A relation on a set $A$ is just a set of ordered pairs $R\subseteq A\times A$. Three tests — reflexive, symmetric, transitive — sort relations into types, and the ones that pass all three split $A$ into equivalence classes.
In Class 11 you met a relation as a link between members of two sets: “is the capital of”, “is a factor of”, “is the square of”. This year we look at relations inside one set and ask what kind of link they are. Is every element linked to itself? If $a$ is linked to $b$, is $b$ automatically linked to $a$? Can links be chained? These three questions decide whether a relation behaves like “is equal to” — and relations that do behave like equality are everywhere: “has the same remainder on division by $5$”, “has the same blood group as”, “is parallel to”, “was born in the same month as”.
Why care? Because a relation that behaves like equality lets us sort a set into non-overlapping groups, and then treat each group as a single object. Clocks do this (13:00 and 1 o'clock are “the same” hour), calendars do it (every Monday is “the same” day of the week), and hospitals do it when they group donors by blood type. By the end of this section you will be able to test any relation for these properties, prove your answer, and find the groups — the equivalence classes.
Relations on a Set: Empty and Universal
Recall that for a set $A$, the Cartesian product $A\times A$ is the set of all ordered pairs $(a,b)$ with $a,b\in A$. If $A$ has $n$ elements then $A\times A$ has $n^2$ elements. (See Class 11 Mathematics, §2.1 Cartesian Products of Sets and §2.2 Relations, if you need a refresher.)
📘 Definition 1.1 — Relation on a set
A relation $R$ on a set $A$ is any subset of $A\times A$. If $(a,b)\in R$ we say “$a$ is related to $b$” and also write $aRb$. If $(a,b)\notin R$ we say $a$ is not related to $b$.
The order inside the pair matters: $(2,4)\in R$ says “$2$ is related to $4$”, which is a different statement from “$4$ is related to $2$”. On $A=\{1,2,3,4\}$, the relation “$a$ divides $b$” contains $(2,4)$ but not $(4,2)$.
Two relations sit at the extreme ends: the one with no pairs at all, and the one with every possible pair.
📘 Definition 1.2 — Empty and universal relations
A relation $R$ on $A$ is the empty relation if no element of $A$ is related to any element of $A$, that is, $R=\varnothing\subseteq A\times A$.
A relation $R$ on $A$ is the universal relation if every element of $A$ is related to every element of $A$, that is, $R=A\times A$.
Both are called trivial relations.
For example, let $A$ be the set of all students in your school. The relation “$a$ and $b$ were born more than $100$ years apart” is empty — no two students qualify. The relation “the sum of the ages of $a$ and $b$ is positive” is universal — every pair qualifies. On $A=\{1,2,3\}$, the universal relation is all $3^2=9$ pairs, and the empty relation has $0$ pairs.
Reflexive, Symmetric and Transitive Relations
Think about the relation “is equal to” on numbers. It has three obvious features: every number equals itself; if $a=b$ then $b=a$; and if $a=b$ and $b=c$ then $a=c$. We give each feature a name and then ask which relations share it.
$R$ is reflexive if $(a,a)\in R$ for every $a\in A$.
$R$ is symmetric if $(a,b)\in R$ implies $(b,a)\in R$, for all $a,b\in A$.
$R$ is transitive if $(a,b)\in R$ and $(b,c)\in R$ together imply $(a,c)\in R$, for all $a,b,c\in A$.
Each property is an “if … then …” promise (reflexive is a promise about every element). To show a property holds you must check it for all elements or pairs — usually with a general argument. To show it fails, one specific counterexample is enough. Keep that asymmetry in mind; it decides how you write every answer in this section.
🧭 Problem-solving strategy — testing a relation
Reflexive? Is $(a,a)$ in $R$ for each $a\in A$? For a finite set, look along the diagonal: $(1,1),(2,2),\dots$ One missing diagonal pair is enough to say “not reflexive”.
Symmetric? Take a general pair $(a,b)\in R$ and use the rule to show $(b,a)\in R$. For a finite set, check that every pair has its “mirror”. One pair without its mirror gives “not symmetric”.
Transitive? Assume $(a,b)\in R$ and $(b,c)\in R$; use both facts to prove $(a,c)\in R$. For a finite set, look for every chain $a\to b\to c$ and check that the shortcut $a\to c$ is there — including chains like $a\to b\to a$, which need $(a,a)$.
Write a conclusion for each property separately, with a reason or a counterexample.
✏️ Example 1 — Three relations on a small set
Let $A=\{1,2,3\}$. Decide which of the three properties each relation has.
