Grade 11 Logic and Set (Chapter-1) - Worked Out Examples

Grade-11 Logic and Set Chapter-1 Worked Out Examples.

Grade 11 Logic and Set (Chapter-1) - Worked Out Examples provides students with step-by-step solutions to the most important questions from the first chapter of the Grade 11 Mathematics syllabus. This resource helps Science, Management, Arts, and other faculty students understand fundamental concepts such as logic, statements, truth tables, logical connectives, sets, subsets, power sets, Venn diagrams, and set operations. By practicing these solved examples, students can strengthen their problem-solving skills, improve conceptual understanding, and prepare effectively for classroom tests, internal assessments, and final board examinations. Grade 11 Logic and Set (Chapter-1) - Worked Out Examples also enables learners to identify common mistakes and apply mathematical reasoning with confidence.

Designed according to the latest Grade 11 curriculum, these worked-out examples explain every solution in a simple and easy-to-follow format. Students can use this study material for self-learning, quick revision, and exam preparation while building a strong foundation for advanced mathematics chapters. Each example follows a logical approach that enhances analytical thinking and encourages accurate problem-solving techniques. Whether you are revising before an exam or learning the chapter for the first time, Grade 11 Logic and Set (Chapter-1) - Worked Out Examples serves as a reliable study companion for achieving better academic performance in Grade 11 Mathematics.

Grade 11 Logic and Set (Chapter-1) - Worked Out Examples | Complete Solutions

Download Grade 11 Logic and Set (Chapter-1) - Worked Out Examples with step-by-step solutions for Science, Management, Arts, and other faculties. Learn logic, sets, Venn diagrams, truth tables, and prepare effectively for Grade 11 Mathematics exams.

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LOGIC

Basic Laws in Logics

The following laws of logic are worth mentioning.

1. Law of Negation: Only one of the statement p or ~ p is true i.e. the statement p and ~ p have the opposite truth value.

2. Law of involution: ~ (~ p) ≡ p i.e. the negation of the negation of a statement is logically equivalent to the same statement.

3. Law of tautology: The statement p ∨ ~ p is a tautology.

4. Law of contradiction: The statement p ∧ ~ p is always false and is known as the contradiction.

5. Law of contra positive: (p ⇒ q) ≡ (~ q ⇒ ~ p)

6. Inverse law: ~ p ⇒ ~ q ≡ q ⇒ p

7. Law of syllogism: If p ⇒ q, q ⇒ r then p ⇒ r which can be written as [(p ⇒ q) ∧ (q ⇒ r)] ⇒ (p ⇒ r)

Algebra of Logic

Using the ideas aforementioned to prove equivalent statements, we can easily prove:

1. Idempotent Laws:     p ∨ p ≡ p,                                     p ∧ p ≡ p

2. Commutative laws:   p ∨ q ≡ q ∨ p,                              p ∧ q ≡ q ∧ p

3. Associative Laws:     p ∨ (q ∨ r) ≡ (p ∨ q) ∨ r,              p ∧ (q ∧ r) ≡ (p ∧ q) ∧ r

4. Distributive Laws:     p ∨ (q ∧ r) ≡ (p ∨ q) ∧ (p ∨ r),      p ∧ (q ∨ r) ≡ (p ∧ q) ∨ (p ∧ r)

5. De Morgan's Laws:   ~ (p ∨ q) ≡ ~ p ∧ ~ q,                   ~ (p ∧ q) ≡ ~ p ∨ ~ q

Negation of a Conditional

A conditional p ⇒ q is equivalent to (~ p) ∨ q.

p	q	~ p	p ⇒ q	(~ p) ∨ q
T	T	F	T	T
T	F	F	F	F
F	T	T	T	T
F	F	T	T	T

Using De Morgan's law, the negation of a conditional p ⇒ q is p ∧ (~ q).

WORKED OUT EXAMPLES

EXAMPLE-1

Construct a truth table for:

a. p ∧ q ⇒ p ∨ q

b. p ∨ ~ q

Solution:

a. p ∧ q ⇒ p ∨ q

p	q	p ∧ q	p ∨ q	p ∧ q ⇒ p ∨ q
T	T	T	T	T
T	F	F	T	T
F	T	F	T	T
F	F	F	F	T

b. p ∨ ~ q

p	q	~ q	p ∨ ~ q
T	T	F	T
T	F	T	T
F	T	F	F
F	F	T	T

EXAMPLE-2

If truth values, of p, q, r and s be T, T, F and F, find the truth value of (p ∧ q) ∨ (r ∧ ~ s).

