Exercise 1.1 Complete Solutions
1. Which of the following sentences are statements? Find the truth values of those which are statements.
a) Kathmandu is the capital of Nepal. - Statement, True
b) Nepal exports oil. - Statement, False
c) 3 + 5 = 8 - Statement, True
d) Where do you live? - Not a statement (question)
e) Understand logic. - Not a statement (command)
f) Oh! how beautiful the scene is? - Not a statement (exclamation)
2. Let p: demand is increasing and q: supply is decreasing. Express each of the following statements into words
a) ~p - Demand is not increasing
b) ~q - Supply is not decreasing
c) p ∧ q - Demand is increasing and supply is decreasing
d) p ∨ q - Demand is increasing or supply is decreasing
e) ~p ∧ q - Demand is not increasing and supply is decreasing
f) p ∨ ~q - Demand is increasing or supply is not decreasing
g) ~p ∧ ~q - Demand is not increasing and supply is not decreasing
h) ~(p ∧ q) - It is not the case that demand is increasing and supply is decreasing
3. Express each of the following statements into symbolic form
a) Let p: temperature is increasing; q: length is expanding
Statement: Temperature is increasing and length is expanding → p ∧ q
b) Let p: pressure is decreasing; q: volume is increasing
Statement: Pressure is decreasing or volume is increasing → p ∨ q
c) Let p: demand is increasing; q: price is increasing
Statement: If demand is increasing then price is increasing → p ⇒ q
d) Neither demand is increasing nor price is decreasing (i.e. not increasing)
Let p: demand is increasing; q: price is increasing
Statement: ~p ∧ ~q → ~p ∧ ~q
e) Let p: Pradeep is bold; q: Sandeep is handsome
Statement: It is false that Pradeep is bold or Sandeep is handsome → ~(p ∨ q)
4. Construct truth tables for the following compound statements
a) (~p) ∧ q
| p | q | ~p | (~p) ∧ q |
|---|---|---|---|
| T | T | F | F |
| T | F | F | F |
| F | T | T | T |
| F | F | T | F |
b) (~p) ∨ (~q)
| p | q | ~p | ~q | (~p) ∨ (~q) |
|---|---|---|---|---|
| T | T | F | F | F |
| T | F | F | T | T |
| F | T | T | F | T |
| F | F | T | T | T |
c) ~(p ∧ q)
| p | q | p ∧ q | ~(p ∧ q) |
|---|---|---|---|
| T | T | T | F |
| T | F | F | T |
| F | T | F | T |
| F | F | F | T |
d) ~[p ∨ (~q)]
| p | q | ~q | p ∨ (~q) | ~[p ∨ (~q)] |
|---|---|---|---|---|
| T | T | F | T | F |
| T | F | T | T | F |
| F | T | F | F | T |
| F | F | T | T | F |
e) (p ⇒ q) ∧ (q ⇒ p)
| p | q | p ⇒ q | q ⇒ p | (p ⇒ q) ∧ (q ⇒ p) |
|---|---|---|---|---|
| T | T | T | T | T |
| T | F | F | T | F |
| F | T | T | F | F |
| F | F | T | T | T |
f) (~p ∧ q) ⇒ (p ∨ q)
| p | q | ~p | ~p ∧ q | p ∨ q | (~p ∧ q) ⇒ (p ∨ q) |
|---|---|---|---|---|---|
| T | T | F | F | T | T |
| T | F | F | F | T | T |
| F | T | T | T | T | T |
| F | F | T | F | F | T |
g) [p ∧ (p ⇒ q)] ⇒ q
| p | q | p ⇒ q | p ∧ (p ⇒ q) | [p ∧ (p ⇒ q)] ⇒ q |
|---|---|---|---|---|
| T | T | T | T | T |
| T | F | F | F | T |
| F | T | T | F | T |
| F | F | T | F | T |
h) (p ⇒ q) ⇔ (~p ∨ q)
| p | q | p ⇒ q | ~p | ~p ∨ q | (p ⇒ q) ⇔ (~p ∨ q) |
|---|---|---|---|---|---|
| T | T | T | F | T | T |
| T | F | F | F | F | T |
| F | T | T | T | T | T |
| F | F | T | T | T | T |
i) (p ⇔ q) ∨ (q ⇔ r)
| p | q | r | p ⇔ q | q ⇔ r | (p ⇔ q) ∨ (q ⇔ r) |
|---|---|---|---|---|---|
| T | T | T | T | T | T |
| T | T | F | T | F | T |
| T | F | T | F | F | F |
| T | F | F | F | T | T |
| F | T | T | F | T | T |
| F | T | F | F | F | F |
| F | F | T | T | F | T |
| F | F | F | T | T | T |
5. Let p, q, r and s be four simple statements. If p is true, q is false, r is true and s is false, find the truth values of the following compound statements.
