NCERT Class 11 Mathematics — Sets, Exercise 1.4. All 12 questions solved.

This is the longest exercise of the chapter and the most mechanical, which makes it the one where care matters most. Three operations do all the work:

$$A \cup B = \{x : x \in A \text{ or } x \in B\}$$ $$A \cap B = \{x : x \in A \text{ and } x \in B\}$$ $$A – B = \{x : x \in A \text{ and } x \notin B\}$$

Two sets are disjoint when $A \cap B = \varphi$ — they share nothing.

Key insight. Union and intersection are symmetric: $A \cup B = B \cup A$ and $A \cap B = B \cap A$. Difference is not. $A – B$ and $B – A$ are usually different sets, and are always disjoint from each other. Question 9 asks for twelve differences precisely so that the asymmetry becomes unmissable — compare its (i) with its (iv), or its (xi) with its (xii).

Question 1

Find the union of each of the following pairs of sets:

(i) $X = \{1, 3, 5\}$, $Y = \{1, 2, 3\}$ (ii) $A = \{a, e, i, o, u\}$, $B = \{a, b, c\}$ (iii) $A = \{x : x \text{ is a natural number and multiple of } 3\}$, $B = \{x : x \text{ is a natural number less than } 6\}$ (iv) $A = \{x : x \text{ is a natural number and } 1 < x \le 6\}$, $B = \{x : x \text{ is a natural number and } 6 < x < 10\}$ (v) $A = \{1, 2, 3\}$, $B = \varphi$

Solution. The union pools both lists, writing each element once even where it appears in both.

(i) $1$ and $3$ are common, so they appear once:

$$X \cup Y = \{1, 2, 3, 5\}$$

(ii) $a$ is common:

$$A \cup B = \{a, b, c, e, i, o, u\}$$

(iii) $A = \{3, 6, 9, 12, \ldots\}$ is infinite and $B = \{1, 2, 3, 4, 5\}$ is finite. The union keeps every multiple of $3$ and adds $1, 2, 4, 5$ (with $3$ already present):

$$A \cup B = \{1, 2, 3, 4, 5, 6, 9, 12, 15, \ldots\} = \{x : x = 1, 2, 4, 5 \text{ or } x \text{ is a multiple of } 3\}$$

(iv) $A = \{2, 3, 4, 5, 6\}$ and $B = \{7, 8, 9\}$. They abut without overlapping, so the union closes the gap:

$$A \cup B = \{2, 3, 4, 5, 6, 7, 8, 9\} = \{x : x \in \mathbf{N},\ 1 < x < 10\}$$

(v) Uniting with the empty set adds nothing:

$$A \cup B = \{1, 2, 3\}$$

Question 2

Let $A = \{a, b\}$, $B = \{a, b, c\}$. Is $A \subset B$? What is $A \cup B$?

Solution. Both $a$ and $b$ belong to $B$, so yes, $A \subset B$. And since everything in $A$ is already in $B$, the union adds nothing new:

$$A \cup B = \{a, b, c\} = B$$

Question 3

If $A$ and $B$ are two sets such that $A \subset B$, then what is $A \cup B$?

Solution. This is question 2 stated in general. If every element of $A$ already lies in $B$, then pooling the two lists produces exactly $B$’s list:

$$A \subset B \quad\Longleftrightarrow\quad A \cup B = B$$

The equivalence runs both ways, and it is one of the four conditions the miscellaneous exercise asks you to prove equivalent.

Question 4

If $A = \{1, 2, 3, 4\}$, $B = \{3, 4, 5, 6\}$, $C = \{5, 6, 7, 8\}$ and $D = \{7, 8, 9, 10\}$; find

(i) $A \cup B$   (ii) $A \cup C$   (iii) $B \cup C$   (iv) $B \cup D$ (v) $A \cup B \cup C$   (vi) $A \cup B \cup D$   (vii) $B \cup C \cup D$

Solution. The four sets form a chain, each overlapping the next in two elements and the one after that in none. Merge and sort:

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

(ii) $A$ and $C$ are disjoint, so nothing collapses: $\{1, 2, 3, 4, 5, 6, 7, 8\}$

(iii) $\{3, 4, 5, 6, 7, 8\}$

(iv) $B$ and $D$ are disjoint: $\{3, 4, 5, 6, 7, 8, 9, 10\}$

(v) Building on (i) and adding $C$: $\{1, 2, 3, 4, 5, 6, 7, 8\}$ — the same as (ii), because $C$’s new elements $7, 8$ were already contributed there.

(vi) $\{1, 2, 3, 4, 5, 6\} \cup \{7,8,9,10\} = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$

(vii) $\{3, 4, 5, 6, 7, 8, 9, 10\}$

Question 5

Find the intersection of each pair of sets of question 1 above.

Solution. Now keep only what appears in both lists.

