

a \neq 0

📝 Practice Questions

Q1: Let $A = \{x, y\}$ , $B = \{1, 2, 3\}$ . Compute $A \times B$ .

Strategy Hint: List all ordered pairs where first coordinate is from A, second from B.

$\boxed{\{(x,1), (x,2), (x,3), (y,1), (y,2), (y,3)\}}$ Q2: Does $R = \{(a,b) \mid a,b \in \mathbb{N},\ a < b\}$ satisfy reflexivity?

Strategy Hint: Check if any natural number is less than itself.

Solution: $a < a$ is false for all $a$ . So $R$ is NOT reflexive.

$\boxed{\text{No}}$ Q3: Is "is a sibling of" an equivalence relation?

Strategy Hint: Check reflexivity, symmetry, transitivity.

Reflexive: A person is not their own sibling → Not reflexive. Symmetric: If X is sibling of Y, then Y is sibling of X → Yes. Transitive: If X is sibling of Y and Y is sibling of Z, then X is sibling of Z → Yes (assuming full siblings).

But since it's not reflexive, it's NOT an equivalence relation.

$\boxed{\text{No — not reflexive}}$ Q4: Find the equivalence classes of $R = \{(a,b) \mid a,b \in \mathbb{Z},\ a \equiv b \pmod{4}\}$ .

Strategy Hint: Group by remainder when dividing by 4.

$\boxed{[0] = \{\ldots,-8,-4,0,4,8,\ldots\},\ [1] = \{\ldots,-7,-3,1,5,9,\ldots\},\ [2] = \{\ldots,-6,-2,2,6,10,\ldots\},\ [3] = \{\ldots,-5,-1,3,7,11,\ldots\}}$ Q5: Is $R = \{(a,b) \mid a,b \in \mathbb{R},\ a^2 = b^2\}$ an equivalence relation?

Strategy Hint: Check the three properties.

Reflexive: $a^2 = a^2$ → ✓ Symmetric: $a^2 = b^2 \implies b^2 = a^2$ → ✓ Transitive: $a^2 = b^2$ and $b^2 = c^2 \implies a^2 = c^2$ → ✓

Yes! Equivalence classes are $\{x, -x\}$ for each $x \in \mathbb{R}$ .

$\boxed{\text{Yes — equivalence classes are pairs } \{x, -x\}}$ Q6: Let $A = \{1,2,3\}$ . How many relations exist on $A$ ?

Strategy Hint: Any relation is a subset of $A \times A$ .

Step 1: $|A \times A| = 3 \times 3 = 9$ Step 2: Number of subsets = $2^9 = 512$

$\boxed{512 \text{ relations}}$ Q7: Draw the directed graph for $R = \{(1,2), (2,3), (3,1), (1,1)\}$ on $\{1,2,3\}$ .

Strategy Hint: Vertices = 1,2,3. Edges = the pairs.
pseudo
1 → 2 → 3
↑    ↓
└────┘
With a self-loop at 1.

$\boxed{\text{Graph representation: } 1 \to 2 \to 3 \to 1 \text{, plus } 1 \to 1}$ Q8: If $|A| = 3$ and $|B| = 4$ , what is $|A \times B \times A|$ ?

Strategy Hint: Multiply cardinalities.

$|A \times B \times A| = 3 \times 4 \times 3 = 36$

$\boxed{36}$ Q9: Let $R = \{(1,1), (1,2), (2,1), (2,2), (3,3)\}$ on $\{1,2,3,4\}$ . Is $R$ reflexive?

Strategy Hint: Reflexive requires $(4,4)$ .

$(4,4) \notin R$ , so NOT reflexive.

$\boxed{\text{No}}$ Q10: Find the matrix representation of $R = \{(a,b) \mid a,b \in \{1,2,3\},\ a \mid b\}$ .

Strategy Hint: $a$ divides $b$ means $b$ is a multiple of $a$ .

(a,b)	1	2	3
1	1 (1\|1)	1 (1\|2)	1 (1\|3)
2	0	1 (2\|2)	0
3	0	0	1 (3\|3)

> M_R = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} >

> \boxed{M_R = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}} >

Q11: If $R$ is symmetric and antisymmetric, prove $R \subseteq \{(a,a) \mid a \in A\}$ .

Strategy Hint: Take any $(a,b) \in R$ and deduce $a = b$ .

Proof: Let $(a,b) \in R$ . By symmetry, $(b,a) \in R$ . Since $R$ is antisymmetric, $(a,b),(b,a) \in R \implies a = b$ . Thus every pair in $R$ has the form $(a,a)$ .

$\boxed{\text{Proved}}$ Q12: Show that "same birthday month" is an equivalence relation on the set of all people. How many equivalence classes are there?

Strategy Hint: The three properties are straightforward. Classes = 12 months.

Reflexive: A person has the same birthday month as themselves. ✓ Symmetric: If A and B share a month, B and A do too. ✓ Transitive: If A∼B (same month) and B∼C (same month), then A∼C. ✓

12 equivalence classes (January through December), possibly with empty classes if no one in the set has that month.

$\boxed{\text{Yes, 12 equivalence classes}}$

🔗 Cross-References

Previous topic: Sets and Set Operations
Next topic: Functions and Their Types — a function is a special type of relation
Across courses: BSMA1003 Maths 2 (linear transformations are relations with special properties); BSMA1002 Stats 1 (random variables are functions from sample space to reals) Join Discord Previous1.1 Sets & OperationsNext1.3 Functions

(a,b)	1	2	3
1	1 (1\|1)	1 (1\|2)	1 (1\|3)
2	0	1 (2\|2)	0
3	0	0	1 (3\|3)