NDA Maths · Sets & Relations

Set Fundamentals, Operations and Algebra

A set is a well-defined collection of distinct objects; the operations (union, intersection, complement, difference, symmetric difference) obey a fixed list of algebraic laws that the NDA tests by asking which identity is NOT correct.

All Sets & Relations notes NDA Maths strategy

Why this matters

Start here — everything else in the chapter is built on these operations. 23 PYQs live in this subtopic, and the dominant shape is identity verification: you are given four set identities and must spot the wrong one. That is pure law-recognition (distributive, De Morgan, absorption) plus careful region-counting on a Venn diagram. All EASY or MODERATE except a handful.

Concept 1 of 4

Sets — notation, empty set, and equal vs equivalent

Intuition

A set is just a collection of distinct objects. Two ideas trip students up and the NDA tests both: the empty set (a set with no elements, written ∅) is still a perfectly good set, and 'equal' sets are not the same as 'equivalent' sets — equal means identical elements, equivalent means merely the same NUMBER of elements.

Definition

Core vocabulary:

Set — a well-defined collection of distinct objects; $x \in A$ means x belongs to A.
Empty (null) set $\emptyset$ — has no elements; it is a subset of every set. A condition with no solutions defines $\emptyset$ (e.g. $\{x \in \mathbb{R} : x^2 + 1 = 0\} = \emptyset$ ).
Subset $A \subseteq B$ — every element of A is in B. Proper subset $A \subset B$ excludes $A = B$ .
Equal sets have exactly the same elements; equivalent sets only have the same number of elements (same cardinality). Equal $\Rightarrow$ equivalent, not the reverse.

Worked example

Are the sets

A = \{1, 2, 3\}

and

B = \{a, e, i\}

equal, equivalent, or neither?

Equal means identical elements — A and B share no elements, so not equal.
Equivalent means the same number of elements — both have 3.
So they are equivalent but not equal.

Answer:Equivalent (both have 3 elements) but not equal.

Practice this conceptself-check · 3 quick reps

Try it yourself

What is the set

S = \{x : x \text{ is real and } x^2 + 1 = 0\}

Practice — Level 1 (3 reps)

Quick reps to lock in the method. Try each, then check.

1.
Is the empty set a subset of every set?
2.
How many elements does $\{x \in \mathbb{R} : x^2 = -4\}$ have?
3.
Are $\{1,2,3\}$ and $\{2,4,6\}$ equal or equivalent?

From the bank · past-year question

Example 1Sets & RelationsEASY

S = \{x : x^2 + 1 = 0,\, x \text{ is real}\}

, then

S

[Q25 · Apr · 2017]

Equal vs equivalent

\{1,3,5\}

and

\{2,4,7\}

are equivalent (both size 3) but not equal (different elements). The NDA pairs these words deliberately — equal is about WHICH elements, equivalent is about HOW MANY.

Drill 2 more on sets — notation, empty set, and equal vs equivalent

Concept 2 of 4

Union, intersection, complement and difference

Intuition

Five operations build every set expression. Union

A \cup B

is 'in A OR B'; intersection

A \cap B

is 'in A AND B'; complement

A'

is 'NOT in A'; difference

A - B

is 'in A but not B'. Picture them as regions of a Venn diagram and most questions become a region count.

Definition

The operations and the identities the bank leans on:

$A \cup B$ (union), $A \cap B$ (intersection), $A'$ (complement, relative to the universal set), $A - B = A \cap B'$ (difference).
Complement is an involution: $(A')' = A$ . A long nested complement like $E-(E-(E-A))$ collapses by cancelling in pairs.
A set can be defined by a condition — solving it gives the set: $(x-a)(x-b) > 0$ (with $a<b$ ) gives $x < a$ or $x > b$ ; multiples of 2 AND 3 are the multiples of 6.

Worked example

E

is the universal set, simplify

E - (E - (E - A))

Innermost: $E - A = A'$ .
Next: $E - A' = A$ (complement of the complement).
Outermost: $E - A = A'$ .
Each $E-\,$ flips the set; an odd number of flips leaves $A'$ .

Answer:

A'

Practice this conceptself-check · 4 quick reps

Try it yourself

A

is the set of multiples of 2 and

B

the set of multiples of 3. What is

A \cap B

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Write $A - B$ using complement.
2.
Simplify $(A')'$ .
3.
Solve $(x-1)(x-4) > 0$ as a set.
4.
Multiples of 4 and multiples of 6 — their intersection is multiples of?