(a) $R_1=\{(1,1),(2,2),(3,3),(1,2)\}$ (b) $R_2=\{(1,2),(2,1)\}$ (c) $R_3=\{(1,1),(2,2),(3,3),(1,2),(2,1),(2,3),(3,2)\}$
💡 Show solution▶
(a)Reflexive: $(1,1),(2,2),(3,3)$ are all present — yes. Symmetric: $(1,2)\in R_1$ but $(2,1)\notin R_1$ — no. Transitive: the chains are $(1,1),(1,2)\Rightarrow(1,2)$ ✓ and $(1,2),(2,2)\Rightarrow(1,2)$ ✓, and chains through the diagonal alone give diagonal pairs ✓. No chain breaks, so yes. $R_1$ is reflexive and transitive but not symmetric.
(b)Reflexive: $(1,1)\notin R_2$ — no. Symmetric: the only pairs are $(1,2)$ and $(2,1)$, and each is the mirror of the other — yes. Transitive: $(1,2)\in R_2$ and $(2,1)\in R_2$ would need $(1,1)\in R_2$, which is missing — no. $R_2$ is symmetric only.
(c)Reflexive: all diagonal pairs present — yes. Symmetric: $(1,2)$ has $(2,1)$ and $(2,3)$ has $(3,2)$ — yes. Transitive: $(1,2)\in R_3$ and $(2,3)\in R_3$, but $(1,3)\notin R_3$ — no. $R_3$ is reflexive and symmetric but not transitive.
Check Load these three relations into Interactive 1 (choose “Example 1 (a)/(b)/(c)”) and watch which light goes red, and which pair causes it.
Interactive 1 — Relation Explorer: Matrix and Arrow Diagram
A relation on $A=\{1,2,3,4\}$ is shown twice: as a $4\times4$ grid (a filled cell in row $a$, column $b$ means $(a,b)\in R$) and as an arrow diagram (a loop at $a$ means $(a,a)\in R$). Click cells to add or remove pairs. Play cycles the checker through the three tests and highlights the evidence: the diagonal for reflexive, a pair and its mirror for symmetric, a chain $a\to b\to c$ and its shortcut $a\to c$ (dashed red if missing) for transitive.
pairs in R—
reflexive?—
symmetric?—
transitive?—
✏️ Example 2 — A relation on ℝ
Let $R$ be the relation on $\mathbb R$ defined by $aRb$ if and only if $a\le b+1$. Is $R$ reflexive, symmetric or transitive?
💡 Show solution▶
Reflexive For every real $a$, $a\le a+1$ is true. So $(a,a)\in R$ for all $a\in\mathbb R$: $R$ is reflexive.
Symmetric Try to find a pair that works one way only. Take $a=3$, $b=1$: is $3\le1+1$? No. Now $a=1$, $b=3$: $1\le3+1$ — yes. So $(1,3)\in R$ but $(3,1)\notin R$: $R$ is not symmetric.
Transitive From $a\le b+1$ and $b\le c+1$ we only get $a\le c+2$, which is weaker than what we need. That suggests a counterexample with a “gap” of almost $2$: take $a=2$, $b=1$, $c=0$. Then $2\le1+1$ ✓ and $1\le0+1$ ✓, but $2\le0+1$ is false. So $(2,1),(1,0)\in R$ while $(2,0)\notin R$: $R$ is not transitive.
Answer $R$ is reflexive, but neither symmetric nor transitive.
✏️ Example 3 — Lines in a plane
Let $L$ be the set of all lines in a plane. Test (a) $R_1=\{(\ell_1,\ell_2):\ell_1\perp\ell_2\}$ and (b) $R_2=\{(\ell_1,\ell_2):\ell_1\parallel\ell_2\}$, where by convention every line is parallel to itself.
💡 Show solution▶
(a) R No line is perpendicular to itself, so $(\ell,\ell)\notin R_1$: not reflexive.
(a) S, T If $\ell_1\perp\ell_2$ then $\ell_2\perp\ell_1$: symmetric. For transitivity take $\ell_1$ the $x$-axis, $\ell_2$ the $y$-axis, $\ell_3$ the $x$-axis again: $\ell_1\perp\ell_2$ and $\ell_2\perp\ell_3$, but $\ell_1\parallel\ell_3$, not perpendicular. In fact two lines perpendicular to the same line are always parallel to each other. Not transitive. So $R_1$ is symmetric only.
(b) Every line is parallel to itself (reflexive); $\ell_1\parallel\ell_2\Rightarrow\ell_2\parallel\ell_1$ (symmetric); and if $\ell_1\parallel\ell_2$ and $\ell_2\parallel\ell_3$ then all three have the same slope (or are all vertical), so $\ell_1\parallel\ell_3$ (transitive). $R_2$ has all three properties.
⚠️ Common mistakes when studying alone
1. “Reflexive” needs every element. On $A=\{1,2,3\}$, the relation $\{(1,1),(2,2)\}$ is not reflexive: $(3,3)$ is missing, even though $3$ does not appear anywhere in $R$.
