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

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

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

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

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