(i) $X \cap Y$: common to $\{1,3,5\}$ and $\{1,2,3\}$ are $1$ and $3$: $\{1, 3\}$

(ii) $A \cap B$: the only shared letter is $a$: $\{a\}$

(iii) Multiples of $3$ among $\{1,2,3,4,5\}$: just $3$, so $\{3\}$

(iv) $\{2,3,4,5,6\} \cap \{7,8,9\} = \varphi$. The two sets abut but do not overlap, which is why the union in question 1(iv) filled a continuous range.

(v) $\{1,2,3\} \cap \varphi = \varphi$. Intersecting with the empty set always empties everything.

Question 6

If $A = \{3, 5, 7, 9, 11\}$, $B = \{7, 9, 11, 13\}$, $C = \{11, 13, 15\}$ and $D = \{15, 17\}$; find

(i) $A \cap B$   (ii) $B \cap C$   (iii) $A \cap C \cap D$   (iv) $A \cap C$ (v) $B \cap D$   (vi) $A \cap (B \cup C)$   (vii) $A \cap D$   (viii) $A \cap (B \cup D)$ (ix) $(A \cap B) \cap (B \cup C)$   (x) $(A \cup D) \cap (B \cup C)$

Solution. Where brackets appear, evaluate the inner operation first — the order genuinely changes the answer.

(i) $\{7, 9, 11\}$

(ii) $\{11, 13\}$

(iii) $A \cap C = \{11\}$ by (iv), and $11 \notin D$, so the result is $\varphi$.

(iv) $\{11\}$

(v) $B$ ends at $13$ and $D$ starts at $15$: $\varphi$

(vi) $B \cup C = \{7, 9, 11, 13, 15\}$; intersecting with $A$ keeps $\{7, 9, 11\}$.

(vii) $\varphi$

(viii) $B \cup D = \{7, 9, 11, 13, 15, 17\}$; intersecting with $A$ gives $\{7, 9, 11\}$ again — the extra elements $15, 17$ were never in $A$ to begin with.

(ix) $A \cap B = \{7, 9, 11\}$ and $B \cup C = \{7, 9, 11, 13, 15\}$. Every element of the first lies in the second, so the intersection is $\{7, 9, 11\}$. (This had to happen: $A \cap B \subset B \subset B \cup C$.)

(x) $A \cup D = \{3, 5, 7, 9, 11, 15, 17\}$ and $B \cup C = \{7, 9, 11, 13, 15\}$. Common: $\{7, 9, 11, 15\}$ — the $15$ arrives from $D$ on the left and $C$ on the right, which is what distinguishes this part from (vi).

Question 7

If $A = \{x : x \text{ is a natural number}\}$, $B = \{x : x \text{ is an even natural number}\}$, $C = \{x : x \text{ is an odd natural number}\}$ and $D = \{x : x \text{ is a prime number}\}$, find

(i) $A \cap B$   (ii) $A \cap C$   (iii) $A \cap D$   (iv) $B \cap C$   (v) $B \cap D$   (vi) $C \cap D$

Solution. $B$, $C$ and $D$ are all subsets of $A$, which settles the first three parts immediately: $X \subset A \Rightarrow A \cap X = X$.

(i) $A \cap B = B$    (ii) $A \cap C = C$    (iii) $A \cap D = D$

(iv) No natural number is both even and odd, so $B \cap C = \varphi$.

(v) An even prime must have $2$ as a divisor and no other, so it is $2$ itself:

$$B \cap D = \{2\}$$

(vi) Every prime except $2$ is odd, so

$$C \cap D = \{x : x \text{ is an odd prime number}\} = \{3, 5, 7, 11, 13, \ldots\}$$

Parts (v) and (vi) together partition the primes: $D = \{2\} \cup (C \cap D)$.

Question 8

Which of the following pairs of sets are disjoint?

(i) $\{1, 2, 3, 4\}$ and $\{x : x \text{ is a natural number and } 4 \le x \le 6\}$ (ii) $\{a, e, i, o, u\}$ and $\{c, d, e, f\}$ (iii) $\{x : x \text{ is an even integer}\}$ and $\{x : x \text{ is an odd integer}\}$

Solution. Disjoint means the intersection is empty, so hunt for a single shared element.

(i) Not disjoint. The second set is $\{4, 5, 6\}$, and $4$ belongs to both. The $\le$ at the lower end is what does it — with $4 < x$ the pair would have been disjoint.

(ii) Not disjoint. $e$ is a vowel and also appears in $\{c, d, e, f\}$.

(iii) Disjoint. Every integer is exactly one of even or odd, never both.

Question 9

If $A = \{3, 6, 9, 12, 15, 18, 21\}$, $B = \{4, 8, 12, 16, 20\}$, $C = \{2, 4, 6, 8, 10, 12, 14, 16\}$, $D = \{5, 10, 15, 20\}$; find

(i) $A – B$   (ii) $A – C$   (iii) $A – D$   (iv) $B – A$ (v) $C – A$   (vi) $D – A$   (vii) $B – C$   (viii) $B – D$ (ix) $C – B$   (x) $D – B$   (xi) $C – D$   (xii) $D – C$

Solution. $A – B$ keeps every element of $A$ that is not in $B$. Work from the first set’s list each time and strike out the shared elements.