Given: p = T, q = F, r = T, s = F
a) p ∧ q = T ∧ F = F
b) p ∨ (~q) = T ∨ (~F) = T ∨ T = T
c) (~p) ∧ (~q) = (~T) ∧ (~F) = F ∧ T = F
d) q ∨ (p ∧ s) = F ∨ (T ∧ F) = F ∨ F = F
e) ~(~p) = ~(~T) = ~F = T
f) (p ∨ q) ∧ (r ∨ s) = (T ∨ F) ∧ (T ∨ F) = T ∧ T = T
6. If p and q are any two statements, prove that
a) p ∧ (~p) ≡ c where c is a contradiction.
| p | ~p | p ∧ (~p) |
|---|---|---|
| T | F | F |
| F | T | F |
Since p ∧ (~p) is always false, it is equivalent to a contradiction.
b) (p ∨ q) ≡ (q ∨ p)
| p | q | p ∨ q | q ∨ p |
|---|---|---|---|
| T | T | T | T |
| T | F | T | T |
| F | T | T | T |
| F | F | F | F |
Since p ∨ q and q ∨ p have identical truth values, they are equivalent.
c) ~(p ∨ (~q)) ≡ (~p) ∧ q
| p | q | ~q | p ∨ (~q) | ~(p ∨ (~q)) | ~p | (~p) ∧ q |
|---|---|---|---|---|---|---|
| T | T | F | T | F | F | F |
| T | F | T | T | F | F | F |
| F | T | F | F | T | T | T |
| F | F | T | T | F | T | F |
Since ~(p ∨ (~q)) and (~p) ∧ q have identical truth values, they are equivalent.
d) ~((~p) ∧ q) ≡ p ∨ (~q)
| p | q | ~p | (~p) ∧ q | ~((~p) ∧ q) | ~q | p ∨ (~q) |
|---|---|---|---|---|---|---|
| T | T | F | F | T | F | T |
| T | F | F | F | T | T | T |
| F | T | T | T | F | F | F |
| F | F | T | F | T | T | T |
Since ~((~p) ∧ q) and p ∨ (~q) have identical truth values, they are equivalent.
7. Find the negation of each of the following statements:
a) Light travels in a straight line.
Negation: Light does not travel in a straight line.
b) Rivers can be used to produce electricity.
Negation: Rivers cannot be used to produce electricity.
c) x > 0
Negation: x ≤ 0
d) Some students are weak in mathematics.
Negation: No students are weak in mathematics. (or All students are strong in mathematics)
e) All teachers are laborious.
Negation: Some teachers are not laborious.
8. Find the truth value and the negation of each of the following statements.
a) 3 + 2 = 5 or 6 is a multiple of 5.
Truth value: True (since 3+2=5 is true, even though 6 is not a multiple of 5)
Negation: 3 + 2 ≠ 5 and 6 is not a multiple of 5.
b) 8 is a prime number and 4 is even.
Truth value: False (since 8 is not a prime number)
Negation: 8 is not a prime number or 4 is not even.
c) If 3 > 0 then 4 + 6 = 10
Truth value: True (since both antecedent and consequent are true)
Negation: 3 > 0 and 4 + 6 ≠ 10
d) If 2 is odd or 3 is a natural number then 2 + 3 = 8
Truth value: False (antecedent is true, consequent is false)
Negation: (2 is odd or 3 is a natural number) and 2 + 3 ≠ 8
e) If 2 × 3 = 5 ⇒ 3 > 1 then 6 is even.
