Relations – NDA Maths Notes

Mathematics needs precise ways to describe how things are connected. The word “related” appears everywhere: one number divides another, two people attend the same university, one quantity is greater than another. A relation is the mathematical tool that captures this idea precisely.

Why does this chapter matter? Chapter 2 builds directly on the ordered pairs and Cartesian products from Chapter 1, and it hands that ordered-pair thinking over to Chapter 3: Functions. Understanding what a relation is and which properties it has is the foundation for understanding functions, mappings, and the probability event structure.

1. Cartesian Products and Ordered Pairs

Before defining a relation formally, we need two building blocks from Chapter 1: ordered pairs and Cartesian products.

1.1  Ordered Pairs

An ordered pair (a, b) is a pair of objects where the order matters. (a, b) and (b, a) are different unless a = b. This is what makes relations directional : “x divides y” is not the same relation as “y divides x”.

1.2  Cartesian Product A×B

A×B is the set of all ordered pairs (a, b) where a ∈ A and b ∈ B.

A×B = {(a, b) : a ∈ A, b ∈ B}

If A = {1, 2} and B = {3, 4}, then A×B = {(1,3), (1,4), (2,3), (2,4)}

Cardinality: |A×B| = |A| · |B| = mn, where m = |A| and n = |B|. So |A×A| = n².

A×B ≠ B×A in general, though |A×B| = |B×A|. The order of the sets in the product matters.  [NDA 2010-I]

1.3  Elements Common to A×B and B×A

An ordered pair (a, b) belongs to both A×B and B×A if and only if a belongs to both A and B, and b belongs to both A and B. So the number of elements common to A×B and B×A is |A∩B|², often written k² where k = |A∩B|.

If A and B share 10 elements, A×B and B×A share 10² = 100 elements.  [NDA 2024-II]

1.4  Determining A from |A×A|

If you know |A×A| = n², you can find n = |A|. Given specific elements in A×A, you can identify the elements of A itself.

If A×A has 16 elements, then |A|² = 16, so |A| = 4. If (0,2) and (1,3) are in A×A, then 0,1,2,3 ∈ A, giving A = {0,1,2,3}.  [NDA 2023-II]

1.5  Counting with Set Differences

When a question involves (X−Y)×(Y−X), compute the set differences first, then multiply.

|X−Y| = |X| − |X∩Y|   and   |Y−X| = |Y| − |X∩Y|

|(X−Y)×(Y−X)| = |X−Y| · |Y−X|

This pattern appeared in three consecutive modern papers.  [NDA 2023-II, NDA 2024-II, NDA 2025-II]

NDA Worked Example 5   Modern Cartesian-Product Counting

Question Question (NDA 2025-II) Set X contains 3n elements and set Y contains 2n elements. They have n elements in common. How many elements does (X−Y)×(Y−X) have?

Solution Compute the set differences before applying the product formula. |X−Y| = |X| − |X∩Y| = 3n − n = 2n |Y−X| = |Y| − |X∩Y| = 2n − n = n |(X−Y)×(Y−X)| = 2n × n = 2n² Same counting structure.  [NDA 2023-II, NDA 2024-II]

Answer: 2n²

The most common mistake is computing |X| × |Y| = 6n² directly, without first removing the shared elements. Apply set-difference cardinality before the product formula.

2. What Is a Relation?

A relation R from A to B is any subset of A×B. It is a collection of ordered pairs, where each first element comes from A and each second element from B.

R ⊆ A×B

A relation on A (also called a relation in A) is any subset of A×A. The relating rule compares elements of A with each other.

Every ordered pair in R is a relation fact: it says “this element is related to that element”. Any element not listed as a pair is not related. The empty set ∅ is a valid relation (nothing is related to anything). The full set A×B is also a valid relation (everything is related to everything).

A relation from A to B is a subset of A×B; a relation in A is a subset of A×A.  [NDA 2013-I, NDA 2021-II]

2.1  Domain, Codomain and Range

Term	Meaning	Who defines it?	Example (R = {(1,2),(3,4)})
Domain	Set of all first elements actually in R	Derived from R	{1, 3}
Codomain	Set from which second elements are drawn	Given in advance	ℕ or specified set
Range	Set of all second elements actually in R	Derived from R	{2, 4}

★   Important

The critical distinction: Range ⊆ Codomain always. Range = Codomain only if every element of the codomain appears as the second element of some pair in R. This is rarely automatic.

