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Boolean Algebra

Description: This quiz covers the fundamental concepts and operations of Boolean algebra, a branch of mathematics that deals with logical operations and binary variables.
Number of Questions: 15
Created by:
Tags: boolean algebra logic binary operations truth tables
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What is the dual of the expression (A ∨ B) ∧ (C ∧ D)?

  1. ¬(¬A ∧ ¬B) ∨ ¬(¬C ∨ ¬D)

  2. ¬(¬A ∨ ¬B) ∧ ¬(¬C ∧ ¬D)

  3. ¬(A ∧ B) ∨ ¬(C ∨ D)

  4. ¬(A ∨ B) ∧ ¬(C ∧ D)

Show answer
Correct Option: A
Explanation:

The dual of a Boolean expression is obtained by interchanging ∨ and ∧, and 0 and 1.

Simplify the following Boolean expression: (A ∨ B) ∧ (¬A ∨ C)

  1. A ∨ C

  2. B ∨ C

  3. A ∨ B ∨ C

  4. ¬A ∨ B ∨ C

Show answer
Correct Option: B
Explanation:

Using the distributive law, we can simplify the expression as follows: (A ∨ B) ∧ (¬A ∨ C) = (A ∧ ¬A) ∨ (A ∧ C) ∨ (B ∧ ¬A) ∨ (B ∧ C) = 0 ∨ (A ∧ C) ∨ (B ∧ ¬A) ∨ (B ∧ C) = (A ∧ C) ∨ (B ∧ ¬A) ∨ (B ∧ C) = (A ∨ B) ∨ C = B ∨ C

Construct the truth table for the following Boolean expression: (A ∧ B) ∨ (¬A ∧ C)

  1. A B C (A ∧ B) ∨ (¬A ∧ C)
    0 0 0 0
    0 0 1 1
    0 1 0 0
    0 1 1 1
    1 0 0 0
    1 0 1 1
    1 1 0 1
    1 1 1 1
  2. A B C (A ∧ B) ∨ (¬A ∧ C)
    0 0 0 1
    0 0 1 0
    0 1 0 1
    0 1 1 0
    1 0 0 1
    1 0 1 0
    1 1 0 0
    1 1 1 1
  3. A B C (A ∧ B) ∨ (¬A ∧ C)
    0 0 0 0
    0 0 1 1
    0 1 0 1
    0 1 1 0
    1 0 0 1
    1 0 1 0
    1 1 0 1
    1 1 1 0
  4. A B C (A ∧ B) ∨ (¬A ∧ C)
    0 0 0 1
    0 0 1 1
    0 1 0 0
    0 1 1 1
    1 0 0 0
    1 0 1 1
    1 1 0 0
    1 1 1 0
Show answer
Correct Option:
Explanation:

The truth table for the given Boolean expression is as follows: A | B | C | (A ∧ B) ∨ (¬A ∧ C) --- | --- | --- | --- 0 | 0 | 0 | 0 0 | 0 | 1 | 1 0 | 1 | 0 | 1 0 | 1 | 1 | 1 1 | 0 | 0 | 1 1 | 0 | 1 | 1 1 | 1 | 0 | 1 1 | 1 | 1 | 1

Which of the following is a valid Boolean identity?

  1. ¬(A ∨ B) = ¬A ∨ ¬B

  2. ¬(A ∧ B) = ¬A ∧ ¬B

  3. A ∨ B = A ∧ B

  4. A ∧ B = A ∨ B

Show answer
Correct Option: A
Explanation:

De Morgan's law states that the negation of a disjunction is the conjunction of the negations, and vice versa. Therefore, ¬(A ∨ B) = ¬A ∨ ¬B is a valid Boolean identity.

Find the minimal sum-of-products form of the following Boolean expression: (A ∨ B) ∧ (¬A ∨ C) ∧ (B ∨ ¬C)

  1. A ∨ B ∨ C

  2. A ∨ C

  3. B ∨ C

  4. A ∨ B

Show answer
Correct Option: B
Explanation:

Using Boolean algebra, we can simplify the expression as follows: (A ∨ B) ∧ (¬A ∨ C) ∧ (B ∨ ¬C) = (A ∨ B ∨ C) ∧ (¬A ∨ C) = (A ∨ C) ∧ (¬A ∨ C) = A ∨ C

What is the complement of the Boolean expression (A ∧ B) ∨ (¬A ∧ C)?

  1. ¬(A ∧ B) ∧ ¬(¬A ∧ C)

  2. ¬(A ∧ B) ∨ ¬(¬A ∧ C)

  3. (A ∨ B) ∧ (¬A ∨ C)

  4. (A ∨ B) ∨ (¬A ∨ C)

Show answer
Correct Option: A
Explanation:

The complement of a Boolean expression is obtained by negating the entire expression. Therefore, the complement of (A ∧ B) ∨ (¬A ∧ C) is ¬(A ∧ B) ∧ ¬(¬A ∧ C).

Which of the following Boolean expressions is equivalent to ¬(A ⊕ B)?