2. Symmetric does not mean “has many pairs”. It means every pair comes with its mirror. The empty relation is symmetric and transitive — there are no pairs to break the promises (we say the conditions hold vacuously).
3. Chains can go back on themselves. If $(1,2)$ and $(2,1)$ are in a transitive relation, then the chain $1\to2\to1$ forces $(1,1)$, and $2\to1\to2$ forces $(2,2)$. Many students forget these chains.
4. “Symmetric + transitive ⇒ reflexive” is false. The tempting argument “$aRb$ gives $bRa$, then transitivity gives $aRa$” only works if $a$ is related to something. On $\{1,2,3\}$, $\{(1,1),(1,2),(2,1),(2,2)\}$ is symmetric and transitive, but $3$ is related to nothing, so $(3,3)$ is missing.
5. A counterexample must use actual elements of $A$. On $\mathbb N$, do not use $0$ or negative numbers as witnesses.
Equivalence Relations
📘 Definition 1.4 — Equivalence relation
A relation $R$ on a set $A$ is an equivalence relation if it is reflexive, symmetric and transitive.
Equality, “is parallel to” (Example 3(b)) and the universal relation are equivalence relations. The most important example in all of school mathematics is “leaves the same remainder”, also called congruence modulo $n$.
✏️ Example 4 — Same remainder on division by 3
Show that $R=\{(a,b)\in\mathbb Z\times\mathbb Z : a-b \text{ is divisible by } 3\}$ is an equivalence relation on $\mathbb Z$.
💡 Show solution▶
Reflexive For any $a\in\mathbb Z$, $a-a=0=3\cdot0$ is divisible by $3$. So $(a,a)\in R$.
Symmetric Suppose $(a,b)\in R$, so $a-b=3k$ for some integer $k$. Then $b-a=-3k=3(-k)$, and $-k$ is an integer. So $(b,a)\in R$.
Transitive Suppose $(a,b)\in R$ and $(b,c)\in R$: $a-b=3k$ and $b-c=3m$ with $k,m\in\mathbb Z$. Adding, $a-c=(a-b)+(b-c)=3(k+m)$, and $k+m\in\mathbb Z$. So $(a,c)\in R$.
Conclusion $R$ is reflexive, symmetric and transitive, hence an equivalence relation. Notice the pattern: write the hypothesis as an equation ($a-b=3k$), do algebra, and show the result has the required form.
Equivalence Classes and Partitions
Under the relation of Example 4, the integer $7$ is related to $1,4,10,-2,\dots$ — exactly the integers that leave remainder $1$ when divided by $3$. Collecting everything related to a given element produces a group.
📘 Definition 1.5 — Equivalence class; partition
Let $R$ be an equivalence relation on $A$ and $a\in A$. The equivalence class of $a$ is the set of all elements of $A$ related to $a$:
$$[a]=\{x\in A : (x,a)\in R\}.$$
A partition of $A$ is a collection of non-empty subsets of $A$ that are pairwise disjoint and whose union is $A$ — every element of $A$ lies in exactly one of the subsets.
📐 Theorem 1.1 — Equivalence classes form a partition
Let $R$ be an equivalence relation on a non-empty set $A$. Then
$a\in[a]$ for every $a\in A$ (so no class is empty);
$[a]=[b]$ if and only if $(a,b)\in R$;
for any $a,b\in A$, either $[a]=[b]$ or $[a]\cap[b]=\varnothing$.
Hence the distinct equivalence classes form a partition of $A$. Conversely, any partition of $A$ defines an equivalence relation: “$a$ and $b$ lie in the same piece”.
Why (3) holds. Suppose $[a]$ and $[b]$ share an element $c$. Then $cRa$ and $cRb$. By symmetry $aRc$, and by transitivity ($aRc$, $cRb$) we get $aRb$. Now if $x\in[a]$ then $xRa$ and $aRb$ give $xRb$, so $x\in[b]$; the same argument the other way round shows $[b]\subseteq[a]$. So two classes that overlap at all are equal. Each of the three properties was used: reflexivity for (1), symmetry and transitivity for (3).
For “same remainder on division by $3$” on $\mathbb Z$ there are exactly three classes:
$$[0]=\{\dots,-6,-3,0,3,6,\dots\},$$
$$[1]=\{\dots,-5,-2,1,4,7,\dots\},$$
$$[2]=\{\dots,-4,-1,2,5,8,\dots\}.$$
They do not overlap, together they make up all of $\mathbb Z$, and $[7]=[1]=[-2]$ are three names for the same class.
✏️ Example 5 — Finding the classes
Let $A=\{1,2,3,\dots,10\}$ and $R=\{(a,b): a \text{ and } b \text{ leave the same remainder on division by } 3\}$. Show that $R$ is an equivalence relation, list its equivalence classes, and find all elements related to $4$.