A relation and a function are not the same. A relation is any subset of A×B. A function is a relation in which every element of the domain is related to exactly one element of the codomain. Every function is a relation; not every relation is a function.  [NDA 2021-II]

2.2  Representing a Relation

Roster form: List the ordered pairs. R = {(1,2), (3,4), (3,6)}.

Set-builder (rule) form: Describe the rule. R = {(x,y) : x divides y, x,y ∈ ℕ}. This is the dominant modern NDA form.

Arrow diagram: Draw elements of A on the left, elements of B on the right, and draw an arrow from a to b whenever (a,b) ∈ R. This makes the properties of a relation visible.

Reflexive Every element has a self-loop 1→1 2→2 3→3 4→4 Every (a,a) ∈ R

Symmetric Every arrow has a reverse a ↔ b b ↔ c a → c If (a,b)∈R then (b,a)∈R

Transitive Chains are closed: x→y, y→z ⇒ x→z x → y y → z x → z  Completing the triangle

3. Number of Relations from A to B

Since a relation from A to B is any subset of A×B, and A×B has mn elements, the total number of possible relations from A to B equals the total number of subsets of a set with mn elements.

Number of relations from A (m elements) to B (n elements) = 2^(mn)

Number of relations on A (n elements) = 2^(n²)

This follows directly from the subset-counting formula from Chapter 1: a set with k elements has 2^k subsets.

What to count	Formula	Example
Relations from A (m) to B (n)	2^(mn)	m=3, n=2: 2⁶ = 64
Relations on A (n elements)	2^(n²)	n=3: 2⁹ = 512
|A×B|	mn	m=3, n=4: 12
|A×A|	n²	n=4: 16
Elements common to A×B and B×A	k²  where k = |A∩B|	k=10: 100  [NDA 2024-II]

★   Important

“On A” means on A×A, which has n² elements. “From A to B” means on A×B, which has mn elements. These are different. The number of relations from A to B is NOT the same as from B to A unless m = n.

NDA Worked Example 4   Number of Relations: The “From B to A” Trap

Question Question (NDA 2015-I) Let A = {x, y, z} and B = {p, q, r, s}. What is the number of distinct relations from B to A?

Solution The question asks for relations from B to A, not from A to B. These are different. |B| = 4  (the “from” set),  |A| = 3  (the “to” set) Relations from B to A are subsets of B×A. |B×A| = 4 × 3 = 12 Number of relations = 2^12 = 4096

Answer: 4096

“From B to A” uses B×A, not A×B. Always identify the “from” set first, then compute its Cartesian product. The distractor 4094 = 2^12 − 2 is the number of proper non-empty subsets: a different question.

Number of relations from A to B = 2^(mn).  [NDA 2010-I, NDA 2013-II]

4. Rule-Defined Relations: Elements, Domain and Range

In most modern NDA questions, a relation is defined by an algebraic rule rather than a roster. The skill is systematic enumeration: list all pairs satisfying the rule, then read off the domain and range.

★   IMPORTANT

Method: (1) Set up the rule. (2) Let x take values one by one. (3) Compute y from the rule; check whether y also belongs to the given set. (4) Collect all valid pairs (x, y). (5) Domain = set of all first elements. Range = set of all second elements. Codomain = the set specified in the question.

Example: R = {(x,y) : 3x + 4y = 5, x,y ∈ ℝ}. Here 0R1 is NOT correct since 3(0)+4(1)=4≠5. Instead, 3(0)+4(5/4)=5, so 0R(5/4). This relation is defined on ℝ, so infinitely many pairs exist. For integer questions, only finitely many pairs are valid.  [NDA 2011-I]

NDA Worked Example :   Domain, Range and Codomain of a Rule-Defined Relation

Question Question (NDA 2023-II) Let A = {1, 2, 3, …, 20}. Define a relation R from A to A by R = {(x,y): 4x − 3y = 1}, where x, y ∈ A. Which statements are correct? Statement 1: The domain of R is {1, 4, 7, 10, 13, 16}. Statement 2: The range of R is {1, 5, 9, 13, 17}. Statement 3: The range of R is equal to the codomain of R.

