NCERT Class 11 Mathematics — Sets, Exercise 1.5. All 7 questions solved.

Everything in this exercise depends on a universal set $U$ being fixed first. The complement of $A$ is what is left of $U$ once $A$ is removed:

$$A’ = \{x : x \in U \text{ and } x \notin A\} = U – A$$

Four laws are used throughout, and questions 4 and 7 exist to make you handle them rather than recite them:

$$A \cup A’ = U, \qquad A \cap A’ = \varphi, \qquad (A’)’ = A, \qquad \varphi’ = U,\ U’ = \varphi$$

De Morgan’s laws are the ones worth committing to memory:

$$(A \cup B)’ = A’ \cap B’, \qquad (A \cap B)’ = A’ \cup B’$$

Key insight. De Morgan’s laws are the statement that negation swaps “or” with “and”. “Not (in $A$ or in $B$)” means “not in $A$ and not in $B$” — which is obvious in ordinary speech and easy to get wrong in symbols. Anyone who writes $(A \cup B)’ = A’ \cup B’$ has kept the connective unchanged, and that is exactly the error the laws exist to prevent.

Question 1

Let $U = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$, $A = \{1, 2, 3, 4\}$, $B = \{2, 4, 6, 8\}$ and $C = \{3, 4, 5, 6\}$. Find

(i) $A’$   (ii) $B’$   (iii) $(A \cup C)’$   (iv) $(A \cup B)’$   (v) $(A’)’$   (vi) $(B – C)’$

Solution. For each part, form the set inside the bracket first, then take whatever of $U$ is left over.

(i) Removing $1, 2, 3, 4$ from $U$: $A’ = \{5, 6, 7, 8, 9\}$

(ii) Removing the evens $2, 4, 6, 8$: $B’ = \{1, 3, 5, 7, 9\}$

(iii) $A \cup C = \{1, 2, 3, 4, 5, 6\}$, so $(A \cup C)’ = \{7, 8, 9\}$

(iv) $A \cup B = \{1, 2, 3, 4, 6, 8\}$, so $(A \cup B)’ = \{5, 7, 9\}$

(v) $A’ = \{5, 6, 7, 8, 9\}$ from (i); complementing again returns everything that was removed:

$$(A’)’ = \{1, 2, 3, 4\} = A$$

(vi) $B – C$ keeps the elements of $B$ that are not in $C$. $B$ and $C$ share $4$ and $6$, so $B – C = \{2, 8\}$ and

$$(B – C)’ = \{1, 3, 4, 5, 6, 7, 9\}$$

Question 2

If $U = \{a, b, c, d, e, f, g, h\}$, find the complements of the following sets:

(i) $A = \{a, b, c\}$   (ii) $B = \{d, e, f, g\}$   (iii) $C = \{a, c, e, g\}$   (iv) $D = \{f, g, h, a\}$

Solution. Each answer is the eight letters of $U$ with the named ones struck out.

A useful check on every part: $A$ and $A’$ between them must account for all eight letters and share none. In (i), $3 + 5 = 8$ ✓; in (iii), $4 + 4 = 8$ ✓.

Question 3

Taking the set of natural numbers as the universal set, write down the complements of the following sets:

(i) $\{x : x \text{ is an even natural number}\}$   (ii) $\{x : x \text{ is an odd natural number}\}$ (iii) $\{x : x \text{ is a positive multiple of } 3\}$   (iv) $\{x : x \text{ is a prime number}\}$ (v) $\{x : x \text{ is a natural number divisible by } 3 \text{ and } 5\}$ (vi) $\{x : x \text{ is a perfect square}\}$   (vii) $\{x : x \text{ is a perfect cube}\}$ (viii) $\{x : x + 5 = 8\}$   (ix) $\{x : 2x + 5 = 9\}$ (x) $\{x : x \ge 7\}$   (xi) $\{x : x \in \mathbf{N} \text{ and } 2x + 1 > 10\}$

Solution. With $U = \mathbf{N}$, the complement of a set is “the natural numbers that fail its condition” — so most parts are answered by negating the condition carefully.