💡 Show solution▶
Equivalence “$a$ and $b$ leave the same remainder” is the same as “$a-b$ is divisible by $3$”, so the proof of Example 4 works word for word on the smaller set $A$ (every element used there stays in $A$).
Classes Sort $1,\dots,10$ by remainder: remainder $1$: $\{1,4,7,10\}$; remainder $2$: $\{2,5,8\}$; remainder $0$: $\{3,6,9\}$. So the classes are $[1]=\{1,4,7,10\}$, $[2]=\{2,5,8\}$, $[3]=\{3,6,9\}$.
Related to 4 The elements related to $4$ form $[4]=[1]=\{1,4,7,10\}$.
Check $4+3+3=10$ elements in total, no element appears twice, and the classes need not have equal sizes. Within a class every pair is related ($10-1=9$ ✓); across classes none is ($5-4=1$ ✗).
Interactive 2 — Sorting a Set into Equivalence Classes
Choose a relation. If it is an equivalence relation, Play sends each element down into its class: the boxes never overlap and every element lands in exactly one box (Theorem 1.1). Change $n$ to see “same remainder on division by $n$” give $n$ classes. The last option is not transitive: watch the sets $[a]$ overlap without being equal, so no partition is possible.
3
elements in A—
number of classes—
partition?—
✏️ Example 6 — Building relations with chosen properties
On $A=\{1,2,3,4\}$ give a relation that is (a) reflexive and symmetric but not transitive; (b) symmetric and transitive but not reflexive; (c) reflexive and transitive but not symmetric.
💡 Show solution▶
(a) Start with the diagonal $\Delta=\{(1,1),(2,2),(3,3),(4,4)\}$ (this makes it reflexive) and add a chain with its mirrors but no shortcut: $R=\Delta\cup\{(1,2),(2,1),(2,3),(3,2)\}$. Then $1\to2\to3$ but $(1,3)\notin R$.
(b) Leave out one diagonal pair, and use a “block” that is complete: $R=\{(1,1),(1,2),(2,1),(2,2)\}$. Every chain stays inside $\{1,2\}$ and every shortcut is present, so it is transitive and symmetric; $(3,3)\notin R$ so it is not reflexive.
(c) $R=\Delta\cup\{(1,2)\}$. Reflexive ✓. Not symmetric: $(2,1)\notin R$. Transitive: every chain involves a diagonal pair, which produces a pair already in $R$.
Check Build each one in Interactive 1 by clicking cells; then press “Make it an equivalence relation” to see the fewest pairs you would have to add.
🚀 Beyond the board syllabus (JEE Main · JEE Advanced)
In JEE Main and JEE Advanced 2026; not examined in the CBSE board exam. JEE problems often ask how many relations of a given type exist on a set $A$ with $n$ elements. Each of the $n^2$ pairs of $A\times A$ is either in or out of $R$, so:
all relations: $2^{n^2}$; reflexive relations: the $n$ diagonal pairs are forced in, the other $n^2-n$ are free: $2^{n^2-n}$;
symmetric relations: the $n$ diagonal pairs are free, and the $\tfrac{n^2-n}{2}$ mirror-pairs $\{(a,b),(b,a)\}$ are in or out together: $2^{n(n+1)/2}$;
reflexive and symmetric: $2^{n(n-1)/2}$; equivalence relations = number of partitions of $A$ (for $n=1,2,3,4$: $1,2,5,15$).
Worked example. For $A=\{1,2,3\}$: there are $2^9=512$ relations, $2^6=64$ reflexive, $2^6=64$ symmetric, $2^3=8$ reflexive-and-symmetric, and $5$ equivalence relations — one for each partition: $\{1\},\{2\},\{3\}$; $\{1,2\},\{3\}$; $\{1,3\},\{2\}$; $\{2,3\},\{1\}$; $\{1,2,3\}$.
Exercises
I
Relations on a finite set. Unless stated otherwise $A=\{1,2,3,4\}$. For each relation say whether it is reflexive, symmetric and/or transitive, giving a counterexample for each property that fails.
1 Reflexive ✓ (all four diagonal pairs). Symmetric ✓ (each pair is its own mirror). Transitive ✓ (any chain $a\to a\to a$ gives $(a,a)$). This identity relation is an equivalence relation.
2 Not reflexive: $(1,1)\notin R$. Not symmetric: $(1,2)\in R$, $(2,1)\notin R$. Not transitive: $(1,2),(2,3)\in R$ but $(1,3)\notin R$. None of the three.
3 Reflexive ✓, symmetric ✓ ($(1,3)$ and $(3,1)$ both present), transitive ✓ (the chains $1\to3\to1$ and $3\to1\to3$ need $(1,1)$ and $(3,3)$, which are present). An equivalence relation with classes $\{1,3\},\{2\},\{4\}$.
