Chapter 1: Mathematical Logic

Exercise 1.1

Balbharati solutions for Mathematics and Statistics 1 (Commerce) 12th Standard HSC Maharashtra State Board

#### EXERCISE 1.1Q 1 PAGE 2

Exercise 1.1 | Q 1 | Page 2

# State which of the following sentences is a statement. Justify your answer if it is a statement. Write down its truth value.

## A triangle has ‘n’ sides

### SOLUTION

It is an open sentence. Hence, it is not a statement.

[Note: Answer given in the textbook is ‘it is a statement’. However, we found that ‘It is not a statement’.]

#### EXERCISE 1.1Q 2 PAGE 2

Exercise 1.1 | Q 2 | Page 2

# State which of the following sentence is a statement. Justify your answer if it is a statement. Write down its truth value.

## The sum of interior angles of a triangle is 180°

### SOLUTION

It is a statement which is true. Hence, it’s truth value is T.

#### EXERCISE 1.1Q 3 PAGE 2

Exercise 1.1 | Q 3 | Page 2

# State which of the following sentence is a statement. Justify your answer if it is a statement. Write down its truth value.

## You are amazing!

### SOLUTION

It is an exclamatory sentence. Hence, it is not a statement.

#### EXERCISE 1.1Q 4 PAGE 2

Exercise 1.1 | Q 4 | Page 2

# State which of the following sentences is a statement. Justify your answer if it is a statement. Write down its truth value.

## Please grant me a loan.

### SOLUTION

It is a request. Hence, it is not a statement.

#### EXERCISE 1.1Q 5 PAGE 2

Exercise 1.1 | Q 5 | Page 2

# State which of the following sentences is a statement. Justify your answer if it is a statement. Write down its truth value.

## √-4 is an irrational number.

### SOLUTION

It is a statement which is false. Hence, it’s truth value is F.

#### EXERCISE 1.1Q 6 PAGE 2

Exercise 1.1 | Q 6 | Page 2

## x2 − 6x + 8 = 0 implies x = −4 or x = −2.

### SOLUTION

It is a statement which is false. Hence, it’s truth value if F.

#### EXERCISE 1.1Q 7 PAGE 3

Exercise 1.1 | Q 7 | Page 3

## He is an actor.

### SOLUTION

It is an open sentence. Hence, it is not a statement.

#### EXERCISE 1.1Q 8 PAGE 3

Exercise 1.1 | Q 8 | Page 3

## Did you eat lunch yet?

### SOLUTION

It is an interrogative sentence. Hence, it is not a statement.

#### EXERCISE 1.1Q 9 PAGE 3

Exercise 1.1 | Q 9 | Page 3

# State which of the following sentence is a statement. Justify your answer if it is a statement. Write down its truth value.

## Have a cup of cappuccino.

### SOLUTION

It is an imperative sentence, hence it is not a statement.

#### EXERCISE 1.1Q 10 PAGE 3

Exercise 1.1 | Q 10 | Page 3

## (x + y)2 = x2 + 2xy + y2 for all x, y ∈ R.

### SOLUTION

It is a statement which is true. Hence, it’s truth value is T.

#### EXERCISE 1.1Q 11 PAGE 3

Exercise 1.1 | Q 11 | Page 3

## Every real number is a complex number.

### SOLUTION

It is a statement which is true. Hence, it’s truth value is T.

#### EXERCISE 1.1Q 12 PAGE 3

Exercise 1.1 | Q 12 | Page 3

## 1 is a prime number.

### SOLUTION

It is a statement which is false. Hence, it’s truth value is F.

#### EXERCISE 1.1Q 13 PAGE 3

Exercise 1.1 | Q 13 | Page 3

## With the sunset the day ends.

### SOLUTION

It is a statement which is true. Hence, it’s truth value is T.

#### EXERCISE 1.1Q 14 PAGE 3

Exercise 1.1 | Q 14 | Page 3

## 1 ! = 0

### SOLUTION

It is a statement which is false. Hence, it’s truth value is F.

[Note: Answer in the textbook is incorrect]

#### EXERCISE 1.1Q 15 PAGE 3

Exercise 1.1 | Q 15 | Page 3

## 3 + 5 > 11

### SOLUTION

It is a statement which is false. Hence, it’s truth value is F.

#### EXERCISE 1.1Q 16 PAGE 3

Exercise 1.1 | Q 16 | Page 3

## The number π is an irrational number.

### SOLUTION

It is a statement which is true. Hence, it’s truth value is T.