(i) Every natural number is even or odd, so the complement of the evens is the odds: $\{x : x \text{ is an odd natural number}\}$

(ii) Symmetrically: $\{x : x \text{ is an even natural number}\}$

(iii) $\{x : x \in \mathbf{N} \text{ and } x \text{ is not a multiple of } 3\}$

(iv) A natural number that is not prime is either composite or equal to $1$ — and $1$ has to be named explicitly, since it is neither prime nor composite:

$$\{x : x \text{ is a positive composite number or } x = 1\}$$

(v) “Divisible by $3$ and $5$” means divisible by $15$. Negating a conjunction turns it into a disjunction — De Morgan again, now applied to conditions rather than sets:

$$\{x : x \in \mathbf{N},\ x \text{ is not divisible by } 3 \text{ or not divisible by } 5\}$$

Equivalently, $\{x \in \mathbf{N} : x \text{ is not a multiple of } 15\}$.

(vi) $\{x : x \in \mathbf{N} \text{ and } x \text{ is not a perfect square}\}$

(vii) $\{x : x \in \mathbf{N} \text{ and } x \text{ is not a perfect cube}\}$

(viii) The condition $x + 5 = 8$ picks out the single number $3$, so the set is $\{3\}$ and its complement is everything else:

$$\{x : x \in \mathbf{N} \text{ and } x \ne 3\}$$

(ix) $2x + 5 = 9$ gives $x = 2$, so the complement is $\{x : x \in \mathbf{N} \text{ and } x \ne 2\}$

(x) The complement of $x \ge 7$ is $x < 7$, which within $\mathbf{N}$ means

$$\{x : x \in \mathbf{N},\ x < 7\} = \{1, 2, 3, 4, 5, 6\}$$

(xi) Solve the inequality first: $2x + 1 > 10 \Rightarrow x > \tfrac92 = 4.5$. In $\mathbf{N}$ that is $\{5, 6, 7, \ldots\}$, so the complement is

$$\left\{x : x \in \mathbf{N},\ x \le \tfrac92\right\} = \{1, 2, 3, 4\}$$

Question 4

If $U = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$, $A = \{2, 4, 6, 8\}$ and $B = \{2, 3, 5, 7\}$, verify that

(i) $(A \cup B)’ = A’ \cap B’$    (ii) $(A \cap B)’ = A’ \cup B’$

Solution. “Verify” means compute both sides separately and compare — not quote the law. Start with the two complements, since both parts need them.

$$A’ = \{1, 3, 5, 7, 9\}, \qquad B’ = \{1, 4, 6, 8, 9\}$$

(i) Left side: $A \cup B = \{2, 3, 4, 5, 6, 7, 8\}$, so

$$(A \cup B)’ = \{1, 9\}$$

Right side: the elements common to $A’$ and $B’$ are $1$ and $9$, so

$$A’ \cap B’ = \{1, 9\}$$

The two agree. ✓

(ii) Left side: $A \cap B = \{2\}$, the only element in both, so

$$(A \cap B)’ = \{1, 3, 4, 5, 6, 7, 8, 9\}$$

Right side: pooling $A’$ and $B’$ gives every number from $1$ to $9$ except $2$:

$$A’ \cup B’ = \{1, 3, 4, 5, 6, 7, 8, 9\}$$

The two agree. ✓

Notice how the sizes behave: $A \cup B$ has $7$ elements and its complement $2$; $A \cap B$ has $1$ and its complement $8$. In each case the totals come to $9$, which is a quick way to catch a dropped element before comparing lists.

