Sets & Relations — NDA Mathematics Formulas and Traps

Set Fundamentals, Operations and Algebra

Union, intersection, complement and difference

Difference and complement identities

A - B = A \cap B' \qquad (A')' = A

Symmetric difference and set-equality conditions

Symmetric difference

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

The laws of set algebra

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.

Common traps

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.

$A - B$ 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

or

A

.

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.

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.

Counting, Subsets and Inclusion-Exclusion

Learn this subtopic in the notes

Power set and counting subsets

Counting subsets

|P(A)| = 2^n \qquad \text{proper subsets} = 2^n - 1 \qquad \text{subsets containing a fixed element} = 2^{n-1}

Inclusion-exclusion for two sets

Inclusion–exclusion (two sets)

|A \cup B| = |A| + |B| - |A \cap B| \qquad |A \cap B| \ge |A| + |B| - |U|

Inclusion-exclusion for three sets

Inclusion–exclusion (three sets)

|A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |B \cap C| - |A \cap C| + |A \cap B \cap C|

Survey problems — exactly one, exactly two, all three

Survey accounting identities

\text{exactly two} = \textstyle\sum\text{pairwise} - 3\,(\text{all three}) \qquad \text{at least two} = \text{exactly two} + (\text{all three})

Common traps

Count the elements before raising 2 to a power

A=\{\lambda, \{\lambda,\mu\}\}

has TWO elements (an element

\lambda

and a SET

\{\lambda,\mu\}

), so

|P(A)|=4

. The brace-inside-brace is one element, not two — miscounting elements is the usual error.

Overlap of multiples uses LCM, not product

Multiples of 4 and 6 in common are multiples of

\mathrm{lcm}(4,6)=12

, NOT

4\times 6 = 24

. Use the LCM whenever 'divisible by both' appears.

Mind the alternating signs

Pairwise intersections are SUBTRACTED, the triple intersection is ADDED. Dropping the

+|A\cap B\cap C|

term is the most common slip — it under-counts the union.

'Exactly two' is not the sum of pairwise intersections

Each all-three person is counted in all three pairwise intersections, so

\sum\text{pairwise}

over-counts them by a factor of 3. Subtract: exactly two

= \sum\text{pairwise} - 3\cdot(\text{all three})

. Mixing up 'exactly' and 'at least' is the single biggest source of wrong answers here.

Relations and the Cartesian Product

Learn this subtopic in the notes

Cartesian product, domain and range

Cartesian product and relation counts

|A \times B| = |A|\,|B| \qquad \text{number of relations} = 2^{|A||B|} \qquad (A \times B) \cap (B \times A) = (A \cap B) \times (A \cap B)

Reflexive, symmetric, transitive, equivalence

Property	Test	Fails when
Reflexive	Is $(a,a)\in R$ for every a?	One element lacks its self-loop (e.g. $(4,4)\notin R$ )
Symmetric	Does $(a,b)\in R$ force $(b,a)\in R$ ?	Some arrow has no reverse
Transitive	Do $(a,b),(b,c)$ force $(a,c)$ ?	A 2-step path with no direct shortcut
Equivalence	All three hold	Any one of R / S / T fails A strict inequality $<$ is transitive ONLY — not reflexive, not symmetric.

Test each property separately; one counterexample kills it.

Common traps

Range can be smaller than the codomain

When a relation is defined by a rule, only some target values are actually hit. The codomain might be all of

\{1,\dots,20\}

while the range is just

\{1,5,9,13,17\}

. 'Range = codomain' is usually FALSE.

Reflexive means EVERY element, not just some

A relation listing

(1,1),(2,2),(3,3)

on the set

\{1,2,3,4\}

is NOT reflexive —

(4,4)

is missing. Reflexivity is an 'all elements' condition; one missing self-pair breaks it.

Factor before you test

A relation like

x^2 - 5xy + 4y^2 = 0

looks intimidating until you factor it to

(x-y)(x-4y)=0

, i.e.

x=y

or

x=4y

. Skipping the factoring step is where the time and the errors go.

Every function is a relation, not the reverse

All functions are relations, but a relation that sends one input to two outputs (or misses an input) is not a function. The statement 'all relations are functions' is FALSE; 'all functions are relations' is true.

Sets & Relations formulas

Set Fundamentals, Operations and Algebra

Union, intersection, complement and difference

Symmetric difference and set-equality conditions

The laws of set algebra

Common traps

Counting, Subsets and Inclusion-Exclusion

Power set and counting subsets

Inclusion-exclusion for two sets

Inclusion-exclusion for three sets

Survey problems — exactly one, exactly two, all three

Common traps

Relations and the Cartesian Product

Cartesian product, domain and range

Reflexive, symmetric, transitive, equivalence

Common traps

More NDA Mathematics formula sheets