# 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.
If A = {3, 5, 7, 9, 11, 12}, determine the truth value of each of the following.

(i) Ǝ 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) Ɐ 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) Ǝ 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) Ɐ 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) Ǝ 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) Ɐ 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) Ɐ n ∈ N, n2 + n + 2 is divisible by 4.
Solution:
The negation of the given statement is :
‘Ǝ n ∈ N, such that n2 + n + 2 is not divisible by 4.’

(ix) Ǝ x ∈ N such that x – 17 < 20
Solution:
The negation of the given statement is :
‘Ɐ 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. If 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.

### 1.3.1 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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