DISCRETE MATHEMATICS/Student Engineering Notebook
Page 03·notes
SETS, RELATIONS, POSETS & LATTICES—~34 MIN READ

Relations, Closures, Equivalence Partitions & Counting

“Reflexive, symmetric, transitive closures, counting relation types on finite sets, and partition Bell numbers.”

1.Binary Relation Axioms on Finite Sets

A binary relation R from set A to set B is a subset of the Cartesian product A times B. When A = B, R is a relation on A.

For a set A with |A| = n elements, the product A times A has n^2 ordered pairs.

Core Properties:

1. Reflexive: forall a in A, (a, a) in R.

2. Irreflexive: forall a in A, (a, a) notin R.

3. Symmetric: forall a, b in A, (a, b) in R implies (b, a) in R.

4. Anti-Symmetric: forall a, b in A, [ (a, b) in R land (b, a) in R ] implies a = b.

5. Asymmetric: forall a, b in A, (a, b) in R implies (b, a) notin R (implies irreflexive).

6. Transitive: forall a, b, c in A, [ (a, b) in R land (b, c) in R ] implies (a, c) in R.

2.Combinatorial Counting of Relation Types

Represent R as an n times n Boolean adjacency matrix:

- There are n main diagonal entries (a_i, a_i).

- There are n^2 - n = n(n - 1) off-diagonal entries, forming n(n - 1) / 2 symmetric pairs { (a_i, a_j), (a_j, a_i) } with i < j.

Derivations:

- Reflexive: Diagonal entries are fixed to 1 (1 choice). The n(n - 1) off-diagonal entries can each be 0 or 1 (2 choices). Total = 2^{n(n - 1)} = 2^{n^2 - n}.

- Symmetric: Diagonal entries can each be 0 or 1 (2^n choices). For each of the n(n - 1)/2 off-diagonal pairs, both must be 0 or both must be 1 (2 choices). Total = 2^n * 2^{n(n - 1)/2} = 2^{n(n + 1)/2}.

- Anti-Symmetric: Diagonal entries can be 0 or 1 (2^n choices). For each of the n(n - 1)/2 pairs, we cannot have both equal 1. Possible choices: (0, 0), (1, 0), (0, 1) (3 choices). Total = 2^n * 3^{n(n - 1)/2}.

WORKED EXAM ANALYSIS · STEP-BY-STEP DERIVATIONOriginal Practice Problem · Modeled on GATE Pattern (2 Marks)

3. Worked Problem: Exact Counting on Finite Set of Size n = 4

⚠️ Pitfall Warning:

Do not mix up Reflexive & Symmetric with just Symmetric. Adding reflexivity forces all n diagonal entries to 1, eliminating 2^n degrees of freedom.

Problem: For a set A containing n = 4 elements, calculate:

1. The number of reflexive relations.

2. The number of symmetric relations.

3. The number of relations that are simultaneously reflexive and symmetric.

4. The number of anti-symmetric relations.

Step 1: Set size parameters: n = 4, n^2 = 16, n(n - 1) / 2 = 4(3) / 2 = 6 pairs.

Step 2: Calculate Reflexive relations:

Formula: 2^{n^2 - n} = 2^{16 - 4} = 2^{12} = 4,096 relations.

Step 3: Calculate Symmetric relations:

Formula: 2^{n(n + 1) / 2} = 2^{4(5) / 2} = 2^{10} = 1,024 relations.

Step 4: Calculate Reflexive & Symmetric relations:

Diagonal has 1 choice (must be 1). Off-diagonal has 6 pairs, each with 2 choices.

Formula: 2^{n(n - 1) / 2} = 2^6 = 64 relations.

Step 5: Calculate Anti-Symmetric relations:

Formula: 2^n 3^{n(n - 1) / 2} = 2^4 3^6 = 16 * 729 = 11,664 relations.

SOURCE TRANSPARENCY & CITATIONS

Primary Academic References & Exam Sources

Every concept in Notes From a B.Tech Brain is synthesized from authoritative academic texts, seminal peer-reviewed research, verified exam archives, and renowned university lectures.

Curated University Lecture Series
Discrete Mathematics and Its Applications (8th Edition) ·
LECTURE SERIES
RECOMMENDED YOUTUBE EDUCATORS & FULL-COURSE PLAYLISTS

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