4 Not reflexive: $(2,2)\notin R$ (also $(3,3)$, $(4,4)$). Symmetric ✓. Not transitive: $(2,1),(1,2)\in R$ but $(2,2)\notin R$.
5 Reflexive ✓. Not symmetric: $(1,2)\in R$ but $(2,1)\notin R$. Transitive ✓: the only chain not using a diagonal pair is $1\to2\to3$, and $(1,3)\in R$.
6 Not reflexive ($(1,1)\notin\varnothing$). Symmetric ✓ and transitive ✓ vacuously — there is no pair that could break either promise.
7 Every pair is present, so all three conditions hold: an equivalence relation with the single class $A$.
8 Not reflexive. Not symmetric: $(3,1)\notin R$. Transitive ✓ vacuously: a chain would need a pair starting at $3$, and there is none.
9 $a+b$ is even exactly when $a$ and $b$ have the same parity. Reflexive ($a+a=2a$ even), symmetric ($a+b=b+a$), transitive (same parity as $b$ is the same parity as each other). Equivalence relation; classes $\{1,3\}$ and $\{2,4\}$. As a set, $R$ has $2^2+2^2=8$ pairs.
10 $R=\{(1,2),(2,3),(3,4)\}$ — the same set as part 2. None of the three properties.
II
Relations on infinite sets. Determine whether each relation is reflexive, symmetric, transitive. Prove each property that holds and give a counterexample for each that fails.
1 On $\mathbb Z$: $aRb$ iff $a-b$ is a multiple of $5$.
2 On $\mathbb Z$: $aRb$ iff $a+b$ is even.
3 On $\mathbb R$: $aRb$ iff $a\lt b$.
4 On $\mathbb N$: $aRb$ iff $a$ divides $b$.
5 On $\mathbb R$: $aRb$ iff $|a-b|\le1$.
6 On $\mathbb Z$: $aRb$ iff $ab\ge0$.
7 On $\mathbb R$: $aRb$ iff $a^2=b^2$. Also find the class of $3$.
8 On $\mathbb N$: $aRb$ iff $a=2b$.
9 On the set of lines in a plane: $\ell_1R\ell_2$ iff $\ell_1\parallel\ell_2$. Describe the class of the line $y=2x$.
10 On the set of all triangles in a plane: $T_1RT_2$ iff $T_1$ and $T_2$ have the same area.
💡 Show solutions▶
1 As in Example 4 with $5$ in place of $3$: $a-a=0=5\cdot0$; $a-b=5k\Rightarrow b-a=5(-k)$; $a-b=5k$, $b-c=5m\Rightarrow a-c=5(k+m)$. Equivalence relation (five classes $[0],\dots,[4]$).
2 $a+a=2a$ is even (R). $a+b=b+a$ (S). If $a+b=2k$ and $b+c=2m$, then $a+c=2k+2m-2b=2(k+m-b)$, even (T). Equivalence relation; the classes are the even and the odd integers.
3 Not reflexive: $1\lt1$ is false. Not symmetric: $1\lt2$ but $2\lt1$ is false. Transitive: $a\lt b$ and $b\lt c$ give $a\lt c$. Transitive only.
4 $a=1\cdot a$, so $a\mid a$ (R). Not symmetric: $2\mid4$ but $4\nmid2$. Transitive: $b=ka$ and $c=mb$ give $c=(mk)a$. Reflexive and transitive, not symmetric.
5 $|a-a|=0\le1$ (R). $|b-a|=|a-b|$ (S). Not transitive: $|0-0.8|\le1$ and $|0.8-1.6|\le1$, but $|0-1.6|=1.6\gt1$. Reflexive and symmetric only.
6 $a\cdot a=a^2\ge0$ (R). $ab=ba$ (S). Not transitive: $1\cdot0\ge0$ and $0\cdot(-1)\ge0$, but $1\cdot(-1)=-1\lt0$. The number $0$ is related to everything, which is what breaks the chain.
7 $a^2=a^2$ (R); $a^2=b^2\Rightarrow b^2=a^2$ (S); $a^2=b^2$ and $b^2=c^2\Rightarrow a^2=c^2$ (T). Equivalence relation; $[3]=\{x: x^2=9\}=\{3,-3\}$. (Any relation of the form “$f(a)=f(b)$” is an equivalence relation, for the same three reasons.)
8 Not reflexive: $1\ne2\cdot1$. Not symmetric: $(2,1)\in R$ but $1\ne2\cdot2$. Not transitive: $(4,2),(2,1)\in R$ but $4\ne2\cdot1$. None.
9 Equivalence relation (Example 3(b)). The class of $y=2x$ is the family of all lines of slope $2$: $\{y=2x+c: c\in\mathbb R\}$.