[Note: Answer in the textbook is incorrect]

#### EXERCISE 1.1Q 17 PAGE 3

Exercise 1.1 | Q 17 | Page 3

## x2 - y2 = (x + y)(x - y) for all x, y ∈ R.

### SOLUTION

It is a statement which is true. Hence, it’s truth value is T.

#### EXERCISE 1.1Q 18 PAGE 3

Exercise 1.1 | Q 18 | Page 3

## The number 2 is the only even prime number.

### SOLUTION

It is a statement which is true. Hence, it’s truth value is T.

#### EXERCISE 1.1Q 19 PAGE 3

Exercise 1.1 | Q 19 | Page 3

## Two co-planar lines are either parallel or intersecting.

### SOLUTION

It is a statement which is true. Hence, it’s truth value is T.

#### EXERCISE 1.1Q 20 PAGE 3

Exercise 1.1 | Q 20 | Page 3

## The number of arrangements of 7 girls in a row for a photograph is 7!.

### SOLUTION

It is a statement which is true. Hence, it’s truth value is T.

#### EXERCISE 1.1Q 21 PAGE 3

Exercise 1.1 | Q 21 | Page 3

## Give me a compass box.

### SOLUTION

It is an imperative sentence. Hence, it is not a statement.

#### EXERCISE 1.1Q 22 PAGE 3

Exercise 1.1 | Q 22 | Page 3

## Bring the motor car here.

### SOLUTION

It is an imperative sentence. Hence, it is not a statement.

#### EXERCISE 1.1Q 23 PAGE 3

Exercise 1.1 | Q 23 | Page 3

## It may rain today.

### SOLUTION

It is an open sentence. Hence, it is not a statement.

#### EXERCISE 1.1Q 24 PAGE 3

Exercise 1.1 | Q 24 | Page 3

## If a + b < 7, where a ≥ 0 and b ≥ 0 then a < 7 and b < 7.

### SOLUTION

It is a statement which is true. Hence, it’s truth value is T.

#### EXERCISE 1.1Q 25 PAGE 3

Exercise 1.1 | Q 25 | Page 3

## Can you speak in English?

### SOLUTION

### It is an interrogative sentence. Hence, it is not a statement.

### Balbharati solutions for Mathematics and Statistics 1 (Commerce) 12th Standard HSC Maharashtra State Board Chapter 1 Mathematical Logic Exercise 1.2 [Page 6]

#### EXERCISE 1.2Q 1.1 PAGE 6

Express the following statement in symbolic form.

e is a vowel or 2 + 3 = 5

#### SOLUTION

Let p : e is a vowel.

q : 2 + 3 = 5

The symbolic form is p ∨ q.

#### EXERCISE 1.2Q 1.2 PAGE 6

Express the following statement in symbolic form.

Mango is a fruit but potato is a vegetable.

#### SOLUTION

Let p : Mango is a fruit.

q : Potato is a vegetable.

The symbolic form is p Λ q.

#### EXERCISE 1.2Q 1.3 PAGE 6

Express the following statement in symbolic form.

Milk is white or grass is green.

#### SOLUTION

Let p : Milk is white.

q : Grass is green.

The symbolic form is p ∨ q.

#### EXERCISE 1.2Q 1.4 PAGE 6

Express the following statement in symbolic form.

I like playing but not singing.

#### SOLUTION

Let p : I like playing.

q : I do not like singing.

The symbolic form is p ∧ q.

#### EXERCISE 1.2Q 1.5 PAGE 6

Express the following statement in symbolic form.

Even though it is cloudy, it is still raining.

#### SOLUTION

Let p : It is cloudy.

q : It is still raining.

The symbolic form is p ∧ q.

#### EXERCISE 1.2Q 2.1 PAGE 6

Write the truth value of the following statement.

Earth is a planet and Moon is a star.

#### SOLUTION

Let p : Earth is a planet.

q : Moon is a star.

The truth values of p and q are T and F respectively.

The given statement in symbolic form is p ∧ q.

∴ p ∧ q ≡ T ∧ F ≡ F

∴ Truth value of the given statement is F.

#### EXERCISE 1.2Q 2.2 PAGE 6

Write the truth value of the following statement.

16 is an even number and 8 is a perfect square.

#### SOLUTION

Let p : 16 is an even number.

q : 8 is a perfect square.

The truth values of p and q are T and F respectively.

The given statement in symbolic form is p ∧ q.

∴ p ∧ q ≡ T ∧ F ≡ F

∴ Truth value of the given statement is F.

#### EXERCISE 1.2Q 2.3 PAGE 6

Write the truth value of the following statement.

