### Mathematical Logic - Ex 1.3 | Maharashtra Board 12th Maths Solutions Chapter 1

### Mathematical Logic - Ex 1.3 | Maharashtra Board 12th Maths Solutions Chapter 1

**Question 1.**

**(i) ÆŽ x ∈ A such that x – 8 = 1**

**Solution:**

Clearly x = 9 ∈ A satisfies x – 8 = 1.

So the given statement is true, hence its truth value is T.

**(ii) â±¯ x ∈ A, x2 + x is an even number**

**Solution:**

For each x ∈ A, x2 + x is an even number.

So the given statement is true, hence its truth value is T.

**(iii) ÆŽ x ∈ A such that x2 < 0**

**Solution:**

There is no x ∈ A which satisfies x2 < 0.

So the given statement is false, hence its truth value is F.

**(iv) â±¯ x ∈ A, x is an even number**

**Solution:**

x = 3 ∈ A, x = 5 ∈ A, x = 7 ∈ A, x = 9 ∈ A, x = 11 ∈ A do not satisfy x is an even number.

So the given statement is false, hence its truth value is F.

**(v) ÆŽ x ∈ A such that 3x + 8 > 40**

**Solution:**

Clearly x = 11 ∈ A and x = 12 ∈ A satisfies 3x + 8 > 40.

So the given statement is true, hence its truth value is T.

**(vi) â±¯ x ∈ A, 2x + 9 > 14**

**Solution:**

For each x ∈ A, 2x + 9 > 14. So the given statement is true, hence its truth value is T.

### Question 2.Write the duals of each of the following.

(i) p ∨ (q ∧ r)

**Solution:**

**The duals of the given statement patterns are :**

**p ∧ (q ∨ r)**

(ii) p ∧ (q ∧ r)

**Solution:**

**p ∨ (q ∨ r)**

(iii) (p ∨ q) ∧ (r ∨ s)

**Solution:**

**(p ∧ q) ∨ (r ∧ s)**

(iv) p ∧ ~q

**Solution:**

**p ∨ ~q**

(v) (~p ∨ q) ∧ (~r ∧ s)

**Solution:**

**(~p ∧ q) ∨ (~r ∨ s)**

(vi) ~p ∧ (~q ∧ (p ∨ q) ∧ ~r)

**Solution:**

**~p ∨ (~q ∨ (p ∧ q) ∨ ~r)**

(vii) [~(p ∨ q)] ∧ [p ∨ ~(q ∧ ~s)]

**Solution:**

**[ ~(p ∧ q)] ∨ [p ∧ ~(q ∨ ~s)]**

(viii) c ∨ {p ∧ (q ∨ r)}

**Solution:**

**t ∧ {p ∧ (q Ar)}**

(ix) ~p ∨ (q ∧ r) ∧ t

**Solution:**

**~p ∧ (q ∨ r) ∨ c**

(x) (p ∨ q) ∨ c

**Solution:**

**(p ∧ q) ∧ t**

### Question 3.Write the negations of the following.

**(i) x + 8 > 11 or y – 3 = 6**

**Solution:**

Let p : x + 8 > 11, q : y — 3 = 6.

Then the symbolic form of the given statement is p ∨ q.

Since ~(p ∨ q) ≡ ~p ∧ ~q, the negation of given statement is :

‘x + 8 > 11 and y – 3 ≠ 6’ OR

‘x + 8 ≮ 11 and y – 3 ≠ 6’

**(ii) 11 < 15 and 25 > 20**

**Solution:**

Let p: 11 < 15, q : 25 > 20.

Then the symbolic form of the given statement is p ∧ q.

Since ~(p ∧ q) ≡ ~p ∨ ~q, the negation of given statement is :

’11 ≮ 15 or 25 > 20.’ OR

’11 ≯ 15 or 25 ≮ 20.’

**(iii) Qudrilateral is a square if and only if it is a rhombus.**

**Solution:**

Let p : Quadrilateral is a square.

q : It is a rhombus.

Then the symbolic form of the given statement is p ↔ q.

Since ~(p ↔ q) ≡ (p ∧ ~q) ∨ (q ∧ ~p), the negation of given statement is :

‘ Quadrilateral is a square but it is not a rhombus or quadrilateral is a rhombus but it is not a square.’

