NCERT Class 12 Mathematics — Relations and Functions, Exercise 1.1. All 16 questions solved.

Exercise 1.1 is entirely about three properties of a relation $R$ on a set $A$:

• Reflexive — $(a, a) \in R$ for every $a \in A$.
• Symmetric — whenever $(a, b) \in R$, also $(b, a) \in R$.
• Transitive — whenever $(a, b) \in R$ and $(b, c) \in R$, also $(a, c) \in R$.

A relation with all three is an equivalence relation.

The exercise is not really testing whether you know those definitions. It is testing whether you understand the asymmetry between proving and disproving them, which is where nearly every mark is lost.

Key insight. To disprove a property you need exactly one counterexample — a single pair that fails. To prove one you need a general argument covering every element of the set. Students routinely try to prove a property by checking a few pairs, which proves nothing, or try to disprove it with a vague statement instead of naming the pair that breaks it. Name the pair.

One more thing worth knowing before starting: a property can hold vacuously. If no pair $(a, b)$ and $(b, c)$ both lie in $R$, then transitivity has nothing to fail on, and the relation is transitive. Questions 1(ii) and 1(v)(d) both turn on this, and it surprises people.

Question 1

Determine whether each of the following relations are reflexive, symmetric and transitive.

Question 1 (i)

Relation $R$ in the set $A = \{1, 2, 3, \dots, 13, 14\}$ defined as $R = \{(x, y) : 3x – y = 0\}$.

Solution. The condition is $y = 3x$, so listing the pairs that stay inside $A$ makes everything visible at once:

$$R = \{(1, 3), (2, 6), (3, 9), (4, 12)\}$$

Reflexive? We would need $(1, 1) \in R$, which requires $3(1) – 1 = 0$. It equals $2$. Not reflexive.

Symmetric? $(1, 3) \in R$, but $(3, 1)$ requires $3(3) – 1 = 0$, and it equals $8$. Not symmetric.

Transitive? $(1, 3) \in R$ and $(3, 9) \in R$, so transitivity would require $(1, 9) \in R$ — but $3(1) – 9 = -6 \neq 0$. Not transitive.

Question 1 (ii)

Relation $R$ in the set $\mathbf{N}$ of natural numbers defined as $R = \{(x, y) : y = x + 5 \text{ and } x < 4\}$.

Solution. The constraint $x < 4$ restricts $x$ to $1, 2, 3$, so

$$R = \{(1, 6), (2, 7), (3, 8)\}$$

Reflexive? $(1, 1)$ would need $1 = 1 + 5$. Not reflexive.

Symmetric? $(1, 6) \in R$, but $(6, 1)$ fails on both counts — $1 \neq 11$, and $6 \not< 4$. Not symmetric.

Transitive? This is the vacuous case. For a failure we would need $(x, y)$ and $(y, z)$ both in $R$, so $y$ would have to appear as a first coordinate. But the first coordinates are $1, 2, 3$ and the second coordinates are $6, 7, 8$, which share no element. No such chain exists, so nothing can break transitivity. Transitive.

Question 1 (iii)

Relation $R$ in the set $A = \{1, 2, 3, 4, 5, 6\}$ as $R = \{(x, y) : y \text{ is divisible by } x\}$.

Solution. Reflexive? Every $x$ divides itself, so $(x, x) \in R$ for all $x \in A$. Reflexive.

Symmetric? $(1, 2) \in R$ because $2$ is divisible by $1$, but $(2, 1) \notin R$ because $1$ is not divisible by $2$. Not symmetric.

Transitive? Suppose $y$ is divisible by $x$ and $z$ is divisible by $y$. Then $y = kx$ and $z = my$ for some integers $k, m$, so $z = mkx$ — meaning $z$ is divisible by $x$. Transitive.

Notice the shape of that argument: it never mentions a particular number, which is exactly what a proof of transitivity requires.

Question 1 (iv)

Relation $R$ in the set $\mathbf{Z}$ of all integers defined as $R = \{(x, y) : x – y \text{ is an integer}\}$.

Solution. The difference of any two integers is always an integer, so every pair is in $R$ — that is, $R = \mathbf{Z} \times \mathbf{Z}$. A relation containing every possible pair automatically satisfies all three properties.