A quadratic equation has two distinct roots or 6 has three prime factors.

#### SOLUTION

Let p : A quadratic equation has two distinct roots.

q : 6 has three prime factors.

The truth values of p and q are F and F respectively.

The given statement in symbolic form is p ∨ q.

∴ p ∨ q ≡ F ∨ F ≡ F

∴ Truth value of the given statement is F.

#### EXERCISE 1.2Q 2.4 PAGE 6

Write the truth value of the following statement.

The Himalayas are the highest mountains but they are part of India in the North East.

#### SOLUTION

Let p : Himalayas are the highest mountains.

q : Himalayas are the part of India in the north east.

The truth values of p and q are T and T respectively.

The given statement in symbolic form is p ∧ q.

∴ p ∧ q ≡ T ∧ T ≡ T

∴ Truth value of the given statement is T.

### Balbharati solutions for Mathematics and Statistics 1 (Commerce) 12th Standard HSC Maharashtra State Board Chapter 1 Mathematical Logic Exercise 1.3 [Page 7]

#### EXERCISE 1.3Q 1.1 PAGE 7

Write the negation of the following statement.

All men are animals.

#### SOLUTION

Some men are not animals.

#### EXERCISE 1.3Q 1.2 PAGE 7

Write the negation of the following statement.

− 3 is a natural number.

#### SOLUTION

– 3 is not a natural number.

#### EXERCISE 1.3Q 1.3 PAGE 7

Write the negation of the following statement.

It is false that Nagpur is capital of Maharashtra

#### SOLUTION

Nagpur is capital of Maharashtra.

#### EXERCISE 1.3Q 1.4 PAGE 7

Write the negation of the following statement.

2 + 3 ≠ 5

#### SOLUTION

2 + 3 = 5

#### EXERCISE 1.3Q 2.1 PAGE 7

Write the truth value of the negation of the following statement.

## √5 is an irrational number.

#### SOLUTION

Truth value of the given statement is T.

∴ Truth value of its negation is F.

#### EXERCISE 1.3Q 2.2 PAGE 7

Write the truth value of the negation of the following statement.

London is in England.

#### SOLUTION

Truth value of the given statement is T.

∴ Truth value of its negation is F.

#### EXERCISE 1.3Q 2.3 PAGE 7

Write the truth value of the negation of the following statement.

For every x ∈ N, x + 3 < 8.

#### SOLUTION

Truth value of the given statement is F.

∴ Truth value of its negation is T.

### Balbharati solutions for Mathematics and Statistics 1 (Commerce) 12th Standard HSC Maharashtra State Board Chapter 1 Mathematical Logic Exercise 1.4 [Pages 10 - 11]

Write the following statement in symbolic form.

If triangle is equilateral then it is equiangular.

#### SOLUTION

Let p : Triangle is equilateral.

q : Triangle is equiangular.

The symbolic form is p → q.

Write the following statement in symbolic form.

It is not true that “i” is a real number.

#### SOLUTION

Let p : i is a real number.

The symbolic form is ~ p.

Write the following statement in symbolic form.

Even though it is not cloudy, it is still raining.

#### SOLUTION

Let p : It is cloudy.

q : It is raining.

The symbolic form is ~p ∧ q.

Write the following statement in symbolic form.

Milk is white if and only if the sky is not blue.

#### SOLUTION

Let p : Milk is white.

q : Sky is blue.

The symbolic form is p ↔ ~ q.

Write the following statement in symbolic form.

Stock prices are high if and only if stocks are rising.

#### SOLUTION

Let p : Stock prices are high.

q : Stock are rising

The symbolic form is p ↔ q.

Write the following statement in symbolic form.

If Kutub-Minar is in Delhi then Taj-Mahal is in Agra.

#### SOLUTION

Let p : Kutub-Minar is in Delhi.

q : Taj-Mahal Is in Agra.

The symbolic form is p → q.

Find the truth value of the following statement.

It is not true that 3 − 7i is a real number.

#### SOLUTION

Let p : 3 – 7i is a real number.

The truth value of p is F.

The given statement in symbolic form is ~p.

∴ ~ p ≡ ~ F ≡ T

∴ Truth value of the given statement is T.

Find the truth value of the following statement.

If a joint venture is a temporary partnership, then discount on purchase is credited to the supplier.

#### SOLUTION

Let p : A joint venture is a temporary partnership. q : Discount on purchase is credited to the supplier.

The truth value of p and q are T and F respectively.

The given statement in symbolic form is p → q.

∴ p → q ≡ T → F ≡ F

∴ Truth value of the given statement is F.