**(iv) It is cold and raining.**

**Solution:**

Let p : It is cold.

q : It is raining.

Then the symbolic form of the given statement is p ∧ q.

Since ~(p ∧ q) ≡ ~p ∨ ~q, the negation of the given statement is :

‘It is not cold or not raining.’

**(v) If it is raining then we will go and play football.**

**Solution:**

Let p : It is raining.

q : We will go.

r : We play football.

Then the symbolic form of the given statement is p → (q ∧ r).

Since ~[p → (q ∧ r)] ≡ p ∧ ~(q ∧ r) ≡ p ∧ (q ∨ ~r), the negation of the given statement is :

‘It is raining and we will not go or not play football.’

**(vi) 2–√ is a rational number.**

**Solution:**

Let p : 2–√ is a rational number.

The negation of the given statement is

‘ ~p : 2–√ is not a rational number.’

**(vii) All natural numbers are whole numers.**

**Solution:**

The negation of the given statement is :

‘Some natural numbers are not whole numbers.’

**(viii) â±¯ n ∈ N, n2 + n + 2 is divisible by 4.**

**Solution:**

The negation of the given statement is :

‘ÆŽ n ∈ N, such that n2 + n + 2 is not divisible by 4.’

**(ix) ÆŽ x ∈ N such that x – 17 < 20**

**Solution:**

The negation of the given statement is :

‘â±¯ x ∈ N, x – 17 ≯ 20.’

### Question 4.Write converse, inverse and contrapositive of the following statements.

**(i) If x < y then x2 < y2 (x, y ∈ R)**

**Solution:**

Let p : x < y, q : x2 < y2.

Then the symbolic form of the given statement is p → q.

Converse : q → p is the converse of p → q.

i.e. If x2 < y2, then x < y.

Inverse : ~p → ~q is the inverse of p → q.

i.e. If x ≯ y, then x2 ≯ y2. OR

If x ≮ y, then x2 ≮ y2.

Contrapositive : ~q → p is the contrapositive of

p → q i.e. If x2 ≯ y2, then x ≯ y. OR

If x2 ≮ y2, then x ≮ y.

**(ii) A family becomes literate if the woman in it is literate.**

**Solution:**

**Let p :**The woman in the family is literate.

**q :**A family become literate.

Then the symbolic form of the given statement is p → q

**Converse**: q → p is the converse of p → q.

i.e. If a family become literate, then the woman in it is literate.

**Inverse**: ~p → ~q is the inverse of p → q.

i.e. If the woman in the family is not literate, then the family does not become literate.

**Contrapositive :**~q → ~p is the contrapositive of p → q. i e. If a family does not become literate, then the woman in it is not literate.

**(iii) If surface area decreases then pressure increases.**

**Solution:**

**Let p :**The surface area decreases.

**q :**The pressure increases.

Then the symbolic form of the given statement is p → q.

**Converse**: q → p is the converse of p→ q.

**i.e. I**f the pressure increases, then the surface area decreases.

**Inverse**: ~p → ~q is the inverse of p → q.

**i.e.**If the surface area does not decrease, then the pressure does not increase.

**Contrapositive**: ~q → ~p is the contrapositive of p → q.

**i.e.**If the pressure does not increase, then the surface area does not decrease.

**(iv) If voltage increases then current decreases.**

**Solution:**

**Let p :**Voltage increases.

**q**: Current decreases.

Then the symbolic form of the given statement is p → q.

**Converse**: q →p is the converse of p → q.

i

**.e.**If current decreases, then voltage increases.**Inverse**: ~p → ~q is the inverse of p → q.

**i.e.**If voltage does not increase, then current does not decrease.

**Contrapositive**: ~q → ~p, is the contrapositive of p → q.

**i.e.**If current does not decrease, then voltage doesnot increase.

### 1.3 QUANTIFIERS, QUANTIFIED STATEMENTS, DUALS, NEGATION OF COMPOUND STATEMENTS, CONVERSE, INVERSE AND CONTRAPOSITIVE OF IMPLICATION.

### 1.3.1 Quantifiers and quantified statements.

Look at the following statements :

p : "There exists an even prime number in the set of natural numbers".

q : "All natural numbers are positive".