Spelling it out: $x – x = 0$ is an integer, so reflexive. If $x – y$ is an integer then so is $y – x = -(x-y)$, so symmetric. If $x – y$ and $y – z$ are integers then so is their sum $x – z$, so transitive.

Question 1 (v)

Relation $R$ in the set $A$ of human beings in a town at a particular time.

(a) $R = \{(x, y) : x$ and $y$ work at the same place$\}$. Everyone works where they work (reflexive); if $x$ works where $y$ does then $y$ works where $x$ does (symmetric); and if $x$ and $y$ share a workplace and $y$ and $z$ share one, all three share it (transitive).

(b) $R = \{(x, y) : x$ and $y$ live in the same locality$\}$. Identical reasoning to (a).

(c) $R = \{(x, y) : x$ is exactly 7 cm taller than $y\}$. Nobody is 7 cm taller than themselves, so not reflexive. If $x$ is 7 cm taller than $y$, then $y$ is 7 cm shorter, so not symmetric. And if $x$ is 7 cm taller than $y$ and $y$ is 7 cm taller than $z$, then $x$ is 14 cm taller than $z$, not 7 — so not transitive.

(d) $R = \{(x, y) : x$ is wife of $y\}$. Not reflexive, and not symmetric (if $x$ is the wife of $y$, then $y$ is the husband of $x$, not the wife). Transitivity is vacuous again: for a chain we would need $y$ to be both a wife and a husband simultaneously, which cannot happen, so no chain exists and nothing fails.

(e) $R = \{(x, y) : x$ is father of $y\}$. Not reflexive, not symmetric. Not transitive either: if $x$ is the father of $y$ and $y$ is the father of $z$, then $x$ is the grandfather of $z$.

Question 2

Show that the relation $R$ in the set $\mathbf{R}$ of real numbers, defined as $R = \{(a, b) : a \leq b^2\}$, is neither reflexive nor symmetric nor transitive.

Solution. All three failures need counterexamples, and the useful trick for this relation is to work with a fraction, where squaring makes a number smaller.

Not reflexive. Take $a = \tfrac{1}{2}$. Then $a^2 = \tfrac{1}{4}$, and $\tfrac{1}{2} \leq \tfrac{1}{4}$ is false, so $\left(\tfrac12, \tfrac12\right) \notin R$.

Not symmetric. Take $(1, 2)$: $1 \leq 4$ is true, so $(1, 2) \in R$. But $(2, 1)$ needs $2 \leq 1$, which is false.

Not transitive. Take $(10, 4)$ and $(4, 3)$: $10 \leq 16$ and $4 \leq 9$ are both true. Transitivity would require $(10, 3)$, needing $10 \leq 9$ — false.

Question 3

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

Solution. Listing $R$ makes all three answers immediate:

$$R = \{(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)\}$$

Reflexive? $(1, 1)$ would need $1 = 2$. Not reflexive.

Symmetric? $(1, 2) \in R$ but $(2, 1)$ would need $1 = 3$. Not symmetric.

Transitive? $(1, 2)$ and $(2, 3)$ are both in $R$, so we would need $(1, 3)$ — which requires $3 = 2$. Not transitive.

Question 4

Show that the relation $R$ in $\mathbf{R}$ defined as $R = {(a, b) : a \leq b}$ is reflexive and transitive but not symmetric.

Solution. Reflexive. $a \leq a$ holds for every real number, so $(a, a) \in R$ always.

Transitive. If $a \leq b$ and $b \leq c$, then $a \leq c$ — this is precisely the transitivity of the order relation on $\mathbf{R}$.

Not symmetric. One counterexample suffices: $(1, 2) \in R$ since $1 \leq 2$, but $(2, 1) \notin R$ since $2 \leq 1$ is false.

Question 5

Check whether the relation $R$ in $\mathbf{R}$ defined by $R = {(a, b) : a \leq b^3}$ is reflexive, symmetric or transitive.

Solution. The same idea as question 2, since cubing a proper fraction also shrinks it.

Not reflexive. $a = \tfrac{1}{2}$ gives $a^3 = \tfrac{1}{8}$, and $\tfrac{1}{2} \leq \tfrac{1}{8}$ is false.