Solution From 4x − 3y = 1, we get y = (4x − 1)/3. For y to be a natural number in A, (4x − 1) must be divisible by 3. x = 1:  y = 3/3 = 1 x = 4:  y = 15/3 = 5 x = 7:  y = 27/3 = 9 x = 10: y = 39/3 = 13 x = 13: y = 51/3 = 17 x = 16: y = 63/3 = 21 : but 21 > 20, so not in A So R = {(1,1), (4,5), (7,9), (10,13), (13,17)}. Statement 1: Domain = {1, 4, 7, 10, 13}. The statement claims {1,4,7,10,13,16}: incorrect, because x=16 gives y=21 ∉ A. Statement 1 is FALSE. Statement 2: Range = {1, 5, 9, 13, 17}. Correct. Statement 2 is TRUE. Statement 3: Codomain = A = {1,2,…,20}. Range = {1,5,9,13,17} ≠ Codomain. Statement 3 is FALSE.

Answer: Statement 2 only: option (b)

Range and codomain are not the same. Codomain is the set you draw the y-values from; range is the set of y-values that actually appear in R. Range ⊆ Codomain; they are equal only by coincidence, not by definition.

For rule-defined relations with a linear condition, enumerate valid pairs by substituting x-values and checking whether y ∈ A.  [NDA 2009-II, NDA 2021-I, NDA 2025-II]

5. Property Verification: Reflexive, Symmetric, Transitive

This is the most important section in the chapter. More than half of all NDA Relations questions ask exactly one thing: classify a relation by its properties. The correct method is to test each property separately, using the definitions below, before drawing any conclusion.

★ IMPORTANT

The standard NDA question form: “The relation R … is: (a) reflexive and symmetric but not transitive; (b) reflexive and transitive but not symmetric; (c) an equivalence relation; (d) none of the above.” You must independently adjudicate each property.

5.1  Reflexive

Reflexive
Formal	∀ a ∈ A : (a, a) ∈ R
Plain English	Every element must appear as both the first and second element of some pair in R. No element of A may be left without its own self-pair. Even one missing (a,a) means R is not reflexive.
How to test	List every element of A. Check that (a,a) ∈ R for each. If any one is missing, R is not reflexive. For algebraic rules: substitute y = x into the rule. If the result is always true, R is reflexive.
Counterexample	x < y on ℝ : no element satisfies x < x. Any strict inequality destroys reflexivity.

★   IMPORTANT

Test reflexivity first. It is often the quickest property to check. If it fails, eliminate any answer choices that require reflexivity. On strict inequalities (x < y, x > y, |x| < y), reflexivity always fails : test x R x once and you are done.

5.2  Irreflexive

Definition: R on A is irreflexive if no element is related to itself: (a,a) ∉ R for every a ∈ A. This is the opposite of reflexive. A relation is NOT irreflexive merely because it is not reflexive: it is irreflexive only if every (a,a) is absent.

The NDA paper does not directly test irreflexivity as a named property, but understanding it prevents the common confusion: a relation that is not reflexive is not automatically irreflexive. It might have some (a, a) pairs, but not all.

5.3  Symmetric

Symmetric
Formal	∀ a, b ∈ A : (a, b) ∈ R → (b, a) ∈ R
Plain English	Every arrow has a reverse arrow. Every pair has its mirror.
How to test	For each pair (a,b) in R where a ≠ b, check that (b,a) is also in R. For algebraic rules: swap x and y. If the rule is unchanged, R is symmetric.
Counterexample	x divides y : 2 divides 6, but 6 does not divide 2. One pair without its reverse is sufficient.

★ IMPORTANT

“a − b divisible by 5” IS symmetric (if 5|(a−b) then 5|(b−a)). “x divides y” is NOT symmetric. These are the most confused pair in the chapter.

5.4  Transitive

Transitive
Formal	∀ a, b, c ∈ A : [(a,b) ∈ R  and  (b,c) ∈ R] → (a,c) ∈ R
Plain English	Chains hold. If you can go from a to b and from b to c, you must be able to go directly from a to c.
How to test	Find every pair of composable pairs: (a,b) and (b,c) both in R. For each such pair, check that (a,c) ∈ R. One missing (a,c) disproves transitivity.
Counterexample	Age difference ≤ 5: persons aged 1, 5, 9. 1↔5 (diff=4≤5) , 5↔9 (diff=4≤5) , but 1↔9 (diff=8>5) . Transitivity fails.

