Chapter 1 Mathematical Logic Miscellaneous Exercise 1 Solutions

Balbharti12th Maharashtra State Board Maths Solutions BookPdf Chapter 1 Mathematical Logic Miscellaneous Exercise 1 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1

(ii) (p ∧ q) → r is logically equivalent to ________.
(A) p → (q → r)
(B) (p ∧ q) → ~r
(C) (~p ∨ ~q) → ~r
(D) (p ∨ q) → r
Solution:
(a) p → (q → r) [Hint: Use truth table.]

(iii) Inverse of statement pattern (p ∨ q) → (p ∧ q) is ________.
(A) (p ∧ q) → (p ∨ q)
(B) ~(p ∨ q) → (p ∧ q)
(C) (~p ∧ ~q) → (~p ∨ ~q)
(D) (~p ∨ ~q) → (~p ∧ ~q)
Solution:
(c) (~p ∧ ~q) → (~p ∨ ~ q)

(iv) If p ∧ q is F, p → q is F then the truth values of p and q are ________.
(A) T, T
(B) T, F
(C) F, T
(D) F, F
Solution:
(b) T, F

(v) The negation of inverse of ~p → q is ________.
(A) q ∧ p
(B) ~p ∧ ~q
(C) p ∧ q
(D) ~q → ~p
Solution:
(a) q ∧ p

(vi) The negation of p ∧ (q → r) is ________.
(A) ~p ∧ (~q → ~r)
(B) p ∨ (~q ∨ r)
(C) ~p ∧ (~q → ~r)
(D) ~p ∨ (~q ∧ ~r)
Solution:
(d) ~p ∨ (q ∧ ~r)

(vii) If A = {1, 2, 3, 4, 5} then which of the following is not true?
(A) Ǝ x ∈ A such that x + 3 = 8
(B) Ǝ x ∈ A such that x + 2 < 9
(C) Ɐ x ∈ A, x + 6 ≥ 9
(D) Ǝ x ∈ A such that x + 6 < 10
Solution:
(c) Ǝ x ∈ A, x + 6 ≥ 9.

(ii) π is an irrational number.
Solution:
It is a statement which is true, hence its truth value is ‘T’.

(iii) India is a country and Himalayas is a river.
Solution:
It is a statement which is false, hence its truth value is ‘F’. ….[T ∧ F ≡ F]

(iv) Please get me a glass of water.
Solution:
It is an imperative sentence, hence it is not a statement.

(v) cos²θ – sin²θ = cos2θ for all θ ∈ R.
Solution:
It is a statement which is true, hence its truth value is ‘T’.

(vi) If x is a whole number the x + 6 = 0.
Solution:
It is a statement which is false, hence its truth value is ‘F’.

(ii) Ɐ n ∈ N, n²+ n is even number while n²– n is an odd number.
Solution:
Let p : Ɐ n ∈ N, n²+ n is an even number.
q : Ɐ n ∈ N, n²– n is an odd number.
Then the symbolic form of the given statement is p ∧ q.
The truth values of p and q are T and F respectively.
∴ the truth value of p ∧ q is F. … [T ∧ F ≡ F].

(iii) Ǝ n ∈ N such that n + 5 > 10.
Solution:
Ǝ n ∈ N, such that n + 5 > 10 is a true statement, hence its truth value is T.
(All n ≥ 6, where n ∈ N, satisfy n + 5 > 10).

(iv) The square of any even number is odd or the cube of any odd number is odd.
Solution:
Let p : The square of any even number is odd.
q : The cube of any odd number is odd.
Then the symbolic form of the given statement is p ∨ q.
The truth values of p and q are F and T respectively.
∴ the truth value of p ∨ q is T. … [F ∨ T ≡ T].

(v) In ∆ ABC if all sides are equal then its all angles are equal.
Solution:
Let p : ABC is a triangle and all its sides are equal.
q : Its all angles are equal.
Then the symbolic form of the given statement is p → q
If the truth value of p is T, then the truth value of q is T.
∴ the truth value of p → q is T. … [T → T ≡ T].

(vi) Ɐ n ∈ N, n + 6 > 8.
Solution:
Ɐ n ∈ N, 11 + 6 > 8 is a false statement, hence its truth value is F.
{n = 1 ∈ N, n = 2 ∈ N do not satisfy n + 6 > 8).

(ii) Ɐ x ∈ A, x + 5 < 12.
Solution:
False

(iii) Ǝ x ∈ A, such that x + 7 ≥ 11.
Solution:
True

(iv) Ɐ x ∈ A, 3x ≤ 25.
Solution:
False

(ii) Ǝ x ∈ A, such that x + 9 ≤ 15.
Solution:
Ɐ x ∈ A, x + 9 > 15.

(iii) Some triangles are equilateral triangle.
Solution:
All triangles are not equilateral triangles.

(ii) (~p ∨ ~q) ↔ [~(p ∧ q)]
Solution:

(iii) ~(~p ∧ ~q) ∨ q
Solution:

(iv) [(p ∧ q) ∨ r] ∧ [~r ∨ (p ∧ q)]
Solution:

(v) [(~p ∨ q) ∧ (q → r)] → (p → r)
Solution:

(ii) [(p ∨ q) ∧ ~p] ∧ ~q
Solution:

All the entries in the last column of the above truth table are F.
∴ [(p ∨ q) ∧ ~p] ∧ ~q is a contradiction.

(iii) (p → q) ∧ (p ∧ ~q)
Solution:

All the entries in the last column of the above truth table are F.
∴ (p → q) ∧ (p ∧ ~q) is a contradiction.

(iv) [p → (q → r)] ↔ [(p ∧ q) → r]
Solution:

All the entries in the last column of the above truth table are T.
∴ [p → (q → r)] ↔ [(p ∧ q) → r] is a tautology.

