Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 12 Discrete Mathematics Ex 12.2 Text Book Back Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 12 Discrete Mathematics Ex 12.2

Question 1.

Let p : Jupiter is a planet and q: India is an island be any two simple statements. Give verbal sentence describing each of the following statements.

(i) ¬P

(ii) P ∧ ¬q

(iii) ¬p v q

(iv) p → ¬q

(v) p ↔ q

Solution:

p: Jupiter is a planet

q: India is an island

(i) ¬P : Jupiter is not a planet

(ii) P ∧ ¬q : Jupiter is a planet and India is not an island.

(iii) ¬p v q : Jupiter is not a planet or India is an island

(iv) p → ¬q : Jupiter is a planet then India is not an island

(v) p ↔ q : Jupiter is a planet if and only if India is an island

Question 2.

Write each of the following sentences in symbolic form using statement variables p and q.

(i) 19 is not a prime number and all the angles of a triangle are equal.

(ii) 19 is a prime number or all the angles of a triangle are not equal.

(iii) 19 is a prime number and all the angles of a triangle are equal.

(iv) 19 is not a prime number.

Solution:

Let p : 19 is a prime number

q : All the angles of a triangle are equal

(i) 19 is not a prime number and all the angles of a triangle are equal ⇒ ¬p ∧ q

(ii) 19 is a prime number or all the angles of a triangle are not equal ⇒ p v ¬q

(iii) 19 is a prime number and all the angles of a triangle are equal ⇒ p ∧ q

(iv) 19 is not a prime number ⇒ ¬p

Question 3.

Determine the truth value of each of the following statements

(i) If 6 + 2 = 5, then the milk is white.

(ii) China is in Europe dr √3 is art integer

(iii) It is not true that 5 + 5 = 9 or Earth is a planet.

(iv) 11 is a prime number and all the sides of a rectangle are equal.

Solution:

(i) If 6 + 2 = 5, then milk is white,

p: 6 + 2 = 5(F)

q: Milk is white (T)

∴ p → q is having the truth value T

(ii) China is in Europe or √3 is an integer

p: China is in Europe (F)

q: √3 is an integer (F)

∴ p v q is having the truth value F

(iii) It is not true that 5 + 5 = 9 or Earth is a planet

p: 5 + 5 = 9 (F)

q: Earth is a planet (T)

∴ ¬p v q is having the truth value T

(iv) 11 is a prime number and all the sides of a rectangle are equal.

p : 11 is a prime number (T)

q: All the sides of a rectangle are equal (F)

∴ p ∧ q is having the truth value F

Question 4.

Which one of the following sentences is a proposition?

- 4 + 7 = 12
- What are you doing?
- 3ⁿ ≤ 81, n ∈ N
- Peacock is our national bird.
- How tall this mountain is!

Solution:

- is a proposition
- not a proposition
- is a proposition
- is a proposition
- not a proposition

Question 5.

Write the converse, inverse, and contrapositive of each of the following implication.

(i) If x and y are numbers such that x = y, then x² = y²

(ii) If a quadrilateral is a square then it is a rectangle.

Solution:

(i) Converse: If x and y are numbers such that x^{2} = y^{2} then x = y.

Inverse: If x and y are numbers such that x ≠ y then x^{2} ≠ y^{2}.

Contrapositive: If x and v are numbers such that x^{2} ≠ y^{2} then x ≠ y.

(ii) Converse: If a quadrilateral is a rectangle then it is a square.

Inverse: If a quadrilateral is not a square then it is not a rectangle.

Contrapositive: If a quadrilateral is not a rectangle then it is not a square.

Question 6.

Construct the truth table for the following statements.

(i) ¬P ∧ ¬q

(ii) ¬(P ∧ ¬q)

(iii) (p v q) v ¬q

(iv) (¬p → r) ∧ (p ↔ q)

Solution:

(i) ¬P ∧ ¬q

(ii) ¬(P ∧ ¬q)

(iii) (p v q) v ¬q

(iv) (¬p → r) ∧ (p ↔ q)

Question 7.

Verify whether the following compound propositions are tautologies or contradictions or contingency.

(i) (P ∧ q) ∧¬ (p v q)

(ii) ((P v q) ∧¬p) → q

(iii) (p → q) ↔ (¬p → q)

(iv) ((p → q) ∧ (q → r)) → (p → r)

Solution:

(i) (P ∧ q) ∧¬ (p v q)

The entries in the last column are only F.

∴ The given statement is a contradiction

(ii) ((P v q) ∧¬p) → q

The entries in the last column are only T.

∴ The given statement is a Tautology

(iii) (p → q) ↔ (¬p → q)

The entries in the last column are a combination of T and F.

∴ The given statement is a contingency.

(iv) ((p → q) ∧ (q → r)) → (p → r)

All the entries in the last column are only T.

∴ The given statement is a tautology.

Question 8.

Show that (i) ¬(p ∧ q) ≡ ¬P v ¬q

(ii) ¬(p → q) ≡ p ∧¬q

Solution:

(i) ¬(p ∧ q) ≡ ¬P v ¬q

The entries in the columns corresponding to ¬(p ∧ q) and ¬P v ¬q are identical and hence they are equivalent.

(ii) ¬(p → q) ≡ p ∧ ¬q

The entries in the columns corresponding to ¬(p → q) and p ∧ ¬q are identical and hence they are equivalent.

Question 9.

Prove that q → p ≡ ¬p → ¬q

Solution:

q → p ≡ ¬p → ¬q

The entries in the columns corresponding to q → p and ¬p → ¬q are identical and hence they are equivalent.

∴ q → q = ¬p → ¬q

Hence proved.

Question 10.

Show that p → q and q → p are not equivalent

Solution:

From the table, it is clear that

p → q ≠ q → P

Question 11.

Show that ¬(p ↔ q) ≡ p ↔ ¬q

Solution:

From the table, it is clear that

¬(p ↔ q) ≡ p ↔ ¬q

Question 12.

Check whether the statement p → (q → p) is a tautology or a contradiction without using the truth table.

Solution:

P → (q → p)

≡ P → (¬q v p) [∵ implication law]

≡ ¬p v (¬q v p) [∵ implication law]

≡ ¬p v (p v ¬q) [∵ Commutative law]

≡ (¬p V p) v (¬p v ¬q) [∵ Distribution law]

≡ T v ¬(p ∧ q) ≡ T [Tautology]

Hence p → (q → p) is a Tautology.

Question 13.

Using the truth table check whether the statements ¬(p v q) v (¬p ∧ q) and ¬P are logically equivalent.

Solution:

From the table, it is clear that ¬P and

¬(p v q) v (¬p ∧ q) are logically equivalent

i.e. ¬(p v q) v (¬p ∧ q) ≡ ¬p

Question 14.

Prove p → (q → r) ≡ (p ∧ q) → r without using the truth table.

Solution:

P → (q → r)

≡ P → (¬q v r) [∵ implication law]

≡ ¬p v (¬q v r) [∵ implication law]

≡ (¬p v ¬q) v r [∵ Associative law]

≡ ¬(p ∧ p) v r [∵ DeMorgan’s law]

≡ (p ∧ q) → r ≡ T [∵ implication law]

Hence Proved.

Question 15.

Prove that p → (¬q v r) ≡ ¬p v (¬q v r) using truth table.

Solution:

From the table, it is clear that the column of p → (¬q v r) and ¬p v (¬q v r) are identical.

∴ p → (¬q v r) ≡ ¬p v (¬q v r)

Hence proved.