Find the truth value of the following statement.

Every accountant is free to apply his own accounting rules if and only if machinery is an asset.

#### SOLUTION

Let p : Every accountant is free to apply his own accounting rules.

q : Machinery is an asset.

The truth values of p and q are F and T respectively.

The given statement in symbolic form is p ↔ q.

∴ p ↔ q ≡ F ↔ T ≡ F

∴ Truth value of the given statement is F.

Find the truth value of the following statement.

Neither 27 is a prime number nor divisible by 4.

#### SOLUTION

Let p : 27 is a prime number.

q : 27 is divisible by 4.

The truth values of p and q are F and F respectively.

The given statement in symbolic form is ~ p ∧ ~ q.

∴ ~ p ∧ ~ q ≡ ~ F ∧ ~ F ≡ T ∧ T ≡ T

∴ Truth value of the given statement is T.

Find the truth value of the following statement.

3 is a prime number and an odd number.

#### SOLUTION

Let p : 3 is a prime number.

q : 3 is an odd number.

The truth values of p and q are T and T respectively.

The given statement in symbolic form is p ∧ q.

∴ p ∧ q ≡ T ∧ T ≡ T

∴ Truth value of the given statement is T.

If p and q are true and r and s are false, find the truth value of the following compound statement.

p ∧ (q ∧ r)

#### SOLUTION

p ∧ (q ∧ r) ≡ T ∧ (T ∧ F)

≡ T ∧ F

≡ F

Hence, truth value if F.

If p and q are true and r and s are false, find the truth value of the following compound statement.

(p → q) ∨ (r ∧ s)

#### SOLUTION

(p → q) ∨ (r ∧ s) ≡ (T → T) ∨ (F ∧ F)

≡ T ∨ F

≡ T

Hence, truth value if T.

If p and q are true and r and s are false, find the truth value of the following compound statement.

~ [(~ p ∨ s) ∧ (~ q ∧ r)]

#### SOLUTION

~ [(~ p ∨ s) ∧ (~ q ∧ r)] ≡ ~[(~T ∨ F) ∧ (~T ∧ F)]

≡ ~[(F ∨ F) ∧ (F ∧ F)

≡ ~ (F ∧ F)

≡ ~ F

≡ T

Hence, truth value if T.

If p and q are true and r and s are false, find the truth value of the following compound statement.

(p → q) ↔ ~(p ∨ q)

#### SOLUTION

(p → q) ↔ ~(p ∨ q) ≡ (T → T) ↔ (T ∨ T)

≡ T ↔ ~ T

≡ T ↔ F

≡ F

Hence, truth value if F.

If p and q are true and r and s are false, find the truth value of the following compound statement.

[(p ∨ s) → r] ∨ ~ [~ (p → q) ∨ s]

#### SOLUTION

[(p ∨ s) → r] ∨ ~ [~ (p → q) ∨ s]

≡ [(T ∨ F) → F] ∨ ~[~ (T → T) ∨ F]

≡ (T → F) ∨ ~ (~ T ∨ F)

≡ F ∨ ~ (F ∨ F)

≡ F ∨ ~ F

≡ F ∨ T

≡ T

Hence, truth value is T.

If p and q are true and r and s are false, find the truth value of the following compound statement.

~ [p ∨ (r ∧ s)] ∧ ~ [(r ∧ ~ s) ∧ q]

#### SOLUTION

~ [p ∨ (r ∧ s)] ∧ ~ [(r ∧ ~ s) ∧ q]

≡ ~ [T ∨ (F ∧ F)] ∧ ~ [(F ∧ ~ F) ∧ T]

≡ ~ (T ∨ F) ∧ ~ [(F ∧ T) ∧ T]

≡ ~ T ∧ ~ (F ∧ T)

≡ F ∧ ~ F

≡ F ∧ T

≡ F

Hence, truth value is F.

Assuming that the following statement is true,

p : Sunday is holiday,

q : Ram does not study on holiday,

find the truth values of the following statements.

Sunday is not holiday or Ram studies on holiday.

#### SOLUTION

Symbolic form of the given statement is ~ p ∨~q

∴ ~ p ∨~ q ≡ ~ T ∨ ~ T

≡ F ∨ F

≡ F

Hence, truth value is F.

Assuming that the following statement is true,

p : Sunday is holiday,

q : Ram does not study on holiday,

find the truth values of the following statements.

If Sunday is not holiday then Ram studies on holiday.

#### SOLUTION

Symbolic form of the given statement is

~ p → ~ q

∴ ~ p → ~ q ≡ ~ T → ~ T

≡ F → F

≡ T

Hence, truth value is T.