Solution:

p	q	r	s	p ∧ q	~ s	r ∧ ~ s	(p ∧ q) ∨ (r ∧ ~ s)
T	T	F	F	T	T	F	T

EXAMPLE-3

If p and q be two statements, prove that ~ (p ∧ q) ≡ ~p ∨ ~ q.

Solution:

p	q	p ∧ q	~ (p ∧ q)	~ p	~ q	~ pv - q
T	T	T	F	F	F	F
T	F	F	T	F	T	T
F	T	F	T	T	F	T
F	F	F	T	T	T	T

EXAMPLE-4

If p and q be statements, then prove that p ∨ (~p ∨ q) is a tautology.

Solution:

p	q	~ p	~ p ∨ q	p ∨ (~ p ∨ q)
T	T	F	T	T
T	F	F	F	T
F	T	T	T	T
F	F	T	T	T

EXAMPLE-5

If r, s and t be statements, then prove that [r ⇒ s) ∧ (s ⇒ t)] ⇒ (r ⇒ t) is a tautology.

Solution:

r	s	t	r ⇒ s	s ⇒ t	r ⇒ t	(r ⇒ s) ∧ (s ⇒ t)	[(r ⇒ s) ∧ (s ⇒ t)] ⇒ (r ⇒ t)
T	T	T	T	T	T	T	T
T	T	F	T	F	F	F	T
T	F	T	F	T	T	F	T
T	F	F	F	T	F	F	T
F	T	T	T	F	T	T	T
F	T	F	T	T	T	F	T
F	F	T	T	T	T	T	T
F	F	F	T	T	T	T	T

EXAMPLE-6

Write the converse, inverse and Contrapositive of the statement 'if 3 is an odd number then 6 is not an odd number".

Solution:

Let p : 3 is an odd number and q : 6 is not an odd number.

Then, the statement is 'p ⇒ q'.

Now,

Converse: q ⇒ p i.e., if 6 is not an odd number then 3 is an odd number.

Inverse: ~ p ⇒ ~ q i.e., if 3 is not an odd number, then 6 is an odd number.

Contrapositive: ~ q ⇒ ~ p i.e., if 6 is an odd number, then 3 is not an odd number.

EXAMPLE-7

Write the negation of the conditional "If Ram is a boy, then Sita is a girl".

Solution:

Let p : Ram is a boy. and q : Sita is a girl.

Thus the given statement is p ⇒ q. It negation is p ∧ ~ q.

i.e., Ram is a boy and Sita is not a girl.

SET

Theorems Based on Sets: Laws of Set Algebra

The laws concerning the set operations are listed as the theorems based on set operations . The theorems with their proofs are given below:

Theorem 1 - IDEMPOTENT LAWS:

Let U be the universal set and A, B, C be the subsets of U. Then,

a. A ⋃ A = A and

b. A ∩ A = A

Proof:

a) A ⋃ A = A

A ⋃ A

= {x: x ∈ A or x ∈ A}

= {x: x ∈ A}

= A

∴ A ⋃ A = A.

b) A ∩ A = A

A ∩ A

= {x: x ∈ A or x ∈ A}

= {x: x ∈ A}

= A

∴ A ∩ A = A.

Theorem 2 - IDENTITY LAWS:

Let A be the subset of a universal set U. Then,

a. A ⋃ ∅ = A

b. A ⋃ U = U

c. A ∩ ∅ = ∅

d. A ∩ U = A

Proof:

a) A ⋃ ∅ = A

Obviously A ⊆ A ⋃ ∅.

For the converse, let x ∈ A ⋃ ∅ ⇒ x ∈ A or x ∈ ∅

⇒ x ∈ A [∵ ∅ ⊆ A]

∴ A ⋃ ∅ = A

b) A ⋃ U = U

U ⊆ A ⋃ U is obvious. For the converse,

Let x A ⋃ U ⇒ x ∈ A or x ∈ U

⇒ x ∈ U [∵A ⊆ U]

∴ A ⋃ U ⊆ U.

