i) Which of the following statement is true
a) 3 + 7 =4 or 3 – 7 = 4
b) If Pune is in Maharashtra, then Hyderabad is in Kerala
c) It is false that 12 is not divisible by 3
d) The square of any odd integer is even.
It is false that 12 is not divisible by 3
ii) Which of the following is not a statement
a) 2+2 =4
b) 2 is the only even prime number
c) Come here
d) Mumbai is not in Maharashtra
Come here
iii) If p is any statement then ( p ? ∼ p) is a
a) Contingency
b) Contradiction
c) Tautology
d) None of these
Tautology
iv) If p and q are two statements then (p → q) ↔ (∼ q → ∼ p) is
a) Contradiction
b) Tautology
c) Neither (i) nor (ii)
d) None of these
v) Negation of p → (p ? ∼ q) is
∼ p → (∼ p ? q)
p ? (∼ p ? q)
∼ p ? (∼ p ? ∼ q)
∼ p → (∼ p → q)
p ? (∼ p ? q)
vi) If p : He is intelligent.
q : He is strong
Then, symbolic form of statement “It is wrong that, he is intelligent or strong” is
∼ p ∨ ∼ p
∼ (p ∧ q)
∼ (p ∨ q)
p ∨ ∼ q
∼ (p ∨ q)
vii)A biconditional statement is the conjunction of two ______ statements
Conditional
ix) The negation of the statement (p ? q) → (r ? ∼ p)
p ? q ? ∼ r
(p ? q) ? r
p ? q ? ∼ r
(p v q) ? (r ? s)
Contrapositive
ix) The negation of the statement (p ? q) → (r ? ∼ p)
p ? q ? ∼ r
(p ? q) ? r
p ? q ? ∼ r
(p v q) ? (r ? s)
p ? q ? ∼ r
x) The false statement in the following is
p ? (∼ p) is contradiction
(p → q) ↔ (∼ q → ∼ p) is a contradiction
∼ (∼ p) ↔ p is a tautology
p ? (∼ p) ↔ p is a tautology
(p → q) ↔ (∼ q → ∼ p) is a contradiction
i) Find the negation of 10 + 20 = 30
The negation of 10 + 20 = 30 is 10 + 20 ≠ 30
ii) State the truth Value of x2 = 25
x2 = 25’ is an open sentence.
It is not a statement in logic.
iii) Write the negation of p → q
∼p → q ≡ (∼p ∨ q) .......[? p → q ≡ ∼p ∨ q]
≡ ∼ (∼p) ∧ ∼q .......[De’Morgan’s Law]
≡ p ∧ ∼q
iv) State the truth value of √3 is not an irrational number
Let p : 3 is irrational number.
∴ Truth value of p is T.
∴ ∼p : 3 is not irrational number.
∴ Truth value of ∼p is F.
v) State the truth value of (p ? ? p)
p | ∼ p | p ∨ ∼p |
T | F | T |
F | T | T |
∴ Truth value of (p ∨ ∼p) is T.
State the truth value of (p ? ∼p)
vi) State the truth value of (p ? ? p)
p | ∼p | p ∧ ∼p |
T | F | F |
F | T | F |
i) If statements p, q are true and r, s are false, determine the truth values of the following.
~ p ∧ (q ∨ ~ r)
a) Answer :
~ p ∧ (q ∨ ~ r)
≡ ∼T ∧ (T ∨ ∼ F)
≡ F ∧ (T ∨ T)
≡ F ∧ T
≡ F
∴ Truth value of ~ p ∧ (q ∨ ~ r) is F
b) Answer :
(p ∧ ~r) ∧ (~q ∨ s)
≡ (T ∧ ∼F) ∧ (∼T ∨ F)
≡ (T ∧ T) ∧ (F ∨ F)
≡ T ∧ F
≡ F
∴ Truth value of (p ∧ ∼r) ∧ (∼q ∨ s) is F.
ii) Write the following compound statements symbolically.
a) Nagpur is in Maharashtra and Chennai is in Tamilnadu.
b) Triangle is equilateral or isosceles.
a) Answer :
Let p: Nagpur is in Maharashtra.
Let q: Chennai is in Tamil Nadu.
