MathsCBSE Class 11 · Sets

Set Operations, Complements and De Morgan's Laws (Class 11)

Updated 17 October 2026 · Fact-checked

A ∪ B holds elements in A or B, A ∩ B those in both, A − B those in A but not B, and A' those of the universal set not in A. De Morgan's laws: (A ∪ B)' = A' ∩ B' and (A ∩ B)' = A' ∪ B'. Counting: n(A ∪ B) = n(A) + n(B) − n(A ∩ B), and for three sets n(A ∪ B ∪ C) = n(A) + n(B) + n(C) − n(A ∩ B) − n(B ∩ C) − n(A ∩ C) + n(A ∩ B ∩ C).

Notice that the region outside both circles, {7, 8}, is (A ∪ B)′ and also A′ ∩ B′, which is De Morgan's first law.

Illustration drawn for QuestionBuddy.

Understand Set Operations, Complements and De Morgan's Laws

Fix a universal set $U$ ; all sets in a problem are subsets of $U$ .

Operations.

Union: $A \cup B = {x : x \in A or x \in B}$ (or includes both).
Intersection: $A \cap B = {x : x \in A and x \in B}$ ; if $A \cap B = \emptyset$ the sets are disjoint.
Difference: $A - B = {x : x \in A, x \in / B}$ . In general $A - B \neq = B - A$ , and $A - B = A \cap B^{'}$ .
Complement: $A^{'} = U - A$ .

Laws. Both $\cup$ and $\cap$ are commutative and associative; each distributes over the other: $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$ and $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$ . Also $A \cup A^{'} = U$ , $A \cap A^{'} = \emptyset$ , $(A^{'})^{'} = A$ , $\emptyset^{'} = U$ .

De Morgan's laws. $(A \cup B)^{'} = A^{'} \cap B^{'}$ and $(A \cap B)^{'} = A^{'} \cup B^{'}$ . In words: the complement of "in A or B" is "in neither", and the complement of "in both" is "missing from at least one".

Proving set identities. (1) *Element method:* show each side is a subset of the other by taking a general element $x$ and tracking membership. (2) *Verification:* for given sets, compute both sides explicitly and compare. For example, with $U = {1, \dots, 8}$ , $A = {1, 2, 3, 4}$ and $B = {3, 4, 5, 6}$ : $(A \cup B)^{'} = {7, 8} = A^{'} \cap B^{'}$ . An example verifies the law for those sets only; it does not prove it in general.

Counting. For finite sets,
$ $n (A \cup B) = n (A) + n (B) - n (A \cap B),$ $
$ $n (A \cup B \cup C) = \sum n (A) - \sum n (A \cap B) + n (A \cap B \cap C) .$ $
Draw a Venn diagram and fill the innermost region first (all three), then the pairs-only regions (pair total minus the triple), then the "only" regions by subtraction, and finally "none" as $n (U)$ minus the union. The practical problems on union and intersection are assessed only formatively in 2026-27, but they are the best check on the laws.

Bounds. Since $A \cap B \subseteq A$ and $n (A \cup B) \leq n (U)$ : $max (0, n (A) + n (B) - n (U)) \leq n (A \cap B) \leq min (n (A), n (B))$ .

Set Operations, Complements and De Morgan's Laws: formulas to remember

Difference as intersection: $A - B = A \cap B^{'}$; Useful in proofs.
De Morgan: $(A \cup B)^{'} = A^{'} \cap B^{'}; (A \cap B)^{'} = A^{'} \cup B^{'}$; Complement swaps union and intersection.
Two sets: $n (A \cup B) = n (A) + n (B) - n (A \cap B)$; Subtract the overlap once.
Three sets: $n (A \cup B \cup C) = Σ n (A) - Σ n (A \cap B) + n (A \cap B \cap C)$; Alternate the signs.
Distributive laws: $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$; And A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C).

How to solve Set Operations, Complements and De Morgan's Laws questions

Use these steps for set operations, proofs and counting.

1Write the universal set and list each set in roster form.
2Compute the operation step by step; for a complement use $U$ .
3To verify an identity, compute both sides separately and compare the lists.
4To prove an identity, take a general element $x$ and show each side is a subset of the other, or use known laws.
5For counting problems, draw a Venn diagram, fill the innermost region first, and work outwards by subtraction.
6Check that the regions add up to $n (U)$ .

Set Operations, Complements and De Morgan's Laws in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Union, intersection and difference of customer lists

Suppose a shop keeps two lists: shoe buyers $S = {Ravi, Meera, Sana, Tom}$ and bag buyers $B = {Meera, Tom, Uma}$ .

Find the customers who bought both, only shoes, and at least one of the two.

Try it first, then show the solutionSolution

Both: $S \cap B = {Meera, Tom}$ .
Only shoes: $S - B = {Ravi, Sana}$ .
At least one: $S \cup B = {Ravi, Meera, Sana, Tom, Uma}$ , five customers (check: $4 + 3 - 2 = 5$ ).

Answer: Both: Meera, Tom. Only shoes: Ravi, Sana. At least one: 5 customers.