(i) $A$ and $B$ share only $12$: $\{3, 6, 9, 15, 18, 21\}$

(ii) $A$ and $C$ share $6$ and $12$: $\{3, 9, 15, 18, 21\}$

(iii) $A$ and $D$ share $15$: $\{3, 6, 9, 12, 18, 21\}$

(iv) Now start from $B$ and remove $12$: $\{4, 8, 16, 20\}$ — a completely different set from (i), which is the asymmetry in action.

(v) From $C$, remove $6$ and $12$: $\{2, 4, 8, 10, 14, 16\}$

(vi) From $D$, remove $15$: $\{5, 10, 20\}$

(vii) $B$ and $C$ share $4, 8, 12, 16$, leaving $\{20\}$

(viii) $B$ and $D$ share $20$: $\{4, 8, 12, 16\}$

(ix) From $C$, remove $4, 8, 12, 16$: $\{2, 6, 10, 14\}$

(x) From $D$, remove $20$: $\{5, 10, 15\}$

(xi) $C$ and $D$ share $10$: $\{2, 4, 6, 8, 12, 14, 16\}$

(xii) From $D$, remove $10$: $\{5, 15, 20\}$

Question 10

If $X = \{a, b, c, d\}$ and $Y = \{f, b, d, g\}$, find

(i) $X – Y$    (ii) $Y – X$    (iii) $X \cap Y$

Solution. The shared elements are $b$ and $d$.

(i) Remove them from $X$: $\{a, c\}$

(ii) Remove them from $Y$: $\{f, g\}$

(iii) $\{b, d\}$

The three answers partition $X \cup Y$ into three disjoint pieces — what is in $X$ only, in $Y$ only, and in both. That is exactly the picture the Venn diagram in the textbook draws.

Question 11

If $\mathbf{R}$ is the set of real numbers and $\mathbf{Q}$ is the set of rational numbers, then what is $\mathbf{R} – \mathbf{Q}$?

Solution. $\mathbf{R} – \mathbf{Q}$ collects the real numbers that are not rational — that is, those which cannot be written as $\dfrac{p}{q}$ with $p, q$ integers and $q \ne 0$. These are precisely the irrational numbers: $\sqrt2$, $\pi$, $e$, and so on.

Question 12

State whether each of the following statements is true or false. Justify your answer.

(i) $\{2, 3, 4, 5\}$ and $\{3, 6\}$ are disjoint sets. (ii) $\{a, e, i, o, u\}$ and $\{a, b, c, d\}$ are disjoint sets. (iii) $\{2, 6, 10, 14\}$ and $\{3, 7, 11, 15\}$ are disjoint sets. (iv) $\{2, 6, 10\}$ and $\{3, 7, 11\}$ are disjoint sets.

Solution.

(i) False. $3$ lies in both, so the intersection is $\{3\} \ne \varphi$.

(ii) False. $a$ is common to both.

(iii) True. The first set holds even numbers, the second odd ones; no element can appear in both.

(iv) True. Same reason — the sets are (iii) with one element dropped from each, and removing elements cannot create an overlap.

Common mistakes

• Repeating shared elements in a union. In question 1(i), writing $\{1, 1, 2, 3, 3, 5\}$ treats the union as a list rather than a set. Each element appears once, however many of the original sets contributed it.
• Question 4(v), recomputing from scratch instead of building up. $A \cup B \cup C$ is $(A \cup B) \cup C$, and part (i) has already produced $A \cup B$. Re-deriving each part independently is where sign-of-fatigue errors creep into a twelve-part question.
• Question 6(iii), reading $A \cap C \cap D$ as $A \cap (C \cup D)$. With three intersections the answer must lie in all three sets, and $\{11\}$ fails at $D$.
• Question 8(i), missing the $\le$. $\{x \in \mathbf{N} : 4 \le x \le 6\}$ includes $4$, so the pair is not disjoint. Had the condition read $4 < x \le 6$ the answer would flip.
• Question 9, computing $B – A$ by negating $A – B$. They are different sets built from different starting lists. Always begin from the set written first.
• Question 12, confusing “disjoint” with “unequal”. $\{2,3,4,5\}$ and $\{3,6\}$ are certainly not equal, but that is not what is being asked. Disjoint is the much stronger claim that they share nothing at all.

Practise next

• Exercise 1.5 — complements and De Morgan’s laws, where $A’ = U – A$ makes the difference operation of question 9 into a named object.
• Miscellaneous Exercise on Chapter 1 — proofs built from these same operations, including the four-way equivalence that question 3 hints at.