Not symmetric. $(1, 2) \in R$ since $1 \leq 8$, but $(2, 1)$ needs $2 \leq 1$ — false.

Not transitive. $(10, 3) \in R$ since $10 \leq 27$, and $(3, \tfrac{3}{2}) \in R$ since $3 \leq \tfrac{27}{8} = 3.375$. But $(10, \tfrac{3}{2})$ would need $10 \leq 3.375$ — false.

Question 6

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

Solution. Symmetric. $R$ contains exactly two pairs, and each is the reverse of the other. There is nothing else to check.

Not reflexive. $(1, 1) \notin R$ — indeed no pair of the form $(a, a)$ is in $R$ at all.

Not transitive. $(1, 2) \in R$ and $(2, 1) \in R$, so transitivity demands $(1, 1) \in R$. It is not. This is the standard trap: a symmetric relation is transitive only if it also contains the resulting “round trip” pairs.

Question 7

Show that the relation $R$ in the set $A$ of all the books in a library of a college, given by $R = \{(x, y) : x$ and $y$ have same number of pages$\}$, is an equivalence relation.

Solution. Write $p(x)$ for the number of pages in book $x$. Then $(x, y) \in R$ means exactly $p(x) = p(y)$, and the three properties follow from the three properties of equality.

Reflexive. $p(x) = p(x)$, so $(x, x) \in R$.

Symmetric. If $p(x) = p(y)$ then $p(y) = p(x)$, so $(y, x) \in R$.

Transitive. If $p(x) = p(y)$ and $p(y) = p(z)$ then $p(x) = p(z)$.

Any relation of the form “$x$ and $y$ have the same something” is an equivalence relation for this reason. Questions 7, 11, 13 and 14 are all the same problem wearing different clothes.

Question 8

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

Solution. The key observation is that $|a – b|$ is even exactly when $a$ and $b$ have the same parity — both odd or both even.

Reflexive. $|a – a| = 0$, which is even.

Symmetric. $|a – b| = |b – a|$, so the condition is unchanged by swapping.

Transitive. If $a, b$ have the same parity and $b, c$ have the same parity, then $a, c$ have the same parity, so $|a – c|$ is even.

For the second part: $\{1, 3, 5\}$ are all odd, so any two of them differ by an even number and are related. $\{2, 4\}$ are both even, likewise. But an odd number and an even number differ by an odd number, so no element of one subset is related to any element of the other.

Question 9

Show that each of the relation $R$ in the set $A = {x \in \mathbf{Z} : 0 \leq x \leq 12}$ is an equivalence relation, and find the set of all elements related to $1$ in each case.

Question 9 (i)

$R = \{(a, b) : |a – b| \text{ is a multiple of } 4\}$.

Solution. Reflexive. $|a – a| = 0 = 4 \times 0$, a multiple of $4$.

Symmetric. $|a – b| = |b – a|$.

Transitive. If $a – b = 4m$ and $b – c = 4n$, then $a – c = 4(m + n)$.

For the elements related to $1$: we need $|a – 1|$ to be a multiple of $4$, so $a \in \{1, 5, 9\}$ once we discard $a = -3$ and $a = 13$, which fall outside $A = \{0, 1, 2, \dots, 12\}$.

Question 9 (ii)

$R = \{(a, b) : a = b\}$.

Solution. Equality is the archetypal equivalence relation — reflexive, symmetric and transitive by definition. The only element equal to $1$ is $1$ itself.

Question 10

Give an example of a relation which is:

(i) Symmetric but neither reflexive nor transitive. On $A = \{1, 2, 3\}$, take $R = \{(1, 2), (2, 1)\}$ — this is question 6.

(ii) Transitive but neither reflexive nor symmetric. On $A = \{1, 2, 3\}$, take $R = \{(a, b) : a < b\} = \{(1, 2), (1, 3), (2, 3)\}$. Strict inequality is transitive, never reflexive, and never symmetric.

(iii) Reflexive and symmetric but not transitive. On $A = \{1, 2, 3\}$, take

$$R = \{(1,1), (2,2), (3,3), (1,2), (2,1), (2,3), (3,2)\}$$

This is reflexive and symmetric by inspection, but $(1, 2)$ and $(2, 3)$ are in $R$ while $(1, 3)$ is not.

