Types of Relations

What a relation is (one-line recap)

A relation $R$ in a set $A$ is simply a subset of $A \times A$ — a collection of ordered pairs of elements of $A$. If $(a, b) \in R$ we say "$a$ is related to $b$" and write $a\,R\,b$. From Class XI you know two ways of describing one: listing the pairs (roster form) or giving the rule (set-builder form), e.g. in $\{1, 2, 3, 4\}$ the rule $b = a + 1$ describes $R = \{(1,2), (2,3), (3,4)\}$.

The two extremes: empty and universal

Since a relation is a subset of $A \times A$, the two extreme cases are the smallest and largest possible subsets.

Empty relation: no element is related to anything — $R = \phi \subset A \times A$.

Universal relation: every element is related to every element — $R = A \times A$.

Both are called trivial relations. Example: in $A = \{1, 2, 3, 4\}$, the rule $a - b = 10$ produces the empty relation (no pair manages a difference of 10), while $\vert a - b \vert \geq 0$ produces the universal relation (every pair qualifies). Similarly, in the set of students of a boys school, "$a$ is sister of $b$" is empty, while "heights of $a$ and $b$ differ by less than 3 metres" is universal.

The three structural properties

Everything in this section revolves around three properties a relation may or may not have.

Definition. A relation $R$ in a set $A$ is called

(i) reflexive if $(a, a) \in R$ for every $a \in A$,

(ii) symmetric if $(a_1, a_2) \in R$ implies $(a_2, a_1) \in R$, for all $a_1, a_2 \in A$,

(iii) transitive if $(a_1, a_2) \in R$ and $(a_2, a_3) \in R$ together imply $(a_1, a_3) \in R$, for all $a_1, a_2, a_3 \in A$.

In arrow language: reflexive means every element carries a self-loop; symmetric means every arrow has a return arrow; transitive means every two-step chain has its one-step shortcut.

How to check the properties (the exam template)

Step 1 — reflexive: take an arbitrary $a \in A$ and test whether $(a, a)$ satisfies the rule. One failing element is enough to destroy reflexivity — the property demands all of $A$.

Step 2 — symmetric: assume $(a, b) \in R$ and test whether the rule forces $(b, a) \in R$. To disprove, exhibit one concrete pair in $R$ whose reverse is not.

Step 3 — transitive: assume $(a, b), (b, c) \in R$ and test whether $(a, c) \in R$ must follow. To disprove, exhibit one concrete broken chain.

Worked check 1. In $\{1, 2, 3\}$, let $R = \{(1,1), (2,2), (3,3), (1,2), (2,3)\}$.

Reflexive: $(1,1), (2,2), (3,3)$ all present ✓. Symmetric: $(1,2) \in R$ but $(2,1) \notin R$ ✗. Transitive: $(1,2) \in R$ and $(2,3) \in R$ but $(1,3) \notin R$ ✗. So $R$ is reflexive but neither symmetric nor transitive.

Worked check 2. In the set $L$ of all lines in a plane, let $R = \{(L_1, L_2) : L_1 \perp L_2\}$.

Reflexive: no line is perpendicular to itself ✗. Symmetric: $L_1 \perp L_2$ certainly gives $L_2 \perp L_1$ ✓. Transitive: if $L_1 \perp L_2$ and $L_2 \perp L_3$, then $L_1$ is parallel to $L_3$, never perpendicular ✗. So perpendicularity is symmetric only.

Note on vacuous truth. If a relation contains no chain $(a,b), (b,c)$ at all, the transitivity condition is never violated, so the relation counts as transitive. Example: $R = \{(1,6), (2,7), (3,8)\}$ in $\mathbb{N}$ (from $y = x + 5$, $x < 4$) has no two pairs that link up, hence it is transitive — while failing reflexivity and symmetry. Watch for this in MCQs.

Equivalence Relations and Equivalence Classes

The star of the section

Definition. A relation $R$ in a set $A$ is an equivalence relation if it is reflexive, symmetric and transitive.

Equivalence relations are the relations that behave like "equality with a theme": congruence of triangles (equal shape and size), similarity of triangles, "same number of pages", "same remainder on division by 3" — each declares two objects interchangeable from one point of view.

The model proof (learn this rhythm). In $\mathbb{Z}$, let $R = \{(a, b) : 2 \text{ divides } a - b\}$.

Step 1 — reflexive: $a - a = 0$ and $2$ divides $0$, so $(a, a) \in R$ for every $a$.