Question 5

Draw appropriate Venn diagrams for each of the following:

(i) $(A \cup B)’$    (ii) $A’ \cap B’$    (iii) $(A \cap B)’$    (iv) $A’ \cup B’$

Solution. The rectangle is the universal set $U$ and the two circles are $A$ and $B$. Shade the region the expression describes. The exercise is really a picture of question 4: (i) and (ii) shade the same region, and so do (iii) and (iv).

(i) $(A \cup B)’$ and (ii) $A’ \cap B’$ — everything outside both circles.

(iii) $(A \cap B)’$ and (iv) $A’ \cup B’$ — everything except the overlap.

Question 6

Let $U$ be the set of all triangles in a plane. If $A$ is the set of all triangles with at least one angle different from $60^\circ$, what is $A’$?

Solution. $A’$ consists of the triangles that fail $A$’s condition. Negating “at least one angle different from $60^\circ$” gives “no angle different from $60^\circ$” — that is, every angle equals $60^\circ$.

A triangle with all three angles $60^\circ$ is equiangular, and an equiangular triangle is equilateral. So

$$A’ = \text{the set of all equilateral triangles}$$

The negation is worth doing slowly: “not (at least one)” is “none”, not “at least one that is equal”.

Question 7

Fill in the blanks to make each of the following a true statement:

(i) $A \cup A’ = \ldots$    (ii) $\varphi’ \cap A = \ldots$ (iii) $A \cap A’ = \ldots$    (iv) $U’ \cap A = \ldots$

Solution.

(i) Every element of $U$ is either in $A$ or outside it, and $A’$ collects exactly those outside. Together they exhaust $U$:

$$A \cup A’ = U$$

(ii) $\varphi’ = U$, and intersecting any subset with the universal set leaves it unchanged:

$$\varphi’ \cap A = U \cap A = A$$

(iii) No element can be both inside and outside $A$:

$$A \cap A’ = \varphi$$

(iv) $U’ = \varphi$, and intersecting with the empty set produces the empty set:

$$U’ \cap A = \varphi \cap A = \varphi$$

Parts (ii) and (iv) are the same manoeuvre with the two roles swapped, which is why they are set together — the work is knowing $\varphi’ = U$ and $U’ = \varphi$, not the intersection itself.

Common mistakes

• Taking a complement without fixing $U$. “The complement of the even numbers” is meaningless until the universal set is named — within $\mathbf{N}$ it is the odd naturals, within $\mathbf{Z}$ the odd integers. Question 3 states $U = \mathbf{N}$ in its stem, and every answer depends on it.
• Question 3(iv), forgetting $1$. The non-primes in $\mathbf{N}$ are the composites and $1$, because $1$ is neither. Writing only “composite numbers” loses an element.
• Question 3(v), negating “and” into “and”. The complement of “divisible by $3$ and $5$” is “not divisible by $3$ or not divisible by $5$”. Keeping the “and” describes a completely different set — the numbers divisible by neither.
• Question 3(xi), forgetting to solve first. $2x + 1 > 10$ has to become $x > 4.5$ before the complement can be written; complementing the inequality in its original form gives $2x + 1 \le 10$, which is correct but leaves the answer unfinished.
• Writing $(A \cup B)’ = A’ \cup B’$. This is the error De Morgan’s laws exist to correct, and question 4 is the arithmetic that refutes it: $A’ \cup B’$ has eight elements where $(A \cup B)’$ has two.
• Question 6, negating “at least one” as “at least one equal”. The opposite of “at least one angle is not $60^\circ$” is “all angles are $60^\circ$” — every triangle in $A’$ is equilateral, with none left over.
• Question 7(ii), reading $\varphi’$ as $\varphi$. The complement of the empty set is the whole universal set. Getting this backwards turns answer (ii) from $A$ into $\varphi$.

Practise next

• Miscellaneous Exercise on Chapter 1 — the proof questions, where the laws verified numerically in question 4 are used as tools rather than checked.
• Exercise 1.4 — worth revising if the difference $B – C$ in question 1(vi) felt awkward, since $A’ $ is simply $U – A$.