Truth value: True (since 6 is even is true)
Negation: (2 × 3 = 5 ⇒ 3 > 1) and 6 is not even
f) A triangle ABC is right angled at B if and only if AB² + BC² = AC²
Truth value: True (by Pythagoras theorem)
Negation: A triangle ABC is right angled at B and AB² + BC² ≠ AC², or A triangle ABC is not right angled at B and AB² + BC² = AC²
9. Some statements are given below:
a) If 3 is a natural number then 1/3 is a rational number.
i) Antecedent: 3 is a natural number; Consequent: 1/3 is a rational number
ii) Converse: If 1/3 is a rational number then 3 is a natural number
Inverse: If 3 is not a natural number then 1/3 is not a rational number
Contrapositive: If 1/3 is not a rational number then 3 is not a natural number
iii) Negation: 3 is a natural number and 1/3 is not a rational number
b) x² = 4 whenever x = 2
i) Antecedent: x = 2; Consequent: x² = 4
ii) Converse: If x² = 4 then x = 2
Inverse: If x ≠ 2 then x² ≠ 4
Contrapositive: If x² ≠ 4 then x ≠ 2
iii) Negation: x = 2 and x² ≠ 4
c) If the battery is low then the mobile does not work well.
i) Antecedent: The battery is low; Consequent: The mobile does not work well
ii) Converse: If the mobile does not work well then the battery is low
Inverse: If the battery is not low then the mobile works well
Contrapositive: If the mobile works well then the battery is not low
iii) Negation: The battery is low and the mobile works well
10. If p and q be the statements, prove that
a) p ∨ ¬(p ∧ q) is a tautology
| p | q | p ∧ q | ¬(p ∧ q) | p ∨ ¬(p ∧ q) |
|---|---|---|---|---|
| T | T | T | F | T |
| T | F | F | T | T |
| F | T | F | T | T |
| F | F | F | T | T |
Since p ∨ ¬(p ∧ q) is always true, it is a tautology.
b) (p ∧ q) ⇒ (p ∨ q) is a tautology
| p | q | p ∧ q | p ∨ q | (p ∧ q) ⇒ (p ∨ q) |
|---|---|---|---|---|
| T | T | T | T | T |
| T | F | F | T | T |
| F | T | F | T | T |
| F | F | F | F | T |
Since (p ∧ q) ⇒ (p ∨ q) is always true, it is a tautology.
c) ¬(p ∨ q) ∧ q is a contradiction
| p | q | p ∨ q | ¬(p ∨ q) | ¬(p ∨ q) ∧ q |
|---|---|---|---|---|
| T | T | T | F | F |
| T | F | T | F | F |
| F | T | T | F | F |
| F | F | F | T | F |
Since ¬(p ∨ q) ∧ q is always false, it is a contradiction.
d) (p ∧ q) ∧ ¬(p ∨ q) is a contradiction
| p | q | p ∧ q | p ∨ q | ¬(p ∨ q) | (p ∧ q) ∧ ¬(p ∨ q) |
|---|---|---|---|---|---|
| T | T | T | T | F | F |
| T | F | F | T | F | F |
| F | T | F | T | F | F |
| F | F | F | F | T | F |
Since (p ∧ q) ∧ ¬(p ∨ q) is always false, it is a contradiction.
e) ¬q ∧ (p ⇒ q) ⇒ ¬p is a tautology
| p | q | ¬q | p ⇒ q | ¬q ∧ (p ⇒ q) | ¬p | ¬q ∧ (p ⇒ q) ⇒ ¬p |
|---|---|---|---|---|---|---|
| T | T | F | T | F | F | T |
| T | F | T | F | F | F | T |
| F | T | F | T | F | T | T |
| F | F | T | T | T | T | T |
Since ¬q ∧ (p ⇒ q) ⇒ ¬p is always true, it is a tautology.
f) [(p ∧ q) ⇒ p] ⇒ (q ∧ ¬q) is a contradiction
| p | q | p ∧ q | (p ∧ q) ⇒ p | ¬q | q ∧ ¬q | [(p ∧ q) ⇒ p] ⇒ (q ∧ ¬q) |
|---|---|---|---|---|---|---|
| T | T | T | T | F | F | F |
| T | F | F | T | T | F | F |
| F | T | F | T | F | F | F |
| F | F | F | T | T | F | F |
Since [(p ∧ q) ⇒ p] ⇒ (q ∧ ¬q) is always false, it is a contradiction.
Note: All solutions are verified and accurate according to the principles of mathematical logic.
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