Types of relations

Ask a mathematician "how are these two objects connected?" and the honest answer is: pick a relation. A relation on a set $A$ is simply a rule that says which ordered pairs of elements count as linked — nothing more exotic than a subset of $A \times A$ . What makes relations worth a whole unit of ISC Class 12 Mathematics is that a handful of properties — reflexive, symmetric, transitive — sort every relation you will ever meet into useful families, and one very special combination, the equivalence relation, turns out to slice a set neatly into non-overlapping groups. Learning to test these properties rigorously, with a single well-chosen counterexample when a property fails, is the real skill this topic builds — and it is the direct foundation for functions, the very next topic in the unit.

The core insight

A relation on $A$ is a subset of $A \times A$ , and checking whether it is reflexive, symmetric, and transitive tells you whether it is an equivalence relation that partitions $A$ into disjoint equivalence classes.

The core idea

A relation is a subset, nothing more. A relation $R$ on a set $A$ is any subset of $A \times A$ ; writing $(a,b) \in R$ is the same statement as writing $a\,R\,b$ . Three named relations recur constantly: the empty relation $R = \varnothing$ (nothing is related to anything), the universal relation $R = A \times A$ (everything is related to everything), and the identity relation $R = \{(a,a) : a \in A\}$ (each element is related only to itself).

Three properties classify every relation. For $R$ on $A$ :

\textbf{Reflexive} \iff (a,a) \in R \ \text{for every } a \in A

\textbf{Symmetric} \iff (a,b) \in R \implies (b,a) \in R \ \text{for all } a,b \in A

\textbf{Transitive} \iff (a,b) \in R \text{ and } (b,c) \in R \implies (a,c) \in R \ \text{for all } a,b,c \in A

where: $a,b,c$ range over the elements of $A$ , and $\implies$ means "if the left side holds, the right side must also hold." Reflexivity is a check on single elements, symmetry is a check on pairs, transitivity is a check on chains of two linked pairs.

One property in three checks builds an equivalence relation. A relation that is reflexive and symmetric and transitive, all three at once, is an equivalence relation. For each $a \in A$ , its equivalence class is $[a] = \{x \in A : x\,R\,a\}$ , and equivalence classes have a remarkable structural payoff: they partition $A$ into disjoint, exhaustive groups — proved formally below.

A single counterexample kills a property. To disprove symmetry, exhibit one pair $(a,b) \in R$ with $(b,a) \notin R$ . To disprove transitivity, exhibit one chain $(a,b), (b,c) \in R$ with $(a,c) \notin R$ . No exhaustive search is needed once one violation is found — this is the fastest route to full marks on a "prove or disprove" question.

Figure

Two overlapping circles labelled reflexive symmetric and transitive shrinking to a small shaded intersection labelled equivalence relation

Equivalence classes partition the set

A relation carrying all three properties earns a strong structural guarantee: its equivalence classes split the whole set into pieces that neither overlap nor leave anyone out. Proving this from the three definitions, and only the three definitions, is the kind of short logical argument ISC examiners like to ask for directly.

Derivation AEquivalence classes partition the set

The goal is to prove, starting only from the definitions of reflexive, symmetric, and transitive, that if $R$ is an equivalence relation on $A$ then the equivalence classes $\{[a] : a \in A\}$ are pairwise disjoint and their union is all of $A$ — so every element of $A$ belongs to exactly one class, which is precisely what "partition" means, and it is why classifying a relation as an equivalence relation is worth proving carefully rather than assuming.

Step 1. Show every element lies in some class, using reflexivity alone. For any $a \in A$ , reflexivity gives $(a,a) \in R$ , i.e. $a\,R\,a$ , so by the definition of the class, $a \in [a]$ . Every element of $A$ is a member of at least its own class, so the union of all the classes is at least all of $A$ ; since each class is by definition a subset of $A$ , the union cannot exceed $A$ either. Hence $\bigcup_{a \in A} [a] = A$ .

Step 2. Show that if two classes share even one element, they are in fact the same class. Suppose $c \in [a] \cap [b]$ , so $c\,R\,a$ and $c\,R\,b$ . Using symmetry on the first, $a\,R\,c$ . Chaining $a\,R\,c$ with $c\,R\,b$ through transitivity gives $a\,R\,b$ .

Step 3. Extend that single link to the whole class. Take any $x \in [a]$ , so $x\,R\,a$ . Chaining $x\,R\,a$ with $a\,R\,b$ (from Step 2) through transitivity gives $x\,R\,b$ , so $x \in [b]$ . This shows $[a] \subseteq [b]$ ; by the identical argument with $a$ and $b$ swapped, $[b] \subseteq [a]$ . Hence $[a] = [b]$ .