10 “Same area” is of the form $f(T_1)=f(T_2)$ with $f$ = area, so it is reflexive, symmetric and transitive: an equivalence relation. (Triangles in one class can have very different shapes.)
III
Equivalence classes and partitions.
1 On $A=\{1,2,\dots,10\}$, $aRb$ iff $a-b$ is divisible by $4$. Show $R$ is an equivalence relation and list its classes.
2 For congruence modulo $3$ on $\mathbb Z$ (Example 4), show that $[7]=[1]$ and $[-1]=[2]$.
3 On $A=\{-3,-2,\dots,3\}$, $aRb$ iff $|a|=|b|$. List the classes. How many are there?
4 On the set of words $\{\text{CAT, DOG, EMU, LION, BEAR, ZEBRA, HORSE, OX}\}$, two words are related if they have the same number of letters. List the equivalence classes.
5 An equivalence relation on $A=\{1,2,3,4,5,6\}$ has classes $\{1,4\}$, $\{2,3,6\}$, $\{5\}$. How many ordered pairs does $R$ contain? List them.
6 $R$ is an equivalence relation on $A$ and $2\in[5]$. Explain why $5\in[2]$ and $[2]=[5]$.
7 Explain why $\{1,2\},\{2,3\},\{4\}$ cannot be the set of equivalence classes of any relation on $\{1,2,3,4\}$.
💡 Show solutions▶
1 Proof as in Example 4 with $4$ in place of $3$. Classes by remainder on division by $4$: $\{1,5,9\}$, $\{2,6,10\}$, $\{3,7\}$, $\{4,8\}$ — four classes, sizes $3+3+2+2=10$.
2 $7-1=6=3\cdot2$, so $(7,1)\in R$ and by Theorem 1.1(2) $[7]=[1]$. $-1-2=-3=3\cdot(-1)$, so $[-1]=[2]$.
3 $[0]=\{0\}$, $[1]=\{-1,1\}$, $[2]=\{-2,2\}$, $[3]=\{-3,3\}$: four classes.
5 Inside a class of size $k$ every ordered pair is in $R$, giving $k^2$ pairs; there are no pairs across classes. Total $2^2+3^2+1^2=14$: $(1,1),(1,4),(4,1),(4,4)$; $(2,2),(2,3),(2,6),(3,2),(3,3),(3,6),(6,2),(6,3),(6,6)$; $(5,5)$.
6 $2\in[5]$ means $(2,5)\in R$. By symmetry $(5,2)\in R$, i.e. $5\in[2]$. By Theorem 1.1(2), $(2,5)\in R$ gives $[2]=[5]$.
7 The sets $\{1,2\}$ and $\{2,3\}$ overlap (both contain $2$) but are not equal. Theorem 1.1(3) says two equivalence classes are either equal or disjoint, so these cannot both be classes.
IV
Constructing, reasoning and modelling.
1 On $A=\{1,2,3\}$ give a relation that is reflexive and symmetric but not transitive.
2 On $A=\{1,2,3\}$ give a relation that is symmetric and transitive but not reflexive.
3 On $A=\{1,2,3\}$ give a relation that is reflexive and transitive but not symmetric.
4 Find the smallest equivalence relation on $\{1,2,3,4\}$ that contains $(1,2)$ and $(2,3)$. How many pairs does it have?
5Telecom. On a set of mobile towers, $aRb$ means “$a$ and $b$ are within $5$ km of each other”. Which properties does $R$ have? Can $R$ be used to split the towers into non-overlapping service zones? Compare with “$a$ and $b$ are managed by the same district office”.
6Medicine. On a set of blood donors, consider (a) “has the same ABO and Rh blood group as” and (b) “can safely give red cells to”, where $X$ can give to $Y$ exactly when every antigen (A, B, Rh) on $X$'s cells is also on $Y$'s. Which properties does each relation have? What is the largest possible number of classes in (a)?
7 A student “proves” that a symmetric, transitive relation is reflexive: “Let $a\in A$. Take $b$ with $aRb$. By symmetry $bRa$; by transitivity $aRa$.” Find the flaw.
💡 Show solutions▶
1 $\{(1,1),(2,2),(3,3),(1,2),(2,1),(2,3),(3,2)\}$: $1\to2\to3$ without $(1,3)$. (Example 1(c).)
2 $\{(1,1),(1,2),(2,1),(2,2)\}$: complete on $\{1,2\}$, but $(3,3)$ is missing. (The empty relation also works.)
3 $\{(1,1),(2,2),(3,3),(1,2)\}$ (Example 1(a)).
4 $1\sim2$ and $2\sim3$ force $1,2,3$ into one class; $4$ can stay alone. So the classes are $\{1,2,3\}$ and $\{4\}$, and $R$ consists of all $9$ pairs from $\{1,2,3\}$ plus $(4,4)$: $10$ pairs. Any equivalence relation containing $(1,2),(2,3)$ must contain these, so this is the smallest.