(iv) Reflexive and transitive but not symmetric. On $A = \{1, 2, 3\}$, take $R = \{(a, b) : a \leq b\}$ — this is question 4 restricted to a finite set.

(v) Symmetric and transitive but not reflexive. On $A = \{1, 2, 3\}$, take $R = \{(1,1), (2,2), (1,2), (2,1)\}$. It is symmetric and transitive, but $(3, 3) \notin R$, so it fails reflexivity — reflexivity demands every element of $A$, including the one the relation ignores entirely.

Question 11

Show that the relation $R$ in the set $A$ of points in a plane given by $R = \{(\mathrm{P}, \mathrm{Q}) : $ distance of the point $\mathrm{P}$ from the origin is same as the distance of the point $\mathrm{Q}$ from the origin$\}$, is an equivalence relation. Further, show that the set of all points related to a point $\mathrm{P} \neq (0, 0)$ is the circle passing through $\mathrm{P}$ with origin as centre.

Solution. Write $d(\mathrm{P})$ for the distance of $\mathrm{P}$ from the origin $\mathrm{O}$. Then $(\mathrm{P}, \mathrm{Q}) \in R$ means $d(\mathrm{P}) = d(\mathrm{Q})$, and as in question 7 the three properties come straight from the properties of equality.

For the second part, fix $\mathrm{P} \neq (0,0)$ and let $r = d(\mathrm{P}) > 0$. A point $\mathrm{Q}$ is related to $\mathrm{P}$ precisely when $d(\mathrm{Q}) = r$ — that is, when $\mathrm{Q}$ lies at distance $r$ from the origin. The set of all such points is by definition the circle of radius $r$ centred at the origin, and $\mathrm{P}$ itself lies on it.

Question 12

Show that the relation $R$ defined in the set $A$ of all triangles as $R = \{(\mathrm{T}_1, \mathrm{T}_2) : \mathrm{T}_1$ is similar to $\mathrm{T}_2\}$, is an equivalence relation. Consider three right angle triangles $\mathrm{T}_1$ with sides $3, 4, 5$; $\mathrm{T}_2$ with sides $5, 12, 13$; and $\mathrm{T}_3$ with sides $6, 8, 10$. Which triangles among $\mathrm{T}_1$, $\mathrm{T}_2$ and $\mathrm{T}_3$ are related?

Solution. Reflexive. Every triangle is similar to itself, with ratio $1$.

Symmetric. If $\mathrm{T}_1 \sim \mathrm{T}_2$ with ratio $k$, then $\mathrm{T}_2 \sim \mathrm{T}_1$ with ratio $\tfrac{1}{k}$.

Transitive. If $\mathrm{T}_1 \sim \mathrm{T}_2$ with ratio $k$ and $\mathrm{T}_2 \sim \mathrm{T}_3$ with ratio $m$, then $\mathrm{T}_1 \sim \mathrm{T}_3$ with ratio $km$.

For the three triangles, similarity means corresponding sides are in a constant ratio. Comparing $\mathrm{T}_1$ and $\mathrm{T}_3$:

$$\frac{6}{3} = \frac{8}{4} = \frac{10}{5} = 2$$

so $\mathrm{T}_1 \sim \mathrm{T}_3$. Comparing $\mathrm{T}_1$ and $\mathrm{T}_2$: $\tfrac{5}{3}$, $\tfrac{12}{4} = 3$ and $\tfrac{13}{5}$ are not equal, so they are not similar. Being right-angled is not enough — the sides must scale together.

Question 13

Show that the relation $R$ defined in the set $A$ of all polygons as $R = \{(\mathrm{P}_1, \mathrm{P}_2) : \mathrm{P}_1$ and $\mathrm{P}_2$ have same number of sides$\}$, is an equivalence relation. What is the set of all elements in $A$ related to the right angle triangle $\mathrm{T}$ with sides $3, 4$ and $5$?

Solution. This is the “same number of pages” relation of question 7 again, with sides in place of pages, so it is an equivalence relation for the same reason.

The triangle $\mathrm{T}$ has three sides, so the polygons related to it are exactly those with three sides. Note that the specific lengths $3, 4, 5$ are irrelevant — the relation only sees the count of sides, so the answer is every triangle, not just those similar to $\mathrm{T}$.