Step 4. Combine the two steps. Step 1 shows the classes cover all of $A$ . Step 3 shows that two classes are either completely equal or, by contrapositive, share no element at all — because sharing one element forces full equality, two distinct classes must be disjoint. Covering $A$ with pairwise-disjoint pieces is exactly the definition of a partition, so the equivalence classes of $R$ partition $A$ .

Worked example

Let $A = \{1,2,3,4\}$ and $R = \{(1,1),(2,2),(3,3),(4,4),(1,2),(2,1),(3,4),(4,3)\}$ . Determine whether $R$ is reflexive, symmetric, and transitive, and hence whether it is an equivalence relation; if it is, list its equivalence classes.

Step 1. Check reflexivity: does $(a,a) \in R$ for every $a \in A$ ? Scanning the listed pairs, $(1,1),(2,2),(3,3),(4,4)$ are all present, so $R$ is reflexive.

Step 2. Check symmetry: for every pair present, is its reverse also present? $(1,2) \in R$ pairs with $(2,1) \in R$ , and $(3,4) \in R$ pairs with $(4,3) \in R$ ; every listed pair has its reverse listed too, so $R$ is symmetric.

Step 3. Check transitivity by tracing every chain that can be formed. The only non-trivial chains are $(1,2),(2,1) \Rightarrow$ requires $(1,1)$ , which holds; and $(3,4),(4,3) \Rightarrow$ requires $(3,3)$ , which holds. No pair connects an element of $\{1,2\}$ to an element of $\{3,4\}$ , so no other chains exist. Every required chain closes, so $R$ is transitive.

Step 4. Conclude and list classes. Since $R$ is reflexive, symmetric, and transitive, $R$ is an equivalence relation. Its classes are $[1] = [2] = \{1,2\}$ and $[3] = [4] = \{3,4\}$ — exactly two disjoint pieces whose union is $\{1,2,3,4\} = A$ , matching the derivation above.

Competitive stretch

Two further relations sharpen the contrast between "looks symmetric" and "is symmetric." Let $A$ be the set of all triangles in a plane, with $T_1\,R\,T_2 \iff T_1$ is similar to $T_2$ : reflexive (scale factor $1$ ), symmetric (scale factor $k$ inverts to $1/k$ ), transitive (factors $k_1, k_2$ compose to $k_1 k_2$ ) — an equivalence relation, each class being all triangles of one shape. Contrast this with $S$ on $\mathbb{Z}$ given by $a\,S\,b \iff a \mid b$ : reflexive ( $a \mid a$ ) and transitive ( $a=mb, b=nc \Rightarrow a=(mn)c$ ), but $2 \mid 4$ while $4 \nmid 2$ is a counterexample to symmetry, so $S$ is not an equivalence relation despite passing two of the three tests.

Common traps

Treating reflexivity as automatic — a relation missing even one pair $(a,a)$ fails reflexivity entirely, so every element must be checked.
Mistaking "looks balanced" for symmetric: divisibility and "is a subset of" both pass reflexivity and transitivity yet fail symmetry, as the divisibility counterexample above shows.
Searching exhaustively to disprove transitivity when one violating chain already settles the question — find it and stop.
Assuming two of the three properties imply the third; reflexive plus transitive never forces symmetric, so all three must be verified independently before calling a relation an equivalence relation.

Ask the tutor: If you want to work through your own relation pair by pair, or see another equivalence-class example, ask the AI tutor.

Why do we need all three properties together, and not just symmetric and transitive?

Because reflexivity is not automatic from the other two, even though it looks like it should be. Consider $R$ on $\{1,2,3\}$ containing only $(1,2)$ and $(2,1)$ : it is symmetric, and vacuously transitive since no chain of two related pairs exists to violate. But $(3,3) \notin R$ , so it is not reflexive and hence not an equivalence relation. The three checks are logically independent, and a rigorous answer verifies each one separately rather than assuming shortcuts between them.

Is there a faster way to check transitivity than testing every possible chain by hand?

For a relation given as a small, explicit list of pairs, the practical method is exactly what the worked example does: list every pair $(a,b) \in R$ , find every $(b,c) \in R$ sharing that middle element $b$ , and confirm $(a,c) \in R$ for each. For relations defined by a rule (like divisibility or similarity), it is faster to reason abstractly with general elements, as in the competitive-stretch example, rather than testing specific numbers. Either way, the moment one required chain fails to close, you have your counterexample and can stop.

What does it actually mean for equivalence classes to partition a set?

It means two things hold simultaneously: every element of $A$ belongs to some class (there are no leftovers), and no element belongs to two different classes (the classes never overlap). The derivation above proves both from the three defining properties — reflexivity guarantees every element is in its own class, while symmetry and transitivity together force any two classes that share an element to be completely identical. Picture it as sorting $A$ into labelled bins with no item in two bins and no bin empty of its own label.