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Boolean Algebra

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Boolean Algebra

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Boolean Algebra operates on the two-element set B = {0, 1} using three operations: OR (+), AND (·), and NOT (′, pronounced “bar” or “complement”). A Boolean variable can only hold 0 or 1 — never 2, never a fraction. The most-tested identity in NDA is De Morgan’s Theorem: (A·B)′ = A′ + B′ and (A + B)′ = A′·B′.

Exam pointer: The trap question asks “what is 1 + 1 in Boolean algebra?” Answer: 1, because OR saturates at 1.

IdentityAND formOR form
IdentityA · 1 = AA + 0 = A
ComplementA · A′ = 0A + A′ = 1
IdempotentA · A = AA + A = A

Remember precedence: NOT binds first, then AND, then OR. NAND/CDS aspirants get 1–2 MCQs from this topic every year, usually worth 3–4 marks.

🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Definition and Huntington’s Postulates

A Boolean Algebra is an algebraic structure (B, +, ·, ′) satisfying: closure under + and ·, commutativity and associativity of both, distributivity of · over + (and + over ·), existence of distinct identity elements 0 and 1, and existence of a complement A′ for every element A. The set {0, 1} with OR, AND, NOT is the simplest instance and is the only Boolean algebra needed for NDA questions.

Core Laws with Variables Defined

Let A, B, C ∈ {0, 1}. All quantities are dimensionless logical values.

  • Identity Law: A + 0 = A and A · 1 = A
  • Complement Law: A + A′ = 1 and A · A′ = 0
  • Idempotent Law: A + A = A and A · A = A
  • Domination Law: A + 1 = 1 and A · 0 = 0
  • Distributive Law: A · (B + C) = A·B + A·C and A + (B·C) = (A + B)·(A + C)
  • Involution Law: (A′)′ = A

De Morgan’s Theorem

The single most heavily tested identity. De Morgan’s Theorem states that the complement of a sum equals the product of complements, and vice versa:

(A + B)′ = A′ · B′ and (A · B)′ = A′ + B′

where A, B ∈ {0, 1}. The rule generalises to n variables: the complement of any Boolean expression is obtained by swapping + with · and complementing every variable individually.

Duality Principle

The dual of any Boolean identity is obtained by simultaneously swapping + ↔ · and 0 ↔ 1, leaving variables untouched. The dual of A + 0 = A is A · 1 = A. If one identity is true, its dual is automatically true — examiners test this by asking the dual form of a given identity.

Switching Circuits

Boolean algebra models switching networks directly. A closed switch = 1, an open switch = 0, series connection = AND (·), parallel connection = OR (+), and an inverter (NOT gate) corresponds to the complement operation. The expression (A + B) · A′ describes a parallel block A+B in series with the complement of A.

Typical NDA Question Types

  1. Evaluate a 2- or 3-variable expression using a given truth-table excerpt.
  2. Apply De Morgan’s theorem to simplify a complemented sum/product.
  3. Identify the dual of a stated identity.
  4. Translate a verbal statement (“at least one of X or Y is false”) into symbols.

Scoring tip: Always build the truth table for unfamiliar expressions — even partial tables eliminate distractors in seconds.

🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Worked Micro-Example

Simplify F = A′B + AB + A′B′.

Step 1 — apply complement law on the first two terms: A′B + AB = (A′ + A)·B = 1·B = B. Step 2 — F reduces to B + A′B′. Step 3 — factor B: B(1 + A′B′) = B · 1 = B. So F = B. A truth table over (A, B) ∈ {0,1}² confirms F equals B for all four input rows.

Minterms, Maxterms, SOP and POS

A minterm is a product (AND) containing every variable exactly once, either complemented or uncomplemented. For 3 variables there are 2³ = 8 minterms: A′B′C′, A′B′C, A′BC′, …, ABC. A maxterm is the corresponding sum (OR) form. Sum of Products (SOP) is a disjunction of minterms selected from a truth-table row where the function equals 1; Product of Sums (POS) is a conjunction of maxterms where the function equals 0. NDA questions occasionally ask students to write the SOP form from a small truth table.

Edge Cases and Operator Traps

  • Precedence pitfall: Without brackets, A + B · C′ parses as A + (B · C′), not (A + B) · C′. Neglecting this yields a wrong truth table.
  • Subtraction fallacy: A′ is not A − 1; it is logical negation. For A = 0, A′ = 1; for A = 1, A′ = 0.
  • Arithmetic contamination: Many aspirants write 1 + 1 = 2 — Boolean OR gives 1 + 1 = 1 because the result must stay in {0, 1}.
  • Dual-of-dual: The dual of a dual is the original expression; this is occasionally framed as a verification MCQ.

Connections to Adjacent NDA Topics

Boolean algebra bridges Set Theory (union ↔ OR, intersection ↔ AND, complement ↔ NOT, with Venn diagrams providing geometric proof) and Propositional Logic (∨, ∧, ¬ mirror +, ·, ′ exactly). It also seeds Probability foundations, where events combine with AND/OR under independence assumptions. Mastery here reduces friction in higher algebra chapters and in reasoning-based questions.

Exam strategy: Boolean Algebra carries ~2% weight in NDA Paper I. Treat it as a 4–6 minute sub-section: memorise De Morgan’s both forms, the four basic laws, and the dual of each, then practise 15 MCQs from previous UPSC papers.

Practice Prompts

  1. Using De Morgan’s theorem, simplify (A + B′)′ and state its dual.
  2. Draw the switching circuit for F = (A + B) · (A′ + C) and identify the input combination (A, B, C) that forces F = 0.

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