★ IMPORTANT

For algebraic relations, try to find a chain a→b→c where the conclusion a→c fails. For social relations, look for a case where the “middle” person connects two extremes that cannot be directly connected.

Property Summary

Property	Formal condition	Standard counterexample	Common Confusion
Reflexive	∀ a ∈ A: (a,a) ∈ R	x < y on ℝ : no element satisfies x < x	Checking only some (a,a) pairs, not all
Symmetric	(a,b) ∈ R → (b,a) ∈ R	x divides y : 2│6 but 6∤2	Confusing “a − b div. by 5” with “x divides y”
Transitive	(a,b),(b,c) ∈ R → (a,c) ∈ R	Age diff. ≤ 5: 1↔5, 5↔9 but 1↔9 fails	Checking a few chains and assuming all hold
Equivalence	All three above hold simultaneously	—	Stopping at two properties; not testing all three

5.5  The Standard Property-Check Sequence

For any given relation R, follow this procedure:

Step 1 : Test reflexivity. Substitute x = x (or y = x) into the rule. If any element fails, record “not reflexive” and move on.

Step 2 : Test symmetry. Swap x and y in the rule, or check that every pair has its reverse. If one pair is one-directional, record “not symmetric”.

Step 3 : Test transitivity. Find a chain a→b→c and check whether a→c follows. Use a specific counterexample if one exists.

Step 4 : Factor algebraic relations before testing. If the rule is a quadratic in x and y, factor it first. The factored form shows exactly which pairs belong to R.

Step 5 : One counterexample is enough to disprove a property. Finding one failure is sufficient: you do not need to show all failures.

NDA Worked Example 1   Roster Property Check: All Three Verdicts

Question Question (NDA 2019-I) Suppose X = {1, 2, 3, 4} and R = {(1,1),(2,2),(3,3),(1,2),(2,1),(2,3),(3,2)}. Which one of the following is correct? (a) R is reflexive and symmetric, but not transitive (b) R is symmetric and transitive, but not reflexive (c) R is reflexive and transitive, but not symmetric (d) R is neither reflexive nor transitive, but symmetric

Solution Step 1 : Reflexive: Check (x,x) for every x ∈ X = {1,2,3,4}. (1,1), (2,2), (3,3). But (4,4) is NOT in R, and 4 ∈ X. Reflexivity fails. Step 2 : Symmetric: Check each non-diagonal pair. (1,2) → (2,1). (2,3) → (3,2). No other non-diagonal pairs exist. R is symmetric. Step 3 : Transitive: Find composable chains. (1,2) and (2,3) are both in R: does (1,3) ∈ R? No. Transitivity fails. Same roster-check pattern.  [NDA 2009-I, NDA 2010-II, NDA 2012-II]

Answer: Neither reflexive nor transitive, but symmetric. Option (d).

The element 4 ∈ X has no self-pair in R. This is the confusion: the three self-pairs (1,1),(2,2),(3,3) make reflexivity look plausible, but X has four elements. Check every element of X, not just the ones listed in R.

NDA Worked Example 2   Divisibility : “Divides” vs “Divisible by 5”

Question Question (NDA 2013-II, NDA 2014-II, NDA 2016-I) Two relations are defined. Classify each. Relation A: R = {(a,b): a − b is divisible by 5}, on ℤ. Relation B: S defined by x S y iff x divides y, on ℤ.

Solution Relation A  (a − b divisible by 5)  [NDA 2013-II] Reflexive: a − a = 0, and 0 is divisible by 5. So (a,a) ∈ R for all a ∈ ℤ. Symmetric: If 5|(a−b), then a − b = 5k, so b − a = 5(−k), also divisible by 5. Transitive: If 5|(a−b) and 5|(b−c), then a − c = 5(k+m), also divisible by 5. Relation A: all three properties hold: equivalence relation (congruence mod 5). Relation B  (x divides y)  [NDA 2014-II, NDA 2016-I] Reflexive: Every integer divides itself (x = 1·x). Symmetric: Does x divides y imply y divides x? No. Take x = 2, y = 6: 2 divides 6, but 6 does not divide 2. Symmetry fails. Transitive: If x|y and y|z, then y = kx and z = my, so z = mkx : x divides z. Relation B: reflexive and transitive, NOT symmetric. Not an equivalence relation. “n is a factor of m” is the same as “n divides m” : same answer: reflexive + transitive, not symmetric.