Memory hook: Intersection is the overlap; difference removes the overlap from one set.

Clubs and the "neither" group

Suppose a club has 40 members. 24 take part in maths events, 18 in science events and 6 in both.

How many take part in neither?

Try it first, then show the solutionSolution

$n (M \cup S) = 24 + 18 - 6 = 36$ .
Neither $= 40 - 36 = 4$ .

Answer: 4 members.

Memory hook: None = Total − (A ∪ B).

Tricks to remember Set Operations, Complements and De Morgan's Laws

Short memory aids. Say each one aloud once, then use it on a question the same day.

Subtract the overlap
Adding n(A) and n(B) counts the overlap twice.
Use it when: Two-set counting.
Complement swaps the sign
De Morgan: break the bar, change ∪ to ∩ (or the reverse).
Use it when: Simplifying complements.
Innermost first
Fill the triple overlap before the pairs and the single regions.
Use it when: Three-set Venn diagrams.
A − B is A ∩ B'
Rewrite every difference as an intersection to apply the laws.
Use it when: Proofs.

Set Operations, Complements and De Morgan's Laws: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Basic operations · 2 marks · Very short answerBoard level

Let $U = {1, 2, \dots, 10}$ , $A = {1, 3, 5, 7, 9}$ , $B = {2, 3, 5, 7}$ . Find $A \cup B$ , $A \cap B$ , $A - B$ , $A^{'}$ and $(A \cup B)^{'}$ .

Try it first, then show the solutionSolution

$A \cup B = {1, 2, 3, 5, 7, 9}$ , $A \cap B = {3, 5, 7}$ . (1 mark)
$A - B = {1, 9}$ , $A^{'} = {2, 4, 6, 8, 10}$ , $(A \cup B)^{'} = {4, 6, 8, 10}$ . (1 mark)

Answer: See working

Example 2: Verifying De Morgan's laws · 3 marks · Short answerBoard level

Verify both De Morgan's laws for $U = {1, \dots, 8}$ , $A = {1, 2, 3, 4}$ , $B = {3, 4, 5, 6}$ .

Try it first, then show the solutionSolution

$A \cup B = {1, \dots, 6}$ , so $(A \cup B)^{'} = {7, 8}$ . $A^{'} = {5, 6, 7, 8}$ , $B^{'} = {1, 2, 7, 8}$ , so $A^{'} \cap B^{'} = {7, 8}$ . (1½ marks)
$A \cap B = {3, 4}$ , so $(A \cap B)^{'} = {1, 2, 5, 6, 7, 8}$ . $A^{'} \cup B^{'} = {1, 2, 5, 6, 7, 8}$ . (1 mark)
Both laws hold for these sets. (½ mark)

Answer: Both laws verified

Example 3: Counting two sets · 3 marks · Short answerHarder board level

In a class of 60 students, 35 like tea, 28 like coffee and 12 like both. Find the number who like (i) only tea, (ii) only coffee, (iii) neither.

Try it first, then show the solutionSolution

(i) Only tea $= 35 - 12 = 23$ . (1 mark)
(ii) Only coffee $= 28 - 12 = 16$ . (1 mark)
(iii) At least one $= 35 + 28 - 12 = 51$ , so neither $= 60 - 51 = 9$ . (1 mark)

Answer: (i) 23 (ii) 16 (iii) 9

Example 4: Proving an identity · 3 marks · Short answerHarder board level

For any sets $A, B, C$ prove that $A \cap (B - C) = (A \cap B) - (A \cap C)$ .

Try it first, then show the solutionSolution

Let $x \in A \cap (B - C)$ . Then $x \in A$ , $x \in B$ and $x \in / C$ . (1 mark)
So $x \in A \cap B$ ; and $x \in / A \cap C$ because $x \in / C$ . Hence $x \in (A \cap B) - (A \cap C)$ . (1 mark)
Conversely, if $x \in (A \cap B) - (A \cap C)$ then $x \in A$ , $x \in B$ and $x \in / A \cap C$ ; since $x \in A$ , this forces $x \in / C$ . So $x \in A \cap (B - C)$ . (1 mark)

Answer: Both sides contain exactly the same elements.

Example 5: Three languages · 5 marks · Long answerHOTS / competency

In a survey of 100 students, 50 study French, 40 German and 30 Spanish; 20 study French and German, 15 German and Spanish, 10 French and Spanish, and 5 study all three. Find the number who study (i) at least one language, (ii) none, (iii) exactly one, (iv) exactly two.

Try it first, then show the solutionSolution

(i) $n (F \cup G \cup S) = 50 + 40 + 30 - 20 - 15 - 10 + 5 = 80$ . (1 mark)
(ii) None $= 100 - 80 = 20$ . (1 mark)
(iv) Exactly two: $(20 - 5) + (15 - 5) + (10 - 5) = 15 + 10 + 5 = 30$ . (1 mark)
(iii) Exactly one $= 80 - 30 - 5 = 45$ (check by regions: $25 + 10 + 10 = 45$ ). (1½ marks)
Total check: $45 + 30 + 5 + 20 = 100$ . (½ mark)

Answer: (i) 80 (ii) 20 (iii) 45 (iv) 30

Where students lose marks in Set Operations, Complements and De Morgan's Laws

The slips examiners see most often, and what to write instead.