Assuming that the following statement is true,

p : Sunday is holiday,

q : Ram does not study on holiday,

find the truth values of the following statements.

Sunday is a holiday and Ram studies on holiday.

#### SOLUTION

Symbolic form of the given statement is p ∧ ~ q

∴ p ∧ ~ q ≡ T ∧ ~ T

≡ T ∧ F

≡ F

Hence, truth value is F.

If p : He swims

q : Water is warm

Give the verbal statement for the following symbolic statement.

p ↔ ~ q

#### SOLUTION

He swims if and only if water is not warm.

If p : He swims

q : Water is warm

Give the verbal statement for the following symbolic statement.

~ (p ∨ q)

#### SOLUTION

It is not true that he swims or water is warm.

If p : He swims

q : Water is warm

Give the verbal statement for the following symbolic statement.

q → p

#### SOLUTION

If water is warm then he swims.

If p : He swims

q : Water is warm

Give the verbal statement for the following symbolic statement.

q ∧ ~ p

#### SOLUTION

Water is warm and he does not swim.

### EXERCISE 1.5PAGE 12

### Balbharati solutions for Mathematics and Statistics 1 (Commerce) 12th Standard HSC Maharashtra State Board Chapter 1 Mathematical Logic Exercise 1.5 [Page 12]

Use quantifiers to convert the following open sentences defined on N, into a true statement.

x^{2} + 3x - 10 = 0

#### SOLUTION

∃ x ∈ N, such that x^{2} + 3x – 10 = 0

It is true statement, since x = 2 ∈ N satisfies it.

Use quantifiers to convert the following open sentences defined on N, into a true statement.

3x - 4 < 9

#### SOLUTION

∃ x ∈ N, such that 3x – 4 < 9

It is true statement, since

x = 2, 3, 4 ∈ N satisfies 3x - 4 < 9.

Use quantifiers to convert the following open sentences defined on N, into a true statement.

n^{2} ≥ 1

#### SOLUTION

∀ n ∈ N, n^{2} ≥ 1

It is true statement, since all n ∈ N satisfy it.

Use quantifiers to convert the following open sentences defined on N, into a true statement.

2n - 1 = 5

#### SOLUTION

∃ n ∈ N, such that 2n - 1 = 5

It is a true statement since all n = 3 ∈ N satisfy 2n - 1 = 5.

Use quantifiers to convert the following open sentences defined on N, into a true statement.

y + 4 > 6

#### SOLUTION

∃ y ∈ N, such that y + 4 > 6

It is a true statement since y = 3, 4, ... ∈ N satisfy y + 4 > 6.

Use quantifiers to convert the following open sentences defined on N, into a true statement.

3y - 2 ≤ 9

#### SOLUTION

∃ y ∈ N, such that 3y - 2 ≤ 9

It is a true statement since y = 1, 2, 3 ∈ N satisfy it.

If B = {2, 3, 5, 6, 7} determine the truth value of ∀ x ∈ B such that x is prime number.

#### SOLUTION

For x = 6, x is not a prime number.

∴ x = 6 does not satisfies the given statement.

∴ The given statement is false.

∴ It’s truth value is F.

If B = {2, 3, 5, 6, 7} determine the truth value of

∃ n ∈ B, such that n + 6 > 12.

#### SOLUTION

For n = 7, n + 6 = 7 + 6 = 13 > 12

∴ n = 7 satisfies the equation n + 6 > 12.

∴ The given statement is true.

∴ It’s truth value is T.

If B = {2, 3, 5, 6, 7} determine the truth value of

∃ n ∈ B, such that 2n + 2 < 4.

#### SOLUTION

There is no n in B which satisfies 2n + 2 < 4.

∴ The given statement is false.

∴ It’s truth value is F.

If B = {2, 3, 5, 6, 7} determine the truth value of

∀ y ∈ B, such that y^{2} is negative.

#### SOLUTION

There is no y in B which satisfies y^{2} < 0.

∴ The given statement is false.

∴ It’s truth value is F.

If B = {2, 3, 5, 6, 7} determine the truth value of

∀ y ∈ B, such that (y - 5) ∈ N

#### SOLUTION

For y = 2, y – 5 = 2 – 5 = –3 ∉ N.

∴ y = 2 does not satisfies the equation (y – 5) ∈ N.

∴ The given statement is false.

∴ It’s truth value is F.