5 Reflexive (distance $0$) and symmetric (distance is the same both ways), but not transitive: towers at $0$, $4$ and $8$ km along a road give $a\sim b$, $b\sim c$, but $a$ and $c$ are $8$ km apart. Because the “neighbourhoods” overlap without being equal, they cannot form non-overlapping zones. “Same district office” is of the form $f(a)=f(b)$, an equivalence relation, and its classes are exactly the zones.
6 (a) Equivalence relation (“same value of blood group”). With $4$ ABO groups and $2$ Rh types there are at most $4\times2=8$ classes. (b) Reflexive (a set of antigens is contained in itself), transitive (if $X$'s antigens are in $Y$'s and $Y$'s are in $Z$'s, then $X$'s are in $Z$'s), but not symmetric: O-negative can give to AB-positive, not the other way round.
7 The step “take $b$ with $aRb$” assumes such a $b$ exists. If $a$ is related to nothing, there is no $b$, and the argument says nothing about $(a,a)$. Counterexample: Exercise IV.2.
How this is examined: the Board paper asks you to prove that a relation is (or is not) an equivalence relation and to write down a class — marks go to each of the three properties separately, so set out all three. JEE turns the same ideas into quick MCQs and counting questions.
E1.Board · 1 mark On $A=\{1,2,3\}$, the relation $R=\{(1,1),(2,2),(1,2),(2,1)\}$ is (A) reflexive and symmetric (B) symmetric and transitive but not reflexive (C) an equivalence relation (D) reflexive and transitive but not symmetric
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Idea Check the diagonal first: $(3,3)\notin R$, so $R$ is not reflexive — this rules out (A), (C), (D). Every pair has its mirror (symmetric), and the chains $1\to2\to1$, $2\to1\to2$ need $(1,1)$, $(2,2)$, which are present (transitive).
Answer (B). The trap is (C): $R$ would be an equivalence relation on $\{1,2\}$, but the question is about $A=\{1,2,3\}$.
E2.Board · 1 markAssertion–Reason. Assertion (A): the relation $aRb$ iff $a-b$ is even is an equivalence relation on $\mathbb Z$. Reason (R): every relation that is symmetric and transitive is also reflexive. (A) Both A and R are true and R is the correct explanation of A (B) Both A and R are true but R is not the correct explanation of A (C) A is true but R is false (D) A is false but R is true
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Idea A is true (Exercise II.2 with “difference” in place of “sum”: $a-a=0$, $b-a=-(a-b)$, $a-c=(a-b)+(b-c)$). R is false: $\{(1,1)\}$ on $\{1,2\}$ is symmetric and transitive but not reflexive.
Answer (C).
E3.Board · 5 marks Show that the relation $R$ on $\mathbb R$ defined by $aRb$ iff $a-b$ is an integer is an equivalence relation. Write down the equivalence class $\left[\tfrac12\right]$, and decide whether $[2.7]=[-0.3]$.
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Reflexive $a-a=0\in\mathbb Z$ for every $a\in\mathbb R$. (1 mark)
Symmetric If $a-b=k\in\mathbb Z$ then $b-a=-k\in\mathbb Z$. (1 mark)
Transitive If $a-b=k$ and $b-c=m$ with $k,m\in\mathbb Z$, then $a-c=k+m\in\mathbb Z$. Hence $R$ is an equivalence relation. (1 mark)
Classes $x\in\left[\tfrac12\right]$ iff $x-\tfrac12=n\in\mathbb Z$, so $\left[\tfrac12\right]=\left\{n+\tfrac12: n\in\mathbb Z\right\}=\{\dots,-1.5,-0.5,0.5,1.5,\dots\}$. Since $2.7-(-0.3)=3\in\mathbb Z$, $(2.7,-0.3)\in R$ and so $[2.7]=[-0.3]$ (Theorem 1.1). (2 marks)
E4.Board · Case study A school in Chennai has $1200$ students, each in exactly one of four houses (Red, Blue, Green, Yellow). On the set $S$ of students consider $R_1$: “is in the same house as”, and $R_2$: “sits next to (on the same bench as)”, where each bench seats exactly two students. (i) Is $R_1$ an equivalence relation? [2 marks] (ii) How many equivalence classes does $R_1$ have? [1 mark] (iii) Is $R_2$ reflexive? symmetric? [1 mark] (iv) (Extra practice.) Suppose at least one bench is shared by two students of the same house. Is $R_3$: “is in the same house as and sits on the same bench as” transitive?
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(i) Yes. Reflexive: every student is in the same house as themselves. Symmetric: if $a$ is in $b$'s house, $b$ is in $a$'s. Transitive: if $a,b$ share a house and $b,c$ share a house, all three are in $b$'s house. (It is “same value of house”.)