From the bank · past-year question

Example 2Sets & RelationsMODERATE

For any three non-empty sets

A,B,C

, what is

(A\cup B)-\{(A-B)\cup(B-A)\cup(A\cap B)\}

equal to?

[Q24 · Apr · 2024]

$A - B$ is not symmetric

A - B

(in A, not B) is different from

B - A

(in B, not A). Only their UNION,

(A-B)\cup(B-A)

, is symmetric — that is the symmetric difference. Don't write

A-B = B-A

The three disjoint pieces rebuild the union

(A-B) \cup (A\cap B) \cup (B-A) = A \cup B

— the only-A, both, and only-B regions tile the whole union. Recognising this collapses many 'simplify the expression' questions to

\emptyset

A

Drill 8 more on union, intersection, complement and difference

Concept 3 of 4

The laws of set algebra

Intuition

Set operations obey a fixed list of laws — the same shape as the laws of logic. The NDA's favourite question gives you four 'identities' and asks which is NOT correct; the wrong one is always a real law with one operation swapped. Knowing the genuine laws makes it a spot-the-impostor exercise.

Definition

The laws you must recognise on sight (A, B, C are any sets):

Distributive: $A \cup (B \cap C) = (A\cup B)\cap(A\cup C)$ and $A \cap (B \cup C) = (A\cap B)\cup(A\cap C)$ .
De Morgan: $(A\cup B)' = A' \cap B'$ and $(A\cap B)' = A' \cup B'$ .
Absorption: $A \cup (A\cap B) = A$ and $A \cap (A\cup B) = A$ .
Idempotent / Identity / Complement: $A\cup A = A$ ; $A\cup\emptyset=A$ ; $A\cup A' = E$ , $A\cap A' = \emptyset$ .

Law	Statement	The impostor to watch for
Distributive	$A\cup(B\cap C)=(A\cup B)\cap(A\cup C)$	Swapping $\cup$ / $\cap$ on one side breaks it
De Morgan	$(A\cup B)'=A'\cap B'$	$(A\cup B)' = A'\cup B'$ is WRONG — the operation flips In predicate form: $x\notin(A\cup B)\Rightarrow x\notin A$ AND $x\notin B$ (not OR).
Absorption	$A\cup(A\cap B)=A$	$A\cup(A\cap B)=A\cup B$ is WRONG — it collapses to just $A$
Subset test 'for all B'	$(A\cap B)\subseteq(C\cap B)\ \forall B \Rightarrow A\subseteq C$	Test such claims by choosing $B=\emptyset$ or $B=E$

Distractors are genuine laws with one operation flipped. Verify a suspect identity on a tiny example or a Venn diagram.

Practice this conceptself-check · 3 quick reps

Try it yourself

Which is NOT a correct identity: (a)

A\cap(A\cup B)=A

, (b)

A\cup(A\cap B)=A\cup B

, (c)

(A\cup B)'=A'\cap B'

Practice — Level 1 (3 reps)

Quick reps to lock in the method. Try each, then check.

1.
State De Morgan for $(A\cap B)'$ .
2.
Simplify $A\cup(A\cap B)$ .
3.
$x\notin(A\cup B)$ means $x\notin A$ ___ $x\notin B$ .

From the bank · past-year question

Example 3Sets & RelationsMODERATE

A

B

and

C

are subsets of a given set, then which one of the following relations is

\textbf{\text{not}}

correct ?

[Q3 · Sep · 2019]

The wrong option is a real law with a flipped operation

These questions never use nonsense — every distractor is a genuine law with

\cup

\cap

swapped or absorption over-simplified. If unsure, plug in

A=\{1\}, B=\{2\}

and compute both sides.

Drill 6 more on the laws of set algebra

Concept 4 of 4

Symmetric difference and set-equality conditions

Intuition

The symmetric difference

A \triangle B

collects elements in exactly one of the two sets — the union minus the overlap. It also gives a clean test for equality: two sets are equal exactly when their symmetric difference is empty. A related trap is that you CANNOT cancel sets like numbers.

Definition

Equality tools:

Symmetric difference: $A \triangle B = (A-B)\cup(B-A) = (A\cup B)-(A\cap B) = (A\cap B')\cup(A'\cap B)$ .
$A \triangle B = \emptyset \iff A = B$ ; also $A\cup B = A\cap B \iff A = B$ .
Cancellation fails: $A\cap B = A\cap C$ does NOT force $B=C$ ; $A\cup C = B\cup C$ forces $A=B$ only when C is disjoint from both.