Each of them asserts a condition for some or all objects in a collection. Words "there exists" and "for all" are called quantifiers. "There exists is called existential quantifier and is denoted by symbol $. "For all" is called universal quantifier and is denoted by ". Statements involving quantifiers are called quantified statements.

Every quantified statement corresponds to a collection and a condition. In statement p the collection is 'the set of natural numbers' and the condition is 'being even prime'. What is the condition in the statement q ?

A statement quantified by universal quantifier " is true if all objects in the collection satisfy the condition. And it is false if at least one object in the collection does not satisfy the conditon. A statement quantified by existential quantifier $ is true if at least one object in the collection satisfies the condition. And it is false if no object in the collection satisfies the condition.

### Ex.1.If A = {1, 2, 3, 4, 5, 6, 7}, determine the truth value of the following.

i) ∃ x ∈ A such that x – 4 = 3

ii) ∀ x ∈ A , x + 1 > 3

iii) ∀ x ∈ A, 8 – x < 7

iv) ∃ x ∈ A, such that x + 8 = 16

Solution :

i) For x = 7, x – 4 = 7 – 4 = 3

\ x = 7 satisfies the equation x – 4 = 3

\ The given statement is true and its truth value is T.

ii) For x = 1, x + 1 = 1 + 1 = 2 which is not greater than or equal to 3

\ For x = 1, x + 1 > 3 is not true.

\ The truth value of given statement is F.

iii) For each x ∈ A 8 – x < 7

\ The given statement is true.

\ Its truth value is T.

iv) There is no x in A which satisfies x + 8 = 16.

\ The given statement is false. \ Its truth value is F

**1.3.2 Dual :**We use letters t and c to denote tautology and contradiction respectively. If two statements contain logical connectives like ∨, ∧ and letters t and c then they are said to be duals of each other if one of them is obtained from the other by interchanging ∨ with ∧ and ,t with c.

The dual of i) p ∨ q is p ∧ q ii) t ∨ p is c ∧ p iii) t ∧ p is c ∨ p

### Ex.1. Write the duals of each of the following :

i) (p ∧ q) ∨ r ii) t ∨ (p ∨ q)

iii) p ∧ [~ q ∨ (p ∧ q) ∨ ~ r]

iv) (p ∨ q) ∧ t

v) (p ∨ q) ∨ r ≡ p ∨ (q ∨ r )

vi) p ∧ q ∧ r

vii) (p ∧ t) ∨ (c ∧ ~ q)

Solution :

i) (p ∨ q) ∧ r ii) c ∧ (p ∧ q)

iii) p ∨ [(~ q ∧ (p ∨ q) ∧ ~ r]

iv) (p ∧ q) ∨ c

v) (p ∧ q) ∧ r ≡ p ∧ (q ∧ r )

vi) p ∨ q ∨ r

vii) (p ∨ c) ∧ (t ∨ ~q)

### 1.3.3 Negations of compound statements :

Negation of conjunction : When is the statment "6 is even and perfect number" is false? It is so, if 6 is not even or 6 is not perfect number. The negation of p ∧ q is ~ p ∨ ~ q. The negation of "6 is even and perfect number" is "6 not even or not perfect number".

**Activity :**Using truth table verify that ~ (p ∧ q) ≡ ~ p ∨ ~ q

**Negation of disjunction :**When is the statement "x is prime or y is even" is false? It is so, if x is not prime and y is not even. The negation of p ∨ q is ~ p ∧ ~ q. The negation of "x is prime or y is even" is "x is not prime and y is not even".

**Activity :**Using truth table verify that ~ (p ∨ q) ≡ ~ p ∧ ~ q

**Note :**'~ (p ∧ q) ≡ ~ p ∨ ~ q' and '~ (p ∨ q) ≡ ~ p ∧ ~ q' are called De'Morgan's Laws

**Negation of implication**: Implication p → q asserts that "if p is true statement then q is true statement". When is an implication a true statement and when is it false? Consider the statement "If bakery is open then I will buy a cake for you." Clearly statement is false only when the bakery was open and I did not buy a cake for you.

The conditional statement "If p then q" is false only in the mcase "p is true and q is false". In all other cases it is true. The negation of the statement "If p then q" is the statement "p and not q". i.e. p does not imply q

**Activity :**Using truth table verify that ~ (p → q) ≡ p ∧ ~ q

Negation of biconditional : The biconditional p ↔ q is the conjuction of statement p → q and q → p.

