Let $A = \{1, 2, 3, 4, 5\}$ and $R = \{(a,b) : a, b \in A,\ |a-b| \leq 1\}$. Determine whether $R$ is reflexive, symmetric and transitive. Is $R$ an equivalence relation?

Let A = \1, 2, 3, 4, 5\ and R = \(a,b) : a, b ∈ A,|a-b| ≤ 1\. Determine whether R is reflexive, symmetric and

Miscellaneous · Q26

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

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Concept understanding — Types of Relations

Types of Relations – From Intuition to Precision

Imagine you have a set of people in a room. A relation is simply a rule that tells you whether two people are connected in some way. "Is the brother of", "lives in the same city as", "is taller than" — each of these is a relation. The question is: what kind of connection is it?

Some relations are very special. They behave in predictable, almost perfect ways. These are the ones we study first.

The Intuition: Three Key Properties

Think about the relation "lives in the same city as" between people.

Reflexive: Does every person live in the same city as themselves? Yes — obviously. A relation is reflexive if every element is related to itself.
Symmetric: If A lives in the same city as B, does B live in the same city as A? Yes — it's mutual. A relation is symmetric if whenever A is related to B, B is also related to A.
Transitive: If A lives in the same city as B, and B lives in the same city as C, does A live in the same city as C? Yes — it chains. A relation is transitive if whenever A is related to B and B is related to C, then A is related to C.

Now think about "is taller than". It is not reflexive (no one is taller than themselves), not symmetric (if A is taller than B, B is not taller than A), but it is transitive (if A > B and B > C, then A > C). Different relations satisfy different combinations.

The Precise Definitions

Let $R$ be a relation on a set $A$ (meaning $R \subseteq A \times A$ ).

R

R

R

The Four Types of Relations You Must Know

These three properties combine to define the most important types:

Type	Reflexive?	Symmetric?	Transitive?	Example
Empty relation	No (unless $A = \emptyset$ )	Yes (vacuously)	Yes (vacuously)	$R = \emptyset$ on $A = \{1,2\}$
Universal relation	Yes	Yes	Yes	$R = A \times A$
Identity relation	Yes	Yes	Yes	$R = \{(a,a) \mid a \in A\}$
Equivalence relation	Yes	Yes	Yes	"Same city", "same remainder mod 3"

Watch out

A common mistake: thinking "symmetric" means "if (a,b) is in R, then (b,a) must also be in R" — that's correct. But it does not require that (a,b) and (b,a) are different pairs. The identity relation is symmetric because (a,a) implies (a,a).

Equivalence Relations — The Most Important Type

An equivalence relation is one that is reflexive, symmetric, and transitive all at once. It captures the idea of "sameness" or "equivalence" in some sense.

Important

An equivalence relation partitions the set into disjoint equivalence classes — groups of elements that are all related to each other. Every element belongs to exactly one class.

Example: On the set of integers, define $a \sim b$ if $a - b$ is divisible by 3. This is an equivalence relation. The classes are:

$\{\dots, -6, -3, 0, 3, 6, \dots\}$ (remainder 0)
$\{\dots, -5, -2, 1, 4, 7, \dots\}$ (remainder 1)
$\{\dots, -4, -1, 2, 5, 8, \dots\}$ (remainder 2)

Other Important Types (for completeness)

Type	Definition	Example
Antisymmetric	If $(a,b) \in R$ and $(b,a) \in R$ , then $a = b$	$\leq$ on real numbers
Asymmetric	If $(a,b) \in R$ then $(b,a) \notin R$	$<$ on real numbers
Partial order	Reflexive + antisymmetric + transitive	$\subseteq$ on sets
Total order	Partial order where every pair is comparable	$\leq$ on real numbers

Tip

To check if a relation is an equivalence relation, test all three properties in order: reflexivity first (easiest to fail), then symmetry, then transitivity. If any fails, it's not an equivalence relation.

Quick Check — Test Yourself

Let $A = \{1,2,3\}$ and $R = \{(1,1), (2,2), (3,3), (1,2), (2,1)\}$ .

Reflexive? Yes — every element has $(a,a)$ .
Symmetric? Yes — $(1,2)$ has $(2,1)$ , and all self-pairs are fine.
Transitive? Check: $(1,2)$ and $(2,1)$ give $(1,1)$ — present. $(2,1)$ and $(1,2)$ give $(2,2)$ — present. No other chains. So yes.

This is an equivalence relation. The classes are $\{1,2\}$ and $\{3\}$ .

Final takeaway: A relation is just a set of ordered pairs. The "types" are simply patterns in which pairs appear. The three properties — reflexivity, symmetry, transitivity — are the building blocks. Master them, and you master the types.

Reflexive: $|a-a|=0\leq1$ for every $a\in A$ , so $(a,a)\in R$ always. Reflexive.

Symmetric: $|a-b|=|b-a|$ always (absolute value is symmetric in its two arguments), so if $(a,b)\in R$ then $(b,a)\in R$ . Symmetric.

Transitive: take $(1,2)\in R$ since $|1-2|=1\leq1$ , and $(2,3)\in R$ since $|2-3|=1\leq1$ . But $(1,3)$ : $|1-3|=2\not\leq1$ , so $(1,3)\notin R$ . Not transitive.

Since transitivity fails, $R$ is not an equivalence relation, even though it is reflexive and symmetric.

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Q.Let A={1,2,3,4,5}A = \{1, 2, 3, 4, 5\}A={1,2,3,4,5} and R={(a,b):a,b∈A, ∣a−b∣≤1}R = \{(a,b) : a, b \in A,\ |a-b| \leq 1\}R={(a,b):a,b∈A, ∣a−b∣≤1}. Determine whether RRR is reflexive, symmetric and transitive. Is RRR an equivalence relation?

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Q.Let $A = \{1, 2, 3, 4, 5\}$ and $R = \{(a,b) : a, b \in A,\ |a-b| \leq 1\}$ . Determine whether $R$ is reflexive, symmetric and transitive. Is $R$ an equivalence relation?

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