Mistake: Adding n(A) and n(B) for the union
The overlap is counted twice.
Wrong working
$n (A \cup B) = 35 + 28 = 63$ in a class of 60.
Right working
$35 + 28 - 12 = 51$ .
Do this: A count above $n (U)$ signals the missing subtraction.
Usually costs: The whole question.
Mistake: Writing (A ∪ B)' = A' ∪ B'
De Morgan changes the operation.
Wrong working
$(A \cup B)^{'} = A^{'} \cup B^{'}$ .
Right working
$(A \cup B)^{'} = A^{'} \cap B^{'}$ .
Do this: Break the bar, change the sign.
Usually costs: The proof.
Mistake: Claiming a law is proved by one example
An example verifies for specific sets only.
Wrong working
"De Morgan holds for A, B above, so it is proved."
Right working
Use a general element, or say "verified for these sets".
Do this: Use the right verb: verify or prove.
Usually costs: The reasoning marks.
Mistake: Filling the Venn diagram from the outside
The overlap values must be known first.
Wrong working
Putting 50 in the French circle as "only French".
Right working
Fill $F \cap G \cap S$ first, then subtract.
Do this: Start from the triple overlap.
Usually costs: Method and answer.

Exam tips

State the universal set at the start of every operation question.
For proofs write "Let $x \in \dots$ " and show both inclusions.
For counting, draw the Venn diagram with the innermost region first and check that the regions add to $n (U)$ .
Use "verify" when you test a law on given sets and "prove" when you argue for any sets.
Rewrite $A - B$ as $A \cap B^{'}$ whenever a proof gets stuck.

Practice questions on Set Operations, Complements and De Morgan's Laws

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
If $A = {1, 2, 3}$ and $B = {3, 4}$ , then $A - B$ is:
(a) ${1, 2}$
(b) ${4}$
(c) ${1, 2, 4}$
(d) ${3}$
Hint
Elements of $A$ not in $B$ .
Answer
(a). ${1, 2}$ .
Q2 · 1 mark · MCQHarder board level
If $n (A) = 20$ , $n (B) = 25$ and $n (A \cup B) = 35$ , then $n (A \cap B)$ is:
(a) 5
(b) 10
(c) 15
(d) 20
Hint
$n (A \cap B) = n (A) + n (B) - n (A \cup B)$ .
Answer
(b). $20 + 25 - 35 = 10$ .
Q3 · 3 marks · Short answerHarder board level
Let $U = {1, 2, \dots, 20}$ , $A$ the multiples of 3 and $B$ the multiples of 4 in $U$ . Find $A \cap B$ , $n (A \cup B)$ and $n ((A \cup B)^{'})$ .
Hint
Multiples of both 3 and 4 are multiples of 12.
Answer
$A = {3, 6, 9, 12, 15, 18}$ (6), $B = {4, 8, 12, 16, 20}$ (5), $A \cap B = {12}$ . $n (A \cup B) = 6 + 5 - 1 = 10$ . $n ((A \cup B)^{'}) = 20 - 10 = 10$ .
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): For any two sets $A$ and $B$ , $A - B = B - A$ .
Reason (R): $A - B$ is the set of elements of $A$ that are not in $B$ .
(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.
Hint
Take $A = {1, 2}$ , $B = {2, 3}$ .
Answer
(d). $A - B = {1}$ but $B - A = {3}$ . R is the correct definition.
Q5 · 4 marks · Short answerHOTS / competency
In a hostel of 200 residents, 120 play chess and 90 play carrom (some residents may play neither). Find the least and the greatest possible number of residents who play both games.
Hint
Use $n (A \cap B) \geq n (A) + n (B) - n (U)$ and $n (A \cap B) \leq min (n (A), n (B))$ .
Answer
Least: $n (A \cap B) \geq 120 + 90 - 200 = 10$ . Greatest: $n (A \cap B) \leq min (120, 90) = 90$ . So between 10 and 90 residents play both.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.What is A − B?Show answer
Elements of A that are not in B.
2.What is (A ∪ B)'?Show answer
A' ∩ B'.
3.What is n(A ∪ B)?Show answer
n(A) + n(B) − n(A ∩ B).
4.What is A ∪ A'?Show answer
The universal set.
5.What is A ∩ A'?Show answer
The empty set.

Set Operations, Complements and De Morgan's Laws: frequently asked questions

What is a universal set?

The set that contains all the objects under discussion in a problem, so complements are taken inside it.

Is A − B the same as B − A?

Not in general. They are equal only when A = B, and then both are empty. Otherwise they are disjoint sets that differ.

Do I need the power set?

No. The NCERT section on the power set has been removed from the 2026-27 syllabus.

Are practical union and intersection problems examined?

The chapter page notes they are assessed only formatively in 2026-27, though they are good practice for the laws.