Then the symbolic form of the given statement is p ∧ q.
b) Answer :
Let p: Triangle is equilateral.
Let q: Triangle is isosceles.
Then the symbolic form of the given statement is p ∨ q.
iii) . Write the converse and contrapositive of the following statements. “If a function is differentiable then it is continuous”.
Let p: A function is differentiable,
q: It is continuous.
∴ The symbolic form of the given statement is p → q.
Converse: q → p
i.e. If a function is continuous then it is differentiable
Contrapositive: ~q → ~p
i.e. If a function is not continuous then it is not differentiable.
iv) Without using truth table prove that:
~ (p ∨ q) ∨ (~ p ∧ q) ≡ ~ p
I | II | III | IV | V | VI | VII |
p | q | ∼p | p ∨ q | ∼(p ∨ q) | ~(p ∧ q) | ~(p ∨ q) ∨ (~p ∧ q) |
T | T | F | T | F | F | F |
T | F | F | T | F | F | F |
F | T | T | T | F | T | T |
F | F | T | F | T | F | T |
From Columns (III) and (VII), we get ∼(p ∨ q) ∨ (∼p ∧ q) ≡ ∼p
i) Write the negation of the statement “ An angle is a right angle if and only if it is of measure 900 ”
Let p: An angle is a right angle.
q: An angle is of measure 90°.
∴ The symbolic form of the above Statement is p ↔ q.
Note that negation of ‘p ↔ q’ is (p ∧ ∼q) ∨ (q ∧ ∼p).
∴ The negation of the given statement is ‘An angle is a right angle and is not of measure 90° or an angle is of measure 90° and not a right angle.
ii) Write the following statements in symbolic form
a) Milk is white if and only if the sky is not blue
b) If Kutab – Minar is in Delhi then Taj- Mahal is in Agra
c) Even though it is not cloudy , it is still raining
a) Answer :
Let p: Milk is white.
q: Sky is blue.
The given statement in symbolic form is p ↔ ∼q.
b) Answer :
Let p: If Kutub − Minar is in Delhi.
q: Taj − Mahal Is in Agra.
The given statement in symbolic form is p → q.
c) Answer :
Let p: It is cloudy.
q: It is still raining.
∴ The symbolic form of the given statement is ~p ∧ q
iii) Use quantifiers to convert the given open sentence defined on N into a true statement
a) n 2 ≥ 1
b) 3x – 4 < 9
c) Y + 4 > 6
a) Answer :
n2 ≥ 1
∀ n ∈ N, n2 ≥ 1
Since, square of all natural numbers is either 1 or greater than 1.
∴ The statement is true.
b) Answer :
3x – 4 < 9
∃ x ∈ N such that 3x – 4 < 9, is a true statement, since x = 2 ∈ N satisfies 3x – 4 < 9
c) Answer :
Y + 4 > 6
∃ Y ∈ N such that Y + 4 > 6, is a true statement, since Y = 3 ∈ N satisfies Y + 4 > 6.
iv) Examine whether the statement pattern is a tautology, contradiction or contingency
( p ? ? q) → ( ? p ? ? q)
p | q | ~p | ~q | p∧~q | ~p∧~q | (p∧~q)→(~p∧~q) |
T | T | F | F | F | F | T |
T | F | F | T | T | F | F |
F | T | T | F | F | F | T |
F | F | T | T | F | T | T |
The truth values in the last column are not identical. Hence, it is contingency.
v) Using truth table prove that ? p ? q ≡ ( p ? q ) ? ? p
I | II | III | IV | V | VI |
p | q | ~p |
∼p ? q |
p v q |
(p v q) ? ∼p |
T | T | F | F | T | F |
T | F | F | F | T | F |
F | T | T | T | T | T |
F | F | T | F | F | F |
From column (IV) and (VI), we get
∴ ∼p ? q ≡ (p ? q) ? ∼p
vi) Write the dual of the following
a) 13 is prime number and India is a democratic country
b ) ( p ? ? q ) ? ( ? p ? q ) ≡ ( p ?q ) ? ? ( p ? q)
a) Answer :
13 is prime number or India is a democratic country.
b) Answer :
(p ? ∼q) ? (∼p ? q) ≡ (p ? q) ? ∼(p ? q)
vii) Write the converse, inverse and contrapositive of the statement “If it snows, then they do not drive the car”
Let p: It snows.
q: They do not drive the car.