Question 14

Let $L$ be the set of all lines in the XY plane and $R$ be the relation in $L$ defined as $R = \{(\mathrm{L}_1, \mathrm{L}_2) : \mathrm{L}_1$ is parallel to $\mathrm{L}_2\}$. Show that $R$ is an equivalence relation. Find the set of all lines related to the line $y = 2x + 4$.

Solution. Two lines are parallel exactly when they have the same slope, so once again this is an equality of some attribute, and the three properties follow as in question 7. (Reflexivity requires treating a line as parallel to itself, which is the convention NCERT adopts here.)

The line $y = 2x + 4$ has slope $2$. A line is related to it precisely when its slope is also $2$, and every such line has the form $y = 2x + c$ for some real constant $c$. Different values of $c$ shift the line up or down without changing its direction.

Question 15

Let $R$ be the relation in the set $\{1, 2, 3, 4\}$ given by

$$R = \{(1,2), (2,2), (1,1), (4,4), (1,3), (3,3), (3,2)\}$$

Choose the correct answer.

Solution. Reflexive? The set has four elements, so we need all of $(1,1), (2,2), (3,3), (4,4)$. All four are present. Reflexive.

Symmetric? $(1, 2) \in R$ but $(2, 1) \notin R$. Not symmetric. That alone eliminates (A), (C) and (D).

Transitive? Confirming (B) properly means checking every chain. The only non-trivial ones are $(1,3)$ with $(3,2)$, requiring $(1,2)$ — present; $(1,2)$ with $(2,2)$, requiring $(1,2)$ — present; and $(3,2)$ with $(2,2)$, requiring $(3,2)$ — present. Chains starting from a pair $(a,a)$ add nothing new. Transitive.

Question 16

Let $R$ be the relation in the set $\mathbf{N}$ given by $R = \{(a, b) : a = b – 2,\ b > 6\}$. Choose the correct answer.

Solution. Both conditions must hold, and the quickest route is to test each option against them.

(A) $(2, 4)$: here $b = 4$, and $4 > 6$ is false. Rejected on the second condition, even though $2 = 4 – 2$ is true.

(B) $(3, 8)$: $b = 8 > 6$ holds, but $a$ would need to be $8 – 2 = 6$, not $3$.

(C) $(6, 8)$: $b = 8 > 6$ holds, and $6 = 8 – 2$ holds. Both conditions satisfied.

(D) $(8, 7)$: $b = 7 > 6$ holds, but $a$ would need to be $7 – 2 = 5$, not $8$.

Common mistakes

• Checking a few pairs and calling a property proved. Verifying that $(1,1)$, $(2,2)$ and $(3,3)$ lie in $R$ proves reflexivity only if those are all the elements. On an infinite set like $\mathbf{R}$ or $\mathbf{Z}$, no amount of checking is a proof — you need a general argument.
• Missing that a property can hold vacuously. In question 1(ii) there is no chain $(x,y), (y,z)$ at all, so transitivity cannot fail and the relation is transitive. Writing “not transitive because there are no such pairs” is exactly backwards.
• Assuming symmetric plus transitive implies reflexive. It is tempting to argue that $(a,b)$ and $(b,a)$ give $(a,a)$ — but that only works for elements that appear in some pair. Question 10(v) is built on precisely this gap: the element $3$ appears in no pair, so $(3,3)$ never arises.
• Forgetting to check every condition in the definition. In question 16, option (A) satisfies $a = b – 2$ and fails only on $b > 6$. Stopping after the first condition gets it wrong.
• Trying to disprove a property without naming the pair. “It is not symmetric because the relation is one-directional” earns nothing. “$(1,2) \in R$ but $(2,1) \notin R$” is the whole answer.
• Reading question 13 as being about similar triangles. The relation counts sides and nothing else, so every triangle qualifies — the lengths $3, 4, 5$ in the question are a deliberate distraction.

Practise next

• Exercise 1.2 — one-one and onto functions, where the same prove-versus-disprove asymmetry applies to injectivity and surjectivity.
• Miscellaneous Exercise on Chapter 1 — mixed questions drawing on both exercises together.