Step 2 — symmetric: if $2$ divides $a - b$, then $b - a = -(a - b)$ is also divisible by $2$, so $(b, a) \in R$.

Step 3 — transitive: if $2$ divides both $a - b$ and $b - c$, then $a - c = (a - b) + (b - c)$ is a sum of two even numbers, hence even, so $(a, c) \in R$.

All three hold, so $R$ is an equivalence relation. $\blacksquare$

Equivalence classes: the partition picture

In the example above, every even integer is related to $0$ and every odd integer is related to $1$. The two sets $$[0] = \{\ldots, -4, -2, 0, 2, 4, \ldots\}, \qquad [1] = \{\ldots, -3, -1, 1, 3, 5, \ldots\}$$ are called equivalence classes: $[a]$ is the set of all elements related to $a$. This always happens — an equivalence relation $R$ in a set $X$ chops $X$ into mutually disjoint nonempty subsets $A_i$ (the classes) such that

(i) all elements within one class are related to each other, (ii) no element of one class is related to any element of a different class, and (iii) the classes together cover $X$: $\cup A_i = X$ with $A_i \cap A_j = \phi$ for $i \neq j$.

Such a family of subsets is called a partition of $X$, and the process reverses: every partition of $X$ arises from exactly one equivalence relation ("belongs to the same piece").

The mod-3 picture. For $R = \{(a,b) : 3 \text{ divides } a - b\}$ in $\mathbb{Z}$, the classes are $$[0] = \{\ldots, -6, -3, 0, 3, 6, \ldots\}, \quad [1] = \{\ldots, -5, -2, 1, 4, 7, \ldots\}, \quad [2] = \{\ldots, -4, -1, 2, 5, 8, \ldots\}$$ and note $[0] = [3] = [-3] = [3r]$ — a class has many names, one for each of its members.

Finding "the set of elements related to $a$" is a standard board sub-question: apply the defining rule with one slot fixed at $a$. For $R = \{(a, b): \vert a - b\vert \text{ is a multiple of } 4\}$ in $\{x \in \mathbb{Z} : 0 \leq x \leq 12\}$, the elements related to $1$ are those differing from 1 by $0, 4, 8, 12$: the set $\{1, 5, 9\}$.

Common mistakes to avoid

Mistake 1 — proving reflexivity from symmetry. "If $(a,b) \in R$ and $(b,a) \in R$ then transitivity gives $(a,a) \in R$" only covers elements that are related to something. Reflexivity must hold for every element of $A$, related or not — this is why symmetric + transitive does not imply reflexive.

Mistake 2 — testing properties on one example. To prove a property you must argue for arbitrary elements; a single verified instance proves nothing. One counterexample, however, disproves a property completely.

Mistake 3 — forgetting vacuous transitivity. A relation with no linking chains is transitive by default (nothing violates the condition).

Mistake 4 — assuming symmetry means "looks symmetric". Test the actual rule: $a \leq b^2$ feels symmetric-ish but $(1, 2) \in R$ (since $1 \leq 4$) while $(2, 1) \notin R$ (since $2 \not\leq 1$).

Mistake 5 — mixing up $[a]$ with $a$. An equivalence class is a set of elements, not a number; and $[1] = [5]$ is a perfectly correct statement when $1$ and $5$ are related.

Solved Examples

Example 1 — The two trivial relations

Let $A$ be the set of all students of a boys school. Show that the relation $R = \{(a, b) : a$ is sister of $b\}$ is the empty relation and $R' = \{(a, b) :$ the difference between heights of $a$ and $b$ is less than 3 metres$\}$ is the universal relation.

Step 1 — test $R$: in a boys school no student can be the sister of another, so no pair whatsoever satisfies the rule: $R = \phi$, the empty relation.

Step 2 — test $R'$: any two students' heights certainly differ by less than 3 metres, so every pair qualifies: $R' = A \times A$, the universal relation.

Answer: $R$ is empty and $R'$ is universal — the two extremes a relation can be.

Example 2 — Congruence is an equivalence relation

Let $T$ be the set of all triangles in a plane and $R = \{(T_1, T_2) : T_1$ is congruent to $T_2\}$. Show that $R$ is an equivalence relation.

Step 1 — reflexive: every triangle is congruent to itself, so $(T_1, T_1) \in R$.

Step 2 — symmetric: if $T_1$ is congruent to $T_2$ then $T_2$ is congruent to $T_1$, so $(T_1, T_2) \in R \Rightarrow (T_2, T_1) \in R$.