∴ The given statement is p → q.
Its converse is q → p.
If they do not drive the car, then it snows.
Its inverse is ~p → ~q.
If it does not snow, then they drive the car.
Its contrapositive is ~q → ~p.
If they drive the car, then it does not snow.
Q5) Answer the following questions
i) Examine whether the statement pattern
[p → (∼q ? r)] ↔ ∼[p → (q → r)] is a tautology, contradiction or contingency.
[p → (~q ∨ r)] ↔ ~[p → (q → r)]
p | q | r | ~q | ~q ∨ r | p → (~q ∨ r) |
q → r | p → (q →r) |
~[p → (q → r)] |
[p → (~q ∨ r)] ↔ ~[p → (q → r)] |
T | T | T | F | T | T | T | T | F | F |
T | T | F | F | F | F | F | F | T | F |
T | F | T | T | T | T | T | T | F | F |
T | F | F | T | T | T | T | T | F | F |
F | T | T | T | T | T | T | T | F | F |
F | T | F | F | F | T | F | T | F | F |
F | F | T | T | T | T | T | T | F | F |
F | F | F | T | T | T | T | T | F | F |
All the truth values in the last column are F.
Hence, it is contradiction.
ii) Using truth table prove that p ? (q ? r) ≡ (p ? q) ? (p ? r)
I | II | II | IV | V | VI | VII | VIII |
p | q | r | q ∧ r | p ∨ q | p ∨ r | p ∨ (q ∧ r) | (p ∨ q) ∧ (p ∨ r) |
T | T | T | T | T | T | T | T |
T | T | F | F | T | T | T | T |
T | F | T | F | T | T | T | T |
T | F | F | F | T | T | T | T |
F | T | T | T | T | T | T | T |
F | T | F | F | T | F | F | F |
F | F | T | F | F | T | F | F |
F | F | F | F | F | F | F | F |
From column (VII) and (VIII), we get p ∨ (q ∧ r) ≡ ( p ∨ q) ∧ ( p ∨ r)
iii) Without using truth table show that
(p ? q) ? (∼p v ∼q) ≡ (p v ∼q) ? (∼p v q)
(p ∨ q) ∧ (∼p ? ∼q)
≡ [(p ∨ q) ∧ ∼p] ∨ [(p ∨ q) ∧ ∼q] .......[Distributive Law]
≡ [(p ∧ ∼p) ∨ (q ∧ ∼p)] ∨ [(p ∧ ∼q) ∨ (q ∧∼q)] .......[Distributive Law]
≡ [F ∨ (q ∧ ∼p)] ∨ [(p ∧ ∼q) ∨ F] .......[Complement Law]
≡ (q ∧ ∼p) ∨ (p ∧ ∼q) .......[Identity Law]
≡ (p ∧ ∼q) ∨ (q ∧ ∼p) .......[Commutative Law]
iv) With proper justification, state the negation of the following.
(p ↔ q) v (~ q → ~r)
~[(p ↔ q) v (~q → ~r)]
≡ ~(p ↔ q) ? (~q → ~r) ....[Negation of disjunction]
≡ [(p ? ~q) v (q ∧ ~p)] ∧ ~(~q → ~r) ....[Negation of double implication]
≡ [(p ? ~q) v (q ? ~p)] ? [~ q ? ~(~r)] ....[Negation of implication]
≡ [(p ? ~q) v (q ? ~p)] ? (~q ? r) ....[Negation of negation]
v) Prepare truth table for (p ? q) ? ~ r
(p ∧ q) ∨ ~ r
(p ∧ q) ∨ ~ r
p | q | r | ~r | p ∧ q | (p ∧ q) ∨ ~ r |
T | T | T | F | T | T |
T | T | F | T | T | T |
T | F | T | F | F | F |
T | F | F | T | F | T |
F | T | T | F | F | F |
F | T | F | T | F | T |
F | F | T | F | F | F |
F | F | F | T | F | T |