Worked example

Sets satisfy

A\cap C = B\cap C

with

C

disjoint from both

A

and

B

. Also

A\cup C = B\cup C

. What can you conclude?

Since $C$ is disjoint from $A$ and $B$ , $A\cap C = \emptyset = B\cap C$ — the first equation gives nothing.
From $A\cup C = B\cup C$ , remove the disjoint $C$ from both sides.
Because $C$ shares no element with $A$ or $B$ , the leftover parts must be equal: $A = B$ .

Answer:

A = B

(but

C

need not be empty).

Practice this conceptself-check · 3 quick reps

Try it yourself

Rewrite

(A\cap B') \cup (A'\cap B)

as a single named operation.

Practice — Level 1 (3 reps)

Quick reps to lock in the method. Try each, then check.

1.
When is $A \triangle B = \emptyset$ ?
2.
Does $A\cap B = A\cap C$ imply $B = C$ ?
3.
$A\cup B = A\cap B$ implies?

From the bank · past-year question

Example 4Sets & RelationsMODERATE

Let A and B be subsets of X and

C = (A \cap B') \cup (A' \cap B)

, where A' and B' are complements of A and B in X. What is C equal to?

[Q6 · Apr · 2018]

You cannot cancel sets like numbers

A\cap B = A\cap C

does NOT give

B=C

, and

A\cup B = A\cup C

does NOT give

B=C

either. Cancellation only works when the cancelled set is disjoint from the rest. Always look for a small counterexample.

Drill 3 more on symmetric difference and set-equality conditions

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Reference tables (1)

The laws of set algebra4 rows

Law	Statement	The impostor to watch for
Distributive	$A\cup(B\cap C)=(A\cup B)\cap(A\cup C)$	Swapping $\cup$ / $\cap$ on one side breaks it
De Morgan	$(A\cup B)'=A'\cap B'$	$(A\cup B)' = A'\cup B'$ is WRONG — the operation flips In predicate form: $x\notin(A\cup B)\Rightarrow x\notin A$ AND $x\notin B$ (not OR).
Absorption	$A\cup(A\cap B)=A$	$A\cup(A\cap B)=A\cup B$ is WRONG — it collapses to just $A$
Subset test 'for all B'	$(A\cap B)\subseteq(C\cap B)\ \forall B \Rightarrow A\subseteq C$	Test such claims by choosing $B=\emptyset$ or $B=E$

Distractors are genuine laws with one operation flipped. Verify a suspect identity on a tiny example or a Venn diagram.

Watch out for (5)

Equal vs equivalent
→ Sets — notation, empty set, and equal vs equivalent
$A - B$ is not symmetric
→ Union, intersection, complement and difference
The three disjoint pieces rebuild the union
→ Union, intersection, complement and difference
The wrong option is a real law with a flipped operation
→ The laws of set algebra
You cannot cancel sets like numbers
→ Symmetric difference and set-equality conditions

Mastery check — 5 interleaved questions

Try each one before clicking. Questions are interleaved across the concepts above, not grouped — interleaving sharpens transfer.

Example 1Sets & RelationsEASY

Consider the following statements: 1.

A = \{1,3,5\}

and

B = \{2,4,7\}

are equivalent sets. 2.

A = \{1,5,9\}

and

B = \{1,5,5,9,9\}

are equal sets. Which of the above statements is/are correct?

[Q35 · Apr · 2021]

Example 2Sets & RelationsEASY

Set

X

contains

3n

elements and set

Y

contains

2n

elements, and they have

n

elements in common. How many elements does

(X-Y)\times(Y-X)

have?

[Q42 · Sep · 2025]

Example 3Sets & RelationsHARD

If A, B and C are subsets of a Universal set, then which one of the following is

\textbf{\text{not}}

correct? (where A' is the complement of A)

[Q5 · Sep · 2018]

Example 4Sets & RelationsMODERATE

Consider the following: 1. A∩B = A∩C ⟹ B = C 2. A∪B = A∪C ⟹ B = C

Which of the above is/are correct?

[Q15 · Sep · 2022]

Example 5Sets & RelationsEASY

Consider the following statements in respect of sets: 1. The union over intersection of sets is distributive. 2. The complement of union of two sets is equal to intersection of their complements. 3. If the difference of two sets is equal to empty set, then the two sets must be equal. Which of the above statements are correct?

[Q37 · Sep · 2021]

Drill every past-year question on this subtopic

23 questions from the bank — paginated, with cart and Word-export support.

Drill the 23 questions