Q.Let $T$ be the set of all triangles in a plane with $R$ a relation in $T$ given by $R = \{(T_1, T_2) : T_1 \sim T_2\}$. Show that $R$ is an equivalence relation.

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.

1. Reflexive: Does every person live in the same city as themselves? Yes — obviously. A relation is reflexive if every element is related to itself.

2. 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.

3. 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 $R \subseteq A \times A$).

The Four Types of Relations You Must Know

These three properties combine to define the most important types:

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.

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)

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.