### EXERCISE 1.6PAGE 16

### Balbharati solutions for Mathematics and Statistics 1 (Commerce) 12th Standard HSC Maharashtra State Board Chapter 1 Mathematical Logic Exercise 1.6 [Page 16]

Prepare truth tables for the following statement pattern.

p → (~ p ∨ q)

#### SOLUTION

p → (~ p ∨ q)

p | q | ~p | ~ p ∨ q | p → (~ p ∨ q) |

T | T | F | T | T |

T | F | F | F | F |

F | T | T | T | T |

F | F | T | T | T |

Prepare truth tables for the following statement pattern.

(~ p ∨ q) ∧ (~ p ∨ ~ q)

#### SOLUTION

(~ p ∨ q) ∧ (~ p ∨ ~ q)

p | q | ~p | ~q | ~p∨q | ~p∨~q | (~p∨q)∧(~p∨~q) |

T | T | F | F | T | F | F |

T | F | F | T | F | T | F |

F | T | T | F | T | T | T |

F | F | T | T | T | T | T |

Prepare truth tables for the following statement pattern.

(p ∧ r) → (p ∨ ~ q)

#### SOLUTION

(p ∧ r) → (p ∨ ~ q)

p | q | r | ~q | p ∧ r | p∨~q | (p ∧ r) → (p ∨ ~ q) |

T | T | T | F | T | T | T |

T | T | F | F | F | T | T |

T | F | T | T | T | T | T |

T | F | F | T | F | T | T |

F | T | T | F | F | F | T |

F | T | F | F | F | F | T |

F | F | T | T | F | T | T |

F | F | F | T | F | T | T |

Prepare truth tables for the following statement pattern.

(p ∧ q) ∨ ~ r

#### SOLUTION

(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 |

Examine whether the following statement pattern is a tautology, a contradiction or a contingency.

q ∨ [~ (p ∧ q)]

#### SOLUTION

p | q | p ∧ q | ~ (p ∧ q) | q ∨ [~ (p ∧ q)] |

T | T | T | F | T |

T | F | F | T | T |

F | T | F | T | T |

F | F | F | T | T |

All the truth values in the last column are T. Hence, it is a tautology.

Examine whether the following statement pattern is a tautology, a contradiction or a contingency.

(~ q ∧ p) ∧ (p ∧ ~ p)

#### SOLUTION

p | q | ~p | ~q | (~q∧p) | (p∧~p) | (~q∧p)∧(p∧~p) |

T | T | F | F | F | F | F |

T | F | F | T | T | F | F |

F | T | T | F | F | F | F |

F | F | T | T | F | F | F |

All the truth values in the last column are F. Hence, it is a contradiction.

Examine whether the following statement pattern is a tautology, a contradiction or a contingency.

(p ∧ ~ q) → (~ p ∧ ~ q)

#### SOLUTION

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.

Examine whether the following statement pattern is a tautology, a contradiction or a contingency.

~ p → (p → ~ q)

#### SOLUTION

p | q | ~p | ~q | p→~q | ~p→(p→~q) |

T | T | F | F | F | T |

T | F | F | T | T | T |

F | T | T | F | T | T |

F | F | T | T | T | T |

All the truth values in the last column are T. Hence, it is tautology.

Prove that the following statement pattern is a tautology.

(p ∧ q) → q

#### SOLUTION

p | q | p ∧ q | (p∧q)→q |

T | T | T | T |

T | F | F | T |

F | T | F | T |

F | F | F | T |

All the truth values in the last column are T. Hence, it is tautology.

Prove that the following statement pattern is a tautology.

(p → q) ↔ (~ q → ~ p)

#### SOLUTION

p | q | ~p | ~q | p→q | ~q→~p | (p→q)↔(~q→~p) |

T | T | F | F | T | T | T |

T | F | F | T | F | F | T |

F | T | T | F | T | T | T |

F | F | T | T | T | T | T |

All the truth values in the last column are T. Hence, it is a tautology.

Prove that the following statement pattern is a tautology.

(~p ∧ ~q ) → (p → q)

#### SOLUTION

p | q | ~p | ~q | ~p∧~q | p→q | (~p∧~q)→(p→q) |

T | T | F | F | F | T | T |

T | F | F | T | F | F | T |

F | T | T | F | F | T | T |

F | F | T | T | T | T | T |

All the truth values in the last column are T. Hence, it is a tautology.

Prove that the following statement pattern is a tautology.

(~ p ∨ ~ q) ↔ ~ (p ∧ q)

#### SOLUTION

p | q | ~p | ~q | ~p∨~q | p∧q | ~p∨~q | (~p∨~q↔~(p ∧ q) |

T | T | F | F | F | T | F | T |

T | F | F | T | T | F | T | T |

F | T | T | F | T | F | T | T |

F | F | T | T | T | F | T | T |

All the truth values in the last column are T. Hence, it is a tautology.

