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.

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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