Answer: Relation A: Equivalence | Relation B: Reflexive + Transitive, NOT symmetric

“a − b divisible by 5” IS an equivalence relation. “x divides y” is NOT symmetric. These are the most commonly confused pair in the chapter.

6. Equivalence Relations and the Property Grid

An equivalence relation is one that is simultaneously reflexive, symmetric and transitive. It is the most important classification in NDA Relations questions.

R is an equivalence relation ⟺ R is reflexive AND symmetric AND transitive

The NDA canonical question gives you a relation and asks which combination of the three properties it has. The options enumerate the most common combinations. Testing all three properties before choosing is essential : stopping at two will lead to the wrong option.

6.1  The Standard Relations Classification Table

This table directly answers the divisibility, inequality and social-relation families that the NDA has tested repeatedly.

Relation	Reflexive?	Symmetric?	Transitive?	Equivalence?
a − b divisible by k, on ℤ	Yes	Yes	Yes	Yes (congruence mod k)
x divides y, on ℕ	Yes	No	Yes	No
n is a factor of m, on ℕ	Yes	No	Yes	No
x = y (equality)	Yes	Yes	Yes	Yes
x ≤ y, on ℝ	Yes	No	Yes	No
x < y (strict), on ℝ	No	No	Yes	No
ab ≥ 0, on ℤ	Yes	Yes	No	No
Same father as, on people	Yes	Yes	Yes	Yes
Is son of, on men	No	No	No	No (none)
Age diff. exactly k	No	Yes	No	No
Age diff. at most k	Yes	Yes	No	No
At least k years older	No	No	Yes	No

★   IMPORTANT

Three consecutive NDA papers (2014-I, 2015-I, 2015-II) tested age-based relations with three different quantifiers: “exactly 5 years”, “at least 5 years older”, and “at most 5 years”. Each gives a different answer. Memorising any one answer is dangerous; understand the properties.

Age relations: three consecutive papers, three different answers.  [NDA 2014-I, NDA 2015-I, NDA 2015-II]

“Born on same day”: equivalence relation.  [NDA 2017-I]

“Same university”: equivalence relation.  [NDA 2010-I]

“Is son of”: none of the standard properties hold. None of the above.  [NDA 2011-I]

“Has same father as”: equivalence relation.  [NDA 2012-II]

ab ≥ 0 on ℤ : reflexive and symmetric, NOT transitive. Counterexample: a=−1, b=0, c=1: (−1)(0)=0≥0 and (0)(1)=0≥0 but (−1)(1)=−1<0.

ab ≥ 0 fails transitivity : counterexample: a=−1, b=0, c=1.  [NDA 2014-I]

log_{1/2} x > log_{1/2} y: since base 1/2 < 1, the logarithm reverses the inequality, making this equivalent to x < y. Transitive only, not reflexive.  [NDA 2020-I & II]

(a,b) R (c,d) iff a+d = b+c on ℕ×ℕ: equivalence relation.  [NDA 2017-II]

NDA Worked Example 3   Algebraic Relation: Factor Before Classifying

Question Question (NDA 2022-I) A relation R is defined on ℕ as x R y ⟺ x² − 5xy + 4y² = 0. Which statements are correct? 1. R is reflexive    2. R is symmetric    3. R is transitive

Solution Step 1 : Factor the relation Never attempt to classify a quadratic relation from its unfactored form. Factor first. x² − 5xy + 4y² = (x − y)(x − 4y) = 0 So x R y if and only if x = y  or  x = 4y. Step 2 : Test reflexivity Substitute y = x: x = x is always true. So (x,x) ∈ R for all x ∈ ℕ. R is reflexive. Step 3 : Test symmetry (find a counterexample) If x R y, must y R x? Suppose x = 4, y = 1: x = 4y = 4(1) = 4 , so 4 R 1. Does 1 R 4? Need 1 = 4 (no) or 1 = 4(4) = 16 (no). So 1 is NOT related to 4. Symmetry fails. Step 4 : Test transitivity (find a counterexample) If x = 4y and y = 4z, then x = 16z. But x R z requires x = z or x = 4z. Neither holds for z ∈ ℕ. Concrete: 16 R 4 (16 = 4×4) and 4 R 1 (4 = 4×1 ).  Does 16 R 1?  Need 16 = 1 or 16 = 4.  Both false. Transitivity fails. Same factoring method.  [NDA 2019-I]