Prove that the following statement pattern is a contradiction.

(p ∨ q) ∧ (~p ∧ ~q)

#### SOLUTION

p | q | ~p | ~q | p∨q | ~p∧~q | (p∨q)∧(~p∧~q) |

T | T | F | F | T | F | F |

T | F | F | T | T | F | F |

F | T | T | F | T | F | F |

F | F | T | T | F | T | F |

All the truth values in the last column are F. Hence, it is a contradiction.

Prove that the following statement pattern is a contradiction.

(p ∧ q) ∧ ~p

#### SOLUTION

p | q | ~p | p∧q | (p∧q)∧~p |

T | T | F | T | F |

T | F | F | F | F |

F | T | T | F | F |

F | F | T | F | F |

All the truth values in the last column are F. Hence, it is a contradiction.

Prove that the following statement pattern is a contradiction.

(p ∧ q) ∧ (~p ∨ ~q)

#### SOLUTION

p | q | ~p | ~q | p∧q | ~p∨~q | (p∧q)∧(~p∨~q) |

T | T | F | F | T | F | F |

T | F | F | T | F | T | F |

F | T | T | F | F | T | F |

F | F | T | T | F | T | F |

All the truth values in the last column are F. Hence, it is a contradiction.

Prove that the following statement pattern is a contradiction.

(p → q) ∧ (p ∧ ~ q)

#### SOLUTION

p | q | ~q | p→q | p∧~q | (p→q)∧(p∧~q) |

T | T | F | T | F | F |

T | F | T | F | T | F |

F | T | F | T | F | F |

F | F | T | T | F | F |

All the truth values in the last column are F. Hence, it is a contradiction.

Show that the following statement pattern is contingency.

(p∧~q) → (~p∧~q)

#### SOLUTION

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.

Show that the following statement pattern is contingency.

(p → q) ↔ (~ p ∨ q)

#### SOLUTION

p | q | ~p | p→q | ~p∨q | (p→q)↔(~p∨q) |

T | T | F | T | T | T |

T | F | F | F | F | T |

F | T | T | T | T | T |

F | F | T | T | T | T |

All the truth values in the last column are T. Hence, it is a tautology. Not contingency.

Show that the following statement pattern is contingency.

p ∧ [(p → ~ q) → q]

#### SOLUTION

p | q | ~q | p→~q | (p→~q)→q | p∧[(p→~q)→q] |

T | T | F | F | T | T |

T | F | T | T | F | F |

F | T | F | T | T | F |

F | F | T | T | F | F |

Truth values in the last column are not identical. Hence, it is contingency.

Show that the following statement pattern is contingency.

(p → q) ∧ (p → r)

#### SOLUTION

p | q | r | p→q | p→r | (p→q)∧(p→r) |

T | T | T | T | T | T |

T | T | F | T | F | F |

T | F | T | F | T | F |

T | F | F | F | F | F |

F | T | T | T | T | T |

F | T | F | T | T | T |

F | F | T | T | T | T |

F | F | F | T | T | T |

The truth values in the last column are not identical. Hence, it is contingency.

Exercise 1.6 | Q 6.1 | Page 16

Using the truth table, verify

p ∨ (q ∧ r) ≡ (p ∨ q) ∧ (p ∨ r)

#### SOLUTION

1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |

p | q | r | q∧r | p∨(q∧r) | p∨q | p∨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 | F | T | F | F |

F | F | T | F | F | F | T | F |

F | F | F | F | F | F | F | F |

The entries in columns 5 and 8 are identical.

∴ p ∨ (q ∧ r) ≡ (p ∨ q) ∧ (p ∨ r)

Using the truth table, verify

p → (p → q) ≡ ~ q → (p → q)

#### SOLUTION

1 | 2 | 3 | 4 | 5 | 6 |

p | q | ~q | p→q | p→(p→q) | ~q→(p→q) |

T | T | F | T | T | T |

T | F | T | F | F | F |

F | T | F | T | T | T |

F | F | T | T | T | T |

In the above truth table, entries in columns 5 and 6 are identical.

∴ p → (p → q) ≡ ~ q → (p → q)

Using the truth table, verify

~(p → ~q) ≡ p ∧ ~ (~ q) ≡ p ∧ q

#### SOLUTION

1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |

p | q | ~q | p→~q | ~(p→~q) | ~(~q) | p∧~(~q) | p∧q |

T | T | F | F | T | T | T | T |

T | F | T | T | F | F | F | F |

F | T | F | T | F | T | F | F |

F | F | T | T | F | F | F | F |

In the above table, entries in columns 5, 7, and 8 are identical.