\ p ↔ q ≡ (p → q) ∧ (q → p)

\ The conditional statement p ↔ q is false if p → q is false or q → p false.

The negation of the statement "p if and only if q " is the statement "p and not q, or q and not p".

\ ~ (p ↔ q) ≡ (p ∧ ~ q) ∨ (q ∧ ~ p)

Activity : Using truth table verify that ~ (p ↔ q) ≡ (p ∧ ~ q) ∨ (q ∧ ~ p)

#### 1.3.4 Converse, inverse and contrapositive

From implication p → q we can obtain three implications, called converse, inverse and contrapositive.

q → p is called the converse of p → q

~ p → ~ q is called the inverse of p → q.

~ q → ~ p is called the contrapositive of p → q.

**Activity :**

Prepare the truth table for p → q, q → p, ~ p → ~q and ~ q → ~p. What is your conclusion

from the truth table ?

i) ........................ ≡ .........................

ii) ........................ ≡ .........................

### Ex.1)Write the negations of the following.

i) 3 + 3 < 5 or 5 + 5 = 9

ii) 7 > 3 and 4 > 11

iii) The number is neither odd nor perfect square.

iv) The number is an even number if and only if it is divisible by 2.

**Solution :**

**i) Let p :**3 + 3 < 5 : q : 5 + 5 = 9

Given statement is p ∨ q and its negation is ~(p ∨ q) and ~(p ∨ q) ≡ ~ p ∧ ~ q

\ The negation of given statement is 3 + 3 > 5 and 5 + 5 ≠ 9

**ii) Let p :**7 > 3; q: 4 > 11

The given statement is p ∧ q

Its negation is ~ (p ∧ q) and ~ (p ∧ q) ≡ ~ p ∨ ~ q

\ The negation of given statement is 7 < 3 or 4 < 11

**iii) Let p :**The number is odd

q : The number is perfect square

Given statement can be written as 'the number is not odd and not perfect square'

Given statement is ~ p ∧ ~ q

Its negation is ~ (~ p ∧ ~ q) ≡ p ∨ q

The negation of given statement is 'The number is odd or perfect square'.

**iv) Let p :**The number is an even number.

q : The number is divisible by 2

Given statement is p ↔ q

Its negation is ~ (p ↔ q)

But ~ (p ↔ q) ≡ (p ∧ ~ q) ∨ (q ∧ ~ p)

\ The negation of given statement is 'A number is even but not divisible by 2 or a number is divisible by 2 but not even'.

**Negation of quantified statement :**while doing the negations of quantified statement we

replace the word 'all' by 'some', "for every" by "there exists" and vice versa.

#### Ex.2.Write the negations of the following statements.

i) All natural numbers are rational.

ii) Some students of class X are sixteen year old.

iii) ∃ n ∈ N such that n + 8 > 11

iv) ∀ x ∈ N, 2x + 1 is odd

Solution :

i) Some natural numbers are not rationals.

ii) No student of class X is sixteen year old.

iii) ∀ n ∈ N, n + 8 < 11

iv) ∃ x ∈ N such that 2x + 1 is not odd

#### Ex.3.Write the converse, inverse and contrapositive of the following statements.

i) If a function is differentiable then it is continuous.

ii) If it rains then the match will be cancelled.

Solution :

**(1) Let p : A function is differentiable**

q : A function is continuous.

\ Given statement is p → q

i) Its converse is q → p

If a function is continuous then it is differentiable.

ii) Its inverse is ~ p → ~ q.

If a function not differentiable then it is not continuous.

iii) Its contrapositive is ~ q → ~ p

If a function is not continuous then it is not differentiable.

#### (2) Let p : It rains, q : The match gets cancelled.

\ Given statement is p → q

i) Its converse is q → p

If the match gets cancelled then it rains.

ii) Inverse is ~ p → ~ q

If it does not rain then the match will not be cancelled.

iii) Its contrapositive is ~ q → ~ p.

If the match is not cancelled then it does not rain

### Mathematical Logic - Ex 1.3 | Maharashtra Board 12th Maths Solutions Chapter 1

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