Answer: Statement 1 only: R is reflexive but not symmetric and not transitive. Option (a).

Factor the quadratic before testing any property. In the unfactored form, reflexivity is not obvious; in the factored form, it is immediate. The same method applies to NDA-2019-I (x² − 4xy + 3y² = (x−y)(x−3y) = 0), which also gives: reflexive only.

x² = y³ on ℕ : Reflexivity: x² = x³ iff x = 0 or x = 1; fails for x=2. Not reflexive. Not symmetric (e.g. x=8, y=4: 64=64, but (4,8): 16≠512). Neither symmetric nor transitive: both statements NOT correct.  [NDA 2022-II]

x = y³ on ℕ : neither symmetric nor transitive.  [NDA 2026-I]

|x+y| < 2 on the open interval (−1,1) : reflexive (|x+x|=2|x|<2 since |x|<1) and symmetric (|x+y|=|y+x|) but not transitive (x=−0.9, y=0, z=0.9 fails).  [NDA 2024-I]

x ≤ y² on positive numbers: reflexive (x ≤ x² for x ≥ 1, but fails for 0 < x < 1). The answer depends on the domain. Check the domain first.  [NDA 2016-I]

7. Equivalence Classes and Partitions

When a relation R on A is an equivalence relation, it partitions A into non-overlapping groups called equivalence classes.

The equivalence class of a (written [a] or [a]_R) is the set of all elements in A that are related to a:

[a] = {b ∈ A: (a, b) ∈ R}

Two fundamental properties of equivalence classes: (1) Any two equivalence classes are either equal or disjoint; they never partially overlap. (2) The union of all equivalence classes is A: every element belongs to exactly one class.

Equivalence Classes : Partition of ℤ by “a − b divisible by 5”
[a₁] …,−10,−5, 0, 5,10,…	[a₂] …, −9,−4, 1, 6,11,…	[a₃] …, −8,−3, 2, 7,12,…	[a₄] …, −7,−2, 3, 8,13,…	[a₅] …, −6,−1, 4, 9,14,…
Any two classes are equal or disjoint  ·  Their union = ℤ  ·  Every integer belongs to exactly one class

For the equivalence relation on ℤ defined by “a − b divisible by 5”, the five classes above partition ℤ completely : as shown in the diagram.  [NDA 2015-I]

8. Inverse Relations and Composition

8.1  Inverse Relation

The inverse relation R⁻¹ is formed by reversing every pair in R:

R⁻¹ = {(b, a) : (a, b) ∈ R}

If R is a relation from A to B, then R⁻¹ is a relation from B to A.

Example: R = {(1,3),(1,5),(2,3),(2,5),(3,5),(4,5)} from A = {1,2,3,4} to B = {1,3,5}. Then R⁻¹ = {(3,1),(5,1),(3,2),(5,2),(5,3),(5,4)}.

Computed R∘R⁻¹ for R = {(a,b): a < b} from {1,2,3,4} to {1,3,5}: R∘R⁻¹ = {(3,3),(3,5),(5,3),(5,5)}.  [NDA 2016-II]

8.2  Composition R∘R⁻¹

(a, c) ∈ R∘R⁻¹ if there exists b ∈ A such that (b, a) ∈ R (i.e. (a, b) ∈ R⁻¹) and (b, c) ∈ R. Here a and c are elements of B, and b is the intermediate element from A through which they are connected.

8.3  Which Properties Survive Inversion?

Property	R⁻¹	P ∩ Q (both have it)	P ∪ Q (both have it)
Reflexive	Always	Always	Always
Symmetric	Always	Always	Always
Transitive	Always	Always	Not always

The table assumes P and Q each possess the property being tested. The only failure is transitivity under union.