Hence, U = A ⋃ U.

c) A ∩ ∅ = ∅

Since empty set is the subset of every set,

∅ ⊆ A ∩ ∅

Conversely, x ∈ A ∩ ∅ ⇒ x ∈ A and x ∈ ∅ ⇒ x ∈ ∅

Hence, ∅ = A ∩ ∅

d) A ∩ U = A

By the definition, A ∩ U ⊆ A

Conversely x ∈ A ⇒ x ∈ A ∩ U [∵ A U]

∴ A ⊆ A ∩ U

Hence, A = A ∩ U or A ∩ U = A.

Theorem 3 - COMMUTATIVE LAWS:

Let A and B the subjects of a universal set U. Then,

a. A ⋃ B = B ⋃ A

b. A ∩ B = B ∩ A

Proof:

a) A ⋃ B = B ⋃ A

A ⋃ B

= {x:x ∈ A or x ∈ B}

= {x:x ∈ B or x ∈ A}

= {x:x ∈ B ⋃ A}

= B ⋃ A

b) A ∩ B = B ∩ A

A ∩ B

= {x:x ∈ A and x ∈ B}

= {x:x ∈ A and x ∈ B}

= B ∩ A

Theorem 4 - ASSOCIATIVE LAWS:

Let A, B and C be subsets of universal set U. Then,

a. A ⋃ (B ⋃ C) = (A ⋃ B) ⋃ C

b. A ∩ (B ∩ C) = (A ∩ B) ∩ C

Proof:

a) A ⋃ (B ⋃ C) = (A ⋃ B) ⋃ C

= A ⋃ (B ⋃ C)

= {x:x ∈ A ⋃ (B ⋃ C)}

= {x:x ∈ A or x ∈ (B ⋃ C)}

= {x:x ∈ A or x ∈ B or x ∈ C)}

= {x:x ∈ (A ⋃ B) or x ∈ C}

= {x:x ∈ (A ⋃ B) ⋃ C}

= (A ⋃ B) ⋃ C

b) A ∩ (B ∩ C) = (A ∩ B) ∩ C

= A ∩ (B ∩ C)

= {x:x ∈ A ∩ (B ∩ C)}

= {x:x ∈ A or x ∈ (B ∩ C)}

= {x:x ∈ A or x ∈ B or x ∈ C)}

= {x:x ∈ (A ∩ B) or x ∈ C}

= {x:x ∈ (A ∩ B) ∩ C}

= (A ∩ B) ∩ C

Theorem 5 - DISTRIBUTIVE LAWS:

Let A, B and C be subsets of universal set U. Then,

a. A ⋃ (B ∩ C) = (A ⋃ B) ∩ (A ⋃ C)

b. A ∩ (B ⋃ C) = (A ∩ B) ⋃ (A ∩ C)

Proof:

a. A ⋃ (B ∩ C) = (A ⋃ B) ∩ (A ⋃ C)

A ⋃ (B ∩ C)

= {x:x ∈ A ⋃ (B ∩ C)}

= {x:x ∈ A or x ∈ (B ∩ C)}

= {x:x ∈A or (x ∈ B and x ∈ C)}

= {x: (x ∈ A or x ∈ B) and (x ∈ A or x ∈ C)}

= {x:x ∈ (A ⋃ B) and x ∈ (A ⋃ C)}

= {x:x ∈ (A ⋃ B) ∩ (A ⋃ C)

= (A ⋃ B) ∩ (A ⋃ C)

b. A ∩ (B ⋃ C) = (A ∩ B) ⋃ (A ∩ C)

A ∩ (B ⋃ C)

= {x:x ∈ A ∩ (B ⋃ C)}

= {x:x ∈ A or x ∈ (B ⋃ C)}

= {x:x ∈A or (x ∈ B and x ∈ C)}

= {x: (x ∈ A or x ∈ B) and (x ∈ A or x ∈ C)}

= {x:x ∈ (A ∩ B) and x ∈ (A ∩ C)}

= {x:x ∈ (A ∩ B) ⋃ (A ∩ C)

= (A ∩ B) ⋃ (A ∩ C)

Theorem 6 - DE-MORGAN'S LAWS:

Let A and B be the subsets of a universal set U. Then,

a. (A ⋃ B) = A ∩ B

b. (A ∩ B) = A ⋃ B

Proof:

a. (A ⋃ B = A ∩ B

= (A ⋃ B)

= {x:x A B}

= {x:x A and x B}

= {x:x A and x B}

= {x:x A B}

= A ∩ B

b. (A ∩ B) = A ⋃ B

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