  1. A ∨ B

  2. A ∧ B

  3. ¬A ∧ ¬B

  4. ¬A ∨ ¬B

Show answer
Correct Option: D
Explanation:

The exclusive OR (⊕) operation is defined as A ⊕ B = (A ∨ B) ∧ ¬(A ∧ B). Therefore, ¬(A ⊕ B) = ¬[(A ∨ B) ∧ ¬(A ∧ B)] = ¬(A ∨ B) ∨ (A ∧ B) = ¬A ∨ ¬B.

Simplify the following Boolean expression using Boolean algebra: (A ∨ B) ∧ (¬A ∨ ¬B)

  1. 0

  2. 1

  3. A ∨ B

  4. ¬A ∨ ¬B

Show answer
Correct Option: A
Explanation:

Using Boolean algebra, we can simplify the expression as follows: (A ∨ B) ∧ (¬A ∨ ¬B) = (A ∨ B) ∧ ¬(A ∧ B) = ¬(A ∧ B) ∧ (A ∨ B) = 0

Which of the following is a valid Boolean law?

  1. Associative law of addition

  2. Associative law of multiplication

  3. Distributive law of addition over multiplication

  4. Distributive law of multiplication over addition

Show answer
Correct Option: C
Explanation:

The distributive law of addition over multiplication states that A ∨ (B ∧ C) = (A ∨ B) ∧ (A ∨ C). This law is valid in Boolean algebra.

What is the dual of the Boolean expression (A ∧ B) ∨ (¬A ∧ C)?

  1. ¬(¬A ∨ ¬B) ∧ ¬(A ∨ ¬C)

  2. ¬(¬A ∨ ¬B) ∨ ¬(A ∨ ¬C)

  3. ¬(A ∨ B) ∧ ¬(¬A ∨ C)

  4. ¬(A ∨ B) ∨ ¬(¬A ∨ C)

Show answer
Correct Option: A
Explanation:

The dual of a Boolean expression is obtained by interchanging ∨ and ∧, and 0 and 1. Therefore, the dual of (A ∧ B) ∨ (¬A ∧ C) is ¬(¬A ∨ ¬B) ∧ ¬(A ∨ ¬C).

Simplify the following Boolean expression: (A ∨ B) ∧ (¬A ∨ C) ∧ (B ∨ ¬C)

  1. A ∨ B ∨ C

  2. A ∨ C

  3. B ∨ C

  4. A ∨ B

Show answer
Correct Option: B
Explanation:

Using Boolean algebra, we can simplify the expression as follows: (A ∨ B) ∧ (¬A ∨ C) ∧ (B ∨ ¬C) = (A ∨ B ∨ C) ∧ (¬A ∨ C) = (A ∨ C) ∧ (¬A ∨ C) = A ∨ C

Which of the following is a valid Boolean identity?

  1. ¬(A ∨ B) = ¬A ∨ ¬B

  2. ¬(A ∧ B) = ¬A ∧ ¬B

  3. A ∨ B = A ∧ B

  4. A ∧ B = A ∨ B

Show answer
Correct Option: A
Explanation:

De Morgan's law states that the negation of a disjunction is the conjunction of the negations, and vice versa. Therefore, ¬(A ∨ B) = ¬A ∨ ¬B is a valid Boolean identity.

Find the minimal sum-of-products form of the following Boolean expression: (A ∨ B) ∧ (¬A ∨ C) ∧ (B ∨ ¬C)

  1. A ∨ B ∨ C

  2. A ∨ C

  3. B ∨ C

  4. A ∨ B

Show answer
Correct Option: B
Explanation:

Using Boolean algebra, we can simplify the expression as follows: (A ∨ B) ∧ (¬A ∨ C) ∧ (B ∨ ¬C) = (A ∨ B ∨ C) ∧ (¬A ∨ C) = (A ∨ C) ∧ (¬A ∨ C) = A ∨ C

What is the complement of the Boolean expression (A ∧ B) ∨ (¬A ∧ C)?

  1. ¬(A ∧ B) ∧ ¬(¬A ∧ C)

  2. ¬(A ∧ B) ∨ ¬(¬A ∧ C)

  3. (A ∨ B) ∧ (¬A ∨ C)

  4. (A ∨ B) ∨ (¬A ∨ C)

Show answer
Correct Option: A
Explanation:

The complement of a Boolean expression is obtained by negating the entire expression. Therefore, the complement of (A ∧ B) ∨ (¬A ∧ C) is ¬(A ∧ B) ∧ ¬(¬A ∧ C).

Which of the following Boolean expressions is equivalent to ¬(A ⊕ B)?

  1. A ∨ B

  2. A ∧ B

  3. ¬A ∧ ¬B

  4. ¬A ∨ ¬B

Show answer
Correct Option: D
Explanation:

The exclusive OR (⊕) operation is defined as A ⊕ B = (A ∨ B) ∧ ¬(A ∧ B). Therefore, ¬(A ⊕ B) = ¬[(A ∨ B) ∧ ¬(A ∧ B)] = ¬(A ∨ B) ∨ (A ∧ B) = ¬A ∨ ¬B.

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