Why transitivity does not survive P∪Q: Consider A = {1,2,3}. Let P = {(1,1),(2,2),(1,2),(2,1)} and Q = {(2,2),(3,3),(2,3),(3,2)}. Both P and Q are symmetric and transitive. P∪Q = {(1,1),(2,2),(3,3),(1,2),(2,1),(2,3),(3,2)}. Now (1,2) and (2,3) are in P∪Q, but (1,3) is not. Transitivity fails for P∪Q. Why all three properties survive R⁻¹: Reflexive: (a,a) ∈ R → (a,a) ∈ R⁻¹ (self-pairs unchanged). Symmetric: if R is symmetric, R⁻¹ = R, so R⁻¹ is symmetric. Transitive: if (a,b),(b,c) ∈ R⁻¹ then (b,a),(c,b) ∈ R; by transitivity of R, (c,a) ∈ R; so (a,c) ∈ R⁻¹.

NDA Worked Example 6   Three-Claim Statement Adjudication: Inverse Relation

Question Question (NDA 2022-I) Consider the following statements in respect of any relation R on a set A: 1. If R is reflexive, then R⁻¹ is also reflexive. 2. If R is symmetric, then R⁻¹ is also symmetric. 3. If R is transitive, then R⁻¹ is also transitive. Which of the above statements are correct?

Solution Claim 1: If (a,a) ∈ R for all a ∈ A (reflexive), then by definition of R⁻¹, (a,a) ∈ R⁻¹ for all a ∈ A. Self-pairs reverse to the same self-pairs. Claim 1 is TRUE. Claim 2: If R is symmetric, then (a,b) ∈ R ↔ (b,a) ∈ R. R⁻¹ = {(b,a):(a,b)∈R} = R (since R is symmetric). A relation equal to R is itself symmetric. Claim 2 is TRUE. Claim 3: Suppose (a,b),(b,c) ∈ R⁻¹. Then (b,a),(c,b) ∈ R. Since R is transitive and (c,b),(b,a) ∈ R, we get (c,a) ∈ R. Therefore (a,c) ∈ R⁻¹. Claim 3 is TRUE. Do properties survive P∩Q and P∪Q? Reflexivity: both. Symmetry: both. Transitivity: P∩Q only, P∪Q. Correct statements: I, II and III for P∩Q; I and II only for P∪Q.  [NDA 2024-II]

Answer: All three statements correct: option (d) 1, 2 and 3.

This is the modern format: three independent claims, each worth one judgement. Adjudicate each separately. Do not assume any one follows from the others.

9. Operations on Relations

When P and Q are both relations on the same set A, we can form P∪Q and P∩Q. The survival table in Section 8 applies here. The most important result: transitivity does not always survive union, even when both relations are transitive.

P and Q both reflexive → P∩Q reflexive; P∪Q reflexive.  P and Q both symmetric → P∩Q symmetric; P∪Q symmetric.  P and Q both transitive → P∩Q transitive; P∪Q NOT always transitive.  [NDA 2024-II]

10. Important Distinctions

Relation vs Function.  A relation is any subset of A×B: no restrictions on how many pairs share the same first element. A function is a special relation where each element of A appears as the first element of exactly one pair. Every function is a relation; not every relation is a function.  [NDA 2021-II]

Domain vs Range vs Codomain.  Domain and range are derived from the pairs actually in R. Codomain is specified in advance. Range ⊆ Codomain always; they are equal only when every codomain element appears in R.

Reflexive vs Irreflexive.  “Not reflexive” means at least one (a,a) is missing. “Irreflexive” means every (a,a) is missing. A relation can be neither reflexive nor irreflexive if some (a,a) pairs exist but not all.

Symmetric vs Antisymmetric.  A symmetric relation requires that (a,b) ∈ R always forces (b,a) ∈ R. Antisymmetry requires that (a,b) ∈ R and (b,a) ∈ R together force a = b. “Antisymmetric” does not mean “not symmetric” : they are different conditions, and both can hold simultaneously.

“Relation on A” vs “relation from A to B”.  “On A” means the relation is a subset of A×A (n² possible pairs). “From A to B” means a subset of A×B (mn possible pairs). These are different domains and give different counting results.  [NDA 2013-I, NDA 2015-I]

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