∴ ~(p → ~q) ≡ p ∧ ~ (~ q) ≡ p ∧ q

Using the truth table, verify

~(p ∨ q) ∨ (~ p ∧ q) ≡ ~ p

#### SOLUTION

1 | 2 | 3 | 4 | 5 | 6 | 7 |

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 |

In the above truth table, the entries in columns 3 and 7 are identical.

∴ ~(p ∨ q) ∨ (~ p ∧ q) ≡ ~ p

Exercise 1.6 | Q 7.1 | Page 16

Using the truth table, verify

p ∨ (q ∧ r) ≡ (p ∨ q) ∧ (p ∨ r)

#### SOLUTION

1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |

p | q | r | q∧r | p∨(q∧r) | p∨q | p∨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 | F | T | F | F |

F | F | T | F | F | F | T | F |

F | F | F | F | F | F | F | F |

The entries in columns 5 and 8 are identical.

∴ ∨ (q ∧ r) ≡ (p ∨ q) ∧ (p ∨ r)

Prove that the following pair of statement pattern is equivalent.

p ↔ q and (p → q) ∧ (q → p)

#### SOLUTION

1 | 2 | 3 | 4 | 5 | 6 |

p | q | p↔q | p→q | q→p | (p→q)∧(q→p) |

T | T | T | T | T | T |

T | F | F | F | T | F |

F | T | F | T | F | F |

F | F | T | T | T | T |

In the above table, entries in columns 3 and 6 are identical.

∴ Statement p ↔ q and (p → q) ∧ (q → p) are equivalent.

Prove that the following pair of statement pattern is equivalent.

p → q and ~ q → ~ p and ~ p ∨ q

#### SOLUTION

1 | 2 | 3 | 4 | 5 | 6 | 7 |

p | q | ~p | ~q | p→q | ~q→~p | ~p∨q |

T | T | F | F | T | T | T |

T | F | F | T | F | F | F |

F | T | T | F | T | T | T |

F | F | T | T | T | T | T |

In the above table, entries in columns 5, 6 and 7 are identical.

∴ Statement p → q and ~q → ~p and ~p ∨ q are equivalent.

Prove that the following pair of statement pattern is equivalent.

~(p ∧ q) and ~p ∨ ~q

#### SOLUTION

1 | 2 | 3 | 4 | 5 | 6 | 7 |

p | q | ~p | ~q | p∧q | ~(p∧q) | ~p∨~q |

T | T | F | F | T | F | F |

T | F | F | T | F | T | T |

F | T | T | F | F | T | T |

F | F | T | T | F | T | T |

In the above table, entries in columns 6 and 7 are identical.

∴ Statement ~(p ∧ q) and ~p ∨ ~q are equivalent.

### EXERCISE 1.7PAGE 17

### Balbharati solutions for Mathematics and Statistics 1 (Commerce) 12th Standard HSC Maharashtra State Board Chapter 1 Mathematical Logic Exercise 1.7 [Page 17]

Write the dual of the following:

(p ∨ q) ∨ r

#### SOLUTION

(p ∧ q) ∧ r

Write the dual of the following:

~(p ∨ q) ∧ [p ∨ ~ (q ∧ ~ r)]

#### SOLUTION

~(p ∧ q) ∨ [p ∧ ~ (q ∨ ~ r)]

Write the dual of the following:

p ∨ (q ∨ r) ≡ (p ∨ q) ∨ r

#### SOLUTION

p ∧ (q ∧ r) ≡ (p ∧ q) ∧ r

Write the dual of the following:

~(p ∧ q) ≡ ~ p ∨ ~ q

#### SOLUTION

~(p ∨ q) ≡ ~ p ∧ ~ q

Write the dual statement of the following compound statement.

13 is a prime number and India is a democratic country.

#### SOLUTION

13 is a prime number or India is a democratic country.

Write the dual statement of the following compound statement.

Karina is very good or everybody likes her.

#### SOLUTION

Karina is very good and everybody likes her.

Write the dual statement of the following compound statement.

Radha and Sushmita cannot read Urdu.

#### SOLUTION

Radha or Sushmita cannot read Urdu.

Write the dual statement of the following compound statement.

A number is a real number and the square of the number is non-negative.

#### SOLUTION

A number is a real number or the square of the number is non-negative.