(v) [(p ∧ (p → q)] → q
Solution:

All the entries in the last column of the above truth table are T.
∴ [(p ∧ (p → q)] → q is a tautology.

(vi) (p ∧ q) ∨ (~p ∧ q) ∨ (p ∨ ~q) ∨ (~p ∧ ~q)
Solution:

All the entries in the last column of the above truth table are T.
∴ (p ∧ q) ∨ (~p ∧ q) ∨ (p ∨ ~q) ∨ (~p ∧ ~q) is a tautology.

(vii) [(p ∨ ~q) ∨ (~p ∧ q)] ∧ r
Solution:

The entries in the last column are neither T nor all F.
∴ [(p ∨ ~q) ∨ (~p ∧ q)] ∧ r is a contingency.

(viii) (p → q) ∨ (q → p)
Solution:

All the entries in the last column of the above truth table are T.
∴ (p → q) ∨ (q → p) is a tautology.

(ii) (p ∨ q) is T and (p ∨ q) → q is F
Solution:

Since the truth values of (p ∨ q) is T and (p ∨ q) → q is F, from the table, the truth values of p and q are T and F respectively.

(iii) (p ∧ q) is F and (p ∧ q) → q is T
Solution:

Since the truth values of (p ∧ q) is F and (p ∧ q) → q is T, from the table, the truth values of p and q are either T and F respectively or F and T respectively or both F.

(ii) (p ∧ q) → r ≡ p → (q → r)
Solution:

The entries in the columns 5 and 7 are identical.
∴ (p ∧ q) → r ≡ p → (q → r).

(ii) ~p ∧ q ≡ (p ∨ q) ∧ ~p
Solution:
(p ∨ q) ∧ ~ p
≡ (p ∧ ~p) ∨ (q ∧ ~p) … (Distributive Law)
≡ F ∨ (q ∧ ~p) … (Complement Law)
≡ q ∧ ~ p … (Identity Law)
≡ ~p ∧ q …(Commutative Law)
∴ ~p ∧ q ≡ (p ∨ q) ∧ ~p.

(iii) ~(p ∨ q) ∨ (~p ∧ q) ≡ ~p
Solution:
~ (p ∨ q) ∨ (~p ∧ q)
≡ (~p ∧ ~q) ∨ (~p ∧ q) … (Negation of disjunction)
≡ ~p ∧ (~q ∨ q) … (Distributive Law)
≡ ~ p ∧ T … (Complement Law)
≡ ~ p … (Identity Law)
∴ ~(p ∨ q) ∨ (~p ∧ q) ≡ ~p.

(ii) p ∧ (q ∨ r)
Solution:
The negation of p ∧ (q ∨ r) is
~ [p ∧ (q ∨ r)]
≡ ~ p ∨ ~(q ∨ r) … (Negation of conjunction)
≡ ~p ∨ (~q ∧ ~r) … (Negation of disjunction)

(iii) (p → q) ∧ r
Solution:
The negation of (p → q) ∧ r is
~ [(p → q) ∧ r]
≡ ~ (p → q) ∨ (~ r) … (Negation of conjunction)
≡ (p ∧ ~q) ∨ (~ r) … (Negation of implication)

(iv) (~p ∧ q) ∨ (p ∧ ~q)
Solution:
The negation of (~ p ∧ q) ∨ (p ∧ ~ q) is
~ [(~p ∧ q) ∨ (p ∧ ~q)]
≡ ~(~p ∧ q) ∧ ~ (p ∧ ~q) … (Negation of disjunction)
≡ [~(~p) ∨ ~q] ∧ [~p ∨ ~(q)] … (Negation of conjunction)
≡ (p ∨ ~ q) ∧ (~ p ∨ q) … (Negation of negation)

(ii)

Solution:
Let p : the switch S₁is closed
q : the switch S₂is closed
r : the switch S₃is closed.
Then the symbolic form of the given statement is : (p ∨ q) ∧ (p ∨ r).

(ii)

Solution:
(ii) Let p : the switch S₁is closed
q : the switch S₂is closed
r : the switch S₃is closed
s : the switch S₄is closed
t : the switch S₅is closed
~ p : the switch S₁‘ is closed or the switch S₁is open
~ q : the switch S₂‘ is closed or the switch S₂is open
~ r : the switch S₃‘ is closed or the switch S₃is open
~ s : the switch S₄‘ is closed or the switch S₄is open
~ t : the switch S₅‘ is closed or the switch S₅is open.
Then the given circuit in symbolic form is
[(p ∧ q) ∨ ~r ∨ ~s ∨ ~t] ∧ [(p ∧ q) ∨ (r ∧ s ∧ t)]
Using the laws of logic, we have,
[(p ∧ q) ∨ ~r ∨ ~s ∨ ~ t] ∧ [(p A q) ∨ (r ∧ s ∧ t)]
= [(p∧ q) ∨ ~(r ∧ s ∧ t)] ∧ [(p ∧ q) ∨ (r ∧ s ∧ t)] … (By De Morgan’s Law)
= (p ∧ q) ∨ [ ~(r ∧ s ∧ t) ∧ (r ∧ s ∧ t)] … (By Distributive Law)
= (p ∧ q) ∨ F … (By Complement Law)
= p ∧ q … (By Identity Law)
Hence, the alternative simplified circuit is :

(B)

Solution:
The symbolic form of the given switching circuits are
(p ∨ q) ∧ (p ∨ r) and p ∨ (q ∧ r)
By Distributive Law,
p ∨ (q ∧ r) ≡ (p ∨ q) ∧ (p ∨ r)
Hence, the given switching circuits are logically equivalent.