Step 3 — transitive: if $T_1$ is congruent to $T_2$ and $T_2$ to $T_3$, then $T_1$ is congruent to $T_3$.

Answer: all three properties hold, so congruence is an equivalence relation. $\blacksquare$

Example 3 — Perpendicularity: symmetric only

Let $L$ be the set of all lines in a plane and $R = \{(L_1, L_2) : L_1$ is perpendicular to $L_2\}$. Show that $R$ is symmetric but neither reflexive nor transitive.

Step 1 — not reflexive: a line is never perpendicular to itself, so $(L_1, L_1) \notin R$.

Step 2 — symmetric: $L_1 \perp L_2$ immediately gives $L_2 \perp L_1$.

Step 3 — not transitive: if $L_1 \perp L_2$ and $L_2 \perp L_3$, then $L_1$ and $L_3$ are parallel, not perpendicular — a concrete broken chain.

Answer: symmetric only. Geometric relations make the cleanest counterexamples; keep this one ready.

Example 4 — Reading properties off a finite list

Show that the relation $R = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)\}$ in the set $\{1, 2, 3\}$ is reflexive but neither symmetric nor transitive.

Step 1 — reflexive: $(1,1), (2,2), (3,3)$ are all in $R$ ✓.

Step 2 — not symmetric: $(1, 2) \in R$ but $(2, 1) \notin R$.

Step 3 — not transitive: $(1, 2) \in R$ and $(2, 3) \in R$, but $(1, 3) \notin R$.

Answer: reflexive only. For a listed relation, the checks are pure inspection — scan for the three diagonal pairs, then hunt for a missing reverse and a missing shortcut.

Example 5 — The parity relation on $\mathbb{Z}$

Show that the relation $R = \{(a, b) : 2$ divides $a - b\}$ in $\mathbb{Z}$ is an equivalence relation, and describe its equivalence classes.

Step 1 — reflexive: $a - a = 0$ is divisible by 2.

Step 2 — symmetric: if $2 \mid (a - b)$ then $2 \mid (b - a)$, since $b - a = -(a - b)$.

Step 3 — transitive: $a - c = (a - b) + (b - c)$ — a sum of two even numbers is even.

Step 4 — the classes: every even integer is related to 0 and every odd integer to 1: $$[0] = \{\text{even integers}\}, \qquad [1] = \{\text{odd integers}\}$$

Answer: $R$ is an equivalence relation partitioning $\mathbb{Z}$ into the evens and the odds — two disjoint classes covering everything.

Example 6 — Equivalence classes inside a finite set

Show that $R = \{(a, b) : \vert a - b \vert$ is even$\}$ in $A = \{1, 2, 3, 4, 5\}$ is an equivalence relation, and show that all elements of $\{1, 3, 5\}$ are related to each other, all elements of $\{2, 4\}$ are related to each other, but no element of $\{1, 3, 5\}$ is related to any element of $\{2, 4\}$.

Step 1 — equivalence: $\vert a - a \vert = 0$ is even (reflexive); $\vert a - b \vert = \vert b - a \vert$ (symmetric); if $a - b$ and $b - c$ are both even, their sum $a - c$ is even, and $\vert a - c\vert$ is even (transitive).

Step 2 — the two blocks: any two odd numbers differ by an even amount, so $1, 3, 5$ are mutually related; likewise $2, 4$. An odd and an even number differ by an odd amount, so no cross-pair is in $R$.

Answer: $R$ is an equivalence relation with classes $\{1, 3, 5\}$ and $\{2, 4\}$ — the partition made visible in a five-element set.

Example 7 — When all three properties fail

Check whether the relation $R = \{(a, b) : b = a + 1\}$ in the set $\{1, 2, 3, 4, 5, 6\}$ is reflexive, symmetric or transitive.

Step 1 — list it: $R = \{(1,2), (2,3), (3,4), (4,5), (5,6)\}$.

Step 2 — not reflexive: $(1, 1) \notin R$ since $1 \neq 1 + 1$.

Step 3 — not symmetric: $(1, 2) \in R$ but $(2, 1) \notin R$ ($1 \neq 2 + 1$).

Step 4 — not transitive: $(1, 2), (2, 3) \in R$ but $(1, 3) \notin R$ ($3 \neq 1 + 1$).

Answer: none of the three properties holds. "Successor" relations are the standard example of a relation with no structure at all.