(ii) Four classes, one per house (assuming every house has at least one student).
(iii) Not reflexive (no one sits next to themselves); symmetric (if $a$ sits next to $b$ then $b$ sits next to $a$).
(iv) Not transitive: take a bench shared by $a$ and $b$ of the same house. Then $(a,b)$ and $(b,a)$ are in the relation, but $(a,a)$ is not (no one sits next to themselves). The chain $a\to b\to a$ breaks.
E5.JEE Main The number of reflexive relations on a set with $4$ elements is (A) $2^{16}$ (B) $2^{12}$ (C) $2^{8}$ (D) $2^{4}$
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Idea $A\times A$ has $16$ pairs. The $4$ diagonal pairs must be in $R$; each of the remaining $12$ may be in or out independently.
Answer (B) $2^{12}=4096$. The trap (A) $2^{16}$ counts all relations, forgetting that the diagonal is forced.
⏱ Fast route Reflexive relations on $n$ elements: $2^{n^2-n}$; here $2^{16-4}$.
E6.JEE Advanced(One or more options correct.) Let $R$ and $S$ be equivalence relations on a set $A$, and for any relation $T$ on $A$ let $T'=\{(b,a):(a,b)\in T\}$ (the pairs reversed). Which statements are always true? (A) $R\cap S$ is an equivalence relation (B) $R\cup S$ is an equivalence relation (C) every symmetric, transitive relation on $A$ is reflexive (D) for every relation $T$ on $A$, $T\cap T'$ is symmetric
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(A) True Diagonal pairs are in both, so in $R\cap S$. If $(a,b)$ is in both, so is $(b,a)$. If $(a,b),(b,c)$ are in both, then $(a,c)$ is in both.
(B) False On $\{1,2,3\}$ take $R=\Delta\cup\{(1,2),(2,1)\}$ and $S=\Delta\cup\{(2,3),(3,2)\}$. Then $R\cup S$ contains $(1,2)$ and $(2,3)$ but not $(1,3)$.
(C) False $\{(1,1)\}$ on $\{1,2\}$ (Common mistake 4).
(D) True If $(a,b)\in T\cap T'$ then $(a,b)\in T$ and $(a,b)\in T'$, i.e. $(b,a)\in T$. Hence $(b,a)\in T'$ and $(b,a)\in T$, so $(b,a)\in T\cap T'$.
Answer (A) and (D).
🔬 Research–Teaching NexusMini project · about 45 minutes
Research question. When public-health teams trace contacts during an outbreak, how do they turn a list of “who met whom” into separate clusters — and why is “met” not enough on its own?
Why it matters. During outbreaks of influenza, COVID-19 or measles, health workers group cases into clusters so they can test and isolate each group. “$a$ met $b$” is reflexive and symmetric but not transitive, so it does not split people into groups. The fix is to use “$a$ and $b$ are linked by a chain of meetings” — which is transitive, and is the smallest equivalence relation containing the meetings (compare Exercise IV.4).
What to do.
With permission, survey $12$–$15$ classmates: “Which of these people did you sit with at lunch this week?” Use first initials only.
Enter the answers as a $0/1$ grid in a spreadsheet (row $a$, column $b$). Make it symmetric and put $1$s on the diagonal.
Find a chain $a\to b\to c$ with no direct link $a\to c$ — evidence that the relation is not transitive.
Grow clusters by hand: start from one person, add everyone linked to them, then everyone linked to those, until nothing new appears. Repeat with someone not yet placed.
Check your clusters are disjoint and cover the whole group (Theorem 1.1). Draw them as an arrow diagram.
Tools: Spreadsheet (Google Sheets, Excel or LibreOffice), paper for the diagram; Interactive 1 for small cases.
Hand in: The grid, your cluster list, one broken chain, and a paragraph explaining why “met” alone could not define clusters.
Going further. Real contact-tracing data is anonymised; open network datasets (for example school-contact studies published by epidemiology groups) let you repeat the analysis on hundreds of people. Search for “connected components” — the computer-science name for these clusters.
Instructor note: classes typically split into $2$–$4$ lunch clusters; a single student who links two friend-groups often merges them, which makes a good discussion of how one chain changes the whole partition.
🧪 Self-check quiz
0/5 correct
Q1. The empty relation on $A=\{1,2\}$ is
Q2. The relation “$a\le b$” on $\mathbb R$ is
Q3. The least number of ordered pairs in a reflexive relation on $\{1,2,3\}$ is
Q4. How many equivalence classes does “$a-b$ is divisible by $5$” have on $\mathbb Z$?
Q5. $R$ is an equivalence relation on $A$ and some element $c$ lies in both $[a]$ and $[b]$. Then